Datasets:

awilliamson

/

mathstack-qa

like 0

<blockquote> <p>Let there are three points <span class="math-container">$(2,5,-3),(5,3,-3),(-2,-3,5)$</span> through which a plane passes. What is the equation of the plane in Cartesian form?</p> </blockquote> <p>I know how to find it in using vector form by computing the cross product to get the normal vector and passing through any one of the given points. But I want to do it a bit differently. </p> <p>We know, the equation of any plane passing through the first point is <span class="math-container">$$a(x-2)+b(y-5)+c(z+3)=0$$</span></p> <p>This equation must satisfy the other two points. However, this given me two equations with three unknowns <span class="math-container">$a,b,c$</span>. So can I not solve by this method?</p>

amd

<p>Remember that the implicit Cartesian equation for a plane is not unique. If you mutiply both sides of such an equation by a nonzero constant, you get another equation for the same plane. So, having an undetermined system for the coefficients of this equation is to be expected. Choose any convenient solution of the system.</p>

<p>I'm trying to prove <span class="math-container">$$\forall z\in\mathbb C-\{-1\},\ \left|\frac{z-1}{z+1}\right|=\sqrt2\iff\left|z+3\right|=\sqrt8$$</span> thus showing that the solutions to <span class="math-container">$\left|(z-1)/(z+1)\right|=\sqrt2$</span> form the circle of center <span class="math-container">$-3$</span> and radius <span class="math-container">$\sqrt8$</span>. But my memories of algebra in <span class="math-container">$\mathbb C$</span> fail me. The simplest I get is writing <span class="math-container">$z=x+i\,y$</span> with <span class="math-container">$(x,y)\in\mathbb R^2-\{(-1,0)\}$</span> and doing the rather inelegant <span class="math-container">$$\begin{align}\left|\frac{z-1}{z+1}\right|=\sqrt2&amp;\iff\left|z-1\right|=\sqrt2\,\left|z+1\right|\\ &amp;\iff\left|z-1\right|^2=2\,\left|z+1\right|^2\\ &amp;\iff(x-1)^2+y^2=2\,((x+1)^2+y^2)\\ &amp;\iff0=x^2+6\,x+y^2+1\\ &amp;\iff(x+3)^2+y^2=8\\ &amp;\iff\left|z+3\right|^2=8\\ &amp;\iff\left|z+3\right|=\sqrt8\\ \end{align}$$</span></p> <p>How can I avoid the steps with <span class="math-container">$x$</span> and <span class="math-container">$y$</span> ?</p>

user376343

<p>For <span class="math-container">$r&gt;0, a,b \in \mathbb{C}, a\neq b\;$</span> the equation <span class="math-container">$\left|\frac{z-a}{z-b}\right|=r$</span> defines a hyperbolic pencil of <a href="https://en.wikipedia.org/wiki/Apollonian_circles" rel="nofollow noreferrer">Apollonian circles</a>. Their centers lie on the line <span class="math-container">$AB,$</span> where <span class="math-container">$A(a), B(b).$</span></p> <p>To find the Apollonian circle (its center and radius) in the particular case <span class="math-container">$$\left|\frac{z-1}{z+1}\right|=\sqrt2,$$</span> it suffices to consider <span class="math-container">$z\in \mathbb{R}$</span> because <span class="math-container">$AB$</span> is the real axis.<br /> We find two real values <span class="math-container">$z=-3\pm 2\sqrt2.$</span> They are limit points of a diameter of the circle.<br /> The midpoint <span class="math-container">$C(-3)$</span> is the center, the radius is <span class="math-container">$2\sqrt2.$</span></p>

<p>1) Can a non-square matrix have eigenvalues? Why?</p> <p>2) True or false: If the characteristic polynomial of a matrix A is p($\lambda$)=$\lambda$^2+1, then A is invertible. Thank you!</p>

user1357015

<p>For 1) No, it has to be a square matrix by definition.</p> <p>To see why, consider the following:</p> <p>Recall that for an eigenvector $v$ and and an eigenvalue $\lambda$, you have that $Av$ = $\lambda v$. </p> <p>Now suppose that dim(v) = n x 1. That means that dim(Av) = n x 1 and dim($\lambda v$) = n x 1. If A is not square then dim(A) must be m x n where $m \neq n$. But then you have that dim($Av$) = (m x n) * (n x1) = m x 1. But we just said that dim(Av) = n x 1 . Thus contradiction.</p> <p>In regards to your second question, this post answers it completely.<br> <a href="https://math.stackexchange.com/questions/166219/is-a-matrix-with-characteristic-polynomial-t2-1-invertible">Is a matrix with characteristic polynomial $t^2 +1$ invertible?</a></p>

<p>Is $2\sqrt{12}$ or $4\sqrt{3}$ a better representation? Also, for $\sqrt{675}$, is $3\sqrt{75}$ or $15\sqrt{3}$ considered more simplified? Why is one more simplified than the others?</p>

Community

<p>You would want to simplify the root as much as possible, removing all squares. You would want to use 4$\sqrt{3}$.</p> <p>Further, the prime factorization of 48 is $2^4*3$, and that is what leads us to 4$\sqrt{3}$.</p>

<p>How to calculate $$\lim_{t\rightarrow1^+}\frac{\sin(\pi t)}{\sqrt{1+\cos(\pi t)}}$$? I've tried to use L'Hospital, but then I'll get</p> <p>$$\lim_{t\rightarrow1^+}\frac{\pi\cos(\pi t)}{\frac{-\pi\sin(\pi t)}{2\sqrt{1+\cos(\pi t)}}}=\lim_{t\rightarrow1^+}\frac{2\pi\cos(\pi t)\sqrt{1+\cos(\pi t)}}{-\pi\sin(\pi t)}$$ and this doesn't get me further. Any ideas?</p>

N. S.

<p>$$\lim_{t\rightarrow1^+}\frac{\sin(\pi t)}{\sqrt{1+\cos(\pi t)}}=\lim_{t\rightarrow1^+}\frac{\sin(\pi t)}{\sqrt{1+\cos(\pi t)}}\frac{\sqrt{1-\cos(\pi t)}}{\sqrt{1-\cos(\pi t)}}=\lim_{t\rightarrow1^+}\frac{\sin(\pi t)\sqrt{1-\cos(\pi t)}}{\sqrt{\sin^2(\pi t)}}$$</p> <p><strong>P.S.</strong> Pay attention to the sign of $\sin(\pi t)$ .</p>

<p>How do I derive the $m$ in the formula: $$I=\left(1+\frac{r}{m}\right)^{mn} -1$$</p> <p>all the values of the variables in the formula except $m$ is given and the question is find $m$. I just don't know how to derive the formula using the knowledge of Algebra I have.</p>

David Holden

<p>this looks like a compound interest question, so we see that the somewhat esoteric Lambert-W function may soon be part of the technical toolbox of chartered accountants! in order to gain a clearer idea of what is happening, using only elementary algebra, OP may find it useful to make an initial substitution, say $ x = \frac{m}{r} $, so that the equation becomes: $$ I=(1+x^{-1})^{nrx} -1 $$ or $$ \sqrt[nr]{1+I} = (1+x^{-1})^x $$ where the LHS is a constant computable from the problem data</p>

<p>If <span class="math-container">$\langle x_n\rangle $</span> is a sequence of positive real numbers such that <span class="math-container">$$x_{(n+2)}=\frac{(x_{n+1}+ x_{n})}{2}$$</span> for all <span class="math-container">$n \in \mathbb{N},\ $</span> let <span class="math-container">$x_1 &lt;x_2$</span></p> <p>then subsequence of odd terms is increasing and subsequence of even terms is decreasing .But how to prove it mathematically? We have </p> <p><span class="math-container">$$x_{(n+2)}-x_n=\frac{(x_{n+1}- x_{n})}{2}= \frac{(x_{n}- x_{n-2})}{4}$$</span> How to proceed from here? Any hint please.</p>

marty cohen

<p>(I'm pretty sure that I and many others have done this before but I'll work it out again.)</p> <p>If <span class="math-container">$x(n+1) = ax(n)+(1-a)x(n-1) $</span> where <span class="math-container">$0 &lt; a &lt; 2$</span> then</p> <p><span class="math-container">$\begin{array}\\ x(n+1)-x(n) &amp;= ax(n)+(1-a)x(n-1)-x(n)\\ &amp;= (a-1)x(n)+(1-a)x(n-1)\\ &amp;= (a-1)(x(n)-x(n-1))\\ \text{so}\\ x(n+k)-x(n+k-1) &amp;= (a-1)^k(x(n)-x(n-1))\\ \text{and}\\ x(n+k)-x(n) &amp;=\sum_{j=1}^{k}(x(n+j)-x(n+j-1)\\ &amp;= \sum_{j=1}^{k}(a-1)^j(x(n)-x(n-1))\\ &amp;= (x(n)-x(n-1))\sum_{j=1}^{k}(a-1)^j\\ &amp;= (x(n)-x(n-1))\dfrac{(a-1)-(a-1)^{k+1}}{1-(a-1)}\\ &amp;= (x(n)-x(n-1))\dfrac{(a-1)-(a-1)^{k+1}}{2-a}\\ \text{so} &amp;\text{putting } n = 1\\ x(k+1)-x(1) &amp;= (x(1)-x(0))\dfrac{(a-1)-(a-1)^{k+1}}{2-a}\\ &amp;= (x(1)-x(0))(\dfrac{a-1}{2-a}-\dfrac{(a-1)^{k+1}}{2-a})\\ \end{array} $</span></p> <p>Since <span class="math-container">$0 &lt; a &lt; 2$</span>, we have <span class="math-container">$-1 &lt; a-1 &lt; 1$</span> so <span class="math-container">$(a-1)^{k+1} \to 0$</span> and</p> <p><span class="math-container">$\begin{array}\\ x(k+1) &amp;\to x(1)+(x(1)-x(0))\dfrac{a-1}{2-a}\\ &amp;=(1+\dfrac{a-1}{2-a})x(1)-\dfrac{a-1}{2-a}x(0)\\ &amp;=\dfrac{2-a+a-1}{2-a}x(1)-\dfrac{a-1}{2-a}x(0)\\ &amp;=\dfrac{1}{2-a}x(1)+\dfrac{1-a}{2-a}x(0)\\ \end{array} $</span></p> <p>This is the case <span class="math-container">$a = \frac12$</span> so</p> <p><span class="math-container">$\begin{array}\\ x(k) &amp;\to \dfrac{1}{2-1/2}x(1)+\dfrac{1-1/2}{2-1/2}x(0)\\ &amp;=\dfrac{1}{3/2}x(1)+\dfrac{1/2}{3/2}x(0)\\ &amp;=\dfrac23 x(1)+\dfrac13 x(0)\\ \end{array} $</span></p>

<p>$$ \mbox{Question: Evaluate}\quad \tan^{2}\left(\pi \over 16\right) + \tan^{2}\left(2\pi \over 16\right) + \tan^{2}\left(3\pi \over 16\right) + \cdots + \tan^{2}\left(7\pi \over 16\right) $$</p> <p>What I did: Well I know that $\tan^{2}\left(7\pi/16\right)$ is the same as $\cot^{2}\left(\pi/16\right)$. Thus this will repeat for all values up to $\tan^{2}\left(4\pi/16\right)$.</p> <p>However, I don't understand where to proceed from there. </p>

Vishwa Iyer

<p><strong>HINT:</strong> Use the half angle formula with $\theta = \pi/4$ to find $\tan(\pi/8)$ and do the same with $\theta = \pi/8$ to find $\tan(\pi/16)$</p> <p>EDIT: The half angle formula is: $$\tan(a) = \frac{2\tan(\frac{a}{2})}{1- \tan^2(\frac{a}{2})}$$ So use this formula for $a = \pi/4$ and $a = \pi/8$</p>

<p>Decide whether the given set of vectors is linearly independent in the indicated vector space:</p> <p>$\{ x_1, x_1 +x_2, x_1 +x_2 +x_3, ..., x_1+\cdots+x_n\} $</p> <p>if $\{x_1, x_2, x_3, ..., x_n\}$ is linearly independent, in some vector space $V$.</p> <hr> <p>If $n=4:$</p> <p>$x_1 - (x_1+x_2) + (x_1+x_2+x_3) - (x_1+x_2+x_3+x_4) = -x_4.$</p> <p>So, if $n$ is even then it's linearly independent right?</p> <hr> <p>If $n=3:$ </p> <p>$x_1 - (x_1+x_2) + (x_1+x_2+x_3) = x_1 + x_3.$</p> <p>What about this situation when $n$ is odd? What can we state from $x_1+x_3$?</p>

Tsemo Aristide

<p>Suppose that such terminal object $e$ exists, you have a morphism of bundles $X\times G\rightarrow e$, implies that $e$ is isomorphic to $X\times G$, for every $G$-principal bundle $P$, you have a morphism $P\rightarrow X\times G$. This implies that $P$ is trivial. The terminal object exists if every $G$-principal bundle is trivial.</p> <p>We have used the fact that a morphism between two principal $G$-bundles over $X$ is an isomorphism.</p> <p><a href="https://math.stackexchange.com/questions/1300173/a-morphism-of-principal-bundles-is-an-isomorphism">A morphism of principal bundles is an isomorphism.</a></p>

<blockquote> <p>Without using a calculator, what is the sum of digits of the numbers from $1$ to $10^n$?</p> </blockquote> <p>Now, I'm familiar with the idea of pairing the numbers as follows:</p> <p>$$\langle 0, 10^n -1 \rangle,\, \langle 1, 10^n - 2\rangle , \dotsc$$</p> <p>The sum of digits of each pair is $9n$. How can I prove this property retains for all pairs?</p>

theREALyumdub

<p>Well, it's not too hard to figure out that the sum of digits of 0 or 1 to 9 is 45. That's very helpful, since that sequence will appear a bunch. Just add 1, and 46 is the sum of digits from 1 to 10 (or 0 to 10).</p> <p>In a way, that and the base shifts are all you need I believe. Let's do 100. We're going to get 10 times 45 in the ones place, for all of the numbers. As for the tens place, it moves ten times slower - but they all still pair up ten times to make 45 (try doing 10 with 20 with 30, 11 with 21 with 31, etc). so the sum from 1 to 99 is 450 + 450 = 900, and then add 1 + 0 + 0 to get 901.</p> <p>In general, it should be $ 45n 10^{n - 1} + 1 $, by the reasoning I have described.</p>

<p>Suppose, for the sake of keeping things as simple as possible, that I have the following equation that I wish to simplify in Mathematica:</p> <p><span class="math-container">$y = x x$</span></p> <p>But suppose further that I also have a restriction, not directly on <span class="math-container">$x$</span>, but on what values <span class="math-container">$y$</span> can have. More specifically, let's suppose <span class="math-container">$y \leq 9$</span>.</p> <p>What Mathematica expression, if any, will allow me to simplify the above expression for <span class="math-container">$y$</span> and get an output that not only simplifies, but also gives me the range of values that <span class="math-container">$x$</span> can take to satisfy my restriction on <span class="math-container">$y$</span>?</p> <p>So, just to be clear, the function or script I am looking for will output something like the following when dealing with the above:</p> <p>y = <span class="math-container">$x^2$</span> and <span class="math-container">$-3\leq x\leq3$</span></p>

Carl Woll

<p>When a function has 2 arguments (not a single list argument), use:</p> <pre><code>f[x_, y_] := x^4 + y^4 Derivative[1, 0][f][x, y] Derivative[0, 1][f][x, y] </code></pre> <blockquote> <p>4 x^3</p> <p>4 y^3</p> </blockquote>

<p>I know basic things about cardinality (I'm only in High School) like that since $\mathbb{Q}$ is countable, its cardinality is $\aleph_0$. Also that the cardinality of $\mathbb{R}$ is $2^{\aleph_0}$.</p> <blockquote> <p>Are there any direct applications of these numbers outside of theoretical math?</p> </blockquote> <p>I know this can be convenient for certain proofs and help understanding sets of numbers, but are there any applications of this?</p>

eggcrook

<p>Quoted from Christian Marks blog(blog seems to be gone now):</p> <blockquote> <p>In an unexpected development for the depressed market for mathematical logicians, Wall Street has begun quietly and aggressively recruiting proof theorists and recursion theorists for their expertise in applying ordinal notations and ordinal collapsing functions to high-frequency algorithmic trading. Ordinal notations, which specify sequences of ordinal numbers of ever increasing complexity, are being used by elite trading operations to parameterize families of trading strategies of breathtaking sophistication.</p> <p>The monetary advantage of the current strategy is rapidly exhausted after a lifetime of approximately four seconds — an eternity for a machine, but barely enough time for a human to begin to comprehend what happened. The algorithm then switches to another trading strategy of higher ordinal rank, and uses this for a few seconds on one or more electronic exchanges, and so on, while opponent algorithms attempt the same maneuvers, risking billions of dollars in the process.</p> <p>The elusive and highly coveted positions for proof theorists on Wall Street, where they are known as trans-quantitative analysts, have not been advertised, to the chagrin of executive recruiters who work on commission. Elite hedge funds and bank holding companies have been discreetly approaching mathematical logicians who have programming experience and who are familiar with arcane software such as the ordinal calculator. A few logicians were offered seven figure salaries, according to a source who was not authorized to speak on the matter.</p> </blockquote>

<p>This is a Question from an Analysis 1 exam. The question is as follows: Decide if the functions <span class="math-container">$f: \mathbb{R} \longrightarrow \mathbb{R}$</span> can be written as the difference of two monotonically increasing functions</p> <p>a) <span class="math-container">$f(x) = \cos(x)$</span></p> <p>b) <span class="math-container">$f(x) = x^2$</span></p> <p>For the moment I’m working on a) my first thought would be to use the MVT and receive something in the form of <span class="math-container">$\cos(x)+2x = -\sin(x)-2x$</span> but as we see -<span class="math-container">$\sin(x)$</span> is not monotonically increasing. Obviously one could also answer with <span class="math-container">$\cos(x) = (\cos(x)+2x) - 2x$</span> but I fear this answer would not be accepted by my professor. If you have any tips or answers for either a) or b) id be grateful</p>

kandb

<p>The support of <span class="math-container">$f$</span> is <span class="math-container">$\mathbb{R}$</span> (note that the support of a function is the subset of its domain on which it does not vanish), which is closed by definition, but <span class="math-container">$\mathbb{R}$</span> is not a compact subset of <span class="math-container">$\mathbb{R}$</span>. In other words, <span class="math-container">$f$</span> vanishes at infinity, but lacks compact support. To see that <span class="math-container">$\mathbb{R}$</span> is not compact, consider an open cover of <span class="math-container">$\mathbb{R}$</span>:</p> <p><span class="math-container">\begin{equation*} \mathbb{R} = \bigcup_{n\in \mathbb{N}}{(-n,n)}. \end{equation*}</span></p> <p>Because this open cover has no finite sub-cover, <span class="math-container">$\mathbb{R}$</span> is not compact (though this is perhaps beyond the scope of your question).</p>

<p>I have a long vector and some of the values (19 out of 64) are complex. I got them using the Mathematica Rationalize function, so the complex ones are written in the a+bi form. Is there a function I can apply to the entire vector, that would change my complex numbers to the form A<em>Exp[I</em>phi]? </p>

bill s

<p>Since the arguments are rational:</p> <pre><code>v = {1/10, -1 - 2 I, 3 - 5/3 I, 7, 9/10 + I}; Abs[v] Exp[I Arg[v]] {1/10, Sqrt[5] E^(I (-π + ArcTan[2])), 1/3 Sqrt[106] E^(-I ArcTan[5/9]), 7, 1/10 Sqrt[181] E^(I ArcTan[10/9])} </code></pre>

<p>I have a question in my paper, Express 4225 as the product of its prime factors in index notation. That was easy to answer, but the next question is express the square root of 42250000 using prime factorisation. Apparently there is a way to use my answer in the first question to do the second, but how do I?</p>

CoffeeCrow

<p>Using that <span class="math-container">$\sqrt{x}=x^{\frac{1}{2}}$</span>, index laws and the prime factorisations of <span class="math-container">$1000$</span> and <span class="math-container">$4255$</span>, as well as that <span class="math-container">$42250000=4225\times10000$</span>, we have;</p> <p><span class="math-container">$\sqrt{42250000}=\sqrt{4225}\sqrt{10000}=\sqrt{5^{2}13^{2}}\sqrt{10^4}=(5^{2}13^{2})^{\frac{1}{2}}({10^4})^{\frac{1}{2}}=5\times 13\times10^2$</span></p> <p>This works because every factor in the prime factorisation of <span class="math-container">$42250000$</span> appears raised to a power which is a multiple of <span class="math-container">$2$</span>, so the square root is guaranteed to be a whole number.</p>

<p>I have a question in my paper, Express 4225 as the product of its prime factors in index notation. That was easy to answer, but the next question is express the square root of 42250000 using prime factorisation. Apparently there is a way to use my answer in the first question to do the second, but how do I?</p>

drhab

<p>If <span class="math-container">$$n=p_1^{2k_1}\cdots p_m^{2k_m}$$</span> where the <span class="math-container">$k_i$</span> are non-negative integers then: <span class="math-container">$$\sqrt n=p_1^{k_1}\cdots p_m^{k_m}$$</span></p>

<p>I have a question in my paper, Express 4225 as the product of its prime factors in index notation. That was easy to answer, but the next question is express the square root of 42250000 using prime factorisation. Apparently there is a way to use my answer in the first question to do the second, but how do I?</p>

Bill Dubuque

<p><strong>Hint</strong> <span class="math-container">$\,f(ab) = f(a)f(b)\,$</span> where <span class="math-container">$\,f(n) := $</span> prime factorization of <span class="math-container">$n$</span> and, <a href="https://math.stackexchange.com/a/21637/242">furthermore,</a> if <span class="math-container">$\,a,b\,$</span> are coprime then <span class="math-container">$\,ab\,$</span> is a square <span class="math-container">$\iff a,b$</span> are squares <span class="math-container">$\iff$</span> every prime occurs to even power in their prime factorizations, all being true by FTA = <em>existence</em> and <em>uniqueness</em> of prime factorizations (<em>all</em> inferences fail lacking such). </p>

<p>Over the past few days, the Chrome browser page with Questions is crashing.</p> <p>It has crashed while typing answers twice and several other times when just viewing and updating the questions listing.</p> <p>Did something change in the past three days that could be affecting this as I have not seen this behavior before and it is very annoying when trying to write answers?</p> <p>Has anyone else seen this or is this a bug?</p>

Pedro

<p>I have been able to diminish the number of crashes by using Robjohn's "rendering off" tab. That seems to fix it a little, since it seems Chrome crashes when compiling and re-compiling the code over and over when we type. It seems every new character entered forces a recompilation of the code that as been already been written.</p>

<p>Is it true that any manifold homotopy equivalent to a k-dimensional CW-complex admits a proper Morse function with critical points all of index &lt;= k? I believe this is not true, so I would like to see a counterexample.</p>

Tom Goodwillie

<p>Suppose that $M$ has a proper Morse function $f\ge 0$ with all critical points of index at most $k$. Then homotopically $M$ can be made of low-dimensional cells: it has homotopical dimension at most $k$. But also $M$, relative to its boundary $M^{\ge c}$, can be made by attaching <i>high</i>-dimensional cells. Thus the pair $(M,M^{\ge c})$ is $(dim(M)-k-1)$-connected. So for example if $k\le dim(M)-3$ and $M$ is simply connected then $M$ must also be "simply connected at infinity".</p>

<p>I am currently trying to show that the sequence of functions defined by <span class="math-container">$f_n(x) = \frac{x}{1 + x^n}$</span> converges pointwise on <span class="math-container">$U = [0, \infty)$</span>. I have found the limits for the three specific cases and they are: <span class="math-container">\begin{equation*} \lim_{n \to \infty} f_n(x) = \begin{cases} x &amp; \text{if $x \in [0, 1)$} \\ \frac{1}{2} &amp; \text{if $x = 1$} \\ 0 &amp; \text{if $x \in (1, \infty)$} \end{cases} \end{equation*}</span> I have started the proof for when <span class="math-container">$x \in [0, 1)$</span> but I am not sure how to proceed. Here is what I have:</p> <p>Let <span class="math-container">$\epsilon &gt; 0$</span>. We want to show that <span class="math-container">$(f_n(x))_n$</span> converges to <span class="math-container">$x$</span> on <span class="math-container">$x \in [0, 1)$</span>, then there exists an <span class="math-container">$N \in \mathbb{N}$</span> such that <span class="math-container">\begin{equation*} \left|\frac{x}{1 + x^n} - x\right| &lt; \epsilon \end{equation*}</span> for all <span class="math-container">$n \geq N$</span>. Then <span class="math-container">\begin{equation*} \left|\frac{x}{1 + x^n} - x\right| = \frac{x^{n + 1}}{1 + x^n} &lt; \epsilon \end{equation*}</span> This is the part where I am lost, but I know that I am suppose to find an <span class="math-container">$N = N(x, \epsilon)$</span>, however, where it confuses me is where we have the <span class="math-container">$n$</span> as the exponent and it is a bit unclear to me how to proceed from here. Would like some assistance for this part.</p>

Anne Bauval

<p>You have <span class="math-container">$x\in[0,1),$</span> <span class="math-container">$\epsilon&gt;0,$</span> and you want <span class="math-container">$x^{n+1}&lt;\left(1+x^n\right)\epsilon,$</span> i.e. <span class="math-container">$$x^n\left(x-\epsilon\right)&lt;\epsilon.$$</span></p> <ul> <li>If <span class="math-container">$x\le\epsilon,$</span> this holds for every <span class="math-container">$n.$</span></li> <li>If <span class="math-container">$x&gt;\epsilon,$</span> this holds iff <span class="math-container">$n\ln x&lt;\ln\frac\epsilon{x-\epsilon},$</span> i.e. <span class="math-container">$n\left|\ln x\right|&gt;\ln\frac{x-\epsilon}\epsilon,$</span> i.e. <span class="math-container">$$n&gt;\frac{\ln\frac{x-\epsilon}\epsilon}{\left|\ln x\right|}.$$</span> For a shorter proof of the pointwise convergence in each of the three cases, see <a href="https://math.stackexchange.com/a/789845/386889">this post</a>.</li> </ul>

<p>Does anybody have suggestions on what to read to learn more about couplings pertaining to statistics?</p> <p>I'm working on a research project on Poisson approximations and am looking to perform a coupling on the unknown distribution. However, I cannot find much material on how to perform a coupling and the general calculations for it. I haven't had formal training on measure theory or intense probability theory, just upper level statistics courses. Any suggestions?</p>

Omer

<p>Have you looked at Lindvall's "Lectures on the Coupling Method"?</p>

<p>Does anybody have suggestions on what to read to learn more about couplings pertaining to statistics?</p> <p>I'm working on a research project on Poisson approximations and am looking to perform a coupling on the unknown distribution. However, I cannot find much material on how to perform a coupling and the general calculations for it. I haven't had formal training on measure theory or intense probability theory, just upper level statistics courses. Any suggestions?</p>

Will Jagy

<p>My friend Marty suggests the Lindvall book as well as</p> <p>H. Thorisson, Coupling, Stationarity, and Regeneration. Springer, New York, 2000. </p> <p><a href="http://www.springer.com/mathematics/probability/book/978-0-387-98779-8" rel="nofollow">http://www.springer.com/mathematics/probability/book/978-0-387-98779-8</a></p> <p>and points out that coupling is used now in basic textbooks in stochastic processes to prove the ergodic theorem for Markov Chains. So he recommends </p> <p>Geoffrey Grimmett and David Stirzaker, Probability and Random Processes, 3rd edition, Oxford University Press, 2001. </p> <p><a href="http://ukcatalogue.oup.com/product/9780198572220.do" rel="nofollow">http://ukcatalogue.oup.com/product/9780198572220.do</a></p> <p>A monograph that presents the 1975 Stein-Chen method: </p> <p>A. D. Barbour, Lars Holst, and Svante Janson, Poisson Approximation </p> <p><a href="http://ukcatalogue.oup.com/product/academic/series/mathematics/osip/9780198522355.do?sortby=bookTitleAscend" rel="nofollow">http://ukcatalogue.oup.com/product/academic/series/mathematics/osip/9780198522355.do?sortby=bookTitleAscend</a></p> <p>A rare accessible discussion is in chapter 2 of:</p> <p>Ross, Sheldon and Peköz, Erol (2007). A second course in probability. www.ProbabilityBookstore.com. ISBN 978-0979570407.</p> <p>Link to page for Peköz, which gives further link for book purchase to Amazon:</p> <p><a href="http://smgpublish.bu.edu/pekoz/" rel="nofollow">http://smgpublish.bu.edu/pekoz/</a></p> <p>A number of useful links at </p> <p><a href="http://www.math.lsa.umich.edu/~fomin/525w07.html" rel="nofollow">http://www.math.lsa.umich.edu/~fomin/525w07.html</a> </p> <p>and see</p> <p><a href="http://en.wikipedia.org/wiki/Stein%27s_method" rel="nofollow">http://en.wikipedia.org/wiki/Stein%27s_method</a> </p> <p><a href="http://en.wikipedia.org/wiki/Coupling_%28probability%29" rel="nofollow">http://en.wikipedia.org/wiki/Coupling_%28probability%29</a></p>

<p>Question: $ \sqrt{x^2 + 1} + \frac{8}{\sqrt{x^2 + 1}} = \sqrt{x^2 + 9}$</p> <p>My solution: $(x^2 + 1) + 8 = \sqrt{x^2 + 9} \sqrt{x^2 + 1}$</p> <p>$=&gt; (x^2 + 9) = \sqrt{x^2 + 9} \sqrt{x^2 + 1}$</p> <p>$=&gt; (x^2 + 9) - \sqrt{x^2 + 9} \sqrt{x^2 + 1} = 0$</p> <p>$=&gt; \sqrt{x^2 + 9} (\sqrt{x^2 + 9} - \sqrt{x^2 + 1}) = 0$</p> <p>So, either $\sqrt{x^2 + 9} = 0$ or $(\sqrt{x^2 + 9} - \sqrt{x^2 + 1}) = 0$</p> <p>From the first expression, I get $x = \pm 3 i$ and from the second expression, I get nothing.</p> <p>Now, notice how in the 2nd step, I could've divided both the sides by $\sqrt{x^2 + 9}$, but I didn't because I learned here that we must never do that and that we should always factor: <a href="https://math.stackexchange.com/questions/1751410/how-to-square-both-the-sides-of-an-equation#comment3574205_1751435">Why one should never divide by an expression that contains a variable.</a></p> <p>So, my question is: is the solution above correct? Would it have been any harm had I divided both the sides by $\sqrt{x^2 + 9}$?</p>

Deepak

<p>A suggested simplification. You should always look for simplifications to the algebra if they're easy to find.</p> <p>Let $y = x^2 + 1$</p> <p>Then you're solving $\sqrt y + \frac{8}{\sqrt y} = \sqrt{y + 8}$</p> <p>$\frac{y + 8}{\sqrt y} = \sqrt{y + 8}$</p> <p>$(y+8)^2 = y(y+8)$</p> <p>$8(y+8) = 0$</p> <p>From this, it should be obvious that $y = -8$ is the only root, giving $x^2 + 1 = -8$ or $x = \pm 3i$.</p>

<p>In Enderton's <em>A Mathematical Introduction to Logic</em>, he defines <span class="math-container">$n$</span>-tuples recursively using ordered pairs, i.e. <span class="math-container">$\langle x_1,\dots,x_{n+1}\rangle=\langle\langle x_1,\dots,x_n\rangle, x_{n+1}\rangle$</span>. But he also notes,</p> <blockquote> <p>Finite sequences are often defined to be certain finite functions, but the above definition is slightly more convenient for us.</p> </blockquote> <p>I believe he's referring to the other definition (which I prefer) of <a href="https://en.wikipedia.org/wiki/Tuple#Tuples_as_functions" rel="nofollow noreferrer"><span class="math-container">$n$</span>-tuples as functions</a>.</p> <p><a href="https://math.stackexchange.com/questions/2122856/endertons-a-mathematical-introduction-to-logic-question-about-n-tuples-in-th">Here</a> is a related question that provides some more background, but I'm not confused about the mathematics of Enderton's definition. I'm curious about why he feels it's slightly more convenient when it feels the opposite to me.</p> <p><strong>In what way might Enderton's choice be slightly more convenient, at least in the context of his logic book?</strong> If you have access to the book, the relevant pages are page 4 and remark 5 on page 15.</p>

Wuestenfux

<p>Actually, you don't need the definition of function by using the above recursive definition. For this, you start with induction base for <span class="math-container">$n=2$</span>:</p> <p><span class="math-container">$(x,y) = \{\{x\},\{x,y\}\}$</span> (ordered pair) is a set and so all <span class="math-container">$n$</span>-tuples are sets.</p>

<p>How can I calculate the following limit epsilon-delta definition?</p> <p>$$\lim_{x \to 0} \left(\frac{\sin(ax)}{x}\right)$$</p> <p>Edited the equation, sorry...</p>

anegligibleperson

<p>you can rewrite $\dfrac{\sin (ax)}{x} = a \dfrac{\sin (ax)}{ax}$, then take limits, as suggested by @rscwieb in the comments</p>

<p>I want to express this polynomial as a product of linear factors:</p> <p>$x^5 + x^3 + 8x^2 + 8$</p> <p>I noticed that $\pm$i were roots just looking at it, so two factors must be $(x- i)$ and $(x + i)$, but I'm not sure how I would know what the remaining polynomial would be. For real roots, I would usually just do use long division but it turns out a little messy in this instance (for me at least) and was wondering if there was a simpler method of finding the remaining polynomial. </p> <p>Apologies for the basic question!</p>

A. Pongrácz

<p>Yo can also do long division for complex polynomials. But in this case, I would suggest to pull out the two factors at once, i.e., divide by their product, which is $(x^2+1)$.</p> <p>$(x^5+x^3+8x^2+8):(x^2+1)= x^3+8$. You can easily factorize this polynomial. (Hint: $8=2^3$.)</p>

<p>According to <a href="http://en.wikibooks.org/wiki/Haskell/Category_theory" rel="noreferrer">the Haskell wikibook on Category Theory</a>, the category below is not a valid category due to the addition of the morphism <em>h</em>. The hint says to &quot; think about associativity of the composition operation.&quot; But I don't don't see why it fails.</p> <p><span class="math-container">$$ f \circ (g \circ h) = (f \circ g) \circ h\\ f \circ (\mathit{id}_B) = (\mathit{id}_A) \circ h\\ $$</span></p> <p>Does this then reduce to <span class="math-container">$f = h$</span> ?</p> <p>And is that not true because <em>f</em> and <em>h</em>, despite both being from B to A, are not equivalent?</p> <p><a href="https://i.stack.imgur.com/gllAK.png" rel="noreferrer"><img src="https://i.stack.imgur.com/gllAK.png" alt="" /></a><br /> _{(source: <a href="https://upload.wikimedia.org/wikibooks/en/6/65/Not-a-cat.png" rel="noreferrer">wikimedia.org</a>)}</p>

Giorgio Mossa

<p>It not so clear what is they mean, but I guess what they mean is that if you consider the graph above (in which edges with different labels are different) then you cannot put on that graph a structure of a category.</p> <p>To prove that you have to use reductio ab absurdum: if there where any category structure on that graph there should be a law of composition such that $g \circ h = \text{id}_B$ and also $f \circ g = \text{id}_A$ (that's follows for what is said in the link you posted above) and so it should also be the case that $$f = f \circ \text{id_B} = f \circ (g \circ h) = (f \circ g) \circ h = \text{id}_A \circ h = h \ .$$</p> <p>This would implies that $f=h$ but by hypothesis $f \ne h$ hence you've arrived to an absurd, so you cannot find any composition law that give to the graph the structure of a category.</p> <p>Hope this helps.</p>

<p>If $\lambda$ is the eigenvalue of matrix $A$,what is the eigenvalue of $A^TA$?I have no clue about it. Can anyone help with that?</p>

Ted Shifrin

<p>The good exercise for you is to prove, in general, that $$\lambda_{\text{max}}(A)\le \max_{\|x\|=1} \|Ax\|=\sqrt{\lambda_{\text{max}}(A^\top A)}.$$</p>

<blockquote> <p><strong>Possible Duplicate:</strong><br> <a href="https://math.stackexchange.com/questions/228080/operatornameimfz-leq-operatornamerefz-then-f-is-constant">$|\operatorname{Im}f(z)|\leq |\operatorname{Re}f(z)|$ then $f$ is constant</a> </p> </blockquote> <blockquote> <p>Let $f\colon\mathbb C \to \mathbb C$ be entire. Show that if $|\operatorname{Im}f(z)|\geqslant |\operatorname{Re}f(z)|$ for all $z \in \mathbb C$, then $f$ is constant on $\mathbb C$. </p> </blockquote> <p>Can I answer this by considering the distance between $f(z)$ and $i$ like in this problem <a href="https://math.stackexchange.com/questions/228080/operatornameimfz-leq-operatornamerefz-then-f-is-constant">$|\operatorname{Im}f(z)|\leq |\operatorname{Re}f(z)|$ then $f$ is constant</a>?</p>

amWhy

<p>$\forall x\in K, (\text{and}\;\;\forall x \in X)$: $x\in F\cup F^c$. $\;x\in F$ or $x\in F^c \implies (x \in F$ or $x \notin F$). That is $x$ is in $F$, or it's not in $F$ (Law of excluded middle). If it's in $F$, it's covered by the open cover of $F$: $\{V_\alpha\}$. If it's in $F^c$ (i.e., if it's not in $F$), it's covered by $F^c$. </p> <p>Whatever the case, it's covered: all of $X$ so all of $K$ is covered, and since K is compact, it can be covered by a finite subcover, and following the proof, $F$ can then be covered by a finite collection of $\{V_\alpha\}$, so it too is compact.</p> <hr> <p>Edit: Since the problem seems to be in understanding $F^c$, the complement of $F$ relative to $X$:</p> <p><strong>Complement clarification</strong>: See image.</p> <p>Let $X = U = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$. </p> <p>Let $F = B = \{0, 1, 4, 5, 6, 7, 8\}.$ </p> <p>Then $F^c = B'= X \setminus F = U \setminus B = \{2, 3, 9\}$</p> <p><img src="https://i.stack.imgur.com/8mTGN.png" alt="enter image description here"></p>

<blockquote> <p><strong>Possible Duplicate:</strong><br> <a href="https://math.stackexchange.com/questions/228080/operatornameimfz-leq-operatornamerefz-then-f-is-constant">$|\operatorname{Im}f(z)|\leq |\operatorname{Re}f(z)|$ then $f$ is constant</a> </p> </blockquote> <blockquote> <p>Let $f\colon\mathbb C \to \mathbb C$ be entire. Show that if $|\operatorname{Im}f(z)|\geqslant |\operatorname{Re}f(z)|$ for all $z \in \mathbb C$, then $f$ is constant on $\mathbb C$. </p> </blockquote> <p>Can I answer this by considering the distance between $f(z)$ and $i$ like in this problem <a href="https://math.stackexchange.com/questions/228080/operatornameimfz-leq-operatornamerefz-then-f-is-constant">$|\operatorname{Im}f(z)|\leq |\operatorname{Re}f(z)|$ then $f$ is constant</a>?</p>

sureshs

<p>The open cover you're dealing with is $\{F^c\} \cup \{V_\alpha\}$. This covers the entire space $X$, not just $K$. To see this, let $x \in X$. Either $x \in F$ or $x \notin F$. If $ x \in F$, then</p> <p>$$x \in F \subseteq \bigcup V_{\alpha}$$</p> <p>Otherwise, $x \in F^c$, so $X \subseteq F^c \cup \bigcup V_{\alpha}$.</p>

<blockquote> <p>How many integer-sided right triangles exist whose sides are combinations of the form $\displaystyle \binom{x}{2},\displaystyle \binom{y}{2},\displaystyle \binom{z}{2}$?</p> </blockquote> <p><strong>Attempt:</strong></p> <p>This seems like a hard question, since I can't even think of one example to this. Mathematically we have,</p> <p>$$\left(\dfrac{x(x-1)}{2} \right)^2+\left (\dfrac{y(y-1)}{2} \right)^2 = \left(\dfrac{z(z-1)}{2} \right)^2\tag1$$ </p> <p>where we have to find all positive <em>integer</em> solutions $(x,y,z)$. </p> <p>I find this hard to do. But here was my idea. Since we have $x^2(x-1)^2+y^2(y-1)^2 = z^2(z-1)^2$, we can try doing $x = y+1$. If we can prove there are infinitely many solutions to,</p> <p>$$(y+1)^2y^2+y^2(y-1)^2 = z^2(z-1)^2\tag2$$ </p> <p>then we are done.</p>

Tito Piezas III

<p>Solving $(1)$ for $z$, we have,</p> <p>$$z = \frac{1\pm\sqrt{1\pm4w}}{2}\tag3$$</p> <p>where,</p> <p>$$w^2 = (x^2-x)^2+(y^2-y)^2\tag4$$</p> <p>It can be shown that $(4)$ has infinitely many integer solutions. (<em>Update</em>: Also proven by Sierpinski in 1961. See link given by MXYMXY, <em><a href="http://www.fq.math.ca/Scanned/17-2/ballew.pdf" rel="nofollow">Pythagorean Triples and Triangular Numbers</a></em> by Ballew and Weger, 1979.)</p> <p>However, the problem is you still have to solve $(3)$. I found with a computer search that with $x&lt;y&lt;1000$, the only integers are $x,y,z = 133,\,144,\,165$, so,</p> <p>$$\left(\dfrac{133(133-1)}{2} \right)^2+\left (\dfrac{144(144-1)}{2} \right)^2 = \left(\dfrac{165(165-1)}{2} \right)^2$$</p> <p><strong>P.S.</strong> If you're curious about <em>rational</em> solutions, then your $(1)$ and $(2)$ have infinitely many.</p>

<p>Let there be $T:R^3 \rightarrow R^3$ <br> $T(0,-1,1)=(3,3,3)$ <br> $T(1,0,-1)=(0,1,1)$ <br> $T(1,1,0)=(1,2,-1)$</p> <p>Is (1,2,3) is the only image of the vector $(1, \frac{-7}{9}, \frac{-8}{9})$?</p> <p>I have thought to create a matrix $[T]^T_E$*$[T]^E_C$=$[T]^T_C$ so I will have a matrix that does the transformation and the I can check the calim, but that did not work </p>

robjohn

<p>As was noted by Adam Hughes, the series for $\frac1{1-z}$ about the point $5i$ is $$ \begin{align} \frac1{1-z} &amp;=\frac1{(1-5i)-(z-5i)}\\ &amp;=\frac1{1-5i}\frac1{1-\color{#C00000}{\frac{z-5i}{1-5i}}}\\ &amp;=\frac1{1-5i}\sum_{k=0}^\infty\left(\color{#C00000}{\frac{z-5i}{1-5i}}\right)^k\\ &amp;=\sum_{k=0}^\infty\frac{(z-5i)^k}{(1-5i)^{k+1}} \end{align} $$ which converges by the ratio test for $|z-5i|\lt|1-5i|=\sqrt{26}$.</p> <p>Another indicator of the radius of convergence of the Taylor series is the distance from the center of the expansion to the nearest singularity. Since the only singularity is at $z=1$, the radius of convergence is $$ |1-5i|=\sqrt{26} $$</p>

<p>There is a whiskey made up of 64% corn, 32% rye, and 4% barley that was made by blending other whiskies together. I am trying to figure out if there is a chance the ratio of this whiskey could be the result of blending two, maybe three whiskies of different ratios.</p> <p>The possible whiskies:</p> <p>Whiskey A is 60% corn, 36% rye and 4% barley.</p> <p>Whiskey B is 81% corn, 15% rye, and 4% barley. </p> <p>Whiskey C is 75% corn, 21% rye, and 4% barley</p> <p>I have a feeling there is a possibility because these whiskies all have 4% barley, but I can't figure out if the other percentages match up in any 1:2:3 ratio in the blend. Any help will be greatly appreciated. Thank you.</p>

Henry

<p>You could any of</p> <ul> <li><p>Whiskey D made up of $\dfrac{17}{21}$ of Whiskey A and $\dfrac{4}{21}$ of Whiskey B </p></li> <li><p>Whiskey E made up of $\dfrac{11}{15}$ of Whiskey A and $\dfrac{4}{15}$ of Whiskey C</p></li> <li><p>Any blend of whiskeys D and E</p></li> </ul> <p>and you would have your 64%, 32% and 4% fractions.</p>

<p>There is a whiskey made up of 64% corn, 32% rye, and 4% barley that was made by blending other whiskies together. I am trying to figure out if there is a chance the ratio of this whiskey could be the result of blending two, maybe three whiskies of different ratios.</p> <p>The possible whiskies:</p> <p>Whiskey A is 60% corn, 36% rye and 4% barley.</p> <p>Whiskey B is 81% corn, 15% rye, and 4% barley. </p> <p>Whiskey C is 75% corn, 21% rye, and 4% barley</p> <p>I have a feeling there is a possibility because these whiskies all have 4% barley, but I can't figure out if the other percentages match up in any 1:2:3 ratio in the blend. Any help will be greatly appreciated. Thank you.</p>

Nick D.

<p>Yes, and in infinitely many different ways!</p> <p>Row reduce $\begin{bmatrix} 60&amp;81&amp;75&amp;64\\ 36&amp;15&amp;21&amp;32\\ 4&amp;4&amp;4&amp;4\\ \end{bmatrix}$ to get $\begin{bmatrix} 1&amp;0&amp;\frac{2}{7}&amp;\frac{17}{21}\\ 0&amp;1&amp;\frac{5}{7}&amp;\frac{4}{21}\\ 0&amp;0&amp;0&amp;0\\ \end{bmatrix}$.</p> <p>Let's say:</p> <p>$A$=ratio of Whiskey A in mixture</p> <p>$B$=ratio of Whiskey B in mixture</p> <p>$C$=ratio of Whiskey C in mixture</p> <p>From the row reduction we get the relations: $$(*)A=\frac{17}{21}-\frac{2}{7}C, B=\frac{4}{21}-\frac{5}{7}C$$</p> <p>Since $A,B,C$ are all between $0$ and $1$, these relations put a restriction on $C$: namely $C\leq\frac{4}{15}$. So choose your favorite value for $C$ in the appropriate range, plug it into the equations in $(*)$. That will give you the amounts of whiskeys A and B you need to get the desired mixture.</p>

<p>These days I have read many descriptions of a noncooperative game like the one below.</p> <p>A noncooperative game is a game in which players are unable to make enforceable contracts outside of the rules/description of such a game.</p> <p>As a graduate student majoring in math, I wonder if there is any mathematically formal definition of a noncooperative( and cooperative) game.</p>

Michael

<p>Two mathematicians who won Nobel prizes in economics:</p> <ol> <li><p>Nash</p></li> <li><p>Shapley</p></li> </ol> <p>for work in non-cooperative and cooperative game theory respectively. Google and JSTOR are your friends.</p>

<p>These days I have read many descriptions of a noncooperative game like the one below.</p> <p>A noncooperative game is a game in which players are unable to make enforceable contracts outside of the rules/description of such a game.</p> <p>As a graduate student majoring in math, I wonder if there is any mathematically formal definition of a noncooperative( and cooperative) game.</p>

Sergio Parreiras

<p>I don't think you will ever find a formal definition. The informal one is very good for all purposes. The closest you can get is the definition of a strategic game (= non-cooperative games) see Osborne and Rubinstein (section 2.1) and the definition of coalitional games with and without transferable payoff ( = cooperative games) see again Osborne and Rubinstein (sections 13.1 and 13.5). By the way, Osborne and Rubinstein are freely available online - to get a copy legally, you only need to register.</p>

<p>when you are checking to see if a sum of say <span class="math-container">$k^2$</span> from <span class="math-container">$k=1$</span> to to <span class="math-container">$k=n$</span> is equal to a sum of <span class="math-container">$(k+1)^2$</span> from <span class="math-container">$k=0$</span> to <span class="math-container">$n−1$</span> can someone explain what is going on here. THanks</p> <p>(looking for a fairly simple way to work the problem without writing out the sums which may help me understand what is going on, )</p>

Johann Birnick

<p>Here is an explicit strategy that works for <span class="math-container">$n \leq 100$</span> cards.</p> <p>As seen in previous answers, we use the order of the 4 transmitted cards to encode a number between 1 and 24. Now if the magician can reduce from the <em>set</em> of the 4 transmitted cards the possible 5th cards to at most 24, we are done.</p> <p>Mathematically speaking, we want to find a map <span class="math-container">$$ f: \mathrm{Comb}_5^n \to \mathrm{Comb}_4^n $$</span> that satisfies <span class="math-container">$c \supseteq f(c)$</span> for all <span class="math-container">$c \in \mathrm{Comb}_5^n$</span> and <span class="math-container">$|f^{-1}(c)| \leq 24$</span> for all <span class="math-container">$c \in \mathrm{Comb}_4^n$</span>. Here <span class="math-container">$\mathrm{Comb}_k^n$</span> denotes all the subsets of <span class="math-container">$\{1,...,n\}$</span> of size 5.</p> <p>My proposed map <span class="math-container">$f$</span> is given as follows: WLOG the cards are numbered from 1 to <span class="math-container">$n$</span>. For <span class="math-container">$c \in \mathrm{Comb}_5^n$</span> take the sum of the card numbers modulo 5, call this <span class="math-container">$1 \leq d \leq 5$</span>, and remove the <span class="math-container">$d$</span>^th smallest card.</p> <p>This works as long as <span class="math-container">$n \leq 100$</span>: Given <span class="math-container">$c \in \mathrm{Comb}_4^n$</span>, what is <span class="math-container">$f^{-1}(c)$</span> ? Go through the cards 1,2,3,...,<span class="math-container">$n$</span>, assume this is the removed card, each time compute the respective <span class="math-container">$d$</span>, and check whether this <span class="math-container">$d$</span> would have actually led to remove this card. One quickly sees that this is the case for roughly every 5th card, maybe slightly more. If you think it through this goes for at most <span class="math-container">$\lceil n/5 \rceil + 4$</span>, which must be <span class="math-container">$\leq 24$</span>, leaving us with <span class="math-container">$n \leq 100$</span>.</p> <p>(If you look at the details, e.g. that one can also exclude the 4 cards given, this works actually for slightly bigger <span class="math-container">$n$</span> as well.)</p> <p>It's not thaat practical on a real table, but certainly doable!</p>

<p>I'm very confused at the following question:</p> <blockquote> <p>Find the basis for the image and a basis of the kernel for the following matrix: $\begin{bmatrix} 7 &amp; 0 &amp; 7 \\ 2 &amp; 3 &amp; 8 \\ 9 &amp; 0 &amp; 9 \\ 5 &amp; 6 &amp; 17 \end{bmatrix}$</p> </blockquote> <p>I just don't know how to do any of this. We find the image by doing the following: $\begin{bmatrix} 7 &amp; 2 &amp; 9 &amp; 5 \\\ 0 &amp; 3 &amp; 0 &amp; 6\\ 7 &amp; 8 &amp; 9 &amp;17 \end{bmatrix}$ Then, after doing RREF, we get: $\begin{bmatrix} 1 &amp; 0 &amp; \frac{9}{7} &amp; \frac{1}{7} \\\ 0 &amp; 1 &amp; 0 &amp; 2\\ 0&amp;0&amp;0&amp;0 \end{bmatrix}$. This gives us an image of {$\begin{bmatrix} 1 \\ 0 \\ \frac{9}{7} \\ \frac{1}{7} \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 0 \\ 2 \end{bmatrix}$}. However, I don't know how to proceed from here. Please help me?</p>

Mathematician 42

<p>So you have to find the basis for the image and kernel of the map $$T_A:\mathbb{R}^3\rightarrow \mathbb{R}^4:X\mapsto AX$$ where $$A=\begin{bmatrix} 7 &amp; 0 &amp; 7 \\ 2 &amp; 3 &amp; 8 \\ 9 &amp; 0 &amp; 9 \\ 5 &amp; 6 &amp; 17 \end{bmatrix}.$$</p> <p>Im not a fan of blindly following an algorithm someone showed you. Instead, let's think about things nice and slowly.</p> <p>So what belongs to $\text{im}(T_A)$? Suppose we take $X_1=\begin{bmatrix} 1\\0\\0\end{bmatrix}$. Then $T_A(X_1)=AX_1=\begin{bmatrix}7\\2\\9\\5 \end{bmatrix}$ belongs to the image by definition. Notice that this simply is the first column of $A$. Similarly, consider $X_2=\begin{bmatrix} 0\\1\\0\end{bmatrix},X_3=\begin{bmatrix} 0\\0\\1\end{bmatrix}$, then $T_A(X_i)$ yields the $i$-th column of $A$.</p> <p>Clearly, $\text{im}(T_A)$ is generated by $T_A(X_1),T_A(X_2)$ and $T_A(X_3)$ (why?). So we have to check whether any of these vectors is redundant. </p> <p>Just by eyeballing this, you can see that $T_A(X_3)=T_A(X_1)+2T_A(X_2)$. Thus $T_A(X_3)$ is redundant and clearly $T_A(X_1),T_A(X_2)$ are linearly independent. It follows that $\text{im}(T_A)$ has $\left\{T_A(X_1),T_A(X_2)\right\}$ as a basis. </p> <p>So looking back at the above, we find the following general procedure: The image of $T_A$ is generated by the columns of $A$. If they are not linearly independent, we have to recursively throw away those columns that arise as linear combinations of the others. This exactly what you are doing when you look at the RREF of $A^T$.</p> <p>Okay, so let's move to the kernel of $T_A$. First of all, general theory tells us that $\dim(\mathbb{R}^3)=\dim(\ker(T_A))+\dim(\text{im}(T_A))$. Hence $\dim(\ker(T_A))=1$. So we only have to find one non-zero vector $X$ such that $T_A(X)=0$. Now consider $X=\begin{bmatrix} a\\b\\c\end{bmatrix}=aX_1+bX_2+cX_3$. Then $T_A(X)=aT_A(X_1)+bT_A(X_2)+cT_A(X_3)$. We are looking for $X$ such that $T_A(X)=0$. We already knew that $T_A(X_3)=T_A(X_1)+2T_A(X_2)$ thus $X=\begin{bmatrix} 1\\2\\-1\end{bmatrix}\in \ker(T_A)$. Putting everything together, we find that $\begin{bmatrix} 1\\2\\-1\end{bmatrix}$ generates $\ker(T_A)$.</p>

<p>I've hit a wall on the above question and was unable to find any online examples that also contain trig in $f(g(x))$. I'm sure I am missing something blatantly obvious but I can't quite get it.</p> <p>$$ g(x)=3x+4 , \quad f(g(x)) = \cos\left(x^2\right)$$</p> <p>So far I've managed to get to the point where I have $f(x+8) = \cos\left(x^2\right)$, by solving $g^{-1}(g(x))$ (loosely based on the last bit of advice <a href="https://www.physicsforums.com/threads/if-f-o-g-16x-7-and-g-x-2x-1-find-f-x.518117/">here</a>), but I can't make that final connection.</p> <p>My best attempt so far was $f(x)=\cos(x^2-16x+64)$, but while that does result in $x^2$, it still ends up wrong due to it being cosine.</p>

ThePortakal

<p><strong>Hint:</strong> $(f \circ g) \circ g^{-1} =f \circ (g \circ g^{-1}) = f$</p>

<blockquote> <p>If $P\left(A\right)=0.8\:$ and $P\left(B\right)=0.4$, find the maximum and minimum values of $\:P(A|B)$.</p> </blockquote> <p>My textbook says the answer is $0.5$ to $1$. But I think the answer should be $0$ to $1$.</p> <p>The textbook claims $P(A∩B)$ is $0.2$ when $P(A'∩B')=0$</p> <p>I think that the minimum value arises when $A$ and $B$ are mutually exclusive. So there isn't a chance of both happening and so you have $\frac{0}{0.4} = P(A|B) = 0$ (right?)</p> <p>I agree with the textbook in saying the maximum value is one.</p>

Rushabh Mehta

<p>For completeness, I'll show both bounds of $P(A|B)$.</p> <h1>Lower Bound</h1> <p>We note that $$P(A)+P(B)-P(A\cup B) = P(A\cap B)$$ which is a simple rearrangement of the standard probability summation formula. Note that the upper bound of $P(A\cup B)=1$, so when we substitute, we get $0.8+0.4-1=P(A\cap B) = 0.2$</p> <p>Hence, $$P(A|B) = \frac{P(A\cap B)}{P(B)} = \frac {0.2}{0.4} = \frac12$$</p> <h1>Upper Bound</h1> <p>We reach an upper bound when $A\rightarrow B$. So, $P(A\cap B)=P(B)$, so $$P(A|B) = \frac{P(A\cap B)}{P(B)} = \frac{P(B)}{P(B)} = 1$$</p> <h1>Intuition for Lower Bound</h1> <p>So why does the formula used to compute the lower bound work? Well, that is because when we want to count the probability of $P(A\cup B)$, we realize that its the sum of $P(A)+P(B)$, but we have double counted the probability where they overlap, notably $P(A\cap B)$. This implies that $$P(A)+P(B)-P(A\cap B) = P(A\cup B)$$</p>

<p><strong>Problem statement:</strong></p> <p>There are three spheres. The one which will roll is $\textbf{X}=(x_1,x_2,x_3)$ with radius $R_X$. The other two spheres are $\textbf{A}=(a_1,a_2,a_3)$ with $R_A$ and $\textbf{B}=(b_1,b_2,b_3)$ with $R_B$. They are both below $X$, meaning $x_3&gt;a_3$ and $x_3&gt;b_3$. They both contact $X$, meaning $|\textbf{X}-\textbf{A}| = R_X+R_A$ and similarly for $B$. I need to find the axis of rotation $\textbf{V}$ and point $\textbf{v}$ about which $\textbf{X}$ can rotate to change its position while preserving its contacts with both $\textbf{A}$ and $\textbf{B}$. With this information actually performing the rotation for my task in a granular physics program is simple. </p> <p><strong>Attempted solution:</strong></p> <p>I thought the axis of rotation would be the vector connecting $A$ and $B$. $$\textbf{V}= \textbf{B}-\textbf{A},$$ suitably normalized. I thought the point about which the rotation should occur would come from analyzing the triangle formed by the locations of the three spheres, and breaking it into two right triangles with height $h$ and bases $d$ and $D$, then solving the system</p> <p>$$(R_A+R_X)^2 = d^2 + h^2,$$ $$(R_B + R_X)^2 = D^2 + h^2,$$ $$d+D = |\textbf{B}-\textbf{A}|,$$</p> <p>which is three equations in three unknowns $h$,$d$, and $D$. I thought $h$ would be the length of the lever arm about which to rotate, and I thought $d$ or $D$ would map $\textbf{A}$ to the pivot point as $\textbf{v} = \textbf{V}d$. When I use this for the rotation I get nonsense results. What am I missing? Any help is appreciated! </p> <p>Here is a figure to indicate a possible configuration before rolling. It is not necessary however that the two lower spheres $\textbf{A}$ and $\textbf{B}$ are touching. </p> <p><a href="https://i.stack.imgur.com/exeOf.png" rel="nofollow noreferrer">It works. Thanks!</a></p> <p><a href="https://i.stack.imgur.com/XXhJ6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XXhJ6.png" alt="enter image description here"></a></p>

Travis Willse

<p>By translating coordinates we may as well take ${\bf B} = {\bf 0}$, so that $\ell = \langle {\bf A} \rangle$. Then, there is a (unique) orthogonal decomposition $${\bf X} = {\bf X}^{\parallel} + {\bf X}^{\perp} ,$$ where ${\bf X}^{\parallel} \parallel \ell$ and ${\bf X}^{\perp} \perp \ell$. (Explicitly, ${\bf X}^{\parallel} = \textrm{proj}_{\bf A} {\bf X}$.)</p> <p>Now, as I pointed out in my comment, any rotation fixing the spheres $A$ and $B$ fixes ${\bf A}, {\bf B}$, so it also fixes the unique line $\ell$ through them and hence fixes ${\bf X}^{\parallel}$. On the other hand, rotations about $\ell$ also fix the plane $\Pi$ orthogonal to $\ell$ and $\bf X$, which by construction intersects $\ell$ at ${\bf X}^{\parallel}\textrm{.}$ The "circular track" along which ${\bf X}$ moves under rotation mentioned in the comments is the circle in $\Pi$ centered at $||{\bf X}^{\parallel}||$ of radius $||{\bf X}^{\perp}||$ (this circle degenerates to a single point in the case that ${\bf X} \in \ell$, that is, when the three centers are collinear.</p>

<p>I am seeking a deeper understanding of the representation of set-based objects in terms of Boolean algebras.</p> <p>Let $\wp(A)$ be the set of subsets of a set $A$. A relation $R \subseteq A \times B$ generates two operators $pre: \wp(B) \to \wp(A)$ and $post: \wp(A) \to \wp(B)$ where $pre$ maps a set $X \subseteq A$ to its preimage with respect to $R$ and $post$ maps $X$ to its image.</p> <p>In the standard Stone duality between the category of sets and Boolean algebras, a function is represented using the preimage operator. The preimage operator generated by a function turns out to be a Boolean algebra homomorphism but the image operator may not be a homomorphism. I see this as one reason to choose the preimage to represent a function. My first question is: Are there other reasons to choose the preimage representation? I feel like there should be something deeper going on. </p> <p>If we leave the setting of functions, the preimage operator generated by a relation isn't necessarily a Boolean algebra homomorphism. So, in the representation of a system of relations over a set by a Boolean algebra with operators (in the sense of Jonsson and Tarski) I see no specific motivation for using the preimage, as opposed to image operator. I see why we want to be consistent in convention, and also use the preimage because of its connection to the semantics of modal logic. However, this appears to be an aesthetic choice. My second question is: Is there a specific reason to choose the preimage, rather than image operator, when representing relations as Boolean algebras with operators?</p>

Nik Weaver

<p>Having a subset of a set is "the same" as having a function from that set into $\{0,1\}$ (namely, the function which is $1$ on the subset and $0$ on its complement). If I have a function $f: X \to Y$ I can compose it with functions from $Y$ to $\{0,1\}$ and thereby turn subsets of $Y$ into subsets of $X$. I guess that's one explanation for why <em>pre</em> is natural in a way that <em>post</em> is not.</p>

<p>I am seeking a deeper understanding of the representation of set-based objects in terms of Boolean algebras.</p> <p>Let $\wp(A)$ be the set of subsets of a set $A$. A relation $R \subseteq A \times B$ generates two operators $pre: \wp(B) \to \wp(A)$ and $post: \wp(A) \to \wp(B)$ where $pre$ maps a set $X \subseteq A$ to its preimage with respect to $R$ and $post$ maps $X$ to its image.</p> <p>In the standard Stone duality between the category of sets and Boolean algebras, a function is represented using the preimage operator. The preimage operator generated by a function turns out to be a Boolean algebra homomorphism but the image operator may not be a homomorphism. I see this as one reason to choose the preimage to represent a function. My first question is: Are there other reasons to choose the preimage representation? I feel like there should be something deeper going on. </p> <p>If we leave the setting of functions, the preimage operator generated by a relation isn't necessarily a Boolean algebra homomorphism. So, in the representation of a system of relations over a set by a Boolean algebra with operators (in the sense of Jonsson and Tarski) I see no specific motivation for using the preimage, as opposed to image operator. I see why we want to be consistent in convention, and also use the preimage because of its connection to the semantics of modal logic. However, this appears to be an aesthetic choice. My second question is: Is there a specific reason to choose the preimage, rather than image operator, when representing relations as Boolean algebras with operators?</p>

Andreas Blass

<p>For a function $f:X\to Y$, the operation "pre-image along $f$" from $\mathcal P(Y)$ to $\mathcal P(X)$ has adjoints on both sides. [I'm viewing the power sets as partially ordered by $\subseteq$ and then viewing these partially ordered sets as categories, so that "adjoint" makes sense.] The left adjoint is the familiar image operator. The right adjoint is its (not so familiar) dual, $A\mapsto Y-f(X-A)$. So from this point of view, "image" has the same level of naturality as its dual, while "pre-image" is "better".</p> <p>Concerning the claim of equal naturality, note that the left adjoint sends $A$ to $\{y\in Y:(\exists x\text{ with }f(x)=y)\ x\in A\}$ while the right adjoint sends $A$ to $\{y\in Y:(\forall x\text{ with }f(x)=y)\ x\in A\}$. So they correspond to the two familiar (mutually dual) quantifiers.</p>

<p><strong>Definition:</strong> An ordinal number $\alpha$ is called a <strong><em>limit ordinal number</em></strong> if there is no ordinal number immediately preceding $\alpha$. </p> <p>Now my lecture notes say that $\omega, 2\omega, \omega^2, \omega^\omega$ are limit ordinal numbers whereas $\omega+3,2^\omega+5$ are not which is intuitively clear. But is there a characterization of limit ordinals that may work when proving the case with some rigor? Or is it "just look and identify" sort of a thing using the ordering of the family of ordinal numbers?</p>

Stefan Mesken

<p>You already have a characterization of limit ordinals. In any specific case you just have to verify that this characterization is fulfilled. Consider for example $\omega^2$:</p> <p>$$ \omega^2 = \omega \cdot \omega = \sup \{ \omega \cdot n \mid n &lt; \omega \}. $$</p> <p>Hence, if $\alpha &lt; \omega^2$, there is some $n &lt; \omega$ such that $\alpha &lt; \omega \cdot n$. But then $$ \alpha + 1 &lt; \omega \cdot n + 1 &lt; \omega \cdot (n+1) \le \omega^2. $$</p> <p>Thus $\omega^2$ is a limit ordinal.</p>

<p>Are there general ways for given rational coefficients <span class="math-container">$a,b,c$</span> (I am particularly interested in <span class="math-container">$a=3,b=1,c=8076$</span>, but in general case too) to answer whether this equation has a rational solution or not?</p>

Max Alekseyev

<p>Multiplying by <span class="math-container">$x^2$</span> and denoting <span class="math-container">$X:=-abx^2$</span>, <span class="math-container">$Y:=ab^2xy$</span>, we get an elliptic curve: <span class="math-container">$$Y^2 = X^3-ab^2cX.$$</span> If it turns out that this curve has zero rank, then the number of solutions is finite and they are given by torsion points.</p> <p><s>Luckily this is the case for the given values <span class="math-container">$a,b,c$</span>, while the torsion points do not give a solution to the original equation.</s></p>

<p>How can I prove that these three sets have no common values:</p> <ul> <li>A: {prime numbers} </li> <li>B: {Fibonacci numbers}</li> <li>C: {8|n+1} </li> </ul> <blockquote> <ul> <li>C: for example 15: 15+1 = 16 => 8|16 &lt;= 16/8 = 2</li> <li>C: for example 23: 23+1 = 24 => 8|24 &lt;= 24/8 = 3</li> </ul> </blockquote>

Dietrich Burde

<p>The claim is that Fibonacci primes $F_n$ (which implies that $n$ is prime, except for $n=4$) do not satisfy $F_p\equiv 7\bmod 8$. Let $a(n)$ denote the sequence of Fibonacci primes. Then they satisfy $a(n)\equiv 1,5\bmod 8$, see <a href="http://oeis.org/A005478" rel="nofollow noreferrer">OEIS</a>, because of $a(n) \equiv 1 \bmod 4$ (Cadwell) for $n&gt;2$.</p>

<p>In writing my senior thesis I met the following problem: Sometimes I have some intuition about some mathematical statement. Yet I find it extremely painful trying to put these intuition into precise form on paper. In particular it is very hard to specify the correct condition for statement.</p> <p>Does anyone have some tips for me in doing so? How do you often do it?</p> <p>Let me elaborate a bit further. I believe it is an experience that any mature mathematician must have went through. We want to go to D, and we need to go through A, B, C. But A can not be stated clearly until one knows B, B can not be stated clearly until one knows A and C, and C can not be stated clearly until one knows B. But we sort of have a vague picture of A, B, C in our mind. It sounds very stupid, but I don't know where to start.</p> <p>I wonder if this question is too vague for MO. So please close it if you see fit.</p>

Angelo

<p>You think about it and try to clarify your ideas, till you can write them up precisely. If there is another way, I am not aware of it.</p>

<p>In writing my senior thesis I met the following problem: Sometimes I have some intuition about some mathematical statement. Yet I find it extremely painful trying to put these intuition into precise form on paper. In particular it is very hard to specify the correct condition for statement.</p> <p>Does anyone have some tips for me in doing so? How do you often do it?</p> <p>Let me elaborate a bit further. I believe it is an experience that any mature mathematician must have went through. We want to go to D, and we need to go through A, B, C. But A can not be stated clearly until one knows B, B can not be stated clearly until one knows A and C, and C can not be stated clearly until one knows B. But we sort of have a vague picture of A, B, C in our mind. It sounds very stupid, but I don't know where to start.</p> <p>I wonder if this question is too vague for MO. So please close it if you see fit.</p>

Igor Pak

<p>I see this in students all the time, and I always give the same advice: talk to somebody. Find a friend who would be willing to listen and challenge you on every point. Sit down with a piece of paper, and try to tell her/him the whole story, explain the theorem you are trying to prove, examples, counter-examples, etc. Even if your friend can't help you formalize the ideas, the act of explaining itself might prove very useful to clarify the matter. Good luck! </p>

<p>In writing my senior thesis I met the following problem: Sometimes I have some intuition about some mathematical statement. Yet I find it extremely painful trying to put these intuition into precise form on paper. In particular it is very hard to specify the correct condition for statement.</p> <p>Does anyone have some tips for me in doing so? How do you often do it?</p> <p>Let me elaborate a bit further. I believe it is an experience that any mature mathematician must have went through. We want to go to D, and we need to go through A, B, C. But A can not be stated clearly until one knows B, B can not be stated clearly until one knows A and C, and C can not be stated clearly until one knows B. But we sort of have a vague picture of A, B, C in our mind. It sounds very stupid, but I don't know where to start.</p> <p>I wonder if this question is too vague for MO. So please close it if you see fit.</p>

Cam McLeman

<p>In addition to the other answers...experiment! Write SAGE (or other) code to look at a hundred or a thousand examples of what you're trying to say something about. You'll probably see the pattern more clearly when it's sitting right in front of you in numerical form, and probably catch a class of exceptions you hadn't thought of before. If your particular intuition isn't easily calculable on a computer, then generate as many examples as you can by hand, and organize the data in such a way that your brain can see any patterns more clearly (and use SAGE or other code to simplify tedious calculations along the way).</p>

<p>For how many positive values of $n$ are both $\frac n3$ and $3n$ four-digit integers?</p> <p>Any help is greatly appreciated. I think the smallest n value is 3000 and the largest n value is 3333. Does this make sense?</p>

fleablood

<p>Some basic thoughts.</p> <p>1) $\frac n3 &lt; n &lt; 3n$ </p> <p>2) If $k$ has four digits then $1000 \le k \le 99999$.</p> <p>So therefore</p> <p>$1000 \le \frac n3 &lt; n &lt; 3n \le 9999$</p> <p>So </p> <p>$1000 \le \frac n3 \implies 3000 \le n$.</p> <p>And $3n \le 9999\implies n \le 3333$.</p> <p>So $3000 \le n \le 3333$.</p> <p>A third thing to note is </p> <p>3) if $\frac n3 \in \mathbb Z \iff 3|n$ so</p> <p>the question becomes: How many multiples of $3$ are there between $3000$ and $3333$ inclusively.</p> <p>One final basic thought</p> <p>4) For any set of consecutive integers every $n$th one of them will be divisible by by $n$.</p> <p>So between $3000$ and $3333$ inclusively, there are $3334$ integers (don't forget to count $3000$ as the <em>first</em> one and $3001$ as the <em>second</em> one) so there are about $\frac {334}3 = 111\frac 13$ multiples of $3$. </p> <p>Of the $333$ integers between $3001$ and $3333$ exactly $\frac 13$ of them, $111$ are divisible by $3$. But $3000$ is also divisible by $3$ so that is $112$ multiples of $3$. </p> <p>They are: $3000, 3003,3006,.......,3327, 3330, 3333$.</p>

<p>Question:</p> <p>Assume $x, y$ are elements of a field $F$. Prove that if $xy = 0$, then $x = 0$ or $y = 0$.</p> <p>My thinking:</p> <p>I am not sure how to prove this. <strong>I can only use basic field axioms.</strong> Should I assume that both x and y are not equal to 0 and then prove by contradiction or should I assume one of x and y is not 0 and then prove the other one has to equal 0?</p> <p>Thanks </p>

fleablood

<p>First prove for any element $a $ that $0*a=0$:</p> <p>$0*a = 0*a +0=$</p> <p>$0a+0a -0a =(0+0)a-0a =$</p> <p>$0a-0a=0$.</p> <p>Let $x\ne 0$ and $y\ne 0$. Then $x^{-1},y^{-1} $ and $g= (y^{-1}x^{-1})$ exist.</p> <p>$(xy)*g=(xy)* (y^{-1}x^{-1})=1\ne 0$.</p> <p>Thus $xy\ne 0$</p> <p>So $xy=0$ only if $x$ and $y $ aren't both non-zero.</p>

<p>How do I integrate $\frac{1}{1-x^2}$ without using trigonometric identities or partial fractions? Thanks!</p>

haqnatural

<p>Hint: $$\frac { 1 }{ 1-{ x }^{ 2 } } =\frac { 1 }{ 2 } \left[ \frac { 1 }{ 1-x } +\frac { 1 }{ 1+x } \right] $$</p>

<p><span class="math-container">$\textbf{Question}$</span>: Show that there exist an uncountable subset <span class="math-container">$X$</span> of <span class="math-container">$\mathbb{R}^{n}$</span> with property that every subset of <span class="math-container">$X$</span> with <span class="math-container">$n$</span> elements is a basis of <span class="math-container">$\mathbb{R}^{n}$</span>.</p> <p><span class="math-container">$\textbf{My Attempt}$</span>: For <span class="math-container">$n= 2$</span>, if we take <span class="math-container">$X = \{(\cos \theta, \sin \theta): 0&lt;\theta &lt; \frac{\pi}{2}\}$</span>, then <span class="math-container">$X$</span> has such property. But for <span class="math-container">$n\geq 3$</span>, I can not find any way. Welcome for answer of this question.</p>

Bart Michels

<p>The set of vectors <span class="math-container">$(1, x, x^2, \ldots, x^{n-1})$</span> with <span class="math-container">$x &gt; 0$</span> works, because the condition amounts to the fact that a Vandermonde determinant is nonzero.</p>

<blockquote> <p>Let <span class="math-container">$V=\{(x,y,z)\in\mathbb{R^3}: x^2+y^2\le z, z\le x+2\}$</span></p> <p>Then the volume of V is:</p> <p>(A) Vol(V) = <span class="math-container">$\frac{75}{8}\pi$</span></p> <p>(B) Vol(V) = <span class="math-container">$\frac{81}{32}\pi$</span></p> <p>(C) Vol(V) = <span class="math-container">$\frac{9}{4}\pi$</span></p> <p>(D) Vol(V) = <span class="math-container">$\frac{45}{8}\pi$</span></p> </blockquote> <p>I already understand how the figure sits on 3D space, is limited by a paraboloid on the bottom and a plane on the top, which is inclined. I guessed it would be easier if I used cylindrical or spherical coordinates but I'm having a hard time with the inclined plane. I would appreciate any hint in or guided resolution in order to understand what thing I'm missing. Thank you.</p>

user2661923

<p>Suppose that the problem was changed to minimizing <span class="math-container">$h(x) = \int_x^2 g(t)dt$</span> and further suppose that you were (temporarily) only considering values of <span class="math-container">$x$</span> such that <span class="math-container">$x &lt; 1$</span>. It is clear that</p> <ol> <li><span class="math-container">$h(-6) = 0$</span>.</li> <li>Because of the symmetry around <span class="math-container">$g(t)$</span>, and because the graph of <span class="math-container">$g(t)$</span> for the area between <span class="math-container">$(-2 \leq t \leq 2)$</span> is completely positive, it is clear that there is no value for <span class="math-container">$x &lt; 1$</span> such that <span class="math-container">$h(x) &lt; 0.$</span></li> </ol> <p>Furthermore, when <span class="math-container">$x$</span> is constrained to <span class="math-container">$x &lt; 1$</span>, minimizing <span class="math-container">$f(x)$</span> is equivalent to minimizing <span class="math-container">$h(x)$</span>. That is, <span class="math-container">$f(x)$</span> will have a minimum at <span class="math-container">$(x = x_0 &lt; 1) \iff h(x)$</span> is similarly minimized at <span class="math-container">$(x = x_0 &lt; 1).$</span></p> <p>Consequently, the problem is reduced to considering only those values of <span class="math-container">$x$</span> such that <span class="math-container">$x &gt; 1$</span>. Further, values of <span class="math-container">$x$</span> such that <span class="math-container">$1 \leq x \leq 2$</span> can be considered at the end of this answer.</p> <p>Given that, it is clear that minimizing <span class="math-container">$f(x)$</span> for <span class="math-container">$x &gt; 2$</span> is (again) similar to minimizing <span class="math-container">$h(x)$</span> for <span class="math-container">$x &gt; 2$</span>. Therefore, the following considerations apply, when <span class="math-container">$x&gt;2$</span>.</p> <ol start="3"> <li><span class="math-container">$\int_x^2 g(t)dt = -\int_2^x g(t)dt.$</span></li> <li>The graph of <span class="math-container">$g(t)$</span> for the region <span class="math-container">$2 \leq x \leq 6$</span> is completely negative.</li> </ol> <p>Therefore, it is immediate that for <span class="math-container">$x &gt; 2$</span>, the minimum achievable value for <span class="math-container">$h(x)$</span> will be <span class="math-container">$0 = h(10) = -\int_2^{10} g(t)dt.$</span></p> <p>Consider <span class="math-container">$\int_2^1 g(t)dt = -\int_ 1^2 g(t)dt &lt; 0$</span>.</p> <p>You can see by the symmetry of the graph of <span class="math-container">$g(t)$</span>, that just as <span class="math-container">$h(-6) = h(2) = h(10),~f(-6) = f(2) = f(10).$</span></p>

<p>I'm finding maximum and minimum of a function $f(x,y,z)=x^2+y^2+z^2$ subject to $g(x,y,z)=x^3+y^3-z^3=3$.</p> <p>By the method of Lagrange multiplier, $\bigtriangledown f=\lambda \bigtriangledown g$ and $g=3$ give critical points. So I tried to solve these equalities, i.e.</p> <p>$\quad 2x=3\lambda x^2,\quad 2y=3\lambda y^2,\quad 2z=-3\lambda z^2,\quad x^3+y^3-z^3=3$. </p> <p>But it is too hard for me. Can anybody solve these?</p>

MJD

<p>Since you suggest that you know how to calculate how many even numbers are in a set that <em>does</em> start with 1, why not calculate how many even numbers are in $\{1, 2, \ldots, 456\}$, then how many even numbers are in $\{1, 2, \ldots, 44\}$, and then subtract?</p>

<p>I know that the domain is: $(-\infty,0) \cup (0,\infty)$, or all Reals except Zero.</p> <p>But if a take the "nearest negative number to zero" and then, the "nearest positive number to zero", my function will hugely increase (Y will approach +infinity).</p> <p>And my domain have all numbers that approach Zero in both sides, except Zero.</p> <p>So, why the function 1/x does not increase between $(-\infty,0)$ and $(0,\infty)$? (Sorry for any english error)</p>

Jacob Manaker

<p>There are two phenomena occurring here. This first is that, in an informal sense, the region where $x\mapsto\frac{1}{x}$ is increasing is also where it is infinite, so we traditionally exclude that from the domain. The second is that this region has all been squashed together, so that $x\mapsto\frac{1}{x}$ has no time to increase. </p> <p>To see this, consider $f:\mathbb{R}\setminus\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\to\tilde{\mathbb{R}}$; $$f(x)=\begin{cases} \frac{1}{x+\frac{\pi}{2}} &amp; x\leq-\frac{\pi}{2} \\ \frac{1}{x-\frac{\pi}{2}} &amp; \frac{\pi}{2}\leq x \end{cases}$$ (I'm using $\tilde{\mathbb{R}}$ to denote the <a href="https://en.wikipedia.org/wiki/Extended_real_number_line" rel="nofollow noreferrer">extended reals</a>, so that $f\left(\pm\frac{\pi}{2}\right)=\pm\infty$.)&nbsp;&nbsp; Both $\tilde{\mathbb{R}}$ and $\mathbb{R}\setminus\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ are still ordered sets under the standard ordering inherited from $\mathbb{R}$, so it makes sense to ask whether $f$ is increasing.&nbsp;&nbsp; And, lo-and-behold, this function does act the way you want it to: it decreases on $(-\infty,-\frac{\pi}{2}]\cup[\frac{\pi}{2},\infty)$ and increases on $\left\{-\frac{\pi}{2},\frac{\pi}{2}\right\}=[-\frac{\pi}{2},\frac{\pi}{2}]\cap\left(\mathbb{R}\setminus\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\right)$. </p> <p>(Note that I've done two things here:</p> <ol> <li><p>I've passed to the extended reals, and </p></li> <li><p>I've shifted apart the two "pieces" of $x\mapsto\frac{1}{x}$.) </p></li> </ol> <p>What is the problem with (1.) above?&nbsp;&nbsp; At two points, $f$ is infinite.&nbsp;&nbsp; Now, $x\mapsto\frac{1}{x}$ is normally not considered on the extended reals (for a reason I will get to below), so we obtain a closer analogy by taking $f$ on everywhere where it is finite.&nbsp;&nbsp; The region wherein $f$ is finite is, of course, $\left(\mathbb{R}\setminus\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\right)\setminus\left\{-\frac{\pi}{2},\frac{\pi}{2}\right\}$.&nbsp;&nbsp; But then we've created a problem: $f$ increases on the very small (two-point!) region $\left\{-\frac{\pi}{2},\frac{\pi}{2}\right\}$, and by cutting out the infinite points, we also just lost the section where $f$ is increasing.&nbsp;&nbsp; So one (trivial) reason why $x\mapsto\frac{1}{x}$ does not increase between your two regions is that you have purposefully cut out the regions on where it is increasing, because those point(s) are infinite. </p> <p>Now, what is the problem with (2.)?&nbsp;&nbsp; But what if we left in the infinite parts, and pulled the two pieces together?&nbsp;&nbsp; Then $f(0)=-\infty$, because the point in the "left" part with rightmost abcissa has ordinate $-\infty$.&nbsp;&nbsp; But we also have $f(0)=\infty$, because the point in the "right" branch with leftmost abcissa has ordinate $\infty$.&nbsp;&nbsp; If $f$ is to be a function (even one into $\tilde{\mathbb{R}}$), $f$ needs to have one value at $0$, not two! We can't have both, can we? Actually, we can, iff $-\infty=\infty$, and that definition is <a href="https://en.wikipedia.org/wiki/Real_projective_line" rel="nofollow noreferrer">one convention for infinities</a> used to extend $x\mapsto\frac{1}{x}$ to $0$.&nbsp;&nbsp; If we do that, then we can't order the extended real line, because then $$\begin{align*} -1&amp;&lt;0&lt;1&lt;\cdots\\ &amp;&lt;1000&lt;\cdots\\ &amp;&lt;\infty=-\infty&lt;\cdots\\ &amp;&lt;-1000&lt;\cdots\\ &amp;&lt;-1\end{align*}$$ So the only way to join our two pieces of $x\mapsto\frac{1}{x}$ involves losing the notion of "increasing" we wanted in the first place. </p>

<p>In reviewing some old homework assignments, I found two problems that I really do not understand, despite the fact that I have the answers.</p> <p>The first is: R(x, y) if y = 2^d * x for some nonnegative integer d. What I do not understand about this relation is how it can possibly be transitive (according to my notes it is). My understanding is that if the relation were transitive, the following would apply: if y = 2^d * x and x = 2^d * z, then y = 2^d * z. That seems impossible unless x = z. Am I missing something?</p> <p>The second is: R(x, y) if x and y are both divisible by 17. What I do not understand about this relation is why it is not reflexive. My understanding is that if the relation is reflexive, if x is divisible by 17 then both x and x are divisible by 17. I think that I am possibly applying the quality of reflexiveness incorrectly to this relation, but I am not quite sure.</p> <p>Thank you for any help in correcting these misunderstandings!</p>

57Jimmy

<p>For the first problem: the integer $d$ is not fixed. And indeed, if $y = 2^d x$ and $x = 2^e z$ you get $y = 2^f z$ for $f=d+e$.</p> <p>For the second problem: reflexivity means that for all $x$, $R(x,x)$. But if $x$ is not divisible by $17$, this does not hold!</p>

<p>In reviewing some old homework assignments, I found two problems that I really do not understand, despite the fact that I have the answers.</p> <p>The first is: R(x, y) if y = 2^d * x for some nonnegative integer d. What I do not understand about this relation is how it can possibly be transitive (according to my notes it is). My understanding is that if the relation were transitive, the following would apply: if y = 2^d * x and x = 2^d * z, then y = 2^d * z. That seems impossible unless x = z. Am I missing something?</p> <p>The second is: R(x, y) if x and y are both divisible by 17. What I do not understand about this relation is why it is not reflexive. My understanding is that if the relation is reflexive, if x is divisible by 17 then both x and x are divisible by 17. I think that I am possibly applying the quality of reflexiveness incorrectly to this relation, but I am not quite sure.</p> <p>Thank you for any help in correcting these misunderstandings!</p>

drhab

<p>What your missing where it concerns transitivity of $R$ is the fact that $d$ is not necessarily the same for each pair $x,y$ with $\langle x,y\rangle\in R$.</p> <p>If $y=2^dx$ and $x=2^{d'}z$ for nonnegative integers $d,d'$ then $y=2^{d+d'}z$ for nonnegative integer $d+d'$.</p> <hr> <p>The second relation you mention is not reflexive because we don not have $\langle x,x\rangle\in R$ for every $x$ (preassuming that $R$ is a relation on the set of integers). </p> <p>E.g. $\langle 3,3\rangle\notin R$ since $3$ is not divisible by $17$.</p>

<p>Let $R$ be an integral domain, and let $r \in R$ be a non-zero non-unit. Prove that $r$ is irreducible if and only if every divisor of $r$ is either a unit or an associate of $r$.</p> <p>Proof. ($\leftarrow$) Suppose $r$ is reducible then $r$ can be expressed as $r = ab$ where $a$, $b$ are not units. This contradicts the fact that every divisor of $r$ is either a unit or an associate of $r$.</p> <p>($\rightarrow$) Suppose every divisor of $r$ is neither a unit nor an associate of $r$. Then $r$ is reducible.</p> <p>Can someone verify if my proof is correct. However, I prefer a direct proof. This proof doesn't look nice. Can someone show me how I can do this directly?</p>

user39082

<p>The Statement is not true. Für example $X=\overline{X}={\mathbb R}^2$, which is simply connected, p the Identity, and $A=S^1$. The Fundamental Group of A is nontrivial, so $i_\sharp$ is not injective, but $p^{-1}A=A$ is connected.</p> <p>By the way, the Statement as above has no meaning anyway. The Path component is path connected anyway, you must mean something else.</p>

<p>$$ \begin{matrix} 1 &amp; 0 &amp; -2 \\ 0 &amp; 1 &amp; 1 \\ 0 &amp; 0 &amp; 0 \\ \end{matrix} $$</p> <p>I am told that the span of vectors equal $R^m$ where $m$ is the rows which has a pivot in it. So when describing the span of the above vectors, is it correct it saying that they don't span $R^3$ but only span $R^2$?.</p> <p>Thanks</p>

Disintegrating By Parts

<p>If $X$ is a real (complex) n-dimensional inner product space, then you choose an orthonormal basis $\{ e_1,e_2,\cdots,e_n \}$ of $X$ and establish a map $$ U_e : X \rightarrow \mathbb{C}^n $$ given by $U_ef = (\langle f,e_1\rangle,\langle f,e_2\rangle,\cdots)$. This will map $e_1$ to $(1,0,0,\cdots)$, $e_2$ to $(0,1,0,\cdots)$, etc.. But you can also do this with another orthonormal basis $\{ e_1',e_2',\cdots \}$ of $X$, and you'll have a different correspondence $U_{e'} : X\rightarrow \mathbb{C}^n$. These maps will preserve the inner product, but the basis mapped to the standard basis in $\mathbb{C}^n$ may vary from $U_e$ to $U_{e'}$. The map $U_{e'}^{-1}U_e : X\rightarrow X$ maps the orthnormal basis $e$ to $e'$; so that map is not the identity. Constructing a map $U_e : X\rightarrow \mathbb{C}^n$ requires choosing an orthonormal basis of $X$; so these maps are not all the same. The map $U_{e'}^{-1}U_{e}$ is not the identity; it is an orthogonal map (or unitary map if the space is complex.) The map $U_{e'}U_{e}^{-1} : \mathbb{C}^n\rightarrow\mathcal{C}^n$ is an orthogonal matrix if the spaces are real, and is a unitary matrix if the spaces are complex.</p>

<p>If the stem of a mushroom is modeled as a right circular cylinder with diameter $1$, height $2$, its cap modeled as a hemisphere of radius $a$ the mushroom has axial symmetry, is of uniform density,and its center of mass lies at center of plane where the cap and stem join, then find $a$.</p> <p>I really need help.</p>

xoxox

<p>$\lim \limits_{n\to\infty}{2^n*Pi/2^n}$</p> <p>This obviously is $Pi$. The circles may seem very small, but can't get $0$ (otherwise, the two points at $-1$ and $1$ wouldn't be connected). Consider their radius' as infinitesimal, so the curve length always will stay $Pi$ and will never get $2$.</p> <p>Also, the the length of $2$ would mean that the curve is a straight line and accordingly that it contains all points with $y=0$ and $-1\le x\le1$. There are simply $2^n+1$ points that touch the x-axis (so the x-coordinates are rational), infinitely many, but not all. This might sound contradictory, however, Cantor has proven that there are more real than rational numbers (<a href="https://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument" rel="nofollow">Cantor's Diagonal Argument</a>).</p>

<p>Given the following <span class="math-container">$\frac{1}{2^2}+ \frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{3^2}+\frac{1}{3^3}+...+$</span></p> <p>Can this be symbolized as: <span class="math-container">$$\sum_{n=2}^{\infty}2^{-n}+(n+1)^{-n}$$</span></p> <p>and if so, are the following values for <span class="math-container">$S_{n}$</span> correct?</p> <p><span class="math-container">$$S_n= \begin{cases} \sqrt[\leftroot{-2}\uproot{2}n]{n},\\ -\frac{1}{4}\\ \end{cases}$$</span></p> <p>The formula for <span class="math-container">$S_n$</span> is <span class="math-container">$$S_n=\frac{n}{2}(a_1+a_n)$$</span></p>

user

<p>Your expression seems uncorrect, we can use a double series</p> <p><span class="math-container">$$\sum_{n=2}^{\infty}\left(\sum_{k=2}^{\infty} \frac1{n^k}\right)=\sum_{n=2}^{\infty}\left(\frac{1}{1-\frac1n}-1-\frac1{n}\right)=\sum_{n=2}^{\infty}\frac{1}{n(n-1)} =\sum_{n=2}^{\infty}\left(\frac1{n-1}-\frac1{n}\right)=1$$</span></p>

<p>In many proofs I see that some variable is "fixed" and/or "arbitrary". Sometimes I see only one of them and I miss a clear guideline for it. Could somebody point me to a reliable source (best a well-known standard book) which explains, when and how to use both in proofs?</p> <p>EDIT: A little add-on to the question: Take a usual induction on natural numbers and assume that you are teaching it to students in their first semester. How do you explain "fixed" and "arbitrary" in this scenario?</p>

ryang

<ol> <li><p>The proof of a universally-quantified (“for each <span class="math-container">$x$</span>”) statement might begin with one of these:</p> <ul> <li>Let <span class="math-container">$x$</span> be <strong>an arbitrary</strong> (i.e., <strong>any</strong>) real number greater than <span class="math-container">$7.$</span></li> <li>Consider <strong>any</strong> (i.e., <strong>an arbitrary</strong>) real number <span class="math-container">$x$</span> greater than <span class="math-container">$7.$</span></li> </ul> <p>The idea is that this <em>representative</em> value is fixed only while applying the remaining steps, and that for each iteration of the proof's flow of logic a value is chosen for <span class="math-container">$x.$</span></p> </li> <li><p>The proof is eventually concluded by invoking <a href="https://en.wikipedia.org/wiki/Universal_generalization" rel="nofollow noreferrer">Universal Introduction</a> to convert that opening sentence to</p> <ul> <li><strong>For each</strong> <span class="math-container">$x{&gt;}7,\ldots,$</span></li> </ul> <p>which finally refers <em>generally</em> to <strong>every</strong> element of <span class="math-container">$(7,\infty).$</span></p> </li> <li><p><a href="https://math.stackexchange.com/a/4053215/21813">How to interpret “let... suppose” in mathematics?</a></p> <p><a href="https://matheducators.stackexchange.com/a/25472/372">An overanalysis of the various synonyms of “for each&quot;</a></p> </li> </ol>

<p>In textbooks and online tutorials I see that the remainder is calculated by using a new unknown variable on the same interval. For example we take the Taylor polynomial $T_n(a)$ but find the remainder $R(x)$ with a new variable $z$ inside it. See <a href="http://www.millersville.edu/~bikenaga/calculus/remainder-term/remainder-term.html" rel="nofollow">this tutorial</a> for an example.</p> <p>Maybe I'm dense but I haven't seen an explanation for this other variable being used. Or at least not one I can understand. I originally thought the remainder was the difference between the value of $f(a)$ and $T_n(a)$ but it appears instead that it is a function like the derivative that can find the error at any point. But I'm not even sure if that's the case.</p> <p>What is the purpose of $z$ in that example? And why do we find the remainder at an arbitrary value instead of the value we are estimating? </p> <p>I'm in Calc II if that's relevant. Thanks.</p>

Christian Blatter

<p>Taylor's theorem with remainder term is not meant to allow the <em>exact</em> computation of $f(x)$ for some $x$ near $a$, using a hell of a detour. The remainder term is just intended to be a help in <em>estimating</em> the error when you replace the exact value $f(x)$ by its $n$th Taylor approximation $j_{\&gt;a}^nf(x)$. The size of the error is connected with the distance $|x-a|$ and the size of $f^{n+1}$ in the neighborhood of $a$ in a particular way. Writing $f^{n+1}(\xi)$, whereby $\xi$ is some unknown point between $a$ and $x$, is just a handy way of referencing the dependence of the error on $f^{(n+1)}$.</p>

<p>If the series $\sum\limits_{n=1}^{\infty} u_n $ is convergent then the sequence $u_n \rightarrow 0$ as $n \rightarrow \infty$. Therefore if the ratio test $R=\frac{u_{n+1}}{u_n}$ gives $R&lt;1$ then we can conclude that $(u_{n})_{n\in \mathbb{N}}$ is convergent, right?</p> <p>Generally, if we can find that a series is convergent then can we always conclude that the sequence that is summed up is convergent, provided that it is not an alternating series?</p>

Logan Clark

<p>If I'm understanding you correctly, then yes, you're right. The ratio test states that if the limit is less than one, the sum converges. I think you might be getting caught up in the terms.</p>

<p>I am trying to solve a modular arithmetic system and I got to the point where I have to solve $22t \equiv 9 \pmod{7}$ for $t$. I researched on the internet and found that there are many ways to solve this, such as using linear diophantine equations, the euclidean algorithm or using inverses.</p> <p>Can someone show me, using which ever method they choose how to solve for $t$.</p>

Phillip Hamilton

<p>Without learning more theory, the basic identity here is that $$ a \equiv b\; (mod\; c)$$ if and only if $$ \; c\; | \;a - b $$ </p> <p>So for $22t\; \equiv 9\;(mod \;7)$, we see $7\;|\;22t - 9$, which means we can rewrite this as a linear combination (where the Euclidean Algorithm would come into play) as $$22t - 9 = 7n$$ for some $n \in \mathbb Z$</p> <p>But this is straightforward enough that we can just quickly look for the smallest smallest $t \in \mathbb N$ such that some $n \in \mathbb Z$ gives us a difference of 9. And we can quickly see that there is no such combination for $t=1$, and that the next smallest possibility $t=2$ and $n=-5$ works.</p> <p>So $t=2$, hence $t \equiv 2\;(mod\;7)$</p>

<p>I haven't been able to come up with a counterexample so far. </p>

Teoc

<p>Take $n=4$, $a=3$, $b=6$. Then you have a counterexample, which shows that the claim is invalid.</p>

<p>Let $(a_{n,k})_{n, k \in \mathbb N} \subset \mathbb C$ be a series satisfying</p> <p>$$ \sum_{n=0}^\infty \left| \sum_{k=0}^\infty a_{n,k}\right| \lt \infty $$ and $$ \sum_{k=0}^\infty \left|a_{n,k}\right| \lt \infty \qquad \forall n\in \mathbb N. $$ Does this imply that $\sum_{k=0}^\infty \sum_{n=0}^\infty a_{n,k}$ converges with $$ \sum_{k=0}^\infty \sum_{n=0}^\infty a_{n,k} = \sum_{n=0}^\infty \sum_{k=0}^\infty a_{n,k}, $$ ie. the order of summation can be rearranged?</p> <p>In a lecture a proof was given (at least I think so!) for the case that instead of the first condition above the series satisfies $$ \sum_{n=0}^\infty \sum_{k=0}^\infty |a_{n,k}| \lt \infty. $$ (Am I right that this last condition itself is enough to justify switching the order because of the Fubini theorem?)</p> <p>Can the series be rearranged in general or is there a (nasty) counter example?</p>

zhw.

<p>Consider the double series</p> <p>$$1-1+0+0+0+0 +\cdots$$ $$0+2-2+0+0+0+\cdots $$ $$0+0+3-3+0+0 +\cdots $$ $$0+0+0 +4 -4+ 0 + \cdots$$ $$0+ 0+0+0+5-5+\cdots$$</p> <p>where of course we continue the rows in this pattern on out to $\infty.$ We have absolutely convergent series in each row and column. But the interated double series sums to $0$ in one order, and $\infty$ in the opposite order.</p>

<p>I have a lot of sum questions right now ... could someone give me the convergence of, and/or formula for, $\sum_{n=2}^{\infty} \frac{1}{n^k}$ when $k$ is a fixed integer greater than or equal to 2? Thanks!!</p> <p>P.S. If there's a good way to google or look up answers to these kinds of simple questions ... I'd love to know it...</p> <p>Edit: Can I solve it by integrating $\frac{1}{x^k}$ ? I can show it converges, but to find the formula? Is my question just the Riemann Zeta function?</p> <p>(edit)</p> <p>Thanks guys! This got me the following result:</p> <p>$\sum_{p} \frac{1}{p} \log \frac{p}{p-1} ~ ~ \leq ~ \zeta(2)$</p> <p>summing over all primes $p$. (And RHS is Riemann zeta function.)</p> <p>First, sum over all integers $p$ instead of primes. Then transform the log into $\sum_{m=1}^{\infty} \frac{1}{m p^{m}}$ (<a href="http://en.wikipedia.org/wiki/Natural_logarithm#Derivative.2C_Taylor_series" rel="nofollow">reference: wikipedia</a>. I know.). Now we have (with rearranging):</p> <p>$\leq ~ \sum_{m=1}^{\infty} \frac{1}{m} \sum_{p=2}^{\infty} \frac{1}{p^{m+1}}$</p> <p>By the result of this question (Arturo's answer), this inner sum, which is $\zeta(m+1)$ is at most $\frac{1}{m+1-1} = \frac{1}{m}$. So we have</p> <p>$\leq ~ ~ \sum_{m=1}^{\infty} \frac{1}{m} \frac{1}{m} = \zeta(2)$</p> <p>I think this is a very pretty little proof. Thanks again, hope you math people enjoyed reading this....</p>

Henry

<p>You are looking for the <em>Riemann zeta function</em> $\zeta(k)$ (or close to it: the sum usually starts at $n=1$). </p> <p>Since you are supposed to be doing research for a class project, perhaps you should search for it.</p>

<p>So I'm a bit confused with calculating a double integral when a circle isn't centered on $(0,0)$. </p> <p>For example: Calculating $\iint(x+4y)\,dx\,dy$ of the area $D: x^2-6x+y^2-4y\le12$. So I kind of understand how to center the circle and solve this with polar coordinates. Since the circle equation is $(x-3)^2+(y-2)^2=25$, I can translate it to $(u+3)^2+(v+2)^2=25$ and go on from there.</p> <p>However I would like to know if I could solve this without translating the circle to the origin. I thought I could, so I simply tried solving $\iint(x+4y)\,dx\,dy$ by doing this: $\int_0^{2\pi}\,d\phi\int_0^5(r\cos\phi + 4r\sin\phi)r\,dr$ but this doesn't work. I'm sure I'm missing something, but why should it be different? the radius is between 0 and 5 in the original circle as well, etc.</p> <p>So my questions are:</p> <ol> <li><p>How can I calculate something like the above integral without translating the circle to the origin? What am I doing wrong?</p></li> <li><p>I would appreciate a good explanation of what are the steps exactly when translating the circle. I kind of "winged it" with just saying "OK, I have to move the $X$ back by 3, so I'll call it $X+3$, the same with the $Y$ etc. If someone could give a clear breakdown of the steps that would be very nice :)</p></li> </ol> <p>Thanks!</p>

rogerl

<p>Figure out where the 13th day of each month falls relative to January 1st (remember there are two possibilities, corresponding to leap year or non leap year). There are only seven possible values: 0, 1, 2, 3, 4, 5, 6 days after the day on which Jan. 1 falls. If each of those values occurs at least once, then at least one of those days must be a Friday. Can you take it from there?</p>

<p>So I'm a bit confused with calculating a double integral when a circle isn't centered on $(0,0)$. </p> <p>For example: Calculating $\iint(x+4y)\,dx\,dy$ of the area $D: x^2-6x+y^2-4y\le12$. So I kind of understand how to center the circle and solve this with polar coordinates. Since the circle equation is $(x-3)^2+(y-2)^2=25$, I can translate it to $(u+3)^2+(v+2)^2=25$ and go on from there.</p> <p>However I would like to know if I could solve this without translating the circle to the origin. I thought I could, so I simply tried solving $\iint(x+4y)\,dx\,dy$ by doing this: $\int_0^{2\pi}\,d\phi\int_0^5(r\cos\phi + 4r\sin\phi)r\,dr$ but this doesn't work. I'm sure I'm missing something, but why should it be different? the radius is between 0 and 5 in the original circle as well, etc.</p> <p>So my questions are:</p> <ol> <li><p>How can I calculate something like the above integral without translating the circle to the origin? What am I doing wrong?</p></li> <li><p>I would appreciate a good explanation of what are the steps exactly when translating the circle. I kind of "winged it" with just saying "OK, I have to move the $X$ back by 3, so I'll call it $X+3$, the same with the $Y$ etc. If someone could give a clear breakdown of the steps that would be very nice :)</p></li> </ol> <p>Thanks!</p>

Rebecca J. Stones

<p>In fact, every year will contain a Friday the 13-th between March and October (so leap years don't enter into it).</p> <p>If March 13 is assigned $0 \pmod 7$, then the other moduli occur as indicated below: $$(\underbrace{\underbrace{\underbrace{\underbrace{\underbrace{\underbrace{\overbrace{31}^{\text{March}}}_{3 \pmod 7},\overbrace{30}^{\text{April}}}_{5 \pmod 7},\overbrace{31}^{\text{May}}}_{1 \pmod 7},\overbrace{30}^{\text{June}},\overbrace{31}^{\text{July}}}_{6 \pmod 7},\overbrace{31}^{\text{August}}}_{2 \pmod 7},\overbrace{30}^{\text{September}}}_{4 \pmod 7})$$</p>

<p>A corollary to the Intermediate Value Theorem is that if $f(x)$ is a continuous real-valued function on an interval $I$, then the set $f(I)$ is also an interval or a single point.</p> <p>Is the converse true? Suppose $f(x)$ is defined on an interval $I$ and that $f(I)$ is an interval. Is $f(x)$ continuous on $I$? </p> <p>Would the answer change if $f(x)$ is one-to-one?</p>

lhf

<p>Here is one converse:</p> <blockquote> <p>If $f$ is monotone and $f(I)$ is an interval, then $f$ is continuous.</p> </blockquote>

<p>If we a sample of <span class="math-container">$n$</span> values from a given population and if <span class="math-container">$X$</span> is the variable of the sample, then the mean of <span class="math-container">$X$</span> is just <span class="math-container">$\dfrac{ \sum x }{n}$</span></p> <p>Now, suppose <span class="math-container">$X$</span> is random variable. For concreteness, let us take <span class="math-container">$X$</span> to be the number of heads in a toss of a coin. Now, <span class="math-container">$X$</span> can be 0 or 1. Therefore, the mean in this case is <span class="math-container">$\dfrac{0+1}{2} = \dfrac{1}{2}$</span> and the formula coincides with the above, the one for samples. </p> <p>In general, for random variable <span class="math-container">$X$</span>, we know the mean is <span class="math-container">$\sum x P(X=x)$</span>.</p> <p>But, how is this different from the mean for samples? What is the motivation for this defition? </p>

Angela Pretorius

<p>This is a birds-eye view of a torus with <span class="math-container">$K_6$</span> embedded on it's surface. You can only see the three vertices on the top of the torus, but I'm sure you can imagine how they connect to the three vertices on the bottom of the torus. <a href="https://i.stack.imgur.com/zF2V9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zF2V9.png" alt="enter image description here"></a></p> <p>The projective plane can be formed by twisting and gluing opposite sides of a square together. When this square is glued together, all the edges should join up to make a copy of <span class="math-container">$K_6$</span>. <a href="https://i.stack.imgur.com/5x9rR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5x9rR.png" alt="enter image description here"></a></p>

<p><span class="math-container">$\lim_{n\to \infty}(0.9999+\frac{1}{n})^n$</span></p> <p>Using Binomial theorem:</p> <p><span class="math-container">$(0.9999+\frac{1}{n})^n={n \choose 0}*0.9999^n+{n \choose 1}*0.9999^{n-1}*\frac{1}{n}+{n \choose 2}*0.9999^{n-2}*(\frac{1}{n})^2+...+{n \choose n-1}*0.9999*(\frac{1}{n})^{n-1}+{n \choose n}*(\frac{1}{n})^n=0.9999^n+0.9999^{n-1}+\frac{n-1}{2n}*0.9999^{n-2}+...+n*0.9999*(\frac{1}{n})^{n-1}+(\frac{1}{n})^n$</span></p> <p>A limit of each element presented above is 0. How should I prove that limit of "invisible" elements (I mean elements in "+..+") is also 0?</p>

farruhota

<p>Alternatively, consider the series: <span class="math-container">$\sum_{n=1}^{\infty} (0.9999+\frac{1}{n})^n$</span>, which converges by the root test: <span class="math-container">$$\lim_{n\to \infty} a_n^{1/n}=\lim_{n\to \infty} (0.9999+\frac1n)=0.9999&lt;1.$$</span> Hence: <span class="math-container">$$\lim_{n\to\infty} a_n=\lim_{n\to\infty} (0.9999+\frac{1}{n})^n=0.$$</span></p>

<p>What is the sum of this series ?</p> <p>$(n-1)+(n-2)+(n-3)+...+(n-k)$ </p> <p>$(n-1)+(n-2)+...+3+2+1 = \frac{n(n-1)}{2}$</p> <p>So how can we find the sum from $n-1$ to $n-k$ ?</p>

Ron Gordon

<p>$ n k $ minus the sum from $1$ to $k$ equals </p> <p>$$ n k - \frac12 k (k+1)$$</p>

<p>What is the sum of this series ?</p> <p>$(n-1)+(n-2)+(n-3)+...+(n-k)$ </p> <p>$(n-1)+(n-2)+...+3+2+1 = \frac{n(n-1)}{2}$</p> <p>So how can we find the sum from $n-1$ to $n-k$ ?</p>

Nick Peterson

<p><strong>Hint:</strong></p> <p>Try writing: $$ \sum_{k=1}^{n-1}k=\sum_{k=1}^{n-k-1}k+\sum_{k=n-k}^{n-1}k. $$ Your formula allows you to find the first two sums; subtraction should do the rest!</p>

<p>I'm considering the original coupon collector problem with a small modification. For the sake of completeness I shall state the original problem again first, where <strong>my question is at the end</strong>. </p> <p>say there is a coupon inside every packet of wafers, for the moment let's assume there are only two distinct coupons $C_1$ and $C_2$ that can be collected. How many times do you need to buy the wafers on average to collect both coupons? </p> <p>The solution to this problem as a classical coupon collector problem is 3. See for example <a href="https://en.wikipedia.org/wiki/Coupon_collector%27s_problem" rel="nofollow noreferrer">Wikipedia</a>.</p> <p><strong>Now my question:</strong> </p> <blockquote> <p>How many times on average, should I buy the wafers if I want at least one $C_2$ to be collected before one $C_1$?</p> </blockquote>

Especially Lime

<p>I'm not sure I understand the problem either, but do you mean "how many packets (on average) do I have to buy to get a $C_2$ followed by a $C_1$"? </p> <p>If so, the answer is $4$. You want the next $C_1$ after the first $C_2$. The expected number of packets up to and including the first $C_2$ is $2$ (expectation of a geometric random variable), and then the expected number of packets after that up to and including the next $C_1$ is also $2$.</p>

<p>For training I have decided to solve this limit of a succession</p> <p><span class="math-container">$$\lim\limits_{n\to \infty }\left(3^{n+1}-3^{\sqrt{n^2-1}}\right).$$</span></p> <p><strong>My first attempt</strong>:</p> <p><span class="math-container">\begin{split} \lim_{n\to \infty }\left(3^{n+1}-3^{\sqrt{n^2-1}}\right) =\\ &amp;=\lim_{n\to \infty } 3^{n+1}\left(1-\frac{3^{n\cdot \sqrt{1-\frac{1}{n^2}}}}{3\cdot 3^{n}}\right)\\ &amp;=\ldots \end{split}</span></p> <p>I have abandoned this approach because I wanted to use a notable limit (if <span class="math-container">$\{b_n\}\to 0$</span>) that could probably be useful; but seeing the rounded brackets I have thought that I have occurred many times:</p> <p><span class="math-container">$$\lim_{n\to \infty} \frac{a^{b_n}-1}{b_n}$$</span></p> <p><strong>My second attempt</strong>:</p> <p><span class="math-container">\begin{split} \lim_{n\to \infty }\left(3^{n+1}-3^{\sqrt{n^2-1}}\right) =\\ &amp;=\lim_{n\to \infty } 3^{\sqrt{n^2-1}}\left(\frac{3^{n+1}}{3^{\sqrt{n^2-1}}}-1\right)\\ &amp;=\ldots \end{split}</span></p> <p>Considering only</p> <p><span class="math-container">$$\frac{3^{n+1}}{3^{\sqrt{n^2-1}}}=3^{(n+1-\sqrt{n^2-1})}$$</span></p> <p>Taking only the exponent I have:</p> <p><span class="math-container">$$n+1-\sqrt{n^2-1}=\frac{((n+1)-\sqrt{n^2-1})\cdot ((n+1)+\sqrt{n^2-1})}{(n+1)+\sqrt{n^2-1}}$$</span> <span class="math-container">$$=\frac{2n+2}{n+1+\sqrt{n^2+1}} \tag 1$$</span></p> <p>If I take the limit of the <span class="math-container">$(1)$</span>, easily:</p> <p><span class="math-container">$$\lim_{n\to \infty}\frac{2n+2}{n+1+\sqrt{n^2+1}}=1$$</span></p> <p>i.e.</p> <p><span class="math-container">$$\lim_{n\to \infty}3^{(n+1-\sqrt{n^2-1})}=3$$</span></p> <p>Definitely I think that this limit is <span class="math-container">$\infty\cdot 2=+\infty$</span>. I think that is correct. Do you users think there are better alternatives? Thanks in advance.</p>

Servaes

<p>For <span class="math-container">$n&gt;1$</span> you have <span class="math-container">$$3^{n+1}-3^{\sqrt{n^2-1}}&gt;3^{n+1}-3^{\sqrt{n^2}}=3^{n+1}-3^n=2\cdot3^n.$$</span></p>

<blockquote> <p>Consider the symmetric group of$S_{20}$ and it's subgroup $A_{20}$ consisting of all even permutations. Let $H$ be a $7$-Sylow subgroup of$A_{20}$. Is $H$ cyclic? And is correct the statement which says that any $7$-Sylow subgroup of $S_{20}$ is subset of $A_{20}$?</p> </blockquote> <p>I know that order of H is 49 and H is not normal subgroup of $A_{20}$. But I don't understand whether it is cyclic or not.</p>

p Groups

<p>In $|S_{20}|$, the highest power of $7$ which divides $20!$ is $7^2$. So it is clear that the Sylow-$7$ subgroup of $S_{20}$ is of order $7^2$. </p> <p>Group of order $7^2$ is either cyclic or isomorphic to $Z_7\times Z_7$. </p> <p>If it is cyclic, then $S_{20}$ will have an element of order $49$, and it should be product of disjoint cycles. Check, whether this is possible? You will reach to your answer.</p>

<p>I've been told that strong induction and weak induction are equivalent. However, in all of the proofs I've seen, I've only seen the proof done with the easier method in that case. I've never seen a proof (in the context of teaching mathematical induction), that does the same proof in both ways, and I can't seem to figure out how to do them myself. It would put my mind at ease if I could see with my own eyes that a proof done with strong induction can be completed with weak induction. Does anyone have a link to proofs proved with both, or could anyone show me a simple proof here? I'm more interested in proofs were strong induction is the easier method.</p>

Arturo Magidin

<p>Statements that say that two propositions are equivalent have to be done carefully, because the background theory is important.</p> <p>Specifically, you are talking about two statements about the natural numbers:</p> <ol> <li><p><strong>Induction</strong> (or "weak" induction): Let $S\subseteq \mathbb{N}$ be such that:</p> <ul> <li>$0\in S$; and</li> <li>For all $n\in\mathbb{N}$, if $n\in S$ then $s(n)\in S$.</li> </ul> <p>Then $S=\mathbb{N}$.</p></li> <li><p><strong>Strong induction</strong>: Let $S\subseteq \mathbb{N}$ be such that:</p> <ul> <li>For all $n\in\mathbb{N}$, if $\{k\in\mathbb{N}\mid k\lt n\}\subseteq S$ then $n\in S$.</li> </ul> <p>Then $S=\mathbb{N}$.</p></li> </ol> <p>Above, $s(n)$ is the successor function.</p> <p>The main difficulty is to establish exactly what our "background" is. The induction Schema makes sense in the context of Peano's postulates; Strong induction requires a defined property of $\lt$.</p> <p>Moreover, it is <strong>not</strong> the case that induction and strong induction are equivalent axioms! That is, if we take the other four Peano axioms,</p> <ol> <li>$0\in\mathbb{N}$.</li> <li>If $n\in\mathbb{N}$, then $s(n)\in\mathbb{N}$.</li> <li>For all $n\in\mathbb{N}$, $0\neq s(n)$. </li> <li>If $s(n)=s(m)$ then $n=m$.</li> </ol> <p>then it is <strong>not</strong> true that Axioms 1-4 + Induction yields a theory equivalent to Axioms 1-4 + Strong induction, even if you throw in an order so that you can state Strong induction!</p> <p>To see this, consider a disjoint union of two copies of the natural numbers; let's call one copy the "green" natural numbers, and the other copy the "purple" natural numbers (I usually use blue and red, but let's avoid politics this year...). We interpret the primitives as follows: $\mathbb{N}$ is the set that contains all green and all purple natural numbers. $0$ corresponds to the green $0$. If $n$ is green, then $s(n)$ is the green $n+1$; if $n$ is purple, then $s(n)$ is the purple $n+1$. The order is defined as follows: if $n$ is green and $m$ is purple, then $n\lt m$. If both $n$ and $m$ are of the same color, then $n\lt m$ if and only if $n$ is smaller than $m$ in the usual order.</p> <p>This model satisfies Peano's axioms 1 through 4; it also satisifies the strong induction postulate.</p> <p>However, in this model, the set $S$ of all green natural numbers satisfies the hypothesis of "Induction" but is <em>not</em> all of $\mathbb{N}$: green $0$ is in $S$, and if $n$ is a green natural number, then so is $s(n)$. This means that this set is <em>not</em> a model for Peano arithmetic. So it is <em>false</em> that weak and strong induction can be swapped with one another and yield equivalent theories with the other four Peano postulates.</p> <p>However, if we add a further property, namely</p> <blockquote> <p>For every $n\in\mathbb{N}$, either $n=0$ or else there exists $m\in\mathbb{N}$ such that $n=s(m)$;</p> </blockquote> <p>then the first four axioms, plus this property, plus strong induction <em>does</em> imply weak induction. </p> <p>I guess the moral is that the statement "weak induction is equivalent to strong induction" has to be made precise before it is true; one has to specify a "background theory" or a set of "background properties" which we may take for granted before the equivalence is established. But in the presence of those "background properties", then Patrick Da Silva's argument is the standard one: any proof (over a suitable theory for $\mathbb{N}$) that uses induction can be reshaped (in a straightforward way) to become a proof that uses strong induction instead; and any proof that uses strong induction can be reshaped (in a more-or-less algorithmic manner) into a proof that uses weak induction.</p>

<p>Let A and B be subsets of $R^n$ Define</p> <p>$A+B=\{a+b\ |\ a\in\, A , b\in \,B\}$</p> <p>Consider the sets $W=\{(x,y) \in\,R^2\ |\ x&gt;0 , y&gt;0\} \\ X=\{(x,y) \in\,R^2\ |\ x\in\,R , y=0\} \\ Y=\{(x,y) \in\,R^2\ |\ xy=1\} \\Z=\{(x,y) \in\,R^2\ |\ |x|\le 1,|y|\le 1\}$<br> Which of the following statements are true </p> <ol> <li>The Set $W+X$ is open</li> <li>The Set $X+Y$ is closed</li> <li>The Set $Y+Z$ is closed</li> </ol> <p>Try I sincerely don't know how to proceed The answer is 1 and 3 I do know that sum of two open set is open in $R^n$ but that doesn't work over here </p>

Aweygan

<p>Here are a few hints for each one.</p> <p>$(1)$ Observe that $A$ is open (Why?), that the translation of any open set is open (Why?), and $$ W+X=\bigcup_{x\in X}(W+x). $$</p> <p>$(2)$ Find a sequence in $X+Y$ convergent in $\mathbb{R}^2$, that does not converge in $X+Y$.</p> <p>$(3)$ Observe that $Z$ is compact (Heine-Borel), and $Y$ is closed (Why?). Try to show that the sum of a closed and a compact set is closed. </p>

<p>Prove that for any primitive Pythagorean triple (a, b, c), exactly one of a and b must be a multiple of 3, and c cannot be a multiple of 3.</p> <p><strong>My attempt:</strong></p> <p>Let a and b be relatively prime positive integers.</p> <p>If $a\equiv \pm1 \pmod{3}$ and $b\equiv \pm1 \pmod{3}$, </p> <p>$c^2=a^2+b^2\equiv 1+1\equiv 2 \pmod{3}$</p> <p>This is impossible as the only quadratic residues modulo 3 are 0 and 1.</p> <p><em>So far, so good.</em></p> <p>If one of a, b is $\equiv 0 \pmod{3}$ and the other is $\equiv \pm1 \pmod{3}$,</p> <p>$c^2=a^2+b^2\equiv 0+1\equiv 1 \pmod{3}$</p> <p><em>This is the part I don't understand. Just because $c^2\equiv 1\pmod{3}$ doesn't mean that $c^2$ must be a perfect square. For example, $a=12$ and $b=13$ satisfy the above conditions but $c^2=a^2+b^2=313$, which isn't a perfect square.</em></p>

Sam Weatherhog

<p>Your proof is complete. You are asked to show that one of $a,b,c$ is divisible by $3$. In the first part you show that $a$ and $b$ can't both be non-divisible by $3$. In the second part, you assume that one of $a,b$ is divisible by $3$ and show that $c^2\equiv 1$ (mod $3$) which implies that $c$ is not divisible by $3$ and hence exactly one of $a,b,c$ is divisible by $3$.</p>

<p>An even graph is a graph all of whose vertices have even degree. </p> <p>A spanning subgraph $H$ of $G$, denoted by $H \subseteq_{sp} G$, is a graph obtained by $G$ by deleting <em>only</em> edges of $G$.</p> <p>I want to show that if $G$ is a connected graph, then $\big|\{H \subseteq_{sp} G | H$ $is$ $even\}\big| = 2^{e-n+1}$, where $e$ is the number of edges and $n$ the number of vertices of $G$.</p> <p>Can anyone give me a solution or a hint? Thanks in advance!</p>

Hw Chu

<p>Let $v_1, \cdots, v_n$ be the vertices of the graph. Fixing the number of vertices, let us do induction on $e$, the number of edges.</p> <p>Since $G$ is connected, The base case is $e = n-1$, which happens when $G$ is a tree. If $H \subseteq_{sp}G$ is even and contains at least one edge, then $H$ contains a cycle, but a tree can have no cycle, this is impossible. So the only even spanning subgraph is the trivial one, i.e., with $n$ vertices and no edges. So $|\{H\subseteq_{sp}G|H \mbox{ is even}\}| = 1$.</p> <p>Suppose the assertion holds when $e = m \geq n-1$. Now suppose that $G$ is a connected graph with $m+1$ edges. Let $G'$ be a connected graph obtained by removing an edge $e_0 = v_1v_2$ from $G$. By induction hypothesis $|\{H\subseteq_{sp}G'|H \mbox{ is even}\}| = 2^{m-n+1}$.</p> <p>How many even spanning subgraoh are there in $G$? There are $2^{m-n+1}$ of them which does not contain $e_0$, and $2^{m-n+1}$ of them containing $e_0$. To see this, let $v_1v_{a_1}v_{a_2}\cdots v_{a_{n-1}}v_2$ be a simple path (no repeated vertices) from $v_1$ to $v_2$. This must exists because $G'$ is connected. Let $\tilde H$ be a even spanning subgraph of $G$ with edges $e_0, v_1v_{a_1}, v_{a_1}v_{a_2}, \cdots, v_{a_{n-1}}v_2$. Then you can show that the following exhausts all even spanning subgraphs containing $e_0$:</p> <p>$$ \{\tilde H \oplus H \ | \ H \subseteq_{sp} G', H \mbox{ is even} \}, $$</p> <p>where $H_1 \oplus H_2$ means the "exclusive or" of $H_1$ and $H_2$, that is, the spanning subgraph whose edges are those which appears exactly once in $H_1$ and $H_2$.</p> <p>This completes the sketch of the proof.</p>

<p>Could somebody please show me how to integrate the following:</p> <p>$dA/dt = -kA$</p> <p>I'm told that the answer is:</p> <p>$A(t) = A(0)e^-kt$</p> <p>but I do not know why. Could you be explicit in your answer and explain precisely why it works? </p>

Dr. Sonnhard Graubner

<p>if we have $f(x)=-5x^2+12x-7$ so we get $f'(x)=-10x+12$ and now must be $-10x+12\geq 0$ or $-10x+12\le 0$</p>

<p>Could somebody please show me how to integrate the following:</p> <p>$dA/dt = -kA$</p> <p>I'm told that the answer is:</p> <p>$A(t) = A(0)e^-kt$</p> <p>but I do not know why. Could you be explicit in your answer and explain precisely why it works? </p>

Prakhar Nagpal

<p>So your first clue should be that it is a function of degree <span class="math-container">$2$</span> so it will have only one maximum or minima, in this case, maxima since it is a quadratic opening downward<span class="math-container">$\rightarrow a \lt 0$</span>. I am including the reason for this as well. Let us say we have a general quadratic <span class="math-container">$$ax^2 + bx + c$$</span> Now taking <span class="math-container">$x^2$</span> common we get <span class="math-container">$$x^2 \Bigl(a + \frac{b}{x} + \frac{c}{x^2}\Bigl)$$</span> Now to determine whether it will open upward or downward we simply see the result of <span class="math-container">$x\rightarrow \infty$</span> and so we get, <span class="math-container">$$\infty^2 \Bigl(a + \frac{b}{\infty} + \frac{c}{\infty^2}\bigl)$$</span> So we get, <span class="math-container">$$\infty^2 \Bigl(a\bigl)=0$$</span> which obviusly says that if <span class="math-container">$a\lt0$</span> then at <span class="math-container">$$x\rightarrow \infty, f(x) \rightarrow -\infty$$</span> and for <span class="math-container">$a\gt0$</span> then at <span class="math-container">$$x\rightarrow \infty, f(x) \rightarrow \infty$$</span> This tells us that this function will be a downward opening parabola as seen in the figure below,<br> <a href="https://www.desmos.com/calculator/w0krov2wrl" rel="nofollow noreferrer">https://www.desmos.com/calculator/w0krov2wrl</a><br> Now to find the value of the maxima we simply set <span class="math-container">$$\frac{df}{dx} = 0$$</span> That is, <span class="math-container">$$\frac{d\Bigl(-5x^2 + 12x -7\bigl)}{dx} = 0$$</span> Which gives us, <span class="math-container">$$-10x + 12 = 0$$</span> Or, <span class="math-container">$$x = \frac{6}{5}$$</span> Now obviously since this is the maxima it means that after this point the function will be strictly decreasing so the interval for which it is strictly increasing is <span class="math-container">$$\forall \; x \lt \frac6 5$$</span></p>

<p>If $X$ is $Beta\left(\dfrac{ \alpha_1}{ 2 }, \dfrac{\alpha_2}{2}\right)$ then $\dfrac{\alpha_2 X}{\alpha_1(1-X)}$ is $F(\alpha_1, \alpha_2)$? </p> <p>Any help is appreciated I don't know where to start. I'm assuming I need the pdf's of each distribution?</p>

Maverick Meerkat

<p>I'm not 100% sure this way is valid, but I'm gonna give it a try using the CDF's:</p> <p><span class="math-container">$Y =\dfrac{\alpha_2 X}{\alpha_1(1-X)}$</span></p> <p>CDF of <span class="math-container">$Y = F_{Y}(y) = \mathbb {P}(Y &lt; y) = \mathbb {P}(\dfrac{\alpha_2 X}{\alpha_1(1-X)} &lt; y) = \mathbb {P}(X &lt;\dfrac{\alpha_1 y}{\alpha_1 y + \alpha_2}) $</span></p> <p>The last stage has some limitation, mainly that <span class="math-container">$ X \neq 1 $</span>, which depends on how you define the support of the Beta distribution. Assuming it does not include 1, we're ok. Note that <span class="math-container">$\alpha_1 y + \alpha_2 &gt; 0$</span>, so we can divide by it without changing signs. Continuing:</p> <p><span class="math-container">$ = F_{X}(\dfrac{\alpha_1 y}{\alpha_1 y + \alpha_2}) = $</span> (according to the CDF definition of a Beta distribution)<br> <span class="math-container">$ = I_{\dfrac{\alpha_1 y}{\alpha_1 y + \alpha_2}}(\dfrac{\alpha_1}{2}, \dfrac{\alpha_2}{2}) $</span></p> <p>But this is exactly the definition of the CDF for the F distribution.</p>

<p>Like in topic, you have 6 dice. You sum their values. What is the probability you get 9? How do I calculate it?</p>

Brian Tung

<p><strong>General approach.</strong> There aren't that many ways to get $9$ from six dice. (I assume these are ordinary six-sided dice.) Up to reordering, there are only the following three combinations:</p> <p>$$ 1, 1, 1, 1, 1, 4 $$</p> <p>$$ 1, 1, 1, 1, 2, 3 $$</p> <p>$$ 1, 1, 1, 2, 2, 2 $$</p> <p>Calculate the number of ways to get each of these (including their reorderings), and divide by $6^6$ to obtain the probability. You should obtain the same value as Matthew Conroy in his answer, hopefully with a lot less computation. (Generating functions are a convenient way to obtain all of the probabilities at once, though.)</p>

<p>Like in topic, you have 6 dice. You sum their values. What is the probability you get 9? How do I calculate it?</p>

N. F. Taussig

<p>Let's assume the dice are six-sided and distinct (each of a different color, say). Then there are $6^6$ possible outcomes in the sample space. Let $x_k$ be the outcome on the $k$th die. Then $$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 9 \tag{1}$$ Equation 1 is an equation in the positive integers subject to the restriction that $x_k \leq 6$ for $1 \leq k \leq 6$. Notice that if $x_k &gt; 6$ for some $k$, then the sum is at least $7 + 5 = 12 &gt; 9$, so we do not need to worry about that restriction (as Brian Tung demonstrated in his answer).</p> <p>A particular solution to equation 1 corresponds to the insertion of addition signs in five of the eight spaces between successive ones in a row of nine ones. For instance, $$1 1 + 1 + 1 1 + 1 + 1 + 1 1$$ corresponds to the outcome $x_1 = 2$, $x_2 = 1$, $x_3 = 2$, $x_4 = 1$, $x_5 = 1$, and $x_6 = 2$, while $$1 1 1 + 1 1 + 1 + 1 + 1 + 1$$ corresponds to the outcome $x_1 = 3$, $x_2 = 2$, $x_3 = x_4 = x_5 = x_6 = 1$. Hence, the number of solutions of equation 1 is the number of ways we can select five of the eight spaces between successive ones in a row of nine ones, which is $\binom{8}{5}$. Hence, the probability of obtaining a sum of $9$ when six six-sided dice are thrown is $$\frac{\binom{8}{5}}{6^6}$$ </p>

<p>I am looking for examples of finitely generated solvable groups that are not polycyclic. In <a href="http://groupprops.subwiki.org/wiki/Finitely_generated_and_solvable_not_implies_polycyclic#Some_examples_based_on_the_general_construction_and_otherwise" rel="nofollow">Wikipedia</a> Baumslag-Solitar group $BS(1,2)$ is an example. But how to prove this fact?</p>

Dietrich Burde

<p>For a polycyclic group, every subgroup is finitely generated, in particular the commutator subgroup is finitely generated. Now assume that $BS(1,2)$ is polycyclic. Hence its derived subgroup is finitely generated. However, the derived subgroup is isomorphic to the group of $2$-adic rationals, i.e., the group of all rationals with denominators powers of $2$. This group is not finitely generated, a contradiction. </p> <p>More generally, all Baumslag-Solitar groups $BS(1,m)$ for $m\ge 2$ are metabelian, but not polycyclic, because the commutator subgroup $\mathbb{Z}[\frac{1}{m}]$ is not finitely generated.</p>

<p><img src="https://ukpstq.bn1.livefilestore.com/y2p5St1yRZbdxBvzMbBTYjqjqtwDvBaoWtc7YRGZXCwBTax0XseUIh_l_O92NO6XAbLeGaqU67bkBI4lroIlcD2ade_rxfDast52B_7ECcMd68/question3.png?psid=1" alt="question"></p> <p>I have attempted these few simple questions, can someone let me know if this is correct please? If not please provide the answer as I learn better that way and if possible explain.</p> <p>i) FA1 Start = -X1 and FA1 End = +X2</p> <p>FA2 Start = -Y1 and FA2 End = +Y3</p> <p>ii) FA1 NOT ACCEPTED and FA2 ACCEPTED</p> <p>iii) In FA1 the string must begin with 'a' and this is in a loop, the next string is a 'b' followed by another 'b' in a loop. it can then go back round through 'a' and start again but must end with a 'b'</p> <p>In FA2 it must start with a 'b' and is in a loop, then it is followed by an 'a' and go back through 'b' or continue onto 'a' and then end or go through 'a' in a loop or start the cycle again through 'b' back to the start but must end with an 'a' at y3.</p>

J.-E. Pin

<p>Your answers:<br> (i) is OK<br> (ii) There are four questions, you need to give four answers.<br> (iii) <strong>Hint</strong>. Give a regular expression for the languages accepted by $FA_1$ and $FA_2$ and then convert them to plain English.</p>

<p>$f(x)=x^4-16x^2+4$, the root of $f(x)$ is $a= \sqrt{3} + \sqrt{5}$</p> <p>Factorise $f(x)$ as a product of irreducible polynomials over $\mathbb{Q}$, over $\mathbb{R}$ and over $\mathbb{C}$.</p> <p>I am really confused as to how to start.</p>

Brenin

<p>Yes. You are asking whether the origin is a nonsingular point of $C=\textrm{Spec}\,A\subset \mathbb A^2_k$. Write the homogeneous decomposition $F=\sum_{d\geq 1}f_d$, where $f_1=aX+bY\neq 0$. Let us show that $P$ is a regular point of $C$. If $P=(0,0)$ were singular, then (by <em>definition</em>) the two partial derivatives of $F$ would vanish at $P$. But then we would find \begin{equation} 0=\frac{\partial F}{\partial X}(P)=a+(\textrm{higher degree terms containing powers of}\, X \,\textrm{and}\, Y) \end{equation} \begin{equation} 0=\frac{\partial F}{\partial Y}(P)=b+(\textrm{higher degree terms containing powers of}\, X \,\textrm{and}\, Y). \end{equation} But this implies $a=0=b$, contradiction. Hence $P$ is regular.</p> <p>Now I claim that saying $P$ is a regular point <em>is equivalent</em> to the assertion that $\mathcal O_{C,P}\,(\,=A_P)$ is regular as a local ring, that is, by <em>definition</em>: \begin{equation} \dim A_P=\dim T_{C,P}\,, \end{equation} where $T_{C,P}$ is the tangent space at $P$. If $P$ is regular then the tangent space at $P$ is a line, so $\dim T_{C,P}=1=\dim A=\dim A_P$. Conversely, if $\dim T_{C,P}=1$ then the partial derivatives of $F$, the generators of $T_{C,P}$, can't both vanish at $P$. Indeed, a point $(\alpha,\beta)\in \mathbb A^2$ is in $T_{C,P}$ if and only if \begin{equation} \frac{\partial F}{\partial X}(P)\cdot\alpha+\frac{\partial F}{\partial Y}(P)\cdot\beta=0. \end{equation} Hence $P$ is regular. </p> <p>So far, we have established that $A_P$ is a <em>regular local ring</em>.</p> <p>Finally, $\dim A_P=\dim A=\dim C=1$. Now, a DVR is a regular local ring of dimension one so your $\mathcal O_{C,P}$ is one such.</p>

<p>$f(x)=x^4-16x^2+4$, the root of $f(x)$ is $a= \sqrt{3} + \sqrt{5}$</p> <p>Factorise $f(x)$ as a product of irreducible polynomials over $\mathbb{Q}$, over $\mathbb{R}$ and over $\mathbb{C}$.</p> <p>I am really confused as to how to start.</p>

Mariano Suárez-Álvarez

<p>Your ring is an integrally closed noetherian local ring with Krull dimension one, and such a thing is a DVR.</p>

<p>$f(x)=x^4-16x^2+4$, the root of $f(x)$ is $a= \sqrt{3} + \sqrt{5}$</p> <p>Factorise $f(x)$ as a product of irreducible polynomials over $\mathbb{Q}$, over $\mathbb{R}$ and over $\mathbb{C}$.</p> <p>I am really confused as to how to start.</p>

Makoto Kato

<p><strong>Lemma</strong> Let $A$ be a Noetherian local domain. Let $\mathfrak{m}$ be its unique maximal ideal. Suppose $\mathbb{m}$ is a non-zero principal ideal. Then $A$ is a discrete valuation ring.</p> <p>Proof: Let $t$ be a generator of $\mathfrak{m}$. We claim that $\bigcap_n \mathfrak{m}^n = 0$. Let $x \in \bigcap_{n&gt;0} \mathfrak{m}^n$. For every integer $n &gt; 0$, there exists $y_n \in A$ such that $x = t^ny_n$. Since $t^ny_n = t^{n+1}y_{n+1}$, $y_n = ty_{n+1}$. Hence $(y_1) \subset (y_2) \subset \cdots$. Since $A$ is Noetherian, there exists $n$ such that $(y_n) = (y_{n+1})$. Hence there exists $a \in A$ such that $y_{n+1} = ay_n$. Hence $y_{n+1} = aty_{n+1}$. Hence $(1 - at)y_{n+1} = 0$. Since $1 - at$ is invertible, $y_{n+1} = 0$. Hence $x = 0$ as desired.</p> <p>Let $x$ be a non-zero element of $\mathfrak{m}$. Since $\bigcap_n \mathfrak{m}^n = 0$. There exists integer $n &gt; 0$ such that $x \in \mathfrak{m}^n - \mathfrak{m}^{n+1}$. Hence there exists $u \in A$ such that $x = t^nu$. Since $u$ is not contained in $\mathfrak{m}$, $u$ is invertible. Hence $A$ is a discrete valuation ring. <strong>QED</strong></p> <p>Let $R=K[X,Y]_{(X,Y)}$. As <a href="https://math.stackexchange.com/questions/239649/quotient-ring-of-a-localization-of-a-ring">this question</a> shows, there exists a canonical isomomorphism $A_{\mathfrak{m}} \cong R/(F)$. Let $F = F_1\cdots F_m$ be a factorization of $F$ into irreducible factors. Since $F(0, 0) = 0$, there exists $i$ such that $F_i(0, 0) = 0$. By the assumption that $F(X, Y) = aX + bY + \cdots$, $F_j(0, 0) \neq 0$ for $j \neq i$. Hence $F_j$ is invertible in $R$ for $j \neq i$. Hence $R/(F) = R(F_i)$. Hence $R/(F)$ is an integral domain. Therefore, by the lemma, it suffices to prove that $\mathfrak{m}$ is principal. By Nakayama's lemma, it suffices to prove that $dim_k \mathfrak{m}/\mathfrak{m}^2 = 1$.</p> <p>Let $I = (X, Y)$ be the ideal generated by $X, Y$ in $k[X, Y]$. It is easy to see that $\mathfrak{m}/\mathfrak{m}^2$ is isomorphic to $I/((F) + I^2)$ as $k[X, Y]$-modules. In particular, it is isomorphic as $k$-vector spaces. Note that $dim_k I/I^2 = dim_k I/((F) + I^2) + dim_k ((F) + I^2)/I^2$. Let $x, y$ be the image of $X, Y$ by the canonical homomorphism $I \rightarrow I/I^2$ respectively. Clearly $x, y$ is a basis of the $k$-vector space $I/I^2$. Hence $dim_k I/I^2 = 2$. On the other hand, $((F) + I^2)/I^2$ is the vector subspace of $I/I^2$ generated by $ax + by$. By the assumption $ax + by \neq 0$. Hence $dim_k ((F) + I^2)/I^2 = 1$. Hence $dim_k \mathfrak{m}/\mathfrak{m}^2 = dim_k I/((F) + I^2) = 1$ as desired.</p>

<p>I am struggling to convert a base64 string to a list of UnsignedInteger16 values. I have limited experience with mathematica, so please excuse me if this should be obvious.</p> <p>I read the base64 string from an XML file and ultimately into a variable base64String. This is a long string with 7057 UnsignedInteger16 values, but it starts off as: &quot;8ynnKdspzinCKbcprSmhKZUpiCl8KW8pYylXKUkpQCk0KSop...&quot;</p> <p>Can convert to ByteArry with n = BaseDecode[base64String] // Normal;</p> <p>Then n has the expected values {243, 41, 231, 41, 219, 41, 206, 41, 194, 41, 183, 41, 173, 41, 161, ...</p> <p>However, what I am looking for is a list with the values {243 + 256 * 41, 231 + 256 * 41, 219 + 256 * 41, ...</p> <p>i.e.: {10739, 10727, 10727, ...</p> <p>Thank you in advance for you suggestions.</p> <p>Edit: Thank you for the suggestion to use Partition. Works well. For the interests in improving my understanding, can any one suggest why I cannot get ImportString[...] to work.</p> <p>My source data is XML file with data along the lines of the following where I truncated the Base64 string for clarity.:</p> <pre><code>&lt;?xml version=&quot;1.0&quot; encoding=&quot;utf-8&quot; ?&gt; &lt;BPplus version=&quot;5.0&quot; filename=&quot;BPplus_00100.xml&quot;&gt; &lt;MeasDataLogger guid=&quot;ee7bee36-ffd6-30ae-53f1-257f0eab7ddd&quot;&gt; &lt;Sys&gt;130&lt;/Sys&gt; &lt;Dia&gt;77&lt;/Dia&gt; &lt;Map&gt;101&lt;/Map&gt; &lt;PressureWaves&gt; &lt;RawPressureWave&gt; &lt;RawPressure&gt;NIBP&lt;/RawPressure&gt; &lt;RawSampleCount&gt;7057&lt;/RawSampleCount&gt; &lt;RawCuffPressureWave&gt;8ynnKdspzinCKbcprSmhKZUpiCl8KW8pYylXKUk... bU1mjV+NWM1RjU=&lt;/RawCuffPressureWave&gt; &lt;/RawPressureWave&gt; &lt;/PressureWaves&gt; &lt;/MeasDataLogger&gt; &lt;/BPplus&gt; </code></pre> <p>The script I then run with the C1, C2 &amp; C3 suggestions.</p> <pre><code>sampleFilenameString = &quot;C:\\BPPdata\\BPplus_00112.xml&quot;; xmldata = Import[sampleFilenameString, &quot;XML&quot;]; RawCuffPressureWave = Cases[xmldata, XMLElement[&quot;RawCuffPressureWave&quot;, _, _], Infinity]; nibpxml = RawCuffPressureWave[[2]]; base64String = ToString[nibpxml[[3]]]; C1 = Partition[Normal@BaseDecode[base64String], 2].{1, 256}; C2 = ImportByteArray[BaseDecode[base64String], &quot;UnsignedInteger16&quot;]; C3 = ImportString[base64String, {&quot;Base64&quot;, &quot;UnsignedInteger16&quot;}, ByteOrdering -&gt; -1] C1 == C2 C1 == C3 C1 </code></pre> <p>produces the following output</p> <pre><code>{17956, 26977, 25964} True False {10739, 10727, 10715, 10702, 10690, 10679, 10669, 10657, ... } </code></pre> <p>I have tried to make the first parameter <code>ToString[base64String]</code>, but that did not change the outcome.</p> <p>If I manually define the string <code>$base64 = &quot;8ynnKdspzinCKbcprSmhKZUpiCl8KW8pYylXKUkpQCk0KSo....</code></p> <p>It works as expected and C1 = C4</p> <pre><code>C4 = ImportString[$base64, {&quot;Base64&quot;, &quot;UnsignedInteger16&quot;}, ByteOrdering -&gt; -1]; C1 == C4 </code></pre> <p>Can anyone explain the difference why <code>$base64</code> variable works but <code>base64String</code> does not?</p> <p>Edit: base64String is not a string! Not sure if this is the best way to convert, but it works in Mathematica 11.x and higher.</p> <pre><code>sampleFilenameString = &quot;C:\\BPPdata\\BPplus_00112.xml&quot;; xmldata = Import[sampleFilenameString, &quot;XML&quot;]; RawCuffPressureWave = Cases[xmldata, XMLElement[&quot;RawCuffPressureWave&quot;, _, _], Infinity]; nibpxml = RawCuffPressureWave[[2]]; (* convert to string then base64 decode *) base64Data = Developer`DecodeBase64ToByteArray[ nibpxml //. XMLElement[_, _, t_] :&gt; t // Flatten // StringJoin] // Normal; C1 = Partition[base64Data, 2].{1, 256}; </code></pre>

kglr

<pre><code>k = {6, 4, 2}; lbls = Row[{Subscript[&quot;N&quot;, &quot;B&quot;] , #}, &quot; = &quot;] &amp; /@ k; cols = {Red, Orange, Black}; </code></pre> <p>You can use the option <code>LegendLayout</code> as follows:</p> <pre><code>linesperlabel = 2; labels = Flatten[Thread[{lbls, ##&amp; @@ ConstantArray[SpanFromAbove, linesperlabel - 1]}]]; plotstyles = Flatten[{#, Directive[#, Dashed]} &amp; /@ cols]; LineLegend[plotstyles, labels, LegendMarkerSize -&gt; {30, 10}, LabelStyle -&gt; Directive[FontSize -&gt; 14, FontColor -&gt; Black, FontFamily -&gt; &quot;Helvetica&quot;], LegendLayout -&gt; (Grid[Transpose[{#[[All, 1]], labels}], Alignment -&gt; {Center, Center}, Spacings -&gt; {1, {Prepend[ConstantArray[0, linesperlabel - 1], 1]}}] &amp;)] </code></pre> <p><a href="https://i.stack.imgur.com/95apE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/95apE.png" alt="enter image description here" /></a></p> <pre><code>functions = k[[#]] Sin[k[[#]] x + b] /. {b -&gt; Range[linesperlabel]} &amp; /@ Range @ 3; Plot[Evaluate @ functions, {x, 0, 2 π}, PlotStyle -&gt; plotstyles, FrameStyle -&gt; Directive[Black, 16], Frame -&gt; {True, True, False, False}, ImageSize -&gt; 450, PlotRange -&gt; {-8, 8}, PlotLegends -&gt; LineLegend[Automatic, labels, LegendMarkerSize -&gt; {30, 10}, LabelStyle -&gt; Directive[FontSize -&gt; 14, FontColor -&gt; Black, FontFamily -&gt; &quot;Helvetica&quot;], LegendLayout -&gt; (Grid[Transpose[{#[[All, 1]], labels}], Alignment -&gt; {Center, Center}, Spacings -&gt; {1, {Prepend[ConstantArray[0, linesperlabel - 1], 1]}}] &amp;)]] </code></pre> <p><a href="https://i.stack.imgur.com/gRGib.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gRGib.png" alt="enter image description here" /></a></p> <p>Use</p> <pre><code>linesperlabel = 3; labels = Flatten[Thread[{lbls, ## &amp; @@ ConstantArray[SpanFromAbove, linesperlabel - 1]}]]; plotstyles = Flatten[{#, Directive[#, Dashed], Directive[#, Dotted]} &amp; /@ cols]; functions = k[[#]] Sin[k[[#]] x + b] /. {b -&gt; Range[linesperlabel]} &amp; /@ Range@3; </code></pre> <p>to get</p> <p><a href="https://i.stack.imgur.com/s0m7Q.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/s0m7Q.png" alt="enter image description here" /></a></p>

<blockquote> <p>Is it possible to find $6$ integers $a_1,a_2,\ldots,a_6 \geq 2$ such that $$a_1+a_1a_2+a_1a_2a_3+a_1a_2a_3a_4+a_1a_2a_3a_4a_5+a_1a_2a_3a_4a_5a_6 = 248?$$</p> </blockquote> <p>I was wondering how we could establish the existence of such numbers. Is there a way to do it without finding the actual $6$ numbers?</p>

HarrySmit

<p>I will provide the first step, from which you can hopefully complete the proof yourself. As @snulty mentioned, we can factor this as follows: $$ a_1(1 + a_2(1 + a_3(1 + a_4(1 + a_5(1 + a_6))))) = 2^3 \cdot 31. $$ This implies that $a_1 = 2$ or $a_1 \geq 4$. However, if $a_1 \geq 4$, we have $$ a_1(1 + a_2(1 + a_3(1 + a_4(1 + a_5(1 + a_6))))) \geq 4(1 + 2(1 + 2(1 + 2(1 + 2(1 + 2))))) = 252, $$ hence this is not an option. </p> <p>We find $a_1 = 2$, hence the problem is reduced to $$a_2(1 + a_3(1 + a_4(1 + a_5(1 + a_6)))) = 123.$$</p> <p>The prime factorization of $123$ is $3 \cdot 41$, so what can we conclude about the value of $a_2$? Finish the rest of the proof!</p>

<p>I've just been studying cyclic quads in geometry at school and I'm thinking see seems pretty interesting, but where would I actually find these in the real world? They seem pretty useless to me...</p>

preferred_anon

<p>I can't think of any applications, and I doubt any satisfactory ones exist - for example, as noted in the comments there may well have been connections to astronomy, but I think it's fair to suggest that almost no-one who is being taught circle theorems is going to use them in their life at any point. </p> <p>Thus, I'm going to interpret this question as:</p> <blockquote> <p>Why would you bother learning a theorem that has no application in real life? </p> </blockquote> <p>And I think there are two good answers to this question:</p> <p><strong>1. It is interesting</strong><br> This is, really, the only reason you're taught anything in your life other than how to pay taxes. Geometry is something that lots of people over a long period of time have found to be intrinsically interesting. The reasons for this are complicated - it's a good intellectual exercise, and for many people intellectual exercises are something they enjoy doing. </p> <p><strong>2. It forces you to think logically</strong><br> The patterns of thought people generally use in mathematics are valuable. Logical arguments are important in all walks of life, and being able to understand and interpret them is an extremely valuable life skill which you really should want to have. </p> <p>I have a lot of sympathy with this question, for the following reason: <strong>you are probably taught mathematics very badly</strong>. The arguments I give above really rely on the idea that you are taught how to prove theorems (and Euclidean geometry is a <em>fantastic</em> exercise in proof). Without that, I would claim that learning geometry <em>really has no value</em>. I would even go so far as to say you shouldn't bother going so far as to learn basic trigonometry (unless you need it to be an engineer or something), <em>unless you study its proof</em>. That really is where all the value, and all the fun, is. </p> <p><em>This is not your fault</em>. But there is something you can do about it. Look up a proof, try to understand it, and if you're lucky you'll get a little intellectual buzz from the 'aha!' moment of it all coming together. But, I'm sorry to say, you'll probably have to do this yourself. Mathematics teaching is woeful in the vast majority of schools, and statistically speaking you are unlikely to even have a teacher capable of explaining to you why these results are <em>true</em>, let alone interesting or useful. </p> <p>So, on the off chance that this answer has spiked your curiosity, I recommend writing another question, called "How do you prove interesting facts about cyclic quadrilaterals?", and you might get a more satisfying answer.</p>

<p>I've just been studying cyclic quads in geometry at school and I'm thinking see seems pretty interesting, but where would I actually find these in the real world? They seem pretty useless to me...</p>

Joseph O'Rourke

<p>Theorem 3 of the Bern-Eppstein paper cited below proves that any polygon of $n$ vertices may be partitioned into $O(n)$ cyclic quadrilaterals. A hint of how this might be achieved can be glimpsed in the figure below, where all the white quadrilaterals are cyclic. <hr /> &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; <img src="https://i.stack.imgur.com/Zi4bX.png" width="400" /> <br /> &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; ^{Fig.5 from cited paper.} <hr /> Quadrilateral meshing is important in many applications. The cyclic quads produced by their algorithm have desirable "quality" characteristics.</p> <blockquote> <p>Bern, Marshall, and David Eppstein. "Quadrilateral meshing by circle packing." <em>International Journal of Computational Geometry &amp; Applications</em> 10.04 (2000): 347-360. (<a href="http://arxiv.org/abs/cs/9908016" rel="nofollow noreferrer">Pre-journal arXiv abstract</a>.) (<a href="http://www.worldscientific.com/doi/abs/10.1142/S0218195900000206" rel="nofollow noreferrer">Journal link</a>.)</p> </blockquote>

<p>Let $K$ and $L$ be two subfields of some field. If a variety is defined over both $K$ and $L$, does it follow that the variety can be defined over their intersection?</p>

Bjorn Poonen

<p>Yes, if varieties are interpreted as <strike>subvarieties</strike> closed subschemes of base extensions of a fixed ambient <strike>variety</strike> scheme (e.g., affine space or projective space).</p> <p>More precisely, suppose that $k \subseteq F$ are fields and <strike>the variety</strike> $X$ is <strike>an $F$-subvariety</strike> a closed subscheme of $\mathbf{P}^n_F$. Say for a field $K$ with $k \subseteq K \subseteq F$ that "<strong>$X$ is defined over $K$</strong>" if $X$ is the base extension of some subvariety of $\mathbf{P}^n_K$. Then $X$ has a minimal field of definition $E$ with $k \subseteq E \subseteq F$, characterized by the property that for any field $K$ with $k \subseteq K \subseteq F$, we have that $X$ is defined over $K$ if and only if $K$ contains $E$.</p> <p>The same statement holds if $\mathbf{P}^n$ is replaced by any fixed <strike>$k$-variety</strike> $k$-scheme $Y$.</p> <p>(Note: this answer does not contradict Pete's. This is just a different interpretation of the question.)</p> <p><strong>EDIT:</strong> As Brian points out, I was indeed assuming that my varieties were closed in the ambient space. The statement about minimal field of definition is not even true for open subschemes in characteristic $p$. For example, if $k=\mathbb{F}_p$ and $F=k(t)$ and $Y=\operatorname{Spec} k[x]$ and $X=\operatorname{Spec} F[x,1/(x-t)]$, then $X$ is the base extension of $\operatorname{Spec} F^{p^n}[x,1/(x^{p^n}-t^{p^n})]$, and hence is definable over $F^{p^n}$ for all $n$, but not over the intersection of all these fields, which is just $k$. </p> <p>On the other hand, the intersection of any <em>finite</em> number of fields of definition is still a field of definition.</p> <p>I have generalized to schemes as suggested by Brian.</p>

<p>Let $A$ be a $m \times n$ matrix. Determine whether or no the set $W= \{y : Ay=0\}$ is a vector space. This proof involves nullspace work and another way of asking it is also proving that $W$ is the nullspace of $A$.</p> <p>I think you can solve this with the use of the subspace theorem being that you can just prove axioms $A1$ ($U+V =$ something in the set) and $M1$ ($kU=$ something in the set) The problem that I am having here is how to set this up so that I can prove this.</p>

Brian Fitzpatrick

<p>We wish to show that $$ W=\{\vec x\in\Bbb R^n:A\vec x=\vec 0\} $$ is a subspace of $\Bbb R^n$. To do so, we use the <a href="https://proofwiki.org/wiki/One-Step_Vector_Subspace_Test" rel="nofollow">one-step vector subspace test</a>. To do so, let $\vec x,\vec y\in W$ and let $\lambda\in\Bbb R$. Then $$ A(\vec x+\lambda\cdot\vec y)=A\vec x+\lambda\cdot A\vec y=\vec 0+\lambda\cdot \vec0=\vec 0 $$ so that $\vec x+\lambda\cdot\vec y\in W$. Hence $W$ is a subspace.</p> <p>Of course, we could alternatively note that $W=\ker T$ where $T$ is the linear transformation $\Bbb R^n\to\Bbb R^m$ defined by $T(\vec x)=A\vec x$, but this probably violates the spirit of the problem.</p>

<p>I tried rewriting $(1+x+x^2)^\frac{1}{x}$ as $e^{\frac{1}{x}\ln(1+x+x^2)}$ and then computing the taylor series of $\frac{1}{x}$ and $\ln(1+x+x^2)$ but I'm still not getting the correct answer..</p>

robjohn

<p>This appears to be the approach you used, but without seeing your work, it is hard to tell where you were having trouble.</p> <p>Use the series $\log(1+x)=x-\frac{x^2}2+\frac{x^3}3+\cdots$ $$ \begin{align} \frac1x\log\left(1+x+x^2\right) &amp;=\frac1x\left(x+x^2-\frac{x^2+2x^3+x^4}2+\frac{x^3+3x^4+3x^5+x^6}3+\cdots\right)\\ &amp;=1+\frac12x-\frac23x^2+\cdots \end{align} $$ Then use the series $e^x=1+x+\frac{x^2}2+\cdots$ $$ \begin{align} \left(1+x+x^2\right)^{1/x} &amp;=e^{1+\frac12x-\frac23x^2+\cdots}\\ &amp;=e\left(1+\left(\frac12x-\frac23x^2+\cdots\right)+\frac12\left(\frac12x-\frac23x^2+\cdots\right)^2+\cdots\right)\\ &amp;=e\left(1+\frac12x-\frac{13}{24}x^2+\cdots\right) \end{align} $$</p>