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 |
|---|---|---|---|---|
733,280 | <p>I cannot understand why $\log_{49}(\sqrt{ 7})= \frac{1}{4}$. If I take the $4$th root of $49$, I don't get $7$.</p>
<p>What I am not comprehending? </p>
| Sujaan Kunalan | 77,862 | <p>$\log_{49}{\sqrt{7}}=\dfrac{1}{4}$ means that $49^{\frac{1}{4}}={\sqrt{7}}$</p>
<p>In other words, when you take the fourth root of $49$, you should get $\sqrt{7}$, not $7$.</p>
<p>Note that $\Large49^{\frac{1}{4}}=(7^2)^{\frac{1}{4}}=7^\frac{2}{4}=7^{\frac{1}{2}}=\sqrt{7}$</p>
|
2,314,327 | <p>I have a quick question here.</p>
<p>For an exercise, I was asked to factor:</p>
<p>$$11x^2 + 14x - 2685 = 0$$</p>
<p>How do I figure this out quickly without staring at it forever? Is there a quicker mathematical way than guessing number combinations, or do I have to guess until I find the right combination of numbers?</p>
<p>The answer is:</p>
<p>$$(11x + 179)(x - 15) = 0 $$</p>
| Janitha357 | 393,345 | <p>Here's one suggestion: Look at $11$ and $-2685$. Factorize $2685$ into primes to get $3.5.179$. Then check the following.</p>
<p>$11.(-3)+(895), 11.(-15)+179, 11.(3)+(-895), 11.(15)+(-179)$ and check which yields $14$. The one that yields $14$ is $11.(-15)+179$. So you get $(11x+179)(x-15)$. </p>
<p>I don't know if this method is standard but it seems to work for me.</p>
|
18,686 | <p>Let us define the following "dimension" of a Borel subet $B \subset \mathbb{R}^k$:</p>
<p>$\dim(B) = \min\{n \in \mathbb{N}: \exists K \subset \mathbb{R}^n, ~{\rm s.t.} ~ B \sim K\}$,</p>
<p>where $\sim$ denotes "homeomorphic to". Obviously, $0 \leq \dim(B) \leq k$.</p>
<p>I have three questions: Given a $B \subset \mathbb{R}$,<br>
1) As $k \to \infty$, how slow can $\dim(B^k)$ grow? Can we choose some $B$ such that $\dim(B^k) = o(k)$ or even $O(1)$?<br>
2) Will it make a difference if we drop the Borel measurability of $B$ or add the condition that $B$ has positive Lebesgue measure?<br>
3) Does this dimension-like notion have a name? The dimension concepts I usually see are Lebesgue's covering dimension, inductive dimension, Hausdorff dimension, Minkowski dimension, etc. I do not think the quantity defined above coincides with any of these, but of course bounds exist.</p>
<p>Thanks!</p>
| Johannes Hahn | 3,041 | <p>As for 1.) $"dim"(\mathbb{Z}^k)=1$ for all $k\in\mathbb{N}$, because all $\mathbb{Z}^k$ are discrete countable and therefore homeomorphic to each other.</p>
|
1,305,257 | <p>I do not understand how to use the following information: If $f$ is entire, then </p>
<p>$$\lim _{|z| \rightarrow \infty} \frac{f(z)}{z^2}=2i.$$</p>
<p>So if $f$ is entire, it has a power series around $z_0=0$, so $f(z)=\Sigma_{n=0}^\infty a_nz^n$, and then we get </p>
<p>$$\lim _{|z| \rightarrow \infty} \frac{\Sigma_{n=0}^\infty a_nz^n}{z^2}=2i.$$ </p>
<p>How do I continue from here? </p>
<p>It is a part of a question. I just want to know how can I use this info. I don't know how I can manipulate summations, and since it's $|z| \rightarrow \infty$ and not $z \rightarrow \infty$ (which is meaningless), I don't really know what I can do here. </p>
<p>Maybe </p>
<p>$$\lim _{|z| \rightarrow \infty} \Sigma_{n=0}^\infty a_nz^{n-2}=2i,$$ but then what?</p>
<p>Thanks in advance for your assistance! </p>
| Hagen von Eitzen | 39,174 | <p>The claim itself is true more generally for any polynomial of even degree with positive leading coefficient.</p>
<p>We do not need that $g(x)<0$ for some $x$ and in fact such $x$ may not exist. What we need is that $\lim_{x\to\pm\infty}g(x)=1$ implies that there exists $M\in \mathbb R$ such that $|x|>M$ implies $|g(x)-1|<\frac12$ (so that $g(x)>\frac12$). Then for $|x|>M$ we conclude $f(x)>\frac12x^4>\frac12M^4$. We may assume wlog. that $\frac12M^4>a_0=f(0)$. Then the minimum the continuous function has on $[-M,M]$ is in fact a global minimum.</p>
|
115,081 | <p>I hope here is the best place to ask this, I will begin my master degree very soon, I've already attended the regular undergraduate courses included Real Analysis, Analysis on manifolds, Abstract Algebra, Field Theory, point-set topology, Algebraic Topology, etc...
I like very much algebraic topology and I found it really beautiful, I would like to know which areas of algebraic topology are the most interesting to begin to work with and which books I can study with my background in order to get the prerequisites to begin to study this subject. </p>
<p>I want as soon as possible has a "taste" of a current research field in algebraic topology, and I know that an algebraic topologist can give me a "shortest way" while I attend the regular courses of my master degree.</p>
<p>Thank you</p>
| Alexandre Eremenko | 25,510 | <p>For your first research problem, I recommend that you find an adviser in your department.
If there is no algebraic topologist in your department, find some other adviser, and
ask to suggest an interesting problem.</p>
<p>It is very unlikely that, as a master student, you will be able to find and solve
a reasonable research
problem yourself, without a help from an experienced adviser.
In this site, people can give you only reading recommendations, and this is probably not enough
to begin your own research.
But of course, there were rare exceptions in history when self-taught mathematicians
did good research.</p>
<p>Here is an outstanding problem in algebraic topology on this site:</p>
<p>fedja (mathoverflow.net/users/1131), Two commuting mappings in the disk, <a href="https://mathoverflow.net/questions/3332">Two commuting mappings in the disk</a> (version: 2009-11-25)</p>
<p>To understand the statement of the problem, little knowledge is required.
What does one need to learn to solve this problem, nobody knows:-)</p>
<p>I wish you luck.</p>
|
3,287,424 | <p>I have a function <span class="math-container">$$f(z)=\begin{cases}
e^{-z^{-4}} & z\neq0 \\
0 & z=0
\end{cases}$$</span></p>
<p>I have to show cauchy riemann equation is satisfied everywhere. I have shown that it isn't differentiable at <span class="math-container">$z=0$</span>. </p>
<p>Usually I will have to convert it in <span class="math-container">$$f(z)=u+iv$$</span> which seems very tedious. Is there some way to do this while keeping it in <span class="math-container">$f(z)$</span> form. </p>
| md2perpe | 168,433 | <p>A function <span class="math-container">$f$</span> is holomorphic at <span class="math-container">$z$</span> if <span class="math-container">$\lim_{|h| \to 0} \frac{f(z+h)-f(z)}{h}$</span> exists. This equation is more fundamental than the Cauchy-Riemann equations, which are derived from this.</p>
<p>In the actual case we get
<span class="math-container">$$
f'(0)
= \lim_{|h| \to 0} \frac{f(h)-f(0)}{h}
= \lim_{|h| \to 0} \frac{e^{-h^{-4}}}{h}
= \{ R = 1/h \}
= \lim_{|R| \to \infty} R e^{-R^{4}}
.
$$</span></p>
<p>When we take partial derivatives, <span class="math-container">$h$</span> follows an axis. Then so does <span class="math-container">$R$</span> and we have <span class="math-container">$R^4 = |R|^4.$</span> This leads to <span class="math-container">$|R e^{-R^{4}}| = |R| e^{-|R|^{4}} \to 0$</span> as <span class="math-container">$|R| \to \infty.$</span> Thus all partial derivatives are <span class="math-container">$0$</span> at <span class="math-container">$z=0$</span>, which implies that the Cauchy-Riemann equations are satisfied.</p>
|
307,701 | <p>Show that if $G$ is a finite group with identity $e$ and with an even number of elements, then there is an $a \neq e$ in $G$, such that $a \cdot a = e$.</p>
<p>I read the solutions here <a href="http://noether.uoregon.edu/~tingey/fall02/444/hw2.pdf" rel="nofollow">http://noether.uoregon.edu/~tingey/fall02/444/hw2.pdf</a></p>
<p>Why do they say $D = \{a, a^\prime\}$? Isn't $D$ not a group? There is no identity and if they include the identity they get 3 elements, which means $|D| = 3 = $ odd.</p>
| Barbara Osofsky | 59,437 | <p>Consider the relation on $G$ given by $g\equiv h\iff g\in\{\ h\ , h^{-1}\ \}$. It is easy to see that this is symmetric, reflexive, and transitive, and so an equivalence relation with equivalence classes $\{\ h\ ,\ h^{-1}\ \}$. The equivalence class of the identity $e$ of $G$ is $\{\ e\ \}$ containing only one element, and all equivalence classes have at most two elements. Since the order of $G$ is even, at least one equivalence class besides $\{\ e\ \}$ must have only one element, and that element is its own inverse. </p>
|
28,568 | <p>Recently, I answered to this problem:</p>
<blockquote>
<p>Given <span class="math-container">$a<b\in \mathbb{R}$</span>, find explicitly a bijection <span class="math-container">$f(x)$</span> from
<span class="math-container">$]a,b[$</span> to <span class="math-container">$[a,b]$</span>.</p>
</blockquote>
<p>using an "iterative construction" (see below the rule).</p>
<p>My question is: is it possible to solve the problem finding a less <em>exotic</em> function?</p>
<p>I mean: I know such a bijection cannot be monotone, nor globally continuous; but my <span class="math-container">$f(x)$</span> has <em>a lot of</em> jumps... Hence, can one do without so many discontinuities?</p>
<hr>
<p>W.l.o.g. assume <span class="math-container">$a=-1$</span> and <span class="math-container">$b=1$</span> (the general case can be handled by translation and rescaling).
Let:</p>
<p>(<strong>1</strong>) <span class="math-container">$X_0:=]-1,-\frac{1}{2}] \cup [\frac{1}{2} ,1[$</span>, and</p>
<p>(<strong>2</strong>) <span class="math-container">$f_0(x):=\begin{cases}
-x-\frac{3}{2} &\text{, if } -1<x\leq -\frac{1}{2} \\ -x+\frac{3}{2} &\text{,
if } \frac{1}{2}\leq
x<1\\ 0 &\text{, otherwise} \end{cases}$</span>,</p>
<p>so that the graph of <span class="math-container">$f_0(x)$</span> is made of two segments (parallel to the line <span class="math-container">$y=x$</span>) and one segment laying on the <span class="math-container">$x$</span> axis; then define by induction:</p>
<p>(<strong>3</strong>) <span class="math-container">$X_{n+1}:=\frac{1}{2} X_n$</span>, and</p>
<p>(<strong>4</strong>) <span class="math-container">$f_{n+1}(x):= \frac{1}{2} f_n(2 x)$</span></p>
<p>for <span class="math-container">$n\in \mathbb{N}$</span> (hence <span class="math-container">$X_n=\frac{1}{2^n} X_0$</span> and <span class="math-container">$f_n=\frac{1}{2^n} f_0(2^n x)$</span>).</p>
<p>Then the function <span class="math-container">$f:]-1,1[\to \mathbb{R}$</span>:</p>
<p>(<strong>5</strong>) <span class="math-container">$f(x):=\sum_{n=0}^{+\infty} f_n(x)$</span></p>
<p>is a bijection from <span class="math-container">$]-1,1[$</span> to <span class="math-container">$[-1,1]$</span>.</p>
<p><em>Proof</em>: <strong>i</strong>. First of all, note that <span class="math-container">$\{ X_n\}_{n\in \mathbb{N}}$</span> is a pairwise disjoint covering of <span class="math-container">$]-1,1[\setminus \{ 0\}$</span>. Moreover the range of each <span class="math-container">$f_n(x)$</span> is <span class="math-container">$f_n(]-1,1[)=[-\frac{1}{2^n}, -\frac{1}{2^{n+1}}[\cup \{ 0\} \cup ]\frac{1}{2^{n+1}}, \frac{1}{2^n}]$</span>.</p>
<p><strong>ii</strong>. Let <span class="math-container">$x\in ]-1,1[$</span>. If <span class="math-container">$x=0$</span>, then <span class="math-container">$f(x)=0$</span> by (<strong>5</strong>). If <span class="math-container">$x\neq 0$</span>, then there exists only one <span class="math-container">$\nu\in \mathbb{N}$</span> s.t. <span class="math-container">$x\in X_\nu$</span>, hence <span class="math-container">$f(x)=f_\nu (x)$</span>. Therefore <span class="math-container">$f(x)$</span> is <em>well defined</em>.</p>
<p><strong>iii</strong>. By <strong>i</strong> and <strong>ii</strong>, <span class="math-container">$f(x)\lesseqgtr 0$</span> for <span class="math-container">$x\lesseqgtr 0$</span> and the range of <span class="math-container">$f(x)$</span> is:</p>
<p><span class="math-container">$f(]-1,1[)=\bigcup_{n\in \mathbb{N}} f(]-1,1[) =[-1,1]$</span>,</p>
<p>therefore <span class="math-container">$f(x)$</span> is surjective.</p>
<p><strong>iv</strong>. On the other hand, if <span class="math-container">$x\neq y \in ]-1,1[$</span>, then: if there exists <span class="math-container">$\nu \in \mathbb{N}$</span> s.t. <span class="math-container">$x,y\in X_\nu$</span>, then <span class="math-container">$f(x)=f_\nu (x)\neq f_\nu (y)=f(y)$</span> (for <span class="math-container">$f_\nu (x)$</span> restrited to <span class="math-container">$X_\nu$</span> is injective); if <span class="math-container">$x\in X_\nu$</span> and <span class="math-container">$y\in X_\mu$</span>, then <span class="math-container">$f(x)=f_\nu (x)\neq f_\mu(y)=f(y)$</span> (for the restriction of <span class="math-container">$f_\nu (x)$</span> to <span class="math-container">$X_\nu$</span> and of <span class="math-container">$f_\mu(x)$</span> to <span class="math-container">$X_\mu$</span> have disjoint ranges); finally if <span class="math-container">$x=0\neq y$</span>, then <span class="math-container">$f(x)=0\neq f(y)$</span> (because of <strong>ii</strong>).
Therefore <span class="math-container">$f(x)$</span> is injective, hence a bijection between <span class="math-container">$]-1,1[$</span> and <span class="math-container">$[-1,1]$</span>. <span class="math-container">$\square$</span></p>
| Asaf Karagila | 622 | <p>It seems that your construction is fine, however coarse and crude. We usually give this question in the introductory course of set theory, the solution is quite elegant too.</p>
<p>Firstly, it is very clear that this function cannot be continuous. Consider a sequence approaching the ends of the interval, the function cannot be continuous there.</p>
<p>Secondly, without the loss of generality assume the interval is $[0,1]$. Define $f(x)$ as following:
$$f(x) = \left\{
\begin{array}{1 1}
\frac{1}{2} & \mbox{if } x = 0\\
\frac{1}{2^{n+2}} & \mbox{if } x = \frac{1}{2^n}\\
x & \mbox{otherwise}
\end{array}
\right.$$</p>
<p>It is relatively simple to show that this function is as needed.</p>
|
28,568 | <p>Recently, I answered to this problem:</p>
<blockquote>
<p>Given <span class="math-container">$a<b\in \mathbb{R}$</span>, find explicitly a bijection <span class="math-container">$f(x)$</span> from
<span class="math-container">$]a,b[$</span> to <span class="math-container">$[a,b]$</span>.</p>
</blockquote>
<p>using an "iterative construction" (see below the rule).</p>
<p>My question is: is it possible to solve the problem finding a less <em>exotic</em> function?</p>
<p>I mean: I know such a bijection cannot be monotone, nor globally continuous; but my <span class="math-container">$f(x)$</span> has <em>a lot of</em> jumps... Hence, can one do without so many discontinuities?</p>
<hr>
<p>W.l.o.g. assume <span class="math-container">$a=-1$</span> and <span class="math-container">$b=1$</span> (the general case can be handled by translation and rescaling).
Let:</p>
<p>(<strong>1</strong>) <span class="math-container">$X_0:=]-1,-\frac{1}{2}] \cup [\frac{1}{2} ,1[$</span>, and</p>
<p>(<strong>2</strong>) <span class="math-container">$f_0(x):=\begin{cases}
-x-\frac{3}{2} &\text{, if } -1<x\leq -\frac{1}{2} \\ -x+\frac{3}{2} &\text{,
if } \frac{1}{2}\leq
x<1\\ 0 &\text{, otherwise} \end{cases}$</span>,</p>
<p>so that the graph of <span class="math-container">$f_0(x)$</span> is made of two segments (parallel to the line <span class="math-container">$y=x$</span>) and one segment laying on the <span class="math-container">$x$</span> axis; then define by induction:</p>
<p>(<strong>3</strong>) <span class="math-container">$X_{n+1}:=\frac{1}{2} X_n$</span>, and</p>
<p>(<strong>4</strong>) <span class="math-container">$f_{n+1}(x):= \frac{1}{2} f_n(2 x)$</span></p>
<p>for <span class="math-container">$n\in \mathbb{N}$</span> (hence <span class="math-container">$X_n=\frac{1}{2^n} X_0$</span> and <span class="math-container">$f_n=\frac{1}{2^n} f_0(2^n x)$</span>).</p>
<p>Then the function <span class="math-container">$f:]-1,1[\to \mathbb{R}$</span>:</p>
<p>(<strong>5</strong>) <span class="math-container">$f(x):=\sum_{n=0}^{+\infty} f_n(x)$</span></p>
<p>is a bijection from <span class="math-container">$]-1,1[$</span> to <span class="math-container">$[-1,1]$</span>.</p>
<p><em>Proof</em>: <strong>i</strong>. First of all, note that <span class="math-container">$\{ X_n\}_{n\in \mathbb{N}}$</span> is a pairwise disjoint covering of <span class="math-container">$]-1,1[\setminus \{ 0\}$</span>. Moreover the range of each <span class="math-container">$f_n(x)$</span> is <span class="math-container">$f_n(]-1,1[)=[-\frac{1}{2^n}, -\frac{1}{2^{n+1}}[\cup \{ 0\} \cup ]\frac{1}{2^{n+1}}, \frac{1}{2^n}]$</span>.</p>
<p><strong>ii</strong>. Let <span class="math-container">$x\in ]-1,1[$</span>. If <span class="math-container">$x=0$</span>, then <span class="math-container">$f(x)=0$</span> by (<strong>5</strong>). If <span class="math-container">$x\neq 0$</span>, then there exists only one <span class="math-container">$\nu\in \mathbb{N}$</span> s.t. <span class="math-container">$x\in X_\nu$</span>, hence <span class="math-container">$f(x)=f_\nu (x)$</span>. Therefore <span class="math-container">$f(x)$</span> is <em>well defined</em>.</p>
<p><strong>iii</strong>. By <strong>i</strong> and <strong>ii</strong>, <span class="math-container">$f(x)\lesseqgtr 0$</span> for <span class="math-container">$x\lesseqgtr 0$</span> and the range of <span class="math-container">$f(x)$</span> is:</p>
<p><span class="math-container">$f(]-1,1[)=\bigcup_{n\in \mathbb{N}} f(]-1,1[) =[-1,1]$</span>,</p>
<p>therefore <span class="math-container">$f(x)$</span> is surjective.</p>
<p><strong>iv</strong>. On the other hand, if <span class="math-container">$x\neq y \in ]-1,1[$</span>, then: if there exists <span class="math-container">$\nu \in \mathbb{N}$</span> s.t. <span class="math-container">$x,y\in X_\nu$</span>, then <span class="math-container">$f(x)=f_\nu (x)\neq f_\nu (y)=f(y)$</span> (for <span class="math-container">$f_\nu (x)$</span> restrited to <span class="math-container">$X_\nu$</span> is injective); if <span class="math-container">$x\in X_\nu$</span> and <span class="math-container">$y\in X_\mu$</span>, then <span class="math-container">$f(x)=f_\nu (x)\neq f_\mu(y)=f(y)$</span> (for the restriction of <span class="math-container">$f_\nu (x)$</span> to <span class="math-container">$X_\nu$</span> and of <span class="math-container">$f_\mu(x)$</span> to <span class="math-container">$X_\mu$</span> have disjoint ranges); finally if <span class="math-container">$x=0\neq y$</span>, then <span class="math-container">$f(x)=0\neq f(y)$</span> (because of <strong>ii</strong>).
Therefore <span class="math-container">$f(x)$</span> is injective, hence a bijection between <span class="math-container">$]-1,1[$</span> and <span class="math-container">$[-1,1]$</span>. <span class="math-container">$\square$</span></p>
| xen | 8,229 | <p>Here you have simpler construction.</p>
<p>Let $(a_n)$ be the sequence in $(a,b)$ defined by $a_n = a + \frac{b-a}{2^n}$. Then let $f\colon [a,b] \to (a,b)$ be given by
$$
f(x) = \begin{cases}
a_1, & x = a,\\
a_2, & x = b,\\
a_{n+2}, & x = a_n, n = 1,2,\dots\\
x, & x \in [a,b] \setminus \{a,b,a_1,a_2,\dots\}.
\end{cases}
$$
Of course definition of $(a_n)$ could be different. The only important thing is that it is a sequence in $(a,b)$.</p>
|
143,274 | <p>I am trying to find the derivative of $\sqrt{9-x}$ using the definition of a derivative </p>
<p>$$\lim_{h\to 0} \frac {f(a+h)-f(a)}{h} $$</p>
<p>$$\lim_{h\to 0} \frac {\sqrt{9-(a+h)}-\sqrt{9-a}}{h} $$</p>
<p>So to simplify I multiply by the conjugate</p>
<p>$$\lim_{h\to0} \frac {\sqrt{9-(a+h)}-\sqrt{9-a}}{h}\cdot \frac{ \sqrt{9-(a+h)}+ \sqrt{9-a}}{\sqrt{9-(a+h)}+\sqrt{9-a}}$$</p>
<p>which gives me </p>
<p>$$\frac {-2a-h}{h(\sqrt{9-(a+h)}+\sqrt{9-a})}$$</p>
<p>I have no idea what to do from here, obviously I can easily get the derivative using other methods but with this one I have no idea how to proceed.</p>
| Argon | 27,624 | <p>Everything you have done is right except for the last step.</p>
<p>$$\begin{align}
&\lim_{h\to0} \frac {\sqrt{9-(a+h)}-\sqrt{9-a}}{h}\cdot \frac{ \sqrt{9-(a+h)}+ \sqrt{9-a}}{\sqrt{9-(a+h)}+\sqrt{9-a}}=\\
&\lim_{h\to0} \frac{9-(a+h)-(9-a)}{h(\sqrt{9-(a+h)}+\sqrt{9-a})}=\\
&\lim_{h\to0} \frac{9-a-h-9+a}{h(\sqrt{9-(a+h)}+\sqrt{9-a})}=\\
&\lim_{h\to0} \frac{h}{h(\sqrt{9-(a+h)}+\sqrt{9-a})}=\\
&\lim_{h\to0} \frac{1}{\sqrt{9-(a+h)}+\sqrt{9-a}}
\end{align}$$</p>
<p>The limit is then easy to evaluate.</p>
|
4,373,055 | <p>We define a bound vector to be a quantity with a defined starting point, magnitude and direction. A free vector has no defined starting point, just magnitude and direction.</p>
<p>So what is a position vector (of a point)? It is defined relative to something else (Origin), so it has a starting point, size and direction like a bound vector. So it makes sense to call position vector a bound vector. However if the point is moving around (say some physical object), then position vector direction and magnitude is changing - so is it still bound vector?</p>
<p>And if it is neither a free vector nor bound, does this mean position vectors is a "3rd type" of a vector?</p>
| ryang | 21,813 | <p>I've not heard of a bound versus free vector, but based on your given definition,</p>
<p>since a position vector <em>does</em> have a <em>defined</em> starting point, magnitude and direction,</p>
<p>then it <em>is</em> a bound vector.</p>
<p>P.S. While its starting point is defined and fixed, its magnitude and direction are defined and varying.</p>
|
288,051 | <p>In enumerative combinatorics, a <i>bijective proof</i> that $|A_n| = |B_n|$ (where $A_n$ and $B_n$ are finite sets of combinatorial objects of size $n$) is a proof that constructs an explicit bijection between $A_n$ and $B_n$.
Bijective proofs are often prized because of their beauty and because of the insight that they often provide. Even if a combinatorial identity has already been proved (e.g., using generating functions), there is often interest in finding a bijective proof.</p>
<p>In spite of the importance of bijective proofs, the process of discovering or constructing a bijective proof seems to be an area that has been relatively untouched by computers. Of course, computers are often enlisted to generate all small examples of $A_n$ and $B_n$, but then the process of searching for a bijection between $A_n$ and $B_n$ is usually done the "old-fashioned" way, by playing around with pencil and paper and using human insight.</p>
<p>It seems to me that the time may be ripe for computers to search directly for bijections. To clarify, I do not (yet) envisage computers autonomously producing full-fledged bijective proofs. What I want computers to do is to search empirically for a <i>combinatorial rule</i>—that says something like, take an element of $A_n$ and do $X$, $Y$, and $Z$ to produce an element of $B_n$—that appears to yield a bijection for small values of $n$.</p>
<p>One reason that such a project has not already been carried out may be that the sheer diversity of combinatorial objects and combinatorial rules may seem daunting. How do we even describe the search space to the computer?</p>
<p>It occurs to me that, now that proof assistants have "come of age," people may have already had to face, and solve (at least partially), the problem of systematically encoding combinatorial objects and rules. This brings me to my question:</p>
<blockquote>
<p>Does there exist a robust framework for encoding combinatorial objects and combinatorial rules in a way that would allow a computer to empirically search for bijections? If not, is there something at least close, that could be adapted to this end with a modest amount of effort?</p>
</blockquote>
<p>In my opinion, Catalan numbers furnish a good test case. There are many different types of combinatorial objects that are enumerated by the Catalan numbers. As a first "challenge problem," a computer program should be able to discover bijections between different kinds of "Catalan objects" on its own. If this can be done, then there is no shortage of more difficult problems to sink one's teeth into.</p>
| JimN | 62,043 | <p>Does the OEIS count?
Three of your papers are listed in <a href="https://oeis.org/wiki/Works_Citing_OEIS" rel="nofollow noreferrer">https://oeis.org/wiki/Works_Citing_OEIS</a> as having used the help of either OEIS or Superseeker , I presume to help find a (potential?) bijection.</p>
|
1,599,467 | <p>Here $f$ is a non-zero linear functional on a vector space $X$. I can show this true for one direction, </p>
<blockquote>
<p>Let $x_1, x_2 \in x + N(f)$</p>
<p>$\implies x_1 = x + y_1, \quad x_2 = x + y_2$, where $y_1, y_2 \in N(f).$</p>
<p>Then $f(x_1) = f(x) + f(y_1) = f(x) = f(x) + f(y_2) = f(x_2)$.</p>
</blockquote>
<p>I'm not sure how to prove the converse holds true too.</p>
| Jendrik Stelzner | 300,783 | <p>As symplectomorphic already pointed out this is just
\begin{align*}
f(x) = f(y)
&\iff f(x) - f(y) = 0
\iff f(x-y) = 0 \\
&\iff x-y \in N(f)
\iff x + N(f) = y + N(f).
\end{align*}</p>
<p>Notice however that this is really just a special case of the more general fact that for every linear map $f \colon V \to W$ we have for all $x,y \in V$ that
$$
f(v) = f(w) \iff v + \ker f = w + \ker f.
$$
The proof of this can be done by the same calculation as before, but I think it is important to realize that this is precisely the statement that the induced linear map
$$
\hat{f} \colon V/\ker f \to W, \quad v + \ker f \mapsto f(v)
$$
is a monomorphism, i.e. that $\hat{f} \colon V / \ker f \to \mathrm{im} f$ is an isomorphism. So this is also see this as a direct consequence of the first isomorphism theorem.</p>
|
73,629 | <p>I want to use <code>Listplot</code> with <code>Tooltip</code>that displays <code>Position</code>of the element I'm hovering over.</p>
<pre><code>data={{0,1},{1,3},{2,2}};
ListPlot[Tooltip[data]]
</code></pre>
<p>This is displaying the value of the element. Can I use the <code>Position</code>function in the tooltip?</p>
| akater | 1,859 | <p><code>Map</code> and <code>MapIndexed</code> solutions are formally valid but <code>Thread</code> is often cleaner. (Poor guy is also underused, I believe.)</p>
<pre><code>data = RandomInteger[10, {10, 2}];
ListPlot@Thread@Tooltip[#, Range@Length@#]&@data
</code></pre>
|
4,358 | <p>I've been reading a bit about how the set of bounds changes for a set depending on what superset one works with. I considered the sets $S\subseteq T\subseteq\mathbb{Q}$ and worked out a few contrived examples:</p>
<p>If $S=T=$ {$x\in\mathbb{Q}\ | \ x^2\lt 2$}, so here $S$ is not bounded above in $T$, but it is bounded about in $\mathbb{Q}$, with $2$ being a possibility.</p>
<p>Also, if $S=$ {$x\in\mathbb{Q}\ | \ x^2\lt 1$} and $T=$ {$x\in\mathbb{Q}\ | \ x\lt 2 \ \text{and}\ x\neq 1$}, then $S$ is bounded in $T$ and $\sup_\mathbb{Q} S=1$ exists, but $\sup_T S$ does not exist.</p>
<p>My question is, is it possible for $S$ to be bounded in $T$ where $\sup_T S$ exists, but $\sup_\mathbb{Q} S$ does not? And moreover, can both $\sup_T S$ and $\sup_\mathbb{Q} S$ exist, but not be equal? Any example of this would be much appreciated.</p>
| Arturo Magidin | 742 | <p>Yes, it is possible for $\sup_T(S)$ to exist, but $\sup_\mathbb{Q}(S)$ not to exist. Yes, it's possible for both to exist and be distinct.</p>
<p>Here is an example of the latter. Take $S=\{x\in\mathbb{Q}|0\leq x\lt 1\}$. Let $T=S\cup{{2}}$. Then $S$ is bounded above in $T$, and has a supremum, namely $2$. Indeed, for every $t\in T$, if $t\lt 2$, then there exists $s\in S$ such that $t\lt s\leq 2$, so $2$ is the supremum of $S$. And of course, the supremum of $S$ in $\mathbb{Q}$ is $1$.</p>
<p>For the former, take $S=\{x\in\mathbb{Q}|0\leq x \lt \sqrt{2}\}$ (or any irrational number you please), and take $T=S\cup\{2\}$ (or any number greater than $\sqrt{2}$). </p>
<p>In general, if you have partially ordered sets $P\subseteq Q$, and a subset $A$ of $P$, you can have that $A$ has a supremum in $P$ but not in $Q$; has a supremum in $Q$ but not in $P$; has suprema in neither; has the same supremum in both; or has suprema in both but they are distinct. The one thing you can say is that if both suprema exist, then $\sup_Q(A)\leq\sup_P(A)$. So in your situation, you do know that if they both exist you will have $\sup_{\mathbb{Q}}(S)\leq\sup_{T}(S)$, but you can have strict inequality.</p>
|
95,314 | <p>To evaluate this type of limits, how can I do, considering $f$ differentiable, and $ f (x_0)> 0 $</p>
<p>$$\lim_{x\to x_0} \biggl(\frac{f(x)}{f(x_0)}\biggr)^{\frac{1}{\ln x -\ln x_0 }},\quad\quad x_0>0,$$</p>
<p>$$\lim_{x\to x_0} \frac{x_0^n f(x)-x^n f(x_0)}{x-x_0},\quad\quad n\in\mathbb{N}.$$</p>
| Did | 6,179 | <p>Hint: write $x=x_0+h$ with $h\to0$ and expand each term up to order $h$.</p>
<p>For example, $\log(x)-\log(x_0)=\log(1+h/x_0)=h/x_0$ and $f(x)=f(x_0)+hf'(x_0)$ hence $f(x)/f(x_0)=1+hf'(x_0)/f(x_0)$, hence...</p>
|
1,336,344 | <p>Given a matrix A of a strongly $k$ regular graph G(srg($n,k,\lambda,\mu$);$\lambda ,\mu >0;k>3$). The matrix A can be divided into 4 sub matrices based on adjacency of vertex $x \in G$.
$A_x$ is the symmetric matrix of the graph $(G-x)$, where $C$ is the symmetric matrix of the graph created by vertices of $(G-x)$ which are not adjacent to $x$ and $D$ is the symmetric matrix of the graph created by vertices of $(G-x)$ which are adjacent to $x$. </p>
<p>$$ A_x = \left(\begin{array}{cccccc|ccc|c}
0&1&0&0&1&0&1&0&0&0\\
1&0&1&0&0&0&0&1&0&0\\
0&1&0&1&0&0&0&0&1&0\\
0&0&1&0&0&1&1&0&0&0\\
1&0&0&0&0&1&0&0&1&0\\
0&0&0&1&1&0&0&1&0&0\\
\hline
1&0&0&1&0&0&0&0&0&1\\
0&1&0&0&0&1&0&0&0&1\\
0&0&1&0&1&0&0&0&0&1\\
\hline
0&0&0&0&0&0&1&1&1&0\\
\end{array}\right)
=
\left( \begin{array}{ccc}
C & E & 0 \\
E^{T} & D & 1\\
0 & 1 & 0\\
\end{array} \right)
$$</p>
<p>It should be noted that</p>
<ol>
<li><p>Interchanging/swapping any two rows (or columns) of $C$ does not affect matrix $D$ (and vice versa).</p></li>
<li><p>Any change in $C$ or $D$ or both $C$ and $D$ changes matrix $E$.</p></li>
</ol>
<p><strong>Problem:</strong> If some vertices of $G$ is rearranged (i.e., permuted), $A$ will be different, say, <strong><em>this new matrix is $B$. Again, matrix $B$ can be divided into 4 sub matrices based on adjacency of vertex $x \in G$ and $ B_x$ can be obtained.</em></strong></p>
<p>Assume:</p>
<ol>
<li>$C$ is always the adjacency matrix of a regular graph and bigger than $D$.</li>
<li>There exists an algorithm that always order $D$ (for a vertex $x \in G$) takes $O(K)$ time (assumed to be polynomial/exponential, does not matter). </li>
<li><p>Each row of E has different permutation, i.e., rows might have same number of 1's but different permutation/arrangement of 1.</p>
<p>For $n$ vertices there will be total $n$ numbers of $C,D$, each of them will take $O(K)$ (assumed to be polynomial) time to sort. If each $C$ takes time $O(f(n))$ to sort, then the total complexity will be $O(n \cdot K \cdot f(n))$. </p></li>
</ol>
<p><strong>Question:</strong> According to the three assumptions above, does there exist a polynomial time algorithm to sort $C$ so that $B=A$? I.e., is there a polynomial $f$ ?</p>
<p>The problem is connected to <a href="https://math.stackexchange.com/questions/1240637/counting-problem-of-combinations-of-symmetric-matrix">this question</a>.</p>
<p><strong><em>If anything is unclear, please ask a specific question, I will try my best to answer.</em></strong> </p>
| vzn | 42,153 | <p>this will not be a <em>direct</em> answer at what does not really seem a <em>direct</em> question, which in its apparently big scope/ ambition pushes outside the bounds of SE questions but is nevertheless aimed to be a scientific and guiding response. you seem to be studying <a href="https://en.wikipedia.org/wiki/Graph_isomorphism_problem" rel="nofollow noreferrer">graph isomorphism</a> in general, and as you are apparently aware the precise complexity of this is a major open problem. and it is known/ utilized generally that regular graphs are the "hard" ones for GI otherwise one can leverage nonuniform vertex degrees to (usually) significantly decrease the search space of vertex permutations. </p>
<p>so the best route to studying known hard scientific problems is to relate your own approach to the literature aka <em>"terra cognita"</em>, which is quite large on this subject. for problems this significant, most straightforward approaches are attempted/ analyzed "somewhere". Here are some fairly close references to your technique, they share many close similarities. & then the task is to try to determine how your technique is different. Your basic idea of assigning columns/ rows of the adjacency matrix to different matrices is basically a partitioning algorithm of vertices into different sets. so:</p>
<ul>
<li><p><a href="https://mathoverflow.net/questions/96858/complexity-of-equitable-partitions">complexity of open partitions</a> / open question by McKay on mathoverflow. he asks about the complexity of partitioning regular graphs into vertex partitions with the "equitability" property that may be similar to yours. he is the writer of <a href="http://www3.cs.stonybrook.edu/~algorith/implement/nauty/implement.shtml" rel="nofollow noreferrer">nauty</a> the top software for graph isomorphism computations.</p></li>
<li><p><a href="http://haralick.org/journals/graph_automorphism_vertex_partitioning.pdf" rel="nofollow noreferrer">Efficient graph automorphism by vertex partitioning</a> / Fowler et al. from abstract, an algorithm that runs in P time for all graphs tested including regular graphs.</p></li>
</ul>
<p>My suggestion is also to try to convert your problem to isomorphism instances and empirically test nauty on it (just as the 1st paper uses empirical testing, which is accepted with conditions in this area), and that may be one of the more efficient known approaches, because nauty encodes vast swathes of state-of-the-art knowledge/ algorithm(s) on graph isomorphism complexity study.</p>
|
688,742 | <p>Given $P\colon\mathbb{R} \to \mathbb{R}$ , $P$ is injective (one to one) polynomial function i need to formally prove that $P$ is onto $\mathbb{R}$</p>
<p>my strategy so far .......
polynomial function is continuous and since it one-to-one function it must be strictly monotonic and now i have no idea what to do .... </p>
<p>there is a theorem saying all continuous and monotonic functions have an inverse function and another theorem saying function have an inverse function if and only if its one-to-one and onto ... </p>
<p>but for formal proof of the on-to function property i think i need to show here that every element in the co-domain/target-set have a "source" in the domain .</p>
<p>i don't think its useful for this proof but only a polynomials of odd degrees have the one-to-one property . </p>
| Federico Poloni | 65,548 | <p>I can guess that your doubt comes from a first step that you do in your mind that goes like this: "$\lim_{x\rightarrow \infty}{e^{-x}} = 0$, so I can replace $e^{-x}$ with $0$ in the text and I remain with
$$
\lim_{x\to ∞} \frac{11-0}{7},
$$
then what happens to the limit in this last equation?"</p>
<p>This reflects a common misunderstanding. Repeat after me: <strong>in a limit, one cannot replace arbitrary subexpressions with their limit</strong>. There is no theorem that says that. If you've been doing that, you are wrong; change your habits.</p>
<p>For an easier example when this doesn't work, $\lim_{x\to 0} x = 0$, but
$$
\lim_{x\to 0} \frac{x+x^3}{x^2} \neq \lim_{x\to 0} \frac{0+x^3}{x^2}.
$$</p>
<p>Clive Newstead's and Ilmari Karonen's answers show some alternative manipulations that are legitimate by theorems.</p>
|
69,590 | <p>Consider the following code.</p>
<pre><code>f[a_,b_]:=x
x=a+b;
f[1,2]
(* a + b *)
</code></pre>
<p>From a certain viewpoint, one might expect it to return <code>3</code> instead of <code>a + b</code>: the symbols <code>a</code> and <code>b</code> are defined during the evaluation of <code>f</code> and <code>a+b</code> should evaluate to their sum.</p>
<p>Why is this viewpoint wrong? What's the right way to make it behave the way I want it to? (Something more clever than <code>f[p_,q_]:=x/.{a->p,b->q};</code>.) </p>
| Mr.Wizard | 121 | <h2>Analysis</h2>
<p>In <em>Mathematica</em> when a definition is <em>applied</em> the expressions (arguments) that match pattern objects on the left-hand-side (LHS) are substituted into the matching right-hand-side (RHS) names before it is evaluated. This is separate from the evaluation that does or does not take place at the time the rule or definition is created. This substitution-before-evaluation is invariant between <code>Set</code> and <code>SetDelayed</code>. </p>
<p>Please consider this example:</p>
<pre><code>ClearAll[f, a, b, x]
f[a_, b_] = (Print["one: ", a, b]; x);
x := Print["two: ", a, b];
f[1, 2]
{a, b} = {3, 4};
f[1, 2]
</code></pre>
<blockquote>
<p>one: ab</p>
<p>two: ab</p>
<p>two: 34</p>
</blockquote>
<ul>
<li><p>The use of <code>Set</code> in the definition creation causes the RHS to evaluate, printing "one:", but this <code>Print</code> is not part of the definition created as it does not remain in the evaluated form of <code>(Print["one: ", a, b]; x)</code>.</p></li>
<li><p>When the definition is used no <code>a</code> or <code>b</code> appear in the explicit unevaluated RHS (<code>x</code>) therefore no substutions are made. The RHS is then evaluated and the <code>Print</code> statement in the global definition of <code>x</code> fires.</p></li>
<li><p>Between the first and second time the function is used <code>a</code> and <code>b</code> are given global values. When the function is next called the same evaluation sequence takes place but when <code>x</code> evaluates to <code>Print["two: ", a, b]</code> this time <code>a</code> and <code>b</code> evaluate to <code>3</code> and <code>4</code>.</p></li>
</ul>
<hr>
<h2>Alternatives</h2>
<p>You asked for "something more clever than <code>f[p_,q_]:=x/.{a->p,b->q};</code>" but that may in essence be what you need. This code effectively performs the substitution <em>after</em> the RHS evaluation each time the definition is applied. If you want to use the argument values <em>during</em> evaluation you will need to use <a href="http://reference.wolfram.com/language/ref/Block.html" rel="nofollow noreferrer"><code>Block</code></a> or another function that works like <code>Block</code>. Here is an example:</p>
<pre><code>ClearAll[f, a, b, x]
f[aa_, bb_] := Block[{a = aa, b = bb}, x]
x = a + b;
f[1, 2]
x = 3 a + b^2;
f[1, 2]
</code></pre>
<blockquote>
<pre><code>3
7
</code></pre>
</blockquote>
<p>All that remains is to make construction of these definitions easier which we can do with a little meta-programming. Here is one method:</p>
<pre><code>SetAttributes[blockSet, HoldFirst]
blockSet[LHS_ := RHS_] :=
With[{subs =
Cases[Unevaluated[LHS],
Verbatim[Pattern][p_, x_] :> Append[Hold[p], Module[{p}, p]], -2]},
{
ReplaceAll[Hold[LHS], HoldPattern[#] :> #2 & @@@ subs],
Set @@@ Hold @@ subs
} /. {Hold[newLHS_], Hold[sets__]} :> (newLHS := Block[{sets}, RHS])
]
</code></pre>
<p>Now:</p>
<pre><code>Remove[f, a, b, x]
blockSet[
f[a_, b_] := x
]
?f
</code></pre>
<blockquote>
<pre><code>Global`f
f[a$752_, b$753_] := Block[{a = a$752, b = b$753}, x]
</code></pre>
</blockquote>
<p>And:</p>
<pre><code>x = a + b;
f[1, 2]
</code></pre>
<blockquote>
<pre><code>3
</code></pre>
</blockquote>
|
1,487,966 | <p>I have been looking at stereographic projections in books, online but they all seem...I don't know how else to put this, but very pedantic yet skipping the details of calculations.</p>
<p>Say, I have a problem here which asks;</p>
<blockquote>
<p>Let <span class="math-container">$n \geq 1$</span> and put <span class="math-container">$S^n=\{(x_0,x_1...,x_n) \in \mathbb{R}^{n+1}|{x_0}^2+...+{x_n}^2=1\}$</span> (So I understand this is a unit sphere in <span class="math-container">$n+1$</span> dimensions). Let <span class="math-container">$P=\{(1,0,...,0)\}$</span> and consider <span class="math-container">$S^n$</span>\ <span class="math-container">$P$</span> and <span class="math-container">$Y=\{(y_0...y_n)\in \mathbb{R}^{n+1}|y_0=0\}$</span> both with Euclidean metric. Thus <span class="math-container">$X$</span> is an <span class="math-container">$n$</span>-sphere with a point removed and <span class="math-container">$Y \cong \mathbb{R}^n$</span>.</p>
</blockquote>
<p>That is the set up. The problem I don't know how to approach is,</p>
<blockquote>
<p><span class="math-container">$i$</span>) For <span class="math-container">$x=(x_0,...,x_n) \in X$</span> and let <span class="math-container">$f(x)$</span> be a unique point of <span class="math-container">$Y$</span> such that <span class="math-container">$P=(1,0...0)$</span> and <span class="math-container">$x$</span>, <span class="math-container">$f(x)$</span> are collinear. So, find <span class="math-container">$ \lambda(x)$</span> such that <span class="math-container">$f(x)=\lambda(x)P+(1-\lambda(x))x$</span> fo some <span class="math-container">$\lambda(x) \in \mathbb{R}$</span>.</p>
<p><span class="math-container">$ii$</span>) For <span class="math-container">$y=(y_0,...y_n) \in Y$</span> let <span class="math-container">$g(y)$</span> be a unique point of <span class="math-container">$X$</span> such that <span class="math-container">$(1,0,...,0)$</span>, <span class="math-container">$y$</span> and <span class="math-container">$g(y)$</span> are collinear, by similar method to <span class="math-container">$i$</span>), find a formula for <span class="math-container">$g(y)$</span>.</p>
<p><span class="math-container">$iii$</span>) Prove that <span class="math-container">$f : X \rightarrow Y$</span> and <span class="math-container">$g : Y \rightarrow X$</span> are inverse to each other and deduce that <span class="math-container">$X$</span> is homeomorphic to <span class="math-container">$Y$</span></p>
</blockquote>
<p>Leaving aside part iii, I don't get how to do i.
I have seen some examples on removing the North Pole(which is NOT my case) but then the equation for <span class="math-container">$f(x)$</span> appears out of nowhere in those cases.
I don't know where and how they were obtained, no explanations and steps were given.</p>
<p>So, <span class="math-container">$\lambda(x)$</span> is a real number or, at least a scalar I understand. Given my conditions, I tried substituting the points to the given collinear form but found only that <span class="math-container">$\frac{y_0-x_0}{1-x_0} = \lambda(x)$</span> and also <span class="math-container">$\frac{x_i-y_i}{x_i} = \lambda(x)$</span> unless <span class="math-container">$i=0$</span>. Which doesn't make sense to me, as I found well, 2 different <span class="math-container">$\lambda(x)$</span>s (haven't I??)</p>
<p>I just don't get this stereographic projection thing "analytically". I have seen pictures and diagrams which visualises it and that's all very nice but algebraically/analytically, I cannot make sense of it.</p>
<p>Would anyone care helping me out at all?? Thank you so much....</p>
| Bernard | 202,857 | <p>For part i) you have to find the point of the line (Px) that lies in the hyperplane $x_0=0$, i. e. you have to solve for
$$\lambda +(1-\lambda)x_0=0,\tag{1}$$
since the straight line $(Px)$ has a parametic representation:
$$\lambda(1 ,0,\dots,0)+(1-\lambda)(x_0,x_1,\dots,x_n)=\bigl(\lambda +(1-\lambda)x_0, (1-\lambda)x_1, \dots,(1-\lambda)x_n\bigr).$$
Solving for $\lambda$ in $(1)$, we get
$$\lambda=\frac{x_0}{x_0-1}.$$</p>
|
3,349,630 | <p>If a,b,c are positive real numbers,prove that
<span class="math-container">$$ \frac{a}{b+2c} + \frac{b}{c+2a} + \frac{c}{a+2b} \ge 1 $$</span>
I tried solving and i have no idea how to proceed I mechanically simplified it it looks promising but im still stuck. This is from the excersice on Cauchy Schwartz Inequality.</p>
| Dr. Sonnhard Graubner | 175,066 | <p>Using Cauchy-Schwarz in Engel form we get
<span class="math-container">$$\frac{a^2}{ab+2ac}+\frac{b^2}{bc+ab}+\frac{c^2}{ac+2ac}\geq \frac{(a+b+c)^2}{3ab+3ac+3bc}\geq 1$$</span> if <span class="math-container">$$(a+b+c)^2\geq 3ab+3ac+3bc$$</span>
and this is <span class="math-container">$$a^2+b^2+c^2\geq ab+bc+ca$$</span></p>
|
190,948 | <p>So I have </p>
<pre><code>Emin[T_, d_] := If[T == 0, 0, 1/(-1 + E^(1/T)) - d/(-1 + E^(d/T))];
</code></pre>
<p>which I can solve wonderfully for </p>
<pre><code>Solve[Emin[T, 10] == 0.5, T, Reals]
{{T -> 0.910427}}
</code></pre>
<p>Now as you see <code>Emin[T_, d_]</code> is also dependent on d which I set to d=10 in the second line.</p>
<p>All I'd like is to now Plot this solve for different d.</p>
<p>I've got so far this:</p>
<pre><code>Tlist = Table[Solve[Emin[T, d] == 0.5, T, Reals] // Values // Flatten // N, {d, 5, 10}]
{{0.936864}, {0.919994}, {0.913886}, {0.911604}, {0.910747}, {0.910427}}
</code></pre>
<p>But as you see I cannot plot this because d is missing in those entries. </p>
<p>So can anyone either help me add the correct d value to the entries or maybe help me plot this even simpler? Thanks in advance !</p>
| Ulrich Neumann | 53,677 | <p>With the additional information <code>d01,2,3...10</code> try</p>
<pre><code>dT = Table[{d, T /. NSolve[Emin[T, d] == 0.5, T, Reals][[1]]}, {d, 1,10}]
(*{{1, T /. {}[[1]]}, {2, T /. {}[[1]]}, {3, 1.19888}, {4,0.988141},
{5, 0.936864}, {6, 0.919994}, {7, 0.913886}, {8,0.911604}, {9, 0.910747}, {10, 0.910427}}*)
Show[ContourPlot[Emin[T, d] == 0.5, {d, 0, 10}, {T, .5, 2},FrameLabel -> {T, d}], Graphics[Point[dT[[3 ;;]]]]]
</code></pre>
<p><a href="https://i.stack.imgur.com/73fcv.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/73fcv.jpg" alt="enter image description here"></a></p>
|
3,869,237 | <p>I know this is quite weird or it does not make much sense, but I was wondering, does <span class="math-container">$\int e^{dx}$</span> has any meaning or whether it makes sense at all? If it does means something, can it be integrated and what is the result?</p>
| zkutch | 775,801 | <p>I agree, that formally it make no sense, but as lunch exercise in mathematical fantasy , if we can give some sense to <span class="math-container">$(dx)^n$</span> as some measure, then we can imagine <span class="math-container">$\int e^{dx}=\int \sum \frac{(dx)^n}{n!}$</span>. Now the point is what is <span class="math-container">$(dx)^n$</span>..</p>
|
3,421,858 | <p><span class="math-container">$\sqrt{2}$</span> is irrational using proof by contradiction.</p>
<p>say <span class="math-container">$\sqrt{2}$</span> = <span class="math-container">$\frac{a}{b}$</span> where <span class="math-container">$a$</span> and <span class="math-container">$b$</span> are positive integers. </p>
<p><span class="math-container">$b\sqrt{2}$</span> is an integer. ----[Understood]</p>
<p>Let <span class="math-container">$b$</span> denote the smallest such positive integer.----[My understanding of this is </p>
<p>that were are going to assume b is the smallest possible integer such that <span class="math-container">$\sqrt{2}$</span> = <span class="math-container">$\frac{a}{b}$</span>, ... Understood]</p>
<p>Then <span class="math-container">$b^{*}$</span> := <span class="math-container">$b(\sqrt{2} - 1)$</span>----[I'm not sure I understand the point that's being made here,</p>
<p>from creating a variable <span class="math-container">$b^{*} = a - b$</span> ]</p>
<p>Next, <span class="math-container">$b^{*}$</span> := <span class="math-container">$b(\sqrt{2} - 1)$</span> is a positive integer such that <span class="math-container">$b^{*}\sqrt{2}$</span> is an integer.----[ I get that (<span class="math-container">$a - b$</span>) has to</p>
<p>be a positive integer, why does it follow that then <span class="math-container">$b^{*}\sqrt{2}$</span> is an integer?]</p>
<p>Lastly, <span class="math-container">$b^{*}<b$</span>, which is a contradiction.----[I can see that given <span class="math-container">$b^{*}$</span> := <span class="math-container">$b(\sqrt{2} - 1)$</span>, we then have </p>
<p><span class="math-container">$b^{*}<b$</span>, I don't get how that creates a contradiction]</p>
<p>Any help is appreciated, thank you. </p>
| Matt Samuel | 187,867 | <p>Simply calculate <span class="math-container">$$b(\sqrt2-1)\sqrt2=b(2-\sqrt2)=2b-b\sqrt2$$</span>
which is an integer. This is actually the heart of the proof. We assumed <span class="math-container">$b$</span> was the smallest integer satisfying <span class="math-container">$b\sqrt2$</span> being an integer, but we contradicted that by constructing a smaller one. </p>
|
4,383,800 | <p>I can already see that the <span class="math-container">$\lim_\limits{n\to\infty}\frac{n^{n-1}}{n!e^n}$</span> converges by graphing it on Desmos, but I have no idea how to algebraically prove that with L’Hopital’s rule or induction. Where could I even start with something like this?</p>
<p>Edit: For context, I came across this limit while studying the series expansion for the Lambert W Function, <span class="math-container">$W(x)= \sum_{n=0}^{\infty}\frac{(-n)^{n-1}x^n}{n!}$</span> . By the ratio test, it is clear that <span class="math-container">$|x|<\frac1e$</span> in order to converge, but I needed to use the Alternating Series Test to see whether this series converges at <span class="math-container">$x= \pm\frac1e$</span>. Finding <span class="math-container">$\lim_\limits{n\to\infty}|a_n|$</span> is the first step of the test.</p>
| Community | -1 | <p><strong>Hint:</strong></p>
<p>Taking the logarithm, you need the limit of</p>
<p><span class="math-container">$$(n-1)\log n-\sum_{k=1}^n\log n-n.$$</span></p>
<p>Then the summation can be estimated by an integral and the second term is</p>
<p><span class="math-container">$$\sim n(\log n-1).$$</span></p>
<p>A more precise bounding of the integral will lead to a formal proof.</p>
|
2,529,682 | <p>Right now I'm stuck on the following problem, since I feel like I should be using total probability, but I dont know what numbers to use as what.</p>
<p>Let's say there's a population of students. In this population:</p>
<p>30% have a bike</p>
<p>10% have a motorcycle</p>
<p>12% have a car.</p>
<p>8% have a bike AND a motorcycle</p>
<p>7% car and a bike</p>
<p>4% have a motorcycle and a car</p>
<p>2% have a bike, a car and a motorcycle</p>
<p>What percentage of students owns no vehicles?</p>
<p>I draw myself a Venn diagram, but I can't get my mind around the problem. My thinking right now is just substracting each percentage off 100%, but that just feels wrong.</p>
<p>Using total probability feels wrong aswell, since I have no idea what to calculate. I want to calculate P(A) = people that own a vehicle, but then P(A|Hi) doesn't really have a value.</p>
<p>Anyone has any ideas?</p>
| Arthur | 15,500 | <p>The derivative is restricted to that domain precisely because the original function is. How would you evaluate the derivative of $\ln x$ at negative $x$? You can't, and therefore the derivative is only defined for positive $x$.</p>
<p>The derivative of $\ln x$ does have a very natural extension to the negative numbers. However, that extension has little to do with the original function.</p>
|
1,921,114 | <p><img src="https://i.stack.imgur.com/D8IcM.jpg" alt="enter image description here"></p>
<p>So I solved this system without using matrices, just by (sort of) reverting to high school math instincts.
$$w+x=-5$$
$$x+y=4$$
$$y+z=1$$
$$w+z=8$$
From this, I got $w=-9,x=4,y=0$ and $z=1$. How would I convert this back to what appears to be a matrix form?</p>
| Sam Waggoner | 1,145,242 | <p>In the equations, you have four variables, <em>x1</em>, <em>x2</em>, <em>x3</em>, and <em>x4</em>. The challenge is to 1) solve for each, and 2) state them in terms of some single constant s, instead of the other variables.
The first step is to solve for each of the variables, which will result in this:</p>
<p><span class="math-container">$x_1 = -5 - x_2$</span></p>
<p><span class="math-container">$x_2 = 4 - x_3$</span></p>
<p><span class="math-container">$x_3 = 1 - x_4$</span></p>
<p><span class="math-container">$x_4 = -8 - x_1$</span></p>
<p>Now we need to put these in terms of <span class="math-container">$s$</span>. So, arbitrarily set <span class="math-container">$x_1$</span> to <span class="math-container">$s$</span>.</p>
<p>Thus, <span class="math-container">$x_4$</span> can be restated:</p>
<p><span class="math-container">$x_4 = 8 - s$</span></p>
<p>And then you can restate <span class="math-container">$x_3$</span> using the updated <span class="math-container">$x_4$</span>:</p>
<p><span class="math-container">$x_3 = 1 - (-8 - s)$</span></p>
<p>And simplify to:</p>
<p><span class="math-container">$x_3 = 9 + s$</span></p>
<p>Now you can restate <span class="math-container">$x_2$</span> using the updated <span class="math-container">$x_3$</span> in the same way:</p>
<p><span class="math-container">$x_2 = 4 - (9 + s)$</span></p>
<p>And simplify to:</p>
<p><span class="math-container">$x_2 = -5 - s$</span></p>
<p>Restate <span class="math-container">$x_1$</span> using the updated <span class="math-container">$x_2$</span> in the same way:</p>
<p><span class="math-container">$x_1 = -5 - (-5 - s)$</span></p>
<p>And simplify to:</p>
<p><span class="math-container">$x_1 = 0 + s$</span></p>
<p>Our four simplified equations in terms of s are:</p>
<p><span class="math-container">$x_1 = 0 + s$</span></p>
<p><span class="math-container">$x_2 = -5 - s$</span></p>
<p><span class="math-container">$x_3 = 9 + s$</span></p>
<p><span class="math-container">$x_4 = 8 - s$</span></p>
<p>Which can be formatted in the way of the solution as:</p>
<p><span class="math-container">$
\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4 \\
\end{bmatrix}
$</span>
<span class="math-container">$=$</span>
<span class="math-container">$
\begin{bmatrix}
0 \\
-5 \\
9 \\
8 \\
\end{bmatrix}
$</span>
<span class="math-container">$ + s$</span>
<span class="math-container">$
\begin{bmatrix}
1 \\
-1 \\
1 \\
-1 \\
\end{bmatrix}
$</span></p>
|
2,489,988 | <p>A sequence of numbers is formed from the numbers $1, 2, 3, 4, 5, 6, 7$ where all $7!$ permutations are equally likely. What is the probability that anywhere in the sequence there will be, at least, five consecutive positions in which the numbers are in increasing order?</p>
<p>I approached this problem in the following way, but I am wondering if there is a better way, since my approach is quite complicated.</p>
<p><strong>My Approach:</strong> There are three possibilities: a sequence have $7$ consecutive positions in which numbers increase, have $6$ consecutive positions in which numbers increase, and $5$ consecutive positions in which numbers increase.</p>
<p>There is only $1$ sequence that have $7$ consecutive positions. Namely, the sequence $(1,2,3,4,5,6,7)$.</p>
<p>There are $12$ sequences that have $6$ consecutive positions. Namely, we choose $1$ number from $(1,2,3,4,5,6,7)$, and move it to either sides. As an illustration, if we choose $3$, then we can get $(3,1,2,4,5,6,7)$ or $(1,2,4,5,6,7,3)$.</p>
<p>Now consider when there are $5$ consecutive positions in which numbers increase. We choose $2$ numbers that are <em>not</em> in the increasing subsequence. </p>
<p>If $1$ and $7$ are not chosen, we can place them in front of the subsequence, of after. For example, if we choose $(2,5)$, then we will have $(2,5,1,3,4,6,7)$,$(5,2,1,3,4,6,7)$, $(1,3,4,6,7,2,5)$ and $(1,3,4,6,7,5,2)$. This is $\binom{5}{2}\times4$.</p>
<p>Then I'm not sure how to proceed when we choose $1$ and/or $7$?</p>
| CogitoErgoCogitoSum | 52,938 | <p>${}_7 C_2 = 21$, are the number of ways of choosing two out of the seven numbers. We pull them out. The remaining five retain their increasing order. So now find the number of ways you can replace those two values at either the beginning and/or end of the sequence, and multiply this result by 21. If they happen to be replaced where they started then the length of the increasing sub-sequence grows, but this is allowed anyway. Two elements can be replaced in the sequence in 6 ordered ways, by my count. That is $2!=2$ orders times 3 ways of splitting them up. In total then I figure there are 126 ways of creating sequences with 5 or more increasing elements anywhere in the sequence. Divide by your $7!=5040$ permutations for a probability.</p>
|
237,142 | <p>I am having a problem with the final question of this exercise.</p>
<p>Show that $e$ is irrational (I did that). Then find the first $5$ digits in a decimal expansion of $e$ ($2.71828$).</p>
<p>Can you approximate $e$ by a rational number with error $< 10^{-1000}$ ? </p>
<p>Thank you in advance</p>
| Belgi | 21,335 | <p>I don't think we have to really know anything about $e$ to say that
we can approximate it with a rational number with an error less then
$10^{-1000}$.</p>
<p>Say $e=a_0.a_1 a_2 \ldots$</p>
<p>There is clearly a rational number $q=a_0.a_1 \ldots a_{1001}$ and
$|q-e|\leq10^{-1000}$ (note that the difference is bounded by a geometric
sum)</p>
|
1,566,111 | <p>prove $(n)$ prime ideal of $\mathbb{Z}$ iff $n$ is prime or zero</p>
<hr>
<p><strong>Defintions</strong></p>
<p>Def of prime Ideal (n)
$$ ab\in (n) \implies a\in(n) \vee b\in(n) $$
Def 1] integer n is prime if $n \neq 0,\pm 1 $ and only divisors are $\pm n,\pm 1$ </p>
<p>Def 2 of n is prime] If $n\neq0,\pm1$ only divisors of n are $\pm1,\pm n$
$$ n|ab \implies n|a \vee n|b $$</p>
<hr>
<p>$\Rightarrow $] (<em>Prime Ideal $(n)$ of $\mathbb{Z} $$\Rightarrow$ $n$ prime or zero</em>)</p>
<p>Now consider the case where $(n)\neq (0)$. that is $n\neq 0$ </p>
<p>Using the def of prime Ideals $$\begin{aligned} ab \in (n) \implies a \in(n) \vee b \in (n) \end{aligned} $$</p>
<p>Well, If an element $x\in(n) \iff x=q*n \iff n|x$ </p>
<p>$$n|ab \implies n|a \vee n|b$$
So, $n$ is prime ,nonzero. </p>
<p>In the case that $(n)=(0)$ clearly $n=0$ since there are no zero divisors in $\mathbb{Z}$</p>
<p>$\Leftarrow$] <em>(n is prime or zero $\Rightarrow $ $(n)$ is a prime ideal of $\mathbb{Z}$)</em></p>
<p>Consider the case where $p is prime$ so $n\neq 0, \pm1$</p>
<p>$$ \begin{aligned}
n|ab &\implies n|a \vee n|b \\
ab \in (n) &\implies a \in (n) \vee b\in(n) \end{aligned}$$</p>
<p>in the case $n=0$, since $\mathbb{Z}$ has no zero divisors
$$ab=0 \implies a=0 \vee b=0 $$
So (0) is a prime ideal. </p>
<hr>
<p>Concern if this prove holds, also would be surprised if this question is not out there in this site. Did a search and clicked on similar questions and could not find it.And of course any other ways to prove it.</p>
| Robert Soupe | 149,436 | <p>Another way would be to show that $\mathbb{Z}$ is a principal ideal domain or that it has unique factorization. Don't you also need the definition that a prime ideal has to be properly contained within the whole ring?</p>
<p>If $n = \pm 1$, then $\langle n \rangle = \mathbb{Z}$ and thus it can't be a prime ideal. If $n$ is composite and divisible by some prime $p$, then $\langle n \rangle$ is properly contained within $\langle p \rangle$ and thus $\langle n \rangle$ is not a prime ideal either.</p>
<p>And then you just proceed with what you have already demonstrated.</p>
|
2,130,658 | <p>How would I go about proving this mathematically? Having looked at a proof for a similar question I think it requires proof by induction. </p>
<p>It seems obvious that it would be even by thinking about the first few cases. As for $n=0$ there will be no horizontal dominoes which is even, and for $n=1$ there can only be one vertical domino so there are $0$ horizontal dominoes, which is again even. Then for $n=2$ you can have either two horizontal or two vertical dominoes which again gives $0$ or $2$ horizontal dominoes which is again an even number. And so on for n greater than $2$.</p>
<p>I would like to prove that the number of way of dividing a 2-high-by-n-wide rectangle into dominoes so that $2j$ dominoes are horizontal is ${n-j\choose j}$ and deduce that $U_n$ (where $U_n$ is the number of ways to divide a 2-high-by-n-wide rectangle into 2-wide-by-1-high dominoes) = $$\sum_j {n-j\choose j}$$ where this sum is over all the integers $j$ with $0\le j\le \frac{n}{2}$.</p>
<p>I understand that trivially for a 2-high-by-n-wide rectangle you can divide it by exactly $2j=n$ horizontal dominoes or by $n$ vertical dominoes or some combination of vertical and horizontal dominoes, but how can I use this knowledge to construct the proof?</p>
| Brian Tung | 224,454 | <p>We assume, given the description of the problem in the OP, that the rectangle is a horizontal strip of height $2$ and width $n$.</p>
<p><strong>Proof by Contradiction.</strong> Suppose we had a covering of the $2 \times n$ rectangle containing an odd number of horizontal dominoes. Then either the upper row or the lower row contains more dominoes than the other; without loss of generality, assume the upper row contains more. Then the upper-row horizontal dominoes cover more squares than the lower-row ones do. But vertical dominoes must cover an equal number of upper and lower dominoes. So more upper-row squares are covered than lower-row squares, violating the definition of the rectangle. Ergo, no such covering is possible.</p>
<p><strong>Proof by Induction.</strong> As you point out the number of horizontal dominoes is clearly zero (and therefore even) for $n = 0$ or $1$. For $n \geq 2$, consider the last two squares on the right. There are two cases: In one case, they are covered by a single vertical domino, in which case the number of horizontal dominoes is equal to the number in the first $2 \times n-1$. In the other case, they are covered by two horizontal dominoes, in which case the number of horizontal dominoes is two more than the number in the first $2 \times n-2$. In either case, the parity of the number of horizontal dominoes is unaffected, so the result obtains.</p>
<hr>
<p>You've added a second question: how to show that the number of ways to cover the $2$-by-$n$ rectangle with $2j$ horizontal dominoes (and therefore $n-j$ vertical ones).</p>
<p>The easiest way I can think of to proceed is to observe that each row must look identical. Vertical dominoes perforce look identical in each row, and if the horizontal ones don't line up one on top of the other, then there will be an extra space left over somewhere. (Actually, two extra spaces, but at any rate—extra spaces.)</p>
<p>Therefore, we can just count the number of ways that we can cover a single strip of $n$ squares with pieces of either one square (those represent half a vertical domino) or two squares (those represent a horizontal domino with a matching domino in the other row). If there are $j$ two-square pieces altogether in the strip, then there must be $n-j$ pieces in the row, because the two-square pieces collectively cover $2j$ squares, leaving the remaining $n-2j$ squares to be covered by $n-2j$ one-square pieces: $n-2j+j = n-j$ pieces in all.</p>
<p>Each arrangement corresponds to choosing $j$ of those $n-j$ pieces to be the two-square pieces, so the number of ways of doing that is</p>
<p>$$
\binom{n-j}{j}
$$</p>
<p>I'm assuming you can handle the rest of the deduction.</p>
|
1,088,338 | <p>There are at least a few things a person can do to contribute to the mathematics community without necessarily obtaining novel results, for example:</p>
<ul>
<li>Organizing known results into a coherent narrative in the form of lecture notes or a textbook</li>
<li>Contributing code to open-source mathematical software</li>
</ul>
<p>What are some other ways to make auxiliary contributions to mathematics?</p>
| Community | -1 | <p>Even people with no mathematical background at all can contribute to mathematics.</p>
<p>One obvious way is by running software such as the <a href="http://www.mersenne.org/">GIMPS</a> client for finding Mersenne primes, though the value of such primes to theoretical mathematics is debatable.</p>
<p>Another, vastly more important one is to contribute anything, literally <em>anything</em> at all to human civilization. In reality, it is that complex civilization which makes mathematics possible in the first place. Because they do not have to scramble for food in the dirt at the risk of their lives, people have the time to explore "non-productive" subjects such as mathematics. The number of great minds who could have become Eulers or Euclids but did not because they had to work as swineherds or died of a flu at age nine surely outnumbers those who actually did ever do any mathematical work. By keeping civil order, hospitals and agriculture going, you keep mathematics going as well.</p>
|
1,088,338 | <p>There are at least a few things a person can do to contribute to the mathematics community without necessarily obtaining novel results, for example:</p>
<ul>
<li>Organizing known results into a coherent narrative in the form of lecture notes or a textbook</li>
<li>Contributing code to open-source mathematical software</li>
</ul>
<p>What are some other ways to make auxiliary contributions to mathematics?</p>
| Jair Taylor | 28,545 | <p>It is useful to contribute to databases of mathematical objects such as <a href="http://oeis.org/">Sloane's Encyclopedia of Integer Sequences</a>, <a href="http://findstat.org/">FindStat</a> and others. These resources have become invaluable for mathematicians searching for references to objects that they don't know the name of. Using these databases, mathematicians are able to find connections between seemingly distant fields if both make use of a certain integer sequence or other object in a database. For more information, see Billey and Tenner's manifesto <a href="http://www.ams.org/notices/201308/rnoti-p1034.pdf">here</a>.</p>
|
2,666,568 | <p>I have a dynamical system: $\dot{\mathbf x}$= A$\mathbf x$ with $\mathbf x$=
$\bigl( \begin{smallmatrix} x \\ y\end{smallmatrix} \bigr)$ and A =
$\bigl( \begin{smallmatrix} 3 & 0 \\ \beta & 3 \end{smallmatrix} \bigr). \beta$ real, time-independent.</p>
<p>I calculated the eigenvalue $\lambda$ = 3 with the algebraic multiplicity of 2.</p>
<p>The <strong>first question</strong> is about eigenvectors when $\beta = 0$ and when $\beta \neq$ 0:</p>
<p>1) when $\beta$ = 0, I have $\bigl( \begin{smallmatrix} 0 & 0 \\ \beta & 0 \end{smallmatrix} \bigr) \bigl( \begin{smallmatrix} x \\ y\end{smallmatrix} \bigr)$ = $\bigl( \begin{smallmatrix} 0 & 0 \\ 0 & 0 \end{smallmatrix} \bigr) \bigl( \begin{smallmatrix} x \\ y\end{smallmatrix} \bigr)$ = $\bigl( \begin{smallmatrix} 0 \\ 0\end{smallmatrix} \bigr)$.
Does this allow for eigenvectors calculation? Does it tell me anything at all?</p>
<p>2) when $\beta \neq$ 0, I have $\bigl( \begin{smallmatrix} 0 & 0 \\ \beta & 0 \end{smallmatrix} \bigr) \bigl( \begin{smallmatrix} x \\ y\end{smallmatrix} \bigr)$ = $\bigl( \begin{smallmatrix} 0 \\ 0\end{smallmatrix} \bigr)$, so my eigenvector is $\bigl( \begin{smallmatrix} 0 \\ 1\end{smallmatrix} \bigr)$ and A is defective? Is there any other eigenvector?</p>
<p>The <strong>other question</strong> is further on case 2) when $\beta$ = 3. I am to find any fixed points + their stability, but first I am wondering whether I did the above correctly. I am not sure how to approach it, thought about a trajectory expression, but I am confused by the defective A.</p>
<p>Thanks!</p>
| gt6989b | 16,192 | <p>When $\beta = 0$, you have $A = 3I$ so the eigenvectors are the standard basis and the eigenvalues are both $1$.</p>
<p>When $\beta \ne 0$, the matrix is not diagonalizable, it only has a Jordan form.</p>
|
4,231,509 | <p>I'm trying to prove that the group <span class="math-container">$(\mathbb{R}^*, \cdot)$</span> is not cyclic (similar to [1]). My efforts until now culminated into the following sentence:</p>
<blockquote>
<p>If <span class="math-container">$(\mathbb{R}^*,\cdot)$</span> is cyclic, then <span class="math-container">$\exists x \in \mathbb{R}^*$</span> such that <span class="math-container">$x \cdot x \neq x \in \mathbb{R}^*$</span>.</p>
</blockquote>
<p>The assumption written above is only not true for the neutral element on <span class="math-container">$\mathbb{R}^*$</span>. Are there any follow-ups that I should do to improve that sentence?</p>
<p>[1] <a href="https://math.stackexchange.com/questions/1491510/show-that-q-and-r-are-not-cyclic-groups">Show that (Q, +) and (R, +) are not cyclic groups.</a></p>
| Infinity_hunter | 826,797 | <p>Suppose <span class="math-container">$(\mathbb{R}^*, \cdot)$</span> is cyclic then there exist <span class="math-container">$x\in \mathbb{R}^*$</span> such that <span class="math-container">$\mathbb{R}^* = \langle x \rangle =\{x^n \vert \, n \in \mathbb{Z}\}$</span>. However we see that the set <span class="math-container">$\{x^n \vert \, n \in \mathbb{Z}\}$</span> is <a href="https://en.wikipedia.org/wiki/Countable_set" rel="nofollow noreferrer">countable</a> but <span class="math-container">$\mathbb{R}^*$</span> is <a href="https://en.wikipedia.org/wiki/Uncountable_set" rel="nofollow noreferrer">uncountable</a> which is a cotradiction.</p>
<p>A similar argument shows that <span class="math-container">$\mathbb{R}^*$</span> can not be generated by a finite number of elements , that is for any reals <span class="math-container">$x_i$</span> (<span class="math-container">$i =1,2,\ldots ,n$</span>) we must have <span class="math-container">$\mathbb{R}^* \ne \langle x_1, x_2 , \ldots , x_n \rangle$</span> for all positive integer <span class="math-container">$n$</span>.</p>
|
1,722,287 | <p>So far I know that when matrices A and B are multiplied, with B on the right, the result, AB, is a linear combination of the columns of A, but I'm not sure what to do with this. </p>
| Tryss | 216,059 | <p>If you know the rank-nullity theorem, it's easy :</p>
<p>$\forall x \in \text{Ker} B, ABx = A(Bx) = A(0) = 0$</p>
<p>So $x\in \text{Ker} AB$, hence $\text{Ker} B \subset \text{Ker} AB$, that imply that </p>
<p>$\text{dim Ker} B \leq \text{dim Ker} AB$ , and by the rank nullity theorem, you get that $\text{rank} B \geq \text{rank} AB$</p>
|
1,007,399 | <p>I came across following problem</p>
<blockquote>
<p>Evaluate $$\int\frac{1}{1+x^6} \,dx$$</p>
</blockquote>
<p>When I asked my teacher for hint he said first evaluate</p>
<blockquote>
<p>$$\int\frac{1}{1+x^4} \,dx$$</p>
</blockquote>
<p>I've tried to factorize $1+x^6$ as</p>
<p>$$1+x^6=(x^2 + 1)(x^4 - x^2 + 1)$$
and then writing</p>
<p>$$I=\int\frac{1}{1+x^6} \,dx=\int\frac{1}{(x^2 + 1)(x^4 - x^2 + 1)} \,dx=\int\frac{1+x^2-x^2}{(x^2 + 1)(x^4 - x^2 + 1)} \,dx$$
$$I=\int\frac{1}{x^4 - x^2 + 1} \,dx-\int\frac{x^2}{(x^2 + 1)(x^4 - x^2 + 1)} \,dx$$</p>
<p>However $$x^4-x^2+1=\left(x^2-\frac12\right)^2+\frac{3}{4}$$
But I can't see how it helps</p>
<p>I've also tried to reverse engineer the <a href="http://www.wolframalpha.com/input/?i=integrate+%281%29%2F%281%2Bx%5E6%29" rel="noreferrer">solution given by Wolfram Alpha</a></p>
<p>And I need to have terms similar to<br>
$$\frac{x^2-1}{x^4-x^2+1} \quad , \quad \frac{1}{1+x^2} \quad , \quad \frac{1}{(x+c)^2+1}\quad , \quad \frac{1}{(x+c)^2+1}$$ in integrand, How can I transform my cute looking integrand into these huge terms?</p>
<p>Since in exams I will neither have access to WA nor time to reverse engineer the solution moreover it does not seem intuitive,is there any way to solve this problem with some nice tricks or maybe substitutions?</p>
| John_dydx | 82,134 | <p>As suggested, you can transform your denominator $1+x^6$ into $(1+x^2)(1-x^2+x^4)$. You can then use partial fractions to split the following into something that can be integrated:</p>
<p>$\Large\int \frac{1}{(1+x^2)(1-x^2+x^4)}dx$ </p>
<p>Hope that helps. It looks like a very nasty and messy integration to do. Best wishes.</p>
|
1,007,399 | <p>I came across following problem</p>
<blockquote>
<p>Evaluate $$\int\frac{1}{1+x^6} \,dx$$</p>
</blockquote>
<p>When I asked my teacher for hint he said first evaluate</p>
<blockquote>
<p>$$\int\frac{1}{1+x^4} \,dx$$</p>
</blockquote>
<p>I've tried to factorize $1+x^6$ as</p>
<p>$$1+x^6=(x^2 + 1)(x^4 - x^2 + 1)$$
and then writing</p>
<p>$$I=\int\frac{1}{1+x^6} \,dx=\int\frac{1}{(x^2 + 1)(x^4 - x^2 + 1)} \,dx=\int\frac{1+x^2-x^2}{(x^2 + 1)(x^4 - x^2 + 1)} \,dx$$
$$I=\int\frac{1}{x^4 - x^2 + 1} \,dx-\int\frac{x^2}{(x^2 + 1)(x^4 - x^2 + 1)} \,dx$$</p>
<p>However $$x^4-x^2+1=\left(x^2-\frac12\right)^2+\frac{3}{4}$$
But I can't see how it helps</p>
<p>I've also tried to reverse engineer the <a href="http://www.wolframalpha.com/input/?i=integrate+%281%29%2F%281%2Bx%5E6%29" rel="noreferrer">solution given by Wolfram Alpha</a></p>
<p>And I need to have terms similar to<br>
$$\frac{x^2-1}{x^4-x^2+1} \quad , \quad \frac{1}{1+x^2} \quad , \quad \frac{1}{(x+c)^2+1}\quad , \quad \frac{1}{(x+c)^2+1}$$ in integrand, How can I transform my cute looking integrand into these huge terms?</p>
<p>Since in exams I will neither have access to WA nor time to reverse engineer the solution moreover it does not seem intuitive,is there any way to solve this problem with some nice tricks or maybe substitutions?</p>
| Quanto | 686,284 | <p>With <span class="math-container">$1+x^6= (1+x^2)(x^4-x^2+1)$</span>, decompose the integrand</p>
<p><span class="math-container">\begin{align}
& \int \frac{dx}{1+x^6}
=\frac13\int \left( \frac{1}{1+x^2}+\frac12\frac{x^2+1}{x^4-x^2+1}- \frac32\frac{x^2-1}{x^4-x^2+1}\right)dx \\
&\hspace{15mm}=\frac13\int \frac{dx}{1+x^2}+\frac16\int \frac{d(x-\frac1x)}{(x-\frac1x)^2+1}dx - \frac12\int \frac{d(x+\frac1x)}{(x+\frac1x)^2-3} dx \\
\end{align}</span></p>
|
637,819 | <p>$$x\in(\cap F)\cap(\cap G)=[\forall A\in F(x\in A)]\land[\forall A\in G(x\in A)]$$</p>
<p>Since the variable $A$ is bounded by universal quantifier, it is regarded as bounded variable, according to the rules, the variable is free to change to other letters while the meaning statement remains unchanged. But,the above statements mention two different families of sets, $F$ and $G$, why is it correct to state the sets of $F$ and $G$ by using the same letter $A$, for the first $A$ in the first part of the conjunction stands for sets in $F$ while the latter stands for sets in $G$? Isn't different letters should be used to refer those sets($A$ for $F$ while $B$ for $G$)? I am extremely confused with the usage of bound variables. Please explain, thanks!</p>
| Mauro ALLEGRANZA | 108,274 | <p>In a quantified expression like $\forall x A(x)$ the occurrences of the variable $x$ into the formuala $A(x)$ are <em>bounded</em> by the quantifier (i.e. $\forall x$) becuase they occur into the <strong>scope</strong> of the quantifier.</p>
<p>So, in the example :</p>
<blockquote>
<p>$\forall x (A(x) \land B(x))$</p>
</blockquote>
<p>both the occurence of $x$ in $A$ <strong>and</strong> the occurrence of $x$ in $B$ are <em>bound</em>.</p>
<p>In the expression :</p>
<blockquote>
<p>$\forall x A(x) \land B(x)$</p>
</blockquote>
<p>I have renoved the parentheses, so "restricting" the scope of the quantifier: now the $x$ in $A(x)$ is <em>bound</em>, but the $x$ in $B(x)$ is <em>free</em>.</p>
<p>When you consder the formula :</p>
<blockquote>
<p>$\forall x A(x) \land \forall x B(x)$</p>
</blockquote>
<p>two different quantifiers act one on $A(x)$ and one on $B(x)$; so, both occurences of $x$ are bound, but the fact that the two quantifiers use both $x$ is immaterial. The "effect" of the quantifier "vanish" outside its scope.</p>
<p>The last formula is equivalent to :</p>
<blockquote>
<p>$\forall x A(x) \land \forall y B(y)$</p>
</blockquote>
<p>because the variable in the quantifier preceding $A(x)$ is not "linked" to the variable in the quantifier preceding $B(x)$.</p>
|
592,560 | <p>Let G be an abelian group. Show that, if G is not cyclic, then for all $x\in G$, there is a divisor $d$ of $n = |G|$ which is strictly smaller than n satisfying $x^d=1$. </p>
<p>I'm guessing that this is a consequence of Lagrange's Theorem. We can have that G is a disjoint union of left cosets that all have the same cardinality. So $|H| < |G|$ since G is composed with more than just one left coset. By Lagrange's Theorem, we have that $|H|=d$ and then $d$ divides $n$. However the "if G is not cyclic" part is bothering me. Does the fact that G is not cyclic put a restriction?</p>
| amWhy | 9,003 | <p>You've shown that $d$ divides $n$, but you haven't shown that $d \neq n$ (i.e., you haven't shown that $d$ is strictly smaller than $n$). </p>
<p>This is where the "non-cyclic" condition on $G$ comes in. Suppose for the sake of contradiction that $H = G$, and that $H$ is the only non-trivial subgroup of $G$, and hence $|H| = d\mid n = |G|$ and $d = n$ for every non-identity element $h\in H$. Then $h^n = 1$, and no $m$, $1\lt m \lt n$ exists such that $h^m = n$. But then $h$ generates $H = G$, and hence $G$ is cyclic. Contradiction.</p>
|
1,798,261 | <p>what is multilinear coefficient? I heard it a couple of times and I tried to google it, all I am getting is multiple linear regression.
I am confused at this point. </p>
| Zelos Malum | 197,853 | <p>By what you write we have that $c=0$ which means we've already lost 1 dimension, then we have $a+b=0$ which means it's a line only, a line is definitionally a subspace of the 3 dimensional space.</p>
|
912,176 | <p>If $y,z$ are elements of an archimedean field $F$ and if $y<z$, then there is a rational element $r$ of $F$ such that $y<r<z$</p>
<p>The proof begins with saying that it is no loss of generality that we assume that $0<y<z$</p>
<p>I don't understand well why this the case. Please guide me .</p>
<p>I think there will be loss of generality because suppose : $z>0$ but still it's possible that $y<0$ i.e. $z \in P ~;~ y \notin P$ where $P$ is a positive class in $F$</p>
<p>Thank you for your help.</p>
| Adam Hughes | 58,831 | <p>It's not because in an ordered field if you have arbitrary elements $a,b,c\in F$ then</p>
<p>$$a<b\iff a+c<b+c$$</p>
<p>Hence if $0<y$, for example, then you can just add $1-y$ to both sides to get</p>
<p>$$0<1<z+1-y$$</p>
<p>and you have $y'=1, z'=z+1-y$ so that your new question is only about positives, and is <strong>equivalent</strong>.</p>
<p>In particular, say that $r$ is rational between <em>any</em> $y,z\in F$ and say $-N<y$ is an <em>integer</em> less than $y$, then</p>
<p>$$0<y+N<r+N<z+N$$</p>
<p>so that</p>
<p>$$\begin{cases} y'=y+N\\
r'=r+N \\
z'=z+N
\end{cases}$$</p>
<p>and $y<r<z\iff y'<r'<z'$, so that things work the same with positives as with any choice. Also, clearly $r'$ is rational.</p>
|
3,756,436 | <p>Recently I was doing a physics problem and I ended up with this quadratic in the middle of the steps:</p>
<p><span class="math-container">$$ 0= X \tan \theta - \frac{g}{2} \frac{ X^2 \sec^2 \theta }{ (110)^2 } - 105$$</span></p>
<p>I want to find <span class="math-container">$0 < \theta < \frac{\pi}2$</span> for which I can later take the largest <span class="math-container">$X$</span> value that solves this equation, i.e. optimize the implicit curve to maximize <span class="math-container">$X$</span>.</p>
<p>I tried solving this by implicit differentiation (assuming <span class="math-container">$X$</span> can be written as a function of <span class="math-container">$\theta$</span>) with respect to <span class="math-container">$\theta$</span> and then by setting <span class="math-container">$\frac{dX}{d\theta} = 0$</span>:</p>
<p><span class="math-container">\begin{align}
0 &= X \sec^2 \theta + \frac{ d X}{ d \theta} \tan \theta - \frac{g}{2} \frac{ 2 \left( X \sec \theta \right) \left[ \frac{dX}{d \theta} \sec \theta + X \sec \theta \tan \theta \right]}{ (110)^2 } - 105 \\
0 &= X \sec^2 \theta - \frac{g}{2} \frac{ 2( X \sec \theta) \left[ X \sec \theta \tan \theta\right] }{ (110)^2 } \\
0 &= 1 - \frac{ Xg \tan \theta}{(110)^2} \\
\frac{ (110)^2}{ g \tan \theta} &= X
\end{align}</span></p>
<p>This is still not an easy equation to solve. However, one of my friends told we could just take the discriminant of the quadratic in terms of <span class="math-container">$X$</span>, and solve for <span class="math-container">$\theta$</span> such that <span class="math-container">$D=0$</span>.</p>
<p>Taking discriminant and equating to 0, I get</p>
<p><span class="math-container">$$ \sin \theta = \frac{ \sqrt{2 \cdot 10 \cdot 105} }{110}$$</span></p>
<p>and, the angle from it is, 24.45 degrees</p>
<p>I tried the discriminant method, but it gave me a different answer from the implicit differentiation method. I ended up with two solutions with the same maximum value of <span class="math-container">$X$</span> but different angles: <span class="math-container">$\theta =24.45^\text{o}$</span> and <span class="math-container">$X=1123.54$</span> (from discriminant method), and
<span class="math-container">$\theta = 47^\text{o}$</span> and <span class="math-container">$X=1123.54$</span> (from implicit differentiation).</p>
<p>I later realized the original quadratic can only have solutions if <span class="math-container">$D(\theta) > 0$</span>, where <span class="math-container">$D$</span> is the discriminant. Using the discriminant, I can find a lower bound on the angle. Once I have the lower bound, if I can prove that <span class="math-container">$X$</span> decreases monotonically as a function of <span class="math-container">$\theta$</span>, then I can use the lower bound for further calculations of <span class="math-container">$\theta$</span>.</p>
<p>So then I used the implicit function theorem and got</p>
<p><span class="math-container">$$ \frac{dX}{ d \theta} =- \frac{X \sec^2 \theta -\frac{g X^2}{2 (110)^2} 2 \sec \theta ( \sec \theta \tan \theta) } {\tan \theta - \frac{g \sec^2 \theta}{2 (110)^2} 2X }$$</span></p>
<p>Now the problem here is that I can't prove this function is in monotonic in terms of <span class="math-container">$\theta$</span> as the implicit derivative is a function of both <span class="math-container">$\theta$</span> and <span class="math-container">$X$</span>.</p>
| Anatoly | 90,997 | <p>I reworded the problem setting <span class="math-container">$X=y$</span> and <span class="math-container">$\theta=x$</span>. Considering <span class="math-container">$g=9.81$</span>, solving the quadratic equation for <span class="math-container">$y$</span> gives the two solutions</p>
<p><span class="math-container">$$\displaystyle y =-\frac{ 1100}{981} \cos x \\ \left( \sqrt{10} \sqrt{121000 \tan^2 x - 20601 \sec^2 x} - 1100 \tan x \right)$$</span></p>
<p><span class="math-container">$$\displaystyle y =\frac{ 1100}{981} \cos x \\ \left( \sqrt{10} \sqrt{121000 \tan^2 x - 20601 \sec^2 x} +1100 \tan x \right)$$</span></p>
<p>The separate plots of the two functions are <a href="https://www.wolframalpha.com/input/?i=y%20%3D%20-1100%2F981%20cos%5E2%28x%29%20%28sqrt%2810%29%20sqrt%28121000%20tan%5E2%28x%29%20-%2020601%20sec%5E2%28x%29%29%20-%201100%20tan%28x%29%29%2C%20%20x%3D0%20to%20%20pi%20" rel="nofollow noreferrer">here</a> and <a href="https://www.wolframalpha.com/input/?i=y%20%3D%201100%2F981%20cos%5E2%28x%29%20%28sqrt%2810%29%20sqrt%28121000%20tan%5E2%28x%29%20-%2020601%20sec%5E2%28x%29%29%20%2B%201100%20tan%28x%29%29%2C%20%20x%3D0%20to%20%20pi%20" rel="nofollow noreferrer">here</a>. The combined plot of the two functions is <a href="https://www.wolframalpha.com/input/?i=y+%3D+-1100%2F981+cos%5E2%28x%29+%28sqrt%2810%29+sqrt%28121000+tan%5E2%28x%29+-+20601+sec%5E2%28x%29%29+-+1100+tan%28x%29%29%2C++++++y+%3D+1100%2F981+cos%5E2%28x%29+%28sqrt%2810%29+sqrt%28121000+tan%5E2%28x%29+-+20601+sec%5E2%28x%29%29+%2B+1100+tan%28x%29%29%2C+++++++++++x%3D0+to++pi+" rel="nofollow noreferrer">here</a>.
By clarity, I also paste them here:</p>
<p><a href="https://i.stack.imgur.com/sj2Lv.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sj2Lv.gif" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/qy0v2.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qy0v2.gif" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/Ixm6n.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ixm6n.gif" alt="enter image description here" /></a></p>
<p>The maximal value is achieved by the second function. Its <a href="https://www.wolframalpha.com/input/?i=y%20%3D%201100%2F981%20cos%5E2%28x%29%20%28sqrt%2810%29%20sqrt%28121000%20tan%5E2%28x%29%20-%2020601%20sec%5E2%28x%29%29%20%2B%201100%20tan%28x%29%29%20derivative" rel="nofollow noreferrer">derivative</a> is quite complicated, and a closed form for the value of <span class="math-container">$x$</span> that maximizes <span class="math-container">$y$</span> may probably not exist. However, as already noteď in the OP and in the comments, the numerical solution is given by <span class="math-container">$x\approx 0.832$</span> radians (corresponding to about <span class="math-container">$47.67$</span> degrees), which leads to a maximum of <span class="math-container">$y\approx 1123.54$</span>, as confirmed <a href="https://www.wolframalpha.com/input/?i=Local%20maximum%20%20%201100%2F981%20cos%5E2%28x%29%20%28sqrt%2810%29%20sqrt%28121000%20tan%5E2%28x%29%20-%2020601%20sec%5E2%28x%29%29%20%2B%201100%20tan%28x%29%29%20%2C%20x%3D0%20to%20pi%2F2" rel="nofollow noreferrer">here</a>. By a similar numerical assessment of the first function, it can be confirmed that the symmetrical minimum value of <span class="math-container">$y\approx -1123.54$</span> is achieved for <span class="math-container">$x\approx \pi-0.832\approx 2.309$</span> radians (corresponding to about <span class="math-container">$132.33$</span> degrees), as shown <a href="https://www.wolframalpha.com/input/?i=Local%20minimum%20%20%20-1100%2F981%20cos%5E2%28x%29%20%28sqrt%2810%29%20sqrt%28121000%20tan%5E2%28x%29%20-%2020601%20sec%5E2%28x%29%29%20-%201100%20tan%28x%29%29%20%2C%20x%3D0%20to%20pi" rel="nofollow noreferrer">here</a>.</p>
<p>Note that setting the determinant equal to zero, as cited by the OP, only gives the values of <span class="math-container">$x$</span> for which the values of the two functions coincide. These values, which have rather tricky <a href="https://www.wolframalpha.com/input/?i=121000%20tan%5E2%20x%20-%2020601%20sec%5E2%20x%3D0" rel="nofollow noreferrer">closed forms</a> that can be approximated by <span class="math-container">$x\approx 0.4253$</span> and <span class="math-container">$\approx 2.7163$</span>, correspond to the points where the two functions meet, in the leftmost and rightmost portion of the combined plot.</p>
|
493,102 | <p>I have a concern with nested quantifiers.</p>
<p>I have: $$ \forall x \exists y \forall z(x^2-y+z=0) $$ such that $$ x,y,z \in \Bbb Z^+$$ </p>
<p>My first question, can it be read like this:</p>
<p>$$ \forall x \forall z \exists y(x^2-y+z=0) $$</p>
<p>The way I did it, is I started off with $x=1, z=1 $ </p>
<p>$$ 2-y = 0 $$ $$ y =2 $$</p>
<p>Is this a good approach?</p>
| MJD | 25,554 | <p>Consider $$\forall y. \exists x. x\text{ is the mother of }y$$ means that for every person $y$, there is some person $x$ who is $y$'s mother, which is true; every person has a mother. But $$\exists x.\forall y. x\text{ is the mother of }y$$ says that there is some person $x$, so that for every person $y$, $x$ is $y$'s mother, which is false; there is no person who is everyone's mother.</p>
<p>The order is important.</p>
|
244,679 | <p>I have a two variable function <code>z[x,y] = f[x,y] + g[x,y]</code>, such that I know the functional form of <code>f[x,y]</code> but not of <code>g[x,y]</code>. I have to do some symbolic calculations with the function <code>z[x,y]</code>, but I would like to keep only the first order in <code>g[x,y]</code> (treating <code>g</code> as small). So, for example, I would like Mathematica to approximate <code>(z[x,y])^3 = (f[x,y] + g[x,y])^3 = f[x,y]^3 + 3*f[x,y]^2*g[x,y]</code>, or <code>(D[z[x,y], x])^2 = (D[f[x,y], x])^2 + 2*D[f[x,y], x]*D[g[x,y], x]</code>. Is there a way to do it? I have tried the most naive way, namely to use <code>Series['exp'[z],{g, 0, 1}]</code>, treating <code>g[x,y]</code> as a parameter rather than a function, but (as expected) it doesn't work. Do you know a way to do it?</p>
<p>Thank you very much in advance for anyone who will reply!</p>
| Henrik Schumacher | 38,178 | <p>First OP's implementation with timing on my machine:</p>
<pre><code>First@AbsoluteTiming[
u1 = input;
Do[f = SparseArray[{{i_} :>
If[MemberQ[lll, i],
0, -2.0*u1[[i - 1]] - 2.0*u1[[i + 1]] - 2.0*u1[[i + dos]] -
2.0*u1[[i - dos]] + (1. + I)*u1[[i]]]}, {dos*dos}];
u1 = LinearSolve[s, f];, {j, 1, steps, 1}];
result0 = u1;
]
</code></pre>
<blockquote>
<p>17.1247</p>
</blockquote>
<p>The computation of <code>f</code> from <code>u1</code> is linear and as such can be represented by the following sparse matrix:</p>
<pre><code>A = Times[
SparseArray[Partition[Complement[Range[dos dos], lll], 1] -> 1. + 0. I, {dos dos}, 0. I],
SparseArray[{
Band[{1, 1}] -> (1. + 1. I)
, Band[{1, 1 + 1}] -> -2.0 + 0. I
, Band[{1, 1 + dos}] -> -2.0 + 0. I
, Band[{1 + 1, 1}] -> -2.0 + 0. I
, Band[{1 + dos, 1}] -> -2.0 + 0. I
},
{dos dos, dos dos}, 0. I
]
]; // AbsoluteTiming // First
</code></pre>
<blockquote>
<p>0.053559</p>
</blockquote>
<p>So, using also one-time factorization of <code>s</code>, OP's implementation can be rewritten into</p>
<pre><code>First@AbsoluteTiming[
sinv = LinearSolve[s];
u1 = input;
Do[u1 = sinv[A.u1];, {j, 1, steps, 1}];
result1 = u1;
]
</code></pre>
<blockquote>
<p>0.141947</p>
</blockquote>
<p>That is almost a 100-fold speedup.</p>
<p>This can also be formulated a bit shorter with <code>Nest</code> (which is not faster than <code>Do</code>):</p>
<pre><code>First@AbsoluteTiming[
sinv = LinearSolve[s];
result2 = Nest[sinv[A.#] &, input, steps];
]
</code></pre>
<blockquote>
<p>0.143743</p>
</blockquote>
<p>Both lead to exactly the same result:</p>
<pre><code>Max[Abs[result1 - result2]]
</code></pre>
<p>Comparison with OP's result:</p>
<pre><code>Max[Abs[result1 - result0]]
</code></pre>
<blockquote>
<p>541.03</p>
</blockquote>
<p>Looks huge, but the relative error is much smaller:</p>
<pre><code>Max[Abs[result1 - result0]]/Max[Abs[result0]]
</code></pre>
<blockquote>
<p>1.64591*10^-14</p>
</blockquote>
<p>What to learn from this: Pattern matching is all nice, but i maximizes rather programming speed, not execution speed.</p>
|
3,773,133 | <p>I have been thinking about this problem for a couple of months, and eventually failed. Could someone help me?</p>
<blockquote>
<p>Let <span class="math-container">$M$</span> and <span class="math-container">$X$</span> be two symmetric matrices with <span class="math-container">$M\succeq 0$</span> and <span class="math-container">$X=X^T$</span>, and let <span class="math-container">$p$</span> be a nonzero real number <span class="math-container">$|p|\le 10$</span>. What is the derivative of <span class="math-container">$f$</span> with respect to <span class="math-container">$M$</span>, where</p>
<p><span class="math-container">$$f(M) = \operatorname{Trace}(M^pX),$$</span></p>
<p>and <span class="math-container">$\operatorname{Trace}(A)$</span> is the trace operator to calculate the sum of elements in diagonal line of <span class="math-container">$A$</span>.</p>
</blockquote>
| rych | 73,934 | <p>Let <span class="math-container">$f(A) = \operatorname{tr}(A)$</span> then</p>
<p><span class="math-container">$df_A[H]=\operatorname{tr}H=(\operatorname{tr}\circ\operatorname{id})[H]=(\operatorname{tr}\circ\, dA_A)[H]$</span></p>
<p>for <span class="math-container">$A=M^pX$</span>, and e.g., <span class="math-container">$p=2$</span>, <span class="math-container">$dA_M=dM_M\,M\,X + M\,dM_M\,X$</span> and the chain rule gives (reusing letter <span class="math-container">$H$</span> for the auxiliary variable):</p>
<p><span class="math-container">$df_M[H]=(\operatorname{tr}\circ\operatorname{dA_M})[H]=(\operatorname{tr}\circ\,(dM_M\,M\,X + M\,dM_M\,X))[H]=\operatorname{tr}(H\,M\,X + M\,H\,X)$</span></p>
<p><a href="https://en.wikipedia.org/wiki/Trace_(linear_algebra)" rel="nofollow noreferrer">If products of three symmetric matrices are considered, any permutation is allowed</a>, therefore</p>
<p><span class="math-container">$df_M[H]=2\operatorname{tr}(M\,X\,H)=2\operatorname{tr}(M\,X\,\circ\,\operatorname{id})[H]$</span></p>
<p>More generally, <span class="math-container">$$d(\operatorname{tr}(M^pX))=p\operatorname{tr}(M^{p-1}X\circ\,\operatorname{id})$$</span></p>
|
2,797,902 | <p>AFAIK, every mathematical theory (by which I mean e.g. the theory of groups, topologies, or vector spaces), started out (historically speaking) by formulating a set of axioms that generalize a specific structure, or a specific set of structures. </p>
<p>For example, when people think of a “field” they AFAIK usually think of $\mathbb R$, or $\mathbb C$. A topology started out as a concept defined on $\mathbb R^n$ if I’m not mistaken. </p>
<p>But I’ve also seen cases where a certain structure has a natural topological structure, such as certain sets of propositions in first order logic. As far as I know, the people who formulated the axioms of a topology had no idea of this application. <strong>And the topological structure of a set of FOL statements is certainly conceptually vastly different from one on $\mathbb R^n$, certainly not two structures I would have expected to have such a deep commonality.</strong></p>
<p><strong>I would like to make a list of examples of mathematical structures that</strong></p>
<ol>
<li><p>Are interesting and well-behaved structures (e.g. not mere pathological counter examples)</p></li>
<li><p>satisfy the axioms of some mathematical theory in an interesting and nontrivial way,</p></li>
<li><p>But whose emergence is (conceptually/historically) very different from the structure of which those axioms were originally intended as a generalization.</p></li>
</ol>
| Ethan Bolker | 72,858 | <p>The useful <a href="https://en.wikipedia.org/wiki/Zariski_topology" rel="nofollow noreferrer">Zariski topology</a> in algebraic geometry satisfies the usual axioms for a topology in a context that doesn't really match "the structure of which those axioms were originally intended as a generalization".</p>
|
1,227,419 | <p>$$\int\limits_6^{16}\left(\frac{1}{\sqrt{x^3+7x^2+8x-16}}\right)\,\mathrm dx=\frac{\pi }{k}$$</p>
<p><strong>Note:</strong> $k$ is a constant.</p>
| mickep | 97,236 | <p>Hint: Since $x^3+7x^2+8x-16=(x-1)(x+4)^2$, I suggest you to do the substitution $u=\sqrt{x-1}$. In the end, you will find that a primitive function is given by
$$
\frac{2}{\sqrt{5}}\arctan(\sqrt{x-1}/\sqrt{5}).
$$
The value of $k$? Well, I leave that fun to you!</p>
|
2,623,324 | <p>Assume that the measure space is finite for this to make sense. Also, we know that $L^p$ spaces satisfy log convexity, that is -
$$\|f\|_r \leq \|f\|_p^\theta \|f\|_q^{1-\theta}$$
where $\frac{1}{r}=\frac{\theta}{p} +\frac{1-\theta}{q}$.
The text which I am following says 'Indeed this is trivial when $q=\infty$, and the general case then follows by convexity'. I understand that it is true when $q=\infty$, however I am unable to use that it is true for $q=\infty$ for proving the general case. I have found a proof which uses log-convexity and the fact that $\lim_{p\rightarrow 0}\|f\|_p^p=\mu(\text{supp}f)$. Is there some way that I can do this by using that it is true for $q=\infty$?</p>
| Dr. Sonnhard Graubner | 175,066 | <p>write your Limit in the form $$e^{\lim_{n\to \infty}\frac{\ln(n+1)}{\sqrt{n}}}$$</p>
|
2,651,054 | <p>I have this expression:
$$(x + y + z’)(x’ + y’ + z)$$ which I am trying to simplify. I decide to multiply it out in order to get, $${\color{red}{(xx')}}+(xy')+(xz)+(yx')+{\color{red}{(yy')}}+(yz)+(z'x')+(z'y')+{\color{red}{(z'z)}}.$$
I know that the $xx', yy'$ and $zz'$ would just be $0$, however, now I am stuck. I can't seem to find something to pull out and simplify further.</p>
| user284331 | 284,331 | <p>For $f(z)=z^{3}+2$, then $f(z)=f(1)+f'(1)(z-1)+\dfrac{f''(1)}{2!}(z-1)^{2}+\dfrac{f^{(3)}(1)}{3!}(z-1)^{3}$.</p>
|
2,651,054 | <p>I have this expression:
$$(x + y + z’)(x’ + y’ + z)$$ which I am trying to simplify. I decide to multiply it out in order to get, $${\color{red}{(xx')}}+(xy')+(xz)+(yx')+{\color{red}{(yy')}}+(yz)+(z'x')+(z'y')+{\color{red}{(z'z)}}.$$
I know that the $xx', yy'$ and $zz'$ would just be $0$, however, now I am stuck. I can't seem to find something to pull out and simplify further.</p>
| David | 119,775 | <p>As suggested in comments, other methods are probably easier, but you certainly can do this by division if you wish. We want
$$z^3+2=a_0+a_1(z-1)+a_2(z-1)^2+a_3(z-1)^3\ .\tag{$*$}$$
Dividing the LHS by $z-1$ gives
$$z^3+2=(z-1)(z^2+z+1)+3\ .$$
Looking at $(*)$, when we divide the RHS by $z-1$ the remainder is obviously $a_0$; for the LHS it is $3$ as we have just shown; so $a_0=3$. Now in $(*)$, subtract $3$ from both sides and divide by $z-1$; we have already done the working so we just write down
$$z^2+z+1=a_1+a_2(z-1)+a_3(z-1)^2\ .$$
Now follow the same procedure to find $a_1$, and so on.</p>
|
269,178 | <p>i would like to know where i could find a plot of</p>
<p>$$ J_{ia}(2\pi i)$$ (1)</p>
<p>using Quantum mechanics i have conjectured that if $ a= \frac{x}{2} $ and $ i= \sqrt{-1} $ then </p>
<p>$$ J_{it}(2\pi i)\approx0=\zeta (1/2+2it)$$ at least for big $ t \to \infty $ (2)</p>
<p>however i do not know how to check or disprove this fact.</p>
<p>the idea is that the Operator $$ -D^{2}y(x)+4\pi ^{2}e^{4x}y(x)=E_{n}y(x) $$ (3)</p>
<p>has a Weyl term for the Eigenvalues as $ N(T)= \frac{\sqrt{T}}{2\pi}log( \frac{\sqrt{T}}{2\pi e}) $</p>
<p>inthe same fashion as the Riemann zeta function</p>
<p>the condition (2) is stablished by imposing that the eigenvalue problem satisfy $ y(0)=0=y(\infty) $</p>
| Ron Gordon | 53,268 | <p>Here is a plot over some sample values in Wolfram Alpha:</p>
<p><a href="http://www.wolframalpha.com/share/clip?f=d41d8cd98f00b204e9800998ecf8427emmfihklhut" rel="nofollow">http://www.wolframalpha.com/share/clip?f=d41d8cd98f00b204e9800998ecf8427emmfihklhut</a></p>
|
324,119 | <p>I've been reading about the Artin Spin operation. It's defined as taking the classical <span class="math-container">$n$</span>-knot (<span class="math-container">$S^n\hookrightarrow S^{n+2}$</span>) to an <span class="math-container">$(n+1)$</span>-knot. For the <span class="math-container">$1$</span>-knot case (in <span class="math-container">$\mathbb{R}^3$</span>), I reproduce the procedure in <a href="https://arxiv.org/pdf/math/0410606.pdf" rel="nofollow noreferrer">knot spinning</a>, p. 8, </p>
<ol>
<li><p>We manipulate a knot <span class="math-container">$K$</span> so that all but a trivial arc <span class="math-container">$a$</span> lie in the upper half space <span class="math-container">$H^3=\{(x,y,z)\>|\>z\geq 0\}$</span>. We remove the interior of <span class="math-container">$a$</span>.</p></li>
<li><p>We rotate <span class="math-container">$H^3$</span> around <span class="math-container">$\mathbb{R}^2$</span> in <span class="math-container">$\mathbb{R}^4$</span>, inducing a parameterization <span class="math-container">$(x,y,z)\mapsto (x, y, z\cos\theta, z\sin\theta)$</span></p></li>
</ol>
<p><strong>-- Question --</strong>
How is this operation similar to the suspension functor (on a topological space <span class="math-container">$X$</span>) <span class="math-container">$\Sigma X\equiv X\wedge S=\frac{X\times S}{X\vee S}$</span>?</p>
| Friedrich Knop | 89,948 | <p>There are various generalizations of the Jordan-Hölder theorem. Beyond groups and groups with operators it holds for any equational theory which contains a Mal'cev operation. This means that from the given operations one can form a ternary operation <span class="math-container">$m(x,y,z)$</span> satisfying <span class="math-container">$m(x,x,z)=z$</span> and <span class="math-container">$m(x,z,z)=x$</span> for all <span class="math-container">$x,y,z$</span>. Every equational theory containing a group operation is an example since <span class="math-container">$m(x,y,z)=xy^{-1}z$</span> has the Mal'cev property. Another example not of that type would be the class of Heyting algebras. A non-example is the class of lattices.</p>
<p>The fact that every equational theory containing a Mal'cev operation satisfies the Jordan-Hölder theorem has been proved by Lambek in <a href="https://doi.org/10.4153/CJM-1958-005-6" rel="noreferrer">Goursat's theorem and the Zassenhaus lemma. Canad. J. Math. 10 (1958) 45–56</a>. It is my impression that the Mal'cev condition is pretty much optimal in this context.</p>
<p>The JH-theorem clearly also holds in abelian categories but that is not covered by Lambek's theorem above. A common generalization has been achieved by Carboni-Lambek-Pedicchio with the notion of a Mal'cev category (in <a href="https://doi.org/10.1016/0022-4049(91)90022-T" rel="noreferrer">Diagram chasing in Mal’cev categories, J. Pure Appl. Algebra 69 (1991) 271–284</a>). These are exact categories (in the sense of Barr, not Quillen!) which have the curious property that every reflexive relation is already an equivalence relation. The category of models of an equationial theory with a Mal'cev operation would be an example (that's an amusing exercise). The category should also have a <span class="math-container">$0$</span>-object in order to be able to define kernels. Then one can show that then all objects of finite length have the JH-property. A proof has been given in <a href="https://doi.org/10.1016/j.aim.2007.03.001" rel="noreferrer">F. Knop: Tensor envelopes of regular categories. Advances in Mathematics 214 (2007) 571-617</a> even though the fact must have been at least folklore.</p>
|
4,218,729 | <p>You have a database of <span class="math-container">$25,000$</span> potential criminals. The probability that this database includes the art thief is <span class="math-container">$0.1$</span>. In a stroke of luck, you found a DNA sample of this thief from the crime scene. You compare this sample with a database of <span class="math-container">$25,000$</span> men. And lo and behold, a match is found! You are well aware that DNA matches are not always perfect: if you pick two different persons at random, the chance that their DNA samples would match using the current testing techniques is <span class="math-container">$1$</span> in <span class="math-container">$10,000$</span>. What's the probability that the database includes the art thief, given that a DNA match has been found?</p>
<p>I'm sure this question uses Bayes' theorem where <span class="math-container">$A=$</span> the database includes the art thief, <span class="math-container">$B=$</span> a DNA match has been found.
I need to find <span class="math-container">$P(A|B)= P(B|A)*P(A)/P(B)$</span>
to calculate <span class="math-container">$P(B)$</span>, there are different cases,
1). the database does include the thief, and a person matches with the DNA from the scene.
2). the database doesn't include the thief, but a person matches with the DNA.</p>
<p>I'm not really sure how to calculate 1) and 2). Please help me.</p>
<p>I also want to know if the probability of two people matching is <span class="math-container">$1/10000$</span>, is the probability of two people not matching <span class="math-container">$9999/10000$</span>?</p>
| true blue anil | 22,388 | <p>** Recast answer(12 Aug) **</p>
<p><em>Answer recast to include possibility that</em> a <strong>wrong match</strong> <em>may be found even if thief is in the database, and to present with greater clarity and simplicity through a contingency table</em></p>
<p>If the probability of a match being wrong is <span class="math-container">$P=1/10 000$</span>, that of it being correct must be <span class="math-container">$P=9999/10 000$</span></p>
<p>I'll also assume that</p>
<ul>
<li>all <span class="math-container">$25,000$</span> are lined up and tested</li>
<li><em>exactly</em> one match is found</li>
</ul>
<p><em>Also, even if the thief is present <span class="math-container">$(T)$</span>,the match might not be with the thief but with someone else, say</em> <span class="math-container">$X$</span></p>
<p>Drawing up a contingency table,</p>
<p><span class="math-container">$\quad\quad\quad\quad\quad\quad T\quad|\quad T^c$</span></p>
<p>Matches <span class="math-container">$T\quad \;\,A \quad\quad\; C$</span></p>
<p>Matches <span class="math-container">$X\quad \;\,B \quad \quad\;D$</span></p>
<p><span class="math-container">$A = 0.1*0.9999*0.9999^{24999}= 0.82075$</span>%</p>
<p><span class="math-container">$B = 0.1*.0001*24999*.0001*.9999^{24998}=0.00021$</span>%</p>
<p><span class="math-container">$C = 0.9*0 = 0$</span>%</p>
<p><span class="math-container">$D = 0.9*0.0001*25000*0.9999^{24999}= 18.46866$</span>%</p>
<p>Finally, P(thief present | match found)</p>
<p><span class="math-container">$=\dfrac{A+B}{A+B+C+D} = \dfrac{0.82075+0.00021}{0.82075+0.00021+0+18.46866}$</span></p>
<p><span class="math-container">$\boxed{\approx 4.26\%}$</span></p>
|
3,936,676 | <p>Well this format of a limit <span class="math-container">$0^0$</span> is an indeterminate form.</p>
<p>I claim that whatever this limit is (which depends on the exact question) should always be in between <span class="math-container">$[0,1]$</span>.</p>
<p>Is my claim correct?</p>
<p>I have no mathematical proof for it but just a basic idea, that any number base <span class="math-container">$0$</span> should try to pull towards <span class="math-container">$0$</span>, whereas any number power <span class="math-container">$0$</span> should pull towards <span class="math-container">$1$</span>.</p>
| Mike | 544,150 | <p>That's not true in general. Take <span class="math-container">$$y(t)=e^{-1/t}$$</span> and <span class="math-container">$$x(t)=-5t$$</span> say.
Then <span class="math-container">$$\lim_{t \rightarrow 0^+} y(t) = \lim_{t \rightarrow 0^+} x(t)=0,$$</span> but <span class="math-container">$$\lim_{t \rightarrow 0^+} (e^{-1/t})^{-5t}= e^5 >1.$$</span></p>
<p>For an even more dramatic example take <span class="math-container">$y(t)=e^{-1/t^2}$</span> and <span class="math-container">$x(t)=-t$</span>.</p>
<p>HOWEVER, if <span class="math-container">$y(t)$</span> and <span class="math-container">$x(t)$</span> as above are both positive for then infact the limit will indeed fall between 0 and 1, as any positive number less than 1 raised to a positive power is in [0,1].</p>
|
957,940 | <p>I'm "walking" through the book "A walk through combinatorics" and stumbled on an example I don't understand. </p>
<blockquote>
<p><strong>Example 3.19.</strong> A medical student has to work in a hospital for five
days in January. However, he is not allowed to work two consecutive
days in the hospital. In how many different ways can he choose the
five days he will work in the hospital?</p>
<p><strong>Solution</strong>. The difficulty here is to make sure that we do not choose two consecutive days. This can be assured by the following
trick. Let $a_1, a_2, a_3, a_4, a_5$ be the dates of the five days of
January that the student will spend in the hospital, in increasing
order. Note that the requirement that there are no two consecutive
numbers among the $a_i$, and $1 \le a_i \le 31$ for all $i$ is equivalent
to the requirement that $1 \le a_i < a_2 — 1 < a_3 — 2 < a_4 — 3 < a_5 — 4 \le 27$. In other words, there is an obvious bijection between the
set of 5-element subsets of [31] containing no two consecutive
elements and the set of 5-element subsets of [27].</p>
<p>*** Instead of choosing the numbers $a_i$, we can choose the numbers $1 \le a_i < a_2 — 1 < a_3 — 2 < a_4 — 3 < a_5 — 4 \le 27$, that is, we can
simply choose a five-element subset of [27], and we know that there
are $\binom{27}{5}$ ways to do that.</p>
</blockquote>
<p>What I don't understand here $1 \le a_i < a_2 — 1 < a_3 — 2 < a_4 — 3 < a_5 — 4 \le 27$:</p>
<ul>
<li>Why do the subtracting numbers increment with every other $a_i$? </li>
<li>Why 27?</li>
</ul>
<p>And the very last sentence (***) is unclear to me.</p>
<ul>
<li>Why is there no talk about "non-consecutive"? Why choosing 5 elements of 27 is equivalent to choosing 5 non-consecutive elements out of 31? I miss the connection. </li>
</ul>
<p>I'd be very thankful if you could help me to understand this example!</p>
| Trold | 180,885 | <p>Henry's answer for the first part is already very good, so this only tries to tackle the sentence ***. </p>
<p>Understanding the correspondence between five element subsets of $[27]$ and the non-consecutive days problem might be easier by going from the five element subsets of 27 to the days problem asked, rather than the other way around. </p>
<p>We know that we'll be forbidden from working on four specific days -- namely, the days after the first four workdays. (We're technically forbidden from working the day after the fifth workday, too, but since we've already filled our monthly quota who cares?) -- so let's throw them out now and worry about which days exactly we've thrown out later. We're now left with days we'll number 1 to 27.</p>
<p>So let's pick five workdays out of the remaining twenty seven (here's our five element subset of $[27]$) and reconstruct the monthly calendar. I'm going to use $a_i$ for the numbers we picked from $[27]$ and $c_i$ for how they'll appear as days of the month a normal human calendar. By way of example I'm going to say we picked $\{3,4,12,15,20\}$</p>
<p>For $a_1$ we needn't do anything, we'll just copy it down as $c_1$ and say our first workday is Jan. 3d. We're forbidden from working the next day, so we put back the first of the four days we discarded and increment our remaining $a_i$ by one. This fixes our problem with $a_2=4$, and informs us we next have work on the 5th. We do the same for the three $a_i$ left and so on. </p>
<p>In the end, our five element set from $[27]$ is mapped to the $a_1$ (3rd), $a_2+1$ (5th), $a_3+2$ (14th), $a_4+3$ (18th), and $a_5+4$ (24th) days of January. (Proof that this is actually a bijection is left out, but not really difficult so eh).</p>
|
1,002,719 | <p>If we have</p>
<p>$f: \{1, 2, 3\} \to \{1, 2, 3\}$</p>
<p>and</p>
<p>$f \circ f = id_{\{1,2,3\}}$</p>
<p>is the following then always true for every function?</p>
<p>$f = id_{\{1,2,3\}}$</p>
| Dustan Levenstein | 18,966 | <p>If $f: S \to S$ is a function satisfying $f(f(x)) = x$ for all $x \in S$, consider what the implications are of having a fixed $x \in S$ so that $f(x) \neq x$; setting $y := f(x)$, we must have:</p>
<p>$$f(x) = y$$ by definition, and $$f(y) = f(f(x)) = x.$$</p>
<p>Given this observation, can you then describe exactly what the nature of such a function $f$ must be?</p>
|
19,253 | <p><strong>Bug introduced in 7.0.1 or earlier and fixed in 10.0.2 or earlier</strong></p>
<hr>
<p>I have access to two versions of Mathematica, version 7.0.1 on Linux and version 8 on Windows. When I try the following two lines on version 7, the kernel quits when it tries to plot. In version 8 it plots just fine. You can import the data using:</p>
<pre><code>temp = Import["http://pastie.org/pastes/6098244/download", "Table"];
</code></pre>
<p>Any idea why this crashes in version 7? </p>
<pre><code>ListContourPlot[temp]
</code></pre>
<p>It may also crash in Mathematica 7 for Windows too, but I don't personally have a copy of that on my machine.</p>
<p>ETA: Other users have reported this also crashes in some versions 8 and 9 across different platforms, so I changed the name of the post accordingly.</p>
<hr>
<p>If the pastebin link fails, please copy the data from here:</p>
<pre><code>temp={{-0.1,11000,1.00214},{-0.1,11050,0.998176},
{-0.1,11100,0.994088},{-0.1,11150,0.990016},{-0.1,11200,0.986111},
{-0.1,11250,0.982526},{-0.1,11300,0.979411},{-0.1,11350,0.976904},
{-0.1,11400,0.97513},{-0.1,11450,0.974195},{-0.1,11500,0.974179},
{-0.1,11550,0.975137},{-0.1,11600,0.977087},{-0.1,11650,0.98002},
{-0.1,11700,0.983885},{-0.1,11750,0.988602},{-0.1,11800,0.994051},
{-0.1,11850,1.00008},{-0.1,11900,1.00652},{-0.1,11950,1.01315},
{-0.1,12000,1.01975},{-0.1,12050,1.02608},{-0.1,12100,1.03189},
{-0.1,12150,1.03694},{-0.1,12200,1.04098},{-0.1,12250,1.04378},
{-0.1,12300,1.04513},{-0.1,12350,1.04486},{-0.1,12400,1.04282},
{-0.1,12450,1.03889},{-0.1,12500,1.03303},{-0.1,12550,1.02522},
{-0.1,12600,1.01549},{-0.1,12650,1.00392},{-0.1,12700,0.990664},
{-0.1,12750,0.975894},{-0.1,12800,0.959842},{-0.1,12850,0.942777},
{-0.1,12900,0.924999},{-0.1,12950,0.906838},{-0.1,13000,0.888643},
{-0.1,13050,0.870773},{-0.1,13100,0.853588},{-0.1,13150,0.837442},
{-0.1,13200,0.822673},{-0.1,13250,0.809592},{-0.1,13300,0.798481},
{-0.1,13350,0.789577},{-0.1,13400,0.783075},{-0.1,13450,0.779116},
{-0.1,13500,0.777787},{-0.098,11000,0.990164},{-0.098,11050,0.986921},
{-0.098,11100,0.983944},{-0.098,11150,0.981344},{-0.098,11200,0.979226},
{-0.098,11250,0.977683},{-0.098,11300,0.976797},{-0.098,11350,0.976633},
{-0.098,11400,0.977237},{-0.098,11450,0.978633},{-0.098,11500,0.980822},
{-0.098,11550,0.983781},{-0.098,11600,0.98746},{-0.098,11650,0.991787},
{-0.098,11700,0.996664},{-0.098,11750,1.00197},{-0.098,11800,1.00756},
{-0.098,11850,1.01329},{-0.098,11900,1.01897},{-0.098,11950,1.02443},
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{-0.076,12600,0.999955},{-0.076,12650,0.999954},{-0.076,12700,0.999954},
{-0.076,12750,0.999954},{-0.076,12800,0.999954},{-0.076,12850,0.999954},
{-0.076,12900,0.999954},{-0.076,12950,0.999954},{-0.076,13000,0.999954},
{-0.076,13050,0.999953},{-0.076,13100,0.999953},{-0.076,13150,0.999953},
{-0.076,13200,0.999953},{-0.076,13250,0.999953},{-0.076,13300,0.999953},
{-0.076,13350,0.999953},{-0.076,13400,0.999953},{-0.076,13450,0.999953},
{-0.076,13500,0.999953},{-0.074,11000,0.999983},{-0.074,11050,0.999983},
{-0.074,11100,0.999982},{-0.074,11150,0.999982},{-0.074,11200,0.999982},
{-0.074,11250,0.999982},{-0.074,11300,0.999981},{-0.074,11350,0.999981},
{-0.074,11400,0.999981},{-0.074,11450,0.999981},{-0.074,11500,0.99998},
{-0.074,11550,0.99998},{-0.074,11600,0.99998},{-0.074,11650,0.99998},
{-0.074,11700,0.999979},{-0.074,11750,0.999979},{-0.074,11800,0.999979},
{-0.074,11850,0.999979},{-0.074,11900,0.999979},{-0.074,11950,0.999978},
{-0.074,12000,0.999978},{-0.074,12050,0.999978},{-0.074,12100,0.999978},
{-0.074,12150,0.999978},{-0.074,12200,0.999977},{-0.074,12250,0.999977},
{-0.074,12300,0.999977},{-0.074,12350,0.999977},{-0.074,12400,0.999977},
{-0.074,12450,0.999977},{-0.074,12500,0.999976},{-0.074,12550,0.999976},
{-0.074,12600,0.999976},{-0.074,12650,0.999976},{-0.074,12700,0.999976},
{-0.074,12750,0.999976},{-0.074,12800,0.999976},{-0.074,12850,0.999976},
{-0.074,12900,0.999976},{-0.074,12950,0.999975},{-0.074,13000,0.999975},
{-0.074,13050,0.999975},{-0.074,13100,0.999975},{-0.074,13150,0.999975},
{-0.074,13200,0.999975},{-0.074,13250,0.999975},{-0.074,13300,0.999975},
{-0.074,13350,0.999975},{-0.074,13400,0.999975},{-0.074,13450,0.999975},
{-0.074,13500,0.999975},{-0.072,11000,0.999986},{-0.072,11050,0.999986},
{-0.072,11100,0.999986},{-0.072,11150,0.999985},{-0.072,11200,0.999985},
{-0.072,11250,0.999985},{-0.072,11300,0.999985},{-0.072,11350,0.999985},
{-0.072,11400,0.999984},{-0.072,11450,0.999984},{-0.072,11500,0.999984},
{-0.072,11550,0.999984},{-0.072,11600,0.999984},{-0.072,11650,0.999984},
{-0.072,11700,0.999983},{-0.072,11750,0.999983},{-0.072,11800,0.999983},
{-0.072,11850,0.999983},{-0.072,11900,0.999983},{-0.072,11950,0.999983},
{-0.072,12000,0.999983},{-0.072,12050,0.999982},{-0.072,12100,0.999982},
{-0.072,12150,0.999982},{-0.072,12200,0.999982},{-0.072,12250,0.999982},
{-0.072,12300,0.999982},{-0.072,12350,0.999982},{-0.072,12400,0.999981},
{-0.072,12450,0.999981},{-0.072,12500,0.999981},{-0.072,12550,0.999981},
{-0.072,12600,0.999981},{-0.072,12650,0.999981},{-0.072,12700,0.999981},
{-0.072,12750,0.999981},{-0.072,12800,0.999981},{-0.072,12850,0.999981},
{-0.072,12900,0.999981},{-0.072,12950,0.999981},{-0.072,13000,0.99998},
{-0.072,13050,0.99998},{-0.072,13100,0.99998},{-0.072,13150,0.99998},
{-0.072,13200,0.99998},{-0.072,13250,0.99998},{-0.072,13300,0.99998},
{-0.072,13350,0.99998},{-0.072,13400,0.99998},{-0.072,13450,0.99998},
{-0.072,13500,0.99998},{-0.07,11000,0.999989},{-0.07,11050,0.999989},
{-0.07,11100,0.999989},{-0.07,11150,0.999989},{-0.07,11200,0.999989},
{-0.07,11250,0.999988},{-0.07,11300,0.999988},{-0.07,11350,0.999988},
{-0.07,11400,0.999988},{-0.07,11450,0.999988},{-0.07,11500,0.999988},
{-0.07,11550,0.999988},{-0.07,11600,0.999988},{-0.07,11650,0.999987},
{-0.07,11700,0.999987},{-0.07,11750,0.999987},{-0.07,11800,0.999987},
{-0.07,11850,0.999987},{-0.07,11900,0.999987},{-0.07,11950,0.999987},
{-0.07,12000,0.999987},{-0.07,12050,0.999987},{-0.07,12100,0.999987},
{-0.07,12150,0.999986},{-0.07,12200,0.999986},{-0.07,12250,0.999986},
{-0.07,12300,0.999986},{-0.07,12350,0.999986},{-0.07,12400,0.999986},
{-0.07,12450,0.999986},{-0.07,12500,0.999986},{-0.07,12550,0.999986},
{-0.07,12600,0.999986},{-0.07,12650,0.999986},{-0.07,12700,0.999986},
{-0.07,12750,0.999986},{-0.07,12800,0.999986},{-0.07,12850,0.999985},
{-0.07,12900,0.999985},{-0.07,12950,0.999985},{-0.07,13000,0.999985},
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{-0.07,13350,0.999985},{-0.07,13400,0.999985},{-0.07,13450,0.999985},
{-0.07,13500,0.999985},{-0.068,11000,0.999992},{-0.068,11050,0.999992},
{-0.068,11100,0.999992},{-0.068,11150,0.999992},{-0.068,11200,0.999992},
{-0.068,11250,0.999992},{-0.068,11300,0.999992},{-0.068,11350,0.999992},
{-0.068,11400,0.999992},{-0.068,11450,0.999992},{-0.068,11500,0.999991},
{-0.068,11550,0.999991},{-0.068,11600,0.999991},{-0.068,11650,0.999991},
{-0.068,11700,0.999991},{-0.068,11750,0.999991},{-0.068,11800,0.999991},
{-0.068,11850,0.999991},{-0.068,11900,0.999991},{-0.068,11950,0.999991},
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{-0.066,11100,0.999995},{-0.066,11150,0.999995},{-0.066,11200,0.999995},
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{-0.066,11550,0.999995},{-0.066,11600,0.999995},{-0.066,11650,0.999995},
{-0.066,11700,0.999995},{-0.066,11750,0.999995},{-0.066,11800,0.999995},
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{-0.066,12300,0.999994},{-0.066,12350,0.999994},{-0.066,12400,0.999994},
{-0.066,12450,0.999994},{-0.066,12500,0.999994},{-0.066,12550,0.999994},
{-0.066,12600,0.999994},{-0.066,12650,0.999994},{-0.066,12700,0.999994},
{-0.066,12750,0.999994},{-0.066,12800,0.999994},{-0.066,12850,0.999994},
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{-0.066,13350,0.999994},{-0.066,13400,0.999994},{-0.066,13450,0.999994},
{-0.066,13500,0.999994},{-0.064,11000,0.999998},{-0.064,11050,0.999998},
{-0.064,11100,0.999998},{-0.064,11150,0.999998},{-0.064,11200,0.999998},
{-0.064,11250,0.999998},{-0.064,11300,0.999998},{-0.064,11350,0.999998},
{-0.064,11400,0.999998},{-0.064,11450,0.999998},{-0.064,11500,0.999998},
{-0.064,11550,0.999998},{-0.064,11600,0.999998},{-0.064,11650,0.999998},
{-0.064,11700,0.999998},{-0.064,11750,0.999998},{-0.064,11800,0.999998},
{-0.064,11850,0.999998},{-0.064,11900,0.999998},{-0.064,11950,0.999998},
{-0.064,12000,0.999998},{-0.064,12050,0.999998},{-0.064,12100,0.999998},
{-0.064,12150,0.999998},{-0.064,12200,0.999998},{-0.064,12250,0.999998},
{-0.064,12300,0.999998},{-0.064,12350,0.999998},{-0.064,12400,0.999998},
{-0.064,12450,0.999998},{-0.064,12500,0.999998},{-0.064,12550,0.999998},
{-0.064,12600,0.999998},{-0.064,12650,0.999998},{-0.064,12700,0.999998},
{-0.064,12750,0.999998},{-0.064,12800,0.999998},{-0.064,12850,0.999998},
{-0.064,12900,0.999998},{-0.064,12950,0.999998},{-0.064,13000,0.999998},
{-0.064,13050,0.999998},{-0.064,13100,0.999998},{-0.064,13150,0.999998},
{-0.064,13200,0.999998},{-0.064,13250,0.999998},{-0.064,13300,0.999998},
{-0.064,13350,0.999998},{-0.064,13400,0.999998},{-0.064,13450,0.999998},
{-0.064,13500,0.999998},{-0.062,11000,0.999999},{-0.062,11050,0.999999},
{-0.062,11100,0.999999},{-0.062,11150,0.999999},{-0.062,11200,0.999999},
{-0.062,11250,0.999999},{-0.062,11300,0.999999},{-0.062,11350,0.999999},
{-0.062,11400,0.999999},{-0.062,11450,0.999999},{-0.062,11500,0.999999},
{-0.062,11550,0.999999},{-0.062,11600,0.999999},{-0.062,11650,0.999999},
{-0.062,11700,0.999999},{-0.062,11750,0.999999},{-0.062,11800,0.999999},
{-0.062,11850,0.999999},{-0.062,11900,0.999999},{-0.062,11950,0.999999},
{-0.062,12000,0.999999},{-0.062,12050,0.999999},{-0.062,12100,0.999999},
{-0.062,12150,0.999999},{-0.062,12200,0.999999},{-0.062,12250,0.999999},
{-0.062,12300,0.999999},{-0.062,12350,0.999999},{-0.062,12400,0.999999},
{-0.062,12450,0.999999},{-0.062,12500,0.999999},{-0.062,12550,0.999999},
{-0.062,12600,0.999999},{-0.062,12650,0.999999},{-0.062,12700,0.999999},
{-0.062,12750,0.999999},{-0.062,12800,0.999999},{-0.062,12850,0.999999},
{-0.062,12900,0.999999},{-0.062,12950,0.999999},{-0.062,13000,0.999999},
{-0.062,13050,0.999999},{-0.062,13100,0.999999},{-0.062,13150,0.999999},
{-0.062,13200,0.999999},{-0.062,13250,0.999999},{-0.062,13300,0.999999},
{-0.062,13350,0.999999},{-0.062,13400,0.999999},{-0.062,13450,0.999999},
{-0.062,13500,0.999999},{-0.06,11000,1.},{-0.06,11050,1.},{-0.06,11100,1.},
{-0.06,11150,1.},{-0.06,11200,1.},{-0.06,11250,1.},{-0.06,11300,1.},
{-0.06,11350,1.},{-0.06,11400,1.},{-0.06,11450,1.},{-0.06,11500,1.},
{-0.06,11550,1.},{-0.06,11600,1.},{-0.06,11650,1.},{-0.06,11700,1.},
{-0.06,11750,1.},{-0.06,11800,1.},{-0.06,11850,1.},{-0.06,11900,1.},
{-0.06,11950,1.},{-0.06,12000,1.},{-0.06,12050,1.},{-0.06,12100,1.},
{-0.06,12150,1.},{-0.06,12200,1.},{-0.06,12250,1.},{-0.06,12300,1.},
{-0.06,12350,1.},{-0.06,12400,1.},{-0.06,12450,1.},{-0.06,12500,1.},
{-0.06,12550,1.},{-0.06,12600,1.},{-0.06,12650,1.},{-0.06,12700,1.},
{-0.06,12750,1.},{-0.06,12800,1.},{-0.06,12850,1.},{-0.06,12900,1.},
{-0.06,12950,1.},{-0.06,13000,1.},{-0.06,13050,1.},{-0.06,13100,1.},
{-0.06,13150,1.},{-0.06,13200,1.},{-0.06,13250,1.},{-0.06,13300,1.},
{-0.06,13350,1.},{-0.06,13400,1.},{-0.06,13450,1.},{-0.06,13500,1.}};
</code></pre>
| Community | -1 | <p>It turns out to indeed be a problem with the triangulation algorithm. I figured it out by plotting the list incrimentally as in <code>ListContourPlot[temp[[;;n]]]</code> and found it would plot for <code>n<103</code> but then gave an error "The data generates an inconsistent triangulation. You can perturb the data to make it valid." Alexey Popkov gave the suggestion of perturbing the data with some artificial noise, but that isn't necessary in this case.</p>
<p>Searching that error message took me to this <a href="http://forums.wolfram.com/mathgroup/archive/2011/Aug/msg00358.html">page</a>
which suggested I could rescale the x and y values to lie between 0 and 1, and then it would plot just fine. Something about the difference in range in the x and y dimensions short-circuited it. I tried the method listed there and it worked just fine.</p>
<pre><code>maxmin={Min[#],Max[#]}&/@Transpose[temp][[{1,2}]];
rescaled=Transpose[{
Rescale[temp[[All,1]]],Rescale[temp[[All,2]]],temp[[All,3]]
}];
</code></pre>
<p>Then to do a contour plot of the rescaled data (which works without error), and apply a rescaling directly to the axis of the resulting plot:</p>
<pre><code>gr=ListContourPlot[rescaled,PlotRange->All];
gr/.GraphicsComplex[a__]:> GeometricTransformation[
GraphicsComplex[a], RescalingTransform[{{0,1},{0,1}},maxmin]]
</code></pre>
<p>A 3D plot like this:</p>
<pre><code>gr=ListPlot3D[rescaled,PlotRange->All];
gr/.GraphicsComplex[a__]:>GeometricTransformation[
GraphicsComplex[a], RescalingTransform[{{0,1},{0,1},{0,1}},Append[maxmin,{0,1}]]]
</code></pre>
|
85,957 | <p>Assume the annually gross salary is $100,000.</p>
<p>Tax brackets:</p>
<ol>
<li>0 - 50K - 10% = 5K in taxes</li>
<li>50K - 70K - 20% = 4K in taxes</li>
<li>70K - $90K - 30% = 6K in taxes</li>
<li>90K and up - 40% = 4K in taxes</li>
</ol>
<p>The income tax on 100K would be 19K.</p>
<p>So the net salary would be 81K.</p>
<p>However, knowing only the net salary (81K) and the tax brackets (1-4) how can I find out what was the gross salary?</p>
<p>Thank you very much.</p>
<p><strong>UPDATE:</strong> Would the same rules apply if I have 2 level of brackets? For instance Federal and State income tax. Lets assume that the brackets are the same and Federal and State charge you the same amount on yeah tax bracket.</p>
<p><strong>UPDATE 2:</strong> Actually I will combine Federal and State tax into one set of brackets and apply the same rule as it was one set of brackets.</p>
| Arturo Magidin | 742 | <p>First, determine the maximum net for each bracket:</p>
<ol>
<li><p>If your gross is 0-50,000, then your net is 0-45,000.</p></li>
<li><p>If your gross is 50,001-70,000, then your net is 45,001-61,000 (the most you will pay in taxes is 5,000 for the first 50,000, and 4,000 for the next 20,000).</p></li>
<li><p>If your gross is 70,001-90,000, then your net is 61,001-75,000 (the most you will pay in taxes is 5,000 for the first 50,000, 4,000 for the next 20,000, and 6,000 for the final 20,000, for a total of 15,000 in taxes).</p></li>
<li><p>If your gross is 90,001 and up, your net is 75,001 and up.</p></li>
</ol>
<p>So, if the net salary is 81,000, then you are in the final bracket. You've earned 75,000 for the first 90,000; and you've earned a further 6,000 more <em>net</em>. </p>
<p>If $x$ is the amount over 90,000 that you earned, then you know that $0.6x=6,000$ (since you keep 60% of the amount over 90,000). Solving for $x$ gives $x=\frac{6000}{0.6} = 10,000$, so the gross salary was $\$$100,000.</p>
<p>Indeed, if your gross is 100,000, then you pay 5,000 of the first 50,000 and keep 45,000; you pay 4,000 of the next 20,000, and keep 16,000 (you've paid 9,000 in taxes and kept 61,000 in net); you pay 6,000 of the next 20,000 in taxes, and keep 14,000 (your running total is 15,000 in taxes and 75,000 in net); and you pay 4,000 in taxes for the next 10,000 and keep 6,000 (your final total: 19,000 in taxes, 81,000 in net).</p>
<p>Once you know what tax bracket the final total lies in, you can figure out how much "after the last dividing line" he originally earned.</p>
<p>For example, say the net salary is 67,500. This corresponds to a gross salary in the 70,000-90-000 bracket; it is exactly 6,500 over the final net if you made 70,000, so if $x$ is the amount over 70,000, as you pay 30% of $x$ in taxes you must have $6,500=0.7x$. Solving for $x$ gives $x=9285.71$, so the gross would be $70,000+x = 79,285.71$.</p>
|
85,957 | <p>Assume the annually gross salary is $100,000.</p>
<p>Tax brackets:</p>
<ol>
<li>0 - 50K - 10% = 5K in taxes</li>
<li>50K - 70K - 20% = 4K in taxes</li>
<li>70K - $90K - 30% = 6K in taxes</li>
<li>90K and up - 40% = 4K in taxes</li>
</ol>
<p>The income tax on 100K would be 19K.</p>
<p>So the net salary would be 81K.</p>
<p>However, knowing only the net salary (81K) and the tax brackets (1-4) how can I find out what was the gross salary?</p>
<p>Thank you very much.</p>
<p><strong>UPDATE:</strong> Would the same rules apply if I have 2 level of brackets? For instance Federal and State income tax. Lets assume that the brackets are the same and Federal and State charge you the same amount on yeah tax bracket.</p>
<p><strong>UPDATE 2:</strong> Actually I will combine Federal and State tax into one set of brackets and apply the same rule as it was one set of brackets.</p>
| dku.rajkumar | 20,185 | <pre><code>consider net salary is x, now we will find out gross salary y
if x < 45000(0.9 * 50000)
y = x/0.9
end
if 45000 < x < 61000 (45000 + 0.8 * 20000)
y = 50000 + (x- 45000)/0.8
end
if 61000 < x < 75000 (61000 + 0.7 * 20000)
y = 70000 + (x - 61000)/0.7
end
if 75000 < x
y = 90000 + (x- 75000)/0.6
end
</code></pre>
|
114,754 | <p>I have several questions concerning the proof. I don't think I quite understand the details and motivation of the proof. Here is the proof given by our professor.</p>
<p>The space of polynomials $F[x]$ is not finite-dimensional.</p>
<p><em>Proof</em>. Suppose
$$F[x] = \operatorname{span}\{f_1,f_2,\dots,f_n\}$$</p>
<p>Let us choose a positive integer $N$ such that $N > deg (f_i)$ for all $i = 1,\dots,n$. As $\{f_1,f_2,\dots,f_n\}$
spans $F[x]$ we can find scalars $a_1, a_2, \dots,a_n$ such that $x^N = a_1f_1 + a_2f_2 + \cdots + a_nf_n$.
Then the polynomial </p>
<p>$$G(x) = x^N - a_1f_1 - a_2f_2 - \cdots - a_nf_n \equiv 0$$</p>
<p>is a polynomial of degree $N$ which is identically zero. This is a contradiction since $G(x)$ cannot
have more than $N$ roots. </p>
<p><strong>Questions</strong></p>
<ul>
<li>Why is $G(x)$ <strong>identically</strong> zero and what does it mean for it to be <strong>identically</strong> zero? </li>
<li>After obtaining that $G(x) = x^N - a_1f_1 - a_2f_2 - \cdots - a_nf_n \equiv 0$, why must we have more than $N$ roots?</li>
<li><p>What is the <strong>motivation</strong> behind choosing $N$ such that $N > deg (f_i)$ for all $i = 1,\dots,n?$ </p></li>
<li><p>What are the <strong>implications</strong> if we chose $N$ less than or equal to $f_i$ such that $f_i$ has the greatest degree? How will the proof fail in this case? </p></li>
<li><p>If there are any other important <strong>details</strong> and key <strong>insights</strong> that are worthy to be pointed out please let me know so that I can better my understanding of the argument presented.</p></li>
</ul>
| Pierre-Yves Gaillard | 660 | <p>I think the proof in your question is incorrect. Let's assume that $F$ is a field and $x$ an indeterminate. A correct proof might (I believe) go as follows:</p>
<p>Suppose
$$
F[x]=\operatorname{span}\{f_1,f_2,\dots,f_n\}.
$$</p>
<p>Let us choose a positive integer $N$ such that $N > \deg (f_i)$ for all $i=1,\dots,n$. As $\{f_1,f_2,\dots,f_n\}$ spans $F[x]$ we can find scalars $a_1, a_2, \dots,a_n$ such that
$$
x^N=a_1f_1+a_2f_2+\cdots+a_nf_n.
$$
This yields the contradiction
$$
0= x^N - a_1f_1 - a_2f_2 - \cdots - a_nf_n\neq0.
$$</p>
|
2,881,673 | <p>I've searched all over the internet and cannot seem to factorise this polynomial.</p>
<p>$x^4 - 2x^3 + 8x^2 - 14x + 7$</p>
<p>The result should be $(x − 1)(x^3 − x^2 + 7x − 7)$</p>
<p>What are the steps to get to that result?
I've tried grouping but doesn't seem to work...</p>
| yakobyd | 569,127 | <p>Considering $(a-b)^2 = a^2 - 2ab + b^2$, notice that you may manipulate the polynomial as follows:</p>
<p>\begin{align}
x^4 - 2x^3 + 8x^2 -14x + 7 &= (x^4 -2x^3 + x^2) + (7x^2 - 14x + 7) \\
&= x^2(x^2 - 2x + 1) + 7(x^2 - 2x + 1) \\
&= (x-1)^2(x^2 + 7)
\end{align}</p>
<p>which yields a complete factorization over the real numbers.</p>
|
375,372 | <p>Using the $\epsilon-M $ definition of the limit, calculate
$$\lim_{x\to\infty}\frac{3x^2+7}{x^2+x+8}.$$</p>
<p>Working so far: </p>
<p>$$\lim_{x\to\infty}\frac{3x^2+7}{x^2+x+8}=3$$</p>
<p>Given $\epsilon>0$, I want M s.t. $x>M \implies \left|\frac{3x^2+7}{x^2+x+8}-3 \right|<\epsilon$</p>
<p>$$\left|\frac{3x^2+7}{x^2+x+8}-3 \right|<\epsilon$$</p>
<p>$$\left|\frac{3x^2+7-3(x^2+x+8)}{x^2+x+8} \right|<\epsilon$$</p>
<p>$$\left|\frac{-3x-17}{x^2+x+8} \right|<\epsilon$$</p>
<p>And now I'm stuck.. Any help would be great, thanks.</p>
| mathemagician | 49,176 | <p>Let $x>\max\{\sqrt{17},\frac{3}{\epsilon-1}\}$. We have $x>0$. This gives us</p>
<p>$$\left|\frac{-3x-17}{x^2+x+8} \right|= \left|\frac{3x+17}{x^2+x+8} \right|=\frac{3x+17}{x^2+x+8}<\frac{3x+17}{x^2}<\frac{3}{x}+1<\epsilon$$</p>
<p>This is incomplete as it assumes $\epsilon\neq 1$. When $\epsilon=1$ you can solve a quadratic and find which $x$'s work.</p>
|
681,608 | <blockquote>
<p>Prop. 6.9: Let $X \to Y$ be a finite morphism of non-singular curves, then for any divisor $D$ on $Y$ we have $\deg f^*D=\deg f\deg D$.</p>
</blockquote>
<p>I can not understand two points in the proof:</p>
<p>(1) (Line 9) Now $A'$ is torsion free, and has rank equal to $r=[K(X):K(Y)]$.</p>
<p>Since it is a torsion-free module over PID $O_Q$, I see it is free, but how to calculate its rank?</p>
<p>(2) (Line 15) Clearly $tA'=\bigcap(tA'_{\mathfrak m_i}\cap A')$ so ... I don't know how to show the claim ? </p>
| Tomo | 62,940 | <p>I like @SomeEE's answer for Line 15. I want to give a different justification for Line 9, because I wasn't able to justify to myself the statement 'Passing to quotient fields.' Unfortunately in my notation below, B = A' and A is the local ring at Q; noting that normalization commutes with localization we see that A ⊂ B in my notation below is indeed an integral extension of rings.</p>
<p>We have an extension of function fields of curves over an algebraically closed field k.
In fact we may reduce to the setting of k ⊂ A ⊂ B where B is the integral closure of the PID A, and is finite as an A-module.
Since k is alg. closed it is perfect hence K(B) is sep./k hence over K(A).
Now a beautiful theorem:</p>
<blockquote>
<p><strong>Thm</strong> If L/K is separable and A is a PID, then every f.g. B-submodule M ≠ 0 of L is a free A-module of rank [L : K]. In particular, B admits an integral basis over A.</p>
</blockquote>
<p>(One can find this theorem, for example, in the first few pages of Neukirch.)</p>
<p>Hence B has rank [K(B) : K(A)].</p>
<p>Thank you!</p>
<p>P.S. I was also at first stumped by the isomorphism</p>
<p>$$A'/(tA_{\mathfrak m_i}\cap A')\cong A_{\mathfrak m_i}/tA_{\mathfrak m_i},$$</p>
<p>but this isn't so mysterious once one takes into account that the ring on the left is already local with maximal ideal $\mathfrak m_i$, by @SomeEE's 'Dedekind' comment.</p>
|
161,029 | <p>I have not seen a problem like this so I have no idea what to do.</p>
<p>Find an equation of the tangent to the curve at the given point by two methods, without elimiating parameter and with.</p>
<p>$$x = 1 + \ln t,\;\; y = t^2 + 2;\;\; (1, 3)$$</p>
<p>I know that $$\dfrac{dy}{dx} = \dfrac{\; 2t\; }{\dfrac{1}{t}}$$</p>
<p>But this give a very wrong answer. I am not sure what a parameter is or how to eliminate it.</p>
| Lemon | 26,728 | <p><strong>Method 1</strong> Eliminating. I think they want to write everything in terms of x first. </p>
<p>$x = 1 + ln(t) \iff e^{x - 1} = t$</p>
<p>$y = t^2 + 2 = e^{2x -2} + 2 \implies y' = 2e^{2x -2}$</p>
<p>At (1,3) $y' = 2$. So the tangent line is $y = 2(x- 1) + 3$ or parametrically let $x - 1 = t
\iff x = 1 + t$ and $y = 2t + 3$</p>
<p><strong>Method 2</strong> No eliminating. </p>
<p>Let $r = (1 + ln(t),2 + t^2) \implies r' = (1/t, 2t)$. At (1,3), $t = 1$, therefore $r'(1) = (1,2)$</p>
<p>$r = (1,3) + s(1,2)$ which gives you $x = 1 + s$ and $y = 3 + 2s$</p>
|
192,883 | <p>Can anyone please give an example of why the following definition of $\displaystyle{\lim_{x \to a} f(x) =L}$ is NOT correct?:</p>
<p>$\forall$ $\delta >0$ $\exists$ $\epsilon>0$ such that if $0<|x-a|<\delta$ then $|f(x)-L|<\epsilon$</p>
<p>I've been trying to solve this for a while, and I think it would give me a greater understanding of why the limit definition is what it is, because this alternative definition seems quite logical and similar to the real one, yet it supposedly shouldn't work.</p>
| Pedro | 23,350 | <p>I'm a little confused when you ask wether it is correct or not. Do you ask as opposed to the well known definition? That is, as opposed to </p>
<blockquote>
<p>We say that $\lim\limits_{x\to a}f(x)=\mathscr L$ if for every $\epsilon >0$ there exists a $\delta >0$ such that, for all $x$, if $0<|x-a|<\delta$ then $|f(x)-\mathscr L|<\epsilon$.</p>
</blockquote>
<p>If so, you can just see $\epsilon$ and $\delta$ are reversed.</p>
<p>However, suppose this definition was not given, and we want to define what we mean by limit. First, we clearly want to understand that we're concerned about what happens <strong>near</strong> $a$, but not <strong>at</strong> $a$. The idea of a "limit" near $a$ is then that of a value $\mathscr L$ a function $f(x)$ approaches when $x$ approaches $a$, So the idea we want to capture is that a function $f(x)$ has a number $\mathscr L$ as a <strong>limit</strong> when $x$ approaches $a$ if we can make $\mathscr L$ and $f(x)$ <strong>as near as we wish</strong>, by taking $x$ <em>sufficiently</em> near $a$. So when do we say that two number are <strong>near</strong>? We need to formalize our idea of proximity of numbers. </p>
<p>$\bf A$. So, we can associate to each $x,y\in \bf R$ the real number $d(x,y)=|x-y|$ and call it the <em>distance from $x$ to $y$</em>. Note that this distance has some properties we really want any notion of distance to have:</p>
<p>$(1)$ The distance is symmetric. $$d(x,y)=d(y,x)$$</p>
<p>$(2)$ It is always positive, <em>unless</em> $x=y$. That is, the distance of two numbers is zero if and only if they are the same number. $$d(x,y)>0 \text{ and } d(x,y)=0\iff x=y$$</p>
<p>$(3)$ The distance from a point $x$ to a point $y$ will always be less than or equal to the distance from $x$ to another $z$ plus the distance from that $z$ to $y$. We're just saying the shortest distance from $x$ to $y$ is precisely the straight line that joints them (this generalizes to higher dimensions). $$d(x,y)\leq d(x,z)+d(z,y)$$</p>
<p>$\bf B$. Now, consider this silly theorem:</p>
<blockquote>
<p>Suppose $x,y\in \Bbb R$. Then, for every $\epsilon>0$, $d(x,y)<\epsilon$ if and only if $x=y$. </p>
</blockquote>
<p>One direction is easy: if $x=y$, then clearly $d(x,y)=0<\epsilon$ for positive $\epsilon$. Now suppose that for any $\epsilon>0$, we have that $d(x,y)=0<\epsilon$. We aim to prove $x=y$. We will argue by contradiction. Since the distance is symmetric, we might assume $x<y$. Then $d(x,y)=|x-y|>0$. So $|x-y|$ is an $\epsilon>0$, which would mean that $|x-y|<|x-y|$, which is absurd. Thus, it must be that $x=y$.</p>
<p>$\bf C$. Now, we have a formal definition of the notion of <em>near</em>, and we can say that two numbers are near if $d(x,y)$ is small. In particular we just showed that if $x=y$, $d(x,y)$ is a smaller than any positive quantity given, since it is zero. It is but only logical to say that making $\mathscr L$ as near as we wish to $f(x)$ is making $|f(x)-\mathscr L|<\epsilon$ , no matter how small $\epsilon$ is, with $\epsilon<0$.</p>
<p>$\bf D$. We can now try and think about a degree of closeness. Given a number $\delta >0$, we can standarize and say that $x$ and $a$ are sufficiently near if $d(x,a)<\delta$. Think of it as the ruler we use to tell wether you can go in a rollercoaster ride or not. If $d(x,a)\geq \delta$, then $x$ is "bad" and we discard it.</p>
<p>$\bf E$. It is important to note this: we are concerned about making $f(x)$ and $\mathscr L$ close, and we want to succeed in doing so by making $x$ close to $a$. It is not our objective to make $x$ and $a$ close, but our means. So our definition must capture this: given a desired proximity of $f$ and $\mathscr L$, there must be a moment in which any $x$ close enough to $a$ will make $f(x)$ be close to $\mathscr L$. We don't forget that $x\neq a$, which only means $0<|x-a|$. So, let's try and write something down, considering what we discussed in $\bf B,C,D$. </p>
<blockquote>
<p>We say that $f(x)$ has $\mathscr L$ as a limit when $x$ approaches $a$ if for any <em>prescribed degree of closeness</em>, making $x$ <em>sufficiently close</em> to $a$, but <em>not equal to $a$</em>, will imply that for all those $x$, $f(x)$ and $\mathscr L$ will be within that prescribed degree of closeness.</p>
</blockquote>
<p>Applying $\bf B,C$, we can write</p>
<blockquote>
<p>We say that $f(x)$ has $\mathscr L$ as a limit when $x$ approaches $a$ if for any $\epsilon >0$, making $x$ <em>sufficiently close</em> to $a$, but <em>not equal to $a$</em>, will imply that for all those $x$, $|f(x)-\mathscr L|<\epsilon$.</p>
</blockquote>
<p>Using $\bf D$, we can write</p>
<blockquote>
<p>We say that $f(x)$ has $\mathscr L$ as a limit when $x$ approaches $a$ if for any $\epsilon >0$, there is a $\delta$ such that for all $x\neq a$, $|x-a|<\delta$ implies $|f(x)-\mathscr L|<\epsilon$.</p>
</blockquote>
<p>Finally, since $x\neq a$ is the same as $0<|x-a|$, we can be more succint and write.</p>
<blockquote>
<p>We say that $f(x)$ has $\mathscr L$ as a limit when $x$ approaches $a$ if for any $\epsilon >0$, there is a $\delta$ such that for all $x$, $0<|x-a|<\delta$ implies $|f(x)-\mathscr L|<\epsilon$.</p>
</blockquote>
<p>You can prove, or find proofs, that if the limit of $f$ exists for some $a$, then it is <strong>unique</strong>. And this is important. As others have pointed out, this definition, giving freedom on $d(f(x),\mathscr L)$,for example, makes limits lose their uniqueness. </p>
<p>Now, your definition clearly doesn't capture our original idea: It is somehow saying $f$ has $\mathscr L$ as a limit if for any prescribed degree of closeness $\delta$, there will exist some positive number $\epsilon$ such that making $d(x,a)<\delta$ will imply $d(f(x),\mathscr L)<\epsilon$. But this is giving us a lot of freedom on $d(f(x),\mathscr L)$, which we are most worried about. <em>Of course</em> we can make $x$ and $a$ as near as we wish, but the question is, can we make $f$ and $\mathscr L$ as close as we wish? </p>
|
3,554,393 | <p>Namely, I need to prove <span class="math-container">${\max\limits_i} |\lambda_i| \leq {\max\limits_i}{\sum\limits_j} |M_{ij}|\mid$</span>, where <span class="math-container">$M$</span> is the matrix and <span class="math-container">$\lambda_i$</span> are its eigenvalues.</p>
<p>I'm not sure if there is any helpful theorem or lemma. Is there any hint on how to solve this?</p>
| Arthur | 15,500 | <p>Hint: Take an eigenvector <span class="math-container">$v$</span> corresponding to the maximum eigenvalue, and scale it so that the largest entry (in absolute value) is <span class="math-container">$1$</span>, at the <span class="math-container">$j$</span>th row. Look at the <span class="math-container">$j$</span>th row of <span class="math-container">$Mv$</span>.</p>
|
440,242 | <p>I'm pretty sure almost all mathematicians have been in a situation where they found an interesting problem; they thought of many different ideas to tackle the problem, but in all of these ideas, there was something missing- either the "middle" part of the argument or the "end" part of the argument. They were stuck and couldn't figure out what to do.</p>
<blockquote>
<ol>
<li>In such a situation what do you do?</li>
<li>Is the reason for the "missing part" the incompleteness in the theory of the topic that the problem is related to? What can be done to find the "missing part"?</li>
</ol>
</blockquote>
<p>For tenure-track/tenure professors, maybe this is not a big deal because they have "enough" time and can let the problem "stew" in the "back-burner" of their mind, but what about limited-time positions, e.g. PhD students, postdocs, etc., where the student/employee has to prove their capability to do "independent" research so that they can be hired for their next position? I think for these people it is quite a bit of a problem because they can't really afford to spend a "lot" of time thinking about the same problem.</p>
| Jon Bannon | 6,269 | <p>A little buddhism goes a long way here, it seems.</p>
<p>There is a slight difference between being stuck and being obsessed.</p>
<p>As the saying goes, doing the same thing over and over again and expecting a different result is the definition of insanity.</p>
<p>So, if by stuck you mean obsessively trying to remove a single obstacle despite it's <em>not</em> being ready to move, I'm not so sure of the value of this.</p>
<p>On the other hand, if by stuck you mean that you are filling in more and more detail around the obstacle in order to "soften it up", then this is quite a healthy process.</p>
<p>I mention the buddhism, because "it is stuck" is much healthier than "I am stuck" (although a buddhist may not even acknowledge that anything is stuck at all). Mathematicians are a bit too obsessed with the "I", and this creates undue emotional distress and paralysis. As you mention, this situation is much worse for those who don't have the luxury of time. This creates a very sad situation, because being all bunched up like this really messes with the ability to relax and perceive what is right in front of you. Rather than naturally growing a point of view, you end up endlessly beating yourself up for trying to repeatedly stuff a square block in a round hole.</p>
<p>If you stop taking things so seriously, you will find that it is enough to attend to the problem and obstacle fluidly, and enjoy the process of exploring around the obstacle. A good problem provides many opportunities to do this.</p>
<p>You might ask what constitutes a good problem, even. Stubbornly chasing the accolades of breaking a hard problem looks a little silly if you stand back far enough, especially when contrasted with enjoying the unfolding of a breathtaking structure that is "ripe", so to speak.</p>
<p>I get the impression that too many people are in love with the idea of being a mathematician, more than with the mathematics itself. If you honestly investigate the quality of your experience in doing math, maybe a break is in order, or a career shift, or a problem shift. Good mathematics is supposed to be fun. I think we forget this much too often!</p>
<p>Regarding the career trajectory, there are economic realities to consider. Everyone feels that their work/research/contribution should be valuable to the development of their subject. This is certainly true, but I think we tend to overvalue our own contributions. In the grand scheme of things, only a few of us will find a fruitful enough direction to support a career in research. (I am not one of these, although I've had some rewarding ideas.) If you put the "I" on the back burner, this is easier to accept. Maybe, despite your interest in obsessing over a problem, your effort would be better spent elsewhere? Maybe you fit into the world differently than you plan to? Ironically, accepting this reality may be the best way to get "unstuck" on your problem.</p>
<p>Once I was talking with Peter Jones from Yale about this sort of thing. I was talking to him about how hard I was working on such and such and how I had to learn to do this and that thing better and change my focus in such a way. Peter said to me, "That's funny, I always just did what I liked!" I realized afterward that there was a sobering reality to this. He just fit into his part of the mathematical world a bit more naturally than the rest of us. All of his experience perhaps led naturally to where he was. Others, like me, were forcing it.</p>
<hr />
<p>Edit:</p>
<p>Here is a Zen Koan:</p>
<blockquote>
<p>A monk asked Tozan, "How can we escape the cold and the heat?" Tozan replied, "Why not go where there is no cold and heat?" "Is there such a place?", the monk asked. Tozan commented, "When cold, be thoroughly cold; when hot be hot through and through."</p>
</blockquote>
<p>So there is an answer the OP's question akin to the above Koan:</p>
<blockquote>
<p>When stuck, be thoroughly stuck.</p>
</blockquote>
|
440,242 | <p>I'm pretty sure almost all mathematicians have been in a situation where they found an interesting problem; they thought of many different ideas to tackle the problem, but in all of these ideas, there was something missing- either the "middle" part of the argument or the "end" part of the argument. They were stuck and couldn't figure out what to do.</p>
<blockquote>
<ol>
<li>In such a situation what do you do?</li>
<li>Is the reason for the "missing part" the incompleteness in the theory of the topic that the problem is related to? What can be done to find the "missing part"?</li>
</ol>
</blockquote>
<p>For tenure-track/tenure professors, maybe this is not a big deal because they have "enough" time and can let the problem "stew" in the "back-burner" of their mind, but what about limited-time positions, e.g. PhD students, postdocs, etc., where the student/employee has to prove their capability to do "independent" research so that they can be hired for their next position? I think for these people it is quite a bit of a problem because they can't really afford to spend a "lot" of time thinking about the same problem.</p>
| Boris Bukh | 806 | <p>Not much changes with tenure -- we do not suddenly change our habits after a decade of working.</p>
<p>Our job is and has always been to advance the science. Since the most important unsolved problems tend to be also the most well-known, it usually means that we are almost (but not always!) perpetually stuck on the problems that we really want to solve.</p>
<p>Practically, this means that we spend most of our time trying to gain insights that would help us on a few important problems. So, when we are stuck, we 1) make toy problems encapsulating some of the difficulties 2) seek generalizations that would remove distractions 3) read more, or talk to others in a hope to broaden our knowledge 4) work on something else to avoid getting into fruitless loops.</p>
<p>However, we always remember the problems that we must, eventually, solve.</p>
|
3,424,259 | <p>The system in question is <span class="math-container">$$\begin{cases} x_1 -x_2 + x_3 = -1 \\ -3x_1 +5x_2 + 3x_3 = 7 \\ 2x_1 -x_2 + 5x_3 = 4 \end{cases}$$</span></p>
<p>After writing this in matrix-form and performing row-operations we can show that</p>
<p><span class="math-container">$$
\begin{matrix}
-1 & 4 & 8 &| 11 \\
0 & 1 & 3 &|6\\
0 & 1 & 3&|10 \\
\end{matrix}
$$</span></p>
<p>Substitute back our variables and we get</p>
<p><span class="math-container">$$\begin{cases} x_2 + 3x_3 = 6 \\ x_2 +3x_3 = 10 \end{cases}$$</span></p>
<p>Which is a contradictory statement that shows our system has no solutions. Is this a suffciently rigorous way of answering the question : 'Does this system have a solution?'.</p>
| G. Gare | 568,973 | <p>You can do one step more and transform the last line to <span class="math-container">$(0, 0, 0 | 4)$</span> so that it's clear that the system is not compatible and hence admits no solution (contradiction <span class="math-container">$0=4$</span> for any value of the variables <span class="math-container">$x_1, x_2, x_3$</span>).</p>
|
126,168 | <p>$$\zeta(it)=2it\pi it−1\sin(i\pi t/2)\Gamma(1−it)\zeta(1−it).$$
Everything on the RHS is never zero,</p>
<p>Does that means LHS has no zeros, since $\sin(s)$ has a simple zero at $s=0$ while $\zeta(1−s)$ has a simple pole at $s=0$ (the Laurent expansion), so the product $\sin(i\pi t/2)\zeta(1−it)$ is finite and nonzero at $t=0$.</p>
<p>My question is that from functional equation one can't get direct answer $\zeta(0)=−1/2$ unless using alternate infinite series !</p>
| Aryabhata | 1,102 | <p>Do you mean the functional equation</p>
<p>$$\zeta(s) = 2(2\pi)^{s-1} \Gamma(1-s) \zeta(1-s) \sin (\frac{1}{2} \pi s)$$</p>
<p>?</p>
<p>which is valid for all $s \neq 1$ and $s \neq 0$.</p>
<p>Even though you cannot directly substitute $s =0$ or $s=1$, what you can do is multiply by $s-1$ to get</p>
<p>$$[\zeta(s)(s-1)] = 2(2\pi)^{s-1} [\Gamma(1-s) (s-1)] \zeta(1-s) \sin (\frac{1}{2} \pi s) \tag{1}$$</p>
<p>$\zeta$ has a simple pole at $1$ with Residue $1$, and $\Gamma$ has a simple pole at $0$ with Residue $1$.</p>
<p>Taking $s \to 1$ in 1) now gives you the result that $\displaystyle \zeta(0) = -\frac{1}{2}$.</p>
|
126,168 | <p>$$\zeta(it)=2it\pi it−1\sin(i\pi t/2)\Gamma(1−it)\zeta(1−it).$$
Everything on the RHS is never zero,</p>
<p>Does that means LHS has no zeros, since $\sin(s)$ has a simple zero at $s=0$ while $\zeta(1−s)$ has a simple pole at $s=0$ (the Laurent expansion), so the product $\sin(i\pi t/2)\zeta(1−it)$ is finite and nonzero at $t=0$.</p>
<p>My question is that from functional equation one can't get direct answer $\zeta(0)=−1/2$ unless using alternate infinite series !</p>
| Max Clifford | 71,661 | <p>The LHS has no zeroes and therein lies the proof of the Prime Number Theorem.</p>
<p>You can get $\zeta(0)=-1/2$ from the functional equation; though you must be comfortable with limit manipulation as the poster above rightly states that the functional equation is not valid for $s=0,1$. The derivation at the end of this page is simple enough.</p>
<p><a href="http://planetmath.org/valueoftheriemannzetafunctionats0" rel="nofollow">http://planetmath.org/valueoftheriemannzetafunctionats0</a>.</p>
<p>Regards</p>
|
1,680,269 | <p>Here $\mathbb{Z}_{n}^{*}$ means $\mathbb{Z}_{n}-{[0]_{n}}$</p>
<p>My attempt:</p>
<p>$(\leftarrow )$</p>
<p>$p$ is a prime, then, for every $[x]_{n},[y]_{n},[z]_{n}$ $\in (\mathbb{Z}_{n}^{*},.)$ are verified the following:</p>
<p>1) $[x]_{n}.([y]_{n}.[z]_{n}) = ([x]_{n}.[y]_{n}).[z]_{n}$, since from the operation . we have $[a]_{n}.[b]_{n}=[a.b]_{n}$ and . is associative in $\mathbb{Z}$.</p>
<p>2) There is an element $e$ such that $[x]_{n}.e = e.[x]_{n} = [x]_{n}$, since $ [x]_{n}.[1]_{n} = [x.1]_{n} = [x]_{n} = [x.1]_{n} = [x]_{n}[1]_{n}$</p>
<p>But I don't know how to check the inverse property, neither how to do the $(\rightarrow)$ part. </p>
<p>Thanks!</p>
| Aloizio Macedo | 59,234 | <p>Consider the function</p>
<p>$$f: K \times L \rightarrow \mathbb{R}$$
$$(x,y) \mapsto |x-y|.$$</p>
<p>$f=\vert\cdot\vert \circ -|_{K \times L},$ where $-: \mathbb{R} \times \mathbb{R}$ is the subtraction and $| \cdot|$ is the module function. Hence, $f$ is the composition of continuous functions, therefore continuous. It is clear that $\inf_{x \in K \times L} f=d$. It is also clear, since those sets are disjoint, that $f>0$. Since $f$ is a real continuous function in a compact set, it achieves its infimum in its domain. Therefore, $d>0$.</p>
|
1,298,971 | <p>Any help on this problem is greatly appreciated! I'm completely stuck</p>
<p>School board officials are debating whether to require all high school seniors to take a proficiency exam before graduating. A student passing all three parts (mathematics, language skills, and general knowledge) would be awarded a diploma; otherwise, he or she would receive only a certificate of attendance. A practice test given to this year’s ninety-five hundred seniors resulted in the
following numbers of failures:
$$\begin{array}{|l|l|}
\hline \\
\textbf{Subject Area} & \textbf{Number Failing} \\\hline \\
\text{Mathematics} & 3325 \\\hline \\
\text{Language Skills} & 1900 \\\hline \\
\text{General Knowledge} & 1425\\ \hline
\end{array}$$</p>
<p>If “Student fails mathematics,” “Student fails language skills,” and “Student fails general knowledge” are independent events, what proportion of next year’s seniors can be expected to fail to qualify for a diploma?</p>
| Barry Cipra | 86,747 | <p>The length of a side, $L$, is analogous to the <em>diameter</em> of a circle, not its radius. So the appropriate variable to use is $\ell=L/2$, in which case the perimeter is $8\ell$, which integrates to $4\ell^2$, and that <em>is</em> correct, since $4\ell^2=(2\ell)^2=L^2$.</p>
<p>(Remark: I posted this before reading Andre Nicolas's answer. It's basically the same. But maybe the first sentence here is helpful.)</p>
|
2,166,917 | <p>$20$ questions in a test. The probability of getting correct first $10$ questions is $1$. The probability of getting correct next $5$ questions is $\frac 13$. The probability of getting correct last $5$ questions is $\frac 15$. What is the probability of getting exactly $11$ questions correctly?</p>
<p>This is the question. I don't know how to calculate this question.</p>
<p>I tried $$1* {}^5C_1*{\frac 13}\left(\frac 23\right)^4*\left(\frac 45\right)^5+1*\left(\frac 23\right)^5*{}^5C_1*\left(\frac 15\right)\left(\frac 45\right)^4$$</p>
<p>But I am not sure the answer.</p>
<p>What if asking futher about expectation and variance? that will be a mess</p>
| Jaroslaw Matlak | 389,592 | <p>For $k<0$:</p>
<p>$P(k)=0.$</p>
<p>For $k\geq10$:</p>
<p>$$P(k)=\sum_{i=\max(k-15,0)}^{\min(k-10,5)}p(i,k-10-i)$$
Where
$$p(i,j)={5 \choose i}\left(\frac{1}{3}\right)^i\left(\frac{2}{3}\right)^{5-i} {5 \choose j}\left(\frac{1}{5}\right)^j\left(\frac{4}{5}\right)^{5-j}$$</p>
<p>Finally: for $k\in \{0,1,2,...,20\}$:
$$P(k)=\begin{cases}0&, \text{for }k<10\\
\sum_{i=\max(k-15,0)}^{\min(k-10,5)}p(i,k-10-i), \text{for }k\geq 10\end{cases}$$</p>
<p>For $k=11$ we have:
$$P(11)=p(0,1)+p(1,0)=5\frac{2+4}{15}\left(\frac{8}{15}\right)^4=\frac{8192}{50625}\approx 0.16$$</p>
|
2,231,949 | <p>To find the minimal polynomial of $i\sqrt{-1+2\sqrt{3}}$, I need to prove that
$x^4-2x^2-11$ is irreducible over $\Bbb Q$. And I am stuck. Could someone please help? Thanks so much!</p>
| dxiv | 291,201 | <p>As noted already, the rational root theorem excludes rational roots, which only leaves a product of rational quadratics as a potential factorization.</p>
<p>By the way the polynomial was constructed, it is known that its roots are $\,\pm i \sqrt{2 \sqrt{3}-1}\,$ and $\,\pm \sqrt{2 \sqrt{3}+1}\,$. Any quadratic factor of the quartic would need to have two of those as roots, and among quadratics with rational (non-complex) coefficients the only possible pairings would be between the two real roots, or between the two complex roots, respectively. However, both cases result in quadratics that have irrational coefficients $\,(x^2 - 2 \sqrt{3} - 1)\cdot(x^2 + 2 \sqrt{3} - 1)\,$, thus no factorization over $\mathbb{Q}$ exists.</p>
|
3,328,387 | <p>Suppose I have two positive semi-definite <span class="math-container">$n$</span>-by-<span class="math-container">$n$</span> matrices <span class="math-container">$A$</span>, <span class="math-container">$B$</span> and an <span class="math-container">$n$</span>-by-<span class="math-container">$n$</span> identity matrix <span class="math-container">$I$</span>, and I'm looking for a way to compute, approximate or bound the following quantity:</p>
<p><span class="math-container">$$(A\otimes I + I \otimes A)^{-1} \text{vec}B$$</span></p>
<p>Concretely I'm dealing with matrices with <span class="math-container">$n$</span> ranging from <span class="math-container">$100$</span> to <span class="math-container">$4000$</span>, so <span class="math-container">$A$</span> is easy to invert, while <span class="math-container">$A \otimes I$</span> is too large, so need a way to compute this using operations on <span class="math-container">$n$</span>-by-<span class="math-container">$n$</span> matrices</p>
<p>Additionally, I found the following to give a decent approximation when <span class="math-container">$A$</span>, <span class="math-container">$B$</span> can be Kronecker-factored, wondering if there's a reason for that.</p>
<p><span class="math-container">$$0.5 A^{-0.5} B A^{-0.5}$$</span></p>
<p>Any tips or literature pointers are appreciated!</p>
| Kwin van der Veen | 76,466 | <p>Your equation is equivalent to solving the following equation for <span class="math-container">$P$</span></p>
<p><span class="math-container">$$
A\,P + P\,A^\top = B. \tag{1}
$$</span></p>
<p>Such equation also solves for the <a href="https://en.wikipedia.org/wiki/Controllability_Gramian#Controllability_Gramian" rel="nofollow noreferrer">infinite time (controllability) Gramian</a>, which can also be calculated with</p>
<p><span class="math-container">$$
P = -\int_0^\infty e^{A\,t}\,B\,e^{A^\top t}\,dt. \tag{2}
$$</span></p>
<p>This integral should also be of size <span class="math-container">$n \times n$</span>. As time goes to infinity <span class="math-container">$e^{A\,t}$</span> is not well defined when <span class="math-container">$A$</span> is positive semi-definite. It can also be <a href="https://en.wikipedia.org/wiki/Sylvester_equation" rel="nofollow noreferrer">noted</a> that in order to be able to solve for <span class="math-container">$P$</span> the spectra of <span class="math-container">$A$</span> and <span class="math-container">$-A$</span> need to be disjoint. In order words <span class="math-container">$A$</span> can't have an eigenvalue of zero, which would imply that <span class="math-container">$A$</span> should be positive definite instead of positive semi-definite. When multiplying both sides of <span class="math-container">$(1)$</span> by minus ones keeps it equivalent to the original problem. This is also equivalent to multiplying both <span class="math-container">$A$</span> and <span class="math-container">$B$</span> by minus one, which transforms <span class="math-container">$(2)$</span> into</p>
<p><span class="math-container">$$
P = \int_0^\infty e^{-A\,t}\,B\,e^{-A^\top t}\,dt. \tag{3}
$$</span></p>
<p>Now the term <span class="math-container">$e^{-A\,t}$</span> does stay bounded and converges to zero as time goes to infinity. The rate of how fast this integral converges should be roughly inversely proportional to the smallest eigenvalue of <span class="math-container">$A$</span>, however any finite time numerical approximation of this integral can be used as a lower bound for <span class="math-container">$P$</span>.</p>
|
92,660 | <p>Let $X$ be a nonsingular projective variety over $\mathbb{C}$, and let $\widetilde{X}$ be the blow-up of X at a point $p\in X$.
What relationships exist between the degrees of the Chern classes of $X$ (i.e. of the tangent bundle of $X$) and the degrees of the Chern classes of $\widetilde{X}$?</p>
<p>Thanks.</p>
| Johannes Nordström | 13,061 | <p>Like Georges says, 15.4 of Fulton's Intersection Theory deals with the general theory. For this special case it's not too hard to work out the Chern classes by hand though.</p>
<p>Let $f : \widetilde X \to X$ be the projection and $E \cong \mathbb{C}P^{n-1}$ the exceptional divisor.
$H^*(\widetilde X) \cong f^*H^*(X) \oplus \langle \textrm{Poincare duals of planes } P_k \textrm{ in } E $ $\textrm{of dimension }k = 1,\ldots, n-1\rangle$. Note that $[P_{n-i}][P_{n-j}] = -[P_{n-i-j}]$, while $(f^*\alpha) [P_k] = 0$ for any $\alpha \in H^i(X)$ ($i, k > 0$).</p>
<p>$f^* c_i(X)$ and $c_i(\widetilde X)$ are equal outside the exceptional divisor, so their difference is Poincare dual to something in $E$. On the other hand the restriction of $f^*c_i(X)$ to $E$ is 0 (for $i > 0$), while the restriction of $c_i(E)$ is $c_i(\mathcal{O}(1)^n \oplus \mathcal{O}(-1)) = \left({n\choose i} - {n \choose i-1}\right)H^i$, where $H \in H^2(E)$ is the hyperplane class. For $0 < i < n$ we deduce that $c_i(\widetilde X) = f^*c_i(X) - \left({n\choose i} - {n \choose i-1}\right)[P_{n-i}]$ by comparing the evaluations on $P_i$.</p>
|
424,694 | <p>Let <span class="math-container">$p$</span> be a prime, and consider <span class="math-container">$$S_p(a)=\sum_{\substack{1\le j\le a-1\\(p-1)\mid j}}\binom{a}{j}\;.$$</span>
I have a rather complicated (15 lines) proof that <span class="math-container">$S_p(a)\equiv0\pmod{p}$</span>. This must be
extremely classical: is there a simple direct proof ?</p>
| Zhi-Wei Sun | 124,654 | <p>Actually, a further extension was given in my paper <a href="http://maths.nju.edu.cn/%7Ezwsun/92a.pdf" rel="nofollow noreferrer">Combinatorial congruences and Stirling numbers</a> [Acta Arith. 126 (2007), 387-398]. Now I state Corollary 1.3 (a consequence of Theorem 1.1) in the 2007 paper of mine.</p>
<p>Let <span class="math-container">$p$</span> be a prime. If <span class="math-container">$l,l',m$</span> are positive integers with <span class="math-container">$l'\ge l>m/p$</span> and
<span class="math-container">$l'\equiv l\pmod {(p-1)p^{\lfloor\log_p m\rfloor}}$</span>, then
<span class="math-container">$$\sum_{j\equiv r\pmod {p-1}}\binom{l'}j S(j,m)
\equiv\sum_{j\equiv r\pmod {p-1}}\binom{l}j S(j,m)\pmod p,$$</span>
where <span class="math-container">$S(j,m)$</span> denotes the Stirling number of the second kind.</p>
<p>In the case <span class="math-container">$m=1$</span>, this yields Glaisher's result.</p>
|
143,655 | <p>According to <a href="http://en.wikipedia.org/wiki/Lipschitz_continuity#Properties" rel="nofollow noreferrer">wikipedia</a> a function <span class="math-container">$f\colon \mathbb{R}^n\to\mathbb{R}^n$</span> that is continuously differentiable, is also locally Lipschitz.</p>
<p>I there someone who knows a good reference which contains a proof of this statement?</p>
| Shuhao Cao | 7,200 | <p>The exact quote on wiki was:</p>
<blockquote>
<p>In particular, any $C^1$ function is locally Lipschitz, as continuous functions on a locally compact space are locally bounded so its gradient is.</p>
</blockquote>
<p>The logic here is we would like to show the gradient of a $C^1$-function is locally bounded on a locally compact space, thus to obtain the Lipschitz continuity. Hence your question boils down to how to prove:</p>
<blockquote>
<p>A continuous function on a locally compact space is locally bounded.</p>
</blockquote>
<p>While I am not one hundred percent sure which real analysis textbook has the exact result, you may refer to this <a href="http://en.wikipedia.org/wiki/Continuous_functions_on_a_compact_Hausdorff_space" rel="nofollow">Wikipedia entry</a>, we could use <a href="http://en.wikipedia.org/wiki/Stone%E2%80%93Weierstrass_theorem" rel="nofollow">Stone-Weierstrass theorem</a> to prove: roughly speaking, since $\mathbb{R}^n$ is locally compact, for any point $x\in \mathbb{R}^n$, choose a compact neighborhood $S$ of $x$, we could use S-W theorem to prove that, for a continuous function on $S$, there exists a sequence of vector-valued polynomials $\{p_n\}$ that converge uniformly to the function of interest $f$ on $S$, and use the triangle inequality in the supreme norm, we could see $f$ is bounded on $S$. Thus the local boundedness of $f$ is proved.</p>
<p>To sum up: Gradient of $C^1$-function is continuous $\to$ A continuous function on a locally compact space(notice the gradient of $f$ you gave is the Jacobian matrix but the argument still applies) is locally bounded $\to$ Gradient of $C^1$-function is locally bounded $\to$ By mean value theorem $C^1$-function is locally Lipschitz.</p>
<hr>
<p><code>EDIT:</code> As Neal suggested, it suffices to argue as follows: For any $x_0\in \mathbb{R}^n$, on a compact set within some neighborhood $S \subset B(x_0,\epsilon)$, for $x,y\in S$:
$$
\frac{\|f(x)-f(y)\|}{\|x-y\|} \leq \sup_{\xi\in S} \| Df\|<L(x_0)
$$
again we rely on <em>continuous functions are bounded on compact set</em>, and the Lipschitz constant depends on $x_0$, this is the local Lipschitz property you need.</p>
|
3,906,920 | <p>A string in <span class="math-container">$\{0, 1\}*$</span> has even parity if the symbol <span class="math-container">$1$</span> occurs in the word an even number of times; otherwise, it has odd parity.</p>
<p>(a) How many words of length <span class="math-container">$n$</span> have even parity?</p>
<p>(b) How many words of length <span class="math-container">$n$</span> have odd parity?</p>
<p>It seems to me that correct approach will be to use summations of combinations, and correct answer will yield <span class="math-container">$2^{n-1}$</span> in both cases. Is there any way to prove separately that each of them is equal to <span class="math-container">$2^{n-1}$</span>? Would be happy to know your ideas!</p>
| ho boon suan | 436,996 | <p>Let <span class="math-container">$X\in\mathfrak{X}(M)$</span> be a smooth vector field on <span class="math-container">$M$</span>. If all you have is a smooth bijection <span class="math-container">$F\colon M\to N$</span> with an inverse <span class="math-container">$F^{-1}$</span> that isn’t smooth, there is no way to guarantee that the resulting vector field <span class="math-container">$F_*X$</span> is smooth. That said, a smooth bijection is enough to define a unique rough (that is, not necessarily smooth) vector field.</p>
<p>Following the suggestion of @Tom Ariel in the comments, consider the (constant) coordinate vector field <span class="math-container">$X$</span> defined by <span class="math-container">$X_p=\partial/\partial x|_p$</span> on <span class="math-container">$\mathbf{R}$</span>. Its pushforward under the map <span class="math-container">$F\colon\mathbf{R}\to\mathbf{R}$</span> defined by <span class="math-container">$F\colon x\mapsto x^3$</span> gives us a rough vector field <span class="math-container">$F_*X$</span> defined by
<span class="math-container">$$\begin{align*}
(F_*X)_p
&=dF_{F^{-1}(p)}(X_{F^{-1}(p)}) \\
&=dF_{p^{1/3}}(\partial/\partial x|_{p^{1/3}}) \\
&=3p^{2/3}(\partial/\partial x|_p).
\end{align*}$$</span>
This vector field is not smooth; the map <span class="math-container">$x\mapsto 3x^{2/3}$</span> isn’t even differentiable at <span class="math-container">$x=0$</span>.</p>
<p>However, if we have a diffeomorphism <span class="math-container">$G\colon M\to N$</span>, then we are guaranteed that <span class="math-container">$G_*$</span> takes smooth vector fields to smooth vector fields, since <span class="math-container">$G_*X\colon N\to TN$</span> is then the composition of smooth maps
<span class="math-container">$$N
\overset{G^{-1}}{\longrightarrow}
M
\overset{X}{\longrightarrow}
TM
\overset{dG}{\longrightarrow}
TN.$$</span></p>
<p>(For references, see for example Propositions 8.1 and 8.19 in John Lee’s <em>Introduction to Smooth Manifolds</em>.)</p>
|
2,067,003 | <p>(Mathematics olympiad Netherlands) Let $A,B$ and $C$ denote chess players in a tournament. The winner of each match plays the next match against the oponent that did not play the current. At the end of the tournament $A$, $B$ and $C$ played $10$, $15$ and $17$ times respectively. Each match only ended up in a win. <em>Question</em>: Which player lost the second match?</p>
<p><em>UPDATE</em>: So I think I got the answer. Denote $n$ as the amount of matches between $A$ and $B$. Since $A$ plays the same amount of matches against $B$ as the otherway around, we have $15 - n = 17 - (10-n) \implies n = 4$. So $A$ plays a total of 10 matches, while there are a total of 21 matches. This is only possible if $A$ plays all the even matches <strong>and</strong> he loses that match (else contradiction to amount of matches played).</p>
| MatheMagic | 397,530 | <p>I got my answer in this way:
$\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n(n+1)}=\sum_{n=1}^{\infty}(-1)^{n+1}\left(\frac1n-\frac1{n+1}\right)=\sum_{n=1}^{\infty}\frac {(-1)^{n+1}}{n}-\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n+1}=\\log2+\log2-1=2\log2-1$</p>
|
2,239,240 | <p>I'm looking to do some independent reading and I haven't been able to find rough prerequisites for Differential Topology at the level of Milnor or Guillemin and Pollack.</p>
<p>Is a semester of analysis (Pugh) and a semester of topology (Munkres) enough to make sense of most of it or should I take a second semester of analysis first?</p>
| Zev Chonoles | 264 | <p><strong>As a general rule, you can do anything you want in any order as long as each step is a valid operation.</strong></p>
<p>Replacing $2+3$ by $5$ is certainly fine. </p>
<p>Expanding an expression $(a+b)^2$ as $a^2+2ab+b^2$ is also correct. </p>
<p>Do whichever you want.</p>
|
2,239,240 | <p>I'm looking to do some independent reading and I haven't been able to find rough prerequisites for Differential Topology at the level of Milnor or Guillemin and Pollack.</p>
<p>Is a semester of analysis (Pugh) and a semester of topology (Munkres) enough to make sense of most of it or should I take a second semester of analysis first?</p>
| Ethan Bolker | 72,858 | <p>This is a very good question.</p>
<p>There are some standard <em>conventions</em> for order of operations - for example, when you see
$2 + 3 \times 4$ you do the multiplication first. This convention is not really a part of arithmetic, it's just an agreed upon way to write the calculation that avoids having to write all the pairs of parentheses. </p>
<p>If we didn't have the convention you would have to write exactly what you meant:
$2 + (3 \times 4)$. You can always do that anyway -
you don't have to rely on the convention. Use as many "extra" parentheses as you wish. (Sometimes that's a good idea when you're writing a computer program and aren't sure the computer is following the same conventions you are.)</p>
<p>Anything surrounded by a pair of parentheses is evaluated as an independent calculation. Think of the parentheses as a circle surrounding the expression, just cut off so it doesn't interfere with the lines above and below it on the page.</p>
<p>In this particular question you are probably confused by (the horrible ugly) PEMDAS, which puts the P before the E. So in $(2+3)^2$ you evaluate the sum in the parentheses first. And you need those parentheses, because PEMDAS says $2+3^2$ is $2+3\times3 = 11$. </p>
<p>But when you rewrite $(2+3)^2$ as $(2+3) \times (2+3)$ there is no longer an exponent. You already know the answer is $5 \times 5$. But there's another way to get there that involves real mathematics, not just a convention. That's the <em>distributive law</em> that tells you how multiplication and addition combine:
$$
a \times (b + c) = (a \times b) + (a \times c).
$$
(I wrote that out with all the parentheses - didn't rely on PEMDAS.)
Using the distributive law twice (and the commutative law too):
$$
\begin{align}
(2+3) \times (2+3) & = (2+3) \times 2 + (2+3) \times 3 \\
& = 2 \times 2 + 3 \times 2 + 2 \times 3 + 3 \times 3 \\
& = \text{and so on}
\end{align}
$$</p>
<p>(Here I relied on the "multiplication before addition" convention and
left out the extra parentheses).</p>
<p>You may recognize that calculation as (the horrible ugly abbreviation) FOIL - which is just a mnemonic for the distributive law twice.</p>
|
107,399 | <p>Let's say we have a set a\of associations:</p>
<pre><code>dataset = {
<|"type" -> "a", "subtype" -> "I", "value" -> 1|>,
<|"type" -> "a", "subtype" -> "II", "value" -> 2|>,
<|"type" -> "b", "subtype" -> "I", "value" -> 1|>,
<|"type" -> "b", "subtype" -> "II", "value" -> 2|>
}
</code></pre>
<p>where every entry is unique in terms of <code>{#type, #subtype}</code>, </p>
<h3>I'd like to build a nested association for more handy querying, e.g. I would like to have:</h3>
<pre><code>nested["a", "II", "value"]
</code></pre>
<blockquote>
<pre><code>2
</code></pre>
</blockquote>
<h3>I can start with</h3>
<pre><code>GroupBy[dataset, {#type &, #subtype &}]
</code></pre>
<blockquote>
<pre><code><|
"a" -> <|
"I" -> {<|"type" -> "a", "subtype" -> "I", "value" -> 1|>},
"II" -> {<|"type" -> "a", "subtype" -> "II", "value" -> 2|>}|>,
"b" -> <|
"I" -> {<|"type" -> "b", "subtype" -> "I", "value" -> 3|>},
"II" -> {<|"type" -> "b", "subtype" -> "II", "value" -> 4|>}
|>|>
</code></pre>
</blockquote>
<p>But <code>nested["a", "I"]</code> points to a <strong>list with one association</strong>, what is expected but I would like to drop that list. </p>
<p>It seems that the third argument of <code>GroupBy</code> isn't generalized to handle nested grouping... </p>
<p>So basically I would like to have <code>... "I" -> <|"type" -> "a", ...</code>.</p>
<h3>What is a generic way to go?</h3>
<p>I can do:</p>
<ul>
<li><p>nested <code>GroupBy</code>: </p>
<pre><code>GroupBy[dataset, #type &, GroupBy[#, #subtype &, First] &]
</code></pre></li>
<li><p><code>Map</code> later:</p>
<pre><code>GroupBy[dataset, {#type &, #subtype &}] // Map[First, #, {-3}] &
</code></pre></li>
</ul>
<p>But the first is not handy in general while the second is ugly (and not general either). </p>
<hr>
<p>Acceptable outputs are:</p>
<pre><code><|
"a" -> <|
"I" -> <|"type" -> "a", "subtype" -> "I", "value" -> 1|>,
...
|>
</code></pre>
<p>or</p>
<pre><code><|
"a" -> <|
"I" -> <|"value" -> 1|>,
...
|>
</code></pre>
<p>or</p>
<pre><code><|
"a" -> <|
"I" -> 1 ,
...|>
</code></pre>
<p>but the first is the most desired one because we may have more that one ("value") key left.</p>
| gwr | 764 | <p>Here is the approach I would take to transform your dataset to a nested <code>Association</code>:</p>
<pre><code>Clear[ makeNested ];
makeNested[ assoc_, keylist_] := GroupBy[
assoc,
keylist
] // Apply[
Association,
#,
{ Length @ keylist }
] &
</code></pre>
<p>Now <code>makeNested[ dataset, { #type &, #subtype &} ]</code> and</p>
<p><code>makeNested[ dataset, { #type &, #subtype &, #type &, #subtype & } ]</code> will work as wanted.</p>
<p>For example:</p>
<pre><code>(nested = makeNested[ dataset, { #type &, #subtype & } ]) // Dataset
</code></pre>
<p><a href="https://i.stack.imgur.com/rHaiK.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rHaiK.jpg" alt="Nested Dataset"></a></p>
<pre><code>nested["a","I","value"]
</code></pre>
<blockquote>
<p>1</p>
</blockquote>
<h2>Update: A more streamlined solution</h2>
<p>Inspired by WReach's solution I tried to streamline this into a compact function that also get's rid of the keys that have already been used in the nesting. So here it is:</p>
<pre><code>Clear[ formatLeaves ];
Options[ formatLeaves ] = {
"DropKeys" -> None
};
formatLeaves/: List[ formatLeaves[ data_ ] ] := Module[
{
keys = OptionValue[ formatLeaves, "DropKeys" ]
},
KeyDrop[ data, keys ]
]
Clear[ makeNested ];
makeNested[ data_ , keylist_?(VectorQ[ #, StringQ ]&) ] := Module[
{
listKeys = Map[Key] @ keylist,
lastKey
},
lastKey = Last @ listKeys;
SetOptions[ formatLeaves, "DropKeys" -> listKeys ];
GroupBy[
data,
ReplaceAll[
listKeys,
Rule[
lastKey,
lastKey -> formatLeaves
]
]
]
]
</code></pre>
<p>Now:</p>
<pre><code>(nested = makeNested[ dataset, { "type", "subtype" } ]) // Dataset
</code></pre>
<p><a href="https://i.stack.imgur.com/d2wFc.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/d2wFc.jpg" alt="Cleaned up nested data"></a></p>
<pre><code>nested[ "a","I","value" ]
</code></pre>
<blockquote>
<p>1</p>
</blockquote>
<p>Naturally this saves space in memory and on disk:</p>
<pre><code>Map[ByteCount] @ { nested, dataset }
</code></pre>
<blockquote>
<p>{2064, 2280}</p>
</blockquote>
|
1,581,161 | <p>Let the triangle $ABC$ and the angle $\widehat{ BAC}<90^\circ$ </p>
<p>Let the perpendicular to $AB$ passing by the point $C$ and the perpendicular to $AC$ passing by $B$ intersect the circumscribed circle of $ABC$ on $D$ and $E$ respectively .
We suppose that $DE=BC$</p>
<p>What is the angle $\widehat{BAC}$ </p>
<p>I tried using law of sines in triangle
Also , let O be center of circle so OD=OE=r</p>
| James Pak | 187,056 | <p><a href="https://i.stack.imgur.com/YElQu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YElQu.png" alt="enter image description here"></a></p>
<p>Note that $\angle BAC=\angle BEC$, and that $\angle BAC=180^\circ-\angle DFE=\angle CFE$. As $DE=BC$, $\angle BEC=\angle DCE$. Therefore, $\angle BAC = 60^\circ$.</p>
<p>Appendix:</p>
<p><a href="https://i.stack.imgur.com/m4iHU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/m4iHU.png" alt="enter image description here"></a></p>
<p>See the comment section below.</p>
|
273,499 | <blockquote>
<p>Show that every group $G$ of order 175 is abelian and list all isomorphism types of these groups. [HINT: Look at Sylow $p$-subgroups and use the fact that every group of order $p^2$ for a prime number $p$ is abelian.]</p>
</blockquote>
<p>What I did was this. $|G| = 175$. Splitting 175 gives us $175 = 25 \cdot 7$. Now we want to calculate the Sylow $p$-groups, i.e we want</p>
<p>$$P= n_7: n_7 \equiv 1 mod 7 \hspace{1.5cm} n_7|25$$
$$Q= n_{25}: n_{25} \equiv 1 mod 25 \hspace{1.5cm} n_{25} | 7$$</p>
<p>After listing all elements that are $\equiv 1 mod 7$ and $1 mod 25$ you see that the only (avaliable) ones are $n_7 = n_{25} = 1$. This tells us that both groups $P,Q$ are normal subgroups of $G$. I think, by definition of a normal subgroup, they are abelian and so this tells us that $G$ is abelian. To list all the isomorphism types, we want the semidirect product (SDP) such that</p>
<p>$$P \rightarrow Aut(Q) = C_7 \rightarrow C_{20}$$</p>
<p>As there are no elements of order 7 in $C_{20}$, the only SDP we have is the trivial SDP, i.e the direct product</p>
<p>$$C_7 \times C_{25} \cong C_{175}$$</p>
<p>We know that $175 = 5^2 \cdot 7$ and so multiplying the powers shows us that there are 2 non-isomorphic groups:</p>
<p>$$C_{25} \times C_7$$
$$C_5 \times C_5 \times C_7$$</p>
<p>My question for this is is my reasoning also correct for things like showing the abelian groups? I saw something which said something about $P \cap Q = I_G$ and they used this but I don't understand what it was.</p>
<p>The next question, assuming that I had to possibilites for my $p$ subgroup, i.e $n_p = 1 or x$, how would I go about answering this question? (I am doing a question like this now and am stuck as I have two Sylow $p$-subgroups).</p>
| DonAntonio | 31,254 | <p>What you did looks fine, albeit slightly messy and overkill: if you already know there's one unique Sylow $\,5-$subgroup $\,P\,$ of order $\,25\,$ and one single Sylow $\,7-$ subgroup $\,Q\,$ of order $\,7\,$ , both of them abelian, and then you already know:</p>
<p>(1) $\,P,Q\triangleleft G\,$</p>
<p>(2) $\,P\cap Q=1\,$</p>
<p>Then we get</p>
<p>$$|PQ|=\frac{|P||Q|}{|P\cap Q|}=|P||Q|=25\cdot 7=175\Longrightarrow G=PQ\cong P\times Q$$</p>
<p>And since direct product of abelian groups is abelian we're done.</p>
|
2,484 | <p>It's been quite a while since I was tutoring a high school student and even longer since not a gifted one.</p>
<p>However, this time, something was amiss. I have asked him to show me how he does some exercise, and then another and the only thing I wanted to do was to shout:</p>
<blockquote>
<p><strong>You are doing it <em>wrong</em>!</strong></p>
</blockquote>
<p>I hope I didn't let it show and tried to work it out in small steps. However, it was like he didn't wanted to learn, he just wanted to get the problem set done.</p>
<p>One mistake he would frequently do, is to mess the signs in trigonometric reductions (e.g. $\cos(90^\circ + x)$ would be $\sin x$ instead of $-\sin x$). In fact, he memorized all the formulas, but couldn't recall them properly. I've tried to teach him how to recover formulas from graphs of $\sin$ and $\cos$, but the response was along the lines of "I don't need to understand it" and "I would like to do it without thinking".</p>
<p>For example, there was a pair of tasks 1) prove that $X(\alpha) = Y(\alpha)$ and 2) calculate $X(30^\circ)$, where $Y$ was simple and $X$ was not. However, in the class they covered such calculations, so he evaluated the $X$ directly. Yet, pointing out the simpler way didn't changed his approach.</p>
<p>Another example could be an expression with non-round numbers, where I suggested to hide them behind $\alpha$ and its variations (esp. since such substitution made the necessary formulas and reductions easily visible). Nevertheless, he tirelessly rewrote them all the way through calculations and clung to their concreteness like to some kind of lifeline.</p>
<p>I am aware that an experienced tutor would do much better than I did, but that's not what I'm after. What shook me most strongly was this strange quality of his approach to mathematics that would produce a strong feeling of dissatisfaction (similar to the one you have when seeing something ugly). What I would like to ask about is:</p>
<p><strong>How can I change such a student's approach to math?</strong></p>
<p>It seems like a Catch-22. He sees math as horrendous, but won't learn otherwise, because he doesn't want to deal with it. On the other hand, when he does, it's so wrong that no wonder he sees math this way.</p>
<p>For a change I wanted to show him something engaging, e.g. how math allows us to build nice things using 3D animation software, but he would not make a connection. It was just a few cool pics useful only as long as they can be used to raise his status (e.g. his friends think them awesome).</p>
<p>I have no illusions on whether I could teach him to enjoy math. But, is it possible to make him not hurt mathematics? Right now, I admit defeat.</p>
| Community | -1 | <p><em>What does he want?</em> The question says more about what he doesn't want.</p>
<p>Maybe he wants to pass an exam to avoid math for a few years, and wants to improve his chance of passing. Maybe he wants to run a business, and would be interested in business algebra.</p>
<p>Starting from his goals can make the relevant activities more appealing for him. That may get him in a frame of mind to appreciate other approaches to math.</p>
|
2,805,192 | <p><a href="https://i.stack.imgur.com/OhGi4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OhGi4.png" alt="enter image description here"></a></p>
<p>Definition:</p>
<p><a href="https://i.stack.imgur.com/mV4NU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mV4NU.png" alt="enter image description here"></a></p>
<p>Call the set in the hint E. I've already proven that E is finite and X can be covered by finitely many neighborhoods of radius $\delta$.</p>
<p>What I'm not sure of is the next part, where I n need to show that E is dense. I want to show that every point of X can be a point of E. This happens because for a point x $\in$ X such that x $\in$ $N_\delta(q)$ for some q $\in$ E, we can than choose another smaller $\delta$ so that E also counts x amongst its member.</p>
<p>So for $\delta = 1/n \ (n = 1,2,3,...)$, every point of X can be a point of E. Hence E is dense.</p>
<p>This argument seems to go in the direction of the hint, but I'm not really sure about it. Since wouldn't we need to prove that E is dense for a fixed $\delta$?</p>
| Igor Sikora | 464,503 | <p>Your argument is nearly valid. It doesn't show that every $x \in X$ belongs to $E$, but rather that every $x$ is in the closure of $E$. And that's ok, because this is the definition of a dense subset: $E$ is dense in X if $\bar{E}=X$.</p>
<p>Also, having said that, you cannot fix $\delta$. That is because of the definition of the closure of $E$: $x\in \bar{E}$ iff $\forall_{\delta >0} N_{\delta}(x)\cap E\neq\emptyset$.</p>
|
2,805,192 | <p><a href="https://i.stack.imgur.com/OhGi4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OhGi4.png" alt="enter image description here"></a></p>
<p>Definition:</p>
<p><a href="https://i.stack.imgur.com/mV4NU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mV4NU.png" alt="enter image description here"></a></p>
<p>Call the set in the hint E. I've already proven that E is finite and X can be covered by finitely many neighborhoods of radius $\delta$.</p>
<p>What I'm not sure of is the next part, where I n need to show that E is dense. I want to show that every point of X can be a point of E. This happens because for a point x $\in$ X such that x $\in$ $N_\delta(q)$ for some q $\in$ E, we can than choose another smaller $\delta$ so that E also counts x amongst its member.</p>
<p>So for $\delta = 1/n \ (n = 1,2,3,...)$, every point of X can be a point of E. Hence E is dense.</p>
<p>This argument seems to go in the direction of the hint, but I'm not really sure about it. Since wouldn't we need to prove that E is dense for a fixed $\delta$?</p>
| Berci | 41,488 | <p>Let indeed $x_1$ be arbitrary, and set $\delta_1=1$. Denote $X_1$ the finite set $\{x_1,\dots, x_{n_1}\}$ obtained as in the hint for $\delta_1$. <br>
Let $x_{n_1+1}, \dots$ be points satisfying $d(x_k, x_j) \ge\delta_2:=\frac12$. Denote their set $X_2$, it's finite again. <br>
Continue with $\delta_n:=\frac1n$. </p>
<p>Now, the dense set $E$ is considered to be $E:=\bigcup_nX_n$. </p>
|
754,012 | <p>Is it possible to show that the harmonic series is divergent by showing that the sequence of partial sums is a monotone increasing sequence that is unbounded?</p>
| marty cohen | 13,079 | <p>Another way,
which is moderately equivalent.</p>
<p>Let
$S_n = \sum_{k=1}^n \frac1{k}$.
$S_n$ is obviously an increasing sequence.</p>
<p>$S_{2n}-S_n
=\sum_{k=1}^{2n} \frac1{k}-\sum_{k=1}^n \frac1{k}
=\sum_{k=n+1}^{2n} \frac1{k}
>\sum_{k=n+1}^{2n} \frac1{2n}
=\frac{n}{2n}
=\frac12
$</p>
<p>so, by induction
$S_{2^mn}-S_n
>\frac{m}{2}
$
so $S_{2^mn}$ can be made
as large as you want
by making $m$ large enough.</p>
<p>Note that this also shows that
$S_{2n}-S_n
=\sum_{k=n+1}^{2n} \frac1{k}
<\sum_{k=n+1}^{2n} \frac1{n}
=\frac{n}{n}
=1
$
so, by induction
$S_{2^mn}-S_n
<m
$
so to make $S_{2^mn}$
large,
you have to go up exponentially.</p>
<p>This is why $S_n$ is about
$\log n$.</p>
|
3,632,576 | <p>Considering that input <span class="math-container">$x$</span> is a scalar, the data generation process works as follows:</p>
<ul>
<li>First, a target t is sampled from {0, 1} with equal probability.</li>
<li>If t = 0, x is sampled from a uniform distribution over the interval
[0, 1]. </li>
<li>If t = 1, x is sampled from a uniform distribution over the
interval [0, 2].</li>
</ul>
<p>I'm trying to find the formula for <span class="math-container">$P(t=1)$</span>, <span class="math-container">$P(t=0)$</span>, <span class="math-container">$P(x|t=1)$</span> and <span class="math-container">$P(x|t=0)$</span> and then find the posterior probability <span class="math-container">$P(t = 0|x)$</span> as a function of <span class="math-container">$x$</span>.</p>
<p>So far I have that <span class="math-container">$P(t=1)$</span> and <span class="math-container">$P(t=0)$</span> <span class="math-container">$=$</span> <span class="math-container">$\frac12$</span> but I wasn't sure how to find <span class="math-container">$P(x|t=1)$</span> and <span class="math-container">$P(x|t=0)$</span>.</p>
<p>I know from there we can just use <span class="math-container">$P(A|B)=\frac{P(B|A)P(A)}{P(B)}$</span> to compute the probability <span class="math-container">$P(t = 0|x)$</span> as a function of <span class="math-container">$x$</span>. Is that correct?</p>
| S.C. | 544,640 | <p>Here is a more detailed/adapted version of Spivak's proposed solution:</p>
<p>By assumption, we know that there exists an <span class="math-container">$N \gt 0: \forall t \geq N: \left|f(t)-a\right|\lt \frac{\varepsilon}{3}$</span>. Consider <span class="math-container">$f$</span> restricted to the domain <span class="math-container">$[N,M]$</span> for some arbitrary <span class="math-container">$M \gt N$</span>. Denote this as <span class="math-container">$f|_{[N,M]}$</span>. For all <span class="math-container">$t \in [N,M]$</span>, we know that <span class="math-container">$\left|f|_{[N,M]}(t)-a\right| \lt \frac{\varepsilon}{3}$</span>. Next, let <span class="math-container">$g(t)=\left|f|_{[N,M]}(t)-a\right|$</span> and be defined on the same domain. Then for all <span class="math-container">$t \in [N,M]: g(t) \lt \frac{\varepsilon}{3} \quad (\dagger_1)$</span>.</p>
<p>By assumption, we know that for any <span class="math-container">$x \gt 0: f|_{[0,x]}$</span> is integrable on <span class="math-container">$[0,x]$</span>. Then we certainly have that <span class="math-container">$f|_{[0,M]}$</span> is integrable on <span class="math-container">$[0,M]$</span>, which tells us that for <span class="math-container">$0 \lt N \lt M: f|_{[N,M]}$</span> is integrable on <span class="math-container">$[N,M]$</span>.</p>
<p>Because <span class="math-container">$f|_{[N,M]}$</span> is integrable on <span class="math-container">$[N,M]$</span>, we know that <span class="math-container">$f|_{[N,M]}(t)-a$</span> is integrable on <span class="math-container">$[N,M]$</span>, and because <span class="math-container">$f|_{[N,M]}(t)-a$</span> is integrable on <span class="math-container">$[N,M]$</span>, we must have that <span class="math-container">$\left|f|_{[N,M]}(t)-a\right|$</span> is integrable on <span class="math-container">$[N,M]$</span>. Of course, then, we are certain that <span class="math-container">$g$</span> is integrable on <span class="math-container">$[N,M]$</span>, which allows us to deduce from <span class="math-container">$(\dagger_1)$</span> that <span class="math-container">$\displaystyle \int_N^M g(t)dt \lt \int_N^M \frac{\varepsilon}{3}=(M-N)\frac{\varepsilon}{3}$</span>.</p>
<p>We can rewrite this as:</p>
<p><span class="math-container">\begin{align}\int_N^M\left(\left|f|_{[N,M]}(t)-a\right|\right)dt \lt(M-N)\frac{\varepsilon}{3}\end{align}</span></p>
<p>Applying the result from exercise 37 (Chapter 13), we then have that:</p>
<p><span class="math-container">$$\left|\int_N^M \left(f|_{[N,M]}(t)-a\right)dt\right| \leq \int_N^M\left(\left|f|_{[N,M]}(t)-a\right|\right)dt \lt(M-N)\frac{\varepsilon}{3} $$</span></p>
<p>From this, we have that:</p>
<p><span class="math-container">$$\left|\int_N^M f|_{[N,M]}(t)dt- \int_N^M adt\right| \lt (M-N)\frac{\varepsilon}{3},$$</span> which we can further simplify to:</p>
<p><span class="math-container">$$\left|\int_N^M f|_{[N,M]}(t)dt- (M-N)a\right|\lt(M-N)\frac{\varepsilon}{3} \quad (\dagger_2)$$</span></p>
<p>Note that <span class="math-container">$M,N \gt 0 \implies \frac{M-N}{M} \lt 1$</span>. This implies that we can modify <span class="math-container">$(\dagger_2)$</span> as follows:</p>
<p><span class="math-container">$$\frac{\left|\int_N^M f|_{[N,M]}(t)dt- (M-N)a\right|}{M}\lt\frac{(M-N)\frac{\varepsilon}{3}}{M}\lt\frac{\varepsilon}{3} \quad (\dagger_3)$$</span></p>
<p><span class="math-container">$0 \lt M=|M|$</span>, so <span class="math-container">$(\dagger_3)$</span> can be rewritten as:</p>
<p><span class="math-container">$$\left|\frac{1}{M}\int_N^M f|_{[N,M]}(t)dt -\frac{M-N}{M}a\right|\lt \frac{\varepsilon}{3} \quad (\dagger_4)$$</span></p>
<p><strong>Importantly</strong>, note that, by construction, <span class="math-container">$(\dagger_4)$</span> is valid for any <span class="math-container">$M \gt N$</span>.</p>
<p>Recall that we are trying to prove: <span class="math-container">$\displaystyle \lim_{x \to \infty}\frac{1}{x}\int_0^xf(t)dt=a \iff \forall \varepsilon \gt 0: \exists M \in \mathbb R: \forall x \gt M: \left|\frac{1}{x}\cdot\int_0^xf(t)dt-a \right|\lt \varepsilon$</span>.</p>
<p>We will make use of the following triangle inequality lemma, applied to three terms:</p>
<p><span class="math-container">$$|a+b+c|\leq |a+b|+|c| \leq |a|+|b|+|c| \implies |a+b+c|\leq |a|+|b|+|c|$$</span></p>
<p>Consider the following two inequalities:</p>
<p><span class="math-container">$\displaystyle\left| \frac{1}{M}\int_0^N f(t) \right|\lt \frac{\varepsilon}{3}$</span> and <span class="math-container">$\displaystyle \left|\frac{M-N}{M}a-a \right| \lt \frac{\varepsilon}{3}$</span></p>
<p>We must select an <span class="math-container">$M \gt N$</span> that satisfies both of these statements simultaneously. Realize that the left side of each inequality, for a sufficiently large <span class="math-container">$M$</span>, is a strictly decreasing positive function. So if one <span class="math-container">$M$</span> satisfies both inequalities simultaneously, any <span class="math-container">$M' \gt M$</span> does, too. Finally, we can add these two inequalities to <span class="math-container">$(\dagger_4)$</span> and apply our lemma, which gives us:</p>
<p><span class="math-container">$$\left |\frac{1}{M}\int_0^N f(t)dt+\frac{1}{M}\int_N^M f|_{[N,M]}(t)dt-\frac{M-N}{M}a+ \frac{M-N}{M}a-a\right| \lt \frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3}=\varepsilon$$</span></p>
<p>This simplifies to our desired result of:</p>
<p><span class="math-container">$$\left|\frac{1}{M}\int_0^M f(t)dt-a\right| \lt \varepsilon$$</span></p>
<p>From our previous comments, we know that for any <span class="math-container">$M' \gt M$</span>, this inequality holds, so we are done.</p>
|
2,528,716 | <p>How can I prove </p>
<blockquote>
<p>$$x^2+y^2-x-y-xy+1≥0$$</p>
</blockquote>
<p>I tried $(x+y)^2-3xy-(x+y)+1≥0 \rightarrow(x+y-1)(x-y)-3xy+1≥0$ I can not continue</p>
| mathlove | 78,967 | <p>Another way :$$\begin{align}x^2+y^2-x-y-xy+1&=x^2+(-1-y)x+y^2-y+1\\\\&=\left(x+\frac{-1-y}{2}\right)^2-\left(\frac{-1-y}{2}\right)^2+y^2-y+1\\\\&=\left(x-\frac{1+y}{2}\right)^2+\frac 34(y-1)^2\ge 0\end{align}$$</p>
|
2,528,716 | <p>How can I prove </p>
<blockquote>
<p>$$x^2+y^2-x-y-xy+1≥0$$</p>
</blockquote>
<p>I tried $(x+y)^2-3xy-(x+y)+1≥0 \rightarrow(x+y-1)(x-y)-3xy+1≥0$ I can not continue</p>
| Michael Rozenberg | 190,319 | <p>We need to prove that
$$x^2-(y+1)x+y^2-y+1\geq0,$$ which is a quadratic inequality of $x$.</p>
<p>Thus, it's enough to prove that
$$(y+1)^2-4(y^2-y+1)\leq0$$ or
$$(y-1)^2\geq0,$$
which is obvious.</p>
|
1,282,486 | <p>Given the function $f(x) = |8x^3 − 1|$ in the set $A = [0, 1].$
Prove that the function is not differentiable at $x = \frac12.$ </p>
<p>The answer in my book is as follows:</p>
<p>$$\lim_{x \to \frac12-} \dfrac{f(x)-f(1/2)}{x-1/2} = -6$$
$$\lim_{x \to \frac12+} \dfrac{f(x)-f(1/2)}{x-1/2} = 6$$ </p>
<p>Can anyone explain how the $6$'s were derived. I understand that as $x$ tends to $\frac12$ from the negative side, the bottom will be negative, so thats why the first one is a minus.</p>
<p>But how do you get to the $6$, what am I missing? Obviously $f(\frac12)=0$ but what do you make $f(x)=$ as $x$ tends to $\frac12$</p>
<p>Thanks</p>
| Przemysław Scherwentke | 72,361 | <p>$(8x^3-1)/(x-1/2)=8(x^3-1/8)/(x-1/2)=8(x^2+(1/2)x+1/4)$, which is equal to 6 at $x=1/2$.</p>
|
1,282,486 | <p>Given the function $f(x) = |8x^3 − 1|$ in the set $A = [0, 1].$
Prove that the function is not differentiable at $x = \frac12.$ </p>
<p>The answer in my book is as follows:</p>
<p>$$\lim_{x \to \frac12-} \dfrac{f(x)-f(1/2)}{x-1/2} = -6$$
$$\lim_{x \to \frac12+} \dfrac{f(x)-f(1/2)}{x-1/2} = 6$$ </p>
<p>Can anyone explain how the $6$'s were derived. I understand that as $x$ tends to $\frac12$ from the negative side, the bottom will be negative, so thats why the first one is a minus.</p>
<p>But how do you get to the $6$, what am I missing? Obviously $f(\frac12)=0$ but what do you make $f(x)=$ as $x$ tends to $\frac12$</p>
<p>Thanks</p>
| zhw. | 228,045 | <p>It's almost easier to prove a more general result: If $f(a) = 0$ and $f'(a) \ne 0,$ then $|f(x)|$ is not differentiable at $a.$</p>
|
232,562 | <p>Ax-Grothendieck Theorem states that if $\mathbf K$ is an algebraically closed field, then any injective polynomial map $P:\mathbf K^n\longrightarrow \mathbf K^n$ is bijective.</p>
<blockquote><b>Question 1.</b> What does the inverse map of $P$ look like ? What kind of map is that ?</blockquote>
<p>$P^{-1}$ need not be polynomial, as the example $x^p$ in $\mathbf F_p^{alg}$ shows.</p>
<blockquote><b>Question 2.</b> Are there conditions under which $P^{-1}$ is polynomial ?</blockquote>
| R. van Dobben de Bruyn | 82,179 | <p>In light of abx's comment, I came up with the following argument. I'm sure some version of this must be in the literature somewhere.</p>
<p>Recall that a dominant morphism $Y \to X$ of varieties is <em>separable</em> if $K(X) \to K(Y)$ is a separable field extension. In characteristic $0$, this is automatic.</p>
<blockquote>
<p><strong>Theorem.</strong> Let $k$ be an algebraically closed field, and let $f \colon Y \to X$ be a morphism of $k$-varieties, with $X$ normal. If $f$ is bijective (in particular, dominant) and separable, then $f$ is an isomorphism.</p>
</blockquote>
<p><em>Proof.</em> By a suitable version of Zariski's main theorem (see e.g. <a href="http://stacks.math.columbia.edu/tag/05K0" rel="nofollow">Tag 05K0</a>), there exists an open immersion $Y \subseteq Z$ and a <em>finite</em> morphism $g \colon Z \to X$ extending $f$. By restricting to the underlying reduced scheme of the unique irreducible component of $Z$ containing $Y$, we may assume $Z$ is integral. In particular $K(Z) = K(Y)$, and $g$ is separable since $f$ is.</p>
<p>Since $g$ is separable, it is generically étale, hence there exists an open $U \subseteq X$ such that
$$g \colon g^{-1}(U) \to U$$
is finite étale. In particular, $g$ is finite flat over $U$ of some rank $r \geq 1$. If $x \in U$ is a closed point, then $f^{-1}(x) \to x$ is finite étale of rank $r$, thus a disjoint union of $r$ copies of $\operatorname{Spec} k$ (since $k$ is algebraically closed). Since $f$ is bijective, we have $r = 1$, so $f$ is birational, i.e.
$$K(X) \stackrel \sim \to K(Z).$$
Since $X$ is normal and $g$ finite, this forces $g$ to be an isomorphism. Since $f$ is bijective, we conclude that $Y = Z$ and $f = g$. $\square$</p>
<p><strong>Remark.</strong> I do not know how to get rid of the algebraically closed condition without weakening the hypotheses to $f$ being <em>universally bijective</em> (in a restricted sense: we only need it for base change along field extensions) and $X$ <em>geometrically normal</em> (when $k$ is perfect, this is equivalent to normal).</p>
|
2,471,633 | <p>Let $B$ an open ball in $\mathbb{R}^{n}$, and $(K_{j})_{j}$ be an increasing sequence of compact subsets of $B$ whose union equals $B$. For each $j$, let $\rho_{j}$ be a cut-off function in $C_{c}^{\infty}(B)$ that equals 1 on a neighborhood of $K_{j}$ and whose support is in $K_{j+1}$. Finally, let $\theta$ be a smooth function whose laplacian $\Delta\theta\equiv1$, and set
$$\phi_{j}:=\rho_{j}\theta.$$
My question: can we conclude that
$$\sup_{(x,j)\in B\times\mathbb{N}}|\Delta\phi_{j}(x)|$$
is bounded?</p>
| Dap | 467,147 | <p>Using the divergence theorem,</p>
<p>$$\int_{K_{j+1}\setminus K_j} \Delta \phi_j + \int_{K_j} \Delta \phi_j=\int_B \Delta \phi_j = \int_{\partial B} \nabla \phi_j \cdot dS=0.$$</p>
<p>But $\int_{K_j} \Delta \phi_j = \operatorname{vol}(K_j),$ so</p>
<p>$$\sup_{K_{j+1}\setminus K_j}(-\Delta\phi_j)\geq \frac{1}{\operatorname{vol}(K_{j+1}\setminus K_j)}\int_{K_{j+1}\setminus K_j} -\Delta \phi_j=\frac{\operatorname{vol}(K_j)}{\operatorname{vol}(K_{j+1}\setminus K_j)}\to \infty.$$</p>
|
1,611,506 | <blockquote>
<p>$$\int (2x^2+1)e^{x^2} \, dx$$</p>
</blockquote>
<p>It's part of my homework, and I have tried a few things but it seems to lead to more difficult integrals. I'd appreciate a hint more than an answer but all help is valued.</p>
| Ben Longo | 137,131 | <p>Start by expanding the integrand.</p>
<p>$$\begin{align}
I&=\int \left(2x^2e^{x^2}+e^{x^2}\right)\,dx\\
&=\int 2x^2e^{x^2}\,dx+\int {e^{x^2}}\,dx\tag{a}\\
&=x e^{x^2}-\int 2x^2 e^{x^2}\,dx+\int 2x^2 e^{x^2}\,dx\\
&=x e^{x^2}+C
\end{align}$$</p>
<p>$(\text{a})$: Use integration by parts on the second integral with $u=e^{x^2}$ and $dv=dx$.</p>
|
58,024 | <p><img src="https://i.stack.imgur.com/cTpA2.jpg" alt="Show that..."></p>
<p>The picture says it all. "Vis at" means "show that". My first thought was that h is 2x, which is not correct. Maybe the formulas for area size is useful? </p>
<p>EDIT: (To make the question less dependent from the <a href="https://math.meta.stackexchange.com/questions/1805/on-the-inclusion-of-pages-of-text-as-images-in-questions">picture</a>.)</p>
<p>A square with side $x$ is placed in the right triangle with legs $g$ and $h$. Show that $x=\frac{gh}{g+h}$.</p>
| robjohn | 13,854 | <p>Consider the area of the whole triangle and the areas of the constituent triangles and square.</p>
|
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