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 |
|---|---|---|---|---|
4,189,417 | <p>The book I am currently reading, is written in a local language, uses the symbol <span class="math-container">$$\gtreqqless$$</span></p>
<p>Noted usage in the book (translated):</p>
<blockquote>
<p>Let <span class="math-container">$f(x) = x^3e^{-3x}$</span> for x > 0. Then what is the maximum value of <span class="math-container">$f(x)$</span> ?</p>
</blockquote>
<blockquote>
<p>Solution : <span class="math-container">$\dotsc$</span> <span class="math-container">$f'(x) = 3e^{-3x}x^2(1-x)$</span>... We have to work explicitly with <span class="math-container">$1 - x$</span> since <span class="math-container">$3e^{-3x}$</span> is positive for x > 0 <span class="math-container">$\therefore$</span> We have to see when 1 - x <span class="math-container">$\geqslant$</span> 0 and when 1 - x <span class="math-container">$\leqslant$</span> 0. We can see that 1 - x <span class="math-container">$\gtreqqless$</span> 0 <span class="math-container">$\Longleftrightarrow$</span> 1 <span class="math-container">$\gtreqqless$</span> x <span class="math-container">$\dotsc$</span></p>
</blockquote>
<p>So how is the symbol read ,or, What does it mean ?</p>
<p>Edit: I have searched through the book and it does not provide any page on symbols used and no past explanation has been provided.</p>
| ultralegend5385 | 818,304 | <p>It is similar to <span class="math-container">$\pm$</span>, so for example, <span class="math-container">$(a\pm b)^2=a^2\pm 2ab+b^2$</span>, if you take a plus you take it everywhere, if you take a minus, you take it everywhere.</p>
<p>Likewise, if you consider <span class="math-container">$1-x>0$</span>, you have <span class="math-container">$1>x$</span>. (Here, we took the topmost symbol!)</p>
<p>If you have <span class="math-container">$1-x=0$</span>, it means <span class="math-container">$1=x$</span>; here we took the second symbol <span class="math-container">$=$</span>. You can see the pattern now.</p>
<p>Hope this helps. Ask anything if not clear :)</p>
|
284,507 | <p>In the textbook "Topology without tears" I found the definition.</p>
<p>$(X, \tau)$ is diconnected iff there exists open sets $A,B$ with $X = A \cup B$ and $A \cap B = \emptyset$.</p>
<p>In Walter Rudin: Principles of Analysis, I found.</p>
<p>$E \subseteq X$ is connected iff it is not the union of two nonempty separated sets. Where two sets $A,B$ are separeted iff $A \cap \overline{B} = \emptyset$ and $\overline{A} \cap B = \emptyset$.</p>
<p>So I am confused, why is in the first defintion nothing about the concept of separted sets said, for these two definitions are not logical negates of one another????</p>
| k.stm | 42,242 | <p>They are equivalent.</p>
<p>Let <span class="math-container">$X$</span> be disconnected as in the first sense.
Then there are (nonempty) open sets <span class="math-container">$A$</span>, <span class="math-container">$B$</span> in <span class="math-container">$X$</span> such that <span class="math-container">$X = A \cup B$</span> and <span class="math-container">$A \cap B = \emptyset$</span>, that is <span class="math-container">$X\setminus A = B$</span>.
Now since <span class="math-container">$A$</span> is open in <span class="math-container">$X$</span>, the set <span class="math-container">$B$</span> is the complement of an open set, so <span class="math-container">$B$</span> is closed and <span class="math-container">$\overline B = B$</span>.
Therefore, <span class="math-container">$A \cap B = \emptyset$</span> becomes <span class="math-container">$A \cap \overline B = \emptyset$</span>. Do the same for <span class="math-container">$A$</span> and <span class="math-container">$\overline A \cap B$</span>.
So the space is the union of two separated sets and therefore not connected.</p>
<p>Let <span class="math-container">$X$</span> be not connected as in the second sense.
Then there are (nonempty) separated sets <span class="math-container">$A$</span>, <span class="math-container">$B$</span> in <span class="math-container">$X$</span> such that <span class="math-container">$X = A \cup B$</span>.
Since <span class="math-container">$A \cap \overline B = \emptyset$</span>, you have <span class="math-container">$A = X\setminus \overline B$</span>, so <span class="math-container">$A$</span> is open in <span class="math-container">$X$</span>. Do the same for <span class="math-container">$B$</span>.
From <span class="math-container">$A \cap \overline B = \emptyset$</span> you also get <span class="math-container">$A \cap B = \emptyset$</span>, since <span class="math-container">$B \subseteq \overline B$</span>.
Therefore, the space is the union of nonempty open sets, so the space is disconnected.</p>
<p>Now, the definition given by Rudin mentions subsets, but extending the first definition by calling a subset <span class="math-container">$E \subseteq X$</span> connected if <span class="math-container">$E$</span> is connected with respect to the subspace topology, and checking that <span class="math-container">$E \subseteq X$</span> is disconnected if and only if <span class="math-container">$E \subseteq E$</span> is disconnected with respect to the subspace topology, you see, that both notions are really the same.</p>
|
4,200,434 | <p>I am having difficulty with what should be a routine question, Exercise 2.2.2 (b) of <em>Understanding Analysis</em> by Stephen Abbott (2015).</p>
<blockquote>
<p><strong>Exercise 2.2.2.</strong> Verify, using the definition of convergence of a sequence, that the following sequences converge to the proposed limit.</p>
<p>(b) <span class="math-container">$$\lim_{n \rightarrow \infty} \frac{2n^2}{n^3 + 3} = 0.$$</span></p>
</blockquote>
<p><strong>My attempt.</strong></p>
<p><em>Due to a transcription error in my question and some answers based on my erroneous transcription, followed by re-edits etc., this has caused some confusion. To avoid any further future confusion, I have reverted this question, together with my attempt to the most recent state before I received correct answers, and upvoted answers accordingly. For the members of the community who took the time to assist, please accept my apologies, and also my gratitude for the time you've spent writing up your answers.</em></p>
<p>I understand that verification using the formal definition (without recourse to theorems occurring later in the text) requires supplying a particular <span class="math-container">$N(\epsilon) \in \mathbb{N}$</span> in response to any <span class="math-container">$\epsilon > 0$</span> such that whenever <span class="math-container">$n \geq N$</span>, then</p>
<p><span class="math-container">$$\left \vert \frac{2n^2}{n^3+3} \space \right \vert < \epsilon.$$</span></p>
<p>Because <span class="math-container">$n > 0$</span>, I can simplify the above to get</p>
<p><span class="math-container">$$\left \vert \frac{2n^2}{n^3+3} \space \right \vert = \frac{2n^2}{n^3 + 3} < \epsilon.$$</span></p>
<p>Which yields the following cubic polynomical in <span class="math-container">$n$</span>, which I am unsure how to solve analytically:</p>
<p><span class="math-container">$$\epsilon n^3 - 2n^2 + 3 \epsilon > 0.$$</span></p>
| Community | -1 | <p>After the edit to the OP which changes the <span class="math-container">$n^2$</span> in the denominator to <span class="math-container">$n^3$</span>:</p>
<p>Observe that
<span class="math-container">$$2n^3 \leq 2n^3 + 6 = 2(n^3 + 3)$$</span>
so
<span class="math-container">$$2n^2 = \frac{2n^3}{n} \leq \frac{2(n^3+3)}{n}$$</span>
Dividing the left and right sides by <span class="math-container">$n^3 + 3$</span> gives
<span class="math-container">$$\frac{2n^2}{n^3 + 3} \leq \frac{2}{n}$$</span>
Thus the LHS is sandwiched between <span class="math-container">$0$</span> and <span class="math-container">$2/n$</span>, and the latter converges to <span class="math-container">$0$</span> as <span class="math-container">$n \to \infty$</span>.</p>
|
449,457 | <p>A desk has three drawers. The first contains two gold coins, the second has two silver coins and the third has one gold and one silver coin. A drawer is selected at random and a coin is drawn at random from the drawer. Suppose that the coin selected was silver. Use Bayes's Theorem to find the probability that the other coin in that drawer is gold. </p>
| angryavian | 43,949 | <p>You select a drawer and select a coin from that drawer.</p>
<p>Let $A$ denote the event that the coin that you select is silver.</p>
<p>Let $B$ denote the event that the other coin in the drawer is gold.</p>
<p>You are equally likely to pick any coin, so</p>
<p>$$P(A) = \frac{3\text{ silver coins}}{6 \text{ coins}} = \frac 12$$</p>
<p>There are 6 different ways you could have selected a coin. Only one of them results in you selecting a silver coin AND the other coin in the drawer being gold, so
$$P(A \cap B) = \frac{1\text{ desired outcome}}{6 \text{ total outcomes}} = \frac 16$$</p>
<p>By the definition of conditional probability,
$$P(B \mid A) = \frac{P(B \cap A)}{P(A)} = \frac{1/6}{1/2} = \frac 13$$</p>
<hr>
<p>As I mentioned in my comment, Bayes's Theorem is useful if you want to know $P(B\mid A)$ but you only know $P(A\mid B)$. Here, we don't know $P(A \mid B)$ either, so Bayes's Theorem doesn't really help.</p>
|
2,139,951 | <p>A set S of integers $\{s_1, s_2, . . . , s_k\}$ such that 0 ∈ S is called a set of digits
modulo m if and only if any integer b can be written as
$$b = f_nm^n + f_{n−1}m^{n−1} + · · · + f_1m + f_0$$
for some $n \geq 0$ and some $f_i$ ∈ S, moreover such expression is unique assuming
that $f_n \neq 0$.
Show that:</p>
<p>a) $k = m$ and the set of $s_i$ is a complete set of representatives modulo m</p>
<p>b) S = {0, 1, −1} is a set of digits modulo 3</p>
<p>c) S = {0, 2, −2} is NOT a set of digits modulo 3</p>
<p>I can't really seem to get anywhere with this. I've tried representing b in a few ways using the division algorithm and then trying to convert to the expression above but this doesn't really seem to get me anywhere. I'd appreciate any help, thanks!</p>
| fleablood | 280,126 | <p>1) $S$ must contain a complete representative of modulo $m$ and $k \ge n$.</p>
<p>Suppose not: Let $a\in \mathbb Z$ be such that $a \not \equiv s_i \mod m$ for any $s_i$ in $S$. Then $a = \sum_{i=0}^n s_i m^i; s_i \in S$ is impossible as $\sum_{i=0}^n s_i m^i \equiv s_0 \mod m$ but $a \not \equiv s_0 \mod m$. So $a$ can not be represented in terms of $S$.</p>
<p>2) $S$ may not contain any two integers, $s \ne t$ but $s \equiv t \mod m$.</p>
<p>Suppose $S$ could: If $s, t \in S$ and $s \ne t$ and $s \equiv t \mod m$. So $s = t + km$ for some k. Let $k = \sum k_i m^i;k_i \in S$ be the unique representation of $k$ using terms of $S$. Then $s = t + \sum k_i m^{i+1}$ is one representation of $s$ using terms of $S$. And $s = s$ is another. That contradicts that all representations are unique.</p>
<p>So 1) and 2) mean $S$ must be a completer modulo $m$ representation.</p>
<p>b) Will show by induction. </p>
<p>We want to find $n = \sum s_i 3^i; s_i \in \{0,1,-1\}$.</p>
<p>So we will need $n \equiv \sum s_i 3^i \mod 3 \equiv s_0 \mod 3$.</p>
<p>Now $n \equiv 0, -1, 1$ so we can and we must choose $s_0 = 0, -1, 1$ so that $n \equiv s_0 \mod 3$.</p>
<p>Now $n - s_0 \equiv 0 \mod 3$. Let $n_1 = n/3$. We can, and must, repeat these steps inductively. To get $n = \sum_{i = 0}^N s_i 3^i$ being a unique representation.</p>
<p>c) Note if $n = \sum_{i=0}^m s_i 3^i; s_i = 0, -2,2$ then $n$ is even. No odd number can be thusly represented. So this fails to be a set of digits modulo $3$.</p>
|
1,791,815 | <p>I'm trying to find an elementary way to see that the 1st de Rham cohomology of the n-sphere is zero for $n>1$, $H^1(S^n) = 0$.</p>
<p>This is part of an attempt to find the de Rham cohomology of the n sphere generally. I have shown that $H^k(S^n) = H^{k-1}(S^{n-1})$ (for $k>1$) but to find the de Rham cohomology generally I need to show that $H^1(S^n) = 0$ and I am struggling to do this.</p>
<p>Is it possible to take a closed one form $\alpha$ on a sphere and explicitly find a function $f$ on $S^n$ that has $df = \alpha$, I've tried writing it out in coordinates and explicitly integrating but I haven't managed to get very far with this!</p>
| R Mary | 309,615 | <p>I think the function you are looking for is $f(x)=\int_{\gamma_x} \alpha$ where $\gamma_x$ is a path from some chosen base point $x_0$ to $x$. But then for this to be well defined you have to say that all paths between $x$ and $x_0$ are homotopic, i.e. that your sphere is simply connected, so the whole argument gets a bit circular...</p>
<p>P.S. Traying to write things out in coordinates will not get you far because a closed form not being exact is a "global" thing, not a local one</p>
|
1,791,815 | <p>I'm trying to find an elementary way to see that the 1st de Rham cohomology of the n-sphere is zero for $n>1$, $H^1(S^n) = 0$.</p>
<p>This is part of an attempt to find the de Rham cohomology of the n sphere generally. I have shown that $H^k(S^n) = H^{k-1}(S^{n-1})$ (for $k>1$) but to find the de Rham cohomology generally I need to show that $H^1(S^n) = 0$ and I am struggling to do this.</p>
<p>Is it possible to take a closed one form $\alpha$ on a sphere and explicitly find a function $f$ on $S^n$ that has $df = \alpha$, I've tried writing it out in coordinates and explicitly integrating but I haven't managed to get very far with this!</p>
| Thomas | 284,057 | <p>Write $S^n$ as the union of two open sets $U= S^n-N$, $V=S^n-S $where $N,S$ are the north and south pole. Note that the intersection $U\cap V$ is connected if $n\geq 2$. $U,V$ are contracile (diffeomorphic to $R^,$, and on each of these sets the form $\alpha$ is exact and andmits primitive $f_U, f_V$. On the intersection $d(f_U-f_V)=0$ and by connexity $f_U-f_V$ must be a constant $c$. So $f_V=f_U+c$ one the intersection. As a primitive of $\alpha$ one can take $f_U$ on $U$ and $f_U +c$ on $V$. The main point is that the intersection is connected.</p>
|
3,738,719 | <p>We got</p>
<p><span class="math-container">$$x'=\frac{t^2x^2+1}{2t^2}$$</span></p>
<p><span class="math-container">$$\frac{dx}{dt}=\frac{t^2x^2+1}{2t^2}$$</span></p>
<p><span class="math-container">$$2t^2dx=\left(t^2x^2+1\right)dt$$</span></p>
<p><span class="math-container">$$0=\left(t^2x^2+1\right)dt+\left(-2t^2dx\right)$$</span></p>
<p>I get the integrating factor</p>
<p><span class="math-container">$$\mu \left(t\right)=\frac{e^{x+2}}{t}$$</span></p>
<p>But it doesn't really work out!</p>
<p>What should I do?</p>
| user577215664 | 475,762 | <p><span class="math-container">$$x'=\frac{t^2x^2+1}{2t^2}$$</span>
<span class="math-container">$$x'=\frac{x^2}{2}+\frac{1}{2t^2}$$</span>
Change the variable
<span class="math-container">$$x=u-\dfrac 1t$$</span>
<span class="math-container">$$u'=\frac{u^2}{2}-\dfrac ut$$</span>
This is Bernoulli 's differential equation. You can put it in the following form for <span class="math-container">$u \ne 0$</span>:
<span class="math-container">$$\dfrac {d(ut)}{(ut)^2}=\dfrac {dt}{2t}$$</span>
Integrate.
<span class="math-container">$$\dfrac {2}{(ut)}=k-\ln t$$</span>
<span class="math-container">$$u=\dfrac 2{t(k-\ln t)}$$</span>
<span class="math-container">$$x(t)=\dfrac 2{t(k-\ln t)}-\dfrac 1t$$</span></p>
<p>For <span class="math-container">$u=0$</span> you have that: <span class="math-container">$$x(t)=-\dfrac 1 t$$</span></p>
|
302,599 | <p><strong>Problem:</strong></p>
<p>For every $n$ in $\mathbb{N}$, we consider the sets $A_{n}:=\left \{ (2n+1)\lambda :\ \lambda \in \mathbb{N}\right \}$. The question is to find $\bigcap_{n=1}^{\infty }A_{n}$, and $\bigcup_{n=1}^{\infty }A_{n}$.</p>
<p>For the intersection, I think it is the empty set, because for every $n$ in $\mathbb{N}$; we have $n\notin A_{n}\Rightarrow n\notin \bigcap_{n=1}^{\infty }A_{n}$. For the union, I tried the first few sets, but I can't see exactly what the union of all the sets should be. Any help is appreciated.</p>
| André Nicolas | 6,312 | <p>These are all the positive integers that have a factor of the shape $2n+1$. That's the collection of all positive integers that have an odd factor $\ge 3$. Every positive integer qualifies, <strong>except</strong> for the powers $1,2,4,8,\dots$ of $2$.</p>
<p>So our set is the complement of the set of powers of $2$. </p>
|
524,387 | <p>Having watched an <a href="http://www.youtube.com/watch?v=NircIRdNUrE" rel="nofollow">integralCALC video lesson</a>, given</p>
<p>$$w=xe^{y/z}, x = t^2, y = 1-t, z=1+2t$$
which could be rewritten as
$$w=t^2 e^\frac{1-t}{1+2t}$$</p>
<p>How does $dw/dt$ differ from $\partial w/\partial t$? Is there some intuition behind the partial derivative (such as a geometric interpretation)?</p>
| dezign | 32,937 | <p>There's no difference. It's just that we use straight $d$s to denote derivatives of functions with respect to one variable. It may seem like the function $w$ depends on more than one variable, but the other variables all depend on $t$, so at the end of the day all the values of the function $w$ are determined by the values of $t$, so we use the straight $d$s to denote its derivative. </p>
|
3,660,825 | <p>I know this question has been asked many times and there is good information out there which has clarified a lot for me but I still do not understand how the addition and multiplication tables for <span class="math-container">$GF(4)$</span> is constructed?</p>
<p>I'm just starting to learn about fields in general, galois fields and the concept of "it can't be 0 or 1 so it must be x"</p>
<p>I've seen;
<a href="https://math.stackexchange.com/q/172980/378927">Galois Field GF(4)</a>;
<a href="https://math.stackexchange.com/q/2777107/378927">Addition and Multiplication in $F_4$</a>;
<a href="https://math.stackexchange.com/q/168135/378927">Explicit construction of a finite field with $8$ elements</a></p>
<p>but none explicity explain the construction and I'm too new to be told "its an extension of <span class="math-container">$GF(2)$</span>"</p>
<p>Thank you in advance</p>
| Wuestenfux | 417,848 | <p>Well, take an irreducible polynomial of degree 2 over <span class="math-container">$GF(2)$</span>. There is exactly one namely <span class="math-container">$f(x)=x^2+x+1$</span>.
Then the quotient ring <span class="math-container">$GF(2)[x]/\langle f(x)\rangle$</span> is a field, <span class="math-container">$GF(4)$</span>, with 4 elements.</p>
<p>To construct the field, note that <span class="math-container">$f(x)$</span> has a zero in <span class="math-container">$GF(4)$</span> namely residue class <span class="math-container">$x+\langle f(x)\rangle$</span>. Call it <span class="math-container">$\alpha$</span>. Then <span class="math-container">$f(\alpha)=\alpha^2+\alpha+1=0$</span>, i.e., <span class="math-container">$\alpha^2=\alpha+1$</span>.
Thus the elements of <span class="math-container">$GF(4)$</span> are <span class="math-container">$0,1,\alpha,\alpha^2=\alpha+1$</span>. From here its simple to construct the addition and multiplication tables.</p>
|
992,125 | <p>If I rolled $3$ dice how many combinations are there that result in sum of dots appeared on those dice be $13$?</p>
| G Cab | 317,234 | <p>the formula introduced by Masacroso applies to a bulk of different schemes in combinatorics and diophantine geometry, all stemming out from finding
<span class="math-container">$$N_{\,b} (s,r,m) = \text{No}\text{. of solutions to}\;\left\{ \begin{gathered}
0 \leqslant \text{integer }x_{\,j} \leqslant r \hfill \\
x_{\,1} + x_{\,2} + \cdots + x_{\,m} = s \hfill \\
\end{gathered} \right.$$</span>
<em>(note that here, for generality, the allowed range for the variables is taken as <span class="math-container">$0\, \ldots \,r$</span>;<br>
the conversion to <span class="math-container">$1\, \ldots \,6$</span> for dice problems is quite straight, leading to the formulas already provided above)</em>. </p>
<p>It is preferable to express <span class="math-container">${N_{\,b} }$</span> as follows</p>
<p><span class="math-container">$$
N_{\,b} (s,r,m)\quad \left| {\;0 \le {\rm integers}\;s,r,m} \right.\quad = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,{s \over r}\, \le \,m} \right)} {\left( { - 1} \right)^{\,k} \left( \matrix{
m \cr
k \cr} \right)\left( \matrix{
s + m - 1 - k\left( {r + 1} \right) \cr
s - k\left( {r + 1} \right) \cr} \right)}
$$</span>
where the binomial coefficient is defined as</p>
<p><span class="math-container">$$\left( \begin{gathered}
x \\
q \\
\end{gathered} \right) = \left\{ {\begin{array}{*{20}c}
{\frac{{x^{\,\underline {\,q\,} } }}
{{q!}}} & {0 \leqslant \text{integer }q} \\
0 & {\text{otherwise}} \\
\end{array} } \right.$$</span></p>
<p>re. [1], [2].</p>
<p>When defined in this way, in fact, the limits of summation are implicit in the summand (that is why they are indicated in brackets) and that greatly simplifies further manipulations.</p>
<p>The o.g.f. , as explained in the precedent answer, is
<span class="math-container">$$
F_{\,b} (x,r,m) = \sum\limits_{0\, \le \,s\,\left( { \le \,m\,r} \right)} {N_{\,b} (s,r,m)\;x^{\,s} } = \left( {1 + x + \, \cdots \, + x^{\,r} } \right)^m = \left( {{{1 - x^{\,r + 1} } \over {1 - x}}} \right)^m
$$</span></p>
<p>Thus <span class="math-container">$Nb$</span> can also be expressed in terms of multinomials ..., and that is why it is also called "r-nomial coefficient" (actually, as defined above, an "r+1-nomial"): eg. in OEIS A008287 [5]. </p>
<p><span class="math-container">$Nb$</span> satisfies many recurrences, one of which is :</p>
<p><span class="math-container">$$\left\{ \begin{gathered}
N_{\,b} (s,r,0) = \left[ {0 = s} \right] \hfill \\
N_{\,b} (s,r,m + 1) = \sum\limits_{0\, \leqslant \,j\, \leqslant \,r} {N_{\,b} (s - j,r,m)} \hfill \\
\end{gathered} \right.$$</span></p>
<p>where:
<span class="math-container">$$\left[ P \right] = \left\{ {\begin{array}{*{20}c}
1 & {P = TRUE} \\
0 & {P = FALSE} \\
\end{array} } \right.\text{ }\;\;\text{is the Iverson bracket}$$</span></p>
<p>and which just corresponds to:</p>
<p><span class="math-container">$$F_{\,b} (x,r,m) = \left( {\frac{{1 - x^{\,r + 1} }}
{{1 - x}}} \right)^m = \left( {\frac{{1 - x^{\,r + 1} }}
{{1 - x}}} \right)\left( {\frac{{1 - x^{\,r + 1} }}
{{1 - x}}} \right)^{m - 1} $$</span></p>
<p>Each way in which <span class="math-container">$F_{\,b}$</span> can be rewritten turns into a relation for <span class="math-container">$N_{\,b}$</span>, for instance<br>
<span class="math-container">$$
F_{\,b} (x,r,m) = \left( {{{1 - x^{r + 1} } \over {1 - x}}} \right)^{\,m} = \left( {{{1 - x^r } \over {1 - x}} + x^r } \right)^{\,m} = \left( {1 + x\left( {{{1 - x^r } \over {1 - x}}} \right)} \right)^{\,m}
$$</span></p>
<p>And to complete the picture you also have the double o.g.f.
<span class="math-container">$$
G_{\,b} (x,r,z) = \sum\limits_{0\, \le \,s,\,m} {N_{\,b} (s,r,m)\;x^{\,s} \;z^{\,m} } = {1 \over {1 - z{{1 - x^{\,r + 1} } \over {1 - x}}}}
$$</span></p>
<p>The applications include:<br>
a) number of ways to roll <span class="math-container">$m$</span> dice, with <span class="math-container">$r+1$</span> facets numbered from 0 to r, and getting a total of <span class="math-container">$s$</span>;<br>
b) number of ways to dispose <span class="math-container">$s$</span> indistinguishable balls into <span class="math-container">$m$</span> distinguishable bins of capacity <span class="math-container">$r$</span> as it is called in many publications,<br>
but, beware that this might be misleading , as is not the model of "throwing balls into bins", rather the reverse of "<em>throwing bins into balls</em>" , in the sense of throwing separators into a row of balls, i.e. the "bars_and_stars" model,
but provided that the <span class="math-container">$m-1$</span> bars are inserted incrementally, and then with the restriction that they shall not encompass more than <span class="math-container">$r$</span> balls ;<br>
c) number of different histograms, with <span class="math-container">$m$</span> bars, each bar of length <span class="math-container">$0\, \ldots \,r$</span>, total length <span class="math-container">$s$</span>;<br>
d) number of points with integer coordinates, lying on the diagonal plane <span class="math-container">$x_{\,1} + x_{\,2} + \cdots + x_{\,m} = s$</span>, within a <span class="math-container">$m$</span>-dimensional cube of side <span class="math-container">$0\, \ldots \,r$</span>;<br>
e) number of 2-D lattice paths, from <span class="math-container">$(0,0)$</span> to <span class="math-container">$(m,s)$</span>, with steps in <span class="math-container">$\left( {1,0\, \ldots \,r} \right)$</span> ;<br>
f) finally note that <span class="math-container">$N_{\,b}$</span> recurrence above entails a <em>"moving-window summation"</em> of fixed width <em>0..r</em>, so that it can be exploited in topics involving that.</p>
<p>The various underlying models provide different perspectives useful to grasp the properties of this function.<br>
It is clear for instance that <span class="math-container">$N_{\,b} (s,r,m) = N_{\,b} (m\,r - s,r,m)$</span>
because distributing <span class="math-container">$s$</span> balls is the same as distributing <span class="math-container">$m r-s$</span> voids,
or by looking at the complement of the histogram, or by viewing at the <span class="math-container">$m$</span>-cube from the opposite diagonal corner.</p>
<p>@PardonMe..<br>
A clear, precise and fundamental basis to Generating Functions (and to much more) is given in [1].<br>
[3] provides a general exposition of how this function may be derived (it also deals with the case of bins with different capacities..).<br>
In [4] then, although it deals with partitions, you get a clear picture of how to derive from the o.g.f. the combinatorial properties that it encapsulates, as Masacroso did in his exposition above.</p>
<hr>
<p>[1] "Concrete Mathematics: a foundation for computer science" R. L. Graham - D.E. Knuth - O. Patashnik - Addison-Wesley 2nd Ed. 1994<br>
[2] <a href="http://en.wikipedia.org/wiki/Binomial_coefficient" rel="nofollow noreferrer">http://en.wikipedia.org/wiki/Binomial_coefficient</a><br>
[3] <a href="https://www.mathpages.com/home/kmath337/kmath337.htm" rel="nofollow noreferrer">https://www.mathpages.com/home/kmath337/kmath337.htm</a><br>
[4] <a href="http://www.math.upenn.edu/~wilf/PIMS/PIMSLectures.pdf" rel="nofollow noreferrer">http://www.math.upenn.edu/~wilf/PIMS/PIMSLectures.pdf</a><br>
[5] <a href="https://oeis.org/A008287" rel="nofollow noreferrer">https://oeis.org/A008287</a><br>
[6] <a href="http://arxiv.org/abs/1202.0228v7" rel="nofollow noreferrer">http://arxiv.org/abs/1202.0228v7</a></p>
|
2,918,497 | <p>From <em>Linear Algebra</em> by Friedberg, Insel, and Spence:</p>
<blockquote>
<p>Given <span class="math-container">$M_1=\begin{pmatrix} 1&0\\0 &1\end{pmatrix}$</span>, <span class="math-container">$M_2=\begin{pmatrix} 0&0\\0 &1\end{pmatrix}$</span> and <span class="math-container">$M_3=\begin{pmatrix}0&1\\1 &0\end{pmatrix}$</span>,</p>
<p>prove that <span class="math-container">$\text{span}\{M_1, M_2, M_3\}$</span> is the set of all symmetric <span class="math-container">$2 \times2$</span> matrices.</p>
</blockquote>
<p>For reference, we just learned about linear combinations/span, but only in terms of vectors, nothing really with matrices.</p>
| gt6989b | 16,192 | <p><strong>HINT</strong></p>
<p>Span is the set of all linear combinations. Note that
$$
aM_1 + bM_2 + cM_3 =
\begin{pmatrix}
a & c \\
c & a+b
\end{pmatrix}
$$</p>
<p>Can you finish finding the span?</p>
<hr>
<p><strong>HINT 2</strong></p>
<p>The way you phrased it, the span does not <em>include</em> all symmetric matrices, but the span <em>is</em> that set. Is $M_2$ symmetric?</p>
|
1,357,488 | <p>Let $E$ together with $g$ be a inner product space(over field $\mathbb R$) , $\text{dim}E=n<\infty$ and $\{e_1,\cdots,e_n\}$ is orthonormal basis of $E$ that $\{e^1,\cdots,e^n\}$ is its dual basis. Now we define $\omega:=e^1\wedge\cdots\wedge e^n$ as element of volume of $E$.</p>
<p>How can I prove that $\omega(u_1,\cdots,u_n)\omega(v_1,\cdots,v_n)=det[g(u_i,v_j)] \qquad \forall u_i,v_j\in E\qquad\text{and } i=1,\cdots,n\ ?$ </p>
<p>Of course, I prove that $\omega(u_1,\cdots,u_n)=det[u_1 \cdots u_n]_{n\times n}$ ($u_i$'s are columns of the matrix)but I do not find out how uses it for my problem.</p>
| Joonas Ilmavirta | 166,535 | <p>The equation you wish to prove is linear in each of the vectors $u_j$ and $v_j$.
Therefore it suffices to show the identity when these vectors are basis vectors.
There are $n$ basis vectors from which we now want to choose the $n$ vectors $u_1,\dots,u_n$.
If we choose any two to be the same, then both sides of the identity vanish (and the identity is true), so we may assume that they are all distinct.
Both sides of the identity change signs in the same way if we permute the vectors $u_1,\dots,u_n$, so way may in fact assume $u_j=e_j$.
Similarly, we may assume $v_j=e_j$.
So it only remains to check this one case, and now the identity is again trivial: both sides equal one.</p>
|
3,822,822 | <p>Four cards are face down on a table. You are told that two are red and two are black, and you need to guess which two are red and which two are black. You do this by pointing to the two cards you’re guessing are red (and then implicitly you’re guessing that the other two are black). Assume that all configurations are equally likely, and that you do not have psychic powers. Find the probability that exactly j of your guesses are correct, for j = 0, 1, 2, 3, 4.
Hint: Some probabilities are 0.</p>
<p>My professor worked out this example in class and I know the answers for j = 1 and j = 3 are zero, j = 0 is 1/6, j = 2 is 2/3, and j = 4 is 1/6 but I do not understand the process or concept behind the question. I do not understand where the numbers are coming from and why only the even js have a probability but not the odd js. The j guesses and pointing is confusing me. Can someone please help explain the question, I would appreciate it.</p>
| Anatoly | 90,997 | <p>HINT: with <span class="math-container">$4$</span> cards, of which two are red and two are black, there are <span class="math-container">$\frac{4!}{2!2!}=6$</span> possible sequences. When you choose two cards, you automatically define the other two cards and then choose a sequence.</p>
<p>Now consider the two cards that you choose: they can be both correct (this implies that the other two are also correct), both wrong (this implies that the other two are also wrong) or one correct and one wrong (this implies that the other two are one correct and one wrong as well). Now you can easily complete the solution.</p>
|
4,634,263 | <p>I have the following problem that asks to find the shock curve for the following IVP
<span class="math-container">$$u_t + (u^2)_x = 0, \quad u(x,0) = \frac{1}{2}e^{-x},$$</span>
to then obtain a weak solution from this curve using Rankine Hugoniot condition. I'm pretty lost on how to raise the problem. In Evans there is an example where it is easy to get u^- and u^+, so we can solve for the shock curve using an ODE. In this case, does anyone have an idea how to raise the problem?</p>
| Ethan Bolker | 72,858 | <p>@FD_bfa offers a <a href="https://math.stackexchange.com/questions/4634210/why-are-two-disjoint-events-defined-to-be-independent-if-one-has-zero-probabilit/4643623#4643623">good answer</a>: a definition should cover edge and degenerate cases with no modification.</p>
<p>Here's another way to look at it. Two events are independent if finding out that one has occurred tells you nothing about the probability of the other. That's clearly the case when one (or both) of the events has probability <span class="math-container">$0$</span>.</p>
|
3,081,076 | <p>In do Carmo, one exercise gives a plane in <span class="math-container">$\mathbb R^3$</span>, <span class="math-container">$ax +by +cz+d = 0$</span>, and tells us to show that <span class="math-container">$|d|/\sqrt{a^2 + b^2 + c^2}$</span> measures the distance from the plane to the origin.</p>
<p>However, this seems a bit ambiguous since we don't know what the plane actually is.</p>
<p><strong>By distance, does he mean minimal distance?</strong></p>
| Alex J Best | 31,917 | <p>Something similar was already asked at <a href="https://math.stackexchange.com/questions/3021423/independence-of-points-on-elliptic-curve">Independence of points on Elliptic curve</a> the answers there are the same as above but maybe a little more extensive.</p>
<p>Edit: I just noticed the question asker is the same as the previous question. So maybe this question really is trying to ask something different. The height of a point will depend on the elliptic curve itself so you will need to put the equation of the curve somewhere.</p>
|
1,023,013 | <blockquote>
<p>Prove:$\forall a_1,b_1,a_2,b_2: \left|\max(a_1,b_1) - \max(a_2,b_2)\right| \le \max(\left|a_1-a_2\right|, \left| b_1-b_2 \right|) $</p>
</blockquote>
<p>How can I prove it easily? Or should I just go over each case?
All those cases make me confused!</p>
| Empy2 | 81,790 | <p>$$\frac12(a_n^2-a_n-a_{n+1}^2+a_{n+1})=\frac12(a_n-a_{n+1})(a_n+a_{n+1}-1)$$<br>
Can you show that $|\frac12(a_n+a_{n+1}-1)|<1$?</p>
|
554,869 | <p>I want to show that the following conditions are equivalent for a nonzero element $a$ in a Boolean algebra $\mathcal{B}$:</p>
<p>1) for all $x\in\mathcal{B},a\leq x$ or $a\leq x'$</p>
<p>2) for all $x,y\in\mathcal{B},a\leq x\sqcup y\Rightarrow a\leq x$ or $a\leq y$</p>
<p>3) $a$ is minimal among nonzero elements of $\mathcal{B}$</p>
<p>I can't show any of the implications. Could you give me some hint?</p>
| egreg | 62,967 | <p>(1)$\implies$(2) If $a\le x\sqcup y$ and $a\not\le x$, then $a\le x'$. Thus $a\le(x\sqcup y)\sqcap x'=\ldots\le\ldots$</p>
<p>(2)$\implies$(3) Let $b\le a$; then $a\le b\sqcup b'$, so $a\le b$ or $a\le b'$. In the first case $a=b$; in the second case $b\le b'$, so …</p>
<p>(3)$\implies$(1) …</p>
|
90,342 | <p>Here $\mathbb{T}^n := (\mathbb{R} / \mathbb{Z})^n$ is the topological group of the n-dimensional torus and $k \in \mathbb{Z}^n$ is a non-null vector, I'm working about the subgroup</p>
<p>$S = \{x \in \mathbb{T}^n : k \cdot x = 0_{\mathbb{T}^n}\}$</p>
<p>where $\cdot$ is the scalar product. I think that $S$ is isomorphic, as topological group, to $\mathbb{T}^{n-1}$, but I could not prove it.</p>
<p>About the case $n=2$, setting $k = (k_1, k_2)$, $k_2 \neq 0$ I found as a possible candidate isomorphism</p>
<p>$\mathbb{T} \to \mathbb{T}^2 : x \mapsto (k_2 x, \lfloor k_2 x \rfloor / k_2 - k_1 x)$</p>
<p>where $\lfloor \;\rfloor$ is the floor function, however this seems too complicated in the general case, I hope you have useful tips, thanks!</p>
| James Cranch | 14,901 | <p>It is known, in fact, that every compact connected abelian Lie group is isomorphic to a torus.</p>
<p>Your construction has a danger of producing disconnected subgroups, however: this occurs when the coordinates of $k$ are all multiples of some fixed $i>1$. (You can see this even in small dimensions).</p>
|
1,990,033 | <p>Suppose I have a point $P(x_1, y_1$) and a line $ax + by + c = 0$. I draw a perpendicular from the point $P$ to the line. The perpendicular meets the line at point $Q(x_2, y_2)$. I want to find the coordinates of the point $Q$, i.e., $x_2$ and $y_2$.</p>
<p>I searched up for similar questions where the coordinates of end points of the line segment are given. But here, I've got an equation for the line. So, I am pretty clueless how to solve this.</p>
<p>Please give me a formula to arrive at my answer (if any) and show me its derivation too. I am a high school student with a basic knowledge of trigonometry. I have no idea of calculus, so please give me a simplified answer.</p>
<p>Any help is highly appreciated. Thanks a lot in advance...</p>
| 655321 | 197,645 | <p>Since
$$ \sin a \cdot \sin b = \frac{1}{2} \left[ \cos (a - b) - \cos (a+b) \right] $$
$$ \sin a \cdot \cos b = \frac{1}{2} \left[ \sin (a + b) + \sin (a - b) \right] $$
and
$$ \sin a - \sin b = 2 \sin\left[ \frac{1}{2} (a-b) \right] \cos \left[ \frac{1}{2} (a+b) \right] $$
we can use these identities to solve
$$ 4 \sin x \cdot \sin 2x \cdot \sin 4x = \sin 3 x $$
$$ \Rightarrow \quad 2 \sin x \cdot \left[ \cos (-2x) - \cos 6x \right] = \sin 3x $$
$$ \Rightarrow \quad 2 \sin x \cdot \cos 2x - 2 \sin x \cdot \cos 6x = \sin 3x $$
$$ \Rightarrow \quad \sin 3x + \sin (-x) - (\sin 7x + \sin(-5x)) = \sin 3x $$
$$ \Rightarrow \quad \sin 3x - \sin x - \sin 7x + \sin 5x = \sin 3x $$
$$ \Rightarrow \quad - \sin x - \sin 7x + \sin 5x = 0 $$
$$ \Rightarrow \quad \sin x = \sin 5x - \sin 7x $$
$$ \Rightarrow \quad \sin x = 2 \sin (-x) \cos 6x $$
$$ \Rightarrow \quad \sin x = - 2 \sin x \cos 6x $$
$$ \Rightarrow \quad \cos 6x = - \frac{1}{2} $$
$$ \Rightarrow \quad 6x = \frac{2}{3} (3 \pi n \pm \pi) $$
$$ \Rightarrow \quad x = \frac{1}{9} (3 \pi n \pm \pi) $$</p>
<p>or
$$\sin x = 0 \quad \Rightarrow \quad x = n \pi \mathrm{.}$$</p>
|
1,143,893 | <p>Struggling with basic AS Statistical maths, Any help would be much appreciated.</p>
<p>The two events $A,B$ are such that $P(A)= 0.65, P(A\cup B)= 0.93$</p>
<p>Evaluate $P(B)$ given that $A$ and $B$ are independent. (4 Marks) </p>
<p>Thank you.</p>
| user 1 | 133,030 | <p>$P(AUB) = P(A) + P(B) - P(A\cap B) $.</p>
|
3,342,761 | <p>I'm trying to get an equation for a solid angle of a segment of octahedron in the same vein as described in this article <a href="http://www.rorydriscoll.com/2012/01/15/cubemap-texel-solid-angle/" rel="nofollow noreferrer">cubemap-texel-solid-angle</a>. I ended up having to integrate
<span class="math-container">$$\int \int \frac{1}{(x^2+y^2+(1-x-y)^2)^\frac{3}{2}} \,\mathrm{d}x \,\mathrm{d}y$$</span> where <span class="math-container">$0 \leq x \leq 1$</span> and <span class="math-container">$0 \leq y \leq 1-x$</span>. That is, integral over a segment of a triangle mapped onto the sphere(one octant). Does anyone know how to integrate that?
<a href="https://i.stack.imgur.com/UQR7z.png" rel="nofollow noreferrer">Subdivided polyhedrons by courtesy of Gavin Kistner</a></p>
<p><strong>Update</strong>
Thanks to the general formula from <a href="https://math.stackexchange.com/questions/1211287/how-to-find-out-the-solid-angle-subtended-by-a-tetrahedron-at-its-vertex/3305625#3305625">this answer</a>:
<span class="math-container">$$\omega=\cos^{-1}\left(\frac{\cos\alpha-\cos\beta\cos\gamma}{\sin\beta\sin\gamma}\right)-\sin^{-1}\left(\frac{\cos\beta-\cos\alpha\cos\gamma}
{\sin\alpha\sin\gamma}\right)-\sin^{-1}\left(\frac{\cos\gamma-\cos\alpha\cos\beta}{\sin\alpha\sin\beta}\right)$$</span></p>
<p>We can calculate a solid angle for all the triangles. Here is the <a href="https://gist.github.com/aschrein/ad2554b9c54a207f6dcfbe431b130d04" rel="nofollow noreferrer">gist</a> and the <a href="https://www.shadertoy.com/view/tlBXDd" rel="nofollow noreferrer">shadertoy</a>. The naive implementation is not numerically stable at small angles.</p>
<p><strong>Update</strong> See <a href="https://math.stackexchange.com/q/3343709">this answer</a></p>
| Jean Marie | 305,862 | <p>All 8 facets of an octaedron are alike and interchangeable. </p>
<p>Then, the looked for solid angle is </p>
<blockquote>
<p><span class="math-container">$$\frac18(4\pi)=\frac{\pi}{2}.$$</span></p>
</blockquote>
|
1,880,150 | <p>I work in a warehouse where we take components and put them together to make a finished good product.</p>
<p>We have these values for each component:</p>
<ol>
<li><p>Quantity Required</p></li>
<li><p>Quantity Used</p></li>
<li><p>Variance (Quantity Used- Quantity Required)</p></li>
<li><p>Overage ((Variance/Quantity Required)) * 100%</p></li>
</ol>
<p>So basically these values tell me how much we needed, how much we used, and how much we wasted.</p>
<p>At the bottom of the report that shows all these values, I take the sum of Quantity Required, Quantity Used, and Variance. I also take the average of Quantity Required and Quantity Used and subtract these averages to find the average Variance.</p>
<p><strong>The problem:</strong> Sometimes the Quantity Required value for a component is zero. How should I adjust my summations/averages accordingly? </p>
<p><strong>What I've tried:</strong> I've simply made the summations/averages <strong>ignore</strong> the rows where the Quantity Required is zero, but I think this is <strong>wrong</strong>. If my boss asks me "What is the average overage for Item X for this year?" my method will ignore the rows with zeros in the Quantity Required column, thus ignoring an amount in Quantity Used that is technically still an amount to consider.</p>
<p><strong>Sample Data of 1 item being used across multiple jobs:</strong></p>
<pre><code> Quantity Required Quantity Used Variance Overage
Job A 1000 1100 100 10%
Job B 900 800 -100 -11.1%
Job C 0 300 300 (N/A)
Summations: 1900 1900 0 0%
Averages: 950 950 0 0%
</code></pre>
| Justthisguy | 81,821 | <p>It's a good rule of thumb to never add quantities that have different units. When you add, for example, the entries in the first column, the first entry has units "number of component A", whereas the second entry has units "number of component B". So what are the units of the sum?</p>
<p>A good way to aggregate this data is to compute the total cost of the goods used and the total cost of goods used under normal usage (i.e. if the quantities used in the "quantity required" column were used). Then you overage cost is </p>
<p>(cost of goods used) - (cost of goods that should have been used), </p>
<p>and your overage percentage is </p>
<p>((cost of goods used) - (cost of goods that should have been used)) / (cost of goods that should have been used).</p>
<p>You can do this analysis item-by-item as well.</p>
|
961,304 | <p>I am studying a book and I am stagnating on a what should be a straightfoward proof:</p>
<p>Show that if $X$ is compact, $V\subset X$ is open and $x\in V$, then there exists an open set $U$ in $X$ with $x\in U\subset \bar U\subset V$</p>
<p>I don't know how to find the appropriate set $U$. I am guessing you need to do something like taking the intersection with $V$ of a finite subcover of X and then show that its closure is contained in $V$ ...</p>
<p>Could someone nudge me in the right direction?</p>
| Ilmari Karonen | 9,602 | <p>Let $X$ be a <a href="//en.wikipedia.org/wiki/Random_variable" rel="nofollow">random variable</a> uniformly distributed over the interval $[a,b]$. If $f : [a,b] \to \mathbb R$ is a (Borel) <a href="//en.wikipedia.org/wiki/Measurable_function" rel="nofollow">measurable function</a>, then $Y = f(X)$ is also a random variable, and its <a href="//en.wikipedia.org/wiki/Probability_distribution" rel="nofollow">distribution</a> can be characterized by all the usual statistical properties, such as the mean and the median.</p>
<p>In particular, the <a href="//en.wikipedia.org/wiki/Mean_of_a_function" rel="nofollow">mean of the function</a> $f$, as defined in your question, corresponds exactly to the <a href="//en.wikipedia.org/wiki/Expected_value" rel="nofollow">mean of the random variable</a> $Y$, as given by the <a href="//en.wikipedia.org/wiki/Law_of_the_unconscious_statistician" rel="nofollow">law of the unconscious statistician</a>.</p>
<p>Similarly, we can naturally define the <a href="http://en.wikipedia.org/wiki/Median" rel="nofollow">median</a> of the function $f$ to be the median of the corresponding random variable $Y$, i.e. a value $m$ such that $\mathrm{Pr}(Y \ge m) \ge \frac12$ and $\mathrm{Pr}(Y \le m) \ge \frac12$.</p>
<p>In particular, since $\mathrm{Pr}(Y \ge m) = \frac{1}{b-a}\mu(\{x \in [a,b]: f(x) \ge m\})$, where $\mu(S) = \int_S 1\, dx$ denotes the <a href="//en.wikipedia.org/wiki/Measure_%28mathematics%29" rel="nofollow">measure</a> of the set $S$, and conversely for $\mathrm{Pr}(Y \le m)$, it follows that the median $m$ of $Y$ (and thus of $f$) is simply the value such that the graph of $f$ lies at or below $m$ for half of the interval $[a,b]$, and at or above $m$ for the other half.</p>
<p>(The median, as defined above, may not be unique if $f$ is not continuous. However, if there are multiple values of $m$ that satisfy the definition above, then they will form a single interval. In this sense, the situation is no different from that of the median of a probability distribution in general.)</p>
|
3,267,153 | <p>I want to understand an equation I saw in a paper. If <span class="math-container">$\theta$</span> is some angle, and <span class="math-container">$\sigma_y = \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix}$</span>, then it is stated that <span class="math-container">$e^{-i \theta \sigma_y } = R_{\theta}$</span> (the rotation matrix of angle <span class="math-container">$\theta$</span>). </p>
<p>Now set <span class="math-container">$A:= -i \theta \sigma_y = \begin{bmatrix} 0 & \theta \\ -\theta & 0 \end{bmatrix}$</span>, so that <span class="math-container">$e^{A}$</span> is defined as <span class="math-container">$\sum_{j=0}^{\infty} \frac {1} {k!} A^k$</span>. Now I know how to exponentiate a diagonal matrix, but I am not sure how to do it in this case (anti-diagonal). Insights appreciated.</p>
| Marc Bogaerts | 118,955 | <p>A simple method (based on the chinese remainder theorem) goes as follows:</p>
<p>To solve <span class="math-container">$f \operatorname{mod} p = u$</span> and <span class="math-container">$f \operatorname{mod} q = v$</span> (where p and q are coprime polynomials):</p>
<p>put <span class="math-container">$\ \bar{p} = p^{-1}\operatorname{mod} q$</span> and <span class="math-container">$\ \bar{q} = q^{-1}\operatorname{mod} p$</span>
then <span class="math-container">$f = vp\bar{p}+uq\bar{q}$</span>. Indeed we have <span class="math-container">$ f = upp^{-1}\operatorname{mod} q = u$</span> and <span class="math-container">$ f = vqq^{-1}\operatorname{mod} p = v$</span>. </p>
<p>In this case we have for <span class="math-container">$p = x^2+4$</span>, <span class="math-container">$q = x+3$</span>, <span class="math-container">$u = 7x-20$</span> and <span class="math-container">$v=-2$</span> that <span class="math-container">$\bar{p} = -\frac{x}{39}$</span> and <span class="math-container">$\bar{q} = -\frac{x}{13}+\frac{3}{13}$</span> so that <span class="math-container">$f = -\frac{19}{39}x^3+\frac{20}{13}x^2+\frac{197}{39}x-\frac{180}{13}$</span>. But this only one of the many solutions, the all differ by a multiple of <span class="math-container">$pq$</span>. In this case <span class="math-container">$f + \frac{58}{39}\ pq\ $</span> yields the given polynomial.</p>
|
1,010,820 | <p>I tried the following :</p>
<p>\begin{align}\sec\theta + \tan\theta&=4\\
\frac1{\cos\theta} + \frac{\sin\theta}{\cos\theta}&=4\\
\frac{1+\sin\theta}{\cos\theta}&=4\\
\frac{1+\sin\theta}4&=\cos\theta\end{align}</p>
<p>now don't know how to evaluate further ?</p>
| user72272 | 72,272 | <p>one way is to use the half angle formule like @Mark Bennet suggested. then you have
$$\sec(\theta)=\frac{1+t^2}{1-t^2},\qquad\tan(\theta)=\frac{2t}{1-t^2}$$ where $t=\tan\left(\frac{\theta}{2}\right)$. This results in the following equationt
$$\frac{1+2t+t^2}{1-t^2}=\frac{(1+t)^2}{(1-t)(1+t)}=\frac{1+t}{1-t}=4\\t=\frac{3}{5}\rightarrow\theta=2\arctan\left(\frac{3}{5}\right)$$
and $$\cos(\theta)=\frac{1-t^2}{1+t^2}=\frac{1-\frac{9}{25}}{1+\frac{9}{25}}=\frac{8}{17}.$$</p>
|
2,717,842 | <p>I want to show that if $X$ is a locally connected topological space, $A\subseteq X$ is a subspace and $f:X \rightarrow A$ is continuous such that $f|_{A} = Id_{A}$, then $A$ must be locally connected as well.</p>
<p>My progress so far:</p>
<p>Take $U\subseteq A$ $A-$open and $x\in U$. Since $f^{-1} (U)$ is $X-$open and $x\in f^{-1} (U)$, there exists a connected, $X-$open subset $V\subseteq X$ such that $x\in V\subseteq f^{-1}(U)$.</p>
<p>Now we have $x\in V\cap A \subseteq f^{-1}(U)\cap A = U$</p>
<p>My guess is that $V\cap A$ should be connected, but I am unsure if this is correct.</p>
<p>Any help would be appreciated!</p>
| William Elliot | 426,203 | <p>Let K be the nonnegative x and y axis.<br>
Project every point of the 1st quadrant at a 45 degree angle o K.<br>
Thus K is a retract of the first quadrant.<br>
In general, every V is retract of the space between the two lines. </p>
<p>Let C be the comb space of all lines from (0,0) to (1,1/n)<br>
for all n in N and the line from (0,0) to (1,0). </p>
<p>In view of the retracts of all the V's of K can the first quadrant be retracted to K? Thus be a counterexample.</p>
|
1,976,001 | <p>The statement goes as following: if $3 \mid 2a$, then $3 \mid a$ and $a$ is an integer. In my approach, I used prime factorization, but is this actually valid? This was my approach:</p>
<p>$$3 \mid 2a \implies 2a = 2 \cdot 3 \cdot k, k \in \mathbb{N}$$</p>
<p>$$\frac{2a}{2} = \frac{2\cdot 3 \cdot k}{2}$$</p>
<p>$$a = 3 \cdot k$$</p>
<p>$$\therefore 3 \mid a$$</p>
<p>Is this valid or am I doing forbidden things here?</p>
| Hagen von Eitzen | 39,174 | <p>Note that $2a=3k$ implies
$$ a=3a-2a=3(a-k)$$</p>
|
2,347,995 | <p>Not sure what I'm doing wrong.</p>
<p>Here's my work:</p>
<p>Expressing the first part $A \setminus (B\setminus C)$ using logical symbols:</p>
<p>$A \land \neg(B \land \neg C)$ becomes</p>
<p>$A \land \neg B\lor C$ (De Morgan's law)</p>
<p>While the second expression $(A \setminus B) \cup (A \cap C)$ is </p>
<p>$(A \land \neg B) \lor (A \land C)$ which becomes</p>
<p>$A \land \neg B \lor A \land C$ (Associative Law)</p>
<p>or $A \land \neg B \land C$</p>
<p>How are these two expressions similar? Thanks for any help in advance! </p>
| Michael Rozenberg | 190,319 | <p>Since $\cos$ is an even function, we can assume $0<x<1$.
we need to prove that $$\frac{2\sin^2\frac{x}{2}}{x^2}\geq\frac{1}{4}$$ or
$$\sin\frac{x}{2}\geq\frac{1}{\sqrt2}\cdot\frac{x}{2}.$$
Let $f(x)=\sin{x}-\frac{1}{\sqrt2}x$, where $0<x<0.5$.</p>
<p>Thus, $$f'(x)=\cos{x}-\frac{1}{\sqrt2}>\cos\frac{1}{2}-\frac{1}{\sqrt2}>0,$$
which says $f(x)\geq f(0)=0$ and we are done!</p>
|
311,950 | <p>I'm trying to show that any subset bounded of $\Bbb{R}^k$ is totally bounded.</p>
<p>Here is what I did: </p>
<p>(1)A subset of a totally bounded Set is bounded:</p>
<p>Proof: Let $X$ be a totally bounded subset and $Y\subset X$ then there exists an $\epsilon /2$-net $\{x_1,x_2,..,x_n\}$ and $X\subset \displaystyle\bigcup_{i=1}^n B(x_i,\epsilon/2)$ . Let $\{x_1,x_2,..,x_m\}$ be the points whose balls contain $Y$ $(m\le n)$. Now $\forall i \in \{1,..,m\} \exists q_i \in A \cap B(x_i,\epsilon/2) $ and $B(x_i,\epsilon/2)\subset B(q_i,\epsilon)$. We have for every $x\in B(x_i,\epsilon/2)$ $$ d(x,q_i)\le d(x,x_i)+d(x_i,q_i) < \frac \epsilon 2 + \frac \epsilon 2 =\epsilon$$</p>
<p>Hence $Y \subset \displaystyle\bigcup_{i=1}^m B(x_i,\epsilon/2) \subset \bigcup_{i=1}^m B(q_i,\epsilon)$ and $q_i \in Y$ for all $i$ hence $Y$ is totally bounded</p>
<p>Back to the problem:</p>
<p>Let $A \subset \Bbb{R}^k$ be a bounded set then $A \subset B(0,R)$ for some $R$ then $A \subset [-R,R] \times [-R,R] \times ...\times [-R,R] $ then $A$ is a subset of a compact set by the Heine-Borel Theorem which is also a totally bounded set, hence by (1) $A$ is totally bounded.</p>
<p>I'm trying to do this problem by only using (1) without invoking the Heine-Borel theorem, can anyone tell me how that can be done? (Is the proof of (1) right for that matter?)</p>
| Emanuele Paolini | 59,304 | <p>If $B$ is bounded in $\mathbb R^n$ it is contained in some cube $[-R,R]^n$. Then for any $\epsilon>0$ just consider all balls of radius $\epsilon$ centered in the set
$$
(\frac \epsilon 2\mathbb Z)^n \cap [-R,R]^n
$$
these balls are a finite number (less then $(4R/\epsilon+1)^n$) and cover the whole cube $[-R,R]^n(edit this)$.</p>
|
1,508,753 | <p>Show that for $x,y\in\mathbb{R}$ with $x,y\geq 0$, the arithmetic mean-quadratic mean inequality
$$\frac{x+y}{2}\leq \sqrt{\frac{x^2+y^2}{2}}$$
holds.</p>
<p>After my calculations I'll get: </p>
<p>$$-x^2+2xy-y^2$$
which can't be $\leq 0$.</p>
| Yes | 155,328 | <p>Initially we may guess that
$$
\frac{1}{1-x} \to -1
$$
as $x \to 2$. To prove this, note that we have $x\neq 1$ only if
$$
\bigg| \frac{1}{1-x} - (-1) \bigg| = \bigg|\frac{2-x}{1-x} \bigg| = \bigg|\frac{x-2}{x-1} \bigg|;
$$
we have $0 < |x-2| < 1/2$ only if $||x-1|-1| \leq |x-2| < 1/2$, only if $1/2 < |x-1|$, and only if
$$
\bigg| \frac{x-2}{x-1} \bigg| < 2|x-2|;
$$
given any $\varepsilon > 0$, we have $0 < |x-2| < \varepsilon/2$ only if $2|x-2| < \varepsilon$; hence
we have proved this:
for every $\varepsilon > 0$, we have $0 < |x-2| < \min \{1/2, \varepsilon/2 \}$ only if
$$
\bigg| \frac{1}{1-x} + 1 \bigg| < \varepsilon,
$$
which shows that
$$
\lim_{x \to 2}\frac{1}{1-x} = -1.
$$</p>
|
1,508,753 | <p>Show that for $x,y\in\mathbb{R}$ with $x,y\geq 0$, the arithmetic mean-quadratic mean inequality
$$\frac{x+y}{2}\leq \sqrt{\frac{x^2+y^2}{2}}$$
holds.</p>
<p>After my calculations I'll get: </p>
<p>$$-x^2+2xy-y^2$$
which can't be $\leq 0$.</p>
| Ian Miller | 278,461 | <p>The definition of convergence says a sequence $\{a_n\}$ converges to a limit $L$ if given an $\epsilon$ however small there exists $N(\epsilon)$ such that $|a_n-L|<\epsilon$ for all $n>N(\epsilon)$.</p>
<p>So let $x_n$ be values getting progressively closer to 2. I.e. For any $\epsilon_2$ there exists $N_2(\epsilon_2)$ such that $|x_n-2|<\epsilon$ for all $n>N_2(\epsilon_2)$.</p>
<p>By triangle inequality $|x_n-1|<|x_n-2|+1=\epsilon_2+1$</p>
<p>Let us show that given $\epsilon$ there does exist $N(\epsilon)$ such that $|\frac{1}{1-x_n}--1|<\epsilon$ for all $n>N$.
$$|\frac{1}{1-x_n}--1|<\epsilon$$
$$\Leftarrow |\frac{2-x_n}{1-x_n}|<\epsilon$$
$$\Leftarrow \frac{|2-x_n|}{|1-x_n|}<\epsilon$$
$$\Leftarrow \frac{\epsilon_2}{\epsilon_2+1}<\epsilon$$
$$\Leftarrow \epsilon_2\epsilon<\epsilon_2+1$$
$$\Leftarrow 1<\epsilon_2(1-\epsilon)$$
$$\Leftarrow \frac{1}{1-\epsilon}<\epsilon_2$$</p>
<p>So given any $\epsilon$ choose $\epsilon_2>\frac{1}{1-\epsilon}$ and we find a find a corresponding value $x_n$ (as we defined $x_n$ as converging) such that
$|\frac{1}{1-x_n}|<\epsilon$.</p>
|
48,679 | <p>I've been going through Fermats proof that a rational square is never congruent. And I've stumbled upon something I can't see why is. Fermat says: ''If a square is made up of a square and the double of another square, its side is also made up of a square and the double of another square'' Im having difficulties understanding why this is. Can anyone help me understand this?</p>
| Phil Wild | 7,842 | <p>Another example from theoretical high-energy physics I've encountered: sometimes when physicists have some equation of motion for an arbitrary number $N$ of particles with positions $x_i$, e.g. something of the form $\frac{1}{N}\sum_i f(x_i) + \frac{1}{N^2}\sum_{ij} g(x_i, x_j) = 0$, they wish to know what the solutions to this equation look like for large $N$. A technique they use is to replace the variables $x_i$ with a probability measure $\mu$ on the space of their possible values, which is supposed to represent the number of $x_i$'s in a given region in the large $N$ limit, and instead of solving the original equation they solve the analogous equation in $\mu$, e.g. $\int f(x) \mathrm{d}\mu(x) + \int g(x, y) \mathrm{d}(\mu \times \mu) (x, y) = 0$. In fact it's not hard to come up with a toy example where the original equation can be solved exactly for all $N$ and the solutions "look like" a particular probability distribution in the large $N$ limit, but that probability distribution fails to satisfy the corresponding equation, and for that reason I have some doubt that this method can be turned into something rigorous.</p>
|
48,679 | <p>I've been going through Fermats proof that a rational square is never congruent. And I've stumbled upon something I can't see why is. Fermat says: ''If a square is made up of a square and the double of another square, its side is also made up of a square and the double of another square'' Im having difficulties understanding why this is. Can anyone help me understand this?</p>
| Peter Shor | 2,294 | <p>The <a href="http://en.wikipedia.org/wiki/Replica_trick" rel="noreferrer">replica method</a> and the <a href="http://en.wikipedia.org/wiki/Cavity_method" rel="noreferrer">cavity method</a> have been used by physicists to calculate thermodynamic quantities in various statistical mechanics settings (including quite a few classes of random combinatorial objects). The results are often exactly right, even though the method is not at all rigorous. Michel Talagrand has recently <a href="http://people.math.jussieu.fr/~talagran/spinglasses/" rel="noreferrer">proven rigorously</a> some of the results that have been obtained by these methods.</p>
|
3,975,162 | <p>Let <span class="math-container">$(X, A, µ)$</span> be a positive metric space. If <span class="math-container">$\mu(X) < \infty$</span> and <span class="math-container">$(A_n)_{(n \in N^*)},A \in X$</span> <br />
show that if <span class="math-container">$\mu(A\bigtriangleup A_n)\rightarrow 0$</span> then <span class="math-container">$μ(A_n)\rightarrow\mu(A)$</span></p>
<p>What I have tried so far is</p>
<p>use that <span class="math-container">$A\bigtriangleup A_n = (A/A_n)\cup(A_n/A)$</span> <br />
and since <span class="math-container">$(A/A_n)\cap(A_n/A)= \emptyset$</span> then <span class="math-container">$\mu(A\bigtriangleup A_n) = \mu(A/A_n)+\mu(A_n/A) \rightarrow 0$</span> <br />
now I am trying to contain <span class="math-container">$μ(A_n)$</span> in an inequality where both side converge to <span class="math-container">$A$</span></p>
<p>EDIT: thanks to Thorgott's point both <span class="math-container">$\mu(A/A_n)$</span> and <span class="math-container">$\mu(A_n/A)$</span> converge to <span class="math-container">$0$</span> and <span class="math-container">$(A/A_n)\cap A_n = \emptyset$</span>
then <span class="math-container">$\mu(A) = \mu(A/A_n) + \mu(A_n) \Rightarrow \mu(A_n) \rightarrow0+\mu(A)$</span></p>
| Infinity_hunter | 826,797 | <p>Just observe that <span class="math-container">$y$</span> does not divide any number in the form of <span class="math-container">$ky+ 1$</span> since it divides <span class="math-container">$ky$</span>. Now note that <span class="math-container">$(y+1)^2 = 1 + ky$</span> where <span class="math-container">$k = y + 2$</span>.</p>
|
1,606,709 | <p>I am studing Kähler differentials and I tried to understand the geometric motivation behind these settings. What I do not understand is the role which plays the diagonal in all these theory. The cotangent sheaf is later defined in terms of the diagonal map. Why is this geometrically interesting? I tried to write a short introduction to Kähler differential to make the geometric nature more available, but I do not now if it make sense. Here it is: </p>
<blockquote>
<p>Differential $ 1 $-forms are linear transformations $ \omega_{p}:T_p X\mapsto K $ assigning an element in $ K $ to a tangent vector of the tangent space $ T_p X $ of a point $ p\in X $ in some differential manifold $ X $. Differential $ 1 $-forms can be viewed as \textit{infinitesimal} direction vectors $ \triangle p $. This means in physical terms, that the scalar $ \omega_p(\triangle t)\in K $ with $ \triangle t $ a tangent vector represents the \textit{work} required to move from $ x_i $ to $ x_{i+1} $ with $ p\in (x_i,x_{i+1}) $ along some curve. In other words, differential forms are cotangent vectors over some field $ K $, which give information about the work which is locally required to move along some curve. However, they can be generalized and captured by sheaf theory. For this attempt, we observe first that $ \triangle p $ is related to Taylor expansions. Indeed, let $ f $ be a smooth function, that is $ f\in C^{\infty}(\mathbb{C}) $, on a differential manifold $ X$ and let $\mathfrak{J}$ be the ideal of smooth functions vanishing at the point $p\in X$. The zero order part of the Taylor series of a smooth function $f$ is the value of $f$ at the point $ p $, let us say, $ f(p)=c $, so that $ f-c\in\mathfrak{J}$. Now the first order derivatives of $f-c$ correspond to the first order terms in the Taylor series and these are given by the image of $f$ in $\mathfrak{J}/\mathfrak{J}^2$. Let us denote this map by $ d(f) $ with $ d: \mathcal{O}_X\rightarrow \mathfrak{J}/\mathfrak{J}^2 $ and where $ \mathcal{O}_X $ denotes the ring of smooth functions on $ X $. Moreover, if $ f $ is constant, this means a fixed vector, then $ d(f)=0 $. Another important input is that $ \triangle p$ is required to be non zero, as there is no direction available for the zero vector. But $ \triangle p=0 $ is satisfied if and only if the two endpoints of the tangent vector $ \triangle p $ are choosen to be the same, which happens if and only if the point $ p\in X $ correspond to an element on the diagonal of $ X\times X $. So we demand that we just consider elements in $ X\times X $ vanishing on the diagonal (or in the complement of the diagonal).</p>
</blockquote>
<p>Summarizing up, my two questions are the following: </p>
<p>1) which geoemtric interpretation have the diagonal in these context? </p>
<p>2) Higher derivations seemed to me a generalization of Kähler differentials, but what is their motivation or geometric nature (analogy to differential geometry), since I cannot see any connection between Kähler differentials and higher derivations ? </p>
| DeepSea | 101,504 | <p><strong>Hint</strong>: Let $a =\dfrac{1}{2}$ and $b$ be the expression you are trying to calculate, then $b^2 = a + b$. You can use quadratic equation to finish.</p>
|
1,022,184 | <p>A game is played as follows: A random number $X$ is chosen uniformly from $[0,
1]$. Then a sequence $Y_1, Y_2,\ldots$ of random numbers is chosen independently
and uniformly from $[0, 1]$. The game ends the first time that $Y_i > X$. You are
then paid $(i-1)$ dollars. What is a fair entrance fee for this game?</p>
| alexjo | 103,399 | <p>Let $Z$ represent the payment. Then
$$\Bbb{P}(Z = k|X = x) = \Bbb{P}(Y_1 \le x, Y_2 \le x, \ldots , Y_k \le x, Y_{k+1} > x)
= x^k(1 − x) .$$
Therefore,
$$\Bbb{P}(Z = k) = \int_0^1 x^k(1 − x)\operatorname{d}x=\left[\frac{x^{k+1}}{k+1}-\frac{x^{k+2}}{k+2}\right]_0^1=\frac{1}{k+1}-\frac{1}{k+2}=\frac{1}{(k+1)(k+2)}.$$</p>
<p>Thus,
$$\Bbb{E}(Z) =\sum_{k=0}^{\infty} k\Bbb{P}(Z = k)=\sum_{k=0}^{\infty}\frac{k}{(k+1)(k+2)},$$
which diverges. </p>
<p>Thus, you should be willing to pay any amount to play this game.</p>
|
3,565,632 | <blockquote>
<p>Given <span class="math-container">$x_1 := a > 0$</span> and <span class="math-container">$x_{n+1} := x_n + \frac{1}{x_n}$</span> for <span class="math-container">$n \in \mathbb{N}$</span>, determine whether <span class="math-container">$(x_n)$</span> converges or diverges.</p>
</blockquote>
<p>Since <span class="math-container">$x_1 > 0$</span>, it seems obvious that the sequence is strictly increasing and always positive because we are always adding a positive number to each subsequent element in the sequence. </p>
<p>The trickier part is to show whether <span class="math-container">$(x_n)$</span> is bounded or not. If <span class="math-container">$(x_n)$</span> is bounded, then <span class="math-container">$\exists M \in \mathbb{R}$</span> such that </p>
<p><span class="math-container">\begin{align}|x_n| \leq M ~\forall n \in \mathbb{N}\tag{1} \end{align}</span></p>
<p>Alternatively, I think this means that <span class="math-container">$M$</span> could be a supremum of the sequence and another way to rewrite <span class="math-container">$(1)$</span> is given any <span class="math-container">$\epsilon > 0$</span></p>
<p><span class="math-container">\begin{align} x_n + \epsilon \leq M
\tag{2}\end{align}</span></p>
<p>Choose <span class="math-container">$\epsilon = \frac{1}{x_n}$</span>. Then </p>
<p><span class="math-container">\begin{align}
x_{n+1} = x_n + \frac{1}{x_n} \leq M \implies x_n + \epsilon \leq M
\tag{3}\end{align}</span></p>
<p>Thus I conclude that the inequality holds <span class="math-container">$\forall n \in \mathbb{N}$</span> and that <span class="math-container">$(x_n)$</span> is convergent. Thus by the Monotone Sequence Convergence Theorem, <span class="math-container">$(x_n)$</span> is convergent.</p>
<hr>
<p>The part I am not sure about is my reasoning to show that <span class="math-container">$x_n$</span> being bounded is correct or not. </p>
| Alex Ortiz | 305,215 | <p>Here is a suggestion:</p>
<p>Try proving the Liebniz rule for differentiating the product of two functions on <span class="math-container">$\mathbb R$</span>:</p>
<p><span class="math-container">$$
(fg)^{(n)}(x) = \sum_{k=0}^n{n\choose k}f^{(k)}(x)g^{(n-k)}(x),
$$</span>
and apply this to <span class="math-container">$f(x) = 1/x$</span> and <span class="math-container">$g(x) = \sin(x)$</span> to show that
<span class="math-container">$$
\left|\left(\frac{\sin x}{x}\right)^{(n)}\right| \le n!|x|^{-n-1}\sum_{k=0}^n\frac{|x|^k}{k!}.
$$</span>
Now try to use this to show the inequality you want.</p>
<hr>
<p><strong>Edit:</strong> This likely isn't enough to show the inequality on its own, as the bound we get still blows up at least like <span class="math-container">$1/|x|$</span> on the right-hand side as <span class="math-container">$x \to 0^+$</span>. We aren't taking advantage of the cancellations in the sum over <span class="math-container">$k$</span>, which evidently matter here.</p>
|
2,164,333 | <p>Let $T$ be linear transformation from $V$ to $W$. I know how to prove the result that nullity$(T) = 0$ if and only if $T$ is an injective linear transformation. But I still don't intuitively understand why the kernel only containing the zero vector means that $T$ is injective, and vice versa. In contrast, the relation between the image of $T$ and condition of being surjective is easy to see, since in order to map to all of the elements of $W$ the image of $T$ must have the same dimension as $W$. This can intuitively be seen with a diagram of the mapping from $V$ to $W$, for example. I can't really imagine a diagram that plainly shows the injective condition.</p>
<p>In short, what about nullity$(T) = 0$ imples that $T$ is a one-to-one function?</p>
| Maczinga | 411,133 | <p>Because if $\ker(T)\ne\{0\}$ then you have several manners to produce the null vector. Hence you cannot be injective.</p>
|
2,164,333 | <p>Let $T$ be linear transformation from $V$ to $W$. I know how to prove the result that nullity$(T) = 0$ if and only if $T$ is an injective linear transformation. But I still don't intuitively understand why the kernel only containing the zero vector means that $T$ is injective, and vice versa. In contrast, the relation between the image of $T$ and condition of being surjective is easy to see, since in order to map to all of the elements of $W$ the image of $T$ must have the same dimension as $W$. This can intuitively be seen with a diagram of the mapping from $V$ to $W$, for example. I can't really imagine a diagram that plainly shows the injective condition.</p>
<p>In short, what about nullity$(T) = 0$ imples that $T$ is a one-to-one function?</p>
| Leif | 413,652 | <p>If you have for example some linear mapping given by the matrix $A$ such that $f(X) = AX$, you can look at the Kernel. If this mapping is injective, the only vector from $X$ that would give me zero would be a trivial vector, otherwise $X$ (it's actually a null space of matrix $A$) contains more vectors that could give me as a result the zero vector (the same result for two different options is a contradiction with the injectivity).</p>
|
395,850 | <p>Is Dirac delta a function? What is its contribution to analysis? </p>
<p>What I know about it:
It is infinite at 0 and 0 everywhere else. Its integration is 1 and I know how does it come.</p>
| fgp | 42,986 | <p>The delta <em>distribution</em> is not a function from <span class="math-container">$\mathbb{R} \to \mathbb{R}$</span>. For <em>any</em> such function <span class="math-container">$f$</span> that is zero except at finitly many points, as the delta distribution is, <span class="math-container">$\int f= 0$</span> (this holds for both the riemann and the lebesgue integral). Yet <span class="math-container">$\int \delta = 1$</span>...</p>
<p>You can distributions as a function that maps <em>functions</em> (well, rather <em>some</em> functions) to real values though, i.e as a function <span class="math-container">$\mathbb{R}^\mathbb{R} \to \mathbb{R}$</span> (since it doesn't may <em>every</em> function to a real value, the domain is not really <span class="math-container">$\mathbb{R}^\mathbb{R}$</span> but a subset thereof).</p>
<p>The delta distribution is particularly simply, it just evaluates the function at <span class="math-container">$0$</span>, i.e. <span class="math-container">$$
\delta(f) = f(0) \text{.}
$$</span> Instead of <span class="math-container">$\delta(f)$</span>, people often write <span class="math-container">$\int f \delta$</span>, and that way you get <span class="math-container">$$
\int f \delta = f(0) \text{.}
$$</span>
But this is just notation - just as the <span class="math-container">$df$</span> and <span class="math-container">$dx$</span> in <span class="math-container">$\frac{df}{dx}$</span> aren't real numbers, yet <em>sometimes</em> you may treat them as if they were, <span class="math-container">$\delta$</span> isn't a function - yet <em>sometimes</em> it behaves like one. For example, you may use partial integration to compute <span class="math-container">$\delta'$</span> by doing <span class="math-container">$$
d'(f) = \int f \delta' = - \int f' \delta \text{.}
$$</span>
Thus, <span class="math-container">$\delta'$</span> is the distribution which, given a function <span class="math-container">$f$</span>, evaluates the <em>derivative</em> of <span class="math-container">$f$</span> at <span class="math-container">$0$</span> (and multiplities by <span class="math-container">$-1$</span>), i.e. returns <span class="math-container">$-f'(0)$</span>. This is, in fact, the very definition the derivative of a distribution btw.</p>
<p>This syntax also allows you to treat every function (well, actually every interable function) <span class="math-container">$h$</span> as a distribution. Following the path laid out by the integral notation for distributions, you get <span class="math-container">$$
h(f) = \int h f = \int h(x) f(x) \, dx \text{.}
$$</span></p>
<p>Note that this allows you to assign a derivative to a lot of functions which aren't otherwise differentiable. If you convert them into a distribution first, you can then set <span class="math-container">$$
h'(f) = \int h' f = -\int h f' \text{.}
$$</span>
provided that <span class="math-container">$f$</span> is suitable smooth.</p>
<p>This leaves the question of which <span class="math-container">$f$</span> are actually allowed here, i.e. which functions are mapped to real values open. You'll generally want them to be differentiable arbitrarily often, but that still leaves multiple choices. You'll have to read up on the theory of distributions to understand all the details.</p>
|
338,638 | <p>There is a wonderful series of articles by Flajolet et. al. about Mellin Transforms and the asymptotic analysis of generating functions. In particular, on page 45 of the article <a href="http://algo.inria.fr/flajolet/Publications/mellin-harm.pdf" rel="nofollow noreferrer">Mellin Transforms and Asymptotics: Harmonic Sums</a>, they state the following result:</p>
<blockquote>
<p>Proposition 6 (Growth of Special Dirichlet Series): Let <span class="math-container">$\lambda_{k}$</span>
and <span class="math-container">$\mu_{k}$</span> admit asymptotic expansions in descending powers of <span class="math-container">$k$</span>
as
<span class="math-container">$$\lambda_{k}\sim\sum_{r=0}^{\infty}\frac{a_{r}}{k^{\alpha_{r}}}$$</span>
<span class="math-container">$$\mu_{k}\sim k^{w}\left(1+\sum_{r=1}^{\infty}\frac{b_{r}}{k^{\beta_{r}}}\right)$$</span>
Then the Dirichlet series <span class="math-container">$\sum_{k}\lambda_{k}\mu_{k}^{-s}$</span> can be
continued to a meromorphic function <span class="math-container">$\Lambda\left(s\right)$</span> in the
whole of the complex plane.</p>
</blockquote>
<p>Now, let <span class="math-container">$V$</span> be an arbitrary set of infinitely many positive integers, and let: <span class="math-container">$$\zeta_{V}\left(s\right)\overset{\textrm{def}}{=}\sum_{v\in V}\frac{1}{v^{s}}$$</span></p>
<p>Enumerating the elements of <span class="math-container">$V$</span> in increasing order as <span class="math-container">$v_{1},v_{2},\ldots$</span>, recall that one way of defining the natural density of <span class="math-container">$V$</span> (denoted <span class="math-container">$d\left(V\right))$</span> is: <span class="math-container">$$d\left(V\right)=\lim_{n\rightarrow\infty}\frac{n}{v_{n}}$$</span> In the case where <span class="math-container">$d\left(V\right)$</span> exists and is positive, this gives the asymptotic <span class="math-container">$d\left(V\right)v_{n}\sim n$</span>. As such, for: <span class="math-container">$$\frac{\zeta_{V}\left(s\right)}{\left(d\left(V\right)\right)^{s}}=\sum_{n=1}^{\infty}\frac{1}{\left(d\left(V\right)v_{n}\right)^{s}}$$</span> I have that <span class="math-container">$$\lambda_{k}=1$$</span> and <span class="math-container">$$\mu_{k}=d\left(V\right)v_{k}\sim k$$</span> which is obtained by taking:<span class="math-container">$$\mu_{k}\sim k^{w}\left(1+\sum_{r=1}^{\infty}\frac{b_{r}}{k^{\beta_{r}}}\right)$$</span> setting <span class="math-container">$w=1$</span>, and letting all the <span class="math-container">$b_{r}$</span>s be <span class="math-container">$0$</span>. Hence, unless I am mistaken, Proposition 6 implies that <span class="math-container">$\zeta_{V}\left(s\right)$</span> extends to a meromorphic function on <span class="math-container">$\mathbb{C}$</span> whenever <span class="math-container">$V$</span> has positive natural density.</p>
<p>However, consider the following. Let <span class="math-container">$\mathbb{P}$</span> denote the set of prime numbers. Then, <span class="math-container">$\zeta_{\mathbb{P}}\left(s\right)$</span> is the so-called Prime Zeta Function, which is known to have a natural boundary on the imaginary axis. On the other hand, since <span class="math-container">$d\left(\mathbb{P}\right)=0$</span>, it follows that <span class="math-container">$\mathbb{N}/\mathbb{P}$</span> has a well-defined natural density of <span class="math-container">$1$</span>, and thus, by Proposition 6, that <span class="math-container">$\zeta_{\mathbb{N}/\mathbb{P}}\left(s\right)$</span> is meromorphic on <span class="math-container">$\mathbb{C}$</span>. However, since I can write:<span class="math-container">$$\zeta_{\mathbb{P}}\left(s\right)=\zeta\left(s\right)-\zeta_{\mathbb{N}\backslash\mathbb{P}}\left(s\right)$$</span> it follows that <span class="math-container">$\zeta_{\mathbb{P}}\left(s\right)$</span> is the difference of two meromorphic functions, which forces <span class="math-container">$\zeta_{\mathbb{P}}\left(s\right)$</span> to be meromorphic on <span class="math-container">$\mathbb{C}$</span>, which is obviously not correct. </p>
<p>So, where's the error, and how (if at all) can it be rectified? In particular, when, if ever, does the existence of <span class="math-container">$d\left(V\right)$</span> imply that <span class="math-container">$\zeta_{V}\left(s\right)$</span> is meromorphic on <span class="math-container">$\mathbb{C}$</span>?</p>
| reuns | 84,768 | <blockquote>
<p><span class="math-container">$$F(s)=\sum_{k}\lambda_{k}\mu_{k}^{-s}$$</span> extends meromorphically with finitely many poles in each strip because <span class="math-container">$$\Gamma(s)F(s)= \int_0^\infty f(t)t^{s-1}dt, \qquad f(t)=\sum_{k}\lambda_{k} e^{-t \mu_{k}}$$</span> where <span class="math-container">$f$</span> has an expansion in powers of <span class="math-container">$t$</span> near <span class="math-container">$0$</span> which makes clear the exact conditions you need for it to hold. </p>
</blockquote>
<ul>
<li>The expansion <span class="math-container">$\mu_k= k^{w}\left(1+\sum_{r\le R}b_{r}k^{-\beta_r} + o(k^{-\beta_R})\right)$</span> leads to (expanding the <span class="math-container">$\exp$</span>) <span class="math-container">$$e^{-\mu_k t}= e^{-k^wt} \exp(- \sum_{r\le R} b_r k^{(w-\beta_r) t}) (1+o(k^{-\beta_R})) \\= e^{-k^wt} \sum_{j\le J} c_j t^{d_j} k^{e_j}+ o(t^{d_J}k^{-e_j} e^{-k^wt})$$</span> </li>
<li>together with the expansion <span class="math-container">$\lambda_k = \sum_{r \le R} a_r k^{-\alpha_r} + o(k^{-\alpha_R})$</span> it leads to <span class="math-container">$$f(t) = \sum_{l\le L} C_l t^{D_l} \sum_k k^{E_l} e^{-k^wt}+ o(t^{d_L} \sum_k k^{e_L} e^{-k^wt})$$</span></li>
</ul>
<p>For every <span class="math-container">$r$</span>, for <span class="math-container">$L$</span> large enough we have <span class="math-container">$t^{d_L} \sum_k k^{e_L} e^{-k^wt} = o(t^r)$</span> so the error term won't be a problem. </p>
<p>And from our knowledge of <span class="math-container">$\Gamma(s) \zeta(Bs+C)$</span>, of Mellin inversion and the residue and Tauberian theorem we know that <span class="math-container">$\sum_k k^{E_l} e^{-k^wt}$</span> has an expansion in powers of <span class="math-container">$t$</span>.</p>
<p>Thus so does <span class="math-container">$f$</span> <span class="math-container">$$f(t) = \sum_{m \le M} u_m t^{v_m} + o(t^{v_M}), \qquad v_m \to \infty$$</span> from which we have our meromorphic continuation to <span class="math-container">$\Re(s) > -v_M$</span> <span class="math-container">$$\Gamma(s)F(s)= \int_0^\infty (f(t)-\sum_{m \le M} u_m t^{v_m} 1_{t < 1})t^{s-1}dt+ \sum_{m \le M}\frac{u_m }{s+v_m}$$</span></p>
|
121,431 | <p>I need a good reference for the basic definitions of the dual of locally compact group (not necessarily abelian), its natural topology, $\sigma$-algebra, and the Plancherel measure on it (when they are defined). This topic seems pretty standard to me, but when I needed a basic reference on this (both to check my memories and to be able to cite it in a paper I am writing), I didn't find one. </p>
<p>By the way, the wikipedia webpage "Plancherel measure" should be completely rewritten.
There is not even a definition, just a list of examples (and the definition given in the finite case is not compatible with the one given in the compact case). I would be happy to rewrite it, when I have a reference to check the details. </p>
| Uwe Franz | 30,364 | <p>Hartmut Fuehr's book (Abstract harmonic analysis of continuous wavelet transforms,
Springer Lecture Notes in Mathematics, Nr. 1863, 2005, X, 193 p., Softcover
ISBN: 3-540-24259-7), contains a "-reasonably self-contained- exposition of Plancherel theory", see also <a href="http://www.matha.rwth-aachen.de/~fuehr/book.html">http://www.matha.rwth-aachen.de/~fuehr/book.html</a>.</p>
<p>Then there is also Dixmier's book: J. Dixmier, Les C*-algèbres et leurs représentations, Gauthier-Villars, 1969.</p>
|
2,069,849 | <p>For a given $x\in\left\{ 0,1\right\} ^{n}$, define $x_{\setminus k}=\left(x_{1},\dots,x_{k-1},x_{k+1},\dots,x_{n}\right)$. [Is there standard notation for this?]</p>
<p><strong>Prove or disprove the following claim:</strong></p>
<p>Let $X$ be some random variable with values in $\left\{ 0,1\right\} ^{n}$. There exists $k\in\left[n\right]$ s.t for all $y\in\left\{ 0,1\right\} ^{n}$,
$$Pr\left(X=y\vert\bigoplus_{i=1}^{n}X_{i}=\bigoplus_{i=1}^{n}y_{i}\right)=Pr\left(X_{\setminus k}=y_{\setminus k}\right)$$</p>
<p>Where $\oplus$ is the XOR function.</p>
<p>Intuitively the claim sounds true, since for $n-1$ boolean coordinates and a parity of 1's there exists a unique $n$-coordinate boolean vector.</p>
| Will Sherwood | 124,497 | <p>Throughout the tune, we receive 12 partridges in pear trees, 22 turtle doves, 30 french hens, etc. So to total everything up, I think this sum is more compact:
$$\sum_{n=1}^{12}(13-n)f(n)$$</p>
|
1,161,183 | <p>Find the number of ways of choosing three numbers from the set ${1,2,3,…,20}$, so that the sum of the three numbers is divisible by $3$. I wrote $a+b+c=3k$, where $k=1,2,...20$, till $k=6$, there were no problems, but then restrictions started($a<21$). Even if I come to know of the correct method from here, the summation of all the answers is a tedious job(the answers come in binomial coefficients). Therefore, s there a shorter way?</p>
| Théophile | 26,091 | <p>The three numbers might all be multiples of 3, in which case there are $6 \choose 3$ ways to pick them.</p>
<p>There are several other possibilities: they might all be congruent to 1 modulo 3, for example. Find all such possibilities and count accordingly.</p>
|
1,161,183 | <p>Find the number of ways of choosing three numbers from the set ${1,2,3,…,20}$, so that the sum of the three numbers is divisible by $3$. I wrote $a+b+c=3k$, where $k=1,2,...20$, till $k=6$, there were no problems, but then restrictions started($a<21$). Even if I come to know of the correct method from here, the summation of all the answers is a tedious job(the answers come in binomial coefficients). Therefore, s there a shorter way?</p>
| Asinomás | 33,907 | <p>We only need concern ourselves with congruence $\bmod 3$.</p>
<p>There are $7$ numbers that are $1\bmod 3$</p>
<p>There are $7$ numbers that are $2\bmod 3$</p>
<p>There are $6$ numbers that are $0\bmod 3$</p>
<p>We now count the ways to add to $0\bmod 3$:</p>
<hr>
<p>Three zeros:</p>
<p>$0+0+0$ (There are $\binom{6}{3}$ of these)</p>
<hr>
<p>Two zeros:</p>
<hr>
<p>One zero:</p>
<p>$0+1+2$ (There are $7\cdot7\cdot 6$ of these )</p>
<hr>
<p>No zeros:</p>
<p>$1+1+1$ (There are $\binom{7}{3}$ of these)</p>
<p>$2+2+2$ (There are $\binom{7}{3}$ of these)</p>
<p>Adding up the answer is $\binom{6}{3}+7\cdot7\cdot6+2\binom{7}{3}=384$</p>
<p>There are $\binom{20}{3}=1140$ ways to choose, $3$ numbers of $20$. A third of that would be $380$. So more than a third of the subsets of size three are $0\bmod 3$</p>
|
4,577,925 | <p>Which expression is larger,
<span class="math-container">$$
99^{50}+100^{50}\quad\textrm{ or }\quad 101^{50}?
$$</span></p>
<p>Idea is to use the Binomial Theorem:</p>
<p>The right hand side then becomes
<span class="math-container">$$
101^{50}=(100+1)^{50}=\sum_{k=0}^{50}\binom{50}{k}1^{50-k}100^k=100^{50}+\sum_{k=0}^{49}\binom{50}{k}100^k
$$</span></p>
<p>The left hand side reads
<span class="math-container">$$
99^{50}+100^{50}=(100-1)^{50}+100^{50}=\sum_{k=0}^{50}\binom{50}{k}(-1)^{50-k}100^k+100^{50}
$$</span></p>
<p>Thus, since both sides have the summand <span class="math-container">$100^{50}$</span>, it remains to compare
<span class="math-container">$$
\sum_{k=0}^{50}\binom{50}{k}(-1)^{50-k}100^k\quad\textrm{and}\quad \sum_{k=0}^{49}\binom{50}{k}100^k
$$</span></p>
| robjohn | 13,854 | <p>The binomial theorem says
<span class="math-container">$$
101^{50}=\sum_{k=0}^{50}(+1)^k\binom{50}{k}100^{50-k}
$$</span>
and
<span class="math-container">$$
99^{50}=\sum_{k=0}^{50}(-1)^k\binom{50}{k}100^{50-k}
$$</span>
Furthermore, <span class="math-container">$100^{50}=2\binom{50}{1}100^{49}$</span>, which is the difference in the <span class="math-container">$k=1$</span> terms. Thus, the difference is the sum of the differences of the odd terms for <span class="math-container">$k\gt1$</span>:
<span class="math-container">$$
101^{50}-\left(100^{50}+99^{50}\right)=2\sum_{j=1}^{24}\binom{50}{2j+1}100^{49-2j}
$$</span></p>
|
4,577,925 | <p>Which expression is larger,
<span class="math-container">$$
99^{50}+100^{50}\quad\textrm{ or }\quad 101^{50}?
$$</span></p>
<p>Idea is to use the Binomial Theorem:</p>
<p>The right hand side then becomes
<span class="math-container">$$
101^{50}=(100+1)^{50}=\sum_{k=0}^{50}\binom{50}{k}1^{50-k}100^k=100^{50}+\sum_{k=0}^{49}\binom{50}{k}100^k
$$</span></p>
<p>The left hand side reads
<span class="math-container">$$
99^{50}+100^{50}=(100-1)^{50}+100^{50}=\sum_{k=0}^{50}\binom{50}{k}(-1)^{50-k}100^k+100^{50}
$$</span></p>
<p>Thus, since both sides have the summand <span class="math-container">$100^{50}$</span>, it remains to compare
<span class="math-container">$$
\sum_{k=0}^{50}\binom{50}{k}(-1)^{50-k}100^k\quad\textrm{and}\quad \sum_{k=0}^{49}\binom{50}{k}100^k
$$</span></p>
| cr001 | 254,175 | <p><span class="math-container">$$\sum_{k=0}^{49}\binom{50}{k}100^k$$</span><span class="math-container">$$=50\cdot100^{49}+1225\cdot100^{48}+ ...$$</span><span class="math-container">$$>62\cdot100^{49}$$</span><span class="math-container">$$>99^{50}$$</span></p>
<p>The last inequality comes from <span class="math-container">$$\ln{62} + 49\ln{100}=229.78...>50\ln{99}=229.75...$$</span></p>
|
19,294 | <p>Instead of taking a one to one correspondence meaning each set has the same number of elements. why not use the concept of coverings of topology? The irrational numbers covers the whole numbers but not vice versa?</p>
<p>A hierarchy of coverings instead of infinities. Wouldn't that make those infinities more manageable in those terms?( yes I know topology can be expressed in set theory)</p>
| Konrad Waldorf | 3,473 | <p>A nice point of view is to consider principal bundles with structure group $B\mathbb{C}^\times$. One can probably take any abelian Lie group instead of $\mathbb{C}^\times$. Principal $B\mathbb{C}^\times$-bundles are one way to give a precise meaning to what you are calling "patching together classifying spaces". </p>
<p>The point is that there is an isomorphism
$$
H^1(X,B\mathbb{C}^\times) = H^2(X,\mathbb{C}^\times),
$$
which explains the relation to the second cohomology group. It is basically saying that principal $B\mathbb{C}^\times$-bundles are the <em>same</em> as $\mathbb{C}^\times$-gerbes. </p>
<p>This isomorphism is induced from the exact sequence
$$
1 \to \mathbb{C}^\times \to E\mathbb{C}^\times \to B\mathbb{C}^\times \to 1
$$
of groups, and the fact that the sheaf of $E\mathbb{C}^\times$-valued functions on a paracompact space $X$ is soft. All this is very nicely explained in Gajer's Inventiones paper "Geometry of Deligne cohomology". </p>
|
2,456,561 | <p>A bowl contains 16 chips, of which 6 are red, 7 are white and 3 are blue. If four chips are taken at random and without replacement, find the probability that there is at least 1 chip of each colour.</p>
<p>Why is the answer not the following:</p>
<p><a href="https://i.stack.imgur.com/25ImM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/25ImM.png" alt="enter image description here"></a></p>
| JMoravitz | 179,297 | <p>Your thought process seems to be the following:</p>
<ul>
<li>Pick the red chip</li>
<li>Pick the white chip</li>
<li>Pick the blue chip</li>
<li>Pick one more chip</li>
</ul>
<p>When you picked your one more chip however, the final chip will be an additional chip of an earlier color. Supposing that the chips were labeled with numbers as well as colors, the sequence of selections red1, white2, blue3, red4 would have resulted in the same collection of chips as red4, white2, blue3, red1.</p>
<p>Remember that to apply multiplication principle, we should have counted every outcome exactly once. Despite this, if instead we overcounted we can still salvage the approach so long as each outcome was counted exactly the same number of times each and we divide by the number of times each each counted as per the "shepherd's principle."</p>
<p>You can reason then that every outcome you will have counted exactly twice and so we should divide by two to account for that. Dividing by $\binom{16}{4}$, the number of ways of selecting four chips, will give the probability.</p>
<p>An alternate corrected approach which avoids division by symmetry arguments:</p>
<p>Recognize that to get at least one of each of the three colors represented in the four drawn chips exactly one of the following will occur:</p>
<ul>
<li>There will be two red chips, one blue, and one white</li>
<li>There will be one red chip, two blue, and one white</li>
<li>There will be one red chip, one blue, and two whites</li>
</ul>
<p>Find the probabilities of each of these respective sub-cases and add.</p>
<p>$\dfrac{\binom{6}{2}\binom{7}{1}\binom{3}{1}+\binom{6}{1}\binom{7}{2}\binom{3}{1}+\binom{6}{1}\binom{7}{1}\binom{3}{2}}{\binom{16}{4}} = \frac{9}{20}$</p>
|
3,056,031 | <p>From the first chapter of Arfken's Mathematical Methods for physicists (rotation of the coordinate axis):</p>
<blockquote>
<p>To go on to three and, later, four dimensions, we find it convenient to use a more compact notation. Let
<span class="math-container">\begin{equation}
x → x_1,
\textbf{ }
y → x_2
\end{equation}</span></p>
<p><span class="math-container">\begin{equation}
a_{11} = cos\phi,\textbf{ } a_{12} = sin\phi
\end{equation}</span></p>
<p><span class="math-container">\begin{equation}
a_{21} = −sin\phi, \textbf{ } a_{22} = cos\phi
\end{equation}</span></p>
<p>Then Eqs. become</p>
<p><span class="math-container">\begin{equation}
x′_1 = a_{11}x_1 + a_{12}x_2
\end{equation}</span></p>
<p><span class="math-container">\begin{equation}
x′_2 = a_{21}x_1 + a_{22}x_2.
\end{equation}</span></p>
<p>The coefficient <span class="math-container">$a_{ij}$</span> may be interpreted as a direction cosine, the cosine of the angle between <span class="math-container">$x'_i$</span> and <span class="math-container">$x_j$</span> ; that is,</p>
</blockquote>
<p>This is all good. Later, the book states:</p>
<blockquote>
<p>From the definition of <span class="math-container">$a_{ij}$</span> as the cosine of the angle between the positive <span class="math-container">$x′_i$</span> direction and the positive <span class="math-container">$x_j$</span> direction we may write (Cartesian coordinates):</p>
<p><span class="math-container">\begin{equation}
a_{ij}=\frac{\partial x'_i}{\partial x_j}
\end{equation}</span></p>
</blockquote>
<p>Where did that come from? I can't find anything about the cosine as a partial derivative, and I don't see how that works.</p>
| Manraj | 584,828 | <p>You can verify-</p>
<p><span class="math-container">$\frac{\partial x_1'}{\partial x_1}=a_{11},\frac{\partial x_1'}{\partial x_2}=a_{12},\frac{\partial x_2'}{\partial x_1}=a_{21}\ and\ \frac{\partial x_2'}{\partial x_2}=a_{22}$</span></p>
|
1,856,171 | <p>Functional</p>
<p>$F(x)=(x^2-x+1)Q(x)+x-1$</p>
<p>$G(x)=(x^2-x+1)T(x)+x+1$</p>
<p>$F(x).G(x)=(x^2-x+1)H(x)+ax+b$</p>
<p>find a and b</p>
| G Cab | 317,234 | <p>Suppose you mean a vector ($\mathbf{v}$) "specular" to the given one ($\mathbf{u}$) with respect to the line $L$.
Since the line is passing through the origin, let's fix our mind on taking the vectors as orientated segments from the origin.<br>
If you do the cross product $\mathbf{w} = \mathbf{u} \times \mathbf{L}
$ you get a vector which is orthogonal either to the line and to $\mathbf{u}$, i.e. it is normal to the plane containing the line and $\mathbf{u}$.
Then $\mathbf{v} \times \mathbf{L}$ shall be equal to $-\mathbf{w}$ : solve such a linear system of three equations in three unknowns. Hope that is of help to you.</p>
|
3,123,681 | <p>let us consider following problem taken from book</p>
<p><em>An appliance store purchases electric ranges from
two companies. From company A, 500 ranges are
purchased and 2% are defective. From company B,
850 ranges are purchased and 2% are defective.
Given that a range is defective, find the probability
that it came from company B</em></p>
<p>so here we are assuming that probability of selection company is equally right? that means that <span class="math-container">$P(A)=P(B)=\frac{1}{2}$</span> , also <span class="math-container">$2$</span> % defective means that probability of selection of defective from ranges is equal to <span class="math-container">$0.02$</span>, for instance in Company A, number of defective ranges is <span class="math-container">$500*0.02=10$</span> there probability is equal to <span class="math-container">$\frac{10}{500}=0.02=2$</span>% </p>
<p>we know probability of selecting defective range is equal to</p>
<p><span class="math-container">$ \frac{1}{2} *2$</span>% + <span class="math-container">$\frac{1}{2} *2$</span>% and probability of selecting defective from company B will be <span class="math-container">$1/2 * 2$</span>% divided by probability of selection of defective range, but book says that answer is <span class="math-container">$0.65 $</span>, how?</p>
| DXT | 372,201 | <p>We have <span class="math-container">$$x-\frac{x^3}{3}<\tan^{-1}(x)<x,x>0$$</span></p>
<p>So <span class="math-container">$$\sum^{n}_{k=1}\bigg[\frac{1}{n+k}-\frac{1}{3(n+k)^3}\bigg]<\sum^{n}_{k=1}\tan^{-1}\bigg(\frac{1}{n+k}\bigg)<\sum^{n}_{k=1}\frac{1}{n+k}$$</span></p>
<p>Now Using Limit <span class="math-container">$$\lim_{n\rightarrow \infty}\sum^{n}_{k=1}\frac{1}{n+k}=\lim_{n\rightarrow \infty}\sum^{n}_{k=1}\frac{1}{n}\cdot \frac{1}{1+\frac{k}{n}}$$</span></p>
<p>Using Riemann sum, <span class="math-container">$$ = \int^{1}_{0}\frac{1}{1+x}dx = \ln(2)$$</span></p>
<p>Same way </p>
<p><span class="math-container">$$\lim_{n\rightarrow \infty}\sum^{n}_{k=1}\frac{1}{(n+k)^3}=\lim_{n\rightarrow \infty}\sum^{n}_{k=1}\frac{1}{n^2}\cdot \frac{1}{n}\cdot \frac{1}{(1+\frac{k}{n})^3}=0$$</span></p>
<p>And using Squeeze principle <span class="math-container">$$\lim_{n\rightarrow \infty}\sum^{n}_{k=1}\tan^{-1}\bigg(\frac{1}{n+k}\bigg)=\ln(2)$$</span></p>
|
629,340 | <p>Is it true that if for each partition of a graph G's vertices into two non empty sets there is an edge with end points in both sides then G is connected? Intuitively this seems true to me. But I cannot prove this. I would very much appreciate some assistance. Thanks </p>
| Nick Peterson | 81,839 | <p>Yes, it is!</p>
<p>A graph is connected precisely when there is a path between any two vertices; however, another characterization is this:</p>
<p>Let $R$ be the relation on the vertex set $V$ defined by $vRw$ iff there is a path between $v$ and $w$. It is not too hard to show that this is an equivalence relation, and so it partitions $V$ in to equivalence classes; a connected graph is a graph for which there is only one equivalence class.</p>
<p>This makes the result obvious: if you have multiple equivalence classes, pick one; the partition obtained by putting this equivalence class on one side and the rest on the other must have an edge crossing it. But then the other end of that edge is in your equivalence class, a contradiction!</p>
|
14,568 | <p>The Poisson summation says, roughly, that summing a smooth $L^1$-function of a real variable at integral points is the same as summing its Fourier transform at integral points(after suitable normalization). <a href="http://en.wikipedia.org/wiki/Poisson_summation_formula">Here</a> is the wikipedia link.</p>
<p>For many years I have wondered why this formula is true. I have seen more than one proof, I saw the overall outline, and I am sure I could understand each step if I go through them carefully. But still it wouldn't tell me anything about why on earth such a thing should be true.</p>
<p>But this formula is exceedingly important in analytic number theory. For instance, in the book of Iwaniec and Kowalski, it is praised to high heavens. So I wonder what is the rationale of why such a result should be true.</p>
| Darsh Ranjan | 302 | <p>In what follows, I'll use the convention $$ \hat{f}(\xi) = \int_{-\infty}^{\infty} f(x)e^{-2\pi i x \xi}dx,$$ so that $$ f(x) = \int_{-\infty}^{\infty} \hat{f}(\xi)e^{2\pi i x \xi}d\xi.$$</p>
<p>I like the following interpretation of Poisson summation, which also gives a generalization: Consider the Dirac comb distribution $C(x) = \sum_{n\in \mathbb{Z}} \delta(x-n)$. This is a tempered distribution, so it has a Fourier transform. In fact, it is its own Fourier transform. To justify this, I'm going to give a very nonrigorous argument. But if intuition is the main goal, then I think it will help. First, note that $C(x)$ is periodic with period 1. Thus, its "Fourier transform" is actually a Fourier series: its support is in $\mathbb{Z}$. This follows by noting that</p>
<p>$$\begin{align}C(x) &= \int_{-\infty}^{\infty} \hat{C}(\xi)e^{2\pi i x \xi}d\xi; \\
C(x) &= \sum_{n\in \mathbb{Z}}a_n e^{2\pi i n x};
\end{align}$$</p>
<p>Where the first line is the Fourier inversion formula and the second line is the Fourier series for $C$. It follows by uniqueness that $\hat{C}(\xi) = \sum_{n\in \mathbb{Z}}a_n \delta(\xi - n)$. On the other hand, the (inverse) Fourier transform of $\hat{C}$ is also supported on $\mathbb{Z}$, so $\hat{C}$ is also periodic with period 1. Thus, all the $a_n$ are the same:</p>
<p>$$\hat{C}(\xi) = a\sum_{n\in \mathbb{Z}}\delta(\xi - n),$$ where $a$ is some scalar. It's not hard to see that the scalar has to be 1. </p>
<p>To derive Poisson summation from this, use the convolution theorem: let $f$ be any function. On the one hand, $$(f*C)(x) = \sum_{n\in \mathbb{Z}} f(x+n).$$ On the other hand, we can use the convolution theorem: $$\widehat{(f*C)}(\xi) = \hat{f}(\xi)\hat{C}(\xi) = \hat{f}(\xi)\sum_{n\in \mathbb{Z}} \delta(\xi-n) = \sum_{n\in \mathbb{Z}} \hat{f}(n)\delta(\xi-n).$$ The last sum gives the Fourier series of the periodic function $f*C$: $$(f*C)(x) = \sum_{n\in \mathbb{Z}} \hat{f}(n)e^{2\pi i n x}.$$ Plugging in $x=0$ gives the Poisson summation formula, QED. But the result for general $x$ is interesting as well: given a function $f$, you can obtain a periodic function $g(x)$ by (a) adding up $f(x+n)$ over all integers $n$, or (b) taking the Fourier transform of $f$ at integer frequencies and making that the Fourier series of $g$. The result is that (a) and (b) give the same function. </p>
|
14,568 | <p>The Poisson summation says, roughly, that summing a smooth $L^1$-function of a real variable at integral points is the same as summing its Fourier transform at integral points(after suitable normalization). <a href="http://en.wikipedia.org/wiki/Poisson_summation_formula">Here</a> is the wikipedia link.</p>
<p>For many years I have wondered why this formula is true. I have seen more than one proof, I saw the overall outline, and I am sure I could understand each step if I go through them carefully. But still it wouldn't tell me anything about why on earth such a thing should be true.</p>
<p>But this formula is exceedingly important in analytic number theory. For instance, in the book of Iwaniec and Kowalski, it is praised to high heavens. So I wonder what is the rationale of why such a result should be true.</p>
| paul garrett | 15,629 | <p>A variant on @Darsh Ranjan's argument: since $u=\sum_{n\in\mathbb Z}\delta(x-n)$ is invariant under translation by $\mathbb Z$, and is annihilated by multiplication by $e^{2\pi inx}-1$ for all $n\in \mathbb Z$ (this captures the order-zero aspect!) it suffices to show that there is a unique (up to scalar multiples) distribution meeting these conditions, and that the Fourier transform of $u$ also does meet them. The latter is immediate, since Fourier transform interchanges the two conditions.</p>
<p>For uniqueness: annihilation by multiplication by $e^{2\pi ix}-1$ implies that the support is $\mathbb Z$. We know the classification of distributions supported at points, hence, supported on a discrete closed subset: Dirac deltas and derivatives. Using smooth cut-offs to examine the behavior at a given integer, since $e^{2\pi ix}-1$ vanishes just to order $1$, the order of the $n$-th component is $0$. Thus, such a distribution is of the form $\sum_n c_n\cdot \delta(x-n)$. Translation invariance implies that all the coefficients are the same.</p>
|
4,509,098 | <p><span class="math-container">$△ABC$</span> - triangle with <span class="math-container">$45°, 105°$</span> and <span class="math-container">$30°.$</span></p>
<p>Perimeter of triangle is <span class="math-container">$\sqrt6 + 2\sqrt3 - \sqrt2.$</span></p>
<p>Find the longest side?</p>
<p>Do you have any ideas? Seems something to do with Law of sines and cosines, something else?</p>
<p>Thank you!</p>
<p>P.S.: how to use Latex?</p>
| abcdefu | 1,080,275 | <p>The longest side will be the side that is opposite to the largest angle of the triangle.</p>
<p>You haven't defined the angles clearly so I can't tell which side is largest but the approach remains same.</p>
<p>No need of any law of sines or cosines.</p>
<p><strong>EDIT</strong></p>
<p>Let the sided opposite to <span class="math-container">$105^o$</span> be <span class="math-container">$a$</span>, opposite to <span class="math-container">$45^o$</span> be <span class="math-container">$b$</span> and opposite to <span class="math-container">$30^o$</span> be <span class="math-container">$c$</span>.</p>
<p>Now by law of sines: <span class="math-container">$$\frac{a}{\operatorname{sin}105}=\frac{b}{\operatorname{sin}45}=\frac{c}{\operatorname{sin}30}$$</span>
or <span class="math-container">$$\frac{a}{\frac{1+\sqrt3}{2\sqrt2}}=\frac{b}{\frac{1}{\sqrt2}}=\frac{c}{\frac{1}{2}}$$</span> or <span class="math-container">$$b=\frac{2a}{1+\sqrt3} $$</span> and <span class="math-container">$$c=\frac{\sqrt2\cdot a}{1+\sqrt3}$$</span> <span class="math-container">$\implies$</span> <span class="math-container">$$a+\frac{2a}{1+\sqrt3}+\frac{\sqrt2\cdot a}{1+\sqrt3}=\sqrt6+2\sqrt3-\sqrt2$$</span> After rigorous solving we get, <span class="math-container">$$a=2$$</span>
Hence the largest side is <span class="math-container">$2$</span></p>
|
3,651,173 | <p>I am trying to use generating function to find closed form formula for this expression:
<span class="math-container">$$\sum_{n \geq 0} \frac{n^2+4n+5}{n!}$$</span>
but I don't how to start this. any suggestion or hints. thank you</p>
| failedentertainment | 583,480 | <p>Recall that <span class="math-container">$e^x = \sum_{n \geq 0} \frac{x^n}{n!}$</span> so that <span class="math-container">$$xe^x = \sum_{n\geq 0} \frac{x^{n+1}}{n!} = \sum_{n \geq 0} (n+1) \frac{x^{n+1}}{(n+1)!} = \sum_{n \geq 1} n \frac{x^n}{n!}$$</span> where the latter equality follows by reindexing. Similarly, <span class="math-container">$$x^k e^x = \sum_{n \geq 0} \frac{x^{n+k}}{n!} = \sum_{n \geq 0} (n+k)(n+k-1) \cdots (n+1)\frac{x^{n+k}}{(n+k)!} = \sum_{n \geq k} n(n-1)\cdots(n-k+1) \frac{x^n}{n!} $$</span></p>
<p>Note that since the coefficient <span class="math-container">$n(n-1)\cdots(n-k+1)$</span> vanishes for <span class="math-container">$n =0,1,\cdots,k-1$</span>, we can include lower indices without changing anything, so that <span class="math-container">$$x^ke^x = \sum_{n \geq 0} n(n-1)\cdots (n-k+1) \frac{x^n}{n!}$$</span></p>
<p>In particular, <span class="math-container">$$x^2e^x = \sum_{n\geq 0} n(n-1) \frac{x^n}{n!}$$</span> So <span class="math-container">$$(x^2 +5x+5)e^x = \sum_{n\geq 0} \frac{(n^2 +4n +5)x^n}{n!}$$</span>
The way of thinking here is to knock out the highest degree terms first. You want to take care of the <span class="math-container">$n^2$</span> in the numerator of your sum, so you subtract off an <span class="math-container">$x^2e^x$</span>, and you gain an extra <span class="math-container">$n$</span>, which you take care of with an <span class="math-container">$xe^x$</span>, etc.</p>
<p>Evaluating the given expression at <span class="math-container">$x=1$</span> we have <span class="math-container">$$11e = \sum_{n\geq 0} \frac{(n^2 +4n +5)}{n!}$$</span> which was what you wanted.</p>
|
2,440,785 | <p>if $f: X \to X$ is continuous where $X$ is a topological space with a cofinite topology, then:</p>
<p>$$(i) \ f^{-1}(x) \text{ is finite for all $x$} \\ \text{or} \\ (ii) \ f \text{ is constant}$$</p>
<p>My approach:</p>
<p>I couldn't build up a proper approach here to be honest. I believe we need to use the fact that inverse functions preserve differences of sets. But couldn't go on. </p>
<p>Any hints?</p>
| Community | -1 | <p>If the pre-image of every point is finite, you are done.</p>
<p>Otherwise there is a point with infinite pre-image. But single points are closed, so the pre-image must be closed. The only closed set that is NOT finite is all of $X$. So now you have a point whose pre-image is all of $X$, which means $f$ is constant.</p>
|
177,915 | <p>I have two inequalities that I can't prove:</p>
<ol>
<li>$\displaystyle{n\choose i+k}\le {n\choose i}{n-i\choose k}$</li>
<li>$\displaystyle{n\choose k} \le \frac{n^n}{k^k(n-k)^{n-k}}$</li>
</ol>
<p>What is the best way to prove them? Induction (it associates with simple problems, but sometimes I find it difficult to use, what is worrying), or maybe combinatorial interpretation?</p>
| Davide Giraudo | 9,849 | <ol>
<li><p>Let $S_j^n:=\{A\subset [n],|A|=j\}$. Let $S'\subset S_k^{n-i}\times S_i^n$ which consists of pairs of disjoint subsets. The map
$$\varphi\colon S'\to S_{i+k}^n,\varphi(A,B)=A\cup B$$
is onto, hence $|S'|\geq |S_{i+k}^n|=\binom n{i+k}$. Since $S'\subset S_k^{n-i}\times S_i^n$, its cardinal is smaller than the cardinal of the product, which gives the result. </p></li>
<li><p>We have to show that
$$k^k(n-k)^{n-k}\binom nk\leq n^n.$$
The LHS is the number of maps from $[n]$ to $[n]$ such that there exists a susbet of $k$ element which is stable and its complement is also stable, and the RHS is the total number of maps from $[n]$ to $[n]$. </p></li>
</ol>
|
3,422,505 | <p>The problem is as follows:</p>
<blockquote>
<p>In a certain shopping mall which is many stories high there is a glass
elevator in the middle plaza. One shopper ridding the elevator notices
a kid drops a spheric toy from the top of the building where is
located the toy store. The shopper riding the elevator labeled <span class="math-container">$A_{1}$</span>
is descending towards the ground with a velocity of
<span class="math-container">$\vec{v}=-5\hat{j}\,\frac{m}{s}$</span>. Find the speed (in meters per
second) and the acceleration in <span class="math-container">$\frac{m}{s^{2}}$</span> which will be seen
by the shopper in the glass elevator in the instant <span class="math-container">$t=3\,s$</span>. You may
use <span class="math-container">$g=10\,\frac{m}{s^{2}}$</span></p>
</blockquote>
<p>The given alternatives on my book are as follows:</p>
<p><span class="math-container">$\begin{array}{ll}
1.&-35\hat{i}-10\hat{j}\frac{m}{s}\\
2.&-25\hat{i}-10\hat{j}\frac{m}{s}\\
3.&-30\hat{i}-10\hat{j}\frac{m}{s}\\
4.&-25\hat{i}+10\hat{j}\frac{m}{s}\\
5.&-40\hat{i}-10\hat{j}\frac{m}{s}\\
\end{array}$</span></p>
<p>For this problem I'm totally lost at how should I understand or calculate the speed as seen from the observer. My first guess is that it might be the sum of the two speeds?. In other words that the speed of the shopper inside the glass elevator is the sum of the sphere as seen by him and the real speed. Or could it be the opposite?.</p>
<p>I'm still confused at this part.</p>
<p>The only thing which I could come up with was to write the position equation for the sphere as shown below:</p>
<p><span class="math-container">$y(t)=y_{0}+v_{oy}t-\frac{1}{2}gt^2$</span></p>
<p>Although <span class="math-container">$v_{oy}=0$</span>, and <span class="math-container">$t=3\,s$</span> there is no given information about how high is the building.</p>
<p>Then I turned my attention to the speed at <span class="math-container">$t=3\,s$</span> this would mean:</p>
<p><span class="math-container">$v_{f}=v_{o}-gt$</span></p>
<p><span class="math-container">$v_{f}=0-10(3)=-30\,\frac{m}{s}$</span></p>
<p>That would be the real speed of the sphere at that instant. My intuition tells me that the observer will see the ball going faster? and how about the acceleration?</p>
<p>Then and more importantly how can I find the acceleration and the velocity as seen by the observer riding in the elevator?. Can somebody help me here?.</p>
| Doug M | 317,162 | <p>If the elevator is moving at a constant velocity, the acceleration due to gravity is the same inside the elevator as outside.</p>
<p><span class="math-container">$a = -10\frac {m}{s^2}$</span></p>
<p>Velocities will be different due to the different frames of refference.</p>
<p>For the person in the elevator the relative velocity is.</p>
<p><span class="math-container">$v = (-10t + 5) \frac{m}{s}$</span> at time <span class="math-container">$t=3, v(t) = -25\frac {m}{s}$</span></p>
|
176,033 | <p>What is the highest dimension for which the space of reduced positive definite quadratic forms (or the fundamental domain of $SL_n(\mathbb{R})/SL_n(\mathbb{Z})$) has been explicitly calculated? I know it's been done for $n \leq 7$ in 1970's. Has there been any progress since then? If not, is it because there is some sort of difficulty, or because, although the computation is very much possible in practice, people thought it's not worth the effort?</p>
| JHM | 20,516 | <p>I very much enjoy Douglas Grenier's "Fundamental Domains for the General linear group", Pacific J. of Math., Vol 132, 2, 1988, wherein fundamental domains $F_n$ for the symmetric spaces $S_n \simeq SL(n,R) / SO(n)$ of $n \times n$ positive definite unimodular symmetric matrices are constructed inductively. That is, $F_n$ is constructed from $F_{n-1}$. The usual Minkowski reduction does not permit such an inductive description. One could therefore, in principal, see for oneself how to push the known results forward. Grenier presents an algorithm in his paper for reducing an arbitrary element in $S_n$ to his fundamental domain $F_n$, and the difficulty of pushing a fundamental domain for $S_7$ onward to $S_8$ may perhaps be phrased in terms of computing time of MAPLE or OCTAVE.</p>
<p>The basic element in Grenier's paper are modified Iwasawa coordinates on $S_n$. Instead of the usual Iwasawa coordinates $KAN$, where $AN$ is the identity component of an $R$-split Borel subgroup of $SL(n,R)$ and $K$ of course being $SO(n,R)$, Grenier works relative to an $R$-split maximal parabolic subgroup $P$, namely the stabilizer of a single $1$-dimensional flag. Of course one knows that $S_n$ is a homogeneous $P(R)^o$-space, and it is an interesting point to determine exactly global bijective coordinates for $S_n$ relative to a connected maximal $R$-split torus of $P(R)^o$ and the unipotent radical of $P(R)^o$.</p>
<p>Now my own opinion (which is negligible) is that one does not want an explicit description, even if one wants an explicit description. Any explicit description is necessarily so explicit and complicated as to be totally useless. More seriously, I don't think there is any question about the action of $SL(n,Z)$ on $S_n$ which requires a fundamental domain to be constructed. The finite presentation of $SL(n,Z)$, finite covolume, etc., can be achieved many other ways. E.g. via the Borel-Serre bordification or Siegel sets (coarse reduction theory) a la Borel/Harish-Chandra. In short, fundamental domains for $S_n$ relative to $SL(n,Z)$ are (to me) totally fictional and useless.</p>
|
2,016,781 | <p>Premise:</p>
<ol>
<li>Let P(n) return the real part of the nth nontrivial root of the zeta function</li>
<li>The first several roots of the zeta function are already known </li>
</ol>
<p>Proof:</p>
<ol>
<li>Pick any integer n. For example, 1.</li>
<li>Solve for P(n)</li>
<li>Solve for P(n+1)</li>
<li>Both (2) and (3) are 1/2 as established by Alan Turing and others</li>
<li>By induction all P(n) are 1/2</li>
</ol>
<p>QED the real part of all nontrivial roots of the riemann zeta function is 1/2</p>
<p>Any problems? </p>
| 5xum | 112,884 | <p>The problem with this proof is that it isn't a proof. Step number 4 claims something that was not proven in previous steps. We don't know what $f$ is, that's the whole point.</p>
<p>Also, even if this "proof" was correct, this is not how a proof by induction. A proof of induction would look like:</p>
<ol>
<li>Prove that the real part of $f(1)$ is $\frac12$.</li>
<li>Assume that the real part of $f(n)$ is $\frac12$.</li>
<li>Prove, from assumption $(2)$, that the real part of $f(n+1)$ is $\frac12$</li>
</ol>
<hr>
<p>An example of your proof, showing something very wrong:</p>
<ol>
<li>Define $P(n) = n$</li>
<li>The first several values of $P(n)$ are already known</li>
</ol>
<p>Claim: for all values $n$, the value $P(n)$ is smaller than $1000000$.</p>
<p>Proof:</p>
<ol>
<li>Pick any integer $n$, for example $1$.</li>
<li>Solve for $P(n)$</li>
<li>Solve for $P(n+1)$</li>
<li>Both (2) and (3) are smaller than $1000000$ as $1<1000000$ and $2<1000000$.</li>
<li>By induction all $P(n)$ are smaller than $1000000$</li>
</ol>
<hr>
<p>An example of a good proof by induction:</p>
<p>Claim: For all values $n\in\mathbb N$, the sum $\sum_{i=1}^n i$ is equal to $\frac{n(n+1)}{2}$</p>
<p>Proof:</p>
<p>First of all, the claim is true for $n=1$ since $$\sum_{i=1}^1 i = 1 = \frac{1\cdot(1+1)}{2}$$</p>
<p>Now, assume that the claim is true for $n$.</p>
<ol>
<li>By assumption, $$\sum_{i=1}^n i = \frac{n(n+1)}{2}.$$</li>
<li>By definition, $$\sum_{i=1}^{n+1}i = \sum_{i=1}^n i + (n+1)$$</li>
<li>From (1) and (2), it follows that
$$\sum_{i=1}^{n+1}i = \sum_{i=1}^n i + (n+1) = \frac{n(n+1)}{2} + n+1$$</li>
<li>Simplifying the expression, we get $$\frac{n(n+1)}{2} + n+1 = \frac{n^2+n}{2} + \frac{2n+2}{2} = \frac{n^2+3n+2}{2} = \frac{(n+1)(n+2)}{2}$$</li>
<li>From (4), it follows that $$\sum_{i=1}^{n+1}i = \frac{(n+1)(n+2)}{2}$$ in other words, the claim is true for $n+1$.</li>
</ol>
<p>So, assuming that the claim is true for $n$, we have proven (in point (5)), that the claim is true for $n+1$. We have also proven that the claim is true for $1$. This allows us to conclude that the claim is true for all values of $n$.</p>
<hr>
<p>Unlike <em>your</em> proof, this valid induction proof cannot be refuted by the same arguments. It is valid for all values of $n$.</p>
|
1,752,106 | <p>You roll $n$ fair dice. Let X be the number of pairs of dice that sum to $7$.
Write $X$ as a sum of indicator variables and find $E[X]$</p>
<p>This is how I approached the problem</p>
<p>Let $S$ be all the strings length n on [6].</p>
<p>Let $X(S)$ be the number of ways to add up the numbers on the n dice that adds up to 7. Then we can write $X$ as $X = X_{1}+X_{2}+.....+X_{n}$</p>
<p>To find the expected value we can use: $$ \sum_{s\in S} X(S)P(S)$$</p>
<p>Let $X(S) = k$ (k is the numbers that show up on the $n$ dice and adds up to 7. Eg: n = 4 and numbers that show up on the dice are $4631$. $4$ and $3$ make $7$ and $6$ and $1$ make $7$, so k = 2.)</p>
<p>Now I have no idea how to find the P(S). I feel like the denominator of the P(S) will be $6^{n}$, but I don't know how to find the numerator.</p>
<p>Thank you.</p>
| Graham Kemp | 135,106 | <p>You wish to find the expected count of pairs of dice whose results add up to seven.</p>
<p>Eg: If the results are <code>1,2,3,4,3,6,2,5,4</code> then there are $7$ ways to select pairs which add up to seven, because there is one <code>1</code> and one <code>6</code>, two <code>2</code> and one <code>5</code>, two <code>3</code> and two <code>4</code>. </p>
<p>Let $N_k$ be the count of dice whose number is $k$, and express $X$ in terms of $N_1, N_2, N_3, N_4, N_5, N_6$. </p>
<p>Now find the expectation.</p>
<hr>
<blockquote class="spoiler">
<p> $$\begin{align}\mathsf E(X) ~=~& \mathsf E(N_1N_6+N_2N_5+N_3N_4)\\[1ex] ~=~ & 3~\mathsf E(N_iN_j; i\neq j) & \textsf{Linearity of Expectation}\\[1ex] ~=~& 3~\mathsf E(N_i \mathsf E(N_j\mid N_i); i\neq j) & \textsf{Iteration of Expectation} \\[1ex] ~=~& 3~\mathsf E(N_i\cdot \tfrac 1 5(n-N_i)) & N_j\mid N_i ~\sim~\mathcal{Bin}(n-N_i, 1/5) \\[1ex] ~=~& \tfrac 3 5(n\mathsf E(N_i)-\mathsf E(N_i^2)) \\[1ex] ~=~& \tfrac 3 5(n\mathsf E(N_i)-\mathsf {Var}(N_i)-\mathsf E(N_i)^2) \\[1ex] ~=~& \tfrac 3 5(\tfrac {n^2}6-\tfrac {5n}{36} -\tfrac{n^2}{36}) & N_i\sim\mathcal{Bin}(n, 1/6) \\[1ex] ~=~& \tfrac{n(n-1)}{12}\end{align}$$</p>
</blockquote>
<p>For which André Nicolas obtained by a much more elegant approach. </p>
|
789,043 | <p>Given a finite sequence of decimal digits $a_1,a_2,...,a_n$ prove that there exists a natural number $m$ such that decimal representation of $2^m$ starts with that sequence of digits.</p>
<p>Thanks for your help :)</p>
| Community | -1 | <p>It is not a proof, just a try to find by approximations the first solutions, without the guarantee that they exist.</p>
<p>I tried a small script to select candidates which were submited next to a <a href="http://www.wolframalpha.com/input/?i=2%5E2621" rel="nofollow">large digital computer</a> for example manually.</p>
<p>The principle is the decimal approximation. The algorithm keeps only the 11 digits of a number hoping that the truncature will not affect a lot the most left digits. To extend the search , it selects also a few approximative matches, always under the target, since the truncature makes the result smaller.</p>
<p>There is an example of implementation for the target 999 , easy to adapt to other ones:</p>
<pre><code>function main()
{
var i, k = 68719476736,res="",res2="" ; /// begin with 2^36
for( i=37 ; i < 50000 ; i++ )
{
k = 2* ((""+k).substr(0,7)) ;
var kk = (""+k).substr(0,3) ;
if( kk=="999" ) // if( kk=="998" || kk=="999" )
{
res +=( k+" ("+i+"), " ) ;
res2 +=( "2^"+i+"," ) ;
}
}
return res+"\n"+res2+"\n" ;
}
// scratchpad formalism to output the result as the last seen variable
// type CTRL L to run
var tto , ttt = main() ;
tto = ttt ;
</code></pre>
<p>Truncated numbers selected with the power in parenthesis :</p>
<p>$99912223808 (2621), 99928497088 (4757), 99944773024 (6893), $
$99961051568 (9029), 99977332768 (11165), 99993616640 (13301), $
$99905846016 (15922), $
$99922118264 (18058), 99938393032 (20194), 99954670632 (22330), $
$99970950824 (24466), 99987233664 (26602), $
$99915739784 (31359), 99932013584 (33495), 99948290144 (35631), $
$99964569264 (37767), 99980851144 (39903), $
$99997135584 (42039), 99909361920 (44660), 99925634840 (46796), $
$99941910296 (48932); ...$</p>
<p>After external tests, $2^{2621}$, $2^{4757}$, $2^{6893}$ , $2^{9029}$ , $2^{11165}$ seem to work well with the target of the test $999$.</p>
<p><strong>edit</strong> : keeping 11 significative digits and testing only the target, I get directly the same begining of the sequence solution. </p>
|
13,937 | <p>Prove that $$\sum_{\substack{x,y \in \mathbb N \\ ax-by \ne 0 }}\frac1{|ax-by|xy}$$ converges, where $a, b \in \mathbb N$
</p>
<p>This 2-D problem can be proved by an integral test. I'm looking for some other proofs that can be easily generalized to higher dimensional cases like below,</p>
<p>Prove that $$\sum_{\substack{x_1,x_2, \cdots, x_d \in \mathbb N \\ \bf a \cdot \bf x \ne 0 }}\frac1{|{\bf a} \cdot {\bf x}|x_1 x_2 \cdots x_d}$$ converges, with ${\bf a} \in \mathbb Z^d$, and $\forall \; 1 \le i \le d, a_i \ne 0$.</p>
| Jack Schmidt | 583 | <p>For abelian groups, the ring End(A) is very important. As far as non-abelian groups A go, End(A) is not even (usually considered) a group.</p>
<p>"Adding" homomorphisms doesn't work in the non-abelian case.</p>
<p>If you define (f+g)(x) = f(x) + g(x), then (f+g)(x+y) = f(x+y) + g(x+y) = f(x) + f(y) + g(x) + g(y), but (f+g)(x) + (f+g)(y) = f(x) + g(x) + f(y) + g(y). To conclude that:</p>
<p> f(y) + g(x) = g(x) + f(y)</p>
<p>are equal, you use that + is commutative, that A is abelian. More precisely, if you take f=g to be the identity endomorphism, then f+g is an endomorphism iff A is abelian.</p>
<p>"Composing" homomorphisms doesn't work to form a group, since they are not invertible.</p>
<p>Aut(A), the group of invertible endomorphisms, does form a group. Aut(A) does not determine if a group is abelian or not: 4×2 and the dihedral group of order 8 have isomorphic automorphism groups.</p>
<p>Instead of a ring, End(A) sits inside the "near-ring" of self-maps. See <a href="http://en.wikipedia.org/wiki/Nearring#Mappings_from_a_group_to_itself">the wikipedia article on nearring</a> for an explanation.</p>
|
1,922,417 | <p>This is not for the Maths part of the General GRE. This is for the GRE <a href="https://www.ets.org/gre/subject/about" rel="noreferrer">Subject Test</a> in <a href="https://www.ets.org/gre/subject/about/content/mathematics" rel="noreferrer">Maths</a>. Feel free to add or comment.</p>
<ul>
<li><p><a href="https://math.stackexchange.com/questions/2967070">How do I know the definition of rings or of anything on the GRE given that definitions can vary?</a></p></li>
<li><p><a href="https://math.stackexchange.com/questions/2470560">What does the subject GRE measure?</a></p></li>
<li><blockquote>
<p>I think trying to relearn an entire undergraduate degree's worth of mathematics purely for this test would be an enormous waste of precious time and energy. You can't make this test your life,you have other priorities. I'd begin talking to graduate students who have been through the test-see what they'd recommend. - <a href="https://math.stackexchange.com/a/792069">Mathemagician1234</a></p>
</blockquote></li>
</ul>
<hr>
<p><a href="http://www.mathematicsgre.com/" rel="noreferrer">http://www.mathematicsgre.com/</a></p>
<p><a href="http://www.physicsgre.com/viewtopic.php?f=13&t=1078" rel="noreferrer">http://www.physicsgre.com/viewtopic.php?f=13&t=1078</a></p>
| BCLC | 140,308 | <h2>Top web resource:</h2>
<p><a href="https://en.wikipedia.org/wiki/DM_Ashura" rel="noreferrer">DM Ashura</a> aka <a href="https://www.mathsub.com/" rel="noreferrer">Bill Shillito</a>: <a href="https://www.mathsub.com/" rel="noreferrer">https://www.mathsub.com/</a></p>
<hr />
<h2>Official resources:</h2>
<p>(Official) 0568</p>
<p><a href="http://www.math.ucla.edu/%7Ecmarshak/GRE1.pdf" rel="noreferrer">http://www.math.ucla.edu/~cmarshak/GRE1.pdf</a></p>
<p><a href="http://web.archive.org/web/20150701191646/http://rambotutoring.com/GR0568-solutions.pdf" rel="noreferrer">http://web.archive.org/web/20150701191646/http://rambotutoring.com/GR0568-solutions.pdf</a>
<a href="http://online.fliphtml5.com/ebvz/nytg/" rel="noreferrer">http://online.fliphtml5.com/ebvz/nytg/</a>
<a href="https://web.archive.org/web/20161023164437/http://online.fliphtml5.com/ebvz/nytg/#p=1" rel="noreferrer">https://web.archive.org/web/20161023164437/http://online.fliphtml5.com/ebvz/nytg/#p=1</a></p>
<p><a href="http://www.math.ucla.edu/%7Eiacoley/gre/Practice%202.pdf" rel="noreferrer">http://www.math.ucla.edu/~iacoley/gre/Practice%202.pdf</a>
<a href="http://www.math.ucla.edu/%7Eiacoley/gre/Practice%202%20solutions.pdf" rel="noreferrer">http://www.math.ucla.edu/~iacoley/gre/Practice%202%20solutions.pdf</a></p>
<hr />
<p>(Official) 1268 aka 1768</p>
<p><a href="http://www.math.ucla.edu/%7Eiacoley/gre/Practice%201.pdf" rel="noreferrer">http://www.math.ucla.edu/~iacoley/gre/Practice%201.pdf</a>
<a href="http://www.math.ucla.edu/%7Eiacoley/gre/Practice%201%20solutions.pdf" rel="noreferrer">http://www.math.ucla.edu/~iacoley/gre/Practice%201%20solutions.pdf</a></p>
<p><a href="http://seansmathematicsblog.blogspot.hk/2016/08/solutions-to-math-subject-gre-sample.html" rel="noreferrer">http://seansmathematicsblog.blogspot.hk/2016/08/solutions-to-math-subject-gre-sample.html</a>
<a href="https://drive.google.com/file/d/0B4qQg_AuKUglUnd6YWIyYWNudWM/view?usp=drive_web" rel="noreferrer">https://drive.google.com/file/d/0B4qQg_AuKUglUnd6YWIyYWNudWM/view?usp=drive_web</a></p>
<p><a href="http://webpages.sou.edu/%7Estonelakb/math/testprep" rel="noreferrer">http://webpages.sou.edu/~stonelakb/math/testprep</a></p>
<p><a href="http://www.rambotutoring.com/GR1268-solutions.pdf" rel="noreferrer">http://www.rambotutoring.com/GR1268-solutions.pdf</a></p>
<p><a href="https://www.scribd.com/doc/315056948/GR1268-Solutions" rel="noreferrer">https://www.scribd.com/doc/315056948/GR1268-Solutions</a>
<a href="https://www.scribd.com/user/285425100/RG" rel="noreferrer">https://www.scribd.com/user/285425100/RG</a></p>
<p><a href="http://www.mathematicsgre.com/viewtopic.php?t=3494" rel="noreferrer">http://www.mathematicsgre.com/viewtopic.php?t=3494</a>
<a href="https://drive.google.com/file/d/0B-uVGGkZosoPcGVRcDNPN0d2ZVU/view" rel="noreferrer">https://drive.google.com/file/d/0B-uVGGkZosoPcGVRcDNPN0d2ZVU/view</a></p>
<hr />
<p>(Official) 9367</p>
<p><a href="http://www.math.ucla.edu/%7Ecmarshak/GRE2.pdf" rel="noreferrer">http://www.math.ucla.edu/~cmarshak/GRE2.pdf</a></p>
<p><a href="http://www.math.ucla.edu/%7Eiacoley/gre/Practice%203.pdf" rel="noreferrer">http://www.math.ucla.edu/~iacoley/gre/Practice%203.pdf</a>
<a href="http://www.math.ucla.edu/%7Eiacoley/gre/Practice%203%20solutions.pdf" rel="noreferrer">http://www.math.ucla.edu/~iacoley/gre/Practice%203%20solutions.pdf</a></p>
<p><a href="http://seansmathematicsblog.blogspot.hk/2016/08/solutions-to-mathematics-subject-gre.html" rel="noreferrer">http://seansmathematicsblog.blogspot.hk/2016/08/solutions-to-mathematics-subject-gre.html</a>
<a href="https://drive.google.com/file/d/0B4qQg_AuKUglY0I1bkZ5azl6NDQ/view" rel="noreferrer">https://drive.google.com/file/d/0B4qQg_AuKUglY0I1bkZ5azl6NDQ/view</a></p>
<hr />
<p>(Official) 8767</p>
<p><a href="http://www.math.ucla.edu/%7Ecmarshak/GRE3.pdf" rel="noreferrer">http://www.math.ucla.edu/~cmarshak/GRE3.pdf</a></p>
<p><a href="http://www.math.ucla.edu/%7Eiacoley/gre/Practice%204.pdf" rel="noreferrer">http://www.math.ucla.edu/~iacoley/gre/Practice%204.pdf</a>
<a href="http://www.math.ucla.edu/%7Eiacoley/gre/Practice%204%20solutions.pdf" rel="noreferrer">http://www.math.ucla.edu/~iacoley/gre/Practice%204%20solutions.pdf</a></p>
<p><a href="http://seansmathematicsblog.blogspot.hk/2016/08/solutions-to-mathematics-subject-gre_7.html" rel="noreferrer">http://seansmathematicsblog.blogspot.hk/2016/08/solutions-to-mathematics-subject-gre_7.html</a>
<a href="https://drive.google.com/open?id=0B4qQg_AuKUgld2hoV3h0OFBaZHM" rel="noreferrer">https://drive.google.com/open?id=0B4qQg_AuKUgld2hoV3h0OFBaZHM</a></p>
<hr />
<p>(Official) 9768</p>
<p><a href="http://www.math.ucla.edu/%7Ecmarshak/GRE4.pdf" rel="noreferrer">http://www.math.ucla.edu/~cmarshak/GRE4.pdf</a></p>
<p><a href="http://www.math.ucla.edu/%7Eiacoley/gre/Practice%205.pdf" rel="noreferrer">http://www.math.ucla.edu/~iacoley/gre/Practice%205.pdf</a>
<a href="http://www.math.ucla.edu/%7Eiacoley/gre/Practice%205%20solutions.pdf" rel="noreferrer">http://www.math.ucla.edu/~iacoley/gre/Practice%205%20solutions.pdf</a></p>
<hr />
<p>(Official) 35-item practice test</p>
<p><a href="http://www.math.ucla.edu/%7Eiacoley/gre/othertests/math97-99.pdf" rel="noreferrer">http://www.math.ucla.edu/~iacoley/gre/othertests/math97-99.pdf</a>
<a href="https://www.geneseo.edu/%7Ejohannes/GRE%20Practice%20Problems.pdf" rel="noreferrer">https://www.geneseo.edu/~johannes/GRE%20Practice%20Problems.pdf</a></p>
<hr />
<p>(Official) ETS Major Field Test in Mathematics Sample Questions</p>
<p><a href="https://www.ets.org/Media/Tests/MFT/pdf/mft_mathII.pdf" rel="noreferrer">https://www.ets.org/Media/Tests/MFT/pdf/mft_mathII.pdf</a></p>
<p><a href="http://www.wou.edu/%7Ebeaverc/404/S14/major%20field%20test.pdf" rel="noreferrer">http://www.wou.edu/~beaverc/404/S14/major%20field%20test.pdf</a></p>
<hr />
<h2>Web resources:</h2>
<p><a href="http://rambotutoring.com/" rel="noreferrer">Rambotutoring.co</a>m practice test and answers</p>
<p><a href="http://www.rambotutoring.com/GREpractice.pdf" rel="noreferrer">http://www.rambotutoring.com/GREpractice.pdf</a>
<a href="http://www.rambotutoring.com/GREpracticeanswers.pdf" rel="noreferrer">http://www.rambotutoring.com/GREpracticeanswers.pdf</a></p>
<hr />
<p>More practice tests (rea i think)</p>
<p><a href="http://www.math.ucla.edu/%7Eiacoley/gre/othertests/math01.pdf" rel="noreferrer">http://www.math.ucla.edu/~iacoley/gre/othertests/math01.pdf</a>
<a href="http://www.math.ucla.edu/%7Eiacoley/gre/othertests/math02.pdf" rel="noreferrer">http://www.math.ucla.edu/~iacoley/gre/othertests/math02.pdf</a>
<a href="http://www.math.ucla.edu/%7Eiacoley/gre/othertests/math03.pdf" rel="noreferrer">http://www.math.ucla.edu/~iacoley/gre/othertests/math03.pdf</a>
<a href="http://www.math.ucla.edu/%7Eiacoley/gre/othertests/math04.pdf" rel="noreferrer">http://www.math.ucla.edu/~iacoley/gre/othertests/math04.pdf</a>
<a href="http://www.math.ucla.edu/%7Eiacoley/gre/othertests/math05.pdf" rel="noreferrer">http://www.math.ucla.edu/~iacoley/gre/othertests/math05.pdf</a>
<a href="http://www.math.ucla.edu/%7Eiacoley/gre/othertests/math06.pdf" rel="noreferrer">http://www.math.ucla.edu/~iacoley/gre/othertests/math06.pdf</a></p>
<p><a href="http://www.math.ucla.edu/%7Eiacoley/gre/othertests/math01e.pdf" rel="noreferrer">http://www.math.ucla.edu/~iacoley/gre/othertests/math01e.pdf</a>
<a href="http://www.math.ucla.edu/%7Eiacoley/gre/othertests/math02e.pdf" rel="noreferrer">http://www.math.ucla.edu/~iacoley/gre/othertests/math02e.pdf</a>
<a href="http://www.math.ucla.edu/%7Eiacoley/gre/othertests/math03e.pdf" rel="noreferrer">http://www.math.ucla.edu/~iacoley/gre/othertests/math03e.pdf</a>
<a href="http://www.math.ucla.edu/%7Eiacoley/gre/othertests/math04e.pdf" rel="noreferrer">http://www.math.ucla.edu/~iacoley/gre/othertests/math04e.pdf</a>
<a href="http://www.math.ucla.edu/%7Eiacoley/gre/othertests/math05e.pdf" rel="noreferrer">http://www.math.ucla.edu/~iacoley/gre/othertests/math05e.pdf</a>
<a href="http://www.math.ucla.edu/%7Eiacoley/gre/othertests/math06e.pdf" rel="noreferrer">http://www.math.ucla.edu/~iacoley/gre/othertests/math06e.pdf</a></p>
<hr />
<p>GRE Subject Test Math Practice (App)</p>
<p><a href="https://play.google.com/store/apps/details?id=com.varsitytutors.learningtools.gresubjecttestmath&hl=en" rel="noreferrer">https://play.google.com/store/apps/details?id=com.varsitytutors.learningtools.gresubjecttestmath&hl=en</a></p>
<hr />
<p>Sheir.Org GRE Mathematics Sample Questions</p>
<p><a href="http://www.sheir.org/gre_mathematics_sample_questions.html" rel="noreferrer">http://www.sheir.org/gre_mathematics_sample_questions.html</a></p>
<p><a href="http://www.sheir.org/gre-mathematics-sample-test.html" rel="noreferrer">http://www.sheir.org/gre-mathematics-sample-test.html</a></p>
<hr />
<p>Sheir.Org GRE Mathematics Practice Test</p>
<p><a href="http://www.sheir.org/gre_math_practice_questions.html" rel="noreferrer">http://www.sheir.org/gre_math_practice_questions.html</a></p>
<p><a href="http://www.sheir.org/gre-math-practice-test.html" rel="noreferrer">http://www.sheir.org/gre-math-practice-test.html</a></p>
<p><a href="http://www.sheir.org/gre-math-subject-test-practice.html" rel="noreferrer">http://www.sheir.org/gre-math-subject-test-practice.html</a></p>
<p><a href="http://www.sheir.org/gre-mathematics-subject-test-practice.html" rel="noreferrer">http://www.sheir.org/gre-mathematics-subject-test-practice.html</a></p>
<hr />
<p>FPSC Mathematics Lecturer Test Sample Questions</p>
<p><a href="http://www.sheir.org/fpsc-math-sample-questions.html" rel="noreferrer">http://www.sheir.org/fpsc-math-sample-questions.html</a></p>
<hr />
<p>FPSC Mathematics Lecturer Test Past Papers</p>
<p><a href="http://www.sheir.org/fpsc-mathematics-past-papers.html" rel="noreferrer">http://www.sheir.org/fpsc-mathematics-past-papers.html</a></p>
<p><a href="http://www.sheir.org/fpsc-mathematics-past-test.html" rel="noreferrer">http://www.sheir.org/fpsc-mathematics-past-test.html</a></p>
<p><a href="http://www.sheir.org/fpsc-mathematics-past-test-papers.html" rel="noreferrer">http://www.sheir.org/fpsc-mathematics-past-test-papers.html</a></p>
<p><a href="http://www.sheir.org/math-lecturer-past-papers.html" rel="noreferrer">http://www.sheir.org/math-lecturer-past-papers.html</a></p>
<p><a href="http://www.sheir.org/mathematics-lecturer-test-past-papers.html" rel="noreferrer">http://www.sheir.org/mathematics-lecturer-test-past-papers.html</a></p>
<hr />
<p>Sample Questions (1-10) FPSC Mathematics Lecturer test BS-17.pdf</p>
<p><a href="https://www.scribd.com/doc/311749932/Mathematics-Subject-test-sample-questions-MCQs-21-30-pdf" rel="noreferrer">Mathematics Subject test sample questions MCQs (21-30).pdf</a></p>
<p><a href="http://documents.mx/documents/mathematics-subject-test-sample-questions-mcqs-31-40.html" rel="noreferrer">Mathematics Subject test sample questions MCQs (31-40).pdf</a></p>
<p>Mathematics Subject test sample questions MCQs (41-50).pdf</p>
<hr />
<p>UCLA Christian Parkinson</p>
<p><a href="http://www.math.ucla.edu/%7Echparkin/gre.html" rel="noreferrer">http://www.math.ucla.edu/~chparkin/gre.html</a></p>
<hr />
<p>UCLA Charlie Marshak</p>
<p><a href="http://www.math.ucla.edu/%7Ecmarshak/GREWorkshop.html" rel="noreferrer">http://www.math.ucla.edu/~cmarshak/GREWorkshop.html</a></p>
<p><a href="https://web.archive.org/web/20160911233702/http://www.math.ucla.edu/%7Ecmarshak/GREworkshopProblems.pdf" rel="noreferrer">https://web.archive.org/web/20160911233702/http://www.math.ucla.edu/~cmarshak/GREworkshopProblems.pdf</a>
<a href="https://web.archive.org/web/20160911233717/http://www.math.ucla.edu/%7Ecmarshak/GREworkshopSolutions.pdf" rel="noreferrer">https://web.archive.org/web/20160911233717/http://www.math.ucla.edu/~cmarshak/GREworkshopSolutions.pdf</a></p>
<p><a href="https://www.math.ucla.edu/%7Ecmarshak/GRE/Problems/GREBootcampProblems.pdf" rel="noreferrer">https://www.math.ucla.edu/~cmarshak/GRE/Problems/GREBootcampProblems.pdf</a>
<a href="https://www.math.ucla.edu/%7Ecmarshak/GRE/Solutions/GREBootcampSolutions.pdf" rel="noreferrer">https://www.math.ucla.edu/~cmarshak/GRE/Solutions/GREBootcampSolutions.pdf</a></p>
<p><a href="https://www.math.ucla.edu/%7Ecmarshak/GRE/GREProb.pdf" rel="noreferrer">https://www.math.ucla.edu/~cmarshak/GRE/GREProb.pdf</a>
<a href="https://www.math.ucla.edu/%7Ecmarshak/GRE/GRESol.pdf" rel="noreferrer">https://www.math.ucla.edu/~cmarshak/GRE/GRESol.pdf</a></p>
<hr />
<p>UCSB</p>
<p><a href="http://web.math.ucsb.edu/%7Epadraic/ucsb_2014_15/math_gre_w2015/" rel="noreferrer">http://web.math.ucsb.edu/~padraic/ucsb_2014_15/math_gre_w2015/</a></p>
<p><a href="http://web.math.ucsb.edu/%7Epadraic/ucsb_2014_15/math_gre_w2015/math_gre_w2015.html" rel="noreferrer">http://web.math.ucsb.edu/~padraic/ucsb_2014_15/math_gre_w2015/math_gre_w2015.html</a></p>
<p><a href="http://web.math.ucsb.edu/%7Epadraic/ucsb_2014_15/math_gre_w2015/math_gre_w2015_syllabus.pdf" rel="noreferrer">http://web.math.ucsb.edu/~padraic/ucsb_2014_15/math_gre_w2015/math_gre_w2015_syllabus.pdf</a></p>
<p><a href="http://web.math.ucsb.edu/%7Epadraic/ucsb_2014_15/math_gre_w2015/math_gre_w2015_lecture1.pdf" rel="noreferrer">http://web.math.ucsb.edu/~padraic/ucsb_2014_15/math_gre_w2015/math_gre_w2015_lecture1.pdf</a></p>
<p><a href="http://web.math.ucsb.edu/%7Epadraic/ucsb_2014_15/math_gre_w2015/math_gre_w2015_lecture2.pdf" rel="noreferrer">http://web.math.ucsb.edu/~padraic/ucsb_2014_15/math_gre_w2015/math_gre_w2015_lecture2.pdf</a></p>
<p><a href="http://web.math.ucsb.edu/%7Epadraic/ucsb_2014_15/math_gre_w2015/math_gre_w2015_lecture3.pdf" rel="noreferrer">http://web.math.ucsb.edu/~padraic/ucsb_2014_15/math_gre_w2015/math_gre_w2015_lecture3.pdf</a></p>
<p><a href="http://web.math.ucsb.edu/%7Epadraic/ucsb_2014_15/math_gre_w2015/math_gre_w2015_lecture4.pdf" rel="noreferrer">http://web.math.ucsb.edu/~padraic/ucsb_2014_15/math_gre_w2015/math_gre_w2015_lecture4.pdf</a></p>
<p><a href="http://web.math.ucsb.edu/%7Epadraic/ucsb_2014_15/math_gre_w2015/math_gre_w2015_lecture5.pdf" rel="noreferrer">http://web.math.ucsb.edu/~padraic/ucsb_2014_15/math_gre_w2015/math_gre_w2015_lecture5.pdf</a></p>
<p><a href="http://web.math.ucsb.edu/%7Epadraic/ucsb_2014_15/math_gre_w2015/lecture%206.pdf" rel="noreferrer">http://web.math.ucsb.edu/~padraic/ucsb_2014_15/math_gre_w2015/lecture%206.pdf</a></p>
<p><a href="http://web.math.ucsb.edu/%7Epadraic/ucsb_2014_15/math_gre_w2015/math_gre_w2015_lecture7.pdf" rel="noreferrer">http://web.math.ucsb.edu/~padraic/ucsb_2014_15/math_gre_w2015/math_gre_w2015_lecture7.pdf</a></p>
<p><a href="http://web.math.ucsb.edu/%7Epadraic/ucsb_2014_15/math_gre_w2015/math_gre_w2015_lecture8.pdf" rel="noreferrer">http://web.math.ucsb.edu/~padraic/ucsb_2014_15/math_gre_w2015/math_gre_w2015_lecture8.pdf</a></p>
<p><a href="http://web.math.ucsb.edu/%7Epadraic/ucsb_2014_15/math_gre_w2015/math_gre_w2015_lecture10.pdf" rel="noreferrer">http://web.math.ucsb.edu/~padraic/ucsb_2014_15/math_gre_w2015/math_gre_w2015_lecture10.pdf</a></p>
<p><a href="http://web.math.ucsb.edu/%7Epadraic/ucsb_2014_15/math_gre_w2015/math_gre_w2015_practice_test.pdf" rel="noreferrer">http://web.math.ucsb.edu/~padraic/ucsb_2014_15/math_gre_w2015/math_gre_w2015_practice_test.pdf</a></p>
<hr />
<p>Clark University Study Group for the GRE Subject Test in Mathematics</p>
<p><a href="http://aleph0.clarku.edu/%7Edjoyce/GREmath/" rel="noreferrer">http://aleph0.clarku.edu/~djoyce/GREmath/</a></p>
<p><a href="http://aleph0.clarku.edu/%7Edjoyce/GREmath/linalg.pdf" rel="noreferrer">http://aleph0.clarku.edu/~djoyce/GREmath/linalg.pdf</a></p>
<p><a href="http://aleph0.clarku.edu/%7Edjoyce/GREmath/linalgDiscuss.pdf" rel="noreferrer">http://aleph0.clarku.edu/~djoyce/GREmath/linalgDiscuss.pdf</a></p>
<p><a href="http://aleph0.clarku.edu/%7Edjoyce/GREmath/highcalc.pdf" rel="noreferrer">http://aleph0.clarku.edu/~djoyce/GREmath/highcalc.pdf</a></p>
<hr />
<p>SubjectMath.Com</p>
<p><a href="http://media.wix.com/ugd/57c670_fb82db77a2134bd6836923eda969cda6.pdf" rel="noreferrer">http://media.wix.com/ugd/57c670_fb82db77a2134bd6836923eda969cda6.pdf</a></p>
<p><a href="http://media.wix.com/ugd/57c670_a4c54c95e72c44aa8944ef3172ec6f25.pdf" rel="noreferrer">http://media.wix.com/ugd/57c670_a4c54c95e72c44aa8944ef3172ec6f25.pdf</a></p>
<p><a href="http://media.wix.com/ugd/57c670_2d7c1c187d394b8aa39d14cc3bf4665e.pdf" rel="noreferrer">http://media.wix.com/ugd/57c670_2d7c1c187d394b8aa39d14cc3bf4665e.pdf</a></p>
<p><a href="http://media.wix.com/ugd/57c670_5deab4f8898941cf8eed1f07bb339fb9.pdf" rel="noreferrer">http://media.wix.com/ugd/57c670_5deab4f8898941cf8eed1f07bb339fb9.pdf</a></p>
<p><a href="https://www.youtube.com/watch?v=qXoZ8n72v5M" rel="noreferrer">https://www.youtube.com/watch?v=qXoZ8n72v5M</a></p>
<p><a href="https://www.youtube.com/watch?v=-LYugb1LO9g" rel="noreferrer">https://www.youtube.com/watch?v=-LYugb1LO9g</a></p>
<p><a href="https://www.youtube.com/watch?v=EPPMAVwz59Y" rel="noreferrer">https://www.youtube.com/watch?v=EPPMAVwz59Y</a></p>
<p><a href="https://www.youtube.com/watch?v=C2xJWoD95_k" rel="noreferrer">https://www.youtube.com/watch?v=C2xJWoD95_k</a></p>
<p><a href="https://www.youtube.com/watch?v=84nZMwr7saU" rel="noreferrer">https://www.youtube.com/watch?v=84nZMwr7saU</a></p>
<hr />
<p>GRE Subject Calculus</p>
<p><a href="https://www.coursehero.com/file/10329376/GRE-Subject-Calculus/" rel="noreferrer">https://www.coursehero.com/file/10329376/GRE-Subject-Calculus/</a></p>
<p>GRE Subject Additional Topics</p>
<p><a href="https://www.coursehero.com/file/10329377/GRE-Subject-Additional-topics" rel="noreferrer">https://www.coursehero.com/file/10329377/GRE-Subject-Additional-topics</a></p>
<hr />
<blockquote>
<p>UChicago has a very large list of unique questions that I don't see linked here (if it is, sorry I just missed it!). See: <a href="http://math.uchicago.edu/%7Emin/GRE/" rel="noreferrer">math.uchicago.edu/~min/GRE</a> – Jonathan Rayner</p>
</blockquote>
<hr />
<blockquote>
<p>Just found a YouTube channel called ProfOmarMath - has a selection of good videos discussing content and dissecting GRE questions. See <a href="https://www.youtube.com/playlist?list=PLBiVnG9A5gceFbk7oTBHALETI2SUmplIL" rel="noreferrer">youtube.com/playlist?list=PLBiVnG9A5gceFbk7oTBHALETI2SUmplIL</a> - Jordan Mitchell Barrett</p>
</blockquote>
<hr />
<h2>Book resources:</h2>
<ol>
<li>Morris Bramson --> I think this is better than REA because Bramson tests more on understanding while REA tests more on knowledge. Idk. Disagreements welcome.</li>
</ol>
<p><a href="https://www.coursehero.com/file/14751549/Morris-Bramson-Mathematics-Subject-Test-AdvanceBookFi-org/" rel="noreferrer">https://www.coursehero.com/file/14751549/Morris-Bramson-Mathematics-Subject-Test-AdvanceBookFi-org/</a></p>
<hr />
<ol start="2">
<li>REA (rated low by <a href="https://en.wikipedia.org/wiki/DM_Ashura" rel="noreferrer">DM Ashura</a> aka <a href="https://www.mathsub.com/" rel="noreferrer">Bill Shillito</a>)</li>
</ol>
<p><a href="https://books.google.com/books?id=Njk8Ea2IAU0C" rel="noreferrer">https://books.google.com/books?id=Njk8Ea2IAU0C</a></p>
<p><a href="https://rads.stackoverflow.com/amzn/click/com/0738608386" rel="noreferrer" rel="nofollow noreferrer">https://www.amazon.com/GRE-Mathematics-Test-Preparation/dp/0738608386</a></p>
<p>REA_GRE_MATH_SUB.pdf</p>
<hr />
<ol start="3">
<li>Princeton</li>
</ol>
<p><a href="https://rads.stackoverflow.com/amzn/click/com/0375429727" rel="noreferrer" rel="nofollow noreferrer">https://www.amazon.com/Cracking-Mathematics-Subject-Test-Edition/dp/0375429727</a></p>
<p>Quote: This book is for GRE Math Subject Test, NOT for GRE General Test.</p>
<p>Cracking The Gre Math Sunject Test.pdf</p>
<p>Cracking+the+GRE+Mathematics+Subject+Test(2010).pdf</p>
<p>Typos</p>
<p><a href="http://www.mathematicsgre.com/viewtopic.php?t=38" rel="noreferrer">http://www.mathematicsgre.com/viewtopic.php?t=38</a>
<a href="http://www.mathematicsgre.com/viewtopic.php?t=1432" rel="noreferrer">http://www.mathematicsgre.com/viewtopic.php?t=1432</a>
<a href="http://www.mathematicsgre.com/viewtopic.php?f=1&t=234" rel="noreferrer">http://www.mathematicsgre.com/viewtopic.php?f=1&t=234</a>
<a href="http://www.mathematicsgre.com/viewtopic.php?t=234" rel="noreferrer">http://www.mathematicsgre.com/viewtopic.php?t=234</a></p>
<p>-</p>
<p>Calculus Practice</p>
<p><a href="https://www.coursehero.com/file/p2an7lo/P-A-21-B-16-C-15-D-12-E-10-78-%D0%9D%D0%B0%D0%BC-%D0%B2%D0%B0%D0%B6%D0%B5/" rel="noreferrer">https://www.coursehero.com/file/p2an7lo/P-A-21-B-16-C-15-D-12-E-10-78-%D0%9D%D0%B0%D0%BC-%D0%B2%D0%B0%D0%B6%D0%B5/</a></p>
<p>-</p>
<p>UCI</p>
<p><a href="http://sites.uci.edu/gremath/files/2013/08/First-Day-Test-.pdf" rel="noreferrer">http://sites.uci.edu/gremath/files/2013/08/First-Day-Test-.pdf</a>
<a href="http://sites.uci.edu/gremath/files/2011/09/multivariable_calculus.pdf" rel="noreferrer">http://sites.uci.edu/gremath/files/2011/09/multivariable_calculus.pdf</a>
<a href="http://sites.uci.edu/gremath/files/2011/09/more_calculus_from_test2.pdf" rel="noreferrer">http://sites.uci.edu/gremath/files/2011/09/more_calculus_from_test2.pdf</a>
<a href="http://sites.uci.edu/gremath/files/2011/09/calculus_practice_test3.pdf" rel="noreferrer">http://sites.uci.edu/gremath/files/2011/09/calculus_practice_test3.pdf</a>
<a href="http://sites.uci.edu/gremath/files/2011/09/ALGEBRA_REVIEW1.pdf" rel="noreferrer">http://sites.uci.edu/gremath/files/2011/09/ALGEBRA_REVIEW1.pdf</a>
<a href="http://sites.uci.edu/gremath/files/2011/09/linear_algebra_from_test2.docx" rel="noreferrer">http://sites.uci.edu/gremath/files/2011/09/linear_algebra_from_test2.docx</a></p>
<p><a href="http://sites.uci.edu/gremath/files/2013/08/chapter-1.pdf" rel="noreferrer">http://sites.uci.edu/gremath/files/2013/08/chapter-1.pdf</a>
<a href="http://sites.uci.edu/gremath/files/2013/08/chapter-2.pdf" rel="noreferrer">http://sites.uci.edu/gremath/files/2013/08/chapter-2.pdf</a>
<a href="http://sites.uci.edu/gremath/files/2013/08/chapter-3.pdf" rel="noreferrer">http://sites.uci.edu/gremath/files/2013/08/chapter-3.pdf</a>
<a href="http://sites.uci.edu/gremath/files/2013/08/chapter-4.pdf" rel="noreferrer">http://sites.uci.edu/gremath/files/2013/08/chapter-4.pdf</a>
<a href="http://sites.uci.edu/gremath/files/2013/08/chapter-5.pdf" rel="noreferrer">http://sites.uci.edu/gremath/files/2013/08/chapter-5.pdf</a>
<a href="http://sites.uci.edu/gremath/files/2013/08/chapter-6.pdf" rel="noreferrer">http://sites.uci.edu/gremath/files/2013/08/chapter-6.pdf</a>
<a href="http://sites.uci.edu/gremath/files/2013/08/Chapter-7.pdf" rel="noreferrer">http://sites.uci.edu/gremath/files/2013/08/Chapter-7.pdf</a></p>
<hr />
<h2>Web resources (Advice)</h2>
<p>GRE advice&strategy - be realistic</p>
<p>Post by j0equ1nn » Sat Apr 04, 2009 4:08 pm</p>
<p><a href="http://www.mathematicsgre.com/viewtopic.php?f=1&t=217" rel="noreferrer">http://www.mathematicsgre.com/viewtopic.php?f=1&t=217</a></p>
<p>Maths SE questions</p>
<p><a href="https://math.stackexchange.com/questions/269549/recommending-books-for-gre-math-subject-test">recommending books for GRE math subject test</a></p>
<p><a href="https://math.stackexchange.com/questions/16514/math-gre-subject-exam">https://math.stackexchange.com/questions/16514/math-gre-subject-exam</a></p>
<p><a href="https://math.stackexchange.com/questions/807285/multivariable-calculus-for-gre">Multivariable Calculus for GRE</a></p>
<p><a href="https://math.stackexchange.com/questions/91170/books-to-study-for-math-gre-self-study-have-some-time">Books to study for Math GRE, self-study, have some time.</a></p>
<p><a href="https://math.stackexchange.com/questions/217697/gre-math-subject-test">GRE Math Subject Test</a></p>
<p><a href="https://math.stackexchange.com/questions/791162/ultimate-gre-prep">Ultimate GRE Prep</a></p>
<hr />
<h2>Maths SE Tags:</h2>
<p><a href="https://math.stackexchange.com/questions/tagged/gre-exam">https://math.stackexchange.com/questions/tagged/gre-exam</a></p>
<p><a href="http://academia.stackexchange.com/questions/tagged/gre">http://academia.stackexchange.com/questions/tagged/gre</a></p>
<p><a href="https://math.stackexchange.com/search?q=gre">https://math.stackexchange.com/search?q=gre</a></p>
<hr />
<h2>Resources suggested to me:</h2>
<ul>
<li><p><a href="http://academicpublishers.in/site/book_details/competitive-mathematics/mathematics-for-competitive-examinations" rel="noreferrer">http://academicpublishers.in/site/book_details/competitive-mathematics/mathematics-for-competitive-examinations</a></p>
</li>
<li><p><a href="https://mualphatheta.org/past_tests#tests2018" rel="noreferrer">https://mualphatheta.org/past_tests#tests2018</a></p>
</li>
<li><p><a href="http://scherk.pbworks.com/w/page/14864231/Quiz%3A%20Linear%20algebra" rel="noreferrer">http://scherk.pbworks.com/w/page/14864231/Quiz%3A%20Linear%20algebra</a></p>
</li>
</ul>
|
1,903,762 | <blockquote>
<p>Largest number that leaves same remainder while dividing 5958, 5430 and 5814 ?</p>
</blockquote>
<hr>
<p>$$5958 \equiv 5430 \equiv 5814 \pmod x$$
$$3\times 17\times 19 \equiv 5\times 181\equiv 3\times 331\pmod x$$
$$969 \equiv 905\equiv 993\pmod x$$</p>
<p>After a bit of playing with the calculator,
I think the answer is $48$ but I don't know how to prove it. </p>
<p>Sorry if the answer is too obvious, I am still trying to wrap my head modular arithmetic and not very successful yet.It would be great help if anybody would give me some hints on how to proceed ahead. Thanks. $\ddot \smile$ </p>
| B. Goddard | 362,009 | <p>Since $5958 \equiv 5430 \mod{x}$, then $x | 5958 -5430 = 528$. Likewise $x |
5958 - 5814 = 144$ and $x | 5814 - 5430 = 384$. So $x$ is a common divisor of
$528,$ $144,$ and $384$. So you want $x=\gcd(528, 144, 384) = 48$.</p>
|
6,534 | <p>I apologize if this isn't the right place to ask this question.</p>
<p>Two features of stackexchange would be very useful for a personal math blog -- Latex works great, and comments and replies can be voted upon.</p>
<p>Is there any way to use the stackexchange functionality in a personal math blog?</p>
| GNUSupporter 8964民主女神 地下教會 | 290,189 | <h1>Short answer</h1>
<p>View my profile for a quick start, and <a href="https://vincenttam.gitlab.io/bhdemo" rel="nofollow noreferrer">my minimal demo site</a> (<a href="https://gitlab.com/vincenttam/bhdemo" rel="nofollow noreferrer">source</a>).</p>
<h1>Complete answer</h1>
<p>I recommend a <em>static blog</em> powered by open-source technologies <a href="https://themes.gohugo.io/beautifulhugo/" rel="nofollow noreferrer">Beautiful Hugo</a> with <a href="https://staticman.net" rel="nofollow noreferrer">Staticman</a>. The former provides a blogging theme with <a href="https://katex.org" rel="nofollow noreferrer">KaTeX</a>, whereas the later provides commenting support for static sites.</p>
<p>To <strong>start writing math online</strong> using Beautiful Hugo <em>without any installation</em>, you may</p>
<h2>Quick start</h2>
<ol>
<li>Clone <a href="https://gitlab.com/pages/hugo" rel="nofollow noreferrer">GitLab's sample</a></li>
<li>Follow the README from <a href="https://gitlab.com/pages/hugo#gitlab-user-or-group-pages" rel="nofollow noreferrer">GitLab User or Group Pages</a>. (<em>Skip</em> the first two sections unless you want to preview your posts before publishing.)</li>
<li>(<em>Optional</em>: commenting support) <strong>Clone <a href="https://gitlab.com/vincenttam/bhdemo" rel="nofollow noreferrer">my demo GitLab site</a> instead</strong>, and follow my guide in <a href="https://math.stackexchange.com/users/290189/gnusupporter-8964%E6%B0%91%E4%B8%BB%E5%A5%B3%E7%A5%9E-%E5%9C%B0%E4%B8%8B%E6%95%99%E6%9C%83?tab=profile">my profile</a> <em>from step 2</em>.</li>
</ol>
<p><img src="https://i.stack.imgur.com/wMOlt.png" alt="sample Hugo site" /><br />
Figure 1: A <a href="https://jwinternheimer.github.io/post/2017-03-05-math-sample/" rel="nofollow noreferrer">data analysis blog</a> showing math equations.</p>
<h2>Technical specifications</h2>
<h3>Abstract comparisons</h3>
<ol>
<li>Advantages of static sites over dynamic sites (e.g. WordPress): the former concentrates on content delivery (the case of a personal blog) <em>without</em> handling logic from clients' requests (counterexample: library catalog search). Unlike dynamic web servers, static ones won't slow down drastically upon massive requests.
Therefore, many data analysts prefer static pages, say <a href="https://www.datascienceblog.net/post/other/staticman_comments/" rel="nofollow noreferrer">https://www.datascienceblog.net/post/other/staticman_comments/</a> Further reading: <a href="https://www.netlify.com/blog/2016/05/18/9-reasons-your-site-should-be-static/" rel="nofollow noreferrer">https://www.netlify.com/blog/2016/05/18/9-reasons-your-site-should-be-static/</a></li>
<li>Advantages of static comments over dynamic comments (e.g. WordPress) and/or comments offload (e.g. Disqus): the former allows search engines to grasp the <em>whole</em> page as site content <em>without</em> an extra database. This boosts both load speed and SEO. A WordPress site requires a database, which bears a considerable maintenance cost (for the service provider and/or the end users). You may see <a href="https://irz.fr/wordpress-jekyll" rel="nofollow noreferrer">IRZ's article about migration from WordPress</a> (in French) for an estimate of the cost of a medium WordPress site.</li>
<li>Advantages of free and open source softwares over proprietary technologies: there are plenty of discussions about this. Long story short, you can avoid being locked by proprietary licenses by switching to open source technologies. As a short example, imagine what you file would become if you saved your work in Microsoft Word <em>ten years ago</em>.</li>
<li>Advantages of decentralized technologies over central ones: in some countries, some large web sites (e.g. Google) are <em>banned</em>. Despite my support of <a href="https://stackoverflow.com/users/895245/ciro-santilli-%E6%96%B0%E7%96%86%E6%94%B9%E9%80%A0%E4%B8%AD%E5%BF%83-%E5%85%AD%E5%9B%9B%E4%BA%8B%E4%BB%B6-%E6%B3%95%E8%BD%AE%E5%8A%9F">Ciro Santilli 新疆改造中心 六四事件 法轮功</a>'s <a href="https://github.com/cirosantilli/china-dictatorship/blob/master/FAQ.adoc" rel="nofollow noreferrer">attack on SE</a>, one has to wait for an indefinitely long period for its victory. Therefore, technologies allowing the use of custom domains are preferred.</li>
</ol>
<h3>Specific comparisons</h3>
<ol>
<li>Hugo vs Jekyll: <a href="https://gohugo.io" rel="nofollow noreferrer">Hugo</a> is a static blog renderer written in <a href="https://golang.org/" rel="nofollow noreferrer">Go</a>, which is one of the fastest programming language developed by Google. I had been a <a href="https://jekyllrb.com" rel="nofollow noreferrer">Jekyll</a> user for three years. As my old blog grew to over three hundred articles, it took me more than a minute for site regeneration due to the speed of Ruby.</li>
<li>KaTeX vs MathJax: personally, speed is king . To see this, load this site's <a href="https://math.meta.stackexchange.com/q/5020">MathJax tutorial</a>. The former is less mature in terms of <span class="math-container">$\rm \LaTeX$</span> syntax support, but it should cover the majority of this site's <span class="math-container">$\rm \LaTeX$</span> code. (known exceptions: AMS-CD)</li>
<li>Staticman vs Disqus: apart from point (2) in the previous section, the later has <em>neither</em> Markdown <em>nor</em> math support. Although the main Staticman API suffers from <a href="https://github.com/eduardoboucas/staticman/issues/227" rel="nofollow noreferrer">an error lasting for several months</a>, this can be overcome by following the next point.</li>
<li>GitLab vs GitHub: apart from point (3) in the above section, the former offers a convenient all-in-one CD/CI service. This allows users to remotely run scripts (i.e. build sites) <em>without</em> installation. Using the later, one either use other third-party CI/CD service or put the generated HTML code in a separate orphan branch. The first option is less convenient, whereas the second option implements a <em>wrong</em> Git model. Git only manages <em>source code, not</em> data, binaries nor generated code.</li>
</ol>
<h2>Variations</h2>
<p>Since <a href="https://staticman.net" rel="nofollow noreferrer">Staticman</a> is an open-source API service converting HTML form data into Git commits to be pushed to the remote Git repository, users can incorporate it with <em>any</em> (or even <a href="https://github.com/sdessus/comments_blog_tdm" rel="nofollow noreferrer">none</a>) static site generator.</p>
<h3>Jekyll + Staticman v3</h3>
<p>Despite Hugo's faster site regeneration, some users might want to stick with <a href="https://jekyllrb.com" rel="nofollow noreferrer">Jekyll</a>. Here're two examples.</p>
<ol>
<li>GitHub Pages: <a href="https://github.com/daattali/beautiful-jekyll#features" rel="nofollow noreferrer">https://github.com/daattali/beautiful-jekyll#features</a></li>
<li>GitLab Pages: <a href="https://staticman-gitlab-pages.frama.io/jekyll/" rel="nofollow noreferrer">https://staticman-gitlab-pages.frama.io/jekyll/</a></li>
</ol>
<h3>Use Framagit instead of GitLab.com</h3>
<p><strong>Framagit generates pages in <a href="https://framagit.org/staticman-gitlab-pages/jekyll/-/jobs/" rel="nofollow noreferrer">half</a> of the time of <a href="https://gitlab.com/VincentTam/gl-jekyll/-/jobs" rel="nofollow noreferrer">GitLab.com</a>.</strong></p>
<p>Since <a href="https://framagit.org" rel="nofollow noreferrer">Framagit</a> is a self-hosted GitLab instance, the setup follows the <em>same</em> logic, except that you have to use a <em>different</em> GitLab bot, Heroku app name and specify your API parameter to be <code>GITLAB_BASE_URL = "https://framagit.org/"</code>.</p>
<p>Note: It's suggested that every Staticman user use his/her own API instance.</p>
|
346,855 | <p>Let's start with the text of the assignment:</p>
<p><em>Let $V$ be a (finite-dimensional) vector space, and let $L<V$ be a subspace of $V$. Prove that $[a] = a + L$ holds for every vector $a\in{V}$.</em></p>
<p>(where $[S]$ in general is the set of all possible linear combinations of vectors from $S$, a.k.a. the <strong>span</strong> of $S$.It can also be a single vector. In that case, you can just put the vector inside brackets instead of the set.)</p>
<p>So, what causes me problems it that it seems to me that this is simply not true. I will show you my reasoning (what I think is a counterexample) and I hope you could help me understand it!</p>
<p>Let's take $L=V^2$ and $V=V^3$. So in that case, we can take $a=\vec{i}$.</p>
<p>In that case, $[a]=[\vec{i}]$, which is a set of all possible linear combinations of $\vec{i}$ over the field $\mathbb{R}$. Clearly, this can only define a <em>line</em> in either $V^2$ or $V^3$, so it cannot be the same as $a+L$, i.e. $[\vec{i} \cup V^2]$, because that would cover the whole $V^2$, not just a line.</p>
<p>Am I missing something here?</p>
<p>P.S.
If you didn't catch it, the plus sign used here is a sum of vector spaces, i.e. for some vector spaces $L$ and $M$, $L+M=[L \cup M]:=\left\{a + b | a \in L, b \in M\right\}$</p>
| lel | 60,013 | <p>To answer my own question... it was a misunderstanding after all. My reasoning is correct but the $[]$ brackets were actually representing simply the equivalence class for a relation mentioned earlier, but I missed it!</p>
|
2,960,054 | <p>We are told to proof the Minkowski inequality specialized to sequences:</p>
<p><span class="math-container">$$
\left(\sum_{i=1}^n|a_i+b_i|^p\right)^{1/p}\leq
\left(\sum_{i=1}^n|a_i|^p\right)^{1/p}
+\left(\sum_{i=1}^n|b_i|^p\right)^{1/p}
$$</span></p>
<p>when <span class="math-container">$p\geq1$</span> </p>
<p>Well, the problem is that all the proofs I have found "use" the Holder's inequality to proof that, but we haven't seen it in class yet so I can't "use" it to proof the Minkowski inequality.</p>
<p>The proof must somehow "elementary", with things we have already learnt in class. "Cauchy-Schwartz inequality" and the "Triangle inequality"are two of the inequalities we have learnt, for example.</p>
<p>Could someone help me? (and the rest of the students in my class, who are in the same situation).</p>
<p>Thanks in advance!</p>
| River Li | 584,414 | <p>Denote
<span class="math-container">$$S := \sum_{i=1}^n |a_i|^p, \quad \mathrm{and}\quad T := \sum_{i=1}^n |b_i|^p.$$</span></p>
<p>If <span class="math-container">$S = 0$</span> or <span class="math-container">$T = 0$</span>, the desired inequality is true.
In the following, assume that <span class="math-container">$S, T > 0$</span>.</p>
<p>Let
<span class="math-container">$$t_0 = \frac{S^{1/p}}{S^{1/p} + T^{1/p}}.$$</span>
Then <span class="math-container">$0 < t_0 < 1$</span>. We have
<span class="math-container">\begin{align*}
(S^{1/p} + T^{1/p})^p
&= \frac{S^{1/p} + T^{1/p}}{(S^{1/p} + T^{1/p})^{1 - p}}\\
&= \frac{S^{1/p}}{(S^{1/p} + T^{1/p})^{1 - p}} + \frac{ T^{1/p}}{(S^{1/p} + T^{1/p})^{1 - p}}\\
&= t_0^{1-p} S
+ (1 - t_0)^{1-p}T\\
&= \sum_{i=1}^n (t_0^{1-p}|a_i|^p + (1 - t_0)^{1-p}|b_i|^p)\\
&= \sum_{i=1}^n \left[t_0\left(\frac{|a_i|}{t_0}\right)^p + (1 - t_0)\left(\frac{|b_i|}{1 - t_0}\right)^p\right]\\\
&\ge \sum_{i=1}^n \left(t_0 \cdot \frac{|a_i|}{t_0} + (1 - t_0)\cdot \frac{|b_i|}{1 - t_0} \right)^p \tag{1}\\
&= \sum_{i=1}^n (|a_i| + |b_i|)^p\\
&\ge \sum_{i=1}^n |a_i + b_i|^p
\end{align*}</span>
where in (1) we have used Jensen's inequality
for convex function <span class="math-container">$g(u) = u^p$</span> on <span class="math-container">$u\ge 0$</span>.</p>
<p>We are done.</p>
<hr>
<p>If Jensen inequality is not allowed, one may prove the following lemma.</p>
<p><em>Lemma</em>. Let <span class="math-container">$1 \le p < \infty$</span> and <span class="math-container">$u, v \ge 0$</span>. Then, for all <span class="math-container">$t \in (0, 1)$</span>,
<span class="math-container">$$(u + v)^p \le t^{1 - p} u^p + (1 - t)^{1 - p}v^p.$$</span>
(<em>Hint</em>: The nontrivial case is <span class="math-container">$p > 1$</span> and <span class="math-container">$u, v > 0$</span>. Let <span class="math-container">$f(t) := t^{1 - p} u^p + (1 - t)^{1 - p}v^p$</span>. Then <span class="math-container">$f'(t) = (1 - p)[t^{-p}u^p - (1 - t)^{-p}v^p]$</span>. Then <span class="math-container">$f(t)$</span> has the global minimum at <span class="math-container">$t = u/(u+v)$</span>.)</p>
<p><em>Reference</em>.</p>
<p>[1] Heinz König, “A simple proof of the Minkowski inequality,” 1990.</p>
|
9,304 | <p>I have the following question: I have a file that has structure:</p>
<pre><code>x1 y1 z1 f1
x2 y2 z2 f2
...
xn yn zn fn
</code></pre>
<p>I can easily visualize it with <em>Mathematica</em> using <code>ListContourPlot3D</code>. But could you please tell me how I can plot contour plot for this surface? I mean with these data I have a set of surfaces corresponding to different isovalues (f) and I want to plot intersection between all these surfaces and some certain plane. I tried to Google but didn't get any results. Any help and suggestions are really appreciated.
Thanks in advance!</p>
| halirutan | 187 | <p>Ok, lets give this a try. @Mr.Wizard already showed you, how you can use <code>Interpolate</code> to make a function from your discrete data and since you didn't provide some test-data, I just assume we are speaking of an isosurface of a function $f(x,y,z)=c$ which is defined in some box in $\mathbb{R}^3$. </p>
<p>For testing we use $$f(x,y,z) = x^3+y^2-z^2\;\;\mathrm{and}\;\; -2\leq x,y,z \leq 2$$ which accidently happens to be the first example of <code>ContourPlot3D</code>. </p>
<p>The idea behind the following is pretty easy: As you may know from school, there is a simple representation of a plane in 3d which uses a point vector $p_0$ and two direction vectors $v_1$ and $v_2$. Every point on this plane can be reached through the $(s,t)$ parametrization</p>
<p>$$p(s,t)=p_0+s\cdot v_1+t\cdot v_2$$</p>
<p>Please note that $p_0, p, v_1, v_2$ are vectors in 3d and $s,t$ are scalars. The other form which we will use only for illustration is called the <em>normal form</em> of a plane. It is give by</p>
<p>$$n\cdot (p-p_0)=0$$</p>
<p>where $n$ is the vector normal to the plane, which can easily be calculated with the cross-product by $v_1\times v_2$. Lets start by looking at our example. To draw the plane inside <code>ContourPlot3D</code> we use the normal form which is <code>plane2</code>:</p>
<pre><code>f[{x_, y_, z_}] := x^3 + y^2 - z^2;
v1 = {1, 1, 0};
v2 = {0, 0, 1};
p0 = {0, 0, 0};
plane1 = p0 + s*v1 + t*v2;
plane2 = Cross[v1, v2].({x, y, z} - p0);
gr3d = ContourPlot3D[{f[{x, y, z}], plane2}, {x, -2, 2}, {y, -2, 2}, {z, -2, 2},
Contours -> {0},
ContourStyle -> {ColorData[22, 3], Directive[Opacity[0.5], ColorData[22, 4]]}]
</code></pre>
<p><img src="https://i.stack.imgur.com/ExNvT.png" alt="Mathematica graphics"></p>
<p>What we do now is, that we try to find the contour value (which is 0 here) of $f(x,y,z)$ for all points, that lie on our plane. This is like doing a normal <code>ContourPlot</code> because our plane is 2d (although placed in 3d space). Therefore, we use the 2d to 3d mapping of <code>plane1</code></p>
<pre><code>gr2d = ContourPlot[f[plane1], {s, -2, 2}, {t, -2, 2}, Contours -> {0},
ContourShading -> None, ContourStyle -> {ColorData[22, 1], Thick}]
</code></pre>
<p><img src="https://i.stack.imgur.com/ymazK.png" alt="Mathematica graphics"></p>
<p>Look at the intersection. It is exactly the loop we would have expected from the 3d illustration. Now you could object something like "ew.. but I really like a curve in 3d..". Again, the mapping from this 2d curve to 3d is given in the plane equation. You can simply extract the <code>Line[..]</code> directives from the above plot and transfer it back to 3d:</p>
<pre><code>Show[{gr3d,
Graphics3D[{Red, Cases[Normal[gr2d], Line[__], Infinity] /.
Line[pts_] :> Tube[p0 + #1*v1 + #2*v2 & @@@ pts, .05]}]
}]
</code></pre>
<p>I extract the <code>Line</code>s with <code>Cases</code> and then use the exact same definition of <code>plane1</code> as pure function to transform the <code>pts</code>. </p>
<p><img src="https://i.stack.imgur.com/HBSIm.png" alt="Mathematica graphics"></p>
<p>When I'm not completely wasted at 5:41 in the morning, than this approach should work for your interpolated data too.</p>
<h2>Apply method on test-data</h2>
<p>I uploaded your test-data to our Git-repository and therefore, the code below should work without downloading anything. The approach is the same as above but some small things have changed, since we work on interpolated data now. I'll explain only the differences.</p>
<p>First we import the data and since we have a long list of <code>{x,y,z,f}</code> values, we transform them to <code>{{x,y,z},f}</code> as required by the <code>Interpolation</code> function. I'm not using the interpolation-function directly. I wrap a kind of protection around it which tests whether a given <code>{x,y,z}</code> is numeric and lies inside the interpolation box. Otherwise I just return 0.</p>
<pre><code>data = {Most[#], Last[#]} & /@
Import["https://raw.github.com/stackmma/Attachments/master/data_9304_187.m"];
ip = Interpolation[data];
fip[{x_?NumericQ, y_?NumericQ, z_?NumericQ}] :=
If[Apply[And, #2[[1]] < #1 < #2[[2]] & @@@
Transpose[{{x, y, z}, First[ip]}]],
ip[x, y, z], 0.0]
</code></pre>
<p>The code below is almost the same. I only adapted the plane that it goes through your interpolation box. Furthermore, if you inspect your data you see that the value run from 0 to 1.2. Therefore I'm plotting the 0.5 contour by subtracting 0.5 from the function value and using <code>Contours -> {0}</code>. Remember that when I would simply plot the 0.5 contour, it would draw me a different plane, since we use one combined <code>ContourPlot3D</code> call.</p>
<p>Furthermore, notice that I normalized the direction vectors of the plane. This makes it easier to adjust the 2d plot of the contour. The rest should be the same.</p>
<pre><code>v1 = Normalize[{30, 30, 0}];
v2 = Normalize[{0, 0, 21}];
p0 = {26, 26, 17};
plane1 = p0 + s*v1 + t*v2;
plane2 = Cross[v1, v2].({x, y, z} - p0);
gr3d = ContourPlot3D[{fip[{x, y, z}] - 0.5, plane2}, {x, 27, 30}, {y,
27, 30}, {z, 17.3, 21}, Contours -> {0},
ContourStyle -> {Directive[Opacity[.5], ColorData[22, 3]],
Directive[Opacity[.8], ColorData[22, 5]]}]
gr2d = ContourPlot[fip[plane1] - 0.5, {s, 2, 5}, {t, 1, 4},
Contours -> {0}, ContourShading -> None,
ContourStyle -> {ColorData[22, 1], Thick}];
Show[{gr3d,
Graphics3D[{Red,
Cases[Normal[gr2d], Line[__], Infinity] /.
Line[pts_] :> Tube[p0 + #1*v1 + #2*v2 & @@@ pts, .05]}]}]
</code></pre>
<p><img src="https://i.stack.imgur.com/T9WjZ.png" alt="Mathematica graphics"></p>
<p>As you can see, your sphere has a whole inside.</p>
<p><img src="https://i.stack.imgur.com/IPLSw.png" alt="Mathematica graphics"></p>
|
2,909,036 | <p>I'm reading a book and there is some strange proof (strange for me) of the theorem that within each interval, no matter how small, there are rational points. Proof: we need only take a denominator $n$ large enough so that the interval $[0, \frac{1}{n}]$ is smaller than the interval $[A, B]$ in question; then at least one of the fractions $\frac{m}{n}$ must lie within the interval.</p>
<p>I can't understand how they figured out that $[A, B]$ contains at least one $\frac{m}{n}$ and how it's connected to the interval $[0, \frac{1}{n}]$.</p>
| hamam_Abdallah | 369,188 | <p>$\Bbb R$ is Archimedian thus</p>
<p>$$\exists n>0 \; : \; n(B-A)>1$$</p>
<p>therefore</p>
<p>$$nA<nA+1<nB$$</p>
<p>Let $m=\lfloor nA+1\rfloor$
then</p>
<p>$$nA<m\le nA+1<nB$$</p>
<p>finally</p>
<p>$$A<\frac{m}{n}<B$$</p>
|
235,145 | <p>I have the following 3D-model:</p>
<pre><code>RevolutionPlot3D[Sqrt[E^-x (1 + E^x)^2], {x, 0, 4},
RevolutionAxis -> {1, 0, 0}]
</code></pre>
<p>Is there a way to make this model into a solid so that I can export it to let it be printed by a 3D printer. Now it is kind of hollow from the inside but I want it to be filled.</p>
| cvgmt | 72,111 | <p>Revolve three parametric curves. <code>{x, f[x]}</code> , <code>{0, y}</code> and <code>{4, y}</code></p>
<pre><code>SetOptions[RevolutionPlot3D, Mesh -> False];
f[x_] := Sqrt[E^-x (1 + E^x)^2];
a = RevolutionPlot3D[{x, f[x]}, {x, 0, 4},
RevolutionAxis -> {1, 0, 0}, PlotStyle -> Red];
b = RevolutionPlot3D[{0, y}, {y, 0, f[0]},
RevolutionAxis -> {1, 0, 0}, PlotStyle -> Yellow];
c = RevolutionPlot3D[{4, y}, {y, 0, f[4]},
RevolutionAxis -> {1, 0, 0}, PlotStyle -> Cyan];
Show[a, b, c, Boxed -> False, Axes -> False,
ViewPoint -> {0.76, -1.39, 2.98}]
</code></pre>
<p><a href="https://i.stack.imgur.com/fmRu0.png" rel="noreferrer"><img src="https://i.stack.imgur.com/fmRu0.png" alt="enter image description here" /></a></p>
|
105,158 | <p>Wigner's D-matrices is defined as $D_{m'm}^j(\phi,\theta,\psi)=\langle jm'|R(\phi,\theta,\psi)|jm\rangle$; it produces a square matrix (indices $m$ and $m'$) of dimension $2j+1$. It is also the case that these matrices (for all positive $j$ multiple of $1/2$) are a representation of the rotation group $SU(2)$, which is a double cover of $SO(3)$. </p>
<p>Now, I can see this for the specific case $j=1$ (3x3 matrices representing rotations in 3-dimensional Euclidean space), but what does it mean for other values of $j$? Say for $j=3/2$, does that represent the subgroup of 3-dimensional rotations given some parameter, in a 4-dimensional ambient geometry?</p>
<p>Thanks.</p>
| J. M. ain't a mathematician | 498 | <p>For $n=-1,0$, finding the roots of the spherical Bessel functions $j_n(x)$ and $y_n(x)$ is somewhat easy, since:</p>
<p>$$\begin{array}{ll}
j_{-1}(x)&=&\frac{\cos\,x}{x}&\quad&y_{-1}(x)&=&\mathrm{sinc}(x)\\
j_0(x)&=&\mathrm{sinc}(x)&\quad&y_0(x)&=&-\frac{\cos\,x}{x}\\
\end{array}$$</p>
<p>where $\mathrm{sinc}(x)=\dfrac{\sin\,x}{x}$ is the sine cardinal. Solving for zeros of other orders results in rather complicated transcendental equations, which I doubt have closed-form solutions. However, you will want to see these <a href="http://dlmf.nist.gov/10.21">DLMF</a> <a href="http://dlmf.nist.gov/10.58">entries</a> for some more information that can help you in numerically determining the zeros (e.g., asymptotic expansions); approximations derived from formulae there can then be subsequently polished with Newton-Raphson or some other iterative method of choice.</p>
|
219,165 | <p>I have the following equation which I want to solve it for r:</p>
<pre><code>TGB5 = (-((2 q^2)/r^5) + 2/r + (16 P \[Pi] r)/3)/(4 \[Pi]);
rlarge5 = Last[r /. Solve[TGB5 == T, r, Reals]] // Normal
</code></pre>
<p>The answer is a root object:</p>
<blockquote>
<p>Root[-3 q^2 + 3 #1^4 - 6 [Pi] T #1^5 + 8 P [Pi] #1^6 &, 4]</p>
</blockquote>
<p>Now, this function of r should be expanded about T=Infinity. But Mathematica gives the error:</p>
<pre><code>rser5 = Series[rlarge5, {T, Infinity, 4}] // Normal
</code></pre>
<blockquote>
<p>Series::nmer: Root[-3 q^2+3 #1^4-6 [Pi] T #1^5+8 P [Pi] #1^6&,4] is not a meromorphic function of T at [Infinity].</p>
</blockquote>
<p>Is there anyway to expand this root object?</p>
| Kheeyal | 1,894 | <p>In Mathematica 12.1, the following code gives the right answer:</p>
<pre><code>r /. Last[AsymptoticSolve[TGB5 == T, r, {T, Infinity, 8}]] // Expand
</code></pre>
<p>which is:</p>
<pre><code> -((4 P^2)/(9 \[Pi]^3 T^5)) + (128 P^4 q^2)/(81 \[Pi] T^5) - P/(
3 \[Pi]^2 T^3) - 1/(2 \[Pi] T) + (3 T)/(4 P)
</code></pre>
<p>I do not know why the answer from <strong>AsymptoticSolve</strong> command is too far from what the <strong>Series</strong> command computes (Please see Hanlon answer.). Does any one know? </p>
|
252,049 | <p>I have seen on the nLab that we can <a href="https://ncatlab.org/nlab/show/loop+space+object" rel="nofollow">view the loop space as a particular homotopy pullback</a>, and that it is even the way a "loop space object" is defined in general (when it exists). </p>
<p>Can someone give me some intuition of why it is so?
To which construction in geometry does this correspond (up to homotopy)? </p>
<p>Any reference would also be welcome.</p>
| Emily | 130,058 | <p>Here are some asides on my other answer.</p>
<p>On general loop space objects: See the discussion around (specially above) [Remark 1.1.2.6, <a href="http://www.math.harvard.edu/~lurie/papers/HA.pdf" rel="nofollow noreferrer">HA</a>] and [Section 2.3.2, <a href="https://drive.google.com/file/d/0ByJtAd7Hyh5XNGxCdVdiRlJfMnM/view" rel="nofollow noreferrer">Mauro Porta's master thesis</a>].</p>
<hr>
<p>Similarly, the suspension of a topological space <span class="math-container">$X$</span> is the homotopy <em>pushout</em> of the diagram:
<span class="math-container">$$*\hookleftarrow X\hookrightarrow *.$$</span></p>
<p>What follows is about the <em>unreduced</em> suspension of <span class="math-container">$X$</span>.</p>
<p>For unpointed topological spaces, we have the following formula<span class="math-container">$^2$</span> for homotopy pushouts of topological spaces:
<span class="math-container">$$\mathrm{hocolim}\left(X\xleftarrow{f}A\xrightarrow{g}Y\right)=\frac{X\coprod(A\times[0,1])\coprod Y}{\left((a,0)\sim f(a)\text{ and }(a,1)\sim g(a))\right)}.$$</span>
Picture from Dugger:</p>
<p><img src="https://i.stack.imgur.com/aC3xt.png" width="400"></p>
<p>(Unreduced!) suspension of <span class="math-container">$S^1$</span> from Wikipedia:</p>
<p><img src="https://upload.wikimedia.org/wikipedia/commons/c/c3/Suspension.svg" width="200"></p>
<hr>
<p><span class="math-container">$^2$</span>See Example 2.2 of <a href="https://pages.uoregon.edu/ddugger/hocolim.pdf" rel="nofollow noreferrer">Dugger's A Primer on Homotopy Colimits</a>.</p>
|
3,334,910 | <p>Question. <span class="math-container">$\square ABCD$</span> is a square with <span class="math-container">$AB = 10$</span>. Circle <span class="math-container">$O$</span> inscribes the <span class="math-container">$\square ABCD$</span>. The center of the arc is <span class="math-container">$A$</span>. What is the area of the colored area?</p>
<p><a href="https://i.stack.imgur.com/1N2eT.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1N2eT.png" alt="enter image description here"></a></p>
<p>Explanation: This problem can be solved by integral. But, this problem is from an elementary math book, which means that integral is not a good solution for this.</p>
<p>Our teacher solved this problem with integral, but she was not able to solve it with elementary math. She asked my school students about this problem, but none of us were able to solve it. How can we solve the problem using only elementary math?</p>
<p>Korean elementary math does not contain:</p>
<ol>
<li>irrationals and imaginary numbers</li>
<li>functions</li>
<li>Of course, angle functions (such as <span class="math-container">$sin$</span>, <span class="math-container">$cos$</span>, <span class="math-container">$tan$</span>, etc)</li>
<li>inscribed angle</li>
<li>non-linear equations</li>
<li>similarity and congruence</li>
</ol>
<p>If you are unsure that you can use certain formulas or something else, comment me and I'll answer you.</p>
<p><strong>Edit 1.</strong> Actually, we do not study <span class="math-container">$\pi$</span> in elementary school. We just assimilate it to <span class="math-container">$3.14$</span>. But it's not so important, so let's use <span class="math-container">$\pi$</span>.</p>
| Quanto | 686,284 | <p><a href="https://i.stack.imgur.com/Wxgzb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Wxgzb.png" alt="enter image description here"></a></p>
<p>Evaluate the areas of the circular sectors ABP and OPE, as well as the quadrilateral APOF and then subtract them from the half square to get the shaded area. (A unit square assumed below.)</p>
<p>With A being the origin, the two circles are</p>
<p><span class="math-container">$$r=1; \space \space r(\sin\theta + \cos\theta)=r^2+\frac{1}{4}.$$</span> </p>
<p>Eliminate <span class="math-container">$r$</span> to obtain,</p>
<p><span class="math-container">$$\sin\beta = \frac{5+\sqrt{7}}{8},\>\>\>\>\>
\sin\alpha = \frac{0.5-\cos\beta}{0.5}=\frac{\sqrt{7}-1}{4}$$</span></p>
<p>Then, the area of the quadrilateral AFOP which is made of two triangles AFP and FPO, is</p>
<p><span class="math-container">$$A_q = \frac{1}{4}\sin\beta + \frac{1}{4}\left(\frac{1}{2}-\cos\beta \right)=\frac{2+\sqrt{7}}{16}
$$</span></p>
<p>And the area of the two sectors is</p>
<p><span class="math-container">$$A_c=\frac{1}{2}\left( \frac{\pi}{2}-\beta\right) + \frac{\alpha}{8}
=\frac{\pi}{8}-\frac{1}{8}\sin^{-1} \left(\frac{23\sqrt{7}+67}{128}\right). $$</span></p>
<p>Therefore, the shaded area is</p>
<p><span class="math-container">$$A_s=\frac{1}{2}-A_c-A_q=\frac{6-\sqrt{7}-2\pi}{16}+\frac{1}{8}\sin^{-1} \left(\frac{23\sqrt{7}+67}{128}\right). $$</span></p>
|
114,278 | <p>If I wanted to show that a group of order $66$ has an element of order $33$, could I just say that it has an element of order $3$ (by Cauchy's theorem, since $3$ is a prime and $3 \mid 66$), and similarly that there must be an element of order $11$, and then multiply these to get an element of order $33$? I'm pretty sure this is wrong, but if someone could help me out I would appreciate it. Thanks. </p>
| Pete L. Clark | 299 | <p>Step 1: Show that any finite group order $4k+2$ has an index two subgroup. (<a href="https://math.stackexchange.com/questions/108796/group-of-order-4k2-prove-that-the-following-permutation-is-odd">Hint</a>.)</p>
<p>Step 2: Show that any group of order $33$ is cyclic. </p>
|
38,746 | <p>Divide $\mathbb{R}^2$ into disjoint unit squares, let Q be a tile (a polyomino, a finite set of connected unit squares). If, for every finite set S of unit squares in $\mathbb{R}^2$, - I can find a finite set of disjoint Q-tiles (a tiling) such that S is a subset of this tiling, how to prove that I can tile entire $\mathbb{R}^2$ with Q-tiles, (without axiom of choice)?</p>
| Alon Amit | 308 | <p>Consider a sequence $S_1, S_2, S_3, \ldots$ of finite regions that exhaust the plane, namely $\cup S_i = \mathbb{R}^2$. For instance, you may take $S_i$ to be the square of edge side $2i+1$ centered at the origin.</p>
<p>Each $S_i$ is coverable with copies of your tile $Q$. Define a tree as follows: The vertices at level $i$ are all the possible coverings of $S_i$, and a vertex at level $i$ is a connected with a vertex at level $i+1$ iff the latter is an extension of the former - namely, a covering of $S_{i+1}$ that contains the covering of $S_i$ and possibly adds a finite number of additional tiles. (to give our tree a root we can just add a vertex, representing "S_0", connected to all the ways to position a single tile $Q$ over $S_1$). </p>
<p>It should be easy to prove that this is indeed an infinite tree. By Kőnig's lemma it contains an infinite branch which defines a tiling of the while plane. Kőnig's lemma is equivalent to a weak form of AC, probably countable-AC or something similar; I don't think you can avoid some form of choice or dependent choice for your proof. </p>
|
38,746 | <p>Divide $\mathbb{R}^2$ into disjoint unit squares, let Q be a tile (a polyomino, a finite set of connected unit squares). If, for every finite set S of unit squares in $\mathbb{R}^2$, - I can find a finite set of disjoint Q-tiles (a tiling) such that S is a subset of this tiling, how to prove that I can tile entire $\mathbb{R}^2$ with Q-tiles, (without axiom of choice)?</p>
| JDH | 413 | <p>I know three different proofs of this.</p>
<ul>
<li><p>First, you can appeal to König's lemma. View the set
of tilings of increasingly large squares based at the
origin (that is, they minimally cover such a square) as a
tree under the subtiling relation. This is a finitely
branching tree, since any tiling of a square has only
finitely many extensions to cover the next larger square.
But your assumption says that this tree is infinite. Thus,
it has an infinite branch, by König's lemma. Such a
branch gives a tiling of the entire plane.</p></li>
<li><p>Second, you can view the previous argument as a
compactness argument, if you consider the right topology.</p></li>
<li><p>Third, one can argue from nonstandard analysis. Take a
nonstandard model of the natural numbers. By the transfer
principle, there must be a tiling of some nonstandard
size square using tiles from your tile set. But a nonstandard
(pseudo)finite square includes many actual copies of the
standard plane inside it, around the nonstandard numbers in the
very center of the nonstandard square. Thus, there is a
tiling of the actual standard plane.</p></li>
</ul>
|
849,486 | <p>Find all <em>distinct</em> integers $x$ and $y$ that satisfy the following equation.</p>
<p>$$
x\log y=y\log x.
$$</p>
<p>Obviously, if $x=y$, the equation is satisfied.
I found $x=2$, and $y=4$. I think we cannot find all solutions (if there are many).</p>
<p>P.S. The base of the $\log(\cdot)$ is not important.</p>
| Answer | 160,449 | <p>Here using some of the $\log$ laws we can find all solution classes.</p>
<p>$$a \log (b)=\log(b^a)$$ </p>
<p>and</p>
<p>$$e^{\log(x)}=x$$ shall be all that is required.</p>
<p>$$x\log y = y \log x$$</p>
<p>$$\log y^x = \log x^y$$
$$e^{\log (y^x)} = e^{\log (x^y)}$$
$$y^x = x^y$$</p>
<p>Once we have reached this point, we have a duplicate of <a href="https://math.stackexchange.com/questions/9505/xy-yx-for-integers-x-and-y">this question.</a></p>
|
849,486 | <p>Find all <em>distinct</em> integers $x$ and $y$ that satisfy the following equation.</p>
<p>$$
x\log y=y\log x.
$$</p>
<p>Obviously, if $x=y$, the equation is satisfied.
I found $x=2$, and $y=4$. I think we cannot find all solutions (if there are many).</p>
<p>P.S. The base of the $\log(\cdot)$ is not important.</p>
| puru | 151,916 | <p>We have $x^y = y^x$, with $0 < x < y$ (WLOG).</p>
<p>Let $y= rx$, where $r > 1$.
Then $x^{rx} = (rx)^x$.
Taking the natural logarithm of both sides, $rx \ln x = x (\ln r + \ln x)$.
Dividing by $x$, and rearranging, $(r−1) \ln x = \ln r.$
(Notice that $r = 1$ would be a solution, from which $y = x$.)
With $r > 1$, $\ln x = (\ln r) / (r−1).$
Taking exponentials,$ x = e^{(\ln r) / (r−1)} = (e^{\ln r})^{1/(r−1)}$.
Therefore we have the parametric solution, $x = r^{1/(r−1)}, y = rx = r^{r/(r−1)}.$</p>
<p>If we set $r = 1 + 1/k$, where $k > 0$, we have $x = (1 + 1/k)^k$, $y = (1 + 1/k)^{k+1}$.</p>
<p>Since $r > 1, 1/(r−1) $is positive, and so $x > 1$, and therefore $y > 1$.
(Notice that, if we allow $0 < r < 1, 1/(r−1) $and $r/(r−1)$ are negative, and so again $x$ and $y$ are greater than $1$. Of course, this must be the case, because $r = a$ and $r = 1/a$ yield essentially the same solution, with $x$ and $y$ swapped.)</p>
<p>We have shown that, for a rational solution, $k$ is an integer; that is $x = [(a + 1)/a]^a$.
If $a > 1$, $a$ and $a + 1$ are relatively prime, and so $[(a + 1)/a]^a$ is a fraction in its lowest terms, with denominator $> 1$, and therefore not an integer.
So the only integer solution is $x = 2, y = 4$.</p>
<p>Hope this helps. Correct me if I'm wrong! </p>
<p>You can refer <a href="http://functionspace.org/topic/2341" rel="nofollow">here</a> for more discussion on the same problem! </p>
|
4,127,468 | <p>Suppose we have two functions <span class="math-container">$f,g:\Bbb R\rightarrow \Bbb R$</span>. The chain rule states the following about the derivative of the composition of these functions, namely that
<span class="math-container">$$
(f \circ g)'(x) = f′(g(x))\cdot g′(x).
$$</span>
However, the equivalent expression using Leibniz notation seems to be saying something different. I know that <span class="math-container">$f'(g(x))$</span> means the derivative of <span class="math-container">$f$</span> evaluated at <span class="math-container">$g(x)$</span>, but when considering the Leibniz equivalent of the chain rule, it appears that it should really mean the derivative of <span class="math-container">$f$</span> with respect to <span class="math-container">$g(x)$</span>. If we let <span class="math-container">$z=f(y)$</span> and y=<span class="math-container">$g(x)$</span>, then
<span class="math-container">$$
{\frac {dz}{dx}}={\frac {dz}{dy}}\cdot {\frac {dy}{dx}}.
$$</span>
Where here the <span class="math-container">$\frac{dz}{dy}$</span> corresponds to <span class="math-container">$f'(g(x))$</span>. Since <span class="math-container">$y=g(x)$</span>, I am tempted to believe that the expression <span class="math-container">$f'(u)$</span> means the derivative of <span class="math-container">$f$</span> with respect to <span class="math-container">$u$</span>; it would make sense in this case as we are treating <span class="math-container">$g(x)$</span> as the independant variable. This leaves me with the question: does <span class="math-container">$f'(g(x))$</span> mean the derivative of <span class="math-container">$f$</span> evaluated at <span class="math-container">$g(x)$</span>, <span class="math-container">$\frac{df}{dx} \Bigr\rvert_{x = g(x)}$</span>, or the derivative of <span class="math-container">$f$</span> with respect to <span class="math-container">$g(x)$</span>, <span class="math-container">$\frac{df}{dg(x)}?$</span></p>
| littleO | 40,119 | <p>In my opinion the usual way of writing the chain rule in Leibniz notation is confusing and, I would say, bad. It's a frequent source of confusion on this website.</p>
<p>The function that is called <span class="math-container">$z$</span> on the left is not the same as the function that is called <span class="math-container">$z$</span> on the right. In other words, two different functions are being called by the same name. It would be better to give the function on the left its own name, such as <span class="math-container">$\hat z(x) = z(y(x))$</span>. Then, using Leibniz notation, the chain rule could be written as <span class="math-container">$\frac{d\hat z}{dx} = \frac{dz}{dy} \frac{dy}{dx}$</span>. This is still a little confusing: <span class="math-container">$\frac{dz}{dy}$</span> is to be interpreted as <span class="math-container">$z'(y(x))$</span>.</p>
<p>In my opinion the notation
<span class="math-container">$$\hat z'(x) = z'(y(x)) y'(x)$$</span>
is far more clear.</p>
<p>To specifically address the final part of your question: <span class="math-container">$f'(g(x))$</span> is the derivative of <span class="math-container">$f$</span> evaluated at <span class="math-container">$g(x)$</span>. I would not use the phrase "derivative of <span class="math-container">$f$</span> with respect to <span class="math-container">$g(x)$</span>".</p>
<hr />
<p><strong>Edit:</strong> Here is the thought process behind the Leibniz notation, and an explanation for why it has become so popular despite the fact that I think it's confusing.</p>
<p>Think about the quantity <span class="math-container">$z(y(x))$</span>, and imagine what happens if <span class="math-container">$x$</span> is perturbed by a small amount <span class="math-container">$\Delta x$</span>. Then the output of <span class="math-container">$y$</span> is perturbed by a small amount <span class="math-container">$\Delta y$</span>, and the output of <span class="math-container">$z$</span> is correspondingly perturbed by a small amount <span class="math-container">$\Delta z$</span>. And we have
<span class="math-container">$$
\frac{\Delta z}{\Delta x} = \frac{\Delta z}{\Delta y} \frac{\Delta y}{\Delta x}
$$</span>
The term on the left is approximately <span class="math-container">$\hat z'(x)$</span>, but you can see the temptation to call it <span class="math-container">$\frac{dz}{dx}$</span>. The term <span class="math-container">$\frac{\Delta z}{\Delta y}$</span> is approximately <span class="math-container">$z'(y(x))$</span>, but you can see the temptation to call it <span class="math-container">$\frac{dz}{dy}$</span>. And the term <span class="math-container">$\frac{\Delta y}{\Delta x}$</span> is approximately <span class="math-container">$y'(x)$</span>, and of course you see the temptation to call it <span class="math-container">$\frac{dy}{dx}$</span>.</p>
|
91,252 | <p>Let $v_1, \ldots, v_n$ be a set of vectors in a vector space $V$. Show that $v_1, \ldots, v_n$ is a basis of $V$ if and only if for any non-zero linear function $f$ on $V$ there is a vector $v$ in $\operatorname{span}(v_1, \ldots, v_n)$ such that $f(v) \neq 0$. </p>
<p>Suppose that $v_1, \ldots, v_n$ is not a basis of $V$. Then the complement $W$ of $\operatorname{span}(v_1, \ldots, v_n)$ in $V$ is not empty. Let $f$ be a function such that $f(x)=1$ for all $x\in W$ and $f(x)=0$ for all $x$ in $\operatorname{span}(v_1, \ldots, v_n)$. My question is: is $f$ linear in $V$?</p>
| Arturo Magidin | 742 | <p>As Henning points out, the statement is false as written; but if you add the assumption that $v_1,\ldots,v_n$ are linearly independent, then the result is true. </p>
<p>As to your idea, it's either completely wrong or <em>almost</em> right, depending on what you mean by "complement."</p>
<p>If by "complement" you mean the set-theoretic complement (that is, $W$ is the set of all vectors in $V$ that are not in $\mathrm{span}(v_1,\ldots,v_n)$), which was my interpretation when I read your post, then your approach is completely wrong: this set is not closed under sums, and in general you cannot expect it to be well-behaved.</p>
<p>If by "complement" you mean a <em>linear</em> complement (as lhs understood your writing), that is, a subspace $W$ such that $W\cap\mathrm{span}(v_1,\ldots,v_n)=\{\mathbf{0}\}$ and $W+\mathrm{span}(v_1,\ldots,v_n) = V$), then you are <em>almost</em> right: you cannot define $f$ as $1$ on <em>all</em> of $W$ (that would not be homogeneous or additive); but you <em>can</em> find a basis for $W$ (say, by extending $v_1,\ldots,v_n$ to a basis of $V$, $v_1,\ldots,v_n,v_{n+1},\ldots,v_{n+m}$ and letting $W=\mathrm{span}(v_{n+1},\ldots,v_{n+m})$), define $f$ as $1$ <em>on that basis</em>, and then "extending linearly". </p>
|
1,188,928 | <p>I know that on a smooth projective variety any coherent sheaf has a finite locally free resolution. I read somewhere that this implies that any object in $D^b(X)$ for $X$ smooth projective is then isomorphic in $D^b(X)$ to a complex of locally free objects. It seems to me it should be a proof by induction on the length of the complex, but I am pretty lost in approaching it.
Could you give me some (even very big) hint or reference for a proof?</p>
<p>What I can show is that every object is quasi-isomorphic to its resolution, since if we have
\begin{equation}
\ldots \rightarrow \mathcal{E}^1 \rightarrow \mathcal{E}^0 \xrightarrow{f} \mathcal{F} \rightarrow 0
\end{equation}
I can see $\mathcal{F}$ as a complex living just in degree 0, the resolution as a complex whose degree 0 is $\mathcal{E}^0$ and the quasi-isomorphism of complexes as being given by $f$. Although, I don't know how to generalize this process to complexes with arbitrary finite length.</p>
<p>Another thing I thought is that I might try to use the finite resoulutions of each sheaf in the complex and put them together in a suitable way. Is there any smart way to do it?</p>
| Jeremy Rickard | 88,262 | <p>Let
$$\mathcal{F}_\bullet=\dots\stackrel{d^\mathcal{F}_3}{\to}\mathcal{F}_2
\stackrel{d^\mathcal{F}_2}{\to}\mathcal{F}_1
\stackrel{d^\mathcal{F}_1}{\to}\mathcal{F}_0
\stackrel{d^\mathcal{F}_0}{\to}0\to\dots$$
be a bounded complex of coherent sheaves (assuming without loss of generality that $\mathcal{F}_i=0$ for $i<0$).</p>
<p>Then you can construct a locally free resolution
$$\begin{array}{ccccccccccc}
\dots&\stackrel{d^\mathcal{E}_3}{\to}&\mathcal{E}_2&
\stackrel{d^\mathcal{E}_2}{\to}&\mathcal{E}_1&
\stackrel{d^\mathcal{E}_1}{\to}&\mathcal{E}_0&
\stackrel{d^\mathcal{E}_0}{\to}&0\to&\dots\\
&&\downarrow&&\downarrow&&\downarrow&&\downarrow&\\
\dots&\stackrel{d^\mathcal{F}_3}{\to}&\mathcal{F}_2&
\stackrel{d^\mathcal{F}_2}{\to}&\mathcal{F}_1&
\stackrel{d^\mathcal{F}_1}{\to}&\mathcal{F}_0&
\stackrel{d^\mathcal{F}_0}{\to}&0\to&\dots
\end{array}$$
by starting with any epimorphism $\mathcal{E}_0\to\mathcal{F}_0$ from a locally free sheaf to $\mathcal{F}_0$, and once you've constructed $\mathcal{E}_i$ and all the relevant maps for $i<k$, taking $\mathcal{E}_k$ to be any locally free sheaf with an epimorphism to the pullback of the obvious diagram
$$\begin{array}{ccc}
&&\ker(d^\mathcal{E}_{k-1})\\
&&\downarrow\\
\mathcal{F}_k&\to&\ker(d^\mathcal{F}_{k-1})
\end{array}$$</p>
|
635,893 | <p>I am trying to prove following statement:</p>
<blockquote>
<p>$[m,n]$ is a set of functions defined as $f \in [m,n] \leftrightarrow f: \{1,...,m\} \rightarrow \{1,...,n\}$. The size of $[m,n]$ is $n^m$ for $m,n \in \mathbb{N}_{\gt0}$.</p>
</blockquote>
<p>I have tried to prove it but I am not entirely sure about its correctness:</p>
<p><strong>1)</strong> For the basic step $m=n=1$.</p>
<p>The size of $\{1\} \rightarrow \{1\}$ is $1$. And it equals $1^1 = 1$.</p>
<p><strong>2)</strong> Then I assume that for some $m,n$ the size $[m,n] = n^m$. Now comes the first problem: should I be proving it for $[m, n+1],[m+1,n],[m+1,n+1]$ or is some of it redundant?</p>
<p>When trying to prove $[m, n+1]$ I rewrite it as $[m,n+1] = (n+1) * (n+1) * (n+1) * ... * (n+1) = (n+1)^m$ but I don't use the induction assumption so is that correct?</p>
<p>Again $[m+1,n] = n*n*...*n = n^{m+1}$</p>
<p>Finally $[m+1,n+1] = (n+1)*(n+1)*...*(n+1) = (n+1)^{m+1}$.</p>
<p>During the process I didn't really used my induction assumption, so I am worried that this wouldn't qualify as a proof by induction. So what would be the correct way to prove this? </p>
| Carsten S | 90,962 | <p>HINT: If $\int_a^b f(x)\,dx\ne0$, then $\int_0^a f(x)\,dx\ne\int_0^b f(x)\,dx$.</p>
|
3,444,500 | <p>I need to prove that <span class="math-container">$3x^6+12x^5+9x^4-24x^3+9x^2-4x+3=0$</span> does not have any real root. I tried analyzing the derivatives to see the maxima and minima but I can't compute them exactly, so I couldn't proceed further. Hints are appreciated.</p>
| Axion004 | 258,202 | <p>One approach would be applying <a href="https://en.wikipedia.org/wiki/Sturm%27s_theorem" rel="nofollow noreferrer">Sturm's Theorem</a> to find the total number of real roots of the polynomial</p>
<p><span class="math-container">$$p(x)=3x^6+12x^5+9x^4-24x^3+9x^2-4x+3$$</span></p>
<p>Sturm's theorem expresses the number of distinct real roots of <span class="math-container">$p$</span> located in an interval in terms of the number of changes of signs of the values of the Sturm sequence at the bounds of the interval. Applied to the interval of all the real numbers, it gives the total number of real roots of <span class="math-container">$p$</span>.</p>
<p>The computation itself becomes slightly messy as you go further down the Sturm sequence. Although, it is still doable with the help of a Computer Algebraic System such as Mathematica. Suppose we wish to find the number of roots in some range for the polynomial</p>
<p><span class="math-container">$$p(x)=3x^6+12x^5+9x^4-24x^3+9x^2-4x+3$$</span></p>
<p>Let</p>
<p><span class="math-container">$$p_0(x)=3x^6+12x^5+9x^4-24x^3+9x^2-4x+3$$</span>
<span class="math-container">$$p_1(x)=p'(x)=18 x^5 + 60 x^4 + 36 x^3 - 72 x^2 + 18 x - 4$$</span>
The remainder of the Euclidean division of <span class="math-container">$p_0$</span> by <span class="math-container">$p_1$</span> is </p>
<p><span class="math-container">$$-\frac{11x^4}{3}-16x^3+14x^2-\frac{16x}{3}+\frac{31}{9}$$</span></p>
<p>multiplying by <span class="math-container">$-1$</span> we obtain</p>
<p><span class="math-container">$$p_2(x)=\frac{11x^4}{3}+16x^3-14x^2+\frac{16x}{3}-\frac{31}{9}$$</span></p>
<p>Next, the remainder of the Euclidean division of <span class="math-container">$p_1$</span> by <span class="math-container">$p_2$</span> is </p>
<p><span class="math-container">$$\frac{22464x^3}{121}-\frac{20448x^2}{121}+\frac{7488x}{121}-\frac{2592}{121}$$</span></p>
<p>multiplying by <span class="math-container">$-1$</span> we obtain</p>
<p><span class="math-container">$$p_3(x)=-\frac{22464x^3}{121}+\frac{20448x^2}{121}-\frac{7488x}{121}+\frac{2592}{121}$$</span></p>
<p>Next, the remainder of the Euclidean division of <span class="math-container">$p_2$</span> by <span class="math-container">$p_3$</span> is </p>
<p><span class="math-container">$$\frac{43439x^2}{18252}-\frac{242x}{351}-\frac{7381}{6084}$$</span></p>
<p>multiplying by <span class="math-container">$-1$</span> we obtain</p>
<p><span class="math-container">$$p_4(x)=-\frac{43439x^2}{18252}+\frac{242x}{351}+\frac{7381}{6084}$$</span></p>
<p>Next, the remainder of the Euclidean division of <span class="math-container">$p_3$</span> by <span class="math-container">$p_4$</span> is </p>
<p><span class="math-container">$$-\frac{1920402432x}{15594601}+\frac{1249896960}{15594601}$$</span></p>
<p>multiplying by <span class="math-container">$-1$</span> we obtain</p>
<p><span class="math-container">$$p_5(x)=\frac{1920402432x}{15594601}-\frac{1249896960}{15594601}$$</span></p>
<p>Next, the remainder of the Euclidean division of <span class="math-container">$p_4$</span> by <span class="math-container">$p_5$</span> is </p>
<p><span class="math-container">$$\frac{8062408717}{12332584368}$$</span></p>
<p>multiplying by <span class="math-container">$-1$</span> we obtain</p>
<p><span class="math-container">$$p_6(x)=-\frac{8062408717}{12332584368}$$</span></p>
<p>As this is a constant, this finishes the computation of the Sturm sequence. To find the number of real roots of <span class="math-container">$p_0$</span> one has to evaluate the sequences of the signs of these polynomials at <span class="math-container">$-\infty$</span> and <span class="math-container">$\infty$</span>. </p>
<p><span class="math-container">$$\text{For $-\infty$, the sequence of signs is: (+, −, +, +, −,−,−)}$$</span>
<span class="math-container">$$\text{For $\infty$, the sequence of signs is: (+, +, +, −,−,+,−)}$$</span></p>
<p>Therefore</p>
<p><span class="math-container">$$V(-\infty)-V(+\infty)=3-3=0$$</span></p>
<p>which shows that <span class="math-container">$p$</span> has no real roots.</p>
|
3,237,234 | <p>Find the maximum and minimum
value of <span class="math-container">$$\cos 2x + 3\sin x.$$</span></p>
| auscrypt | 675,509 | <p>I think you mean
<span class="math-container">$$\frac{\mathrm d}{\mathrm dx}\int_2^{x^4}\tan(t^2)\,\mathrm dt$$</span>
as the integral is usually done with respect to a "dummy variable"; reusing variables like that is bad practice.</p>
<p>In this case, using <a href="https://en.wikipedia.org/wiki/Leibniz_integral_rule" rel="nofollow noreferrer">Leibniz's integral rule</a>, we get </p>
<p><span class="math-container">$$4x^3\tan(x^8)$$</span></p>
|
3,237,234 | <p>Find the maximum and minimum
value of <span class="math-container">$$\cos 2x + 3\sin x.$$</span></p>
| HK Lee | 37,116 | <p>When <span class="math-container">$F(x)=\int^x_0\ \tan\ t^2\ dt$</span>, then <span class="math-container">$$ \frac{d}{dx}\ F =_{{\rm Fundamental \ Theorem\ of\ Calculus}} \tan\
x^2$$</span> so that <span class="math-container">$$ \frac{d}{dx}\int_2^{x^4}\ \tan\ t^2\ dt
=\frac{d}{dx}\{F(x^4) +C\} =_{{\rm Chain\ Rule}} F'(x^4)(4x^3)=\tan\ x^8\cdot (4x^3)$$</span></p>
|
1,605,100 | <p>$$\lim_{n\to\infty}\sum_{k=1}^n \frac{n}{n^2+k^2}$$</p>
<p>I tried this using powerseries just putting as $x=1$ there ,even tried thinking subtracting $s_{n+1} - s_{n}$ would be of some help, also I thought of writing the denominator as product of two complex numbers and then doing the partial fractions but it did not help. Any method guys?</p>
<p>Thanks in advance for guiding to think about these kind of problems.</p>
| Tsemo Aristide | 280,301 | <p>Since $H$ is a subgroup of finite index, $G/H$ has a finite numbers of elements. Let $S_H=\{ aHa^{-1},a\in H\}$, consider the map $f:S_H\rightarrow G/H$ such that $f(a)$ is the class of $a$. We show that $f$ is injective. Suppose that $f(a)=f(b)$, this implies that $b\in aH$, thus $b=ah, h\in H$, $bHb^{-1}=ahH{(ah)}^{-1}=a(hHh^{-1})a^{-1}=aHa^{-1}$, this implies that $aHa^{-1}=bHb^{-1}$. Thus $f$ is injective, since $G/H$ is finite, $S_H$ is finite.</p>
|
2,287,312 | <p>So the 12 marbles are distinct with each marble being a different color. Therefore I have to take into account, the different colors and different amount each person have. For example, person 1 can have 2 marbles, person 2 can have 7 marbles, and person 3 can have 3 marbles with each marble being a different color. At first, I thought it was:</p>
<p>$$\binom{12}{1}\binom{11}{1}\binom{10}{1}$$</p>
<p>but I don't think it covers the case of each person having a different amount of marbles for each case. </p>
| RRL | 148,510 | <p>Hint:</p>
<p>If $x$ is irrational then by density of rationals there is a sequence of rationals $(x_n)$ converging to $x$ such that</p>
<p>$$x = \lim_n x_n =\lim_n f(x_n) \neq f(x) = 0.$$</p>
<p>Now consider rational $x \neq 0$.</p>
|
2,287,312 | <p>So the 12 marbles are distinct with each marble being a different color. Therefore I have to take into account, the different colors and different amount each person have. For example, person 1 can have 2 marbles, person 2 can have 7 marbles, and person 3 can have 3 marbles with each marble being a different color. At first, I thought it was:</p>
<p>$$\binom{12}{1}\binom{11}{1}\binom{10}{1}$$</p>
<p>but I don't think it covers the case of each person having a different amount of marbles for each case. </p>
| Alex Ortiz | 305,215 | <p>For continuity at the origin, let $\epsilon>0$ be given. Then we claim that if $|x| < \epsilon$, then $|f(x) - f(0)| =|f(x)| < \epsilon$. Indeed, if $x$ is rational, then $|f(x)| = 0 < \epsilon$ and if $x$ is irrational, then $|f(x)| = |x| < \epsilon$.</p>
<p>If $a$ is irrational, then by density of the rationals, there is some sequence $r_n\to a$ with $r_n$ rational for every $n$. By our definition of $f$, $|f(a) - f(r_n)| \ge \eta$ for some $\eta>0$ and all $n$, so $f$ is not continuous at $a$.</p>
<p>If $a$ is rational, then by the density of the irrationals, there is a sequence $\alpha_n\to a$ with $\alpha_n$ irrational for every $n$. By our definition of $f$, $|f(a) - f(\alpha_n)| \ge \eta'>0$ for some $\eta'$ and all $n$, so $f$ is not continuous at $a$.</p>
|
1,224,692 | <p>What is the most computationally efficient way to find the layer on which a ball (i) belongs when arranged in a tetrahedron or 3 dimensional triangle with a triangular base. The ball on the top layer is numbered one. The balls on the second layer are numbered 2 - 4. The fifth layer 4-10 and so on. </p>
| Mastrel | 152,513 | <p>Suppose there exists a non-zero solution to $(A-\lambda M)x=0$. Then that means that $A-\lambda M$ is not one-to-one (as $(A-\lambda M)0 =0 = (A-\lambda M)x$ but $x \not = 0$), thus $A - \lambda M$ is not invertible and hence $A- \lambda M$ has determinant $0$. </p>
|
1,890,496 | <p>I am so confused on how to find the domain of this function $3^\sqrt{x^2-3x}$ without graphing it. I have no idea what to do in this situation.</p>
| SquirtleSquad | 339,227 | <p>Hint: Think of this as the composition of $3^x$, $\sqrt{x}$, and $x^2-3x$. Which of these has a domain that isn't all real numbers? And what is the domain of a composition?</p>
|
2,382,304 | <p>Find the general solution of the equation $$ \frac{dy}{dx}=1+xy $$</p>
<p>My attempt: Arranging it in standard linear equation form $\frac{dy}{dx}+Py=Q$, we get
$$\frac{dy}{dx}+(-x)y=1$$
Hence, the integrating factor(I.F.) = $e^{\int-xdx}=e^{-x^2/2}$
Hence, the solution is $$y(e^{-x^2/2})=\int{e^{-x^2/2}dx}+c$$</p>
<p>How do I solve this integral now? </p>
| hamam_Abdallah | 369,188 | <p>The incomplete equation is
$$\frac {y'}{y}=x $$</p>
<p>thus
$$\ln (\frac {y}{\lambda})=\frac {x^2}{2} $$
and
$$y_h=\lambda e^{\frac {x^2}{2}} $$</p>
<p>the constante variation method gives</p>
<p>$$\lambda'(x)=e^{\frac {-x^2}{2}} $$</p>
<p>the general solution is</p>
<p>$$y_g=\Bigl(\lambda+\int e^{\frac {-x^2}{2}}dx\Bigr)e^{\frac {x^2}{2}}$$</p>
|
423,938 | <p>I don't know if I apply for this case sin (a-b), or if it is the case of another type of resolution, someone with some idea without using derivation or L'Hôpital's rule? Thank you.</p>
<p>$$\lim_{x\to0}\frac{\sin(x^2+\frac{1}{x})-\sin\frac{1}{x}}{x}$$</p>
| Community | -1 | <p>We have
$$\frac{\sin(x^2+\frac{1}{x})-\sin\frac{1}{x}}{x}=\frac{\sin x^2\cos\frac{1}{x}+\sin\frac{1}{x}\cos x^2-\sin\frac{1}{x}}{x}$$
and since $\sin x^2\sim_0 x^2$ and $\left|\cos \frac{1}{x}\right|\leq 1$ and since $\cos x^2\sim_0 1-\frac{x^4}{2}$ we can see easily that the given limit is $0$.</p>
|
104,368 | <p>I can show that there infinitely many solutions to this equation. Is it possible that the set
of rational solutions is dense?</p>
| Ramsey | 12,107 | <p>Over $\mathbb{C}$, elliptic curves with, say, a point of order $N$ can be identified with the quotient of the upper half plane $\mathbb{H}$ by $\Gamma_1(N)$ just by associating to the class of $\tau\in \mathbb{H}$ the isomorphism class of the air $(\mathbb{C}/(\mathbb{Z} \oplus \mathbb{Z}\tau), 1/N)$. Under this identification, modular forms of level $N$ can be realized as sections of a certain line bundle on "the space" (using the term loosely) of such isomorphism classes that is natural in a sense because is has only to do with this moduli interpretation (roughly speaking, the fiber at each point is a tensor power of the space of differentials on the associated elliptic curve).</p>
<p>One can take this observation a lot further to get a good notion of modular forms over base rings other than $\mathbb{C}$ by studying sections of these natural invertible sheaves on modular curves over these more general bases. In particular, one gets a good notion of $p$-adic analytic modular forms by looking at rigid-analytic moduli spaces of elliptic curves.</p>
<p>ADDED IN EDIT:</p>
<p>Regarding your second question, here is perhaps one reason related to my answer above to "believe" that such a moduli interpretation shouldn't exist. I'm not sure how convincing it is...</p>
<p>If there were such an interpretation, then one should be able to mimic the stuff in my answer above to get a geometric notion of modular forms for non-congruence subgroups over a more general class of rings, including, say, $\mathbb{Z}$. Then basic facts from algebraic geometry would kick in and tell you that the Fourier coefficients of such forms should have bounded denominators, as they do for congruence subgroups. This, however, is false for modular forms on non-congruence subgroups. I'm not sure of the history behind these results, but I know that Winnie Li and her collaborators have proven theorems in this area.</p>
|
3,587,990 | <p>I am facing the following combination problem : </p>
<p>I do have <span class="math-container">$B$</span> colored boxes, each one containing <span class="math-container">$N_b$</span> balls of the given color.
The total number of balls is thus <span class="math-container">$\sum_{b=1}^{B} N_b = N$</span>.</p>
<p>I want to distribute the balls to P players with the two following aims:</p>
<ul>
<li>The number of balls received by each player must be the same (<em>ie</em> <span class="math-container">$N/P$</span>)</li>
<li>The sum of the number of different colors hold by each player must be minimal : if each player get <span class="math-container">$c_p$</span> different colors, I want <span class="math-container">$\sum_{p=1}^{P} c_p$</span> as small as possible.</li>
</ul>
<p>Is there any formulae or algorithm to respect both constraints ?</p>
<p><strong>Edit</strong></p>
<p>What is this problem useful ?</p>
<p>This problem arises eg in scientific computing : you have a available number of processus (<span class="math-container">$P$</span>) and you want to perform arithmetics on several (<span class="math-container">$B$</span>) arrays of different sizes (<span class="math-container">$N_b$</span>). Obviously you want the arrays to be uniformly distributed (otherwise one process will have much more work and will slow the whole procedure), but you also want to minimize the splits in the original arrays since each one implies a communication between processus, which are also time consuming. </p>
<p>What to start with : </p>
<p>I guess we can easily fall back to the case of a distribution <span class="math-container">$\tilde D=\{\tilde N_1, \tilde N_2, ... \tilde N_{B}\}$</span> to distribute on P' players, with <span class="math-container">$P' \le P$</span> and <span class="math-container">$\tilde N_b < N/P$</span> : indeed, any boxe in the original distribution having more than <span class="math-container">$N/P$</span> ball will fill up 100% capacity of one (or more) player.
We can then try to do pairs (or triplets, etc.) with the remaining boxes to reach N/P, but we may not have a perfect match without cuts in the general case.</p>
| RobPratt | 683,666 | <p>You can solve the problem via integer linear programming as follows. Let nonnegative integer decision variable <span class="math-container">$x_{b,p}$</span> be the number of balls of color <span class="math-container">$b$</span> distributed to player <span class="math-container">$p$</span>. Let binary decision variable <span class="math-container">$y_{b,p}$</span> indicate whether <span class="math-container">$x_{b,p}>0$</span>. The problem is to minimize <span class="math-container">$\sum_{b,p} y_{b,p}$</span> subject to:
<span class="math-container">\begin{align}
\sum_p x_{b,p} &=N_b &&\text{for all $b$}\\
\sum_b x_{b,p} &=N/P &&\text{for all $p$}\\
x_{b,p} &\le M_{b,p} y_{b,p} &&\text{for all $b$ and $p$}
\end{align}</span>
Here <span class="math-container">$M_{b,p}$</span> is a "big-M" constant upper bound on <span class="math-container">$x_{b,p}$</span>. A good choice is <span class="math-container">$M_{b,p}=\min(N_b,N/P)$</span>.</p>
|
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