qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
886,935 | <p>Let $A_1,A_2,A_3,\dots$ be a sequence of sequences where each
$$A_i = a_{i,1},a_{i,2},a_{i,3},\dots$$</p>
<p>Each sequence $A_i$ converges and in particular as $t \rightarrow \infty$, $a_{i,t} \rightarrow L_i$ for every $i$. We also have that in the limit as $i \rightarrow \infty$ the limits of these sequences converge to 1. I.e. as $i \rightarrow \infty$, $L_i \rightarrow 1$.</p>
<p>QUESTION: When does the following occur?</p>
<p>In the limit as $t \rightarrow \infty$
$$\prod_{i=1}^{\infty} a_{i,t} \rightarrow \prod_{i=1}^{\infty}L_i $$</p>
<p>I know how to prove that the limit of the product of two (convergent) sequences is the product of the limit of those sequences. I cannot see whether this argument extends to infinite products. Also I am not sure how to use the fact that the $L_i$ converge to 1.</p>
| Chen Jiang | 167,817 | <p>Usually this is not true. You must take care when consider the limit of a limit thing.</p>
<p>Here is a counterexample.</p>
<p>We define $a_{i,t}$ inductively. Let $a_{1,1}=1$ and $a_{1,t}=2$ if $t>1$. Assume that we defined $a_{j,t}$ for all $j<i$. Then we define
$$
a_{i,t}=
\begin{cases}
1 &t<i\\
\prod_{j<i}a_{j,i}&t=i\\
1+\frac{1}{i}&t>i
\end{cases}
$$</p>
<p>Then $$L_i=1+\frac{1}{i},$$
and $$\prod_{i=1}^\infty a_{i,t}=1.$$</p>
|
1,761,527 | <p>Let $f,g,h:X\to\mathbb{R}$ such that $f(x)\leq g(x)\leq h(x)$ for all $x\in X$. If $f$ and $h$ are differentiable in $a$ and $h(a)=f(a)$. Then $g$ is differentiable and $g'(a)=f'(a).$</p>
<p>How can I can prove that $g$ is differentiable in $a$.? Thanks</p>
| Tryss | 216,059 | <p>Without loss of generality, let $a=0$ and $f(0) = g(0) = h(0) = 0$</p>
<p>Then</p>
<p>$$\frac{f(\epsilon)}{\epsilon} \leq \frac{g(\epsilon)}{\epsilon} \leq \frac{h(\epsilon)}{\epsilon} $$</p>
<p>Now, take the limit when $\epsilon \to 0$, and as $f$ and $h$ are differentiable at $0$, you have that $\lim_{\epsilon \to 0} \frac{g(\epsilon)}{\epsilon}$ exists and is equal to $f'(0)$. So we have the result</p>
|
2,650,634 | <p>EDIT: I know how to integrate the last part. I'm just try to find mistake in converting Sum to integral</p>
<p>Question: </p>
<blockquote>
<p>$$a_n=\left(\left(1+\left(\frac1n\right)^2\right)\left(1+\left(\frac2n\right)^2\right)\cdots\left(1+\left(\frac{n}n\right)^2\right)\right)^n$$ find<br>
$$\lim_{n\to\infty}a_n^{-1/n^2}$$</p>
</blockquote>
<p>My Approach:
Let $$y=\lim_{n\to\infty}a_n^{-1/n^2}$$ I Converted this to
$$\ln y=\lim_{n\to\infty}{-1\over n}\sum^n_{k=1}\left(\ln\left(1+{k^2\over n^2}\right) \right)$$</p>
<p>and then to (here's where i think the mistake is in conversion, not integration):</p>
<p>$$\ln y=-\int_0^1\ln(1+x^2)dx$$ </p>
<p>However, the answer is $\ln y=1/2-\ln2$</p>
<p>Please the help me find mistake</p>
| Wouter | 89,671 | <p>You're right, your book is wrong.</p>
<p><a href="https://i.stack.imgur.com/R4hyF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/R4hyF.png" alt="enter image description here"></a></p>
|
2,725,839 | <p>The question is below.<a href="https://i.stack.imgur.com/k3UMf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/k3UMf.png" alt="enter image description here"></a></p>
<p>I was able to solve part (a) because the $x$-coordinate would just be the circumference of the circle, which is $2\pi$. Therefore, $P = (2\pi, 0)$. </p>
<p>I am confused with parts (b) and after, but I know that the $y$-coordinate will remain to be $1$ because that's the radius of the circle and how far it is off from the $x$-axis. Any help with how to find the $x$-coordinate will help for part (b). Thank you in advance. </p>
| Narasimham | 95,860 | <p>The parametric equation of point $P$ of cycloid on rolling wheel circle radius $a=1$</p>
<p>$$ x= a (\theta- \sin \theta),\quad y= a(1-\cos \theta) $$</p>
<p>With a bit of calculus horizontal slope is </p>
<p>$$ \frac{dy/d \theta}{dx/d \theta}= \tan \phi_h = \cot \phi =\frac {\sin \theta }{1- \cos \theta}= \cot \frac{\theta}{2} \rightarrow \boxed{\theta = 2 \phi} $$</p>
<p>So amount crank radius rotates is double of what the cycloid vertical tangent tardily rotates.</p>
<p><a href="https://i.stack.imgur.com/ORNIJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ORNIJ.png" alt="Cycloid"></a></p>
<p>This parametric form I think answers all your questions.</p>
<p>The y-cordinate of center is always $a$ no matter what the amount of roll $\theta $ is. The question is designed <em>to confuse you</em>, it appears.</p>
|
1,396,449 | <p>How does the Newton Interpolation work? The definition can be found here: <a href="http://www.nptel.ac.in/courses/122104018/node109.html" rel="nofollow">http://www.nptel.ac.in/courses/122104018/node109.html</a></p>
<p>Not how it's defined since that's mathematically clear, but I'm trying to grasp the general intuition about how it was formulated.</p>
| Olivier Oloa | 118,798 | <p><strong>Hint.</strong> Here is an approach using the <a href="http://dlmf.nist.gov/5.12">Euler beta function</a>. </p>
<p>By the change of variable, $u=e^{-x}$, you get
$$
\begin{align}
\int_0^{\infty} \frac{\log^2(1 - e^{-x})x^5}{e^x - 1} \, dx&=-\int_0^1\frac{\log^2(1-u)\log^5u}{1-u}\:du\\\\
&=-\left.\partial_b^2\partial_a^5\left(\int_{0}^{1} {u}^{a-1}(1-u)^{b-1}du\right)\right|_{a=1,b=0}\\\\
&=-\left.\partial_b^2\partial_a^5\left(\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}\right)\right|_{a=1,b=0}\\\\
&=\frac{4}{3} \pi ^4 \zeta(3)+20 \pi ^2 \zeta(5)-360 \zeta (7).
\end{align}
$$</p>
|
1,396,449 | <p>How does the Newton Interpolation work? The definition can be found here: <a href="http://www.nptel.ac.in/courses/122104018/node109.html" rel="nofollow">http://www.nptel.ac.in/courses/122104018/node109.html</a></p>
<p>Not how it's defined since that's mathematically clear, but I'm trying to grasp the general intuition about how it was formulated.</p>
| Leucippus | 148,155 | <p>Given
$$I = \displaystyle \int\limits_0^{\infty} \frac{\log^2(1 - e^{-x})x^5}{e^x - 1} \, dx $$
then let $t = e^{-x}$ to obtain
\begin{align}
I &= - \int_{1}^{0} ( - \ln t)^{5} \, \ln^{2}(1-t) \, \frac{dt}{1-t} \\
&= - \int_{0}^{1} \frac{\ln^{5}t \, \ln^{2}(1-t)}{1-t} \, dt \\
&= \int_{0}^{1} \ln^{5}t \, \left( \frac{1}{3} \, \frac{d}{dt} \ln^{3}(1-t)\right) \, dt \\
&= \left[ \frac{1}{3} \ln^{5}t \, \ln^{3}(1-t) \right]_{0}^{1} - \frac{5}{3} \, \int_{0}^{1} \frac{\ln^{3}(1-t) \, \ln^{4}t}{t} \, dt \\
&= - \frac{5}{3} \, \int_{0}^{1} \frac{\ln^{3}(1-t) \, \ln^{4}t}{t} \, dt
\end{align}</p>
<p>From here consider the Beta function in the form
\begin{align}
B(x,y) = \int_{0}^{1} t^{x-1} \, (1-t)^{y-1} \, dt
\end{align}
for which
\begin{align}
I &= - \frac{5}{3} \, \partial_{x}^{4} \partial_{y}^{3} \, \left. B(x,y) \right|_{x=0, y=1} \\
&= 5! \, (\zeta(4) \, \zeta(3) + \zeta(2) \, \zeta(5) - 3 \, \zeta(7))
\end{align}</p>
|
794,842 | <p>Let the statement $?PQR$ be determined by the following truth-table.</p>
<pre><code>P Q R ?PQR
T T T T
T T F F
T F T F
T F F T
F T T T
F T F T
F F T F
F F F T
</code></pre>
<ol>
<li>After ‘Answer:’ below, give a logically equivalent sentence of ?PQR in FOL. But here’s the catch: you may only use the Boolean connectives (i.e. ¬, Ʌ, and V) in the sentence you give.
I am trying to figure this assignment out; but cannot figure out the formula for determining the sentence. Any ideas?</li>
</ol>
| Peter Smith | 35,151 | <p>@AsafKaragila, being a nice guy, has given you a hint, and patiently explained more in his comments. Being not such a nice guy, can I point out that your question rather suggests that you haven't been to the library and done your basic groundwork? More or less any elementary logic text will have a section, under the title "expressive completeness" or some such, which will explain this idea. Faced with such a very basic question like this, your first reaction should be to check out two or three texts about the relevant principles at stake, till you find one whose explanations give you that "Aha! Got it!!" feeling. </p>
|
131,842 | <p>Let $X,Y$ be normed linear spaces. Let $T: X\to Y$ be linear. If $X$ is finite dimensional, show
that $T$ is continuous. If $Y$ is finite dimensional, show that $T$ is continuous if and only if $\ker T$ is closed. </p>
<p>I am able to show that $X$, finite dimensional $\implies$ $T$ is bounded, hence continuous. </p>
<p>For the second part: This is what I have: </p>
<p>Suppose $T$ is continuous. By definition $\ker T = \{ x\in X : Tx = 0 \}$ , and so $\ker T$ is the continuous inverse of a closed set. Hence $\ker T $ is closed. </p>
<p>First, is what I have attempted okay. How about the other direction? </p>
| azarel | 20,998 | <p>If $\ker(T)$ is closed then $X/\ker(T)$ is a normed vector space. Observe that the map $\overline{T}:X/\ker(T)\to Y$ given by $\overline{T}(x+\ker(T))=T(x)$ is a well-defined linear map by the first part $\overline T$ is continuous since $X/\ker(T)$ is a finite dimensional vector space (as it is isomorphic to a subspace of $Y$). Let $\pi: X \to X/\ker(T)$ denote the quotient map. Note that $T=\overline{T}\circ \pi$ hence $T$ is continuous since it is a composition of continuous functions.</p>
|
241,210 | <p>I am confused with the concept of topology base. Which are the properties a base has to have?</p>
<p>Having the next two examples for $X=\{a,b,c\}$:</p>
<p>1) $(X,\mathcal{T})$ is a topological space where $\mathcal{T}=\{\emptyset,X,\{a\},\{b\},\{a,b\}\}$. Which is the general procedure to follow in order to get a base for $(X,\mathcal{T})$? Can $\mathcal{B}=\{\{a\},\{b\}\}$ be a base?</p>
<p>2) Having just $X$ and no topology $\mathcal{T}$ defined for $X$, is $\mathcal{A}=\{X,\{a\},\{c\}\}$ a base? What is the topology that it generates? </p>
<p>Thank you very much.</p>
| murad.ozkoc | 301,250 | <p>Let $(X,\tau)$ be a topological space and $\mathcal{B}\subseteq 2^X$.
$$\mathcal{B} \text{ is a base for topology } \tau \text { on } X$$
$$:\Leftrightarrow$$
$$1) \mbox{ } \mathcal{B}\subseteq \tau$$
$$2) \mbox{ } (\forall A\in\tau)(\exists\mathcal{A}\subseteq\mathcal{B})(A=\cup\mathcal{A})$$</p>
|
307,458 | <p>Let <span class="math-container">$\mathbf{x}_1, \dots, \mathbf{x}_n \in \mathbb{R}^d$</span> be <span class="math-container">$n$</span> given vectors. Define the function</p>
<p><span class="math-container">$$ \mathcal{K}(\mathbf{x},\mathbf{y}) := \alpha\exp\left(-\frac{\|\mathbf{x}-\mathbf{y}\|^2}{2\sigma^2}\right) $$</span></p>
<p>where <span class="math-container">$\alpha$</span> and <span class="math-container">$\sigma$</span> are given constants. Now define the <span class="math-container">$n\times 1$</span> vector</p>
<p><span class="math-container">$$ \mathcal{K}_n(\mathbf{x}) := \begin{bmatrix} \mathcal{K}(\mathbf{x},\mathbf{x}_1) & \dots & \mathcal{K}(\mathbf{x},\mathbf{x}_n) \end{bmatrix}^\top $$</span></p>
<p>Let <span class="math-container">$\mathbf{A}$</span> be any <span class="math-container">$n \times n$</span> positive definite matrix and <span class="math-container">$\mathbf{b}$</span> be any <span class="math-container">$n \times 1$</span> real vector. Let <span class="math-container">$\lambda > 0$</span> be given. Consider the optimization problem</p>
<p><span class="math-container">$$ \max_{\mathbf{x} \in \mathbb{R}^d} \quad\mathcal{K}_n(\mathbf{x})^\top\mathbf{b}-\lambda\mathcal{K}_n(\mathbf{x})^\top\mathbf{A}\mathcal{K}_n(\mathbf{x}) $$</span></p>
<p>Is this optimization problem known in literature? How do I do gradient ascent on this?</p>
| Arash | 124,260 | <p>You are looking for a stochastic optimizer <a href="https://en.wikipedia.org/wiki/Stochastic_optimization" rel="nofollow noreferrer">[1]</a>.</p>
<p>This optimization is stochastic. Thus, it requires you optimize the expectation $E[.]$ of a particular value.</p>
<p>In an engineering problem, they often have some constraint on top of optimization which form a stochastic programming problem <a href="https://en.wikipedia.org/wiki/Stochastic_programming" rel="nofollow noreferrer">[2]</a><a href="https://www2.isye.gatech.edu/people/faculty/Alex_Shapiro/TutorialSP.pdf" rel="nofollow noreferrer">[3]</a><a href="http://users.iems.northwestern.edu/~jrbirge/html/dholmes/StoProIntro.html" rel="nofollow noreferrer">[4]</a>. In such problems, the constraints also become stochastic.</p>
<p>I refer you to a great publication on Stochastic Model Predictive Control:</p>
<ul>
<li>Mesbah, A., 2016. Stochastic model predictive control: An overview and perspectives for future research. IEEE Control Systems, 36(6), pp.30-44. <a href="https://cloudfront.escholarship.org/dist/prd/content/qt1wt3d4vr/qt1wt3d4vr.pdf" rel="nofollow noreferrer">[5]</a></li>
</ul>
<p>To make the life easy, you can use a <em>Monte Carlo</em> method with a bunch of scenarios to perform an optimization. I refere you to a few publications:</p>
<ul>
<li><p>Janson, L., Schmerling, E. and Pavone, M., 2018. Monte Carlo motion planning for robot trajectory optimization under uncertainty. In Robotics Research (pp. 343-361). Springer, Cham. <a href="https://arxiv.org/pdf/1504.08053.pdf" rel="nofollow noreferrer">[6]</a></p></li>
<li><p>Maldonado, D.A., 2017. Sequential Monte Carlo Methods for Parameter Estimation, Dynamic State Estimation and Control in Power Systems (Doctoral dissertation). <a href="http://repository.iit.edu/handle/10560/4176" rel="nofollow noreferrer">[7]</a></p></li>
<li>Oldewurtel, F., Jones, C.N., Parisio, A. and Morari, M., 2014. Stochastic model predictive control for building climate control. IEEE Transactions on Control Systems Technology, 22(3), pp.1198-1205. <a href="https://infoscience.epfl.ch/record/190572/files/06573317.pdf" rel="nofollow noreferrer">[8]</a></li>
</ul>
|
804,871 | <p>Prove that if x and y are odd natural numbers, then $x^2+y^2$ is never a perfect square.</p>
<p>Let $x=2m+1$ and $y=2l+1$ where m,l are integers.</p>
<p>$x^2+y^2=(2m+1)^2+(2l+1)^2=4(m^2+m+l^2+l)+2$</p>
<p>Where do I go from here?</p>
| Achari S Ganesha | 86,508 | <p>When a perfect square number is divided by 4 it doesn't leave a remainder of 2- GANESHA A S</p>
|
1,811,081 | <blockquote>
<p>Let $1,x_{1},x_{2},x_{3},\ldots,x_{n-1}$ be the $\bf{n^{th}}$ roots of unity. Find: $$\frac{1}{1-x_{1}}+\frac{1}{1-x_{2}}+......+\frac{1}{1-x_{n-1}}$$</p>
</blockquote>
<p>$\bf{My\; Try::}$ Given $x=(1)^{\frac{1}{n}}\Rightarrow x^n=1\Rightarrow x^n-1=0$</p>
<p>Now Put $\displaystyle y = \frac{1}{1-x}\Rightarrow 1-x=\frac{1}{y}\Rightarrow x=\frac{y-1}{y}$</p>
<p>Put that value into $\displaystyle x^n-1=0\;,$ We get $\displaystyle \left(\frac{y-1}{y}\right)^n-1=0$</p>
<p>So we get $(y-1)^n-y^n=0\Rightarrow \displaystyle \left\{y^n-\binom{n}{1}y^{n-1}+\binom{n}{2}y^2-......\right\}-y^n=0$</p>
<p>So $\displaystyle \binom{n}{1}y^{n-1}-\binom{n}{2}y^{n-1}+...+(-1)^n=0$ has roots $y=y_{1}\;,y_{2}\;,y_{3}\;,.......,y_{n-1}$</p>
<p>So $$y_{1}+y_{2}+y_{3}+.....+y_{n-1} = \binom{n}{2} =\frac{n(n-1)}{2}\;,$$ Where $\displaystyle y_{i} = \frac{1}{1-x_{i}}\;\forall i\in \left\{1,2,3,4,5,.....,n-1\right\}$</p>
<p>My Question is can we solve it any less complex way, If yes Then plz explain here, Thanks</p>
| Nate | 91,364 | <p>Let $f(x) = \frac{(x^n-1)}{(x-1)} = x^{n-1}+x^{n-2}+...+x+1$</p>
<p>Then $\frac{f'(x)}{f(x)} = \frac{1}{x-x_1} + \frac{1}{x-x_2}+ ... + \frac{1}{x-x_{n-1}}$</p>
<p>So what you want is just $\frac{f'(1)}{f(1)}$, which is easy to compute as $\frac{n-1}{2}$.</p>
|
3,424,720 | <p>I want to calculate the above limit. Using sage math, I already know that the solution is going to be <span class="math-container">$-\sin(\alpha)$</span>, however, I fail to see how to get to this conclusion.</p>
<h2>My ideas</h2>
<p>I've tried transforming the term in such a way that the limit is easier to find:
<span class="math-container">\begin{align}
\frac{\cos(\alpha + x) - \cos(\alpha)}{x}
&= \frac{\cos(x)\cos(\alpha)-\sin(x)\sin(\alpha)-\cos(\alpha)}{x} & (1) \\
&= \frac{\cos(x)\cos(\alpha)-\cos(x)}{x} - \frac{\sin(x)\sin(\alpha)}{x} & (2) \\
&= \frac{\cos(\alpha)(\cos(x)-1)}{x} - \sin(\alpha) & (3) \\
\end{align}</span></p>
<p>However, I'm not sure whether term <span class="math-container">$(3)$</span> is enough to solve the problem. Surely, for <span class="math-container">$x \to 0$</span>, this evaluates to <span class="math-container">$\frac{0}{0} - \sin(\alpha)$</span>, which is not enough to determine the limit.</p>
<p>Another try was to transform the term in such a way that I can use <span class="math-container">$\lim_{x \to 0} \frac{\sin x}{x} = 1$</span>. For instance, I found that
<span class="math-container">$$
\frac{\cos(\alpha + x) - \cos(\alpha)}{x} = \frac{\sin(\alpha - 90^° + x) - \sin(\alpha-90^°)}{x} \qquad (4)
$$</span></p>
<p>However, this only seems to lead to more complicated term manipulations that I did not manage to bring to a useful point.</p>
<p>What can I do to solve this problem?</p>
| user | 505,767 | <p>You are right and from here </p>
<p><span class="math-container">$$ x\frac{\cos(\alpha)(\cos(x)-1)}{x^2} - \sin(\alpha) \to0\cdot \left(-\frac{\cos \alpha}2\right) -\sin(\alpha)=-\sin \alpha$$</span></p>
<p>indeed recall </p>
<p><span class="math-container">$$\frac{\cos(x)-1}{x^2} \to -\frac12$$</span></p>
<p>As an alternative refer to derivative definition if you are allowed tu use that.</p>
|
1,569,476 | <p>We throw fair dice until $6$ will appear. Let $X$ denote total number of throws and $Y$ - number of $5$ we received.</p>
<ol>
<li>Find distribution $(X,Y)$</li>
<li>Are variables $X$ and $Y$ independent?</li>
</ol>
<p>I have to say that I have utterly no idea how to proceed with this question, detailed explanation appreciated.</p>
| Alain Remillard | 278,299 | <p>$X$ follow a distribution called geometric
$$P(X= x) = \left(\frac56\right)^{x-1} \times \frac16$$
Having something other than 6 for the first $x - 1$ throw, then a 6.</p>
<p>Of course, $Y$ and $X$ are dependant. If you get a 6 on the first throw ($X = 1$), the $Y = 0$, you can't have any five.</p>
<p>Knowing the number of throws you needed to get a 6, the number of 5 is a binomial distribution.
$$P(Y=y|X=x) = Binomial\left(x-1,\frac15\right) = \binom{x-1}{y}\left(\frac15\right)^y\left(\frac45\right)^{x-1-y}$$ if $y\leq x-1$, $0$ otherwise.</p>
<p>The combine distribution will be, if $y \leq x-1$
$$P(X=x, Y=y) = P(X = x) \times P(Y=y | X=x) = \left(\frac56\right)^{x-1} \frac16\binom{x-1}{y}\left(\frac15\right)^y\left(\frac45\right)^{x-1-y}$$
$$P(X=x,Y=y) = \binom{x-1}{y}\left(\frac15\right)^y\left(\frac45\right)^{x-1-y}\left(\frac56\right)^{x-1} \frac16$$
The probability is $0$ otherwise.</p>
|
3,797,724 | <blockquote>
<p>Find all real continuous functions that verifies :
<span class="math-container">$$f(x+1)=f(x)+f\left(\frac{1}{x}\right) \ \ \ \ \ \ (x\neq 0) $$</span></p>
</blockquote>
<p>I found this result <span class="math-container">$\forall x\neq 1 \ \ f(x)=f\left(\frac{x}{x-1} \right)$</span> and I tried to study the behaviour of the function <span class="math-container">$g$</span> defined as <span class="math-container">$g(x)=\frac{x}{x-1}$</span> and compare it with <span class="math-container">$x$</span> in order to use fixed point theorem but it won't work.</p>
<p>I need a hint and thanks.</p>
| md2perpe | 168,433 | <h3>Some results</h3>
<p>Assume that <span class="math-container">$f$</span> is defined and continuous on <em>all</em> of <span class="math-container">$\mathbb{R}$</span> even if the equation <span class="math-container">$f(x+1) = f(x) + f(\frac{1}{x})$</span> is not defined for <span class="math-container">$x=0.$</span></p>
<p>From the given identity
<span class="math-container">$
f(x+1) = f(x) + f(\frac1x)
$</span>
we get
<span class="math-container">$
f(\frac1x) = f(x+1) - f(x) \to f(1) - f(0)
$</span>
as <span class="math-container">$x\to0$</span>. Thus <span class="math-container">$\lim_{R\to\infty}f(R) = \lim_{R\to\infty}f(-R) = f(1) - f(0).$</span></p>
<p>Furthermore, we have
<span class="math-container">$$\begin{align}
f(2) &= f(1+1) = f(1) + f(\frac{1}{1}) = f(1) + f(1) \\
f(3) &= f(2+1) = f(2) + f(\frac{1}{2}) = f(1) + f(1) + f(\frac{1}{2}) \\
f(4) &= f(3+1) = f(3) + f(\frac{1}{3}) = f(1) + f(1) + f(\frac{1}{2}) + f(\frac{1}{3}) \\
\vdots \\
f(n) &= f(1) + \sum_{k=1}^{n-1} f(\frac{1}{k}) \\
\end{align}$$</span>
Taking limits as <span class="math-container">$n\to\infty$</span> gives
<span class="math-container">$
\lim_{n\to\infty} f(n) = f(1) + \sum_{k=1}^{\infty} f(\frac{1}{k})
,
$</span>
but since <span class="math-container">$\lim_{n\to\infty} f(n) = f(1) - f(0)$</span> this implies
<span class="math-container">$$
f(0) = -\sum_{k=1}^{\infty} f(\frac{1}{k})
.
$$</span>
However, the convergence of the series implies that <span class="math-container">$|f(\frac1k)|\to 0$</span> as <span class="math-container">$k\to\infty.$</span> By continuity of <span class="math-container">$f$</span> this means that <span class="math-container">$f(0)=\lim_{k\to\infty}f(\frac1k)=0.$</span> And from that follows that
<span class="math-container">$
\sum_{k=1}^{\infty} f(\frac{1}{k}) = 0
$</span>
and that
<span class="math-container">$
\lim_{R\to\infty}f(\pm R) = f(1)
.
$</span></p>
<p>We also get that
<span class="math-container">$$
f(0) = f((-1)+1) = f(-1) + f(\frac1{-1}) = 2 f(-1)
,
$$</span>
i.e. <span class="math-container">$f(-1) = 0.$</span></p>
|
621,109 | <p>I need to find all the numbers that are coprime to a given $N$ and less than $N$.
Note that $N$ can be as large as $10^9.$ For example, numbers coprime to $5$ are $1,2,3,4$.</p>
<p>I want an efficient algorithm to do it. Can anyone help? </p>
| Robert Israel | 8,508 | <p>Once you find the prime factors $p_1, \ldots, p_k$ of $N$, you could use a sieve: start with $1 \ldots N-1$, delete all multiples of $p_1$, then all multiples of $p_2$, etc. What's left is coprime to $N$.</p>
|
96,657 | <p>I'm trying to understand an alternative proof of the idea that if $E$ is a dense subset of a metric space $X$, and $f\colon E\to\mathbb{R}$ is uniformly continuous, then $f$ has a uniform continuous extension to $X$.</p>
<p>I think I know how to do this using Cauchy sequences, but there is this suggested alternative. For each $p\in X$, let $V_n(p)$ be the set of $q\in E$ such that $d(p,q)<\frac{1}{n}$. Then prove that the intersections of the closures
$$
A=\bigcap_{n=1}^\infty\overline{f(V_n(p))}
$$
consists of a single point, $g(p)$, and so $g$ is the desired continuous extension of $f$. Why is this intersection a single point, and why is $g$ continuous?</p>
<hr>
<p>This is what I did so far. Since $f$ is uniformly continuous, for given $\epsilon>0$, there is $\delta>0$ such that $\text{diam }f(V)<\epsilon$ whenever $\text{diam }V<\delta$. Since $V_n(p)$ has diameter at most $\frac{2}{n}$, taking $n>2/\delta$ would imply
$$
\text{diam }f(V_n(p))=\text{diam }\overline{f(V_n(p))}<\epsilon
$$
So I think $\lim_{n\to\infty}\text{diam }\overline{f(V_n(p))}=0$, which would imply $A$ consists of at most one point. I noticed that the closures form a descending sequence of closed sets, but I couldn't tell if they are bounded since $X$ is an arbitrary metric space, in order to conclude that the intersection is nonempty, and hence a single point.</p>
<p>Lastly, why is $g$ continuous at points $p\in X\setminus E$? I was trying to think of an argument with sequences converging to $p$ since $p$ is a limit point of $E$, but got stumping on how to show $g$ is actually continuous. Thanks.</p>
| yunone | 1,583 | <p>I had a lot of help on this question in chat from users Srivatsan and t.b. the other day. I tried my best to write up what was said as an answer here.</p>
<hr>
<p>Notice that the sets $\overline{f(V_n(p))}\supseteq\overline{f(V_{n+1}(p))}\supseteq\cdots$ form a nested sequence of closed sets. Moreover, let $\epsilon>0$ be given. Since $f$ is uniformly continuous, there exists $\delta>0$ such that $d(p,q) < \delta$ implies $d(f(p),f(q))<\epsilon$ for $p,q\in E$. Taking $n$ large enough so that $\frac{2}{n}<\delta$, then for $q,r\in V_n(p)$,
$$
d(q,r)<d(q,p)+d(p,r)<\frac{2}{n}<\delta
$$
so $d(f(q),f(r))<\epsilon$. Thus $f(V_n(p))$ is bounded in $\mathbb{R}$, so $\overline{f(V_n(p))}$ is bounded as well. Hence for large enough $n$ the sets form a compact nested sequence. Since $V_n(p)$ has diameter at most $\frac{2}{n}$, taking $n>2/\delta$ would imply
$$
\operatorname{diam } f(V_n(p))=\operatorname{diam }\overline{f(V_n(p))}<\epsilon
$$
So $\lim_{n\to\infty}\operatorname{diam }\overline{f(V_n(p))}=0$, and thus their intersection consists of a single point. Also, since $\operatorname{diam }f(V_n(p))\to 0$ as $n\to\infty$, and so by choosing points arbitrarily close to $p$, their images under $g$ are arbitrarily close to $g(p)$. (To be more explicit, letting $\delta$ be small enough such that for $x,y\in E$, then $d(x,y)<2\delta$ implies $d(f(x),f(y))<\epsilon/3$, choose $n$ large enough that $\frac{1}{n}<\delta$, and thus for any $x,y\in V_n(p)$, $d(x,y)<2/n<2\delta$, so $\operatorname{diam }f(V_n(p))<2\epsilon/3$, so $d(f(x),g(p))<2\epsilon/3$. Note also that this can be done for any $p$.)</p>
<p>I contend that $g$ is uniformly continuous. Let $\epsilon>0$ be given. Since $f$ is uniformly continuous, there exists $\delta>0$ such that $d(r,s)<\delta$ implies $d(f(r),f(s))<\epsilon/3$. Now let $p,q\in X$ be any points such that $d(p,q)<\delta/3$. By the above reasoning, choose $n$ large enough so that $n>\frac{3}{\delta}$, and both $d(g(r),g(p))<\epsilon/3$ and $d(g(s),g(q))<\epsilon/3$ for $r\in V_n(p)$ and $s\in V_n(q)$. Also,
$$
d(r,s)<d(r,p)+d(p,q)+d(q,s)<\delta
$$
so $d(f(r),f(s))=d(g(r),g(s))<\epsilon/3$. By the triangle inequality, $d(g(p),g(q))<\epsilon$, so $g$ is uniformly continuous, and thus continuous on $X$, and of course $g|_E=f$.</p>
|
2,289,777 | <p>I have a function $f(x)=(8x^2+7)^3(x^3-7)^4$</p>
<p>I have differentiated it using the chain rule and arrived at:</p>
<p>$3(8x^2+7)^2 \cdot 16x \cdot 4(x^3-7)^3 \cdot 3x^2$ And apparently this is wrong?</p>
<p>What am I missing here? </p>
| Mike Miller | 67,263 | <p>If $p(x)=(8x^2+7)^3(x^3-7)^4$ then by the product rule $(fg)' = f'g + fg'$ we have $f(x) = (8x^2+7)^3$ and $g(x) = (x^3-7)^4$ so by the product and chain rule</p>
<p>$$p'(x) = 48x(8x^2+7)^2(x^3-7)^4 + 12x^2(8x^2+7)^3(x^3-7)^3\\=12x(8x^2+7)^2(x^3-7)^3(12x^3+7x-28).$$</p>
|
3,927,845 | <p>n is a natural number. Prove <span class="math-container">$6^n \geq n3^n$</span> holds for every natural number.</p>
<p><span class="math-container">$n = 1:$</span></p>
<p><span class="math-container">$$6 \geq 3 $$</span></p>
<p><span class="math-container">$n \rightarrow n + 1:$</span></p>
<p><span class="math-container">$$6 ^{n+1} = (3*2)^{n+1} = 3^{n+1}2^{n+1} \geq 3^{n+1}(n+1)2\geq 3^{n+1}(n+1) $$</span></p>
<p>Is my proof correct?</p>
<p>EDIT:
<span class="math-container">$$2^{n+1} \geq(n+1)2 \text{ because of Bernoulli inequality}$$</span></p>
| hamam_Abdallah | 369,188 | <p>Let us prove that <span class="math-container">$$(\forall n\ge 1)\;\; P_n : 2^n\ge n$$</span></p>
<p><strong>first step</strong>
<span class="math-container">$$2^1=2\ge 1 \implies P_1$$</span></p>
<p><strong>second step</strong></p>
<p>Let <span class="math-container">$ n\ge 1 $</span> such that <span class="math-container">$ 2^n\ge n$</span>.</p>
<p><span class="math-container">$$2^n\ge n \implies 2^{n+1}\ge 2n$$</span></p>
<p><span class="math-container">$$\implies 2^{n+1}\ge n+ n$$</span>
<span class="math-container">$$\implies 2^{n+1}\ge n+ 1$$</span></p>
<p>Now, just replace <span class="math-container">$$2^n=\frac{6^n}{3^n}$$</span></p>
|
59,965 | <p>If I have a function $f(x,y)$, is it always true that I can write it as a product of single-variable functions, $u(x)v(y)$?</p>
<p>Thanks.</p>
| André Nicolas | 6,312 | <p>Unfortunately, the answer is no. For example, let $f(x,y)=x-y$.</p>
<p>Note that $f(1,1)=0$. If $f(x,y)=u(x)v(y)$ for all $x$, $y$, then $0=f(1,1)= u(1)v(1)$. But if $u(1)v(1)=0$, then $u(1)=0$ or $v(1)=0$. </p>
<p>Suppose for example that $u(1)=0$. Then $u(1)v(y)=0$ for all $y$. If $f(x,y)=u(x)v(y)$ for all $x$, $y$, then $1-y=0$ for all $y$. This is clearly not true. </p>
<p>The example $f(x,y)=x-y$ is not really special. "Most" functions $f(x,y)$ are not expressible as $u(x)v(y)$.</p>
|
2,496,114 | <p>Show that $ a \equiv 1 \pmod{2^3 } \Rightarrow a^{2^{3-2}} \equiv 1 \pmod{2^3} $</p>
<p>Show that$ a \equiv 1 \pmod{2^4 } \Rightarrow a^{2^{4-2}} \equiv 1 \pmod{2^4} $</p>
<p><strong>Answer:</strong></p>
<p>$ a \equiv 1 \pmod{2^3} \\ \Rightarrow a^2 \equiv 1 \pmod{2^3} \\ \Rightarrow a^{2^{3-2}}=a^{2^1} \equiv 1 \pmod{2^3} $</p>
<p><strong>Am I right?</strong></p>
| Guy Fsone | 385,707 | <p>See that, $\displaystyle\frac{\sin x}{x} =f(x) = \frac{1}{2}\int_{-1}^{1} e^{-itx} dt$ Then
$$|f^{(n)}(x)| =\left|\frac{1}{2}\int_{-1}^{1} (-it)^ne^{-itx} dt\right| \le\frac{1}{2}\int_{-1}^{1} |t|^n dt=\int_0^1t^n\,dt=\frac1{n+1}.$$</p>
|
134,523 | <p>Is there any recursively axiomized system with infinitely many proofs for some propositions or a proposition? So we will have at least one proposition which is deduced from the recursively axiomatic system in infinite ways.Could any one give an example or proof?</p>
<p><strong>EDIT</strong>:We define a proof or a reduced proof as one which could not omit any proposition in the arguments,otherwise would be invalid. Or we can view it as question of formal language,that is :is there a formal language with at least one word which can be parsed in infinitely many ways?Of course there have to be no rewriting like "irrelevent padding or stupid detours"</p>
| Andreas Blass | 6,794 | <p>In any of the usual axiomatic systems (ZF set theory, Peano arithmetic, etc.), any proposition that has a proof at all will have infinitely many. The reason is that you can add irrelevant padding or stupid detours to any argument. To make a real question along these lines, you'd have to specify some notion of "genuinely different" proofs that prohibits such trivial variations.</p>
|
134,523 | <p>Is there any recursively axiomized system with infinitely many proofs for some propositions or a proposition? So we will have at least one proposition which is deduced from the recursively axiomatic system in infinite ways.Could any one give an example or proof?</p>
<p><strong>EDIT</strong>:We define a proof or a reduced proof as one which could not omit any proposition in the arguments,otherwise would be invalid. Or we can view it as question of formal language,that is :is there a formal language with at least one word which can be parsed in infinitely many ways?Of course there have to be no rewriting like "irrelevent padding or stupid detours"</p>
| none | 35,233 | <p>The question of whether two proofs are essentially the same is sometimes called "proof identity". It is an active topic in proof theory and there is an old MO thread with some references:</p>
<p><a href="https://mathoverflow.net/questions/3776/when-are-two-proofs-of-the-same-theorem-really-different-proofs">When are two proofs of the same theorem really different proofs</a></p>
|
908,083 | <p>I'd like to know what methods can I apply to simplify the fraction $\frac{4x + 2}{12 x ^2}$ </p>
<p>Is it valid to divide above and below by 2? (I didn't know it but Geogebra's Simplify aparantly does this)</p>
<p>Thanks in advance</p>
| 2'5 9'2 | 11,123 | <p>If you are looking for a formula that can be evaluated with a minimal number of bit operations for any given $x$, it might be best to multiply by $\frac{\frac14}{\frac14}$ and get $$\frac{x+\frac12}{3x^2}$$ which in binary would be $$\frac{x+0.1}{11\cdot x^{10}}$$</p>
|
4,013,796 | <p>If we make a regular polygon with n vertices (n edges) and triangulate on the inside with n-3 edges, then triangulate on the outside with (n-3) edges (or draw dotted lines inside again), a Maximal Planar Graph is formed. Edges shouldn't be repeated and there's no loops or directions.</p>
<p>How many distinct graphs of this type are there?</p>
<p>It's connected to an earlier question where it was asked 'How many Distinct Maximal Planar Graphs are there? <a href="https://math.stackexchange.com/questions/4009628/how-many-distinct-maximal-planar-graphs-exist-with-n-vertices">How many distinct Maximal Planar Graphs exist with $n$ vertices?</a></p>
<p>Will Orrick gave the OEIS numbers A000109. It was then wondered if there was a known formula for those numbers or bounds on them. It's conjectured that graphs of the type in this question might constitute most Maximal Planar Graphs, so a formula for the 'polygon' types might be an approximate formula or a lower bound for all types of Maximal Planar Graphs.</p>
<p>Dividing the polygon on the inside can be done in C(n-2) ways, where C(n) are the Catalan numbers A000108, so a connection was looked for - and A000109 seem to be close to C(n-2)*2^(n-13) at least for n=9 to 23.</p>
<p>So, the answer to this question would be of interest and any thoughts on connecting the A000108 and A000109 numbers. Lots of ways have been tried so far, e.g. since the next Catalan number can be formed by adding the product of ones before, e.g. C(4) = C(3)*C(1) + C(2)*C(2) + C(1)*C(3), perhaps something similar happens for A000109, or by incorporating numbers from both sequences. Lots of coincidences (probably) have been found including the 233 number in A000109 say X(10) is C(9-2) - the sum of the Catalan numbers before it.</p>
<p>Think I'm going to go crazy looking for patterns any longer! Any suggestions please!</p>
| Will Orrick | 3,736 | <p>This is too long to be a comment, but only provides a modest reduction in the <span class="math-container">$C_{n-2}^2$</span> upper bound.</p>
<p>For <span class="math-container">$n\ge3$</span> let <span class="math-container">$a_n$</span> be the number of unordered pairs of triangulations of the <span class="math-container">$n$</span>-gon with no common diagonals. Let <span class="math-container">$g(x)=a_3x^3+a_4x^4+a_5x^5+a_6x^6+\ldots$</span> be the generating function. The first few coefficients are <span class="math-container">$a_3=\frac{1}{2}$</span>, <span class="math-container">$a_4=1$</span>, <span class="math-container">$a_5=5$</span>, <span class="math-container">$a_6=34$</span> (see <a href="https://oeis.org/A257887" rel="nofollow noreferrer">A257887</a>). The non-integer value of <span class="math-container">$a_3$</span> makes sense in light of the property that there is one unordered pair of triangulations for every set of two ordered pairs of triangulations in which the triangulations trade places, <em>except</em> in the case of the triangle, where the only choice is for both triangulations to be empty. The OEIS omits the <span class="math-container">$a_3$</span> term. To understand why <span class="math-container">$a_6=34$</span> let <span class="math-container">$T$</span>, <span class="math-container">$Z$</span>, and <span class="math-container">$F$</span> denote the triangle, Z-shaped, and fan triangulations described in one of your comments. Then there is one <span class="math-container">$TT$</span> pair, six <span class="math-container">$TF$</span> pairs, <span class="math-container">$12$</span> <span class="math-container">$ZF$</span> pairs, nine <span class="math-container">$ZZ$</span> pairs, and six <span class="math-container">$FF$</span> pairs. The <span class="math-container">$TT$</span> pair and three of the <span class="math-container">$ZZ$</span> pairs give the octahedron; all others give the graph <span class="math-container">$K_6$</span> minus a path of length <span class="math-container">$4$</span> described by Dan Uznanski in a comment <a href="https://math.stackexchange.com/questions/4009628/how-many-distinct-maximal-planar-graphs-exist-with-n-vertices">here</a>. This example shows that one must account, not only for rotations and reflections of a pair of triangulations, but also for different choice Hamiltonian circuit, which can result in different triangulation pairs for the same graph.</p>
<p>Danièle Huguet and Dov Tamari in <em>La structure polyédrale des complexes de parenthésages</em>, J. Combin. Inform. System Sci., 3(2):69–81, 1978 showed that the generating function satisfies
<span class="math-container">$$
x=\sum_{j=0}^\infty C_j^2\left(x-\frac{2}{x}g(x)\right)^{j+1}.
$$</span>
In other words, <span class="math-container">$g(x)$</span> can be obtained by inverting the formal power series
<span class="math-container">$$
\phi(x)=\sum_{j=0}^\infty C_j^2x^{j+1},
$$</span>
which means finding the formal power series <span class="math-container">$u(x)$</span> such that <span class="math-container">$\phi(u(x))=x$</span>. Then <span class="math-container">$g(x)=\frac{x}{2}(x-u(x))$</span>. I have not been able to find a copy of Huguet and Tamari's paper, but a derivation is also given in</p>
<blockquote>
<p>Dimbinaina Ralaivaosaona, Jean Bernoulli Ravelomanana, and Stephan Wagner, <a href="https://drops.dagstuhl.de/opus/volltexte/2018/8925/pdf/LIPIcs-AofA-2018-32.pdf" rel="nofollow noreferrer"><em>Counting Planar Tanglegrams</em></a>, in James Allen Fill and Mark Daniel Ward eds., <em>29th International Conference on Probabilistic, Combinatorial and Asymptotic Methods for the Analysis of Algorithms (AofA 2018)</em>, Leibniz International Proceedings in Informatics (LIPIcs) 110:32:1–32:18, 2018.</p>
</blockquote>
<p>Ralaivaosaona, Ravelomanana, and Wagner point out that <span class="math-container">$\phi(x)$</span> is the series expansion of a complete elliptic integral and use this to do an asymptotic analysis. They find that <span class="math-container">$a_n$</span> grows like a constant times <span class="math-container">$n^{-3}\left(\frac{4\pi}{4-\pi}\right)^n$</span>. The upper bound <span class="math-container">$C_{n-2}^2$</span> grows like a constant times <span class="math-container">$n^{-3}\cdot16^n$</span>. The revised upper bound has slower growth since <span class="math-container">$\frac{4\pi}{4-\pi}\approx14.64$</span>.</p>
|
1,982,216 | <p>Consider the operator $B: L^1\left(\mathbb{R^+} \right)\to L^1\left(\mathbb{R^+} \right)$ defined for each $f\in L^1\left(\mathbb{R^+} \right)$ by
$$(Bf)(t)=\int_0^\infty\alpha (t,s)f(s)ds, \ \ \ \text{for} \ \ t\geq 0$$
where $\alpha:\mathbb{R^+} \times \mathbb{R^+}\to\mathbb{R}$ is a real function satisfying
$$\left|\alpha(t,s)\right| \leq\beta(t) \ \ \ \ \text{for all} \ \ t,s\geq 0,$$</p>
<p>where $\beta:\mathbb{R^+} \to\mathbb{R^+}$ is a positive integrable function.</p>
<p>Then $B$ defines a bounded operator on $L^1\left(\mathbb{R^+} \right)$.
Now if we suppose that the function $\alpha$ is constant with respect to the second argument, i.e. $\alpha(t,s)=\gamma(t)$ for all $t,s\geq0$, then one can see that $B$ is a finite rank operator ans thus compact.</p>
<p>Now in the general case ($\alpha$ not constant with respect to the second argument), can we say that $B$ is a compact operator ? how can we prove this if it's true ? Is there a reference which deals with the compactness of such operators ?</p>
| Davide Giraudo | 9,849 | <p>Following the references in <a href="https://math.stackexchange.com/questions/35115/classifying-the-compact-subsets-of-lp">this thread</a> about the characterization of compact subsets of $L^p\left(\mathbb R^n\right) $, $1\leqslant p\lt \infty$, the condition
$$\lim_{\delta\to 0}\sup_{\substack{ t,t'\geqslant 0\\ |t'-t|\lt \delta}} \sup_{ s\geqslant 0}\left|\alpha\left(t',s\right)- \alpha\left(t,s\right) \right|=0\quad \mbox{ and } \quad \left|\alpha(t,s)\right| \leqslant\beta(t) \ \ \ \ \text{for all} \ \ t,s\geq 0,$$
for an integrable function $\beta$ is sufficient for compactness of $B$.</p>
|
536,073 | <p>I came across several questions like this in the problem section of a book on coding theory & cryptography and I have no idea how to tackle them. There must be a certain trick that allows for efficiently solving such problems by hand.</p>
| Ekaveera Gouribhatla | 31,458 | <p>$$2^9\equiv-1(mod)19$$ $\implies$
$$2^{162}\equiv1(mod)19$$ $\implies$
$$2^{170}\equiv256(mod)19=9(mod)19$$ Also
$$3^3\equiv 2^3(mod)19$$ So
$$3^{63}\equiv2^{63}(mod)19=-1(mod)19$$ So
$$2^{170}+3^{63}\equiv 8(mod)19$$</p>
|
2,832,614 | <blockquote>
<p>Prove by induction that
$$\lim_{x \to a} \frac{x^n-a^n}{x-a}=na^{n-1}.$$</p>
</blockquote>
<p>I did a strange proof using two initial results: We know that result is true for $n=1$ and $n=2$. Assuming the result is true for $n=k-1$ and $n=k$, I can prove the result for $n=k+1$. For this I used my assumption for the case of $n=k$, and used a random multiplier to get the desired result:
$$\lim_{x \to a} \frac{x^{k}-a^{k}}{x-a}·\frac{x+a}{x+a}.$$
During the simplification I had to use my $n=k-1$ case result as well.</p>
<p>Is this proof OK in terms of induction principle? I can show my whole working if needed. </p>
<p>Thank you.</p>
<p><strong>Edit</strong></p>
<p>whole proof:</p>
<p>$\lim_{x \to a} \frac{x^{k}-a^{k}}{x-a}·\frac{x+a}{x+a}=k.a^{k-1}$ (I'll omit limit notation for clarity)</p>
<p>$$\frac{x^{k}-a^{k}}{x-a}·\frac{x+a}{x+a}=k.a^{k-1}$$</p>
<p>$$\frac{1}{2a}\frac{x^{k+1}+ax^k-a^kx-a^{k+1}}{x-a}=k.a^{k-1}$$</p>
<p>$$\frac{1}{2a}(\frac{x^{k+1}-a^{k+1}}{x-a}+\frac{ax(x^{k-1}-a^{k-1})}{x-a})=k.a^{k-1}$$</p>
<p>$$\frac{1}{2a}[\frac{x^{k+1}-a^{k+1}}{x-a}+{a^2(k-1).a^{k-2}}]=k.a^{k-1}$$</p>
<p>Hence, $$\lim_{x \to a} \frac{x^{k+1}-a^{k+1}}{x-a}=(k+1)a^{k}.$$</p>
| Doug M | 317,162 | <p>My algebra would be:</p>
<p>Suppose $\lim_\limits{x\to a} \frac {x^n-a^n}{x-a} = na^{n-1}$</p>
<p>Show that $\lim_\limits{x\to a} \frac {x^{n+1}-a^{n+1}}{x-a} = na^{n}$</p>
<p>$\frac {x\cdot x^{n} - a\cdot a^{n}}{x-a}$</p>
<p>Add and subtract the same term such that your expression will factor into something that lest you use the inductive hypothesis.</p>
<p>$\frac {x\cdot x^{n} -xa^n+xa^n- a\cdot a^{n}}{x-a}\\
\frac {x(x^{n} -a^n)}{x-a} +\frac{x -a}{x-a}a^n$</p>
<p>Then invoke the inductive hypothesis. Evaluate the limits for what remains and you are done.</p>
|
3,436,891 | <p>I have the following stupid question in my mind while i am studying for exams.
Does <span class="math-container">$X<\infty \ a.s$</span>, implies that <span class="math-container">$\mathbb E(X)<\infty$</span>? </p>
<p>Further on this, is the converse of the above statement true? Do give me a bit summary on this. Thanks very much.</p>
<p>I thought this was true until I realize the following example: Let's consider a simple symmetric random walk, we know that each state is null-recurrent. Let <span class="math-container">$\tau_L$</span> be a stopping time when the walk first hits <span class="math-container">$L$</span> started from <span class="math-container">$0$</span>. then
<span class="math-container">$$\mathbb P(\tau_L<\infty)=1$$</span>
so <span class="math-container">$\tau_L<\infty \ a.s.$</span>
but we also know that
<span class="math-container">$$\mathbb E(\tau_L)=\infty$$</span>
Is this a counter-example? Thanks, i am a bit weak on measure theory. </p>
| Simon Segert | 497,898 | <p>An even simpler counterexample is to take <span class="math-container">$P(X=n)=C/n^2$</span> where n is a positive integer and <span class="math-container">$C$</span> is a normalizing constant.</p>
<p>However, the converse is true. Usually, the expectation is only defined when <span class="math-container">$X$</span> is absolutely integrable. We have <span class="math-container">$P(|X|>M)M\leq E(|X|)<\infty$</span> for any <span class="math-container">$M>0$</span>. If <span class="math-container">$X$</span> is infinite with positive probability, then <span class="math-container">$P(|X|>M)\ge P(|X|=\infty)>0$</span> for any <span class="math-container">$M$</span>, so the left hand side of the inequality diverges as <span class="math-container">$M\to\infty$</span>, a contradiction.</p>
|
115,269 | <p>I'm studying the remainder class $\mathbb{Z}_n$, I've grabbed something, but something else is unfocused. Let
$$20x \equiv 4\pmod{34}$$
then GCD(20,34)=2 so I rewrite as:
$$10x \equiv 2\pmod{17}$$
and successively:
$$10x \equiv 1\pmod{17}$$
Now I know $\gcd(10, 17)=1$</p>
<blockquote>
<p>Question 1: Why? Is this cause I've divided both $20$ and $34$ for the
same $\gcd=2$? If $d$ is a $\gcd$ for $a$ and $b$, then $\gcd(a/d, b/d)=1$?</p>
</blockquote>
<p>At this point I can:
$$1=10\alpha + 17\beta$$
that will be:
$$1=10(-5)+17\cdot 3$$</p>
<blockquote>
<p>Question 2: I know that $-5$ is the solution of equation, but why I have to choose $-5$ more than $3$?</p>
</blockquote>
<p>Now $-5$ is a solution for:
$$10x \equiv 1\pmod{17}$$
and $-5\cdot2$ is a solution for:
$$10x \equiv 2\pmod{17}$$
$[-10]_{17}$ is class, the entire set is $\{-10+17k: k\in\mathbb{Z}\}$.</p>
<blockquote>
<p>Question 3: This set will contain all invertible elements. Right? Now
If I took $[8]_{17}$ and $[7]_{17}$ why the class $[56]_{17}$ is not
in the same class of $[1]_{17}$ since it is invertible, matter of
fact, $\gcd(56,17)=1$? Is $\gcd(x,y)=1$ the only way to test if $[x]_y$ is
invertible? I know that element $a$ was invertible if $ba=ab=1$, but this
contradicts my knowledge.</p>
</blockquote>
| Manolito Pérez | 13,293 | <p>More generally, you can set $f(k) = \sum_{i=1}^k a_i$, and then you get the recurrence equation $f(k) = f(k-1) + a_k$, with $f(1)$ known. Now solve the recurrence equation, and you have the formula you're looking for. </p>
|
115,269 | <p>I'm studying the remainder class $\mathbb{Z}_n$, I've grabbed something, but something else is unfocused. Let
$$20x \equiv 4\pmod{34}$$
then GCD(20,34)=2 so I rewrite as:
$$10x \equiv 2\pmod{17}$$
and successively:
$$10x \equiv 1\pmod{17}$$
Now I know $\gcd(10, 17)=1$</p>
<blockquote>
<p>Question 1: Why? Is this cause I've divided both $20$ and $34$ for the
same $\gcd=2$? If $d$ is a $\gcd$ for $a$ and $b$, then $\gcd(a/d, b/d)=1$?</p>
</blockquote>
<p>At this point I can:
$$1=10\alpha + 17\beta$$
that will be:
$$1=10(-5)+17\cdot 3$$</p>
<blockquote>
<p>Question 2: I know that $-5$ is the solution of equation, but why I have to choose $-5$ more than $3$?</p>
</blockquote>
<p>Now $-5$ is a solution for:
$$10x \equiv 1\pmod{17}$$
and $-5\cdot2$ is a solution for:
$$10x \equiv 2\pmod{17}$$
$[-10]_{17}$ is class, the entire set is $\{-10+17k: k\in\mathbb{Z}\}$.</p>
<blockquote>
<p>Question 3: This set will contain all invertible elements. Right? Now
If I took $[8]_{17}$ and $[7]_{17}$ why the class $[56]_{17}$ is not
in the same class of $[1]_{17}$ since it is invertible, matter of
fact, $\gcd(56,17)=1$? Is $\gcd(x,y)=1$ the only way to test if $[x]_y$ is
invertible? I know that element $a$ was invertible if $ba=ab=1$, but this
contradicts my knowledge.</p>
</blockquote>
| Brian M. Scott | 12,042 | <p>There are indeed methods of dealing with many more complicated summations; the excellent book <em>Concrete Mathematics</em>, by Graham, Knuth, & Patashnik, is full of such techniques. One of them is <em>finite calculus</em>; <a href="http://www.cs.purdue.edu/homes/dgleich/publications/finite-calculus.pdf">this PDF</a> contains a decent brief introduction using the notation of <em>Concrete Mathematics</em>.</p>
<p>One of the basic results of finite calculus is that $$\sum_{k=a}^bk^{\underline{m}}=\left[\frac{k^{\underline{m+1}}}{m+1}\right]_a^{b+1}\;,$$ where $x^{\underline{m}}=x(x-1)(x-2)\cdots(x-m+1)$ is called a <em>falling power</em> of $x$. </p>
<blockquote>
<p>Note the similarity with $$\int_a^bx^m dx=\left[\frac{x^{m+1}}{m+1}\right]_a^b\;,$$ though there is a difference in the upper limit of evaluation.</p>
</blockquote>
<p>In this case we write $k^2=k(k-1)+k=k^{\underline{2}}+k^{\underline{1}}$, so that $$\sum_{k=1}^nk^2=\sum_{k=1}^n(k^{\underline{2}}+x^{\underline{1}})\;,$$ and </p>
<p>$$\begin{align*}\sum_{k=1}^n(k^2+3k)&=\sum_{k=1}^n(k^{\underline{2}}+4x^{\underline{1}})\\
&=\sum_{k=1}^nk^{\underline{2}}+4\sum_{k=1}^nx^{\underline{1}}\\
&=\left[\frac{k^{\underline{3}}}3\right]_1^{n+1}+4\left[\frac{k^{\underline{2}}}2\right]_1^{n+1}\\
&=\frac13\left((n+1)^{\underline{3}}-1^{\underline{3}}\right)+2\left((n+1)^{\underline{2}}-1^{\underline{2}}\right)\\
&=\frac13(n+1)^{\underline{3}}+2(n+1)^{\underline{2}}\\
&=\frac13(n+1)n(n-1)+2(n+1)n\\
&=n(n+1)\left(\frac{n-1}3+2\right)\\
&=\frac{n(n+1)(n+5)}3\;.
\end{align*}$$</p>
<p>This may look a little complicated, but once you have the basic tools of finite calculus, which aren’t actually very hard, you can generalize this to all sorts of polynomial sums and even beyond that.</p>
|
115,269 | <p>I'm studying the remainder class $\mathbb{Z}_n$, I've grabbed something, but something else is unfocused. Let
$$20x \equiv 4\pmod{34}$$
then GCD(20,34)=2 so I rewrite as:
$$10x \equiv 2\pmod{17}$$
and successively:
$$10x \equiv 1\pmod{17}$$
Now I know $\gcd(10, 17)=1$</p>
<blockquote>
<p>Question 1: Why? Is this cause I've divided both $20$ and $34$ for the
same $\gcd=2$? If $d$ is a $\gcd$ for $a$ and $b$, then $\gcd(a/d, b/d)=1$?</p>
</blockquote>
<p>At this point I can:
$$1=10\alpha + 17\beta$$
that will be:
$$1=10(-5)+17\cdot 3$$</p>
<blockquote>
<p>Question 2: I know that $-5$ is the solution of equation, but why I have to choose $-5$ more than $3$?</p>
</blockquote>
<p>Now $-5$ is a solution for:
$$10x \equiv 1\pmod{17}$$
and $-5\cdot2$ is a solution for:
$$10x \equiv 2\pmod{17}$$
$[-10]_{17}$ is class, the entire set is $\{-10+17k: k\in\mathbb{Z}\}$.</p>
<blockquote>
<p>Question 3: This set will contain all invertible elements. Right? Now
If I took $[8]_{17}$ and $[7]_{17}$ why the class $[56]_{17}$ is not
in the same class of $[1]_{17}$ since it is invertible, matter of
fact, $\gcd(56,17)=1$? Is $\gcd(x,y)=1$ the only way to test if $[x]_y$ is
invertible? I know that element $a$ was invertible if $ba=ab=1$, but this
contradicts my knowledge.</p>
</blockquote>
| Norbert | 19,538 | <p>My approach will look like overkill for your particular problem, but you asked for the general method of summation such series.</p>
<p>If you can find explicit formula for series of the form $S_r(k)=\sum\limits_{i=0}^k i^r$, $r\in\mathbb{Z}_+$ then you can find formula for the series of the form $\sum\limits_{i=1}^k P_n(i)$, where $P_n(x)$ is the polynomial of degree $n$. Indeed, if $P_n(x)=\sum\limits_{r=1}^n a_r x^r$, then
$$
\sum\limits_{i=1}^k P_n(i)=
\sum\limits_{i=1}^k \sum\limits_{r=1}^n a_r i^r=
\sum\limits_{r=1}^n a_r \sum\limits_{i=1}^k i^r=
\sum\limits_{r=1}^n a_r S_r(k)
$$
So the question is how to compute $S_r(k)$. This can be made by recurrence formula. For the begining
$$
S_0(k)=\sum\limits_{i=0}^k i^0=k+1
$$
Now to derive recurrence formula note that
$$
(i+1)^{r+1}-i^{r+1}=
\left(\sum\limits_{p=0}^{r+1} {r+1 \choose p} 1^{r+1-p} i^p\right) -i^{r+1}=
$$</p>
<p>$$
\left(\sum\limits_{p=0}^{r+1} {r+1 \choose p} i^p\right) -i^{r+1}=
\sum\limits_{p=0}^{r} {r+1 \choose p} i^p
$$
Now lets sum this equalities by $i$ from $0$ to $k$. Then we get
$$
\sum\limits_{i=0}^{k} \left((i+1)^{r+1}-i^{r+1}\right)=
\sum\limits_{i=0}^{k} \sum\limits_{p=0}^{r} {r+1 \choose p} i^p=
\sum\limits_{p=0}^{r} {r+1 \choose p}\sum\limits_{i=0}^{k} i^p=
\sum\limits_{p=0}^{r} {r+1 \choose p}S_p(k)=
$$
$$
{r+1 \choose r}S_{r}(k)+\sum\limits_{p=0}^{r-1} {r+1 \choose p}S_p(k)=
(r+1)S_{r}(k)+\sum\limits_{p=0}^{r-1} {r+1 \choose p}S_p(k)
$$
Note that $\sum\limits_{i=0}^{k} \left((i+1)^{r+1}-i^{r+1}\right)$ is a telescopic sum and it is equals to $(k+1)^{r+1}$. So we get the following equality
$$
(k+1)^{r+1}=(r+1)S_{r}(k)+\sum\limits_{p=0}^{r-1} {r+1 \choose p}S_p(k)
$$
This gives us
$$
S_r(k)=\frac{1}{r+1}\left((k+1)^{r+1}-\sum\limits_{p=0}^{r-1} {r+1 \choose p}S_p(k)\right)
$$
From this formula one can show by induction that $S_r(k)$ is a polynomial of variable $k$ of degree $r+1$. Coefficients of this polynomial is strongly related to so called <a href="http://en.wikipedia.org/wiki/Bernoulli_number" rel="nofollow">Bernoulli numbers</a>, denoted by $B_n$. In fact,
$$
S_r(k)=\frac{1}{r+1}\sum\limits_{i=0}^{r}{r+1\choose i}B_i k^{r+1-i}
$$
The first Bernoulli numbers are: $B_0=1$, $B_1=-1/2$, $B_2=1/6$, $B_3=-1/30$ et cetera.</p>
<p>Now we return to the original problem
$$
S_1(k)=
\frac{1}{2}\left((k+1)^2-\sum\limits_{p=0}^{0} {2 \choose p}S_p(k)\right)=
\frac{1}{2}\left((k+1)^2-{2 \choose 0}S_0(k)\right)=
\frac{1}{2}\left(k^2+2k+1-(k+1)\right)=\frac{k^2+k}{2}=\frac{k(k+1)}{2}
$$
and
$$
S_2(k)=
\frac{1}{3}\left((k+1)^{3}-\sum\limits_{p=0}^{1} {3 \choose p}S_p(k)\right)=
\frac{1}{3}\left((k+1)^{3}-{3 \choose 0}S_0(k)-{3 \choose 1}S_1(k)\right)=
\frac{1}{3}\left((k+1)^{3}-(k+1)-3\frac{k^2+k}{2}\right)=
\frac{2k^3+3k^2+k}{6}=\frac{k(2k+1)(k+1)}{6}
$$
Finally
$$
\sum\limits_{i=1}^k(i^2+3i)=
\sum\limits_{i=1}^k i^2+3\sum\limits_{i=1}^k i=
$$
$$
S_2(k)+3S_1(k)=
\frac{2k^3+3k^2+k}{3}+\frac{k^2+k}{2}=
\frac{k^3+6 k^2+5 k}{3}=
\frac{k (k+1) (k+5)}{3}
$$</p>
|
88,159 | <p>Define $\omega=e^{i \pi /4}$. Is there an elegant way of showing that $20^{1/4} \omega^3$ is not inside $\mathbb{Q}(20^{1/4} \omega)$?</p>
<p>The way i am doing it is by observing that $20^{1/4} \omega$ is algebraic over $\mathbb{Q}$ with minimal polynomial $x^4+20$ and i assume that $20^{1/4} \omega^3$ is in the span over $\mathbb{Q}$ of $1, \alpha, \alpha^2,\alpha^3$, with $\alpha=20^{1/4} \omega$. But it is very messy to arrive at a contradiction.</p>
<p>Thanks.</p>
| Georges Elencwajg | 3,217 | <p>Let $K= \mathbb{Q}(20^{1/4}\omega) $, <em>a field of dimension $4$ over $\mathbb Q$</em> .<br>
Clearly $\omega^2\in K \iff K \;$is the splitting field of $f(X)=X^4+20$ over $\mathbb Q$ .<br>
It suffices to show that this is not the case.<br>
The resolvent of $f(X)$ is $r(Y)=Y^3-80Y $ . This implies that the Galois group of $X^4+20$ is the dihedral group $D_4$ of order $8$ , (cf. for example Morandi's <a href="http://books.google.com/books/about/Field_and_Galois_theory.html?id=HALy09o35TQC" rel="nofollow"><em>Field and Galois Theory</em></a>, Theorem 13.4, page 126).<br>
Thus the splitting field of $f(X)$ over $\mathbb Q$ has <em>dimension $8$</em>
and is thus not equal to $K$.</p>
|
64,977 | <p>Suppose I had a complete bipartite graph with edges each given some numerical "cost" value. Is there a way to select a subset of those edges such that each vertex on each side of the graph is mapped to each vertex on the other (one to one) and the total "costs" is maximized (or minimized)?</p>
<p>Has anyone ever formulated something equivalent to this?</p>
| Or Zuk | 1,778 | <p>You can find a set $\Omega$ with size $O(d)$. Assume for simplicity that $d=2^k-1$ for an integer $k$. Let $y_1,..,y_k$ be binary variables and define
$x_j = \sum_i y_i \alpha_i(j)$ where $\alpha_i(j)$ is simply the $i$-th digit of $j$ when represented in binary. (For example, if $n=7$, you'll get $y_1, y_2, y_3, y_1+y_2, y_1+y_3, y_2+y_3, y_1+y_2+y_3$). You can get $2^k-1$ such linear combinations and they're all balanced and independent - that is $Pr(x_i=1)=1/2$ and $Pr(x_i=a,x_j=b)=1/4, \: \forall a,b=0,1$.
This will satisfy the condition $p(\Omega,i,j,A) = |A|/4$. When $d=2^k-1$ this gives $f(d)\leq d+1$. For general $d$ (not be a power of two minus one) you should get $f(d) \leq 2d$ taking $d=2^k$ as the worst case.</p>
|
2,995,495 | <p>I'm trying to prove that, for every <span class="math-container">$x \geq 1$</span>:</p>
<p><span class="math-container">$$\left|\arctan (x)-\frac{π}{4}-\frac{(x-1)}{2}\right| \leq \frac{(x-1)^2}{2}.$$</span> </p>
<p>I could do it graphically on <span class="math-container">$\Bbb R$</span>, but how to make a formal algebraic proof?</p>
| Barry Cipra | 86,747 | <p>It is convenient to let <span class="math-container">$x=u+1$</span> and rewrite the inequality to be proved as a pair of inequalities:</p>
<p><span class="math-container">$${u\over2}-{u^2\over2}\le\arctan(u+1)-\arctan1\le{u\over2}+{u^2\over2}$$</span></p>
<p>for <span class="math-container">$u\ge0$</span>. Now</p>
<p><span class="math-container">$$\arctan(u+1)-\arctan1=\int_1^{u+1}{dt\over1+t^2}=\int_0^u{dt\over1+(1+t)^2}=\int_0^u{dt\over2+2t+t^2}$$</span></p>
<p>while</p>
<p><span class="math-container">$${u\over2}-{u^2\over2}=\int_0^u{1-2t\over2} dt\qquad\text{and}\qquad{u\over2}+{u^2\over2}=\int_0^u{1+2t\over2}dt$$</span></p>
<p>It suffices, therefore, to show that, for <span class="math-container">$t\ge0$</span>, we have</p>
<p><span class="math-container">$${1-2t\over2}\le{1\over2+2t+t^2}\le{1+2t\over2}$$</span></p>
<p>The first inequality follows from <span class="math-container">$(1-2t)(2+2t+t^2)=2-2t-3t^2\le2$</span>; the second inequality is even easier:</p>
<p><span class="math-container">$${1\over2+2t+t^2}\le{1\over2}\le{1+2t\over2}$$</span></p>
|
480,195 | <p>Three friends brought 3 pens together each 10 dollars. Next day they got 5 dollars cash back so they shared each 1 dollar and donated 2 dollars. Now the pen cost for each guy will be 9 dollars (\$10 -\$1).</p>
<p>But if you add all 9+9+9 = 27 dollars and donated amount is 2 dollars so total 29 dollars. </p>
<p>Where is the other \$1?</p>
| stig | 92,473 | <p>10 x 3 = 30.
30 - 5 = 25.
25 + 3 = 28.
28 + 2 = 30.
so, no missing $.</p>
|
526,837 | <p>Let $(\Omega, {\cal B}, P )$ be a probability space, $( \mathbb{R}, {\cal R} )$ the usual
measurable space of reals and its Borel $\sigma$- algebra, and $X : \Omega \rightarrow \mathbb{R}$ a random variable.</p>
<p>The meaning of $ P( X = a) $ is intuitive when $X$ is a discrete random variable, because it's the definition of the probability mass function. I am not sure if my question makes sense, but how should I think of $ P( X = a) $ when $X$ is a continuous random variable? </p>
| lab bhattacharjee | 33,337 | <p>As $$\frac{-4x}{1+2x}=\frac{-2(1+2x)+2}{1+2x}=-2+\frac2{2x+1},$$</p>
<p>$$\int\frac{-4x}{1+2x}dx=-2\int dx+\int\frac2{1+2x}dx=-2x+\ln|1+2x|+C$$ </p>
<p>putting $1+2x=u$ in the second integral
and where $C$ is an arbitrary constant of indefinite integral</p>
<p>In your case $C=-1$</p>
<p>In fact, if $$f'(x)=g'(x)\iff \int df(x)=\int d g(x)\implies f(x)=g(x)+K$$ where $K$ is an arbitrary constant of indefinite integral</p>
|
2,032,711 | <p>In a triangle $ABC$, if $\sin A+\sin B+\sin C\leq1$,then prove that $$\min(A+B,B+C,C+A)<\pi/6$$
where $A,B,C$ are angles of the triangle in radians.</p>
<p>if we assume $A>B>C$,then $\sum \sin A\leq 3 \sin A$,and $ A\geq \frac{A+B+C}{3}=\pi/3$.also $\sum \sin A\geq 3\sin C$ and $ C\leq \frac{A+B+C}{3}=\pi/3$.But I could not proceed with this. Please help me in this regard.Thanks.</p>
| Djura Marinkov | 361,183 | <p>A must be larger than $\frac {5\pi} 6 $ cos if it's not, then $\sin A\ge\frac 1 2$ so $\sin B+\sin C\le\frac 1 2$ but $B+C\ge\frac {\pi} 6 $ which is not possible since $\sin B+\sin C>\sin(B+C)\ge\frac 1 2$</p>
|
878,115 | <p>Question1:
I found 30 boxes. In 10 boxes i found 15 balls. In 20 boxes i found 0 balls.
Afer i collected all 15 balls i put them randomly inside the boxes.</p>
<p>How much is the chance that all balls are in only 10 boxes or less?</p>
<p>Question2:
I found 30 boxes. In 10 boxes i found 15 balls. In 20 boxes i found 0 balls. In two of the boxes i could find 3 balls. (So in one box has to be 2 balls and in the other seven boxes have to be 1 ball.)
Afer i collected all 15 balls i put them randomly inside the boxes.</p>
<p>How much is the chance that i find in only 2 boxes 6 balls or more?</p>
<p>I wrote a c# programm and tried it 1 million times.
My solution was: With a chance of 12,4694% all balls are in 10 boxes or less.</p>
| Mr.Spot | 155,516 | <p><strong>Solution of Question 1:</strong></p>
<p>This is an occupancy problem with $n=30$ boxes and $k=15$ balls.</p>
<p>Let's first consider the expected number of empty boxes. That is much easier to obtain. The exact answer is $30(1-1/30)^{15}=18.04.$ This is approximately $30/\sqrt e.$ See the answer by Mr.Spot to this question:</p>
<p><a href="https://math.stackexchange.com/questions/873798/making-400k-random-choices-from-400k-samples-seems-to-always-end-up-with-63-dis/">Making 400k random choices from 400k samples seems to always end up with 63% distinct choices, why?</a></p>
<p>The probability of exactly $j$ empty boxes is, for $n-k\le j\le n-1$:</p>
<p>$$P(j)={n \choose j}\sum_{m=0}^{n-j}(-1)^m {n-j \choose m}\left(1-\frac{j+m}n\right)^k $$</p>
<p>See this:</p>
<p><a href="http://probabilityandstats.wordpress.com/2010/04/04/a-formula-for-the-occupancy-problem/" rel="nofollow noreferrer">http://probabilityandstats.wordpress.com/2010/04/04/a-formula-for-the-occupancy-problem/</a></p>
<p>For $n=30$ and $k=15:$</p>
<p>$P(20)$ to $P(24)$ is
$0.096,0.024,0.0036,0.00029,0.000013,....$</p>
<p>and the probability of at least $20$ empty cells = 0.124371</p>
<p><strong>Solution of Question 2:</strong></p>
<p>You throw 15 balls into 30 boxes. What is the probability of the following result:</p>
<p>2 triple, 1 double and 7 single occupancy boxes and the rest empty.</p>
<p>$${30 \choose 2} {28 \choose 1}{27 \choose 7}\frac{15!}{3!3!2!1!^7}\frac{1}{30^{15}} $$
$$=\frac{(30)(29)...(21)}{30^{15}2!1!7!}\frac{15!}{3!3!2!1!^7} $$
First we select the 2 different boxes for the triples; of the 28 remaining boxes we select 1 for the double; of the 27 remaining boxes we select 7 for the singles. Then the multinomial coefficient gives the number of ways to assign the 15 balls to those particular boxes and there are $30^{15}$ equally likely ways to throw the 15 balls into the 30 boxes.</p>
|
3,371,302 | <p>trying to find all algebraic expressions for <span class="math-container">${i}^{1/4}$</span>.</p>
<p>Using. Le Moivre formula , I managed to get this : </p>
<blockquote>
<p><span class="math-container">${i}^{1/4}=\cos(\frac{\pi}{8})+i \sin(\frac{\pi}{8})=\sqrt{\frac{1+\frac{1}{\sqrt{2}}}{2}} + i \sqrt{\frac{1-\frac{1}{\sqrt{2}}}{2}}$</span></p>
</blockquote>
<p>What's about other expressions.</p>
| J. W. Tanner | 615,567 | <p><strong>Hint:</strong>. If <span class="math-container">$\exp(i\theta)^4=\exp(i4\theta)=i$</span>, then <span class="math-container">$4\theta+2\pi n=\dfrac {\pi}2$</span>, where <span class="math-container">$n\in\Bbb Z$</span>. If you solve for <span class="math-container">$\theta, $</span> you should get the other solutions besides <span class="math-container">$\theta=\dfrac{\pi}8$</span>.</p>
|
3,371,302 | <p>trying to find all algebraic expressions for <span class="math-container">${i}^{1/4}$</span>.</p>
<p>Using. Le Moivre formula , I managed to get this : </p>
<blockquote>
<p><span class="math-container">${i}^{1/4}=\cos(\frac{\pi}{8})+i \sin(\frac{\pi}{8})=\sqrt{\frac{1+\frac{1}{\sqrt{2}}}{2}} + i \sqrt{\frac{1-\frac{1}{\sqrt{2}}}{2}}$</span></p>
</blockquote>
<p>What's about other expressions.</p>
| Rhys Hughes | 487,658 | <p>If <span class="math-container">$(a+bi)^2=c+di$</span>, then:</p>
<p><span class="math-container">$$a=\pm\sqrt{\frac{c\pm |c+di|}{2}},b=\pm\sqrt{\frac{-c\pm |c+di|}{2}}$$</span></p>
<p>First we get the result: <span class="math-container">$$\sqrt{\lambda i}=\pm\sqrt{\frac\lambda 2}(1+i)$$</span></p>
<p>and then <span class="math-container">$$\sqrt[4]{\lambda i}=\sqrt[4]{\frac\lambda 2}\bigg(\pm\sqrt{\frac{1\pm\sqrt2}{2}}\pm i\sqrt{\frac{-1\pm\sqrt 2}{2}}\bigg)$$</span></p>
<p>What you seek is when <span class="math-container">$\lambda=1$</span> (the solutions are more exhaustive than necessary because this set of solutions considers <span class="math-container">$(p)+(q)i$</span> and <span class="math-container">$(qi)+(-pi)i$</span> as distinct)</p>
|
1,650,204 | <p>I was given this problem and I can't seem to think of a solution.</p>
<p>Here is a possibly helpful graphic:</p>
<p><a href="https://i.stack.imgur.com/VKZkv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VKZkv.png" alt="Here is a possibly helpful link:" /></a></p>
<blockquote>
<p>Given two parallel lines (representing the banks of a river) and two arbitrary points <span class="math-container">$A$</span> and <span class="math-container">$B$</span> outside of the river (one above the top parallel line and one below the bottom parallel line). A bridge is to be constructed connecting the two sides of the river at point <span class="math-container">$P$</span> and <span class="math-container">$Q$</span> are to be an equal distance between points <span class="math-container">$A$</span> and <span class="math-container">$B$</span>, i.e. <span class="math-container">$\overline{AP}$</span> = <span class="math-container">$\overline{BQ}$</span>.</p>
<p>Where should the bridge be placed, assuming that it runs at right angles to the banks?</p>
</blockquote>
| P Vanchinathan | 28,915 | <p>One talks of eigenvalues etc for linear transformations that are functions from a vector space to itself. $T(u+v)= T(u)+T(v),\ T(av)= aT(v)$.</p>
<p>Having studied physics you must have a good understanding of the 2 and 3-dimensional (euclidean) spaces. I'll stick to them. Any three dimensional rotation around origin is a linear transformation (check it using parallelogram law for addition of vectors). Same for a refelection about a plane passing through origin.</p>
<p>A linear transformation on this vector space (in co-ordinates) is simply a function that is given by three linear homogeneous polynomials.
(or two homogeneous poly in two variables in the 2-dimensional case)</p>
<p>$$(x,y,z)\mapsto (a_1x+b_1y+c_1z, a_2+b_2y+c_2z, a_3x+b_3y+c_3z)$$</p>
<p>This is often written in matrix form as $$\left(\begin{array}{lll}
a_1 &a_2&a_3\\ b_1&b_2&b_3 \\ c_1&c_2&c_3 \\ \end{array}\right)$$</p>
<p>A vector is called an eigenvector for such a transformation if it moves in its own line (connecting it to the origin) by this transformation. The ratio of the lengths of the vector after and before transformation is the eigenvalue of that eigenvector.</p>
<p>Clear that in 2d there are no eigenvectors for rotations (except the zero degree one!). In 3d, we have axis of rotation: they are fixed and so they are eigenvectors of eigenvalue 1. </p>
<p>For reflections there will always be eigenvectors of eigenvalue $-1$ those vectors perpendicular to reflecting plane (or line in the case of 2d).</p>
<p>I have explained in geometric terms. There are lots and lots of linear transformation other than rotations and reflections. For them, eigenvalues and eigenvectors are computed by matrix calculations and solving linear systems.</p>
|
3,381,939 | <p>Be T tree order n given:</p>
<p><span class="math-container">$(a)$</span> <span class="math-container">$ 95 < n < 100$</span></p>
<p><span class="math-container">$(b)$</span> Just have vertices with degree 1,3,5</p>
<p><span class="math-container">$(c)$</span> T has twice the vertices of degree 3 that degree 5</p>
<p><span class="math-container">$n$</span> is...?</p>
<p>Okay, i think that can use handshake to sum degree to obtain <span class="math-container">$2e$</span> and solve for algebra, but I don't see how propose equation. Hope that can help me with it </p>
| J.G. | 56,861 | <p>If such a formula existed, one with a minimal number of implications would contain at least one <span class="math-container">$\to$</span> (because you've already checked the case with none), and we could write some such contradiction as <span class="math-container">$a\to b$</span>. (If <span class="math-container">$(a\to b)\lor c$</span> is a contradiction, so is <span class="math-container">$a\to b$</span>; if <span class="math-container">$(a\to b)\land c$</span> is a contradiction, so is <span class="math-container">$(c\land a)\to b$</span>.) We then need <span class="math-container">$a$</span> to be true and <span class="math-container">$b$</span> false in all interpretations. But then <span class="math-container">$b$</span> is an option with one fewer <span class="math-container">$\to$</span>, a contradiction.</p>
|
2,437,983 | <p>What is the chance that at least two people were born on the same day of the week if there are 3 people in the room?</p>
<p>I know how to get the answer which is 19/49 when considering all 3 people <strong>not being born on the same day</strong>. However, when I try to calculate the answer directly I seem to get it wrong.</p>
<p>Considering exactly 2 people being born on the same day I get 1*1/7*6/7. And then, exactly 3 people is 1*1/7*1/7. Thus, the total is 6/49 + 1/49 = 7/49. This must be something fairly simple, but I was just wondering where I'm going wrong. </p>
<p>Thanks</p>
| mechanodroid | 144,766 | <p>Let $\mathcal{I}$ be the collection of all open intervals, and $\mathcal{A}$ be the collection of all half open-intervals $[a,b)$. We shall prove $\mathcal{B}_\mathbb{R} = \sigma(\mathcal{I})$ and $\sigma(\mathcal{I}) = \sigma(\mathcal{A})$.</p>
<p>Open intervals are open sets so $\mathcal{I} \subseteq \mathcal{B}_\mathbb{R} \implies \sigma(\mathcal{I}) \subseteq \mathcal{B}_\mathbb{R}$.</p>
<p>Take an open set $U \subseteq \mathbb{R}$. For $x \in U$ there exists an open interval $I_x \subset U$ such that $x \in U$. Take $I_x$ to be the largest such interval. We have $U = \bigcup_{x\in U} I_x$. The intervals $I_x$ are disjoint, otherwise they wouldn't be the largest. Since a union of open intervals can be disjoint only if it is at most countable (since each of them contains a distinct rational number), there exists countably many intervals $I_{x_{n}}$ such that $U = \bigcup_{n=1}^\infty I_{x_{n}}$. Thus, $U \in \sigma(\mathcal{I})$.</p>
<p>Hence, $\mathcal{B}_\mathbb{R} = \sigma(\mathcal{I})$.</p>
<p>Now, take an open interval $(a, b) \in \mathcal{I}$ and notice
$$ (a, b) = \bigcup_{n=1}^\infty \underbrace{\left[a +\frac{1}{n}, b\right)}_{\in \mathcal{A}} \in \sigma(\mathcal{A})$$</p>
<p>so $\mathcal{I} \subseteq \sigma(\mathcal{A})\implies \sigma(\mathcal{I}) \subseteq \sigma(\mathcal{A})$.</p>
<p>Similarly, take $[a,b) \in \mathcal{A}$ and notice:</p>
<p>$$ [a, b) = \bigcap_{n=1}^\infty \underbrace{\left(a -\frac{1}{n}, b\right)}_{\in \mathcal{I}} \in \sigma(\mathcal{I})$$</p>
<p>so $\mathcal{A} \subseteq \sigma(\mathcal{I})\implies \sigma(\mathcal{A}) \subseteq \sigma(\mathcal{I})$.</p>
<p>Hence, $\sigma(\mathcal{I}) = \sigma(\mathcal{A})$.</p>
<p>Finally, we can conclude $\mathcal{B}_\mathbb{R} = \sigma(\mathcal{A})$.</p>
|
1,058,831 | <p>Let $A$ and $B$ be two Hermitian matrices. I wanted to know if there is any relation between the maximum eigenvalue of $AB$ and that of $A$ and $B$. Is the following relation true?
$\lambda_{\text{max}}\left(AB\right)\le \lambda_{\text{max}}\left(A\right)\lambda_{\text{max}}\left(B\right)$</p>
<p>If it is true, then is this relation valid when $A$ and $B$ are real.</p>
| copper.hat | 27,978 | <p>Let $A=B=\operatorname{diag}(-2,1)$, then $AB = \operatorname{diag}(4,1)$, but
$\lambda_\max (AB) = 4$, $\lambda_\max (A) = \lambda_\max (B)= 1$.</p>
<p>If the matrices are positive semi-definite, then $\|A\|=\lambda_\max(A)$ (since
$A$ is unitarily diagonalisable) and the
spectral norm is submultiplicatve, hence the result holds.</p>
|
1,058,831 | <p>Let $A$ and $B$ be two Hermitian matrices. I wanted to know if there is any relation between the maximum eigenvalue of $AB$ and that of $A$ and $B$. Is the following relation true?
$\lambda_{\text{max}}\left(AB\right)\le \lambda_{\text{max}}\left(A\right)\lambda_{\text{max}}\left(B\right)$</p>
<p>If it is true, then is this relation valid when $A$ and $B$ are real.</p>
| Algebraic Pavel | 90,996 | <p>If $A$ and $B$ are $\color{blue}{\rm HPD}$, then
$$
\begin{split}
\color{red}{\lambda_{\max}(AB)}
&=
\lambda_{\max}(B^{1/2}AB^{1/2})
\\&=
\max_{x\neq 0}\frac{x^*B^{1/2}AB^{1/2}x}{x^*x}
\\&=
\max_{x\neq 0}\frac{x^*Ax}{x^*B^{-1}x}
\\&=
\max_{x\neq 0}\left(\frac{x^*Ax}{x^*x}\right)\left(\frac{x^*x}{x^*B^{-1}x}\right)
\\&\color{red}{\leq}
\max_{x\neq 0}\left(\frac{x^*Ax}{x^*x}\right)\max_{y\neq 0}\left(\frac{y^*y}{y^*B^{-1}y}\right)
\\&=
\max_{x\neq 0}\left(\frac{x^*Ax}{x^*x}\right)\max_{y\neq 0}\left(\frac{y^*By}{y^*y}\right)
\\&=
\color{red}{\lambda_{\max}(A)\lambda_{\max}(B)}.
\end{split}
$$
Otherwise, as already indicated, the inequality does not need to be true if $A$ or $B$ is indefinite. By the continuity argument, it can also be extended for the case when $A$ and $B$ are only semi-definite.</p>
|
2,849,017 | <p>\begin{align}
dA & = 2RR\,dv = 2R^2\,dv \\[8pt]
A & = \int_0^\pi 2R^2\,dv \\[8pt]
\text{arclength} & = R\,dv
\end{align}</p>
<p><a href="https://i.stack.imgur.com/YBIX5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YBIX5.png" alt="enter image description here"></a></p>
<p>Area of a circle with radius $R$ is $\pi R^2$.</p>
<p>I am trying to discover new correct methods for finding the area of a circle. Using rectangles is not something new per say, but summing infinitesimal rectangles going through the center with the angle as a parameter I have not seen before. Anyway, I have encountered a problem:</p>
<p>Let $dA$ be the infinitesimal rectangle of width $R\,dv$ and length $2R$. As seen in the picture, the position of the rectangle is determined by the angle $V$. If, for example, $V=0$ the rectangle will be positioned "horisontal", and if $V=\frac{\pi}{2}$ the rectangle will be positioned "vertical". </p>
<p>It does seem logical to me that if we add all the rectangles going from $V: 0$ to $\pi$ the sum will be the area of the circle. But we get $2\pi R^2$, double the area, why?</p>
<p><strong>Update:</strong>
WOW! I have not been given the reason for why my method failed, but I figured it out myself! So by drawing the rectangle corresponding to $v$ and $v+dv$ I saw and realized that they overlapped each other, which is bad if we want the exact result. But by symmetry I found that the formed overlap is a parallelogram with $base=R$ and $height=R$, thus every overlap is worth $Area= base*height = R^2$. Subtracting this in the final integral gives us A $= \int_0^\pi 2R^2 - R^2\,dv = \pi R^2$</p>
<p><strong>Update2</strong></p>
<p>I may be right for the wrong reasons here. The area of the overlap can't possible be $R^2$, that seems too large considering that the area of the rectangle itself is smaller than that when dv is infinitesmal. </p>
<p><a href="https://i.stack.imgur.com/IlZGw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IlZGw.png" alt="enter image description here"></a></p>
| Michael Hardy | 11,667 | <p>The length of the red line parallel to the diameter is $2R\cos v.$</p>
<p>Its distance from that diameter is $R\sin v.$ Therefore its infinitesimal change is $R\cos v \,dv.$</p>
<p>And $v$ needs to go from $-\pi/2$ to $\pi/2,$ for reasons that should be apparent from the picture.</p>
<p>So the area is $\displaystyle \int_{-\pi/2}^{\pi/2} 2R^2 \cos^2 v\, dv. $</p>
|
686,981 | <p>So I have to solve the equation $$y^2=4\tag{1.9.88 unit 3*}$$</p>
<p>I did this: $$y^2=4 \text{ means } \sqrt{y^2}=\sqrt{4}=>y=2$$</p>
<p>But I have a problem, $y$ can be either negative or positive so I need to do: $$\sqrt{y^2}=|y|=2=>y=2- or- y=-2$$</p>
<p>Is it right?</p>
| Hakim | 85,969 | <p>Never forget this rule: <em>For all $x\in\mathbb{R}$, the following holds:</em> $$\sqrt{x^2}=|x|=\begin{cases}x & x\gt 0\\ -x & x\leqslant0\end{cases}$$
Applying that to the equation $y^2=4:$ $$\begin{align}
\sqrt{y^2}=\sqrt{4}&\iff |y|=2\\ &\iff y=2\,\,\mathrm{or}\,\,y=-2
\end{align}$$
So you're right, keep up the good work!</p>
<p>I hope this helps.<br>
Best wishes, $\mathcal H$akim.</p>
|
995,159 | <p>I have a matrix $(a_{j,k})_{j,k\in\mathbb{N}}$ given by:</p>
<p>$ a_{j,k} = \dfrac{1 -e^{-jk}}{jk + 1}$</p>
<p>and I need to show that this induces a bounded operator on $\ell^2$. I'm pretty sure Schur's test is inconclusive. So my guess is to use the Hilber-Schmidt test, which states that if,</p>
<p>$\sum\limits_{j=1}^{\infty}\sum\limits_{k=1}^{\infty} \left|\dfrac{1 -e^{-jk}}{jk + 1}\right|^2 < \infty$, then $(a_{j,k})$ induces a bounded operator. </p>
<p>However, I'm not sure how to do this summation - can anyone give me a hint?</p>
<p>Many thanks </p>
| Winther | 147,873 | <p>We have</p>
<p>$$\left|\frac{1-e^{-jk}}{jk+1}\right| < \frac{1}{jk}$$</p>
<p>so</p>
<p>$$\sum_{j,k=1}^\infty \left|\frac{1-e^{-jk}}{jk+1}\right|^2 < \sum_{j,k=1}^\infty \frac{1}{j^2k^2} = \left(\sum_{j=1}^\infty \frac{1}{j^2}\right)^2 < \infty$$</p>
<p>since $\sum_{j=1}^\infty \frac{1}{j^2} = \frac{\pi^2}{6}$.</p>
|
3,101,286 | <p>I would like to get this text translated from Dutch to English:</p>
<p><a href="https://i.stack.imgur.com/0IWlQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0IWlQ.png" alt="enter image description here"></a></p>
<p>I tried using Google translator but the result is confusing me:</p>
<p><a href="https://i.stack.imgur.com/4SSrE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4SSrE.png" alt="enter image description here"></a></p>
| drhab | 75,923 | <p>(a) Let <span class="math-container">$(y_n)_n$</span> be a bounded sequence in <span class="math-container">$\mathbb C$</span>. Show that for every sequence <span class="math-container">$(x_n)_n$</span> in <span class="math-container">$\mathbb C$</span> for which the series <span class="math-container">$\sum_nx_n$</span> converges absolutely also the series <span class="math-container">$\sum_n(x_ny_n)$</span> will converge absolutely.</p>
<p>(b) Now assume that <span class="math-container">$(y_n)_n$</span> is a sequence in <span class="math-container">$\mathbb C$</span> such that for every sequence <span class="math-container">$(x_n)_n$</span> for which series <span class="math-container">$\sum_nx_n$</span> converges absolutely also the series <span class="math-container">$\sum_n(x_ny_n)$</span> converges absolutely. Does this imply that <span class="math-container">$(y_n)_n$</span> is bounded? Give an argumentation of your answer.</p>
<p>So actually in (b) you are asked whether the converse of (a) is true or is not true.</p>
|
3,583,475 | <p>Write as a single fraction:</p>
<p><span class="math-container">$(4x+2y)/(3x) - (5x+9y)/(6x) + 4$</span></p>
<p>Simplify your answer as much as possible.</p>
<p>The answer that I got from when I did the math was: (27x-5y)/(6x). But I have asked some of my friends who some got a different answer from mine. Please let me know if this seems right, and if it isn't please help me. Thanks in advance.</p>
| trula | 697,983 | <p>your answer is right, don't believe the others.</p>
|
2,777,982 | <p>I was asked to describe the surface described by</p>
<p>$${\bf r}^\top {\bf A} {\bf r} + {\bf b}^\top {\bf r} = 1,$$</p>
<p>where $3 \times 3$ positive definite matrix ${\bf A}$ and vector $\bf b$ are given.</p>
<p>My intuition tells me that it is a rotated ellipsoid with a centre that is off the origin. However, I am told to show this via the substitution ${\bf r} = {\bf x} + {\bf a}$, with $\bf a$ being a constant vector, and dictate the conditions on this vector to obtain a new quadric surface ${\bf x}^\top{\bf A}{\bf x} = C$. However, upon substitution, I get a ridiculously messy answer involving combinations of position and constant vectors. Is there a trick I am missing out on? Thank you!</p>
| mvw | 86,776 | <blockquote>
<p>However, I am told to show this via the substitution
${\bf r} = {\bf x} + {\bf a}$, with $\bf a$ being a constant vector, and dictate the
conditions on this vector to obtain a new quadric surface
${\bf x}^\top{\bf A}{\bf x} = C$.</p>
</blockquote>
<p>Following your advice:
\begin{align}
1
&= r^\top A r + b^\top r \\
&= (x + a)^\top A(x+a) + b^\top(x+a) \\
&= x^\top A x + \underbrace{x^\top A a + a^\top A x + b^\top x}_{f(x)} + a^\top A a + b^\top a
\end{align}
We need to choose $a$ such that the terms depending on $x$ vanish:
$$
0 = x^\top A a + a^\top A x + b^\top x
$$
The "positive definite" property holds for symmetric matrices, which fulfill
$$
A^\top = A
$$
so we have
$$
x^\top A a = x^\top A^\top a = (A x)^\top a = a^\top A x
$$
using symmetry and the rules for transposition, which gives
$$
0 = a^\top A x + a^\top A x + b^\top x = (2 a^\top A + b^\top) x
$$
so for all vectors $x$ to hold, we need
$$
0 = 2 a^\top A + b^\top \iff \\
0 = 0^\top = 2 A^\top a + b = 2 A a + b \iff \\
a = -(A^{-1}b)/2
$$
Justification of the last step: </p>
<p>For a positive definite matrix $A$ all eigenvalues $\lambda_i$ are positive, so its determinant (the product of its eigenvalues) is positive as well, and thus non-vanishing. Therefore it has an inverse $A^{-1}$.</p>
<p>So we end up with
$$
x^\top A x = C
$$
where
\begin{align}
C
&= 1 - a^\top A a - b^\top a \\
&= 1 - (A^{-1} b)^\top A (A^{-1} b) / 4 + b^\top (A^{-1} b)/2 \\
&= 1 - b^\top (A^{-1})^\top b / 4 + b^\top A^{-1} b /2 \\
&= 1 - b^\top (A^\top)^{-1} b / 4 + b^\top A^{-1} b /2 \\
&= 1 - b^\top A^{-1} b / 4 + b^\top A^{-1} b /2 \\
&= 1 + \underbrace{b^\top A^{-1} b}_{> 0} / 4
\end{align}
because matrix transposition and inverting operations commute and using symmetry again.</p>
<p>$A^{-1}$ has eigenvalues $1/\lambda_i$ which are all positive as well, so $A^{-1}$ is positive definite too. This assures that the constant number $C$ is positive and we deal with an <a href="https://en.wikipedia.org/wiki/Ellipsoid#As_quadric" rel="nofollow noreferrer">ellipsoid surface</a>.</p>
<p>Here is an example for
$$
A =
\begin{pmatrix}
-2 & -1 & 0 \\
-1 & 2 & -1 \\
0 & -1 & 2
\end{pmatrix}
\quad\quad
b =
\begin{pmatrix}
4 \\ -1 \\ -3
\end{pmatrix}
$$
<a href="https://i.stack.imgur.com/qC2w4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qC2w4.png" alt="Example"></a></p>
<p><strong>Update:</strong></p>
<p>Here is a <a href="https://ggbm.at/cMbcMAhN" rel="nofollow noreferrer">link</a> to an interactive version of the above scene.</p>
|
1,723,942 | <p>The four color theorem declares that any map in the plane (and, more generally, spheres and so on) can be colored with four colors so that no two adjacent regions have the same colors. </p>
<p>However, it's not clear what constitutes a map, or a region in a map. Is this actually a theorem in graph theory, something about every planar graph with some properties admitting a certain coloring? </p>
<p>Or does one really prove it using regions in the plane (which I guess we take to have non-fractal boundaries, or something). What is the precise definition? </p>
| Robert Israel | 8,508 | <p>The theorem is a statement about planar graphs. In the application to geography, you have a finite collection of "countries" (disjoint connected open subsets of the plane with, let's say, piecewise-smooth boundaries). Two countries are considered to be adjacent if the intersection of their boundaries has nonzero length. You then consider the graph with vertices corresponding to countries and edges between adjacent countries. </p>
|
1,944,628 | <p>Let $(X,{\mathcal T}_X)$ and $(Y,{\mathcal T}_Y)$ be topologiclal spaces, and let $f,g:(X,{\mathcal T}_X)\to(Y,{\mathcal T}_Y)$ be continuous maps. </p>
<p>Define the equality set as $$E(f,g) = \{x\in X \ | \ f(x) = g(x) \}$$</p>
<p>I have worked out that if $(Y,{\mathcal T}_Y)$ is Hausdorff, then $E(f,g)$ is ${\mathcal T}_X$-closed (see <a href="https://math.stackexchange.com/questions/199617/the-set-of-points-where-two-maps-agree-is-closed">this answer</a>).</p>
<p>In order to get a better understanding I am trying to find examples of continuous maps $f,g$ with $E(f,g)$ not closed. My understanding is that this only occurs for some maps where the target space $(Y,{\mathcal T}_Y)$ is not Hausdorff.</p>
| Hermès | 127,149 | <p>Define the <strong>Sierpinski space S</strong> to be the set $X = \{0,1\}$, together with the topology $\mathscr{T}_X = \{\emptyset, \{1\}, \{0,1\}\}$. There clearly $S$ is not Hausdorff, as we cannot isolate $0$ and $1$ using two disjoint open sets.
Let $\mathbb{R^*_+}$ be equipped with its natural topology. Define $f,g : \mathbb{R^*_+} \rightarrow S$ such that:</p>
<p>(1) $f(x) = 1, \forall x \in \mathbb{R^*_+}$</p>
<p>(2) $g(x) = 1$ if $x \in ]0,1[$ and $g(x) = 0$ if $x \geq 1$.</p>
<p>$f$ is constant therefore continuous, and g is continuous as well. If fact, it amounts to prove that the inverse image of any open set of $S$ is open in $\mathbb{R^*_+}$. Now, we have: $g^{-1}(\{1\}) = ]0,1[$ (which is open in $\mathbb{R^*_+})$, $g^{-1}(S) = \mathbb{R^*_+}$ (open) and $g^{-1}(\emptyset) = \emptyset$ (open as well). Therefore $g$ is indeed continuous, and we have:</p>
<p>$E(f,g) = ]0,1[$, which is not closed in $\mathbb{R^*_+}$, therefore providing us a counterexample.</p>
|
669,582 | <p>Let $X,τ$ be a topological space. Prove that a subset $A$ of $X$ is dense if and only if every open subset of $ X$ contain some point of $A$</p>
<p>this is what I got</p>
<p>Let $X,τ$ be a topological space</p>
<p>Part 1: Assume that a subset $A$ of $X$ is dense, show that every open subset of $X$ contain some point of $A$
Let $a∈A$, by axiom i) of the closure of set, $A⊂Cl A$, so $a∈Cl A$. Since $A$ is dense, $Cl A=X$, so $a∈X$. Let $ℵ_x $be the the collection of open neihborhood in $X$, by the axiom ii) of the open neighborhood system, $a∈X$ then $a∈N$ for each $N∈ℵ_x$. In other word, every open subset of $X$ contain some point of $A$.</p>
<p>Part 2: Assume that every open subset of $X$ contain some point of $A$, show that subset $A$ of $X$ is dense</p>
<p>I'm kinda stuck on how to show $Cl A=X$. I know that I need to show $Cl A⊂X$ and the other way around, but I'm not sure how.</p>
| aphorisme | 127,196 | <p>Since every (non-empty) open subset of $X$ contains a point from $A$ it has to hold that </p>
<p>(1.) $\forall U \in \mathcal{O}(X)\setminus\{\emptyset\}: U \cap A \neq \emptyset$. </p>
<p>Because of $A \subseteq \mathcal{Cl}A$ it holds also that </p>
<p>(2.) $\complement \mathcal{Cl}A \cap A = \emptyset$</p>
<p>So let's take a look at the given complement:
\begin{align*}
\complement\mathcal{Cl}{A} &= \complement\bigcap_{U\in\mathcal{O}(X),A\subseteq \complement U}\complement U
\\&= \bigcup_{U\in\mathcal{O}(X),A\subseteq \complement U} U =: U_0.
\end{align*}</p>
<p>By definition, $U_0$ has to be an open set. By (2.) it follows that $U_0\cap A = \emptyset$. Then with (1.), $U_0$ has to be the empty set. </p>
<p>Therefore $\complement \mathcal{Cl}A = \emptyset$ and so $\mathcal{Cl}A = X$. </p>
|
2,130,911 | <p>I'm unsure how to compute the following : 3^1000 (mod13)</p>
<p>I tried working through an example below,</p>
<p>ie) Compute $3^{100,000} \bmod 7$
$$
3^{100,000}=3^{(16,666⋅6+4)}=(3^6)^{16,666}*3^4=1^{16,666}*9^2=2^2=4 \pmod 7\\
$$</p>
<p>but I don't understand why they divide 100,000 by 6 to get 16,666. Where did 6 come from? </p>
| Mark | 310,244 | <p><a href="https://en.wikipedia.org/wiki/Fermat%27s_little_theorem" rel="nofollow noreferrer">Fermat's Little Theorem</a> says that:
$$a^p \equiv a\pmod{p}$$
Or, that:
$$a^{p-1}\equiv 1\pmod{p}$$
You're looking at this mod $7$, so $3^{7-1} = 3^6\equiv 1\pmod{p}$.
So, we're trying to split $100,000$ into $6k+r$ where $0\leq r < 5$, which is what writing $100,000 = 16,666\times 6+4$.</p>
|
1,447,852 | <p>Compute this sum:</p>
<p><span class="math-container">$$\sum_{k=0}^{n} k \binom{n}{k}.$$</span></p>
<p>I tried but I got stuck.</p>
| Brian Cheung | 248,555 | <p>$$\sum_{k=0}^nk\binom{n}{k}=\sum_{k=1}^nk\binom{n}{k}=\sum_{k=1}^nn\binom{n-1}{k-1}=n\sum_{k=0}^{n-1}\binom{n-1}{k}=n2^{n-1}$$<br>
Identities used:<br>
1)$$\binom{n}{k}=\dfrac{n!}{k!(n-k)!}=\dfrac nk\dfrac{(n-1)!}{(k-1)!(n-k)!}=\dfrac nk\binom{n-1}{k-1}$$<br>
2)$$\sum_{k=0}^n\binom nk=2^n$$
which can be proved by expanding $(1+1)^n$ with binomial theorem.</p>
|
1,082,390 | <p>$$\lim_{x \to \infty} \left(\sqrt{4x^2+5x} - \sqrt{4x^2+x}\ \right)$$</p>
<p>I have a lot of approaches, but it seems that I get stuck in all of those unfortunately. So for example I have tried to multiply both numerator and denominator by the conjugate $\left(\sqrt{4x^2+5x} + \sqrt{4x^2+x}\right)$, then I get $\displaystyle \frac{4x}{\sqrt{4x^2+5x} + \sqrt{4x^2+x}}$, but I can conclude nothing out of it. </p>
| Bumblebee | 156,886 | <p>As $x\to\infty$ $$\displaystyle \frac{4x}{\sqrt{4x^2+5x} + \sqrt{4x^2+x}}=\displaystyle \frac{4}{\sqrt{4+\dfrac{5}{x}} + \sqrt{4+\dfrac{1}{x}}}\to1$$</p>
|
2,426,897 | <p>Let $\mathbb{H}$ be the ring of real quaternions and $Z(\mathbb{H})$ be its center. Of course $Z(\mathbb{H})=\mathbb{R}$. </p>
<p>Suppose $a+bi+cj+dk$, $x+yi+zj+wk \in \mathbb{H}$ such that $(a+bi+cj+dk)(x+yi+zj+wk) \in Z(\mathbb{H})$. </p>
<p>Does it imply $(x+yi+zj+wk)(a+bi+cj+dk) \in Z(\mathbb{H})$?</p>
| Dr. Sonnhard Graubner | 175,066 | <p>it is $$44<\sqrt{2017}<45$$ since $$44^2=1936$$ and $$45^2=2025$$</p>
|
2,426,897 | <p>Let $\mathbb{H}$ be the ring of real quaternions and $Z(\mathbb{H})$ be its center. Of course $Z(\mathbb{H})=\mathbb{R}$. </p>
<p>Suppose $a+bi+cj+dk$, $x+yi+zj+wk \in \mathbb{H}$ such that $(a+bi+cj+dk)(x+yi+zj+wk) \in Z(\mathbb{H})$. </p>
<p>Does it imply $(x+yi+zj+wk)(a+bi+cj+dk) \in Z(\mathbb{H})$?</p>
| Will Jagy | 10,400 | <p>As to the next question, Pell -1 and Pell +1..Hmm, seems to be showing the exponents incorrectly.</p>
<p>$$ {106515299132603184503844444}^2 - 2017 \cdot 2371696115380807559791481^2 = -1 $$</p>
<p>$$ 22691017898615873418283839489716246568157231499338273^2 - 2017 \cdot 505243842362839347335084683756885179819279000763128^2 = 1 $$</p>
<p>So, $\sqrt {2017}$ is above
$$ \frac{106515299132603184503844444}{2371696115380807559791481} $$
and below
$$ \frac{22691017898615873418283839489716246568157231499338273}{505243842362839347335084683756885179819279000763128} $$</p>
<pre><code>parisize = 4000000, primelimit = 500509
? a = 106515299132603184503844444
%1 = 106515299132603184503844444
? b = 2371696115380807559791481
%2 = 2371696115380807559791481
? a^2 - 2017 * b^2
%3 = -1
?
? c = 22691017898615873418283839489716246568157231499338273
%4 = 22691017898615873418283839489716246568157231499338273
? d = 505243842362839347335084683756885179819279000763128
%5 = 505243842362839347335084683756885179819279000763128
? c^2 - 2017 * d^2
%6 = 1
?
?
? a^2 + 2017 * b^2 - c
%7 = 0
? 2 * a * b - d
%8 = 0
?
?
</code></pre>
|
2,426,897 | <p>Let $\mathbb{H}$ be the ring of real quaternions and $Z(\mathbb{H})$ be its center. Of course $Z(\mathbb{H})=\mathbb{R}$. </p>
<p>Suppose $a+bi+cj+dk$, $x+yi+zj+wk \in \mathbb{H}$ such that $(a+bi+cj+dk)(x+yi+zj+wk) \in Z(\mathbb{H})$. </p>
<p>Does it imply $(x+yi+zj+wk)(a+bi+cj+dk) \in Z(\mathbb{H})$?</p>
| kingW3 | 130,953 | <p>Knowing the answer lies between $40$ and $50$ the rational thing to do would be to try the number in the middle of $40$ and $50$ i.e $45$. </p>
<p>If you don't feel comfortable with multiplying $45\cdot 45$ you can use the
formula for $(40+5)^2=40^2+2\cdot 40\cdot 5+5^2=1600+400+25=2025$</p>
<p>Now you can use that $45^2-44^2=(45-44)(45+44)=89$ so one can see that $45^2>45^2-7=2017>45^2-89$.</p>
<p>It helps speed up a little of calculation for contests and such.</p>
|
2,426,897 | <p>Let $\mathbb{H}$ be the ring of real quaternions and $Z(\mathbb{H})$ be its center. Of course $Z(\mathbb{H})=\mathbb{R}$. </p>
<p>Suppose $a+bi+cj+dk$, $x+yi+zj+wk \in \mathbb{H}$ such that $(a+bi+cj+dk)(x+yi+zj+wk) \in Z(\mathbb{H})$. </p>
<p>Does it imply $(x+yi+zj+wk)(a+bi+cj+dk) \in Z(\mathbb{H})$?</p>
| mathreadler | 213,607 | <p>You can use the Newton square root method <a href="https://en.wikipedia.org/wiki/Methods_of_computing_square_roots" rel="nofollow noreferrer">wikipedia</a> for integers:</p>
<p>$$x_{n+1} = \frac 1 2 \left(x_n+ \frac S {x_n}\right)$$</p>
<p>Let us start with a crappy guess </p>
<ol>
<li>$x_1 = 500$:</li>
<li>$x_2 = \frac{1}{2} (500 + 2017/500) = 252$</li>
<li>$x_3 = \frac 1 2 (252 + 2017/252) = 130$</li>
<li>$x_4 = \frac 1 2 (130 + 2017/130) = 72.7$</li>
<li>$x_5 = \frac 1 2 (73 + 2017/73) = 50$</li>
<li>$x_6 = \frac 1 2 (50 + 2017/50) = 45$</li>
</ol>
<p>Now as the iterations seem to converge we can try $45^2 = 2025$ and we have our answer.</p>
|
2,426,897 | <p>Let $\mathbb{H}$ be the ring of real quaternions and $Z(\mathbb{H})$ be its center. Of course $Z(\mathbb{H})=\mathbb{R}$. </p>
<p>Suppose $a+bi+cj+dk$, $x+yi+zj+wk \in \mathbb{H}$ such that $(a+bi+cj+dk)(x+yi+zj+wk) \in Z(\mathbb{H})$. </p>
<p>Does it imply $(x+yi+zj+wk)(a+bi+cj+dk) \in Z(\mathbb{H})$?</p>
| Simon Fraser | 717,270 | <p><span class="math-container">$\sqrt{1600} = 40$</span> and <span class="math-container">$\sqrt{2500} = 50$</span>. <span class="math-container">$(40+4)^2 = 40^2 + 8\cdot 40 + 16 = 1936$</span> and <span class="math-container">$(40+5)^2 = 40^2 + 10\cdot 40 +25 = 2025$</span>. Hence the desired answer is <span class="math-container">$44$</span> and <span class="math-container">$45$</span>.</p>
|
317,160 | <p>If $ f(x) = \frac{1}{(1+x)\sqrt x} $ how to find all $ p > 0 $ such that
$$ \int^{\infty}_0 |f(x)|^p dx < \infty $$
The integral is with respect to lebesgue measure. Any solution or hints would be helpful. The answer is the integral converges iff $ p\in (\frac{2}{3}, 2) $.</p>
| Mhenni Benghorbal | 35,472 | <p>Using the change of variables $ z=\frac{1}{1+x} $ and the <a href="http://en.wikipedia.org/wiki/Beta_function" rel="nofollow">beta function</a></p>
<p>$$ \beta(x,y) = \int_{0}^{1}t^{x-1}(1-t)^{y-1}dt,\quad Re(x),\,Re(y)>0, $$</p>
<p>we have</p>
<p>$$ \int_{0}^{\infty} \frac {1}{(1+x)^p x^{\frac{p}{2}}} dx = \int _{0}^{1}\! \left( 1-z \right) ^{-\frac{p}{2}}{z}^{\frac{3p}{2}-2}{dz}.$$</p>
<p>Comparing with the existence conditions for the beta function implies that </p>
<p>$$ 1-\frac{p}{2}>0,\quad\frac{3p}{2}-1 > 0. $$</p>
<p>Solving the two inequalities gives the desired result.</p>
|
3,393,466 | <p>I am in final year of my undergraduate in mathematics from a prestigious institute for mathematics. However a thing that I have noticed is that I seem to be slower than my classmates in reading mathematics. As in, how muchever I try, I seem to finish my works at the last moment and I rarely find any time for extra reading. Is there any suggestions or tips that I could try that you know of? Or is it advisable to skip details in favour of saving time?</p>
| Toney Leung | 645,097 | <p>I'm similar to you.While reading mathematics,I always think it over deeply until I thoroughly understand the full meaning.Knowing the connections between different math concepts clearly will be good for your further study in mathematics.Because undergraduate mathematics is just the beginning of real mathematics，and strength your basic knowledge will help you understand the graduate courses easily.If you wanna speed up your reading,I think you can skip some details when you don't understand them ,leave them behind for now and read it again when you are inspiring.</p>
|
1,905,863 | <p>I'm on the section of my book about separable equations, and it asks me to solve this:</p>
<p>$$\frac{dy}{dx} = \frac{ay+b}{cy+d}$$</p>
<p>So I must separate it into something like: $f(y)\frac{dy}{dx} + g(x) = constant$</p>
<p>*note that there are no $g(x)$</p>
<p>but I don't think it's possible. Is there something I'm missing?</p>
| Claude Leibovici | 82,404 | <p>For your specific example, extracting the horizontal asymptote you can also write $$\frac{cy+d}{ay+b}=\frac c a+\frac{a d-b c}{a (a y+b)}$$ Then $$I=\int\frac{cy+d}{ay+b}\,dy=\frac c a \int dy+\frac{a d-b c} a\int \frac {dy} { (a y+b)}=\frac c a \int dy+\frac{a d-b c} {a^2}\int \frac {a\,dy} { (a y+b)}$$ I am sure that you can take it from here.</p>
|
522,289 | <p>It is an exercise on the lecture that i am unable to prove.</p>
<p>Given that $gcd(a,b)=1$, prove that $gcd(a+b,a^2-ab+b^2)=1$ or $3$, also when will it equal $1$?</p>
| lab bhattacharjee | 33,337 | <p>HINT:</p>
<p>Let prime $p$ divides $a+b, a^2-ab+b^2$</p>
<p>$\implies p$ divides $\{(a+b)^2-(a^2-ab+b^2)\}=3ab$</p>
<p>If $p|a,$ as $p|(a+b),p$ must divide $(a+b)-a=b\implies p|(a,b)$ </p>
<p>But as $(a,b)=1,p$ can not divide $a$</p>
<p>Similarly, $p$ can not divide $b$</p>
<p>$\implies p|3$</p>
<p>$\implies (a+b,a^2-ab+b^2)|3$</p>
<p>and $(a+b,a^2-ab+b^2)=3$ if $3|(a+b)\iff 3|(a^2-ab+b^2)$</p>
|
580,974 | <blockquote>
<p>Show that the following equalities hold true for every $n$ from $\mathbb{N}$ and
every $x$ from $\mathbb{R}$</p>
<p>$$\sin^{(n)}(x)=\sin(x+n\fracπ2)$$</p>
</blockquote>
<p>How do I solve this?</p>
| Tim Ratigan | 79,602 | <p>$$\begin{align}
\sin\left(x+\frac{(2n+1)\pi}{2}\right)&=\sin x\cos\left(\frac{(2n+1)\pi}{2}\right)+\cos x\sin\left(\frac{(2n+1)\pi}{2}\right) \\
&=\sin x\cos(n\pi+\pi/2)+\cos x\sin(n\pi+\pi/2) \\
&=-\sin x\sin(n\pi)+\cos x \cos(n\pi) \\
\end{align}$$</p>
<p>You can clearly see that this is not $\sin^n(x)$.</p>
|
580,974 | <blockquote>
<p>Show that the following equalities hold true for every $n$ from $\mathbb{N}$ and
every $x$ from $\mathbb{R}$</p>
<p>$$\sin^{(n)}(x)=\sin(x+n\fracπ2)$$</p>
</blockquote>
<p>How do I solve this?</p>
| hhsaffar | 104,929 | <p><strong>Hint:</strong></p>
<p>Show $sin^{(1)}(x)=sin(x+\frac\pi2)$</p>
<p>Then suppose for $n$ we have $sin^{(n)}(x)=sin(x+n\fracπ2)$, show that $sin^{(n+1)}(x)=sin(x+(n+1)\fracπ2)$. Do this by differentiating $sin^{(n)}(x)$ once. </p>
|
201,381 | <p>I have basic training in Fourier and Harmonic analysis. And wanting to enter and work in area of number theory(and which is of some interest for current researcher) which is close to analysis. </p>
<blockquote>
<p>Can you suggest some fundamental papers(or books); so after reading these I can have, hopefully(probably), I will have some thing to work on(I mean, chance of discovering something new)?</p>
</blockquote>
| arsmath | 3,711 | <p>An older book that I really liked was Ayoub's <i>An Introduction to the Analytic Theory of Numbers</i>. It describes several famous classical theorems proved by analytical methods. It requires very little background beyond analysis. You can read the introduction <a href="http://www.ams.org/books/surv/010/surv010-endmatter.pdf">here</a>.</p>
|
201,381 | <p>I have basic training in Fourier and Harmonic analysis. And wanting to enter and work in area of number theory(and which is of some interest for current researcher) which is close to analysis. </p>
<blockquote>
<p>Can you suggest some fundamental papers(or books); so after reading these I can have, hopefully(probably), I will have some thing to work on(I mean, chance of discovering something new)?</p>
</blockquote>
| Bobby Grizzard | 37,644 | <p>An area where analysis (especially the Fourier kind) is used quite heavily is in the study of so-called "Beurling-Selberg maximal functions," which have applications to several areas in number theory, such as counting lattice points and studying the behavior of zeta- and $L$-functions. The basic idea is that there are some naturally occurring functions, such as characteristic functions of intervals, balls, etc., which obviously don't have nice analytic properties, and it can be very fruitful to find functions which majorize and minorize your "bad" function that have "nice" properties with respect to their fourier transforms.</p>
<p>The paper entitled</p>
<p><em>A survey on Beurling-Selberg majorants and some consequences of the Riemann hypothesis</em></p>
<p>by Emanuel Carneiro seems like it would be a good starting point, but I can't seem to find a full text online. Carneiro and others have written quite a few papers on this topic, such as this one:</p>
<p><a href="https://www.ndsu.edu/pubweb/~littmann/research/Gaussian_2_4.pdf" rel="nofollow">https://www.ndsu.edu/pubweb/~littmann/research/Gaussian_2_4.pdf</a></p>
<p>by Carneiro, Littmann, and Vaaler, which I would describe as mostly analysis, with some discussion of applications to number theory. You might find this stuff interesting.</p>
|
2,062,398 | <p>May you tell me if my translation to symbolic logic is correct? </p>
<p>Thank you so much! Here is the problem:</p>
<p>To check that a given integer $n > 1$ is a prime, prove that it is enough to show that $n$ is not divisible by any prime $p$ with $p \le \sqrt{n}$.</p>
<p>$$\forall p \in P ~\forall n \in N ~(p \nmid n \land p\le \sqrt{n} \land n>1 \rightarrow n \in P )$$</p>
| Community | -1 | <p><strong>Hint.</strong> Try to prove by contradiction. i.e., assume that $\sqrt{2} + \sqrt[3]{5}$ is of the form $p/q$ and do some algebraic manipulations to obtain a contradiction.</p>
|
4,294,577 | <p>If I have a function with all positive integer for the coefficients, is there a way to have a lower bound? Zero isn't an option, because I've done the rational root theorem and found all possible roots. If you need, I can provide the function and its list of possible roots below:</p>
<p><span class="math-container">$70x^{4}+163x^{3}+109x^{2}+37x+6$</span></p>
<p><span class="math-container">$±1, ±1/2, ±1/5, ±1/7, ±1/10, ±1/14, ±1/35, ±1/70,$</span>
<span class="math-container">$±2, ±2/5, ±2/7, ±2/35,$</span>
<span class="math-container">$±3, ±3/2, ±3/5, ±3/7, ±3/10, ±3/14, ±3/35, ±3/70,$</span>
<span class="math-container">$±6, ±6/5, ±6/7, ±6/35$</span></p>
| Torsten Schoeneberg | 96,384 | <p>This is part of proposition 4 in Bourbaki's books on <em>Lie Groups and Algebras</em>, ch. VIII §5 no. 2. Since Bourbaki's setting is slightly more general (they are looking at a semisimple <span class="math-container">$\mathfrak g$</span> with a splitting Cartan subalgebra <span class="math-container">$\mathfrak h$</span> over a general characteristic <span class="math-container">$0$</span> field <span class="math-container">$k$</span>, where the definition of <span class="math-container">$Aut^0$</span> is those automorphisms which become elementary (i.e. products of automorphisms of the form <span class="math-container">$e^{ad(x)}$</span>) after scalar extension to an algebraic closure), and their overall statement is also more general, it is a bit hard to track down what exactly we would need for the proof of the inclusion <span class="math-container">$\psi(N) \subseteq W$</span> you are after. All the more so since, as usual, Bourbaki's proof can lead us down multiple rabbitholes of references and lemmata before (in particular lemma 2 before the proposition seems to be crucial here). But I think the approach is:</p>
<p>As you say in a comment, via twisting <span class="math-container">$\phi \in N$</span> with an element which lands in the Weyl group, it suffices to show that if <span class="math-container">$\psi(\phi)$</span> stabilizes (as a set) a base <span class="math-container">$B$</span> of our root system <span class="math-container">$R$</span>, then it must be the identity on <span class="math-container">$R$</span> (i.e. fix it pointwise).</p>
<p>So assume we have such a <span class="math-container">$\phi$</span>. We can now twist it further, without changing its image under <span class="math-container">$\psi$</span>, via certain elementary automorphisms (which act as identity on <span class="math-container">$\mathfrak h$</span> but scale the elements of certain root spaces) to ensure that on the entire space <span class="math-container">$\sum_{\alpha \in R} \mathfrak g_\alpha$</span> (the sum of all root spaces, a vector space complement of <span class="math-container">$\mathfrak h$</span> in <span class="math-container">$\mathfrak g$</span>), <span class="math-container">$\phi$</span> does not have the eigenvalue <span class="math-container">$1$</span>. (The exact construction of such a twisting takes up most of Bourbaki's proof of the proposition and looks very technical at first, but the idea is kind of straightforward.)</p>
<p>Now the above mentioned lemma 2 says that this last statement implies that <span class="math-container">$\phi$</span> is the identity on <span class="math-container">$\mathfrak h$</span>. To see this, they use that on the one hand, the nilspace of <span class="math-container">$\phi -id$</span> in <span class="math-container">$\mathfrak g$</span> has dimension <span class="math-container">$\ge \dim\mathfrak h$</span> (that comes from a general fact about CSA's as minimal Engel subalgebras proven in an earlier volume), so by that assumption of no eigenvalue <span class="math-container">$1$</span> on that complement, this nilspace is <span class="math-container">$\mathfrak h$</span>, i.e. <span class="math-container">$\phi-id$</span> is nilpotent on <span class="math-container">$\mathfrak h$</span>; on the other hand, because the restriction of <span class="math-container">$\phi$</span> to <span class="math-container">$\mathfrak h$</span> has finite order (it stabilizes the finite set of coroots, which span <span class="math-container">$\mathfrak h$</span>), by Maschke's Theorem it is semisimple (in our case <span class="math-container">$k=\mathbb C$</span>, diagonalizable, and there might be an easier way to see this); putting this together, when restricted to <span class="math-container">$\mathfrak h$</span>, the endomorphism <span class="math-container">$\phi - id$</span> is semisimple and nilpotent, hence identically zero.</p>
|
2,596,098 | <p>For a square matrix $A$ and identity matrix $I$, how does one prove that $$\frac{d}{dt}\det(tI-A)=\sum_{i=1}^n\det(tI-A_i)$$ Where $A_i$ is the matrix $A$ with the $i^{th}$ row and $i^{th}$ column vectors removed?</p>
| A.Γ. | 253,273 | <p>One way to prove this claim is to take the matrix $tI-A$ and replace the first $t$ on the main diagonal with $t_1$, the second one with $t_2$ etc. Let the resulting determinant be $p(t_1,t_2,\ldots,t_n)$. Then
$$
\det(tI-A)=p(t,t,\ldots,t),
$$
and by the chain rule
$$
\frac{d}{dt}\det(tI-A)=\frac{\partial}{\partial t_1}p(t,t,\ldots,t)+\ldots+\frac{\partial}{\partial t_n}p(t,t,\ldots,t).
$$
The partial derivative w.r.t $t_i$ can be simply calculated as $\det(tI-A_i)$ e.g. using the determinant expansion along the $i$-th column.</p>
|
2,431,548 | <p>Okay, so, my teacher gave us this worksheet of "harder/unusual probability questions", and Q.5 is real tough. I'm studying at GCSE level, so it'd be appreciated if all you stellar mathematicians explained it in a way that a 15 year old would understand. Thanks!</p>
<p>So, John has an empty box.
He puts some red counters and some blue counters into the box. </p>
<p>The ratio of the number of red counters to blue counters is 1:4</p>
<p>Linda takes out, at random, 2 counters from the box.</p>
<p>The probability that she takes out 2 red counters is 6/155</p>
<p>How many red counters did John put into the box?</p>
| MEK MEK g | 622,454 | <p>John has a empty box. He puts some red counters and some blue counters. The ratio of the number of red counters to blue counters is 1:4. Linda takes at random 2 counters. The probability that she took 2 red counters is 6/155. How many red counters did John put in the box?</p>
<p>Let there be <span class="math-container">$x$</span> red counters, then there will be <span class="math-container">$4x$</span> blue counters which equals <span class="math-container">$5x$</span> counters altogether in the box.</p>
<p>so the probability that the first counter will be red is </p>
<p>Now there are <span class="math-container">$x-1$</span> red counters left out of <span class="math-container">$5x-1$</span> counters in total so </p>
<p>the probablility that the second counter is red is </p>
<p>So there were 25 red marbles in the box :)</p>
|
853,774 | <blockquote>
<p>If $(G,*)$ is a group and $(a * b)^2 = a^2 * b^2$ then $(G, *)$ is abelian for all $a,b \in G$.</p>
</blockquote>
<p>I know that I have to show $G$ is commutative, ie $a * b = b * a$</p>
<p>I have done this by first using $a^{-1}$ on the left, then $b^{-1}$ on the right, and I end up with and expression $ab = b * a$. Am I mixing up the multiplication and $*$ somehow?</p>
<p>Thanks</p>
| Ivo Terek | 118,056 | <p>In one side, $(a \ast b)^2 = a^2 \ast b^2$ (hypothesis). On the other side, $(a \ast b)^2 = a \ast b \ast a \ast b $. So: $$a^2 \ast b^2 = a \ast b \ast a \ast b \\ a \ast b = b \ast a $$
In groups there is only <strong>one</strong> operation, which we often <em>think</em> as multiplication. It could be well addition, or composition of functions, etc.</p>
|
3,407,368 | <p>Please help me to think through this.</p>
<p>Take Riemann, for example. Finding a non-trivial zero with a real part not equal to <span class="math-container">$\frac{1}{2}$</span> (i.e., a counterexample) would disprove the conjecture, and also so it to be decidable.</p>
<p>How about demonstrating that Riemann is undecidable? Would that not imply that we can check zeros ad infinitum without resolving the hypothesis? But, checking zeros can only provide a counterexample, i.e., a disproof. </p>
<p>How (if at all) do these statements differ?</p>
<p>Any non-trivial zeros that we can find through brute force checking will have a real part of <span class="math-container">$\frac{1}{2}$</span>.</p>
<p>All non-trival zeros have a real part of <span class="math-container">$\frac{1}{2}$</span>.</p>
<p>Is my assumption that all non-trivial zeros is in the infinite set of zeros that can be checked by brute force correct, or even relevant? Or meaningful?</p>
<p>Please be kind. I'm not sure if my question even makes sense.</p>
| Shiranai | 580,826 | <p>It is important to differentiate two aspects of mathematics, the deductive system (which is about what can be proved or not) and the model (about what is true or what is false). They are related, as everything that can be proved is true, but the converse does not hold: being true does not imply it can be proved.</p>
<p>Let's take RH for example. Assume it is true, does that mean we can prove it? Maybe yes, maybe not. There is no easy way to tell beforehand. Assume RH is false, does that mean we can disproove it? <strong>Yes</strong>, as it being false then there is a non trivial zero, and therefore we can prove that it is infact a zero.</p>
<p>So yes, RH could be undecidable, but only in the case that it is true. This generalizes to every existential question (with a sufficiently powerful deductive system).</p>
|
440,583 | <p><strong>Question.</strong> Is there an entire function <span class="math-container">$F$</span> satisfying first two or all three of the following assertions:</p>
<ul>
<li><span class="math-container">$F(z)\neq 0$</span> for all <span class="math-container">$z\in \mathbb{C}$</span>;</li>
<li><span class="math-container">$1/F - 1\in H^2(\mathbb{C}_+)$</span> -- the classical Hardy space in the upper half-plane;</li>
<li><span class="math-container">$F$</span> is bounded in every horizontal half-plane <span class="math-container">$\{z\colon \text{Im}(z) > \delta\}$</span>?</li>
</ul>
<p><strong>Thoughts</strong>. Let <span class="math-container">$G= 1/F$</span>. Then we have <span class="math-container">$G(z) = 1 + \int_0^{\infty} h(x)e^{izx}\, dx$</span> for some <span class="math-container">$h\in L^2[0,\infty)$</span> and all <span class="math-container">$z\in \mathbb{C}_+$</span>. For nice functions <span class="math-container">$h$</span> (e.g., for super-exponentially decreasing) this integral representation can be extended to the whole complex plane and probably the example can be constructed in terms of <span class="math-container">$h$</span>. However, I don't know if it is possible to find <span class="math-container">$h$</span> such that <span class="math-container">$G$</span> is non-zero for every <span class="math-container">$z$</span>.</p>
| Pavel Gubkin | 498,423 | <p>Here is a slightly different construction, from <a href="https://math.stackexchange.com/a/4642796/1048496">this answer</a> by @reuns , the function built this way has a <span class="math-container">$\exp(\exp(|z|))$</span> growth compared to triple exponent in the <a href="https://mathoverflow.net/a/440610/498423">wonderful example by @AlexandreEremenko</a>.</p>
<p>For <span class="math-container">$z\in\mathbb{C}$</span>, consider
<span class="math-container">$$
g(z) = \int_{i\infty}^z\frac{e^{is}(e^{is} - 1)}{s}ds = - \frac{e^{iz}(e^{iz} - 2)}{2iz} + \int_{i\infty}^z\frac{e^{is}(e^{is} - 2)}{2is^2}ds,
$$</span>
where the equality is from integration by parts. For <span class="math-container">$z\neq 0$</span>, let the contour in the last integral consist of two components: the vertical half-line and the arc of the circle with center in <span class="math-container">$0$</span> (as in the picture).<a href="https://i.stack.imgur.com/mXhSL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mXhSL.png" alt="contour" /></a></p>
<p>After integration by parts the integral over the half-line is exponentially small and the integral over the arc is <span class="math-container">$O(1/|z|)$</span> in every upper half-plane.</p>
<p>It follows that <span class="math-container">$|g(z)| = O(1/|z|)$</span> as <span class="math-container">$z\to\infty$</span> uniformly in every upper half-plane and <span class="math-container">$f(z) = e^{g(z)}$</span> will be the answer to the initial question.</p>
|
1,648,587 | <blockquote>
<p><strong>Problem.</strong> Consider two arcs <span class="math-container">$\alpha$</span> and <span class="math-container">$\beta$</span> embedded in <span class="math-container">$D^2\times I$</span> as shown in the figure. The loop <span class="math-container">$\gamma$</span> is obviously nullhomotopic in <span class="math-container">$D^2\times I$</span>, but show that there is no nullhomotopy of <span class="math-container">$\gamma$</span> in the complement of <span class="math-container">$\alpha\cup \beta$</span>.<br />
<a href="https://i.stack.imgur.com/0GU17.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0GU17.png" alt="enter image description here" /></a></p>
</blockquote>
<p>I tried to use van Kampen's theorem to find the fundamental group of <span class="math-container">$X=D^2\times I-\alpha\cup \beta$</span>. Let <span class="math-container">$A=D^2\times I-\alpha$</span> and <span class="math-container">$B=D^2\times I-\beta$</span>. To find the fundamental group of <span class="math-container">$A$</span>, we note that after a homeomorphism, <span class="math-container">$A$</span> looks like a cylinder with it's axis removed. The fundamental group of <span class="math-container">$A$</span> is thus <span class="math-container">$\mathbf Z$</span>. Similarly for <span class="math-container">$B$</span>. Now we need to find the normal subgroup in <span class="math-container">$\pi_1(A)\sqcup \pi_1(B)$</span> generated by words of the form <span class="math-container">$i_{AB}(\omega)i_{BA}(\omega)^{-1}$</span> and quotient by it. Here <span class="math-container">$i_{AB}:A\cap B\to A$</span> and <span class="math-container">$i_{BA}:A\cap B\to B$</span> are the inclusion maps and <span class="math-container">$\omega$</span> is a loop in <span class="math-container">$A\cap B$</span>. Intuitively, the only loops in <span class="math-container">$A\cap B$</span> whose image in <span class="math-container">$A$</span> in nontrivial in <span class="math-container">$A$</span> are the ones which link with <span class="math-container">$\alpha$</span>. I am not sure how to say this precisely and I will be grateful if someone can help me with this. Similarly for <span class="math-container">$B$</span>. I am not able to make progress from here.</p>
| Ian Mateus | 17,751 | <p>If you picture the solid cylinder as a solid ball, you will be able to deform $X$ into a solid ball without two parallel chords, which deformation retracts to the figure eight.</p>
|
2,472,746 | <p>I have that $f_n$ is measurable on a finite measure space.</p>
<p>Define $F_k=\{\omega:|f_n|>k \}$</p>
<p>$F_k$ are measurable and have the property $F_1 \supseteq F_2\supseteq\cdots$</p>
<p>Can I then claim that $m\left(\bigcap_{n=1}^\infty F_n\right) = 0$?</p>
| Tsemo Aristide | 280,301 | <p>If $f$ is real valued, $\cap F_k$ is empty. so its measure is zero.</p>
|
1,437,287 | <p>On <a href="https://en.wikipedia.org/wiki/Geometric_series#Geometric_power_series" rel="nofollow">Wikipedia</a> it is stated that by differentiating the following formula holds:</p>
<p>$$ \sum_n n q^n = {1\over (1-q)^2}$$</p>
<p>Does this not require a proof? It seems to me because the series is infinite it is not clear that differentiation commutes with taking the limit. </p>
<blockquote>
<p>How to prove this?</p>
</blockquote>
| ncmathsadist | 4,154 | <p>Power series are termwise differentiable in their open interval of convergence. This is a basic property of power series.</p>
|
3,344,728 | <p>Let <span class="math-container">$S$</span> be the set of all real numbers except <span class="math-container">$-1$</span>. Define <span class="math-container">$*$</span> on <span class="math-container">$S$</span> by
<span class="math-container">$$a*b=a+b+ab.$$</span></p>
<p>Goal: Show that <span class="math-container">$*$</span> gives a binary operation on S.</p>
<p>In order to prove that <span class="math-container">$*$</span> is a binary operation, I need to prove that <span class="math-container">$S$</span> is closed under <span class="math-container">$*$</span>, so I tried to prove that <span class="math-container">$a+b+ab$</span> never equals <span class="math-container">$-1$</span>. I cannot figure out, however, how to do this. </p>
| Community | -1 | <p>Hint: Show that <span class="math-container">$a*b=(a+1)(b+1)-1$</span></p>
|
4,467,841 | <p>For a complex number <span class="math-container">$z=a+bi$</span> and a positive real value <span class="math-container">$R$</span>, we have <span class="math-container">$e^{Rbi}=\cos(Rb)+i\sin(Rb)$</span>. I am struggling to understand this since no matter how large <span class="math-container">$b$</span> or <span class="math-container">$R$</span> is, we have <span class="math-container">$|e^{Rbi}| \in [-1, 1]$</span>. What is the best way to understand this intuitively? For instance, in an applied sense, is it true that
<span class="math-container">$$\bigg|\sum_{z: \ a, b \geq 0}e^{Rbi}f(z)\bigg|\leq \bigg|\sum_{z: \ a, b \geq 0}f(z)\bigg|,$$</span>
where <span class="math-container">$f$</span> is some generic function and <span class="math-container">$\sum_{z: \ a, b \geq 0}f(z)>0$</span>? This makes sense to me because each <span class="math-container">$|e^{Rbi}|$</span> is no larger than <span class="math-container">$1$</span>.</p>
| Nuke_Gunray | 1,060,681 | <p>Starting with the basics, <span class="math-container">$t\mapsto e^{it} = \cos t +i\sin t$</span> just describes the complex unit circle, which for <span class="math-container">$t\in\mathbb{R}$</span> is run through counter-clockwise with constant speed <span class="math-container">$1$</span>. This just means that one cycle around the circle takes <span class="math-container">$2\pi$</span> time units, which is the length of the unit circle.</p>
<p>So what happens if we plug in <span class="math-container">$2$</span>, i.e. <span class="math-container">$e^{i2t}$</span>? Now the speed is <span class="math-container">$2$</span>, which means one cycle takes <span class="math-container">$\pi$</span> time units.
So by increasing the factor <span class="math-container">$R\in\mathbb{R}$</span>, the geometric shape of the curve traced out by <span class="math-container">$t\mapsto e^{iRt}$</span> is always the unit circle. The only thing that changes is the speed with which this circle is orbited.
This means that <span class="math-container">$|e^{Rit}| = $</span> for all <span class="math-container">$t,R\in\mathbb{R}$</span>, since we never are able to leave the circle.</p>
|
4,304 | <p>I am trying to understand <a href="http://en.wikipedia.org/wiki/All-pairs_testing" rel="nofollow noreferrer"><strong>pairwise testing</strong></a>.</p>
<p>How many combinations of tests would be there for example, if</p>
<blockquote>
<p><code>a</code> can take values from 1 to m</p>
<p><code>b</code> can take values from 1 to n</p>
<p><code>c</code> can take values from 1 to p</p>
</blockquote>
<p><code>a</code>, <code>b</code> and <code>c</code> can take m, n and p distinct values respectively. <strong>What are the total number of pairwise combinations possible?</strong></p>
<hr />
<p>With a pairwise testing tool that I am testing, I am getting 40 results for m = n = p = 6. I am trying to mathematically understand how I get 40 values.</p>
| castal | 1,123 | <p>After reading <a href="http://www.satisfice.com/tools.shtml" rel="nofollow">this page</a>, it seems that pairwise testing requires a set of test cases in which every pair of values from any two of the n categories occurs at least once among the test case n-tuples. In the present case, the problem is to find a minimal subset of the 6x6x6 = 216 total triples (a,b,c) such that</p>
<ul>
<li><p>each pair of values for a and b<br>
occurs at least once, i.e. (a,b,*),</p></li>
<li><p>each pair of a and c values occurs<br>
at least once, i.e. (a,*,c)</p></li>
<li><p>each pair of b and c values occurs
at least once, i.e. (*,b,c)</p></li>
</ul>
<p>Any subset satisfying these requirements must have at least 36 elements just to satisfy the (a,b,*) requirement. In the present case I think 36 test cases are also sufficient, as in the following set of triples:</p>
<p>(1, 1, 1), (1, 2, 2), (1, 3, 3), (1, 4, 4), (1, 5, 5), (1, 6, 6)</p>
<p>(2, 1, 6), (2, 2, 1), (2, 3, 2), (2, 4, 3), (2, 5, 4), (2, 6, 5)</p>
<p>(3, 1, 5), (3, 2, 6), (3, 3, 1), (3, 4, 2), (3, 5, 3), (3, 6, 4)</p>
<p>(4, 1, 4), (4, 2, 5), (4, 3, 6), (4, 4, 1), (4, 5, 2), (4, 6, 3)</p>
<p>(5, 1, 3), (5, 2, 4), (5, 3, 5), (5, 4, 6), (5, 5, 1), (5, 6, 2)</p>
<p>(6, 1, 2), (6, 2, 3), (6, 3, 4), (6, 4, 5), (6, 5, 6), (6, 6, 1)</p>
<p>In this example each of the three kinds of pairs occurs once and only once, i.e. there is no overlap. I don't think this will be possible in general, so it might not always be easy to come up with minimal subsets that cover all the cases.</p>
|
2,735,984 | <p>I tried to solve this recurrence by taking out $n+1$ as a common in the RHS, but still have $n \cdot a_n$ and $a_n$</p>
| Tal-Botvinnik | 331,471 | <p><strong>HINT</strong></p>
<p>Assuming you meant $\frac{n+1}{n}$ (since you stated that you factored $n+1$ from the rhs), you can divide by $n+1$ and define
$$b_n=\frac{a_n}{n+1}.$$</p>
<p>Now, what is the equation satisfied by $b_n$? Do you know how to solve such an equation?</p>
|
4,030,359 | <p>Consider the abstract von Neumann algebra
<span class="math-container">$$M:= \ell^\infty-\bigoplus_{i \in I} B(H_i)$$</span>
which consists of elements <span class="math-container">$(x_i)_i$</span> with <span class="math-container">$\sup_i \|x_i\| < \infty$</span> and <span class="math-container">$x_i \in B(H_i)$</span>.</p>
<p>Let <span class="math-container">$N$</span> be the algebraic direct sum <span class="math-container">$\bigoplus_i B(H_i)$</span>, thus it consists of elements <span class="math-container">$(x_i)_i$</span> such that only finitely many <span class="math-container">$x_i$</span> are non-zero. I want to show that <span class="math-container">$N$</span> is <span class="math-container">$\sigma$</span>-weakly dense in <span class="math-container">$M$</span>.</p>
<p>I guess my main problem is that I don't understand the <span class="math-container">$\sigma$</span>-weak topology on <span class="math-container">$M$</span>. By a result of Sakai, it is the unique topology on <span class="math-container">$M$</span> coming from a weak<span class="math-container">$^*$</span>-topology when we realise <span class="math-container">$M$</span> as the dual of some other Banach space.</p>
<p>Maybe, we can do the following
<span class="math-container">$$M \cong \ell^\infty-\bigoplus_{i \in I} T(H_i)^* \cong \left(\ell^1-\bigoplus_{i \in I} T(H_i)\right)^*$$</span>
But this does not seem very practical! Any insight in the matter will be appreciated!</p>
| daw | 136,544 | <p>Does this functions fit to your conditions?
<span class="math-container">$$
f(x) = \sqrt{|x|}
$$</span></p>
|
4,030,359 | <p>Consider the abstract von Neumann algebra
<span class="math-container">$$M:= \ell^\infty-\bigoplus_{i \in I} B(H_i)$$</span>
which consists of elements <span class="math-container">$(x_i)_i$</span> with <span class="math-container">$\sup_i \|x_i\| < \infty$</span> and <span class="math-container">$x_i \in B(H_i)$</span>.</p>
<p>Let <span class="math-container">$N$</span> be the algebraic direct sum <span class="math-container">$\bigoplus_i B(H_i)$</span>, thus it consists of elements <span class="math-container">$(x_i)_i$</span> such that only finitely many <span class="math-container">$x_i$</span> are non-zero. I want to show that <span class="math-container">$N$</span> is <span class="math-container">$\sigma$</span>-weakly dense in <span class="math-container">$M$</span>.</p>
<p>I guess my main problem is that I don't understand the <span class="math-container">$\sigma$</span>-weak topology on <span class="math-container">$M$</span>. By a result of Sakai, it is the unique topology on <span class="math-container">$M$</span> coming from a weak<span class="math-container">$^*$</span>-topology when we realise <span class="math-container">$M$</span> as the dual of some other Banach space.</p>
<p>Maybe, we can do the following
<span class="math-container">$$M \cong \ell^\infty-\bigoplus_{i \in I} T(H_i)^* \cong \left(\ell^1-\bigoplus_{i \in I} T(H_i)\right)^*$$</span>
But this does not seem very practical! Any insight in the matter will be appreciated!</p>
| Adam Rubinson | 29,156 | <p>An example similar to daw's is:</p>
<p><span class="math-container">$$f(x)=\begin{cases}\arcsin(-2x-1) + \frac{\pi}{2} & -1\leq x< 0\\ \arcsin(2x-1) + \frac{\pi}{2} & 0\leq x\leq 1 \end{cases}$$</span></p>
|
1,569,543 | <blockquote>
<p>Prove or disprove: $(\ln n)^2 \in O(\ln(n^2)).$ </p>
</blockquote>
<p>I think I would start with expanding the left side. How would I go about this?</p>
| Casteels | 92,730 | <p>No. The path on $4$ vertices is known to have (Laplacian) eigenvalues $2-2\cos\left(\frac{k\pi}{4}\right)$ for $k\in\{0,1,2,3\}$. </p>
<p>The second smallest eigenvalue is $2-\sqrt{2}\neq 1$. In fact none of the eigenvalues are $1$.</p>
|
206,825 | <p>Let's say i have</p>
<p>N1 = -584</p>
<p>N2 = 110</p>
<p>Z = 0.64 </p>
<p>How do i calculate from Z which value is it in range of N1..N2? Z is range from 0 to 1.</p>
| André Nicolas | 6,312 | <p>We want to map numbers in the interval $[0,1]$ to numbers in your interval $[-584,110]$ by a function $fx)$ of the shape $f(x)=px+q$.</p>
<p>We want $f(0)=-584$, and $f(1)=110$, so $q=-584$ and $p=694$.</p>
<p>So our function is $694x-584$. Plug in $x=0.64$.</p>
<p><strong>Remark:</strong> Exactly the same idea will work if our target interval is, say, $[a,b]$. We get $q=a$ and $p=b-a$. So the linear function $f(x)$ that maps $[0,1]$ to $[a,b]$ is given by $f(x)=(b-a)x+a$. </p>
|
2,808,159 | <p><strong>The question is:</strong> </p>
<blockquote>
<p>A half cylinder with the square part on the $xy$-plane, and the length $h$ parallel to the $x$-axis. The position of the center of the square part on the $xy$-plane is $(x,y)=(0,0)$.
<img src="https://i.stack.imgur.com/fB5le.jpg" alt="Image description"> </p>
</blockquote>
<p>$S_1$ is the curved portion of the half-cylinder $z=(r^2-y^2)^{1/2}$ of length $h$.<br>
$S_2$ and $S_3$ are the two semicircular plane end pieces.<br>
$S_4$ is the rectangular portion of the $xy$-plane </p>
<p>Gauss' law:
$$\iint_S\mathbf E\cdot \mathbf{\hat{n}}\,dS=\frac{q}{\epsilon_0}$$
$\mathbf E$ is the electric field $\left(\frac{\text{Newton}}{\text{Coulomb}}\right)$.<br>
$\mathbf{\hat{n}}$ is the unit normal vector.<br>
$dS$ is an increment of the surface area $\left(\text{meter}^2\right)$.<br>
$q$ is the total charge enclosed by the half-cylinder $\left(\text{Coulomb}\right)$.<br>
$\epsilon_0$ is the permitivity of free space, a constant equal to $8.854\times10^{-12}\,\frac{\text{Coulomb}^2}{\text{Newton}\,\text{meter}^2}$. </p>
<p>The electrostatic field is:
$$\mathbf{E}=\lambda(x\mathbf{i}+y\mathbf{j})\;\text{,}$$
where $\lambda$ is a constant.</p>
<p>Use this formula to calculate the part of the total charge $q$ for the curved portion $S_1$ of the half-cylinder:
$$\iint_S\mathbf E\cdot \mathbf{\hat{n}}\,dS=\frac{q}{\epsilon_0}=\iint_R\left\{-E_x[x,y,f(x,y)]\frac{\partial f}{\partial x} -E_y[x,y,f(x,y)]\frac{\partial f}{\partial y} +E_z[x,y,f(x,y)] \right\}\,dx\,dy$$</p>
<p>The goal is to find the total charge $q$ enclosed by the half-cylinder, expressed in terms of $\lambda$, $r$ and $h$. </p>
<p><strong>The solution should be:</strong><br>
$$\pi r^2\lambda h\epsilon_0$$</p>
<p><strong>This is what I've tried:</strong><br>
First calculate Gauss' law for $S_1$:
\begin{align}
f(x,y)&=z=(r^2-y^2)^{1/2}=\sqrt{(r^2-y^2)} \\
\frac{\partial f}{\partial x}&=\frac12(r^2-y^2)^{-\frac12}\cdot 0=0 \\
\frac{\partial f}{\partial y}&=\frac12(r^2-y^2)^{-\frac12}\cdot -2y=-\frac{y}{\sqrt{(r^2-y^2)}}=-\frac yz \\
\\
\mathbf{E}&=\lambda(x\mathbf{i}+y\mathbf{j}) \\
E_x[x,y,f(x,y)]&=\lambda x \\
E_y[x,y,f(x,y)]&=\lambda y \\
E_z[x,y,f(x,y)]&=0 \\
\\
\text{length}&=h \\
\end{align}</p>
<p>Using the formula<br>
$$\iint_R\left\{-E_x[x,y,f(x,y)]\frac{\partial f}{\partial x} -E_y[x,y,f(x,y)]\frac{\partial f}{\partial y} +E_z[x,y,f(x,y)] \right\}\,dx\,dy$$
we get:<br>
\begin{align}
&\iint_R\left\{-\lambda x\cdot 0-\lambda y\cdot -\frac{y}{z} + 0\right\}\,dx\,dy \\
&=\iint_R\frac{\lambda y^2}{z}\,dx\,dy \\
&=\lambda\iint_R\frac{y^2}{\sqrt{r^2-y^2}}\,dx\,dy \\
\end{align}</p>
<p>Since the length is $h$ and the length is parallel to the $x$-axis:
\begin{align}
&\lambda \int_R\int_0^h\frac{y^2}{\sqrt{r^2-y^2}}\,dx\,dy \\
&=\lambda\int_R\left[\frac{y^2x}{\sqrt{r^2-y^2}}\right]_0^h\,dy \\
&=\lambda\int_R\frac{y^2h}{\sqrt{r^2-y^2}}\,dy \\
\end{align}</p>
<p>Subsitute:<br>
\begin{align}
y&=r\sin\theta \\
\theta&=\arcsin\left(\frac1r y\right) \\
\frac{dy}{d\theta}&=\frac{d}{d\theta}\left(r\sin\theta\right)=r\cos\theta \\
dy&=r\cos(\theta)\,d\theta \\
\\
&\lambda\int\frac{hr^2\sin^2\theta}{\sqrt{r^2-r^2\sin^2\theta}}\cdot r\cos(\theta)\,d\theta \\
&=\lambda h\int\frac{r^3\sin^2\theta\cos\theta}{r\sqrt{1-\sin^2\theta}}\,d\theta \\
&=\lambda hr^2\int\frac{\sin^2\theta\cos\theta}{\sqrt{\cos^2\theta}}\,d\theta \\
&=\lambda hr^2\int\frac{\sin^2\theta\cos\theta}{\cos\theta}\,d\theta \\
&=\lambda hr^2\int\sin^2\theta\,d\theta \\
&=\lambda hr^2\int\frac{1-\cos2\theta}{2}\,d\theta \\
&=\frac12\lambda hr^2\int1-\cos2\theta\,d\theta \\
&=\frac12\lambda hr^2\int1\,d\theta-\int\cos2\theta\,d\theta \\
&=\frac12\lambda hr^2\left[\theta-\frac12\sin2\theta\right] \\
\\
\text{substitute back } \theta=\arcsin\left(\frac1r y\right)\text{:} \\
&=\frac12\lambda hr^2\left[\arcsin\left(\frac1r y\right)-\frac12\sin\left(2\arcsin\left(\frac1r y\right)\right)\right] \\
\text{the boundaries of }y\text{ are }-r\text{ and }r\text{:} \\
&=\frac12\lambda hr^2\left[\arcsin\left(\frac{y}{r}\right)-\frac12\sin\left(2\arcsin\left(\frac{y}{r}\right)\right)\right]_{-r}^r \\
&=\frac12\lambda hr^2\left(\frac{\pi}{2}-0\right) - \left(-\frac{\pi}{2}-0\right) \\
&=\frac12\pi\lambda hr^2
\end{align}</p>
<p>Calculate Gauss' law for $S_2$ and $S_3$:<br>
The surfaces of $S_2$ and $S_3$ are equal. </p>
<p>Since:<br>
$\bullet$ the position of the center of the square part on the $xy$-plane is $(x,y)=(0,0)$, the direction of the electrostatic field at both surfaces is opposite: $(\lambda x \mathbf{i})$,<br>
$\bullet$ and the unit normal vectors are in opposite direction,<br>
the addition of the result of Gauss' law will not be equal to 0. </p>
<p>The surface of each of the surfaces is $\frac12 \pi r^2$.<br>
The electric field in the $x$-direction is $\lambda x\mathbf{i}$.<br>
$x$ for $S_2$ = $\frac12 h$.<br>
$x$ for $S_3$ = $-\frac12 h$.<br>
$\mathbf{\hat{n}}$ for $S_2$ = $\mathbf{i}$.<br>
$\mathbf{\hat{n}}$ for $S_3$ = $-\mathbf{i}$. </p>
<p>Therefore for $S_2$:
\begin{align}
\mathbf{E}\cdot \mathbf{\hat{n}} \times \text{surface area}&=\lambda x\mathbf{i} \cdot \mathbf{i} \times \frac12 \pi r^2 \\
&=\lambda \frac12 h\mathbf{i} \cdot \mathbf{i} \times \frac12 \pi r^2 \\
&=\frac14 \pi\lambda hr^2 \\
\end{align}</p>
<p>And for $S_3$:
\begin{align}
\mathbf{E}\cdot \mathbf{\hat{n}} \times \text{surface area}&=\lambda x\mathbf{i} \cdot -\mathbf{i} \times \frac12 \pi r^2 \\
&=\lambda (-\frac12 h)\mathbf{i} \cdot -\mathbf{i} \times \frac12 \pi r^2 \\
&=\frac14 \pi\lambda hr^2 \\
\end{align}</p>
<p>Calculate Gauss' law for $S_4$:<br>
Since $S_4$ lies in the $xy$-plane, the electrostatic field $\mathbf{E}=\lambda(x\mathbf{i}+y\mathbf{j})\;\text{,}$ lies parallel to the surface, thus the result of Gauss' law is $0$.</p>
<p>The net result is:<br>
\begin{align}
\iint_S\mathbf E\cdot \mathbf{\hat{n}}\,dS&=\frac{q}{\epsilon_0} \\
&=\frac12 \pi\lambda hr^2 + \frac14 \pi\lambda hr^2 + \frac14 \pi\lambda hr^2 +0 \\
&=\pi\lambda hr^2 \\
\end{align}</p>
<p>The total charge $q$ enclosed in the half-cylinder is thus:<br>
$$q=\pi\lambda hr^2 \epsilon_0$$. </p>
<p>This solves the problem I was having.</p>
| Arthur Sauer | 563,844 | <p>The lower and upper boundaries for $y$ in $S_1$ should be $-r$ and $r$ instead of $0$ and $r$. I was thinking in polar coordinates instead of in Cartesian coordinates... </p>
<p>I'll edit the original post accordingly</p>
|
109,754 | <p>Please help me get started on this problem:</p>
<blockquote>
<p>Let <span class="math-container">$V = R^3$</span>, and define <span class="math-container">$f_1, f_2, f_3 ∈ V^*$</span> as follows:<br />
<span class="math-container">$f_1(x,y,z) = x - 2y$</span><br />
<span class="math-container">$f_2(x,y,z) = x + y + z$</span><br />
<span class="math-container">$f_3(x,y,z) = y-3z$</span></p>
<p>Prove that <span class="math-container">$\{f_1,f_2,f_3\}$</span> is a basis for <span class="math-container">$V^*$</span>, and then find a basis for <span class="math-container">$V$</span> for which it is the dual basis.</p>
</blockquote>
| KnifeWrench | 61,233 | <p>Proving $\lbrace f_1,f_2,f_3\rbrace$ is a basis for $V^*$ can be done by row reducing the coefficients of $\lbrace f_1,f_2,f_3 \rbrace$ and showing that it has a rank of $3$.</p>
<p>The dual basis of $ \lbrace f_1,f_2,f_3 \rbrace $ is found by calculating the inverse of coefficients of $f_i$ which is:</p>
<p>$x_{1}=
\
\begin{bmatrix}
4/10 \\ -3/10 \\ -1/10
\end{bmatrix},
x_{2}=
\begin{bmatrix}
6/10 \\ 3/10 \\ 1/10
\end{bmatrix},
x_{3}=
\begin{bmatrix}
2/10 \\ 1/10 \\ -3/10
\end{bmatrix}
]\
$</p>
<p>Checking that $f_{i}(x_{j})=$ $1,$ if $i=j$ and $0$ if $i \ne j $</p>
<p>$f_{1}(x_1)=(4/10)-2(-3/10)=1\\
f_{1}(x_2)=(6/10)-2(3/10)=0\\
f_{1}(x_3)=(2/10)-2(1/10)=0$</p>
<p>$f_{2}(x_1)=(4/10)+(-3/10)+(-1/10)=0\\
f_{2}(x_2)=(6/10)+(3/10)+(1/10)=1\\
f_{2}(x_3)=(2/10)+(1/10)+(-3/10)=0$</p>
<p>$f_{3}(x_1)=(-3/10)-3(-1/10)=0\\
f_{3}(x_2)=(3/10)-3(1/10)=0\\
f_{3}(x_3)=(1/10)-3(-3/10)=1$</p>
|
3,281,828 | <p>I am new to permutation and combination and am looking for guidance in the following example:</p>
<p>We have 3 people - A, B, C</p>
<p>How many ways are there to arrange them into Rank 1,2,3</p>
<p>Looking at the example, it is clear that No repetitions are allowed and that ordering is not important (in the sense - Rank 1 - A, Rank 2 - B, Rank 3 - C is the same as Rank 2 - B, Rank 3 - C, Rank 1 - A).</p>
<p>So as a permutation problem we have answer as 3! = 6</p>
<p>Where as a variation problem we have the answer as 3!/0! = 3! = 6 (but if ordering is not important then it would be 3^3 = 27)</p>
<p>Please can you help me understand how me decide between Permutation or variation? and whether ordering is important or not?</p>
| Vedvart1 | 354,933 | <p>A <em>permutation</em> asks how many ways there are to <em>permute</em>, or order, some number of elements from a larger set of elements. For instance, suppose you're organizing a photo with family members. Only <span class="math-container">$5$</span> of them will fit in the photo, but you have <span class="math-container">$9$</span> family members present. Then how many ways are there to take this photo, if you care about the order they're in? Well, imagine we're filling the <span class="math-container">$5$</span> photo slots from left to right. For the very first slot, we have <span class="math-container">$9$</span> options, from all the family members. Now for the second slot, we have <span class="math-container">$8$</span> options; the next we have <span class="math-container">$7$</span>, then <span class="math-container">$6$</span>, and the last slot has <span class="math-container">$5$</span> options to fill with. This means we must have <span class="math-container">$9\cdot8\cdot7\cdot6\cdot5$</span> ways to take the picture. In basic combinatorics, this is denoted the permutation, <span class="math-container">$P(n,k)$</span>, and is the number of ways to <em>permute</em> <span class="math-container">$k$</span> objects from a group of <span class="math-container">$n$</span>:</p>
<p><span class="math-container">$$
P(n,k)=n\cdot(n-1)\cdot(n-2)\cdots(n-k+1)=\frac{n!}{(n-k)!}
$$</span></p>
<p>A <em>combination</em> asks how many ways there are to <em>combine</em> elements from a larger group, <strong>not including order</strong>. For this, we can use our friend, the permutation. Imagine the same situation, lining <span class="math-container">$k$</span> of your <span class="math-container">$n$</span> family members up for a photo, but this time you don't care about order. There are <span class="math-container">$P(n,k)$</span> ways to do it including order; for each group of <span class="math-container">$k$</span> family members, there are <span class="math-container">$k!$</span> ways to put them in order. Thus, we want to know the muber of ways to choose family members in order <em>divided by</em> the number of ways to order them, which leaves us with simply the number of ways to choose them. This is the combination <span class="math-container">$C(n,k)$</span>, or more often denoted <span class="math-container">$n\choose k$</span>:</p>
<p><span class="math-container">$$
C(n,k)= {n\choose k} = \frac{P(n,k)}{k!} = \frac{n!}{k!(n-k)!}
$$</span></p>
<p>To return to your initial question: suppose we have people <span class="math-container">$A,B,C$</span>, and want to arrange them into ranks <span class="math-container">$1,2,3$</span>. Ordering <em>is</em> actually important here - to see this imagine the ranks are places in an Olympic race. Clearly <span class="math-container">$A→1,B→2,C→3$</span> is different from <span class="math-container">$C→1,A→2,B→3$</span>, as <span class="math-container">$C$</span> and their team might argue very strongly! In this case, we have <span class="math-container">$3$</span> options for picking rank <span class="math-container">$1$</span>. We then have <span class="math-container">$2$</span> options for rank <span class="math-container">$2$</span>, and then only <span class="math-container">$1$</span> option for the final rank. This gives us <span class="math-container">$3\cdot2\cdot1=3!$</span> ways to do it; this is a permutation in disguise, as we are arranging <span class="math-container">$3$</span> objects from a pool of <span class="math-container">$3$</span>, which there is <span class="math-container">$P(3,3)$</span> ways to do. And if we check we find that <span class="math-container">$P(3,3)=\frac{3!}{(3-3)!}=\frac{3!}{0!}=3!$</span> - exactly the result we were looking for!</p>
<p>The <span class="math-container">$3^3=27$</span> answer comes in a slightly different variation which involves neither permutations <em>or</em> combinations. This occurs when you allow replacement of the objects! Imagine you're pulling names from a hat, but every time you pull a name and read it, you put it back. How many ways are there to pull the names? Say we have <span class="math-container">$3$</span> names. For the first name we pull, we have <span class="math-container">$3$</span> possible names to pull. But since we put it back, on the next draw we also have <span class="math-container">$3$</span> possibilities! Thus, if we draw <span class="math-container">$k$</span> times, we have <span class="math-container">$3^k$</span> ways to pull the names. Applied to your problem, imagine a variation in which instead of ranks we were giving <span class="math-container">$A,B,$</span> and <span class="math-container">$C$</span> awards. If we can give multiple awards to each person, we have replacement. With <span class="math-container">$3$</span> people and <span class="math-container">$3$</span> awards, there would be <span class="math-container">$3^3=27$</span> ways to give these awards to the people.</p>
|
4,454,630 | <p>Is it true that for integers <span class="math-container">$i+j+k= 3m = n$</span> where <span class="math-container">$i , j, k , m , n\ge 0$</span> the inequality holds ?
<span class="math-container">$$
\binom{n}{i\;j\;k} \le \binom{n}{m\;m\;m}
$$</span>
I tried to show
<span class="math-container">$$
\frac{n!}{m!m!m!} \Big/ \frac{n!}{i!j!k!} = \frac{i!j!k!}{m!m!m!} \ge 1
$$</span>
but I have no idea how to proceed , are there any related theorems ?</p>
| user26857 | 121,097 | <p>It is well known that <span class="math-container">$$\bigcap_{i=1}^\infty I^n=\{x\in R\mid\exists a\in I\text{ such that }x=ax\} $$</span>
(See <a href="https://math.stackexchange.com/a/18850/121097">here</a>.)</p>
<p>Let <span class="math-container">$\mathrm{Min}(R)=\{P_1,\dots,P_m\}$</span>. If <span class="math-container">$J\not\subseteq P_i$</span> for all <span class="math-container">$i=1,\dots,m$</span>, there exists <span class="math-container">$x_i\in J$</span> such that <span class="math-container">$x_i\notin P_i$</span> for all <span class="math-container">$i=1,\dots,m$</span>. Let <span class="math-container">$a_i\in I$</span> such that <span class="math-container">$x_i(1-a_i)=0$</span>. It follows that <span class="math-container">$1-a_i\in P_i$</span> for all <span class="math-container">$i=1,\dots,m$</span>. Then <span class="math-container">$(1-a_1)\cdots(1-a_m)$</span> belongs to the product of all minimal prime ideals, hence it belongs to the intersection of all minimal prime ideals, so it is nilpotent. Now write <span class="math-container">$(1-a_1)\cdots(1-a_m)=1-a$</span> with <span class="math-container">$a\in I$</span>. Since <span class="math-container">$1-a$</span> is nilpotent, <span class="math-container">$a=1-(1-a)$</span> is invertible, so <span class="math-container">$I=R$</span>, a contradiction.</p>
|
1,242,075 | <p>I have some function $g$. I know that $g \in C^1[a,b]$, so $g'(x)$ exists. I want to know, if $g: [a,b] \to [a,b]$ is onto. How can I find out if this is true or not?</p>
<p>P.S. I am not saying all $g$ have the said property, I want to have some kind of test to distinguish functions with this property from functions without it.</p>
| Blind | 207,277 | <p>We provide a class of functions $g\in C^1[a,b]$ such that $g(x)\in [a,b]$ for all $x\in[a,b]$.</p>
<p>(1) $g(a), g(b)\in[a,b]$;</p>
<p>(2) $|g^\prime(x)|\leq \alpha \; \forall x\in (a,b)$, where
$$
\alpha\leq\min\left\{\frac{b-g(a)}{b-a};\frac{g(b)-a}{b-a}\right\}.
$$</p>
<p>Indeed, if $x\in [a, b]$ then there exists $\xi_1, \xi_2\in(a,b)$ such that
$$
|g(x)-g(a)|=|g^\prime(\xi_1)(x-a)|\leq \alpha(b-a),
$$
$$
|g(b)-g(x)|=|g^\prime(\xi_2)(b-x)|\leq \alpha(b-a).
$$</p>
<p>It follows that
$$
g(x)\leq\alpha(b-a)+g(a)\leq b-g(a)+g(a)=b,
$$
$$
g(x)\geq g(b)-\alpha(b-a)\geq g(b)-[g(b)-a]=a.
$$
Therefore, $g(x)\in [a,b]$ for all $x\in[a,b]$.</p>
<p><strong>Example</strong> We can choose $g(x)=\frac{1}{2}x^2\in C^1[0,1]$. We have</p>
<p>(1) $g(0)=0\in[0,1], g(1)=1/2\in[0,1]$;</p>
<p>(2) We have
$$
\alpha:=\min\left\{\frac{1-g(0)}{1-0};\frac{g(1)-0}{1-0}\right\}=1,
$$
and
$$
|g^\prime(x)|=|x|\leq \alpha\quad \forall x\in(0,1).
$$</p>
|
3,236,205 | <p>I'm studying for a qualifying exam in algebra and I've come across the following problem:</p>
<blockquote>
<p>Let <span class="math-container">$G$</span> be a finite group with a subgroup <span class="math-container">$N$</span>. Let <span class="math-container">$Aut(G)$</span> be the group of automorphisms of <span class="math-container">$G$</span>. Prove that if <span class="math-container">$|Aut(G)|$</span> and <span class="math-container">$|N|$</span> are relatively prime, then <span class="math-container">$N$</span> is contained in the center of <span class="math-container">$G$</span>.</p>
</blockquote>
<p>I'm struggling to find a good way to approach this. I'm able to prove a similar result that if <span class="math-container">$|G|$</span> and <span class="math-container">$|Aut(G)|$</span> are relatively prime, then <span class="math-container">$G$</span> is abelian:</p>
<p>Let <span class="math-container">$\theta$</span> be a homomorphism from <span class="math-container">$G$</span> to the inner automorphism group of <span class="math-container">$G$</span>, denoted <span class="math-container">$\theta:G\rightarrow Inn(G)$</span>. Then because inner automorphisms are conjugations by fixed elements, the kernel of <span class="math-container">$\theta$</span> is the center of <span class="math-container">$G$</span>, denoted <span class="math-container">$ker(\theta)=Z(G)$</span>. By the first isomorphism theorem, it follows that <span class="math-container">$G/Z(G)\cong Inn(G)\subseteq Aut(G)$</span>. Hence, by Lagrange's theorem, <span class="math-container">$|G/Z(G)|$</span> divides <span class="math-container">$|Aut(G)|$</span>. Similarly, by consequence of Lagrange's theorem, <span class="math-container">$|G/Z(G)||Z|=|G|$</span>, as <span class="math-container">$Z(G)$</span> is a normal subgroup of <span class="math-container">$G$</span>. Since <span class="math-container">$|G/Z(G)|$</span> divides <span class="math-container">$|G|$</span> and <span class="math-container">$|Aut(G)|$</span>, and <span class="math-container">$|G|$</span> and <span class="math-container">$|Aut(G)|$</span> are relatively prime, it follows that <span class="math-container">$|G/Z(G)|=1$</span>, implying that <span class="math-container">$G=Z(G)$</span>. The desired result follows. <span class="math-container">$\blacksquare$</span></p>
<p>It's not obvious to me how to transform this proof into one of the desired problem (or if there even is a good way to simply modify what I already have). If I knew that <span class="math-container">$N$</span> were a normal subgroup, I could potentially apply the same kind of argument using Lagrange's theorem, but I don't have that assumption. Without more information on the structure of <span class="math-container">$G$</span>, it doesn't seem likely that I'll be able to use an element argument to show that <span class="math-container">$N\subseteq Z(G)$</span>.</p>
| Santana Afton | 274,352 | <p>Hint:</p>
<p>First, note that <span class="math-container">$|N|$</span> is coprime to <span class="math-container">$|\!\operatorname{Aut}(G)|$</span> if and only if the order of every <span class="math-container">$n\in N$</span> is coprime to <span class="math-container">$|\!\operatorname{Aut}(G)|$</span>. </p>
<p>Second, let <span class="math-container">$\varphi_n$</span> be the inner automorphism corresponding to any <span class="math-container">$n\in N$</span>. What happens if <span class="math-container">$\varphi_n$</span> has nontrivial order? What does it mean if <span class="math-container">$\varphi_n$</span> has trivial order?</p>
|
186,726 | <p>Just a soft-question that has been bugging me for a long time:</p>
<p>How does one deal with mental fatigue when studying math?</p>
<p>I am interested in Mathematics, but when studying say Galois Theory and Analysis intensely after around one and a half hours, my brain starts to get foggy and my mental condition drops to suboptimal levels.</p>
<p>I would wish to continue studying, but these circumstances force me to take a break. It is truly a case of "the spirit is willing but the brain is weak"?</p>
<p>How do people maintain concentration over longer periods of time? Is this ability trainable or genetic? (Other than taking illegal drugs like Erdős.)</p>
<p>I know this is a really soft question, but I guess asking mathematicians is the best choice since the subject of Mathematics requires the most mental concentration compared to other subjects.</p>
| unclejamil | 11,743 | <p>Often just a change in the way you are thinking about the subject is enough to wake you up and continue learning so let me pass on my own trick to keeping fresh when I get burned out reading through a book. Take a break for 10 minutes and then switch over to learning how to use the software associated with what you were just studying for a bit. There's a ton of good software out there that implements what you are learning about and it's all fascinating stuff. </p>
<p>Some examples:</p>
<ul>
<li><a href="http://www.gap-system.org/" rel="nofollow">GAP</a> (for group theory)</li>
<li><a href="http://www.singular.uni-kl.de/" rel="nofollow">Singular</a> or <a href="http://www.math.uiuc.edu/Macaulay2/" rel="nofollow">Macaulay2</a> (for algebraic geometry)</li>
<li><a href="http://pari.math.u-bordeaux.fr/" rel="nofollow">PARI</a> (for number theory)</li>
<li><a href="https://www.gnu.org/software/mit-scheme/" rel="nofollow">MIT-Scheme</a> and <a href="http://groups.csail.mit.edu/mac/users/gjs/6946/" rel="nofollow">scmutils</a> (for mechanics using <a href="https://mitpress.mit.edu/index.php?q=books/structure-and-interpretation-classical-mechanics" rel="nofollow">SCIM</a> and differential geometry using <a href="http://amzn.com/0262019345" rel="nofollow">functional differential geometry</a>)</li>
<li><a href="http://www.geogebra.org/" rel="nofollow">Geogebra</a> (for straight up geometry)</li>
</ul>
<p>General Tools that are useful for general mathematics:</p>
<ul>
<li><a href="https://www.gnu.org/software/gsl/" rel="nofollow">GNU Scientific Library</a> </li>
<li><a href="http://www.sagemath.org/" rel="nofollow">Sage</a></li>
</ul>
<p>When you take what you are reading about and write a program that uses it or try to understand how such things are implemented in a piece of software then you will lock in the details. The abstract becomes concrete and your excitement level with the material goes up. After you've put in some time with the software, take another break, and then turn back to the books. </p>
<p>It's hard to get bored or fatigued when you're super into what you are studying. If you are a computer person then this might help. Anything to make it more exciting is worth considering in my book.</p>
|
4,576,565 | <p>How would you decide whether a tridiagonal matrix with all ones in the diagonals has only a trivial solution (as matrix b is zero in the equation Ax=b)?</p>
<p>Edit: So, a general solution to an n by n matrix of the following appearance:</p>
<p><span class="math-container">$\begin{bmatrix}
1& 1& 0& 0\\
1& 1& 1& 0\\
0& 1& 1& 1\\
0& 0& 1& 1\\
\end{bmatrix}$</span></p>
| P.S. Dester | 386,763 | <p>The reason <span class="math-container">$\alpha$</span> can't be greater or equal than <span class="math-container">$2$</span> is analogous to the reason <span class="math-container">$\beta$</span> can't be greater or equal than <span class="math-container">$1$</span> in this other one-dimensional integral:
<span class="math-container">$$\int_0^1 \frac{1}{r^\beta}dr = \frac{1^{1-\beta}}{1-\beta} - \lim_{r\to 0^+} \frac{r^{1-\beta}}{1-\beta}. $$</span>
The limit only exists if <span class="math-container">$\beta < 1$</span> because <span class="math-container">$\lim_{r\to0^+}r^c = \infty$</span> when <span class="math-container">$c<0$</span>.</p>
<p>We need to evaluate the limit of the integral through a limit because the function is not continuous at <span class="math-container">$r=0$</span>. In such cases, we call the integral an <a href="https://en.wikipedia.org/wiki/Improper_integral" rel="nofollow noreferrer">improper integral</a>.</p>
<p>In the original problem, we have that
<span class="math-container">$$
\begin{align*}
\iint_{D}\frac{1}{(x+y)^{\alpha}}dxdy&=\int _0^1\left(\:\int _0^{1-x}\:\frac{1}{\left(x+y\right)^{\alpha \:}}dy\right)dx\\
&= \int_0^1 \frac{1-x^{1-\alpha}}{1-\alpha} dx\\
&= \frac{1}{1-\alpha} - \left(\frac{1^{2-\alpha}}{(1-\alpha)(2-\alpha)} - \lim_{x\to 0^+}\frac{x^{2-\alpha}}{(1-\alpha)(2-\alpha)}\right).
\end{align*}
$$</span>
Again, the limit exists only if <span class="math-container">$\alpha<2$</span>.</p>
|
2,223,577 | <p>$\mathbb{Q}[e^{\frac{2\pi i}{5}}]$ is an extension of $\mathbb{Q}$ of degree 4, since $x^4+x^3+x^2+x+1$ is the irreducible polynomial of $\theta=e^{\frac{2\pi i}{5}}$ over $\mathbb{Q}$.</p>
<p>I'm asked if there is a quadratic extension $K$ of $\mathbb{Q}$ inside $\mathbb{Q}[e^{\frac{2\pi i}{5}}]$.
I suspect that the answer is no.</p>
<p>Since a quadratic extension is always of the form $\mathbb{Q}[\sqrt{k}]$ for an integer $k$, a naive way is to show that the equality $$(a+b\theta+c\theta^2+d\theta^3)^2=k$$ for an integer $k$ is impossible.
But that seems tedious. </p>
<p>Another approach would be that in such case, $\mathbb{Q}[e^{\frac{2\pi i}{5}}]$ is itself a quadratic extension, so we are expected to find an element that looks like $\sqrt{a+b\sqrt{k}}$ inside (for rational $a,b$). But then I'm stuck again.</p>
<p>Any ideas?</p>
| Paramanand Singh | 72,031 | <p>We can get an answer in this simple case by solving the equation $$x^{4}+x^{3}+x^{2}+x+1=0$$ Dividing by $x^{2}$ and putting $y=x+1/x$ we get $y^{2}+y-1=0$ so that $$y=\frac{-1\pm\sqrt{5}}{2}$$ and $$x=\frac{y\pm\sqrt{y^{2}-4}} {2}$$ and then clearly we can see that there is a tower of field extensions $$\mathbb{Q} \subset \mathbb{Q} (y) \subset \mathbb{Q} (x) $$ and the degree of each extension is $2$.</p>
<p>More generally Gauss proved that if $p=2^{2^{k}}+1$ is prime then the equation $x^{p} - 1=0$ can be solved by a series of $2^k$ quadratic equations so that the extension $\mathbb{Q} \subset \mathbb{Q} (\zeta_{p}) $ of degree $p-1=2^{2^{k}}$ can be realized as a tower of $2^{k}$ quadratic extensions of $\mathbb{Q}$. This development by Gauss is not so popular as compared to the more abstract work of Galois. See <a href="http://paramanands.blogspot.com/2009/12/gauss-and-regular-polygons-gaussian-periods.html" rel="nofollow noreferrer">this</a> and <a href="http://paramanands.blogspot.com/2009/12/gauss-and-regular-polygons-gaussian-periods-contd.html" rel="nofollow noreferrer">this</a> for more details. </p>
|
59,888 | <p>Very similar to </p>
<p><a href="https://mathematica.stackexchange.com/questions/11638/parameters-in-plot-titles">Parameters in plot titles</a></p>
<p>in which I want to call a parameter from an array using <code>PlotLabel</code> in my plot using <code>Manipulate</code>. I've tried all of the suggestions in the above post and keep getting the same error. Simple code:</p>
<pre><code>f = {1, 2, 3, 4, 5, 6};
Manipulate[
Plot[Sin[f[[g]] x], {x, -2, 2}, PlotLabel -> Text["f =" f[[g]]]], {g,
1, 3, 1}]
</code></pre>
<p><img src="https://i.stack.imgur.com/Ebrf4.gif" alt="enter image description here"></p>
<p>Curiously, the first <code>f</code> value does not show up incorrectly to the left of the <code>"f="</code> string. Instead of <code>Text[]</code> I've also tried <code>HoldForm[]</code>, <code>TraditionalForm[]</code>, <code>Defer[]</code>. I'm not too picky on where <code>"f ="</code> shows up, but this is pretty confusing to me as why this doesn't work as it should. Thanks</p>
| ubpdqn | 1,997 | <p>Just another way using v10 <code>StringTemplate</code></p>
<pre><code>Manipulate[
Plot[Sin[f[[g]] x], {x, -2, 2},
PlotLabel -> TemplateApply[s, g]], {g, Range[6]},
Initialization :> (f = Range[6];
s = StringTemplate["f=<*f[[`1`]]*>"])]
</code></pre>
<p><img src="https://i.stack.imgur.com/llgXk.gif" alt="enter image description here"></p>
<p>If you prefer the traditional form of expressions rather than the version in graphics one way (there are almost certainly better):</p>
<pre><code>Manipulate[
With[{lab =
Rasterize[TraditionalForm[TemplateApply[st, g] // DisplayForm],
RasterSize -> 100, ImageSize -> 50]},
Plot[Sin[f[[g]] x], {x, -2, 2}, PlotLabel -> lab]], {g, Range[6],
SetterBar},
Initialization :> (f = Range[6];
st = StringTemplate["f=<*f[[`1`]]*>", CombinerFunction -> RowBox])]
</code></pre>
<p><img src="https://i.stack.imgur.com/MWRPX.gif" alt="enter image description here"></p>
|
2,319,126 | <p>I have to find dimension of $V+W,$ where$ V$ is a vector subspace given by solutions of the linear system:</p>
<p>$$x+2y+z=0$$
$$3y+z+3t=0$$</p>
<p>and $W$ is the subspace generated from vectors
$(4,0,1,3)^T,(1,0,-1,0)^T$.</p>
<p>I don't know how to combine the two subspaces and calculate the dimension.</p>
| Siong Thye Goh | 306,553 | <p>Hint:</p>
<p>The dimension of the basis of the solution space is $2$. </p>
<p>Also, none of the vectors in the basis of $W$ is a solution to the linear system.</p>
<p>Can you compute the dimension of $V+W$ now?</p>
|
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