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 |
|---|---|---|---|---|
31,451 | <p>I have created a notebook, with functions. The only thing that I want to change in the notebook later is the input file. I therefor want to have a button that runs every input step-by-step so that I do not have to go into every input and press enter for it to update. How could this be done. </p>
| bill s | 1,783 | <p>Perhaps the simplest thing is to command-A and enter, which selects all the cells and evaluates. Or you can use a button</p>
<pre><code>evalStart := Module[{},
FrontEndExecute[FrontEndToken[InputNotebook[], "EvaluateInitialization"]];
FrontEndExecute[FrontEndToken[InputNotebook[], "SelectAll"]];
FrontEndExecute[FrontEndToken[InputNotebook[], "EvaluateCells"]];];
Button["do it", evalStart]
</code></pre>
<p>When you evaluate this, a button appears. Every time you press the button, it evaluates everything by first evaluating the initialization cells (if any), then selecting all, then evaluating all the cells. </p>
<p><img src="https://i.stack.imgur.com/YEml6.png" alt="enter image description here"></p>
|
1,074,809 | <p>What is a non-decreasing sequence of sets and how come it can have a limit?</p>
<p>It appear in a probability theory book</p>
| Swapnil Tripathi | 117,387 | <p><strong>Hint:</strong> $\sqrt{y}$ is defined for $y\ge 0$. So $2x-1\ge 0$ is the domain of the function. (as $\sin$ is defined on whole $\mathbb{R}$)</p>
|
325,860 | <p>Let $a,b \in \mathbb{R}$, $a<b$ and let $f$ be a differentiable real-valued function on an open subset of $\mathbb{R}$ that contains [a,b]. Show that if $\gamma$ is any real number between $f'(a)$ and $f'(b)$ then there exists a number $c\in(a,b)$ such that $\gamma=f'(c)$.</p>
<p>Hint: Combine mean value theorem with the intermediate value theorem for the function $\frac{(f(x_1)-f(x_2))}{x_1-x_2}$ on the set $\{(x_1,x_2)\in E^2: a\leq x_1 < x_2 \leq b\}$.</p>
<p>This is question number 7 on page 109 of Rosenlicht (introduction to Analysis).</p>
<p>I am having a lot of trouble trying to start on this problem.</p>
| Pedro M. | 21,628 | <p>Hint: Let $$g(x_1,x_2) = \frac{f(x_1) - f(x_2)}{x_1 - x_2}$$</p>
<p>Then $$\lim_{x_2 \searrow a} g(a,x_2) = f'(a) \text{ and } \lim_{x_1 \nearrow b} g(x_1,b) = f'(b)$$ </p>
<p>So if $\varepsilon$ is sufficiently small, what can you say about $g(a,a+\varepsilon), \gamma$ and $g(b-\varepsilon,b)$?</p>
<p>Now if you define $h(t) = g(a + (b-\epsilon-a)t, a+\epsilon + (b-a-\epsilon)t)$, what does the Intermediate Value Theorem tell you?</p>
|
2,640,909 | <p>I encountered a problem with 4 variables and I was wondering if anyone knows how to solve this:
This is what is known:</p>
<p>$$ \left\lbrace
\begin{align}
a+b &= 1800 \\
c+d &= 12 \\
a/c &= 100 \\
b/d &= 250 \\
(a+b)/(c+d) &= 150
\end{align}
\right.$$</p>
<p>Below is a screenshot from a spreadsheet. The red numbers are the 4 unknowns that I'm trying to figure out how to solve for (I happen to know them, but would love to understand how to solve for them when I do not know them).
<a href="https://i.stack.imgur.com/rlb4A.png" rel="nofollow noreferrer">screenshot</a>
Any help would be greatly appreciated! Thank you!</p>
| Yash Jain | 522,158 | <p>$b=1800-a$</p>
<p>$d=12-c$</p>
<p>$\cfrac ac=100$</p>
<p>$a=100c$</p>
<p>$\cfrac {1800-a}{12-c}=250$</p>
<p>$\cfrac {2.5a}c=250$</p>
<p>$\cfrac {1800-a}{12-c}=\cfrac {2.5a}c$</p>
<p>$\cfrac {1800-100c}{12-c}=\cfrac {2.5(100c)}c$</p>
<p>$1800c-10c^2=250c(12-c)$</p>
<p>Which gives you the value of $c$</p>
|
342,491 | <p>How to prove the following:</p>
<p>$a_n = \left\{\left(1+\frac{1}{n}\right)^n\right\}$ is bounded sequence, $ n\in\mathbb{N}$</p>
| Aryabhata | 1,102 | <p>There was a comment that you can use $\text{AM} \ge \text{GM}$ to prove the boundedness of this.</p>
<p>Here is a proof. As a side effect, we also prove the convergence.</p>
<p>First we show that $x_n = \left(1 + \frac{1}{n}\right)^n$ is monotonically increasing. We prove this, using $\text{AM} \ge \text{GM}$.</p>
<p>We have that, by taking $n$ copies of $\left(1 + \frac{1}{n}\right)$ and one copy of $1$ that,</p>
<p>$$\frac{\left(1 + \frac{1}{n}\right) + \dots + \left(1 + \frac{1}{n}\right) + 1}{n+1} \gt \sqrt[n+1]{\left(1 + \frac{1}{n}\right)^n}$$</p>
<p>i.e.</p>
<p>$$ \frac{n+2}{n+1} \gt \sqrt[n+1]{\left(1 + \frac{1}{n}\right)^n}$$</p>
<p>and so</p>
<p>$$\left(1 + \frac{1}{n+1}\right)^{n+1} \gt \left(1 + \frac{1}{n}\right)^n$$</p>
<p>Now we use $\text{AM} \ge \text{GM}$ (again!). </p>
<p>We take $n$ copies of $1$ and one copy of $\frac{1}{2}$ to get</p>
<p>$$\frac{n + \frac{1}{2}}{n+1} \ge \sqrt[n+1]{\frac{1}{2}}$$</p>
<p>i.e.</p>
<p>$$ 2^{\frac{1}{n+1}} \ge \frac{2n+2}{2n+1}$$
i.e</p>
<p>$$ 2 \ge \left(1 + \frac{1}{2n+1}\right)^{n+1} $$</p>
<p>And so
$$ 4 \ge \left(1 + \frac{1}{2n+1}\right)^{2n+1} $$</p>
<p>Since the sequence is monotonically increasing, this bound applies to the whole sequence.</p>
<p>(Side effect: Since the sequence is monotonic, and bounded, it is convergent)</p>
|
58,710 | <p>I've got a bunch of readings at various dates and times. I'd like to be able to interpolate and then integrate over, say days. I can do this by converting my date/time to <code>AbosluteTime</code> and get the interpolating function that way. Then by integrating and plugging in seconds as my limits of integration I get what I want.</p>
<p>Also, since seconds in really finer than I need, I convert all of the <code>AbsoluteTimes</code> to the number of hours since the first reading.</p>
<p>I was wondering if there might be a way to do this directly using dates. For what it's worth, my readings last for about 2 months and the values are between about 90 and 300.</p>
<p><strong>SAMPLE DATA</strong></p>
<pre><code> x[[1 ;; 5]]
</code></pre>
<blockquote>
<p>{{{2014, 8, 4, 10, 36, 0.}, 257.},{{2014, 8, 4, 16, 28, 0.}, 385.}, </p>
<p>{{2014, 8, 4, 22, 53, 0.}, 176.},
{{2014, 8, 5, 6, 52, 0.}, 148.}, {{2014, 8, 5, 11, 19, 0.}, 192.}}</p>
</blockquote>
| Dr. belisarius | 193 | <p>Some sample data: </p>
<pre><code>r = # - #[[1]] &@(Range[##, 3690] & @@ (AbsoluteTime[{#, {"Day", "Month", "YearShort"}}] & /@
{"05/01/14", "05/03/14"}));
hours = N[r/3600];
data = Transpose[{hours, hours^2}];
</code></pre>
<p>Integrate: </p>
<pre><code>f = Interpolation@data;
Integrate[f@x, {x, 0, Last@hours}]
(*9.45434*10^8*)
</code></pre>
|
505,067 | <p>I am writing a computer program which generate 5 digit alpha-numeric codes. Each character can contain a-z,A-Z and 0-9 (62 different possibility per character).</p>
<p>So if there are 5 character how many possibility are there?</p>
| Old John | 32,441 | <p>If it is not an integer, there must be some prime which divides the denominator but not the numerator. The same will be true of all integer powers $n\ge 1$, so it cannot be an integer.</p>
|
2,442,233 | <p>let $A= \{x^2 \mid 0 < x <1\}$ and $B =\{x^3 \mid 1 < x < 2 \}$.</p>
<p>Which of the statement is true?</p>
<p>1.there is a one to one, onto function from $A$ to $B$.</p>
<p>2.there is no onto function from $A$ to $B$</p>
<p>my attempt ; there will be no onto function from $A$ to $B$ because order of $A$ is $2$ and order $B$ is $3$ so there is no one one function from $A$ to $B$.</p>
<p>So my answer is option 2,,
is my answer is correct or not ...pliz tell me the solution</p>
| operatorerror | 210,391 | <p>Note that
$$
B=(1,2^3)=(1,8)\\
A=(0,1)
$$
Then, take
$$
f:A\to B\\
f(x)=7x+1
$$
which is linear and thus invertible. </p>
|
3,497,554 | <blockquote>
<p>Show that <span class="math-container">$\frac{(2n-1)!}{(n)!(n-1)!}$</span> is odd or even according as <span class="math-container">$n$</span> is or is not a power of <span class="math-container">$2$</span>.</p>
</blockquote>
<p>I know that the index of the highest power of <span class="math-container">$2$</span> contained in <span class="math-container">$n!$</span> is <span class="math-container">$n-1$</span> when <span class="math-container">$n$</span> is a power of <span class="math-container">$2$</span> and <span class="math-container">$n-r$</span> when <span class="math-container">$n$</span> is equal to <span class="math-container">$2^r-1$</span>. </p>
<p>I've expanded the above to the result that the index of the highest power of <span class="math-container">$2$</span> contained in <span class="math-container">$n!$</span> is <span class="math-container">$2^r-1$</span> when <span class="math-container">$n$</span> is equal to <span class="math-container">$2^r+1$</span>. <em>(If you're wondering how I derived it, refer to the formula given in <a href="http://math.stackexchange.com/q/3497392/710663">this</a> question and correct me if I'm wrong)</em>. Using that, I've got that the highest power of <span class="math-container">$2$</span> in the term when <span class="math-container">$n$</span> is a power of <span class="math-container">$2$</span> is <span class="math-container">$\frac{2^{r+1}-(r+2)}{(2^r-1)(2^r-(r+1))}$</span>.</p>
<p>Now here, if <span class="math-container">$r$</span> is odd, then the numerator is odd, so the whole term is odd and everything is fine. But if <span class="math-container">$r$</span> is odd, then the numerator is even and the denominator is odd, so the term is even which contradicts the above statement.</p>
<p>Putting <span class="math-container">$n=2^r+1$</span> the term equals <span class="math-container">$\frac{(2^{r+1}+1)}{(2^r+1)(2^r)}$</span>. Finding the highest powers of <span class="math-container">$2$</span> I've got <span class="math-container">$\frac{(2^{r+1}-1)}{(2^r-1)(2^r-1)}$</span> which is odd.</p>
<p>I'm getting nearly opposite results.</p>
<p><strong>Am I doing something very wrong?</strong></p>
<p>Any help would be highly appreciated.</p>
| ViHdzP | 718,671 | <p>Let <span class="math-container">$\nu_2(x)$</span> be the exponent of the highest power of <span class="math-container">$2$</span> that divides <span class="math-container">$x$</span>. By <a href="https://en.wikipedia.org/wiki/Legendre's_formula" rel="nofollow noreferrer">Legendre’s Formula</a>, </p>
<p><span class="math-container">\begin{equation}\nu_2\left(\frac{(2n-1)!}{n!(n-1)!}\right)=\sum_{i=0}^\infty\left\lfloor\frac{2n-1}{2^i}\right\rfloor-\left\lfloor\frac{n}{2^i}\right\rfloor-\left\lfloor\frac{n-1}{2^i}\right\rfloor\Rightarrow\label{1}\tag{1}\end{equation}</span> </p>
<p><span class="math-container">\begin{equation}\nu_2\left(\frac{(2n-1)!}{n!(n-1)!}\right)=\sum_{i=0}^\infty2^{-i}\left(r_{2^i}(n)+r_{2^i}(n-1)-r_{2^i}(2n-1)\right),\label{2}\tag{2}\end{equation}</span> where <span class="math-container">$\lfloor\cdot\rfloor$</span> represents the floor function, and <span class="math-container">$r_{2^i}$</span> denotes the least non-negative residue modulo <span class="math-container">$2^i$</span>.</p>
<p>Every term in <span class="math-container">$(\ref{2})$</span> is a positive number times a multiple of <span class="math-container">$2^i$</span> greater than <span class="math-container">$-2^i+1$</span>, and therefore non-negative. Therefore, the problem’s expression will be odd iff every term in <span class="math-container">$(\ref{2})$</span> is in fact <span class="math-container">$0$</span>. That is, iff either <span class="math-container">$2^i\mid n-1$</span>, or <span class="math-container">$r_{2^i}(n)\leq 2^{i-1}$</span> for all <span class="math-container">$i$</span>.</p>
<p>Any power of two, clearly, always withholds the second condition. Any number that is not a power of two, we can write as <span class="math-container">$n=k2^\alpha$</span>, for <span class="math-container">$k>1$</span> odd. If <span class="math-container">$2^{\beta-1}<k<2^\beta$</span>, <span class="math-container">$n$</span> will withhold neither condition for <span class="math-container">$i=\alpha+\beta$</span>. This proves what we wanted. <span class="math-container">$\blacksquare$</span></p>
|
830,755 | <p>I would like to compute the following,
$$
\int_0^{\infty}\int_0^{\infty}e^{-x^2-2xy-y^2}\ dx\,dy
$$
It is obvious that we can rewrite the integral above to,
$$
\int_0^{\infty}\int_0^{\infty}e^{-(x+y)^2}\ dx\,dy
$$
so we are ending up with something looking like a gaussian integral. I think that a smart substitution would help but all I tried ended up to be something I am not able to compute...</p>
<p>I really would appreciate any hint.</p>
<p>Thanks in advance!</p>
| Pranav Arora | 117,767 | <p>Switch to polar coordinates i.e $x=r\cos\theta$, $y=r\sin\theta$ and $dx\,dy=r\,dr\,d\theta$ to obtain:
$$\int_0^{\pi/2} \int_0^{\infty} re^{-r^2(1+\sin(2\theta))}dr\,d\theta=\int_0^{\pi/2} \frac{1}{2(1+\sin(2\theta))}\,d\theta$$
Write $\sin(2\theta)=\frac{2\tan\theta}{1+\tan^2\theta}$ to get:
$$\int_0^{\pi/2} \frac{1}{2(1+\sin(2\theta))}\,d\theta=\frac{1}{2}\int_0^{\pi/2} \frac{\sec^2\theta}{1+\tan^2\theta+2\tan\theta}\,d\theta$$
Use the substitution $\tan\theta=t \Rightarrow \sec^2\theta\,d\theta=dt$:
$$\frac{1}{2}\int_0^{\pi/2} \frac{\sec^2\theta}{1+\tan^2\theta+2\tan\theta}\,d\theta=\frac{1}{2}\int_0^{\infty} \frac{dt}{(1+t)^2}=\boxed{\dfrac{1}{2}}$$</p>
|
2,190,508 | <p>Let $f:[a,b] \to \mathbb R$ be nice differentiable function. By Fundamental Theorem of Calculus (FTC), we have $$\int_a^{b} f'(t) dt = f(b) -f(a).$$</p>
<p>By Cauchy-Schwartz inequality, we have, $\int_{a}^{b} |f(t)| dt \leq (b-a)^{1/2}(\int_a^b |f(t)|^2 dt)^{1/2} .$ Using this, and FTC, we have $\int_{a}^{b} (f'(t))^2 dt \geq (b-a)^{-1} (f(b)-f(a))^2.$ </p>
<blockquote>
<p>Question: (1) Can compute $\int_a^{b}(f'(t))^2 dt$ precisely? (2) Can we improve inequality $\int_{a}^{b} (f'(t))^2 dt \geq (b-a)^{-1} (f(b)-f(a))^2$ ?</p>
</blockquote>
| Micah | 30,836 | <p>In answer to (1), let $f(x)=e^{x^2/2}$. Then $(f'(x))^2=(xe^{x^2/2})^2=x^2e^{x^2}$. This has no elementary antiderivative, as we have
$$
\int x^2e^{x^2} \, dx=\int x(xe^{x^2} \, dx)= \frac{1}{2}xe^{x^2}-\frac{1}{2}\int e^{x^2}\, dx
$$
and this last integral is well-known not to be elementary.</p>
<p>It follows that there can be no general elementary formula for $\int (f'(x))^2 \, dx$.</p>
|
3,810,944 | <p>Consider a positive integer <span class="math-container">$n$</span> and the function
<span class="math-container">$f:\mathbb{N}\to \mathbb{N}$</span> (<span class="math-container">$\mathbb N$</span> includes <span class="math-container">$0$</span>) by</p>
<p><span class="math-container">$$f(x) = \begin{cases} \frac{x}{2}
& \text{if } x \text{ is even} \\ \frac{x-1}{2} + 2^{n-1} & \text{if } x \text{ is odd} \end{cases} $$</span>
Determine the set</p>
<p><span class="math-container">$$ A = \{ x\in \mathbb{N} \mid \underbrace{\left( f\circ f\circ ....\circ f \right)}_{n\ f\text{'s}}\left( x \right)=x \}. $$</span></p>
<p>(Romania NMO 2013)</p>
<p>The solution starts by stating that <span class="math-container">$f(x)<x, \quad\forall x\ge 2^n-1$</span>. This was easy enough to understand. However, they continue by saying this implies that <span class="math-container">$A\subset\{0,1,\dots,2^n-1\}$</span>. Why is that?</p>
<p>Please help me understand! Thanks in advance!</p>
| Community | -1 | <p><strong>Hint:</strong></p>
<p>Letting all divisions be integer and splitting the cases after the low order bits of <span class="math-container">$x$</span>, we can expand the function iterates for, say, <span class="math-container">$n=4$</span>, and the pattern becomes obvious.</p>
<p><span class="math-container">$$f(x)=\begin{cases}\frac x2&\text{ if } x \text{ is even}\\\frac{x}2+8&\text{ if } x \text{ is odd}\end{cases}$$</span></p>
<p><span class="math-container">$$f(f(x))=\begin{cases}\frac x4&\\\frac{x}4+4\\\frac{x}4+8\\\frac{x}4+12\end{cases}$$</span></p>
<p><span class="math-container">$$f(f(f(x)))=\begin{cases}
\frac x8&\\\frac{x}8+2\\\frac{x}8+4\\\frac{x}8+6\\
\frac x8+8&\\\frac{x}8+10\\\frac{x}8+12\\\frac{x}8+14\end{cases}$$</span></p>
<p><span class="math-container">$$f(f(f(f(x))))=\begin{cases}
\frac x{16}&\\\frac{x}{16}+1\\\frac{x}{16}+2\\\frac{x}{16}+3\\
\frac x{16}+4&\\\frac{x}{16}+5\\\frac{x}{16}+6\\\frac{x}{16}+7\\
\frac x{16}+8&\\\frac{x}{16}+9\\\frac{x}{16}+10\\\frac{x}{16}+11\\
\frac x{16}+12&\\\frac{x}{16}+13\\\frac{x}{16}+14\\\frac{x}{16}+15\end{cases}$$</span></p>
|
875,924 | <p>For what values of $x\in\mathbb{R}$ is $f$ continuous?</p>
<p>$f(x) = \left\{
\begin{array}{lr}
0 & \text{if}\, x \in \Bbb Q\\
1 & \text{if}\, x \notin \Bbb Q
\end{array}
\right.$</p>
<p>The solution I found:</p>
<p>$f(x)$ is continuous nowhere. For, given any number $a$ and any $\delta>0$, the interval $(a-\delta, a+ \delta)$ contains infinitely many rational numbers and infinitely many irrational numbers.</p>
<p>Since $f(a) = 0$ or $1$, there are infinitely many numbers $x$ with $0<|x-a|< \delta $, and $|f(x) - f(a)| = 1$</p>
<p>Thus, $\lim_{x \to a}f(x) \neq f(a)$</p>
<p>My question: I'm having a hard time visualizing that you can't have a rational number $x$ with rational numbers on both sides of that $x$ such that $\lim_{x \to a}f(x) = f(a)$(is that because you can go as 'deep' in to the interval as you like to reach the irrational number?). Can anybody attempt to give me some insight in to how this works? </p>
<p>I have never given it much thought that you can have infinitely many numbers even in the smallest of intervals, and it's pretty overwhelming for me to even imagine.</p>
| Mhenni Benghorbal | 35,472 | <p>You can use <a href="http://www.math.washington.edu/~morrow/334_13/continuity.pdf" rel="nofollow">sequential continuity</a>.</p>
|
322,448 | <p>I'm supposed to show that: </p>
<p>$$y=\frac{5(x-1)(x+2)}{(x-2)(x+3)} = P + \frac{Q}{(x-2)} + \frac{R}{(x+3)}$$ </p>
<p>and the required answers are: $$ P=5, Q=4, R=-4 $$ </p>
<p>I tried to solve this with partial fractions like so: </p>
<p>$$5(x-1)(x+2) = A(x+3) + B(x-2)$$ </p>
<p>$\implies$ $A$=4, $B$=-4<br>
$\implies$ $Q$=4, R=-4 </p>
<p>But where does $P$=5 come from?</p>
<p>Or should I have first multiplied out the numerator and denominator and then used long division to solve?</p>
| Community | -1 | <p>$$\frac{5(x-1)(x+2)}{(x-2)(x+3)} = P + \frac{Q}{(x-2)} + \frac{R}{(x+3)}$$ gives<br>
$$5(x-1)(x+2)=P(x-2)(x+3)+Q(x+3)+R(x-2)........(1)$$<br>
Now put $x=2, x=-3,x=0$ respectively on both sides of $(1)$ to get, $Q=4,R=-4,P=5$ respectively. Hence you can reach the desired result.</p>
|
2,207,260 | <p>I have to find the asymptotics of the following integral</p>
<p>$$\int^1_0 \frac{\sin(x)}{x(1+x)^n} dx$$</p>
<p>as $n\to\infty$.</p>
<p>I know I am supposed to use the lebesgue dominated convergence theorem and create a sequence $b_n$ and aim for $\lim_{n\to\infty} a_n/b_n =1$ but I can't seem to get my head round it.</p>
<p>Any help I would be very much grateful for</p>
| tired | 101,233 | <p>Denote the integral in question by $I_n$ and squeeze it baby!</p>
<p><a href="https://math.stackexchange.com/questions/390899/prove-that-x-fracx33-sin-x-x-for-all-x0">Notice</a> that the inequalities $x-x^3/3!<\sin(x)<x$ hold on $x\in(0,1)$</p>
<p>so</p>
<p>$$
\int_0^1dx\frac{1-x^2/3!}{(1+x)^n}<I_n<
\int_0^1dx\frac{1}{(1+x)^n}
$$</p>
<p>doing the algebra (which is elementary) you get</p>
<p>$$
\frac{1}{n-1}+\mathcal{O}(n^{-2})<I_n<\frac{1}{n-1}+\mathcal{O}( n^{-2})
$$</p>
<p>which shows that $I_n\sim n^{-1}$ or $\lim_{n\rightarrow\infty}nI_n=1$</p>
<hr>
<p>For the record: </p>
<p>Integration by parts also works nicely here: use $u(x)=\text{sinc}(x),\,v'(x)=(1+x)^{-n}$</p>
|
882,540 | <p>Does anyone know of a non-trivial (i.e. cardinality $\geq 2)$ algebraic structure $(X,+,-)$ satisfying the following identities?</p>
<ol>
<li><p>$(x+a)-a=x$</p></li>
<li><p>$(x-a)+a=x$</p></li>
<li><p>$(x+y)+a = (x+a)+(y+a)$</p></li>
<li><p>$(x-y)+a = (x+a)-(y+a)$</p></li>
</ol>
<p><em>Remark.</em> The Abelian group of order $2$ doesn't satisfy the last two conditions.</p>
<p><strong>Motivation.</strong> I think its cool that if $X$ is such an algebraic structure, then for every $a \in X$, the functions $$x \mapsto x+a, \qquad x \mapsto x-a$$</p>
<p>are automorphism of $X$. This mean that if $a \in X$ and $f \in \mathrm{Aut}(X)$, then $f+a \in \mathrm{Aut}(X)$ and $f-a \in \mathrm{Aut}(X).$</p>
| Jack Yoon | 141,253 | <p>Well I get one stupid example where $X$ is any set and $+$ and $-$ act trivially. What I mean by trivially is:</p>
<p>$x+a=x-a=x$ for all $x, a \in X$. Which is clearly not interesting.</p>
|
102,280 | <p>What are the usual tricks is proving that a group is not simple? (Perhaps a link to a list?)</p>
<p>Also, I may well be being stupid, but why if the number of Sylow p groups $n_p=1$ then we have a normal subgroup?</p>
| edgar | 23,702 | <p>The summation
$$ \sum\limits_{i=0}^n \frac{i}{2^{i}} $$</p>
<p>is equal to
$$ \sum\limits_{i=0}^n i(\frac{1}{2})^i $$</p>
<p>which is of the form
$$ \sum\limits_{i=0}^n ix^i, x = \frac{1}{2} $$</p>
<p>Consider the summation $$ \sum\limits_{i=0}^n x^i = \frac{x^{n+1} - 1}{x - 1} $$</p>
<p>By differentiating with respect to $x$ we have
$$ \frac{d}{dx}\left( \sum\limits_{i=0}^n x^i \right) = \frac{d}{dx} \left( \frac{x^{n+1} - 1}{x - 1} \right)$$
$$ \Rightarrow \sum\limits_{i=0}^n ix^{i-1} = \frac{(x - 1)(n + 1)x^n - (x^{n+1} -1)(1)}{(x - 1)^2} $$
$$ \Rightarrow \sum\limits_{i=0}^n ix^{i-1} = \frac{(x^{n+1} - x^n)(n + 1) - x^{n+1} + 1}{(x - 1)^2} $$
$$ \Rightarrow \sum\limits_{i=0}^n ix^{i-1} = \frac{(nx^{n+1} - nx^n) + (x^{n+1} - x^n) - x^{n+1} + 1}{(x - 1)^2} $$
$$ \Rightarrow \sum\limits_{i=0}^n ix^{i-1} = \frac{nx^{n+1} - nx^n + x^{n+1} - x^n - x^{n+1} + 1}{(x - 1)^2} $$
$$ \Rightarrow \sum\limits_{i=0}^n ix^{i-1} = \frac{nx^{n+1} - nx^n - x^n + 1}{(x - 1)^2} $$
By multiplying $x$ to both sides,
$$ x\sum\limits_{i=0}^n ix^{i-1} = x\left( \frac{nx^{n+1} - nx^n - x^n + 1}{(x - 1)^2} \right)$$
$$ \Rightarrow \sum\limits_{i=0}^n ix^{i} = \frac{nx^{n+2} - nx^{n+1} - x^{n+1} + x}{(x - 1)^2} $$
By substitution $ \left( x = \frac{1}{2} \right) $,
$$ \sum\limits_{i=0}^n i\left(\frac{1}{2}\right)^{i} = \frac{n\left(\frac{1}{2}\right)^{n+2} - n\left(\frac{1}{2}\right)^{n+1} - \left(\frac{1}{2}\right)^{n+1} + \left(\frac{1}{2}\right)}{(\left(\frac{1}{2}\right) - 1)^2} $$</p>
<p>$$ \Rightarrow \sum\limits_{i=0}^n i\left(\frac{1}{2}\right)^{i} = \frac{n\left(\frac{1}{2^{n+2}}\right) - n\left(\frac{1}{2^{n+1}}\right) - \left(\frac{1}{2^{n+1}}\right) + \left(\frac{1}{2}\right)}{\left(-\frac{1}{2}\right)^2} $$</p>
<p>$$ \Rightarrow \sum\limits_{i=0}^n i\left(\frac{1}{2}\right)^{i} = \frac{\frac{n}{2^{n+2}} - \frac{n + 1}{2^{n+1}} + \frac{1}{2}}{\left(\frac{1}{4}\right)} $$</p>
<p>$$ \Rightarrow \sum\limits_{i=0}^n i\left(\frac{1}{2}\right)^{i} = 4 \left( \frac{n}{2^{n+2}} - \frac{n + 1}{2^{n+1}} + \frac{1}{2} \right) $$</p>
<p>$$ \Rightarrow \sum\limits_{i=0}^n i\left(\frac{1}{2}\right)^{i} = 4 \left( \frac{n}{2^{n+2}} - \frac{2(n + 1)}{2^{n+2}} + \frac{2^{n+1}}{2^{n+2}} \right) $$</p>
<p>$$ \Rightarrow \sum\limits_{i=0}^n i\left(\frac{1}{2}\right)^{i} = \frac{n - 2(n + 1) + 2^{(n+1)}}{2^{n}} $$</p>
<p>$$ \Rightarrow \sum\limits_{i=0}^n \frac{i}{2^{i}} = \frac{n - 2(n + 1) + 2^{(n+1)}}{2^{n}} $$</p>
|
3,374,024 | <p>Basicly, i am bad at mathematics very hard to me got score 70. I try learn trigonometry now, starting from the very basic.
I practice to find the value of <span class="math-container">$$\sin(135^\circ)$$</span> using my knowing about <span class="math-container">$\sin(45^\circ)={\sqrt2\over2}$</span>, <span class="math-container">$\cos(45^\circ)={\sqrt2 \over 2}$</span>, <span class="math-container">$\sin(90^\circ)=1$</span>, <span class="math-container">$\cos(90^\circ)=0$</span>, the quadrant rule and sum angle formula. I hope the expert can help me to correcting and giving a tips.</p>
<p>Using the quadrant rule :
Since <span class="math-container">$135^\circ$</span> on second quadrant the sine sign is positive</p>
<p>Using sum formula of sin :
<span class="math-container">$$\sin(135^\circ)=\sin(90^\circ+45^\circ)$$</span>
<span class="math-container">$$=\sin(90^\circ)×\cos(45^\circ)+\cos(90^\circ)×\sin(45^\circ)$$</span>
<span class="math-container">$$=1×\cos(45^\circ)+0×\sin(45^\circ)$$</span>
<span class="math-container">$$=\cos(45^\circ)+0$$</span>
<span class="math-container">$$=\cos(45^\circ)$$</span>
<span class="math-container">$$={\sqrt2\over2}$$</span>
For your informatiom, i am just economic faculty at second semester, and my college tutor giving us a calculus, that contain trigonometry. I need to learn this because 2 weeks again i got calculus examination. Thank you for your support and tips and explanation.</p>
<p>Is my practice is correct ? Can i get the simple way to remember the sum formula of sine and cosine ? What suitable practice for me after i practicing with <span class="math-container">$135^\circ$</span> ?</p>
| Taroccoesbrocco | 288,417 | <p>'True' refers to the English statement (2.2), 'correct' refers to the sequent <span class="math-container">$\Gamma \vdash \Psi$</span>. The authors aim to define when a sequent is <em>correct</em> in a technical sense, and they use the common sense of 'true' in natural language (the meta-level).</p>
<p>In other words, the meaning of the paragraph is that if the situation described by (2.2) holds (i.e. the statement <span class="math-container">$\Psi$</span> is provable from the undischarged assumptions <span class="math-container">$\Gamma$</span>) then we say that the sequent <span class="math-container">$\Gamma \vdash \Psi$</span> is correct.</p>
<p>Note that it is not true that all sequents are correct: for instance, <span class="math-container">$X \vdash
\lnot X$</span> is not correct (if your system is consistent) because there is no proof of <span class="math-container">$\lnot X$</span> from <span class="math-container">$X$</span>.</p>
|
56,804 | <p>I know the Galois group is $S_3$. And obviously we can swap the imaginary cube roots. I just can't figure out a convincing, "constructive" argument to show that I can swap the "real" cube root with one of the imaginary cube roots. </p>
<p>I know that if you have a 3-cycle and a 2-cycle operating on three elements, you get $S_3$. I have a general idea that based on the order of the group there's supposed to be at least a 3-cycle. But this doesn't feel very "constructive" to me. </p>
<p>I wonder if I've made myself understood in terms of what kind of argument I'd like to see?</p>
| lhf | 589 | <p>What you're probably missing is that, although the real cube root of 2 stands out, <em>all</em> three complex cube roots of 2 are algebraically the same, that is, they cannot be told apart, and so can be permuted at will. Hence $S_3$.</p>
<p>Contrast this with the seemingly analogous situation of, say, the 6-th roots of unity. Here the roots are <em>not</em> the same algebraically, because $x^6-1$ is not irreducible.</p>
|
1,939,107 | <p>An unbiased coin is tossed repeatedly and outcomes are recorded. What is the expected no of toss to get HT ( one head and one tail consecutively) ?</p>
<p>I tried to solve above question using the following probability tree diagram.</p>
<p><a href="https://i.stack.imgur.com/10cO3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/10cO3.png" alt="enter image description here"></a></p>
<p>Probability in each branch is = $0.5$. I double circled the satisfying toss events.
While observing the diagram I noticed that, from 2nd toss onward our required event starts showing up. Additionally,</p>
<ol>
<li>in the $\text{2nd}$ toss (or the 3rd level) we have one satisfying case.</li>
<li>in the $\text{3rd}$ toss (or the 4th level) we have two satisfying case.</li>
<li>in the $\text{4th}$ toss (or the 5th level) we have three satisfying case.</li>
<li>in the $\text{5th}$ toss (or the 6th level) we have four satisfying case.</li>
<li>etc.</li>
</ol>
<p>i.e. in the $\text{kth}$ toss we would have $(k-1)$ satisfying case.
So,</p>
<p>$\\ E = \sum_{k=2}^{\infty } k.P(k)\
\\ E = \sum_{k=2}^{\infty } k.\left \{ (k-1)*(0.5)^k \right \}\\
\\ E = \sum_{k=2}^{\infty } \left \{ (k^2-k)*(0.5)^k \right \}\\$</p>
<p>I understand the series summation would converge because we have the $0.5^{k}$ term which eventually becomes zero.</p>
<p>But could not solve this polynomial and power series summation.</p>
<p>Please suggest how to solve this and or any other method.</p>
| JMoravitz | 179,297 | <p>You can treat this as a <a href="https://en.wikipedia.org/wiki/Markov_chain" rel="nofollow">Markov Chain</a> with the following states:</p>
<p>End, Head, Tail (not end) (<em>or equivalently worded, HT has occured, heads is most recent result, and no heads have occured respectively</em>)</p>
<p>This has the transition matrix $\begin{bmatrix}1&0.5&0\\0&0.5&0.5\\0&0&0.5\end{bmatrix}$, which is in the standard form for an absorbing chain: $\left[\begin{array}{c|c}I&S\\\hline0&R\end{array}\right]$</p>
<p>We look to the <a href="https://en.wikipedia.org/wiki/Absorbing_Markov_chain#Fundamental_matrix" rel="nofollow">fundamental matrix</a>: $(I-R)^{-1}$ which in this case is $\begin{bmatrix}0.5&-0.5\\0&0.5\end{bmatrix}^{-1} = \frac{1}{1/4}\begin{bmatrix}0.5&0.5\\0&0.5\end{bmatrix}=\begin{bmatrix}2&2\\0&2\end{bmatrix}$</p>
<p>The fundamental matrix holds the information about how many steps are expected until we reach an absorbing state. Since our initial state can be thought of as being a tail (<em>or more correctly worded, as being that no head has yet occurred</em>), we look to the corresponding column and add the entries to find the expected time.</p>
<p>In this case, we see it will take an expected number of $2+2=4$ flips to have flipped a tails after a heads.</p>
|
3,238,306 | <p>Hello I have the differential equation <span class="math-container">$$x'=\frac{8t+10x}{17t+x}$$</span>
Brought it in a eqation where I can substitute <span class="math-container">$$ u=\frac{x}{t}$$</span>
and after some transformations I got the equation
<span class="math-container">$$ \frac{17+u}{8-7u-u^2}du=\frac{1}{t}dt $$</span>
with the following equation
<span class="math-container">$$ -2ln(u-1)+ln(u+8)=ln(t)+c$$</span>
Then I tried to multiply it with <span class="math-container">$$e$$</span> to make the <span class="math-container">$$ln()$$</span> disappear
then I got the equation <span class="math-container">$$ \frac{u+8}{(u-1)^2}=te^c$$</span> but now I dont know how to resubstitute or to break the fracture.Thanks,Ciwan.</p>
| Community | -1 | <p>Let <span class="math-container">$a\equiv x+ 2,\,b\equiv y+ 2$</span>
<span class="math-container">$$\therefore\,ab- a- b= xy+ x+ y\geqq 0$$</span>
<span class="math-container">$$\because\,x,\,y\geqq 0$$</span></p>
|
290,173 | <p>Suppose $d$ is a bounded metric on $X$, i.e. $d(x,y)< K<\infty$ for all $x,y\in X$. Is there a standard way to convert $d$ into another metric $\widetilde{d}$ on $X$ with the property that $\widetilde{d}(x,y)\to\infty$ if and only if $d(x,y)\to K$? One way would be to find some function $f$ such that $\widetilde{d}(x,y)=f(d(x,y))$ satisfies the given conditions, but it is not obvious that this is always possible.</p>
<p>The following properties are additionally useful, but not necessary: </p>
<ul>
<li>$\widetilde{d}$ preserves the topology of $(X,d)$</li>
<li>$d(x,y)>d(x',y')\implies \widetilde{d}(x,y)>\widetilde{d}(x',y')$</li>
</ul>
<p>Also, if this is possible when $(X,d)$ satisfies certain extra assumptions but not in general, answers in this direction are welcome.</p>
| Adrián González Pérez | 12,604 | <p>I do not know of any procedure for constructing such metric space in the general setting. On the contrary, if you restrict your class of metric spaces to the smaller class of complete locally compact path-metric spaces then the Hopf-Rinow theorem, see the First Chapter of [Gr], will prevent the construction of your $(X,\tilde{d})$.</p>
<p>If $(X,d)$ is in the intermediate class of (non-complete and non-compact) locally compact path-metric spaces, then I recall having seen the construction of a $(X,\tilde{d})$ satisfying</p>
<ol>
<li>$(X,\tilde{d})$ is a complete locally compact path-metric space</li>
<li>The topology of $X$ is preserved.</li>
<li>$d(x,y) \leq \tilde{d}(x,y)$.</li>
</ol>
<p>The procedure will be something like completing $(X,d)$, embedding $X$ inside its metric completion $\bar{X}$ and then enlarging the metric of $X$ around the points in $\bar{X} \setminus X$ (think of this as sending the extra points "to infinity"). I ignore whether your second condition can be preserved under this construction</p>
<p>[Gr] <em>Gromov, Misha</em>, Metric structures for Riemannian and non-Riemannian spaces. Progress in Mathematics (Boston, Mass.). 152. Boston, MA: Birkhäuser. xix, 585 p. (1999). <a href="https://zbmath.org/?q=an:0953.53002" rel="nofollow noreferrer">ZBL0953.53002</a>.</p>
|
114,340 | <p>Let $M\subset \mathbb C^2$ be a hypersurface defined by $F(z,w)=0$. Then for some point $p\in M$, I've
$$\text{ rank of }\left(
\begin{array}{ccc}
0 &\frac{\partial F}{\partial z} &\frac{\partial F}{\partial w} \\
\frac{\partial F}{\partial z} &\frac{\partial^2 F}{\partial ^ 2z} &\frac{\partial^2 F}{\partial z\partial w} \\
\frac{\partial F}{\partial w} &\frac{\partial^2 F}{\partial w\partial z} & \frac{\partial^2 F}{\partial w^2} \\
\end{array}
\right)_{\text{ at p}}=2.$$</p>
<p>What does it mean geometrically? Can anyone give a geometric picture near $p$? </p>
<p>Any comment, suggestion, please.</p>
<p>Edit: Actually I was reading about Levi flat points and Pseudo-convex domains. I want to understand the relation between these two concepts. A point p for which the rank of the above matrix is 2 is called Levi flat. If the surface is everywhere Levi flat then it is locally equivalent to $(0,1)\times \mathbb{C}^n$, so I have many examples....but what will happen for others for example take the three sphere in $\mathbb{C}^2$ given by $F(z,w)=|z|^2+|w|^2−1=0$. This doesn't satisfy the rank 2 condition. Can I have precisely these two situations?</p>
| Qiaochu Yuan | 232 | <p>The story extends to non-square matrices. Consider a collection $A$ of $a$ vertices, another collection $B$ of $b$ vertices, and a collection of directed edges from the first collection to the second (in other words, a bipartite graph, but where we choose explicitly a $2$-coloring of the vertices). We can organize this data into a $b \times a$ matrix whose entries describe the number of edges from a vertex in the first collection to a vertex in the second. </p>
<p>Now consider a third collection $C$ of $c$ vertices and some directed edges from vertices in $B$ to vertices in $C$. We can organize this data into a $c \times b$ matrix as above. </p>
<p>Then the product of the two matrices above is a $c \times a$ matrix describing how to get from vertices in $A$ to vertices in $C$ by following edges that go through $B$. </p>
<p>This idea is the basis of a combinatorial approach to linear algebra which is described, for example, in Brualdi and Cvetkovic's <a href="http://books.google.com/books?id=pwx6t8QfZU8C&dq=brualdi+combinatorial+linear+algebra&source=gbs_navlinks_s">A combinatorial approach to matrix theory and its applications</a>. </p>
<hr>
<p>For the initiated, the above admits the following nice interpretation: we have described composition of morphisms of a particularly nice <a href="http://en.wikipedia.org/wiki/Category_theory">category</a>. In fact, if I'm not mistaken, it's the free category with finite <a href="http://en.wikipedia.org/wiki/Biproduct">biproducts</a> on one object.</p>
|
1,359,743 | <p>If $\tan \theta = 3\cfrac{15}{16}$, then find $\sin \theta$.</p>
| Rick | 251,424 | <p>Let us rewrite $3 \frac{15}{16}$ as an improper fraction. So we get $3 \dfrac{15}{16} = \dfrac{16\cdot3+15}{16} = \dfrac{63}{16}$</p>
<p>We know that $\tan(\theta) = \dfrac{o}{a}$, where $o$ and $a$ are the opposite and adjacent sides of $\theta$. We also know through the Pythagorean theorem that the hypotenuse is $\sqrt{o^2+a^2} = \sqrt{63^2+16^2} = 65$. Recall that:
$$\sin(\theta) = \frac{o}{h}$$ where $h$ is the hypotenuse and $o$ is the opposite side. Subbing in the values we obtained and get:
$$\sin(\theta) = \frac{63}{65}$$</p>
|
1,403,228 | <blockquote>
<p><strong>Question:</strong></p>
<p>Let <span class="math-container">$m, n, q, r \in \mathbb Z$</span>. If <span class="math-container">$m = qn + r$</span>, show that <span class="math-container">$\gcd(m, n) = \gcd(n, r)$</span>. Hence justify the Euclidean Algorithm.</p>
</blockquote>
<p>I found this question in a past test paper, but cannot seem to find a reference in my textbook that indicates how I can go about "proving" the above statement. Can anyone please point me in the right direction?</p>
| Goob | 953,211 | <p>Let <span class="math-container">$m=qn+r$</span>. We have <span class="math-container">$gcd(m,n)=gcd(qn+r,n)=gcd(r,n)$</span>, where the last equality follows that if <span class="math-container">$d=gcd(qn+r,n)$</span> then since <span class="math-container">$d|n$</span> we know <span class="math-container">$d|qn$</span> so we can sort of ignore that term.</p>
|
2,871,084 | <p>Consider that we have a circle drawn a round the <strong>origin (0,0)</strong>. That circle has some points drawn on its circumference. Each of those points has range and azimuth $(r,\theta)$, the <strong>r</strong> and <strong>$\theta$</strong> values of these points are calculate with responding to the <strong>origin (0,0)</strong>. </p>
<p>I want to move or translate that circle and its (on-circumference) points to a new center (x,y), the new center can be in any quadrant. Here is an image for more demonstration (consider that it was translated to the first quadrant):<a href="https://i.stack.imgur.com/slaiD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/slaiD.png" alt="enter image description here"></a></p>
<p><strong>The question is what is the new $(r, \theta)$ of each point with responding to the origin (0,0) after translation to the new center?</strong></p>
| Batominovski | 72,152 | <p>Assume that $\mathcal{C}\big([a,b]\big)$ is taken modulo constant functions. Any function in the image of $R$ is differentiable everywhere except on a set of Lebesgue measure $0$, but not every continuous function has this property. A properly shifted and rescaled version of Weierstrass function as suggested by Quintic_Solver is a continuous function which is differentiable nowhere, so it does not lie in $\text{im}(R)$. </p>
<p>However, you can shift and rescale the <a href="https://en.wikipedia.org/wiki/Cantor_function" rel="nofollow noreferrer">Cantor function</a> to fit your interval $[a,b]$ to get a continuous function on $[a,b]$ that is not in the image of $R$, but the Cantor function is differentiable everywhere outside the <a href="https://en.wikipedia.org/wiki/Cantor_set" rel="nofollow noreferrer">Cantor set</a>, which has measure $0$. Therefore, the condition that a function is differentiable outside a set of Lebesgue measure $0$ is also insufficient to be in $\text{im}(R)$. Indeed, $\text{im}(R)$ consists of only (but not all) <a href="https://en.wikipedia.org/wiki/Absolute_continuity" rel="nofollow noreferrer">absolutely continuous functions</a> on $[a,b]$ vanishing at $a$ (and the Cantor function is not absolutely continuous).</p>
<p>Being absolutely continuous is also insufficient. As zhw. illustrated, the function $g(x):=\sqrt{x-a}$ is absolutely continuous, but any of its <a href="https://en.wikipedia.org/wiki/Weak_derivative" rel="nofollow noreferrer">weak derivative</a> is unbounded. Therefore, $g'$ is not Riemann-integrable. Every absolutely continuous function has a weak derivative. We know that every function in $\text{im}(R)$ is absolutely continuous. Thus, each function in $\text{im}(R)$ has a bounded weak derivative, whence <a href="https://en.wikipedia.org/wiki/Lipschitz_continuity" rel="nofollow noreferrer">Lipschitz continuous</a>.</p>
<p>To get all the image of $R$, we note that a (real- or complex-valued) function $f$ on $[a,b]$ is Riemann-integrable if and only if it is bounded and continuous outside a set of Lebesgue measure $0$. Hence, $F\in\text{im}(R)$ iff a weak derivative $F'$ of $F$ is bounded and continuous outside a set of Lebesgue measure $0$ ($F'$ exists because $F$ is absolutely continuous). This translates as follows: $\text{im}(R)$ consists of all functions $F$ that is Lipschitz continuous and continuously differentiable everywhere outside a set of Lebesgue measure $0$.</p>
|
3,170,368 | <p>The limit of <span class="math-container">$f(x,y)=\frac{xy^2}{x^2+y^4}$</span> as <span class="math-container">$(x,y) \longrightarrow(0,0)$</span> is doesn't exist, because if we take two paths:</p>
<p>1) Along the path, <span class="math-container">$x=0$</span>, <span class="math-container">$y\longrightarrow0$</span>:
<span class="math-container">$$\lim_{(x,y)\longrightarrow(0,0)}\frac{xy^2}{x^2+y^4}=0$$</span>
2) Along the path, <span class="math-container">$x=y^2$</span>, <span class="math-container">$y\longrightarrow0$</span>:
<span class="math-container">$$\lim_{(x,y)\longrightarrow(0,0)}\frac{xy^2}{x^2+y^4}=\lim_{y\longrightarrow0}\frac{y^2y^2}{(y^2)^2+y^4}=\frac{1}{2}$$</span></p>
<p>But if we use polar coordinate methods of evaluating limits of functions of two variable, we get the limit of the above function becomes zero, i.e.,
<span class="math-container">$x=r\cos(\theta),y=r\sin(\theta)$</span>, we know that <span class="math-container">$x^2+y^2=r^2$</span> and this indicates that <span class="math-container">$r\longrightarrow0$</span> as <span class="math-container">$(x,y)\longrightarrow(0,0)$</span>, therefore
<span class="math-container">$$\lim_{(x,y)\longrightarrow(0,0)}\frac{xy^2}{x^2+y^4}=\lim_{r\longrightarrow0}\frac{r^3\cos(\theta)\sin^2(\theta)}{r^2\cos^2(\theta)+r^4\sin^4(\theta)}$$</span>
<span class="math-container">$$=\lim_{r\longrightarrow0}\frac{r\cos(\theta)\sin^2(\theta)}{\cos^2(\theta)+r^2\sin^4(\theta)}=0$$</span>
Please someone help me on this contradictory, why this is so happened?.</p>
| Kavi Rama Murthy | 142,385 | <p>When you fixing <span class="math-container">$\theta$</span> and take limit as <span class="math-container">$ r \to 0$</span> you are taking limits along straight lines. It is not enough to take limits along these lines. When you take limit along the parabola <span class="math-container">$x=y^{2}$</span> we get a different value for the limit so the limit of the function does not exist. </p>
|
3,170,368 | <p>The limit of <span class="math-container">$f(x,y)=\frac{xy^2}{x^2+y^4}$</span> as <span class="math-container">$(x,y) \longrightarrow(0,0)$</span> is doesn't exist, because if we take two paths:</p>
<p>1) Along the path, <span class="math-container">$x=0$</span>, <span class="math-container">$y\longrightarrow0$</span>:
<span class="math-container">$$\lim_{(x,y)\longrightarrow(0,0)}\frac{xy^2}{x^2+y^4}=0$$</span>
2) Along the path, <span class="math-container">$x=y^2$</span>, <span class="math-container">$y\longrightarrow0$</span>:
<span class="math-container">$$\lim_{(x,y)\longrightarrow(0,0)}\frac{xy^2}{x^2+y^4}=\lim_{y\longrightarrow0}\frac{y^2y^2}{(y^2)^2+y^4}=\frac{1}{2}$$</span></p>
<p>But if we use polar coordinate methods of evaluating limits of functions of two variable, we get the limit of the above function becomes zero, i.e.,
<span class="math-container">$x=r\cos(\theta),y=r\sin(\theta)$</span>, we know that <span class="math-container">$x^2+y^2=r^2$</span> and this indicates that <span class="math-container">$r\longrightarrow0$</span> as <span class="math-container">$(x,y)\longrightarrow(0,0)$</span>, therefore
<span class="math-container">$$\lim_{(x,y)\longrightarrow(0,0)}\frac{xy^2}{x^2+y^4}=\lim_{r\longrightarrow0}\frac{r^3\cos(\theta)\sin^2(\theta)}{r^2\cos^2(\theta)+r^4\sin^4(\theta)}$$</span>
<span class="math-container">$$=\lim_{r\longrightarrow0}\frac{r\cos(\theta)\sin^2(\theta)}{\cos^2(\theta)+r^2\sin^4(\theta)}=0$$</span>
Please someone help me on this contradictory, why this is so happened?.</p>
| Dagnachew Jenber | 627,123 | <p>I asked this question a couple of days ago. I want to answer my question myself, I'm not fully satisfied with your answers. Let me get started by stating the following Theorem with out proof. </p>
<p><strong>Theorem 1.</strong> If the limits of <span class="math-container">$f(x,y)$</span> and <span class="math-container">$g(x,y)$</span> exists as <span class="math-container">$(x,y)\rightarrow(0,0)$</span>, then the limit of the product <span class="math-container">$f(x,y)g(x,y)$</span> exists as <span class="math-container">$(x,y)\rightarrow(0.0)$</span> and is equal to the product of individual limits.</p>
<p><strong>Theorem 2.</strong> If the limits of <span class="math-container">$f(x,y)$</span> exists and the limits of <span class="math-container">$g(x,y)$</span> doesn't exist other than <span class="math-container">$\pm \infty$</span>, then the limit of the product <span class="math-container">$f(x,y)g(x,y)$</span> doesn't exist.</p>
<p>Theorem 1 states that to be the limit of the product exists the limits of the individual functions must exist.</p>
<p>Now let me apply the <strong>Theorem 1</strong> and <strong>2</strong> to see the above example has no contradiction with the two methods, using paths and using polar coordinate method. As we have seen, the limit doesn't exist since we got two different limits for the same function along two different paths. Now let's check why the limit is doesn't exist when we use polar coordinate. </p>
<p>Let <span class="math-container">$x=r\cos(\theta), y=r\sin(\theta)$</span>, we know that <span class="math-container">$x^2+y^2=r^2$</span>, this indicates that <span class="math-container">$r\rightarrow 0$</span> as <span class="math-container">$(x,y)\rightarrow(0,0)$</span>. Therefore
<span class="math-container">$$\lim_{(x,y)\rightarrow(0,0)}\frac{xy^2}{x^2+y^4}=\lim_{r\rightarrow 0}\frac{r^3\cos(\theta)\sin^2(\theta)}{r^2\cos^2(\theta)+r^4\sin^4(\theta)}$$</span>
<span class="math-container">$$=\lim_{r\rightarrow 0}\frac{r\cos(\theta)\sin^2(\theta)}{\cos^2(\theta)+r^2\sin^4(\theta)}\ne\bigg(\lim_{r\rightarrow 0}r\bigg)\bigg(\lim_{r\rightarrow 0}\frac{\cos(\theta)\sin^2(\theta)}{\cos^2(\theta)+r^2\sin^4(\theta)}\bigg)$$</span>
As you see the right hand equation, which is the product of two limits <span class="math-container">$\lim_{r\rightarrow 0}r$</span> which exists, but the limit <span class="math-container">$\lim_{r\rightarrow 0}\frac{\cos(\theta)\sin^2(\theta)}{\cos^2(\theta)+r^2\sin^4(\theta)}$</span> doesn't exist, that is why I put <span class="math-container">$\ne$</span> in between according to <strong>Theorem 1</strong>. Therefore in this case we can't apply <strong>Theorem 1</strong> and by <strong>Theorem 2</strong> the limit doesn't exist. </p>
|
1,870,675 | <blockquote>
<p>Let $R$ be a commutative ring with unity, and let $R^{\times}$ be the group of units of $R$. Then is it true that $(R,+)$ and $(R^{\times},\ \cdot)$ are not isomorphic as groups ? </p>
</blockquote>
<p>I know that the statement is true in general for fields. And it is trivially true for any finite ring (as $|R^{\times}| \le |R|-1<|R|$, so they are not even bijective).</p>
<p>I can show that the groups are not isomorphic whenever $\operatorname{char} R \ne 2$ , but I am unable to deal with $\operatorname{char} R=2$ case ... Please help. Thanks in advance.</p>
| arctic tern | 296,782 | <p>Counterexample: $R=\mathbb{R}\times\mathbb{Z}_2$ satisfies $(R,+)\cong(R^\times,\cdot)$.</p>
|
221,403 | <p>I have a series of functions <span class="math-container">$f_d$</span> in <span class="math-container">$d$</span> variables and would like compute the sum of each one evaluated at each lattice point within the <span class="math-container">$d$</span>-sphere of radius <span class="math-container">$R$</span>; that is, at each point <span class="math-container">$(x_1,x_2 \dots x_d) \in \mathbb{Z}^d:\sqrt{x_1^2+x_2^2 \dots +x_d^2} \leq R$</span>.</p>
<p>I have no problem evaluating the functions <span class="math-container">$f_d$</span> at these points; I think the best way would be to use <code>Part</code> within <code>Sum</code> on the list of lattice points. <strong>What I am unsure of</strong> is how to generate <span class="math-container">$P_R$</span>, the list of lattice points not outside the <span class="math-container">$d$</span>-sphere of radius <span class="math-container">$R$</span>. </p>
| Vitaliy Kaurov | 13 | <h2>Method 1: select after building the grid</h2>
<p>The cubic lattice with side size <span class="math-container">$2r$</span> in <span class="math-container">$d$</span>-dimensions symmetric wrt origin can be defined as:</p>
<pre><code>latt[r_,d_]:=Tuples[Range[-r,r],d]
</code></pre>
<p>Then your points are selected as:</p>
<pre><code>pts[r_,d_]:=Select[latt[r,d],Norm[#]<=r&]
</code></pre>
<p>Example for 3-sphere:</p>
<pre><code>Graphics3D[{Point[pts[10,3]],{Opacity[.2],Sphere[{0,0,0},10]}}]
</code></pre>
<p><a href="https://i.stack.imgur.com/8iVNg.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8iVNg.gif" alt="enter image description here"></a></p>
<h2>Method 2: filter during building the grid</h2>
<p>Another way of doing this is to check points not after grid is build, but during the process:</p>
<pre><code>sphrPTS[r_,d_]:=
Flatten[Array[If[Norm[{##}]<=r,{##},Nothing]&,1+2Table[r,d],-r],d-1]
</code></pre>
<p>To test let's see now <span class="math-container">$2$</span>-sphere:</p>
<pre><code>Graphics[Point[sphrPTS[30, 2]]]
</code></pre>
<p><a href="https://i.stack.imgur.com/ZUvvj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZUvvj.png" alt="enter image description here"></a></p>
|
1,450,669 | <p>Find the value of $$\lim\limits_{x\to 0^+}[1+[x]]^{\frac2x}$$where $[x]$ denotes greatest integer function less than or equal to $x$. <br></p>
<hr>
<p><strong>My attempt:</strong> <br>
I calculated $[1+[x]]$ to be $1$ as $x\to 0^+$. <br> Now I am stuck. <br> Please help me.</p>
| Calvin Khor | 80,734 | <p>For $0<x\ll 1$, $[1+[x]] = 1$. Thus $[1+[x]]^{2/x} = 1^{2/x} = 1$.This gives
$$ \lim_{x→ 0^+}[1+[x]]^{2/x} = \lim_{x→ 0^+} 1 = 1$$</p>
|
3,493,729 | <p>Although I looked up the answer on integral calculator com but I still have little to no idea as to how one would proceed to solve this integral.
Integrate <span class="math-container">$\dfrac{e^x(x^4+2)}{(1+x^2)^{5/2}}$</span> wrt <span class="math-container">$x.$</span> I initially tried to convert it to the form <span class="math-container">$e^x\cdot(f(x)+f'(x)).$</span> However, I wasn't successful in spite of struggling for more than half an hour :(</p>
<p>Any help would be appreciated :)</p>
| Kavi Rama Murthy | 142,385 | <p>If it is onto then <span class="math-container">$f^{-1}(0)$</span> and <span class="math-container">$f^{-1}(1)$</span> are disjoint closed non-empty subsets of <span class="math-container">$[0,1]$</span> which contradicts connectedness of <span class="math-container">$[0,1]$</span>. Hence such a function does not exist. </p>
|
3,425,780 | <p>Let <span class="math-container">$H_i$</span> be a <span class="math-container">$\mathbb C$</span>-Hilbert space and <span class="math-container">$T$</span> be a densely-defined linear operator from <span class="math-container">$H_1$</span> to <span class="math-container">$H_2$</span>.</p>
<blockquote>
<p>How can we show that if <span class="math-container">$T$</span> is injective and <span class="math-container">$\operatorname{im}T$</span> is dense, then <span class="math-container">$T^\ast$</span> is injective as well? I've read that the reason is that <span class="math-container">$$\ker T^\ast=(\operatorname{im}T)^\perp=\{0\}\tag1,$$</span> but I don't get why <span class="math-container">$(1)$</span> holds.</p>
</blockquote>
<p>I know that for a general densely-defined <span class="math-container">$T$</span>, <span class="math-container">$\ker T^\ast=(\operatorname{im}T)^\perp$</span> and hence <span class="math-container">$(\ker T^\ast)^\perp=\overline{\operatorname{im}T}$</span>. On the other hand, the identity <span class="math-container">$\ker T=(\operatorname{im}T^\ast)^\perp$</span> can only be concluded, when <span class="math-container">$T$</span> is closable (since this is equivalent to <span class="math-container">$T^\ast$</span> being densely-defined).</p>
<p>So, assuming <span class="math-container">$\operatorname{im}T$</span> is dense, the only thing I was able to infer is that <span class="math-container">$$(\ker T^\ast)^\perp=\overline{\operatorname{im}T}=H_2\tag2.$$</span> Now if<span class="math-container">$T^\ast$</span> would be continuous, then <span class="math-container">$\ker T^\ast$</span> would be closed and hence <span class="math-container">$H_2=\ker T^\ast\oplus(\ker T^\ast)^\perp$</span>, which would immediately yield <span class="math-container">$\ker T^\ast=\{0\}$</span> and hence the claim.</p>
| Toby Mak | 285,313 | <p>The equation of the parabola is:</p>
<p><span class="math-container">$$a \left(x-\frac{1}{4}\right)^2 - \frac{9}{8}$$</span>
<span class="math-container">$$=a \left(x^2-\frac{1}{2}x+\frac{1}{16} \right)- \frac{9}{8}$$</span>
<span class="math-container">$$=ax^2-\frac{a}{2}x+ \left(\frac{a}{16}-\frac{9}{8} \right)$$</span></p>
<p>Then we find that: <span class="math-container">$$a+b+c = \frac{9}{16}(a-2).$$</span> </p>
<p>When <span class="math-container">$a=0$</span>, <span class="math-container">$a+b+c = -\frac{9}{8}$</span>. Since the gradient of the line is positive, if we try <span class="math-container">$a+b+c=-1$</span>, we will get a positive value of <span class="math-container">$a$</span>.</p>
<p>Therefore we have that: <span class="math-container">$-1 = \frac{9}{16}(a-2) \Rightarrow a = \frac{2}{9}$</span>.</p>
|
2,341,817 | <blockquote>
<p>If <span class="math-container">$ \log_p q + \log_q r + \log_r p = 0 $</span></p>
<p>Then what is the value of,</p>
<p><span class="math-container">$$(\log_p q)^3 + (\log_q r)^3 + (\log_r p)^3$$</span></p>
<p>given that <span class="math-container">$p,q,r \neq 1$</span></p>
<p>A. It is odd prime</p>
<p>B. It is even prime</p>
<p>C. Odd composite</p>
<p>D. Irrational</p>
</blockquote>
<p>I have tried using the identity that, if <span class="math-container">$ a + b + c = 0$</span>, <span class="math-container">$a^3+b^3+c^3= 3abc$</span> , but it gives me the answer <span class="math-container">$0$</span>.</p>
| Tianlalu | 394,456 | <p><strong>HINT</strong></p>
<p><br>
$x^3+y^3+z^3=3xyz$ when $x+y+z=0$.</p>
<p>AND</p>
<p><br>
$\log_p q=\frac{\log q}{\log p}$, $\log_q r=\frac{\log r}{\log q}$, $\log_r p=\frac{\log p}{\log r}$.</p>
|
2,341,817 | <blockquote>
<p>If <span class="math-container">$ \log_p q + \log_q r + \log_r p = 0 $</span></p>
<p>Then what is the value of,</p>
<p><span class="math-container">$$(\log_p q)^3 + (\log_q r)^3 + (\log_r p)^3$$</span></p>
<p>given that <span class="math-container">$p,q,r \neq 1$</span></p>
<p>A. It is odd prime</p>
<p>B. It is even prime</p>
<p>C. Odd composite</p>
<p>D. Irrational</p>
</blockquote>
<p>I have tried using the identity that, if <span class="math-container">$ a + b + c = 0$</span>, <span class="math-container">$a^3+b^3+c^3= 3abc$</span> , but it gives me the answer <span class="math-container">$0$</span>.</p>
| John D | 389,906 | <p>Based on the identity </p>
<p>$$a+b+c=0 \Rightarrow a^3+b^3+c^3 =3abc,$$ the answer is 3. Just note that </p>
<p>$\log_p(q)=\frac{ln(q)}{ln(p)},$ etc, so your hypothesis is just </p>
<p>$$\frac{ln(q)}{ln(p)}+\frac{ln(r)}{ln(q)}+\frac{ln(p)}{ln(r)}=0.$$ Put now </p>
<p>$$a= \frac{ln(q)}{ln(p)} , \; b=\frac{ln(r)}{ln(q)},\; c=\frac{ln(p)}{ln(r)}$$ and note that $abc=1.$</p>
<p>Hope this helps</p>
|
1,214,042 | <p>"A particle of mass m is attracted towards a fixed point 0 with a force inversely proportional to its instantaneous distance from 0. If the particle is released from rest, at a distance L from 0, find the time for it to reach 0"</p>
<p>My attempt at this question:
Noting the fact $x(0)=L$ and $x'(0)=0$, I tried to set the equation to this problem, $m\frac {d^2x}{dt^2}=\frac {-k}{x}$. Since we are covering laplace transform and gamma functions in my class right now, my attempt was to apply laplace transform to the equation above, however I am having trouble finding the answer. Please help me out. Thank you very much in advance for your help.</p>
| JJacquelin | 108,514 | <p>$$2m\frac {d^2x}{dt^2}\frac {dx}{dt}=\frac {-2k}{x}\frac {dx}{dt}$$
$$m(x')^2=-2k\ln(x)+constant$$
With conditions $x(0)=L$ and $x'(0)=0$ :
$$m(x')^2=-2k\ln(x)+2k\ln(L)$$
$x$ is decreassing, hense $x'\leq 0$
$$x'=-\sqrt{-\frac{2k}{m}\ln{\frac{x}{L}}}$$
$$\frac{dx}{\sqrt{\ln(L)-\ln(x)}}=-\sqrt{\frac{2k}{m}}dt$$
$$\int_{x=L}^{x=0}\frac{dx}{\sqrt{\ln(L)-\ln(x)}}=-\sqrt{\frac{2k}{m}}T$$
where $x(T)=0$</p>
<p>The change of variable $\sqrt{\ln(L)-\ln(x)}=y$ transforms the integral into the well-known Gauss integral.
$$L\sqrt{\pi}=\sqrt{\frac{2k}{m}}T$$
$$T=L\sqrt{\frac{\pi m}{2k}}$$</p>
|
2,462,852 | <p>I want to prove that $f(x)=x^3+x+2$, $f: \mathbb R \rightarrow \mathbb R$ is bijective without calculus. My attempts at showing to prove that it' injective and surjective are written below:</p>
<p>$1)$ Injectivity: </p>
<p>I want to show that $\forall a,b \in \mathbb R$ $f(a)=f(b) \implies a=b$.</p>
<p>I started like this:
$$f(a)=f(b) \implies a^3+a+2=b^3+b+2$$
$$\implies a(a^2+1)=b(b^2+1)$$
$$\implies \frac{a}{b}=\frac{b^2+1}{a^2+1}$$
Then I said since $\frac{b^2+1}{a^2+1}>0$ $\forall a,b \in \mathbb R$ then either $a \land b < 0$ or $a \land b > 0$. (For the case when $b=0 \land a \in \mathbb R$ it would be easy to prove that $a=b$.) From there it seemed pretty obvious that $\frac{a}{b}=\frac{b^2+1}{a^2+1} \implies a=b$ so I couldn't really draw a logical argument to show that $a=b$.</p>
<p>$2)$ Surjectivity: </p>
<p>I want to show that $\forall b \in \mathbb R$ $\exists a \in \mathbb R$ s.t. $f(a)=b$. </p>
<p>I started like this:</p>
<p>Let $b \in \mathbb R$ and set $f(a)=b$ then we have:</p>
<p>$$a^3+a+2=b$$
$$\implies a^3+a=b-2$$
$$\implies a(a^2+1)=b-2$$</p>
<p>But then I couldn't find an expression for $a \in \mathbb R$ in terms of $b$.
So I'm wondering if anyone can tell me how I can proceed with my surjectivity and injectivity proofs.</p>
| Piquito | 219,998 | <p>Injectivity is clear and so is for the fact that there is not a positive root. </p>
<p>Applying now Descartes's Rule we have
$$f(x)=x^3+x+2 \text{ have no change signs }\\f(-x)=-x^3-x+2 \text{ has one change sign }$$
Consequently the maximum number of negative roots is $0+1=1$ and $f^{-1}$ exists.</p>
|
284,995 | <p>There is a theorem that if a Hopf algebra $H$ is commutative or cocommutative, then $S^2=id_H$, where $S$ denotes the antipode.</p>
<p>May I know if the converse is true?</p>
<p>(i.e. if $S^2=id_H$, does it follow that $H$ is commutative or cocommutative?)</p>
<p>If no, what would be the simplest counter-example?</p>
<p>Sincere thanks for any help!</p>
| Bogdan | 13,737 | <p>The converse is not true, Sweedler's 4-dimensional Hopf algebra (first example of a non-comm and non-cocomm Hopf algebra) is a counter example. This example is completely taken from section 4.3.6 of "Hopf Algebras" by Dascalescu, Nastasescu and Raianu. </p>
<p>Let $k$ be a field of characteristic $\neq 2$. As a vector space, define $H = \langle 1, c, x, cx \rangle$ to be 4-dimensional over $k$. Define the multiplication by $c^2 = 1, x^2=0, xc = -cx$ and the unit to be $1$. Define the comultiplication $\Delta(c) = c \otimes c$ and $\Delta(x) = c \otimes x + x \otimes 1$, and the counit $\epsilon(c) = 1$ and $\epsilon(x) = 0$. Finally, let the antipode be $S(c) = c^{-1} = c$ and $S(x) = -cx$.</p>
<p>By using that $S$ is an antimorphism we have $S(cx) = S(x)S(c) = -cxc = xcc = x$ and thus $S^2(cx) = S(x) = -cx \neq cx$ because $k$ is of characteristic $\neq 2$.</p>
|
165,235 | <p>I have a weighted, directed graph with 100 vertices and the maximal number of edges, 9900. Are there <em>Mathematica</em> tools or packages available to visualize which edges have large weights? (If you know of non-<em>Mathematica</em> software, that's fine, too.)</p>
<p>Background: My institution has 100 majors, and I'm interested in studying how students change majors, where they start compared to where they end up.</p>
| Tugrul Temel | 60,365 | <p>To draw the subgraph with large weights, you might want to apply the following function:</p>
<pre><code>select[matrix_, lB_, uB_] :=
matrix*Map[Boole[lB <= # <= uB] &, matrix, {-1}];
sa = SparseArray[select[mat, .1, .6]];
weightedG = Graph[sa["NonzeroPositions"], EdgeWeight -> sa["NonzeroValues"], DirectedEdges -> True];
GraphPlot[weightedG]
</code></pre>
<p>You can apply the <code>select[...]</code> function for various domains <code>{lB, uB}</code> on matrix <code>mat</code> and draw the directed graph only for the desired domain.</p>
|
2,939,605 | <p>The given task goes as follows:</p>
<blockquote>
<p>Show that <span class="math-container">$ f: \mathbb{R} \longrightarrow \mathbb{R}$</span> defined by <span class="math-container">$f(x) = \sqrt{1 + x^2} $</span> is not a polynomial function.</p>
</blockquote>
<p>I tried this approach - if <span class="math-container">$f(x)$</span> is a <span class="math-container">$n$</span>-degree polynomial function, then the <span class="math-container">$(n+1)$</span>-st derivative equals to 0 and I was trying to determine the <span class="math-container">$k$</span>-th derivative of <span class="math-container">$f(x)$</span> (and show it differs from 0 for any <span class="math-container">$k$</span>) but without success. Since <span class="math-container">$f(x)$</span> is continuous and defined over whole R domain, I have no idea how to carry on. Any ideas? </p>
| nonuser | 463,553 | <p>Suppose it is a polynomial, then <span class="math-container">$$\sqrt{1+x^2}= ax^n+...+b$$</span></p>
<p>so <span class="math-container">$$1+x^2= (ax^n+...+b)^2 = a^2x^{2n}+...$$</span></p>
<p>Since polynomial are equal iff they have same degree we have <span class="math-container">$2=2n$</span> so <span class="math-container">$n=1$</span> and now we have
<span class="math-container">$$ 1+x^2 = (ax+b)^2 = a^2x^2+2abx+b^2 \implies ab = 0$$</span>
a contradiction, since <span class="math-container">$|a|=|b|= 1$</span>.</p>
|
2,846,700 | <p>Consider the functions $f(x)$, $g(x)$, $h(x)$, where $f(x)$ is neither odd nor even, $g(x)$ is even and $h(x)$ is odd. Is it possible for $f(x) + g(x)$ to be </p>
<ol>
<li>even;</li>
<li>odd?</li>
</ol>
<p>For the second case I can imagine for example $f(x) = x - 1$ and $g(x) = 1$. Then $f$ is neither even nor odd and $g$ is even but their sum is odd, hence it's possible to get odd function from the sum of neither odd nor even and even function.</p>
<p>It feels like $f(x) + g(x)$ can never be even, but I couldn't manage to prove that.</p>
<p>I've tried to do it the following way:
Let $f(x) = - g(x) - h(x)$, which doesn't contradict the initial statement. Then we can express $g(x)$ and $h(x)$ and see whether the facts that they are either even or odd holds, but this always leads to valid equations:</p>
<p>$$
h(x) = \frac{f(-x) - f(x)}{2} \;\;\; \text{is an odd function} \\
g(x) = \frac{-f(x) - f(-x)}{2} \;\;\; \text{is an even function}
$$</p>
<p>I'm stuck at that point.</p>
<blockquote>
<p>How can I prove/disprove that $f(x) + g(x)$ may be even?</p>
</blockquote>
| Cato | 357,838 | <p>f(x) = f(-x)</p>
<p>$g(x) \ne g(-x)$ for some x</p>
<p>z(x) = f(x) + g(x)</p>
<p>then $z(-x) = f(-x) + g(-x) \ne f(x) + g(x) $ for some x</p>
|
1,718,391 | <p>The sequence is </p>
<p>$$\sum_{n=1}^{\infty} \frac{n-1}{n!}$$</p>
<p>I could show that it converges using the Ratio test, but evaluating it seems to be hard. I'm trying to avoid power series and Taylor series and such. Is it possible to solve this some other, simpler way? </p>
<p>I tried writing out the first few terms, but since it's non-geometric, it doesn't seem to help. </p>
| DonAntonio | 31,254 | <p>Your sum is only</p>
<p>$$\sum_{n=1}^\infty\frac{n-1}{n!}=\sum_{\color{red}{n=0}}^\infty\overbrace{\frac1{n!}}^{=\frac n{n!}=\frac1{(n-1)!}}-\sum_{\color{red}{n=1}}^\infty\frac1{n!}=e^1-(e^1-1)=1$$</p>
|
701,110 | <p>A particular combinatorial sequence I was looking at turned out to obey the following pair of recurrence relations:</p>
<p>$$N_{2n+1}=\sum^n_{k=0}N_{2k}$$
$$N_{2n}=\left(\sum^{n-1}_{k=0}N_{2k+1}\right)+1=\sum^n_{k=0}N_{2k-1}$$</p>
<p>For the second form of the second equation, I'm using the convention $N_{-1}=1$.</p>
<p>The even terms depend on the odd terms and vice versa, but by applying these relations twice and rearranging sums, we can express the odd members only in terms of previous odd members, and similarly for even members. I'll skip the details and focus on the odd terms as an example. Setting $x_n=N_{2n+1}$, we have (if my calculations are correct):</p>
<p>$$x_{n+1}=\sum^n_{i=-1}(n-i)x_i$$</p>
<p>Note that by definition $x_{-1}=N_{-1}=1$.</p>
<p>Are there any general methods for solving a recurrence such as this one, where each term depends on <em>all</em> previous terms?</p>
| Ross Millikan | 1,827 | <p>Yes, this is preserved as we go to higher dimensions as long as the number of terms is the same as the dimension of the space. The Pythagorean theorem (iterated) says this sum of squares is the square of the distance from the origin to the point.</p>
|
701,110 | <p>A particular combinatorial sequence I was looking at turned out to obey the following pair of recurrence relations:</p>
<p>$$N_{2n+1}=\sum^n_{k=0}N_{2k}$$
$$N_{2n}=\left(\sum^{n-1}_{k=0}N_{2k+1}\right)+1=\sum^n_{k=0}N_{2k-1}$$</p>
<p>For the second form of the second equation, I'm using the convention $N_{-1}=1$.</p>
<p>The even terms depend on the odd terms and vice versa, but by applying these relations twice and rearranging sums, we can express the odd members only in terms of previous odd members, and similarly for even members. I'll skip the details and focus on the odd terms as an example. Setting $x_n=N_{2n+1}$, we have (if my calculations are correct):</p>
<p>$$x_{n+1}=\sum^n_{i=-1}(n-i)x_i$$</p>
<p>Note that by definition $x_{-1}=N_{-1}=1$.</p>
<p>Are there any general methods for solving a recurrence such as this one, where each term depends on <em>all</em> previous terms?</p>
| copper.hat | 27,978 | <p>Sure, if $Q$ is a rotation matrix ($Q^T Q = I$), then with the Euclidean norm we have
$\|x\|^2 = \langle x , x \rangle = \langle x , Q^T Qx \rangle = \langle Qx , Qx \rangle = \|Qx\|^2$.</p>
|
873,224 | <p>I'm not mathematically inclined, so please be patient with my question.</p>
<p>Given </p>
<ul>
<li><p>$(x_0, y_0)$ and $(x_1, y_1)$ as the endpoints of a cubic Bezier curve.</p></li>
<li><p>$(c_x, c_y)$ and r as the centerpoint and the radius of a circle.</p></li>
<li><p>$(x_0, y_0)$ and $(x_1, y_1)$ are on the circle.</p></li>
<li><p>if it makes the calculation simpler, it's safe to assume the arc is less than or equal to $\frac{\pi}{2}$.</p></li>
</ul>
<p>How do I calculate the two control points of the Bezier curve that best fits the arc of the circle from $(x_0, y_0)$ to $(x_1, y_1)$?</p>
| me22 | 875,771 | <p>As a different way to choose where the error falls, one could pick the control points such that the <em>enclosed area</em> matches that of the circle.</p>
<p>If we consider the first quarter of a circle (the same as the first image in the other answer), then the area should be <span class="math-container">$\pi/4$</span>, so we need the integral</p>
<p><span class="math-container">$$
\begin{align}
\pi/4
&= \int_{0}^{1} y(t)x'(t)dt \\
&= \int_{0}^{1} ( (1-t)^3 + 3(1-t)^2t + 3(1-t)t^2c )( 3(1-t)^2c + 6(1-t)t(1-c) ) dt \\
&= -\frac{3c^2}{20}+\frac{3c}{5} + \frac12 \\
c &= 2 - \sqrt{\frac13(22-5\pi)} \\
&≈ 0.5517784778...
\end{align}
$$</span></p>
<p>That's slightly smaller than the <a href="https://spencermortensen.com/articles/bezier-circle/" rel="nofollow noreferrer">other two answers</a>, as it attempts to even out the error between over- and under-shooting the circle, rather than minimizing the maximum error.</p>
<p><em>Aside: Edits appreciated to add a solution for angles other than 90°. My maths are rusty.</em></p>
|
1,390,593 | <p>"Since the operation of left multiplication is faithful, $G$ is isomorphic to its image in $\operatorname{Perm}(G)$. If $G$ has order $n$, $\operatorname{Perm}(G)$ is isomorphic to $S_n$." - Artin's Algebra</p>
<p>So left multiplication as an action means $\phi_g(x)=gx$</p>
<p>So proving isomorphism, means bijective homomorphism, but I don't understand:</p>
<p>Injective: $ga_1=b$ and $ga_2=b$ $\implies a_1=g^{-1}b=a_2,\square$</p>
<p>Surjective: $\forall b\in G, ga=b$ where you can let $a=g^{-1}b$ where clearly $g^{-1}b\in G$ whenever $g,b\in G$. $\square$</p>
<p>But then homomorphism fails since $\phi_g(ab)=gab\ne\phi_g(a)\phi_g(b)=gagb$</p>
| mich95 | 229,072 | <p>$\phi(g)$ is bijective as you said, so it belongs to $Perm(G) \cong S_{n}$. Now the map, used in Cayley's theorem, is $f:G \to Perm(G)$, $f(g) =\phi_{g}$. $\phi$ is not necessarly a homomorphism, but $f$ is a homomorphism, in fact injective. Let me know if you need more details!</p>
|
3,166,201 | <blockquote>
<p>If <span class="math-container">$x$</span> and <span class="math-container">$y$</span> are prime numbers which satisfy <span class="math-container">$x^2-2y^2=1$</span>, solve for <span class="math-container">$x$</span> and <span class="math-container">$y$</span>.</p>
</blockquote>
<p><strong>My attempt</strong>: </p>
<p><span class="math-container">$x^2-2y^2=1$</span></p>
<p><span class="math-container">$\implies (x+\sqrt{2}y)(x-\sqrt{2}y)=1$</span></p>
<p><span class="math-container">$\implies (x+\sqrt{2}y)=1$</span> and <span class="math-container">$(x-\sqrt{2}y)=1$</span></p>
<p><span class="math-container">$\implies x=1$</span> and <span class="math-container">$y=0$</span></p>
<p>Clearly <span class="math-container">$x$</span> and <span class="math-container">$y$</span> are not prime numbers . Why is my solution not working. I have been able to solve similar type of equations by factorizing and then listing down the integer factors and the different cases. Why is it not working here?</p>
| Mann | 126,204 | <p>The fault is that irrationals can also produce the product to <span class="math-container">$1$</span>.<br>
Consider <span class="math-container">$x=3$</span> and <span class="math-container">$y=2$</span> then we get, <span class="math-container">$(3+\sqrt{2} \cdot 2)(3-\sqrt{2}\cdot 2)=1$</span><br>
Hence, the fault is moving from step 2 to step 3. You should look under the ring of <span class="math-container">$a+b\sqrt{2}$</span> in that step. </p>
|
1,897,163 | <p>Let $\frac{x^2}{y}\to c\neq 0$ as $(x,y)\to (0,0)^+$. Why does this mean that $y$ goes faster to $0$ as $x$ does?</p>
<p>For values of x and y that are small enough, we surely do not have $x^2/y=1$ so they cannot be identical, okay. Moreover, if the quotient would tend to 0, this would mean that $x$ would be faster decaying to 0 than y. </p>
| guestDiego | 338,527 | <p>Of course $f_n(0)=0$ for any $n$ and there is nothing to prove. Suppose $x\neq 0$. Then $1+x^2=:q>1$ and
$$
\lim_{n\to +\infty} \frac{1}{q^n}=0,
$$
therefore
$$
\lim_{n\to +\infty}f_n(x)=x\cdot \lim_{n\to +\infty} \frac{1}{q^n}=x\cdot 0=0,
$$</p>
|
3,575,015 | <p>This question is taken out of the text book High Dimensional Probability by Roman. I am unsure from the context if it means locally or globally Lipschitz. But the question is as follows:</p>
<p>Every differentiable function <span class="math-container">$f:\mathbb{R}^n\rightarrow{\mathbb{R}}$</span> is Lipschitz, and </p>
<p><span class="math-container">$\|f\|_{Lip}\le{∥∇f∥_∞}$</span> .</p>
<p>I have shown that if a function is continuously differentiable, it is locally Lipchitz. But since the question uses <span class="math-container">$\|f\|_{Lip}$</span> I believe I am to show that if a function is only differentiable, it is globally Lipchitz.</p>
<p>Any hints / clues?</p>
| Cesareo | 397,348 | <p>Considering</p>
<p><span class="math-container">$$
f''+\alpha f'+\beta f = 0
$$</span></p>
<p>with <span class="math-container">$\alpha = \frac ba, \beta = \frac ca$</span> after solving the ODE with the established initial conditions we get</p>
<p><span class="math-container">$$
\frac{f_1}{f_t} = \frac{\left(\alpha \left(e^{\sqrt{\alpha ^2-4 \beta }}-1\right)+\sqrt{\alpha ^2-4 \beta }
\left(e^{\sqrt{\alpha ^2-4 \beta }}+1\right)\right) e^{\frac{1}{2} (t-1) \left(\sqrt{\alpha ^2-4 \beta
}+\alpha \right)}}{\alpha \left(e^{t \sqrt{\alpha ^2-4 \beta }}-1\right)+\sqrt{\alpha ^2-4 \beta }
\left(e^{t \sqrt{\alpha ^2-4 \beta }}+1\right)}
$$</span></p>
<p>now making <span class="math-container">$\beta=0$</span> we have</p>
<p><span class="math-container">$$
\frac{f_1}{f_t}{\Big |}_{\beta=0}=\frac{\left(\alpha \left(e^{\alpha }-1\right)+\alpha \left(e^{\alpha }+1\right)\right) e^{\alpha
(t-1)}}{\alpha \left(e^{\alpha t}-1\right)+\alpha \left(e^{\alpha t}+1\right)}=1
$$</span></p>
<p>Follows a plot showing in light blue in the plane <span class="math-container">$\alpha\times\beta$</span>, the stable region such that <span class="math-container">$\lim_{t\to \infty}\frac{f_1}{f_t}=0$</span>. In red is the stable region associated to <span class="math-container">$\alpha > 0, \beta = 0$</span></p>
<p><a href="https://i.stack.imgur.com/r4tnB.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/r4tnB.jpg" alt="enter image description here"></a></p>
<p>Note the parabolic shape associated to <span class="math-container">$\alpha^2-4\beta =0$</span></p>
|
651,027 | <p><img src="https://i.stack.imgur.com/SIdAQ.jpg" alt="enter image description here"></p>
<p>I'm doing this Solving Abstract Problem but I'm not sure which one it is. I mean from the Series I can see there's a pattern but in the Options I don't see images that link with the Series. Do you have any idea - unless it's B but it's note very clear.</p>
| lhf | 589 | <p>I think the answer is E.</p>
<p>At each step in the series, X goes round 45 degrees counterclockwise while the circle goes round 45 degrees clockwise. </p>
|
1,041,623 | <p>I have been suggested to read the <a href="http://press.princeton.edu/chapters/gowers/gowers_VIII_6.pdf"><em>Advice to a Young Mathematician</em> section </a> of the <em>Princeton Companion to Mathematics</em>, the short paper <a href="http://alumni.media.mit.edu/~cahn/life/gian-carlo-rota-10-lessons.html"><em>Ten Lessons I wish I had been Taught</em></a> by Gian-Carlo Rota, and the <a href="http://terrytao.wordpress.com/career-advice/"><em>Career Advice</em></a> section of Terence Tao's blog, and I am amazed by the intelligence of the pieces of advice given in these pages. </p>
<p>Now, I ask to the many accomplished mathematicians who are active on this website if they would mind adding some of their own contributions to these already rich set of advice to novice mathematicians. </p>
<p>I realize that this question may be seen as extremely opinion-based. However, I hope that it will be well-received (and well-answered) because, as Timothy Gowers put it,</p>
<blockquote>
<p>"The most important thing that a young mathematician needs to learn is
of course mathematics. However, it can also be very valuable to learn
from the experiences of other mathematicians. The five contributors to
this article were asked to draw on their experiences of mathematical
life and research, and to offer advice that they might have liked to
receive when they were just setting out on their careers."</p>
</blockquote>
| Adam Ledger | 504,652 | <p>If I could give a younger me advice, I would tell myself to lower my pride and accept as much help as I could in the times it was available to me, and shed my fear of humiliation and inferiority that kept me from even considering talking to others entirely let alone other mathematicians.</p>
<p>While you should strive to be more receptive to constructive criticism every day, I also think it's important to disregard criticism when it's source is from an individual that considers their authority in the field in question to be a conclusive presumption that must be held by yourself and all others.</p>
|
331,543 | <blockquote>
<p>Given a fair six-sided die. Find the probability generating functions for the number of the throw on which the rth six appears. Hence find the probability that the fifth six occurs on the 20th throw. </p>
</blockquote>
<p>For the rth six is it true that $X~Geo((1/6)^r)$, so PGF is $(1/6)^rt+(1/6)^r\times(1-(1/6)^r)t^2$ etc? And for the 20th throw do I look for the coefficient of $t^{20}$?</p>
| Marko Riedel | 44,883 | <p>We will calculate the PGF as asked for in the problem. The probability that the $r$-th six occurs after $n$ rolls of the die is
$$ \frac{1}{6} \binom{n-1}{r-1}
\left(\frac{5}{6}\right)^{n-r} \left(\frac{1}{6}\right)^{r-1}.$$
Therefore the PGF is
$$p(z) = \sum_{n\ge r} \binom{n-1}{r-1}
\left(\frac{5}{6}\right)^{n-r} \left(\frac{1}{6}\right)^r z^n =
\left(\frac{1}{6}\right)^r \left(\frac{6}{5}\right)^r
\sum_{n\ge r} \binom{n-1}{r-1}
\left(\frac{5}{6}\right)^n z^n \\=
\left(\frac{1}{5}\right)^r
\sum_{n\ge r} \binom{n-1}{r-1}
\left(\frac{5}{6}\right)^n z^n =
\left(\frac{1}{5}\right)^r
\sum_{n\ge 0} \binom{n+r-1}{r-1}
\left(\frac{5z}{6}\right)^{n+r} \\=
\left(\frac{z}{6}\right)^r
\sum_{n\ge 0} \binom{n+r-1}{r-1}
\left(\frac{5z}{6}\right)^n.$$</p>
<p>Now note that $$ \sum_{n\ge 0} \binom{n+r-1}{r-1} q^n = \frac{1}{(1-q)^r}$$
so that $$ p(z) = \left(\frac{z}{6}\right)^r \frac{1}{(1-5z/6)^r} =
\left(\frac{z}{6-5z} \right)^r$$</p>
<p>It follows that the desired probability is
$$ [z^{20}] \left(\frac{z}{6}\right)^5 \frac{1}{(1-5z/6)^5} =
\left(\frac{1}{6}\right)^5 [z^{15}] \frac{1}{(1-5z/6)^5} =
\left(\frac{1}{6}\right)^5 \binom{15+4}{4}
\left( \frac{5}{6}\right)^{15} \sim
0.0323525. $$</p>
|
979,947 | <p>If $(A, B, C)$ are distinct integers $> 1$, and $$f(A, B, C) = \frac{\frac{A^2-1}{A} + \frac{B^2-1}{B}}{\frac{C^2-1}{C}},$$ then for what (if any) triplets $(A, B, C)$ is $f(A, B, C)$ an integer?</p>
<p>UPDATE: I've done some more work and have come up with the following: Based on the equations I combined to arrive at $f(A,B,C)$, it follows that: $$f(A,B,C)>2\Rightarrow A>B>C>1$$ and $$f(A,B,C)=2\Rightarrow A>C>B>1$$</p>
<p>My further work: </p>
<p>Let $f(A,B,C)=k$ where $k>1$ is an integer. Rewrite $f(A,B,C)$ as $$k=\frac{A-\frac{1}{A}+B-\frac{1}{B}}{C-\frac{1}{C}}$$</p>
<p>Thus $$A-\frac{1}{A}+B-\frac{1}{B}-k(C-\frac{1}{C})=A-\frac{1}{A}+B-\frac{1}{B}-kC+\frac{k}{C}=0$$</p>
<p>Note that $\frac{1}{A}+\frac{1}{B}<1$ for all integer $A,B$ such that $A>B>1$.</p>
<p>Now we consider three cases: $1)$ $k<C$, $2)$ $k=C$, $3)$ $k>C$.</p>
<p>Case $1$: $k<C$</p>
<p>Since $\frac{k}{C}<1$ it follows that these equations hold:$$A+B=kC$$ $$\frac{1}{A}+\frac{1}{B}=\frac{A+B}{AB}=\frac{k}{C}$$ </p>
<p>By substitution $$\frac{kC}{AB}=\frac{k}{C}\Rightarrow AB=C^2$$</p>
<p>This is only true if $A=np^2$, $B=nq^2$, and $C=npq$ for integers $n,p,q\geq 1$ and $p>q$. From this we have $$p^2>pq; pq>q^2\Rightarrow p^2>pq>q^2\Rightarrow np^2>npq>nq^2\Rightarrow A>C>B>1\Rightarrow k\leq2$$ This last condition, $k\leq2$, follows from the conditions stated above and leaves us with two possibilities: $k=2$ or $k=1$.</p>
<p>If $k=1$ we have $A+B=C$ which violates the condition $A>C$.</p>
<p>If $k=2$ we have $A+B=2C\Rightarrow \frac{A+B}{2}=C$. By substitution we have $$\frac{A+B}{AB}=\frac{2}{\frac{A+B}{2}}$$ $$4AB=(A+B)^2\Rightarrow A^2-2AB+B^2=0\Rightarrow (A-B)^2=0$$</p>
<p>However, this leads to the result that $A=B$ which violates that stated conditions.</p>
<p>Case $1$, $k<C$, fails.</p>
<p>Case $2$: $k=C$</p>
<p>Since $\frac{k}{C}=1$ it follows that $$A+B+1=C^2$$ $$\frac{1}{A}+\frac{1}{B}=0$$</p>
<p>This later equation is clearly false and thus Case $2$, $k=C$ fails.</p>
<p>Case $3$: $k>C$</p>
<p>Let $p,q$ be integers with $C>q\geq0$ and $p>1$. From this we can write $k$ as $k=pC+q$.</p>
<p>Note that $q=0 \Rightarrow A+B+p=C^2$ and $\frac{1}{A}+\frac{1}{B}=0$ (as above) which is clearly a contradiction, so $q\geq1$.</p>
<p>Substituting in $k$ above we have $$A-\frac{1}{A}+B-\frac{1}{B}-pC^2-qC+p+\frac{q}{C}=0$$ </p>
<p>Applying the same logic as above we have the following equations $$\frac{A+B}{AB}=\frac{q}{C}$$ $$A+B=pC^2+qC-p$$</p>
<p>By substitution we have the following system of equations (one cubic and one quadratic) $$pC^3+qC^2-pC=qAB\Rightarrow pC^3+qC^2-pC-qAB=0$$ $$pC^2+qC-p-A-B=0$$</p>
<p>Since at least one $C$ that works must solve both equations we denote the roots of the cubic as $x_1,x_2,x_3$ and the roots of the quadratic as $x_1,y_2$.</p>
<p>By Vieta's formulas we have $$x_1+x_2+x_3=-\frac{q}{p}$$ $$x_1 x_2+x_1 x_3+x_2 x_3=-1$$ $$x_1 x_2 x_3=\frac{qAB}{p}$$</p>
<p>and $$x_1+y_2=-\frac{q}{p}$$ $$x_1 y_2= -\frac{p+A+B}{p}$$</p>
<p>From the first equations (setting $x_1+x_2+x_3=x_1+y_2$) we have $y_2=x_2+x_3$.</p>
<p>Thus $$x_1 y_2=x_1 x_2+x_1 x_3=-\frac{p+A+B}{p}\Rightarrow x_2 x_3-\frac{p+A+B}{p}=-1\Rightarrow x_2 x_3=\frac{A+B}{p}$$ $$x_2 x_3=\frac{A+B}{p}\Rightarrow x_1\frac{A+B}{p}=\frac{qAB}{p}\Rightarrow x_1=\frac{qAB}{A+B}$$</p>
<p>Since $x_1=C$ and $B>C$ we have $$B>\frac{qAB}{A+B}\Rightarrow B^2>(q-1)AB$$</p>
<p>Since $A>B$ it follows $AB>B^2$ and so $q-1<1\Rightarrow q<2$. Since $q\neq0$ (see above) we have $q=1$ and so $C=\frac{AB}{A+B}$.</p>
<p>Substituting for $C$ in the quadratic (although the cubic has the same result) we have $$0=p(\frac{AB}{A+B})^2+\frac{AB}{A+B}-p-A-B$$ Which, after a fair bit of rearranging, becomes $$p=\frac{(A+B)^3-AB(A+B)}{(AB)^2-(A+B)^2}$$ Thus $f(A,B,C)$ is an integer (with $A,B,C$ constrained as above) iff $$p=\frac{(A+B)^3-AB(A+B)}{(AB)^2-(A+B)^2}$$ has integer solutions $p>0$ and $A>B>2$.</p>
<p>Any help with solving this last bit would be greatly appreciated. So far guesswork and Wolfram Alpha have failed to produce results.</p>
| Pieter21 | 170,149 | <p>Always take the educated guess. You often can exclude one or two silly answers, and with chance 1 out of 3 you beat leaving it blank with a few points.</p>
<p>For example: what is the main chemical element on the moon?</p>
<ol>
<li>Water</li>
<li>Cheese</li>
<li>Oxygen</li>
<li>Iron</li>
<li>Silicon</li>
</ol>
<p>Well, if you know it's not cheese, and even Water is not an element, you can take a random pick out the 3 other options to beat the odds.</p>
|
979,947 | <p>If $(A, B, C)$ are distinct integers $> 1$, and $$f(A, B, C) = \frac{\frac{A^2-1}{A} + \frac{B^2-1}{B}}{\frac{C^2-1}{C}},$$ then for what (if any) triplets $(A, B, C)$ is $f(A, B, C)$ an integer?</p>
<p>UPDATE: I've done some more work and have come up with the following: Based on the equations I combined to arrive at $f(A,B,C)$, it follows that: $$f(A,B,C)>2\Rightarrow A>B>C>1$$ and $$f(A,B,C)=2\Rightarrow A>C>B>1$$</p>
<p>My further work: </p>
<p>Let $f(A,B,C)=k$ where $k>1$ is an integer. Rewrite $f(A,B,C)$ as $$k=\frac{A-\frac{1}{A}+B-\frac{1}{B}}{C-\frac{1}{C}}$$</p>
<p>Thus $$A-\frac{1}{A}+B-\frac{1}{B}-k(C-\frac{1}{C})=A-\frac{1}{A}+B-\frac{1}{B}-kC+\frac{k}{C}=0$$</p>
<p>Note that $\frac{1}{A}+\frac{1}{B}<1$ for all integer $A,B$ such that $A>B>1$.</p>
<p>Now we consider three cases: $1)$ $k<C$, $2)$ $k=C$, $3)$ $k>C$.</p>
<p>Case $1$: $k<C$</p>
<p>Since $\frac{k}{C}<1$ it follows that these equations hold:$$A+B=kC$$ $$\frac{1}{A}+\frac{1}{B}=\frac{A+B}{AB}=\frac{k}{C}$$ </p>
<p>By substitution $$\frac{kC}{AB}=\frac{k}{C}\Rightarrow AB=C^2$$</p>
<p>This is only true if $A=np^2$, $B=nq^2$, and $C=npq$ for integers $n,p,q\geq 1$ and $p>q$. From this we have $$p^2>pq; pq>q^2\Rightarrow p^2>pq>q^2\Rightarrow np^2>npq>nq^2\Rightarrow A>C>B>1\Rightarrow k\leq2$$ This last condition, $k\leq2$, follows from the conditions stated above and leaves us with two possibilities: $k=2$ or $k=1$.</p>
<p>If $k=1$ we have $A+B=C$ which violates the condition $A>C$.</p>
<p>If $k=2$ we have $A+B=2C\Rightarrow \frac{A+B}{2}=C$. By substitution we have $$\frac{A+B}{AB}=\frac{2}{\frac{A+B}{2}}$$ $$4AB=(A+B)^2\Rightarrow A^2-2AB+B^2=0\Rightarrow (A-B)^2=0$$</p>
<p>However, this leads to the result that $A=B$ which violates that stated conditions.</p>
<p>Case $1$, $k<C$, fails.</p>
<p>Case $2$: $k=C$</p>
<p>Since $\frac{k}{C}=1$ it follows that $$A+B+1=C^2$$ $$\frac{1}{A}+\frac{1}{B}=0$$</p>
<p>This later equation is clearly false and thus Case $2$, $k=C$ fails.</p>
<p>Case $3$: $k>C$</p>
<p>Let $p,q$ be integers with $C>q\geq0$ and $p>1$. From this we can write $k$ as $k=pC+q$.</p>
<p>Note that $q=0 \Rightarrow A+B+p=C^2$ and $\frac{1}{A}+\frac{1}{B}=0$ (as above) which is clearly a contradiction, so $q\geq1$.</p>
<p>Substituting in $k$ above we have $$A-\frac{1}{A}+B-\frac{1}{B}-pC^2-qC+p+\frac{q}{C}=0$$ </p>
<p>Applying the same logic as above we have the following equations $$\frac{A+B}{AB}=\frac{q}{C}$$ $$A+B=pC^2+qC-p$$</p>
<p>By substitution we have the following system of equations (one cubic and one quadratic) $$pC^3+qC^2-pC=qAB\Rightarrow pC^3+qC^2-pC-qAB=0$$ $$pC^2+qC-p-A-B=0$$</p>
<p>Since at least one $C$ that works must solve both equations we denote the roots of the cubic as $x_1,x_2,x_3$ and the roots of the quadratic as $x_1,y_2$.</p>
<p>By Vieta's formulas we have $$x_1+x_2+x_3=-\frac{q}{p}$$ $$x_1 x_2+x_1 x_3+x_2 x_3=-1$$ $$x_1 x_2 x_3=\frac{qAB}{p}$$</p>
<p>and $$x_1+y_2=-\frac{q}{p}$$ $$x_1 y_2= -\frac{p+A+B}{p}$$</p>
<p>From the first equations (setting $x_1+x_2+x_3=x_1+y_2$) we have $y_2=x_2+x_3$.</p>
<p>Thus $$x_1 y_2=x_1 x_2+x_1 x_3=-\frac{p+A+B}{p}\Rightarrow x_2 x_3-\frac{p+A+B}{p}=-1\Rightarrow x_2 x_3=\frac{A+B}{p}$$ $$x_2 x_3=\frac{A+B}{p}\Rightarrow x_1\frac{A+B}{p}=\frac{qAB}{p}\Rightarrow x_1=\frac{qAB}{A+B}$$</p>
<p>Since $x_1=C$ and $B>C$ we have $$B>\frac{qAB}{A+B}\Rightarrow B^2>(q-1)AB$$</p>
<p>Since $A>B$ it follows $AB>B^2$ and so $q-1<1\Rightarrow q<2$. Since $q\neq0$ (see above) we have $q=1$ and so $C=\frac{AB}{A+B}$.</p>
<p>Substituting for $C$ in the quadratic (although the cubic has the same result) we have $$0=p(\frac{AB}{A+B})^2+\frac{AB}{A+B}-p-A-B$$ Which, after a fair bit of rearranging, becomes $$p=\frac{(A+B)^3-AB(A+B)}{(AB)^2-(A+B)^2}$$ Thus $f(A,B,C)$ is an integer (with $A,B,C$ constrained as above) iff $$p=\frac{(A+B)^3-AB(A+B)}{(AB)^2-(A+B)^2}$$ has integer solutions $p>0$ and $A>B>2$.</p>
<p>Any help with solving this last bit would be greatly appreciated. So far guesswork and Wolfram Alpha have failed to produce results.</p>
| David | 175,745 | <p>Yes, if you have bad luck guessing then it would be better in that case to leave it blank. Since it is a SAT and not a SGT, they should penalize more for a wrong answer such as -1/2 so that if someone doesn't know an answer, they are less inclined to guess. Then they might get a better assessment of aptitude rather than guessing "skills" (actually luck). +1 for each correct answer, -1/2 for each wrong answer, and 0 for each blank answer would be my recommendation.</p>
|
4,380,862 | <p>How can I solve the tangent point and <span class="math-container">$a$</span> when <span class="math-container">$f(x)=\left(\log_{a}{x}\right)^2$</span> is tangent to <span class="math-container">$g(x)=-ax+2$</span>?</p>
<p>Although this can be solved by substituting <span class="math-container">$a=e^2$</span> and <span class="math-container">$x=e^{-2}$</span>, then</p>
<p><span class="math-container">$f\left(e^{-2}\right)=g\left(e^{-2}\right)$</span> and <span class="math-container">$f^\prime\left(e^{-2}\right)=g^\prime\left(e^{-2}\right)$</span> can be proved,</p>
<p>is there any general solution for this, rather than just substituting random numbers? Maybe using Lambert-W could help?</p>
| Dstarred | 955,900 | <p>One cannot continuously attack this question algebraically (there is no general solution for <span class="math-container">$x$</span> without <span class="math-container">$a$</span>), seeing that <span class="math-container">$(\log_a{x})^2 = -ax+2$</span>, will eventually end up in a form similar to<span class="math-container">$f(x)e^{f(x)} = g(a)$</span> or <span class="math-container">$h(x)^{t(x)} = r(a)^{p(a)}$</span>, where both need applications of Lambert-W functions to continue.</p>
<p>Instead, we have to restrict the domain and analyse the nature of solutions. This goes as such:</p>
<h1>Analysing domain and nature of functions</h1>
<p>Let <span class="math-container">$f(x) = \log_a^2{x} = \big(\frac{\ln{x}}{\ln{a}}\big)^2$</span>, then <span class="math-container">$f'(x) = \frac{2\ln{x}}{x\cdot\ln^2{a}}, a>0, a ≠ 1$</span></p>
<p>and</p>
<p><span class="math-container">$g(x) = -ax+2$</span>, then <span class="math-container">$g'(x) = -a$</span></p>
<p>By equating <span class="math-container">$f'(x) = 0$</span>, we see that function has a minimum at <span class="math-container">$x = 1$</span> regardless of <span class="math-container">$a$</span>, and this is a minimum because of the +parabolic nature of <span class="math-container">$f(x)$</span> in the form <span class="math-container">$t^2$</span>.</p>
<p>From this we can judge <span class="math-container">$f(x)$</span> is <strong>strictly decreasing</strong> for <span class="math-container">$x \in (0, 1)$</span> and <strong>strictly increasing</strong> for <span class="math-container">$x > 1$</span> <span class="math-container">$[1]$</span></p>
<p>Since <span class="math-container">$g'(x) = -a$</span> and <span class="math-container">$a > 0, \implies g'(x) < 0 \implies g(x)$</span> is <strong>strictly decreasing</strong> <span class="math-container">$[2]$</span></p>
<p>Comparing <span class="math-container">$[1]$</span> and <span class="math-container">$[2]$</span> we observe solutions for <span class="math-container">$\color{red}{x \in (0, 1)}$</span> only.</p>
<h1>Analysing nature of solutions to <span class="math-container">$f(x) = g(x)$</span></h1>
<p>We observed that <span class="math-container">$f'(x), g'(x) < 0, \text{ } x \in (0,1)$</span>. This suggests only <span class="math-container">$1$</span> solution to</p>
<p><span class="math-container">$$\{x\} \begin{cases} f(x) = g(x) \\ f'(x) = g'(x) \end{cases}$$</span></p>
<p>which is the point of tangency.</p>
<p>Now, observe <span class="math-container">$g(x) = f(x) = 1$</span> is always a solution of <span class="math-container">$x$</span> and it is regardless of <span class="math-container">$a$</span>.</p>
<blockquote>
<p><span class="math-container">$1 = -ax + 2 \to x = \frac{1}{a} \text{ and } f(\frac{1}{a}) = 1$</span></p>
</blockquote>
<p>Hence, <span class="math-container">$\exists\{x_T, a_T\}$</span> such that <span class="math-container">$f(x_T) = g(x_T) = 1$</span> and <span class="math-container">$f'(x_T) = g'(x_T)$</span> and <span class="math-container">$\{x_T, a_T\}$</span> is the only set of tangent solution, with <span class="math-container">$x_T = \frac{1}{a_T}$</span></p>
<p>Therefore,</p>
<p><span class="math-container">$$f'(\frac{1}{a}) = g'(\frac{1}{a})$$</span>
<span class="math-container">$$\frac{2\ln{\frac{1}{a}}}{\frac{1}{a}\big(\ln{a}\big)^2} = -a$$</span>
<span class="math-container">$$-\frac{2\ln{a}}{\big(\ln{a}\big)^2} = -1$$</span>
<span class="math-container">$$\frac{2}{\ln{a}} = 1$$</span>
<span class="math-container">$$\therefore a = e^2 \text{ and } x = \frac{1}{e^2} = e^{-2} \in (0, 1) \text{ as required }$$</span></p>
|
549,411 | <p>How would I go about generalizing the product rule to the product of $n$ functions $\psi_1(x), \ \psi_2(x), ..., \ \psi_n(x)$? That is, I'm hoping to obtain an expression for</p>
<p>$$
\frac{d}{dx} \prod_{j = 1}^n \psi_j(x)
$$</p>
| Marv Turner | 104,945 | <p>Their intersection is the empty set</p>
|
245,789 | <p>Let $f$ be a function defined on the interval $(−1, 1)$ such that for all $x, y \in (−1, 1)$, $$f(x + y) = \frac{{f(x) + f(y)}}{{1 - f(x)f(y)}}.$$ Suppose that $f$ is differentiable at $x = 0$, show that $f$ is differentiable on $(−1, 1)$.</p>
<p>Need your kind guidance and help. Thanks in advance!</p>
| Hagen von Eitzen | 39,174 | <p>Letting $x=y=0$, we obtain $f(0)=\frac{2f(0)}{1-f(0)^2}$, hence $f(0)=0$ or $1-f(0)^2=2$. Since the latter is impossible, $f(0)=0$.
Then we are given that the limit
$$\tag1\lim_{h\to0}\frac{f(h)}h=\lim_{h\to0}\frac{f(h)-f(0)}h=f'(0)$$
exists and of course even more so we have
$$\tag2\lim_{h\to0}f(h)=0.$$
We find
$$\begin{align}f(x+h)-f(x)&=\frac{f(x)+f(h)}{1-f(x)f(h)}-f(x)\\
&=\frac{f(x)+f(h)-f(x)(1-f(x)f(h))}{1-f(x)f(h)}\\
&=f(h)\cdot\frac{1+f(x)^2}{1-f(x)f(h)}\end{align}$$
and thus see (using (1) and (2)) that the derivative at $x$ exists:
$$ f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{f(h)}h\cdot \lim_{h\to0}\frac{1+f(x)^2}{1-f(x)f(h)}=f'(0)\cdot (1+f(x)^2).$$</p>
|
1,248,517 | <p>I try to solve the following problem: <br>
Given a stream of symmetric matrices $A_0, A_1, ...,A_n$ such that $A_i$ is different from $A_{i-1}$ only in one place, I want to compute the eigenvalues of $A_i$. <br>
Since the matrices are very large, computing the eigenvalues from scratch isn't efficient (and since the matrices are different only in one place, that's also not very smart..), and I try to find how to calculate the eigenvalues of $A_i$ using the eigenvalues of $A_{i-1}$. <br></p>
<p>Any help will be welcomed, <br></p>
<p>Thanks</p>
| Adhvaitha | 228,265 | <p>We have
$$3n^2+7n+4 = (n+2)(3n+1) + 2$$
Hence, if $d$ is a common divisor of $3n^2+7n+4$ and $n+2$, then $d$ must divide $2$. Hence, $\gcd(3n^2+7n+4,n+2) = 1 \text{ or }2$. $3n^2+7n+4$ is always even. Further, $n+2$ has the same parity as $n$.</p>
<p>Hence, if $n$ is odd, $$\gcd(3n^2+7n+4,n+2) = 1$$
while if $n$ is even $$\gcd(3n^2+7n+4,n+2) = 2$$</p>
|
1,248,517 | <p>I try to solve the following problem: <br>
Given a stream of symmetric matrices $A_0, A_1, ...,A_n$ such that $A_i$ is different from $A_{i-1}$ only in one place, I want to compute the eigenvalues of $A_i$. <br>
Since the matrices are very large, computing the eigenvalues from scratch isn't efficient (and since the matrices are different only in one place, that's also not very smart..), and I try to find how to calculate the eigenvalues of $A_i$ using the eigenvalues of $A_{i-1}$. <br></p>
<p>Any help will be welcomed, <br></p>
<p>Thanks</p>
| Mark Fischler | 150,362 | <p>Apply Euclid's algorithm:
$$3n^2 + 7n + 4 = (3n+1)(n+2) + 2$$
$$ \text{gcd }(3n^2 + 7n + 4) = \text{ gcd }(n+2,2)
$$
So the answer is $1$ if $n$ is odd, but $2$ if $n$ is even. </p>
|
3,584,927 | <p>If I have two isomorphic groups, can I write <span class="math-container">$A \xrightarrow{\sim} B$</span> rather than <span class="math-container">$A \cong B$</span> to mean "A is isomorphic to B", or is the arrow notation only used if I have a map <span class="math-container">$\varphi : A \xrightarrow{\sim} B$</span> ?</p>
| PrincessEev | 597,568 | <p>If <span class="math-container">$A$</span> and <span class="math-container">$B$</span> are isomorphic, then it implies the existence of an isomorphism from one to the other. Though, that said, I wouldn't use that notation since it seems to imply you're talking about an isomorphism rather than the fact they're isomorphic.</p>
|
1,501,717 | <p>Imagine a real-valued semilog function $\DeclareMathOperator{\sog}{sog}\sog$ with the property that</p>
<blockquote>
<p>$$\sog(\sog(x)) = \log(x)$$
for all real $x>0$.</p>
</blockquote>
<p>My questions: </p>
<ul>
<li>Does such a function exist?</li>
<li>How do I compute it?</li>
</ul>
| Will Jagy | 10,400 | <p>EDIT: I got my copy of Hellmuth Kneser (1950) from the Göttingen digital repository, click to download just the article itself (rather than the whole journal issue) <a href="http://gdz.sub.uni-goettingen.de/dms/load/pdf/?PPN=GDZPPN002175851" rel="nofollow noreferrer">LINK HELLMUTH KNESER 1950 Reele analytische</a>
or straight to <a href="http://wwwuser.gwdg.de/~subtypo3/gdz/pdf/PPN243919689_0187/PPN243919689_0187___LOG_0008.pdf" rel="nofollow noreferrer">DIRECT TO PDF</a>. If anyone tries and has trouble with the link(s), I have now made a nice pdf that is small enough for email, I sent it to myself and gmail says 2MB, no trouble. I put the journal cover, contents page, then a third page after the article itself. </p>
<p>There are two issues, both of which are non-problems for the logarithm. If you are considering a function on the reals that is also real valued, as long as there are no fixpoints, we can expect to produce real analytic half-iterates, in an open set around the relevant portion of the real axis. The open set may vary in width. To be specific, the open set cannot be expected to include complex fixpoints of the function. </p>
<p>It will generally turn out that fractional iterates will not extend to the entire complex plane, even when the original function is entire and single valued. </p>
<p>If you have a fixpoint on the real line, where the derivative of the function is negative, it is easy to see that the half iterate cannot stay real valued. For example, a half iterate of $-x$ is $ix,$ because $i(ix) = -x.$</p>
<p>In the presence of fixpoints with derivative larger than $1$ or strictly between $0$ and $1,$ we get to solve a Schöder equation. With $x^2$ at $1$ we arrive at $x^{\sqrt 2},$ because $$ \left( x^{\sqrt 2} \right)^{\sqrt 2} = x^2 $$ for positive $x.$ As soon as we try to include $0,$ we are stuck with $|x|^{\sqrt 2},$ which is $C^1$ but not $C^2$ at the origin. So, fixpoints are a problem.</p>
<p>Finally, the hardest bit is when the fixpoint has derivative $1.$ I spent quite a bit of time finding a half-iterate for $\sin x.$ It works, the result is in between the sine wave and a sawtooth curve (of line segments) that is tangent to the sine curve at multiples of $\pi.$ And, for $0 < x < \pi,$ it is real analytic. I wrote to Jean Ecalle, it turns out that the function really is $C^\infty$ when we expand to include $x=0,$ therefore $C^\infty$ on the entire real line. Let's see, amplitude a little larger than $\sin x$ itself, not as large as the sawtooth; $\pi/2 \approx 1.570796,$ and we get $f_{1/2} (1.570796) \approx 1.140179.$</p>
<p><a href="https://i.stack.imgur.com/dZVE8.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dZVE8.jpg" alt="enter image description here"></a> </p>
<p>The (substantial) work is summarized at <a href="https://mathoverflow.net/questions/45608/formal-power-series-convergence/46765#46765">https://mathoverflow.net/questions/45608/formal-power-series-convergence/46765#46765</a></p>
<p>As commented, Helmuth Kneser constructed a real analytic half-iterate for $e^x.$ At about the same time I did $\sin x,$ I also did $e^x-1,$ which has a fixpoint with derivative $1$ at $x=0.$ Once again, real analytic for $x > 0,$ also for $x < 0,$ but just $C^\infty$ on the whole real line. Having trouble posting the picture, partly because it is a jpeg instead of a pdf. Anyway, the outcome is that $\log x$ works, but $\log (1+x)$ is more difficult.</p>
|
1,501,717 | <p>Imagine a real-valued semilog function $\DeclareMathOperator{\sog}{sog}\sog$ with the property that</p>
<blockquote>
<p>$$\sog(\sog(x)) = \log(x)$$
for all real $x>0$.</p>
</blockquote>
<p>My questions: </p>
<ul>
<li>Does such a function exist?</li>
<li>How do I compute it?</li>
</ul>
| Franklin Pezzuti Dyer | 438,055 | <p>This question already has a <em>great</em> answer, but I thought that you might like to see an actual piecewise construction of such a function (defined on $(-\ln(2),\infty)$). My function is not analytic, but it is continuous and differentiable on $\mathbb R$.</p>
<p>First notice that if we find a function $h$ with the property
$$(h\circ h)(x)=e^x$$
then it follows that
$$(h^{-1}\circ h^{-1})(x)=\log(x)$$
and we may use $h^{-1}=\text{sog}$.</p>
<p>Okay, this piecewise definition gets a bit gross, so bear with me.</p>
<p>Define a sequence $a_n$ as follows:
$$a_{-1}=-\log(2),\space\space a_0=0$$
$$a_{n+2}=\exp(a_n)$$
and define a sequence of functions $H_n$ as follows:
$$H_{-2}(x)=\log(e^x+0.5)$$
$$H_{n+1}(x)=\exp(H^{-1}_n(x))$$
Then one may show verify rather simply that the function
$$h(x) = \left\{
\begin{array}{lr}
H_{-2}(x) & : -\infty \lt x \le -\log(2)\\
H_n(x) & : a_n\lt x \le a_{n+1}
\end{array}
\right.\\$$
satisfies $(h\circ h)(x)=e^x$. If it helps, I can write the definition out for you this way:
$$h(x) = \left\{
\begin{array}{lr}
\ln(e^x+0.5) & : -\infty \lt x \le -\log(2)\\
e^x-0.5 & : -\log(2)\lt x \le 0\\
x+0.5 & : 0\lt x \le 0.5\\
e^{x-0.5} & : 0.5\lt x \le 1\\
\text{etc}...\\
\end{array}
\right.\\$$
With a bit more difficulty (and induction), one may prove that it is differentiable on $\mathbb R$ (unfortunately, it is <em>not</em> twice differentiable). Here is a its graph (in orange) alongside a graph of $y=e^x$ (in purple):</p>
<p><a href="https://i.stack.imgur.com/UNTOO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UNTOO.png" alt="enter image description here"></a></p>
<p>Now, for $\text{sog}(x)$, all we need to do is graph $y=h^{-1}(x)$:</p>
<p><a href="https://i.stack.imgur.com/8pxJk.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8pxJk.png" alt="enter image description here"></a></p>
|
2,367,875 | <blockquote>
<p>Let $K$ be a field and $x$ be trnacendental over $K$. Compute
$[K(x):K(\frac{x^5}{1+x})]$.</p>
</blockquote>
<p>I've never came across questions like these. It's easy to see that this degree is at most $5$, since: $$x^5-\alpha(x+1)=0$$ ($\alpha$ being $\frac{x^5}{1+x}$). But how can I show the other direction? Why can't there be any $f\in K(\alpha)[X]$ with $\deg f<5$ and $f(x)=0$? Or can there be?</p>
<p>Help would be appriciated. Thank you!</p>
| Ennar | 122,131 | <p>What you have to do is show that $f(X)=X^5-\alpha(X+1)$ is irreducible over $K(\alpha)$ and you are done.</p>
<p>This is because we would have $[K(\alpha)(x):K(x)][K(x):K(\alpha)] = [K(\alpha)(x):K(\alpha)] = 5$ and $K(x)\neq K(\alpha)$ would imply $[K(x):K(\alpha)]=5$ by $5$ being prime.</p>
<p>Now, $f$ is primitive polynomial in $K[\alpha][X]$, so by Gauss's lemma, it is irreducible over $K(\alpha)$ if and only if it is irreducible over $K[\alpha]$.</p>
<p>First of all, $f$ can't have a root in $K[\alpha]$ since it would mean that $$p(\alpha)^5-\alpha(p(\alpha)+1)=0$$ and that $\alpha$ is algebraic over $K$.</p>
<p>Thus, if $f$ weren't irreducible, it would have to allow factorization of the form</p>
<p>$$f(X) = (X^3 +a(\alpha)X^2+b(\alpha)X+c(\alpha))(X^2+d(\alpha)X+e(\alpha))$$ which would lead to system</p>
<p>\begin{align}
a(\alpha)+d(\alpha)&=0\\
b(\alpha)+a(\alpha)d(\alpha)+e(\alpha)&= 0\\
c(\alpha)+b(\alpha)d(\alpha)+a(\alpha)e(\alpha) &= 0\\
c(\alpha)d(\alpha)+b(\alpha)e(\alpha)&= -\alpha\\
c(\alpha)e(\alpha)&=-\alpha
\end{align}</p>
<p>and now you can use that $K[X]\cong K[\alpha]$ since $\alpha$ is transcedental and degree arguments to show that the above system is impossible to solve in $K[X]$.</p>
|
2,367,875 | <blockquote>
<p>Let $K$ be a field and $x$ be trnacendental over $K$. Compute
$[K(x):K(\frac{x^5}{1+x})]$.</p>
</blockquote>
<p>I've never came across questions like these. It's easy to see that this degree is at most $5$, since: $$x^5-\alpha(x+1)=0$$ ($\alpha$ being $\frac{x^5}{1+x}$). But how can I show the other direction? Why can't there be any $f\in K(\alpha)[X]$ with $\deg f<5$ and $f(x)=0$? Or can there be?</p>
<p>Help would be appriciated. Thank you!</p>
| pisco | 257,943 | <p>We only need to show that the polynomial $x^5-\alpha x - \alpha$ is irreducible over $K(\alpha)[x]$. Note that the polynomial is primitive, thus it suffices to show it is irreducible over $K[\alpha][x]$.</p>
<p>We use the idea of Eisenstein criterion, let $\varphi:K[\alpha]\to K$ be the evaluation map sending $\alpha \mapsto 0$. Extend this homomorphism to get $\pi: K[\alpha][x]\to K[x]$. If $x^5-\alpha x -\alpha = fg$, with $f,g \in K[\alpha][x]$, $f,g$ non-unit. Then
$$\pi(f)\pi(g) = x^5$$
Note that $\ker \pi = (\alpha)$, the ideal generated by $\alpha$. </p>
<p>Since $K[x]$ is a UFD, $f = x^r + \alpha f_0, g = x^{5-r} + \alpha g_0$, where $f_0,g_0\in K[\alpha][x]$. But this would imply that the constant term of $fg$ is a multiple of $\alpha^2$. Hence $x^5-\alpha x -\alpha$ is irreducible over $K[\alpha][x]$.</p>
|
3,178,648 | <blockquote>
<p>We assign to every element <span class="math-container">$i$</span> from <span class="math-container">$N=\{1,2,...,n\}$</span> a positive integer <span class="math-container">$a_i$</span>. Suppose <span class="math-container">$$a_1+a_2+...+a_n = 2n-2$$</span> then prove that map <span class="math-container">$T: \mathcal{P}(N) \to \{1,2,...,2n-2\}$</span> defined with <span class="math-container">$$T(X) = \sum _{i\in X}a_i$$</span> is surjective. </p>
</blockquote>
<hr>
<p>We can assume that <span class="math-container">$a_1\leq a_2\leq ...\leq a_n$</span>. </p>
<p>Clearly, <span class="math-container">$a_1 = a_2 = 1$</span> and thus <span class="math-container">$1,2,2n-3,2n-4$</span> are in a range. </p>
<p>Also, if <span class="math-container">$a_i=2$</span> for some <span class="math-container">$i$</span> then we could easily apply induction. </p>
<p>Say <span class="math-container">$b_1< b_2<...<b_k$</span> are all different values that appear among <span class="math-container">$a_i$</span>. </p>
<p>Then we have <span class="math-container">$n _1\cdot b_1+n_2\cdot b_2+...+n_k \cdot b_k = 2n-2$</span> and <span class="math-container">$n_1+n_2+..+n_k = n$</span>. We have to prove that for each <span class="math-container">$l\leq 2n-2$</span> we have <span class="math-container">$$n' _1\cdot b_1+n'_2\cdot b_2+...+n'_k \cdot b_k = l$$</span></p>
<p>for some <span class="math-container">$n'_i\leq n_i$</span>. And here it stops. I have no idea how to find all those <span class="math-container">$n_i'$</span>. Any ideas?</p>
| Dong-gyu Kim | 561,790 | <p>How about using an induction on <span class="math-container">$n$</span>?</p>
<p>From the assumption, <span class="math-container">$a_1\leq a_2\leq \cdots \leq a_n$</span>, we derive some facts: (you mentioned)</p>
<ul>
<li><p><span class="math-container">$a_1=a_2 = 1$</span></p></li>
<li><p><span class="math-container">$a_3 = 1$</span> or <span class="math-container">$2$</span> if <span class="math-container">$n\geq3$</span></p></li>
</ul>
<p>The second fact makes us possible to use an induction!</p>
<p>The initial case, <span class="math-container">$n=2$</span>, is trivial. Now, assume <span class="math-container">$n\geq 3$</span>.</p>
<p>Let define a new sequence <span class="math-container">$\{a_i'\}_{i=1}^{n-1}$</span> such that</p>
<ul>
<li><span class="math-container">$a_1'=a_2' = 1$</span></li>
<li>Case 1. <span class="math-container">$a_3=2$</span>. For <span class="math-container">$i\geq 3$</span>, <span class="math-container">$a_i' = a_{i+1}$</span></li>
<li>Case 2. <span class="math-container">$a_3=1$</span>. <span class="math-container">$a_3' = a_4 -1$</span> and <span class="math-container">$a_i'=a_{i+1}$</span> for <span class="math-container">$i\geq4$</span>.</li>
</ul>
<p>Then, <span class="math-container">$a_1'+\cdots+a_{n-1}'=2n-4$</span>. By the induction hypothesis, there is a surjection <span class="math-container">$T_{n-1}:\mathcal{P}([n-1])\rightarrow [2n-4]$</span>. (here, <span class="math-container">$[n]=\{1,2,\cdots,n\}$</span>)</p>
<p>For the Case 1, we can extend <span class="math-container">$T_{n-1}$</span> to <span class="math-container">$T_n$</span>.</p>
<p>However, an extension of <span class="math-container">$T_{n-1}$</span> is not easy for the Case 2.
(I have no idea for the exentsion for this case, unfortunately.)</p>
|
3,178,648 | <blockquote>
<p>We assign to every element <span class="math-container">$i$</span> from <span class="math-container">$N=\{1,2,...,n\}$</span> a positive integer <span class="math-container">$a_i$</span>. Suppose <span class="math-container">$$a_1+a_2+...+a_n = 2n-2$$</span> then prove that map <span class="math-container">$T: \mathcal{P}(N) \to \{1,2,...,2n-2\}$</span> defined with <span class="math-container">$$T(X) = \sum _{i\in X}a_i$$</span> is surjective. </p>
</blockquote>
<hr>
<p>We can assume that <span class="math-container">$a_1\leq a_2\leq ...\leq a_n$</span>. </p>
<p>Clearly, <span class="math-container">$a_1 = a_2 = 1$</span> and thus <span class="math-container">$1,2,2n-3,2n-4$</span> are in a range. </p>
<p>Also, if <span class="math-container">$a_i=2$</span> for some <span class="math-container">$i$</span> then we could easily apply induction. </p>
<p>Say <span class="math-container">$b_1< b_2<...<b_k$</span> are all different values that appear among <span class="math-container">$a_i$</span>. </p>
<p>Then we have <span class="math-container">$n _1\cdot b_1+n_2\cdot b_2+...+n_k \cdot b_k = 2n-2$</span> and <span class="math-container">$n_1+n_2+..+n_k = n$</span>. We have to prove that for each <span class="math-container">$l\leq 2n-2$</span> we have <span class="math-container">$$n' _1\cdot b_1+n'_2\cdot b_2+...+n'_k \cdot b_k = l$$</span></p>
<p>for some <span class="math-container">$n'_i\leq n_i$</span>. And here it stops. I have no idea how to find all those <span class="math-container">$n_i'$</span>. Any ideas?</p>
| CopyPasteIt | 432,081 | <p>The motivation for this work was thinking about Calvin Lin's bonus problems. </p>
<hr>
<p>Proposition 1: Let <span class="math-container">$(a_i)_{1\le i\le n}$</span> be a family of positive integers such that </p>
<p><span class="math-container">$$\tag 1 \sum_{i=1}^n a_i = 2n - k \; \text{ with } \,1 \le k \lt n$$</span></p>
<p>Then at least one term of the family is equal to one. <br>
Proof<br>
The proof is extremely easy and is left as an exercise. <span class="math-container">$\quad \blacksquare$</span></p>
<p>We call a family <span class="math-container">$(a_i)_{1\le i\le n}$</span> satisfying <span class="math-container">$\text{(1)}$</span> a <span class="math-container">$2n -k$</span> system.</p>
<p>Proposition 2: For any <span class="math-container">$2n -k$</span> system <span class="math-container">$(a_i)_{1\le i\le n}$</span> suppose <span class="math-container">$a_\iota \ge 2$</span> for some term. Then there are at least <span class="math-container">$a_\iota - 1$</span> terms that are equal to <span class="math-container">$1$</span>.<br>
Proof<br>
We can always decrement a term greater than one and increment another term and still have a <span class="math-container">$2n -k$</span> system. If <span class="math-container">$a_\iota$</span> was two greater than the number of units we could construct a system with no units, contradicting proposition 1. <span class="math-container">$\quad \blacksquare$</span></p>
<p>We can now prove the main result.</p>
<p>Proposition 3: Let <span class="math-container">$(a_i)_{1\le i\le n}$</span> be a <span class="math-container">$2n -k$</span> system.<br>
Then for any positive integer <span class="math-container">$m$</span> such that <span class="math-container">$1 \le m \le 2n -k$</span>, there exist a subset of <span class="math-container">$K$</span> of <span class="math-container">$\{1, \dots, n\}$</span> such that </p>
<p><span class="math-container">$$\tag 2 \sum _{k \in K}a_k = m$$</span>
Proof<br>
We begin with a 'ledger' with all of our 'chips' on the right side. Now start by moving the units, one by one, over to left side while 'shouting out' the left side total. If we run out of units and anything remains on the right side, take any 'chip' and move it over to the left side while 'giving unit change' of <span class="math-container">$1$</span> less back over to the right (see proposition 2); then call out the new total, an increment by <span class="math-container">$1$</span> over the prior summation. This is what our algorithm does, and it can resume by moving the (transferred) unit chips back over to the left side. Now just keep repeating these rules.</p>
<p>When this process is complete (nothing remains on the right side of the ledger), all the numbers in <span class="math-container">$1 \le m \le 2n -k$</span> have been 'called out' by continually bumping up the left side of the ledger. <span class="math-container">$\quad \blacksquare$</span></p>
<p>Here is an alternate set theoretic proof of the result.</p>
<p>Proposition 4: Let <span class="math-container">$(a_s)_{s \in S}$</span> be a <span class="math-container">$2n -k$</span> system, where <span class="math-container">$|S| = n$</span>.<br>
Then for any positive integer <span class="math-container">$m$</span> such that <span class="math-container">$1 \le m \le 2n -k$</span>, there exist a subset of <span class="math-container">$K$</span> of <span class="math-container">$S$</span> such that </p>
<p><span class="math-container">$$\tag 3 \sum _{k \in K}a_k = m$$</span>
Proof<br>
We can partition <span class="math-container">$S$</span> into two sets <span class="math-container">$U$</span> and <span class="math-container">$V$</span> so that <span class="math-container">$a_u = 1$</span> for <span class="math-container">$u \in U$</span> and <span class="math-container">$a_v \gt 1$</span> for <span class="math-container">$v \in V$</span>.</p>
<p>If <span class="math-container">$U$</span> has <span class="math-container">$\alpha$</span> elements construct subsets <span class="math-container">$U_i$</span> of <span class="math-container">$U$</span> for <span class="math-container">$1 \le i \le \alpha$</span> satisfying</p>
<p><span class="math-container">$\quad |U_i| = i$</span><br>
<span class="math-container">$\quad U_i \subset U_{i+1}$</span></p>
<p>If <span class="math-container">$V$</span> has <span class="math-container">$\beta$</span> elements construct subsets <span class="math-container">$V_i$</span> of <span class="math-container">$V$</span> for <span class="math-container">$1 \le i \le \beta$</span> satisfying</p>
<p><span class="math-container">$\quad |V_i| = i$</span><br>
<span class="math-container">$\quad V_i \subset V_{i+1}$</span></p>
<p>Consider the integers between <span class="math-container">$\sum _{k \in V_i}a_k$</span> and <span class="math-container">$\sum _{k \in V_{i+1}}a_k$</span>. Let <span class="math-container">$v_{i+1}$</span> be the integer in <span class="math-container">$V_{i+1}$</span> that is not in <span class="math-container">$V_{i}$</span>.
Then the sums of the form</p>
<p><span class="math-container">$$\quad \sum _{k \in V_i}a_k + \sum _{k \in U_j}a_k \text{ with } |U_j| \lt v_{i+1}$$</span></p>
<p>completely cover these integers (see proposition 2).</p>
<p>It is easy to explain why the sums</p>
<p><span class="math-container">$$\quad \sum _{k \in V_\beta}a_k + \sum _{k \in U_j}a_k \text{ with } 1 \le j \le \alpha$$</span></p>
<p>take out all integers between <span class="math-container">$\sum _{k \in V_\beta}a_k$</span> and <span class="math-container">$2n -k$</span> and also must take out <span class="math-container">$2n -k$</span> itself.</p>
<p>Finally, it is a 'mop up' job to show that the sums</p>
<p><span class="math-container">$$\quad \sum _{k \in U_j}a_k \text{ with } |U_j| \lt V_{1}$$</span></p>
<p>cover everything remaining in the interval <span class="math-container">$[1, v_1-1]$</span>. <span class="math-container">$\quad \blacksquare$</span></p>
|
2,566,546 | <p>Show that for any $x_1 < x_2$ and $y_1 < y_2$ one has
$P(x_1 < X ≤ x_2, y_1 < Y ≤ y_2) = F(x_2, y_2) + F(x_1, y_1) − F(x_1, y_2) − F(x_2, y_1)$.</p>
<p>Would I just need to split the LHS to something that gives me the right?</p>
| Elias Costa | 19,266 | <p>If $P^2=P$ then $P\in\mathbb{R}^{n\times n}$ is root of equation
$$
P^2-P=0.
$$
But we have a bit more. If $P^2=P$ then
$$
\cdots=P^{2\cdot n}=\cdots=P^{4}=P^{2}=P^{}
$$
Then for any fixed $n$ and $n$ real numbers $a_1,a_2,\ldots, a_n$ such that $ a_1+a_2+\ldots+a_n=1$ we have that $P$ is root of equation
$$
a_n\cdot P^{2\cdot n}+a_{n-1}\cdot P^{2\cdot (n-1)}+\ldots+a_1\cdot P=P
$$</p>
|
2,912,401 | <p>I'm trying to prove that the sequence $\left(\frac{1}{2},\frac{1}{3},\frac{2}{3},\frac{1}{4},\frac{2}{4},\frac{3}{4},\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5},\cdots\right)$ is not a Cauchy sequence. </p>
<p>I know that a sequence of real numbers is <em>not</em> Cauchy if there exists an $\epsilon>0$ such that, for all $N\in\mathbb{N}$, there exists $m,n>N$ such that $|x_{m}-x_{n}|\geq\epsilon$. It intuitively makes sense to me that the sequence cannot be Cauchy, as the distance between points where the denominator changes (like $\cdots\frac{99}{101},\frac{100}{101},\frac{1}{102},\cdots$) keeps growing larger. However, I'm not sure how to find indices $m$ and $n$ in general with $|x_{m}-x_{n}|\geq\epsilon$. Thanks in advance for any help!</p>
| José Carlos Santos | 446,262 | <p>Let $(a_n)_{n\in\mathbb N}$ be your sequence. Take $\varepsilon=\frac12$. Given $N\in\mathbb N$, take $n\geqslant N$ such that $a_n$ is of the form $\frac k{k+1}$ for some $k\in\mathbb N$ and let $m=n+1$. Then $a_m=\frac1{k+2}$. Therefore,$$\left\lvert a_m-a_n\right\rvert=\frac k{k+1}-\frac1{k+2}\geqslant\frac12=\varepsilon.$$
<hr />
Or you can say that your sequence diverges, since the subsequence$$\frac12,\frac13,\frac14,\ldots$$converges to $0$, whereas the subsequence$$\frac12,\frac23,\frac34,\ldots$$converges to $1$. So, the sequence diverges and therefore it is not a Cauchy sequence.</p>
|
1,429,563 | <blockquote>
<p>Let <span class="math-container">$\{F_n\}_{n\geq 0}$</span> be the Fibonacci sequence.</p>
<p>Prove that the number of primes <span class="math-container">$p$</span> so that <span class="math-container">$p\mid F_{p-1}$</span> is infinite.</p>
</blockquote>
<hr />
<p>I tried to use induction, to no avail.</p>
| Jack D'Aurizio | 44,121 | <p>Assume that $p$ is a prime for which $5$ is a quadratic residue.</p>
<p>That is equivalent, by quadratic reciprocity, to $p\equiv\pm 1\pmod{10}$.</p>
<p>Since the explicit formula for Fibonacci numbers gives:
$$ F_n = \frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n\right),$$
provided that $\sqrt{5}\in\mathbb{F}_p$, from <a href="https://en.wikipedia.org/wiki/Fermat's_little_theorem" rel="nofollow">Fermat's little theorem</a> it follows that:
$$ F_{p-1}\equiv F_0 \equiv 0\pmod{p}$$
as wanted. We may recover an infinitude of odd primes $\equiv 1\pmod{5}$ by factoring $\Phi_5(n)=\frac{n^5-1}{n-1}=n^4+n^3+n^2+n+1$ for different values of $n$, for instance.</p>
|
1,696,713 | <p>I am solving exact differential equation, but I am stuck on the step on how to simplify this term or how to rewrite it. </p>
<p>$e^{-2\ln{\sin{x}}}$</p>
| Ittay Weiss | 30,953 | <p>If you are interested in causality, then the book "Causality: Models, Reasoning, and Inference" by Judea Pearl should be of interest to you. It's a little difficult to judge from your question what you are actually after, but my guess is you will find plenty of illumination in that book. </p>
|
209,760 | <p>Is it possible to transform this equation to give <span class="math-container">$R$</span>?
<span class="math-container">$$y=x\left[\frac{\left(1+\frac{R}{12}\right)^{12\times{25}}}{\frac{R}{12}}-1\right]$$</span> </p>
| Yoni Rozenshein | 36,650 | <p>Letting $w = \frac y x + 1$ and $r = \frac R {12}$, we are left with inverting
$$ w = \frac {(1+r)^{300}} r $$
$$ w = \frac {(1+r)^{300}} {(r^{1/300})^{300}} $$
$$ w^{1/300} = r^{-1/300} + r^{299/300} $$
Letting $z = r^{1/300}$ and multiplying by $z$ we have
$$ z^{300} - w^{1/300}z + 1 = 0$$</p>
<p>This is a trinomial equation of degree 300 in a form similar to Glasser's form, <a href="http://en.wikipedia.org/wiki/Bring%E2%80%93Jerrard_normal_form#Glasser.27s_derivation" rel="nofollow">which you can read about here</a>.
(You can get Glasser's form exactly by substituting $z$ with $cz$ with an appropriate constant $c$, dependent on $w$ of course.)</p>
<p>The answer's not pretty. You might be better off solving it numerically.</p>
|
3,672,814 | <p><span class="math-container">$g_n(y)=\sum_{k=1}^{n} \frac{\sin ky}{k},x \in [0,\pi]$</span>. So i try to find the pointwise limit function first, following this approach in this post <a href="https://math.stackexchange.com/questions/13490/proving-that-the-sequence-f-nx-sum-limits-k-1n-frac-sinkxk-is">Proving that the sequence $F_{n}(x)=\sum\limits_{k=1}^{n} \frac{\sin{kx}}{k}$ is boundedly convergent on $\mathbb{R}$</a>. I find it to be <span class="math-container">$\lim\limits_{n\to\infty}\ \sum\limits_{k=1}^{n}\frac{\sin kx}{k}=\frac{\pi-x}{2}=g(x) ,\, x\in(0,2\pi).$</span>
I tried to prove it by Dirichlet test:<br />
take <span class="math-container">$h_n(x)=\frac{1}{n}\quad \text{and} \quad u_n(x)=\sin(nx) ,
\text{then} |\sum_{k=1}^n\sin(kx)| \leq \frac{1}{\sin(\frac{x}{2})} \leq \frac{1}{\sin(\frac{\delta}{2})}$</span>.<br />
I want to prove or disprove that <span class="math-container">$(g_n)$</span> converge uniformly on <span class="math-container">$[0,\pi]$</span>.</p>
| RRL | 148,510 | <p>You have shown correctly with the Dirichlet test that <span class="math-container">$(g_n)$</span> converges uniformly on the interval <span class="math-container">$[\delta, \pi]$</span> for any <span class="math-container">$\delta > 0$</span>. This relies on the fact that <span class="math-container">$\left|\sum_{k=1}^n \sin kx \right| \leqslant 1 / \sin (\delta/2)$</span> is uniformly bounded for all <span class="math-container">$n \in \mathbb{N}$</span> and <span class="math-container">$x \in [\delta,\pi]$</span>. </p>
<p>However, if <span class="math-container">$\delta = 0$</span> the partial sums are not bounded and the Dirichlet test is not applicable. This is a clue that convergence is not uniform on <span class="math-container">$[0,\pi]$</span>.</p>
<p>If the convergence were uniform, then by the Cauchy criterion for any <span class="math-container">$\epsilon > 0$</span> there exists <span class="math-container">$N \in \mathbb{N}$</span> such that for all <span class="math-container">$m > n \geqslant N$</span> and all <span class="math-container">$x \in [0,\pi]$</span> we would have</p>
<p><span class="math-container">$$\tag{1}\left|\sum_{k=n+1}^{m}\frac{\sin kx }{k} \right| < \epsilon$$</span></p>
<p>We can show that this criterion is violated and the convergence is not uniform. Take <span class="math-container">$\epsilon = 1/(4\sqrt{2})$</span> and, for any integer <span class="math-container">$N$</span>, no matter how large, choose <span class="math-container">$n = N$</span>, <span class="math-container">$m = 2n$</span> and <span class="math-container">$x_n = \pi/(4n) \in [0,\pi]$</span>. For any <span class="math-container">$k > n$</span>, we have <span class="math-container">$kx_n > \pi/4$</span> and <span class="math-container">$\sin kx_n > \sin (\pi/4) = 1/\sqrt{2}$</span>.</p>
<p>Hence,</p>
<p><span class="math-container">$$\tag{2}\left|\sum_{k=n+1}^{2n}\frac{\sin kx_n }{k} \right| > \frac{1}{\sqrt{2}}\sum_{k=n+1}^{2n}\frac{1 }{k} > \frac{1}{\sqrt{2}}\cdot n \cdot \frac{1}{2n} = \frac{1}{2\sqrt{2}} > \epsilon,$$</span></p>
<p>and we have a contradiction to condition (1).</p>
|
2,657,571 | <p>Apologies if this is a duplicate. I searched and didn't find anything quite like it.</p>
<p>Suppose I have a drawer with an equal number of N black socks and N white socks. They're all mixed up. So, my chances of picking a matching pair in the first two selections is (N-1)/(2N-1), right? Well, what if, <em>before</em> I pick the first sock, I randomly (so I don't know the colors of the socks I'm moving) <em>partition</em> the drawer so that there are N socks on each side, and I draw one sock from each side. <em>Do the chances of drawing a matching pair change</em>?</p>
<p>On the one hand, we can see that selection from one side doesn't <em>change</em> the composition of socks on the other side of the partition. However, whichever color I choose from the "first" side, it's likely that there are more of that color on that side. On other words, if I draw a black sock from one side, it's more likely that that side had N-1 blacks and 1 white than it is that that side had 1 black and N-1 whites.</p>
<p>My suspicion is that I need to do some kind of hypothesis testing, where I consider the chances of every possible partitioning, but that's way above my skill level.</p>
| G Tony Jacobs | 92,129 | <p>It doesn't matter if the partitioning happens before or after the first draw. Suppose it happens after. Suppose also that the first sock drawn was black.</p>
<p>Now, we partition off $N$ from the remaining $2N-1$, to obtain our pool for the second draw. On average, the composition of this pool is $\frac{N-1}{2N-1}$ black and $\frac{N}{2N-1}$ white. Drawing from it, we have a matching pair if we draw from the portion of it that is black, i.e., $\frac{N-1}{2N-1}$.</p>
<p>The trick to simplifying the work is to work with <em>expected values</em> for the number of socks of each color in the second part of the partition, not actual values.</p>
|
2,575,156 | <p>I thought of this question through an excercise in algebraic geometry where the rings were $\Gamma(V)$ and $\mathcal{O}_P(V)$, although my question is more general. If $R\subset S$ two (commutative with 1) rings (R subring of S) and I prime ideal in R, then consider I'=(I) the ideal in S, the ideal generated by I. Is that ideal prime? I would think that the answer is positive even in this general case but i can't seem to find any obvious and fast proof. <br/>Any help would be appreciated. <br/>Just for reference the question came up from excercise 2.18 in fulton algebraic curves in which long story short i had to prove that there is a 1-1 correspondence between prime ideals in $\mathcal{O}_P(V)$ and prime ideals in $\Gamma(V)$.</p>
| A Square | 497,575 | <p>As I don't have enough reputation for writing comments, I'll post it here.</p>
<p>It is not true in the general case. Take for example $\mathbb{Z}$ and the ideal generated by $(2)$ then this ideal in $\mathbb{Q}$ is just $\mathbb{Q}$.</p>
<p>You can find a plenty of examples in algebraic number theory.
For instance $\mathbb{Z} \subset \mathbb{Z}[i]$ but $(5)$ is not a prime ideal in $\mathbb{Z}[i]$ as it can be factored $(5) = (1+2\cdot i)\cdot (1-2\cdot i)$</p>
|
153,999 | <p>How to solve: $|\sqrt{x-1}-2| + |\sqrt{x-1}-3|=1$.</p>
<p>I would like to know how to solve an absolute value equation when there is a square root sign inside.</p>
| Community | -1 | <p>Let us call $\sqrt{x-1}$ as $y$. Note that by definition $y \geq 0$. Now we need to find $y$ such that $$\lvert y-2 \rvert + \lvert y - 3 \rvert = 1$$
To solve this lets split into three cases.</p>
<ol>
<li>$y < 2$. This gives us that $$\lvert y-2 \rvert + \lvert y - 3 \rvert = (2-y) + (3-y) = 5 -2y > 5 -2 \times 2 =1$$ Hence, $y < 2$ is not possible.</li>
<li>$2 \leq y \leq 3$. This gives us that $$\lvert y-2 \rvert + \lvert y - 3 \rvert = (y-2) + (3-y) = 1$$ Hence, all $y \in [2,3]$ satisfies this.</li>
<li>$y > 3$. This gives us that $$\lvert y-2 \rvert + \lvert y - 3 \rvert = (y-2) + (y-3) = 2y-5 > 2 \times 3 -5 = 1$$ Hence, $y > 3$ is not possible.</li>
</ol>
<p>This means that $y \in [2,3]$. Hence, we get that $\sqrt{x-1} \in [2,3]$ i.e. $x - 1 \in [4,9]$. Hence, $$x \in [5,10]$$</p>
|
1,675,411 | <p>So far I have this:</p>
<p>First consider $n = 5$. In this case $(5)^2 < 2^5$, or $25 < 32$. So the inequality holds for $n = 5$.</p>
<p>Next, suppose that $n^2 < 2^n$ and $n \geq 5$. Now I have to prove that $(n+1)^2 < 2^{(n+1)}$.</p>
<p>So I started with $(n+1)^2 = n^2 + 2n + 1$. Because $n^2 < 2^n$ by the hypothesis, $n^2 + 2n + 1$ < $2^n + 2n + 1$. As far as I know, the only way I can get $2^{n+1}$ on the right side is to multiply it by $2$, but then I get $2^{n+1} + 4n + 2$ on the right side and don't know how to get rid of the $4n + 2$. Am I on the right track, or should I have gone a different route?</p>
| Saikat | 230,452 | <p>First, we prove a simple lemma. </p>
<p>$$
\begin{align}
n & \gt 5\\
\implies n-1 &\gt 4\\
\implies (n-1)^2 &\gt 16 \gt 2 \\
\implies n^2 - 2n + 1 &\gt 2\\
\implies n^2 &\gt 2n +1\\
\end{align}
$$ </p>
<p>Now, we start with our induction step.
$$
\begin{align}
n^2 &\lt 2^n \text{ From hypothesis}\\
\implies 2n^2 & \lt 2^{n+1}\\
\implies n^2 + n^2 &\lt 2^{n+1}\\
\implies n^2 + 2n + 1 &\lt 2^{n+1} (n^2 \gt 2n+1)\\
\implies (n+1)^2 &\lt 2^{n+1} \\
\end{align}$$</p>
|
2,056,979 | <p>I am attempting:
<a href="https://i.stack.imgur.com/FM80Z.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FM80Z.png" alt="enter image description here"></a></p>
<p>My solution is: But I am not sure where I am going wrong. The answer I get is not divisible by 7.
<a href="https://i.stack.imgur.com/9onWu.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9onWu.jpg" alt="enter image description here"></a></p>
| Arnaldo | 391,612 | <p><strong>Hint</strong></p>
<p>Write like this:</p>
<p>$$4^{k+2}-4^{k+1}+5^{2k+1}-5^{2k-1}=3\cdot4^{k+1}+24\cdot5^{2k-1}=3\cdot(4^{k+1}+5^{2k-1})+21.5^{2k-1}$$</p>
<p>Can you finish? </p>
<p>P.S: Use hyp induction for $k$ and note that $7|21$.</p>
<p>P.S: You can also prove that without use induction.</p>
|
2,736,596 | <p>I want to proof this equation below.
$$
f^{-1}(\bigcup B_{j})=\bigcup(f^{-1}B_{j})
$$
I know it's true but I don't have idea how to proof and explain.
Please help.</p>
| bsbb4 | 337,971 | <p>Let $x \in f^{-1}(\bigcup B_j)$. This means that there exists $y \in \bigcup B_j$ such that $f(x)=y$. For some $i$ in the index set of $\bigcup$, $y \in B_i \subset \bigcup B_j$. This gives that $ x \in f^{-1}(B_j)$, so we also get $x \in \bigcup f^{-1}(B_j)$.</p>
<p>Conversely, assume $x \in \bigcup f^{-1}(B_j)$. There must exist an $i$ such that $x \in f^{-1} (B_i)$. Again we have $B_i \subset \bigcup B_j$, so $x \in f^{-1}(\bigcup B_j)$.</p>
<p>Therefore $\bigcup f^{-1}(B_j) = f^{-1}(\bigcup B_j)$.</p>
|
4,264,808 | <p>So we know that <span class="math-container">$\frac{n^2}{2} \geq \frac{n}{2}$</span>, but I'm stuck proving that <span class="math-container">$\frac{n^2}{2} \geq \frac{n^2}{2^{\sqrt{\log n}}}\geq \frac{n}{2}$</span>. Am I missing something?</p>
| lone student | 460,967 | <p>Let <span class="math-container">$\sqrt {\log n}=u,\thinspace u≥0$</span>, then we have</p>
<p><span class="math-container">$$\begin{align}&\frac{n^2}{2^{\sqrt{\log(n)}}} ≥ \frac{n}{2}\\
\iff &\frac{n^2}{2^{\sqrt{\log(n)}}} ≥ \frac{n^2}{2n}\\
\iff &n≥2^{\sqrt{\log(n)}-1}\\
\iff &e^{u^2}≥2^{u-1}\\
\iff &e^{u^2}>e^{u-1}>2^{u-1}.\end{align}$$</span></p>
<p>This means, the equality is impossible.</p>
<p>Therefore, we have</p>
<p><span class="math-container">$$\frac{n^2}{2^{\sqrt{\log(n)}}} > \frac{n}{2}.$$</span></p>
|
3,245,854 | <p>Hi I am trying to solve an integral problem that involves trig substitution. First I tried completing the square, which gave me <span class="math-container">$1/\sqrt{(x+3)^2+2^2}$</span>. I know I am supposed to use <span class="math-container">$x = \arctan(\theta)$</span>. Does that mean it should be: </p>
<p><span class="math-container">$x + 3 = \arctan(\theta)$</span></p>
<p><span class="math-container">$x = \arctan(\theta) - 3$</span></p>
<p>and then Integrate from there? I am not sure if this is a good way of thinking about this problem. I would appreciate any help!
<a href="https://i.stack.imgur.com/N2IV6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/N2IV6.png" alt="enter image description here"></a></p>
| José Carlos Santos | 446,262 | <p>Do <span class="math-container">$x=2\tan(\theta)-3$</span> and <span class="math-container">$\mathrm dx=2\sec^2(\theta)\,\mathrm d\theta$</span>. You will get<span class="math-container">$$\int\frac{2\sec^2(\theta)}{\sqrt{4\tan^2(\theta)+4}}\,\mathrm d\theta=\int\sec(\theta)\,\mathrm d\theta.$$</span>Now, you can use the formula for the <a href="https://math.stackexchange.com/q/2340800/446262">primitive of the secant</a>.</p>
|
2,661,004 | <p>I have the following given as an answer to a questions, but I don't understand how I get $1+ \log n$ in the second step. I think I am to use L'Hospital's, but deriving the logarithm is $1/(n\ln(b))$, so I don't know why I'm keeping $\log(n)$. I know it's probably something basic, but what am I missing about that step? </p>
<p><a href="https://i.stack.imgur.com/rQY1W.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rQY1W.png" alt="enter image description here"></a></p>
| Henno Brandsma | 4,280 | <p>You just ignore the part of the ball that sticks out. By definition of the subspace metric we just use $B_{(1,y]}(7,r) = \{x \in (1,7] : |x - 7| < r\} = (7-r, 7]$ for $r<6$. So instead of all points closer then $r$ to $7$ in all of the reals we just take those inside the subspace in question. </p>
<p>This corresponds to open sets in the subspace $Y$ of $X$ as sets of the form $O \cap Y$ for open $O$ (open in $X$). Outside the subspace the subspace "sees" no points, essentially.</p>
|
131,583 | <p>My question is: usually, a partial differential equation, for example, those coming from physics, is written in a language of vector calculus in a local coordinate. Is there any way (or any <strong>algorithm</strong>) that we can use to rewrite it using language of differential forms, tensor, exterior calculus, Hodge star and other operators which are coordinate independent? An example, the <strong>Grad f</strong> can be rewritten as a geometric form: (df)#, where # is a sharp operator turning a one-form into a vector. I am currently facing this problem to turn a partial differential equation into its coordinate-independent form, which involves forms, tensors, exterior calculus and other operators. </p>
<p>Thank you for anyone who help me about this problem!</p>
| Liviu Nicolaescu | 20,302 | <p>To write the equations in invariant form you first need to define the various objects involved in invariant form. Sometime this is not so obvious and new insight is needed. Here are two examples that hopefully will illustrate ways in which things can get tricky.</p>
<p>1.The Euler-Lagrange equations are coordinate independent, but if the Lagrangian is a bit more sophisticated it is difficult to come-up with an invariant form. One convenient way to do this is via the Legendre transform leading to Hamiltonian equations. However, this can be used only for certain classes of Lagrangians. </p>
<p>2.The equations arising in gauge theory require a bit of finesse since they necessitate the introduction of a new object (e.g. a principal bundle) whose topological type hides some subtle physical quantity. The difficulties do not stop here and this topic is still being investigated from a rather sophisticated point of view. Andy Manion post, <a href="http://liviusmathblog.blogspot.com/2013/01/gauge-theory-and-variational-bicomplex.html" rel="nofollow"><strong><em>Gauge theory and the variational bicomplex</em></strong></a> is a good place to start. </p>
|
131,583 | <p>My question is: usually, a partial differential equation, for example, those coming from physics, is written in a language of vector calculus in a local coordinate. Is there any way (or any <strong>algorithm</strong>) that we can use to rewrite it using language of differential forms, tensor, exterior calculus, Hodge star and other operators which are coordinate independent? An example, the <strong>Grad f</strong> can be rewritten as a geometric form: (df)#, where # is a sharp operator turning a one-form into a vector. I am currently facing this problem to turn a partial differential equation into its coordinate-independent form, which involves forms, tensors, exterior calculus and other operators. </p>
<p>Thank you for anyone who help me about this problem!</p>
| Bazin | 21,907 | <p>Let $\mathcal M$ be a smooth manifold. A linear PDE on $\mathcal M$ is a sum of terms
$$
X_1\dots X_N u,\quad\text{where the $X_j$ are smooth vector fields.}
$$
We may use the convention that if $N=0$, this is just the multliplication by a smooth function.
In other words a linear PDE on $\mathcal M$ is a (non-commutative) polynomial with constant coefficients of vector fields. This PDE is said to be of order $m$ when the polynomial is of degree $m$. A general PDE on $\mathcal M$ is a function $F(u,P_1 u,\dots, P_m u)$ where $P_j$ is a linear differential operator of order $j$ defined above.</p>
<p>Let me give a simple celebrated example. Euler system describing non-viscous incompressible fluids is usually written as
$$
\partial_t u+u\cdot \nabla u=-\nabla p,\quad \text{div}u=0,\quad u_{\vert t=0}=u_0.
$$
The function $u$ is defined on $\mathbb R_t\times\mathbb R^3_x$ and valued in $\mathbb R^3_x$. A geometric version on a $3D$ manifold is shedding light on this system of equations.
In the first place $u$ should be considered as a 1-form, the pressure $p$ is a function and $dp$ is its differential, also a 1-form. Now suppose that we have on our manifold an additional structure, such as a Riemannian structure allowing to identify 1-forms to vector fields. In the case of a Riemannian structure given by a metric tensor $g$, we may identify a 1-from $u$ with a vector field $v$ through the pairing
$$
g(v,w)=\langle u,w\rangle,\quad \text{for any vector field $w$. We write this as $gv=u$}.
$$
We may then consider the Lie derivative $\mathcal L_v$ and in coordinates in $\mathbb R^3$
$$
\sum_{1\le j\le 3}\mathcal L_v(u_j) dx_j=\mathcal L_v(u)-\sum_{1\le j\le 3}u_j \mathcal L_v(dx_j)=\mathcal L_v(u)-\sum_{1\le j\le 3}u_j dv_j=\mathcal L_v(u)-\frac12 d(\vert v\vert^2).
$$
Euler system expresses the exactness of a 1-form and reads in a simply connected 3D Riemannian manifold
$$
\partial_t u+\mathcal L_v(u)-\frac12 d(g(v,v))=-dp, gv=u, \text{div $v$=0}
$$
so that (note that $\mathrm{curl}\, u$ appears simply as the exterior differential $du$: two-forms in 3D are also three dimensional), since the Lie derivative commutes with the exterior differentiation, with $\omega =du$,
$$
\partial_t \omega +\mathcal L_v(\omega)=0,\quad gv=u,\omega =du,\quad \text{div $v$=0}.
$$
This last expression provides a geometric view on Euler's system. Similar things could be done for the Navier-Stokes system.</p>
|
551,471 | <p>In a proof I am reading, the author considers the situation where $G$ is a reductive algebraic group (variety) over the complex numbers $\mathbb C$ and $N\trianglelefteq G$ is a closed, normal subgroup of $G$ which is also finite. The author concludes that <em>"because $G$ is reductive"</em>, $N$ must be contained in the center of $G$. I can't seem to make that same deduction, though. Is this some well-known result? Can you give a proof?</p>
<p>Btw, if $G$ is connected then this does not seem to have anything to do with reductivity at all. Let $u\in N$, then $G.u\subseteq N$ is finite. Since $0=\dim(G.u)=\dim(G)-\dim(G_u)$, the stabilizer of $u$ is a closed subgroup of $G$ which has maximal dimension, therefore $G=G_u$. However, the author explicitly states that the reductivity of $G$ is the reason for his deduction, and I am curious what that would be.</p>
| Community | -1 | <p>I recall writing about this a while ago. This holds for connected algebraic groups over an algebraically closed field $k$, not necessarily reductive: Let $H$ be the normal subgroup. Let $y\in H$ and let $G\to G$ be given by $x\mapsto xyx^{-1}$. Since $H$ is normal, this sends $G$ into $H$. The image of a connected group is connected, so the image is a single point. In particular, taking $x = 1_G$, we see that the image of $G$ is actually $y$. Hence $y$ commutes with every element of $G$.</p>
<p>If you take the definition of $G$ being reductive as just trivial unipotent radical, this is surely false. For instance, $N = G$, and $G$ a nonabelian finite group that is an algebraic group.</p>
|
3,441,647 | <p>I was asked to find the domain of <span class="math-container">$$\arcsin[\frac{x^2+1}{2x}]$$</span>
My first step was <span class="math-container">$$-1\leq\frac{x^2+1}{2x} \leq1$$</span></p>
<p>What I don't understand is why I cannot cross multiply to get <span class="math-container">$$-2x\le {1+x}^{2} \le 2x$$</span> and then solve the inequality? I tried doing this and got the wrong answer.</p>
| Axion004 | 258,202 | <p>In order to find the domain, you need to check every possible real value of <span class="math-container">$x$</span>. This includes values of <span class="math-container">$x$</span> where <span class="math-container">$x\ge 0$</span> and where <span class="math-container">$x<0$</span>. When you multiplied all three terms of</p>
<p><span class="math-container">$$-1\leq\frac{x^2+1}{2x} \leq1$$</span></p>
<p>by <span class="math-container">$2x$</span> to form</p>
<p><span class="math-container">$$-2x\le {1+x}^{2} \le 2x$$</span></p>
<p>you only considered the case where <span class="math-container">$x\ge 0$</span>. Since an inequality flips sign when multiplied by a negative number, multiplying all of the terms of </p>
<p><span class="math-container">$$-1\leq\frac{x^2+1}{2x} \leq1$$</span></p>
<p>by <span class="math-container">$2x$</span> when <span class="math-container">$x<0$</span> forms </p>
<p><span class="math-container">$$-2x\ge {1+x}^{2} \ge 2x$$</span></p>
<p>Therefore, you need to break up the analysis into two cases and consider what happens when <span class="math-container">$x\ge 0$</span> and <span class="math-container">$x<0$</span>.</p>
<p>Case 1: Suppose <span class="math-container">$x\ge 0$</span>. Then</p>
<p><span class="math-container">$$-2x\le {1+x}^{2} \le 2x$$</span></p>
<p>by the second inequality</p>
<p><span class="math-container">$${1+x}^{2} \le 2x$$</span></p>
<p>we see that <span class="math-container">$x=1$</span> is the only solution.</p>
<p>Case 2: Suppose <span class="math-container">$x < 0$</span>. Then</p>
<p><span class="math-container">$$-2x\ge {1+x}^{2} \ge 2x$$</span></p>
<p>where from the first inequality</p>
<p><span class="math-container">$$1+x^2 \le -2x$$</span></p>
<p>we have that <span class="math-container">$x=-1$</span> is the only solution.</p>
<p>So, the domain of <span class="math-container">$\arcsin\left(\frac{x^2+1}{2x}\right)$</span> is <span class="math-container">$x\in\{-1,1\}$</span>.</p>
|
2,077,053 | <p>Please check my proof</p>
<p>If $H\triangleleft G$ then $gHg^{-1}=H $ for every $g\in G $</p>
<p>If $K\triangleleft G$ then $gKg^{-1}=K$ for every $g\in G$</p>
<p>Then $H\cap K = gHg^{-1}gKg^{-1}=gHKg^{-1}$</p>
<p>Then $H\cap K\triangleleft G$</p>
| George Law | 141,584 | <p>The third line is wrong: you seem to be claiming that $H\cap K=HK$, which isn't true (e.g. try $G=\mathbb Z/6\mathbb Z$, $H=\langle2\rangle$, $K=\langle3\rangle$).</p>
<p>What you should be doing is proving that $g(H\cap K)g^{-1}=H\cap K$. Hint: Take $a\in H\cap K$; then $a\in H$ and $a\in K$, so what can you say about $gag^{-1}$?</p>
|
2,782,338 | <blockquote>
<p>Let $X_1, X_2, \cdots$ be independent and identically distributed random variables with expectation $\mu$. Let $N$ be a positive integer-valued random variable such that $E[N] < \infty$ and such that $I_{N≥n}$ is independent of $X_n$ for all $n$. Prove that $$E\!\left[\sum_{i=1}^NX_i\right]=\mu\, E[N]$$</p>
</blockquote>
<p>This is a question from an exam a few years ago. I don’t even know where to start here. What is meant by $I_{N≥n}$?</p>
| Stefan Hansen | 25,632 | <p>First of all, $I_{N\geq n}$ is the random variable defined by
$$
I_{N\geq n}(\omega)=I_{N(\omega)\geq n}=\begin{cases} 1\quad\text{if } N(\omega)\geq n\\0\quad \text{if } N(\omega)<n\end{cases}
$$
and is often called an indicator function (for the set $\{N\geq n\}$). Now, led by the assumption that $I_{N\geq n}$ is independent of $X_n$ for all $n$, we have that
$$
\mathrm{E}\Big[\sum_{n=1}^NX_n\Big]=\mathrm{E}\Big[\sum_{n=1}^\infty X_n I_{N\geq n}\Big]=\sum_{n=1}^\infty \mathrm{E}[X_nI_{N\geq n}]=\sum_{n=1}^\infty \mu P(N\geq n),
$$
where we in the last equality used the independence, and in the second-to-last equality used dominated convergence. All there is left now is to conclude that
$$
\sum_{n=1}^\infty P(N\geq n)=\mathrm{E}[N].
$$</p>
|
76,474 | <p>Use Newton's method to apporzimate the indicated root of the equation correct to six decimal places. The negative root of $e^x = 4-x^2$</p>
<p>I do not know what a negative root is nor do I really know what I am supposed to do. I am guessing raise everything by loge.</p>
| Ross Millikan | 1,827 | <p>The negative root is the one where $x<0$. This equation has two roots, one near $x=-2$ and one near $x=1$. You could look at <a href="http://www.wolframalpha.com/input/?i=plot%204-x%5E2-exp%28x%29" rel="nofollow">this plot</a></p>
<p>You are solving $0=4-x^2-e^x$. For Newton's method, you need to take the derivative, find a starting value like $x_0=-2$, then iterate $x_{i+1}=x_i-\frac{f(x_i)}{f'(x_i)}$ until the change in $x$ gets smaller than $10^{-6}$</p>
<p>Newton's method works well when you start close to the root as long as the derivative doesn't get too small. In this case, I find convergence within three cycles.</p>
|
2,254,694 | <p>I have been stuck on this question for a while now. I have tried many attempts. Here are two that I thought looked promising but lead to a dead end:</p>
<p>Attempt 1:</p>
<p>Write out the terms of $b_n$:</p>
<p>$$b_1=a_{2}-\frac{a_{1}}{2}$$
$$b_2=a_{3}-\frac{a_{2}}{2}$$
$$b_3=a_{4}-\frac{a_{3}}{2}$$
$$\cdots$$
$$b_n=a_{n+1}-\frac{a_{n}}{2}$$</p>
<p>Adding up the terms you get:</p>
<p>$$\sum_{i = 1}^n b_i=a_{n+1}+\frac{a_n}{2}+\frac{a_{n-1}}{2}+\cdots+\frac{a_2}{2}-\frac{a_1}{2}.$$</p>
<p>But a dead end here.</p>
<p>Attempt 2:</p>
<p>For $ε=\dfrac{1}{2}$, $\exists K$ such that $\forall n>K$, $$\left|a_{n+1}-\frac{a_n}{2}\right|<\frac{1}{2}.$$</p>
<p>Now I attempt to prove $\{a_n\}$ is Cauchy and hence converges. </p>
<p>For $m>n>K$,
\begin{align*}
|a_m-a_n|&=\left|a_m-\frac{a_{m-1}}{2}+\frac{a_{m-1}}{2}-\frac{a_{m-2}}{2^2}+\cdots -+\frac{a_{n+1}}{2^{m-n-1}}-a_n\right|\\
&\leq \left|a_m-\frac{a_{m-1}}{2}\right|+\frac{1}{2}\left|a_{m-1}-\frac{a_{m-2}}{2}\right|+\cdots+\left|\frac{a_n}{2^{m-n}}-a_n\right|\\
&\leq \frac{1}{2}+\frac{1}{2} × \frac{1}{2}+\cdots+\left|\frac{a_n}{2^{m-n}}-a_n\right|\\
&<1+\left|\frac{a_n}{2^{m-n}}-a_n\right|,
\end{align*}
and a dead end. </p>
| Lakshya Gupta | 439,351 | <p>Hint :
If $L_t n > \infty$ $\frac{A_{n+1}}{A_{n}}$ is less than $1$. Then the sequence $A_n$ must converge.</p>
|
2,765,570 | <p>Let $E = \{\frac{1}{n} | n \in \mathbb{N}\}$. Show that the function $f(x) = 1 \text{ if } x \in E \text{ and } f(x) = 0 \text{ if } x \notin E$ is integrable on $[0,1]$.</p>
<p>I know of the definition of integrable but I'm having a hard time applying it to this function.</p>
| orangeskid | 168,051 | <p>Let $N \ge 1$. Consider a division of $[0,1]$ with $\|\Delta\| = \delta$. Let $(\xi_i)$ a system of intermediate points for $\Delta$. Let us estimate $S(f,\Delta, \xi)$. Consider $S_1$ the sum corresponding to the intervals of $\Delta$ that intersect $[0,\frac{1}{N}$. We have $|S_1|\le 1/N + \delta$. Let $S_2$ be the corresponding to the other intervals. Note that a non-zero contribution can comeonly the intervals that contain a point of form $\frac{1}{n}$. Now, outside the interval $[0, \frac{1}{N}]$ there are $N-1$ such points. Therefore, for the sum $S_2$ we have
$|S_2|\le 2(N-1)\cdot \delta$. Therefore,
$$|S|\le |S_1|+ |S_2|\le 1/N + \delta + 2(N-1)\cdot \delta$$ </p>
<p>Note that $N\ge 2$ was chosen arbitrary. So choose $N$ first, then take $\delta$ so that $\delta< \frac{1}{2N(N-1)}$. We get
$$S< \frac{3}{N}$$
We see that $S(f, \delta, \xi)$ converges to $0$ as $\|\Delta\|\to 0$. We conclude that $f$ is integrable and has integral $0$</p>
<p>One can consider for $E$ any sequence with limit $0$, and $f$ any bounded function that is $0$ outside $E$. The idea is the same, start with some $N$, consider the interval $[0, \frac{1}{N}]$ and notice that the number of points $a_n$ outside $[0, \frac{1}{N}]$ is finite. </p>
|
495,064 | <p>I'm currently tackling the following problems:</p>
<p>a) Let $a,b \in \mathbb{Z}$ and let $m$ be a nonnegative integer. Prove that $(a,b)=1$ if and only if $(a^m,b)=1$.</p>
<p>If $(a^m,b)=1$ then $pa^m+qb=1$ for some integers $p$ and $q$. It follows that $(pa^{m-1})a+qb=1$, which shows that $(a,b)=1$. This took care of the ''$\Leftarrow$'' direction. I can't think of a way for the ''$\Rightarrow$'' direction. Can you help me here?</p>
<p>b) In addition to the above, let now $n$ be a nonnegative integer as well. Prove that $(a,b)=1$ if and only if $(a^m,b^n)=1$.</p>
<p>The ''$\Leftarrow$'' direction is completely analogous. How about the ''$\Rightarrow$'' direction? </p>
<p><strong>Edit:</strong> Wow, this really wasn't that hard. Here is what I ended up with:</p>
<p>(a) We prove this by contradiction. First assume that $(a,b)=1$ and suppose that $(a^m,b)=d>1$. Then there is a prime $p$ in the prime factorization of $d$ such that $p \mid a^m$ and $p \mid b$. But if $p \mid a^m$, then $p \mid a$, which implies that $p \mid a$ and $p \mid b$, which contradicts the fact that $(a,b)=1$. Thus, $(a^m,b)=1$.</p>
<p>Now assume that $(a^m,b)=1$ and $(a,b)=d>1$. Then we have a prime $p$ such that $p \mid a$ and $p \mid b$, whence $p \mid a^m$ and therefore $p \mid (a^m,b)$, contradiction. Thus, $(a,b)=1$. This completes the proof.</p>
<p>(b) This proof is analogous to the one in (a). Assume that $(a,b)=1$ and suppose that $(a^m,b^n)=d>1$. Then we have a prime $p$ such that $p \mid b^n$ and $p \mid a^n$. But then $p \mid a$ and $p \mid b$, from which it follows that $p \mid (a,b)$, which contradicts our initial assumption. Thus, $(a^m,b^n)=1$.</p>
<p>Now assume that $(a^m,b^n)=1$ and that $(a,b)=d>1$. Then we have a prime $p$ such that $p \mid a$ and $p \mid b$, whence $p \mid a^m$ and $p \mid b^n$. It follows that $p \mid (a^m,b^n)$, contradiction. Thus, $(a,b)=1$. This completes the proof.</p>
| Pedro | 23,350 | <p>This really boils down to two important facts:</p>
<p>$(1)$ If $p$ is prime, then $p\mid ab\implies p\mid a $ or $p\mid b$.</p>
<p>$(2)$ We have $(a,b)>1$ iff there exists a prime $p$ such that $p\mid a $ and $p\mid b$ (, if and only if $p\mid (a,b))$.</p>
<p>Consequently, if $(a^m,b)>1, p\mid a^m\implies p\mid a$, also $p\mid b$ so $p\mid (a,b)$, and $(a,b)>1$. Conversely, if $p\mid (a,b)$ then trivially $p\mid a^m$, thus $p\mid (a^m,b)$.</p>
<p>The other claim's proof is completely analogous. For ease, assume say $n\leqslant m$. </p>
|
2,033,834 | <blockquote>
<p>Evaluate the work integral where $F(x,y)=\langle-y,x\rangle$ over a triangle with vertices $A(-2,-2)$, $B(2,-2)$, $C(0,1)$.</p>
</blockquote>
<p>I am not sure how to approach this problem.
I tried setting $AB(4,0)$, $BC(-2,3)$ and $CA(-2,-3)$ but I am not sure how to proceed.</p>
<p>Without using Green's theorem</p>
| John Wayland Bales | 246,513 | <p>The three segments $AB,BC,CA$ can each be parameterized on the interval $[0,1]$ as follows:</p>
<p>\begin{eqnarray}
(1-t)A+tB=(1-t)\langle-2,-2\rangle+t\langle2,-2\rangle&=&\langle-2+4t,-2\rangle\\
(1-t)B+tC=(1-t)\langle2,-2\rangle+t\langle0,1\rangle&=&\langle2-2t,-2+3t\rangle\\
(1-t)C+tA=(1-t)\langle0,1\rangle+t\langle-2,-2\rangle&=&\langle-2t,1-3t\rangle
\end{eqnarray}</p>
<p>The differential vectors $\langle dx,dy\rangle$ for the three respective segments are $\langle4,0\rangle dt,\langle-2,3\rangle dt,\langle-2,-3\rangle dt$</p>
<p>Thus</p>
<p>\begin{equation}
\int_{AB}-y\,dx+x\,dy=\int_0^1(2)(4)+(2-4t)(0)dt=\int_0^18\,dt=8
\end{equation}
\begin{equation}
\int_{BC}-y\,dx+x\,dy=\int_0^1(2-3t)(-2)+(2-2t)(3)dt=\int_0^12\,dt=2
\end{equation}
\begin{equation}
\int_{CA}-y\,dx+x\,dy=\int_0^1(-1+3t)(-2)+(-2t)(-3)dt=\int_0^12\,dt=2
\end{equation}</p>
<p>Therefore,</p>
<p>\begin{equation}
\int_{ABC}-y\,dx+x\,dy=8+2+2=12
\end{equation}</p>
|
5,581 | <p>(Wade's rewording)</p>
<p>There is a phrase that contains $4$ words and $22$ total letters (not counting the three spaces between the words). One letter appears four times, no letter appears three times, three letters appear twice, and $12$ letters appear once each. What is this phrase?</p>
<blockquote>
<p>@@$$##&&&%%%%????????????</p>
</blockquote>
<p>There are $4$ words and the & represents a whitespace.<br>
@,#,\$,% represent one letter each ($4$ total letters).<br>
Each ? represents one of the other $22$ letter of the alphabet ($12$ total letters).<br>
$16$ different letters used.<br>
ciphertext also needs to be rearranged - it is 'scrambled.' </p>
<p>How do I approach this?<br>
Not looking for the answer really.<br>
I just need a tool to unscramble the phrase and decrypt the ciphertext simultaneously.<br>
Decrypto only does the later. Thank you. </p>
<p>@Noldorin: This is not homework. From elsewhere. </p>
<p>@Kenny: How are encryption/decryption methodologies not math? This is an applied math problem, whereas your preferences tend to be theoretical/ pure math?</p>
| Arturo Magidin | 742 | <p>As David Speyer notes, the problem is too unconstrained for a solution as it stands. You can make some guesses using frequency tables for letters in English, but that's about the only place where some math will come in.</p>
<p>If it were a "straight cryptogram" (the letters are placed in the correct order, whether with the correct word spacing or in blocks), then you are a little closer to a problem in which mathematics can come in. You would have frequency tables, plus also tables on digraphs or trigraphs (blocks of 2 or 3 letters, when certain letters tend to occur next to each other); the two together can give you a lot of information (though the shorter the text, the less useful information they give: you need larger cryptograms to avoid the information getting lost in the "noise"). </p>
<p>If you want to see some of the general procedures to attack "classic" cryptographic systems like monoalphabetic ciphers (where each letter in the original is replaced by another letter, the same one throughout, and the order of the letters is not modified), and polyalphabetic ciphers (where the substitutions depend on the location of the letter; these range from simple <a href="http://en.wikipedia.org/wiki/Vigen%C3%A8re_cipher" rel="nofollow">Vigenère substitution</a> all the way to substitutions like the <a href="http://en.wikipedia.org/wiki/Enigma_machine" rel="nofollow">Engima cipher</a>), or transposition systems (where the order of the letters is changed, but the letters themselves are not), then I can recommend <em>Elementary Cryptanalysis: A mathematical approach</em> by Abraham Sinkov (Mathematical Association of America, New Mathematical Library, ISBN 0883856220). Pretty much all of them are obsolete as far as modern cryptography goes, but it can give you a taste of how (and when) mathematics comes in. </p>
<p>For your problem as stated, however, the role of mathematics is so limited as to become nearly useless. </p>
|
2,963,560 | <blockquote>
<p>Given <span class="math-container">$a,b,c\in \Bbb Z$</span>, pairwise distinct, and <span class="math-container">$n\in \Bbb N\setminus\{0\}$</span> prove that
<span class="math-container">$$S(n)=\frac{a^n}{(a-b)(a-c)}+\frac{b^n}{(b-a)(b-c)}+\frac{c^n}{(c-a)(c-b)}\in \Bbb Z.$$</span></p>
<p>Source: tagged as Kurshar 1959 in a text with problems from math contests</p>
</blockquote>
<p><strong>My attempt</strong>: I approached the problem trying first to solve a particular instance, such as <span class="math-container">$n=1$</span>, to get some insight. </p>
<p>In this particular case (<span class="math-container">$n=1$</span>), the proof is straightforward:
<span class="math-container">$$S(1)=\frac{a(b-c)-b(a-c)+c(a-b)}{(a-b)(a-c)(b-c)}$$</span>
<span class="math-container">$$S(1)=\frac{ab-ac-ab+bc+ac-bc}{(a-b)(a-c)(b-c)}=0\in \Bbb Z$$</span>
Then, I tried the avenue of an induction proof, considering that the proposition is true for <span class="math-container">$S(1)$</span> so that assuming that it is also true for <span class="math-container">$S(n-1)$</span> it would imply it is true for <span class="math-container">$S(n)$</span>. But I couldn't make this step to work. </p>
<p>Hints and answers, not necessarily with induction will be appreciated. But if possible with induction, that would be nice. Sorry if this is a dup.</p>
| Batominovski | 72,152 | <p>Let <span class="math-container">$f(x)$</span> denote the remainder of <span class="math-container">$x^n$</span> when divided by <span class="math-container">$g(x):=(x-a)(x-b)(x-c)$</span>. Because <span class="math-container">$g(x)\in\mathbb{Z}[x]$</span> is monic and <span class="math-container">$x^n\in\mathbb{Z}[x]$</span>, we conclude that <span class="math-container">$f(x)\in\mathbb{Z}[x]$</span>. Observe that <span class="math-container">$f(x)$</span> has degree at most <span class="math-container">$2$</span> and <span class="math-container">$f(t)=t^n$</span> for <span class="math-container">$t\in\{a,b,c\}$</span>. We apply Lagrange interpolation to get
<span class="math-container">$$f(x)=\frac{(x-b)(x-c)}{(a-b)(a-c)}a^n+\frac{(x-c)(x-a)}{(b-c)(b-a)}b^n+\frac{(x-a)(x-b)}{(c-a)(c-b)}c^n\,.$$</span>
Since <span class="math-container">$f(x)\in\mathbb{Z}[x]$</span>, the coefficient of <span class="math-container">$x^2$</span>, which is
<span class="math-container">$$\frac{a^n}{(a-b)(a-c)}+\frac{b^n}{(b-c)(b-a)}+\frac{c^n}{(c-a)(c-b)}\,,$$</span>
must be an integer.</p>
<p>More generally, we also have
<span class="math-container">$$\sum_{j=1}^k\,\frac{a_j^n}{\prod\limits_{i\in\{1,2,\ldots,k\}\setminus\{j\}}\,\left(a_j-a_i\right)}\in\mathbb{Z}$$</span>
for all pairwise distinct integers <span class="math-container">$a_1,a_2,\ldots,a_k$</span>. The prove is similar, by considering <span class="math-container">$g(x)=\prod\limits_{j=1}^k\,\left(x-a_j\right)$</span> instead. Then, the required expression is the coefficient of <span class="math-container">$x^{k-1}$</span> in the polynomial <span class="math-container">$f(x)$</span> obtained as the remainder when dividing <span class="math-container">$x^n$</span> by <span class="math-container">$g(x)$</span>.</p>
|
1,179,959 | <p>taking $v$ as $x$ axis and $u$ as $y$ axis I would like to know whether ${1\over v}+{1\over u}={1\over f}$ has graph of the form $xy=c^2$?</p>
| Rory Daulton | 161,807 | <p>Not quite.</p>
<p>The shape is indeed a hyperbola with horizontal and vertical asymptotes, as does $xy=c^2$, but the center is moved. The center of $xy=c^2$ is at the origin, while your function has the center at the point $(f,f)$. The vertices of your graph are at the origin and $(2f,2f)$.</p>
<p>We can show this by rewriting your equation as</p>
<p>$$(x-f)(y-f)=f^2$$</p>
<p>And, as @Blue points out in a comment, there is a "hole" at the origin: the origin is on the hyperbola but not in the graph of your original equation.</p>
<p>This can also be shown by the standard conic-section technique of a change of variables causing a rotation of the axes of $x+y=xy/f$ by $45^\circ$. Here is a graph where $f=1$.</p>
<p><img src="https://i.stack.imgur.com/fMZOC.jpg" alt="enter image description here"></p>
|
554,003 | <p>How can I find a closed form for the following sum?
$$\sum_{n=1}^{\infty}\left(\frac{H_n}{n}\right)^2$$
($H_n=\sum_{k=1}^n\frac{1}{k}$).</p>
| Sangchul Lee | 9,340 | <p><strong>EDITED.</strong> Some simplifications were made.</p>
<hr>
<p>Here is a solution.</p>
<p><strong>1. Basic facts on the dilogarithm.</strong> Let $\mathrm{Li}_{2}(z)$ be the <em>dilogarithm</em> function defined by</p>
<p>$$ \operatorname{Li}_{2}(z) = \sum_{n=1}^{\infty} \frac{z^{n}}{n^{2}} = - \int_{0}^{z} \frac{\log(1-x)}{x} \, dx. $$</p>
<p>Here the branch cut of $\log $ is chosen to be $(-\infty, 0]$ so that $\operatorname{Li}_{2}$ defines a holomorphic function on the region $\Bbb{C} \setminus [1, \infty)$. Also, it is easy to check (by differentiating both sides) that the following identities hold</p>
<p>\begin{align*}
\operatorname{Li}_{2}\left(\tfrac{z}{z-1}\right)
&= -\mathrm{Li}_{2}(z) - \tfrac{1}{2}\log^{2}(1-z); \quad z \notin [1, \infty) \tag{1} \\
\operatorname{Li}_{2}\left(\tfrac{1}{1-z}\right)
&= \color{blue}{\boxed{\operatorname{Li}_{2}(z) + \zeta(2) - \tfrac{1}{2}\log^{2}(1-z)}} + \color{red}{\boxed{\log(-z)\log(1-z)}}; \quad z \notin [0, \infty) \tag{2}
\end{align*}</p>
<p>Notice that in (2), the blue-colored part is holomorphic on $|z| < 1$ while the red-colored part induces the branch cut $[-1, 0]$.</p>
<p><strong>2. A useful power series.</strong> Now let us consider the power series</p>
<p>$$ f(z) = \sum_{n=0}^{\infty} \frac{H_n}{n} z^n. $$</p>
<p>Then $f(z)$ is automatically holomorphic inside the disc $|z| < 1$. Moreover, it is easy to check that</p>
<p>$$ \sum_{n=1}^{\infty} H_{n} z^{n-1}
= \frac{1}{z} \left( \sum_{n=1}^{\infty} \frac{z^{n}}{n} \right)\left( \sum_{n=0}^{\infty} z^{n}\right)
= -\frac{\log(1-z)}{z(1-z)}. $$</p>
<p>thus integrating both sides, together with the identity $\text{(1)}$, we obtain the following representation of $f(z)$.</p>
<p>$$f(z)
= \operatorname{Li}_{2}(z) + \tfrac{1}{2}\log^{2}(1-z)
= -\operatorname{Li}_{2}\left(\tfrac{z}{z-1}\right). \tag{3}$$</p>
<p><strong>3. Integral representation and the result.</strong> By the Parseval's identity, we have</p>
<p>$$ \sum_{n=1}^{\infty} \frac{H_{n}^{2}}{n^{2}}
= \frac{1}{2\pi} \int_{0}^{2\pi} f(e^{it})f(e^{-it}) \, dt
= \frac{1}{2\pi i} \int_{|z|=1} \frac{f(z)}{z} f\left(\frac{1}{z}\right) \, dz \tag{4} $$</p>
<p>Since $\frac{1}{z}f(z)$ is holomorphic inside $|z| = 1$, the failure of holomorphy of the integrand stems from the branch cut of</p>
<p>\begin{align*}
f\left(\tfrac{1}{z}\right)
&= -\operatorname{Li}_{2}\left(\tfrac{1}{1-z}\right) \\
&= -\color{blue}{\left( \operatorname{Li}_{2}(z) + \zeta(2) - \tfrac{1}{2}\log^{2}(1-z) \right)} - \color{red}{\log(-z)\log(1-z)},
\end{align*}</p>
<p>which is $[0, 1]$. To resolve this, we utilize the identity $\text{(2)}$. Note that the blue-colored portion does not contributes to the the integral $\text{(4)}$, since it remains holomorphic inside $|z| < 1$. That is, only the red-colored portion gives contribution to the integral. Consequently we have</p>
<p>\begin{align*}
\sum_{n=1}^{\infty} \frac{H_{n}^{2}}{n^{2}}
&= -\frac{1}{2\pi i} \int_{|z|=1} \frac{f(z)}{z} \color{red}{\log(-z)\log(1-z)} \, dz. \tag{5}
\end{align*}</p>
<p>Since the integrand is holomorphic on $\Bbb{C} \setminus [0, \infty)$, we can utilize the keyhole contour wrapping around $[0, 1]$ to reduce $\text{(5)}$ to</p>
<p>\begin{align*}
\sum_{n=1}^{\infty} \frac{H_{n}^{2}}{n^{2}}
&=-\frac{1}{2\pi i} \Bigg\{ \int_{0^{-}i}^{1+0^{-}i} \frac{f(z)\log(-z)\log(1-z)}{z} \, dz \\
&\qquad \qquad + \int_{1+0^{+}i}^{+0^{+}i} \frac{f(z)\log(-z)\log(1-z)}{z} \, dz \Bigg\} \\
&=-\frac{1}{2\pi i} \Bigg\{ \int_{0}^{1} \frac{f(x)(\log x + i\pi)\log(1-x)}{x} \, dx \\
&\qquad \qquad - \int_{0}^{1} \frac{f(x)(\log x - i\pi)\log(1-x)}{x} \, dx \Bigg\} \\
&=-\int_{0}^{1} \frac{f(x)\log(1-x)}{x} \, dx. \tag{5}
\end{align*}</p>
<p>Plugging $\text{(3)}$ to the last integral and simplifying a little bit, we have</p>
<p>\begin{align*}
\sum_{n=1}^{\infty} \frac{H_{n}^{2}}{n^{2}}
&= - \int_{0}^{1} \frac{\operatorname{Li}_2(x)\log(1-x)}{x} \, dx - \frac{1}{2}\int_{0}^{1} \frac{\log^{3}(1-x)}{x} \, dx \\
&= \left[ \frac{1}{2}\operatorname{Li}_2(x)^2 \right]_0^1 - \frac{1}{2} \int_{0}^{1} \frac{\log^3 x}{1-x} \, dx \\
&= \frac{1}{2}\zeta(2)^{2} + \frac{1}{2} \Gamma(4)\zeta(4) \\
&= \frac{17\pi^{4}}{360}
\end{align*}</p>
<p>as desired.</p>
|
167,228 | <p>I have shown several plots by Show function:</p>
<pre><code>Show[p1, p2, p3, p4, p5, p6, p7, p8, PlotRange -> All]
</code></pre>
<p>I need to save data of all plots in one <code>txt</code> format, so that they can be recognizable from each other. How can I do that?</p>
<p>Each plot is a 2D plot say, $f_i$ as a function of $x$.</p>
| Akku14 | 34,287 | <p><strong>Improved answer:</strong></p>
<p>Find all roots and add conjugate complex roots to get all real factorisations.</p>
<pre><code>sr = Select[Root[x^5 + 1, #] & /@ Range[5], Im[#] == 0 &]
(* {-1} *)
si = Select[Root[x^5 + 1, #] & /@ Range[5], Im[#] != 0 &]
(* {Root[1 - #1 + #1^2 - #1^3 + #1^4 &, 1],
Root[1 - #1 + #1^2 - #1^3 + #1^4 &, 2],
Root[1 - #1 + #1^2 - #1^3 + #1^4 &, 3],
Root[1 - #1 + #1^2 - #1^3 + #1^4 &, 4]} *)
pa = Plus @@ # & /@ Partition[si, 2]
Factor[x^5 + 1, Extension -> {#}] & /@ Union[sr, pa] //ToRadicals // Union
(* {-(1/4) (1 + x) (-2 + (1 + Sqrt[5]) x - 2 x^2) (2 + (-1 + Sqrt[5]) x + 2 x^2),
(1 + x) (1 - x + x^2 - x^3 + x^4)} *)
</code></pre>
|
4,531,507 | <p><a href="https://i.stack.imgur.com/rJmjH.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rJmjH.jpg" alt="enter image description here" /></a></p>
<blockquote>
<p>Find the area of <span class="math-container">$ABCD$</span> if <span class="math-container">$PSC$</span> is a semicircle.</p>
</blockquote>
<p>I have used the tangent properties of circles. I also assumed many variables to get a relationship between product of sides of the rectangle <span class="math-container">$ABCD$</span> but eventually and unfortunately, I proved <span class="math-container">$$1=1$$</span></p>
<p>Any help is greatly appreciated.</p>
| timon92 | 210,525 | <p>Let <span class="math-container">$E$</span> be the projection of <span class="math-container">$D$</span> on <span class="math-container">$CQ$</span>. Move triangle <span class="math-container">$BCQ$</span> by <span class="math-container">$\vec{BA}$</span>. The question boils down to finding the area of a parallelogram whose area is equal to <span class="math-container">$CQ\cdot DE =10\cdot DE$</span>.</p>
<p>Let <span class="math-container">$M$</span> be the midpoint of <span class="math-container">$PC$</span>, and <span class="math-container">$R$</span> be the midpoint of <span class="math-container">$CQ$</span>. Can you prove that triangles <span class="math-container">$CDE$</span> and <span class="math-container">$MCN$</span> are congruent? This will lead to <span class="math-container">$DE=5$</span>, hence the area equals <span class="math-container">$50$</span>.</p>
|
2,379,466 | <p>Let there be a set A ,where A = {0,1,2,3,4,5,6,7,8,9} and we are to choose a two member subset {k,m} from A such that k,m are elements of A</p>
<p>What is the probability that |k-m| = 2 ?</p>
| Graham Kemp | 135,106 | <p>Noting that the members in set $A$ are all of the integers between 0 and 9 inclusive</p>
<p>Thus we are being asked for the probability of selecting two members in the list that are seperated by one other member.</p>
<p>So how many ways are there to do this? How many equally-likely ways are there to select <em>any</em> two from the ten members? Divide and calculate.</p>
|
2,379,466 | <p>Let there be a set A ,where A = {0,1,2,3,4,5,6,7,8,9} and we are to choose a two member subset {k,m} from A such that k,m are elements of A</p>
<p>What is the probability that |k-m| = 2 ?</p>
| farruhota | 425,072 | <p>$1)$ All possible two member subsets: combination of $10$ digits taken $2$ at a time.</p>
<p>$2)$ All possible two member subsets satisfying $|k-m|=2: \{0,2\}, \{1,3\}, \cdots, \{7,9\}.$</p>
<p>$3)$ The probability is to divide $2)$ by $1)$ to get $8/45$.</p>
<p>Note: $\{0,2\}=\{2,0\}$. If insisted on distinct $k$ and $m$, then $1)$ permutation, $2)$ twice more, $3)$ the same answer.</p>
|
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