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>How to rigorously evaluate $L:=\lim\limits_{n \to \infty } n \int_0^\frac{\pi}{2}1- \sqrt[n]{\sin(t)}dt$?
I tried to use the beat function since $$B(x,y)= 2\int_0^\frac{\pi}{2}\sin^{2x-1}(t)\cos^{2y-1}(t)dt$$
$$\int_0^\frac{\pi}{2}\sqrt[n]{\sin(t)}dt= 0.5 B\left(\frac{1}{2n} + \frac{1}{2},\frac{1}{2}\right)= \frac{\Gamma\left(\frac{1}{2n} + \frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{2\Gamma\left(\frac{1}{2n} + 1\right)}=\frac{n\sqrt{\pi}\ \Gamma\left(\frac{1}{2n} + \frac{1}{2}\right)}{ \Gamma\left(\frac{1}{2n} \right)}$$
$$L =\lim\limits_{n \to \infty } n \left( \frac{\pi}{2} - \frac{n\sqrt{\pi}\ \Gamma\left(\frac{1}{2n} + \frac{1}{2}\right)}{ \Gamma\left(\frac{1}{2n} \right)} \right)$$
This limit is very hard to deal with, I tried to use Gauss product of the gamma function to try to simplify this mess but it didn't work.
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I noticed that if we define $f(x)=\int_0^\frac{\pi}{2}{\sin^x (t)}dt$ assuming $f'(0) $ exist at $0$
$$-f'(0)= \lim_{h \to 0 } \frac{\int_0^\frac{\pi}{2}{1-\sin^h (t)}dt }{h}= L$$
$$ \dfrac{d}{d x}\int_0^\frac{\pi}{2}{\sin^x (t)}dt=\int_0^\frac{\pi}{2} \dfrac{\partial}{\partial x} {\sin^x (t)}dt \ \ \ $$
Assuming we can interchange the derivative and the integral.
$$L = -\lim_{x \to 0}x \int_0^\frac{\pi}{2}{\sin^{x-1} (t)}dt $$
Assuming all of these steps are true and could be justified $ \lim\limits_{x \to 0}x \int_0^\frac{\pi}{2}{\sin^{x-1} (t)}dt $ is a hard limit and using the beta function will create the same mess as the first approach.
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How to rigorously evaluate $\lim\limits_{n \to \infty } n \int_0^\frac{\pi}{2}1- \sqrt[n]{\sin(t)}dt$?
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Is the starting value $k$ ensure convergence for almost all polynomials?
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Sobolev space on the unit sphere is usually defined with the help of spherical harmonics, but is there any reference where the equivalence to the classical fractional sobolev space definition is shown? And preferably also the construction of the classical fractional sobolev space.
I would greatly appreciate it, thanks.
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I have this question and it seems like I am stuck somewhere. Could someone help me with this? It involves parametric curves with the parameter of the curve in terms of $\pi$.
>Find the cartesian equation of the curve, sketch the curve and indicate with an arrow the direction in which the curve is traced as the parameter increases. [*Verify using Mathematica*]<br/>
i). $\;\;\;x = \sin \frac{1}{2}\theta, \;\; y = \cos \frac{1}{2}\theta, \;\; -\pi \leq \theta \leq \pi.$<br/>
ii). $\;\;x = \frac{1}{2}\cos \theta, \;\; y = 2 \sin \theta, \;\; 0 \leq \theta \leq \pi.$
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Let $f:\mathbb{R}^n_+\mapsto\mathbb{R}$ be concave and homogeneous of degree $\lambda\in(-\infty,0)\cup(0,1)$. Does the inequality $\lambda f(x)>0$ necessarily hold for all $x>0$? I suspect that this is the case, because I can't find a counter example. I know that for $\lambda=1$ the answer is NO as can be seen from the example $f(x_1,x_2)=\sqrt{x_1x_2}-x_2$. I was trying to find a reference that deals with this kind of question but to no avail.
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I am trying to understand if there is a way we can define the $S_n$ group, just like we do for $D_n$. We know that $D_n$ is defined as $D_n = <r,s | s^2 = e, r^n = e, srs = r^{-1}>$.
Can we do a similar definition for $S_n$>?
I know that a transposition $(a,b)$ and a n-cycle (1,2,3...,n) generate $S_n$ if and only if $gcd(b-a, n) = 1$.
Using this can we write something on the lines of
$S_n = <(a,b),r | |(a,b)| = 2, |r| = n, gcd(b-a,n) = 1>$?
I suppose we could add a relation among the two elements mentioned.
For example , for $S_4$, let (a,b) = (12) and r = (1234), then the relation
$r(a,b)r(a,b)r = (a,b)$ holds.
So, can I write $S_4 = <(a,b),r | |(a,b)| = 2, |r| = n, gcd(b-a,n) = 1, r(a,b)r(a,b)r = (a,b)>$.
Thank you for your help.
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Definition of Symmetric Group $\S_n$ like $D_n$?
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I am reading the paper [1]" P. S. Kenderov, I. S. Kortezov and W. B. Moors, Continuity points of quasi-continuous mappings, Topology Appl. 109 (2001), 321–346." I am confused with the statement (iii) of the Theorem 7 which says "For every minimal non-empty-valued mapping $F : Z \rightarrow X$ where Z is an $\alpha$-favorable space the set $\{ z: \tilde{F}(z)\subset X \}$ is of the second Baire category in any open subset of $Z$." But, in my opinion, they only proved that it is a subset of a second category in any open subset of $Z$. Also, this theorem is similar to Theorem 2 of the same paper, and there, they said that it is a subset of a set which is second category in any open subset of $Z$. Am I missing something? We know that a dense subset of a second category set may not be second category. So how are they claiming that it should be of second category?
[1]: https://core.ac.uk/download/pdf/81985595.pdf
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Is the statement of Theorem 7 is correct?
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I am trying to understand if there is a way we can define the $S_n$ group, just like we do for $D_n$. We know that $D_n$ is defined as $D_n =\langle r,s\mid s^2 = e, r^n = e, srs = r^{-1}\rangle$.
Can we do a similar definition for $S_n$?
I know that a transposition $(a,b)$ and a $n$-cycle $(1,2,3...,n)$ generate $S_n$ if and only if $\gcd(b-a, n) = 1$.
Using this, can we write something on the lines of
$S_n =\langle (a,b),r\mid|(a,b)| = 2, |r| = n,\gcd(b-a,n) = 1\rangle$?
I suppose we could add a relation among the two elements mentioned.
For example , for $S_4$, let $(a,b) = (12)$ and $r = (1234)$, then the relation
$r(a,b)r(a,b)r = (a,b)$ holds.
So, can I write $S_4 = \langle(a,b),r\mid|(a,b)| = 2, |r| = n,\gcd(b-a,n) = 1, r(a,b)r(a,b)r = (a,b)\rangle$?
Thank you for your help.
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Presentation of the symmetric roup $S_n$ like $D_n$?
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What are the uses of defining the derivative as $\lim \limits_{h\to 0} \frac{f(x+h)-f(x-h)}{2h}$?
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I was just messing with numbers when I created this formula.
But to understand the formula, you have to understand this: Multiplication is repeated addition, as $8*6$ is more or less $8+8+8+8+8+8$, or vice-versa. In the same way, exponents are repeated multiplication, as $5^3$, can be written as $5*5*5$, which can be written as $(5+5+5+5+5)*5$, which can be written as $5+5 25$ times, or $25+25+25+25+25$, which is $125$
For the mathematicians out there, you would know what is tetration. $3↑↑3$ is $7,625,597,484,987$.
This formula is for calculating by decimals.
$(x^\dfrac{a}{b}=\sqrt[b]{x^a})$ For example, if want to calculate what $10^{1.5}$, it would be $10*(\sqrt10)$
For $10*1.5$, it would be $10+(10/2)$
Do you see the pattern yet?
I want to test if this works for tetration, pentation, sextation, and so on. But I couldn't find a calculator capable of doing these. So can the answerer please add the link to one of these calcs? I prefer online, but if there is none online, then a downloadable one will also work.
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Motivation: In $\langle\mathbb{Z}/7\mathbb{Z},\times\rangle,\ \langle 3\rangle = (3,2,6,4,5,1).$
> Given a $k-$tuple of distinct integers, $q_1, q_2, \ldots, q_k,$ (all
> nonzero) does $\exists$ integers $a,m,n$ with $n>\max\{ q_1,q_2,\ldots,q_k\},\ $ such that $a m^i \equiv q_{i}
\pmod n\ \forall\ i\in \{1,2,\ldots, k\}\ $
> [In other words, $q_i$ is the remainder when $am^i$ is divided by $n.]\ ?$
Maybe there is a counter-example like $(100,99,98, 3,2,1).$ But I doubt it. I imagine it is true and we can even take $a=1,$ for example. I think it could be true by Chinese remainder theorem and/or Fermat's Little theorem?
It should be true for $k=2,$ so maybe then we can use induction on $k$?
I think that without the "nonzero" and "all distinct" conditions, there would be easy counter-examples.
Edit: I think $n$ must be prime for this to work, right? Well, I think that if $n$ is prime then every number other than $0$ and $1$ "generates" all of $\{1,2,\ldots,n-1\},$ although I don't recall the proof for this. But is it true that if n is not prime, then in $\langle\mathbb{Z}/n\mathbb{Z},\times\rangle,\ $ then for any $u,\ \langle u\rangle\ $ does not generate all of $\{1,2,\ldots,n-1\}\ ?$ Can someone link to proofs of these two theorems? I'm sure they are in some Elementary Number/Group Theory pdf. I also have the book, "Elementary Number Theory" by Springer, but I'm not sure if these theorems/proof are in there...
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Uses of the symmetric derivative $\lim \limits_{h\to 0} \frac{f(x+h)-f(x-h)}{2h}$?
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I am trying to understand the conditions for time series $\{y_t\}$ to be stationary, i.e.:
- $E(y_t)=\mu$ is constant for all $t >0$ and
- $Var(y_t) = V$ is constant for all $t >0.$
I came across a [youtube video by ritvikmath](https://www.youtube.com/watch?v=oY-j2Wof51c&ab_channel=ritvikmath) explaining this concept and it shows this example:
[![enter image description here][1]][1]
In 3:00 he says that the mean of plotted series is **constant** (but the series is not stationary because it displays seasonality, but I don't care about this part right now) - and I don't understand it completely.
If there were multiple time series, expected value of each variable $Y_t$ could be estimated as mean of $y_t$s. But how is it done if there is only one series? For each variable $Y_t$ we have only one datapoint, I don't understand how it's possible to know $E(y_t)$ here. Same for variance. In general, I don't understand how we check these two conditions in real life.
[1]: https://i.stack.imgur.com/G19Vbt.png
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Having trouble understanding this proof from Pugh's Real Mathematical Analysis 2nd Edition (p. 288).
Theorem. A function $f:U → \mathbb{R}^m$ is differentiable at $p ∈ U$ if and only if
each of its components $f_i$ is differentiable at $p$. Furthermore, the derivative of its $i^{th}$
component is the $i^{th}$ component of the derivative.
Proof. Assume that $f$ is differentiable at $p$ and express the $i^{th}$ component of $f$ as
$f_i$ = $\pi_if$ where $\pi_i : \mathbb{R}^m → \mathbb{R}$ is the projection that sends a vector $w = (w_1, . . . , w_m)$ to $w_i$. Since $\pi_i$ is linear it is differentiable. By the Chain Rule, $f_i$ is differentiable at $p$ and $(Df_i)_p = (D\pi_i) \circ (Df)_p = \pi_i \circ (Df)_p$.
(1) Is $\pi_if$ a composition of two functions (the lack of $\circ$ is confusing me)? If not, what is it?
(2) On the last equality, how does $(D\pi_i)$ become $\pi_i$? I understand this to mean that the derivative of a function, $\pi_i$, is equal to itself? I thought this was only possible for the exponential function?
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A 2-dimensional simplex is just a triangle, and a 3-dimensional simplex is just a tetrahedron… therefore for convenience, I simply use the term triangle and tetrahedron in the following words.
We know that an arbitrary (Euclidean) triangle is either right or oblique, and an oblique triangle is either acute or obtuse. In accordance with the remarkable [de Gua's theorem](https://en.wikipedia.org/wiki/De_Gua%27s_theorem "The square of the area of the base (i.e., the face opposite the right trihedron) of a trirectangular tetrahedron is equal to the sum of the squares of the areas of its other three faces. "), the three-dimensional analog of the right triangle is far and away the trirectangular tetrahedron. However, there does not appear to be an generally acknowledged three-dimensional analogue of the acute triangle as yet; different scholars adopt distinct extensions in their papers. For instance, according to [this article](https://zbmath.org/?q=an:1424.65225 "Simplices rarely contain their circumcenter in high dimensions"), [this article](https://zbmath.org/?q=an:1430.51027 "Duality of isosceles tetrahedra"), and [this article](http://arxiv.org/abs/1005.1033 "Random Gaussian Tetrahedra"), the following definitions (about a (non-degenerate) tetrahedron) are mentioned:
1. all [of] its dihedral angles are smaller than $\fracπ2$, or the orthogonal projection of each vertex onto the plane of the opposite face lies within that face;
2. it contains its circumcenter in the interior, or its vertices do not lie on a hemisphere of the circumscribed sphere;
3. the circumcenter of each face lies inside the face, or all facets of it are acute;
4. each of its four [solid angle](https://en.wikipedia.org/wiki/Solid_angle#Tetrahedron "trihedral angle")s' measure is less than $\fracπ2$.
It is noticeable that all angles of a triangle are smaller than $\frac\pi2$ iff its circumcenter lies inside the triangle, and hence both ⒈ and ⒉ can be viewed as practical generalizations of “acuteness” to three dimensions (although the two are not equivalent).
Still, analogizing those [conditions for being acute triangles](https://en.wikipedia.org/wiki/Acute_and_obtuse_triangles "necessary and sufficient conditions of acute triangles"), one may consider the following definitions (about a tetrahedron $ABCD$):
5. its [Monge point](https://mathworld.wolfram.com/MannheimsTheorem.html "Monge's Tetrahedron Theorem") $M$ (the reflection of the circumcenter $O$ in the centroid; alternatively, $\overrightarrow{OM}=\frac{\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}}2$) lies in the interior;
6. $S_{\triangle{ABC}}^2+S_{\triangle{ABD}}^2+S_{\triangle{ACD}}^2>S_{\triangle{BCD}}^2$ and three other similar inequalities (cf. the [law of cosines for the tetrahedron](https://en.wikipedia.org/wiki/Trigonometry_of_a_tetrahedron#Law_of_cosines_for_the_tetrahedron " denotes the area of a triangle."));
7. ${\left\lvert AB\right\rvert}^{-2}+{\left\lvert AC\right\rvert}^{-2}+{\left\lvert AD\right\rvert}^{-2}>h_A^{-2}$ and three other similar inequalities, where $h_A$ denotes the length of the altitude from vertex $A$;
8. the distance between $O$ and $M$ is less than the tetrahedron's circumradius;
9. et cetera.
All of them (perhaps apart from ⒊) can be viewed as the necessary and sufficient conditions that a tetrahedron is “acute”, but which condition is the most substantial generalization of that for the acute triangle? (Aren't there any convincing criteria?)
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What is the substantial generalization of the “acuteness” to 3-simplex?
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What this formula from econometrics textbook calculates?
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Let $f\in C^{2}[0,1]$ such that $f(0)=0,f(1/2)=f(1)=1$. Show that
$$\int_{0}^{1}[f''(x)]^2dx\ge 12.$$
My idea: try to find a function $f(x)$ such that the equality holds and use Cauchy-Schwarz. So I tried some "good" functions to achieve the minimum value.
1. Assume $f(x)=ax^3+bx^2+cx$ is a polynomial. Then I find that when $f(x)=-2x^2+3x$, the integral $\int_{0}^{1}[f''(x)]^2dx$ has minimum value $16$. It seems it doesn't help to solve the problem.
2. Since $f''(x)=(f(x)-ax-b)''$ and replace $f(x)$ by $f(x)-x$, we can change the condition to $f(0)=f(1)=0, f(1/2)=1/2$. I find that $f(x)=\frac{1}{2}\sin(\pi x)$ satisfies the condition and $\int_{0}^{1}[f''(x)]^2dx=\frac{\pi^2}{8}=12.1761...$ which is very closed to $12$. But it still doesn't help.
It seems that it does't easy to find such $f(x)$ such that the equality holds. So how to prove this inequality? And what is the minumum value of this integral? May be not 12? Thank you!
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How to prove this integral inequality $\int_{0}^{1}[f''(x)]^2dx\ge 12$?
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How to calculate Gal$(F(\mu_{p^\infty})/F(\mu_p))$ for a number field $F$?
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Let $f\in C^{2}[0,1]$ such that $f(0)=0,f(1/2)=f(1)=1$. Show that
$$\int_{0}^{1}[f''(x)]^2dx\ge 12.$$
My idea: try to find a function $f(x)$ such that the equality holds and use Cauchy-Schwarz. So I tried some "good" functions to achieve the minimum value.
1. Assume $f(x)=ax^3+bx^2+cx$ is a polynomial. Then I find that when $f(x)=-2x^2+3x$, the integral $\int_{0}^{1}[f''(x)]^2dx$ has minimum value $16$. It seems it doesn't help to solve the problem.
2. Since $f''(x)=(f(x)-ax-b)''$ and replace $f(x)$ by $f(x)-x$, we can change the condition to $f(0)=f(1)=0, f(1/2)=1/2$. I find that $f(x)=\frac{1}{2}\sin(\pi x)$ satisfies the condition and $\int_{0}^{1}[f''(x)]^2dx=\frac{\pi^2}{8}=12.1761...$ which is very closed to $12$. But it still doesn't help.
It seems that it isn't easy to find such $f(x)$ such that the equality holds. So how to prove this inequality? And what is the minumum value of this integral? May be not 12? Thank you!
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alright, this question is philosophical and somewhat fuzzy. i also admit to knowing little about logic. all in all, this question can possibly be easily resolved by either pointing to (perhaps even well-known) literature i haven’t found or by pointing out a fault in my reasoning.
joel david hamkins is a proponent of multiverse interpretation of ZFC, where we should see ZFC and other formalisations of the concept of sets as theories of not one universe of sets, but of a multiverse of sets, in which there are many “universes” of sets, .. or let’s call them *aliverses*.
the reason he gives is that since we can perfectly model – within ZFC – different theories ZFC, so one in which CH is true and one in which it is false, we *know* how such “places” look like, so to say that .. we are in only “one true universe” in which CH is either true or it is false and we just don’t know .. would be to disregard either of these “places” as unreal – even though we can readily visit them. instead, he regards these “places” as just aliverses of a multiverse of sets.
that position is known as a *pluralist view* and i find his arguement compelling. but that’s not quite the way we think of set theory, is it? of course, prof. hamkins would propose we adapt our way of thinking here, but then, when doing any mathematics founded on set theory, we would never be in “one definite” of these aliverses, but rather always in “some arbitrary” aliverse, make an argument in them and arrive at statements, which by the arbitrariness of the aliverse, holds in any of them. just as when we're doing group theory, we are never in one definite group, but rather awalys in some arbitrary group. in fact, just as we prepend “let $G$ be a group” before argueing in some arbitrary group, we should prepend “let $U$ be a set-theoretic universe fulfilling ZFC” before doing any mathematics founded on set-theory at all.
however, i still think of sets as belonging to *one universe of sets*, just as i think of natural numbers as belonging to *one structure of natural numbers*. there may be, in some abstract sense, multipile competing definite concepts of sets or natural numbers, but i can’t ever quite get to know them. rather i accept that the one concept of sets and the one concept of natural numbers that i have in mind are amorphous and somewhat fuzzy. taking that seriously, i would have to say that the independency of CH from ZFC, which i regard to be a fitting formalisation of my amorphous concept of sets, tells me that, with respect to my concept of sets, CH *just simply is neither true nor false*, that is to say: *classical logic is not appropriate*.
ok, so i’m probably not the first to think of this. but i haven’t seen this argument so far. so i have to wonder: am i missing something? doesn’t the independency phenomenon make a case for non-classical logics?
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doesn't the independency phenomenon make a case for non-classical logic?
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What does this formula (from an econometrics textbook) calculate?
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alright, this question is philosophical and somewhat fuzzy. i also admit to knowing little about logic. all in all, this question can possibly be easily resolved by either pointing to (perhaps even well-known) literature i haven’t found or by pointing out a fault in my reasoning.
joel david hamkins is a proponent of multiverse interpretation of ZFC, where we should see ZFC and other formalisations of the concept of sets as theories of not one universe of sets, but of a multiverse of sets, in which there are many “universes” of sets, .. or let’s call them *aliverses*.
the reason he gives is that since we can perfectly model – within ZFC – different theories ZFC, so one in which CH is true and one in which it is false, we *know* how such “places” look like, so to say that .. we are in only “one true universe” in which CH is either true or it is false and we just don’t know .. would be to disregard either of these “places” as unreal – even though we can readily visit them. instead, he regards these “places” as just aliverses of a multiverse of sets.
that position is known as a *pluralist view* and i find his arguement compelling. but that’s not quite the way we think of set theory, is it? of course, prof. hamkins would propose we adapt our way of thinking here, but then, when doing any mathematics founded on set theory, we would never be in “one definite” of these aliverses, but rather always in “some arbitrary” aliverse, make an argument in them and arrive at statements, which by the arbitrariness of the aliverse, holds in any of them. just as when we're doing group theory, we are never in one definite group, but rather awalys in some arbitrary group. in fact, just as we prepend “let $G$ be a group” before argueing in some arbitrary group, we should prepend “let $U$ be a set-theoretic universe fulfilling ZFC” before doing any mathematics founded on set-theory at all.
however, i still think of sets as belonging to *one universe of sets*, just as i think of natural numbers as belonging to *one structure of natural numbers*. there may be, in some abstract sense, multipile competing definite concepts of sets or natural numbers, but i can’t ever quite get to know them. rather i accept that the one concept of sets and the one concept of natural numbers that i have in mind are amorphous and somewhat fuzzy. taking that seriously, i would have to say that the independency of CH from ZFC and then from ZF, which i regard to be a fitting formalisation of my amorphous concept of sets, tells me that, with respect to my concept of sets, CH *just simply is neither true nor false*, that is to say: *classical logic is not appropriate*.
ok, so i’m probably not the first to think of this. but i haven’t seen this argument so far. so i have to wonder: am i missing something? doesn’t the independency phenomenon make a case for non-classical logics?
*remark*. i realize that the axiom of choice implies LEM within ZF from say intuitionistic logic. so i’m ready give up full choice in favor of dependent choice in order to drop LEM.
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I'm looking at a problem I believe is combinatorial.
Find **all** possible solutions $\mathbf{x}$ to:
$$\mathbf{a} = [a_1, a_2, ..., a_n], a_k \in \mathbb{N^+}$$
$$\mathbf{l} = [l_1, l_2, ..., l_n], l_k \in \mathbb{N}$$
$$\sum a_k \cdot x_k = B, x_k \le l_k, x_k \in \mathbb{N}$$
$\mathbf{a}$, $\mathbf{l}$ (limit) and $B$ are given constants.
A few trivial examples would be:
$$\mathbf{a}=[2, 3, 5, 7], \mathbf{l}=[2,1,0,1]$$
$B=15$ gives no solutions, $\mathbf{x} \in []$
$B=10$ gives 1 solution, $\mathbf{x} \in [[0,1,0,1]]$
$B=5$ gives 2 solutions, $\mathbf{x} \in [[1,1,0,0], [0,0,1,0]]$
$$\mathbf{a}=[2, 3, 5, 7], \mathbf{l}=[3,3,3,3]$$
$B=15$ gives 4 solutions, $\mathbf{x} \in [[0, 0, 3, 0], [0, 1, 1, 1], [1, 1, 2, 0], [3, 3, 0, 0]]$
$B=10$ gives 3 solutions, $\mathbf{x} \in [[0, 0, 2, 0], [1, 1, 1, 0], [2, 2, 0, 0]]$
$B=5$ gives 2 solutions, $\mathbf{x} \in [[1,1,0,0], [0,0,1,0]]$
NOTE: I solved above equations by hand so please keep it in mind in case I missed something. In this example the 4 first entries from the prime numbers. This is not a feature of the problem but just to give an example. The numbers are however unique in $\mathbf{a}$ but it could equally well be: $\mathbf{a}=[1,2,7,10,22], \mathbf{l}=[...], B=123$
In the problem domain where I want to apply this the dimensionality of $\mathbf{a}$ will be around 25-30 or so, but it is the same formulation of the problem as above.
Does anyone know if there exists efficient solver(s) for this, preferably without licenses, or do I have to roll my own?
If there are, do you know if it is possible to run them on i.e. pause/resume assuming you need to run it in batches?
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alright, this question is philosophical and somewhat fuzzy. i also admit to knowing little about logic. all in all, this question can possibly be easily resolved by either pointing to (perhaps even well-known) literature i haven’t found or by pointing out a fault in my reasoning.
joel david hamkins is a proponent of multiverse interpretation of ZFC, where we should see ZFC and other formalisations of the concept of sets as theories of not one universe of sets, but of a multiverse of sets, in which there are many “universes” of sets, .. or let’s call them *aliverses*.
the reason he gives is that since we can perfectly model – within ZFC – different theories ZFC, so one in which CH is true and one in which it is false, we *know* how such “places” look like, so to say that .. we are in only “one true universe” in which CH is either true or it is false and we just don’t know .. would be to disregard either of these “places” as unreal – even though we can readily visit them. instead, he regards these “places” as just aliverses of a multiverse of sets.
that position is known as a *pluralist view* and i find his arguement compelling. but that’s not quite the way we think of set theory, is it? of course, prof. hamkins would propose we adapt our way of thinking here, but then, when doing any mathematics founded on set theory, we would never be in “one definite” of these aliverses, but rather always in “some arbitrary” aliverse, make an argument in them and arrive at statements, which by the arbitrariness of the aliverse, holds in any of them. just as when we're doing group theory, we are never in one definite group, but rather awalys in some arbitrary group. in fact, just as we prepend “let $G$ be a group” before arguing in some arbitrary group, we should prepend “let $U$ be a set-theoretic universe fulfilling ZFC” before doing any mathematics founded on set-theory at all.
however, i still think of sets as belonging to *one universe of sets*, just as i think of natural numbers as belonging to *one structure of natural numbers*. there may be, in some abstract sense, multipile competing definite concepts of sets or natural numbers, but i can’t ever quite get to know them. rather i accept that the one concept of sets and the one concept of natural numbers that i have in mind are perhaps *amorphous and vague*. taking that seriously, i would have to say that the independency of CH from ZFC and then from ZF, which i regard to be a fitting formalisation of my amorphous concept of sets, tells me that, with respect to my concept of sets, CH *just simply is neither true nor false*, that is to say: *classical logic is not appropriate*.
ok, so i’m probably not the first to think of this. but i haven’t seen this argument so far. so i have to wonder: am i missing something? doesn’t the independency phenomenon make a case for non-classical logics?
*remark*. i realize that the axiom of choice implies LEM within ZF from say intuitionistic logic. so i’m ready give up full choice in favor of dependent choice in order to drop LEM.
|
i have been working with generating functions and i am fed up with always doing the partial fraction decomposition. i have found the following trick that can (hopefully) skip
pfd.
$$
\frac{z^j}{(1-z)^k} \iff \sum_{n\geq 0} \binom{n+k-1-j}{k-1} \cdot z^n
$$
i would like to prove it, however i seem to lack the skills to do so. can anyone confirm whether the proposed trick is actually valid?
thanks in advance
|
It is well known that handles of indices k and (k+1) may be cancelled if the attaching sphere of (k+1)-handle *intersect* the belt sphere of k-handle *at exaclty one point*. Is it also necessary condition for the cancelling? In Rourke and Sanderson book I found that for dimension $n\geq 6$ and for sufficiently small k the requirement of the uniqeness of the point of the intersection may be replaced by more weak condition that the belt and attached spheres have only intersection number equals to one. But since isotopic attaching maps give homeomorhic manifolds, it seems to very plausible that we can slightly move an attaching map to get more that one intersection points, but keep the manifold in any dimensions...
|
alright, this question is philosophical and somewhat fuzzy. i also admit to knowing little about logic. all in all, this question can possibly be easily resolved by either pointing to (perhaps even well-known) literature i haven’t found or by pointing out a fault in my reasoning.
joel david hamkins is a proponent of multiverse interpretation of set theory, where we should see ZFC and other formalisations of the concept of sets as theories of not one universe of sets, but of a multiverse of sets, in which there are many “universes” of sets, .. or let’s call them *aliverses*.
the reason he gives is that since we can perfectly model – within ZFC – different theories ZFC, so one in which CH is true and one in which it is false, we *know* how such “places” look like, so to say that .. we are in only “one true universe” in which CH is either true or it is false and we just don’t know .. would be to disregard either of these “places” as unreal – even though we can readily visit them. instead, he regards these “places” as just aliverses of a multiverse of sets.
that position is known as a *pluralist view* and i find his arguement compelling. but that’s not quite the way we think of set theory, is it? of course, prof. hamkins would propose we adapt our way of thinking here, but then, when doing any mathematics founded on set theory, we would never be in “one definite” of these aliverses, but rather always in “some arbitrary” aliverse, make an argument in them and arrive at statements, which by the arbitrariness of the aliverse, holds in any of them. just as when we're doing group theory, we are never in one definite group, but rather awalys in some arbitrary group. in fact, just as we prepend “let $G$ be a group” before arguing in some arbitrary group, we should prepend “let $U$ be a set-theoretic universe fulfilling ZFC” before doing any mathematics founded on set-theory at all.
however, i still think of sets as belonging to *one universe of sets*, just as i think of natural numbers as belonging to *one structure of natural numbers*. there may be, in some abstract sense, multipile competing definite concepts of sets or natural numbers, but i can’t ever quite get to know them. rather i accept that the one concept of sets and the one concept of natural numbers that i have in mind are perhaps *amorphous and vague*. taking that seriously, i would have to say that the independency of CH from ZFC and then from ZF, which i regard to be a fitting formalisation of my amorphous concept of sets, tells me that, with respect to my concept of sets, CH *just simply is neither true nor false*, that is to say: *classical logic is not appropriate*.
ok, so i’m probably not the first to think of this. but i haven’t seen this argument so far. so i have to wonder: am i missing something? doesn’t the independency phenomenon make a case for non-classical logics?
*remark*. i realize that the axiom of choice implies LEM within ZF from say intuitionistic logic. so i’m ready give up full choice in favor of dependent choice in order to drop LEM.
|
I have the following basic question regarding continuous functions.
Is the following statement correct?
> Given functions
>
> \begin{align*} h & : A \times C \to B \times D ,\\ f &: A \to B ,\\ g &: C
\to D , \end{align*}
>
> if $h$ and $f$ are continuous, then $g$ is continuous.
I know that if $f$ and $g$ are continuous, then $h$ is continuous, while if $h$ is continuous it is not necessarily the case that both $f$ and $g$ are continuous. What about this 'intermediate' case?
Thanks a lot in advance for any feedback.
|
I'm trying to calculate this integral using residue theorem:
$$
\int_{-\infty}^{\infty}\text{d}\omega\frac{f(\omega)}{(\omega+a+i\epsilon)(\omega-a-i\epsilon)}
$$
where $f(\omega)$ is bounded in the whole complex plane and $f(a+i\epsilon)\neq f(-a-i\epsilon)$.
It seems to me that the integrand is proportional to $\frac{1}{\omega^2}$, so it should be convergent at infinity in both upper complex plane and lower complex plane. But if I close the contour in the upper half plane, I got
$$
i\pi\frac{f(a+i\epsilon)}{a+i\epsilon}
$$
and when I close the contour in the lower half plane, I got
$$
i\pi\frac{f(-a-i\epsilon)}{a+i\epsilon}
$$
My question is why these two results do not coincide?
|
I have an integral which I have used contour integration to calculate it. But it seems that I have some contradiction here?
|
Let $A$,$B$ be two positive semidefinite symmetric matrices on $\mathbb{R}^d$ with the same eigenvectors. Let $P$ be an orthogonal projection. Does it hold that
$$
\|APB\| \leq \|AB\|,
$$
where $\|\cdot\|$ denotes the operator norm. If $P$ was on the left or right we would directly have $\|ABP\| \leq \|AB\|\|P\| \leq \|AB\|$. However here, trying to show it from the definition of the operator norm does not help me to conclude as for all $x$, we have
$$
\|APBx\|^2 = x^{\top}BPA^2PBx,
$$
and I don't know how to drop $P$ from this expression.
Does it still hold if we drop the assumption that $A$,$B$ share the same eigenvectors?
|
I have the following basic question regarding continuous functions.
Is the following statement correct?
> Given functions
>
> \begin{align*} h & : A \times C \to B \times D ,\\ f &: A \to B ,\\ g &: C
\to D , \end{align*}
>
> with $h := (f, g)$, if $h$ and $f$ are continuous, then $g$ is continuous.
I know that if $f$ and $g$ are continuous, then $h$ is continuous, while if $h$ is continuous it is not necessarily the case that both $f$ and $g$ are continuous. What about this 'intermediate' case?
Thanks a lot in advance for any feedback.
|
I'm looking at a problem I believe is combinatorial.
Find **all** possible solutions $\mathbf{x}$ to:
$$\mathbf{a} = [a_1, a_2, ..., a_n], a_k \in \mathbb{N^+}$$
$$\mathbf{l} = [l_1, l_2, ..., l_n], l_k \in \mathbb{N}$$
$$\sum a_k \cdot x_k = B, x_k \le l_k, x_k \in \mathbb{N}$$
$\mathbf{a}$, $\mathbf{l}$ (limit) and $B$ are given constants.
A few trivial examples would be:
Example 1)
$$\mathbf{a}=[2, 3, 5, 7], \mathbf{l}=[2,1,0,1]$$
$B=15$ gives no solutions, $\mathbf{x} \in []$
$B=10$ gives 1 solution, $\mathbf{x} \in [[0,1,0,1]]$
$B=5$ gives 1 solution, $\mathbf{x} \in [[1,1,0,0]]$
Example 2)
$$\mathbf{a}=[2, 3, 5, 7], \mathbf{l}=[3,3,3,3]$$
$B=15$ gives 4 solutions, $\mathbf{x} \in [[0, 0, 3, 0], [0, 1, 1, 1], [1, 1, 2, 0], [3, 3, 0, 0]]$
$B=10$ gives 3 solutions, $\mathbf{x} \in [[0, 0, 2, 0], [1, 1, 1, 0], [2, 2, 0, 0]]$
$B=5$ gives 2 solutions, $\mathbf{x} \in [[1,1,0,0], [0,0,1,0]]$
NOTE: I solved above equations by hand so please keep it in mind in case I missed something. In the examples $\mathbf{a}$ contains the 4 first entries from the prime numbers. This is not a feature of the problem but just to give an example. The numbers are however unique in $\mathbf{a}$ but it could equally well be: $\mathbf{a}=[1,2,7,10,22], \mathbf{l}=[...], B=123$
In the problem domain where I want to apply this the dimensionality of $\mathbf{a}$ will be around 25-30 or so, but it is the same formulation of the problem as above.
Does anyone know if there exists efficient solver(s) for this, preferably without licenses, or do I have to roll my own?
If there are, do you know if it is possible to run them on i.e. pause/resume assuming you need to run it in batches?
|
Is the statement of Theorem 7 is correct? The points of some subset which is of the second Baire category in every non-empty open subset of Z means?
|
Let $A$ be a Noetherian commutative ring. Suppose that there exists a maximal ideal $\mathfrak{m}$ of $A$ such that there exists an isomorphism $A/\mathfrak{m}\cong A_{\mathfrak{m}}$ of rings where $A_{\mathfrak{m}}$ is the localization of $A$ at $\mathfrak{m}$. **Is it true that $\mathfrak{m}$ is the zero ideal? If not, can additional conditions be added to make it true?**
|
When do we have $A/\mathfrak{m}\cong A_{\mathfrak{m}}$?
|
I'm looking at a problem I believe is combinatorial.
Find **all** possible solutions $\mathbf{x}$ to:
$$\mathbf{a} = [a_1, a_2, ..., a_n], a_k \in \mathbb{N^+}$$
$$\mathbf{l} = [l_1, l_2, ..., l_n], l_k \in \mathbb{N}$$
$$\sum a_k \cdot x_k = B, x_k \le l_k, x_k \in \mathbb{N}$$
$\mathbf{a}$, $\mathbf{l}$ (limit) and $B$ are given constants.
A few trivial examples would be:
Example 1)
$$\mathbf{a}=[2, 3, 5, 7], \mathbf{l}=[2,1,0,1]$$
$B=15$ gives no solutions, $\mathbf{x} \in []$
$B=10$ gives 1 solution, $\mathbf{x} \in [[0,1,0,1]]$
$B=5$ gives 1 solution, $\mathbf{x} \in [[1,1,0,0]]$
Example 2)
$$\mathbf{a}=[2, 3, 5, 7], \mathbf{l}=[3,3,3,3]$$
$B=15$ gives 4 solutions, $\mathbf{x} \in [[0, 0, 3, 0], [0, 1, 1, 1], [1, 1, 2, 0], [3, 3, 0, 0]]$
$B=10$ gives 3 solutions, $\mathbf{x} \in [[0, 0, 2, 0], [1, 1, 1, 0], [2, 2, 0, 0]]$
$B=5$ gives 2 solutions, $\mathbf{x} \in [[1,1,0,0], [0,0,1,0]]$
NOTE: I solved above equations by hand so please keep it in mind in case I missed something. In the examples $\mathbf{a}$ contains the 4 first entries from the prime numbers. This is not a feature of the problem but just to give an example. The numbers are however unique in $\mathbf{a}$ but it could equally well be: $\mathbf{a}=[1,2,7,10,22], \mathbf{l}=[...], B=123$
In the problem domain where I want to apply this the dimensionality of $\mathbf{a}$ will be around 25-30 or so, but it is the same formulation of the problem as above.
Does anyone know if efficient solver(s) exists for this, preferably without licenses, or do I have to roll my own?
|
Presentation of the symmetric group $S_n$ like $D_n$?
|
I am reading Rudin's "Principles of Mathematical Analysis" and I am stuck at the following theorem
> Theorem 8.14:
>
> If, for some $x$, there are constants $\delta > 0$ and $M < \infty$ such that:
> $$\left| f(x + t) - f(x) \right| \leq M|t|$$
>
> for all $t \in (-\delta, \delta)$, then
> $$\lim_{N\rightarrow\infty} s_N(f; x) = f(x)$$
A few Definitions that needed:
> Dirichlet kernel:
> $$D_N(x) = \sum_{n=-N}^Ne^{inx} = \frac{\sin\left( (N+\frac12)x \right)}{\sin(x/2)} \qquad (77)$$
> $$s_N(f; x) = \frac{1}{2\pi}\int_{-\pi}^{\pi} f(x - t) D_N(t) dt \qquad (78)$$
And the proof going as follows:
> Define $g(t) = \frac{f(x-t) - f(x)}{\sin(t/2)}$
>
> for $0 < |t| \leq \pi$, and put $g(0) = 0)$. By the definition (77),
> $$\frac{1}{2\pi} \int_{-\pi}^{\pi}D_N(x) dx = 1$$
>
> Hence (78) shows that:
> $$s_N(f; x) - f(x) = \frac{1}{2\pi} \int_{-\pi}^{\pi} g(t) \sin\left((N +\frac12)t\right)$$
> $$= \frac{1}{2\pi} \int_{-\pi}^{\pi} \left[ g(t)\cos\frac{t}2 \right]\sin(Nt) \space dt + \frac{1}{2\pi} \int_{-\pi}^{\pi} \left[ g(t)\sin\frac{t}2 \right]\cos(Nt) \space dt$$
>
> $g(t)\cos(t/2)$ and $g(t)\sin(t/2)$ are bounded. The last two integrals thus tend to $0$ as $N \rightarrow \infty$.
Can anyone explain why they tend to $0$?
|
Numbers are represented using binary digits "bits" which are integers in the set $\{0,1\}$.
Positive integers for example, are usually encoded as a sequence of bits so that which power of 2 they encode correspond to the order which they are stored. For example the number $a$ represented by a binary vector $$[b_{N-1},\cdots,b_1,b_0]$$:
$$a = \sum_{i=0}^{N-1} b_k \cdot 2^k$$
Now if we know beforehand the range of numbers to be stored, $[0,a_{max}]$. In many cases $a_{max}$ may not be $2^k-1$ for some $k$. In other words... to be able to store any such number we will have to select a number of bits so that a certain set of combination of bits will be unused. From a data compression stand-point, this is unfortunate. Especially bothersome it can be if $a_{max}$ is a small number and we have very many numbers to store.
Take for example $a_{max}$ = 2, this will require us to use $2$ bits, although 1/4 for each such pair of bits of what we can represent will go unused (the number 3 represented by the two bits 11 never occurs).
How can we in this case find a way to compress a set of numbers in the range $\{ 0,\cdots,a_{max} \}$ so that we "on average" (in some sense) don't need more bits than theoretically necessary?
|
How can I design a non-integral-power efficient range coding that is statistically "fair" over the whole range?
|
I'm looking at a problem I believe is combinatorial.
Find **all** possible solutions $\mathbf{x}$ to:
$$\mathbf{a} = [a_1, a_2, ..., a_n], a_k \in \mathbb{N^+}$$
$$\mathbf{l} = [l_1, l_2, ..., l_n], l_k \in \mathbb{N}$$
$$\sum a_k \cdot x_k = B, x_k \le l_k, x_k \in \mathbb{N}$$
$\mathbf{a}$, $\mathbf{l}$ (limit) and $B$ are given constants.
A few trivial examples would be:
Example 1)
$$\mathbf{a}=[2, 3, 5, 7], \mathbf{l}=[2,1,0,1]$$
$B=15$ gives no solutions, $\mathbf{x} \in []$
$B=10$ gives 1 solution, $\mathbf{x} \in [[0,1,0,1]]$
$B=5$ gives 1 solution, $\mathbf{x} \in [[1,1,0,0]]$
Example 2)
$$\mathbf{a}=[2, 3, 5, 7], \mathbf{l}=[3,3,3,3]$$
$B=15$ gives 4 solutions, $\mathbf{x} \in [[0, 0, 3, 0], [0, 1, 1, 1], [1, 1, 2, 0], [3, 3, 0, 0]]$
$B=10$ gives 3 solutions, $\mathbf{x} \in [[0, 0, 2, 0], [1, 1, 1, 0], [2, 2, 0, 0]]$
$B=5$ gives 2 solutions, $\mathbf{x} \in [[1,1,0,0], [0,0,1,0]]$
NOTE: I solved above equations by hand so please keep it in mind in case I missed something. In the examples $\mathbf{a}$ contains the 4 first entries from the prime numbers. This is not a feature of the problem but just to give an example. The numbers are however unique in $\mathbf{a}$ but it could equally well be: $\mathbf{a}=[1,2,7,10,22], \mathbf{l}=[...], B=123$
In the problem domain where I want to apply this the dimensionality of $\mathbf{a}$ will be around 25-30 or so, but it is the same formulation of the problem as above.
Does anyone know if efficient solver(s) exists for this, or do I have to roll my own?
|
Doing my homework on functional analysis I faced with the following problem.
Let $\displaystyle (Uf) (s)=\int\limits_{-1}^{1}\frac{f(t)}{|s-t|^{5/6}}dt$.
Using Young inequality it can be shown that $U$ is bounded operator from $L^3(-1,1)$ to $L^r(-1,1)$ with $r<6$ and it is compact. If $r>6$ then it is easy to see (by considering functions of the form $f(t)=1/t^\beta$) that $U$ is not bounded from $L^3(-1,1)$ to $L^r(-1,1)$. But I don't know how to investigate the case when $r=6$.
Is $U$ bounded or even compact from $L^3(-1,1)$ to $L^6(-1,1)$?
|
I want to find the strong form of Euler-Lagrange equation of the following variational problem( $\partial\Omega$ smooth ):
$$\min_{u\in H^2(\Omega)} I[u]:=\int_\Omega (\partial_x^2u)^2 + 2(\partial_x\partial_yu)^2+(\partial_y^2u)^2 \,dx\,dy$$
I can get the weak form:
$$\forall v\in H^2(\Omega),\int_\Omega \partial_x^2u\ \partial_x^2v + 2\partial_x\partial_y u\ \partial_x\partial_y v + \partial_y^2u\ \partial_y^2 v \,dx\,dy=0,$$
i.e. $$\int_\Omega \nabla^2 u:\nabla^2v\,dx\,dy=0.$$
By integration by parts,
$$-\int_\Omega div(\nabla^2 u)\cdot \nabla v\,dx\,dy+\int_{\partial\Omega} (\nabla^2u\nabla v)\cdot n\,d\sigma=0$$
($n$ is outward unit vector, and we assume $\nabla v$ is a column vector)
$$\Rightarrow \int_\Omega \Delta^2 u\ v\,dx\,dy - \int_{\partial\Omega} div(\nabla^2 u)\cdot n\ v\,d\sigma+\int_{\partial\Omega} (\nabla^2u\nabla v)\cdot n\,d\sigma=0.$$
Take $v|_\Omega=0,\nabla v|_\Omega=0$ we get $\Delta^2 u=0$; Similarly we get:
$$div(\nabla^2 u)\cdot n=\partial_n(\Delta u)=0\ on\ \partial\Omega.$$
But the last boundary condition is confusing me. What can be implied from "for all $v\in H^2(\Omega),$"
$$\int_{\partial\Omega} (\nabla^2u\nabla v)\cdot n\,d\sigma=0?$$
Apparently it shouldn't be "$\nabla^2 u\ n=0$",since it contains 2 equations, and it times $\nabla v$ which is a gradient instead of an arbituary vector field. Can anyone help with the last boundary condition? Thanks!
|
For a strongly regular simple undirected graph to be disconnected, does each of its component has to be complete (aka $\mu=0$)?
|
You can split the integral in to two separate ones
$$
\int_0^{\frac12} (f''(x))^2 \, \mathrm{d}x+\int_{\frac12}^{1} (f''(x))^2 \, \mathrm{d}x
$$
and find the smallest function for both using calculus of variations and combine them piecewise. The reason that you have to do this is because calculus of variations only makes statements about the boundary.
The solution is
$$
f(x)=
\begin{cases}
-2x^3+\frac52 x& x \in [0,\frac12]\\
2x^3-6x^2+\frac{11}2x -\frac12& x \in (\frac12,1].
\end{cases}
$$
This achieves the lower bound of 12.
From there on it is indeed Cauchy-Schwartz, similar to this https://math.stackexchange.com/questions/499416/how-prove-this-int-abfx2dx-ge-dfrac4b-a?rq=1 .
|
I am reading the [Paper][1] " P. S. Kenderov, I. S. Kortezov and W. B. Moors, Continuity points of quasi-continuous mappings, Topology Appl. 109 (2001), 321–346." Just before the Theorem 2 of the paper, the authors of the paper mentioned "The set of points of continuity $C(f)$ is not necessarily residual in Z. It is however of the second Baire category in every non-empty open subset of Z. I.e., for every non-empty open subset $V \subset Z$ the set $C(f )\cap V$ is not of the first Baire category." Which they proved in the theorem, but I am not getting it. I can see that there exists a first category set H such that for any open set V, $C(f )\cap V\subset V\setminus H$. What am I missing?
[1]: https://core.ac.uk/download/pdf/81985595.pdf
|
How the set $C(f)\cap V$ is a second category in $V$?
|
How to proof a sigma algebra is equal to the union of two given subsets?
|
I cannot give a satisfying answer, but I had a look in Neustrev's book which convinced me that the author's exposition in not really clear.
Quote from Chapter 3:
> Here we give a detailed answer to the following fundamental question: Given an abstract $\mathbb R$-algebra $\mathcal F$, find a set (smooth manifold) $M$ whose $\mathbb R$-algebra of (smooth) functions can be identified with $\mathcal F$.
I guess he considers two questions:
1. Find a **set** $M$ such that $\mathcal F(M) =$ algebra of functions $f : M \to \mathbb R$ can be identified with $\mathcal F$.
2. Find a **smooth manifold** $M$ such that $C^\infty(M)$ can be identified with $\mathcal F$.
In the first case one can regard $M$ as a discrete topological space so that $\mathcal F(M) = C^0(M) =$ algebra of continuous functions $f : M \to \mathbb R$. We may also regard $M$ as $0$-dimensional smooth manifold, at least if $M$ is countable.
Neustrev's approach to smooth manifold is a bit unusual. He introduces them by an algebraic definiton (Chapter 4). In Chapter 5 he gives the "standard" definition (5.8. Coordinate definition of manifolds).
Now consider Example 3.2. The author asks
> Can this algebra $\mathcal F$ be realized as an algebra of nice functions on some set $M$?
His answer is
> Thus any sequence $\{a_i\} \in \mathcal F$ may be viewed as the function on $M = \mathbb R$ given by $x \mapsto \sum_{i=0}^\infty a_ix^i$.
This is **false**. It can be regarded as an element of the polynomial algebra $\mathbb R[x]$, but in general not as function $\mathbb R \to \mathbb R$ because $\sum_{i=0}^\infty a_ix^i$ does in general not convergence (only for $x = 0$ we can be sure).
But even if it were true, we would only get an isomorphism between $\mathcal F$ and the subalgebra of $C^\infty(\mathbb R)$.
In Exercise 3.3 the answer is clear for $\mathcal F_1$. Take a two-point set $M = \{a, b\}$. Then $\mathcal F_1 \approx \mathcal F(M)$. Concerning $\mathcal F_2$ : I did not check whether multiplication is associative and distributive, but let us assume it (you can do the checkings). I think you wanted to have the complex multiplication here, but also other multiplications producing an algebra are legit.
In 3.4 and 3.5 a general method is developped how to find candidates for solving the general problem. In Example 3.6 he comes back to $\mathcal F_2$ and shows that there is no solution.
|
I cannot give a satisfying answer, but I had a look in Neustrev's book which convinced me that the author's exposition in not really clear.
Quote from Chapter 3:
> Here we give a detailed answer to the following fundamental question: Given an abstract $\mathbb R$-algebra $\mathcal F$, find a set (smooth manifold) $M$ whose $\mathbb R$-algebra of (smooth) functions can be identified with $\mathcal F$.
I guess he considers two questions:
1. Find a **set** $M$ such that $\mathcal F(M) =$ algebra of functions $f : M \to \mathbb R$ can be identified with $\mathcal F$.
2. Find a **smooth manifold** $M$ such that $C^\infty(M)$ can be identified with $\mathcal F$.
In the first case one can regard $M$ as a discrete topological space so that $\mathcal F(M) = C^0(M) =$ algebra of continuous functions $f : M \to \mathbb R$. We may also regard $M$ as $0$-dimensional smooth manifold, at least if $M$ is countable.
Neustrev's approach to smooth manifold is a bit unusual. He introduces them by an algebraic definiton (Chapter 4). In Chapter 5 he gives the "standard" definition (5.8. Coordinate definition of manifolds).
Now consider Example 3.2. The author asks
> Can this algebra $\mathcal F$ be realized as an algebra of nice functions on some set $M$?
His answer is
> Thus any sequence $\{a_i\} \in \mathcal F$ may be viewed as the function on $M = \mathbb R$ given by $x \mapsto \sum_{i=0}^\infty a_ix^i$.
This is **false**. It can be regarded as an element of the polynomial algebra $\mathbb R[x]$, but in general not as function $\mathbb R \to \mathbb R$ because $\sum_{i=0}^\infty a_ix^i$ does in general not convergence (only for $x = 0$ we can be sure).
But even if it were true, we would only get an isomorphism between $\mathcal F$ and a subalgebra of $C^\infty(\mathbb R)$. But perhaps one could accept it as a positive answer in a generalized sense to "Can this algebra $\mathcal F$ be realized as an algebra of nice functions on some set $M$?"
In Exercise 3.3 the answer is clear for $\mathcal F_1$. Take a two-point set $M = \{a, b\}$. Then $\mathcal F_1 \approx \mathcal F(M)$. Concerning $\mathcal F_2$ : I did not check whether multiplication is associative and distributive, but let us assume it (you can do the checkings). I think you wanted to have the complex multiplication here, but also other multiplications producing an algebra are legit.
In 3.4 and 3.5 a general method is developped how to find candidates for solving the general problem. In Example 3.6 he comes back to $\mathcal F_2$ and shows that there is no solution.
|
The Constant Elasticity of Variance (CEV) process is a one dimensional diffusion process given by the following stochastic differential equation.
\begin{equation}
d X_t = \mu X_t \cdot dt + \sigma X_t^\beta \cdot d B_t \tag{1}
\end{equation}
where $\mu,\sigma,\beta$ are positive real parameters and $B_t$ is the Brownian motion.
In what follows we assume that the starting value of the process reads $X_0 = x > 0$ and that $\beta > 1 $.
The infinitesimal generator of this process reads $ {\mathfrak G}_z := \mu z d/d z + \sigma^2/2 z^{2 \beta} d^2/d z^2$ . The eigen-functions of this operator $ {\mathfrak G}_z \phi^{\pm} (z) = \lambda \phi^{\pm}(z) $ to the eigenvalue $\lambda > 0 $ are given below:
\begin{eqnarray}
\phi^{+}(z) &=&U\left(\frac{\lambda}{2(-1+\beta) \mu}, 1+ \frac{1}{2(-1+\beta)}, \frac{\mu z^{2-2\beta} }{(-1+\beta) \sigma^2} \right) \tag{2a} \\
\phi^{-}(z) &=& L_{-\frac{\lambda}{2(-1+\beta) \mu}}^{(\frac{1}{2(-1+\beta)})} \left( \frac{\mu z^{2-2\beta} }{(-1+\beta) \sigma^2}\right) \tag{2b}
\end{eqnarray}
where $U$ is the confluent hypergeometric function and $L$ is the generalized Laguerre polynomial. We have checked that $U(z)$ is a strictly increasing function of $z$.
---------------------------
Now, by using the theory of diffusion processes, see section 4.6 pages 128-134 in
<cite authors="Ito, K.; McKean, H. P. jun.">_Ito, K.; McKean, H. P. jun._, Diffusion processes and their sample paths, Berlin-Heidelberg-New York: Springer-Verlag. XVII, 321 p. (1965). [ZBL0127.09503](https://zbmath.org/?q=an:0127.09503).</cite>
, we have found the Laplace transform of the first hitting time $\tau_y := inf(s>0:X_s = y)$ of a horizontal barrier $b$ by this process. The quantity in question reads:
\begin{eqnarray}
E_x \left[ e^{-\lambda \tau_y} \right] = \frac{\phi^{+}(x)}{\phi^{+}(y)} \quad \mbox{for $x \le y$} \tag{3}
\end{eqnarray}
Now, by inverting the Laplace transform in $(3)$, by using the Bromwich integral and then the Cauchy theorem, we have expressed the probability density function of the first hitting time $n_x(t;y) := P_x\left( \tau_y \in dt\right)/dt $ as follows:
\begin{eqnarray}
n_x(t;y) = \sum\limits_{p=0}^\infty
\underbrace{
\frac{U\left( -\zeta_p^{(y;\mu,\sigma,\beta)},
1+ \frac{1}{2(-1+\beta)}, \frac{\mu x^{2-2\beta}}{(-1+\beta) \sigma^2} \right)}{
\zeta_p^{(y;\mu,\sigma,\beta)}
U^{(1,0,0)}\left( -\zeta_p^{(y;\mu,\sigma,\beta)},
1+ \frac{1}{2(-1+\beta)}, \frac{\mu y^{2-2\beta}}{(-1+\beta) \sigma^2} \right)
}
}_{{\mathfrak w}_p^{(y;\mu,\sigma,\beta)}}
\cdot
\underline{(2 (-1+\beta) \mu \zeta_p^{(y;\mu,\sigma,\beta)} ) \cdot e^{-2 (-1+\beta) \mu \cdot \zeta_p^{(y;\mu,\sigma,\beta)} \cdot t}} \tag{5}
\end{eqnarray}
As we can see the quantity in $(5)$ is an infinite linear combination of exponential distributions with weights $ \left( {\mathfrak w}_p^{(y;\mu,\sigma,\beta)} \right)_{p=1}^\infty $ that sum up to unity.
Here $ \left( \zeta_p^{(y;\mu,\sigma,\beta)} \right)_{p=1}^\infty $ are zeros of the function $ {\mathbb R}_+ \ni \lambda \rightarrow U(-\lambda,1+ \frac{1}{2(-1+\beta)}, \frac{\mu y^{2-2\beta}}{(-1+\beta) \sigma^2}) \in {\mathbb R} $.
------------------------
Now we took the following process parameters $\mu,\sigma,\beta = 3/2,1/2,5/2$
and the first value and the barrier $x,y = 3/2, 5/2 $ and we plotted the quantity $(5)$ below. We also verified the normalization numerically. Here we go:
{\[Mu], \[Sigma], \[Beta]} = {3/2, 1/2, 5/2};
(*Here x\[LessEqual]y*)
{x, y} = {3/2, 5/2};
SetOptions[FindRoot, WorkingPrecision -> mprec, PrecisionGoal -> prec];
mzeros = \[Lambda] /.
Table[FindRoot[
HypergeometricU[-\[Lambda],
1 + 1/(2 (-1 + \[Beta])), (\[Mu] y^(
2 - 2 \[Beta]))/((-1 + \[Beta]) \[Sigma]^2)] == 0, {\[Lambda],
n}], {n, 0, 50}];
mzeros = Sort[#[[1]] & /@ Tally[mzeros]];
ts = Array[# &, {300}, {1/100, 3}];
vals = {#,
Total[Table[
HypergeometricU[-mzeros[[p]],
1 + 1/(2 (-1 + \[Beta])), (\[Mu] x^(
2 - 2 \[Beta]))/((-1 + \[Beta]) \[Sigma]^2)]/
\!\(\*SuperscriptBox[\(HypergeometricU\),
TagBox[
RowBox[{"(",
RowBox[{"1", ",", "0", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[-mzeros[[p]],
1 + 1/(2 (-1 + \[Beta])), (\[Mu] y^(
2 - 2 \[Beta]))/((-1 + \[Beta]) \[Sigma]^2)] (2 (-1 + \
\[Beta]) \[Mu]) Exp[-2 (-1 + \[Beta]) \[Mu] mzeros[[p]] #], {p, 1,
Length[mzeros]}]]} & /@ ts;
ListPlot[vals, PlotRange :> All,
AxesLabel -> {"t", "\!\(\*SubscriptBox[\(n\), \(x\)]\)(t,y)"}]
f = Interpolation[vals];
NIntegrate[f[xi], {xi, 0.01, 3}]
[![enter image description here][2]][2]
As you can see the distribution in question has a correct shape and a correct normalization. The negative values close to the origin are due to a numerical error.
--------------------------
Having said all this my question would be how do we evaluate the limit of $\beta \rightarrow 1_+$. In this case the process tends towards the geometric Brownian motion and as such we should have:
\begin{equation}
\lim_{\beta \rightarrow 1_+} n_x(t;y) \stackrel{(??)}{=}
\frac{\left| \log(\frac{y}{x} )\right|}{\sqrt{2 \pi t^3} \sigma} e^{-\frac{1}{2 \sigma^2 t} \left[ \log(\frac{y}{x} - (\mu - \frac{\sigma^2}{2} ) t\right]^2}
\end{equation}
as shown in [a previous question on a similar topic][1].
How do we work out this limit analytically in our framework?
[1]: https://math.stackexchange.com/questions/4643792/the-distribution-of-the-first-hitting-time-for-a-generic-linear-diffusion
[2]: https://i.stack.imgur.com/dYZ1G.png
|
$\newcommand{\ga}{\gamma}$Let
\begin{equation*}
L(s):=\int_0^{\pi/2} \frac{\sin^2(sx)}{\sin^2x}\,f(x)\,dx- \frac\pi2f(0)s
-\frac{f'(0)}2\ln s,
\end{equation*}
\begin{equation*}
R:=\int_{0}^{\pi/2} \frac{f(x)-f(0)-f'(0)x}{2\sin^2x}\,dx+\frac{f'(0)}2\,(1+\ln2+\ga).
\end{equation*}
We have to show that
\begin{equation*}
L(s)\overset{\text{(?)}}\to R \tag{10}\label{10}
\end{equation*}
(as $s\to\infty$).
Let
\begin{equation*}
g(x):=f(x)-f(0)-f'(0)x,
\end{equation*}
so that
\begin{equation*}
L(s)=I_1(s)+f(0)(I_2(s)-\pi s/2)+f'(0)(I_3(s)-\tfrac12\,\ln s),
\end{equation*}
\begin{equation*}
R=\tfrac12\,J_1+f'(0)J_3,
\end{equation*}
where
\begin{equation*}
I_1(s):=\int_0^{\pi/2} \frac{\sin^2(sx)}{\sin^2x}\,g(x)\,dx,
\end{equation*}
\begin{equation*}
I_2(s):=\int_0^{\pi/2} \frac{\sin^2(sx)}{\sin^2x}\,dx, \quad
I_3(s):=\int_0^{\pi/2} \frac{\sin^2(sx)}{\sin^2x}\,x\,dx,
\end{equation*}
\begin{equation*}
J_1:=\int_0^{\pi/2} \frac{g(x)\,dx}{\sin^2x},\quad
J_3:=\frac{1+\ln2+\ga}2.
\end{equation*}
So, it is enough to show that
\begin{equation*}
I_1(s)\overset{\text{(?)}}\to \tfrac12\,J_1, \tag{20}\label{20}
\end{equation*}
\begin{equation*}
I_2(s)-\pi s/2\overset{\text{(?)}}\to 0, \tag{30}\label{30}
\end{equation*}
\begin{equation*}
I_3(s)-\tfrac12\,\ln s\overset{\text{(?)}}\to J_3. \tag{40}\label{40}
\end{equation*}
Since $f\in C^2[0,\frac{\pi}{2}]$, we have $\int_0^{\pi/2} \frac1{\sin^2x}\,|g(x)|\,dx<\infty$. Also, $\sin^2(sx)=\frac12\,(1-\cos2sx)$. So, \eqref{20} follows by the Riemann--Lebesgue lemma (RL).
To prove \eqref{30} and \eqref{40}, the key is to consider
\begin{equation*}
h(x):=\frac1{\sin^2 x}-\frac1{x^2}
\end{equation*}
for $x\in(0,\pi/2]$. Indeed, note that $\int_0^{\pi/2}|h|<\infty$.
So, again by RL,
\begin{align*}
I_2(s)&=\int_0^{\pi/2} h(x)\sin^2(sx)\,dx \\
& +\int_0^\infty \frac{\sin^2(sx)}{x^2}\,dx - \int_{\pi/2}^\infty \frac{\sin^2(sx)}{x^2}\,dx \\
&=\frac12\int_0^{\pi/2} h(x)\,dx+o(1) \\
& +\frac{\pi s}2 -\frac12 \int_{\pi/2}^\infty \frac1{x^2}\,dx+o(1) \\
&=\frac{\pi s}2+o(1),
\end{align*}
so that \eqref{30} is proved as well.
Finally,
\begin{equation*}
\begin{aligned}
I_3(s)&=\int_0^{\pi/2} h(x)x\sin^2(sx)\,dx +\int_0^{\pi/2} \frac{\sin^2(sx)}{x}\,dx \\
&=\frac12\int_0^{\pi/2} h(x)x\,dx +o(1)
+\frac12\, (-\text{Ci}(\pi s)+\ln s+\ga +\ln\pi) \\
&=\frac12\,(1+\ln2-\ln\pi)
+\frac12\, (\ln s+\ga +\ln\pi)+o(1),
\end{aligned}
\tag{50}\label{50}
\end{equation*}
where $\text{Ci}(\pi s):=-\int_{\pi s}^\infty\frac{dt}t\,\cos t$, the value of the [cosine integral][1] at $\pi s$; for the penultimate equality in \eqref{50}, use again the identity $\sin^2(sx)=\frac12\,(1-\cos2sx)$, RL, and the latter expression in the definition of $\text{Ci}$ in that Wikipedia article, and for the last equality in \eqref{50} use the first expression in the definition of $\text{Ci}$ in the Wikipedia article.
So, \eqref{40} is proved, too. $\quad\Box$
[1]: https://en.wikipedia.org/wiki/Trigonometric_integral#Cosine_integral
|
I was just messing with numbers when I created this formula.
But to understand the formula, you have to understand this: Multiplication is repeated addition, as $8*6$ is more or less $8+8+8+8+8+8$, or vice-versa. In the same way, exponents are repeated multiplication, as $5^3$, can be written as $5*5*5$, which can be written as $5+5+5+5+5*(5)$, which can be written as $5+5 25$ times, or $25+25+25+25+25$, which is $125$
For the mathematicians out there, you would know what is tetration. $3↑↑↑3$ is $7,625,597,484,987$.
This formula is for calculating by decimals. For example, if want to calculate what $10^{1.5}$, it would be $10*(\sqrt10)$
For $10*1.5$, it would be $10+(10/2)$
Do you see the pattern yet?
I want to test if this works for tetration, pentation, sextation, and so on. But I couldn't find a calculator capable of doing these. So can the answerer please add the link to one of these calcs? I prefer online, but if there is none online, then a downloadable one will also work.
|
I am a bit new to functional analysis and I stumbled upon this problem that confuses me.
Consider the space $X$ of bounded, non-decreasing, right-continuous functionals $F$ on $[0,1]$. Specifically, the space of cumulative distribution functions on $[0,1]$ is a subset of $X$. I have a linear functional $\varphi$ defined on $X$ through a set of properties. Intuitively (based on my knowledge of the specific function $\varphi$), it seems obvious that it should be of the form
$$ \varphi(F) = \int_0^1 G_\varphi(z) dF(z) $$
for some function $G_\varphi$. Also, it seems obvious that $G_\varphi \in X$.
Now, I am trying to prove this using the Riesz representation theorem. For this, we need an inner product on $X$. My first attempt was to define an "inner product" $\langle \cdot, \cdot \rangle : X \times X \to \mathbb{R}$ of the form
$$ \langle F, G \rangle = \int_0^1 G(z) dF(z)$$
If I blindly apply the Riesz representation theorem using this "inner product" and recover "Riesz representer" $G_\varphi$ using properties of $\varphi$, then I get exactly the answer I expect. The problem, however, is that $\langle \cdot, \cdot \rangle$ defined above is *not* an inner product, as it is not symmetric.
An alternative would be to move to a space $Y$ in which the derivatives of $F \in X$ live (i.e., define $\varphi$ as a function of the pdf $f \in Y$ corresponding to the original cdf $F \in X$ instead), which has the suitable inner product
$$ \langle f, g \rangle = \int_0^1 f(z) g(z) dz.$$
The problem is that the cdf $F(z)$ in my problem does not necessarily have a derivative $f
(z)$, especially at $z = 1$.
Does anyone know whether there exists a version of the Riesz representation theorem that can deal with the stieltjes integral above? Or are there other creative suggestions? Thanks in advance!
|
I was just messing with numbers when I created this formula.
But to understand the formula, you have to understand this: Multiplication is repeated addition, as $8*6$ is more or less $8+8+8+8+8+8$, or vice-versa. In the same way, exponents are repeated multiplication, as $5^3$, can be written as $5*5*5$, which can be written as $(5+5+5+5+5)*5$, which can be written as $5+5 25$ times, or $25+25+25+25+25$, which is $125$
For the mathematicians out there, you would know what is tetration. $3↑↑↑3$ is $7,625,597,484,987$.
This formula is for calculating by decimals. For example, if want to calculate what $10^{1.5}$, it would be $10*(\sqrt10)$
For $10*1.5$, it would be $10+(10/2)$
Do you see the pattern yet?
I want to test if this works for tetration, pentation, sextation, and so on. But I couldn't find a calculator capable of doing these. So can the answerer please add the link to one of these calcs? I prefer online, but if there is none online, then a downloadable one will also work.
|
If I have two matrices of type $n\times m,$ is there a way to compare their singular values?
For example, if $A$ has singular values $\lambda_1,\cdots,\lambda_n$ and $B$ has singular values $\sigma_1,\ldots,\sigma_n,$ is there a way to compare $\lambda_1+\cdots+\lambda_n$ and $\sigma_1+\cdots+\sigma_n$ or $\lambda_1+\cdots+\lambda_n$ and $\sigma_1^2+\cdots+\sigma_n^2?$
|
If I have two matrices of type $n\times m,$ is there a way to compare their singular values?
For example, if $A$ has singular values $\lambda_1,\cdots,\lambda_n,$ $\lambda_1\geqslant\lambda_2\geqslant\cdots\geqslant\lambda_n\geqslant0,$ and $B$ has singular values $\sigma_1,\ldots,\sigma_n,$ $\sigma_1\geqslant\sigma_2\geqslant\cdots\geqslant\sigma_n\geqslant0,$ is there a way to compare $\lambda_1+\cdots+\lambda_n$ and $\sigma_1+\cdots+\sigma_n$ or $\lambda_1+\cdots+\lambda_n$ and $\sigma_1^2+\cdots+\sigma_n^2?$
|
I'm studying from a Deep Learning book (Ian Goodfellow et al).
At page 256 the text explains that, considering a set of $k$ regression models, each produces an error $ϵ_i$ for every example, drawn from a zero-mean multivariate normal distribution with variances $E[ϵ_i^2]=v$ and covariances $E[ϵ_iϵ_j]=c$. Then the error made by the average prediction of all the ensemble models is $\frac{1}{k} \sum_{i=1}^k \epsilon_i$.
The expected squared error of the ensemble predictor is:
$
\begin{align*}
E\left[\left(\frac{1}{k} \sum_{i=1}^k \epsilon_i\right)^2\right] &= \frac{1}{k^2} E\left[\sum_{i=1}^k (\epsilon_i)^2 + \sum_{i \neq j} \epsilon_i \epsilon_j\right] \\
&= \frac{1}{k} v + \frac{k-1}{k} c.
\end{align*}
$$
So, the question is: How do we obtain the first quadratic decomposition? I'm having trouble with it.
Could someone explain?
Thank you so much!
|
I am reading the Eisenbud's commutative Algebra book, proof of Theorem 13.9 and stuck at some statement.
Let me first raise a question that I think is relevant.
**Q.** Let $R$ be a normal domain with finite extension of fields $K' \subseteq K'' \subseteq K(R) $ ( $K(R)$ is the quotient field ), possibly simple extension. Consider $R':=K'\cap R$ and $R'':=K''\cap R$.
Then, 1) $R' \hookrightarrow R''$ is an integral extension?
2) The field extension $K(R'')/ K(R')$ is finite?
I want to understand proof of next theorem in the Eisenbud's Commutative Algebra book.
> Theorem 13.9 (Going Down for integral extensions of normal rings). Let $S$ be a normal domain, and let $R$ be a domain containing $S$. If $R$ is integral over $S$, then **going down** holds between $R$ and $S$ : Given primes $Q_1 \supset Q$ and a prime $P_1$ of $R$ lying over $Q_1$, there exists a prime $P$ of $R$ lying over $Q$ and contained in $P_1$ ( Figure 13.1 ). This results holds even if $S$ and $R$ are not Noetherian.
[![enter image description here][1]][1]
**For the proof, he first proved the theorem for case that $K(R)$ is a finite extension of $K(S)$.**
And completion of the proof of the theorem 13.9 is as folllows :
[![enter image description here][2]][2]
Here, Figure 13.3 is,
[![enter image description here][3]][3]
I want to understand the underlined statement. Return to the FIgure 13.1. above. Let's make substitution :
$$ S \to R':=K'\cap R \\ R \to R'':=K''\cap R \\ Q_1 \to P_1':= P_1 \cap R' \\ P_1 \to P_1'':=P_1 \cap R'' \\ Q \to P' \\ \exists P \to \exists P'' $$
[![enter image description here][4]][4]
Then, if our question is true, then as Eisenbud argues ( by the theorem 13.9 for the case that $K(R)/K(S)$ is finite ), there exists $P'' \subseteq R''$ contained in $P_1'':=P_1 \cap R''$ and contracting to $Q$ in $S$ ( $\because$ $S\cap P''= ( S\cap R') \cap P'' = S \cap ( R' \cap P'') \stackrel{\text{!}}{=} S \cap P' = Q$ ).
And my origial question is really true? Can anyone help?
[1]: https://i.stack.imgur.com/mbHrb.png
[2]: https://i.stack.imgur.com/JZUg0.png
[3]: https://i.stack.imgur.com/LT8hs.png
[4]: https://i.stack.imgur.com/uM1P3.png
|
Is there a good way to solve this problem simply?
I set $t=x+\frac{1}{x}$ but this will make the equation very complicated.
$$f\left( x+\frac{1}{x} \right)=\frac{x+x^3}{1+x^4}$$ find $f(x)$
|
For a strongly regular simple undirected graph to be disconnected, does each of its component has to be complete (aka $\mu=0$)?
My reasoning is that two vertices disconnected components have 0 common neighbours, thus that $\mu$ has to be zero. Then by $${\displaystyle (v-k-1)\mu =k(k-\lambda -1)}$$
either $k=0$ (disjoint vertices) or $\lambda =k-1$. The second case implies for a vertex $u$, a neighbour $v$ can only connect to $u$ and all other neighbours of $u$, which seems to me that it can only be a clique of size $k+1$.
|
Let a,b,c≥0. Prove that:
>a/b+b/c+c/a+3³√abc/(a+b+c)≥4
This is a problem from Samin Riasat's Olympiad Inequalities worksheet. The worksheet gives a hint as to prove and use a/b+b/c+c/a≥(a+b+c)/³√abc. I've done it, substituted it in the original inequality and now I have to prove:
>(9AM²+3GM²)/(3AM•GM)≥4
Where AM and GM are the arithmetic and geometric means of a, b and c. How can I finish the problem?
|
Can you please tell me where you found the above formula?\
Seems to be something wrong with your formula.\
You may use this formula:
$$f(\tilde{f},T,\rho,n)\approx \frac{\rho ^T}{n}\cdot \sum _{k=0}^{n-1} \Re\left(\tilde{f}\left(\rho\cdot \exp \left(\frac{i ((2 \pi ) k)}{n}\right)\right)\cdot \exp \left(\frac{i ((2 \pi ) k T)}{n}\right) \right)$$
import numpy as np
def abate_whitt(ftilde, T, rho, n):
sum = 0.0
for k in range(0, n):
exp1 = np.exp(1j * 2*np.pi * k / n)
exp2 = exp1**T
sum += np.real(ftilde(rho*exp1) * exp2)
return (rho**T / n) * sum
def ftilde(z): return z / ((z + 1) * (z + 2))
def f(T): return (-1)**T - (-2)**T
T = 10
rho= 3
n = 81
f1 = abate_whitt(ftilde, T, rho, n)
f2 = f(T)
print("value=", f1, "exact=", f2, "error=", f1-f2)
value= -1022.9999999999935 exact= -1023 error= 6.480149750132114e-12
|
$$\int_0^{+\infty}\frac{(\sin x)^a}{x^b}dx=?,\,\text{suppose} \ a,b \in \mathbb{Z^+},a-b\geq 0,b\geq 2$$
I know how to calculate if $a-b\equiv 0 \pmod 2$，but I don't know
what to do when $a-b \equiv 1\pmod 2$.
|
Let $a,b,c\ge0$. Prove that:
$$\frac ab+\frac bc+\frac ca+\frac{3\sqrt[3]{abc}}{a+b+c}\ge4$$
This is a problem from Samin Riasat's Olympiad Inequalities worksheet. The worksheet gives a hint as to prove and use $$\frac ab+\frac bc+\frac ca\geq\frac{a+b+c}{\sqrt[3]{abc}}$$
I've done it, substituted it in the original inequality and now I have to prove:
$$\frac{9\mathrm{AM}^2+3\mathrm{GM}^2}{3\mathrm{AM}\cdot\mathrm{GM}}\geq4$$
where $\mathrm{AM}$ and $\mathrm{GM}$ are the arithmetic and geometric means of $a$, $b$ and $c$. How can I finish the problem?
|
It can be shown that when X, Y, and Z are all Banach spaces (or at least when X or Y are Banach spaces) over the number fields R or C, and when B : X×Y →Z is a bilinear function, the continuity of B for each component implies its overall continuity.
However, I'm curious about the scenario when X, Y, and Z are merely normed spaces, not necessarily complete. Is it possible for a bilinear mapping to be continuous per component, but not continuous as a whole?
The abovementioned proposition can be proven using the Uniform Boundedness Principle. Therefore, it seems natural to search for an example that fails to satisfy the UBP as a typical counterexample to this statement. Despite this, I'm finding it challenging to conceive a meaningful counterexample.
Could anyone provide an instance that illustrates this, or give some guidance on how to think about this problem?
|
What is the geometrical interpretation of $1 - i \cdot \text{Im}\left(\frac{1}{z}\right) = z$ in the Gaussian number plane?
I get that the first part is the real part 1 with no imaginary part so it lies on the real axis, so it basically means I have a point that is at 1 on the real axis and dependent on the value of z I transform it so it is equal to point z? The last part is what I do not get.
|
What is the geometrical interpretation of $1 - i \cdot \text{Im}\left(\frac{1}{z}\right) = z$ in the Gaussian number plane?
|
I'm trying to calculate this integral using residue theorem:
$$
\int_{-\infty}^{\infty}\text{d}\omega\frac{f(\omega)}{(\omega+a+i\epsilon)(\omega-a-i\epsilon)}
$$
where $f(\omega)$ is bounded at infinity and $f(a+i\epsilon)\neq f(-a-i\epsilon)$.
It seems to me that the integrand is proportional to $\frac{1}{\omega^2}$, so it should be convergent at infinity in both upper complex plane and lower complex plane. But if I close the contour in the upper half plane, I got
$$
i\pi\frac{f(a+i\epsilon)}{a+i\epsilon}
$$
and when I close the contour in the lower half plane, I got
$$
i\pi\frac{f(-a-i\epsilon)}{a+i\epsilon}
$$
My question is why these two results do not coincide?
|
For a strongly regular simple undirected graph to be disconnected, does each of its component has to be complete (aka $\mu=0$)?
My reasoning is that two vertices disconnected components have 0 common neighbours, thus that $\mu$ has to be zero. Then by $${\displaystyle (v-k-1)\mu =k(k-\lambda -1)}$$
either $k=0$ (disjoint vertices) or $\lambda =k-1$. The second case implies for a vertex $u$, a neighbour $v$ can only connect to $u$ and all other neighbours of $u$, which seems to me that it can only be a clique of size $k+1$.
Edit: The proof was found in the book Algebraic Graph Theory[proof][1].
[1]: https://i.stack.imgur.com/wDMts.png
|
Does this show $\bar{Y} is a sufficient statistic?
|
Does this show $\bar{Y}$ is a sufficient statistic?
|
Kindly help me with the following integral :
$
I_l(a) = \int_{-1}^{+1} dx\, e^{a x} P_l(x) \quad
$
($a$ is real and positive).
I thought to use the following relation given in Gradshteiyn and also in Prudnikov,
$\int_{-1}^{+1}\, dx\, e^{ipx} P_l(x) = \sqrt{\frac{2 \pi}{p}} \, e^{\frac{i l \pi}{2}} J_{l + \frac{1}{2}} (p) $.
But it seems that the formula cited there holds for real $p$. Can someone please enlighten me if $p$ becomes complex? I am unable to trace the original source.
|
Can you please tell me where you found the above formula?\
Seems to be something wrong with your formula.\
You may use this formula:
$$f(\tilde{f},T,\rho,n)\approx \frac{\rho ^T}{n}\cdot \sum _{k=0}^{n-1} \Re\left(\tilde{f}\left(\rho\cdot \exp \left(\frac{i ((2 \pi ) k)}{n}\right)\right)\cdot \exp \left(\frac{i ((2 \pi ) k T)}{n}\right) \right)$$
**IMPORTANT**: The radius $\rho$ must include all poles of $\tilde{f}(z)$
import numpy as np
def abate_whitt(ftilde, T, rho, n):
sum = 0.0
for k in range(0, n):
exp1 = np.exp(1j * 2*np.pi * k / n)
exp2 = exp1**T
sum += np.real(ftilde(rho*exp1) * exp2)
return (rho**T / n) * sum
def ftilde(z): return z / ((z + 1) * (z + 2))
def f(T): return (-1)**T - (-2)**T
T = 10
rho= 3
n = 81
f1 = abate_whitt(ftilde, T, rho, n)
f2 = f(T)
print("value=", f1, "exact=", f2, "error=", f1-f2)
value= -1022.9999999999935 exact= -1023 error= 6.480149750132114e-12
|
I have the following basic question regarding continuous functions.
Is the following statement correct?
> Given functions
>
> \begin{align*} h & : A \times C \to B \times D ,\\ f &: A \to B ,\\ g &: C
\to D , \end{align*}
>
> with $h := (f, g)$, both product spaces endowed with the product topology and all sets nonempty, if $h$ and $f$ are continuous, then $g$ is continuous.
I know that if $f$ and $g$ are continuous, then $h$ is continuous, while if $h$ is continuous it is not necessarily the case that both $f$ and $g$ are continuous. What about this 'intermediate' case?
Thanks a lot in advance for any feedback.
|
A conical buoy weighing $B$ lbs with its vertex $A$ ft below the surface has its top sticking up $\frac{A}{3}$ ft above the surface. How much work to lower the top of the buoy to surface level??
If I add $\frac{37B}{27}$ lbs of weight to the top it causes a displacement of $\frac{A}{3}$. Work = (force)*(displacement) = $\frac{37AB}{81}$ ft-lbs. But I have the wrong answer. Any ideas?
**Non-OP edit**: This is problem $65$ of Additional Problems for Chapter $7$ of Simmons Calculus, with answer given of $\frac{67}{324}AB$.
|
I'm trying to calculate this integral using residue theorem:
$$
lim_{\epsilon \rightarrow 0}\int_{-\infty}^{\infty}\text{d}\omega\frac{\omega^2-k^2v_F^2}{(\omega+kv+i\epsilon)(\omega-kv-i\epsilon)}\frac{1}{(\omega-v_q+i\epsilon)(\omega-v_q-i\epsilon)}
$$
It seems to me that the integrand is proportional to $\frac{1}{\omega^2}$, so it should be convergent at infinity in both upper complex plane and lower complex plane. But if I close the contour in the upper half plane, I got
$$
\frac{i\pi(v_F^2-v^2)}{kv(v-v_q)^2}+\frac{v_F^2-v_q^2}{v_q^2-v^2}\frac{\pi}{\epsilon}
$$
and when I close the contour in the lower half plane, I got
$$
\frac{i\pi(v_F^2-v^2)}{kv(v+v_q)^2}+\frac{v_F^2-v_q^2}{v_q^2-v^2}\frac{\pi}{\epsilon}
$$
My question is why these two results do not coincide?
|
I have the following basic question regarding continuous functions.
Is the following statement correct?
> Fix nonempty toplogical spaces $A, B, C, D$ with $A \times C$ and $B \times D$ endowed with the product topology. Also, fix functions
>
> \begin{align*} h & : A \times C \to B \times D ,\\ f &: A \to B ,\\ g &: C
\to D , \end{align*}
>
> with $h := (f, g)$. Thus, if $h$ and $f$ are continuous, then $g$ is continuous.
I know that if $f$ and $g$ are continuous, then $h$ is continuous, while if $h$ is continuous it is not necessarily the case that both $f$ and $g$ are continuous. What about this 'intermediate' case?
Thanks a lot in advance for any feedback.
|
I was going through the proof of Fundamental theorem of algebra using 'Liouvilles theorem'. **I have two queries/doubt in a proof**.
1. In the proof we assume that, $p(z) =0$ has no roots in $\mathbb{C}$ and then defined $f(z) =\frac{1}{p(z)}$ for all $z\in\mathbb{C}$. From this, **doesn't it directly follow that**, $$|f(z)|=\frac{1}{|p(z)|}≤1$$ so that $f(z)$ is bounded on $\mathbb{C}$.
2. Once we get that, $f(z) =\frac{1}{p(z)}$ is constant say $$f(z) =\frac{1}{p(z)}=k$$ for all $z\in\mathbb{C}$. where k is constant. **How, does it follows that**, $p(z)$ is constant? Isn't $k$ may be equals to $0$? If $k=0$ then what?
Please help me
|
You can split the integral in to two separate ones
$$
\int_0^{\frac12} (f''(x))^2 \, \mathrm{d}x+\int_{\frac12}^{1} (f''(x))^2 \, \mathrm{d}x
$$
and find the smallest function for both using calculus of variations and combine them piecewise. The reason that you have to do this is because calculus of variations only makes statements about the boundary.
The solution is
$$
f(x)=
\begin{cases}
-2x^3+\frac52 x& x \in [0,\frac12]\\
2x^3-6x^2+\frac{11}2x -\frac12& x \in (\frac12,1].
\end{cases}
$$
This achieves the lower bound of 12.
From there on it is indeed Cauchy-Schwarz, similar to this https://math.stackexchange.com/questions/499416/how-prove-this-int-abfx2dx-ge-dfrac4b-a?rq=1 .
|
I've got this theorem right here. It involves two arbitrary lines $AB$ and $CD$ and two transversals $AC$ and $BD$, intersecting **in between these lines**.
The theorem (or lemma, I don't know) in question states that $\alpha+\beta = \varphi + \theta$.
[![The drawing of the theorem / lemma][1]][1]
The simplest proof of which I know is based on the [Sum of Angles of Triangle equals Two Right Angles][2]: two transversals intersecting in between two lines form two triangles with these lines. Then sums of angles of these triangles are equal to one another. One pair of angles is equal as vertical angles, so the sum of two remaining angles in one triangle should be equal to the sum of two remaining angles, which was to be proven.
I've got three questions in regards to this fact:
- Does this fact (is it a lemma or a theorem?) **has a name**? I thought of something related to bowtie, butterfly, or hourglass, but there are numerous facts names after these objects.
- What is the correct way **to formulate this fact** by using parallel lines and transversals?
- If there's any proof of it only through parallel lines and transversals, then **what is that proof**?
[1]: https://i.stack.imgur.com/jXre3.png
[2]: https://proofwiki.org/wiki/Sum_of_Angles_of_Triangle_equals_Two_Right_Angles
|
Yoni Rozenshein (https://math.stackexchange.com/users/36650/yoni-rozenshein), Currently, what is the largest publicly known prime number such that all prime numbers less than it are known?, URL (version: 2013-03-14): https://math.stackexchange.com/q/330221
"Given current computing abilities, I'd guess your prime is somewhere between 2^50 and 2^60."
Since autumn 2020 that "time dependend prime" is well above 2^78.
See the b-file of OEIS https://oeis.org/A033844 ("a(n) = prime(2^n)").
Line 78 in this textfile reads:
78 17254990129969542495182251
It may take some more years to update https://oeis.org/A095124 ("a(n) = prime(2^(2^n))") beyond a(6) = 870566678511500413493.
|
I have the following partitioning condition,
$$\text{p(n|parts in N)}=\text{p(n|distinct parts in M)}, \text{for n}\ge1$$
Suppose I construct a set covered by the theorem $N=(1,3,6)$, for $$ \begin{align} n &= 1 \space 2 \space 3 \space 4 \space 5 \space 6 \\ p(n|\text{parts in (1,3,6)}) &= 1 \space 1 \space 2 \space 2 \space 2 \space 4\end{align}$$
I need your help in understanding how these parts were derived, my initial logic:
$$\begin{align}1 &\to 1 \\ 2 &\to 1+1 \\ 3 &\to 3, 1+1+1 \\ 4 &\to 3+1, 1+1+1+1 \\ 5 &\to 3+1+1, 1+1+1+1+1 \\ 6 &\to 6, 3+3, 3+1+1+1, 1+1+1+1+1+1 \end{align}$$
So essentially, all parts that partition the integer into either 1, 3 or 6. Would this be the correct approach?
|
What do parts in partition refer to?
|
I am not sure if the complete answer is known, but the interior and the exterior of the image $E$ of the exponential map are known:
1. The interior of $E$ consists of all matrices in $SL(n,\mathbb R)$ which have no negative eigenvalues.
2. The exterior of $E$ consists of all matrices in $A\in SL(n,\mathbb R)$ which have at least one negative eigenvalue of odd multiplicity.
3. Thus, the boundary of $E$ consists of matrices $A\in SL(n,\mathbb R)$ such that some eigenvalues of $A$ are negative but all such eigenvalues have even multiplicity. For instance, $A=Diag(-1/2,-1/2,-2, -2, 1)$ is an example.
This result is due to M.Nishikawa but his paper does not seem to be accessible. Another proof can be found in
<cite authors="Đoković, Dragomir Ž.">_Đoković, Dragomir Ž._, [**The interior and the exterior of the image of the exponential map in classical Lie groups**](https://doi.org/10.1016/0021-8693(88)90134-2), J. Algebra 112, No. 1, 90-109, Corrigendum 115, No. 2, 521 (1988). [ZBL0638.22006](https://zbmath.org/?q=an:0638.22006).</cite>
He also gives a description of the image $E$ of the exponential map for $SL(n,\mathbb C)$ but it is a bit too hard to state and you can find it in his paper. The set $E$ is open and dense (for some reason, he does not say "open" but it is open because the exponential map is holomorphic).
|
The finite-dimensional representations of $\frak{sl}_2$ can be constructed explicitly as partial differentiation of polynomials in two variables. Does a similar explicit construction exist for the finite-dimensional representations of $\frak{sl}_3$?
|
Let $(X_i, Y_i)_{i=1}^{\infty}$ be iid continuous random vectors with continuous joint density, where $X_1$ have support $\mathcal{X}$. Let $B_n\subset \mathcal{X}$ be decreasing subsets such that $\cap B_n= x_0\in\mathcal{X}$.
Let $S = \{i\leq n: X_i\in B_n\}$.
I want to show that
$$
\frac{1}{|S|}\sum_{i\in S}Y_i \overset{P}{\to} \mathbb{E}[Y_1\mid X_1=x_0].
$$
I assume that the necessary condition for this convergence is $|S|\to\infty$ or that $nP(X_i\in B_n)\to\infty$. Is it sufficient? Is there some theory that describes this?
|
For region $\Omega$ with smooth boundary, $f$ a vector field smooth enough, if $$\int_{\partial\Omega} \vec{f}\cdot\nabla v=0,\forall v\in H^2(\Omega),$$
then what can we obtain from it about $f$?
This is actually from <https://math.stackexchange.com/questions/4889518/the-boundary-condition-of-strong-form-of-a-4th-order-pde-in-a-variational-proble> and I don't know how to get the boundary condition. I don't think the above implies $f=0$ on $\partial\Omega$, since $v$ is a gradient and not an arbituary vector field. Can anyone help me with it? Thanks!
|
I have the following partitioning condition,
$$\text{p(n|parts in N)}=\text{p(n|distinct parts in M)}, \text{for n}\ge1$$
*Where N is any set of integers such that no element of N is a power of two times an element of N, and M is the set containing all elements of N together with all their multiples of powers of two.*
Suppose I construct a set covered by the theorem $N=(1,3,6)$, for $$ \begin{align} n &= 1 \space 2 \space 3 \space 4 \space 5 \space 6 \\ p(n|\text{parts in (1,3,6)}) &= 1 \space 1 \space 2 \space 2 \space 2 \space 4\end{align}$$
I need your help in understanding how these parts were derived, my initial logic:
$$\begin{align}1 &\to 1 \\ 2 &\to 1+1 \\ 3 &\to 3, 1+1+1 \\ 4 &\to 3+1, 1+1+1+1 \\ 5 &\to 3+1+1, 1+1+1+1+1 \\ 6 &\to 6, 3+3, 3+1+1+1, 1+1+1+1+1+1 \end{align}$$
So essentially, all parts that partition the integer into either 1, 3 or 6. Would this be the correct approach?
|
Calculate: $I = \int_{\substack{x_1 \geq 0, \ldots, x_n \geq 0 \\ , x_1+\cdots+x_n \leq 2}} \sqrt[n]{x_1 \cdots x_n\left(x_1+\cdots+x_n\right)} d x_1 \cdots d x_n$
Attempt: Perform the substitution $ x_i = 2y_i $, the jacobian will be $ 2^n $ and so $I=2^{n+1+\frac{1}{n}} \cdot \int_{\substack{x_1 \geq 0, \ldots, x_n \geq 0 \\ , x_1+\cdots+x_n \leq 1}} \prod x_i^{\frac{1}{n}} \cdot \sqrt[n]{\left(\sum_j x_j\right)}$
I don't know how to continue, any ideas?
Thanks for the help!
|
>In a given grocery store, apples have an average weight of $194$g and standard deviation of $40$g. Suppose that their weights are independent and follow a normal distribution.
If a customer requests $1$ kg of apples and given that the employer chooses randomly each apple and continues until the total weight is the intended one; calculate the probability of being necessary exactly $6$ apples $(1kg=1000g)$.
Each apple's weight is a random variable $X \sim N(194,40^2)$; and defining $S_n = X_1+\dots+ X_n$;I should calculate $P(S_5<1000\leq S_6)$; where $X_i \sim N(194,40^2); i=1, \dots,6$; but how could I compute this probability?
|
$f(x,y)$ is a real polynomial such that in the equation $f(x,y)=0$ we can express $x$ in $y$ with composition of polynomial functions and square roots, and can express $y$ in $x$ with composition of polynomial functions and square roots.
The curve $f(x,y)=0$ on $\Bbb R^2$ has $n$ connected components.
Is it true that we can always find $f_1,\dots,f_n$, each of which is a composition of polynomial functions and square roots, such that $f=\prod_{i=1}^nf_i$ and the curve $f_i=0$ has one-to-one correspondence with the connected components?
---------
Example: $f(x,y)=x^2-y^2-1$.
Clearly in the equation $f(x,y)=0$ we can express $x$ in $y$ with $x=\pm\sqrt{1+y^2}$ and express $y$ in $x$ with $y=\pm\sqrt{x^2-1}$.
$f(x,y)=0$ has $2$ connected components.
and we can find $f_1,f_2$:
$$x^2-y^2-1=f_1f_2=(x+\sqrt{y^2+1})(x-\sqrt{y^2+1})$$
and the curves $f_1=0,f_2=0$ are the two components.
[![enter image description here][1]][1]
------------
Example: $f(x,y)=2 x^2 y^2 + 1.5-(x^2 - 1)^2 - (y^2 - 1)^2$.
It is quadratic in $x^2,y^2$ so in the equation $f(x,y)=0$ we can express $x$ in $y$ with composition of polynomial functions and square roots, and can express $y$ in $x$ with composition of polynomial functions and square roots.
The curve $f(x,y)=0$ has $5$ connected components:
[![enter image description here][2]][2]
I checked that $2 x^2 y^2 + 1.5-(x^2 - 1)^2 - (y^2 - 1)^2=f_1f_2f_3f_4f_5$ by [WolframAlpha][3].
$f_1=\frac{\sqrt{x^{2}+\frac{1}{8}}+\sqrt{y^{2}+\frac{1}{8}}-1}{\sqrt{x^{2}+\frac{1}{8}}+\sqrt{y^{2}+\frac{1}{8}}+1}$
$f_2=x-\sqrt{\left(1+\sqrt{y^{2}+\frac{1}{8}}\right)^{2}-\frac{1}{8}}$
$f_3=x+\sqrt{\left(1+\sqrt{y^{2}+\frac{1}{8}}\right)^{2}-\frac{1}{8}}$
$f_4=y-\sqrt{\left(1+\sqrt{x^{2}+\frac{1}{8}}\right)^{2}-\frac{1}{8}}$
$f_5=y+\sqrt{\left(1+\sqrt{x^{2}+\frac{1}{8}}\right)^{2}-\frac{1}{8}}$
and the curves $f_1=0,f_2=0,f_3=0,f_4=0,f_5=0$ are the five components.
[1]: https://i.stack.imgur.com/Bf5RX.gif
[2]: https://i.stack.imgur.com/bZP6i.gif
[3]: https://www.wolframalpha.com/input?i=FullSimplify%5BExpand%5B%28x%5E2-%281%2Bsqrt%28y%5E2%2B1%2F8%29%29%5E2%2B1%2F8%29%28y%5E2-%281%2Bsqrt%28x%5E2%2B1%2F8%29%29%5E2%2B1%2F8%29%28sqrt%5Bx%5E2%2B1%2F8%5D%2Bsqrt%5By%5E2%2B1%2F8%5D-1%29%2F%28sqrt%5Bx%5E2%2B1%2F8%5D%2Bsqrt%5By%5E2%2B1%2F8%5D%2B1%29%5D%5D
|
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