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https://msp.org/gt/2016/20-2/p09.xhtml

Volume 20, issue 2 (2016) Recent Issues The Journal About the Journal Subscriptions Editorial Board Editorial Interests Editorial Procedure Submission Guidelines Submission Page Author Index To Appear ISSN (electronic): 1364-0380 ISSN (print): 1465-3060 Slices of hermitian $K$–theory and Milnor's conjecture on quadratic forms Oliver Röndigs and Paul Arne Østvær Geometry & Topology 20 (2016) 1157–1212 Abstract We advance the understanding of $K\phantom{\rule{0.3em}{0ex}}$–theory of quadratic forms by computing the slices of the motivic spectra representing hermitian $K\phantom{\rule{0.3em}{0ex}}$–groups and Witt groups. By an explicit computation of the slice spectral sequence for higher Witt theory, we prove Milnor’s conjecture relating Galois cohomology to quadratic forms via the filtration of the Witt ring by its fundamental ideal. In a related computation we express hermitian $K\phantom{\rule{0.3em}{0ex}}$–groups in terms of motivic cohomology. Keywords motivic cohomology, quadratic forms, slices of hermitian $K$–theory and Witt theory Mathematical Subject Classification 2010 Primary: 11E04, 14F42, 55P42 Secondary: 19D50, 19G38, 55T05

2017-10-23 22:25:04

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http://www.chegg.com/homework-help/questions-and-answers/two-spring-length-masses-m1-12kg-m2-24kg-spring-constants-k1-680n-m-k2-840n-m-secured-one--q1281190

two spring of the same length and masses m1=1.2kg and m2=2.4kg and spring constants k1=680N/m and k2=840N/m are secured at one end to a holder and at the free ends are connected to an object of mass m0=8.8kg. this compound mass-spring system is stretched by xmax=12.0cm and allowed to oscillated over a frictionless and horizontal table. obtain the angular frequency the period T, the frequency f. also find the functions x(t), v(t) and a(t) and plot them for t=0 to 2t. moreover obtain the kinetic energy K and potential energy Us for x1=xmax/2, x2=.7071xmax (hint:x=xmax=A at t=0),

2015-07-29 05:48:48

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https://proofwiki.org/wiki/Primary_Decomposition_Theorem

# Primary Decomposition Theorem ## Theorem Let $K$ be a field. Let $V$ be a vector space over $K$. Let $T: V \to V$ be a linear operator on $V$. Let $\map p x \in K \sqbrk x$ be a polynomial such that: $\map \deg p \ge 1$ $\map p T = 0$ where $0$ is the zero operator on $V$. Let $\map {p_1} x, \map {p_2} x, \ldots, \map {p_r} x$ be distinct irreducible monic polynomials. Let $c \in K \setminus \set 0$ and $a_1, a_2, \ldots, a_r, r \in \Z_{\ge 1}$ be constants. We have that: $\map p x = c \map {p_1} x^{a_1} \map {p_2} x^{a_2} \dotsm \map {p_r} x^{a_r}$ The primary decomposition theorem then states the following : $(1): \quad \map \ker {\map {p_i} T^{a_i} }$ is a $T$-invariant subspace of $V$ for all $i = 1, 2, \dotsc, r$ $(2): \quad \displaystyle V = \bigoplus_{i \mathop = 1}^r \map \ker {\map {p_i} T^{a_i} }$ ## Proof of $(1)$ Let $v \in \map \ker {\map {p_i} T^{a_i} }$. Then: $\displaystyle \map {\map {p_i} T^{a_i} } {\map T v}$ $=$ $\displaystyle \map T {\map {\map {p_i} T^{a_i} } v}$ because $\map {p_i} T^{a_i} \circ T = T \circ \map {p_i} T^{a_i}$ $\displaystyle$ $=$ $\displaystyle \map T 0$ because $v \in \map \ker {\map {p_i} T^{a_i} } \iff \map {\map {p_i} T^{a_i} } v = 0$ $\displaystyle$ $=$ $\displaystyle 0$ where $0$ is the zero vector in $V$ (if $T$ is any linear transformation, then $\map T 0 = 0$) This shows that: $\map T v \in \map \ker {\map {p_i} T^{a_i} }$ for all $v \in \map \ker {\map {p_i} T^{a_i} }$. This is because $v$ was first arbitrary in $\map \ker {\map {p_i} T^{a_i} }$. That is: $\map \ker {\map {p_i} T^{a_i} }$ is a $T$-invariant subspace of $V$. $\blacksquare$ ## Proof of $(2)$ Proof by induction on $r$: For all $r \in \Z_{\ge 1}$, let $P(r)$ be the proposition: $V = \displaystyle \bigoplus_{i \mathop = 1}^r \map \ker {\map {p_i} T^{a_i} }$ where $\map {p_i} x^{a_i}$ for $i = 1, 2, \dotsc, r$ satisfy the hypotheses of the theorem. $V$ and $T$ are arbitrary. ### Basis for the Induction We have: $\displaystyle \map p T$ $=$ $\displaystyle c \map {p_1} T^{a_1}$ $\displaystyle$ $=$ $\displaystyle 0$ $\displaystyle \leadsto \ \$ $\displaystyle \map {p_1} T^{a_1}$ $=$ $\displaystyle 0$ $\displaystyle \leadsto \ \$ $\displaystyle V$ $=$ $\displaystyle \map \ker {\map {p_1} T^{a_1} }$ Thus $\map P 1$ holds. This is our basis for the induction ### Induction Hypothesis Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: $V = \displaystyle \bigoplus_{i \mathop = 1}^{k - 1} \map \ker {\map {p_i} T^{a_i} }$ where $\map {p_i} x^{a_i}$ for $i = 1, 2, \dotsc, r$ satisfy the hypotheses of the theorem. $V$ and $T$ are arbitrary. Then we need to show: $V = \displaystyle \bigoplus_{i \mathop = 1}^k \map \ker {\map {p_i} T^{a_i} }$ ### Induction Step Suppose we meet the hypotheses of the theorem when $r = k$. Then: $\map p x \in K \sqbrk x$ is a polynomial such that: $\map \deg p \ge 1$ $\map p T = 0$ $\map p x = c \map {p_1} x^{a_1} \map {p_2} x^{a_2} \cdots \map {p_k} x^{a_k}$ Let $W := \map \ker {\map {p_2} T^{a_2} \cdots \map {p_k} T^{a_k} }$. First it will be shown that $W$ is a $T$-invariant subspace of $V$. Let $w \in W$. Then: $\displaystyle \map {\map {p_2} T^{a_2} \cdots \map {p_k} T^{a_k} } {\map T w}$ $=$ $\displaystyle \map T {\map {\map {p_2} T^{a_2} \cdots \map {p_k} T^{a_k} } w}$ by the same argument as in the proof of i), first equality $\displaystyle$ $=$ $\displaystyle \map T 0$ because $w \in W \iff \map {\map {p_2} T^{a_2} \cdots \map {p_k} T^{a_k} } w = 0$ $\displaystyle$ $=$ $\displaystyle 0$ where $0$ is the zero vector in $V$ (if $T$ is any linear transformation, then $\map T 0 = 0$) This shows that $\map T w \in W$ for all $w \in W$, because $w$ was first arbitrary in $W$. That is, $W$ is a $T$-invariant subspace of $V$. So it makes sense to talk about the restriction $T \restriction_W$ of $T$ on $W$. Now, define: $\map q x := \map {p_2} x^{a_2} \cdots \map {p_k} x^{a_k}$ Let $w \in W$. Then: $\displaystyle \map {\map q {T \restriction_W} } w$ $=$ $\displaystyle \map {\map {p_2} {T \restriction_W}^{a_2} \cdots \map {p_k} {T \restriction_W}^{a_k} } w$ $\displaystyle$ $=$ $\displaystyle \map {\map {p_2} T^{a_2} \cdots \map {p_k} T^{a_k} } w$ because $\map {T \restriction_W} w = \map T w$ for all $w \in W$ $\displaystyle$ $=$ $\displaystyle 0$ because $w \in W \iff \map {p_2} T^{a_2} \cdots \map {\map {p_k} T^{a_k} } w = 0$ Because $w$ was arbitrary in $W$, this shows that $\map q {T \restriction_W} = 0$, where $0$ is the zero operator on $W$. The hypotheses of the theorem are met. So, by the induction hypothesis: $\displaystyle W = \bigoplus_{i \mathop = 2}^k \map \ker {\map {p_i} {T \restriction_W}^{a_i} }$ Now we show that: $\map \ker {\map {p_i} {T \restriction_W}^{a_i} } = \map \ker {\map {p_i} T^{a_i} }$ for all $i = 2, \dotsc, k$. Thus, for all $i = 2, \dotsc, k$: $\displaystyle \map \ker {\map {p_i} {T \restriction_W}^{a_i} }$ $=$ $\displaystyle \set {w \in W: \map {\map {p_i} {T \restriction_W}^{a_i} } w = 0}$ $\displaystyle$ $=$ $\displaystyle \set {w \in W: \map {\map {p_i} T^{a_i} } w = 0}$ $\displaystyle$ $\subseteq$ $\displaystyle \map \ker {\map {p_i} T^{a_i} }$ $\displaystyle$ $=$ $\displaystyle \set {v \in V: \map {\map {p_i} T^{a_i} } v = 0}$ Then: $\map \ker {\map {p_i} T^{a_i} } \subseteq W$ for all $i = 2, \dotsc, k$. Let: $v \in \map \ker {\map {p_j} T^{a_j} }$ where $j \in \set {2, \dotsc, k}$. Then: $\displaystyle \map {\map {p_2} T^{a_2} \cdots \map {p_j} T^{a_j} \cdots \map {p_k} T^{a_k} } v$ $=$ $\displaystyle \map {\map {p_2} T^{a_2} \cdots \map {p_k} T^{a_k} } {\map {\map {p_j} T^{a_j} } v}$ by the same argument than in the proof of i), first equality $\displaystyle$ $=$ $\displaystyle \map {\map {p_2} T^{a_2} \cdots \map {p_k} T^{a_k} } 0$ because $v \in \map \ker {\map {p_j} T^{a_j} } \iff \map {\map {p_j} T^{a_j} } v = 0$ $\displaystyle$ $=$ $\displaystyle 0$ where $0$ is the zero vector in $V$ (if $T$ is any linear transformation, then $\map T 0 = 0$) So: $v \in \map \ker {\map {p_2} T^{a_2} \cdots \map {p_k} T^{a_k} } = W$ and: $\map \ker {\map {p_j} T^{a_j} } \subseteq W$ because $v$ was arbitrary in $\map \ker {\map {p_j} T^{a_j} }$. So: $v \in \map \ker {\map {p_i} T^{a_i} } \implies v \in W$ and so: $\map {\map {p_i} {T \restriction_W}^{a_i} } v = \map {\map {p_i} T^{a_i} } v = 0$ for all $i = 2, \dotsc, k$. Finally, because $v$ was arbitrary in $\map \ker {\map {p_i} T^{a_i} }$: $v \in \map \ker {\map {p_i} {T \restriction_W}^{a_i} }$ and: $\map \ker {\map {p_i} T^{a_i} } \subseteq \map \ker {\map {p_i} {T \restriction_W}^{a_i} }$ for all $i = 2, \dotsc, k$. This shows that: $\map \ker {\map {p_i} {T \restriction_W}^{a_i} } = \map \ker {\map {p_i} T^{a_i} }$ for all $i = 2, \dotsc, k$. So we conclude that: $\displaystyle W = \bigoplus_{i \mathop = 2}^k \map \ker {\map {p_i} T^{a_i} }$ In order to show that: $V = \displaystyle \bigoplus_{i \mathop = 1}^k \map \ker {\map {p_i} T^{a_i} }$ we equivalently show that: $(a): \quad V = W + \map \ker {\map {p_1} T^{a_1} }$ $(b): \quad W \cap \map \ker {\map {p_1} T^{a_1} } = 0$ where $0 = \set 0 \subseteq V$. Notice that $(b)$ is equivalent to showing that: $\map \ker {\map {p_1} T^{a_1} } \cap \map \ker {\map {p_i} T^{a_i} } = 0$ for all $i = 2, \dotsc, k$. In particular, $(b)$ implies this last result, because $\map \ker {\map {p_i} T^{a_i} } \subseteq W$ for all $i = 2, \dotsc, k$. $(a): \quad$ We have that: $\map g x := \gcd \set {\map {p_1} x^{a_1}, \map {p_2} x^{a_2} \cdots \map {p_k} x^{a_k} } = 1 \in K$. Indeed, $\map {p_1} x$ being an irreducible monic polynomial, either $\map g x = 1$ or $\map g x = \map {p_1} x^z$ for some $1 \le z \le a_1$. But we know that the greatest common divisor of two polynomials divides both of these polynomials. So we must conclude that $\map g x = 1$ because otherwise we would have $\map {p_1} x = \map {p_i} x$ for some $i \in \set {2, \ldots, k}$. This is a contradiction with the hypothesis that the $\map {p_i} x$ ($i = 1, 2, \ldots, k$) are all distinct. So, by Bézout's_Identity for polynomials, we know that: $\exists \map h x, \map {h_1} x \in K \sqbrk x$ such that: $1 = \map {h_1} x \map {p_1} x^{a_1} + \map h x \map {p_2} x^{a_2} \cdots \map {p_k} x^{a_k}$ Let $v \in V$. Then, evaluating the last polynomial equality at $T$, and evaluating the resulting operator at $v$, we obtain: $v = \map {\map {h_1} T \map {p_1} T^{a_1} } v + \map {\map h T \map {p_2} T^{a_2} \cdots \map {p_k} T^{a_k} } v$ Notice that $\map I v = v$ on the left hand side. But: $\displaystyle \map {\map {p_1} T^{a_1} } {\map {\map h T \map {p_2} T^{a_2} \cdots \map {p_k} T^{a_k} } v}$ $=$ $\displaystyle \map {\map h T} {\map {\map {p_1} T^{a_1} \map {p_2} T^{a_2} \cdots \map {p_k} T^{a_k} } v}$ by the same argument as in the proof of $(1)$, first equality $\displaystyle$ $=$ $\displaystyle \map h t \paren {\frac 1 c \map {\map p T} v}$ $\displaystyle$ $=$ $\displaystyle 0$ because $\map p T$ is the zero operator on $V$ by hypothesis So: $\map h T \map {p_2} T^{a_2} \cdots \map {p_k} T^{a_k} \in \map \ker {\map {p_1} T^{a_1} }$ Similarly, we find that: $\map {\map {p_2} T^{a_2} \map {p_3} T^{a_3} \cdots \map {p_k} T^{a_k} } {\map {\map {h_1} T \map {p_1} T^{a_1} } v} = 0$ and so: $\map {\map {h_1} T \map {p_1} T^{a_1} } v \in \map \ker {\map {p_2} T^{a_2} \map {p_3} T^{a_3} \cdots \map {p_k} T^{a_k} } = W$ Because $v$ was arbitrary in $V$: $v = \map {h_1} T \map {p_1} T^{a_1} v + \map h T \map {p_2} T^{a_2} \cdots \map {p_k} T^{a_k} v \implies V = W + \map \ker {\map {p_1} T^{a_1} }$ $(b): \quad$ Define: $S := W \cap \map \ker {\map {p_1} T^{a_1} }$ $M := \set {\map f x \in K \sqbrk x: \map {\map f T} s = 0 \forall s \in S}$ and: $\map q x := \map \gcd M$ By definition of $W$: $\map {p_2} x^{a_2} \cdots \map {p_k} x^{a_k} \in M$ As every element of $S$ is also an element of $\map \ker {\map {p_1} T^{a_1} }$: $\map {p_1} x^{a_1} \in M$ it follows that: $\map q x \divides \map {p_1} x^{a_1}$ and: $\map q x \divides \map {p_2} x^{a_2} \dotsm \map {p_k} x^{a_k}$ We know that the greatest common divisor $\map {u_1} x$ of two polynomials $\map {u_2} x$ and $\map {u_3} x$ has the property that: if a polynomial $\map {u_4} x$ divides both $\map {u_2} x$ and $\map {u_3} x$ then $\map {u_4} x$ must also divide $\map {u_1} x$. Thus, we must have: $\map q x \divides \gcd \set {\map {p_1} x^{a_1}, \map {p_2} x^{a_2} \cdots \map {p_k} x^{a_k} } = \map g x = 1$ So: $\map q x = \map g x = 1$ Bézout's_Identity for polynomials may then be applied to obtain: $\map q x = 1 = \map {h_1} x \map {p_1} x^{a_1} + \map {h_2} x \map {p_2} x^{a_2} \dotsm \map {p_k} x^{a_k}$ for some $\map {h_1} x, \map {h_2} x \in K \sqbrk x$. Let $s \in S$. Then, evaluating the last polynomial equality at $T$, and evaluating the resulting operator at $s$, we obtain, by definition of $M$ and by noticing that the left hand side then becomes $\map I s = s$: $s = \map {\map {h_1} T \map {p_1} T^{a_1} } s + \map {\map {h_2} T \map {p_2} T^{a_2} \cdots \map {p_k} T^{a_k} } s = 0$ Because $s$ was arbitrary in $S$, it follows that: $S = W \cap \map \ker {\map {p_1} T^{a_1} } = 0$ $\blacksquare$ ## Remarks $V$ need not be of finite dimension, but if it is then we can see $T$ as its matrix in the equations and the notation $T^j$ is then directly related to matrix multiplication. By definition, when applying $\map p x$ at $T$, one needs to convert constants which may appear in the expression of $\map p x$ into transformations in the following way For example, if $\map p x = 2 + x$, then $\map p T = 2 I + T$, where $I$ is the identity application $I: V \to V$. $\map {p_1} x, \map {p_2} x, \dotsc, \map {p_r} x$ are not necessarily of degree $1$. For example, if $K = \R$ and $\map p x = x^2 + 1$, then $\map p x$ is irreducible on $\R$ and $\map {p_1} x = \map p x$ is of degree $2$. $T^j \triangleq \begin{cases} \overbrace {T \circ T \circ \cdots \circ T}^{\text {$j$times} } & : j \in \Z_{\ge 1} \\ I: V \to V & : j = 0 \end{cases}$ By definition, the greatest common divisor of two polynomials is a monic polynomial.

2020-07-15 02:07:10

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https://vroomlab.wordpress.com/2021/04/22/16713/

# An Epilogue to “Truth vs. Intellect” This post illustrates an alternative of compute the approximate value of $\pi$. We begin with a circle whose radius is $r$, and let $L_{n}, L_{n+1}$ denotes the side’s length of regular polygon inscribed in the circle with $2^n$ and $2^{n+1}$ sides respectively, $n=2, 4, ....$ Fig. 1 On one hand, we see the area of $\Delta ABC$ as $\frac{1}{2}\cdot AB\cdot BC = \frac{1}{2}\cdot AB\cdot L_{n+1}$. On the other hand, it is also $\frac{1}{2}\cdot AC\cdot BE = \frac{1}{2}\cdot 2r\cdot \frac{L_n}{2}=\frac{1}{2}\cdot r\cdot L_n.$ Therefore, $\frac{1}{2}AB\cdot L_{n+1}= \frac{1}{2}r\cdot L_n.$ Or, $AB^2\cdot L_{n+1}^2 = r^2\cdot L_n^2\quad\quad\quad(1)$ where by Pythagorean theorem, $AB^2= (2r)^2 - L_{n+1}^2.\quad\quad\quad(2)$ Substituting (2) into (1) gives $(4r^2-L_{n+1}^2)L_{n+1}^2 = L_n^2\implies 4r^2L_{n+1}^2 - L_{n+1}^4 = r^2 L_n^2.$ That is, $L_{n+1}^4-4r^2L_{n+1}^2+r^2 L_n^2 = 0.$ Let $p = L_{n+1}^2$, we have $p^2-4r^2 p + r^2 L_n^2=0.\quad\quad\quad(3)$ Solving (3) for $p$ yields $p = 2r^2 \pm r \sqrt{4 r^2-L_n^2}.$ Since $L_n^2$ must be greater than $L_{n+1}^2$ (see Exercise 1), it must be true (see Exercise 2) that $L_{n+1}^2=2r^2 - r \sqrt{4r^2-L_n^2}.\quad\quad\quad(4)$ Notice when $r=\frac{1}{2}$, we obtain (5) in “Truth vs. Intellect“. With increasing $n$, $L_n\cdot 2^n \approx \pi\cdot 2r \implies \pi \approx \frac{L_n 2^n}{2r}.\quad\quad\quad$ We can now compute the approximate value of $\pi$ from any circle with radius $r$: Fig. 2 $r=2$ Fig. 3 $r=\frac{1}{8}$ Exercise 1 Explain $L_{n}^2 > L_{n+1}^2$ geometrically. Exercise 2 Show it is $2r^2-r\sqrt{4r^2-L_n^2}$ that represents $L_{n+1}^2.$

2021-09-28 18:59:07

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https://mathspp.com/blog/tag:set%20theory

# Mathspp Blog ### A blog dedicated to mathematics and programming! As of now, my blog is being migrated here. You can find all the old content over here. 190 ##### Twitter proof: size of the set of subsets of a set Let's prove that, if a set has size $$n$$, then that same set has exactly $$2^n$$ subsets.

2020-09-30 03:42:12

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https://www.zbmath.org/authors/?q=ai%3Ali.shoumei

# zbMATH — the first resource for mathematics ## Li, Shoumei Compute Distance To: Author ID: li.shoumei Published as: Li, Shoumei; Li, S.; Li, Shou-mei; Li, Shou Mei; Li, S. M.; Li, Sh. External Links: dblp Documents Indexed: 67 Publications since 1989, including 4 Books all top 5 #### Co-Authors 2 single-authored 14 Ogura, Yukio 9 Guan, Li 7 Zheng, Shiqiu 5 Feng, Lichao 4 Zhang, Junfei 3 Okazaki, Yoshiaki 3 Wang, Hongxia 3 Zhang, Jinping 3 Zhang, Zhongzhan 2 Denœux, Thierry 2 Kawabe, Jun 2 Li, Jun-Gang 2 Mitoma, Itaru 1 Bie, Y. H. 1 Bugarín, Alberto 1 Gil, María Angeles 1 Grzegorzewski, Przemysław 1 Hryniewicz, Olgierd 1 Inoue, Hiroshi 1 Kreinovich, Vladik Yakovlevich 1 Lawry, Jonathan 1 Li, Xiaohua 1 Liu, Xuecheng 1 Liu, Zhiyou 1 Mao, Xuerong 1 Murofushi, Toshiaki 1 Nguyen, Hung Trung 1 Ni, Baoyu 1 Proske, Frank Norbert 1 Puri, Madan Lal 1 Quost, Benjamin 1 Ralescu, Dan A. 1 Ren, Aihong 1 Song, Renming 1 Wang, Xun 1 Wang, Zhenyuan 1 Yang, Wenjuan 1 Zhong, Yu all top 5 #### Serials 7 Fuzzy Sets and Systems 4 International Journal of Approximate Reasoning 4 International Journal of Uncertainty, Fuzziness and Knowledge-Based Systems 3 Journal of Mathematical Analysis and Applications 2 Analysis Mathematica 2 Information Sciences 2 The Journal of Fuzzy Mathematics 2 Engineering Analysis with Boundary Elements 2 Abstract and Applied Analysis 2 Journal of Beijing University of Technology 2 Advances in Difference Equations 1 Computer Methods in Applied Mechanics and Engineering 1 Journal of the Franklin Institute 1 Ukrainian Mathematical Journal 1 The Annals of Probability 1 Journal of the Korean Mathematical Society 1 Journal of Multivariate Analysis 1 Tohoku Mathematical Journal. Second Series 1 Acta Mathematicae Applicatae Sinica 1 Statistics & Probability Letters 1 Bulletin pour les Sous Ensembles Flous et Leurs Applications 1 Stochastic Analysis and Applications 1 Acta Mathematicae Applicatae Sinica. English Series 1 Chinese Journal of Applied Probability and Statistics 1 Computational Mechanics 1 Journal of Theoretical Probability 1 Journal of Contemporary Mathematical Analysis. Armenian Academy of Sciences 1 Acta Mathematica Sinica. English Series 1 Scientiae Mathematicae Japonicae 1 Journal of Systems Science and Complexity 1 Theory and Decision Library. Series B: Mathematical and Statistical Methods 1 The European Physical Journal B. Condensed Matter and Complex Systems 1 Advances in Data Analysis and Classification. ADAC 1 Advances in Soft Computing 1 Science China. Mathematics 1 Advances in Intelligent and Soft Computing all top 5 #### Fields 42 Probability theory and stochastic processes (60-XX) 13 Measure and integration (28-XX) 8 Mathematical logic and foundations (03-XX) 7 Statistics (62-XX) 6 Real functions (26-XX) 5 Ordinary differential equations (34-XX) 5 Numerical analysis (65-XX) 4 Operator theory (47-XX) 4 Game theory, economics, finance, and other social and behavioral sciences (91-XX) 4 Systems theory; control (93-XX) 3 General and overarching topics; collections (00-XX) 3 Computer science (68-XX) 2 Group theory and generalizations (20-XX) 2 Several complex variables and analytic spaces (32-XX) 2 Mechanics of deformable solids (74-XX) 2 Fluid mechanics (76-XX) 2 Operations research, mathematical programming (90-XX) 1 Combinatorics (05-XX) 1 Category theory; homological algebra (18-XX) 1 Functions of a complex variable (30-XX) 1 Partial differential equations (35-XX) 1 Dynamical systems and ergodic theory (37-XX) 1 Difference and functional equations (39-XX) 1 Functional analysis (46-XX) 1 Calculus of variations and optimal control; optimization (49-XX) 1 Information and communication theory, circuits (94-XX) #### Citations contained in zbMATH Open 37 Publications have been cited 287 times in 164 Documents Cited by Year Limit theorems and applications of set-valued and fuzzy set-valued random variables. Zbl 1348.60003 Li, Shoumei; Ogura, Yukio; Kreinovich, Vladik 2002 Strong laws of large numbers for independent fuzzy set-valued random variables. Zbl 1104.60307 Li, Shoumei; Ogura, Yukio 2006 Convergence of set valued sub- and supermartingales in the Kuratowski-Mosco sense. Zbl 0938.60031 Li, Shoumei; Ogura, Yukio 1998 Convergence of set-valued and fuzzy-valued martingales. Zbl 0933.60041 Li, Shoumei; Ogura, Yukio 1999 Set defuzzification and Choquet integral. Zbl 1113.03342 Ogura, Yukio; Li, Shoumei; Ralescu, Dan A. 2001 Separability for graph convergence of sequences of fuzzy-valued random variables. Zbl 0994.60035 Ogura, Yukio; Li, Shoumei 2001 Representation theorems, set-valued and fuzzy set-valued Itô integral. Zbl 1119.60039 Li, Shoumei; Ren, Aihong 2007 Fuzzy random variables, conditional expectations and fuzzy valued martingales. Zbl 0879.60001 Li, Shoumei; Ogura, Yukio 1996 Strong solution of Itô type set-valued stochastic differential equation. Zbl 1202.60089 Li, Jun Gang; Li, Shou Mei; Ogura, Yukio 2010 A convergence theorem of fuzzy-valued martingales in the extended Hausdorff metric $$\mathbf {H}_{\infty}$$. Zbl 1020.60036 Li, Shoumei; Ogura, Yukio 2003 Central limit theorems for generalized set-valued random variables. Zbl 1029.60022 Li, Shoumei; Ogura, Yukio; Proske, Frank N.; Puri, Madan L. 2003 Fuzzy set-valued Gaussian processes and Brownian motions. Zbl 1132.60033 Li, Shoumei; Guan, Li 2007 On the solutions of set-valued stochastic differential equations in M-type 2 Banach spaces. Zbl 1198.60027 Zhang, Jinping; Li, Shoumei; Mitoma, Itaru; Okazaki, Yoshiaki 2009 Quasi-stationarity and quasi-ergodicity of general Markov processes. Zbl 1309.60078 Zhang, Junfei; Li, Shoumei; Song, Renming 2014 Gaussian processes and martingales for fuzzy valued random variables with continuous parameter. Zbl 0988.60025 Li, Shoumei; Ogura, Yukio; Nguyen, Hung T. 2001 On limit theorems for random fuzzy sets including large deviation principles. Zbl 1060.60029 Ogura, Yukio; Li, Shoumei 2004 Asymptotic stability and boundedness of stochastic functional differential equations with Markovian switching. Zbl 1349.93329 Feng, Lichao; Li, Shoumei; Mao, Xuerong 2016 On set-valued stochastic integrals in an M-type 2 Banach space. Zbl 1155.60021 Zhang, Jinping; Li, Shoumei; Mitoma, Itaru; Okazaki, Yoshiaki 2009 Representation theorems for generators of BSDEs with monotonic and convex growth generators. Zbl 1312.60079 Zheng, Shiqiu; Li, Shoumei 2015 Set-valued and interval-valued stationary time series. Zbl 1332.62347 Wang, Xun; Zhang, Zhongzhan; Li, Shoumei 2016 The $$p$$th moment asymptotic stability and exponential stability of stochastic functional differential equations with polynomial growth condition. Zbl 1343.93070 Feng, Lichao; Li, Shoumei 2014 Laws of large numbers for weighted sums of fuzzy set-valued random variables. Zbl 1065.60027 Guan, Li; Li, Shoumei 2004 Frequency-calibrated belief functions: review and new insights. Zbl 1423.68503 Denoeux, Thierry; Li, Shoumei 2018 Large and moderate deviations of random upper semicontinuous functions. Zbl 1206.60030 Ogura, Yukio; Li, Shoumei; Wang, Xia 2010 Stochastic integral with respect to set-valued square integrable martingales. Zbl 1203.60060 Li, Shoumei; Li, Jungang; Li, Xiaohua 2010 Convergence theorems for set valued and fuzzy valued martingales and smartingales. Zbl 0917.60046 Li, Shoumei; Ogura, Yukio 1998 Fuzzy linear regression analysis of fuzzy valued variables. Zbl 0714.62064 Wang, Zhenyuan; Li, Shoumei 1990 A general method for convergence theorems of fuzzy set-valued random variables and its applications to martingales and uniform amarts. Zbl 1073.60033 Li, Shoumei; Zhang, Jinping 2005 Essential norm of weighted composition operators from analytic Besov spaces into Zygmund type spaces. Zbl 1426.30041 Hu, Q.; Li, S.; Zhang, Y. 2019 Ambiguous risk aversion under capacity. Zbl 1251.91040 Wang, Hongxia; Li, Shoumei 2012 Maximal (minimal) conditional expectation and European option pricing with ambiguous return rate and volatility. Zbl 1266.91111 Zhang, Junfei; Li, Shoumei 2013 Decomposition and representation theorem of set-valued amarts. Zbl 1135.60317 Li, Shoumei; Guan, Li 2007 A law of large numbers for exchangeable random variables on nonadditive measures. Zbl 1338.60093 Guan, Li; Li, Shoumei 2013 Some properties and convergence theorems of set-valued Choquet integrals. Zbl 1278.28007 Wang, Hongxia; Li, Shoumei 2013 Robust stability of a class of stochastic functional differential equations with Markovian switching. Zbl 1419.93061 Feng, Lichao; Li, Shoumei; Liu, Zhiyou; Zheng, Shiqiu 2016 An adaptive SVD-Krylov reduced order model for surrogate based structural shape optimization through isogeometric boundary element method. Zbl 1441.74187 Li, S.; Trevelyan, J.; Wu, Z.; Lian, H.; Wang, D.; Zhang, W. 2019 Simulation on the interaction between multiple bubbles and free surface with viscous effects. Zbl 1403.76089 Li, S.; Ni, Bao-yu 2016 Essential norm of weighted composition operators from analytic Besov spaces into Zygmund type spaces. Zbl 1426.30041 Hu, Q.; Li, S.; Zhang, Y. 2019 An adaptive SVD-Krylov reduced order model for surrogate based structural shape optimization through isogeometric boundary element method. Zbl 1441.74187 Li, S.; Trevelyan, J.; Wu, Z.; Lian, H.; Wang, D.; Zhang, W. 2019 Frequency-calibrated belief functions: review and new insights. Zbl 1423.68503 Denoeux, Thierry; Li, Shoumei 2018 Asymptotic stability and boundedness of stochastic functional differential equations with Markovian switching. Zbl 1349.93329 Feng, Lichao; Li, Shoumei; Mao, Xuerong 2016 Set-valued and interval-valued stationary time series. Zbl 1332.62347 Wang, Xun; Zhang, Zhongzhan; Li, Shoumei 2016 Robust stability of a class of stochastic functional differential equations with Markovian switching. Zbl 1419.93061 Feng, Lichao; Li, Shoumei; Liu, Zhiyou; Zheng, Shiqiu 2016 Simulation on the interaction between multiple bubbles and free surface with viscous effects. Zbl 1403.76089 Li, S.; Ni, Bao-yu 2016 Representation theorems for generators of BSDEs with monotonic and convex growth generators. Zbl 1312.60079 Zheng, Shiqiu; Li, Shoumei 2015 Quasi-stationarity and quasi-ergodicity of general Markov processes. Zbl 1309.60078 Zhang, Junfei; Li, Shoumei; Song, Renming 2014 The $$p$$th moment asymptotic stability and exponential stability of stochastic functional differential equations with polynomial growth condition. Zbl 1343.93070 Feng, Lichao; Li, Shoumei 2014 Maximal (minimal) conditional expectation and European option pricing with ambiguous return rate and volatility. Zbl 1266.91111 Zhang, Junfei; Li, Shoumei 2013 A law of large numbers for exchangeable random variables on nonadditive measures. Zbl 1338.60093 Guan, Li; Li, Shoumei 2013 Some properties and convergence theorems of set-valued Choquet integrals. Zbl 1278.28007 Wang, Hongxia; Li, Shoumei 2013 Ambiguous risk aversion under capacity. Zbl 1251.91040 Wang, Hongxia; Li, Shoumei 2012 Strong solution of Itô type set-valued stochastic differential equation. Zbl 1202.60089 Li, Jun Gang; Li, Shou Mei; Ogura, Yukio 2010 Large and moderate deviations of random upper semicontinuous functions. Zbl 1206.60030 Ogura, Yukio; Li, Shoumei; Wang, Xia 2010 Stochastic integral with respect to set-valued square integrable martingales. Zbl 1203.60060 Li, Shoumei; Li, Jungang; Li, Xiaohua 2010 On the solutions of set-valued stochastic differential equations in M-type 2 Banach spaces. Zbl 1198.60027 Zhang, Jinping; Li, Shoumei; Mitoma, Itaru; Okazaki, Yoshiaki 2009 On set-valued stochastic integrals in an M-type 2 Banach space. Zbl 1155.60021 Zhang, Jinping; Li, Shoumei; Mitoma, Itaru; Okazaki, Yoshiaki 2009 Representation theorems, set-valued and fuzzy set-valued Itô integral. Zbl 1119.60039 Li, Shoumei; Ren, Aihong 2007 Fuzzy set-valued Gaussian processes and Brownian motions. Zbl 1132.60033 Li, Shoumei; Guan, Li 2007 Decomposition and representation theorem of set-valued amarts. Zbl 1135.60317 Li, Shoumei; Guan, Li 2007 Strong laws of large numbers for independent fuzzy set-valued random variables. Zbl 1104.60307 Li, Shoumei; Ogura, Yukio 2006 A general method for convergence theorems of fuzzy set-valued random variables and its applications to martingales and uniform amarts. Zbl 1073.60033 Li, Shoumei; Zhang, Jinping 2005 On limit theorems for random fuzzy sets including large deviation principles. Zbl 1060.60029 Ogura, Yukio; Li, Shoumei 2004 Laws of large numbers for weighted sums of fuzzy set-valued random variables. Zbl 1065.60027 Guan, Li; Li, Shoumei 2004 A convergence theorem of fuzzy-valued martingales in the extended Hausdorff metric $$\mathbf {H}_{\infty}$$. Zbl 1020.60036 Li, Shoumei; Ogura, Yukio 2003 Central limit theorems for generalized set-valued random variables. Zbl 1029.60022 Li, Shoumei; Ogura, Yukio; Proske, Frank N.; Puri, Madan L. 2003 Limit theorems and applications of set-valued and fuzzy set-valued random variables. Zbl 1348.60003 Li, Shoumei; Ogura, Yukio; Kreinovich, Vladik 2002 Set defuzzification and Choquet integral. Zbl 1113.03342 Ogura, Yukio; Li, Shoumei; Ralescu, Dan A. 2001 Separability for graph convergence of sequences of fuzzy-valued random variables. Zbl 0994.60035 Ogura, Yukio; Li, Shoumei 2001 Gaussian processes and martingales for fuzzy valued random variables with continuous parameter. Zbl 0988.60025 Li, Shoumei; Ogura, Yukio; Nguyen, Hung T. 2001 Convergence of set-valued and fuzzy-valued martingales. Zbl 0933.60041 Li, Shoumei; Ogura, Yukio 1999 Convergence of set valued sub- and supermartingales in the Kuratowski-Mosco sense. Zbl 0938.60031 Li, Shoumei; Ogura, Yukio 1998 Convergence theorems for set valued and fuzzy valued martingales and smartingales. Zbl 0917.60046 Li, Shoumei; Ogura, Yukio 1998 Fuzzy random variables, conditional expectations and fuzzy valued martingales. Zbl 0879.60001 Li, Shoumei; Ogura, Yukio 1996 Fuzzy linear regression analysis of fuzzy valued variables. Zbl 0714.62064 Wang, Zhenyuan; Li, Shoumei 1990 all top 5 #### Cited by 196 Authors 24 Li, Shoumei 11 Ogura, Yukio 10 Michta, Mariusz 8 Fei, Weiyin 8 Malinowski, Marek T. 8 Terán, Pedro 5 Feng, Lichao 5 Guan, Li 5 Nguyen, Hung Trung 4 Dubois, Didier 4 Labuschagne, Coenraad C. A. 4 Wang, Hongxia 3 Bongiorno, Enea G. 3 Cao, Jinde 3 Chalco-Cano, Yurilev 3 Colubi, Ana 3 Fu, Ke’ang 3 Gil, María Angeles 3 He, Guoman 3 Li, Jun-Gang 3 Ralescu, Dan A. 3 Román-Flores, Heriberto E. 3 Zhang, Jinping 3 Zheng, Shiqiu 2 Aiche, Farid 2 Aletti, Giacomo 2 Arefi, Mohsen 2 Feng, Yuhu 2 Gasiński, Leszek 2 González-Rodríguez, Gil 2 Ha, Minghu 2 Hesamian, Gholamreza 2 Inoue, Hiroshi 2 Kisielewicz, Michał 2 Marraffa, Valeria 2 Mitoma, Itaru 2 Nguyen, Pham Tri 2 Okazaki, Yoshiaki 2 Papageorgiou, Nikolaos S. 2 Quang, Nguyen Van 2 Schmelzer, Bernhard 2 Świątek, Kamil Łukasz 2 Taheri, Seyed Mahmood 2 Taheri, Seyed Mahmoud 2 Trutschnig, Wolfgang 2 Wang, Xizhao 2 Wang, Yangeng 2 Wei, Guo 2 Wu, Rangquan 2 Wu, Zhihui 2 Zhang, Hanjun 2 Zhang, Li-Xin 2 Zhang, Zhongzhan 2 Zhu, Yixia 1 Abbas, Moncef 1 Abbas, Umber 1 Agliardi, Elettra 1 Agliardi, Rossella 1 Alonso de la Fuente, Miriam 1 Alsaadi, Fuad Eid S. 1 Al-saedi, Ahmed Eid Salem 1 Annamalai, Anguraj 1 Arumugam, Vinodkumar 1 Balch, Michael Scott 1 Băleanu, Dumitru I. 1 Bandyopadhyay, Abhirup 1 Beck, Jan B. 1 Beranger, Boris 1 Beresteanu, Arie 1 Bilokon, Paul Alexander 1 Bogdan, Krzysztof 1 Candeloro, Domenico 1 Capasso, Vincenzo 1 Castaing, Charles 1 Cattaneo, Marco E. G. V. 1 Chang, Chia-Wen 1 Chen, Jiqiang 1 Choi, Gyeong Suk 1 Colonius, Fritz 1 Couso, Inés 1 Denœux, Thierry 1 Destercke, Sébastien 1 Driouchi, Tarik 1 Edalat, Abbas 1 Fan, Shengjun 1 Fei, Chen 1 Feinstein, Zachary 1 Figueroa-García, Juan Carlos 1 Gao, Yongling 1 Gashi, Bujar 1 Ghasemi, Reza 1 Godet-Thobie, Christiane 1 Hao, Zhiqi 1 Hoang, Thi Duyen 1 Hong, Dug Hun 1 Hong, Jiarong 1 Joo, Sang Yeol 1 Kandasamy, Banupriya 1 Kar, Samarjit 1 Katagiri, Hideki ...and 96 more Authors all top 5 #### Cited in 64 Serials 38 Fuzzy Sets and Systems 13 Information Sciences 9 Journal of Mathematical Analysis and Applications 8 International Journal of Approximate Reasoning 7 Stochastic Analysis and Applications 7 International Journal of Uncertainty, Fuzziness and Knowledge-Based Systems 6 Statistics & Probability Letters 3 Tohoku Mathematical Journal. Second Series 3 Computational Statistics and Data Analysis 2 Journal of the Franklin Institute 2 Kybernetika 2 Systems & Control Letters 2 Optimization 2 Soft Computing 2 Stochastics and Dynamics 2 Fuzzy Optimization and Decision Making 2 Advances in Difference Equations 1 Applicable Analysis 1 Computer Methods in Applied Mechanics and Engineering 1 International Journal of Control 1 International Journal of General Systems 1 Mathematical Methods in the Applied Sciences 1 Metrika 1 The Annals of Probability 1 Applied Mathematics and Computation 1 Czechoslovak Mathematical Journal 1 Illinois Journal of Mathematics 1 Journal of Computational and Applied Mathematics 1 Journal of Differential Equations 1 Journal of Econometrics 1 Journal of Multivariate Analysis 1 Kybernetes 1 Nonlinear Analysis. Theory, Methods & Applications. Series A: Theory and Methods 1 Operations Research 1 Proceedings of the American Mathematical Society 1 Ricerche di Matematica 1 Theoretical Computer Science 1 Acta Mathematicae Applicatae Sinica. English Series 1 Asia-Pacific Journal of Operational Research 1 Applied Mathematics Letters 1 Mathematical and Computer Modelling 1 Communications in Statistics. Theory and Methods 1 European Journal of Operational Research 1 Linear Algebra and its Applications 1 Stochastic Processes and their Applications 1 Applied Mathematics. Series B (English Edition) 1 Statistical Papers 1 Electronic Journal of Probability 1 Electronic Communications in Probability 1 Finance and Stochastics 1 Abstract and Applied Analysis 1 Journal of Inequalities and Applications 1 Acta Mathematica Sinica. English Series 1 Quantitative Finance 1 Journal of Systems Science and Complexity 1 OR Spectrum 1 Central European Journal of Mathematics 1 Mediterranean Journal of Mathematics 1 Nonlinear Analysis. Hybrid Systems 1 Advances in Fuzzy Systems 1 Set-Valued and Variational Analysis 1 Symmetry 1 Statistics and Computing 1 Open Mathematics all top 5 #### Cited in 25 Fields 104 Probability theory and stochastic processes (60-XX) 27 Measure and integration (28-XX) 22 Statistics (62-XX) 19 Mathematical logic and foundations (03-XX) 16 Computer science (68-XX) 16 Systems theory; control (93-XX) 14 Real functions (26-XX) 13 Ordinary differential equations (34-XX) 9 Game theory, economics, finance, and other social and behavioral sciences (91-XX) 8 Functional analysis (46-XX) 8 Operations research, mathematical programming (90-XX) 6 Dynamical systems and ergodic theory (37-XX) 6 Numerical analysis (65-XX) 5 Operator theory (47-XX) 4 Partial differential equations (35-XX) 3 Calculus of variations and optimal control; optimization (49-XX) 3 General topology (54-XX) 3 Information and communication theory, circuits (94-XX) 2 Integral equations (45-XX) 2 Global analysis, analysis on manifolds (58-XX) 1 Linear and multilinear algebra; matrix theory (15-XX) 1 Group theory and generalizations (20-XX) 1 Potential theory (31-XX) 1 Mechanics of deformable solids (74-XX) 1 Geophysics (86-XX)

2021-07-28 01:44:16

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https://www.amplifiedparts.com/products/vacuum_tubes?sort=recent&filters=1a646c2341a2391b2383

# Vacuum Tubes The first electronic amplification of sound was done with Vacuum Tubes. We have pre-amp, power and rectifier vacuum tubes among other types in New Old Stock from the golden era of American & European manufacturing as well as current production tubes from today. #### JJ Electronics JJ vacuum tubes and capacitors are widely used in guitar and audio amplifiers, recording studio equipment and in a variety of applications for audiophiles. #### Vacuum Tubes The first electronic amplification of sound was done with Vacuum Tubes. We have pre-amp, power and rectifier vacuum tubes among other types in New Old Stock from the golden era of American & European manufacturing as well as current production tubes from today. Vacuum Tube - E83CC Frame Grid, JJ Electronics Alternative to the 12AX7 or ECC83. The JJ Electronic E83CC is a high quality dual triode that rivals the NOS version of this tube. The E83CC was originally designed as an audio specific dual triode with very low noise and microphonics. The JJ E83CC stays true to the original design with the signature thick frame grid and boxed plate. The rigid frame allows for tighter tolerances and closer construction which results in lower noise and increased stability. The JJ Electronic E83CC is a great choice for high fidelity amplifiers as well as guitar amps and studio equipment. The E83CC has the same pinout and amplification factor as a standard 12AX7 and can be used in any 12AX7 or ECC83 position. Guitarist Description: This frame grid tube has an incredibly rich yet balanced tone. The E83CC works well for clean sounds but also breaks up with a satisfying warmth and plenty of articulation. $21.95 On Backorder Vacuum Tube Set - 12AX7 Sampler, 10 total tubes Package of 10 Ten 12AX7s. One of each of the following: JJ Electronic 12AX7, Mullard 12AX7, Tube Amp Doctor premium selected 12AX7, Ruby Tubes 12AX7, Sovtek 12AX7WA, JJ Electronic ECC803S, Chinese high-grade 12AX7B, Groove Tubes 12AX7, Electro-Harmonix 12AX7, Tung-Sol 12AX7$139.94 Vacuum Tube Sampler Set - EL84, 9 tubes total Package of 9 Sampler set of EL84 vacuum tubes. Contains one each of the following: Electro-Harmonix EL84, Genalex Gold Lion EL84, JJ Electronic EL84, Mullard Reissue EL84, Sovtek EL84, Tung-Sol EL84, JJ Electronic EL844, Sovtek EL84M, Tube Amp Doctor EL84 $139.75 Vacuum Tube - 6SN7, JJ Electronics The JJ Electronic 6SN7 is the classic double triode with a rugged construction the offers the high reliability that JJ is known for. The JJ 6SN7 features the outstanding linearity and sound of the original tubes. This tube is an excellent replacement for Hi-Fi amplifiers and preamps and will replace any 6SN7, 6SN7GTA, 6SN7GTB, or 6SN7GT.$16.50 Vacuum Tube - 12BH7-A, JJ Electronics The JJ Electronic 12BH7-A is a faithful reproduction of the original tube. Its rugged construction offers high reliability. The JJ 12BH7-A can be used as a replacement for any 12BH7 tube. $16.95 Vacuum Tube - 5881, JJ Electronics The JJ Electronic 5881 is very similar to the 6L6GC but with a plate dissipation of 23W rather than 30W. Another physical difference is the shorter bulb. These tubes have a soft and balanced response that sounds great in both guitar amps and hifi equipment. The JJ 5881 will work in any amp that uses 6L6 tubes. Starting at$18.95 Vacuum Tube - 12AX7S, JJ Electronics, Mid Gain A new take on the classic 12AX7 preamp tube. While it uses the same rugged build and thick glass as other JJ preamp tubes, this tube differs in construction from the typical 12AX7 in that it features a medium-long plate, which provides a dynamic and open sound with a rich forward sounding mid range. This tube offers low noise and low microphonics. $12.45 Vacuum Tube - 12AX7 / ECC83, JJ Electronics, package of 3 Package of 3 This tube has a well balanced, colorful tone with strongly defined lows, mids and smooth highs. It allows for more clean head-room than higher gain 12AX7s. In overdrive, it is smooth and strong with well defined lows and mids. When pushed into overdrive it offers clean distortion with well balanced lows and mids. The JJ 12AX7 is well suited for all types of music and playing styles. It is also highly recommended for studio pre-amps and hi-fi gear$32.85 Vacuum Tube - 5751, JJ Electronics A dual triode that can be used in place of most 12AX7 circuits. The rugged build of this tube and reduced gain (30% less than a 12AX7) can improve headroom and give better control in high gain stages allowing users to more easily dial in sweet spots. The JJ 5751 features a smooth and balanced tone and a detailed response with low noise and low microphonics. $13.75 Vacuum Tube - 5Y3 S, JJ Electronics, Rectifier The JJ Electronic 5Y3S is a ruggedly built octal rectifier tube with a directly heated cathode. This tube’s solid construction and thick glass envelope offer high reliability. The JJ 5Y3S will work in any 5Y3 position.$15.50 Vacuum Tube - 6L6 CHP, Groove Tubes, Black Plate, Matched Pair Octal power tube (Max Plate Watts = 30W) EXCLUSIVE GT design. This black-plate (carbonized nickel) design features extended heatsink wings. Has strong and focused output, and is highly recommended for enlivening any stock Fender® or Marshall® that uses a standard 6L6. High-performance design delivers 30 watts, with very high performance/headroom. An outstanding replacement for any tube amp with 6L6 power tubes. $50.00 Vacuum Tube - 6L6GC, JJ Electronics The JJ Electronic 6L6GC is the classic pentode built with the rugged construction that only JJ could deliver. At 30W, this tube delivers plenty of headroom. The highs are clear and responsive with warm mids and tight solid lows. Known for their reliability and articulation, the JJ 6L6GC is a great option guitar amplifiers and Hi-Fi stereo equipment. Starting at$17.50 Vacuum Tube - 6L6 GE Reissue, Groove Tubes, Matched Pair Octal power tube (Max Plate Watts = 30W) EXCLUSIVE GT design. The GE is a faithful reproduction of the original "clear top" 6L6. The flagship of the GT 6L6 line, it looks and sounds exactly like the original G.E. tube used by Fender® in the '50s and '60s. It took more than four years to develop, and has stunning clarity and power. $52.00 Vacuum Tube - 6V6 C, Groove Tubes, Matched Pair Octal power tube (Max Plate Watts = 14W) Classic design and affordable. Warm overdrive and soft distortion character. The 6V6 is a low-output power tube (12 - 18 watts) often used in Fender® amplifiers, including the Champ®, Deluxe™, Dual Professional, Vibro-King®, Princeton®, Vibrolux® and others. It has a very warm, round and soft response with rich harmonics, and distorts easily. One of the best choices ever produced for rehearsal and recording amplifiers.$40.00 Vacuum Tube - 6V6, JJ Electronics This 6V6 has the standard rugged construction we’ve come to expect from JJ which allows it handle much higher plate voltages than the typical 6V6 tube. The JJ 6V6 can even be used in place of a 6L6 in some amplifiers. Overall this tube has a warm and balanced tone with incredible separation and response. These tubes can cover a spectrum of sounds and styles all within the same amplifier. Starting at $14.95 Vacuum Tube - 6V6-SD-M, Groove Tubes, Matched Pair The 6V6 is a low-output power tube (12-18 watts) often used in Fender® amplifiers, including the Champ®, Deluxe™, Dual Professional, Vibro-King®, Princeton®, Vibrolux® and others. It has a very warm, round and soft response with rich harmonics, and distorts easily. One of the best choices ever produced for rehearsal and recording amplifiers. Our most premium design, for long life and professional performance. The choice for many pro players who use Groove Tubes in their favorite amps for recording and rehearsal; fully captures the spirit and sound of tube amps from the '50s and '60s to the present.$55.00 On Backorder Vacuum Tube - 7027A, JJ Electronics The JJ Electronic 7027A offers the same reliability and tone as the JJ 6L6GC but with the correct pinout for vintage amplifiers that use 7027 tubes like older Ampegs. At 30W, this tube delivers plenty of headroom. The highs are clear and responsive with warm mids and tight solid lows. Known for their reliability and articulation, the JJ 7027A is a great option guitar amplifiers and Hi-Fi stereo equipment. Starting at $21.95 Vacuum Tube - 7591, JJ Electronics The JJ Electronic 7591 is a reliable and accurate reproduction of the classic tube. This 19 Watt pentode has been used in a variety of amplifiers including Hi-Fi amps and Ampegs. The JJ 7591 is a great replacement for any 7591 tube position. Starting at$17.95 Vacuum Tube - E34L, JJ Electronics The JJ Electronic E34L adds more power to a standard EL34 due to its higher grid voltage rating. This increase gives the tube more headroom and a tighter response on the low end. The wide frequency response of this tube offers excellent harmonics that work in a variety of genres ranging from the cleanest cleans to rich and aggressive breakup. Starting at $13.95 Vacuum Tube - E34L, JJ Electronics, Blue Glass Blue glass envelope. The JJ Electronic E34L adds more power to a standard EL34 due to its higher grid voltage rating. This increase gives the tube more headroom and a tighter response on the low end. The wide frequency response of this tube offers excellent harmonics that work in a variety of genres ranging from the cleanest cleans to rich and aggressive breakup. Starting at$18.95 Vacuum Tube - ECC802 / 12AU7, JJ Electronics, Long Plate A high performance, long plate version of the ECC82 / 12AU7. Like all JJ preamp tubes, these are built into a rugged package using the spiral filament and robust plate structure resulting in low microphonics and high reliability. The long plate version of this tube differs from the standard version with its stronger and fuller sound which offers a wider and more balanced frequency response. Starting at $13.45 Vacuum Tube - ECC803 / 12AX7, JJ Electronics, Long Plate A high performance, long plate version of the ECC83 / 12AX7. Like all JJ preamp tubes, these are built into a rugged package using the spiral filament and robust plate structure resulting in low microphonics and high reliability. The long plate version of this tube differs from the standard version with its stronger and fuller sound. This tube will work in any 12AX7 or ECC83 position. Starting at$13.95 Vacuum Tube - ECC823, JJ Electronics, Dual Triode In the ECC823, Section 1 of the tube (pins 6, 7, and 8) is the low gain 12AU7/ECC82 spec (gain factor 17) and Section 2 (pins 1, 2, and 3) is the high gain 12AX7/ECC83 spec (gain factor 100). $12.95 Vacuum Tube - ECC99, JJ Electronics The JJ Electronic ECC99 is a noval based dual triode. This tube features a rugged construction and is also available with gold pins. The ECC99 is typically used as a driver for power triodes in 2A3 and 300B amplifiers but can also be used as a power tube in small amps and headphone amps. This tube can also be a functional replacement for 5687, E182CC, 6840, and 6BL7 tubes. Starting at$15.95 Don't see what you're looking for? Send us your product suggestions!

2019-11-15 03:05:59

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https://studydaddy.com/question/what-is-the-magnetic-quantum-number-of-the-19th-electron-of-calcium

Waiting for answer This question has not been answered yet. You can hire a professional tutor to get the answer. QUESTION # What is the magnetic quantum number of the 19th electron of Calcium? 0 of Calcium "Ca" is 20. We know that there are four which can be assigned to "describe" any electron in an atom. Each electron has a unique set of these four quantum numbers. n = principal quantum number, which defines the energy level of the electron. l = azimuthal (angular) quantum number defining the energy sub-level. Its values range from 0 " to " (n-1); 0=s, 1=p, 2=d, 3=f m(l) = magnetic quantum number indicating orbital within a sublevel. Range from -l" through "0" to "+l m(s) = spin quantum number. It identifies an electron within an orbital and can have either of the two values +1/2 and -1/2 ================================ The electronic configuration of calcium is: 1"s"^2; 2"s"^2, 2"p"^6; 3"s"^2, 3"p"^6; 4"s"^2 or ["Ar"] 4"s"^2 it is evident that the 19th electron goes to 4"s" energy level. As such for it n=4, l=0 We know that for l=0, m(l)=0 only. Hence magnetic quantum number m(l) of 19th electron of calcium is =0

2019-05-20 21:24:27

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http://mathhelpforum.com/advanced-algebra/274021-please-solve.html

Let S be a set of positive real numbers. If S contains at least four distinct elements show that there are elements x,y ∈ S such that $Given\ z(x,\ y) = \dfrac{x- y}{1 + xy}\ and\ x,\ y \in S,$ $\text{where S is a set of at least 4 distinct positive real numbers}$ $Prove\ \exists\ x,\ y \in\ S\ such\ that\ 0 < z(x,\ y) < \dfrac{\sqrt{3}}{3}.$ What can you say if S has at least 7 elements? What if S has at least n elements, where n > 2?

2017-06-26 05:45:38

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http://symplio.com/30pqg/316045-bfs-spanning-tree

In this tutorial, you will understand the spanning tree and minimum spanning tree with illustrative examples. Breadth-First Search ( or Traversal) also know as Level Order Traversal. BFS spanning tree March 30, 2020 Cinda Heeren / Andy Roth / Geoffrey Tien 11 b c h i a f d g e Root the BFS spanning tree at any vertex, e.g. Breadth-First Search Traversal Algorithm. What is Breadth First Search: Breadth-first search (BFS) is an algorithm for traversing or searching tree or graph data structures. In other words, Spanning tree is a non-cyclic sub-graph of a connected and undirected graph G that connects all the vertices together. }, it only requires $\Theta(1)$ bits per edge. In general, a graph may have more than one spanning tree. (a) Find a spanning tree in the graph below using DFS and BFS starting from vertex a and from vertex f. (b) Using Prims's and Kruskal's algorithm find the minimal spanning tree in the graph below (show the steps). A pseudocode version of the algorithm is as follows: 1: procedure bfs(V,E) 2: S := (v1) {ordered list of vertices of a fix level} 3: V’ := {v1} {v1 is the root of the spanning tree} 4: E’ := {} {no edges in the spanning tree yet} 5: while true Shortest Path And Minimum Spanning Tree In The Un-weighted Graph: BFS technique is used to find the shortest path i.e. Graph and tree traversal using Breadth First Search (BFS) algorithm. If we get one back-edge during BFS, then there must be one cycle. }, it only requires Θ(1) bits per edge. I think I understood the definition of BFS and DFS spanning tree, but I'm not sure my answer is right. But when you have more than 1 connected component , like in the picture, the only thing you have to do is start another BFS or DFS from an unvisited vertex . To visualize a spanning tree, first picture an undirected graph: for instance, a random collection of points connected by lines. • It finds a minimum spanning tree for a weighted undirected graph. Figure 3.12 shows the spanning tree that corresponds to the extended LAN shown in Figure 3.10.In this example, B1 is the root bridge, since it has the smallest ID. In this visualization, we also show that starting from the same source vertex s in an unweighted graph, BFS spanning tree of the graph equals to its SSSP spanning tree. 1 6 4 d с 2 5 3 6 3 5 f 3 9 2 e h 2 1 2 4 4 ] 1 k 5 2 3 6 4 4 3 2 6 6 0 n т P Similarly, we can also find a minimum spanning tree using BFS in the un-weighted graph. To avoid processing a node more than once, we use a … (Reference – Wiki) Example: Breadth first traversal or Breadth first Search is a recursive algorithm for searching all the vertices of a graph or tree data structure. Conclusion When there is only 1 connected component, your BFS or DFS will terminate visiting all the vertices and you will have a spanning tree (which in this case is equal to spanning forest). We use Stack data structure with maximum size of total number of vertices in the graph to implement DFS traversal. for storing the visited nodes of the graph / tree. Breadth First Search (BFS) is an algorithm for traversing an unweighted Graph or a Tree. Usually, a spanning tree formed by branching out from one of the inner points, which is why it is described as a tree. BFS (Breadth First Search) BFS (Breadth First Search) BFS traversal of a graph produces a spanning tree as final result. However, it has been proven assuming a weakly fair daemon. depth- rst: nextEdge selects a frontier edge whose tree endpoint was most recently discovered. The following figure shows a graph with a spanning tree. The main purpose of STP is to ensure that you do not create loops when you have redundant paths in your network. 1st row, then 2nd row, and so on. P2P Networks: BFS can be implemented to locate all the nearest or neighboring nodes in a peer to peer network. Graph Traversal (BFS & DFS), Single Source Shortest Path, Minimum Spanning Tree, RB Trees, B-Trees - addy689/DataStructuresLab SPANNING TREES 117 5. When you enter the command, all STP instances are stopped for the previous mode and are restarted in the new mode. DFS (Depth First Search ) − It is a tree traversal algorithm that traverses the structure to its deepest node. GRAPH THEORY { LECTURE 5: SPANNING TREES 7 DFS and BFS as Tree-Growing Preview of x4.2: The depth- rst and breadth- rst searches use opposite versions of nextEdge. f Mark vertices and edges used in BFS traversal b c h i a f d g e a b b a, c, h c b, d, f d c, f, g e g f c, d, i g d, e i h b, i i f, g, h Loops are deadly to a network. Find a DFS,BFS spanning tree. Breadth First Traversal (or Search) for a graph is similar to Breadth First Traversal of a tree (See method 2 of this post).The only catch here is, unlike trees, graphs may contain cycles, so we may come to the same node again. This algorithm constructs a BFS spanning tree in any connected rooted network. This algorithm constructs a BFS spanning tree in any connected rooted network. A spanning tree of G is a subgraph of G that is a tree containing every vertex of G. Theorem 1 A simple graph is connected if and only if it has a spanning tree. If we perform DFS on unweighted graph, then it will create minimum spanning tree for all pair shortest path tree; We can detect cycles in a graph using DFS. 2. The proof that this produces a spanning tree (the depth first search tree) is essentially the same as that for BFS, so I won't repeat it. However while the BFS tree is typically "short and bushy", the DFS tree is typically "long and stringy". BFS spanning tree constructions have a space complexit y in Ω(log (D)) bits per process. Take a look at the following graph: If we start from node a and want to visit every other … Breadth-First search (BFS) Bipartite test Suppose that the spanning tree generated by BFS algorithm is: What are the two set V1 and V2 Active 4 years, 3 months ago. En dit is het hoofd doel van het Spanning Tree Protocol. Het spanning Tree protocol is een layer twee techniek die netwerk loops ook wel broadcast storm genaamd voorkomt op je netwerk. Notice that both B3 and B5 are connected to LAN A, but B5 is the designated bridge since it is closer to the root. Viewed 2k times -2 $\begingroup$ Is my answer right? The specification for STP is IEEE 802.1D. BFS also uses a Boolean array of size V vertices to distinguish between two states: visited and unvisited vertices (we will not use BFS to detect back edge(s) as with DFS). BFS (Breadth First Search) − It is a tree traversal algorithm that is also known as Level Order Tree Traversal.In this traversal we will traverse the tree row by row i.e. The edges of the spanning tree are in red: 3. In contrast, the solution given in [Joh97] does not compute any distance v alue (actually, 2 Dit alles gebeurd op layer twee niveau. It starts at the tree root and explores the neighbor nodes first, before moving to the next level neighbors. Minimum Spanning Tree Spanning Tree is a graph without loops. Nowadays, it is still the best existing self-stabilizing BFS spanning tree construction in terms of memory requirement, {\em i.e. 07/18/19 - We present results on the last topic we collaborate with our late friend, Professor Ajoy Kumar Datta (1958-2019). We want to show that we will get a spanning... Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 4 Creating a Random Maze We can use the algorithm to compute a spanning tree for creating a random maze. In spanning_tree, the parent and rank dictionaries can be initialized directly: parent = {v: v for v in self._g} rank = {v: 0 for v in self._g} The parent data structure is known as a disjoint-set forest. spanning-tree-mode. Spanning Tree Protocol (STP) is a Layer 2 protocol that runs on bridges and switches. Using DFS we can find path between two given vertices u and v. Spanning Tree is a graph without loops. A spanning tree of an undirected graph G is a connected subgraph that covers all the graph nodes with the minimum possible number of edges. • Prim's algorithm is a greedy algorithm. Ask Question Asked 4 years, 3 months ago. Spanning Tree. Just like we did for BFS, we can use DFS to classify the edges of G into types. 8.5. Depth-first search (DFS) is an algorithm for traversing or searching tree or graph data structures. BFS makes use of Queue. Go to step 2. We use Queue data structure with maximum size of total number of vertices in the graph to implement BFS traversal. 11.4 Spanning Trees Spanning Tree Let G be a simple graph. spanning tree, which we know because we have added n 1 = 4 edges. A graph G can have multiple spanning trees. Met het Spanning Tree Protocol kan je ook redundantie tussen verschillende switchen gaan opbouwen. A spanning tree is a sub-graph of an undirected and a connected graph, which includes all the vertices of the graph having a minimum possible number of edges. Minimum Spanning Tree (Unweighted Graph) A minimum spanning tree describes a path that contains the smallest number of edges that are needed to visit every node in the graph. The spanning tree is complete. Implementation of Prim's algorithm for finding minimum spanning tree using Adjacency list and min heap with time complexity: O(ElogV). the path with the least number of edges in the un-weighted graph. BFS starts with the root node and explores each adjacent node before exploring node(s) at the next level. breadth- rst: nextEdge selects a frontier edge whose tree endpoint BFS algorithm can easily create the shortest path and a minimum spanning tree to visit all the vertices of the graph in the shortest time possible with high accuracy. Caution Be careful when using the spanning-tree mode command to switch between Rapid PVST+ and MST modes. BFS (Breadth First Search) DFS (Depth First Search) DFS traversal of a graph produces a spanning tree as final result. Depth-First Search A spanning tree can be built by doing a depth-first search of the graph. In this tutorial, you will understand the working of bfs algorithm with codes in C, C++, Java, and Python. Detailed explanation. Spanning tree can be defined as a sub-graph of connected, undirected graph G that is a tree produced by removing the desired number of edges from a graph. Nowadays, it is still the best existing self-stabilizing BFS spanning tree construction in terms of memory requirement, {\em i.e. However, it has been proven assuming a weakly fair daemon. B readth-first search is a way to find all the vertices reachable from the a given source vertex, s.Like depth first search, BFS traverse a connected component of a given graph and defines a spanning tree. View session 1.pptx from LEBANESE U 1 at Université Libanaise. If we tried to continue, the next edge BE could not be added because it does not connect two trees, and neither can CE. Back-Edge during BFS, we can also find a minimum spanning tree Let G be a simple graph total of. The BFS tree is a Layer 2 bfs spanning tree that runs on bridges and switches by lines DFS traversal edge... 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Using Breadth First traversal or Breadth First Search ( BFS ) is a non-cyclic of... − it is still the best existing self-stabilizing BFS spanning tree and minimum spanning tree is typically and! The working of BFS and DFS spanning tree Protocol kan je ook redundantie tussen verschillende gaan. Command, all STP instances are stopped for the previous mode and are in! We get one back-edge during BFS, then there must be one cycle rst: nextEdge selects a edge... A BFS spanning tree Protocol kan je ook redundantie tussen verschillende switchen gaan opbouwen to the! With our late friend, Professor Ajoy Kumar Datta ( 1958-2019 ) short and bushy '' the. As final result: BFS technique is used to find the shortest path i.e finds a minimum tree! ( log ( D ) ) bits per process ensure that you do not create loops when have! A weighted undirected graph: if we start from node a and want to visit every other instances are for. A spanning tree are in red: 3 BFS in the un-weighted graph as... $\Theta ( 1 ) bits per edge ask Question Asked 4 years, 3 ago. Then 2nd row, and so on using BFS in the new mode C++ Java... On the last topic we collaborate with our late friend, Professor Ajoy Datta... In this tutorial, you will understand the spanning tree as final result locate all the vertices together traversal... Is my answer is right min heap with time complexity: O ( )... Of total number of edges in the graph / tree Maze we use! Of memory requirement, { \em i.e on the last topic we collaborate with our late,. Search ( or traversal ) also know as level Order traversal traversal of a and. Search: breadth-first Search ( or traversal ) also know as level Order traversal ) − it is the! Order traversal C++, Java, and so on in your network and stringy.! ( ElogV ) it starts at the next level neighbors, C++ Java.: breadth-first Search ( DFS ) is an algorithm for traversing or searching tree or data. 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A recursive algorithm for finding minimum spanning tree for Creating a random.. Nowadays, it is still the best existing self-stabilizing BFS spanning tree in any connected rooted network {., Professor Ajoy Kumar Datta ( 1958-2019 ) node ( s ) at the following graph: for instance a... Tree using BFS in the un-weighted graph structure to its deepest node of Prim 's algorithm for searching the! It is still the best existing self-stabilizing BFS spanning tree construction in of... Every other '', the DFS tree is typically short and bushy '', DFS... A non-cyclic sub-graph of a graph or a tree -2$ \begingroup $is my answer is right tree spanning! This algorithm constructs a BFS spanning tree construction in terms of memory requirement, { \em i.e ensure that do! Of Prim 's algorithm for traversing or searching tree or graph data structures graph. A depth-first Search of the graph to implement DFS traversal of a and... Mode command to switch between Rapid PVST+ and MST modes path with the least number bfs spanning tree... Met het spanning tree is typically short and bushy '', the DFS tree is non-cyclic! Points connected by lines non-cyclic sub-graph of a connected and undirected graph G that connects all the vertices a... Be a simple graph we use Stack data structure with maximum size of total number of vertices the... Long and stringy '' requires Θ ( 1 )$ bits per edge bushy '', the DFS tree typically. Redundantie tussen verschillende switchen gaan opbouwen First, before moving to the next level neighbors any rooted... In a peer to peer network tree, First picture an undirected graph G that connects all the nearest neighboring. Asked 4 years, 3 months ago instance, a graph with a spanning tree can be implemented locate! Path with the root node and explores each adjacent node before exploring node ( s ) at the figure. Are stopped for the previous mode and are restarted in the new mode searching tree graph., Professor Ajoy Kumar Datta ( 1958-2019 ) will understand the spanning tree all STP instances are stopped for previous. Be built by doing a depth-first Search of the spanning tree, First an! As final result the tree root and explores the neighbor nodes First, before moving to the next level used... First, before moving to the next level PVST+ and MST modes data. Root node and explores each adjacent node before exploring node ( s ) at the next level my answer?. Produces a spanning tree construction in terms of memory requirement, { \em i.e an graph. 1 = 4 edges one cycle terms of memory requirement, { i.e... You will understand the spanning tree Protocol ( STP ) is an algorithm for traversing or searching tree or data! Total number of vertices in the un-weighted graph to locate all the of! Answer right sure my answer is right restarted in the un-weighted graph we can use DFS to the! Typically long and stringy '' structure with maximum size of total number vertices! Use Stack data structure with maximum size of total number of edges in the un-weighted graph peer network the mode! Bfs ) is an algorithm for finding minimum spanning tree in the un-weighted graph je ook tussen. Are stopped for the previous mode and are restarted in the graph to implement DFS traversal a! The visited nodes of the spanning tree for Creating a random Maze we can find! It starts at the next level neighbors ( Breadth First Search ) DFS traversal of a produces... Doel van het spanning tree in any connected rooted network unweighted graph tree! Understand the working of BFS algorithm with codes in C, C++ Java! Understand the spanning tree for Creating a random collection of points connected by lines your network not sure answer! You have redundant paths in your network bridges and switches long stringy. Structure with maximum size of total number of vertices bfs spanning tree the un-weighted graph: for instance, graph. Enter the command, all STP instances are stopped for the previous mode and are restarted in the graph... Time complexity: O ( ElogV ) understood the definition of BFS algorithm with codes in C, C++ Java... First, before moving to the next level sub-graph of a connected and graph. The working of BFS and DFS spanning tree, which we know we... To the next level neighbors requirement, { \em i.e long and stringy.... If we get one back-edge during BFS, we can use DFS classify! Or tree data structure ( or traversal ) also know as level Order traversal a recursive for. Least number of edges in the un-weighted graph: for instance, a graph produces a spanning tree using list. More than one spanning tree bfs spanning tree G be a simple graph the vertices together locate all the vertices.... Stringy '' Java, and Python − it is still the best existing self-stabilizing BFS spanning Protocol! Traverses the structure to its deepest node to visit every other Layer 2 Protocol runs! Professor Ajoy Kumar Datta ( 1958-2019 ) the new mode van het spanning tree, First picture an undirected:...

2021-03-04 04:08:15

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https://forum.bebac.at/mix_entry.php?id=21483&order=time&view=mix

Helmut ★★★ Vienna, Austria, 2020-05-26 23:47 (182 d 21:46 ago) Posting: # 21483 Views: 1,597 Simulations 101 [Power / Sample Size] Dear all, off-list I was asked about some discrepancies between sample sizes estimated by PowerTOST’s functions sampleN.scABEL(), sampleN.RSABE() and the tables of the Two Lászlós1 ( Charlie). • Short answer Not relevant in practice. • Longer answer Unless the same  (pseudo-random) number generator with the same  seed and the same number of simulations is used, the outcome will always differ. The Two Lászlós simulated 10,000 studies, whereas in the functions of PowerTOST 100,000 are simulated by default. Excursion: An introduction to Monte Carlo Simulations. With a reasonably large number of simulations the empiric value will converge to the true value. How fast a sufficiently accurate result is obtained, depends on the problem. Hence, it is a good idea to assess that before you set the number of simulations. See IV and V below why we opted for 100,000 as the default in PowerTOST. 1. Rolling two dice Here it’s easy to get the average sum of pips. If you remember probability and combinatorics, you will get the answer 7. If you are a gambler, you simply know it. If you have two handy, roll them. Of course, you will get anything between the possible extremes (2 and 12) in a single roll. But you will need many rolls to come close to the average 7. Since this is a boring excercise (unless you find an idiot betting against 7), let do the job: 2. Estimating π We generate a set of uniformly distributed coordinates {x|y} in the range {–1, +1}. According to Mr Pythagoras all x2 + y2 ≤ 1 are at or within the unit circle and coded in blue. Since $$A = \frac{\pi \, d^2}{4}$$ and $$d = 1$$, we get an estimate of π by 4 times the number of coded points (the “area”) divided by the number of simulations. Here for 50,000: This was a lucky punch since the convergence is lousy. To get a stable estimate with just four significant digits, we need at least 100 million (‼) simulations. Not very efficient. 3. Type I Error of TOST We know that α (probability of the Type I Error) never exceeds the nominal level of the test (generally 0.05). In e.g., a 2×2×2 crossover with CV 20% and 20 subjects it is 0.049999895… We can simulate studies with a GMR at one of the borders of the acceptance range and assess them for ABE as well. We expect ≤5% to pass. The number of passing studies divided by the number of simulations gives the empiric Type I Error. I used the function power.TOST.sim(). The filled circles are runs with a fixed seed of 123456 and open circles ones with random seeds. We see that convergence of this problem is bad. Therefore, e.g., in Two-Stage Designs 106 simulations are common. 4. Sample size for ABEL (EMA) Table A21 gives for a 4-period full replicate study (EMA method), GMR 0.85, CV 50%, and ≥80% power a sample size of 54, whilst sampleN.scABEL() gives only 52. Five runs. As before filled circles are with a fixed seed (default) and open circles with random seeds. With sampleN.scABEL() and only 10,000 simulations I got* five times 52 and once 54 (like The Lászlós). If you have more runs, you will get in some of them more extreme ones (in 1,000 runs of 10,000 I got x̃ 52 and a range of 48–56). However, this range shows that the result is not stable. Hence, we need more simulations to end up with 52 as a stable result. Although you are free to opt for more than 100,000 it’s a waste of time. 5. Sample size for RSABE (FDA) Table A41 gives for a 4-period full replicate study (FDA method), GMR 0.90, CV 70%, and ≥90% power a sample size of only 46, whilst sampleN.RSABE() gives 48. Filled circles are with a fixed seed (default) and open circles with random seeds. With sampleN.RSABE() and 10,000 simulations I got* once 46 (like The Lászlós), once 48, and thrice 50. In 1,000 runs of 10,000 I got x̃ 48 and a range of 44–52. With more simulations we end up with 48 as a stable result. • Long answer for geeks László Tóthfalusi coded in MatLab. Though – like in  – the Mersenne Twister is the standard (pseudo-random) number generator, its implementation might differ. Even with the same seed and the same number of simulations different results may be obtained. For a comparison with the tables, execute the scripts test_ABEL.R and test_RSABE.R in the folder /library/PowerTOST/tests. Patience, please; esp. the first one takes ages to complete. If you want to get a faster result, change in test_ABEL.R lines 71 and 80 from sdsims = TRUE to sdsims = FALSE. Another discrepancy arises from the fact that all sample size functions of PowerTOST return balanced sequences, whereas the Two Lászlós reported the smallest sample size which gives at least the desired power. Therefore, in Tables A1/A3 you find some even numbers (for three sequences) and in Tables A2/A4 some odd numbers (for two sequences), whereas sampleN.RSABE() and sampleN.scABEL() will round up to the next multiple of three and two, respectively. The Two Lászlós simulated subjects, whereas in sampleN.RSABE()/sampleN.scABEL() the ‘key’ statistics2 are used (GMR lognormal-distributed and s2 χ2-distributed). The latter approach is easily 100 times faster than the former. • For ABEL subject simulations are provided by the function sampleN.scABEL.sdsims(). Only if you suspect heterogenicity (i.e., give CV as a vector with two elements, where the first one is CVwT and the second CVwR), you plan a partial replicate design, and CVwT > CVwR, this is the method of choice. If CVwT ≤ CVwR and for full replicate designs we recommend sampleN.scABEL() for speed reasons. See also the vignette. • Sorry, subject simulations are not  implemented  possible for the FDA’s RSABE. Since the FDA’s mixed-effects model for ABE (applicable if swR < 0.294) cannot be set up in , intra-subject contrasts are used. However, it is a reasonably close approximation. For a comparison with SAS see page 16 of the file Implementation_scaledABE_sims.pdf in the folder /library/PowerTOST/doc. Furthermore, likely they used the rounded regulatory constant $$\small{0.893}$$ and not the exact $$\small{\log_{e}(1.25)/0.25=0.8925742\ldots}$$ like sampleN.RSABE(). You can try reg         <- reg_const("FDA") reg$r_const <- 0.893 reg$name    <- "The Laszlos\u2019 likely" and call sampleN.RSABE(..., regulator = reg). In re-calculating the entire tables with 100,000 sim’s I found five of the 352 scenarios where the sample size would be by one sequence smaller. However, I would not go that way because if you will evaluate the study with the SAS-code of the progesterone guidance or the template for Phoenix/WinNonlin, the exact regulatory constant will be used. Another story is at which CV (on the average) we will switch in the simulations. You can try the following code to compare sample sizes for ABE and RSABE: library(PowerTOST) fun <- function(x) {   n.1 <- sampleN.TOST(CV = x, theta0 = theta0, targetpower = target,                       design = design, details = FALSE,                       print = FALSE)[["Sample size"]]   n.2 <- sampleN.RSABE(CV = x, theta0 = theta0, targetpower = target,                        design = design, details = FALSE,                        print = FALSE)[["Sample size"]]   return(n.1- n.2) } theta0 <- 0.9 target <- 0.8 design <- "2x2x4" cat("Equal sample sizes for ABE and RSABE at CV \u2264",     signif(uniroot(fun, interval=c(0.2, 0.35),     extendInt="yes")$root, 3), "\n") Given that, it hardly matters for true HVD(P)s. Note that the tables of the Two Lászlós don’t reach below 30% and they stated in the discussion: • In view of the consequences of the mixed approach, it could be judicious to consider larger numbers of subjects at variations fairly close to 30%. If you want to reproduce simulations, keep the default argument sampleN.foo(..., setseed = TRUE). Recommended, unless you want to assess the performance and convergence. However, results with 100,000 simulations are extremely stable. For the ABEL example I got in 1,000 runs with random seeds 995 times 52 and just five times 54. 1. Tóthfalusi L, Endrényi L. Sample Sizes for Designing Bioequivalence Studies for Highly Variable Drugs. J Pharm Pharmaceut Sci. 2011;15(1):73–84. Open access. 2. Zheng C, Wang J, Zhao L. Testing bioequivalence for multiple formulations with power and sample size calculations. Pharmaceut. Stat. 2012;11(4):334–341. doi:10.1002/pst.1522. • If you perform five runs with 10,000 simulations, you might get different numbers. That’s the effect of random seeds and a low number of simulations. Dif-tor heh smusma 🖖 Helmut Schütz The quality of responses received is directly proportional to the quality of the question asked. 🚮 Science Quotes d_labes ★★★ Berlin, Germany, 2020-05-27 19:30 (182 d 02:03 ago) @ Helmut Posting: # 21485 Views: 1,048 Simulations 101 Dear Helmut! Can't enough! One small correction: The graphics under "Sample size for RSABE" is apparent not correct here. Regards, Detlew Helmut ★★★ Vienna, Austria, 2020-05-27 21:06 (182 d 00:27 ago) @ d_labes Posting: # 21486 Views: 1,044 Simulations 101 Dear Detlew! » Can't enough! THX! » One small correction: The graphics under "Sample size for RSABE" is apparent not correct here. Hhm, why? library(PowerTOST) library(magicaxis) runs <- 5 nsims <- as.integer(10^(seq(3, 6, 0.1))) res <- data.frame(nsims = nsims) for (k in seq_along(nsims)) { for (j in 1:runs) { ifelse (j == 1, setseed <- TRUE, setseed <- FALSE) res[k, j+1] <- sampleN.RSABE(theta0 = 0.9, CV = 0.7, nsims = nsims[k], setseed = setseed, design = "2x2x4", targetpower = 0.9, print = FALSE, details = FALSE)[["Sample size"]] } } ylim <- range(res[, 2:(runs+1)], na.rm = TRUE) dev.new(width = 5, height = 5) op <- par(no.readonly = TRUE) par(pty = "s") plot(res$nsims, res[, 1], type = "n", log = "x", ylim = ylim, axes = FALSE,      xlab = "Number of simulations", ylab = "Sample size") grid(); box() magaxis(1) magaxis(2, minorn = 2, las = 1) for (j in 1:runs) {   ifelse (j == 1,  pch <- 16, pch <- 1)   points(res\$nsims, res[, j+1], pch = pch, col = "blue", cex = 1.15) } points(1e5, 48, pch = 16, col = "#00AA00", cex = 1.15) # default points(1e4, 46, pch = 16, col = "red", cex = 1.15)     # Lászlós par(op) print(res, row.names = FALSE)    nsims V2 V3 V4 V5 V6     1000 50 48 54 52 50     1258 50 48 52 50 52     1584 50 50 52 50 56     1995 50 48 50 50 50     2511 48 48 48 48 52     3162 48 50 50 50 50     3981 46 48 50 50 48     5011 46 50 50 50 48     6309 46 48 48 48 50     7943 46 50 48 48 50    10000 46 50 48 50 50    12589 46 50 48 50 48    15848 48 48 48 48 48    19952 48 48 50 50 48    25118 48 48 50 48 48    31622 48 48 48 48 48    39810 48 48 50 48 50    50118 48 48 48 48 48    63095 48 48 48 48 48    79432 48 48 48 48 48   100000 48 48 48 48 48   125892 48 48 48 48 48   158489 48 48 48 48 48   199526 48 48 48 48 48   251188 48 48 48 48 48   316227 48 48 48 48 48   398107 48 48 48 48 48   501187 48 48 48 48 48   630957 48 48 48 48 48   794328 48 48 48 48 48  1000000 48 48 48 48 48 Dif-tor heh smusma 🖖 Helmut Schütz The quality of responses received is directly proportional to the quality of the question asked. 🚮 Science Quotes d_labes ★★★ Berlin, Germany, 2020-05-28 18:34 (181 d 02:59 ago) @ Helmut Posting: # 21487 Views: 950 Simulations 101 Dear Helmut! » » One small correction: The graphics under "Sample size for RSABE" is apparent not correct here. » » Hhm, why? Today it looks reasonable. Maybe I had a Halunkination ). Regards, Detlew Helmut ★★★ Vienna, Austria, 2020-05-29 13:30 (180 d 08:02 ago) @ Helmut Posting: # 21490 Views: 934 More about the tables Dear sample size geeks, nerds, and simul-ants (weirdos for short), for each combination of design, power, GMR, and CV: Red circles = sample sizes of the Two Lászlós (if >201 none). Green bars = sampleN.scABEL(..., nsims = 1e5) or sampleN.RSABE(..., nsims = 1e5); always balanced sequences. Blue circles = twenty runs simulated with 10,000 sim’s each (random seeds). Starting from the respective result, I decreased the sample size whilst maintaining power. This might give unbalanced sample sizes (even sample sizes for the partial replicate and odd ones for the full replicate). ABEL (EMA) • 3-period 3-sequence partial replicate [TRR|RTR|RTT] • Power 80% • Power 90% • 4-period 2-sequence full replicate [TRTR|RTRT] • Power 80% • Power 90% Now a comparison of the reported sample sizes (y-axis) vs. PowerTOST’ (x-axis). Blue from the tables, and red up-rounded to the next complete sequence. Do the linear fits closer to the unity line? Sometimes yes, sometimes not. • 3-period 3-sequence partial replicate [TRR|RTR|RTT] • Power 80% • Power 90% • 4-period 2-sequence full replicate [TRTR|RTRT] • Power 80% • Power 90% RSABE (FDA) • 3-period 3-sequence partial replicate [TRR|RTR|RTT] • Power 80% • Power 90% • 4-period 2-sequence full replicate [TRTR|RTRT] • Power 80% • Power 90% Comparison of the reported sample sizes (y-axis) vs. PowerTOST’ (x-axis). Blue from the tables, and red up-rounded to the next complete sequence. Do the linear fits get closer to the unity line? Generally yes. • 3-period 3-sequence partial replicate [TRR|RTR|RTT] • Power 80% • Power 90% • 4-period 2-sequence full replicate [TRTR|RTRT] • Power 80% • Power 90% The upper part of the RSABE sample sizes looks disturbing first. Explanation: Since the sample sizes are generally smaller than for ABEL, the many imbalanced sequences have a larger impact. However, rounding up to the next complete sequence (red circles) in the lower part gives a better match (closer to the unity line). Maybe you can make sumfink out of it. I still hold that 10,000 simulations are too few. Dif-tor heh smusma 🖖 Helmut Schütz The quality of responses received is directly proportional to the quality of the question asked. 🚮 Science Quotes Admin contact 21,210 posts in 4,426 threads, 1,481 registered users; online 6 (0 registered, 6 guests [including 4 identified bots]). Forum time: Wednesday 20:33 CET (Europe/Vienna) You can’t fix by analysis what you bungled by design.    Richard J. Light, Judith D. Singer, John B. Willett The Bioequivalence and Bioavailability Forum is hosted by Ing. Helmut Schütz

2020-11-25 19:33:34

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http://mathhelpforum.com/algebra/26525-finding-slope-2-points-print.html

# Finding slope from 2 points • January 21st 2008, 02:02 PM XIII13Thirteen Finding slope from 2 points I have points (-3,-4) and (1,4) working out the problem as $4+4/1+3$ leaves me with slope $m=2$ is this correct? When I graph it it just doesn't look right. • January 21st 2008, 10:22 PM Jhevon Quote: Originally Posted by XIII13Thirteen I have points (-3,-4) and (1,4) working out the problem as $4+4/1+3$ leaves me with slope $m=2$ is this correct? When I graph it it just doesn't look right.

2014-04-19 15:23:15

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https://ltwork.net/a-spinal-injury-victim-who-is-in-water-should-be-group-of--13559864

# A spinal injury victim who is in water should be: Group of answer choices left in the water as is until EMS personnel arrive. ###### Question: A spinal injury victim who is in water should be: Group of answer choices left in the water as is until EMS personnel arrive. immobilized and kept in the water. floated to shore and then stabilized on a spine board. floated to shore and carefully removed from the water. ### Make this into a smaller word problem7.4z−5(−1.6z+2.4):D Make this into a smaller word problem 7.4z−5(−1.6z+2.4) :D... ### Write an equation with the points (-2,-7) and (4,-7) Write an equation with the points (-2,-7) and (4,-7)... ### A verbal statement that represents the expression 4 + k A verbal statement that represents the expression 4 + k... ### Using a simple machine, a student is able to lift a 500n weight by applying only 100n. what is the mechanical Using a simple machine, a student is able to lift a 500n weight by applying only 100n. what is the mechanical advantage of the simple machine?... ### During John Marshall's term on the Supreme Court, federal and Court power expanded. Write a paragraph During John Marshall's term on the Supreme Court, federal and Court power expanded. Write a paragraph contrasting these decisions with Thomas Jefferson's stated views on the proper size and power of the federal government. right answers only please this is important... ### What is magnetic field???List its properties!! ​ What is magnetic field???List its properties!! ​... ### The graph of y=-0.2x² is ? the graph of y=x².there are options to answer this question.choose one of the best answers.a. The graph of y=-0.2x² is ? the graph of y=x².there are options to answer this question.choose one of the best answers.a. narrower than and opens in the same direction as b. wider than and opens in the same direction as.c. narrower than and opens in the opposite direction ofd. wider than and opens... ### The magnitude of the electrical force acting between a +2.4 × 10–8 c charge and a +1.8 × 10–6 The magnitude of the electrical force acting between a +2.4 × 10–8 c charge and a +1.8 × 10–6 c charge that are separated by 0.008 m is n, rounded to the tenths place.... ### To which group did the pilgrims belong? To which group did the pilgrims belong?... ### Suppose that an individual has a body fat percentage of 19.7% and weighs 131 pounds. How many pounds of his weight is made up of fat? Round your answer Suppose that an individual has a body fat percentage of 19.7% and weighs 131 pounds. How many pounds of his weight is made up of fat? Round your answer to the nearest tenth.... ### Let Z be a categorical variable with four levels: 1, 2, 3 and 4, and let X and Y be continuous. Using the following R output, Let Z be a categorical variable with four levels: 1, 2, 3 and 4, and let X and Y be continuous. Using the following R output, estimate the average predicted value of Y in each category of Z when X is 10.... ### There are 20 girls and 16 boys in a classroom. complete the table of ratios below. ​ There are 20 girls and 16 boys in a classroom. complete the table of ratios below. ​ $There are 20 girls and 16 boys in a classroom. complete the table of ratios below. ​$... ### Suppose ^ABC =^XYZ. What is the corresponding congruent part for each segment or angle ? Drag the answers into the boxes to match Suppose ^ABC =^XYZ. What is the corresponding congruent part for each segment or angle ? Drag the answers into the boxes to match each segment or angle... Look at the picture to answer please!! Need help asp :)) $Look at the picture to answer please!! Need help asp :))$... ### Airline passengers arrive randomly and independently at the passenger-screening facility at a major Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The mean arrival rate is 10 passengers per minute. 1. Compute the probability of no arrivals in a one-minute period. 2. Compute the probability that three or fewer passengers ar... ### The temperature to which a mass of air must cool in order to be saturated is The temperature to which a mass of air must cool in order to be saturated is... ### A. a loud crashing sound causes everyone in the room to quickly turn their heads toward the source of A. a loud crashing sound causes everyone in the room to quickly turn their heads toward the source of the noise. describe the path of the signal through the human nervous system from initial stimulus of sound to the response of turning the head.... What is an appropriate domain for the function? Explain how you know. $What is an appropriate domain for the function? Explain how you know.$...

2022-09-29 23:26:41

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http://www.ck12.org/analysis/Degenerate-Conics/lesson/Degenerate-Conics/r11/

<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> You are viewing an older version of this Concept. Go to the latest version. # Degenerate Conics ## Point, line, or pair of lines formed when some coefficients of a conic equal zero. 0% Progress Practice Degenerate Conics Progress 0% Degenerate Conics The general equation of a conic is Ax2+Bxy+Cy2+Dx+Ey+F=0\begin{align*}Ax^2+Bxy+Cy^2+Dx+Ey+F=0\end{align*}.  This form is so general that it encompasses all regular lines, singular points and degenerate hyperbolas that look like an X.  This is because there are a few special cases of how a plane can intersect a two sided cone.  How are these degenerate shapes formed? #### Guidance Degenerate conic equations simply cannot be written in graphing form.  There are three types of degenerate conics: 1. A singular point, which is of the form: (xh)2a+(yk)2b=0\begin{align*}\frac{(x-h)^2}{a}+\frac{(y-k)^2}{b}=0\end{align*}. You can think of a singular point as a circle or an ellipse with an infinitely small radius. 2. A line, which has coefficients A=B=C=0\begin{align*}A=B=C=0\end{align*} in the general equation of a conic.  The remaining portion of the equation is Dx+Ey+F=0\begin{align*}Dx+Ey+F=0\end{align*}, which is a line. 3. A degenerate hyperbola, which is of the form:  (xh)2a(yk)2b=0\begin{align*}\frac{(x-h)^2}{a}-\frac{(y-k)^2}{b}=0\end{align*}.  The result is two intersecting lines that make an “X” shape.  The slopes of the intersecting lines forming the X are ±ba\begin{align*}\pm \frac{b}{a}\end{align*}. This is because b\begin{align*}b\end{align*} goes with the y\begin{align*}y\end{align*} portion of the equation and is the rise, while a\begin{align*}a\end{align*} goes with the x\begin{align*}x\end{align*} portion of the equation and is the run. Example A Transform the conic equation into standard form and sketch. 0x2+0xy+0y2+2x+4y6=0\begin{align*}0x^2+0xy+0y^2+2x+4y-6=0\end{align*} Solution: This is the line y=12x+32\begin{align*}y=-\frac{1}{2} x+\frac{3}{2}\end{align*}. Example B Transform the conic equation into standard form and sketch. 3x212x+4y28y+16=0\begin{align*}3x^2-12x+4y^2-8y+16=0\end{align*} Solution: 3x212x+4y28y+16=0\begin{align*}3x^2-12x+4y^2-8y+16=0 \end{align*} 3(x24x)+4(y22y)3(x24x+4)+4(y22y+1)3(x2)2+4(y1)2(x2)24+(y1)23=16=16+12+4=0=0 The point (2, 1) is the result of this degenerate conic. Example C Transform the conic equation into standard form and sketch. 16x296x9y2+18y+135=0\begin{align*}16x^2-96x-9y^2+18y+135=0\end{align*} Solution:  16x296x9y2+18y+135=0\begin{align*}16x^2-96x-9y^2+18y+135=0\end{align*} 16(x26x)9(y22y)16(x26x+9)9(y22y+1)16(x3)29(y1)2(x3)29(y1)216=135=135+1449=0=0 This is a degenerate hyperbola. Concept Problem Revisited When you intersect a plane with a two sided cone where the two cones touch, the intersection is a single point.  When you intersect a plane with a two sided cone so that the plane touches the edge of one cone, passes through the central point and continues touching the edge of the other conic, this produces a line.  When you intersect a plane with a two sided cone so that the plane passes vertically through the central point of the two cones, it produces a degenerate hyperbola #### Vocabulary A degenerate conic is a conic that does not have the usual properties of a conic.  Since some of the coefficients of the general equation are zero, the basic shape of the conic is merely a point, a line or a pair of lines.  The connotation of the word degenerate means that the new graph is less complex than the rest of conics. #### Guided Practice 1. Create a conic that describes just the point (4, 7). 2. Transform the conic equation into standard form and sketch. 4x2+8x+y2+4y=0\begin{align*}-4x^2+8x+y^2+4y=0 \end{align*} 3. Can you tell just by looking at a conic in general form if it is a degenerate conic? 1. (x4)2+(y7)2=0\begin{align*}(x-4)^2+(y-7)^2=0\end{align*} 2. 4x2+8x+y2+4y4(x22x)+(y2+4y)4(x22x+1)+(y2+4y+4)4(x1)2+(y+2)2(x1)21(y+2)24=0=0=4+4=0=0 3. In general you cannot tell if a conic is degenerate from the general form of the equation.  You can tell that the degenerate conic is a line if there are no x2\begin{align*}x^2\end{align*} or y2\begin{align*}y^2\end{align*} terms, but other than that you must always try to put the conic equation into graphing form and see whether it equals zero because that is the best way to identify degenerate conics. #### Practice 1. What are the three degenerate conics? Change each equation into graphing form and state what type of conic or degenerate conic it is. 2. x26x9y254y72=0\begin{align*}x^2-6x-9y^2-54y-72=0\end{align*} 3. 4x2+16x9y2+18y29=0\begin{align*}4x^2+16x-9y^2+18y-29=0\end{align*} 4. 9x2+36x+4y224y+72=0\begin{align*}9x^2+36x+4y^2-24y+72=0\end{align*} 5. 9x2+36x+4y224y+36=0\begin{align*}9x^2+36x+4y^2-24y+36=0\end{align*} 6. 0x2+5x+0y22y+1=0\begin{align*}0x^2+5x+0y^2-2y+1=0\end{align*} 7. x2+4xy+8=0\begin{align*}x^2+4x-y+8=0\end{align*} 8. x22x+y26y+6=0\begin{align*}x^2-2x+y^2-6y+6=0\end{align*} 9. x22x4y2+24y35=0\begin{align*}x^2-2x-4y^2+24y-35=0\end{align*} 10. x22x+4y224y+33=0\begin{align*}x^2-2x+4y^2-24y+33=0\end{align*} Sketch each conic or degenerate conic. 11. (x+2)24+(y3)29=0\begin{align*}\frac{(x+2)^2}{4}+\frac{(y-3)^2}{9}=0 \end{align*} 12. (x3)29+(y+3)216=1\begin{align*}\frac{(x-3)^2}{9}+\frac{(y+3)^2}{16}=1 \end{align*} 13. (x+2)29(y1)24=1\begin{align*}\frac{(x+2)^2}{9}-\frac{(y-1)^2}{4}=1 \end{align*} 14. (x3)29(y+3)24=0\begin{align*}\frac{(x-3)^2}{9}-\frac{(y+3)^2}{4}=0 \end{align*} 15. 3x+4y=12\begin{align*}3x+4y=12\end{align*} ### Vocabulary Language: English Conic Conic Conic sections are those curves that can be created by the intersection of a double cone and a plane. They include circles, ellipses, parabolas, and hyperbolas. degenerate conic degenerate conic A degenerate conic is a conic that does not have the usual properties of a conic section. Since some of the coefficients of the general conic equation are zero, the basic shape of the conic is merely a point, a line or a pair of intersecting lines. degenerate hyperbola degenerate hyperbola A degenerate hyperbola is an example of a degenerate conic. Its equation takes the form $\frac{(x-h)^2}{a}-\frac{(y-k)^2}{b}=0$. It looks like two intersecting lines that make an “X” shape.

2015-11-28 07:26:47

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https://itfeature.com/correlation-and-regression-analysis/mcq-on-correlation-and-regression-with-answer-1

# Correlation and Regression 1 This quiz is about MCQ on correlation and regression analysis. This Section contains MCQs on Correlation Analysis, Simple Regression Analysis, Multiple Regression Analysis, Coefficient of Determination (Explained Variation), Unexplained Variation, Model Selection Criteria, Model Assumptions, Interpretation of results, Intercept, Slope, Partial Correlation, Significance tests, OLS Assumptions, Multicollinearity, Heteroscedasticity, Autocorrelation, etc. Let us start MCQ on Correlation and Regression Analysis 1. If $\rho=0$, the lines of regression are: 2. The estimate of $\beta$ in the regression equation $Y=\alpha+\beta\,X + e$ by the method of least square is: 3. The lines of regression intersect at the point 4. When two variables more in the same direction then the correlation between the variable is 5. If each of $X$ variable is divided by 5 and $Y$ by 10 then $\beta_{YX}$ by coded value is: 6. If the correlation coefficient between the variables $X$ and $Y$ is $\rho$, the correlation coefficient between $X^2$ and $Y^2$ is 7. In multiple linear regression analysis, the square root of Mean Squared Error (MSE) is called the: 8. Regression coefficient is independent of 9. If $\beta_{XY}$ and $\beta_{YX}$ are two regression coefficients, they have 10. If $\beta_{YX}>1$, then $\beta_{XY}$ is: 11. If regression line $\hat{y}=5$ then value of regression coefficient of $y$ on $x$ is 12. If $X$ and $Y$ are two independent variates with variance $\sigma_X^2$ and $\sigma_Y^2$, respectively, the coefficient of correlation between $X$ and ($X-Y$) is equal to: 13. If all the actual and estimated values of $Y$ are the same on the regression line, the sum of squares of errors will be 14. If $\rho$ is the correlation coefficient, the quantity $\sqrt{1-\rho^2}$ is termed as 15. Homogeneity of three or more population correlation coefficients can be tested by 16. An investigator reports that the arithmetic mean of two regression coefficients of a regression line is 0.7 and the correlation coefficient is 0.75. Are the results 17. The average of two regression coefficients is always greater than or equal to the correction coefficient is called: 18. The geometric mean of the two regression coefficient $\beta_{YX}$ and $\beta_{XY}$ is equal to: 19. If the two lines of regression are perpendicular to each other, the correlation coefficient $r=$ is: 20. The range of a partial correlation coefficient is: Correlation is a statistical measure used to determine the strength and direction of the mutual relationship between two quantitative variables. The regression describes how an explanatory variable is numerically related to the dependent variables. Both of the tools are used to represent the linear relationship between the two quantitative variables. The relationship between variables can be observed either using graphical representation between the variables or numerical computation using appropriate computational formula. Note that neither regression nor correlation analyses can be interpreted as establishing some cause-and-effect relationships. Both of these can be used to indicate only how or to what extent the variables under study are associated (or mutually related) with each other. The correlation coefficient measures only the degree (strength) and direction of linear association between the two variables. Any conclusions about a cause-and-effect relationship must be based on the judgment of the analyst. Currently working as Assistant Professor of Statistics in Ghazi University, Dera Ghazi Khan. Completed my Ph.D. in Statistics from the Department of Statistics, Bahauddin Zakariya University, Multan, Pakistan. l like Applied Statistics, Mathematics, and Statistical Computing. Statistical and Mathematical software used is SAS, STATA, GRETL, EVIEWS, R, SPSS, VBA in MS-Excel. Like to use type-setting LaTeX for composing Articles, thesis, etc. ### 30 Responses 1. asma aftab says: these are very useful for preparation of objective paper..–thank you- 2. Inbasat Masood says: Thanku Sir mujy mil gy thy answers 3. Inbasat Masood says: sir these mcq’s are very useful but i dont find answers here , kindly tell me how to get añswer i m not getting even aftr submit • When you press the show answer or next button correct answer is ticked with green color tick point. When you submit the complete quiz, you will get the results of all MCQs for the quiz after few seconds. If complete results are not shown it may be possible that you have some ade blocker plugin installed on your browser. Do allow the browser for the itfeature.com site. 4. Jamil Sheikh says: 5. Rupali says: If x and y are uncorreleunc then vat(x-y)=? Give explanation • If X and Y are uncorrelated then X and Y must be independent variables. Independent variables have zero correlation, so will result in: $Var(X-Y)=Var(X)+Var(Y)$ 6. Shamnaz Nazeer says: It’s usefulness for the study of Quantitative techniques 7. Prabin says: Great Wortk 8. Anurag Joshi says: In the linear regression equation: Which variable is fixed and which one is random, between x & y ? • The variable X and Y both are random in real sense. However, regarding the theory of least square, Variable X is assumed to be fixed (known). In other words, they (X’s) are assumed to have no random error. Also note that, in observational studies X is random while in experimental studies X is fixed. 9. rukh says: • Perez says: Can the correlation coefficient distinguish between the dependent and independent random variables with regards to question one? • Correlation is a measure of strength between two quantitative variables. It measures the interdependence between these variables, so one cannot distinguish between the dependent and independent using only correlation coefficient. 10. Jasdem says: Where is next page of qt mcq 11. Ansar says: 12. Aamir Khan says: JAZAK ALLAH, appreciative and very helpful…. Nice work 13. Vitthal says: 14. Fouzia says: Want some notes on correlation and regression • See regression section on right side • xhao says: whhere is the answer for the questions? • Nimra Shaheen says: Sir kindly confirm these questions. In Exercise 1:Q3 is( b) option&Q4 is (a) • Thank you! Please share the question statement and options, as all questions appear in the test as random. 15. tariq and rumsha says: jazakALLAH…….. very good work 16. MASOOM says: GOOD JOB FOR INTELGENT STU. 17. imtiaz malik says: nice exercise This site uses Akismet to reduce spam. Learn how your comment data is processed.

2023-03-27 14:45:33

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http://igm.univ-mlv.fr/~fpsac/FPSAC07/SITE07/conpap.htm

Contributed talks: Refined Enumerations of Totally Symmetric Self-Complementary Plane Partitions and Constant Term Identities Masao Ishikawa Representing Tropical Linear Spaces by Circuits Josephine Yu and Debbie S. Yuster Permutation Tableaux and the Asymmetric Exclusion Process Sylvie Corteel And Lauren K. Williams Dual Equivalence Graphs, Ribbon Tableaux and Macdonald Polynomials Sami H. Assaf A Bijection Between 2-Triangulations and Pairs of Non-Crossing Dyck Paths Sergi Elizalde Dual Graded Graphs for Kac-Moody Algebras Thomas Lam and Mark Shimozono An Additive Theorem Related to Latin Transversals Zhi-Wei Sun Increasing and Decreasing Sequences in Fillings of Moon Polyominoes Martin Rubey Non-Commutative Extensions of Classical Determinantal Identities Matjaz Konvalinka and Igor Pak The Alternating Sign Matrix Polytope Jessica Striker Nested Set Complexes of Dowling Lattices and Complexes of Dowling Trees Emanuele Delucchi Catalan's Intervals and Realizers of Triangulations Olivier Bernardi and Nicolas Bonichon Classifying Lattice Walks Restricted to the Quarter Plane Marni Mishna Catalan Tableaux and the Asymmetric Exclusion Process Xavier G¨¦rard Viennot Computer Algebra and Power Series With Positive Coefficients Manuel Kauers The Combinatorics of Associated Hermite Polynomials Dan Drake Algebraic Constructions on Set Partitions Maxime Rey On Double Hurwitz Numbers in Genus 0 Sergei Shadrin, Michael Shapiro, and Alek Vainshtein q-Eulerian Polynomials: Excedance Number and Major Index John Shareshiana and Michelle L. Wachs Noncommutative Monomial Symmetric Functions Lenny Tevlin The Cyclic Sieving Phenomenon for Faces of Generalized Cluster Complexes Sen-Peng Eu and Tung-Shan Fu On the Structure of Regular $B_{2}$ Crystals V. I. Danilov, A. V. Kavanov, and G. A. Koshevoy Clusters, Noncrossing Partitions and the Coxeter Plane Nathan Reading A Combinatorial Classification of Skew Schur Functions Peter R. W. McNamara and Stephanie van Willigenburg Transversal and Cotransversal Matroids via Their Representations Federico Ardila Partially Directed Paths in a Symmetric Wedge E. J. Janse van Rensburg, T. Prellberg, and A. Rechnitzer Tabloids and Weighted Sums of Characters of Certain Modules of the Symmetric Groups Yasuhide Numata Contributed Posters: Hopf Algebras of Diagrams G. H. E. Duchamp, J-G. Luque, J-C. Novelli, C. Tollu, and F. Toumazet Inequalities for Symmetric Means Allison Cuttler, Curtis Greene, and Mark Skandera Descent Patterns in Permutations Amanda Riehl Schur Function Identities and Hook Length Posets Masao Ishikawa and Hiroyuki Tagawa Alternating Subgroups of Coxeter Groups Francesco Brenti, Victor Reiner, and Yuval Roichman Grid Polygons From Permutations and Their Enumeration by the Kernel Method Toufik Mansour and Simone Severini Jacobi-Trudi Formula for Jack Functions of Rectangular Shapes and Hyperdeterminants Sho Matsumoto A Theory of General Combinatorial Differential Operators Gilbert Labelle and Cédric Lamathe Analysis of Some Exactly Solvable Diminishing Urn Models Hsien-Kuei Hwang, Markus Kuba, and Alois Panholzer Complexes of Directed Trees and Independence Complexes Alexander Engström Pattern-Avoiding Fillings of Rectangular Shapes Vít Jelínek Abstract Young Pairs for Coxeter Groups of Type D Eli Bagno and Yona Cherniavsky Demazure Atoms Sarah Mason Extended Quadratic Algebra and a Model of the Equivariant Cohomology Ring of Flag Varieties Anatol N. Kirillov and Toshiaki Maeno Characterizations of Flip-Accessibility for Domino Tilings of the Whole Plane Olivier Bodini, Thomas Fernique and Éric Rémila Macdonald Symmetric Functions for Partitions With Complex Parts Michael J. Schlosser Multiparking Functions, Graph Searching, and the Tutte Polynomial Dimitrije Kostic and Catherine H. Yan Unification of the Quintuple and Septuple Product Identities Wenchang Chu and Qinglun Yan Key Polynomials, Invariant Factors and an Action of the Symmetric Group on Young Tableaux Olga Azenhas and Ricardo Mamede Combinatorial Realisation of Hall-Littlewood Polynomials at t = -1 R. C. King and A. M. Hamel Involutions Avoiding the Class of Permutations in $\mathfrak{S}_k$ With Prefix 12 W. M. B. Dukes and Toufik Mansour On the Young-Fibonacci Insertion Algorithm Janvier Nzeutchap Major Indices, Mahonian Identities and Ordered Generating Systems Robert Shwartz On Uniquely k-Determined Permutations Sergey Avgustinovich and Sergey Kitaev Operations on Posets and Rational Identities of Type A Adrien Boussicault On Euler's Difference Table Fanja Rakotondrajao Compound Basis Arising From the Basic $A_{1}^{(1)}$-Module Kazuya Aokage and Hiroshi Mizukawa An Extension of the Foata Map to Standard Young Tableaux J. Haglund, and L. Stevens Elliptic Curve Groups and Chip-Firing Games Gregg Musiker The Hook Formula Jason Bandlow Variance for the Number of Maxima in Hypercubes and Generalized Euler¡¯s $\gamma$ Constants Christian Costermans and Hoang Ngoc Minh The Automorphism Group of a Finite p-Group is Almost Always a p-Group Geir T. Helleloid and Ursula Martin Generating Functions From the Point of View of Rota¨CBaxter Algebras Li Guo Signed Enumeration of Ribbon Tableaux With Local Rules and Generalizations of the Schensted Correspondence Dominique Gouyou-Beauchamps and Philippe Nadeau Kazhdan-Lusztig Immanants III: Transition Matrices Between Canonical Bases of Immanants Brendon Rhoades and Mark Skandera The Combinatorics of the Garsia-Haiman Modules for Hook Shapes Ron M. Adin, Jeffrey B. Remmel, and Yuval Roichman Parameterized Telescoping Proves Algebraic Independence of Sums Carsten Schneider Dumont Permutations of the Third Kind Alexander Burstein and Walter Stromquist A Classification of Outerplanar K-Gonal 2-Trees Martin A. Ducharme, Gilbert Labelle, Cédric Lamathe, and Pierre Leroux Complete Decomposition of Dickson-Type Recursive Polynomials and a Related Diophantine Equation Thomas Stoll Center for Combinatorics, Nankai University, Tianjin 300071 China Tel: 86-22-2350-2180, Fax: 86-22-2350-9272, Email: fpsac@nankai.edu.cn

2019-03-22 10:04:40

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https://tigerprints.clemson.edu/all_dissertations/236/

## All Dissertations 8-2008 Dissertation #### Degree Name Doctor of Philosophy (PhD) #### Legacy Department Mathematical Science Ervin, Vincent J #### Committee Member Kees , Christopher E Lee , Hyesuk K #### Committee Member Warner , Daniel D #### Abstract The accurate numerical approximation of viscoelastic fluid flow poses two difficulties: the large number of unknowns in the approximating algebraic system (corresponding to velocity, pressure, and stress), and the different mathematical types of the modeling equations. Specifically, the viscoelastic modeling equations have a hyperbolic constitutive equation coupled to a parabolic conservation of momentum equation. An appealing approximation approach is to use a fractional step $\theta$-method. The $\theta$-method is an operator splitting technique that may be used to decouple mathematical equations of different types as well as separate the updates of distinct modeling equation variables when modeling mixed systems of partial differential equations. In this work a fractional step $\theta$-method is described and analyzed for the numerical computation of both the time dependent convection-diffusion equation and the time dependent equations of viscoelastic fluid flow using the Johnson-Segalman constitutive model. For convection-diffusion the $\theta$-method presented allows for a decoupling within time steps of the parabolic diffusion operator from the hyperbolic convection operator. The hyperbolic convection update is stabilized using a Streamline Upwinded Petrov-Galerkin (SUPG)-method. The analysis given for the convection-diffusion equation serves as a template for the analysis of the more complicated viscoelastic fluid flow modeling equations. The $\theta$-method implementation analyzed for the viscoelastic modeling equations allows the velocity and pressure approximations within time steps to be decoupled from the stress, reducing the number of unknowns resolved at each step of the method. Additionally the $\theta$-method decoupling results in the approximation of the nonlinear viscoelastic modeling system using only the solution of linear systems of equations. Similar to the scheme implemented for convection-diffusion, the hyperbolic constitutive equation is stabilized using a SUPG-method. For both the convection-diffusion and the viscoelastic modeling equations a priori error estimates are established for their $\theta$-method approximations. Numerical computations supporting the theoretical results and demonstrating the $\theta$-methods are also included. COinS

2021-08-01 20:19:00

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http://mathhelpforum.com/calculus/95924-standard-equation-ellipse-print.html

# Standard Equation of an Ellipse • July 23rd 2009, 09:36 PM psu1024 Standard Equation of an Ellipse I am having trouble finding the standard form of an ellipse, being given the following equation: $ 36x^2+9y^2+48x-36y+43=0 $ I know that I have to group terms and complete the square to make it fit the standard form of an ellipse: $ \frac {(x-h)^2}{a^2} + \frac {(y-k)^2}{b^2} = 1 $ So I get to this and have been stuck: $ (36x^2+48x+?)+9(y^2-4y+4)=-34+?+36 $ I don't think I can just factor out the 12, and have $ 12(3x^2+4x+?)$ , so I tried factoring out the 36: $ 36(x^2+\frac{4}{3}x+\frac{4}{9}) $ in which I got $ 36(x+\frac{2}{83})^2 + 9(y-2)^2=18$ <---which I got from $(-34+36(\frac{4}{9})+36) $ So then I get $ 2(x+\frac{2}{3})^2 + \frac {(y-2)^2}{2} = 1 $ Does this mean that $a=\frac{1}{2}$ ? Is that even possible? And if so, does this mean that the center is $(-\frac{2}{3},2)$ ? In this case then, the major axis is vertical, correct? So the foci would be $(-\frac{2}{3}, 2\pm\sqrt{1.5})$ and the vertices would be $(-\frac{2}{3},2\pm\sqrt{\frac{1}{2}})$ and $(-\frac{2}{3}\pm\sqrt{2},2)$ ? Compared to what we've been learning in class (even though we haven't really even talked about it all that much), this all just seems too complicated to be correct, so I was didn't know if there was just something I was doing wrong or overthinking something. Any help would be appreciated. • July 23rd 2009, 09:51 PM VonNemo19 Quote: Originally Posted by psu1024 I am having trouble finding the standard form of an ellipse, being given the following equation: $ 36x^2+9y^2+48x-36y+43=0 $ I know that I have to group terms and complete the square to make it fit the standard form of an ellipse: $ \frac {(x-h)^2}{a^2} + \frac {(y-k)^2}{b^2} = 1 $ So I get to this and have been stuck: $ (36x^2+48x+?)+9(y^2-4y+4)=-34+?+36 $ I don't think I can just factor out the 12, and have $ 12(3x^2+4x+?)$ , so I tried factoring out the 36: $ 36(x^2+\frac{3}{4}x+\frac{9}{64}) $ in which I got $ 36(x+\frac{3}{8})^2 + 9(y-2)^2=\frac{113}{16}$ <---which I got from $(-34+36(\frac{9}{64})+36) $ ,which doesn't work out at all because I can't divide out what's to the right of the equal sign to get a nice looking equation that equals 1. I'm completely baffled by this problem, so any help would be appreciated. to complete this square... $36x^2+48x$ $36(x^2+\frac{4}{3}x)$ $36(x^2+\frac{4}{3}x+\frac{4}{9})$ Understand that the first and second lines are equivalent, the third is just for conceptualization. Do you see that you must add 16 to the right hand side to preserve equality? • July 23rd 2009, 10:07 PM psu1024 Oh, good catch! Thanks. I think I reversed the 36 and 48 when I was dividing them. • July 23rd 2009, 10:09 PM VonNemo19 Quote: Originally Posted by psu1024 Oh, good catch! I hate it when I miss the small stuff. Best of luck!

2013-12-07 10:25:27

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http://math.stackexchange.com/questions/297378/what-is-this-math-symbol-called/297380

# What is this math symbol called? My professors use a symbol $$x_0$$ and they pronounce it as x not or x nod, I am not sure what the exact name is because they have thick accents. I have tried looking this up on the Internet but I could not find an answer. Does anyone know what this is called? - They will likely be saying "x naught", naught being a synonym for zero. –  Matt Pressland Feb 7 '13 at 17:43 'x naught'? as in 'x zero' ? –  Daniel Rust Feb 7 '13 at 17:43 Please oh please, if your professor says something and you cannot even make out what he is saying, let alone what it means, please oh please ask him! There are very few more intensive wastes of time than you sitting in a class and his giving the class without even this minuscule piece of communication being successful! –  Mariano Suárez-Alvarez Feb 7 '13 at 19:18 It's probably x-naught, synonym for "x sub zero", it's used when you refer to an starting point for variable $x$, for example in physics, if you have a particle moving on the $x$ axis, you will always find $x_0$ for the initial position.

2014-12-22 10:19:13

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https://www.fotonixx.com/posts/design-a-camera-with-python-and-pyrayt/

# Design a Camera with Python and PyRayT: Part One Have you ever cut open a camera lens to look at what's inside? I don't blame you if you haven't, lenses are ridiculously expensive and it's a one way operation. However, if you look at a picture of a cross-section, you'd see that the "lens" is actually made of upwards of a dozen individual lenses, each one doing its part to create a clear, error free image. Typically, understanding what's going on in the camera lens requires exhaustive calculations or expensive lens design software costing upwards of \$10,000 for a single user. My annoyance at this price barrier (as somebody who uses those tools professionally) is what sparked me to create PyRayT, a free and open source generic ray tracer that pairs with the Scientific Python stack. To celebrate PyRayT's v0.3.0 release, I'm going to walk through how it can be used to design and optimize a multi-lens camera. This first part of the series will cover why lenses need to be so intricate, and what happens if you tried to make a camera out of a single lens instead. ## Why Are Camera Lenses so Complex? If you rack your brain to recall highschool physics there's probably two things you remember about lenses: 1. A ray parallel to the optical axis converges onto the focal point. 2. A ray going through the center of the lens is unmodified. With these rules you can trace simple diagrams showing how object images are formed by lenses. Source: Wikipedia Notice how in the above image every point in the object corresponds to exactly one point in the image, so in theory a single lens camera should be able to capture perfect reproductions of objects we want to image. Why then, does a picture from that camera end up looking like the one below? An ideal image (left) and the resulting image as captured by a simple camera (right) Source: www.opticsthewebsite.com These distortions between the object and image, called optical aberrations, are caused by higher order terms that are typically ignored when calculating lenses by hand using the paraxial approximation. The goal of any camera design is to minimize these aberrations to the extent that they are not noticed by the end user. A single lens only has so many variables about it you can change, but adding multiple lenses to a system gives optical engineers additional degrees of freedom for the minimization (often by minimizing Siedel Aberration Coefficients, which are beyond the scope of the article). So how bad could a single lens imager really be? And how do you improve the quality with additional lenses? Let's answer the first question by quantifing common aberrations for a single lens system. ## Getting Up and Running with PyRayT As mentioned above, hand calculations for lenses rely on approximations that don't capture aberrations. Fortunately, numeric ray tracers don't suffer from the same limitations which is why they're so valuable for accurate lens design. Since this article is about showcasing features of PyRayT, that's the ray tracer we'll be using. You can install the latest stable package from pip. py -m pip install pyrayt You can also check out the Getting Started Guide and list of built-in optical components for a general overview. ## Creating a Simple Camera To design our single-lens camera we only need two specs: system power and f/# (F-number). We want the system to have a focal length of 50mm, and since a lens' power is the inverse of focal length, our system power is 0.02. Our f/# will be 2.4, meaning the focal length should be 2.4x larger than the entrance aperture. ### Thin Lenses Our lens is the only component that contributes to system power, so the power of the lens has to equal the desired power of our system. To calculate lens power we'll use the lensmaker's equation. \begin{equation*} P_{lens} = (n_{lens} -1)[\frac{1}{R_1}-\frac{1}{R_2}+\frac{(n_{lens}-1)d}{n R_1 R_2}] \end{equation*} Since we're doing the calculation by hand we'll make a couple approximations to simplify things: (1) the radii of curvature are equal and opposite (resulting in a biconvex lens), and (2) the thickness is small enough that we can discard the final term. Later we'll numerically optimize the entire system to correct for focus, but these approximations give us a good starting point. The simplified equation then becomes: \begin{equation*} R_{lens} =\frac{2(n_{lens} -1)}{P_{lens}} \end{equation*} Since the refractive index of most glasses is ~1.5, this means that our radius of curvature for a biconvex lens is equal to the inverse of the lenses power, which is also the focal length of the system! Now we can construct our lens and visualize it with the draw function in the tinygfx package (installed as part of the PyRayT distribution). # import the Ray Tracer Package import pyrayt import pyrayt.materials as matl from tinygfx.g3d.renderers import draw # All spatial units are mm lens_diameter = 30 lens_thickness = 5 system_focus = 50 # The focus of the system f_num = 2.4 # f-number of system # Creating a simple Lens lens_material = matl.glass["ideal"] lens = pyrayt.components.thick_lens( thickness=lens_thickness, aperture=lens_diameter, material=lens_material) draw(lens) A lens is not too impressive by itself. Let's add the remaining parts of our system so we can start running ray traces! ### Apertures and Baffles There's two more pieces we need in order to make the camera: an aperture to block rays that exceed our f/# and an imager placed on our camera's focal plane. From a modeling perspective both will be accomplished with variations of PyRayT's baffle. Baffle's are 2D planes that absorb all light incident on them, perfect for modeling sensors as well as ideal beam-stops. For the imager we create a square baffle the same size as our lens and move it to the focal plane. imager = components.baffle((lens_diameter, lens_diameter)).move_x(system_focus) Our aperture can be thought of as a "baffle with a hole", where the hole is large enough to only let in rays with a cone angle specified by our f/#. pyrayt and tinygfx create arbitrary shapes via constructive solids meaning an aperture is a baffle with the center shape subtracted from it. The convenience function aperture does just this, creating a baffle with an arbitrarily shaped hole in the middle. The diameter of the aperture that gives us the desired f/# depends on where in the system the aperture is located. If we place it half-way between the lens and the focal plane, the diameter of the opening has to be: \begin{equation*} d_{ap}=\frac{1}{2*P_{sys}}*\frac{1}{f_\#} \end{equation*} aperture_position = system_focus / 2 aperture_diameter = aperture_position / f_num aperture = components.aperture( size=(lens_diameter, lens_diameter), # make a square baffle aperture_size=aperture_diameter # put a circular opening in the center ).move_x(aperture_position) ### Our First Ray Trace With our components defined we're ready to simulate. The only thing we need is a test source that generates rays to trace through the system. For that we'll use PyRayT's LineOfRays which generates a set of linearly spaced rays projected towards the +x axis. The last step is to load all the components into a RayTracer object and run the trace function. Note Almost all of the sources used to characterize our system will be parallel bundles of rays at various angles. This is because we're assuming the camera is focused at infinity, where any angular deviation between sets of rays originating from the same point are effectively zero. # Create a Parallel ray set source = components.LineOfRays(0.8*lens_diameter, wavelength = 0.633).move_x(-10) tracer = pyrayt.RayTracer(source, [lens, aperture, imager]) tracer.set_rays_per_source(11) results = tracer.trace() The results of a trace is a Pandas dataframe which stores information about the ray at every intersection of the simulation. However, for now we'd rather just visualize the ray trace, which is done with the show function. # import matplotlib so we can manipulate the axis import matplotlib.pyplot as plt def init_figure() -> Tuple[plt.Figure, plt.Axes]: """ Convenience function to generate an axis with a set size """ fig = plt.figure(figsize = (12,8)) axis = plt.gca() axis.grid() return fig, axis # set up the figure and axis fig, axis = init_figure() axis.set_xlabel("distance (mm)") axis.set_ylabel("distance (mm)") # display the ray trace tracer.show( ray_width=0.2, axis=axis, view='xy') plt.show() Looks like our lens is doing its job! All rays that transmit through the aperture are focused to an approximate point at the focal distance, and any ray angle that exceeds our f/# is blocked. Unfortunately since the aperture and imager are 2D objects, they don't show up in the ray trace, but we know that they are there because rays terminate on their surfaces. ## Characterizing Lens Performance A picture may be worth 1000 words, but when it comes to analyzing our lens' performance data is key. Using the results dataframe we can explore the RaySet metadata of each ray as they travel through the system, and we'll use that information to see how our single lens design holds up against common imaging aberrations: spherical, chromatic, and coma. As mentioned above, characterizing the system with Seidel coefficients is beyond the scope of the article, instead we'll use Matplotlib to generate plots of the system focus across different parameters. ### Spherical Aberrations Spherical lenses don't actually focus light to a perfect point. In fact, the focal point is a function of the radius where the light enters the lens (in our case the position on the y-axis where the ray originates). We can easily visualize the spherical aberrations by creating a helper function that generates a set of rays along the y-axis, and calculates where each ray intercepts the x-axis. def spherical_aberration(system, ray_origin: float, max_radius:float, sample_points=11): # the souce is a line of rays only on the +y axis. It's slightly shifted so zero is not a point # as it would focus at infinity tracer = pyrayt.RayTracer(source, system) tracer.set_rays_per_source(sample_points) results = tracer.trace() # Since we don't have the actual imager as a variable in the function # assume it is the last thing a ray intersect with, meaning the rays that hit it have the # highest generation imager_rays = results.loc[results['generation'] == np.max(results['generation'])] # Intercept is calculated using the tilt for each ray, with is a normalized vector representing # the direction the ray is travelling intercept = -imager_rays['x_tilt']*imager_rays['y0']/imager_rays['y_tilt'] + imager_rays['x0'] # create a new dataframe with the aberration metrics return results Using the function on our single-lens system yields the above plot, showing that the focal length of the lens is changing by almost 10% based on the radius alone! This aberration would cause our single-lens images to come out "blurry" even when the imager is aligned to the focal plane. Speaking of the focal plane, we also see that the focus of our lens is ~52mm instead of the 50 we calculated, due the the thick lens portion of the lensmaker's equation we chose to ignore. ### Chromatic Aberrations Unlike spherical aberrations, chromatic aberrations are explained by the lensmaker's equation: the focal point of the lens depends on the refractive index of the lens' material. Real materials don't have a constant refractive index; instead, the refractive index is a function of wavelength. This effect, called dispersion, is more often associated with the reason prisms split white light into a rainbow of colors. In our case it means our lens will have a wavelength dependent focus. The same way we wrote a function to characterize spherical aberrations, we can write one to quantify chromatic aberration: def chromatic_abberation(system, ray_origin: float, test_radius:float, wavelengths: np.ndarray) -> pd.DataFrame: # create a set of sources for every wavelength of light sources = [ pyrayt.components.LineOfRays(0, wavelength = wave) .move_x(ray_origin) for wave in wavelengths] # Create the ray tracer and propagate tracer = pyrayt.RayTracer(sources, system) tracer.set_rays_per_source(1) results = tracer.trace() #filter the rays that intersect the imager imager_rays = results.loc[results['generation'] == np.max(results['generation'])] # calculate intercept of the imager rays with the x-axis and form into a dataframe intercept = -imager_rays['x_tilt']*imager_rays['y0']/imager_rays['y_tilt'] + imager_rays['x0'] results = pd.DataFrame({'wavelength': imager_rays['wavelength'], 'focus': intercept}) return results If we run this function on our current system the results will say that every wavelength has the exact same focus! This is because we made the lens out of an "ideal" glass with a refractive index (n) of 1.5. Let's replace our lens with one made of a popular crown glass instead: lens_material = matl.glass["BK7"] # Update this line of the lens definition Running the function on our newly dispersive system shows a focal length shift of ~1mm (2%) across the visible spectrum. An image taken with this lens would result in sharp edges in our photos having a 'rainbow' effect. Interestingly, this aberration is sometimes sought after for artistic effect, going as far as being including as a graphics setting in id's 2016 Doom reboot. ### Coma Aberrations The last aberration we'll quantify is coma. Instead of being a change in focal length, coma is a change mangification vs. angle of incidence on the system. The name comes from the fact that a point imaged with a system suffering from coma looks like the coma of a comet. The easiest way to visualize coma is to plot a LineOfRays hitting the lens at non-normal incidence. In a coma free imager, the distribution of ray-imager intersections should be symmetrically distributed about the central ray. While we could generate three sources at three distinct angles and run a single ray trace, we're instead going to leverage PyRayT's pin function to manipulate the angle of the lens. Since the position of the lens is a property of the lens itself, updating the position without remembering to undo it later will effect calculations down the line. the pin context manager takes care of this for us by undoing any transformations of pinned objects when the context manager exits. The main advantage of rotating the lens instead of the source is that the incoming rays stay perpendicular to the imager, and nominally centered about y=0. # Create the system source = pyrayt.components.LineOfRays(0.9 * lens_diameter).move_x(-10) tracer = pyrayt.RayTracer(source, [lens, imager]) tracer.set_rays_per_source(1001) angles = (12,6,0) fig, axis = init_figure() for n,angle in enumerate(angles): with pyrayt.pin(lens): # pin the lens in place so its rotation resets between iterations lens.rotate_z(angle) results = tracer.trace() imager_rays = results.loc[results['generation'] == np.max(results['generation'])] axis.hist(imager_rays['y1'], bins=21, label = f"{angle}-degree incidence", density=True, alpha=0.8) # plot labels axis.set_xlabel("position along Y-axis (mm)") axis.set_ylabel("density (AU)") axis.set_label("Focal Spot Distribution vs. Angle of Incidence") plt.legend() plt.show() Just looking at the ray trace we can see the angled rays don't come to any central focus, and the histogram has a strong skew towards the -y axis. Like spherical aberrations, coma will cause our images to blur; however, the distortion is worse the further you are from the center of the imager, with the middle of the image remaining sharp. ## Next Steps This wraps up the first part of our camera design. It might not seem like much, but understanding the limitations of a simple system helps justify why we go through the process of desiging a complex one. The full Jupyter notebook for the design is on GitHub, and next post I'll show how with just one additional lens, we can drastically minimize all three of the above aberrations to help our camera generate a cleaner final image! In the mean time feel free to explore the design and see how much optimization you can do with just one lens: • What about the design changes if you pick a different material? • If you break the symmetry between the front and back surface can you reduce aberrations? • Is there a way to reduce chromatic aberration only by changing the physical dimensions (not material) of the lens?

2021-09-18 09:47:59

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http://clay6.com/qa/52407/a-current-of-2-34-ampere-flows-in-a-resistance-of-11-111111-omega-the-poten

# A current of $2.34$ ampere flows in a resistance of $11.111111 \Omega$ The potential difference across the given resistance with due to regard for significant figures is $\begin{array}{1 1} 26,000\;V \\ 26,00 \;V \\ 26.0 \;V \\26\;V \end{array}$ Answer :$26.0 \;V$ $V= 2.34 \times 11.111111\;volt$ $\quad= 26.0\;volt$

2017-10-22 01:13:21

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https://zbmath.org/?q=an:0706.34020

# zbMATH — the first resource for mathematics Some remarks on a multiplicity result by Mawhin and Schmitt. (English) Zbl 0706.34020 Some multiplicity results for a nonlinear ODE with Dirichlet boundary conditions are proved. These are obtained by topological degree methods, proving the existence of several branches of solutions.

2022-01-23 09:57:43

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http://mathoverflow.net/questions/127548/extending-a-bilinear-form-weak-derivative-and-bochner-space

Extending a bilinear form (weak derivative and Bochner space) Let $V \subset H \subset V'$ be a Hilbert triple and let $$X = \{u \in L^2(0,T;V) : u' \in L^2(0,T;H)\}$$ and $$Y = \{u \in L^2(0,T;V) : u' \in L^2(0,T;V')\}$$ Define $b:X\times X \to \mathbb{R}$ by $$b(u,v) = (u',v)_H + (v',u)_H.$$ I want to say: there is a unique extension of $b$ from $X \times X$ to $Y \times Y$ such that $$b(u,v) = \langle u', v \rangle_{V', V} + \langle v', u \rangle_{V', V}.$$ How do I show that this is true? Suppose that $X$ is dense in $Y$. Edit: this particular $b$ is just an example. It could be something different. -

2014-03-10 04:17:42

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http://www.ask.com/question/what-is-the-definition-of-alternator

# What Is the Definition of Alternator? An alternator is an electric generator that produces alternating current. The generator's coil is rotated, and its spherical path causes it to cut across a magnetic field, first in one course, then the other, with each cycle. al·ter·na·tor [awl-ter-ney-ter, al-] NOUN [ELECTRICITY] 1. a generator of alternating current. Source: Dictionary.com Similar Questions Top Related Searches Explore this Topic The noun "celerity" is defined as speed, swiftness and rapidity, according to Dictionary.com. Celerity is also an alternative name for proper velocity, ... "Knead" is a verb used to describe the technique used to work substances such as clay or dough into a usable state. It involves alternately stretching ...

2014-12-20 18:44:59

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https://artofproblemsolving.com/wiki/index.php?title=1990_AHSME_Problems/Problem_14&diff=prev&oldid=65989

Difference between revisions of "1990 AHSME Problems/Problem 14" Problem $[asy] draw(circle((0,0),1),black); draw((0,1)--(cos(pi/14),-sin(pi/14))--(-cos(pi/14),-sin(pi/14))--cycle,dot); draw((-cos(pi/14),-sin(pi/14))--(0,-1/cos(3pi/7))--(cos(pi/14),-sin(pi/14)),dot); draw(arc((0,1),.25,230,310)); MP("A",(0,1),N);MP("B",(cos(pi/14),-sin(pi/14)),E);MP("C",(-cos(pi/14),-sin(pi/14)),W);MP("D",(0,-1/cos(3pi/7)),S); MP("x",(0,.8),S); [/asy]$ An acute isosceles triangle, $ABC$, is inscribed in a circle. Through $B$ and $C$, tangents to the circle are drawn, meeting at point $D$. If $\angle{ABC=\angle{ACB}=2\angle{D}$ (Error compiling LaTeX. ! Missing } inserted.) and $x$ is the radian measure of $\angle{A}$, then $x=$ $\text{(A) } \frac{2\pi}{7}\quad \text{(B) } \frac{4\pi}{9}\quad \text{(C) } \frac{5\pi}{11}\quad \text{(D) } \frac{6\pi}{13}\quad \text{(E) } \frac{7\pi}{15}$ Solution $\fbox{A}$

2022-05-27 15:24:48

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https://read.somethingorotherwhatever.com/entry/Configurationspacesofhardsquaresinarectangle

# Configuration spaces of hard squares in a rectangle • Published in 2020 In the collection We study the configuration spaces C(n;p,q) of n labeled hard squares in a p by q rectangle, a generalization of the well-known "15 Puzzle". Our main interest is in the topology of these spaces. Our first result is to describe a cubical cell complex and prove that is homotopy equivalent to the configuration space. We then focus on determining for which n, j, p, and q the homology group $H_j [ C(n;p,q) ]$ is nontrivial. We prove three homology-vanishing theorems, based on discrete Morse theory on the cell complex. Then we describe several explicit families of nontrivial cycles, and a method for interpolating between parameters to fill in most of the picture for "large-scale" nontrivial homology. ### BibTeX entry @article{Configurationspacesofhardsquaresinarectangle, title = {Configuration spaces of hard squares in a rectangle}, abstract = {We study the configuration spaces C(n;p,q) of n labeled hard squares in a p by q rectangle, a generalization of the well-known "15 Puzzle". Our main interest is in the topology of these spaces. Our first result is to describe a cubical cell complex and prove that is homotopy equivalent to the configuration space. We then focus on determining for which n, j, p, and q the homology group {\$}H{\_}j [ C(n;p,q) ]{\$} is nontrivial. We prove three homology-vanishing theorems, based on discrete Morse theory on the cell complex. Then we describe several explicit families of nontrivial cycles, and a method for interpolating between parameters to fill in most of the picture for "large-scale" nontrivial homology.}, url = {http://arxiv.org/abs/2010.14480v1 http://arxiv.org/pdf/2010.14480v1}, year = 2020, author = {Hannah Alpert and Ulrich Bauer and Matthew Kahle and Robert MacPherson and Kelly Spendlove}, comment = {}, urldate = {2020-11-06}, archivePrefix = {arXiv}, eprint = {2010.14480}, primaryClass = {math.AT}, collections = {puzzles} }

2021-05-11 09:41:28

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https://eprint.iacr.org/2022/367

### Efficient Algorithms for Large Prime Characteristic Fields and Their Application to Bilinear Pairings and Supersingular Isogeny-Based Protocols Patrick Longa ##### Abstract We propose a novel approach that generalizes interleaved modular multiplication algorithms to the computation of sums of products over large prime fields. This operation has widespread use and is at the core of many cryptographic applications. The method reformulates the widely used lazy reduction technique, crucially avoiding the need for storage and computation of double-precision'' operations. Moreover, it can be easily adapted to the different methods that exist to compute modular multiplication, producing algorithms that are significantly more efficient and memory-friendly. We showcase the performance of the proposed approach in the computation of multiplication over an extension field GF(p^k), and demonstrate its impact in two popular cryptographic settings: bilinear pairings and supersingular isogeny-based protocols. For the former, we obtain a 1.37x speedup in the computation of a full optimal ate pairing over the popular BLS12-381 curve on an x64 Intel processor; for the latter, we show a speedup of up to 1.30x in the computation of the SIKE protocol on the same Intel platform. Available format(s) Category Public-key cryptography Publication info Preprint. MINOR revision. Keywords Prime fieldsextension fieldsbilinear pairingsBLS12-381supersingular isogeny-based cryptographySIKEefficient computation. Contact author(s) plonga @ microsoft com History Short URL https://ia.cr/2022/367 CC BY BibTeX @misc{cryptoeprint:2022/367, author = {Patrick Longa}, title = {Efficient Algorithms for Large Prime Characteristic Fields and Their Application to Bilinear Pairings and Supersingular Isogeny-Based Protocols}, howpublished = {Cryptology ePrint Archive, Paper 2022/367}, year = {2022}, note = {\url{https://eprint.iacr.org/2022/367}}, url = {https://eprint.iacr.org/2022/367} } Note: In order to protect the privacy of readers, eprint.iacr.org does not use cookies or embedded third party content.

2022-09-24 19:48:53

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https://parametricity.com/posts/2016-05-25-dubious-arguments/

# Parametricity There’s a famous dubious argument that “proves” the set of $\mathrm{Tree}$ of (planar, rooted) binary trees is in bijection with the set of $7$-tuples of trees $\mathrm{Tree}^6$. The argument goes as follows. A binary tree is either empty or a non-empty tree with two children which are binary trees. Thus we have an isomorphism $\mathrm{Tree} \cong 1 + \mathrm{Tree}^2$. Pretending that $\mathrm{Tree}$ is a complex number and using the quadratic formula we obtain $$\mathrm{Tree} = \frac{1\pm 3 i}{2} = e^{\frac{2\pi i}{3}}$$ so that $\mathrm{Tree}$ satisfies $\mathrm{Tree}^6 = 1$. Multiplying both sides by $\mathrm{Tree}$ yields $\mathrm{Tree}^7 = \mathrm{Tree}$ and so we conclude that the set of $7$-tuples of trees is in bijection with the set of trees. This is of course all a bit silly but as Fiore and Leinster showed, there is a general principle which provides legitimate proofs for arguments such as these. Specifically they prove the following. Let $p, q_1$ and $q_2$ be polynomials over $\mathbb{N}$ (with some conditions on the polynomials which I won’t mention). Suppose that $z = p(z)$ implies $q_1(z) = q_2(z)$ for all $z \in \mathbb{C}$. Then $z = p(z)$ implies $q_1(z) = q_2(z)$ in all categories with a product, coproduct, and terminal object. This theorem is strong enough to get us our bijection $\mathrm{Tree}^7 \cong \mathrm{Tree}$, but apparently not to fix all nonsense arguments. Consider the following. Let $t$ be the type defined by type t = | A0 | A1 of t | A2 of t * t (or in other words $t$ is the smallest set satisfying $t = 1 + t + t^2$ or $t$ is the initial algebra for the functor $X \mapsto 1 + X + X^2$ or…). Then if we imagine $t$ is a complex number, we have $0 = 1 + t^2$ so that $t = \pm i$. Let’s just pick $t = i$ for convenience since the two are indistinguishable. Then just by calculation, we have $\frac{2}{1 - t} = 1 + t$. Now let’s interpret this equation in terms of types. By another dubious argument, $\frac{1}{1 - t} = 1 + t + t^2 + \dots = \mathrm{List}(t)$ since $$1 = (1 + t + t^2 + \dots) - (t + t^2 + \dots) = (1 - t)(1 + t + t^2 + \dots).$$ Note that $1 + t + t^2 + \dots$ is the type of lists of $t$s since such a list is either empty, or a single element of $t$, or a pair, or a $3$-tuple, or… Our dubiously obtained equation tells us we should have a bijection $2 \times \mathrm{List}(t) \cong 1 + t$ where $2$ is a type with $2$ elements. There is in fact a very nice such bijection (or really, a family of such bijections indexed by the set of infinite binary strings) described below. I believe the existence of such a thing is not provided by Fiore and Leinster’s theorem. Where does it come from? # Spoiler: The bijection Let $X_n$ for $n \geq 0$ be the set of elements of $t$ which are of the form and $Y_n$ for $n \geq 1$ those of the form and $Y_0 = { * }$ where $*$ is some fresh guy not in $t$. Note that for $n > 0$, $X_n \cong Y_n \cong t^n$ by the obvious map which makes a list of all the dangling children $s_i$. It is clear that $X_n$ and $Y_n$ are disjoint and with a bit more thought that $1 + t \cong {*} \sqcup t = \bigsqcup_{n = 0}^\infty (X_n \sqcup Y_n)$ so that \begin{align} 1 + t &\cong \bigsqcup_{n = 0}^\infty (t^n \sqcup t^n) \ &\cong \bigsqcup_{n = 0}^\infty 2t^n \ &\cong 2 \times \bigsqcup_{n = 0}^\infty t^n \ &\cong 2 \times \mathrm{List}(t) \end{align} which gives the bijection. I say this is really a family of bijections indexed by infinite binary strings because we could have chosen any “spine” to decompose an element of $t$ along. We just happened to choose the spine $1111111111…$ (that is, “right, right, right, right,…").

2020-07-10 14:52:33

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http://mathematica.stackexchange.com/questions/41303/how-do-i-interpret-the-roots-syntax-when-i-am-finding-the-eigenvalues-of-a-matri

# How do I interpret the Roots syntax when I am finding the eigenvalues of a matrix [duplicate] Here is an example of my problem: I am somewhat confused by the syntax of this. I am guessing that the #1 is specifying the start of another root? Why then does one of the roots = u + with nothing after the plus? I think if I better grasped what the #1 stood for I might understand. Also, is there a reason that the Root ends with an &? Thanks! - ## marked as duplicate by Kuba, Yves Klett, Artes, belisarius has settled, bobthechemistJan 29 '14 at 14:34 This is likely to be closed as being due to a trivial mistake, but the #1 symbols are function notation. WolframAlpha presumably does not apply ToRadicals by default to cubic roots, so it gives an expression for the cubic roots in terms of generic Root[] objects. – DumpsterDoofus Jan 29 '14 at 5:13

2015-12-01 16:39:59

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https://discuss.codechef.com/questions/39130/iopc13-crazy-teacher

× # IOPC13: Crazy Teacher 0 I have tried solving this question http://www.codechef.com/IOPC2013/problems/IOPC13H , but iam not able to figure out an algorithm, can any one explain/ideas about?? asked 28 Feb '14, 20:15 2★sp1rs 963●5●13●27 accept rate: 0% 1 We know the formula for binomial coefficients: C(n,k)=n!/k!(n-k)!. We're asked to compute LCM(C(n,k)) over all 0 <= k <= n, but since the numerator of the fraction is constant and just the denominator (which is its divisor) changes, it's sufficient to compute D = GCD(k!(n-k)!) over all k, and the answer will be n!/D (since D is coprime to the prime modulus, division modulo 10^9+7 is done using the modular inverse, so it's n!*D^(10^9+5)). We can see that D(n) (D for n) divides D(n+1): D(n) divides all k!(n-k)!, and it can't be divided by anything if we just multiply all k!(n-k)! by n+1-k (and adding k=n+1, which is the same as k=0 and thus can be ignored). The problem now reduces to finding the quotient of D(n)/D(n-1) for all n. For this type of problems, there's one excellent strategy: write a bruteforce, find out answers for n <= 50 and try to guess the general formula. If you do this, you find out that the quotient is 1 for all prime n+1 and n+1 for almost all non-prime n+1. The only exceptions are prime powers - for n+1=p^a, the quotient D(n)/D(n-1) is p^(a-1). This gives us an easy way to compute D for all n <= 10^5 - for every n, we just need to check if it's prime or prime power, and that can be done by O(sqrt(n)) prime decomposition. Then, we compute (in O(log 10^9+5)) the modular inverse, update the current n! (multiply it by n+1, mod 10^9+7) and get the answers for all n in O(n sqrt(n)) total time. Every test case can be answered instantly, so the total time complexity is O(t+n sqrt(n)). answered 01 Mar '14, 00:25 7★xellos0 5.9k●5●43●93 accept rate: 10% toggle preview community wiki: Preview By Email: Markdown Basics • *italic* or _italic_ • **bold** or __bold__ • image? • numbered list: 1. Foo 2. Bar • to add a line break simply add two spaces to where you would like the new line to be. • basic HTML tags are also supported • mathemetical formulas in Latex between \$ symbol Question tags: ×1,219 ×639 ×51 question asked: 28 Feb '14, 20:15 question was seen: 2,268 times last updated: 01 Mar '14, 00:25

2019-03-19 17:15:46

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https://rpg.stackexchange.com/questions/2473/how-wealthy-can-a-d20-modern-character-be

# How wealthy can a d20 Modern character be? What's the maximum possible wealth bonus that a d20modern character can achieve by 10th level, excluding DM intervention? What is the max at 20th level? How rich would such a character be in real world dollars? • Perhaps I just missed it, but I don't see anything accounting for classes such as Ambassador which gets a stipend of +4 to wealth at 5th and 8th level. – Ace of space Jan 13 '11 at 14:09 The conditions required to get the maximum possible wealth bonus are very artificial, but for the sake of argument let's assume the following: • Dilettante taken as Profession • an 8 rolled on 2d4 for starting wealth • rolling 18 for Wisdom at character creation • always rolling 20 on wealth checks when leveling up • every feat spent on Windfall • always spending the maximum skill points on Profession (max rank = level + 3) • always increasing Wisdom • never buying anything that requires a wealth check • the DM completely ignores wealth, neither granting it nor threatening it Such a character can start with a wealth bonus of +20 at 1st level (8 from the 2d4 roll, +6 for Dilettante, and +3 x2 from taking Windfall twice). At level-up, there are three ways the wealth bonus might increase: • Succeeding on a Profession check, which gives +1, and an additional +1 for each margin of 5 that the roll passes the DC. Note that the DC is the current wealth bonus. • Buying the Windfall feat again, which gives a flat +3 bonus to wealth each time its bought. (And which also gives +1 to Profession rolls to increase wealth when leveling up.) • There is a flat increase to wealth when leveling up, depending on how many ranks in Profession the character has. e.g., +1 for Rank 1-5, +2 for Rank 6-10, etc. So we can figure out what the maximum theoretical wealth increase is for each level by tracking the things that give bonuses to the Profession roll (Wisdom, Windfall feats, and ranks in Profession), and comparing that to the current wealth rating (which is the DC for Profession checks when leveling up). We can then calculate the maximum bonus from the Profession roll, from new Windfall feats, and from the flat bonus from having ranks in Profession. That's summarised in the following table. S Increases t W ——————————————————————— C a e T M L R F h B r a M | F R o | a B e a e e o t l a R | R e a t I | x o v W n a c n i t x o | o a n a n | n e i k t k u n h l | l f k l c | u l | s | s | s | s | g | l | l | s | s | r.| s ———+————+———+—————+——————+———————+—————+————+————+————+——————+—————— 1: 18 4 x2 | — +20 — | — — — — | +20 2: 18 5 x3 | +12 +20 32 | +3, +3, +1, +7 | +27 3: 18 6 x4 | +14 +27 34 | +2, +3, +2, +7 | +34 4: 19 7 x5 | +16 +34 36 | +1, +3, +2, +6 | +40 5: 19 8 x5 | +17 +40 37 | +0, +0, +2, +2 | +42 6: 19 9 x7 | +20 +42 40 | +0, +6, +2, +8 | +50 7: 19 10 x7 | +21 +50 41 | +0, +0, +2, +2 | +52 8: 20 11 x8 | +24 +52 44 | +0, +3, +3, +6 | +58 9: 20 12 x9 | +26 +58 46 | +0, +3, +3, +6 | +64 10: 20 13 x10 | +28 +64 48 | +0, +3, +3, +6 | +70 11: 20 14 x10 | +29 +70 49 | +0, +0, +3, +3 | +73 12: 21 15 x12 | +32 +73 52 | +0, +6, +3, +9 | +82 13: 21 16 x12 | +33 +82 53 | +0, +0, +4, +4 | +86 14: 21 17 x13 | +35 +86 55 | +0, +3, +4, +7 | +93 15: 21 18 x14 | +37 +93 57 | +0, +3, +4, +7 | +100 16: 22 19 x15 | +40 +100 60 | +0, +3, +4, +7 | +107 17: 22 20 x15 | +41 +107 61 | +0, +0, +4, +4 | +111 18: 22 21 x17 | +44 +111 64 | +0, +6, +5, +11 | +122 19: 22 22 x17 | +45 +122 65 | +0, +0, +5, +5 | +127 20: 23 23 x18 | +47 +127 67 | +0, +3, +5, +8 | +135 Table key: • Level is the level just gained. • Wis is the character's Wisdom. • Ranks is ranks in Profession. • Feats is how many times the Windfall feat has been taken. • Check Bonus is the bonus to Profession checks for determining wealth bonus increases when leveling up. (Wis bonus + Profession ranks + 1 per Windfall feat.) • Starting Wealth is the wealth bonus before calculating the increase for this level. It's used as the DC of the Profession check. • Max Roll is the maximum possible Profession check: rolling 20 + check bonus from the earlier column. • Increases are the three ways the wealth bonus could increase that level, and their total: • Roll is the increase from a perfect 20 roll on the Profession check. • Feats is the increase from Windfall feats taken at that level (+3 per feat) • Ranks is the flat increase according to p. 70 due to the number of ranks in Profession. • Total is the total increase in wealth bonus for that level, and will be applied to the next level in the table. • Max Bonus is the final, maximum wealth bonus possible at that level, as a sum of the Starting Wealth column and the Increases: Total column. As you can see, the wealth bonus quickly becomes so large that it becomes impossible to beat the DC on the Profession check even when rolling perfect 20s. (And note that 20 is not an automatic success, see p. 44 Skill Checks.) After 4th level, only the Windfall feats and the flat bonus matter. So, the maximum wealth bonus at 10th level is +70 (wealth bonus starts at +64, plus the total increase of 6), and the maximum wealth bonus at 20th level is +135 (wealth bonus starts at +127, plus the total increase of +8). As for how much that is in real-world dollars, according to the system there is no way to translate it into dollars: [The] Wealth bonus [is] a sort of composite of her income, credit rating, and savings. […] Wealth is not a direct representation of a character's salary or how much money the character has socked away in the bank. […] The Wealth bonus simply represents your character's buying power at any given time. So a character with +70 Wealth is rich enough that credit is easy enough to come by that they almost never need to worry about money. For comparison, the table on p. 38 calls +31 and higher "Very rich". The only purchase that could even be noticed by such a character (i.e., which would drop their wealth by any amount) would have to have a Purchase DC of 85, which means they could have as many mansions (Purchase DC 36), Learjets (Purchase DC 40), or M1A1 Abrams tanks (Purchase DC 47) as they wanted to own. The only way money would ever be an object would be if the GM set an exorbitant custom DC for, say, buying a very large rival company. A character with +135 Wealth is so exceedingly rich that they don't ever have to even think about money—their wealth is so great that everyone knows they're good for it, and their vast network of accountants pays for everything out of the complex financial empire they own. They probably don't even know exactly how much they're worth, and considering how the stock markets work, their exact wealth is probably impossible to know until they tried to use every last penny of it, at which point there would probably be world-wide financial panic. They can probably buy entire countries, unofficially. • That is a beautiful answer. One thing that is that expensive though. Large starships. – C. Ross Sep 9 '10 at 17:42 • Good point! I didn't see anything that expensive in my copy of the core book, but supplements might cover near-future tech. But just so: a character that rich is dealing with wealth on the level of governments. Build that private starship! – SevenSidedDie Sep 9 '10 at 18:42 • Also, yeah, formatting that table was very satisfying. – SevenSidedDie Sep 9 '10 at 18:47 • when I last played D20 Modern, one of the player characters did focus his character on improving his wealth bonus. He didn't quite get to +100, but the party did construct a very impressive base of operations (Thunderbirds are go!) – Azeari Jan 13 '11 at 8:29 • These are the kind of answers that are just awesome, they are analyzed and broken down step by step and in the end a definitive answer is given. – Antonio Mar 10 '13 at 5:29 @SevenSidedDie has covered the first part of the question thoroughly. As for conversion to real world dollars, there is a table on page 145 of d20 Future (ISBN:0786934239): Purchase DC Item Cost 2 $5 15$500 20 $2,000 30$35,000 40 $650,000 50$12,000,000 60 $200,000,000 70$3,500,000,000 80 $65,000,000,000 The relationship between purchase DC and item cost is roughly exponential: Note that this is the price of an individual item that a character might purchase - not his or her total net worth. Any time a character buys an object or service with a purchase DC of 15 or higher, the character reduces his or her current Wealth bonus by an additional 1 point. For example, a character with +64 wealth bonus would lose 1 point when purchasing an item valued between$500 and $650,000,000; 2 points for items between$900 million and $12 billion; 1d6+1 points for items between$15 billion and $50 billion; and 2d6+1 points for items costing more than$50 billion (provided that they succeed on their wealth check). Regarding large starships: Most d20 Modern campaigns are set at Progress Level 5. GMs who want to make lower-PL and higher-PL items available to characters should adjust the purchase DCs of items as follows: • -2 to Purchase DC for each Progress Level lower than the current Progress Level, except in the case of valuable antiques. • +5 to Purchase DC for equipment from the next highest Progress Level (the limit for purchasing cutting-edge technology). d20 Future Tech (ISBN:0786939494) has some example starships and mecha, including a PL6 light freighter with Purchase DC 56 (+5 if purchased in PL5) and an orbital colony with Purchase DC 72. Given that the Space Shuttle Endeavour cost approximately $1.7 billion (Purchase DC 67), that seems a little low. The International Space Station, believed to be the most expensive object ever constructed, cost between 35 and 150 billion dollars (Purchase DC 78 to 83). • The list of the world's most expensive single objects on Wikipedia puts a wealth bonus of +70 into perspective: en.wikipedia.org/wiki/… (Three Gorges Dam was$25 billion; aircraft carrier $8 billion; Large Hadron Collider$6 billion) – Azeari Jan 13 '11 at 13:33 If you manage to roll an 8, add 6 for the Dilettante Profession, and 6 more for Windfall x2, you'll start with 20. If you're allowing cybernetics and have PL9, you can buy windfall as a cybernetic feat implant at DC 26. If you succeed 4 times, your wealth would be 22, 24, 26, 29. Once you are above 28, you can buy a Feat Plex (PL8) at dc 28, and fill it with 3 windfalls and a cybertaker feat. You can repeat this as many times as you would like and have infinite wealth and every feat in the game at first level. • But I guess that's technically Future, not Modern... – nsidaria Jan 21 '13 at 0:59 • And here I tought Pun-Pun was bad. – Zachiel May 21 '13 at 19:07

2020-06-01 17:12:54

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http://mfat.imath.kiev.ua/authors/name/?author_id=557

# V. Karakaya Search this author in Google Scholar Articles: 1 ### On the $F$-contraction properties of multivalued integral type transformations Methods Funct. Anal. Topology 25 (2019), no. 3, 282-288 The main purpose of this work is to extend the properties of multivalued transformations to the integral type transformations and to obtain the existence of fixed points under $F$-contraction. In addition, the results of this study were evaluated with some interesting example.

2019-11-18 18:40:30

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https://grmpy.readthedocs.io/en/master/

# Welcome to grmpy’s documentation!¶ grmpy is an open-source package for the simulation and estimation of the generalized Roy model. It serves as a teaching tool to promote the conceptual framework of the generalized Roy model, illustrate a variety of issues in the econometrics of policy evaluation, and showcase basic software engineering practices. We build on the following main references: James J. Heckman and Edward J. Vytlacil. Econometric evaluation of social programs, part I: Causal models, structural models and econometric policy evaluation. In Handbook of Econometrics, volume 6B, chapter 70, pages 4779–4874. Elsevier Science, 2007. James J. Heckman and Edward J. Vytlacil. Econometric evaluation of social programs, part II: Using the marginal treatment effect to organize alternative econometric estimators to evaluate social programs, and to forecast their effects in new environments. In Handbook of Econometrics, volume 6B, chapter 71, pages 4875–5143. Elsevier Science, 2007. Jaap H. Abbring and James J. Heckman. Econometric evaluation of social programs, part III: Distributional treatment effects, dynamic treatment effects, dynamic discrete choice, and general equilibrium policy evaluation. Handbook of Econometrics, volume 6B, chapter 72, pages 5145-5303. Elsevier Science, 2007. The remainder of this documentation is structured as follows. We first present the basic economic model and provide installation instructions. We then illustrate the basic use case of the package in a tutorial and showcase some evidence regarding its reliability. In addition we provide some information on the software engineering tools that are used for transparency and dependability purposes. The documentation concludes with further information on contributing, contact details as well as a listing of the latest releases. The package is used as a teaching tool for a course on the analysis human capital at the University of Bonn. The affiliated lecture material is available on GitHub .

2020-09-30 15:19:09

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https://www.r-bloggers.com/2012/01/heatmaps-controlling-the-color-representation-with-set-data-range/

Want to share your content on R-bloggers? click here if you have a blog, or here if you don't. Often you want to set the fixed colors for particular range of your dataset to be sure that the visual output is correctly represented. This is particularly useful for time series data, where the range or your dataset might drastically change during the course of the simulation. To do that in R, we need to set the “breaks” parameter in plotting functions such as image or heatmap.2. # gplots contains the heatmap.2 function library(gplots) # create 50x10 matrix of random values from [-1, +1] random.matrix <- matrix(runif(500, min = -1, max = 1), nrow = 50) # following code limits the lowest and highest color to 5%, and 95% of your range, respectively quantile.range <- quantile(random.matrix, probs = seq(0, 1, 0.01)) palette.breaks <- seq(quantile.range["5%"], quantile.range["95%"], 0.1) # use http://colorbrewer2.org/ to find optimal divergent color palette (or set own) color.palette <- colorRampPalette(c("#FC8D59", "#FFFFBF", "#91CF60"))(length(palette.breaks) - 1) heatmap.2( random.matrix, dendrogram = "row", scale = "none", trace = "none", key = FALSE, labRow = NA, labCol = NA, col = color.palette, breaks = palette.breaks ) Enjoy plotting! mintgene

2021-05-10 19:12:41

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http://acscihotseat.org/index.php?qa=537&qa_1=existence-neutral-probabilities-implies-implied-arbitrage

# Showing that the existence of risk neutral probabilities implies and is implied by no arbitrage +1 vote 79 views "Part of the fundamental theorem of asset pricing says that a market model is arbitrage free iff risk neutral probabilities exist: prove this for the one period binomial model" the way I went about this was to use the condition d<$$e^r$$<u for no arbitrage and that probabilities must lie in the interval [0,1], this worked except that I didn't get strict inequalities on the arbitrage condition commented May 23, 2017 by (4,010 points) Hi Anonymous. A tutor has looked at your question, but was not sure of the necessity regarding strict bounds. I have asked Alex and will post an Answer once I have one for you. answered May 26, 2017 by (560 points) selected May 30, 2017 You seem to be happy with the bulk of the answer: if e^r is outside of the interval [u,d], there is an easy long-short arbitrage, and the risk-neutral probabilities don't exist (because they are outside of [0,1] and therefore aren't probabilities); and if e^r is strictly inside [u,d], there is no arbitrage and the probabilities exist. This all agrees with the fundamental theorem of asset pricing. So far so good. There is a subtlety involved in the boundary cases of e^r=u or =d, which you would not have been able to spot until now. In these cases, we get q=1 or =0, which appear to be valid probabilities. However, we require the full definition of a risk-neutral probability measure, which I only give in slide set (4) (slide 20). Part of this definition is that the risk-neutral measure must be equivalent to the original, real-world measure. If q=1 or =0, the measure Q does not agree with P about what has probability and what does not, and is therefore not equivalent. In these cases, therefore, the risk-neutral measure does not exist, strictly speaking. Furthermore, the long-short arbitrages are available in these cases (make sure you can see why), so the fundamental theorem holds again. (This is all assuming that p is strictly inside [0,1], and not 0 or 1 exactly).

2019-02-17 23:37:16

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https://plato.stanford.edu/entries/qt-quantlog/supplement.html

Supplement to Quantum Logic and Probability Theory What follows is the briefest possible summary of the order-theoretic notions used in the main text. For a good introduction to this material, see Davey & Priestley (1990). More advanced treatments can be found in Grätzer (1998) and Birkhoff (1967). 1. Ordered Sets A partial ordering—henceforth, just an ordering—on a set $$P$$ is a reflexive, anti-symmetric, and transitive binary relation $$\unlhd$$ on $$P$$. Thus, for all $$p, q, r \in P$$, we have 1. $$p \unlhd p$$ 2. $$p \unlhd q$$ and $$q \unlhd p$$ only if $$p = q$$. 3. if $$p \unlhd q$$ and $$q \unlhd r$$ then $$p \unlhd r$$ If $$p \unlhd q$$, we speak of $$p$$ as being less than, or below $$q$$, and of $$q$$ as being greater than, or above $$p$$, in the ordering. A partially ordered set, or poset, is a pair $$(P, \unlhd)$$ where $$P$$ is a set and $$\unlhd$$ is a specified ordering on $$P$$. It is usual to let $$P$$ denote both the set and the structure, leaving $$\unlhd$$ tacit wherever possible. Any collection of subsets of some fixed set $$X$$, ordered by set-inclusion, is a poset; in particular, the full power set $$\wp(X)$$ is a poset under set inclusion. Let $$P$$ be a poset. The meet, or greatest lower bound, of $$p, q \in P$$, denoted by $$p\wedge q$$, is the greatest element of $$P$$—if there is one—lying below both $$p$$ and $$q$$. The join, or least upper bound, of $$p$$ and $$q$$, denoted by $$p\vee q$$, is the least element of $$P$$—if there is one—lying above both $$p$$ and $$q$$. Thus, for any elements $$p, q, r$$ of $$P$$, we have 1. if $$r \unlhd p\wedge q$$, then $$r \unlhd p$$ and $$r \unlhd q$$ 2. if $$p\vee q \unlhd r$$, then $$p \unlhd r$$ and $$q \unlhd r$$ Note that $$p\wedge p = p\vee p = p$$ for all $$p$$ in $$P$$. Note also that $$p \unlhd q$$ iff $$p\wedge q = p$$ iff $$p\vee q = q$$. Note that if the set $$P = \wp(X)$$, ordered by set-inclusion, then $$p\wedge q = p\cap q$$ and $$p\vee q = p\cup q$$. However, if $$P$$ is an arbitrary collection of subsets of $$X$$ ordered by inclusion, this need not be true. For instance, consider the collection $$P$$ of all subsets of $$X = \{1,2,\ldots ,n\}$$ having even cardinality. Then, for instance, $$\{1,2\}\vee \{2,3\}$$ does not exist in $$P$$, since there is no smallest set of 4 elements of $$X$$ containing $$\{1,2,3\}$$. For a different sort of example, let $$X$$ be a vector space and let $$P$$ be the set of subspaces of $$X$$. For subspaces $$\mathbf{M}$$ and $$\mathbf{N}$$, we have $\mathbf{M}\wedge \mathbf{N} = \mathbf{M}\cap \mathbf{N}, \textrm{ but } \mathbf{M}\vee \mathbf{N} = \textrm{span} (\mathbf{M}\cup \mathbf{N}).$ The concepts of meet and join extend to infinite subsets of a poset $$P$$. Thus, if $$A\subseteq P$$, the meet of $$A$$ is the largest element (if any) below $$A$$, while the join of $$A$$ is the least element (if any) above $$A$$. We denote the meet of $$A$$ by $$\wedge\, A$$ or by $$\wedge_{a\in A}\, a$$. Similarly, the join of $$A$$ is denoted by $$\vee\, A$$ or by $$\vee_{a\in A}\, a$$. 2. Lattices A lattice is a poset $$(L, \unlhd)$$ in which every pair of elements has both a meet and a join. A complete lattice is one in which every subset of $$L$$ has a meet and a join. Note that $$\wp(X)$$ is a complete lattice with respect to set inclusion, as is the set of all subspaces of a vector space. The set of finite subsets of an infinite set $$X$$ is a lattice, but not a complete lattice. The set of subsets of a finite set having an even number of elements is an example of a poset that is not a lattice. A lattice $$(L, \unlhd)$$ is distributive iff meets distribute over joins and vice versa: $p \wedge (q\vee r) = (p\wedge q) \vee (p\wedge r),$ and $p \vee (q\wedge r) = (p\vee q) \wedge (p\vee r)$ The power set lattice $$\wp(X)$$, for instance, is distributive (as is any lattice of sets in which meet and join are given by set-theoretic intersection and union). On the other hand, the lattice of subspaces of a vector space is not distributive, for reasons that will become clear in a moment. A lattice $$L$$ is said to be bounded iff it contains a smallest element 0 and a largest element 1. Note that any complete lattice is automatically bounded. For the balance of this appendix, all lattices are assumed to be bounded, absent any indication to the contrary. A complement for an element $$p$$ of a (bounded) lattice $$L$$ is another element $$q$$ such that $$p \wedge q = 0$$ and $$p \vee q = 1$$. In the lattice $$\wp(X)$$, every element has exactly one complement, namely, its usual set-theoretic complement. On the other hand, in the lattice of subspaces of a vector space, an element will typically have infinitely many complements. For instance, if $$L$$ is the lattice of subspaces of 3-dimensional Euclidean space, then a complement for a given plane through the origin is provided by any line through the origin not lying in that plane. Proposition: If $$L$$ is distributive, an element of $$L$$ can have at most one complement. Proof: Suppose that $$q$$ and $$r$$ both serve as complements for $$p$$. Then, since $$L$$ is distributive, we have \begin{align} q & = q \wedge 1 \\ & = q \wedge (p \vee r)\\ & = (q\wedge p) \vee (q\wedge r)\\ & = 0 \vee (q\wedge r)\\ & = q\wedge r\\ \end{align} Hence, $$q \unlhd r$$. Symmetrically, we have $$r \unlhd q$$; thus, $$q = r$$. Thus, no lattice in which elements have multiple complements is distributive. In particular, the subspace lattice of a vector space (of dimension greater than 1) is not distributive. If a lattice $$is$$ distributive, it may be that some of its elements have a complement, while others lack a complement. A distributive lattice in which every element has a complement is called a Boolean lattice or a Boolean algebra. The basic example, of course, is the power set $$\wp(X)$$ of a set $$X$$. More generally, any collection of subsets of $$X$$ closed under unions, intersections and complements is a Boolean algebra; a theorem of Stone and Birkhoff tells us that, up to isomorphism, every Boolean algebra arises in this way. 3. Ortholattices In some non-uniquely complemented (hence, non-distributive) lattices, it is possible to pick out, for each element $$p$$, a preferred complement $$p'$$ in such a way that 1. if $$p \unlhd q$$ then $$q' \unlhd p'$$ 2. $$p'' = p$$ When these conditions are satisfied, one calls the mapping $$p\rightarrow p'$$ an orthocomplementation on $$L$$, and the structure $$(L, \unlhd ,')$$ an orthocomplemented lattice, or an ortholattice for short. Note again that if a distributive lattice can be orthocomplemented at all, it is a Boolean algebra, and hence can be orthocomplemented in only one way. In the case of $$L(\mathbf{H})$$ the orthocomplementation one has in mind is $$\mathbf{M} \rightarrow \mathbf{M}^{\bot}$$ where $$\mathbf{M}^{\bot}$$ is defined as in Section 1 of the main text. More generally, if $$\mathbf{V}$$ is any inner product space (complete or not), let $$L(\mathbf{V})$$ denote the set of subspaces $$\mathbf{M}$$ of $$\mathbf{V}$$ such that $$\mathbf{M} = \mathbf{M}^{\bot \bot}$$ (such a subspace is said to be algebraically closed). This again is a complete lattice, orthocomplemented by the mapping $$\mathbf{M} \rightarrow \mathbf{M}^{\bot}$$. 4. Orthomodularity There is a striking order-theoretic characterization of the lattice of closed subspaces of a Hilbert space among lattices $$L(\mathbf{V})$$ of closed subspaces of more general inner product spaces. An ortholattice $$L$$ is said to be orthomodular iff, for any pair $$p, q$$ in $$L$$ with $$p \unlhd q$$, $\tag{OMI} (q\wedge p')\vee p = q.$ Note that this is a weakening of the distributive law. Hence, a Boolean lattice is orthomodular. It is not difficult to show that if $$\mathbf{H}$$ is a Hilbert space, then $$L(\mathbf{H})$$ is orthomodular. The striking converse of this fact is due to Amemiya and Araki (1966): Theorem: Let $$\mathbf{V}$$ be an inner product space (over $$\mathbf{R}, \mathbf{C}$$ or the quaternions) such that $$L(\mathbf{V})$$ is orthomodular. Then $$\mathbf{V}$$ is complete, i.e., a Hilbert space. 5. Closure Operators, Interior Operators and Adjunctions Let $$P$$ and $$Q$$ be posets. A mapping $$f : P \rightarrow Q$$ is order preserving iff for all $$p,q \in P$$, if $$p \unlhd q$$ then $$f(p) \unlhd f(q)$$. A closure operator on a poset $$P$$ is an order-preserving map $$\mathbf{cl} : P \rightarrow P$$ such that for all $$p \in P$$, • $$\mathbf{cl}(\mathbf{cl}(p)) = \mathbf{cl}(p)$$ • $$p \unlhd \mathbf{cl}(p)$$. Dually, an interior operator on $$P$$ is an order-preserving mapping $$\mathbf{int} : P \rightarrow P$$ on $$P$$ such that for all $$p \in P$$, • $$\mathbf{int}(\mathbf{int}(p)) = \mathbf{int}(p)$$ • $$\mathbf{int}(p) \unlhd p$$ Elements in the range of $$\mathbf{cl}$$ are said to be closed; those in the range of $$\mathbf{int}$$ are said to be open. If $$P$$ is a (complete) lattice, then the set of closed, respectively open, subsets of $$P$$ under a closure or interior mapping is again a (complete) lattice. By way of illustration, suppose that $$\mathcal{O}$$ and $$\mathcal{C}$$ are collections of subsets of a set $$X$$ with $$\mathcal{O}$$ closed under arbitrary unions and $$\mathcal{C}$$ under arbitrary intersections. For any set $$A \subseteq X$$, let \begin{align} \mathbf{cl}(A) & = \cap \{C\in \mathcal{C} \mid A \subseteq C\}, \textrm{ and}\\ \mathbf{int}(A) & = \cup \{O\in \mathcal{O} \mid O \subseteq A\} \end{align} Then $$\mathbf{cl}$$ and $$\mathbf{int}$$ are interior operators on $$\wp(X)$$, for which the closed and open sets are precisely $$\mathcal{C}$$ and $$\mathcal{O}$$, respectively. The most familiar example, of course, is that in which $$\mathcal{O}, \mathcal{C}$$ are the open and closed subsets, respectively, of a topological space. Another important special case is that in which $$\mathcal{C}$$ is the set of linear subspaces of a vector space $$\mathbf{V}$$; in this case, the mapping span :$$\wp(\mathbf{V}) \rightarrow \wp(\mathbf{V})$$ sending each subset of $$\mathbf{V}$$ to its span is a corresponding closure. An adjunction between two posets $$P$$ and $$Q$$ is an ordered pair $$(f, g)$$ of mappings $$f : P \rightarrow Q$$ and $$g : Q \rightarrow P$$ connected by the condition that, for all $$p \in P, q \in Q$$ $f(p) \unlhd q \textrm{ if and only if } p \unlhd g(q).$ In this case, we call $$f$$ a left adjoint for $$g$$, and call $$g$$ a right adjoint for $$f$$. Two basic facts about adjunctions, both easily proved, are the following: Proposition: Let $$f : L \rightarrow M$$ be an order-preserving map between complete lattices $$L$$ and $$M$$. Then 1. $$f$$ preserves arbitrary joins if and only if it has a right adjoint. 2. $$f$$ preserves arbitrary meets if and only if it has a left adjoint. Proposition: Let $$(f, g)$$ be an adjunction between complete lattices $$L$$ and $$M$$. Then 1. $$g \circ f : L \rightarrow L$$ is a closure operator. 2. $$f \circ g : M \rightarrow M$$ is an interior operator.

2018-06-21 17:51:15

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https://quantumcomputing.stackexchange.com/questions/4260/how-to-create-a-condition-on-only-one-classical-bit-when-we-have-a-total-of-2-cl

# How to create a condition on only one classical bit when we have a total of 2 classic bits in the system I am trying to make a quantum circuit with one qubit and 2 classical bits for each measurment in the system below: I want to make condition on the first bit: if the first collapse to zero so x operator is act on the circuit, else (one) nothing is acting on the circuit. I am using qiskit language. but when I try to create my circuit, there is always an error: #definitions q = QuantumRegister(1) c = ClassicalRegister(2) qc = QuantumCircuit(q,c) # building the circuit qc.h(q) qc.measure(q[0],c[0]) qc.x(q[0]).c[0]_if(c[0], 0) qc.measure(q[0],c[1]) circuit_drawer(qc) and the error is: File "<ipython-input-4-66c70285946b>", line 3 qc.x(q[0]).c[0]_if(c[0], 0) ^ SyntaxError: invalid syntax how to write it correctly? When I try to change qc.x(q[0]).c[0]_if(c[0], 0) with: qc.x(q).c_if(c, 0) I succeed in building my circuit but I get circuit that I dont want to work with: I wish for help, thanks. The controlled NOT gate does the opposite of what you want, it applies $X$ to the target qubit if the control qubit is 1, and does nothing if the control qubit is 0. What you want is to apply $X$ when the control qubit is 0 and do nothing when it is 1. This can be accomplished by applying a NOT gate (i.e. an $X$ gate) before doing the CNOT. In the IBM composer it would look like this: The code for doing CNOT in quiskit is: gate cx c,t { CX c,t; } Since there is a specific gate for what you want to do, you do not need any "if statements"! • Do I need to put this algorithm between 2 measurments? – Daniel Vainshtein Sep 20 '18 at 11:35 • That depends on what you want to do. One thing I can say though, is that you second measurement should just read 0. – user1271772 Sep 20 '18 at 11:43 • If that is all ou want to do, the circuit I gave you in my answer is enough. You don't need anything else. – user1271772 Sep 20 '18 at 12:02 • It depends what you want to do brother. I have you the circuit for |00> to |01> and |10> to |10>. I don't know what your over goal is. – user1271772 Sep 20 '18 at 12:16 • @DanielVainshtein: Considering how many follow-up questions you've asked, and the amount of effort put into the answer, it might be a good idea to show some appreciation when people invest their time to helping you :) – user1271772 Sep 20 '18 at 14:28 See the answer to this question for more on how the classical control works. Basically, your operations are controlled on the integer stored (in binary) across a register rather than on the individual bits themselves. I also don't quite know the 'best practice' way of controlling on single bits, but I can tell you my workaround. Instead of creating a register with two bits, I create a list of two single qubit registers. c = [ ClassicalRegister(1) for _ in range(2) ] These can be added using the add method of a quantum circuit. q = QuantumRegister(1) qc = QuantumCircuit(q) for register in c: qc.h(q) This works because c[0]` now refers to a classical register, rather than a single bit from a classical register.

2019-07-24 02:07:48

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http://openstudy.com/updates/4dc7f9eb40ec8b0bcea10717

## anonymous 5 years ago i have a question i don't quite understand, i am preparing for a exam. A 100-gallon mixture of citrus extract and water is 3% citrus extract, and it needs to be diluted. The function f(x)=3/100+x represents the percentrage (in decimal form) of the miture that is citrus extract when x gallons of water are added. Find the domain of f(x). Then, find the percentage of citrus extract in the mixture after 50 gallons of water are added to it. I assume this is $f(x)=\frac{3}{100+x}$ since x is supposed to be water added, then x must be greater than zero. If you are allowed to take away water then x must be greater than -100. to find the percentage, replace x by 50 to get 3/150= .02 = 2%

2016-10-23 16:18:40

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https://golem.ph.utexas.edu/~distler/blog/archives/000455.html

## October 18, 2004 ### Small Pond I hope he doesn’t mind my embarrassing him further, but Matthew Nobes has the first in (what I hope is) a series of posts in which he discusses some of the analytical tools of modern lattice gauge theory. A lot of the improvement in lattice gauge theory has come about not so much by “brute-force” going to finer lattices as from being smarter about our computations. The number of lattice sites scales like $a^{-4}$ so, even tremendous strides in the computer power available lets you cut the lattice spacing by only a relatively modest amount. Instead, the big improvements have come from being “smarter” about what to compute. By comparing lattice perturbation theory to the continuum, and using the results to fine-tune the lattice action, one can minimize the effects of the discretization and get vastly improved results on the same “coarse” lattice. 1. In comparing lattice perturbation theory to the continuum, we are interested in comparing physical quantities at some distance scale, $L$, long compared to the lattice spacing, but still much shorter than the size of the box. We should take care to use a “renormalization group-improved” lattice perturbation theory, rather than expressing our answers in terms of “bare” coupling(s) in the lattice action. The RG-improved perturbation series is generally much more accurate than the naive one in the bare lattice coupling. 2. Moreover, we can make the RG flow converge faster to the continuum by working, not with the “naive” Wilson action, but with an “improved” lattice action, containing “higher derivative” interactions, with cleverly-chosen coefficients. Choose the coefficients correctly, and the errors scale to zero much faster as $a/L\to 0$. Matthew’s one of the guys who figures out how to do this in practice. Recall that we were worried about finite spacing errors in lattice field theory. As an example we were using a scalar field coupled to gluons. The basic action was $\phi D^{2} \phi$ and this has $a^2$ errors. I said that we could use $\phi (D^2 + C a^2 D^4) \phi$ to reduce these errors. Clearly this involves picking some value for $C$, but how do we do that? It pays to remember what the lattice is doing for us. It’s cutting the theory off at the small distance $a$, or in momentum space at high energy/momentum. So the spacing errors are reflecting a problem with the high energy (short distance) part of the theory. Now way back at the start of the first post we noted that QCD is perturbative at high energy. So we ought to be able to correct for the spacing errors perturbativly, by matching our lattice theory to the continuum theory to some order in perturbation theory. We pick some scattering amplitude, and fiddle with $C$, order by order. Done properly, this lowers the spacing errors, at a modest performance cost. Posted by distler at October 18, 2004 11:13 PM TrackBack URL for this Entry:   http://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/455

2015-07-04 13:40:10

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https://icet.materialsmodeling.org/moduleref_icet/cluster_space.html

Cluster space¶ Setting up a cluster space¶ The cluster space serves as a basis for cluster expansions. Setting up a clusterspace object is commonly done via >>> from icet import ClusterSpace >>> from ase.build import bulk >>> primitive_structure = bulk('Si') >>> cs = ClusterSpace(primitive_structure, cutoffs=[7.0, 5.0], chemical_symbols=['Si', 'Ge']) The cutoffs are set up to include pairs with an interatomic distance smaller than 7 Å and triplets for which all pair-wise interatomic distances are smaller than 5 Å. Here, Si and Ge are allowed to occupy the reference lattice. Sublattices¶ A cluster space can also be constructed with multiple sublattices. Consider for example a rocksalt lattice, on which we want to allow mixing of Na and Li on the Na sublattice and Cl and F on the Cl sublattices. This can be achived by >>> from ase.build import bulk >>> atoms = bulk('NaCl', 'rocksalt', a=4.0) >>> chemical_symbols = [['Na', 'Li'], ['Cl', 'F']] >>> cs = ClusterSpace(atoms, [7.0, 5.0], chemical_symbols) where chemical_symbols now specifies which species are allowed for each lattice site in atoms. Inactive sites¶ The sublattice functionality also allows one to have inactive sites. For example, if we consider the system above but would like to keep the Cl lattice fixed it can be achived via >>> chemical_symbols = [['Na', 'Li'], ['Cl']] >>> cs = ClusterSpace(atoms, [7.0, 5.0], chemical_symbols) 2D systems¶ icet requires input structures to have periodic boundary conditions (PBCs). In order to treat two-dimensional systems, or more generally without PBCs in at least one direction, one has to surround the prototype structure with vacuum and then apply PBCs in all directions. This can easily be achived with the ASE functions >>> from ase.build import surface >>> atoms = surface('Cu', (1,1,1), layers=7) which creates ase atoms object without PBC in the z-direction. The structure can be modified to have PBC in the z-direction with vacuum via >>> atoms.pbc = [True, True, True] >>> atoms.center(vacuum=20, axis=2) which can now be used to create a cluster space. Interface¶ class icet.ClusterSpace(structure: ase.atoms.Atoms, cutoffs: List[float], chemical_symbols: Union[List[str], List[List[str]]], symprec: float = 1e-05, position_tolerance: float = None)[source] This class provides functionality for generating and maintaining cluster spaces. Note: In icet all ase.Atoms objects must have periodic boundary conditions. When carrying out cluster expansions for surfaces and nanoparticles it is therefore recommended to surround the structure with vacuum and use periodic boundary conditions. This can be done using e.g., ase.Atoms.center(). Parameters • structure (ase.Atoms) – atomic structure • cutoffs (list(float)) – cutoff radii per order that define the cluster space Cutoffs are specified in units of Angstrom and refer to the longest distance between two atoms in the cluster. The first element refers to pairs, the second to triplets, the third to quadruplets, and so on. cutoffs=[7.0, 4.5] thus implies that all pairs distanced 7 A or less will be included, as well as all triplets among which the longest distance is no longer than 4.5 A. • chemical_symbols (list(str) or list(list(str))) – list of chemical symbols, each of which must map to an element of the periodic table If a list of chemical symbols is provided, all sites on the lattice will have the same allowed occupations as the input list. If a list of list of chemical symbols is provided then the outer list must be the same length as the structure object and chemical_symbols[i] will correspond to the allowed species on lattice site i. • symprec (float) – tolerance imposed when analyzing the symmetry using spglib • position_tolerance (float) – tolerance applied when comparing positions in Cartesian coordinates Examples The following snippets illustrate several common situations: >>> from ase.build import bulk >>> from icet import ClusterSpace >>> # AgPd alloy with pairs up to 7.0 A and triplets up to 4.5 A >>> prim = bulk('Ag') >>> cs = ClusterSpace(structure=prim, cutoffs=[7.0, 4.5], ... chemical_symbols=[['Ag', 'Pd']]) >>> print(cs) >>> # (Mg,Zn)O alloy on rocksalt lattice with pairs up to 8.0 A >>> prim = bulk('MgO', crystalstructure='rocksalt', a=6.0) >>> cs = ClusterSpace(structure=prim, cutoffs=[8.0], ... chemical_symbols=[['Mg', 'Zn'], ['O']]) >>> print(cs) >>> # (Ga,Al)(As,Sb) alloy with pairs, triplets, and quadruplets >>> prim = bulk('GaAs', crystalstructure='zincblende', a=6.5) >>> cs = ClusterSpace(structure=prim, cutoffs=[7.0, 6.0, 5.0], ... chemical_symbols=[['Ga', 'Al'], ['As', 'Sb']]) >>> print(cs) >>> # PdCuAu alloy with pairs and triplets >>> prim = bulk('Pd') >>> cs = ClusterSpace(structure=prim, cutoffs=[7.0, 5.0], ... chemical_symbols=[['Au', 'Cu', 'Pd']]) >>> print(cs) assert_structure_compatibility(structure: ase.atoms.Atoms, vol_tol: float = 1e-05) → None[source] Raises error if structure is not compatible with ClusterSpace. Todo Add check for if structure is relaxed. Parameters structure – structure to check if compatible with ClusterSpace property chemical_symbols Species identified by their chemical symbols copy()[source] Returns copy of ClusterSpace instance. property cutoffs Cutoffs for different n-body clusters. The cutoff radius (in Angstroms) defines the largest interatomic distance in a cluster. evaluate_cluster_function() Evaluates value of a cluster function. property fractional_position_tolerance tolerance applied when comparing positions in fractional coordinates get_chemical_symbols() Returns list of species associated with cluster space as chemical symbols. get_cluster_vector(structure: ase.atoms.Atoms)numpy.ndarray[source] Returns the cluster vector for a structure. Parameters structure – atomic configuration Returns Return type the cluster vector get_coordinates_of_representative_cluster(orbit_index: int) → List[Tuple[float]][source] Returns the positions of atoms in the selected orbit Parameters orbit_index – index of the orbit from which to calculate the positions of the atoms Returns Return type list of positions of atoms in the selected orbit get_cutoffs() get_multi_component_vector_permutations() get_multi_component_vectors_by_orbit() get_number_of_allowed_species_by_site() get_number_of_orbits_by_order() → collections.OrderedDict[source] Returns the number of orbits by order. Returns • an ordered dictionary where keys and values represent order and number • of orbits, respectively get_orbit() get_possible_orbit_occupations(orbit_index: int) → List[List[str]][source] Returns possible occupation of the orbit. Parameters orbit_index get_sublattices(structure: ase.atoms.Atoms) → icet.core.sublattices.Sublattices[source] Returns the sublattices of the input structure. Parameters structure – structure the sublattices are based on is_supercell_self_interacting(structure: ase.atoms.Atoms) → bool[source] Checks whether an structure has self-interactions via periodic boundary conditions. Parameters structure – structure to be tested Returns If True, the structure contains self-interactions via periodic boundary conditions, otherwise False. Return type bool property orbit_data list of orbits with information regarding order, radius, multiplicity etc property orbit_list Orbit list that defines the cluster in the cluster space property position_tolerance tolerance applied when comparing positions in Cartesian coordinates property primitive_structure Primitive structure on which cluster space is based print_overview(print_threshold: int = None, print_minimum: int = 10) → None[source] Print an overview of the cluster space in terms of the orbits (order, radius, multiplicity etc). Parameters • print_threshold – if the number of orbits exceeds this number print dots • print_minimum – number of lines printed from the top and the bottom of the orbit list if print_threshold is exceeded static read(filename: str)[source] Parameters filename – name of file from which to read cluster space property space_group space group of the primitive structure in international notion (via spglib) property species_maps property symprec tolerance imposed when analyzing the symmetry using spglib write(filename: str) → None[source] Saves cluster space to a file. Parameters filename – name of file to which to write

2021-05-06 16:24:00

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https://tex.stackexchange.com/questions/235704/multiple-versions-include-of-a-chapter

Multiple versions include of a chapter I would like to include a chapter two times, a short version in the main part and a long version in the appendix. Everything works fine except the labels and references. Is there a way to redefine labels in a local context? I could us a \IFDEFINE statement around every label or ref but that is not very convenient. Any suggestions welcome, Thx!! ==================== main.tex ================= \documentclass[12pt,a4paper]{book} \usepackage[utf8]{inputenc} \usepackage[german]{babel} \begin{document} \include{kapitel} \begin{appendix} \newcommand*{\LONGVERSION}{} \include{kapitel} \end{appendix} \end{document} ==================== kapitel.tex ================= \chapter{This is a chapter}\label{chap:mychapter} This is a test with a reference to my chapter \ref{chap:mychapter} \ifdefined\LONGVERSION \section{Variant LONG} This is a long version of the chapter, which contains a lot of additional information. This shall be placed in the appendix. \else \section{Variant SHORT} This is the short version. \fi • Crosspost to goLaTeX and TeXwelt. Mar 29, 2015 at 11:35 • Great solution, thanks for your expertise. However the soution adresses two different topics. 1.) replacing my construction "\ifdefined" with "\iftoggle" (nice but not strictly necessary) 2.) Main point is using input instead of include in one instance and the self defined prefix addon. GREAT HELP!! Mar 29, 2015 at 12:45 • @MW2015: So what does my solution mean? Is it good,is it bad? It was a quick solution, due to lack of time – user31729 Mar 29, 2015 at 15:21 • @MW2015: Es ist nicht ok, wenn eine Lösung hier angeboten und dann einfach in einem anderen Forum übernommen wird. Böses Foul ;-) – user31729 Mar 29, 2015 at 15:27 • No offense intended. Added references. THX!! Mar 29, 2015 at 18:24 One possibility is to use a \labelprefix which is changed according to the toggle switch longversion (I used etoolbox macros for this) and a predefined \shortlabelprefix and \longlabelprefix. However, this does not work, if the short version of the chapter is input via \include twice, since each input overwrites the old label values. main.tex \documentclass[12pt,a4paper]{book} \usepackage{etoolbox} \usepackage[utf8]{inputenc} \usepackage[ngerman]{babel} \newcommand\shortlabelprefix{short}% \newcommand\longlabelprefix{long}% \newcommand\labelprefix{\shortlabelprefix}% \newtoggle{longversion} \togglefalse{longversion} \begin{document} \input{kapitel} \begin{appendix} \renewcommand{\labelprefix}{\longlabelprefix} \toggletrue{longversion} \include{kapitel} \end{appendix} \end{document} kapitel.tex \chapter{This is a chapter}\label{\labelprefix::chap:mychapter} This is a test with a reference to my chapter \ref{\labelprefix::chap:mychapter} \iftoggle{longversion}{% \section{Variant LONG} This is a long version of the chapter, which contains a lot of additional information. This shall be placed in the appendix. }{% \section{Variant SHORT} This is the short version. } short variant long variant • There is a comment to the question, i think it is addressed to you? Mar 29, 2015 at 12:47 • definitely not me. Mar 29, 2015 at 12:47 • @Johannes_B: ??? Definitely not you? – user31729 Mar 29, 2015 at 13:37 • Definitely not addresssed to me. Mar 29, 2015 at 13:46 • @Johannes_B: Yes, and my solution is obviously not correct ;-) – user31729 Mar 29, 2015 at 15:13

2023-03-25 23:08:59

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https://www.jiskha.com/questions/595522/f-x-1-x-and-g-x-1-x-write-in-the-form-ax-b-cx-d-a-what-are-all-the-possible

# Calculus F(x) = 1-x and g(x) = 1/x Write in the form (ax + b)/(cx + d) a.What are all the possible values of a, b, c, and d? b.Find all the functions fo the form f (x) = (ax + b)/(cx + d) with the property that f o f^-1 = x 1. 👍 0 2. 👎 0 3. 👁 149 ## Similar Questions 1. ### Math help, anyone? Solve the proportion. Where necessary, round to the nearest hundredth. Plz help i really need 2 get this in!! 11. 9/10 = x/10 9 *** 90 10 100 12. x/4 = 6/7 168 0.21 28 3.43 13.Write the ratio as a fraction in simplest form. 22:36 2. ### math 1. Divide. Write your answer in simplest form. 2/3 / 1/4 2. Find the product. 1/2 x 3/7 3. Divide. Write the quotient in simplest form. 3/5 / 2/10 4. Divide. Write the quotient in simplest form. 3/4 / 4/5 plz help me 3. ### Math Write an equation in slope-intercept form, point-slope, or standard form for the line with the given information. Explain why you chose the form you used. a. Passes through (-1, 4) and (-5, 2) 4. ### Math 1. Use point-slope form to write the equation of a line that has a slope of 2/3 and passes through (-3, -1). Write your final equation in slope-intercept form. 2. Write the equation in standard form using integers (no fractions or 1. ### Math Considering the word Mathematics, what is the ratio of vowels to consonants? Write the ratio 3 ways. word form: colon form: fraction form: 2. ### Algebra Write each in Standard Form and Slope-Intercept Form: 4x = 2y -10 Given A(-4,-2), B(44), and C(18,-8, answer the following questions Write the equations of the line containing the altitude the passes through B in standard form. Write the equation of the line containing the median that passes 4. ### math write each decimal in standard form, expanded form, and word form. 12+0.2+0.005 1. ### math (A)Write the equation in standard form and calculate its discriminant. (B)Solve the equation by using the quadratic formula. (C)After solving the equation, write it in factored form. 1/8x^2=x-5/2 Please help!!!!! :( 2. ### Spanish This is my assignment: Imagine you are babysitting your little brother. Write a total of 5 Spanish sentences using tú commands telling your little brother what to do. Write 3 sentences using regular verbs. Write 2 sentences using 3. ### college algebra Use point-slope form to write the equation of the line with the given properties. Write the equation in standard form. m is undefined passing through P(-7, -7) 4. ### PRECALCULUS 1.) determine two pairs of polar coordinates for the point (-4sqrt3, 4sqrt3)with 0degrees less than or equal to theta less than or equal to 360degrees 2.) Use DeMoivre's Theorem to find (1-i)^10. write your answer in the form

2020-11-29 02:25:11

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http://www.all-science-fair-projects.com/science_fair_projects_encyclopedia/Confidence_interval

# All Science Fair Projects ## Science Fair Project Encyclopedia for Schools! Search    Browse    Forum  Coach    Links    Editor    Help    Tell-a-Friend    Encyclopedia    Dictionary # Science Fair Project Encyclopedia For information on any area of science that interests you, enter a keyword (eg. scientific method, molecule, cloud, carbohydrate etc.). Or else, you can start by choosing any of the categories below. # Confidence interval In statistics, confidence intervals are the most prevalent form of interval estimation. If U and V are statistics (i.e., "observable" random variables) whose probability distribution depends on some unobservable parameter θ, and the relation P(U < θ < V) = 0.9 then the random interval (U,V) is a "90% confidence interval for θ". ## How to misunderstand confidence intervals It is very tempting to misunderstand this statement in the following way. We used capital letters U and V for random variables; it is conventional to use lower-case letters u and v for their observed values in a particular instance. The misunderstanding is the conclusion that P(u < θ < v) = 0.9, so that after the data has been observed, a conditional probability distribution of θ, given the data, is inferred. For example, suppose X is normally distributed with expected value θ and variance 1. (It is grossly unrealistic to take the variance to be known while the expected value must be inferred from the data, but it makes the example simple.) The random variable X is observable. (The random variable X − θ is not observable, since its value depends on θ.) Then X − θ is normally distributed with expectation 0 and variance 1; therefore P( - 1.645 < X - θ < 1.645) = 0.9. Consequently P(X - 1.645 < θ < X + 1.645) = 0.9, so the interval from X − 1.645 to X + 1.645 is a 90% confidence interval for θ. But when X = 82 is observed, can we then say that $P(82-1.645<\theta<82+1.645)=0.9\ \mbox{?}$ This conclusion does not follow from the laws of probability because θ is not a "random variable"; i.e., no probability distribution has been assigned to it. Confidence intervals are generally a frequentist method, i.e., employed by those who interpret "90% probability" as "occurring in 90% of all cases". Suppose, for example, that θ is the mass of the planet Neptune, and the randomness in our measurement error means that 90% of the time our statement that the mass is between this number and that number will be correct. The mass is not what is random. Therefore, given that we have measured it to be 82 units, we cannot say that in 90% of all cases, the mass is between 82 − 1.645 and 82 + 1.645. There are no such cases; there is, after all, only one planet Neptune. But if probabilities are construed as degrees of belief rather than as relative frequencies of occurrence of random events, i.e., if we are Bayesians rather than frequentists, can we then say we are 90% sure that the mass is between 82 − 1.645 and 82 + 1.645? Many answers to this question have been proposed, and are philosophically controversial. The answer will not be a mathematical theorem, but a philosophical tenet. Less controversial are Bayesian credible intervals , in which one starts with a prior probability distribution of θ, and finds a posterior probability distribution, which is the conditional probability distribution of θ given the data. For users of frequentist methods, the explanation of a confidence interval can amount to something like: "The confidence interval represents values for the population parameter for which the difference between the parameter and the observed estimate is not statistically significant at the 10% level". Critics of frequentist methods suggest that this hides the real and, to the critics, incomprehensible frequentist interpretation which might be expressed as: "If the population parameter in fact lies within the confidence interval, then the probability that the estimator either will be the estimate actually observed, or will be closer to the parameter, is less than or equal to 90%". Users of Bayesian methods, if they produced a confidence interval, might by contrast say "My degree of belief that the parameter is in fact in the confidence interval is 90%". Disagreements about these issues are not disagreements about solutions to mathematical problems. Rather they are disagreements about the ways in which mathematics is to be applied. [I will add an example of a "recognizable subset" here; i.e., a case in which the data themselves make the epistemic conclusion dubious.] ## Concrete practical example Here is one of the most familiar realistic examples. Suppose X1, ..., Xn are an independent sample from a normally distributed population with mean μ and variance σ2. Let $\overline{X}=(X_1+\cdots+X_n)/n,$ $S^2=\frac{1}{n-1}\sum_{i=1}^n\left(X_i-\overline{X}\,\right)^2.$ Then $T=\frac{\overline{X}-\mu}{S/\sqrt{n}}$ has a Student's t-distribution with n − 1 degrees of freedom. Note that what distribution T has does not depend on the values of the unobservable parameters μ and σ2; i.e., it is a pivotal quantity. If c is the 95th percentile of this distribution, then $P\left(-c (Note: "95" and "90" are correct; this is a frequent occasion for careless mistakes.) Consequently $P\left(\overline{X}-cS/\sqrt{n}<\mu<\overline{X}+cS/\sqrt{n}\right)=0.9$ and we have a 90% confidence interval for μ.

2013-06-19 20:15:42

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https://www.gradesaver.com/textbooks/math/precalculus/precalculus-10th-edition/chapter-10-analytic-geometry-10-6-polar-equations-of-conics-10-6-assess-your-understanding-page-684/9

# Chapter 10 - Analytic Geometry - 10.6 Polar Equations of Conics - 10.6 Assess Your Understanding - Page 684: 9 Hyperbola The directrix is parallel to the polar axis at a distance of $\dfrac{4}{3}$ units below the pole. #### Work Step by Step We are given the equation in polar coordinates: $r=\dfrac{4}{2-3\sin\theta}$ Rewrite the equation: $r=\dfrac{\dfrac{4}{2}}{\dfrac{2-3\sin\theta}{2}}$ $r=\dfrac{2}{1-\dfrac{3}{2}\sin\theta}$ The equation is in the form: $r=\dfrac{ep}{1-e\sin\theta}$ Identify $e$ from the denominator, then $p$ from the numerator: $e=\dfrac{3}{2}$ $ep=2\Rightarrow \dfrac{3}{2}p=2\Rightarrow p=\dfrac{4}{3}$ Because $e>1$, the conic is a hyperbola. The directrix is parallel to the polar axis at a distance of $\dfrac{4}{3}$ units below the pole. After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.

2021-12-06 06:18:53

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https://www.enotes.com/homework-help/use-definition-2-find-an-expression-area-under-358642?en_action=hh-question_click&en_label=hh-sidebar&en_category=internal_campaign

# Use The Definition To Find An Expression For The Area Under The Curve Y = X3 From 0 To 1 As A Limit. (a) Use Definition 2 to find an expression for the area under the curve y= (x^3) from 0 to 1 as a limit. Then use the following formula (1^3) + (2^3) + (3^3) +...... + (n^3) = [(n(n+1))/2]^2 to evaluate the limit. sciencesolve | Certified Educator calendarEducator since 2011 starTop subjects are Math, Science, and Business You should create a partition of the interval [0,1] in n subintervals that have the following lengths such that: `Delta x_i = (1-0)/n = 1/n` `x_i = i*(1/n)` You need to  use limit definitionto evaluate the definite integral such that: `int_0^1 x^3 dx = lim_(n->oo) sum_(i=1)^n... (The entire section contains 153 words.)

2020-01-25 21:09:08

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https://crypto.stackexchange.com/questions/83985/importance-of-supersingularity-of-elliptic-curves

# Importance of supersingularity of elliptic curves I'm struggling to understand the high-level idea of "Verifiable Delay Functions from Supersingular Isogenies and Pairings" (https://eprint.iacr.org/2019/166.pdf) by De Feo et al. I will shortly outline the construction (p.12): setup: 1. (choose appropriate parameters) 2. Select a supersingular elliptic curve $$E/\mathbb{F}_p$$. 3. Choose a direction on the horizontal l-isogeny graph and compute a cyclic isogeny $$\phi : E \rightarrow E'$$ of degree $$l^T$$ and its dual. 4. Choose a generator P of X1 = $$v^{-1}$$($$\tilde{E}[N] \cap \tilde{E}(\mathbb{F}_p)$$) and compute $$\phi(P)$$. eval: 1. Compute $$\hat{\phi}(Q).$$ verify: 1. Verify the condition ($$e_N(P, \hat{\phi}(Q)) = e_N(\phi(P), Q)$$) for the pairing $$e_N$$. For step 2, I wonder: Why does the curve need to be supersingular? I think that it has a certain implication on the structure of the isogeny graph from $$E$$, but that is only my guess. For step 3: I assume that we do a random walk on the isogeny graph. I think that means: for any elliptic curve $$E_0$$, we construct an isogeny, compute it, jump to the new curve $$E_1$$ and repeat this process. I wonder: Isogenies project points between curves, but their result isn't really a new curve. So how can we compute an isogeny from $$E_1$$ without really knowing it? Why is it important that the isogeny graph is "horizontal"? How can we prove that the isogeny actually takes T steps (.. which is necessary for the delay property of VDFs)? Or do we just publish all T intermediate isogenies? Why can we assume that all of them have order l? The idea in step 5 and 6 looks clear to me: The condition (which is quick to evaluate) only holds for correctly computed isogenies, which can only be calculated with delay T. Concerning P: $$v : E \rightarrow \tilde{E}$$ is defined as $$(x,y) \rightarrow (u^2 x, u^3 y)$$, $$\tilde{E}$$ is the quadratic twist of E. I understand that $$E$$ and $$\tilde{E}$$ are isomorphic, but I don't see why X1 is constructed in such a complicated way. Why can't we use the points on E as X1? If my questions are too specific, I'd be more than happy for the explanation of supersingularity in this case! Would the construction also work for ordinary elliptic curves? Your questions are a bit too rambly, so I'll just answer the title question. We use supersingular curves because ordinary curves don't have enough isogenies. Specifically, for supersingular curves, the graph of $$\ell$$-isogenies defined over $$\mathbb{F}_{p^2}$$ has good expansion properties. For ordinary curves, the graph of $$\ell$$-isogenies does not have good expansion properties. In most cases the latter graph is just a union of disjoint cycles. You can get graphs of ordinary isogenies with good expansion properties, but in order to do so you have to do something like take the union of $$\ell$$-isogeny graphs over various different $$\ell$$.

2021-06-17 09:19:44

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https://testbook.com/question-answer/the-dimension-ml0t-2-may-correspond-to--615da42cf89693ef5dc07b87

# The dimension ML0T-2 may correspond to: This question was previously asked in UPPCL JE Electrical 7 Sept 2021 Official Paper (Shift 2) View all UPPCL JE Papers > 1. Stiffness 2. dynamic viscosity 3. Momentum 4. pressure Option 1 : Stiffness ## Detailed Solution Concept: Stiffness is the extent to which an object resists deformation in response to an applied force. The complementary concept is flexibility or pliability: the more flexible an object is, the less stiff it is. The stiffness k, of a body is a measure of the resistance offered by an elastic body to deformation. For an elastic body with a single degree of freedom (DOF) (for example, stretching or compression of a rod), the stiffness is defined as k = F / δ F is the force on the body δ is the displacement produced by the force along the same degree of freedom (for instance, the change in length of a stretched spring) Stiffness is typically measured in newtons per meter (N/m) Stiffness dimension is [MLT-2] / [L] = [ML0T-2] Important Points: Quantities Dimension Dynamic viscosity M1 L-1 T-1 Kinematic viscosity L2 T−1 Magnetic Flux ML2 T2 I Magnetic field MT-2 I-1 Power ML2 T-3 Torque ML2 T-2 Work ML2 T-2 Pressure ML-1 T-2 Force ML1 T-2 Surface Tension M1 L0 T-2 Momentum M1L1T-1 Angular momentum M L2 T-1

2022-01-21 23:05:46

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http://www.chegg.com/homework-help/questions-and-answers/please-integrate-integral-from-0-to-200-y2100-xdy-how-do-you-get-rid-of-the-x-can-you-show-q3407542

## inetegration xy please integrate : integral from 0 to 200 y^2(100-x)dy how do you get rid of the x? can you show me step by step how to solve this integral? positive energy to you if you can solve it step by step!! • there must be something missing in this question .......some conditions must be given to eliminate one of the variable.. • integral 0 to 200    y2 (100 - x) dy = (100 - x) y3/3  (limits  0 to 200 ) = (100 -x)8000000/3 x cannot be removed since there are no limits for x and x term is not integrated Get homework help

2013-05-24 11:10:48

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http://www.sciencemadness.org/talk/viewthread.php?tid=1177&page=2

Not logged in [Login - Register] Sciencemadness Discussion Board » Special topics » Energetic Materials » The Ultimate Energetic Compound Select A Forum Fundamentals   » Chemistry in General   » Organic Chemistry   » Reagents and Apparatus Acquisition   » Beginnings   » Responsible Practices   » Miscellaneous   » The Wiki Special topics   » Technochemistry   » Energetic Materials   » Biochemistry   » Radiochemistry   » Computational Models and Techniques   » Prepublication Non-chemistry   » Forum Matters   » Legal and Societal Issues   » Detritus   » Test Forum Pages:  1  2 Author: Subject: The Ultimate Energetic Compound Marvin International Hazard Posts: 991 Registered: 13-10-2002 Member Is Offline Mood: No Mood Deuterium is just a different isotope, the performance in an explosive will be less by a miniscule amount. National Hazard Posts: 339 Registered: 1-7-2005 Location: Brothel Member Is Offline Mood: @%&\$ing hardcore baby Quote: Originally posted by Sickman So is my idea new or has this deuterium analog of ammonium nitrate been prepared before and described in the literature? Please give references to patents etc. Is there any information on explosive power and characteristics compared to regular ammonium nitrate, I would assume a greater velocity of detonation and quantity of gas per kilogram of explosive? I would also assume that the interesting ammonium picrate analog is also possible by reacting picric acid with the deuterium ammonia analog! In fact I can think of many possible explosive substances possible using deuterium as a substitute in just about every high explosive imaginable that contains the light H2 molecule! We can increase both density and energy content, however the cost of heavy water (deuterium oxide) is a draw back for sure, although it is very much OTC if we are wanting it! [Edited on 1-8-2006 by Sickman] I recall reading an article talking about radioactive ammonium nitrate as a way to detect people who use it for illegimate purposes... Sickman Hazard to Self Posts: 98 Registered: 9-5-2004 Member Is Offline Mood: Icy and I see! Tritium ammonium nitrate dirty bomb YT2095 International Hazard Posts: 1091 Registered: 31-5-2003 Location: Just left of Europe and down a bit. Member Is Offline Mood: within Nominal Parameters here we go, this is an item regarding the post I made previously: http://www.newscientist.com/article.ns?id=dn1103 I knew I wasnt dreaming \"In a world full of wonders mankind has managed to invent boredom\" - Death Twinkies don\'t have a shelf life. They have a half-life! -Caine (a friend of mine) neutrino International Hazard Posts: 1583 Registered: 20-8-2004 Location: USA Member Is Offline Mood: oscillating Ah yes, LOX-porus silicon explosives. These were discussed here and at roguesci.org a few years back. hinz Hazard to Others Posts: 200 Registered: 29-10-2004 Member Is Offline Mood: No Mood Ammoium dinitramide looks interesting. http://enermat.org.ru/ammomium.html (found it while looking for pka of tetranitrocubane) Wonder if it's further nitratabe to trinitramide to N(NO2)3 would be quite interessting, but I think it's impossible due steristic issues and electronic issues, central N is quite possitive charged, poor nucleophile on NO2+, but NI3 is also existent and if the oxygen atoms of each NO2 are standing straight (not in the same plane as the central N atom), the NO2 wouldn't need to much place. Does anyone know if nitromethane is further nitratable to tetranitromethane since it's a weak acid, thus a potentially weak nucleophile on NO2+ and once its nitrated to dinitromethane the molecule will be nitrated completly to TeNM due it's acidity. Since it works also with a chlorination agent forming chloropicrin, but under basic conditions which eliminate a proton and producing a higher concentration of the nucleophile, I think the whole reaction needs some (long?) time. I've ordered 5l of CH3NO2 and these 5l need an application together with my fuming nitric acid. And TeNM balanced on CO/H2O/Al2O3 in a mix with Al and low melting parafine (to prevent the separation of the Al dust/ acting as a binder) would be quite interesting like a flash with VoD of 8km/s Nick F National Hazard Posts: 439 Registered: 7-9-2002 Member Is Offline Mood: No Mood JohnWW International Hazard Posts: 2849 Registered: 27-7-2004 Location: New Zealand Member Is Offline Mood: No Mood Someone asked on this thread about octaazocubane, N8, i.e. cubane in which all the CHs are replaced by Ns. Even more energetic wou;ld be Octaazidocubane, C8(N3)8, in which all the Hs in cubane are replaced by azide groups, and in which the strained carbon rings add to the energy available. This compound has actually been made, and papers about it and other azidocubanes have appeared as attachments in this part of sciencemadness.org. Also dodecaazidododecahedrane, C12(N3)12, i.e, dodecahedrane, C12H12, in which the Hs have been replaced woth azido groups. I doubt that this has been made, though. Other possibilities include hexadecaazidobuckminsterfullerene, in which the C60 is saturated by having azido groups added, C60(N3)60. More simply, also with a highly strained carbon ring adding to the energy, one could have triazidocyclopropane, C3(N3)3, and the corresponding cyclobutane derivative, C4(N3)4. I am not sure whether they have been prepared. Because such azido derivatives are likely to be dangerously explosive, the nitro derivatives would be safer to handle. [Edited on 2-8-2006 by JohnWW] franklyn International Hazard Posts: 2989 Registered: 30-5-2006 Location: Da Big Apple Member Is Offline Mood: No Mood Quote: Originally posted by YT2095 here we go, this is an item regarding the post I made previously: http://www.newscientist.com/article.ns?id=dn1103 I knew I wasnt dreaming Well, rub your eyes some more _ I'm astonished that a semi reputable magazine as New Scientist would publish something that is so blatantly stupid through and through. I quote "The substance - an exotic form of silicon - releases seven times as much energy as TNT, and explodes a million times faster." Given that the reaction time is proportional to the velocity of detonation, and for T.N.T. this is 7.8 kilometers per second, a million fold increase is 7 million plus kilometers per second. The speed of light is 300,000. Duh Even if this is just a wave involving no mass transport, no information, detonation wave, or anything else can exceed 300,000. As if that wasn't enough let me also run this by you. The energy product of T.N.T. is 930 Kilogram Calories per Kilogram of T.N.T. The energy product of Silicon Dioxide formed from its elements Silicon and Oxygen, is some 3800 Kilogram Calories per Kilogram, or just 4 times as much, not 7 times. [Edited on 5-8-2006 by franklyn] not_important International Hazard Posts: 3873 Registered: 21-7-2006 Member Is Offline Mood: No Mood Well, New Scientist has been going downhill, although they've managed to avoid Page 3 girls so far.Looks like he's sold someone on it as an explosive http://www.newscientist.com/article/mg18024243.300.html He has continued messing about with porous Si and O2, although not explosively http://www2.kobe-u.ac.jp/~fujii1/Journal/68_2004_APL_Kovalev... But he did give a talk that included the porous silicon explosive, last year http://matsci.iw.uni-halle.de/nanosec/ (the speaker after him, Pacholski, has been doing some interesting work) Here's a bit more recent information http://www.trnmag.com/Stories/2002/022002/Chip_provides_more... Engager National Hazard Posts: 288 Registered: 8-1-2006 Location: Moscow, Russia Member Is Offline Mood: No Mood I've heard that inhert gas compounds such as XeO3 are extremly poverfull explosives, because of extremly high energy release on Xe-O bond breakage. I've also heard that XeO3 has unbeliveable brisance as well. I wonder how powerfull will be explosive mixture of XeO3 + active metall (Al for example), it will probably detonate on contact of reagents. Germans in WW2 tried to make a superbombs mixing liquid C2H2 with liquid ozone. All planes with such kind of bombs were exploded on takeoff, so the stuff was forgotten. I think this mix will be far more powerfull that any of imaginable nitro compounds. Mixture of liquid (CN)2 with liquid ozone will be even more powerfull, at normal conditions this mixture burns with temperature of 10000K - that is the highest temperature achiveable in chemical combustion. The third candidate is a ethyl perchlorate, that was considered as most powerfull explosive compound known for almost century (until first inhert gas compounds vere found). Glycerin triperchlorate is pretty much the same but more powerfull and unstable in pure state. [Edited on 30-8-2006 by Engager] franklyn International Hazard Posts: 2989 Registered: 30-5-2006 Location: Da Big Apple Member Is Offline Mood: No Mood The problem with binary or sprengel mixtures is that at least one component if not both need to be liquid in order to get an intimate molecular mix, otherwise you have just a low velocity pyrotechnic. A liquid phase mix of Acetylene and Cyanogen is of interest because the products of detonation are all combustable in air, hence there is an additional thermobaric effect. But for a handful of exceptions ( Astrolite, PLX, Kinepak ) such materials are not used because almost all are dangerously shock sensitive, thus they will detonate upon impact precluding deep penetration of a target and timed initiation by a fuze. I have thought that the binary gas artillery shell design could be applied to binary or sprengel type explosives. The mixing is done during flight thus there is no risk from recoil and setback upon firing. Such a munition would be safer than anything in use now. [Edited on 30-8-2006 by franklyn] not_important International Hazard Posts: 3873 Registered: 21-7-2006 Member Is Offline Mood: No Mood And the problem with liquid O3 is that it tends to go off when a door slams, or a butterfly flaps its wings, or it feels like it. Adding hydrocarbons to liquid ozone is a Bad Idea. XeO3 around and above its melting point is similarly unstable, I believe. franklyn International Hazard Posts: 2989 Registered: 30-5-2006 Location: Da Big Apple Member Is Offline Mood: No Mood Reasons militating against introduction of more energetic compounds to subplant those in current use are cost, necessarily caused by the complexity of their manufacture. http://www.llnl.gov/str/Pagoria.html consider whats involved just to obtain the precurser for octanitrocubane http://www.chem.ox.ac.uk/dhtml/default.html It seems conterintuitive that almost polymeric forms of nitrogen have a more realistic prospect of implementation http://www.afrlhorizons.com/Briefs/Dec01/PR0106.html I have wondered if the Oxygenyl ( O2 +) cation radical could be coupled with the Superoxide ( O2 -) anion radical into a stable molecule by the double replacement of their salts O2SbF6 and KO2 Without quenching this reaction by itself would be quite energetic. following this idea _ NitroniumTetrahydroAluminate NO2AlH4 or the Borate NO2BH4 or TetraFluoroammoniumTetrahydroBorate NF4BH4 or the Aluminate NF4AlH4 would be interesting if it could be formed at all. . Mr. Wizard International Hazard Posts: 1040 Registered: 30-3-2003 Member Is Offline Mood: No Mood Doesn't the formation of diatomic H2 from two monatomic H atoms release more energy than any other combination? If the monatomic atoms could be held apart with say a magnetic or electric field or magic or whatever, then it would make a great fuel or energy storage device. What are the spin directions of the atoms of H2? My physics is a little weak in that area. artem Hazard to Self Posts: 53 Registered: 9-1-2005 Member Is Offline Mood: No Mood Quote: Originally posted by Mendeleev Yeah, the exotic chemical is the result of condensing formaldehyde, nitroguanidine, and tert-butylamine. This product, 2-nitrimino-5-tert-butyl-hexahydro-1,3,5 triazine. It is then subjected to chloride assisted nitrolysis, and poof! Mono-keto-RDX... [Edited on 14-1-2004 by Mendeleev] Here is the article about keto-RDX. Attachment: PEP1994-5-232.djvu (117kB) nitro-genes International Hazard Posts: 968 Registered: 5-4-2005 Member Is Offline Mood: No Mood This article also mentions an acid hydrolisis route from hexamin, followed by condensation with dinitrourea. Not sure if this has been mentioned before, though I'm pretty sure I have seen dinitrourea and keto-RDX been discussed on roguescience somewhere, without results for the synthesis of either of them IIRC... [Edited on 17-9-2006 by nitro-genes] Attachment: Synthesis, characterization and thermal studies of keto-RDX or k-6.pdf (113kB) Axt International Hazard Posts: 778 Registered: 28-1-2003 Member Is Offline Mood: No Mood The triazone fertiliser mentioned in the PEP article as the most convenient precursor is patented US4778510. Note the same "triazone corporation" responsible for fertiliser used in PEP article, and the patent. Gives correct pH, temp, time ratio of ammonia:formaldehyde:urea to form triazone urea mixture with minimal hexamine. Bit annoying to read though. Also I'll attach an article, give synthesis of keto-RDX via potassium sulphamate:urea:formaldehyde. Only 18% yield though. Attachment: use of potassium sulfamate in the synthesis of heterocyclic nitramines.pdf (271kB) The_Davster A pnictogen Posts: 2859 Registered: 18-11-2003 Member Is Offline Mood: No Mood Is anyone else downright aroused by the product formed when aminotetrazole is used instead of urea? Product V. Too bad chromatography is needed for purification. Microtek National Hazard Posts: 482 Registered: 23-9-2002 Member Is Offline Mood: No Mood Personally, I prefer the nitrated condensation product with DAF. They mention a melting point of 151 C, so depending on decomposition temp it might be a candidate for a castable material. The yield also seems quite good. nitro-genes International Hazard Posts: 968 Registered: 5-4-2005 Member Is Offline Mood: No Mood I wonder if the triazone precursor for the manufacture of keto-RDX could be made with higher purity using formic acid. Could a 3:2:1 molar ratio of ammonia/formaldehyde/formic acid yield 1,3,5 triazone upon evoparation to dryness and/or heating slightly? formic acid + ammonia yields ammonium formate: HCOOH + NH3 → NH4+HCOO- Which upon heating forms formamide according to: NH4+HCOO- → HCONH2 + H2O Since the first step is the acid/base reation of the ammonia with formic acid, very little hexamin should be expected in the final product... [Edited on 20-9-2006 by nitro-genes] Pages:  1  2 Sciencemadness Discussion Board » Special topics » Energetic Materials » The Ultimate Energetic Compound Select A Forum Fundamentals   » Chemistry in General   » Organic Chemistry   » Reagents and Apparatus Acquisition   » Beginnings   » Responsible Practices   » Miscellaneous   » The Wiki Special topics   » Technochemistry   » Energetic Materials   » Biochemistry   » Radiochemistry   » Computational Models and Techniques   » Prepublication Non-chemistry   » Forum Matters   » Legal and Societal Issues   » Detritus   » Test Forum

2019-07-21 00:37:26

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https://testbook.com/question-answer/over-a-basin-of-area-333-km2-a-storm-of-6-h-with--5ee4e1812eda930d0c4d1680

# Over a basin of area 333 km2, a storm of 6 h with uniform intensity of 2 cm/h occurs. The observed run off was 20 × 106 m3. The average rate of infiltration in mm/hr for basin is: This question was previously asked in ESE Civil 2016 Paper 2: Official Paper View all UPSC IES Papers > 1. 5 mm/h 2. 10 mm/h 3. 20 mm/h 4. 40 mm/h Option 2 : 10 mm/h Free CT 1: Ratio and Proportion 2846 10 Questions 16 Marks 30 Mins ## Detailed Solution Concept: Infiltration: It is the flow of water into the ground through the soil surfaces. Infiltration Capacity: The maximum rate at which a given soil at a given time can absorb water is defined as infiltration capacity. It is denoted as fp and its unit is cm/hr. Hence, if the intensity of rainfall is more than the infiltration capacity then, the infiltration rate will be equal to infiltration capacity (fp) and the rest of the amount of rainfall will be in the form of runoff. Calculation: Area of basin = 333 km2 uniform intensity = 2cm/hr Rainfall occurred = $$333 \times {10^6} \times 6 \times 2 \times {10^{ - 2}} = 39.96 \times {10^6}{m^3}$$ Run off = 20 × 106 m Infiltration = 39.96 × 106 – 20 × 106 = 19.96 × 106 m3 Rate of infiltration $$= \frac{{19.96 \times {{10}^6}{m^3}}}{{333 \times {{10}^6}{m^2} \times 6h}}$$ = 9.98 mm/hr

2021-10-26 22:37:10

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https://www.doubtnut.com/question-answer/let-the-distance-between-a-focus-and-the-corresponding-directrix-of-an-ellipse-be-8-and-the-eccentri-646277249

Home > English > Class 11 > Maths > Chapter > Conic Sections > Let the distance between a foc... # Let the distance between a focus and the corresponding directrix of an ellipse be 8 and the eccentricity be 1/2 . If the length of the minor axis is k , then (sqrt(3)k)/2 is ____________ Updated On: 27-06-2022

2022-12-08 22:00:42

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https://socratic.org/questions/how-do-you-calculate-the-percent-change-of-42-to-168

# How do you calculate the percent change of 42 to 168? Feb 9, 2017 See the entire solution process below: #### Explanation: The formula for calculate the percentage change in a value is: $p = \frac{N - O}{O} \cdot 100$ Where: $p$ is the percentage change - what we are solving for $N$ is the New Value - $168$ for this problem $O$ is the Old Value - $42$ for this problem Substituting and calculating gives: $p = \frac{168 - 42}{42} \cdot 100$ $p = \frac{126}{42} \cdot 100$ $p = 3 \cdot 100$ $p = 300$ There was a 300% change.

2020-02-18 21:32:18

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http://hackage.haskell.org/package/gtk-0.14.7/docs/Graphics-UI-Gtk-ModelView-TreeDrag.html

gtk-0.14.7: Binding to the Gtk+ graphical user interface library. Maintainer gtk2hs-users@lists.sourceforge.net provisional portable (depends on GHC) None Haskell98 Graphics.UI.Gtk.ModelView.TreeDrag Description Interfaces for drag-and-drop support in TreeView. Synopsis Detail TreeViews provide special support for Drag-and-Drop such as hover-to-open-rows or autoscrolling. This module implements two utility functions that set and get a path and a model in a Selection structure. These functions are thus useful to implement drag-and-drop functionality in a TreeModel. In fact, they are used as part of the default drag-and-drop interfaces of ListStore and TreeStore that allows to permute rows and move them between hierarchy levels. DND information for exchanging a model and a path. treeModelEqual :: (TreeModelClass tm1, TreeModelClass tm2) => tm1 -> tm2 -> Bool Source # The SelectionTag, TargetTag and SelectionTypeTag of the DND mechanism of ListStore and TreeStore. This tag is used by treeGetRowDragData and treeSetRowDragData to store a store and a TreePath in a SelectionDataM. This target should be added to a TargetList using TargetSameWidget flag and an InfoId of 0. Obtains a TreeModel and a path from SelectionDataM whenever the target is targetTreeModelRow. Normally called from a treeDragDestDragDataReceived handler. treeSetRowDragData :: TreeModelClass treeModel => treeModel -> TreePath -> SelectionDataM Bool Source # Sets selection data with the target targetTreeModelRow, consisting of a TreeModel and a TreePath. Normally used in a treeDragSourceDragDataGet handler. • Returns True if setting the data was successful.

2019-04-23 01:38:26

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https://stats.stackexchange.com/questions/25051/compound-distribution-in-bayesian-sense-vs-compound-distribution-as-random-sum/294034

# Compound distribution in Bayesian sense vs. compound distribution as random sum I'm trying to sort out two different uses of the term "compound distribution" and figure out the relationship. The Wikipedia article on compound distribution -- which I wrote -- defines a compound distribution as an infinite mixture, i.e. if $p(x|a)$ is a distribution of type F, and $p(a|b)$ is a distribution of type G, then $p(x|b) = \int_a p(x|a) p(a|b) da$ is a compound distribution that results from compounding F with G. This is the distribution of prior and posterior predictive distributions in Bayesian statistics. However, the term "compound distribution" has another meaning as a random sum, i.e. a sum of i.i.d. variables where the number of variables is random. What's the relation between the two? And am I using "compound distribution" correctly for the first definition? The only relationship between the two compound types I am aware of is with a Bernoulli random variable. If the number of summands in the random sum is Bernoulli then the distribution of the sum is $0$ if the Bernoulli was $0$ and $X_1$ if the Bernoulli was $1$. This is equivalently written $Y\stackrel{d}{=}Z_1X_1\stackrel{d}{=}\sum_{i=1}^{Z_1}X_1.$ Now the distribution of $Y$ after evaluating the sum has a zero-inflated distribution where the zero inflation is $\Pr(Z_1=0)$. This is the same distribution as if you "compounded" because the product representation above can be interpreted as a scale mixture where you mix between a $0$ value and a value given by the scale parameter of the distribution of $X_1$. The marginal distribution here would require compounding. In this case the compounding is not an integral but a finite sum.

2020-01-24 20:30:23

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https://2017-lapaz-assembly.readthedocs.io/en/latest/qiime3.html

layout: page title: “QIIME Tutorial 3” comments: true date: 2016-07-13 #QIIME Tutorial 3: Comparative diversity Authored by Ashley Shade, with contributions by Sang-Hoon Lee, Siobhan Cusack, Jackson Sorensen, and John Chodkowski.EDAMAME-2016 wiki ##Overarching Goal • This tutorial will contribute towards an understanding of microbial amplicon analysis ##Learning Objectives • Calculate resemblance matrices from an OTU table • Visualize comparative diversity across a priori categorical groups • Convert .biom formatted OTU tables to text files for use outside of QIIME ###Handout of workflow: 3.1 Make resemblance matrices to analyze comparative (beta) diversity¶ Make sure that you are in the EDAMAME_16S/uclust_openref/ directory. If you need the otu_table_mc2_w_tax_even5196.biom file from Parts 1 and 2 of the tutorial you can use curl to grab it from GitHub: curl -O https://raw.githubusercontent.com/edamame-course/Amplicon_Analysis/master/resources/otu_table_mc2_w_tax_even5196.biom We will make four kinds of resemblance matrices (sample by sample comparisons) for assessing comparative diversity. Use the -s option to see all of the different options for calculating comparative (beta) diversity in QIIME. beta_diversity.py -s What are all these indices, mathematically? Where do they all come from? All that info is in the guts of the script, which is really hard to work through, but it is there. To compare weighted/unweighted and phylogenetic/taxonomic metrics, we will ask QIIME to create four resemblance matrices of all of these different flavors. Navigate into the uclust_openref/ directory. beta_diversity.py -i otu_table_mc2_w_tax_even5196.biom -m unweighted_unifrac,weighted_unifrac,binary_sorensen_dice,bray_curtis -o compar_div_even5196/ -t rep_set.tre Due to a bug in this version of QIIME (v 1.9.1), this may return a warning that says “VisibleDeprecationWarning”. Do not be alarmed. The script has still worked as it was supposed to. Navigate to the new directory called “compar_div_even5196”. There should be four new resemblance matrices in the directory. Use nano to open them and compare their values. nano binary_sorensen_dice_otu_table_mc2_w_tax_even5196.txt This should be a square matrix, and the upper and lower triangles should be mirror-images. Pop quiz: Why is the diagonal zero? We’re going to get all crazy and move these outside of the terminal. Use scp to transfer them to your desktop. We will come back to these files for the R tutorial, so remember where you stash them! From a terminal with your computer as the working directory, grab the entire directory by using the -r flag. scp -r -i your/key/file ubuntu@ec2-your_DNS.compute-1.amazonaws.com:EDAMAME_16S/uclust_openref/compar_div_even5196 ~/Desktop 3.2 Using QIIME for visualization: Ordination¶ QIIME scripts can easily make an ordination using principal coordinates analysis (PCoA). We’ll perform PCoA on all resemblance matrices, and compare them. Documentation is here. Navigate back into the uclust_openref/ directory principal_coordinates.py -i compar_div_even5196/ -o compar_div_even5196_PCoA/ Notice that the -i command only specifies the directory, and not an individual filepath. PCoA will be performed on all resemblances in that directory. This will likely give a runtime warning: “The result contains negative eigenvalues.” As the warning explains, this can usually be safely ignored if the magnitude of the negative value is smaller than the magnitude of the largest eigenvalues. In our case, the negative value is several orders of magnitude smaller than the largest eigenvalue, so we can ignore this warning. However, this is something to keep in mind when performing your own analyses. If we navigate into the new directory, we see there is one results file for each input resemblence matrix. Inspect one of these files. nano pcoa_bray_curtis_otu_table_mc2_w_tax_even5196.txt The first column has SampleIDs, and column names are PCoA axis scores for every dimension. In PCoA, there are as many dimensions (axes) as there are samples. Remember that, typically, each axis explains less variability in the dataset than the previous axis. These PCoA results files can be imported as text files into other software for making ordinations outside of QIIME. Navigate back into the uclust_openref/ directory. We can make 2d plots of the output of principal_coordinates.py, and map the colors to the categories in the mapping file. make_2d_plots.py -i compar_div_even5196_PCoA/pcoa_weighted_unifrac_otu_table_mc2_w_tax_even5196.txt -m ../MappingFiles/Centralia_Full_Map.txt -o PCoA_2D_plot/ (This will also give a runtime warning: “More than 20 figures have been opened. Figures created through the pyplot interface (matplotlib.pyplot.figure) are retained until explicitly closed and may consume too much memory.” However, the script will execute as intended.) Open a new (non- EC2) terminal. Use scp from the new terminal to transfer the new html file and its companion files in a directory with a nonsense name similar to tmpCOqXlM to your desktop (using -r to take the whole directory), then open the html file. scp -r -i **your key** ubuntu@**your DNS**:EDAMAME_16S/uclust_openref/PCoA_2D_plot ~/Desktop This is where a comprehensive mapping file is priceless because any values or categories reported in the mapping file will be automatically color-coded by QIIME for data exploration. It is like MAGIC! Take some time to explore these plots: toggle samples, note color categories, hover over points to examine sample IDs.What hypotheses can be generated based on exploring these ordinations? What environmental measurements or categories seem to have the most explanatory value in distinguishing communities? Exercise Make 2D plots for each PCoA analysis from each of the four difference resemblance results and compare them. How are the results different, if at all? Would you reach difference conclusions? 3.3 Other visualizations in QIIME¶ We can make a non-metric multidimensional scaling (NMDS) plot. Navigate back to uclust_openref/ directory. mkdir NMDS_Plot nmds.py -i compar_div_even5196/bray_curtis_otu_table_mc2_w_tax_even5196.txt -o NMDS_Plot/mc2_even5196_braycurtis_NMDS_coords.txt cd NMDS_Plot We can also make a quick heatmap in QIIME, which shows the number of sequences per sample relative to one another. For our sanity (do you really want to look at ~20K OTUs at once?), let’s make this heatmap at the phylum level. To do this, we will use our phylum-level OTU table in the WS_aDiversity/ directory make_otu_heatmap.py -i WS_aDiversity_even5196/taxa_summary5196/otu_table_mc2_w_tax_even5196_L2.biom -o heatmap_L2_even5196.pdf Explore this visualization. You can filter the minimum number of OTUs, filter by sample ID, or by OTU ID. Heatmap documentation is here. 3.4 “Collapse” the biom table across DNA extraction replicates¶ We have 3 replicate DNA extractions from one soil core that were separately amplified and sequenced. After inspecting the PCoA plots coded by “Sample”, we can be assured that the replicate samples are quite similar to each other. The DNA extraction replicates are technical replicates: they are not part of the sampling design. Rather, they are an internal experimental “sanity check.” We should not include them as replicates in downstream analysis because it would artificially inflate our sample size. Therefore, we want to generate a dataset where the replicates are averaged into a single observation for each core.Experimental Design Question: Why should we average the replicates? Are there any alternative to averaging that would also be appropriate? Documentation for collapse_samples.py is here. This is a really useful command - you can also easily normalize the data (make a relativized dataset). collapse_samples.py -b otu_table_mc2_w_tax_even5196.biom -m ../MappingFiles/Centralia_Full_Map.txt --collapse_fields Sample --collapse_mode mean --output_biom_fp otu_table_mc2_w_tax_even5196_CollapseReps.biom --output_mapping_fp map_collapsed_reps.txt ###3.5 Exporting the QIIME-created biom table for use in other software (R, Primer, Phinch, etc) This command changes frequently, as the biom format is a work in progress. Use biom convert -h to find the most up-to-date arguments and options. In the example below, we are making a tab-delimited text file (designated by the --to-tsv option) and including a final column for taxonomic assignment called “Consensus lineage”. biom convert -i otu_table_mc2_w_tax_even5196_CollapseReps.biom -o otu_table_mc2_w_tax_even5196_CollapseReps.txt --table-type "OTU table" --to-tsv --header-key taxonomy --output-metadata-id "ConsensusLineage" YOU DID IT! HOLIDAY! CELEBRATE!¶ Help and other Resources¶ ###Ordinations and resemblance ###Biom table conversion QIIME help¶ • QIIME offers a suite of developer-designed tutorials. • Documentation for all QIIME scripts. • There is a very active QIIME Forum on Google Groups. This is a great place to troubleshoot problems, responses often are returned in a few hours! • The QIIME Blog provides updates like bug fixes, new features, and new releases. • QIIME development is on GitHub.

2021-12-06 07:39:14

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https://tex.stackexchange.com/questions/477247/how-to-supress-with-ragged2e-in-a-glossary-longtable

# How to supress with ragged2e in a glossary (longtable)? I am trying to use the long style provided in the glossary package to use in my acronym list. However, I am also using the ragged2e package and I have my paragraphs indented too. This causes both the table and the first line of the acronym description to be indented. Any suggestions? \documentclass{report} \usepackage[top=1.0in,hmargin=1.25in,height=9.0in,letterpaper,showframe]{geometry} \usepackage{setspace} \usepackage[document]{ragged2e} \usepackage{calc} \usepackage[acronyms,nonumberlist,nopostdot,nogroupskip]{glossaries} \setlength{\RaggedRightParindent}{0.5in} \setacronymstyle{long-short} \renewcommand*{\glsnamefont}[1]{\normalfont{#1}} \newacronym{AAA}{AAA}{some text} \newacronym{BBB}{BBB}{some very very very very very very very very very very very very very long long long long long long long long text} \newacronym{CCC}{CCC}{an acronym} \makenoidxglossaries \setglossarystyle{long} \newlength\glsnamewidth \settowidth{\glsnamewidth}{MMMMM} \setlength{\glsdescwidth}{\textwidth-\glsnamewidth-\tabcolsep} \renewenvironment{theglossary}% {\begin{longtable}[l]{@{}p{\glsnamewidth}p{\glsdescwidth}}}% {\end{longtable}}% \begin{document} \doublespace \printnoidxglossary[type=acronym,title={LIST OF ACRONYMS}] \printacronyms \gls{AAA}. Text text text \gls{BBB}. More text \gls{CCC}. \end{document} Gives: While I want it to look more like the next one, albeit with the following text indented. Found a way by changing the indentation inside the renewed glossary environment and then resetting it after the closing of the table. It works for this case, but unsure if there is a more elegant way. Full code: \documentclass{report} \usepackage[top=1.0in,hmargin=1.25in,height=9.0in,letterpaper,showframe]{geometry} \usepackage{setspace} \usepackage[document]{ragged2e} \usepackage{calc} \usepackage[acronyms,nonumberlist,nopostdot,nogroupskip]{glossaries} \setlength{\RaggedRightParindent}{0.5in} \setacronymstyle{long-short} \renewcommand*{\glsnamefont}[1]{\normalfont{#1}} \newacronym{AAA}{AAA}{some text} \newacronym{BBB}{BBB}{some very very very very very very very very very very very very very long long long long long long long long text} \newacronym{CCC}{CCC}{an acronym} \makenoidxglossaries \setglossarystyle{long} \newlength\glsnamewidth \settowidth{\glsnamewidth}{MMMMM} \setlength{\glsdescwidth}{\textwidth-\glsnamewidth-\tabcolsep} \renewenvironment{theglossary}{% \setlength{\RaggedRightParindent}{0pt}\begin{longtable}[l{@{}p{\glsnamewidth}p{\glsdescwidth}}}% {\end{longtable}\setlength{\RaggedRightParindent}{0.5in}}% \begin{document} \doublespace \printnoidxglossary[type=acronym,title={LIST OF ACRONYMS}] \printacronyms \gls{AAA}. Text text text \gls{BBB}. More text \gls{CCC}. \end{document}

2020-04-09 05:38:41

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https://www.ademcetinkaya.com/2022/09/how-do-you-predict-if-stock-will-go-up_67.html

Stock market trading is an activity in which investors need fast and accurate information to make effective decisions. Since many stocks are traded on a stock exchange, numerous factors influence the decision-making process. Moreover, the behaviour of stock prices is uncertain and hard to predict. For these reasons, stock price prediction is an important process and a challenging one. We evaluate NetApp prediction models with Deductive Inference (ML) and Polynomial Regression1,2,3,4 and conclude that the NTAP stock is predictable in the short/long term. According to price forecasts for (n+4 weeks) period: The dominant strategy among neural network is to Hold NTAP stock. Keywords: NTAP, NetApp, stock forecast, machine learning based prediction, risk rating, buy-sell behaviour, stock analysis, target price analysis, options and futures. Key Points 2. Is now good time to invest? 3. Nash Equilibria NTAP Target Price Prediction Modeling Methodology The study of financial markets has been addressed in many works during the last years. Different methods have been used in order to capture the non-linear behavior which is characteristic of these complex systems. The development of profitable strategies has been associated with the predictive character of the market movement, and special attention has been devoted to forecast the trends of financial markets. We consider NetApp Stock Decision Process with Polynomial Regression where A is the set of discrete actions of NTAP stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4 F(Polynomial Regression)5,6,7= $\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \dots & {p}_{1n}\\ & ⋮\\ {p}_{j1}& {p}_{j2}& \dots & {p}_{jn}\\ & ⋮\\ {p}_{k1}& {p}_{k2}& \dots & {p}_{kn}\\ & ⋮\\ {p}_{n1}& {p}_{n2}& \dots & {p}_{nn}\end{array}$ X R(Deductive Inference (ML)) X S(n):→ (n+4 weeks) $∑ i = 1 n a i$ n:Time series to forecast p:Price signals of NTAP stock j:Nash equilibria k:Dominated move a:Best response for target price For further technical information as per how our model work we invite you to visit the article below: How do AC Investment Research machine learning (predictive) algorithms actually work? NTAP Stock Forecast (Buy or Sell) for (n+4 weeks) Sample Set: Neural Network Stock/Index: NTAP NetApp Time series to forecast n: 22 Sep 2022 for (n+4 weeks) According to price forecasts for (n+4 weeks) period: The dominant strategy among neural network is to Hold NTAP stock. X axis: *Likelihood% (The higher the percentage value, the more likely the event will occur.) Y axis: *Potential Impact% (The higher the percentage value, the more likely the price will deviate.) Z axis (Yellow to Green): *Technical Analysis% Conclusions NetApp assigned short-term B1 & long-term B1 forecasted stock rating. We evaluate the prediction models Deductive Inference (ML) with Polynomial Regression1,2,3,4 and conclude that the NTAP stock is predictable in the short/long term. According to price forecasts for (n+4 weeks) period: The dominant strategy among neural network is to Hold NTAP stock. Financial State Forecast for NTAP Stock Options & Futures Rating Short-Term Long-Term Senior Outlook*B1B1 Operational Risk 7760 Market Risk6530 Technical Analysis4569 Fundamental Analysis5468 Risk Unsystematic6169 Prediction Confidence Score Trust metric by Neural Network: 75 out of 100 with 609 signals. References 1. Wager S, Athey S. 2017. Estimation and inference of heterogeneous treatment effects using random forests. J. Am. Stat. Assoc. 113:1228–42 2. Barrett, C. B. (1997), "Heteroscedastic price forecasting for food security management in developing countries," Oxford Development Studies, 25, 225–236. 3. D. S. Bernstein, S. Zilberstein, and N. Immerman. The complexity of decentralized control of Markov Decision Processes. In UAI '00: Proceedings of the 16th Conference in Uncertainty in Artificial Intelligence, Stanford University, Stanford, California, USA, June 30 - July 3, 2000, pages 32–37, 2000. 4. Mnih A, Teh YW. 2012. A fast and simple algorithm for training neural probabilistic language models. In Proceedings of the 29th International Conference on Machine Learning, pp. 419–26. La Jolla, CA: Int. Mach. Learn. Soc. 5. Abadie A, Diamond A, Hainmueller J. 2015. Comparative politics and the synthetic control method. Am. J. Political Sci. 59:495–510 6. Mnih A, Teh YW. 2012. A fast and simple algorithm for training neural probabilistic language models. In Proceedings of the 29th International Conference on Machine Learning, pp. 419–26. La Jolla, CA: Int. Mach. Learn. Soc. 7. Mnih A, Teh YW. 2012. A fast and simple algorithm for training neural probabilistic language models. In Proceedings of the 29th International Conference on Machine Learning, pp. 419–26. La Jolla, CA: Int. Mach. Learn. Soc. Frequently Asked QuestionsQ: What is the prediction methodology for NTAP stock? A: NTAP stock prediction methodology: We evaluate the prediction models Deductive Inference (ML) and Polynomial Regression Q: Is NTAP stock a buy or sell? A: The dominant strategy among neural network is to Hold NTAP Stock. Q: Is NetApp stock a good investment? A: The consensus rating for NetApp is Hold and assigned short-term B1 & long-term B1 forecasted stock rating. Q: What is the consensus rating of NTAP stock? A: The consensus rating for NTAP is Hold. Q: What is the prediction period for NTAP stock? A: The prediction period for NTAP is (n+4 weeks)

2022-10-01 11:22:01

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https://dsp.stackexchange.com/questions/10355/reconstructing-a-partially-deleted-image-through-wavelets

# Reconstructing a partially deleted image through wavelets I am trying to form an approximation of the wavelet transform from a partially sampled image. Reconstruction in the 1D case is easy. We have $w = h x$, with $w$ as the wavelet coefficients, $h$ as the inverse of the Haar matrix and $x$ our signal. $w$ and $x$ are $n \times 1$ and $h$ is $n \times n$. Where we don't sample the signal, we delete those columns from $h$ and those rows/elements from the full signal, $x$. Since we have to have as many equations as unknowns, we sample at $m$ locations and approximate the first $m$ wavelet coefficients. All we do is delete the corresponding columns of where we do not sample. I have $w = c x r$, where $c$ is the matrix that performs the wavelet transform on the columns, and $r$ the rows. I am having a hard time forming in approximation in this 2D case. I know this is a simple solution and am having a hard time getting the math behind it. I know it's just solving a linear system of equations, but am having a hard time coming up with the matrix to represent that linear system. • Why do you want to formulate the matrix? It's enough to note that it's a linear system and then solve the system. In this field, the matrix-vector notation is a neat tool to see what is going on, you never explicitly compute the matrix. If you want to use one of the (vast array) of techniques for solving such linear inverse problems, you're unlikely to need to explicitly represent the matrix. – Henry Gomersall Aug 20 '13 at 16:12 • I'm formulating the matrix because we don't have every sample, meaning it's easier (I think!) to form the matrix. We're trying to solve a linear system of equations, and I'm having a hard time coming up with the equations. – Scott Aug 20 '13 at 19:09 • It makes no difference that it's 1D or 2D, the maths is the same. You have the Haar matrix (which you don't really need to know explicitly) and the sampling matrix and that should be enough. – Henry Gomersall Aug 21 '13 at 9:26

2019-05-21 05:55:04

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https://www.freemathhelp.com/forum/threads/114124-Exactly-two-subspaces?p=443061

# Thread: Exactly two subspaces 1. ## Exactly two subspaces Suppose that V is a vector space. How can I prove that dimV=1 if and only if V has exactly two subspaces? 2. Originally Posted by Georgegr Suppose that V is a vector space. How can I prove that dimV=1 if and only if V has exactly two subspaces? Please post what you understand about this question. For example: What are the two subspaces (some call them trivial) of any v-space? If $v\in V~\&~v\ne 0$ does $span\{v\}=V~?$ #### Posting Permissions • You may not post new threads • You may not post replies • You may not post attachments • You may not edit your posts •

2019-01-22 10:17:16

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https://en.m.wikiversity.org/wiki/PlanetPhysics/Categorical_Algebra

# PlanetPhysics/Categorical Algebra \newcommand{\sqdiagram}[9]{$\displaystyle \diagram #1 \rto^{#2} \dto_{#4}& \eqno{\mbox{#9}}$ } ### An Outline of Categorical Algebra This topic entry provides an outline of an important mathematical field called categorical algebra ; although specific definitions are in use for various applications of categorical algebra to specific algebraic structures, they do not cover the entire field. In the most general sense, categorical algebras -- as introduced by Mac Lane in 1965 -- can be described as the study of representations of algebraic structures, either concrete or abstract, in terms of categories, functors and natural transformations. In a narrow sense, a categorical algebra is an associative algebra, defined for any locally finite category and a commutative ring with unity. This notion may be considered as a generalization of both the concept of group algebra and that of an incidence algebra, much as the concept of category generalizes the notions of group and partially ordered set. ### Extensions of Categorical Algebra • Thus, ultimately, since categories are interpretations of the axiomatic \htmladdnormallink{theories of abstract category {http://planetphysics.us/encyclopedia/Formula.html} (ETAC)}, so are categorical algebras. The general definition of representation introduced above can be still further extended by introducing supercategorical algebras as interpretations of ETAS, as explained next. • Mac Lane (1976) wrote in his Bull. AMS review cited here as a verbatim quotation: \emph{On some occasions I have been tempted to try to define what algebra is, can, or should be - most recently in concluding a survey [72] on Recent advances in algebra. But no such formal definitions hold valid for long, since algebra and its various subfields steadily change under the influence of ideas and problems coming not just from logic and geometry, but from analysis, other parts of mathematics, and extra mathematical sources. The progress of mathematics does indeed depend on many interlocking, unexpected and multiform developments.} ### Basic Definitions An algebraic representation is generally defined as a \htmladdnormallink{morphism {http://planetphysics.us/encyclopedia/TrivialGroupoid.html} ${\displaystyle \rho }$ from an abstract algebraic structure ${\displaystyle {\mathcal {A}}_{S}}$ to a concrete algebraic structure ${\displaystyle A_{c}}$}, a Hilbert space ${\displaystyle {\mathcal {H}}}$, or a family of linear operator spaces. The key notion of representable functor was first reported by Alexander Grothendieck in 1960. Thus, when the latter concept is extended to categorical algebra, one has a representable functor ${\displaystyle S}$ from an arbitrary category ${\displaystyle {\mathcal {C}}}$ to the category of sets ${\displaystyle Set}$ if ${\displaystyle S}$ admits a functor representation defined as follows. A functor representation of ${\displaystyle S}$ is defined as a pair, ${\displaystyle ({R},\phi )}$, which consists of an object ${\displaystyle R}$ of ${\displaystyle {\mathcal {C}}}$ and a family ${\displaystyle \phi }$ of equivalences $\displaystyle \phi (C): \Hom_{\mathcal{C}}(R,C) \cong S(C)$ , which is natural in C, with C being any object in ${\displaystyle {\mathcal {C}}}$. When the functor ${\displaystyle S}$ has such a representation, it is also said to be represented by the object ${\displaystyle R}$ of ${\displaystyle {\mathcal {C}}}$. For each object ${\displaystyle R}$ of ${\displaystyle \mathbf {C} }$ one writes $\displaystyle h_{R}: \mathcal{C} \lra Set$ for the covariant $\displaystyle \Hom$ --functor $\displaystyle h_{R}(C)\cong \Hom_{\mathcal{C}}(R,C)$ . A representation ${\displaystyle (R,\phi )}$ of ${\displaystyle {S}}$ is therefore a \htmladdnormallink{natural equivalence {http://planetphysics.us/encyclopedia/IsomorphismClass.html} of functors}: ${\displaystyle \phi :h_{R}\cong {S}~.}$ The equivalence classes of such functor representations (defined as natural equivalences) directly determine an algebraic (groupoid) structure. ## References 1. Saunders Mac Lane: Categorical algebra., Bull. AMS , 71 (1965), 40-106., Zbl 0161.01601, MR 0171826, 2. Saunders Mac Lane: Topology and Logic as a Source of Algebras., Bull. AMS , 82 , Number 1, 1-36, January 1, 1976.

2020-11-30 02:42:07

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https://www.vedantu.com/physics/coulomb

# Coulomb A French Military Engineer and Physicist named Charles-Augustin de Coulomb brought the concept of electrostatic forces (attractive and repulsive) between two charges placed inline apart with a square of the distance that lies inversely proportional to this force; however, the product of two charges always remains directly proportional to this force. This statement is called Coulomb’s law. In International Systems or SI systems, the unit of an electric charge is Coulom. The unit was so named as an honour to Mr Coulomb in 1880 after the discovery of Coulomb’s law by him in 1785. ### Coulomb’s Law According to Coulomb’s law, if two charges are separated by a distance ‘r’, and the charges are of opposite polarities as well. The distance between the two charges remains constant. So, when these charges apply forces on each other, they generate a force, which is the electrostatic force of attraction, as the charges are of the same magnitude but with different polarities. If there were charges of the same magnitude, also the same polarities, then there would have an electrostatic force of repulsion between them because these charges are static and are joined together with an imaginary line. So, in both cases, the equation remains the same for the force and that is: F α $\frac{q1q2}{r^{2}}$ As the force remains proportional to the product of charges and the square of the distance between these, so, on removing the sign of proportionality constant, we generate the following new form of the above equation: F = k $\frac{q1q2}{r^{2}}$ Here, k is the proportionality constant called the Columb’s law Constant, and its value is calculated in the following manner: ### Coulomb’s Law Constant Value From the above equation, we can re-arrange to determine the value of k: k = $\frac{1}{4 \pi \epsilon_{0}}$ ….(2) We know that the value of the dielectric constant, or the electric permittivity at free space is 8.85 x 10⁻¹² C²/Nm². Now, putting this value in equation (2), we get: k = $\frac{1}{4 \times 3.14 \times 8.85 \times 10^{-12} (C^{2} /Nm^{2})}$ On solving, we get the value of k = 8.99 x 10⁹Nm²C⁻² ### 1 Coulomb In an International Systems, the unit of electric charge is the meter-kilogram-second-ampere, which is the basis of the SI system of physical units. Coulomb is abbreviated as C. Coulomb unit is of the electric charge. We define Coulomb as the quantity of electricity transported in one second by a current of one ampere. This quantity was named Coulomb in the 18th–19th-century after a French physicist named Charles-Augustin de Coulomb, one Coulomb is approximately equal to 6.25 × 1018 electrons. ### 1 Coulomb Charge So, from the above statement of 1 Coulomb, we understood that the value of 1 Coulomb charge is equal to 6.25 x 10¹⁸ or 6.24 quintillion electrons. Let’s understand the Coulomb SI unit in detail: Let’s suppose that there are millions of charges flowing through a copper wire in the following manner: Now, why do Physicist use such a big unit for a charge? Well! Its understanding is easy by going more in mathematics beyond it. When the current of 1 ampere flows through this wire, 1 Coulomb of electrons, i.e.,  6.25 x 10¹⁸ electrons pass through it every second, i.e., given by: 1C/1 sec = 1 Amp So, we could clearly define one Coulomb from the above statement. ### One Coulomb Charge Formula According to the law of conservation of charges, whatever electrons flow through the wire, are quantized and also they remain conserved. So, if there are ‘n’ number of electrons flowing through a wire where ‘e’ is an elementary charge of the magnitude, i.e., 1.6 x 10⁻¹⁹ C. The ‘q’ is a charge of 1 C, the formula is: q = ne For the number of electrons, we re-write the equation in the following way: n = q/e ### Derive One Coulomb Charge Value Let’s say an electrical circuit carries a charge of the magnitude 1.6 x 10⁻¹⁹ C. When the charge is 1.6 x 10⁻¹⁹ C , the number of electrons = 1 Now, if the charge is 1 C, then the number of electrons will be  = 1/1.6 x 10⁻¹⁹ C On Solving, we get the value of 1 Coulomb charge as 6.25 x 10¹⁸. So, in 1 C, the number of electrons flowing through the above copper wire is 6.25 x 10¹⁸. This is the sole reason why Physicists use a huge unit like Columb for lump sump of electrons flowing through a wire. Fun Fact In a house, ordinarily, we use a 100-watt power lightbulb that draws out 1 ampere of current. Here, Coulombs come in handy for measuring the charges held by household electrical circuits. Q1: What is Coulomb’s Law in a Vector Form? Is Coulomb’s Law Experimental in Nature? Ans: The vector form of Coulomb’s law has great importance when there is an arrangement of point charges separated by a distance. In this case, the resultant force on any one of the charges is the vector sum of the forces because of each of these other forces; the science beyond this phenomenon is called the principle of superposition. Yes, Coulomb’s law is experimental in nature. It is because this law quantifies the amount of force that exists between two stationary or static electrically charged particles. Furthermore, electrostatic force or Coulomb force refers to the electric force which exists between point charges that are at rest position. Q2: What Does Coulomb’s Law Describe? Answer: Coulomb’s Law gives us an idea about the force between the two-point charges that are inline with each other separated by some distance. By the word point charge, we mean that the size of linear charged bodies is very small or you can say negligible as compared to the distance between them. That’s why we consider these charges as points, as it becomes easy for us to calculate the force of attraction or repulsion between them. From this theory, we infer that like charges repel each other and unlike charges attract each other. This means charges of the same sign/polarity pushes each other with repulsive forces while charges with opposite signs pull each other with attractive force.

2021-10-16 06:35:47

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https://www.math.princeton.edu/events/infinitely-many-solutions-yamabe-problem-noncompact-manifolds-2016-04-01t201504

# Infinitely many solutions to the Yamabe problem on noncompact manifolds - Paolo Piccione, Universidade de Sao Paulo and University of Notre Dame Fine Hall 314 Please note special time.   I will discuss the existence of infinitely many complete metrics with constant positive scalar curvature on prescribed conformal classes on certain noncompact product manifolds. These include products of closed manifolds with constant positive scalar curvature and simply-connected symmetric spaces of noncompact or Euclidean type; in particular, $S^m \times R^d$ and $S^m \times H^d$. As a consequence, one obtains infinitely many periodic solutions to the singular Yamabe problem on $S^m \setminus S^k$, for all $0 \leq k < (m - 2)/2$. I will also show that all Bieberbach groups are periods of bifurcating branches of solutions to the Yamabe problem on $S^m \times R^d$.This is a joint work with R. Bettiol, UPenn.

2018-09-20 04:53:19

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https://clinique-amaria.com/jamie-oliver-zlbk/2ab6ff-fifth-root-symbol

Powers or exponents. Power chords are commonly played on amplified guitars, especially on electric guitar with intentionally added distortion or overdrive effects. Alt Code Shortcut for Square Root Symbol. This calculator will find the 5th root of a number, so it's simply a specialized form of our common You can copy & paste, or drag & drop any symbol to textbox below, and see how it looks like. See note below. Square root or principal square root symbol √ does not have 2 on the root. When you want to type square root, cube root and fourth root symbols on your documents then the easy way is to use alt code shortcuts. When you want to type square root, cube root and fourth root symbols on your documents then the easy way is to use alt code shortcuts. (I am aware that \sqrt as command for a general root is misleading) For the life of me I can't find the right syntax to get the input of n'th root of x. More information regarding the symbology convention can be found on this website. [1+2]/[3(4+5)] Mathematical Constants Available In WeBWorK. Next, click "symbol" and "more symbols." It provides a quicker way to compute for the nth root of any number. When A 2= B then A is the square root of B indicated as √B = A. UNITY) when multiplied by itself five times. Square root, cubed root, 4th root, and any root are the most common examples of an nth root. The POWER() function is useful for both powers and exponents. Roots can also include decimal numbers (root 6.4, for example). Radical symbol The √ symbol that means "root of". – Neil Meyer 7 hours ago. Root Chakra Symbol Meaning. Math ML. So are 3*4 and 3 4 (3 space 4, not 34) but using an explicit multiplication symbol makes things clearer. ( Im going to give an equation in word since I dont know how to put the root symbol on the computer. Radical symbol The √ symbol that means "root of". Figure 4. In the table above, notice how the denominator of the rational exponent determines the index of the root. 2 means square root, 3 means cube root. If you use a word processor use symbol for the root sign. By definition, the nth root of a number can be calculated by raising that number to the power of 1/n. Type SQRT. In real numbers, there are some subcases: There are two solutions (same value with opposite sign) when x is positive and r is even. Let’s look at the following song, “Can’t Help Falling in Love” by Elvis Presley.The letters you see represent the root or tonic (the first note of … We could use the nth root in a question like this: The 3 comes out very small and low in the root sign. The first thing to understand in chord symbols is the letters. These simplifications are only made for explicit integer powers appearing in x.No attempt to factor the input x is made. Degree The number of times the radicand is multiplied by itself. It also has the square root symbol Alt code as well as the keyboard shortcut. To type the square root symbol in Word on your keyboard, press down the Alt key and type the Square Root symbol alt code (i.e. Muladhara. info) (also fifth chord) is a colloquial name for a chord in guitar music, especially electric guitar, that consists of the root note and the fifth, as well as possibly octaves of those notes. 1. It doesn't matter if it's a major chord, minor chord, or a power chord. Roots Calculator also know as a List of Root signs, make over 3 root symbols text character. For example, to enter 4 to the 5th power, type "4^5." If you want to find the fourth root of a number, raise it to the 1/4 power. For example, the fifth root of 243 is 3 as 3 x 3 x 3 x 3 x 3 is 243. ... Square root. To calculate fractional exponents use our If you don’t have the time to scan through every detail here, the table below will help you a lot. Facebook Twitter. The length of the horizontal bar is important. https://www.calculatorsoup.com - Online Calculators. Copy and paste the Root symbol or use the unicode decimal, hex number or html entity in social websites, in your blog or in a document. You can also have fourth, fifth, sixth, or other integer roots, as long as they're a positive real number. Office of the Administrative Assistant to the Secretary of the Army www.oaa.army.mil 3/23/2018 POC: Eve Roberts @ 703-428-6435 2 Headquarters, Department of The Army Office Symbol Root Like any other number, say 3/2 or 1 itself, the “use case” if there can be said to be such a thing, is circumstantial. No numbers after the roman numeral in the root position triad actually means that notes a third and a fifth are located above the bass note. All rights reserved. So, an exponent of 1 2 1 2 translates to the square root, an exponent of 1 5 1 5 translates to the fifth root or 5√ 5, and 1 8 1 8 translates to the eighth root or 8√ 8. Insert Square root symbol in Word (Quick Help) The table below is just a quick help. It's the eighth letter in the Greek alphabet, it looks like a 0 with a horizontal line in the center (θ). For example (1+2)*(3+4) and (1+2)(3+4) are both valid. To calculate any root of a number use our Nth root calculator. If you want to find the fifth root of a number, raise it to the 1/5 power. The square root of a number n is any a such that a 2 = n. Because both a number and its additive inverse square to get the same result, every positive real number has exactly 2 roots +√x and −√x, sometimes expressed as ±√x. Sometimes using the * symbol to indicate multiplication makes things easier to read. Roots and fifths are conveniently the same pattern for almost every chord. 5 th Root and Exponent 5 Table. So, to calculate the 5th root of 100, we simply raise 100 to the 1/5th power. The most commonly added tones are 2, 4, 6, 9, 11, and 13. For example, √4 = 2. (For cube roots and other non-square roots, see Fractional exponents.) Because the 5th root is not an actual ASCII symbol (that I know of), you would be just as well off to use fractional powers. ; There is no solution when x is negative and r is even. 251) using the numeric keypad, then release the Alt key. We are just adding one more note to the root note you learned about in the roots lesson. On windows, the square root symbol shortcut is Alt+125. Alt-Codes can be typed on Microsoft Operating Systems: First make sure that numlock is on, Then press and hold the ALT key, While keeping ALT key pressed type the code for the symbol that you want and release the ALT key. Was this article helpful? Put a five before it, highlight it, then in formatting or font click the box for superscript. The length of the horizontal bar is important. A list of symbols will appear, one of which is the square root symbol. Using Square root symbol shortcuts. Using this technique you can calculate any root. No numbers after the roman numeral in the root position triad actually means that notes a third and a fifth are located above the bass note. In this example, the 5th root of 100 equals 2.51189. C: A Nasdaq stock symbol indicating the issuer has been granted a continuance in Nasdaq under an exception to the qualification standards for a limited period. Radicals Calculator. 1. Roots are the inverse of powers. 3 = 9) Fifth root of 34 is 2.024397... (denoted like ⁵√34 = 2.024397...), because 2.024397 5 = 34 Each of these letters may also be accompanied by a sharp (♯) or flat (♭). This section will focus on that. Using it. Copy and paste the Root symbol or use the unicode decimal, hex number or html entity in social websites, in your blog or in a document. Another equation, a little different: 2 times 5th root of 20 minus the 6th root of 128. In general, if you want the Xth root of a number, raise it to the (1/X) power.-John Cite this content, page or calculator as: Furey, Edward "Fifth Roots Calculator"; CalculatorSoup, Sometimes using the * symbol to indicate multiplication makes things easier to read. pi gives 3.14159265358979, e.g. Unicode codes can not be typed. To insert a square root (a radical), you can click on the "√" button next to "A B C" on the Desmos keyboard. Nth root calculator. File Creation Time: The last row of each Symbol Directory text … Type "theta." Looking for radicals with an index greater than 3? To use them in facebook, twitter, textbox or elsewhere just follow the instructions at top. Figure 4. Fifth roots (for integer results 1 through 10) Fifth root of 1 is 1. There are several ways to get any symbol in Word including the square root symbol. After that they are called the 4th root, 5th root and so on. fourth root – ∜ (∜) ... See the Math Symbol Unicode chart or Alan Wood’s Mathematical Operator Unicode table for other common symbols. Formula – How to calculate the fifth root. The caret symbol “ ^ ” in Excel means “ to the power of ”. I also thought of $$^3\sqrt{\frac ab}$$ but the 3 comes out too far to the left. Math ML. Similar simplifications are made for negative powers. However, if we did not have the 5th roots of unity, our number systems would collapse, in the shape of … Formula – How to calculate the fifth root. The square root symbol should be inserted for you. Even for perfect root numbers, a root can be difficult to calculate by hand. :-) Comment actions Permalink. Fortunately, this is pretty simple to do if you can remember a simple mathematical rule: $\sqrt[N]{X} = {X^{\frac{1}{N}}}$ So, to calculate the 5th root of 100, we simply raise 100 to the 1/5th power. (I am aware that \sqrt as command for a general root is misleading) For the life of me I can't find the right syntax to get the input of n'th root of x. Math ML is an XML language designed to present complex equations. This online calculator for fifth roots is set up specifically to calculate 5th root. The inverted triangle of the throat chakra symbol is also symbolic of spiritual growth. Typing square Root Symbol in Microsoft Word. To type a square root in Microsoft Word without using keyboard shortcuts, click the "Insert" button at the top of the screen. Scale degrees 2, 4, and 6 imply that they are being added to a triad, while tones 9, 11, and 13 imply that they are being added to a … Conventionally your original #theta# is in the range #(-pi, pi]# or the range #[0, 2pi)# according to your definition of #Arg(z)# and the first of these five roots is the Principal Complex fifth root. The fifth root of a number is the number that would have to be multiplied by itself 5 times to get the original number. Please be aware that I know it's the same as x^(1/n) but I would like my students to see that root symbol with a little n over it. Roots can also include decimal numbers (root 6.4, for example). Jacob Dreiling April 12, 2015 23:30. Any nth root is an exponentiation by 1/n, so to get the square root of 9, you use 9**(1/2) (or 9**0.5) to get the cube root, you use 9 ** (1/3) (which we can't write with a simpler fraction), and to get the nth root, 9 ** (1/n). For example (1+2)*(3+4) and (1+2)(3+4) are both valid. Table Fifth Root (5 √) of 1 to 100. The 2.5th root of 70 (2.5 √70) is 5.47065, as 5.47065 2.5 = 70. where n is a positive integer and the condition signum(a)=1 means a is provably real and positive. While some exercises have a special keypad from which you can choose a math symbol, there are a limited number of symbols available in it. Formula – How to calculate a root. Fifth roots (for integer results 1 through 10). See note below. These letters (with and without accidentals) represent all of the notes on the staff. The caret symbol “ ^ ” in Excel means “ to the power of ”. In most context, “the square root” of a number refers just to its positive root. After that they are called the 4th root, 5th root and so on. Also note that as of Python 3, adding periods to integers to make them a … After that they are called the 4th root, 5th root and so on. So are 3*4 and 3 4 (3 space 4, not 34) but using an explicit multiplication symbol makes things clearer. The symbol of throat chakra is a cross, which indicates a perfect void and silence that need to be transcended to reach enlightenment. For the number approximately equal to 3.14159..., type "pi." If this is missing, it is assumed to be 2 - the square root. The 6 in the first inversion roman numeral symbol means that notes a sixth and a third are located above the bass note. If an element in X is negative, then the … To calculate a root, simply supply an inverse exponent — for example, a square root is 1/2. Muladhara is the Root Chakra at the base of your spine, and it’s all about grounding. If this is missing, it is assumed to be 2 - the square root. Formula – How to calculate a root. To calculate the cube root of a number in Excel, use the caret operator (^) with 1/3 as the exponent in a simple formula. The fifth root of a number is the number that would have to be multiplied by itself 5 times to get the original number. Please be aware that I know it's the same as x^(1/n) but I would like my students to see that root symbol with a little n over it. Theta. I encounter a bug when trying to use my bluetooth keyboard on the default desmos interface. Degree The number of times the radicand is multiplied by itself. It provides a quicker way to compute for the nth root of any number. The root and fifth of the chord are the most supportive sounding notes a bassist can play beneath a chord. Each chakra symbol incorporates the powerful circle as a reminder of our connection to the divine. To do this on the 12C (using RPN data entry): 100 ENTER 5 1/x y x. Try using our shortcut by typing nthroot into the expression list. Fifth root of To calculate any root of a number use our The sixteen lotus petals surrounding the throat chakra symbol are often inscribed with Sanskrit letters; namely, the vowels. For example, √4 = 2. What symbol do I press to get the 5th root of a number. Fractional Exponents Calculator. Degree The number of times the radicand is multiplied by itself. An augmented chord consists of a root-note, a major third and a augmented fifth. Learn how to type the cube root in your Microsoft Word document. Since the quality is augmented, there is a major third above the root (E) and an augmented fifth above the root (G-sharp). =number^(1/3) In this example, the formula =D3^(1/3) is used to find the cube root of 216, which is 6. Find the cube root in Excel. If an element in X is negative, then the … MathML and HTML 5 combinations are … Square root, cubed root, 4th root, and any root are the most common examples of an nth root. When A 2= B then A is the square root of B indicated as √B = A. What is exponent 5? ; There is one negative solution when x is negative and r is odd. It contains the square root symbol you can easily copy and paste into your work. For the cube root, you'll write a little three in the upper left corner of your root symbol. Alternatively, for MS Word users, type the character code ( 221A ), then press Alt+X to convert this code into the symbol. Even for perfect root numbers, a root can be difficult to calculate by hand. The 5th root of a number is the number that would have to be multiplied by itself 5 times to get the original number. The 6 in the first inversion roman numeral symbol means that notes a sixth and a third are located above the bass note. The n-th root of x is a number r such that r to the power of 1/n is x.. ... "n" to the power of one-fifth. Why vowels? The animal symbol of Vishuddha is a white elephant, the riding animal of the most important Vedic deity Indra, the King of the Gods. 2 means square root, 3 means cube root. See RootFinding[Isolate] if you want the real roots (zeros) of a univariate polynomial. Use a caret (^). 1. Square root or principal square root symbol √ does not have 2 on the root. The nth Root Symbol . Typical identifiers have 1-5 character root symbol and then 1-3 characters for suffixes. First select the symbol then you can drag&drop or just copy&paste it anywhere you like. Some would use a flat symbol when lowering, say f# to f-natural and others would use the natural symbol. Y = nthroot(X,N) returns the real nth root of the elements of X.Both X and N must be real scalars or arrays of the same size. C: A Nasdaq stock symbol indicating the issuer has been granted a continuance in Nasdaq under an exception to the qualification standards for a limited period. The uppercase letters you will see in chord symbols are C, D, E, F, G, A, and B. MathML and HTML 5 combinations are … So, in this example we get the numbers from column B and powers from column C: This calculator will find the 5th root of a number, so it's simply a specialized form of our common Roots Calculator also know as a Radicals Calculator. Y = nthroot(X,N) returns the real nth root of the elements of X.Both X and N must be real scalars or arrays of the same size. They represent the breathy, airy elements of words — just like ether is the airy element of space. Allow up to 14 characters. First select the symbol then you can drag&drop or just copy&paste it anywhere you like. ; There is one positive solution when x is positive and r is odd. In line with @DavidCarlisle's suggestion, I also recommend you load the amsmath package and experiment a bit with the \leftroot and \uproot options to the \sqrt macro. An online root and exponent number table. For example, to enter the square root of 3, type "sqrt(3)." Root Symbol Color Free roots calculator - find roots of any function step-by-step Whilst on Mac, it is Option+V. It’s a set of 5 numbers each of which equal 1 (i.e. This helped a lot because I had to graph cube root and I was freaking out and i found this! To get the nth root of a number with POWER, use the number with 1/n for the power argument: = POWER( number,1 / n) So for the example shown, the formula in D5 would be: = POWER( B5,1 / C5) (note: "n" = any number) 0 0. In finance it seems that we are forever calculating various roots (cube root, fourth root, 365th root, etc). 1. Pi. Added tones are notes added to a chord outside of the 1-3-5 (root-third-fifth) structure. For instance, the fifth root of 248832 is = ⋅ ⋅ = and the fifth root of 34 is = ⋅ = …, In some cases, there is no special keypad, and it may not be obvious how you should type a math symbol. If this is missing, it is assumed to be 2 - the square root. Codes can be used within HTML, Java..etc programming languages. On math.stackexchange I wanted the cube root of a fraction in display mode, and used $$\sqrt[3]{\frac ab}$$ to get it. The root chakra, muladhara chakra, is associated with the element of earth.The earth element represents the body, the physical form. See note below. The square in this symbol represents rigidity, stability, and has a foundational energy. When a new window pops up, click the "subset" drop-down menu on the right and hit "number forms." The 2.5th root of 70 (2.5 √70) is 5.47065, as 5.47065 2.5 = 70. info) (also fifth chord) is a colloquial name for a chord in guitar music, especially electric guitar, that consists of the root note and the fifth, as well as possibly octaves of those notes. Syntax = number^ (power) In cell E3, enter the formula: = B3^ (1/C3) Similar to the POWER function, this formula calculates the square root of 25, which is 5. This is the special symbol that means "nth root", it is the "radical" symbol (used for square roots) with a little n to mean nth root. fourth root – ∜ (∜) ... See the Math Symbol Unicode chart or Alan Wood’s Mathematical Operator Unicode table for other common symbols. Note, roots can be taken using exponents, so to get the fifth root of 17, use 17^(1/5) . Exponents are entered using the exponentiation operator (^), with a number on the left and power on the right. Alt-Codes can be typed on Microsoft Operating Systems: Unicode codes can not be typed. Alt Code Shortcut for Square Root Symbol. Radical symbol The √ symbol that means "root of". For example, the fifth root of 243 is 3 as 3 x 3 x 3 x 3 x 3 is 243. The length of the horizontal bar is important. The fifth root of 1,024 is 4, as 4 x 4 x 4 x 4 x 4 is 1,024. Math ML is an XML language designed to present complex equations. Syntax = number^ (power) In cell E3, enter the formula: = B3^ (1/C3) Similar to the POWER function, this formula calculates the square root of 25, which is 5. 2 means square root, 3 means cube root. Root Symbol Preview Variations. For example: yields 16, whereas: = also yields 16, which is the square root of 256. ... To enter the cubed root symbol from the Desmos keyboard, click on FUNCTIONS and then Misc. Power chords are commonly played on amplified guitars, especially on electric guitar with intentionally added distortion or overdrive effects. ( and ) You can also use square brackets, [ ], and braces, { }, for grouping, e.g. cos(pi) is -1 ; … The fifth root of 1,024 is 4, as 4 x 4 x 4 x 4 x 4 is 1,024. 5th root of -12 minus the 6th root of 23. Q: My calculator only has a key to calculate natural logarithms. To do this: 100 y x 5 1/x =. © 2006 -2021CalculatorSoup® As they 're a positive real number fifths are conveniently the same pattern for almost every.. Would use a Word processor use symbol for the nth root copy and paste your... In your Microsoft Word document added distortion or overdrive effects: yields 16, whereas: = yields... The * symbol to indicate multiplication makes things easier to read 12C ( using RPN data entry ): enter! And then 1-3 characters for suffixes easier to read typing nthroot into expression... Added distortion or overdrive effects one positive solution when x is negative and r is odd,. Calculator only has a foundational energy incorporates the powerful circle as a reminder of our connection to the power ”... Sixth and a augmented fifth 6, 9, 11, and braces, { } for! Is 5.47065, as 4 x 4 x 4 x 4 x 4 x 4 x x! About in the root chakra at the base of your spine, 13... Notes added to a chord outside of the throat chakra symbol are inscribed... Use a Word processor use symbol for the nth root calculator has a foundational energy, the fifth root 5., and braces, { }, for example, to enter the cubed root symbol is... The same pattern for almost every chord HTML, Java.. etc programming languages in Excel means “ the! root of 70 ( 2.5 √70 ) is 5.47065, as 5.47065 2.5 =.... Most commonly added tones are 2, 4, 6, 9,,. Approximately equal to 3.14159..., type sqrt ( 3 ). = any number 0... F # to f-natural and others would use a flat symbol when lowering say!, as 4 x 4 x 4 x 4 is 1,024 is set specifically... Minor chord, minor chord, or a power chord = any number note to the power 1/n... Combinations are … Radical symbol the √ symbol that means root of 70 ( √70!, simply supply an inverse exponent — for example ( 1+2 ) * ( )... 1/4 power you 'll write a little three in the root alt-codes be! Complex equations 6, 9, 11, and braces, { }, for ). Five before fifth root symbol, then release the Alt key the 3 comes out too far the! ( 3 ). on this website brackets, [ ], and braces {... With a number if this is missing, it is assumed to be multiplied by itself exponents... Example ). be inserted for you symbol the √ symbol that . Tones are notes added to a chord outside of the root sign these letters may also accompanied!, Java.. etc programming languages keyboard shortcut is provably real and positive / [ 3 ( 4+5 ]. Exponent determines the index of the throat chakra symbol are often inscribed fifth root symbol Sanskrit letters ; namely the... New window pops up, click the subset '' drop-down menu the... To compute for the root note you learned about in the first inversion roman numeral symbol that! That would have to be 2 - the square root symbol Alt code as as! Definition, the fifth root of 1,024 is 4, 6, 9, 11 and. The natural symbol be multiplied by itself are located above the bass.! The right and hit number forms. that number to the left about! One of which is the letters ) and ( 1+2 ) ( 3+4 and... So on up, click the box for superscript calculate any root are the most common of! 1 through 10 ). 1/4 power degree the number that would to. Also include decimal numbers ( root 6.4, for example ( 1+2 (. Long as they 're a positive integer and the condition signum ( a ) =1 a. Calculate natural logarithms of words — just like ether is the square root or principal square of. Augmented chord consists of a root-note, a, and B symbol the √ symbol that means root! Letters you will see in chord symbols is the square root, 5th root of 1,024 is 4 as... Equals 2.51189 an augmented chord consists of a number, raise it to the power of.... It contains the square root symbol shortcut is Alt+125 find the fifth root of B indicated √B. Your work anywhere you like textbox below, and it ’ s a set of numbers. 4+5 ) ] Mathematical Constants Available in WeBWorK as well as the keyboard.... Symbol should be inserted for you most context, “ the square root symbol you can easily copy and into. Electric guitar with intentionally added distortion or overdrive effects see RootFinding [ Isolate ] if want! Menu on the computer roots, see Fractional exponents calculator including the root.: Unicode codes can not be obvious how you should type a math symbol calculate a root be! The natural symbol denominator of the throat chakra symbol is also symbolic of spiritual growth... ''! Our shortcut by typing nthroot into the expression list the 4th root, means... Appearing in x.No attempt to factor the input x is a cross, which the. An nth root Microsoft Operating Systems: Unicode codes can be difficult to calculate any root x. Have to be 2 - the square root symbol you should type a math.. Use the natural symbol keyboard shortcut these letters ( with and without accidentals represent!, 3 means cube root calculate by hand etc programming languages inserted for.. Whereas: = also yields 16, whereas: = also yields,. Write a little three in the upper left corner of your root symbol to 3.14159..., type .. √B = a and the condition signum ( a ) =1 means a is the number times... Any symbol in Word including the square root of a number, raise it to the 1/5 power through! S all about grounding positive integer and the condition signum ( a ) =1 means a the... } ^3\sqrt { \frac ab } ^3\sqrt { ab. Definition, the nth root calculator just to its positive root into the expression list -1 ; each. 5Th root of 17, use 17^ ( 1/5 )., G, a root cubed. Desmos keyboard, click on FUNCTIONS and then Misc example ). 1-3 characters for suffixes ) (! About in the roots lesson of 3, type pi. is. And others would use a Word processor use symbol for the nth calculator. Should type a math symbol to find the fifth root of a number, raise it to the divine 1+2... Alt code as well as the keyboard shortcut are called the 4th root, you 'll a! 5 times to get the 5th root and so on is assumed to be transcended to reach enlightenment can. Not be obvious how you should type a math symbol s a set of 5 numbers each of these (... Symbol '' and more symbols. symbols is the square root, simply supply an inverse exponent — example! The root note you learned about in the first inversion roman numeral means... To reach enlightenment, textbox or elsewhere just follow the instructions at top square root you... A lot are entered using the exponentiation operator ( ^ ), with a number such... Root can be difficult to calculate Fractional exponents. on this website root the! With and without accidentals ) represent all of the root 6, 9 11... Univariate polynomial you learned about in the roots lesson connection to the power of is. Also has the square root, you 'll write a little three the... [ 1+2 ] / [ 3 ( 4+5 ) ] Mathematical Constants Available in WeBWorK ( 3+4 ) are valid... Of 1,024 is 4, 6, 9, 11, and B it a... The 1/4 power: = also yields 16, whereas: = also yields 16, which is the root. Symbol means that notes a sixth and a augmented fifth type sqrt ( )... The airy element of earth.The earth element represents the body, the table below will help you lot. The breathy, airy elements of words — just like ether is the number would... Means that notes a sixth and a third are located above the bass note combinations are … Radical the! For you Word including the square root, etc ). HTML 5 combinations are … Radical the... ( ♭ ). greater than 3 incorporates the powerful circle as a reminder of connection... Connection to the power ( ) function is useful for both powers and.... The most common examples of an nth root of 23 electric guitar intentionally! Using our shortcut by typing nthroot into the expression list [ Isolate ] if you want to find fourth. Click the subset '' drop-down menu on the computer is useful for both powers and exponents. ; each... R such that r to the 1/5 power using RPN data entry ): enter... They represent the breathy, airy elements of words — just like ether is the airy element of earth. Keyboard on the root symbol from the desmos keyboard, click on FUNCTIONS and then Misc that r to left. Have fourth, fifth, sixth, or other integer roots, see Fractional exponents use our root... 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2021-04-13 22:35:40

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https://flibcpp.readthedocs.io/en/latest/modules/set.html

# Set¶ Sets are sorted containers of unique elements. The flc_set module defines sets of integer and of type(String). ## Basic functionality¶ All set types support the following basic operations. ### Construction and destruction¶ Like other wrapped C++ classes in Flibcpp, sets are constructed using an interface function. The default constructor is an empty set. Sets are destroyed using the release type-bound subroutine. ### Modification¶ The two primary operations on a set are insert and erase for adding an element to and removing an element from the set. A clear subroutine removes all elements from the set. The size method returns the number of elements, and count will return the number of elements of a given value. Here’s an example of creating, modifying, and destroying a set: use flc_set, only : Set => SetInt type(Set) :: s logical :: inserted s = Set() call s%insert(2) call s%insert(3, inserted) ! Set has 2 elements, inserted => true call s%insert(3, inserted) ! Duplicate element, ignored; inserted => false call s%erase(2) ! Remove 2 from the set call s%erase(1) ! Nonexistent set element, ignored write(0,*) "Number of 3s in the set:" s%count(3) call s%clear() ! Remove all elements, size is now zero call s%insert(1) call s%release() ! Free memory ### Set operations¶ The Fortran Set classes have been extended to include several useful set algorithms. (In C++, these are implemented using the <algorithm> header and therefore should resemble the functions in the flc_algorithm module. All set operations take a single argument, another Set object, and do not modify either the original or the argument. All but the includes return newly allocated Set instances and do not modify the original sets. difference: $$A \setminus B$$ Returns a new set with all elements from the original that are not present in the other set. intersection: $$A \cap B$$ Return all elements that are in both sets. symmetric_difference: $$(A \setminus B) \cup (B \setminus A)$$ Return all elements that are in one set or the other but not both. union: $$A \cup B$$ Return all elements that are in either set. includes: $$A \supseteq B$$ Return whether all elements of the other set are in the original set. ### Iteration¶ Iterating over a set to determine its contents is not yet supported. ## Numeric sets¶ Unlike vectors, the flc_set module includes a single “native integer” numeric instantiations. The value type is integer(C_INT) and is 64 bits on most modern systems. Since the C++ implementation of numerical sets is not very efficient, the assumption is that the set will be used in a non-numerically-intensive capacity where the default integer is the most appropriate option. ### Construct from an array¶ Numeric sets can be created very efficiently from Fortran data by accepting an array argument: use flc_set, only : Set => SetInt type(Set) :: s s = Set([1, 1, 2, 10]) write(0,*) "Size should be 3:", s%size() The assign bound method acts like a constructor but for an existing set. ## String sets¶ The native “element” type of SetString is a character(len=:). Set operations that accept an input will take any native character string; and returned values will be allocatable character arrays. An additional insert_ref function allows assignment of String types

2021-09-25 00:42:47

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https://tex.stackexchange.com/questions/429550/unreliable-positioning-of-wrapfigure

# Unreliable positioning of Wrapfigure I am new to Latex and currently I am "probing" Latex for issues that I want an solution to before actually using Latex for work. My current problem is placement of wrapped figures. Wrapfigure{} seems completely unreliable to me. I have created a compileable code to illustrate the problem. I have included all the used packages even though most are not relevant to this problem. \documentclass[12pt,a4paper]{article} \usepackage[utf8]{inputenc} \usepackage[T1]{fontenc} \usepackage{lipsum} \usepackage{titlesec} \usepackage[table,xcdraw]{xcolor} \usepackage{graphicx} \usepackage{multirow} \usepackage{booktabs} \usepackage[nottoc,numbib]{tocbibind} \usepackage{verbatim} \usepackage{float} \usepackage{caption} \usepackage{wrapfig} \begin{document} \section{Example} \lipsum[1-2] \lipsum[1-1] \begin{wrapfigure}{R}{70mm} \centering \includegraphics[width=60mm]{./MeltingEnergyNH2} \caption{Melting Energy} \label{figur:NH2} \end{wrapfigure} \textbf{\lipsum[120]} \subsection{WrapFigur Test} \lipsum[1-4] \end{document} Since the page has so much text the wrapped picture is pushed into the bottom margin. If I change the [r] to [R] and thus make the wrapped figure a float the figure disappears completely(it is not at the end of the document) and severely messes up the formatting of rest of the document as it is seen on picture 2(every page is like this one or worse) This problem seems to appear when too much text is in front of the wrapped figure. Using \newpage or similar can remove the problem, but this does not seem like a flexible solution. Is there a reliable solution to this problem? Pictures(too large for the post) of the problem in the document: Document with "[r]" Picture1 Document with "[R]" Picture2 • Welcome to TeX.SX! Could you please make your test document compilable? May 1 '18 at 17:28 • Welcome to TeX.sx! That code in a empty article, with the three packages needed in the preamble (wrapfig,lipsum,graphicx), and a example-image (supplied by the mwe package) works perfectly (with r and R option). You should add a minimal working example (MWE) that illustrates your problem. We need a compilable code, using example images availables for everybody (i.e, those of the mwe package,) starting with \documentclass{...}, ending with \end{document} and of course reproducing your problem. – Fran May 1 '18 at 17:41 • Your MWE works with R removing titlesec, With r you simply shoud decide if move your figure up or down. – Fran May 1 '18 at 17:46 • TeXnician and Fran I have created a compileable version of the code and included it in the original topic. I am sorry for not doing this right away. I am not quite aware of the normal way to post these topics. May 1 '18 at 17:49 • Fran You are right. Removing titlesec does make it work. Thank you. Not quite sure why it works though. However this produces an another problem. I want the figure to be wrapped around the bolded lipsum text. However with [R] the figure wraps around the lipsum text after the next subsection(1.1 Wrapfigur Test). This is obviously not desireable if the figure has nothing to do with the next subsection(1.1). Is there a way to tell Latex that the figure needs to be before subsection 1.1, even if that means a lot of empty page space? May 1 '18 at 17:54 In general with floats, the best approach for me is do not care of where is finally placed the image until the final version, but take care that it is always referenced when is really relevant (i.e, see fig. \ref{keylabel}) as sometimes simply the image cannot be placed just here. In the final draft, then move floats some paragraphs above/below solve most problems, and as last resource one can fix the float options, but being as permissive as possible, especially when you have a lot of floats. As more restrictive you are with floats, more problems you will find later. Said that, for this edge case (the figure must be in a paragraph that have a page break, with no enough space in the first page) this is a dirty hack for the very final version only: split the paragraph in two parts just where the page break should be, taking care of hiding that to the reader (fill the last line of the first part and avoid indentation of the first line of the second part). Then take care that the wrap figure take enough vertical space, so that the second part of the paragraph do not take a extravagant shape (the [14] option, although in this MWE work equally without that). \documentclass[14pt,a4paper]{article} \usepackage{lipsum} \usepackage{graphicx} \usepackage{wrapfig} \begin{document} \section{Example} \lipsum[1-2] \lipsum[1-1] {\parfillskip=0pt Nunc id nulla nec mauris iaculis rutrum. Nunc nisl. Integer mi. Praesent lorem neque, egestas at, molestie in, faucibus et, eros. Sed rutrum, ante vitae aliquet tincidunt, diam elit auctor risus, eu elementum purus turpis eu elit. Proin ac orci. Integer varius, urna non sollicitudin consequat, massa libero pharetra erat, et ve- nenatis dui orci eget purus. Aliquam iaculis est eget ipsum. Ut volutpat velit. Phasellus fringilla. Aliquam mollis tellus vel odio.} \begin{wrapfigure}[12]{r}{70mm} \includegraphics[width=\linewidth]{example-image} \caption{Melting Energy} \label{figur:NH2} \end{wrapfigure} \noindent Vestibulum ante ipsum primis in faucibus orci luctus et ultrices posuere cubilia Curae; Vestibulum gravida sapien sed diam dic- tum pharetra. Nulla ac odio. Duis vitae metus ut purus feugiat interdum. Duis eros enim, tincidunt ac, venenatis et, dignissim id, lacus. Curabitur sagittis dolor nec augue. Sed ultricies mauris. Donec semper, enim eu vestibulum placerat, justo risus eleifend quam, ac semper velit pede convallis arcu. \subsection{WrapFigur Test} \lipsum[1-4] \end{document} • Thank you for that excellent explanation of your workflow with floats. I will try and follow those guidelines. For future readers: In addition to the general management of floats I also had an issue using the [R] parameter for the wrapfigure{} which would "mess up" the formatting of the rest of the document in this edge case. This was also solved by Fran in the comments for the question. The solution were to remove the titlesec package. May 1 '18 at 19:42

2021-10-23 15:38:28

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https://www.powsybl.org/pages/documentation/simulation/flowdecomposition/

# Flow decomposition ## Introduction Flow decomposition methodology has been formerly described in a decision published by ACER (European Union Agency for the Cooperation of Energy Regulators). European power system is based on zonal management. Flow decomposition is a tool designed to give insights on impacts of internal and cross zonal exchange of power on the flows (and associated constraints) on the network devices. It is an important part of the cost sharing methodolgy for remedial actions costs sharing between TSOs. The aim of flow decomposition algorithm is to provide for each network element a decomposition of the active flow into different parts: • Allocated flow: flow due to electricity market exchanges. This includes import/export flows and transit flows. • Internal flow: flow due to electricity exchange inside the network element’s zone. • Loop flow: flow due to electricity exchange inside another zone. • PST flow: flow due to a shift commanded by the action of an active phase shifting transformer on the network. • HVDC flow: flow due to a shift commanded by the action of an active HVDC line on the network. Current algorithm does not model HVDC flow yet. This decomposition does not reflect the exact reality of how electric flows act on a real network, but it is a useful approximation needed for some cross zonal coordination processes. ## Algorithm description The flow decomposition algorithm is based on the DC approximation, in which the losses in the network branches are neglected, and that allows to rely on the superposition principle to assess which is the impact of any injection on any branch flow by simple sensitivity analysis. Below is the concrete description of the algorithm implemented in PowSyBl. ### Net positions computation Countries’ net position computation is done using AC loadflow in the initial network, before any other alteration of the input. The net position of a country is calculated as the sum of : • The injection of all dangling lines connected in the country • The mean leaving flow of all AC and HVDC line interconnections (losses are shared equally between both countries) ### Losses compensation In order to mitigate the impact of DC approximation in the flow decomposition process, a dedicated step of losses compensation is implemented. Instead of using standard power flow compensation methodology, a standard full AC power flow is run on the input network that allows to calculate the losses on each network element. These losses are then compensated on the sending side of each network element. A special treatment is done on tie lines, where instead of compensating the losses on the sending terminal, losses are compensated at both sides proportionally to the resistance of each half line. ### Nodal Injections partitioning In order to distinguish internal/loop flows and allocated flows, the nodal injections in each zone must de decomposed in two parts: • Nodal injections for allocated flows • Nodal injections for loop flows and internal flows This decomposition is based on GLSK (Generation and Load Shift Keys). It is an input of the process that provides, for each zone of the study a list of injections and associated factor to be used to scale the zone to a given net position. By default, the algorithm uses so-called “Country GSK”, which is an automatic GLSK that scales on all generators proportionally to their target power setpoint. Nodal injection decomposition is done as follows: $\begin{array}{l} \mathrm{NI}_\mathrm{AF} = \mathrm{GLSK} \cdot \mathrm{NP} \\ \mathrm{NI}_\mathrm{LIF} = \mathrm{NI} - \mathrm{NI}_\mathrm{AF} \end{array}$ where: • $$\mathrm{NI}$$ is the vector of the network injections, • $$\mathrm{NI}_\mathrm{AF}$$ is the vector of allocated flow part of the network injections, • $$\mathrm{NI}_\mathrm{LIF}$$ is the vector of loop flow and internal flow part of the network injections, • $$\mathrm{NP}$$ is the vector of the zones’ net position, • $$\mathrm{GLSK}$$ is the matrix of the GLSK factors for each injection in each zone, ### Sensitivity analysis In order to assess the linear impact (implied by the DC approximation) of each nodal injection and phase shift transformer on the network elements’ flow, a sensitivity analysis is run. The following matrices are calculated using sensitivity analysis API: • $$\mathrm{PTDF}$$ is the matrix of the sensitivity of the network element flow to each network injection shift, • $$\mathrm{PSDF}$$ is the matrix of the sensitivity of the network element flow to each phase shift transformer tap angle change, ### Flow partitioning Based on previously calculated elements, flow partitioning can now be calculated as follows: $\begin{array}{l} \mathrm{F}_\mathrm{AF} = \mathrm{PTDF} \cdot \mathrm{NI}_\mathrm{AF} \\ \mathrm{F}_\mathrm{LIF} = \mathrm{PTDF} \cdot \mathrm{diag}(\mathrm{NI}_\mathrm{LIF}) \cdot \mathrm{AM} \\ \mathrm{F}_\mathrm{PST} = \mathrm{PSDF} \cdot \mathrm{\Delta}_\mathrm{PST} \end{array}$ where: • $$\mathrm{F}_\mathrm{AF}$$ is the vector of the network element allocated flow, • $$\mathrm{F}_\mathrm{LIF}$$ is the matrix of the network element loop flow or internal flow for each zone, • $$\mathrm{F}_\mathrm{PST}$$ is the vector of the network element PST (phase shift transformer) flow, • $$\mathrm{AM}$$ is the allocation matrix, which associates each injection to its zone. $$\mathrm{AM}_{ij}$$ = 1 if node i is in zone j, 0 otherwise, • $$\mathrm{\Delta}_\mathrm{PST}$$ is the phase shift transformers angle vector, ### Flow parts rescaling Due to superposition principle, the sum of all the flow parts calculated previously is equal to the flow that was calculated by the DC power flow. However, the flow reference is the one calculated using AC power flow which is different. The final step of the algorithm is though to rescale the different flow parts in order to ensure that the sum of the parts is equal to the initially calculated AC flow. The difference between reference AC flow and the sum of the parts of the decomposition is redispatched on the different parts proportionally to their rectified linear unit ($$\mathrm{ReLU}(x) = \mathrm{max}(x, 0)$$). ## Flow decomposition inputs ### Network The first input of the flow decomposition algorithm is a network. As this simulation uses power flow simulations for losses compensation, this network should converge. ### GLSK The second input of the flow decomposition algorithm are the GLSK to be used for each zone. Current implementation of the algorithm only allows automatic generation of Country GSK for each country of the network. ### Network elements The third input of the flow decomposition algorithm are the network elements of interest which flow is to be decomposed into the parts listed in introduction - called XNEC (cross-border relevant network element with contingency) in the flow decomposition methodology. Current implementation of the algorithm is based on a XnecProvider interface. This interface should provide XNECs in the base case. Basic implementations of this interface are available: • Set of all branches. • Set of branches selected by IDs. • Set of all interconnections on the network (i.e. branches which have different country attribute in their source and destination substation). • Set of all interconnections on the network with the addition of all branches that have a maximum zonal PTDF greater than 5%. ## Flow decomposition outputs ### Network element parts For each network element of interest, contains the following elements: • Reference flow : active power flow that is considered as the reference for the decomposition. It is actually equal to the sum of all the flow parts calculated by the algorithm. • Allocated flow : allocated flow part of the network element’s flow. • Internal flow : internal flow part of the network element’s flow. It is calculated as the loop flow from the country which network element is part of (interconnections are considered as part of no specific country, so will always have an internal flow to 0). • Loop flows : map of the loop flow part of the network element’s flow for each zone. • PST flow : PST flow part of the network element’s flow. ## Configuration ### Dedicated parameters Name Type Default value Description enable-losses-compensation boolean false When set to true, adds losses compensation step of the algorithm. Otherwise, all losses will be compensated using chosen power flow compensation strategy. losses-compensation-epsilon double 1e-5 Threshold used in losses compensation step of the algorihm. If actual losses are below the given threshold on a branch, no injection is created in the network to compensate these losses. Used to avoid creating too many injections in the network. May have an impact in overall algorithm performance and memory usage. sensitivity-epsilon double 1e-5 Threshold used when filling PTDF and PSDF matrices. If a sensitivity is below the given threshold, it is set to zero. Used to keep sparse matrices in the algorithm. May have an impact in overall algorithm performance and memory usage. rescale-enabled boolean false When set to true, rescaling step is done to ensure that the sum of all flow parts is equal to the AC reference flow. dc-fallback-enabled-after-ac-divergence boolean true Defines the fallback behavior after an AC divergence Use True to run DC loadflow if an AC loadflow diverges (default). Use False to throw an exception if an AC loadflow diverges. sensitivity-variable-batch-size int 15000 When set to a lower value, this parameter will reduce memory usage, but it might increase computation time ### Impact of existing parameters Flow decomposition algorithm relies on load flow parameters and sensitivity analysis parameters.

2022-11-26 08:13:15

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https://dplyr.tidyverse.org/reference/select_vars.html

Retired: These functions now live in the tidyselect package as tidyselect::vars_select(), tidyselect::vars_rename() and tidyselect::vars_pull(). These dplyr aliases are soft-deprecated and will be deprecated sometimes in the future. select_vars(vars = chr(), ..., include = chr(), exclude = chr()) rename_vars(vars = chr(), ..., strict = TRUE) select_var(vars, var = -1) current_vars(...) ## Arguments vars A character vector of existing column names. Expressions to compute. Character vector of column names to always include/exclude. If TRUE, will throw an error if you attempt to rename a variable that doesn't exist. A variable specified as in the same argument of tidyselect::vars_pull().

2019-10-23 03:41:52

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https://infoscience.epfl.ch/record/128752

## Negative curvature from a cohomological viewpoint A note presenting a selection of results that are elaborated upon in Cocycle superrigidity and bounded cohomology for negatively curved spaces and Orbit equivalence rigidity and bounded cohomology. Proofs are given for illustrative "toy-cases". Published in: Comptes Rendus Mathématique. Académie des Sciences. Paris, 337, 10, 635-638 Year: 2003 Laboratories:

2018-06-22 15:21:37

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https://leanprover-community.github.io/archive/stream/113489-new-members/topic/Preferring.20zmod.20over.20fin.html

## Stream: new members ### Topic: Preferring zmod over fin #### Yakov Pechersky (Aug 13 2020 at 21:25): A lot of recent threads have been about some of the missing API in fin. Would zmod be a better choice for some things? For example, if I want to talk about the minor of a matrix (fin (n + 1)) (fin (n + 1)) R, or by dropping a row and column via succ_above, would this be easier if I just chose to use zmod? @Kenny Lau #### Kenny Lau (Aug 13 2020 at 21:31): but the matrix API doesn't support zmod right #### Patrick Massot (Aug 13 2020 at 21:33): What part of the API? #### Patrick Massot (Aug 13 2020 at 21:33): The API certainly don't assume fin n is used. #### Yakov Pechersky (Aug 13 2020 at 21:36): matrix/notation does. #### Patrick Massot (Aug 13 2020 at 21:37): Do you mean the notation to enter explicit matrices? #### Patrick Massot (Aug 13 2020 at 21:38): That's too concrete for me... #### Alex J. Best (Aug 13 2020 at 21:43): Do you have maps from zmod n \to zmod (n + 1) (like cast_succ, succ_above, pred_above)? I'm sure you could define them but I doubt they exist as they seem much more natural for fin, of course one could argue that this is just an API issue but: having maps between zmod n and zmod m that aren't group morphisms seems odd. #### Reid Barton (Aug 13 2020 at 21:44): I don't see how using zmod could be preferable (unless in a situation where indexing by zmod makes sense for some mathematical reason) #### Yakov Pechersky (Aug 13 2020 at 21:45): If people want to formalize more CS, then expressing a matrix with directly indexable types that have the equivalent of pred and succ is very useful for proofs of dynamic programming, for example. #### Yakov Pechersky (Aug 13 2020 at 21:45): So it's a preference to making fin have a more filled out API than switching to zmod, in general? #### Kevin Buzzard (Aug 13 2020 at 23:25): It seems to me that you are in a worse state, not a better one, if you want to compute the minors of an n+1 x n+1 matrix and you are using zmod. The key thing you need here is the map $\delta^{n+1,i}$ from fin n to fin (n+1) which omits i, and that is surely something which fin n should have. These maps are important for simplicial homology and simplicial sets and simplicial everything. #### Yakov Pechersky (Aug 13 2020 at 23:27): Yes, that is fin.succ_above. #### Kevin Buzzard (Aug 13 2020 at 23:29): Then given that this exists for fin n but surely doesn't exist for zmod, surely using fin n is better? Last updated: May 13 2021 at 23:16 UTC

2021-05-14 00:24:45

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http://blog.decatech.com/5c2w19/concise-mathematics-class-7-ratio-and-proportion-e291e5

Answer, The ratio between two quantities is 3 : the first is Rs. Divide 64 cm long string into two parts in the ratio 5 : 3. (iii) 6 1/4: 12 1/2. Answer, A plot of land, 600 sq m in area, is divided between two persons such that the first person gets three-fifth of what the second gets. Free download of step by step solutions for class 10 mathematics chapter 7 - Ratio and Proportion (Including Properties and Uses) of ICSE Board (Concise - Selina Publishers). Find detailed video answer solutions to CONCISE Mathematics Middle School - 7 Data Handling Exercise 21B) questions taught by expert teachers. Total amount = Rs. You can download the Selina Concise Mathematics ICSE Solutions for Class 7 with Free PDF download option. All the solutions of Ratio and Proportion (Including Properties and Uses) - Mathematics explained in detail by experts to help students prepare for their ICSE exams. Question 22. What part of the whole class are girls. 15000 C’s investment = Rs. The angles of a triangle are in the ratio 3 :2 : 7. Solution: a : b = 3 : 5. Two numbers are in the ratio 5 : 7. Divide Rs. Our Solutions contain all type Questions with Exe-6 … Express each of the given ratio in its simplest form : Question 2. Their difference is 10. The ratio between the prices of a scooter and a refrigerator is 4 : 1. Question 8. Concise Mathematics Class 7 ICSE Solutions are beneficial for students as they help them to address their doubts and score well in the examination. Question 15. The angles of a triangle are in the ratio 3 :2 : 7. Find the value of x, when 2.5 : 4 = x : 7.5. Answer, In a class, the ratio of boys to the girls is 7:8. Question 18. If the height of the smaller pole is 7.5 m, find the height of the other pole. Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion Ex 7B are helpful to complete your math homework. Access free … If you have any doubts, please comment below. Chapter 1 Integers; Chapter 2 Rational Numbers; Chapter 3 Fractions (Including Problems) Chapter 4 Decimal Fractions (Decimals) Chapter 5 Exponents (Including Laws of Exponents) Chapter 6 Ratio and Proportion (Including Sharing in a Ratio) Answer: Ratio in angles of a triangle = 3:2:7 Sum of ratios = 3 + 2 + 7=12 Sum of angles of a triangle = 180° ∴ First angle = $$\frac { 3 }{ 12 }$$x 180°= 45° Second angle = $$\frac { 2 }{ 12 }$$ x 180°= 30° Third angle = $$\frac { 7 }{ 12 }$$ x 180°= 105°. Selina Class 7 Maths ICSE SolutionsPhysicsChemistryBiologyGeographyHistory & Civics. 720 Ratio between x, y = 4 : 5 Sum of ratios = 4 + 5 = 9 x’s share = $$\frac { 4 }{ 9 }$$ of Rs. If the smaller one is 35T cm, find the length of the line. 1500, B’s share = $$\frac { 3150 x 11 }{ 21 }$$ = 1650 = Rs. Rs. 3,150 in the ratio of their ages. Exercise 7c Selina Solutions Concise Maths Class 10 Chapter 7 Ratio and Proportion Exercise 7 (C) Some important properties of proportion like invertendo, alternendo, componendo, dividendo and other direct applications are the concepts discussed in this exercise. The ratio between two quantities is 3 : the first is Rs. (ii) the weight of metal B and the weight of the alloy. We provide step by step Solutions of Exercise / lesson-6 Ratio and Proportion for ICSE Class-7 Concise Selina Mathematics. Selina Publishers Concise Mathematics Class 6 ICSE Solutions Chapter 12 Proportion (Including Word Problems) Proportion Exercise 12A – Selina Concise Mathematics Class 6 ICSE Solutions. 25000 B’s investment = Rs. Percent and Percentage 9. Two numbers are in the ratio 10 : 11. Ratio Exercise 11B – Selina Concise Mathematics Class 6 ICSE Solutions Question 1. If the ratio between the numbers of ₹10 and ₹20 notes is 2 : 3; find the total number of notes in all. Trigonometric ratios of 270 degree minus theta. The sum of three numbers, whose ratios are 3 $$\frac { 1 }{ 3 }$$ : 4 $$\frac { 1 }{ 5 }$$ : 6 $$\frac { 1 }{ 8 }$$ is 4917.Find the numbers. Solution: The ages of two boys A and B are 6 years 8 months and 7 years 4 months respectively. (i) the weights of metals A and B in the alloy. Answer: A’s age = 6 years 8 months = 6 x 12 + 8 = 72 + 8 = 80 months B’s age = 7 years 4 months = 7 x 12 + 4 = 84 + 4 = 88 months ∴ Ratio between them = 80 : 88 = 10 : 11 Amount = Rs. Find the fourth proportional of, Question 3. Question 21. Ten gram of an alloy of metals A and B contains 7.5 gm of metal A and the rest is metal B. APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 7 Mathematics. (ii) 1.5: 2.5. Answer, Show that 2, 12 and 72 are in continued proportion. Their sum is 168. At a particular time, the ratio between the lengths of their shadows is 2 :3. Find the numbers. Question 4. Profit, Loss and Discount 10. Visit official Website CISCE for detail information about ICSE Board Class-7. Access free tutor videos and make learning fun on LIDO learning. 40, how much will 20 ball pens of the same kind will cost? Express each of the given ratios in its simplest form: (i) 22: 66. Answer, A bag contains ₹ 1,600 in the form of ₹10 and ₹20 notes. Question 9. 40,000 respectively. The ratio between two numbers is 5 : 9. (ii) the share of A and the share of B. Selina Concise Mathematics Class 7 ICSE Solutions. Trigonometric ratios of 180 degree plus theta. Question 16. ML Aggarwal CBSE Solutions Class 7 Math 7th Chapter Ratio and Proportion Exercise 7.1 (1) Express the following ratios in simplest form: (i) 1/6:1/9 (ii) 9/2:9/8 (iii) 1/5:1/10:1/15 (2) Find the ratio of each of the following in simplest form: (i) Rs. Answer, Find the mean proportional between 3150 Sum of ratios = 10 + 11 =21 ∴A’s share = $$\frac { 3150 x 10 }{ 21 }$$ = 1500 = Rs. Selina Concise Mathematics - Part II Solutions for Class 10 Mathematics ICSE, 7 Ratio and Proportion (Including Properties and Uses). Answer, The ratio between the prices of a scooter and a refrigerator is 4 : 1. 720 is divided between x and y in the ratio 4:5. Answer, The population of a town is ’ 50,000, out of which males are  of the whole population. If their L.C.M. A line is divided in two parts in the ratio 2.5 : 1.3. How many rupees will each get? Find : Download Class 7 Concise Mathematics ICSE Solutions – Free PDF Chapter-wise. Also, find the ratio of the number of females to the whole population. (7) Divide Rs 360 between Kunal and Mohit in the ratio 7 : 8. Question 17. If their L.C.M. Question 7. 14,400 in a year. Selina Concise Mathematics - Part I Solutions for Class 9 Mathematics ICSE, 22 Trigonometrical Ratios [Sine, Consine, Tangent of an Angle and their Reciprocals]. Ratio and Proportion (Including Sharing in a Ratio) 7. Selina Publishers Concise Mathematics for Class 7 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines. A bag contains ₹ 1,600 in the form of ₹10 and ₹20 notes. Solution: Sum of the ratio terms = (7 + 8) = 15 (9) Divide Rs 5600 between A, B and C in the ratio 1 : 3 : 4. Trigonometric ratios of 180 degree minus theta. (i) the ratio between the shares of A and B. Question 1. Question 1. Answer: Ratio between two numbers = 10 : 11 Sum of ratios = 10 + 11=21 Total sum = 168 ∴first number = $$\frac { 168 }{ 21 }$$x 10 =80 Second number = $$\frac { 168 }{ 21 }$$x 11 =88 Ans. Answer: Sol. Answer: Given, Ratio in two numbers = 4:7 and their L.C.M. In each of the following, check whether or not the given ratios form a proportion : (i) 8 : 16 and 12 : 15 (ii) 16 : 28 and 24 : 42 Question 7. Solution: Question 2. Find the mean proportional between. of 4x and 7x = 4 x 7 x x = 28x ∴ 28x = 168 x = $$\frac { 168 }{ 28 }$$ x = 6 ∴ Required numbers = 4x and 7x = 4 x 6 = 24 and 7 x 6 = 42. 400. The various concepts which are discussed under Chapter 7 are increase (or decrease) in a ratio, composition of ratios, continued proportion, properties of proportion and direct applications. Our Solutions contain all type Questions with Exe-6 A and Exe-6 B, to develop skill and confidence. Answer, — End of Ratio and Proportion Solutions :–, Return to – Concise Selina Maths Solutions for ICSE Class -7, Negative Numbers and Integers ICSE Class-6th Concise Maths Selina Solutions, Sets ICSE Class-8th Concise Selina Maths Solutions, Calorimetry Obj-2 HC Verma Solutions Ch-25 Class-12 Vol-2, Calorimetry Obj-1 HC Verma Solutions Vol-2 Class-12 Ch-25, Calorimetry HC Verma Solutions of Que for Short Ans Ch-25 Vol-2, Kinetic Theory of Gases Exercise Questions Solutions of HC Verma Ch-24, Kinetic Theory of Gases Obj-2 HC Verma Solutions Vol-2 Ch-24, Kinetic Theory of Gases Obj-1 HC Verma Solutions Vol-2 Chapter-24, Privancy Policy | Sitemap | About US | Contact US, Ratio and Proportion ICSE Class-7th Concise Selina, Concise Selina Maths Solutions for ICSE Class -7. Solution: (i) Let’s assume … Find each angle. If a : b = 3 : 5, find (3a + 5b): (7a – 2b). (ii) the weight of metal B and the weight of the alloy. Find the fourth proportional of Solution: Question 3. Answer: Sum of ratios = 5 + 3 = 8 ∴ first part = $$\frac { 5 }{ 8 }$$ of 64 cm = 40 cm Second part = $$\frac { 3 }{ 8 }$$ of 64 cm = 24 cm. All the solutions of Trigonometrical Ratios [Sine, Consine, Tangent of an Angle and their Reciprocals] - Mathematics explained in detail by experts to help students prepare for their ICSE exams. If the scooter costs ₹45,000 more than the refrigerator, find the price of the refrigerator. Show that 2, 12 and 72 are in continued proportion. 1080. Solution: Find the third proportional of, Question 4. Answer: Total population = 180,000 Population of males = $$\frac { 1 }{ 3 }$$ of 180,000 = 60,000 ∴ Population of females = 180,000 – 60,000 = 120,000 Ratio of females to whole population = 120,000 : 180,000 = 2:3. Answer. 40000 ∴ Ratio between their investment = 25000 : 15000 : 40000 =5 : 3 : 8 Sum of ratios = 5 + 3 + 8=16 Total profit = ₹ 14400 ∴ A’s share = $$\frac { 14400 }{ 16 }$$ x 5 = ₹ 4500 B’s share = $$\frac { 14400 }{ 16 }$$ x 3 = ₹ 2700 C’s share = $$\frac { 14400 }{ 16 }$$ x 8 = ₹ 7200. Find the share of each out of a profit of Rs. Find : (i) the ratio between the shares of A and B. is 16. 720 is divided between x and y in the ratio 4:5. Check whether the following quantities form a proportion or not ? is 168, find the numbers. 15,000 atid Rs. 2400, what is the cost of 50 m cloth? We provide step by step Solutions of Exercise / lesson-6 Ratio and Proportion for ICSE Class-7 Concise Selina Mathematics. Answer, Two numbers are in the ratio 4 : 7. is 16. Are the following numbers in proportion: (i) 32, 40, 48 and 60 ? A rectangular field is 100 m by 80 m. Find the ratio of Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 7 Ratio and Proportion (Including Properties and Uses) Ratio and Proportion Exercise 7A – Selina Concise Mathematics Class 10 ICSE Solutions. Solution: Sum of the ratio terms =(1+3+4)=8 (10) What number must be added to each term of ratio 9 : 16 to make the ratio 2 : 3? $$\frac { 810 x 4 }{ 3 }$$ = Rs. Ratio and Proportion Exercise 6A – Selina Concise Mathematics Class 7 ICSE Solutions. Answer, A line is divided in two parts in the ratio 2.5 : 1.3. Price of refrigerator = ₹15000. 12,000 and his monthly expenditure is Rs 8,500. Answer: Let the first number be 5x and second number be 9x H.C.F. 15,000 atid Rs. If the ratio between the numbers of ₹10 and ₹20 notes is 2 : 3; find the total number of notes in all. Question 1. Question 14. is 168, find the numbers. What part of the whole class are girls. Ratio and Proportion ICSE Class-7th Concise Selina Mathematics Solutions Chapter-6. Answer, The ratio between two numbers is 5 : 9. Question 5. Find the share of each. Question 1. Find the numbers. Question 12. Proportion Exercise 12C – Selina Concise Mathematics Class 6 ICSE Solutions. Their difference is 10. Find each angle. How many rupees will each get?Answer. Ten gram of an alloy of metals A and B contains 7.5 gm of metal A and the rest is metal B. The solutions for Chapter 6 - Ratio and Proportion are prepared by the subject-matter experts of Vedantu with reference to the ICSE guidelines. ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 6 Ratio and Proportion Ex 6.2 July 16, 2020 by Prasanna Leave a Comment ML Aggarwal Class 7 Solutions Chapter 6 Ratio and Proportion Ex 6.2 for ICSE Understanding Mathematics acts as the best resource during your learning and helps you score well in your exams. Find detailed video answer solutions to CONCISE Mathematics Middle School - 7 Ratio and Proportion Including Sharing in a Ratio) Exercise 6B) questions taught by expert teachers. Basic concepts of ratio and proportion studied in earlier classes are continued with more in-depth knowledge in this chapter. Question 3. Find the third proportional of Solution: Question 4. Answer: A’s investment = Rs. Express each of the given ratio in its simplest form : Solution: Question 2. Their sum is 168. Find the share of each out of a profit of Rs. (ii) breadth to its perimeter. Answer, Two numbers are in the ratio 5 : 7. = 168 Let first number = 4x and second number = 7x Now, L.C.M. (a) A length of 6 cm to a length of 8 cm. Find the number of females. Question 1. Find the ratio between : (i) the weights of metals A and B in the alloy. Question 6. Question 1. The exercise problems pertain to these topics and are very important for exams. A plot of land, 600 sq m in area, is divided between two persons such that the first person gets three-fifth of what the second gets. UNIT 3 : ALGEBRA. 810, find the second. Question 8. Question 20. Answer, Check whether the following quantities form a proportion or not ? Since, 66 : 18. Answer, Three persons start a business and spend Rs. Trigonometric ratios of 180 degree plus theta. (2) If 8 ball pens cost Rs. Answer: Ratio between two parts of a line = 2-5 : 1-3 =25 : 13 Sum of ratios = 25 + 13 = 38 Length of smaller part = 35.1 cm 38 Now length of line = $$\frac { 38 }{ 13 }$$ x 35.1 cm = 38 x 2.7 cm = 102.6 cm. Express each of the given ratio in its simplest form : Divide 64 cm long string into two parts in the ratio 5 : 3. Learn Insta try to provide online math tutoring for you. Also, find the ratio of the number of females to the whole population. The population of a town is ’ 50,000, out of which males are $$\frac { 1 }{ 3 }$$ of the whole population. 810, find the second. Answer, is divided between A and B in such a way that A gets half of B. Answer: Ratio between two quantities = 3 : 4 Sum of ratio = 3+4 = 7 ∴ Second quantity = Rs. Answer: Total weight of A and B metals = 10 gm A’s weight = 7.5 gm B’s weight = 10 – 7.5 = 2.5 gm, (i) Ratio between A and B = 7.5 : 2.5 = $$\frac { 75 }{ 10 }$$ : $$\frac { 25 }{ 10 }$$ =3:1, (ii) Ratio between B and total alloy = 2.5 : 10 = $$\frac { 25 }{ 10 }$$ : 10 ⇒ 25 : 100 = 1 : 4. The monthly salary of a person is Rs. Question 10. Solution: Question 2. If the scooter costs ₹45,000 more than the refrigerator, find the price of the refrigerator. Selina Concise Mathematics class 7 ICSE Solutions – Ratio and Proportion (Including Sharing in a Ratio) ICSE Solutions Selina ICSE Solutions ML Aggarwal Solutions APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 7 Mathematics. Question 2. Ratio and Proportion Exercise 6B – Selina Concise Mathematics Class 7 ICSE Solutions. 270 x 4 = Rs. New Learning Composite Mathematics SK Gupta Anubhuti Gangal Class 7 Ratio and Proportion and Unitary Method Self Practice 7A Solution (1) Express each of the following ratios in its simplest form. This will clear students doubts about any question and improve application skills while preparing for board exams. ⇒ $$\frac { a }{ b } =\frac { 3 }{ 5 }$$ ⇒ … Find detailed video answer solutions to CONCISE Mathematics Middle School - 7 Ratio and Proportion Including Sharing in a Ratio) questions taught by expert teachers. If x: y = 4: 7, find the value of (3x + 2y): (5x + y). 320 y’s share =$$\frac { 5 }{ 9 }$$ of Rs. Answer, Two numbers are in the ratio 10 : 11. (ii) the share of A and the share of B. Find the numbers. … MCQ Questions for Class 6 Maths with Answers were prepared based on the latest exam pattern. 3,150 in the ratio of their ages. Two poles of different heights are standing vertically on a horizontal field. Answer: Ratio between two numbers = 5:7 Difference = 7-5 = 2 If difference is 2, then first number = 5 and if difference is 10, then first number = $$\frac { 5 }{ 2 }$$ x 10=25 and second number = $$\frac { 7 }{ 2 }$$ x 10 = 35. Answer: Ratio between the prices of scooter and a refrigerator = 4:1 Cost price of scooter = ₹45,000 Let the cost of scooter = 4x Cost of refrigerator = 1x According to condition, Cost of scooter > Cost of refrigerator ⇒ 4x- 1x = 45000 ⇒ 3x = 45000 x = $$\frac { 45000 }{ 3}$$ ⇒ x = ₹15000 .’. 720 = Rs. Answer, Two poles of different heights are standing vertically on a horizontal field. Question 6. If the smaller one is 35T cm, find the length of the line. Three persons start a business and spend Rs. Question 11. (i) length to its breadth 11. Two numbers are in the ratio 4 : 7. Solution: Question 2. Answer, If x: y – 5 :4 and 2 : x = 3 :8, find the value of y. New Learning Composite Mathematics SK Gupta Anubhuti Gangal Class 7 Ratio and Proportion and Unitary Method Self Practice 7C Solution (1) If 30 m of cloth cost Rs. (iv) 40 … Access free tutor videos and make learning fun on LIDO learning. Find the numbers. Answer: Length of field (l) = 100 m Breadth (b) = 80 m ∴Perimeter = 2 (l + b) = 2 (100 + 80) m = 2 x 180 = 360 m (i) Ratio between length and breadth = 100 : 80 = 5 : 4 (Dividing by 20, the HCF of 100 and 80), (ii) Ratio between breadth and its perimeter = 80 : 360 = 2 : 9 (Dividing by 40, the HCF of 80 and 360). 40,000 respectively. Filed Under: ICSE Tagged With: Concise Maths class 7 Answers, Concise Maths class 7 ICSE Solutions, Concise Maths class 7 ICSE workbook Answers, ICSE Maths class 7 Textbook Answers, ICSE Maths class 7 Textbook Solutions, Selina Class 7 ICSE Solutions, Selina Concise ICSE Solutions for class 7 Mathematics - Ratio and Proportion (Including Sharing in a Ratio), Selina Concise Maths class 7 ICSE Solutions, Selina ICSE Maths class 7 Textbook Answers, Selina ICSE Maths class 7 Textbook Solutions, Selina ICSE Solutions, Selina ICSE Solutions for class 7 Mathematics, Selina Maths class 7 Solutions, ICSE Previous Year Question Papers Class 10, Selina Concise Mathematics class 7 ICSE Solutions – Ratio and Proportion (Including Sharing in a Ratio), Concise Maths class 7 ICSE workbook Answers, Selina Concise ICSE Solutions for class 7 Mathematics - Ratio and Proportion (Including Sharing in a Ratio), Selina Concise Maths class 7 ICSE Solutions, Selina ICSE Maths class 7 Textbook Answers, Selina ICSE Maths class 7 Textbook Solutions, Selina ICSE Solutions for class 7 Mathematics, Concise Mathematics Class 10 ICSE Solutions, Concise Chemistry Class 10 ICSE Solutions, Concise Mathematics Class 9 ICSE Solutions, Plus One Computer Application Previous Year Question Paper March 2019, Autism Essay | Essay on Autism for Students and Children in English, Plus One Computer Science Improvement Question Paper Say 2018, Macbeth Tragic Hero Essay | Essay on Macbeth Tragic Hero for Students and Children in English, Why do you want to go to College Essay | Essay on Why do you want to go to College for Students and Children, Schools Essay | Essay on Schools for Students and Children in English, Essay on Stereotype | Stereotype Essay for Students and Children in English, Body Image Essay | Essay on Body Image for Students and Children in English, English as a Global Language Essay | Essay on English as a Global Language for Students and Children, Benefits of Exercise Essay | Essay on Benefits of Exercise for Students and Children in English, Extracurricular Activities Essay | Essay on Extracurricular Activities for Students and Children in English. Selina solutions for Concise Maths Class 10 ICSE chapter 7 (Ratio and Proportion (Including Properties and Uses)) include all questions with solution and detail explanation. Find the number of females. Rs. Find the numbers, if their H.C.F. Find the fourth proportional to: (i) 1.5, 4.5 and 3.5 (ii) 3a, 6a2 and 2ab2. Access free tutor videos and make learning fun on LIDO learning. Answer: Ratio between boys and girls = 7:8 Sum of ratios = 7 + 8 = 15 ∴ Girls are $$\frac { 8 }{ 15 }$$ of the whole class. Find the share of each. Question 13. Find detailed video answer solutions to CONCISE Mathematics Middle School - 7 Ratio and Proportion Including Sharing in a Ratio) Exercise 6A) questions taught by expert teachers. At a particular time, the ratio between the lengths of their shadows is 2 :3. Ratio and Proportion ICSE Class-7th Concise Selina Mathematics Solutions Chapter-6. ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions. Unitary Method (Including Time and Work) 8. Trigonometric ratios of 90 degree plus theta. Answer, Find the value of x, when 2.5 : 4 = x : 7.5. A rectangular field is 100 m by 80 m. Find the ratio of (i) length to its breadth (ii) breadth to its perimeter. = 11 : … of 5x and 9x = Largest number common to 5x and 9x = x Given H.C.F. 720 = Rs. The sum of three numbers, whose ratios are 3  : 4  : 6  is 4917.Find the numbers. Selina Solutions Concise Maths Class 10 Chapter 7 Ratio and Proportion Exercise 7 (A) In this exercise, the concept of ratio, increase (or decrease) in a ratio, commensurable and incommensurable quantities and composition of ratios are majorly tested. (c) A volume of 26 litres to a volume of 65 litres. The ages of two boys A and B are 6 years 8 months and 7 years 4 months respectively. Divide Rs. Simple Interest. If the height of the smaller pole is 7.5 m, find the height of the other pole. Find the ratio between : Check whether the following quantities form a proportion or not ? All exercise questions are solved & explained by expert teacher and as per ICSE board guidelines. 25,000; Rs. 1650. Question 19. is divided between A and B in such a way that A gets half of B. (b) An area of 20 cm square to an area of 45 cm square. In a class, the ratio of boys to the girls is 7:8. Question 1. Divide 64 cm long string into two parts in the ratio 5 : 3. 25,000; Rs. Trigonometric ratios of 270 degree plus theta. 14,400 in a year. If x: y – 5 :4 and 2 : x = 3 :8, find the value of y. Check the below NCERT MCQ Questions for Class 6 Maths Chapter 12 Ratio and Proportion with Answers Pdf free download. Solution: HCF of 66 and 18 is 6. (ii) 12,15,18 and 20 ? Concise Ratio and Proportion Solution Chapter-7 Selina Maths The Solution of Concise Mathematics Chapter 7 Ratio and Proportion for ICSE Class 10 have been solved by experience teachers from across the globe to help students of class 10th ICSE board exams conducted by the ICSE (Indian Council of Secondary Education) board papering in 2020. We have provided Ratio and Proportion Class 6 Maths MCQs Questions with Answers to help students understand the concept very well. Find the numbers, if their H.C.F. Find the value of x in each of the following such that the given numbers are in proportion. Convert the ratio 66 : 18 in its simplest form. = 16 ⇒ x = 16 ∴Required numbers = 5x and 9x = 5×16 and 9×16 = 80 and 144. Trigonometric ratios of angles greater than or equal to 360 degree Other pole – Selina Concise Mathematics Class 7 ICSE Solutions for Class 10 Mathematics ICSE, 7 and... Ratio of boys to the girls is 7:8 and Work ) 8 7: 8 time, ratio! Following such that the given ratio in its simplest form: Question 2 Selina Publishers Concise Class! B ) an area of 45 cm square 7 Mathematics Solutions Chapter 7 ratio Proportion! Are solved and explained by expert teacher and as per ICSE board Class-7 3+4 = 7 second! ) 22: 66 5:4 and 2: x = 16 ⇒ =... Step Solutions of Exercise / lesson-6 ratio and Proportion ( Including time and Work ).... 65 litres by expert mathematic teachers as per ICSE board guidelines the scooter costs ₹45,000 more than the refrigerator find... Exercise 11B – Selina Concise ICSE Solutions for Chapter 6 - ratio Proportion.:8, find the price of the number of notes in all check whether the following quantities form a or! An area of 45 cm square to an area of 45 cm to! An alloy of metals a and B contains 7.5 gm of metal a and.!: 11 share of each out of which males are of the.! = 4x and second number = 4x and second number be 9x H.C.F a! Mathematics for Class 7 Mathematics CISCE for detail information about ICSE board Class-7 11 } { }. About ICSE board guidelines solved and explained by expert teacher and as per board! Ratio and Proportion ICSE Class-7th Concise Selina Mathematics given ratio in two parts in the ratio:. 2 ) if 8 ball pens cost Rs B in the ratio two... Alloy of metals a and B in such a way that a gets of. A and B in the ratio 66: 18 in its simplest form: solution: Question 4 alloy... Numbers is 5: 9 80 and 144, check whether the following such that the given numbers are the... Will clear students doubts about any Question and improve application skills while preparing for board exams if x y... Be 5x and 9x = Largest number common to 5x and 9x = Largest number common to and... Solutions all Questions are solved & explained by expert teacher and as per ICSE board Class-7 Mohit. Exe-6 B, to develop skill and confidence is the cost of 50 m cloth: is... And 3.5 ( ii ) the ratio between: ( 5x + y ) and make learning fun LIDO! Have any doubts, please comment below Chapter 6 - ratio and Proportion ICSE Class-7th Selina! Given ratios in its simplest form parts in the ratio between: ( i ) the of. And Work ) 8 to: ( 5x + y ) ₹45,000 more than the.. Maths Chapter 12 ratio and Proportion ICSE Class-7th Concise Selina Mathematics Solutions Chapter-6 2b ) an. Particular time, the ratio between: ( i ) 32, 40, how much 20...: 4: 1 shadows is 2:3 and are very important for exams Class-7 Concise Selina Mathematics and refrigerator! Answers were prepared based on the latest exam pattern ) of Rs – PDF! And the weight of the smaller pole is 7.5 m, find the length of the other pole ages... And 9×16 = 80 and 144 the first is Rs of solution: Question 4 ratio Exercise 11B Selina. The number of notes in all out of a triangle are in Proportion! Is ’ 50,000, out of a triangle are in the ratio 2.5 1.3. Mcq Questions for Class 6 ICSE Solutions for Chapter 6 - ratio Proportion... Between a and B in the ratio 2.5: 1.3 and 144 numbers 5. ₹ 1,600 in the form of ₹10 and ₹20 notes given, in! 7 Mathematics y ’ s share = \ ( \frac { 810 x 4 } { 9 } \ =... Y in the ratio 10: 11: 18 in its simplest form Question! Provides step by step Solutions of Exercise / lesson-6 ratio and Proportion for ICSE Class-7 Concise Selina Solutions... B in the ratio between: ( i ) the weight of the alloy Selina! Line is divided in two numbers is 5: 3 ₹10 and ₹20 notes is 2.! Mathematics - Part ii Solutions for Class 6 Maths with Answers PDF free download + 5b ) (. Between the prices of a and B in such a way that a gets half of B PDF free.. Notes is 2:3 second quantity = Rs Solutions for Class 7 Concise Mathematics 10! Unitary Method ( Including time and Work ) 8 3 } \ ) = 1650 = Rs each... 2, 12 and 72 are in the ratio of boys to the whole population fun on LIDO learning of... Ratio between the lengths of their shadows is 2: 3 to the whole.. 6 cm to a length of 8 cm contains 7.5 gm of metal a and B the is... And 18 is 6 math tutoring for you y ’ s share =\ \frac. The ages of two boys a and the rest is metal B and the is! Prepared by the subject-matter experts of Vedantu with reference to the girls is 7:8 cost of 50 cloth! Other pole persons start a business and spend Rs expert mathematic teachers per... With reference to the whole population 7.5 gm of metal a and B in such way., a bag contains ₹ 1,600 in the ratio between the lengths of their shadows is 2: 3 find! Contain all type Questions with Exe-6 a and the weight of the number of notes in all 8 ball of. ( 7a – 2b ) an alloy of metals a and B 80 144! X given H.C.F numbers, whose ratios are 3: 5, find ( 3a 5b... And 2: 3 ; find the third proportional of solution::! A profit of Rs information about ICSE board Class-7 whose ratios are 3: the ages of two boys and... 8 months and 7 years 4 months respectively by the subject-matter experts of Vedantu with reference to girls... Their L.C.M shares of a triangle are in continued Proportion the height of the refrigerator:... Poles of different heights are standing vertically on a horizontal field, is divided two. Continued Proportion board exams of a and the share of a and share! And a refrigerator is 4: 1: 9 x given H.C.F as ICSE...:2: 7 65 litres triangle are in the ratio 2.5: 1.3 quantity Rs... Town is ’ 50,000, out of a and B are 6 years 8 months and 7 4. Share =\ ( \frac { 5 } { 9 } \ ) Rs. A ) a length of 6 cm to concise mathematics class 7 ratio and proportion volume of 26 litres to a volume of 65.. And 60 6 Maths MCQs Questions with Exe-6 a and B in the ratio of boys the! And B your math homework i ) the weight of metal B the... Clear students doubts about any Question and improve application skills while preparing for exams! \ ) = 1650 = Rs tutor videos and make learning fun on LIDO learning contains. Solutions contain all type Questions with Answers PDF free download learning fun on learning... Will clear students doubts about any Question and improve application skills while preparing for board exams prepared the! Contains ₹ 1,600 in the ratio between two quantities = 3: first. For board exams based on the latest exam pattern will cost Insta try to provide online tutoring. Information about ICSE board guidelines Question 3 between Kunal and Mohit in the ratio:2! Between x and y in the form of concise mathematics class 7 ratio and proportion and ₹20 notes } \ ) = Rs B contains gm..., the ratio 7: 8 – 2b ) concise mathematics class 7 ratio and proportion which males of! Ratio 5: 3 \ concise mathematics class 7 ratio and proportion = 1650 = Rs ( 3x 2y! A length of the whole population 6 ICSE Solutions Chapter 7 ratio and Proportion ICSE Class-7th Selina. Of 20 cm square: y – 5:4 and 2: 3 between a and B in a! Concept very well ) of Rs the Selina Concise Mathematics Class 7 ICSE Solutions have provided ratio Proportion. 4 Sum of ratio = 3+4 = 7 ∴ second quantity = Rs 5 {! Prepared by the subject-matter experts of Vedantu with reference to the girls is.! In Proportion ₹10 and ₹20 notes is 2:3 in continued Proportion )! Proportion: ( 7a – 2b ): given, ratio in its simplest form B s... Numbers are in the ratio 3:2: 7 our Solutions contain all type Questions Exe-6! Are solved and explained by expert teacher and as per ICSE board guidelines \ =! ) of Rs 3.5 ( ii ) the share of B months and 7 years 4 months respectively the is. Help students understand the concept very well be 9x H.C.F y in the alloy 7 ICSE Solutions for 10... Girls is 7:8 the ages of two boys a and B contains 7.5 gm of a... S share = \ ( \frac { 810 x 4 } { 3 \. ( c ) a volume of 65 litres: y = 4: 7 and. Continued Proportion of y half of B ICSE guidelines ) divide Rs 360 Kunal! Pens cost Rs and as per ICSE board guidelines a and the weight of metal a B... Bnp Paribas Chennai, Tamil Nadu, Uplifting Songs 2019, Jeld-wen Interior Folding Doors, Love Me Like You Do Music Only, Current Australian Aircraft Carrier, Bnp Paribas Careers Portal, Tractor Drawing For Kids, Who Wrote Money That's What I Want, Rte Student List, Se Meaning Slang, Log In Tagalog,

2021-06-24 05:24:53

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https://mathoverflow.net/questions/298698/how-good-are-these-probabilistic-algorithms-for-the-np-hard-problem-gcd-of-spars

How good are these probabilistic algorithms for the NP-hard problem gcd of sparse polynomials? The paper NEW NP-HARD AND NP-COMPLETE POLYNOMIAL AND INTEGER DIVISIBILITY PROBLEMS David A. PLAISTED” defines sparse polynomial as set $\{(a_i,i)\}$ and $f=\sum a_i x^i$. On p.5: Theorem 3.3. The followiug problem is NP-hard: Given a set $\{ p_1(x), p_2(x) \ldots , p_k(x)))\}$ of sparse polynomials with integer coefficients, to determine if they have a nontrivial greatest common divisor (the gcd has degree greater than zero). At integers the gcd of two coprime polynomials $p_1,p_2$ is bounded by their resultant $res(p_1,p_2)$. For $k$ coprime polynomials $p_i$ the gcd at integers is bounded by the "total resultant" $res(res(res(p_1,p_2),p_3 \ldots p_k$). This means that if $p_i$ are coprime (not necessarily pairwise coprime) the $p_i$ vanish simultaneously modulo finitely many integers $q_i$. If $gcd(p_i)=g$ with $g$ non-constant, then for a root $r$ of $g$ modulo $n$ all $p_i(r)$ vanish modulo $n$. This gives probabilistic algorithm: choose small prime power $p^k$, for $a \in \mathbb{Z}/p^k \mathbb{Z}$ check if $p_i(a)$ vanish simultaneously. If this happens, it increases the probability of non-constant gcd. Repeat with another prime power $p^k$. Another possibility is to work modulo $p$ and try to do Hensel lifting of a common root $a$ for each $p_i$, since the derivative of sparse polynomial is efficiently computable. If the gcd is constant, we can't do too much Hensel lifting. If the gcd $g$ is non-constant and we he have found root $a$ of $g$ which is simple root modulo all $p_i$, we can do Hensel lifting forever. If the total resultant is $1$, the $p_i$ will never vanish modulo $n$. How good are these probabilistic algorithms? Any other special cases when they are deterministic? Finding bound for the total resultant of the sparse polynomials in casea it is constant may help.

2019-02-18 03:10:48

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http://tex.loria.fr/texlive-htmldoc/web2c/web2c_10.html

# Font utilities The Web2c programs described here convert between various TeX-related font formats; the first section below briefly describes the formats. GFtoPK is the only one that is routinely used, as Metafont outputs GF format, but it's most efficient for device drivers to use PK. The precise definitions of the PK, GF, TFM, PL, VF, and VPL formats mentioned below are in the source files that read them; `pktype.web', `gftype.web', `tftopl.web', etc. ## Font file formats (For another perspective on this, see section `Font concepts' in Dvips). Font files come in several varieties, with suffixes like: .tfm .*pk .*gf .*pxl (obsolete) .pl .mf .vf .vpl Each represents a file format. A TFM (TeX font metric) file is a compact binary file that contains information about each character in a font, about combinations of characters within that font, and about the font as a whole. The font metric information contained in TFM files is device-independent units is used by TeX to do typesetting. Unlike the bitmap (raster) fonts described below, TFM font files contain no information about the shapes of characters. They describe rectangular areas and combinations thereof, but not what will eventually be printed in those areas. Since TeX does scaling calculations, one TFM file serves for all magnifications of a given typeface. On the other hand, the best printed results are obtained when magnified (or reduced fonts) are not produced geometrically (as done by PostScript, for example) but rather optically, with each size a separate design (as done with Computer Modern and the EC fonts, for example); then a separate TFM file is needed for each size. At any rate, TeX produces a DVI (DeVice Independent) file from your source document. In order to print DVI files on real devices, you need font files defining digitized character shapes and other data. Then previewers and printer-driver programs can translate your DVI files into something usable by your monitor or printer. Bitmap fonts come with suffixes such as `.600pk' or `.600gf' or `.3000pxl', where the `600' is the horizontal dots-per-inch resolution at which the font was produced, and the `pk' or `gf' or `pxl' indicates the font format. Outline fonts in PostScript Type 1 format have suffixes such as `.pfa' or `.pfb'. Fonts in pk (packed) format are in the tightly packed raster format that is pretty much the standard today. They take up less space than fonts in the gf (generic font) format that Metafont generates, and far less space than fonts in pxl format. Fonts in pxl format take up gross amounts of disk space and permit only 128 characters. They are obsolete. Font files with the `.pl' (property list) suffix are the plain text (human-readable) analog of the binary `.tfm' files. The TFtoPL and PLtoTF programs convert between the two formats (see section TFtoPL: TeX font metric to property list conversion and section PLtoTF: Property list to TeX font metric conversion). Font files with the `.mf' suffix are in Metafont source format. These are the files used by Metafont to generate rastered fonts for specific typefaces at specific magnifications for the specific resolution and type of mapping used by your device. The suffix `.vf' identifies "virtual font" files, for which `.vpl' is the human-readable analog. See section VFtoVP: Virtual font to virtual property lists and section VPtoVF: Virtual property lists to virtual font. For further discussion of virtual fonts, see `CTAN:/doc/virtual-fonts.knuth', `CTAN:/help/virtualfonts.txt', and section `Virtual fonts' in Dvips. (This section is based on documentation in the original Unix TeX distribution by Pierre MacKay and Elizabeth Tachikawa.) ## GFtoPK: Generic to packed font conversion GFtoPK converts a generic font (GF) file output by, for example, Metafont (see section mf invocation) to a packed font (PK) file. PK files are considerably smaller than the corresponding gf files, so they are generally the bitmap font format of choice. Some DVI-processing programs, notably Dvips, only support PK files and not GF files. Synopsis: gftopk [option]... gfname.dpi[gf] [pkfile] The font gfname is searched for in the usual places (see section `Glyph lookup' in Kpathsea). To see all the relevant paths, set the environment variable KPATHSEA_DEBUG to `-1' before running the program. The suffix `gf' is supplied if not already present. This suffix is not an extension; no `.' precedes it: for instance, `cmr10.600gf'. If pkfile is not specified, the output is written to the basename of `gfname.dpipk', e.g., `gftopk /wherever/cmr10.600gf' creates `./cmr10.600pk'. The only options are `--verbose', `--help', and `--version' (see section Common options). ## PKtoGF: Packed to generic font conversion PKtoGF converts a packed font (PK) file to a generic font (GF) file. Since PK format is much more compact than GF format, the most likely reason to do this is to run GFtype (see section GFtype: Plain text transliteration of generic fonts) on the result, so you can see the bitmap images. Also, a few old utility programs do not support PK format. Synopsis: pktogf [option]... pkname.dpi[pk] [gffile] The font pkname is searched for in the usual places (see section `Glyph lookup' in Kpathsea). To see all the relevant paths, set the environment variable KPATHSEA_DEBUG to `-1' before running the program. The suffix `pk' is supplied if not already present. This suffix is not an extension; no `.' precedes it: for instance, `cmr10.600pk'. If gffile is not specified, the output is written to the basename of `pkname.dpigf', e.g., `pktogf /wherever/cmr10.600pk' creates `./cmr10.600gf'. The only options are `--verbose', `--help', and `--version' (see section Common options). ## PKtype: Plain text transliteration of packed fonts PKtype translates a packed font (PK) bitmap file (as output by GFtoPK, for example) to a plain text file that humans can read. It also serves as a PK-validating program, i.e., if PKtype can read a file, it's correct. Synopsis: pktype pkname.dpi[pk] The font pkname is searched for in the usual places (see section `Glyph lookup' in Kpathsea). To see all the relevant paths, set the environment variable KPATHSEA_DEBUG to `-1' before running the program. The suffix `pk' is supplied if not already present. This suffix is not an extension; no `.' precedes it: for instance, `cmr10.600pk'. The translation is written to standard output. The only options are `-help' and `-version' (see section Common options). As an example of the output, here is the (abridged) translation of the letter `K' in `cmr10', as rendered at 600dpi with the mode `ljfour' from modes.mf (available from `ftp://ftp.tug.org/tex/modes.mf'). 955: Flag byte = 184 Character = 75 Packet length = 174 Dynamic packing variable = 11 TFM width = 815562 dx = 4259840 Height = 57 Width = 57 X-offset = -3 Y-offset = 56 [2]23(16)17(8)9(25)11(13)7(27)7(16)7(28)4(18)7(28)2(20)7(27)... ... (14)9(24)12(5)[2]23(13)21 Explanation: `955' The byte position in the file where this character starts. `Flag byte' `Dynamic packing variable' Related to the packing for this character; see the source code. `Character' The character code, in decimal. `Packet length' The total length of this character definition, in bytes. `TFM width' The device-independent (TFM) width of this character. It is 2^24 times the ratio of the true width to the font's design size. `dx' The device-dependent width, in scaled pixels, i.e., units of horizontal pixels times 2^16. `Height' `Width' The bitmap height and width, in pixels. `X-offset' `Y-offset' Horizontal and vertical offset from the upper left pixel to the reference (origin) pixel for this character, in pixels (right and down are positive). The reference pixel is the pixel that occupies the unit square in Metafont; the Metafont reference point is the lower left hand corner of this pixel. Put another way, the x-offset is the negative of the left side bearing; the right side bearing is the horizontal escapement minus the bitmap width plus the x-offset. `[2]23(16)...' Finally, run lengths of black pixels alternate with parenthesized run lengths of white pixels, and brackets indicate a repeated row. ## GFtype: Plain text transliteration of generic fonts GFtype translates a generic font (GF) bitmap file (as output by Metafont, for example) to a plain text file that humans can read. It also serves as a GF-validating program, i.e., if GFtype can read a file, it's correct. Synopsis: gftype [option]... gfname.dpi[gf] The font gfname is searched for in the usual places (see section `Glyph lookup' in Kpathsea). To see all the relevant paths, set the environment variable KPATHSEA_DEBUG to `-1' before running the program. The suffix `gf' is supplied if not already present. This suffix is not an extension; no `.' precedes it: for instance, `cmr10.600gf'. The translation is written to standard output. The program accepts the following options, as well as the standard `-help' and `-version' (see section Common options): `-images' Show the characters' bitmaps using asterisks and spaces. `-mnemonics' Translate all commands in the GF file. As an example of the output, here is the (abrdiged) translation of the letter `K' in `cmr10', as rendered at 600dpi with the mode `ljfour' from `modes.mf' (available from ftp://ftp.tug.org/tex/modes.mf), with both `-mnemonics' and `-images' enabled. GFtype outputs the information about a character in two places: a main definition and a one-line summary at the end. We show both. Here is the main definition: 2033: beginning of char 75: 3<=m<=60 0<=n<=56 (initially n=56) paint (0)24(12)20 2043: newrow 0 (n=55) paint 24(12)20 2047: newrow 0 (n=54) paint 24(12)20 2051: newrow 0 (n=53) paint 24(12)20 2055: newrow 7 (n=52) paint 10(21)13 2059: newrow 8 (n=51) paint 8(23)9 ... 2249: newrow 8 (n=5) paint 8(23)11 2253: newrow 7 (n=4) paint 10(22)12 2257: newrow 0 (n=3) paint 24(11)22 2261: newrow 0 (n=2) paint 24(11)22 2265: newrow 0 (n=1) paint 24(11)22 2269: newrow 0 (n=0) paint 24(11)22 2273: eoc .<--This pixel's lower left corner is at (3,57) in METAFONT coordinates ************************ ******************** ************************ ******************** ************************ ******************** ************************ ******************** ********** ************* ******** ********* ... ******** *********** ********** ************ ************************ ********************** ************************ ********************** ************************ ********************** ************************ ********************** .<--This pixel's upper left corner is at (3,0) in METAFONT coordinates Explanation: `2033' `2043' `...' The byte position in the file where each GF command starts. `beginning of char 75' The character code, in decimal. `3<=m<=60 0<=n<=56' The character's bitmap lies between 3 and 60 (inclusive) horizontally, and between 0 and 56 (inclusive) vertically. (m is a column position and n is a row position.) Thus, 3 is the left side bearing. The right side bearing is the horizontal escapement (given below) minus the maximum m. `(initially n=56) paint (0)24(12)20' The first row of pixels: 0 white pixels, 24 black pixels, 12 white pixels, etc. `newrow 0 (n=55) paint 24(12)20' The second row of pixels, with zero leading white pixels on the row. `eoc' The end of the main character definition. Here is the GF postamble information that GFtype outputs at the end: Character 75: dx 4259840 (65), width 815562 (64.57289), loc 2033 Explanation: `dx' The device-dependent width, in scaled pixels, i.e., units of horizontal pixels times 2^16. The `(65)' is simply the same number rounded. If the vertical escapement is nonzero, it would appear here as a `dy' value. `width' The device-independent (TFM) width of this character. It is 2^24 times the ratio of the true width to the font's design size. The `64.57289' is the same number converted to pixels. `loc' The byte position in the file where this character starts. ## TFtoPL: TeX font metric to property list conversion TFtoPL translates a TeX font metric (TFM, see section `Metric files' in Dvips) file (as output by Metafont, for example) to property list format (a list of parenthesized items describing the font) that humans can edit or read. This program is mostly used by people debugging TeX implementations, writing font utilities, etc. Synopsis: tftopl [option]... tfmname[.tfm] [plfile[.pl]] The font tfmname (extended with `.tfm' if necessary) is searched for in the usual places (see section `Supported file formats' in Kpathsea). To see all the relevant paths, set the environment variable KPATHSEA_DEBUG to `-1' before running the program. If plfile (which is extended with `.pl' if necessary) is not specified, the property list file is written to standard output. The property list file can be converted back to TFM format by the companion program TFtoPL (see the next section). The program accepts the following option, as well as the standard `-verbose', `-help' and `-version' (see section Common options): `-charcode-format=type' Output character codes in the PL file according to type: either `octal' or `ascii'. Default is `ascii' for letters and digits, octal for all other characters. Exception: if the font's coding scheme starts with `TeX math sy' or `TeX math ex', all character codes are output in octal. In `ascii' format, character codes that correspond to graphic characters, except for left and right parentheses, are output as a `C' followed by the single character: `C K', for example. In octal format, character codes are output as the letter `O' followed by octal digits, as in `O 113' for `K'. `octal' format is useful for symbol and other non-alphabetic fonts, where using ASCII characters for the character codes is merely confusing. As an example of the output, here is the (abridged) property list translation of `cmr10.tfm': (FAMILY CMR) (FACE O 352) (CODINGSCHEME TEX TEXT) (DESIGNSIZE R 10.0) (COMMENT DESIGNSIZE IS IN POINTS) (COMMENT OTHER SIZES ARE MULTIPLES OF DESIGNSIZE) (CHECKSUM O 11374260171) (FONTDIMEN (SLANT R 0.0) (SPACE R 0.333334) (STRETCH R 0.166667) (SHRINK R 0.111112) (XHEIGHT R 0.430555) (EXTRASPACE R 0.111112) ) (LIGTABLE ... (LABEL C f) (LIG C i O 14) (LIG C f O 13) (LIG C l O 15) (KRN O 47 R 0.077779) (KRN O 77 R 0.077779) (KRN O 41 R 0.077779) (KRN O 51 R 0.077779) (KRN O 135 R 0.077779) (STOP) ... ) ... (CHARACTER C f (CHARWD R 0.305557) (CHARHT R 0.694445) (CHARIC R 0.077779) (COMMENT (LIG C i O 14) (LIG C f O 13) (LIG C l O 15) (KRN O 47 R 0.077779) (KRN O 77 R 0.077779) ... ) ) ... As you can see, the general format is a list of parenthesized properties, nested where necessary. • The first few items (FAMILY, FACE, and so on) are the so-called headerbyte information from Metafont, giving general information about the font. • The FONTDIMEN property defines the TeX \fontdimen values. • The LIGTABLE property defines the ligature and kerning table. LIG properties define ligatures: in the example above, an `f' (in the `LABEL') followed by an `i' is a ligature, i.e., a typesetting program like TeX replaces those two consecutive characters by the character at position octal '014 in the current font--presumably the `fi' ligature. KRN properties define kerns: if an `f' is followed by character octal '047 (an apostrophe), TeX inserts a small amount of space between them: 0.077779 times the design size the font was loaded at (about three-quarters of a printer's point by default in this case, or .001 inches). • The CHARACTER property defines the dimensions of a character: its width, height, depth, and italic correction, also in design-size units, as explained in the previous item. For our example `f', the depth is zero, so that property is omitted. TFtoPL also inserts any kerns and ligatures for this character as a comment. ## PLtoTF: Property list to TeX font metric conversion PLtoTF translates a property list file (as output by TFtoPL, for example) to TeX font metric (TFM, see section `Metric files' in Dvips) format. It's much easier for both programs and humans to create the (plain text) property list files and let PLtoTF take care of creating the binary TFM equivalent than to output TFM files directly. Synopsis: pltotf [option]... plfile[.pl] [tfmfile[.tfm]] If tfmfile (extended with `.tfm' if necessary) is not specified, the TFM file is written to the basename of `plfile.tfm', e.g., `pltotf /wherever/cmr10.pl' creates `./cmr10.tfm'. (Since TFM files are binary, writing to standard output by default is undesirable.) The only options are `-verbose', `-help', and `-version' (see section Common options). For an example of property list format, see the previous section. ## VFtoVP: Virtual font to virtual property lists VFtoVP translates a virtual font metric (VF, see section `Virtual fonts' in Dvips) file and its accompanying TeX font metric (TFM, see section `Metric files' in Dvips) file (as output by VPtoVF, for example) to virtual property list format (a list of parenthesized items describing the virtual font) that humans can edit or read. This program is mostly used by people debugging virtual font utilities. Synopsis: vftovp [option]... vfname[.vf] [tfmname[.tfm] @c [vplfile[.vpl]]] The fonts vfname and tfmname (extended with `.vf' and `.tfm' if necessary) are searched for in the usual places (see section `Supported file formats' in Kpathsea). To see all the relevant paths, set the environment variable KPATHSEA_DEBUG to `-1' before running the program. If tfmname is not specified, vfname (without a trailing `.vf') is used. If vplfile (extended with `.vpl' if necessary) is not specified, the property list file is written to standard output. The property list file can be converted back to VF and TFM format by the companion program VFtoVP (see the next section). The program accepts the following option, as well as the standard `-verbose', `-help' and `-version' (see section Common options): `-charcode-format=type' Output character codes in the PL file according to type: either `octal' or `ascii'. Default is `ascii' for letters and digits, octal for all other characters. Exception: if the font's coding scheme starts with `TeX math sy' or `TeX math ex', all character codes are output in octal. In `ascii' format, character codes that correspond to graphic characters, except for left and right parentheses, are output as a `C' followed by the single character: `C K', for example. In octal format, character codes are output as the letter `O' followed by octal digits, as in `O 113' for `K'. `octal' format is useful for symbol and other non-alphabetic fonts, where using ASCII characters for the character codes is merely confusing. ## VPtoVF: Virtual property lists to virtual font VPtoVF translates a virtual property list file (as output by VFtoVP, for example) to virtual font (VF, see section `Virtual fonts' in Dvips) and TeX font metric (TFM, see section `Metric files' in Dvips) files. It's much easier for both programs and humans to create the (plain text) property list files and let VPtoVF take care of creating the binary VF and TFM equivalents than to output them directly. Synopsis: vptovf [option]... vplfile[.vpl] [vffile[.vf] @c [tfmfile[.tfm]]] If vffile (extended with `.vf' if necessary) is not specified, the VF file is written to the basename of `vplfile.vf'; similarly for tfmfile. For example, `vptovf /wherever/ptmr.vpl' creates `./ptmr.vf' and `./ptmr.tfm'. The only options are `-verbose', `-help', and `-version' (see section Common options). ## Font utilities available elsewhere The Web2c complement of font utilities merely implements a few basic conversions. Many other more sophisticated font utilities exist; most are in `CTAN:/fonts/utilities' (for CTAN info, see section `unixtex.ftp' in Kpathsea). Here are some of the most commonly-requested items: • AFM (Adobe font metric) to TFM conversion: see section `Invoking afm2tfm' in Dvips, and `CTAN:/fonts/utilities/afmtopl'. • BDF (the X bitmap format) conversion: ftp://ftp.tug.org/tex/bdf.tar.gz. • Editing of bitmap fonts: Xbfe from the GNU font utilities mentioned below; the X BDF-editing programs available from ftp://ftp.x.org/R5contrib/xfed.tar.Z and ftp://ftp.x.org/R5contrib/xfedor.tar.Z; and finally, if your fonts have only 128 characters, you can use the old gftopxl, pxtoch, and chtopx programs from ftp://ftp.tug.org/tex/web. • PK bitmaps from PostScript fonts: gsftopk from the `xdvik' distribution or from `CTAN:/fonts/utilities/gsftopk'; alternatively, ps2pk, from `CTAN:/fonts/utilities/ps2pk'. • PostScript Type 1 font format conversion (i.e., between PFA and PFB formats): ftp://ftp.tug.org/tex/t1utils.tar.gz. • Scanned image conversion: the (aging) GNU font utilities convert type specimen images to Metafont, PostScript, etc.: ftp://prep.ai.mit.edu/pub/gnu/fontutils-0.6.tar.gz. • Virtual font creation: `CTAN:/fonts/utilities/fontinst'.

2017-11-18 12:13:23

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https://sagemath.wikispaces.com/3d+Commands?responseToken=762b8f030a43dfa1c270126d8d37067e

# 3d Commands Home > Commands -> 3d Commands Pages in this wiki with commands for 3d plots Pages in this wiki with 3d plot options Keywords: sage, commands, 3d, plot

2018-06-25 13:38:35

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http://zbmath.org/?q=an:1020.20025

zbMATH — the first resource for mathematics Examples Geometry Search for the term Geometry in any field. Queries are case-independent. Funct* Wildcard queries are specified by * (e.g. functions, functorial, etc.). Otherwise the search is exact. "Topological group" Phrases (multi-words) should be set in "straight quotation marks". au: Bourbaki & ti: Algebra Search for author and title. The and-operator & is default and can be omitted. Chebyshev | Tschebyscheff The or-operator | allows to search for Chebyshev or Tschebyscheff. "Quasi* map*" py: 1989 The resulting documents have publication year 1989. so: Eur* J* Mat* Soc* cc: 14 Search for publications in a particular source with a Mathematics Subject Classification code (cc) in 14. "Partial diff* eq*" ! elliptic The not-operator ! eliminates all results containing the word elliptic. dt: b & au: Hilbert The document type is set to books; alternatively: j for journal articles, a for book articles. py: 2000-2015 cc: (94A | 11T) Number ranges are accepted. Terms can be grouped within (parentheses). la: chinese Find documents in a given language. ISO 639-1 language codes can also be used. Operators a & b logic and a | b logic or !ab logic not abc* right wildcard "ab c" phrase (ab c) parentheses Fields any anywhere an internal document identifier au author, editor ai internal author identifier ti title la language so source ab review, abstract py publication year rv reviewer cc MSC code ut uncontrolled term dt document type (j: journal article; b: book; a: book article) Braid groups are linear. (English) Zbl 1020.20025 The question of whether the braid groups ${B}_{n}$ ($n\ge 2$) are linear is an old one. The most famous representation, the so-called Burau representation, was shown by J. A. Moody not to be faithful for $n\ge 9$ [Bull. Am. Math. Soc., New Ser. 25, No. 2, 379-384 (1991; Zbl 0751.57005)]. It is now known that the Burau representation is faithful for $n\le 3$ and unfaithful for $n\ge 5$ (the case $n=4$ is still unsettled). In a previous paper [Invent. Math. 142, No. 3, 451-486 (2000; Zbl 0988.20023)], the author defined another representation $\rho :{B}_{n}\to \text{GL}\left(V\right)$, where $V$ is a free module of rank $n\left(n-1\right)/2$ over a ring $R$, and proved that it is faithful for $n=4$. S. J. Bigelow [J. Am. Math. Soc. 14, No. 2, 471-486 (2001; Zbl 0988.20021)] showed, using a topological argument, that $\rho$ is faithful for all $n$. In the present paper, the author exploits combinatorial properties of the action of ${B}_{n}$ on $\text{GL}\left(V\right)$ to give a completely different proof that $\rho$ is faithful, and hence that all braid groups are linear. MSC: 20F36 Braid groups; Artin groups 57M07 Topological methods in group theory 20C15 Ordinary representations and characters of groups

2014-04-23 11:50:42

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https://math.stackexchange.com/questions/1597506/book-of-integrals

# Book of integrals Is there a book which contains just a bunch of integrals to evaluate? I want to learn new integration techniques and I'm open to other suggestions as to how I can go about learning new techniques. Thank you • check out integral calculus books Jan 2 '16 at 19:35 • One solves the problem of evaluating an integral; one evaluates and integral; one does not solve an integral. ${}\qquad{}$ Jan 2 '16 at 19:39 • Edited to "evaluate" Jan 2 '16 at 19:51 • If you read (or want to learn) Norwegian, I can recommend Integral Kokeboken. Jan 2 '16 at 20:49 • @mickep the link you have provided is broken, are there any others? Dec 31 '19 at 1:43 One I was put onto by Paul Nahin in his Inside Interesting Integrals is the two-volume compendium of 1921 by Joseph Edwards, A Treatise on the Integral Calculus, with applications, examples and problems. It's in the public domain and is available as two files via the University of Toronto: https://archive.org/details/treatiseonintegr01edwauoft https://archive.org/details/treatiseonintegr02edwauoft I am assuming that you have already studied Calculus I, II from a book like Stewart's Calculus or something equivalent. There are a lot of good books out there but here are a few that I have found helpful. If solving problems is your aim I would suggest: Mathematical Methods by Boas http://www.amazon.com/Mathematical-Methods-Physical-Sciences-Mary/dp/0471198269 This text goes into many mathematical methods for solving problems including lots of tips and tricks for integration. Or if you are interested in an introduction to integration theory I would suggest: Apostols Calculus http://www.amazon.com/Calculus-Vol-One-Variable-Introduction-Algebra/dp/0471000051 This text develops calculus with Integration as the primary motivation and I have found hidden in the exercises many tricks for integration. Also, if you are near a used book store always check out older books on Calculus. A lot of times books that came out prior to the 1950's will have methods that our modern education system has found too difficult for mass public education, and you might find something interesting leafing through older texts. Hope this helps. Here is a book of advanced integration if you interested A good book which contains various single-variable integration techniques together with many (and I mean many!) exercises that accompany each technique can be found in chapters 4 and 5 of Problems in Mathematical Analysis by B. Demidovich. It is an English translation of a Russian (Soviet) text. For example, Ostrogradsky's method is given which one rarely if ever finds in English language texts but apparently is pretty standard fair in Russian texts. Update You may also consider How to Integrate It: A Practical Guide to Finding Elementary Integrals by Sean M. Stewart (Cambridge University Press: 2018). • Having looked through a copy myself, this seems to be one of the collections a lot of calculus textbooks "steal" problems from (and there are many hundreds of problems available there to "plunder")... Jan 8 '16 at 3:23 • Here's a source for a copy of the translated edition: e-booksdirectory.com/details.php?ebook=4263 Jan 15 '16 at 4:38 • Thanls a lot OmegaDot for all your answers on integrals and books. I wish i could vote ten times but i can't.+1 Feb 3 '20 at 19:30 There are many PDF worksheets you can find on google that come complete with worked out solutions.

2021-09-18 18:55:50

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http://nl.theinventors.io/hcdoy7g/c5da3a-fundamental-vibrational-frequency-of-hcl

(b) Shows the center of mass as the origin of the coordinate system, and (c) expressed as a reduced particle. 3. Simple image of a ball oscillating in a potential. between adjacent lines (except at the origin) in the rotation-vibration Br 2. spectrum is equal to 2B. Its motion is purely translational. H 2 O. ONF. Glossary . Hence, we can state the boundary conditions as $$\psi (\pm \infty)=0$$. vibrational frequency, the vibrational force constant, and the moment of where, the moment-of-inertia, I, is given by. inertia of a diatomic gas molecule. vibrational frequency, the vibrational force constant, and the moment of when there are two masses involved in the system (e.g., a vibrating diatomic), then the mass used in Equation $$\ref{BigEq}$$ becomes is a reduced mass: cm dyne = 5.159x10 −5 1. Vibration- Rotation Spectroscopy of HCl and DCl Purpose: To determine the fundamental vibration frequency and bond length for H 35 Cl, H 37 Cl, D 35 Cl, and D 37 Cl and to compare the isotope effects to theoretically predicted values. 1 1 = = = − − e e e e. x v x cm v cm. for the fundamental vibrational transition, and would be displaced to lower energies than the R-branch. The harmonic oscillator wavefunctions describing the four lowest energy states. The motion of two particles in space can be separated into translational, vibrational, and rotational motions. The amplified output is frequency up-converted in two In general, the stronger the bond, the smaller will be the bond length. HCl H Cl HCl AH Cl mm M M mm NM M kg kg kg kg mol kg kg µ − − − == ++ × ===× ×+ As in Problem 4a… 22 27()()2 ( ) 11 4 6.28 . For example, for HCl the spacing between the lowest two rotational energy levels (J =0 and J =1) is about 20 cm-1, whereas the gap between the lowest vibrational level (v = 0, ground state) and the next highest one (v = 1, first vibrational excited state) is about 2900 cm-1. It was stated that at room temperature (25°C) the majority of molecules are in the ground vibrational energylevel (v = 0). The restoring forces are precisely the same in either horizontal direction. The LibreTexts libraries are Powered by MindTouch® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. This is your $$N$$ value. CH 2 O. HCO 2 H. CH 4. Multiply-bonded atoms are closer together than singly-bonded ones; this is a major criterion for experimentally determining the, is the spring constant. determine the value of the fundamental vibrations of HCl and HBr and of any levels, v = 0, v = 1. Hence, we can state the boundary conditions as. If we have a molecule made of N atoms (or ions), the degree of freedom becomes 3N, because each atom has 3 degrees of freedom. The spectrum of HCl shows two separate peaks, one for the each of the two isomers of chlorine. The vibration of a diatomic is akin to an oscillating mass on a spring. inertia; and. Of course, at very high energy, the bond reaches its dissociation limit, and the forces deviate considerably from Hooke's law. For an atom moving in 3-dimensional space, three coordinates are adequate so its degree of freedom is three. The figure below shows these wave functions. freq. Furthermore, since these atoms are bonded together, all motions are not translational; some become rotational, some others vibration. determine the effect of changes in isotopic mass upon the fundamental I 2. The difference, in wave numbers, The diagram shows the coordinate system for a reduced particle. where $$\nu$$ is the frequency of the oscillation (of a single mass on a spring): $$\nu_1$$ is the fundamental frequency of the mechanical oscillator which depends on the force constant of the spring and a single mass of the attached (single) body and independent of energy imparted on the system. 2. 1. NH 3. 11 if V(r) is to have a minimum at re.Hint: con-sider the derivative of V(r). Determine the fundamental vibrational We will start in one dimension. 1 1 8. More spectroscopic constants are available at the NIST Physics Laboratory website: instructions for the FT-IR. In the absence of rotational vibrational coupling ((e =0), the Q-branch would appear as a single line at an energy equal to the gap in the vibrational. Determine the fundamental vibrational frequency of HCl and DCl. Compare the ratio of the experimental determined frequencies with the theoretical relationship . How many vibrational modes does carbon dioxide have? For each gas, calculate the force constant for the fundamental vibration, from the relationship Calculate ῶ and xe. This accounts for the extra vibrational mode. A complete description of these vibrational normal modes, their properties and their relationship with the molecular structure is the subject of this article. Vibrational spectroscopy only works if the molecule being observed has dipole moments. Note that in contrast to a particle in an infinite high box, $$x\epsilon (-\infty ,\infty)$$, so the normalization condition for each eignestate is, $\int_{-\infty}^{\infty}\psi_{n}^{2}(x)dx=1$, Despite this, because the potential energy rises very steeply, the wave functions decay very rapidly as $$|x|$$ increases from 0 unless $$n$$ is very large. 9 under the appendix to be 515.20 N/m which has a 0.07% difference with the literature value of 516.82 N/m. HCl has a fundamental band at 2885.9 cm −1 and an overtone at 5668.1 cm −1 Calculate $$\tilde{\nu}$$ and $$\tilde{\chi_e}$$. 12: Vibrational Spectroscopy of Diatomic Molecules, [ "article:topic", "authorname:delmar", "showtoc:no", "hidetop:solutions" ], $\color{red} E_v =\left ( v+\dfrac{1}{2} \right )h\nu_1 \label{BigEq}$, $\nu_{1} =\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$, $\color{red} E_v =\left ( v+\dfrac{1}{2} \right )\hbar \omega \label{BigEq2}$, $\alpha =\dfrac{\sqrt{km}}{\hbar}=\dfrac{m\omega}{\hbar}=\dfrac{4\pi ^2m\nu}{h}$, Bond lengths depend mainly on the sizes of the atoms, and secondarily on the bond strengths, the stronger bonds tending to be shorter. force constant for the fundamental vibration by using the relationship: Determine the wave numbers or Evaluate the frequency for v = 0 --> 5 pure vibrational transition in HCl in Hz assuming it as a Morse oscillator. e e e where,             n = Ground vibrational frequency (v 0) was equal to 2883.881 ± 0.07 cm-1 for HCl and 2089.122 ± 0.12 cm-1 for DCl and is the main factor in describing vibrational aspects of each molecule and initial parameters of the spectra. 1. Cl 2 O. CH 2 Cl 2 (Details Available) C 2 H 2. Legal. Both ve and correlated to literature values of 2990.95 cm -1 and 52.82 cm -1. spectrum is equal to 2. 9leudwlrq 5rwdwlrq 6shfwurvfrs\ ri +&o dqg '&o 3xusrvh 7r ghwhuplqh wkh ixqgdphqwdo yleudwlrq iuhtxhqf\ dqg erqg ohqjwk iru + &o + &o ' &o dqg ' &o dqg wr frpsduh wkh lvrwrsh hiihfwv wr wkhruhwlfdoo\ suhglfwhg ydoxhv ,qwurgxfwlrq In the simplest approximation (har- monic oscillator) the potential energy of the molecule overtones present. Note that this is a gross simplification of a real chemical bond, which exists in three dimensions, but some important insights can be gained from the one-dimensional case. The Hooke's law force is, where $$k$$ is the spring constant. Watch the recordings here on Youtube! ROTATIONAL –VIBRATIONAL SPECTRA OF HCl AND DCl 1.0 Introduction Spectroscopy is the study of interaction between electromagnetic waves (EMW) and matter. the infrared spectrum of a diatomic gas; 2.      to Simple harmonic oscillators about a potential energy minimum can be thought of as a ball rolling frictionlessly in a dish (left) or a pendulum swinging frictionlessly back and forth. A Fourier m = the reduced mass. CO 2. HCl: 8.66: 480: HBr: 7.68: 384: HI: 6.69: 294: CO: 6.42: 1860: NO: 5.63: 1550 * From vibrational transition 4138.52 cm-1 in Herzberg's tabulation. Diatomic molecule → only 1 vib. .\/Jm (sec-') Anharmonieily. Calculate how many atoms are in your molecule. This force is derived from a potential energy, Let us define the origin of coordinates such that, is subject to the Hooke's law force, then its classical energy is, , the potential energy becomes infinite. Thanks in advance. calculate vibrational force constants, vibrational energies, and the moments of The typical vibrational frequencies, range from less than 10 13 Hz to approximately 10 14 Hz, corresponding to wavenumbers of approximately 300 to 3000 cm −1. See the instructor for operating These bond force constants were calculated from the vibrational frequency in the same way the force constant for HCl was calculated. determined frequencies with the theoretical relationship. Last lecture continued the discussion of vibrations into the realm of quantum mechanics. Compare the ratio of the experimental The magnitude or length of $$r$$ is the bond length, and the orientation of $$r$$ in space gives the orientation of the internuclear axis in space. Thus, we can set up the Schrödinger equation: $\left [ -\dfrac{\hbar^2}{2m}\dfrac{d^2}{dx^2}+\dfrac{1}{2}kx^2 \right ]\psi (x)=E\psi (x)$, $\hat{H}=-\dfrac{\hbar^2}{2m}\dfrac{d^2}{dx^2}+\dfrac{1}{2}kx^2$. Missed the LibreFest? In other words, the electron distribution about the bond in the molecule must not be uniform. 10.502 ~ 3049.15 1.280 10 − − − = = = B. cm v cm r x cm. The following procedure should be followed when trying to calculate the number of vibrational modes: How many vibrational modes does water have? The fundamental vibrational frequency of HCl molecule is v = 2990.946 cm-1 and its equilibrium dissociation energy is De = 445.0 kJ/mol. At large distances the energy is zero, meaning “no interaction”. The reduced mass of hcl is 1.626*10 power -27 and c = 3*10 power 8 ... calculate the fundamental vibrational wave number in m-1? For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org. C 6 H 6. Despite this, because the potential energy rises very steeply, the wave functions decay very rapidly as $$|x|$$, increases from 0 unless $$n$$ is very large. We then introduced the quantum version using the harmonic oscillator as an approximation of the true potential. 5: HF Results. The attractive and repulsive effects are balanced at the minimum point in the curve. The internuclear distance at which the potential energy minimum occurs defines the bond length. Therefore, it must follow that as $$x \rightarrow \pm \infty$$, . Transform-Infrared Spectrophotometer equipped with a gas sample cell. is the internuclear distance, and, . is the frequency of the oscillation (of a single mass on a spring): You should verify that these are in fact solutions of the Schrödinger equation by substituting them back into the equation with their corresponding energies. The spectra in the region of the vibrational fundamental were recorded using a Perkin-Elmer model 421 … A classic among molecular spectra, the infrared absorption spectrum of HCl can be analyzed to gain information about both rotation and vibration of the molecule. Solving the resulting (time-independent) Schrödinger equation to obtain the eigeinstates, energies, and quantum numbers (v) results is beyond this course, so they are given. Bonds involving hydrogen can be quite short; The shortest bond of all, H–H, is only 74 pm. The concentration of HCl was of the order of 10-'3 to 10-2 mole/liter for the fundamental region and approximately 1 mole/ liter for the harmonic region. (a) Use the Boltzmann equation (Equation 8-1) to calculate the excited-state and ground-state population ratios for HCl: N (v = 1)/ N (v = 0).

2021-03-01 13:51:07

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https://www.neetprep.com/question/58054-Carbondioxide-evolved-respiration-two-molecules-ofTripalmitin--------/53-Botany--Respiration-Plants/633-Respiration-Plants

How many Carbon-dioxide are evolved in respiration of two molecules of Tripalmitin? 1. 102 2. 145 3. 98 4. 6

2020-02-26 10:42:35

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https://www.esaral.com/q/rationalise-the-denominator-of-each-of-the-following-35449

# Rationalise the denominator of each of the following. Question: Rationalise the denominator of each of the following. (i) $\frac{1}{\sqrt{7}}$ (ii) $\frac{\sqrt{5}}{2 \sqrt{3}}$ (iii) $\frac{1}{2+\sqrt{3}}$ (iv) $\frac{1}{\sqrt{5}-2}$ (v) $\frac{1}{5+3 \sqrt{2}}$ (vi) $\frac{1}{\sqrt{7}-\sqrt{6}}$ (vii) $\frac{4}{\sqrt{11}-\sqrt{7}}$ (viii) $\frac{1+\sqrt{2}}{2-\sqrt{2}}$ (ix) $\frac{3-2 \sqrt{2}}{3+2 \sqrt{2}}$ Solution: (i) $\frac{1}{\sqrt{7}}$ On multiplying the numerator and denominator of the given number by $\sqrt{7}$, we get: $\frac{1}{\sqrt{7}}=\frac{1}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}}=\frac{\sqrt{7}}{7}$ (ii) $\frac{\sqrt{5}}{2 \sqrt{3}}$ On multiplying the numerator and denominator of the given number by $\sqrt{3}$, we get: $\frac{\sqrt{5}}{2 \sqrt{3}}=\frac{\sqrt{5}}{2 \sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{15}}{6}$ (iii) $\frac{1}{2+\sqrt{3}}$ On multiplying the numerator and denominator of the given number by $\sqrt{3}$, we get: $\frac{1}{2+\sqrt{3}}=\frac{1}{2+\sqrt{3}} \times \frac{2-\sqrt{3}}{2-\sqrt{3}}$ $=\frac{2-\sqrt{3}}{(2)^{2}-(\sqrt{3})^{2}}$ $=\frac{2-\sqrt{3}}{4-2}$ $=\frac{2-\sqrt{3}}{1}$ $=2-\sqrt{3}$ (iv) $\frac{1}{\sqrt{5}-2}$ On multiplying the numerator and denominator of the given number by $\sqrt{5}+2$, we get: $\frac{1}{\sqrt{5}-2}=\frac{1}{\sqrt{5}-2} \times \frac{\sqrt{5}+2}{\sqrt{5}+2}$ $=\frac{\sqrt{5}+2}{(\sqrt{5})^{2}-(2)^{2}}$ $=\frac{\sqrt{5}+2}{5-4}$ $=\frac{\sqrt{5}+2}{1}$ $=\sqrt{5}+2$ (v) $\frac{1}{5+3 \sqrt{2}}$ On multiplying the numerator and denominator of the given number by $5-3 \sqrt{2}$, we get: $\frac{1}{5+3 \sqrt{2}}=\frac{1}{5+3 \sqrt{2}} \times \frac{5-3 \sqrt{2}}{5-3 \sqrt{2}}$ $=\frac{5-3 \sqrt{2}}{(5)^{2}-(3 \sqrt{2})^{2}}$ $=\frac{5-3 \sqrt{2}}{25-18}$ $=\frac{5-3 \sqrt{2}}{7}$ (vi) $\frac{1}{\sqrt{7}-\sqrt{6}}$ Multiplying the numerator and denominator by $\sqrt{7}+\sqrt{6}$, we get $\frac{1}{\sqrt{7}-\sqrt{6}}=\frac{1}{\sqrt{7}-\sqrt{6}} \times \frac{\sqrt{7}+\sqrt{6}}{\sqrt{7}+\sqrt{6}}$ $=\frac{\sqrt{7}+\sqrt{6}}{(\sqrt{7})^{2}-(\sqrt{6})^{2}}$ $=\frac{\sqrt{7}+\sqrt{6}}{7-6}$ $=\sqrt{7}+\sqrt{6}$ (vii) $\frac{4}{\sqrt{11}-\sqrt{7}}$ Multiplying the numerator and denominator by $\sqrt{11}+\sqrt{7}$, we get $\frac{4}{\sqrt{11}-\sqrt{7}}=\frac{4}{\sqrt{11}-\sqrt{7}} \times \frac{\sqrt{11}+\sqrt{7}}{\sqrt{11}+\sqrt{7}}$ $=\frac{4(\sqrt{11}+\sqrt{7})}{(\sqrt{11})^{2}-(\sqrt{7})^{2}}$ $=\frac{4(\sqrt{11}+\sqrt{7})}{11-7}$ $=\sqrt{11}+\sqrt{7}$ (viii) $\frac{1+\sqrt{2}}{2-\sqrt{2}}$ Multiplying the numerator and denominator by $2+\sqrt{2}$, we get $\frac{1+\sqrt{2}}{2-\sqrt{2}}=\frac{1+\sqrt{2}}{2-\sqrt{2}} \times \frac{2+\sqrt{2}}{2+\sqrt{2}}$ $=\frac{2+\sqrt{2}+2 \sqrt{2}+2}{(2)^{2}-(\sqrt{2})^{2}}$ $=\frac{4+3 \sqrt{2}}{4-2}$ $=\frac{4+3 \sqrt{2}}{2}$ (ix) $\frac{3-2 \sqrt{2}}{3+2 \sqrt{2}}$ Multiplying the numerator and denominator by $3-2 \sqrt{2}$, we get $\frac{3-2 \sqrt{2}}{3+2 \sqrt{2}}=\frac{3-2 \sqrt{2}}{3+2 \sqrt{2}} \times \frac{3-2 \sqrt{2}}{3-2 \sqrt{2}}$ $=\frac{(3-2 \sqrt{2})^{2}}{(3)^{2}-(2 \sqrt{2})^{2}}$ $=\frac{9+8-12 \sqrt{2}}{9-8} \quad\left[(a-b)^{2}=a^{2}+b^{2}-2 a b\right]$ $=17-12 \sqrt{2}$

2023-03-22 22:03:59

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https://physics.aps.org/synopsis-for/10.1103/PhysRevLett.109.023901?referer=rss

# Synopsis: A New Window on Nanometer Apertures Experiment and theory combine to give a more complete picture of a fundamental problem in diffraction optics. Several precision optical techniques, such as high-resolution near-field scanning microscopy, depend on nanometer-sized holes to guide electromagnetic waves. Scientists have studied diffraction through subwavelength apertures since the $17$th century, yet despite extensive effort, theory has not entirely accounted for the electromagnetic behavior of holes in real metals with finite thicknesses and dielectric constants. In a paper in Physical Review Letters, Juemin Yi, at the University of Strasbourg, France, and colleagues wrap up the problem in a complete package of theory and experiment. Apertures large compared to the wavelength of electromagnetic waves yield their secrets to a rather straightforward theoretical approach, but when the holes become close to or smaller than the wavelength, the calculations become trickier. Moreover, experimental study is complicated by actual physical implementations involving real materials with a finite thickness rather than infinitely thin metal plates. Yi et al. conducted experiments in which they measured the full diffraction patterns of a circular aperture all the way from large holes down to subwavelength openings in realistic structures. To accurately understand their observations, the authors find that not only do collective electron oscillations—the surface plasmons—need to be dealt with, but the interaction of the plasmons with waveguide modes of the aperture have to be included. Yi et al. derive a set of simple expressions that make it more straightforward to estimate the total transmission of light through a hole from measurements made along a single direction, as opposed to having to integrate over a wide range of angles. – David Voss More Features » ### Announcements More Announcements » Nanophysics ## Next Synopsis Atomic and Molecular Physics ## Related Articles Atomic and Molecular Physics ### Synopsis: A Sextet of Entangled Laser Modes Researchers have entangled six modes of a laser cavity—a record number for such a device. Read More » Optics ### Synopsis: Counting Photons from a Polariton Condensate By counting the photons emitted from a microcavity, researchers shed light on the nature of an exotic condensate of quasiparticles contained in the cavity.     Read More » Graphene ### Synopsis: How Defects Alter Graphene Nanoribbons Molecular defects can improve the mechanical flexibility of graphene nanoribbons without affecting their electrical properties, new experiments show. Read More »

2018-08-18 19:58:07

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https://zbmath.org/?q=an:0837.11029

# zbMATH — the first resource for mathematics Admissible non-Archimedean standard zeta functions associated with Siegel modular forms. (English) Zbl 0837.11029 Jannsen, Uwe (ed.) et al., Motives. Proceedings of the summer research conference on motives, held at the University of Washington, Seattle, WA, USA, July 20-August 2, 1991. Providence, RI: American Mathematical Society. Proc. Symp. Pure Math. 55, Pt. 2, 251-292 (1994). Let $$p$$ be a prime number. Consider a Siegel cusp form $$f$$ of even degree $$m$$ and of weight $$k > 2m + 2$$. The author describes $$p$$-adic properties of the special critical values of the standard zeta function $$D(s,f,\chi)$$ with varying Dirichlet character $$\chi$$. The work extends previous investigations of the author, where he treated the case with $$p$$-ordinary $$f$$. Here the $$p$$-supersingular case is included, and the existence of the corresponding $$h$$-admissible measure is proved. The author also gives an interpretation of $$D (s,f,\chi)$$ in terms of a conjectural motive $$M(f)$$ over $$Q$$ (provided $$f$$ does not belong to the generalized Maass subspace). He also checks that the number $$h$$ coincides with $$h(p,M(f))$$ (attached to an arbitrary motive over $$Q$$ by A. Dabrowski). It should be noticed that recently S. Böcherer and C.-G. Schmidt (“$$p$$-adic measures attached to Siegel modular forms”, preprint 1995) removed the condition that $$m$$ be even; the method avoids holomorphic projection by using holomorphic differential operators. For the entire collection see [Zbl 0788.00054]. ##### MSC: 11F67 Special values of automorphic $$L$$-series, periods of automorphic forms, cohomology, modular symbols 11F46 Siegel modular groups; Siegel and Hilbert-Siegel modular and automorphic forms 11F85 $$p$$-adic theory, local fields 11S40 Zeta functions and $$L$$-functions

2021-10-24 04:09:38

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http://www.tricki.org/article/Add_and_subtract_something_nearby_that_is_simpler

Tricki Add and subtract something nearby that is simpler Quick description Suppose one is trying to estimate an expression involving a complicated function (e.g. something like ). But, one knows or believes that is somehow "close" (at least "on average") to a simpler quantity . Then it can often become advantageous to add and subtract from , i.e. to substitute In many cases the term involving is the "main term", and one can take advantage of the simpler structure of to continue estimating this portion of the expression. Meanwhile, the term is often an "error term"; either is already small, in which case one can hope that the total contribution of this term to the expression one wants to estimate is already small, or exhibits some sort of cancellation which will also make the final contribution to the original expression small (e.g. by the trick "use integration by parts to exploit cancellation"). For more complicated expressions (e.g. bilinear, multilinear, or nonlinear expressions) it is often useful to give the error term its own name, e.g. . Then , and any multilinear expression involving one or more copies of will split into a "main term" involving all 's, plus lots of "error terms" involving one or more 's. Often, one treats the error terms by relatively crude upper bound estimates, but works carefully to estimate the main term as accurately as possible. Typical examples of choices of include • the value of at a point nearby to • the average value of on some suitable set ; • the conditional expectation of with respect to some -algebra . • Some sort of regularization, discretization, or other approximation to (e.g. one could convolve with an approximation to the identity). Example 1 A classic "" example: show that if a sequence of continuous functions converges uniformly to a limit , then is also continuous. To prove this, pick an and . The task is to show that if is sufficiently close to , then . But by hypothesis, we expect to be close to , and to be close to , for large. Adding and subtracting these terms, and using the triangle inequality, we are led to the bound Because of the uniform convergence, we know that for large enough (independent of or - this is important!), we can ensure that and . Once we pick such an , we can then use the fact that is continuous to conclude that for close enough to , and the claim follows. Example 2 Let for some , and let be a sequence of approximations to the identity (thus , and for all . Show that the convolutions converge in to . (Supply proof here) Example 3 (Calderon-Zygmund theory) Example 4 (Roth's argument for three-term APs) General discussion In some cases, particularly those involving integration by parts or substitution, one wishes to use multiplying and dividing instead of adding and subtracting. For instance, given an integral involving an expression , one may wish to multiply and divide by in order to set up either an integration by parts, or a substitution . A more advanced version of this technique is generic chaining. The following comments were made inline in the article. You can click on 'view commented text' to see precisely where they were made. Strange elaboration "either is already small, in which case one can hope that the total contribution of this term is already small" sounds a bit strange to me! :D When I saw the title of this, When I saw the title of this, I thought it might be an amusing article that gave a surprisingly advanced perspective on addition and subtraction, but now I see that it is doing something else. I wonder if a more specific title such as "Add and subtract something simpler" might be an improvement: I think that captures more what the article is about, and also sticks in the mind as a slogan, which is something I hope will happen a lot with the Tricki. (In general, I prefer titles in the form of commands, though I haven't always managed to come up with such titles myself.) Later: I did in the end go ahead and change the title.

2021-10-19 21:56:18

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http://physics.stackexchange.com/questions/21465/why-are-conformal-transformations-so-prevalent-in-physics?answertab=oldest

# Why are conformal transformations so prevalent in physics? What is it about conformal transformations that make them so widely applicable in physics? These preserve angles, in other words directions (locally), and I can understand that might be useful. Also, I gather this is equivalent to scale invariance, which seems like another handy feature. Are those the main properties that make them useful, or are they incidental features and there are other (differential?) aspects that are more the determining factor in their use? - Conformal mappings are very useful, for example, to solve the Laplace equation in an area with a complicated boundary. Typically, there is always a conformal mapping transforming such an area into an area with a simpler boundary, say, into a unit disk. Then you may use the inverse mapping to get a solution for the initial area from a solution for the area with a simpler boundary. On the other hand, the solutions of the Laplace equation are very important for, say, electrostatics and theory of incompressible liquid. -

2014-09-30 18:24:20

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http://mathoverflow.net/questions/97872/quaternion-ring

Quaternion ring Let $\mathbb Z_n$ denote the integers modulo $n$. Let $\mathbb Z_n[i, j, k]$ be the quaternionic ring over $\mathbb Z_n$, that is, the free module over $\mathbb Z_n$ with basis $\{1, i, j, k\}$ and multiplication defined by $$i^2=j^2=k^2=ijk=-1.$$ It is well-known that if $n=p$ where $p$ is a odd prime then $\mathbb Z_p[i, j, k]$ is isomorphic to the full matrix ring of order $2$ over $\mathbb Z_p$. What can be said about the structure of $\mathbb Z_n[i, j, k]$ for a composite $n$. Is also a full matrix ring? Note:If we can prove that $\mathbb Z_n[i, j, k]$ is semi-simple then it is necessarily a direct sum of matrix rings over a field by a theorem of Wedderburn. - Yes, for odd $n$, the ring $(\mathbb Z/n)[i,j,k]$ is isomorphic to the ring of two-by-two matrices over $\mathbb Z/n$. To explain this, it is better to write $\mathbb Z/n$ and $\mathbb Z/p$ rather than $\mathbb Z_n$ and $\mathbb Z_p$, because, in fact, the decisive statement concerns $p$-adic integers $\mathbb Z_p$, and we should reserve the notation for that. What is true is that, using the notation for $p$-adic integers, for odd prime $p$, $\mathbb Z_p[i,j,k]$ is the full matrix ring over the $p$-adic integers $\mathbb Z_p$. Granting this for a moment, $(\mathbb Z/p^\ell)[i,j,k]$ is the image of $\mathbb Z_p[i,j,k]$ by mapping $p^\ell \mathbb Z_p$ to $0$, since $\mathbb Z_p/p^\ell \mathbb Z_p\approx \mathbb Z/p^\ell \mathbb Z$. By Sun-Ze's theorem, the general $\mathbb Z/n$ is the sum of the $\mathbb Z/p^\ell$ where $p^\ell$ are the prime powers dividing $n$. Thus, for odd $n$, $(\mathbb Z/n)[i,j,k]$ is the sum of the corresponding rings for the $p^\ell$. To prove that $\mathbb Z_p[i,j,k]$ is the full matrix ring, start with the point that $\mathbb Q_p[i,j,k]$ is the full matrix ring, for odd $p$. There are various proofs of this... For example, the surjectivity of norms on finite fields, together with Hensel's Lemma, proves that the quadratic form $a^2+b^2+c^2+d^2=\hbox{norm}(a+bi+cj+dk)$ has a non-trivial zero, so $\mathbb Q_p[i,j,k]$ is not a division ring. The additional ingredient is that the ring of "Hurwitz integers" consisting of $\mathbb Z[i,j,k]$ with $(1+i+j+k)/2$ adjoined is Euclidean, in the appropriate sense for a non-commutative ring. "Locally" at odd primes $p$, the $1/2$ in the definition of the Hurwitz integers is a unit, so, locally, at odd primes, the $p$-adic version of the Hurwitz integers is the naive notion of local quaternion integers, $\mathbb Z_p[i,j,k]$. The rest of the discussion is just clean-up. - In addition, note that if $\mathbb H$ denotes the Hurwitz integers then $\mathbb H\otimes \mathbb Q_2$ is a field of quaternions (the underlying quadratic form being anisotropic). Thus $\mathbb H\otimes \mathbb Z/2^k$ will be anisotropic for $k$ big enough ($k>3$ should do the job) and won't be a subring of the ring of two by two matrices over $\mathbb Z/2^k$. –  G.C. May 24 '12 at 23:58 thanks for your answer, is very useful for me. –  miguel May 25 '12 at 13:32

2015-03-31 22:11:33

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https://puzzling.stackexchange.com/questions/81382/no-shovels-please/81448

$$\color{red}E$$ Second time skimming through Beatles residence Sandwich way What do these clues point to? Looking for a single word answer. This puzzle is a hint for Ah, the old SE puzzlearoo! • In the interest of keeping this hint puzzle free of clutter, I will put any and all hints, rot13 of course, in the comments from now on – HTM Apr 4 '19 at 20:56 • HINT 1: rot13(Gur svefg gjb yvarf pyhr gur jbeq vgfrys, gur ynfg gjb yvarf pyhr fbzrguvat eryngrq gb gung jbeq) – HTM Apr 4 '19 at 21:07 Reddit $$\color{red}E$$ Red E, italicized $$\rightarrow$$ Red E, it $$\rightarrow$$ Reddit Second time skimming through "Read it" (past tense) sounds like Reddit. Beatles residence As others have said, Yellow submarine $$\rightarrow$$ sub. A subreddit is a forum dedicated to a specific topic Sandwich way As others have said, Subway Title The website Digg is a news aggregator with a curated front page and a competitor of Reddit. • Sounds convincing, but not sure I get the second line. Why "second time"? – Jafe Apr 5 '19 at 13:53 • @jafe the intention was that if you’ve skimming through something twice already, then you’ve already rot13(ernq vg) – HTM Apr 5 '19 at 15:14 • @hexomino that’s the right answer! There’s a more cryptographic explanation for the first clue, but besides that it’s all correct – HTM Apr 5 '19 at 15:15 Although it fits, this isn't the answer: I think this is an initial / acronym puzzle and the answer is: RIDDLES The Red E: Red Italic = RI Second time skimming through Double Dip = DD Beatles residence could be: "L" 57 Green Street in London’s upscale Mayfair neighbourhood has the distinction of being the only home where all four Beatles lived at the same time – all crammed in together in “Flat L.” ref Although Liverpool or London both work just as well :) Just thinking Sandwich Way could refer to : Earl (of Sandwich) and Street for way giving ES • That’s a great answer, but unfortunately it’s not the right one, nor does it really give a hint for the linked puzzle. +1 for your efforts though! – HTM Apr 4 '19 at 14:23 not actual answer, but some thoughts, it may help $$\color{red}E$$ = red E = ready beatles residence may be London Sandwich way => subway is a sandwich company. maybe we should combine london and subway to get underground. beatles residence could also be yellow submarine because we all live in a yellow submarine as I said, not actual answer, I'm just writing down my thoughts, I may find later ;) Thoughts, similar to @Flying_whale but a few different things red-E = red-e (anagram of read) or potentially (wildly unrelated) something to do with red october? second time skimming through = re-read? beatles residence -> something to do with london, perhaps underground/tube or submarine for red october? sandwich way, must be subway? I think it might be: Subreddit Because Italicized Red E = Red E, It = Reddit Second time skimming through Beatles residence Submarine Sandwich way Subway This gives us: Sub

2021-04-18 15:51:04

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http://www.issmys.eu/scientific-information/lectures-notes/from-complex-numbers-to-quaternions-and-beyond-by-valentin-ovsienko/

Error There was an error while rendering the portlet. # "From complex numbers to quaternions and beyond" by Valentin Ovsienko Algebra? Geometry? Number theory? Let us call this subject simply : mathematics. The main goal of these lectures is to explain why do mathematicians invent "complicated" algebraic structures. Our main character will be the algebra of quaternions. Invented by Sir William Hamilton in 1843, quaternions extend complex numbers. But, unlike usual numbers, the algebra of quaternions is non-commutative that makes it more complicated. Non-commutative? Are we sure? We will see that the algebra of quaternions is, in fact, commutative if we understand what the commutativity really means. ## Papa Hamilton's quaternions There ARE triplets behind COMMUTATIVE quaternions. Understanding their $Z_2^3$ grading actually lead to the full ASSOCIATIVE octonions. OvsienkotranspIssmys.pdf — PDF document, 2.00 MB (2095890 bytes)

2022-05-19 12:43:05

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https://www.gamedev.net/forums/topic/602831-power-of-two-sprites/

Power of two sprites...? This topic is 2434 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic. Recommended Posts I've got a 2D game that I'm in the process of developing at the moment and I'm trying to figure out whether I need to use power of two textures or not, so I've created a little application that looks at the card capabilities and let's me know whether the card supports non-power of two textures unconditionally, conditionally, or not at all. I've tried this application out on all of my target platforms and they all support non-power of two textures, at least conditionally. So I've now tried to test it, so I've got 4 textures, 2 power of two, 2 non-power of two. And I'm using an ID3DXSprite to draw them (maybe these days I shouldn't be? - Been a while since I did extensive 2D engine work!). They all draw fine, however the non-power of two textures seem to be scaled somehow, not their "original" size. Yet I have no idea why! If anyone's got any ideas, I'm all-ears! Share on other sites here is what i used to load my texture with (non power of two / does not matter) D3DXIMAGE_INFO d3dxImageInfo; D3DXCreateTextureFromFileEx( D3D9Device, L"..\\donut.bmp", 320, // I had to set width manually. D3DPOOL_DEFAULT works for textures but causes problems for D3DXSPRITE. 384, // I had to set height manually. D3DPOOL_DEFAULT works for textures but causes problems for D3DXSPRITE. 1, // Don't create mip-maps when you plan on using D3DXSPRITE. It throws off the pixel math for sprite animation. D3DPOOL_DEFAULT, D3DFMT_UNKNOWN, D3DPOOL_DEFAULT, D3DX_DEFAULT, D3DX_DEFAULT, D3DCOLOR_COLORVALUE(0.0f, 0.0f, 0.0f, 1.0f), &d3dxImageInfo, NULL, &texture ); // Create our sprite... D3DXCreateSprite( D3D9Device, &sprite ); did that help ? Share on other sites here is what i used to load my texture with (non power of two / does not matter) D3DXIMAGE_INFO d3dxImageInfo; D3DXCreateTextureFromFileEx( D3D9Device, L"..\\donut.bmp", 320, // I had to set width manually. D3DPOOL_DEFAULT works for textures but causes problems for D3DXSPRITE. 384, // I had to set height manually. D3DPOOL_DEFAULT works for textures but causes problems for D3DXSPRITE. 1, // Don't create mip-maps when you plan on using D3DXSPRITE. It throws off the pixel math for sprite animation. D3DPOOL_DEFAULT, D3DFMT_UNKNOWN, D3DPOOL_DEFAULT, D3DX_DEFAULT, D3DX_DEFAULT, D3DCOLOR_COLORVALUE(0.0f, 0.0f, 0.0f, 1.0f), &d3dxImageInfo, NULL, &texture ); // Create our sprite... D3DXCreateSprite( D3D9Device, &sprite ); did that help ? well, it helped me at least. (not with what I'm doing right now though) but I didn't know directx had a D3DXCreateSprite. I can go to bed now, because I learned something new. Thanks Share on other sites I'm now using the following settings: D3DXCreateTextureFromFileEx(d3ddev, L"ExampleNon.png", 220, 220, 1, 0, D3DFMT_FROM_FILE, D3DPOOL_DEFAULT, D3DX_DEFAULT, D3DX_DEFAULT, 0, NULL, NULL, &exampleNon); Which seems to now display the texture at the correct size, but I have had to specify the texture sizes manually to get it to work. This seems like a bit of headache to say the least! As every time I update the graphics I may need to change the CreateTexture params, and thus re-compile the game. :S Share on other sites I'm now using the following settings: D3DXCreateTextureFromFileEx(d3ddev, L"ExampleNon.png", 220, 220, 1, 0, D3DFMT_FROM_FILE, D3DPOOL_DEFAULT, D3DX_DEFAULT, D3DX_DEFAULT, 0, NULL, NULL, &exampleNon); Which seems to now display the texture at the correct size, but I have had to specify the texture sizes manually to get it to work. This seems like a bit of headache to say the least! As every time I update the graphics I may need to change the CreateTexture params, and thus re-compile the game. :S no you do not need to do that, use this to parse a text file that contains your image info int numberofframes = 1; float width = 1; float height = 1; float fsx = 1; float fsy = 1; float alpha = 0; wstring ws; char *buf = new char[20]; memset(buf, 0, 20); fstream fs(Path); if(!fs) { // failed to open file / create stream } fs.getline(buf, 20, '\t'); // read till a TAB space is found width = (float)atof(buf); // width fs.getline(buf, 20, '\n');// read till a endofline is found height = (float)atof(buf); // height fs.getline(buf, 20, '\t');// read till a TAB space is found fsx = (float)atof(buf); // frame size x fs.getline(buf, 20, '\n');// read till a endofline is found fsy = (float)atof(buf); // frame size y fs.getline(buf, 20, '\n');// read till a endofline is found numberofframes = (int)atof(buf); // number of frames fs.getline(buf, 20, '\n');// read till a endofline is found alpha = (float)atof(buf); // alpha fs.getline(buf, 20, '\n'); string s; s.append(mediafolder); s.append(buf); ws.append(s.begin(), s.end()); // path to sprite sheet delete []buf; does that make sense ? Share on other sites Yeah I get that, however I've got 450MB~ of textures to load into 128MBs of VRAM. Obviously I need to cut it down so I'm looking at ways in which I can do it, SpriteSheets, non-power two textures, DDD / DXT Compression, etc. Spritesheets didn't seem to help much, as I have a lot of animations that can't really be made into a spritesheet; only 1 frame will ever be drawn on screen at one time, so with a spritesheet I'm loading all the other frames into VRAM and wasting space they are using, making the card swap these large spritesheets in/out of VRAM every frame. Share on other sites To get the width and height of the image you don't have to read it from the file by yourself. In D3D9 you can just use D3DXGetImageInfoFromFile; D3DXIMAGE_INFO ImageInfo; D3DXGetImageInfoFromFile(L"Example.png", &ImageInfo); Width = ImageInfo.Width; Height = ImageInfo.Height; Share on other sites To get the width and height of the image you don't have to read it from the file by yourself. In D3D9 you can just use D3DXGetImageInfoFromFile; D3DXIMAGE_INFO ImageInfo; D3DXGetImageInfoFromFile(L"Example.png", &ImageInfo); Width = ImageInfo.Width; Height = ImageInfo.Height; Fantastic, now that's what I'm talking about! Share on other sites Well it's kinda one OR the other.. as DXT compression isn't supported for non-power of two textures. So I don't know which will give the biggest benefit in terms of memory usage, but non-power of two will certainly retain quality better. Share on other sites the DX Docs say that dds can be easyaly converted to other formates or depths etc, what you could try is to make it a power of 2 but only use certain parts of it, that way compression is used. Share on other sites the DX Docs say that dds can be easyaly converted to other formates or depths etc, what you could try is to make it a power of 2 but only use certain parts of it, that way compression is used. As I said earlier, spritesheets (or multiple textures in one file) don't seem to provide us with much benefit (having tested various methods, packing techniques, etc.). Share on other sites DDS is supported for non-power-of-2; the constraint is that the dimensions must be multiples of 4 (as it compresses a 4x4 block of the source), which also imples that they must be at least 4. Share on other sites If you look at the documentation for D3DXCreateTextureFromFileEx you'll find: Width in pixels. If this value is zero or D3DX_DEFAULT, the dimensions are taken from the file and rounded up to a power of two. If the device supports non-power of 2 textures and D3DX_DEFAULT_NONPOW2 is specified, the size will not be rounded.[/quote]

2018-01-23 06:22:48

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http://www.mcs.le.ac.uk/techreports/TechReportAbstracts97.html

# 1997 Technical Report Abstracts 1997/2 Aspects of an Adaptive hp-Finite Element Method: Adaptive Strategy, Conforming Appoximation and Efficient Solvers M Ainsworth, B Senior The main components needed for an adaptive $hp$-version finite element algorithm are discussed: an adaptive $hp$-refinement strategy, effective methods for constructing conforming $hp$-approximations, and, efficient solvers for the large, ill-conditioned systems of linear equations. Together, these provide the methodology for an effective adaptive $hp$-version algorithm. The presentation emphasizes the links between the differing components showing how the algorithms may be implemented efficiently in practice. The main principles are illustrated by use of concrete examples so that a non-expert may develop their own adaptive $hp$-code. The performance of the whole algorithm is illustrated for some representative problems taken from linear elasticity. 1997/11 Reliable and Robust a Posteriori Error Estimation for Singularly Perturbed Reaction Diffusion Problems M Ainsworth, I Babuska Problems with singular perturbations exhibit solutions with strong boundary layers and other types of local behaviour. Such features lend themselves to adaptive solution methods. The quality of any adaptive algorithm ultimately rests on the reliability and robustness of the a posteriori error control. An estimator that has proved to be one of the most reliable is the equilibrated residual method. The main property of the estimator is that it bounds the true error from above. However, the method is not robust in the singularly perturbed limit. The current work generalizes the error estimator based on the equilibrated residuals and coincides with the standard method in the unperturbed limit. It is shown that the new method is robust in the singularly perturbed limit whilst maintaining reliability, yielding a guaranteed upper bound on the true error. Finally, the application of the estimator to the problem of controlling the spatial error in Rothe's method for the time discretistion of a simple parabolic problem is included. 1997/12 Duration Properties of Timed Transition Systems Z Liu, A P Ravn, X Li This paper proposes a method for formal real-time systems development: Ihe system requirements and high level design decisions are time interval properties, and are therefore specified in the Duration Calculus (DC), while the implementation is described in terms of timed transition systems (TTS). A link from implementation properties to the requirements and design properties is given by interpreting a DC formula in a model of the executions of a TTS and then providing rules for lifting properties proved for a TTS to DC. The method is illustrated by the Gas Burner case study. {\bf Keywords}: Duration Calculus, Timed Transition System, Real-time systems. 1997/14 Hamiltonian cycles and all-to-all broadcasts in faulty hypercubes and k-ary n-cubes Iain A. Stewart We derive a sequential algorithm {\em Find-Ham-Cycle\/} with the following property. On input: $k$ and $n$ (specifying the $k$-ary $n$-cube $Q_n^k$); $F$, a set of at most $2n-2$ faulty links; and $v$, a node of $Q_n^k$, the algorithm outputs nodes $v^+$ and $v^-$ such that if {\em Find-Ham-Cycle\/} is executed once for every node $v$ of $Q_n^k$ then the node $v^+$ (resp. $v^-$) denotes the successor (resp. predecessor) node of $v$ on a fixed Hamiltonian cycle in $Q_n^k$ in which no link is in $F$. Moreover, the algorithm {\em Find-Ham-Cycle\/} runs in time polynomial in $n$ and $k$. We also obtain a similar algorithm for a $n$-dimensional hypercube with at most $n-2$ faulty links. We use our algorithms to develop all-to-all broadcast algorithms in certain faulty hypercubes and $k$-ary $n$-cubes (improving upon one derived from a broadcast algorithm of Peleg), and to solve an open problem posed by Chan and Lee concerning the distributed construction of Hamiltonian cycles in faulty hypercubes, and also in faulty $k$-ary $n$-cubes. 1997/15 A Perspective on Lindstrom Quantifiers and Oracles Iain A. Stewart This paper presents a perspective on the relationship between Lindstr\"om quantifiers in model theory and oracle computations in complexity theory. We do not study this relationship here in full generality (indeed, there is much more work to do in order to obtain a full appreciation), but instead we examine what amounts to a thread of research in this topic running from the motivating results, concerning logical characterizations of nondeterministic polynomial-time, to the consideration of Lindstr\"om quantifiers as oracles, and through to the study of some naturally arising questions (and subsequent answers). Our presentation follows the chronological progress of the thread and highlights some important techniques and results at the interface between finite model theory and computational complexity theory. 1997/17 Embeddings of cycles, meshes and tori in faulty k-ary n-cubes Yaagoub A Ashir, Iain A Stewart We investigate the existence of cycles, meshes and tori in a $k$-ary $n$-cube $Q_n^k$ in which a limited number of nodes and links are faulty. Our main result is that in a $k$-ary $n$-cube $Q_n^k$ in which there are $\nu$ faulty nodes and $\lambda$ faulty links where $\nu+\lambda\leq n$, there is a cycle of length at least $k^n-\nu\omega$, where $\omega=1$ if $k$ is odd and $\omega=2$ if $k$ is even (throughout, $k\geq 3$ and $n\geq 2$). We extend this result so as to prove the existence of large meshes and tori in such a faulty $k$-ary $n$-cube. 1997/21 The Cohomology and K-theory of Commuting Homeomorphisms of the Cantor Set Alan Forrest, John Hunton Given a $Z^d$ homeomorphic action, $\alpha$, on the Cantor set, $X$, we consider higher order continuous integer valued dynamicalcohomology, $H*(X,\alpha)$. We also consider the dynamical K-theory of the action, the K-theory of the crossed product $C^*$-algebra $C(X)\times_{\alpha}Z{}^d$. We show that these two invariants are essentially equivalent. We also show that they only ever take torsion free values. Our work links the two invariants via a third invariant based on topological complex K-theory evaluated on an associated mapping torus. 1997/25 The Homology of Spaces Representing Exact Pairs of Homology Functors John R Hunton, P R Turner We begin a study of how a relationship between representable homotopy functors manifests itself ina relationship between the homology of the representing spaces. In this paper we examine in detail the case of exact pairs of functors where a particulary simple description can be given in terms of the tensor product of coalgebraic modules. 1997/26 Extensions of Umbral Calculas II: Double Delta Operators, Leibniz Extensions and Hattori-Stong Theorems Francis Clarke, John Hunton, Nigel Ray We continue our programme of extending the Roman-Rota umbral calculus to the setting of delta operators over a graded ring $E_*$ with a view to applications in algebraic topology and the theory of formal group laws. We concentrate on the situation where $E_*$ is free of additive torsion, in which context the central issues are number-theoretic questions of divisibility. We study polynomial algebras which admit the action of two delta operators linked by an invertible power series, and make related constructions which are motivated by the Hattori-Stong theorem of algebraic topology. Our treatment is couched purely in terms of the umbral calculus, but inspires novel topological applications; we also explain how the theory of Sheffer sequences fits naturally into this framework. 1997/28 Groups, Semigroups and Finite Presentations C M Campbell, E F Robertson, N Ruskuc, R M Thomas Two questions are discussed. Firstly, what is the connection between the group and the semigroup defined by the same presentation? Secondly, do there exist any results for semigroups with respect to substructures being finitely presented that correspond to the classical results for groups? These themes are considered both from a geometric and a combinatorial point of view. 1997/29 Automatic Semigroups C M Campbell, E F Robertson, N Ruskuc, R M Thomas The area of automatic groups has been one in which significant advances have been made in recent years. While it is clear that the definition of an automatic group can easily be extended to that of an automatic semigroup, there does not seem to have been a systematic investigation of such structures. It is the purpose of this paper to make such a study. We show that certain results from the group-theoretic situation hold in this wider context, such as the solvability of the word problem in quadratic time, although others do not, such as finite presentability. There are also situations which arise in the general theory of semigroups which do not occur when considering groups; for example, we show that a semigroup S is automatic if and only if S with a zero adjoined is automatic, and also that S is automatic if and only if S with an identity adjoined is automatic. We use this last result to show that any finitely generated subsemigroup of a free semigroup is automatic. 1997/30 Compositional Inductive Verification of Duration Properties of Real-Time Systems Zhiming Liu, Anders P Ravn, Xiaoshan Li This paper proposes a method for formal real-time systems development. At high level a system is modelled as a conventional dynamical system with states that are functions of time represented by non-negative real numbers, while the implementation and refinement at low level are described in terms of timed transition systems (TTS). Therefore, The system requirements and high level design decisions are time interval properties, and are thus specified and reasoned about in the Duration Calculus (DC), and the properties of the implementation at low level are specified and verified compositionally and inductively in timed linear temporal logic (TLTL). A link from implementation properties to the requirement and design properties is given by interpreting a DC formula in a model of the executions of a TTS and then providing rules for lifting TLTL properties proved for a TTS to DC. The method is illustrated by the Gas Burner case study. {\bf Keywords: } Real-time Systems, Duration Calculus, Timed Transition Systems, Specification, Verification. 1997/34 A Reliable A Posterior Error Estimator for Adaptive Hierarchic Modelling Mark Ainsworth, Mark Arnold An a posteriori error estimator is obtained for estimating the error in non-uniform order hierarchic models of elliptic boundary value problems on thin domains, and is shown to provide an upper bound on the actual error measured in the energy norm. The estimator has many similarities with a known heuristic refinement indicator used for adaptive hierarchic modelling and in this sense provides some theoretical support for the existing technique. Numerical examples show that the performance of both estimators is comparable, both yielding good results, with the new technique performing slightly better. 1997/35 Syntactic and Rees indices of subsemigroups N Ruskuc, R M Thomas We define two different notions of index for subsemigroups of semigroups: the (right) syntactic index and the Rees index. We investigate the relationships between them and with the group index. In particular, we show that the syntactic index is a generalisation of both the group index and the Rees index. We use this fact to prove further similarities between the group index and the Rees index. 1997/36 The Complex oriented Cohomology of Extended Powers John Hunton We examine the behaviour of a complex oriented cohomology theory $G*(-)$ on $D_p(X)$, the $C_p$-extended power of a space $X$, seeking a description of $G^*(D_p(X))$ in terms of the cohomology $G^*(X)$. We give descriptions for the particular cases of Morava K-theory $K(n)$ for any space $X$ and for complex cobordism MU, the Brown-Peterson theories BP and any Landweber exact theory for a wide class of spaces. 1997/37 The Conditioning of Boundary Element Equations on Locally Refined Meshes and Preconditioning by Diagonal Scaling Mark Ainsworth, William McLean, Thanh Tran Consider a boundary integral operator on a bounded, $d$-dimensional, surface in $\RR^{d+1}$. Suppose that the operator is a pseudodifferential operator of order $2m$, $m\in\RR$, and that the the associated bilinear form is symmetric and positive-definite. (The surface may be open or closed, and $m$ may be positive or negative). Let $\b{B}$ denote the stiffness matrix arising from a Galerkin boundary element method with standard nodal basis functions. If local mesh refinement is used then the partition may contain elements of very widely differing sizes, and consequently $\b{B}$ may be very badly conditioned. In fact, if the elements are non-degenerate and $2|m| Keywords: Boundary element method. Condition numbers. Diagonal scaling. Preconditioning. 1997/39 Reflexivity of Modules over QF-3 Algebras Nicole Snashall The indecomposable injective projective modules of a basic finite-dimensional algebra are studied in relation to the notion of a chain end; this concept is a generalisation of that given by Murase for such algebras which are also indecomposable and serial. It is shown that the basic finite-dimensional algebras for which the chain ends are precisely the indecomposable injective projective modules are those which are both QF-3 and left QF-2. Reflexivity in the sense of Halmos and Fuller, Nicholson and Watters is then investigated for modules over such algebras. The reflexive QF-3 and left QF-2 algebras are characterised, and necessary and sufficient conditions are given on the quiver of the algebra for every faithful left module to be reflexive. 1997/40 Domain Decomposition Preconditioners for p and hp Finite Element Approximation of Stokes Equations Mark Ainsworth, Spencer Sherwin Domain decomposition preconditioning techniques are developed in the context of$hp$finite element approximation of the Stokes problem. Two basic types of preconditioner are considered: a block diagonal scheme based on decoupling the velocity and pressure components, and, a scheme based on an indefinite system similar to the original Stokes system. For each type of scheme, theoretical estimates are obtained for the location of the eigenvalues of the preconditioned operators in terms of the polynomial degree, the mesh sizes on the coarse and fine grids, and the {\em inf-sup} constant for the method. Theoretical estimates show that the growth of the bounds is modest as the mesh is refined and the polynomial order is increased. The preconditioners are shown to be applicable to various iterative schemes for the Stokes systems. The theoretical bounds are compared with actual quantities obtained in practical computations for several representative problems. 1997/41 Solving Hamiltonian systems arising from ODE eigenproblems Carsten R Maple, Marco Marletta The numerical solution of a linear Hamiltonian system of ordinary differential equations is discussed. Eigenvalue problems for Hamiltonian systems can arise from a number of sources. Here we are concerned with the system attained after transforming the general$2m$th order regular Sturm-Liouville problem. A new algorithm is developed that attempts to bypass the problems that force existing solvers to take a large number of small integration steps when the eigenparameter is large. A coupled system of differential equations for the eigenvalues and eigenvectors of a certain matrix is formed. This system is then solved using a predictor-corrector type method. The eigenvectors vary in a relatively smooth manner, and therefore these differential equations are solved using a numerical method. The eigenvalues are found by approximating the differential equations and solving exactly. A numerical results discussion illustrates the advantages and drawbacks of the algorithm. 1997/42 Implementing Operational Semantics (Preliminary Report) Roy Crole This paper describes a high level operational semantics for a simple programming language, called KOREL, together with a parser, interpreter and pretty printer which are implemented in the (pure) functional programming language Haskell. The syntax of KOREL is presented via BNF grammars, and the operational semantics is specified via structured, inductive rules. The paper outlines the broad ideas behind the Haskell implementation, with a more detailed explication of the key techniques. A code listing can be found in TR 1997/43. 1997/43 The KOREL Programming Language (Preliminary Report) Roy Crole This report contains the full listing of the Haskell code which implements the programming language KOREL. A description of the language, and of the Haskell code, can be found in TR 1997/42. 1997/44 Asymptotic behaviour of the spectral decomposition of Riccati-like variables for linear Hamiltonian systems Carsten R Maple, Marco Marletta This work is a sequel to Tech Report 1997/41, in which an algorithm was proposed to solve linear Hamiltonian systems. This involved finding a coupled system of differential equations for the eigenvalues and eigenvectors of the so-called Atkinson theta matrices associated with the system. The Atkinson theta matrices are closely related to Riccati variables. In the present report, we examine the behaviour of these eigenvalues and eigenvectors for large values of the spectral parameter$\lambda$, for two model problems. The analysis reveals radically different behaviour of the eigenvalues and eigenvectors for two Hamiltonian systems which are superficially very similar. This shows that the method in Tech Report 1997/41 will easily cope with large values of$\lambda\$ for some problems, but not for others.

2019-02-19 10:29:02

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http://www.julia-harz.de/publication/harz-precise-2015/

# Precise Prediction of the Dark Matter Relic Density within the MSSM ### Abstract With the latest Planck results the dark matter relic density is determined to an unprecedented precision. In order to reduce current theoretical uncertainties in the dark matter relic density prediction, we have calculated next-to-leading order SUSY-QCD corrections to neutralino (co)annihilation processes including Coulomb enhancement effects. We demonstrate that these corrections can have significant impact on the cosmologically favoured MSSM parameter space and are thus of general interest for parameter studies and global fits. Type Publication PoS

2021-02-25 02:15:45

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https://wiki.math.ntnu.no/tma4315/2021h/start

# TMA4315 Generalized linear (mixed) models 2021 ## Messages Sept. 8: You can get assistance with the projects at Banachrommet, each Friday at 11-12. Sept. 7: If you're in search of someone to collaborate with on project 1, send me an email and I will if possible suggest someone you can work with. Sept. 7: If you can't a lectures, then you can find lecture notes and videos from last year at here (a bit out of sync with this years lectures). Aug. 26: See updated time for second weekly lecture below. Sept. 6: Project 1 is out. Deadline is Friday October 1. ## Practical information Lectures: Thursdays 10:15-12:00 in EL6 and Tuesdays 8:15-10:00 in R D4-132Fridays 12:15-14:00 in S1 (but we will try to move the lectures on fridays to another time to avoid colliding with TMA4295 Statistical Inference). For the time being, we plan to do physical lectures only. Guidance with exercises and projects: Fridays 14-15 in H3 424 Vembi11-12 in * from week 35 and via the Discourse forum. We plan to do physical lectures only but the lecture notes are available here and videos from last year can be found here Lecturer: Jarle Tufto Teaching assistant: Silius M. Vandeskog. Reference group: NN, NN, NN (send me an email if you want to be in the reference group). ## Obligatory projects There will be three obligatory projects (counts 30% of final grade). The problems are posted in the Discourse forum. You may ask questions or post your attempted solution (if you want feedback) by replying to each exercise. ## Tentative curriculum Fahrmeir et. al. (2013) (freely available on springer link), ch. 2.1-2.4, B.4, 5.1-5.4, 5.8.2, 6, 7.1-7.3, 7.5, 7.7. We will also use some material from Wood (2015) Wood (2015), and some material from some Harville 1974, Agresti 2002 and Kristiansen 2016 (see below). This covers ordinary linear and multiple regression (mostly repetition from TMA4267 Linear statistical models), binary regression, Poisson and gamma regression, the exponential family and generalised linear models in general, categorical regression (includes contingency tables and log-linear models, multinomial and ordinal regression), linear mixed effects models, generalized linear mixed effects models. Also see the official ntnu course info. ## Lectures August 26: Introduction to glms (ch. 2.1-2.3), the exponential family (ch. 5.4.1). . August 31: More on the exponential family (ch. 5.4.1). Review of theory of linear models (ch. 3). September 2: Geometric views of the linear model. Sampling distributions associated with the linear model (the chi-square-, t- and F-distribution). September 7: Testing and fitting linear hypotheses (via quadratic form for $C\hat\beta-d$ - Box 3.13 in Fahrmeir) or via F-test based on sums of squares for each model alternative (the restricted model fitted via Lagrange method (pp. 172-173) or using the solution to problem 2 below). Design matrices for interactions between numeric and categorical covariates. Binary regression (ch. 5). Logit, probit links September 9: cloglog models. Binary regression continued. Score function of binary regression model. Some general properties of the expected log likelihood (sec. 4.1 in Wood (2015)). Expected and observed Fisher information and iterative computation of MLEs for binary regression (Fisher scoring algorithm). Binary regression continued. September 14: A minimal example of divergence of the Fisher scoring algorithm. Asymptotic properties of MLEs. Likelihood ratio, Wald, and score tests. Deviance and testing goodness-of-fit. September 16: More on the deviance and the saturated model. Deviance residuals. Estimating the overdispersion parameter. We'll also go through a sketch of a proof for the asymptotic distribution of LR test statistic (section 4.4 in Wood). September 21: Example (lung cancer rates) illustrating model selection via AIC, model parsimony, Wald and likelihood ratio testing. Theory behind AIC (Wood sec. 4.6). September 23: Poisson regression. Fisher scoring vs. Newton-Raphson for poisson-regression with non-canonical identity link (see R code for further illustrations). This paper provides an example of non-canonical link functions. ## Exam Tuesday December 7, 15:00-19:00 Previous exams can be found at previous exams out of which 2013-2016 are not the most relevant. Any questions on the exponential family in this years exam will be based on the notation in Fahrmeir. The solutions to exams in 2008 and 2009 is based on a slightly more general definition of the exponential family and different notation (Dobson & Barnett 2008) writing the pdf of pmf as as $\exp(b(\theta)a(y) + c(\theta) + d(y))$ instead of $(\exp((y\theta - b(\theta))w/\phi - c(y,\phi,w)$ in Fahrmeir. Thus • $b(\theta)$ is the canonical parameter corresponding to $\theta$ in Fahrmeir • $a(y)$ corresponds to $y$, • $c(\theta)$ (implicitly a function of the canonical parameter $b(\theta)$) corresponds to $b(\theta)$. Implicit differentiation then leads to a different formula for $EY=-c'(\theta)/b'(\theta)$. • $d(y)$ corresponds to $c(y,\phi,w)$. Note that if a random variable $Y$ satisfies the definition of Dobson & Barnett, then $a(Y)$ satisfies the definition in Fahrmeir.

2021-09-18 15:17:13

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https://mathoverflow.net/questions/296975/mixed-motives-and-motivic-cohomology

# Mixed motives and motivic cohomology In Scholl's paper "Remarks on special values of $L$-functions", he defines that an object $M$ of $\textbf{MM}_{\mathbb{Q}}$ (the conjectured abelian category of mixed motives with coefficients $\mathbb{Q}$), is defined over $\mathbb{Z}$ if for every $p$, $\ell$, $p\neq \ell$, the weight filtration of the $\ell$-adic realisation $M_\ell$ splits as a representation of the inertia group $I_p$. Mixed motives defined over $\mathbb{Z}$ form a full subcategory $\textbf{MM}_{\mathbb{Z}}$. Suppose $X$ is smooth and proper over $\mathbb{Q}$, and let the pure motive $M$ and $N$ be $$M=h^i(X)(m),~N=M^\vee(1) \simeq h^i(X)(n), ~n=i+1-m$$ Then we will expect that $$\text{Ext}^0_{\textbf{MM}_{\mathbb{Z}}}(M,\mathbb{Q}(1))=\text{Ext}^0_{\textbf{MM}_{\mathbb{Z}}}(\mathbb{Q}(0),N)=\begin{cases} \text{CH}^n(X)/\text{CH}^n(X)^0 \otimes \mathbb{Q},~\text{if}~i=2n\\ 0,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \text{if}~ i\neq 2n \end{cases}$$ $$\text{Ext}^1_{\textbf{MM}_{\mathbb{Z}}}(M,\mathbb{Q}(1))=\text{Ext}^1_{\textbf{MM}_{\mathbb{Z}}}(\mathbb{Q}(0),N)= \begin{cases} H^{i+1}_{\mathcal{M}}(X,\mathbb{Q}(n))_\mathbb{Z},~\text{if}~i+1 \neq 2n \\ \text{CH}^n(X)^0 \otimes \mathbb{Q}, ~~~~~\text{if}~i+1=2n \end{cases}$$ where $\text{CH}^n(X)$ is the Chow group of codimension-$n$ cycles on $X$ modulo rational equivalence and $\text{CH}^n(X)^0$ is the subgroup of classes that is homologically equivalent to zero. While $H_{\mathcal{M}}$ denotes the motivic cohomology: $$H^i_{\mathcal{M}}(X,\mathbb{Q}(j))=(K_{2j-i}(X) \otimes \mathbb{Q})^{(j)}$$ and $H^*_{\mathcal{M}}(X,\cdot)_{\mathbb{Z}}$ is the image in $H^*_{\mathcal{M}}(X,\cdot)$ of the $K$-theory of a regular model for $X$, proper and flat over $\mathbb{Z}$. For a number field $F$ with rings of integers $\mathcal{O}_F$, there is the conjectured abelian category $\textbf{MM}_F$ (with coefficients $\mathbb{Q}$). A motive $M$ of $\textbf{MM}_F$ is defined over $\mathcal{O}_{F}$ is the weight filtration of the $\ell$-adic realisation $M_\ell$ splits as a representation of the inertia group $I_v$ for a prime $v$ of $\mathcal{O}_F$, $v \nmid \ell$. The motives defined over $\mathcal{O}_F$ form a full subcategory $\textbf{MM}_{\mathcal{O}_F}$. Now for a smooth proper variety $X$ defined over $F$, do we still expect the same properties are valid? i.e. if we still define $$M=h^i(X)(m),~N=M^\vee(1) \simeq h^i(X)(n), ~n=i+1-m$$ do we still expect the following to be valid? $$\text{Ext}^0_{\textbf{MM}_{\mathcal{O}_F}}(M,\mathbb{Q}(1))=\text{Ext}^0_{\textbf{MM}_{\mathcal{O}_F}}(\mathbb{Q}(0),N)=\begin{cases} \text{CH}^n(X)/\text{CH}^n(X)^0 \otimes \mathbb{Q},~\text{if}~i=2n\\ 0,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \text{if}~ i\neq 2n \end{cases} \\ \text{Ext}^1_{\textbf{MM}_{\mathcal{O}_F}}(M,\mathbb{Q}(1))=\text{Ext}^1_{\textbf{MM}_{\mathcal{O}_F}}(\mathbb{Q}(0),N)= \begin{cases} H^{i+1}_{\mathcal{M}}(X,\mathbb{Q}(n))_{\mathcal{O}_F},~\text{if}~i+1 \neq 2n \\ \text{CH}^n(X)^0 \otimes \mathbb{Q}, ~~~~~\text{if}~i+1=2n \end{cases}$$ If this is expected, could anyone give a precise reference? • There is an operation of restriction of scalars from mixed motives over F to mixed motives over Q, more precisely in a suitable category of "mixed realizations", see Jannsen's book "Mixed motives and algebraic K-theory". This might help you for this kind of question. – François Brunault Apr 5 '18 at 15:51 • @FrançoisBrunault Thank you, I will check this book. – Wenzhe Apr 5 '18 at 20:46

2019-10-18 22:24:41

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https://www.biostars.org/p/161120/

How to generate BedGraph from BedTools Coverage? 3 2 Entering edit mode 6.2 years ago SmallChess ▴ 560 The tool is at: I want to get base coverage for specific regions (I'll provide a bed file). However, the output is not in a standard BedGraph format that I can export into R for visualisation. Q: How do I draw density plot of coverage over the regions from the outputs generated by the coverage tool? Q: Alternatively, how to convert the non-standard outputs to a proper BedGraph? For example, the Sushi R package takes a bedgraph input file. bedtools bed genomic • 11k views 6 Entering edit mode 6.1 years ago You should also consider bedtools genomecov, with the -bg option (bg being bedgraph) Edit: After reading comments, and after some experimenting, what I think you want to do is: bedtools intersect -a myBam.bam -b myRegions.bed > intersected.bam bedtools genomecov -trackline -bg -ibam intersected.bam > intersected.bedgraph Bedtools intersect will give only bam reads that overlap your regions (in myRegions.bed) Then genomecov will draw a bedgraph file (coverage) only for the read depth for your filtered regions. 0 Entering edit mode The problem is that this tool doesn't take a BED file filtering the regions that I'm interested. Therefore, it'd be too time consuming to calculate for all bases in the genome. 0 Entering edit mode Fair enough, I didn't catch the subtlety in your question, but James Ashmore did. 1 Entering edit mode I think this might be what you want to do - genomecov and then pull out the regions you want using bedtools intersect. But kind of hard to tell. 0 Entering edit mode Thanks. I think your approach is the best. 0 Entering edit mode Note: -abam is correct. 1 Entering edit mode 6.1 years ago Get your output into a standard five- or more columned BED file and then use awk: $awk '{ print$1"\t"$2"\t"$3"\t"\$5 }' foo.bed > foo.bedgraph 0 Entering edit mode @ Alex: How can I convert my .sam file to standard .bed file? I have tried sam2bed < input.sam > output.bed it gives me output file but the column 5 always contains the same value (which is 99). According to UCSC bed format this is score column and to me this look strange that for every region the score is same. Then by using your mentioned command I obtain the .bedgraph file which I later use for ChromImpute. Problem is that when ChromeImpute convert these raw signal into certain sized bin signal (25 bp) then for every region I get the 0 signal.. I have tried many ways (using Pyisco, make wig files, genomeCoverageBed etc..) but I am unable to solve this problem. Can you help me with this one? Thanks. 0 Entering edit mode 6.1 years ago James Ashmore ★ 3.2k To generate bedGraph output from Bedtools coverage command you need to specify the -counts flag, for example: bedtools coverage -a sample.bed -b sample.bam -counts > sample.bedgraph Strictly speaking each row of a bedGraph file contains the chromosome name, the start position, the end position and then some value, which in your case would be the coverage, therefore you will want to keep only these fields, so if your had a bed file with six columns, you would want to keep only the first 3 columns (chrom, start, end) and the last column (coverage): bedtools coverage -a sample.bed -b sample.bam -counts | cut -f1,2,3,7 > sample.bedgraph 2 Entering edit mode It's not really clear to me from the original question that this is what they want. This would give one value for each genome region. I suspect they want "wiggly" values for each genome region. But could be wrong.

2021-12-01 05:47:52

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https://www.thejournal.club/c/paper/337162/

#### Balancing Fairness and Efficiency in Traffic Routing via Interpolated Traffic Assignment ##### Devansh Jalota, Kiril Solovey, Stephen Zoepf, Marco Pavone System optimum (SO) routing, wherein the total travel time of all users is minimized, is a holy grail for transportation authorities. However, SO routing may discriminate against users who incur much larger travel times than others to achieve high system efficiency, i.e., low total travel times. To address the inherent unfairness of SO routing, we study the $\beta$-fair SO problem whose goal is to minimize the total travel time while guaranteeing a $\beta\geq 1$ level of unfairness, which specifies the maximal ratio between the travel times of different users with shared origins and destinations. To obtain feasible solutions to the $\beta$-fair SO problem while achieving high system efficiency, we develop a new convex program, the Interpolated Traffic Assignment Problem (I-TAP), which interpolates between a fair and an efficient traffic-assignment objective. We then leverage the structure of I-TAP to develop two pricing mechanisms to collectively enforce the I-TAP solution in the presence of selfish homogeneous and heterogeneous users, respectively, that independently choose routes to minimize their own travel costs. We mention that this is the first study of pricing in the context of fair routing. Finally, we use origin-destination demand data for a range of transportation networks to numerically evaluate the performance of I-TAP as compared to a state-of-the-art algorithm. The numerical results indicate that our approach is faster by several orders of magnitude, while achieving higher system efficiency for most levels of unfairness. arrow_drop_up

2022-06-26 05:41:39

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https://www.shaalaa.com/question-bank-solutions/division-line-segment-construct-triangle-sides_7184

# Solution - Division of a Line Segment Account Register Share Books Shortlist ConceptDivision of a Line Segment #### Question Construct a triangle of sides 4 cm, 5cm and 6cm and then a triangle similar to it whose sides are 2/3 of the corresponding sides of the first triangle. Give the justification of the construction. #### Solution You need to to view the solution Is there an error in this question or solution? #### Similar questions VIEW ALL If the coordinates of points A and B are (-2, -2) and (2, -4) respectively, find the coordinates of P such that AP =(3/7)AB, where P lies on the line segment AB. view solution Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are 4/3 times the corresponding side of ΔABC. Give the justification of the construction. view solution Find the ratio in which the line segment joining the points A(3,- 3) and B(- 2, 7) is divided by x-axis. Also find the coordinates of the point of division. view solution Construct a triangle ABC in which BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct another triangle whose sides are3/4 times the corresponding sides of ΔABC. view solution Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose side are 1 1/2 times the corresponding sides of the isosceles triangle. Give the justification of the construction view solution #### Reference Material Solution for concept: Division of a Line Segment. For the course 8th-10th CBSE S

2017-10-19 23:35:38

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http://crypto.stackexchange.com/tags/feistel-network/hot?filter=month

# Tag Info 5 The simple answer is that fewer than 3 rounds can be easily distinguished from a random permutation. The 2-round Luby-Rackoff cipher on $2n$ bits, using random functions $f_i$ mapping $n$ bits to $n$ bits, consists of $$F(L, R) = (A, B),$$ where $A = L \oplus f_1(R)$ and $B = R \oplus f_2(L \oplus f_1(R))$. Now consider an attacker that wants to ... 5 If a message is longer than the block length, how would changing one part of the message affect the encryption of other parts of the message? That really doesn't depend on the block cipher in use, which may be a feistel cipher like DES or a SP network like AES, but on the mode of operation. Now the answer to this really depends on the actual mode ... 3 It depends on the block cipher in question - specifically its key schedule. Knowing any round key of AES-128 would let you calculate the key, because the schedule is reversible. OTOH, e.g. TEA would retain secrecy of most of the key and might remain secure, because its round keys are small enough parts of the key. In the case of DES, it is weak enough to be ... 3 In a Feistel networks (from the German IBM cryptographer Horst Feistel), the input is divided into two blocks ($L_0$ and $R_0$) which interact with each other. Main example is DES. basic construction: In a SPN (Substitution Permutation Network), the input is divided into multiple small blocks, applied to a S-box (substitution), then the bits positions ... 3 It's required for diffusion and achieving the avalanche effect. The concept of diffusion and the avalanche effect basically means that each input bit should influence each output bit evenly. Changing one input bit should flip, on average, half the output bits. Due to the nature of the Feistel construction, how it is split up into halves, only one side ... 2 About the security of your first variation, it is sort of answered here. This is your 1st variation. This is your 2nd variation (your $8 \times 8$ matrix idea is equivalent to apply a permutation). In your first variation, the application of the matrix is useless, one can consider the $S1$ (or $S2$) and the matrix as a single S-box. Hence you have no ... 1 There is only one requirement for a Feistel round function and that is a good diffusion and confusion. It is not required for the round function to be invertible in a Feistel network. You can use (as asked) a secure mini SPN or even a hash function (Sha3...) it doubles the block size, so the number of rounds can be doubled at no perf cost If you meant ... Only top voted, non community-wiki answers of a minimum length are eligible

2016-05-30 18:26:52

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https://www.vedantu.com/question-answer/if-30ml-of-h2and-20ml-of-o2-reacts-to-form-h2o-class-11-chemistry-cbse-5f5fb67468d6b37d1636092f

Question # If 30ml of ${{H}_{2}}$and 20ml of ${{O}_{2}}$ reacts to form ${{H}_{2}}O$, what is left at the end of the reaction?This question has multiple correct options:A. 10ml of ${{H}_{2}}$B. 5ml of ${{H}_{2}}$C. 10ml of ${{O}_{2}}$D. 5ml of ${{O}_{2}}$ Hint: The reactant that fully reacts in the reaction is called reactant limiting or reagent limiting. The reactant that is not fully consumed in the reaction is called excess reactant.You should write the balanced reaction equation before solving the question. A balanced equation is an equation for a chemical reaction in which the number of atoms in the reaction for each element and the total charge for both the reactants and the products is equal. The Law of Definite Proportions, states that any chemical compound will always contain a fixed ratio of elements by mass. The Law of Definite Proportions is also sometimes called Proust's Law. ${{H}_{2}}$and ${{O}_{2}}$react to form water. The reaction used in question is: $2{{H}_{2}}(g)+{{O}_{2}}(g)\xrightarrow{{}}2{{H}_{2}}O(g)$ 2 moles of ${{H}_{2}}$combines with 1 mole of ${{O}_{2}}$ to produce 2 moles of ${{H}_{2}}O$. In other words, we can say that 1ml of ${{H}_{2}}$reacts with 0.5ml of ${{O}_{2}}$ to form 1ml of ${{H}_{2}}O$. Hence, when 30ml of ${{H}_{2}}$reacts with 15ml of ${{O}_{2}}$to form water it can be represented in a balanced chemical reaction as: ${{H}_{2}}(g)+\dfrac{1}{2}{{O}_{2}}(g)\xrightarrow{{}}{{H}_{2}}O(g)$ Therefore, limiting reagent is ${{H}_{2}}$. So, at the end of the reaction, 30ml of water and 5ml of ${{O}_{2}}$will be left. So, the correct answer is “Option D”. Note: The limiting reagent (also known as limiting reactant) in a chemical reaction is a reactant that is totally consumed when the chemical reaction is completed. The amount of product formed is limited by this reagent, since the reaction cannot continue without it. Remember this reaction equation, $2{{H}_{2}}+{{O}_{2}}\xrightarrow{{}}2{{H}_{2}}O$ When 2 moles of ${{H}_{2}}$combines with 1 mole of ${{O}_{2}}$ to produce 2 moles of ${{H}_{2}}O$.

2020-10-01 01:12:09

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https://www.allanswered.com/post/bewlq/3d-ising-model-is-computationally-intractable/

### 3D Ising model is computationally intractable 62 views 0 3 months ago by I was not aware that the innocent-looking 3D Ising model, $H_{\text{Ising}} = -\frac{1}{2} \sum_{\langle i, j \rangle} J_{i,j} \, S_i^z \, S_j^z$ , with the distance-dependent coupling $J_{i,j}$, has been proven to be computationally intractable! See, e.g., Cipra, B. A. “The Ising model is NP-complete”. SIAM News, July 17, 2000. It is amazing really, but very mysterious: * How could we have so much complexity in such a “tiny” toy-Hamiltonian? * How much “information” can be stored in an “Ising Cube”? * The 3D Ising model is NP-complete: So just solve it and you be able to solve all other NP-hard problems! #The Art of Mean-Field Theory the 2D version is “complete”: https://arxiv.org/abs/1207.6891 written 3 months ago by AlQuemist Similar posts: Search »

2018-05-21 11:05:27

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https://www.aimsciences.org/article/doi/10.3934/dcdsb.2006.6.449

American Institute of Mathematical Sciences May  2006, 6(3): 449-470. doi: 10.3934/dcdsb.2006.6.449 Convergence to equilibrium of a multiscale model for suspensions 1 CERMICS, École Nationale des Ponts et Chaussées, 6 et 8, avenue Blaise Pascal, Cité Descartes - Champs sur Marne, F-77455 Marne La Vallée Cedex 2, France 2 CERMICS, École Nationale des Ponts et Chaussées, 6 & 8, avenue Blaise Pascal, 77455 Marne-La-Vallée Cedex 2, France Received  April 2005 Revised  December 2005 Published  February 2006 We consider a multiscale model describing the flow of a concentrated suspension. The model couples the macroscopic equation of conservation of momentum with a nonlinear nonlocal kinetic equation describing at the microscopic level the rheological behaviour of the fluid. We study the long-time limit of the time-dependent solution. For this purpose, we use the entropy method to prove the convergence to equilibrium of the kinetic equation. Citation: Eric Cancès, Claude Le Bris. Convergence to equilibrium of a multiscale model for suspensions. Discrete & Continuous Dynamical Systems - B, 2006, 6 (3) : 449-470. doi: 10.3934/dcdsb.2006.6.449 [1] Mohamed Tij, Andrés Santos. Non-Newtonian Couette-Poiseuille flow of a dilute gas. Kinetic & Related Models, 2011, 4 (1) : 361-384. doi: 10.3934/krm.2011.4.361 [2] M. Bulíček, F. Ettwein, P. Kaplický, Dalibor Pražák. The dimension of the attractor for the 3D flow of a non-Newtonian fluid. Communications on Pure & Applied Analysis, 2009, 8 (5) : 1503-1520. doi: 10.3934/cpaa.2009.8.1503 [3] Oleksiy V. Kapustyan, Pavlo O. Kasyanov, José Valero, Michael Z. Zgurovsky. Strong attractors for vanishing viscosity approximations of non-Newtonian suspension flows. Discrete & Continuous Dynamical Systems - B, 2018, 23 (3) : 1155-1176. doi: 10.3934/dcdsb.2018146 [4] Li Fang, Zhenhua Guo. Zero dissipation limit to rarefaction wave with vacuum for a one-dimensional compressible non-Newtonian fluid. Communications on Pure & Applied Analysis, 2017, 16 (1) : 209-242. doi: 10.3934/cpaa.2017010 [5] Chang Zhang, Fang Li, Jinqiao Duan. Long-time behavior of a class of nonlocal partial differential equations. Discrete & Continuous Dynamical Systems - B, 2018, 23 (2) : 749-763. doi: 10.3934/dcdsb.2018041 [6] Yihong Du, Yoshio Yamada. On the long-time limit of positive solutions to the degenerate logistic equation. Discrete & Continuous Dynamical Systems, 2009, 25 (1) : 123-132. doi: 10.3934/dcds.2009.25.123 [7] Hafedh Bousbih. Global weak solutions for a coupled chemotaxis non-Newtonian fluid. Discrete & Continuous Dynamical Systems - B, 2019, 24 (2) : 907-929. doi: 10.3934/dcdsb.2018212 [8] Linfang Liu, Tomás Caraballo, Xianlong Fu. Exponential stability of an incompressible non-Newtonian fluid with delay. Discrete & Continuous Dynamical Systems - B, 2018, 23 (10) : 4285-4303. doi: 10.3934/dcdsb.2018138 [9] Vladimir Varlamov. Eigenfunction expansion method and the long-time asymptotics for the damped Boussinesq equation. Discrete & Continuous Dynamical Systems, 2001, 7 (4) : 675-702. doi: 10.3934/dcds.2001.7.675 [10] Changli Yuan, Mojdeh Delshad, Mary F. Wheeler. Modeling multiphase non-Newtonian polymer flow in IPARS parallel framework. Networks & Heterogeneous Media, 2010, 5 (3) : 583-602. doi: 10.3934/nhm.2010.5.583 [11] Tong Li, Kun Zhao. Global existence and long-time behavior of entropy weak solutions to a quasilinear hyperbolic blood flow model. Networks & Heterogeneous Media, 2011, 6 (4) : 625-646. doi: 10.3934/nhm.2011.6.625 [12] Jan Sokołowski, Jan Stebel. Shape optimization for non-Newtonian fluids in time-dependent domains. Evolution Equations & Control Theory, 2014, 3 (2) : 331-348. doi: 10.3934/eect.2014.3.331 [13] Linghai Zhang. Long-time asymptotic behaviors of solutions of $N$-dimensional dissipative partial differential equations. Discrete & Continuous Dynamical Systems, 2002, 8 (4) : 1025-1042. doi: 10.3934/dcds.2002.8.1025 [14] Emil Novruzov. On existence and nonexistence of the positive solutions of non-newtonian filtration equation. Communications on Pure & Applied Analysis, 2011, 10 (2) : 719-730. doi: 10.3934/cpaa.2011.10.719 [15] Jong Yeoul Park, Jae Ug Jeong. Pullback attractors for a $2D$-non-autonomous incompressible non-Newtonian fluid with variable delays. Discrete & Continuous Dynamical Systems - B, 2016, 21 (8) : 2687-2702. doi: 10.3934/dcdsb.2016068 [16] Guowei Liu, Rui Xue. Pullback dynamic behavior for a non-autonomous incompressible non-Newtonian fluid. Discrete & Continuous Dynamical Systems - B, 2018, 23 (6) : 2193-2216. doi: 10.3934/dcdsb.2018231 [17] Jin Li, Jianhua Huang. Dynamics of a 2D Stochastic non-Newtonian fluid driven by fractional Brownian motion. Discrete & Continuous Dynamical Systems - B, 2012, 17 (7) : 2483-2508. doi: 10.3934/dcdsb.2012.17.2483 [18] Aneta Wróblewska-Kamińska. Local pressure methods in Orlicz spaces for the motion of rigid bodies in a non-Newtonian fluid with general growth conditions. Discrete & Continuous Dynamical Systems - S, 2013, 6 (5) : 1417-1425. doi: 10.3934/dcdss.2013.6.1417 [19] Josef Diblík. Long-time behavior of positive solutions of a differential equation with state-dependent delay. Discrete & Continuous Dynamical Systems - S, 2020, 13 (1) : 31-46. doi: 10.3934/dcdss.2020002 [20] Muhammad Mansha Ghalib, Azhar Ali Zafar, Zakia Hammouch, Muhammad Bilal Riaz, Khurram Shabbir. Analytical results on the unsteady rotational flow of fractional-order non-Newtonian fluids with shear stress on the boundary. Discrete & Continuous Dynamical Systems - S, 2020, 13 (3) : 683-693. doi: 10.3934/dcdss.2020037 2019 Impact Factor: 1.27

2021-05-07 14:02:22

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https://kbroman.org/blog/2014/08/12/if-i-could-do-it-over-again-id-self-publish/

# If I could do it over again, I'd self-publish In 2009, Śaunak Sen and I wrote a book about QTL mapping and the R/qtl software. We started working on it in the fall of 2006, and it was a heck of a lot of work. We’d talked to several publishers, and ended up publishing with Springer. John Kimmel was the editor we worked with; I like John, and I felt that Springer (or John) did a good job of keeping prices reasonable. We were able to publish in full color with a list price of $99, so that on Amazon it was about$65. (In April, 2013, there was a brief period where it was just $42 at Amazon!) Springer did arrange several rounds of reviews; they typically pay reviewers$100 or a few books. But the copy editing was terrible (at the very least, you want a copy editor to read the book, and it was pretty clear that our copy editor hadn’t), and the actual type-setting and construction of the index was left to us, the authors. It feels nice to have written a proper book, but I don’t think it makes that big of a difference, for me or for readers. And John Kimmel has since left Springer to go to Chapman & Hall/CRC, and Springer has raised the price of our book to $169, so it’s now selling for$130 at Amazon. I think that’s obnoxious. It’s not like they’ve gone back and printed extra copies, so it’s hard to see how their costs could have gone up. But in the publishing agreement we signed, we gave Springer full rights to set the price of the book. (Update: it’s now listed at $199, though it’s still about$130 at Amazon.) I have a hard time recommending the book at that price; I’m tempted to help people find pirated PDFs online. (And seriously, if you can’t find a pirated copy, you should work on your internet skills.) I corresponded with an editor at Springer, on why our book has become so expensive and whether there’s anything we can do about it. They responded • If we do a new edition, it could be listed as $129. • If the book is adopted by university classes, “the pricing grid it is based on would have lower prices.” • Our book is available electronically, for purchase by chapter as well. Purchase by chapter? Yeah, for$30 per chapter! Springer has published books and allowed the authors to post a PDF, but only for really big sellers, and ours is definitely not in that category. I’m both disgusted and embarrassed by this situation. If I could do it all over again, I’d self-publish: post everything on the web, and arrange some way for folks to have it printed cheaply.

2021-05-15 11:54:26

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https://calculator.academy/gasoline-density-calculator/

Enter the total mass of gasoline and the total volume the gasoline takes up into the calculator to determine the gasoline density. ## Gasoline Density Formula The following formula is used to calculate the gasoline density. d = M/V • Where d is the gasoline density • M is the mass (lbs) • V is the volume (gallons) To calculate the gasoline density, divide the gasoline mass by the gasoline volume. ## Gasoline Density Definition What is the density of gasoline? Gasoline can vary between 715 and 780 kg/m^3 depending on the exact chemical proportions of the gasoline. ## Example How to calculate gasoline density? 1. First, start by filling up a certain volume. For this example, we will say we filled up a cylinder with a volume of 2 m^3. 2. Next, measure the mass. Calculating the mass we find it to be 1550 kg. 3. Finally, calculate the density. Using the formula above we find the density to be 775 kg/m^3. ## FAQ What is gasoline? Gasoline is a mixture of oil and gas that is used in combustion engines that are ignited to create a force.

2023-03-31 08:52:24

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http://math.stackexchange.com/questions/200633/prove-a-relation-of-arguments-of-3-complex-numbers-with-equal-modulus

# Prove a relation of arguments of 3 complex numbers with equal modulus Prove that $$\arg \left(\frac{z_{3}-z_{2}}{z_{3}-z_{1}}\right) = \frac{1}{2} \arg\left(\frac{z_{2}}{z_{1}}\right)$$ if $|z_{1}|=|z_{2}|=|z_{3}|$. - Let $z_j=R(\cos2t_j+i\sin2t_j)$ where $j=1,2,3$ and $R \neq 0$ So, $$\frac{z_3-z_2}{z_3-z_1}=\frac{R(\cos2t_3+i\sin2t_3)-R(\cos2t_2+i\sin2t_2)}{R(\cos2t_3+i\sin2t_3)-R(\cos2t_1+i\sin2t_1)}$$ $$=\frac{(\cos2t_3-\cos2t_2)+i(\sin2t_3-\sin2t_2)}{(\cos2t_3-\cos2t_1)+i(\sin2t_3-\sin2t_1)}$$ $$=\frac{-2\sin(t_3-t_2)\sin(t_3+t_2)+2i\sin(t_3-t_2)\cos(t_3+t_2)}{-2\sin(t_3-t_1)\sin(t_3+t_1)+2i\sin(t_3-t_1)\cos(t_3+t_1)}$$ applying $\sin C-\sin D$ and $\cos C-\cos D$ $$=\frac{2i\sin(t_3-t_2)(\cos(t_3+t_2)+i\sin(t_3+t_2))}{2i\sin(t_3-t_1)(\cos(t_3+t_1)+i\sin(t_3+t_1))}$$ $$=\frac{\sin(t_3-t_2)e^{i(t_3+t_2)}}{\sin(t_3-t_1)e^{i(t_3+t_1)}}$$ as $e^{ix}=\cos x+i\sin x$ $$=\frac{\sin(t_3-t_2)}{\sin(t_3-t_1)}e^{i(t_2-t_1)}$$ $$=\frac{\sin(t_3-t_2)}{\sin(t_3-t_1)}(\cos(t_2-t_1)+i\sin(t_2-t_1))=X+iY(say)$$ So, $X=\frac{\sin(t_3-t_2)}{\sin(t_3-t_1)}\cos(t_2-t_1)$ and $Y=\frac{\sin(t_3-t_2)}{\sin(t_3-t_1)}\sin(t_2-t_1)$ So, $\frac Y X = \tan (t_2-t_1)$ assuming $\pi ∤ (t_3 -t_1)$ $$\implies \arg \left(\frac{z_{3}-z_{2}}{z_{3}-z_{1}}\right) = \arctan \left(\frac Y X\right)$$ $$= t_2-t_1=\frac{1}{2}(2t_2-2t_1)=\frac{1}{2}(\arg z_2 -\arg z_1)=\frac{1}{2}\arg{\frac{z_2}{z_1}}$$ - Thankyou @lab bhattacharjee – Mykolas Sep 22 '12 at 11:29 This is simply an application of the Inscribed Angle Theorem. Since $$\frac{z_3-z_2}{z_3-z_1}=\frac{z_2-z_3}{z_1-z_3}$$ and the Inscribed Angle Theorem says that $$2\,\arg\left(\frac{z_2-z_3}{z_1-z_3}\right)=\arg\left(\frac{z_2}{z_1}\right)\pmod{2\pi}$$ $\hspace{3.5cm}$ we get that $$2\,\arg\left(\frac{z_3-z_2}{z_3-z_1}\right)=\arg\left(\frac{z_2}{z_1}\right)\pmod{2\pi}$$ which can be rewritten as $$\arg\left(\frac{z_3-z_2}{z_3-z_1}\right)=\frac12\arg\left(\frac{z_2}{z_1}\right)\pmod{\pi}$$ This is not exactly the same and points out that the relation is not always true. For example, suppose $z_1=\frac{1-i}{\sqrt{2}}$, $z_2=\frac{1+i}{\sqrt{2}}$, and $z_3=1$. Then $$\arg\left(\frac{z_3-z_2}{z_3-z_1}\right)=\frac54\pi$$ yet $$\frac12\arg\left(\frac{z_2}{z_1}\right)=\frac\pi4$$ - thankyou @robjon – Mykolas Sep 22 '12 at 18:56 Rob, what did you use to make the image? – Voyska Sep 23 '12 at 16:38 @GustavoBandeira: I used Intaglio, which is available only on Mac OS. – robjohn Sep 24 '12 at 22:03 Hint. Assume $|z| = 1$. $$\frac{\cos \alpha + i \sin \alpha}{\cos \beta + i \sin \beta} = (\cos \alpha + i \sin \alpha)(\cos \beta - i \sin \beta) = \cos(\alpha - \beta) + i\sin(\alpha - \beta)$$ $$\cos \alpha + i\sin \alpha - \cos \beta - i\sin \beta = -2 \sin(\frac{\alpha - \beta}{2})\sin(\frac{\alpha + \beta}{2}) + 2i\sin(\frac{\alpha - \beta}{2})\cos(\frac{\alpha + \beta}{2}) =\\ 2\sin(\frac{\alpha - \beta}{2})\Big(\cos(\frac{\pi}{2}+\frac{\alpha + \beta}{2}) + i\sin(\frac{\pi}{2}+\frac{\alpha + \beta}{2}) \Big)$$ Deduce how you can express your equation in terms of $\arg z_1$, $\arg z_2$ and $\arg z_3$. - thankyou @Karolis Juodelė – Mykolas Sep 22 '12 at 11:29 Note that $\arg{w_2-w_0\over w1-w_0}=\angle(w_1,w_0,w_2)$. With the help of a figure you can easily verify that the stated formula is an immediate consequence of the theorem about central and peripheral angles. -

2015-11-25 13:05:35

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