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 |
|---|---|---|---|---|
1,254,189 | <p>I know that I have to study the order of every element in $\mathbb{Q/Z}$.
But what do I do?
I've been struggling of what to do for this question</p>
| k1.M | 132,351 | <p>In the group $\mathbb Q/\mathbb Z$ we have the element $\frac 12+\mathbb Z$ which is of order two. But non of the elements of additive group $\mathbb Q$ is of order two. This shows that there is no such isomorphism. </p>
|
950,485 | <p>I have been trying to solve the following limit but am completely stuck.</p>
<p>$$\lim_{\alpha \rightarrow \infty} 1-\left( \frac{y+\alpha}{\alpha-1} \right)^{-\alpha}$$</p>
<p>I have tried inverting the ratio and came up with the following expression:</p>
<p>$$ 1 - \lim_{\alpha \rightarrow \infty} \left( 1-\frac{y+1}{y+\alpha}\right)^\alpha$$</p>
<p>Which roughly resembles the exponential function:</p>
<p>$$\lim_{\alpha \rightarrow \infty} \left( 1- \frac{x}{\alpha} \right)^\alpha = \exp(-x)$$</p>
<p>Except for the additive term in the denominator. Is there a u-substitution type trick to this?</p>
| Timbuc | 118,527 | <p>If you're already a little advanced in functions and their limits, you could try</p>
<p>$$\lim_{\alpha\to\infty}\frac{\log\left(1-\frac{y+1}{y+\alpha}\right)}{\frac1\alpha}\stackrel{\text{l'Hospital}}=\lim_{\alpha\to\infty}\frac{\frac{y+\alpha}{\alpha-1}\cdot\frac{y+1}{(y+\alpha)^2}}{-\frac1{\alpha^2}}=-\lim_{\alpha\to\infty}\frac{\alpha^2}{\alpha-1}\frac{y+1}{\alpha+y}=-(y+1)$$</p>
<p>So the original limit is $\;e^{-(y+1)}\;$ (why?)</p>
|
1,992,256 | <p>I have to prove</p>
<p>$\sqrt{1} + \sqrt{2} +...+\sqrt{n} \le \frac{2}{3}*(n+1)\sqrt{n+1}$</p>
<p>by using math induction. </p>
<p>First step is to prove that it works for n = 1 , which is true.
Next step is to prove it for n + 1. We can rewrite the formula using</p>
<p>$\sum_{i=1}^{n+1} \sqrt{i}= \sum_{i=1}^{n}\sqrt{i} + \sqrt{i+1}$</p>
<p>and we can substitute sum </p>
<p>$\frac{2}{3}(n+1)\sqrt{n+1} +\sqrt{n+1} \le \frac{2}{3}(n+2)\sqrt{n+2}$</p>
<p>we can transform the left side into </p>
<p>$\sqrt{n+1}(\frac{2}{3}(n+1)+1)$</p>
<p>but how to I further transform the formula in order to find if the sentence is true?</p>
<p>Thanks for all help!</p>
| Jam | 161,490 | <p>You started off well and almost got it. Here's a hint, where LHS and RHS are respectively the left and right hand sides of the formula we want:</p>
<p>$$\begin{align}\mathrm{LHS}^2&=\left(\frac23\right)^2(n+1)\left(n+\frac52\right)^2\\
&=\left(\frac23\right)^2(n^3+6 n^2+11.25n+6.25)
\\
&<\left(\frac23\right)^2(n^3+6n^2+12n+8)&&=\left(\frac23\right)^2(n+2)^3\\
&&&=\mathrm{RHS}^2\\
\mathrm{LHS}&<\mathrm{RHS}
\end{align}$$</p>
<hr>
<p>An alternative approach, with some calculus.</p>
<p>$$\begin{align}
\ln(\mathrm{LHS})&=\ln\frac23+\frac12\ln(n+1)+\ln(n+\frac52)\\
\ln(\mathrm{RHS})&=\ln\frac23+\frac32\ln(n+2)\\
\\
f(n)&=\ln(\mathrm{RHS})-\ln(\mathrm{LHS})\\
&=\frac32\ln(n+2)-\frac12\ln(n+1)-\ln(n+\frac52)\\
f'(n)&=\frac{\frac32}{n+2}-\frac{\frac12}{n+1}-\frac1{n+\frac52}\\
&=\frac{-\frac34}{(n+2)(n+1)(n+\frac52)}\\
\end{align}$$</p>
<p>So $f'(n)$ is never $0$, which means that $f$ has no turning points but we can show that $f(1)=0.049\ldots$ and hence $\ln(\mathrm{RHS})-\ln(\mathrm{LHS})>0$</p>
<hr>
<p>Let me know if you've spotted an error.</p>
<p>-Jam</p>
|
2,804,495 | <p>I was asked to solve this double integral:
Compute the area between $y=2x^2$ and $y=x^2$ and the hyperbolae $xy=1$ and $xy=2$ in </p>
<p>$$ \iint dx \,dy$$</p>
<p>I tried to solve it starting with considering that </p>
<p>$$x^2 \leq y \leq 2x^2 $$</p>
<p>suitabile for integration interval in $y$, obtaining the incomplete form</p>
<p>$$ \int^{x^2}_{2x^2} \int_\ldots^\ldots dx \,dy$$</p>
<p>but I also have
$$1 \leq xy \leq 2$$ and I would obtain a result in which I still have one independent variabile. </p>
<p>Please, can anyone help me? Thanks in advance.</p>
| SlipEternal | 156,808 | <p>Plot the four functions. See where they intersect. Once you do, you will find that you can rewrite this integral as:</p>
<p>$$\displaystyle \int_{ \tfrac{1}{ \sqrt[3]{2} } }^1\left( 2x^2-\dfrac{1}{x}\right)dx + \int_1^{\sqrt[3]{2}}\left(\dfrac{2}{x}-x^2\right)dx$$</p>
|
3,653,979 | <p>Let A = <span class="math-container">$\begin{bmatrix}r_1 & r_2 & r_3 & r_4 & r_5\end{bmatrix}^T$</span> have rows <strong><span class="math-container">$r_1$</span></strong>, <strong><span class="math-container">$r_2$</span></strong>, <strong><span class="math-container">$r_3$</span></strong>, <strong><span class="math-container">$r_4$</span></strong>, <strong><span class="math-container">$r_5$</span></strong> <span class="math-container">$\in$</span> <span class="math-container">$\mathbb{R}^5$</span>. Assume det(A) = -3.</p>
<p>Compute </p>
<p>det <span class="math-container">$\begin{bmatrix}2r_1 + 3r_2 + 4r_3 + 4r_4\\ r_1 + 2r_2\\ r_2+3r_3\\r_3+4r_4\\r_1\end{bmatrix}$</span> and justify your answer.</p>
<p><em>(the part before this was the same thing but with this matrix)</em></p>
<p>det <span class="math-container">$\begin{bmatrix}5r_1 + 5r_2 + 5r_3 + 5r_4 + 5r_5\\ 4r_1 + 4r_2 + 4r_3 + 4r_4\\ r_1\\2r_1+2r_2\\3r_1+3r_2+3r_3\end{bmatrix}$</span> </p>
<p>and I think I figured this one out (you just Gauss-Jordan it and it ends up being equal to A so it's just -3)</p>
<p>But my main confusion is with the first matrix, because even with Gauss-Jordan it doesn't equal A so how can I even find the determinant because it's not a square matrix.</p>
<p>Any help is appreciated. Thanks in advance!</p>
| Ben Grossmann | 81,360 | <p><strong>Hint:</strong> Note that the two matrices under consideration can be written as <span class="math-container">$M_1A$</span> and <span class="math-container">$M_2A$</span> respectively, where
<span class="math-container">$$
M_1 =
\pmatrix{2&3&4&4&0\\ 1&2&0&0&0\\ 0&1&3&0&0\\0&0&1&4&0\\1&0&0&0&0}, \quad
M_2 =
\pmatrix{
5&5&5&5&5\\ 4&4&4&4&0\\ 1&0&0&0&0\\2&2&0&0&0\\3&3&3&0&0
}.
$$</span></p>
|
2,653,483 | <p>Let $a =111 \ldots 1$, where the digit $1$ appears $2018$ consecutive times.</p>
<p>Let $b = 222 \ldots 2$, where the digit $2$ appears $1009$ consecutive times.</p>
<p>Without using a calculator, evaluate $\sqrt{a − b}$.</p>
| hamam_Abdallah | 369,188 | <p>$$a=\sum_{i=0}^{2017}10^i $$
$$=\frac {10^{2018}-1}{10-1} $$
$$b=2\sum_{i=0}^{1008}10^i $$
$$=2\frac {10^{1009}-1}{10-1}$$</p>
<p>$$9 (a-b)=10^{2018}-2.10^{1009}+1$$
$$=(10^{1009}-1)^2$$</p>
<p>then</p>
<p>$$\sqrt {a-b}=\frac {10^{1009}-1}{3} $$</p>
<p>$=333...333.$ (1009 consecutive times).</p>
|
3,705,539 | <p>I am trying to learn more about probability and came across an interesting question that I am stuck on and can no longer find online. There are 20 numbered balls and 10 bins. Someone is trying to assign the balls to the bins, but does it with replacement on accident.</p>
<p>So they did the following: Place a ball in bin 1, record it, then remove ball (with replacement remember). Place a ball in bin 2, record it, then remove ball. Place a ball in bin 3, record it, then remove ball. So for each bin, you have put in 1 ball. There are ten bins, therefore you do that process once for every bin. Once you have done that the experiment is over.</p>
<p>What is the probability exactly 1 ball was assigned to exactly 4 bins? What is the probability at least 2 bins received the same ball?</p>
<p>A) 1 Ball in 4 Bins:</p>
<p>We have <span class="math-container">${20 \choose 1}$</span> being the different ways we can choose the 1 ball that was assigned. Also, we have <span class="math-container">${19 \choose 6}$</span> being the different ways the other 19 balls can be picked for assignment. However, what is the sample size? Would it be <span class="math-container">$20^{10}$</span>? Thus the answer would be <span class="math-container">$\frac{{20 \choose 1}{19 \choose 6}}{20^{10}}$</span>.</p>
<p>B) Probability of at least 2 repeated can be represented as <span class="math-container">$1-P(\text{Zero Repeated})- P(\text{One Repeated})$</span>. So <span class="math-container">$P(0) = {20 \choose 10}/20^{10}$</span> and <span class="math-container">$P(1) = \frac{{20 \choose 1}{19 \choose 9}}{20^{10}}$</span>. Then we can plug and chug.</p>
<p>Are these right? Is this how to think about this type of problem?</p>
| Phicar | 78,870 | <p>The way I understand from the comments is that you are modeling this is by a function <span class="math-container">$f:\{\text{bins}\}\longrightarrow \{\text{balls}\}$</span> in which you take a bin and you assign a ball to it, they can have the same ball(replacement). so there are indeed <span class="math-container">$20^{10}$</span> possible functions.<br><br> Now, for A, you pick the ball in <span class="math-container">$\binom{20}{1}=20$</span> ways and then you choose the other <span class="math-container">$6$</span> bins. But you are assuming those are going each to a different bin and you are not considering different orderings. This contradicts the supposition of replacement. This seems a little more involved that what you propose, because you need to know that you are not overcounting possibilities. I would use the <strong>Inclusion Exclusion Principle</strong> to compute <span class="math-container">$\left |\bigcup _{i=1}^{20}A_i\right |,$</span> where <span class="math-container">$A_i = \{\text{The i-th ball was assigned to 4 bins}\}.$</span> Notice that <span class="math-container">$|A_i|=\binom{10}{4}19^{6}.$</span> For 2 <span class="math-container">$|A_i\cap A_j|=\binom{10}{4}\binom{6}{4}(20-2)^{2}.$</span> Can you have <span class="math-container">$3$</span> balls going to <span class="math-container">$4$</span> bins? No, so you just have to combine this two possibilities. <br><br>
For B, You want to take out the functions that are one to one because if a function is not one to one then 2 bins were going to the same ball. So the probability would be <span class="math-container">$$1-\frac{\binom{20}{10}}{20^{10}}$$</span></p>
|
2,209,438 | <p>I am trying to find this limit,</p>
<blockquote>
<p>$$\lim_{x \rightarrow 0} \frac{1}{x^4} \int_{\sin{x}}^{x} \arctan{t}dt$$</p>
</blockquote>
<p>Using the fundamental theorem of calculus, part 1,
$\arctan$ is a continuous function, so
$$F(x):=\int_0^x \arctan{t}dt$$
and I can change the limit to
$$\lim_{x \rightarrow 0} \frac{F(x)-F(\sin x)}{x^4}$$</p>
<p>I keep getting $+\infty$, but when I actually integrate $\arctan$ (integration by parts) and plot the function inside the limit, the graph tends to $-\infty$ as $x \rightarrow 0+$.</p>
<p>I tried using l'Hospital's rule, but the calculation gets tedious.</p>
<p>Can anyone give me hints?</p>
<p><strong>EDIT</strong></p>
<p>I kept thinking about the problem, and I thought of power series and solved it, returned to the site and found 3 great answers. Thank You!</p>
| Nick Peterson | 81,839 | <p>Power series can help, if you know them.</p>
<p>We know that
$$
\arctan t=\sum_{n=0}^{\infty}(-1)^n\frac{t^{2n+1}}{2n+1},\qquad \lvert t\rvert<1.
$$
Therefore an antiderivative for $\arctan t$ is
$$
F(t):=\sum_{n=0}^{\infty}(-1)^n\frac{t^{2n+2}}{(2n+1)(2n+2)},\qquad \lvert t\rvert<1.
$$
So, as $x\to0$,
$$
F(x)=\frac{x^2}{2}-\frac{x^4}{12}+O(x^6).
$$
Now, recalling that
$$
\sin x=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!}=x-\frac{x^3}{6}+O(x^5)=x+O(x^3),
$$
we see that as $x\to0$ (and therefore $\sin x\to0$),
$$
\begin{align*}
F(\sin x)&=\frac{1}{2}\left(x-\frac{x^3}{6}+O(x^5)\right)^2-\frac{1}{12}\left(x+O(x^3)\right)^4+O[(x+O(x^3))^6]\\
&=\frac{1}{2}\left[x^2-\frac{x^4}{3}+O(x^6)\right]-\frac{1}{12}\left[x^4+O(x^6)\right]+O(x^6)\\
&=\frac{x^2}{2}-\frac{x^4}{4}+O(x^6)
\end{align*}
$$
So, all told, our limit is
$$
\lim_{x\to0}\frac{F(x)-F(\sin x)}{x^4}=\lim_{x\to0}\frac{\,\frac{x^4}{6}\,}{x^4}=\frac{1}{6}
$$</p>
|
2,209,438 | <p>I am trying to find this limit,</p>
<blockquote>
<p>$$\lim_{x \rightarrow 0} \frac{1}{x^4} \int_{\sin{x}}^{x} \arctan{t}dt$$</p>
</blockquote>
<p>Using the fundamental theorem of calculus, part 1,
$\arctan$ is a continuous function, so
$$F(x):=\int_0^x \arctan{t}dt$$
and I can change the limit to
$$\lim_{x \rightarrow 0} \frac{F(x)-F(\sin x)}{x^4}$$</p>
<p>I keep getting $+\infty$, but when I actually integrate $\arctan$ (integration by parts) and plot the function inside the limit, the graph tends to $-\infty$ as $x \rightarrow 0+$.</p>
<p>I tried using l'Hospital's rule, but the calculation gets tedious.</p>
<p>Can anyone give me hints?</p>
<p><strong>EDIT</strong></p>
<p>I kept thinking about the problem, and I thought of power series and solved it, returned to the site and found 3 great answers. Thank You!</p>
| davidlowryduda | 9,754 | <p>Here's one way to do this. The Taylor expansion for $\arctan(x) = x - x^3/3 + x^5/5 + \cdots$. So then
$$ \int_{\sin(x)}^x \arctan(t)dt \sim (x^2/2 - x^4/12 + x^6/30) - \frac{(\sin x)^2}{2} + \frac{(\sin x)^4}{12} - \frac{(\sin x)^6}{30}.$$</p>
<p>Using another Taylor expansion, $\sin x = x - x^3/3! + x^5/5! + \cdots$, so the plan is to pay attention only to those terms of degree up to $4$ in these expansions. Then (dropping all terms of degree greater than $4$),
$$ (\sin x)^2/2 = (x - x^3/3!)^2/2 = x^2/2 - x^4/6$$
and
$$ (\sin x)^4/12 = (x - x^3/3!)^4/4 = x^4/12.$$
Thus
$$ \int_{\sin(x)}^x \arctan(t)dt \sim (x^2/2 - x^4/12) - (x^2/2 - x^4/6) + x^4/12 = x^4/6.$$</p>
<p>So multiplying by $x^{-4}$ and taking the limit gives $1/6$.</p>
<p>As an aside, l'Hopital's rule would also work.</p>
|
9,918 | <p>I recently flagged as "rude or offensive" the comment </p>
<blockquote>
<p>As the tone should suggest, he’s a crank. It’s a hysterical screed with a few nuggets of fact surrounded by a great deal of nonsense. E.g., he may find that set theory ‘doesn’t make sense’, but a great many of us have no trouble making sense of it. – <a href="https://math.stackexchange.com/questions/356264#comment765805_356264">Brian M. Scott</a></p>
</blockquote>
<p>to the question</p>
<p><a href="https://math.stackexchange.com/questions/356264">Infinite sets don't exist!?</a></p>
<p>My complaint is very narrow: I am not arguing that the subject of the comment is right in his critique of infinite sets, only that the word "crank" is inappropriate in this context. In particular, none of the answers to the question provide evidence of crankdom, but rather disagreement over philosophy. If the comment had simply been </p>
<p>"He may find that set theory ‘doesn’t make sense’, but a great many of us have no trouble making sense of it."</p>
<p>I would have no objection.</p>
<p>I could also have flagged it as "not constructive/off topic" or as "too chatty". But to my mind it is the unwarranted rudeness that is the real problem here. More generally, I am quite upset that a productive mathematician is being called a crank. I don't know the subject of the comment personally, but I am familiar with his work and he is most definitely <em>not</em> a crank. </p>
<p>Comments like this reflect quite poorly on math.stackexchange and I hope not to see any more of them. What is the best way to encourage commenters to refrain from baseless accusations of crankdom? Is flagging the appropriate course of action?</p>
| Community | -1 | <p>I'll answer this from a bit of an outside perspective. I'm a moderator on <a href="https://skeptics.stackexchange.com/">Skeptics</a> where this issue was a significant concern. The topics on that site often evoke very strong reactions and it is not unusual for comments to get rather heated. </p>
<p>My observations are that once a comment thread starts to get personal, and the comments are more about the users than about specific issues, there is a high chance the whole thing escalates. In minor cases this just results in long and useless comment threads without direct insults. But if not caught early, some users were happy to escalate to direct insults and other suspension-worthy offenses.</p>
<p>You example is a special case as it wasn't directed at a user of this site, so there is a much lower chance for this to escalate. Still, this comment sparked a rather long and pretty much off-topic comment thread.</p>
<p>You also don't need to make such issues personal, keeping the criticism focused on the actual issues is often even more convinving. Your specific example would not lose anything by omitting the first sentence in my opinion. </p>
<p>In my experience, focusing on specific issues instead of people avoids a lot of moderation problems. A general rule to avoid any personal attacks makes a moderators job a lot easier, and it also makes for a much friendlier experience for all users.</p>
|
1,368,988 | <p>I was thinking about different ways of finding $\pi$ and stumbled upon what I'm sure is a very old method: dividing a circle of radius $r$ up into $n$ isosceles triangles each with radial side length $r$ and central angle $$\theta=\frac{360^\circ}{n}$$ Use $s$ for the side opposite to $\theta$.</p>
<p><img src="https://i.stack.imgur.com/vueMQ.png" alt="An example for n=8."></p>
<p>Then we can approximate the circumference as $sn$. By the law of cosines: $$s=\sqrt{2r^2-2r^2 \cos{\theta}}=r\sqrt{2-2\cos{\left(\frac{360^\circ}{n}\right)}}$$</p>
<p>We know $\pi=\text{circumference}/\text{diameter} \approx \frac{sn}{2r}=\frac{n}{2}\sqrt{2-2\cos\left(\frac{360^\circ}{n}\right)}$. This becomes exact at the limit:
$$\pi=\lim_{n \to \infty}\frac{n}{2}\sqrt{2-2\cos\left(\frac{360^\circ}{n}\right)}$$</p>
<p>Now for my question: <strong>How would you solve the opposite problem?</strong> To make my meaning more clear, above I used the definition of $\pi$ to determine a limit that gives its value. But if I had just been given the limit, what technique(s) could I have used to determine that it evaluates to $\pi$?</p>
| Jack D'Aurizio | 44,121 | <p>$$\sqrt{2-2\cos x}=2\left|\sin\frac{x}{2}\right|$$
hence everything boils down to:
$$ \lim_{x\to 0}\frac{\sin x}{x}=1.$$</p>
|
2,934,028 | <blockquote>
<p>A particle moves along the top of the
parabola <span class="math-container">$y^2 = 2x$</span> from left to right at a constant speed of 5 units
per second. Find the velocity of the particle as it moves through
the point <span class="math-container">$(2, 2)$</span>. </p>
</blockquote>
<p>So I isolate <span class="math-container">$y$</span>, giving me <span class="math-container">$y=\sqrt{2x}$</span>. I then find the derivative of <span class="math-container">$y$</span>, which is <span class="math-container">$1/\sqrt{2x}$</span>. And <span class="math-container">$\sqrt{2x}=y$</span> so the derivative of also equal to <span class="math-container">$1/y$</span>. At <span class="math-container">$(2,2)$</span> the derivative is 0.5. Not sure where to go from there though. Any help is appreciated. </p>
| Community | -1 | <p>in cartesian co-ordinate sysytem :</p>
<p><span class="math-container">$(speed)^2=\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2=25$</span> also, </p>
<p><span class="math-container">$\left(\dfrac{dy}{dt}\right)=\dfrac{1}{y}\left(\dfrac{dx}{dt}\right) $</span> put it in above equation </p>
<p>to get ,</p>
<p><span class="math-container">$\left(\dfrac{dx}{dt}\right)=\dfrac{5y}{\sqrt{1+y^2}}$</span> and <span class="math-container">$\left(\dfrac{dy}{dt}\right)=\dfrac{5}{\sqrt{1+y^2}}$</span></p>
<p>thus, at point (2,2) speed will be </p>
<p><span class="math-container">$\sqrt{ (2\sqrt5)^2+(\sqrt 5)^2}=5 $</span> unit per second</p>
<p>in above calculations gravity is not taken into account </p>
|
2,934,028 | <blockquote>
<p>A particle moves along the top of the
parabola <span class="math-container">$y^2 = 2x$</span> from left to right at a constant speed of 5 units
per second. Find the velocity of the particle as it moves through
the point <span class="math-container">$(2, 2)$</span>. </p>
</blockquote>
<p>So I isolate <span class="math-container">$y$</span>, giving me <span class="math-container">$y=\sqrt{2x}$</span>. I then find the derivative of <span class="math-container">$y$</span>, which is <span class="math-container">$1/\sqrt{2x}$</span>. And <span class="math-container">$\sqrt{2x}=y$</span> so the derivative of also equal to <span class="math-container">$1/y$</span>. At <span class="math-container">$(2,2)$</span> the derivative is 0.5. Not sure where to go from there though. Any help is appreciated. </p>
| user | 505,767 | <p>In that case is useful to use parametric equation that is</p>
<p><span class="math-container">$$y^2=2x \implies \vec s(t)=(2c^2t^2,2ct) \implies \vec v(t)=s'(t)=(4c^2t,2c)$$</span></p>
<p>and since the point <span class="math-container">$(2,2)$</span> is reached at <span class="math-container">$t=1/c$</span>, that is <span class="math-container">$\vec s(1/c)=(2,2)$</span>, the velocity at that point is <span class="math-container">$\vec v(1/c)=(4c,2c)$</span> and since the speed is <span class="math-container">$5$</span> we have</p>
<p><span class="math-container">$$\|v(1/c)\|=\sqrt{16c^2+4c^2}=5 \implies c=\frac{\sqrt 5}2$$</span></p>
<p>therefore the velocity at <span class="math-container">$(2,2)$</span> is</p>
<p><span class="math-container">$$v(t)=\left(2\sqrt 5,\sqrt 5\right)$$</span></p>
<p>As an alternative</p>
<p><span class="math-container">$$y^2=2x \implies 2ydy=2dx \implies \frac{dy}{dx}=\frac1y$$</span></p>
<p>therefore at the point <span class="math-container">$(2,2)$</span> the tangent vector is <span class="math-container">$(4,2)$</span> and then the velocity vector is <span class="math-container">$k(4,2)$</span> and form the condition on the speed we find </p>
<p><span class="math-container">$$5=k\sqrt{16+4}\implies k=\frac{\sqrt 5}2$$</span></p>
|
2,609,537 | <p>Is the following Proof Correct? In particular please comment on the correctness of the given formulas.</p>
<p><strong>Theorem.</strong> Given that $x$ is a real number, $x\neq 0$, and $x + \frac{1}{x}$ is an integer. For all $n\ge 1$, $x^n+\frac{1}{x^n}$ is an integer.</p>
<p><strong>Proof.</strong> We construct the proof by recourse to <em>Strong-Induction</em>. Assume for an arbitrary $n\in\mathbf{Z^+}$ that $x^k+\frac{1}{x^k}$ is an integer for any positive integer $k$ strictly less than $n$. Now Consider the following cases.</p>
<p><em>Case-1:</em> If $n$ is even then for some $l\in\mathbf{Z^+}$, $n = 2l$ thus the <em>Binomial-Theorem</em> implies that
$$(x+\frac{1}{x})^n - \sum_{j=1}^{l-1}\binom{n}{2j}\left(x^{2j}+\frac{1}{x^{2j}}\right)-\binom{n}{l} = x^n+\frac{1}{x^n}$$
from the inductive hypothesis we know that $x+\frac{1}{x}\in\mathbf{Z}$ which implies that $(x+\frac{1}{x})^n \in\mathbf{Z}$ in addition it also follows from the inductive hypothesis that $x^{2j}+\frac{1}{x^{2j}}\in\mathbf{Z}$ for $j\in\{1,2,3,...,l-1\}$ moreover we also that $\binom{n}{r}\in\mathbf{Z^+}$ is always a positive integer implying that $x^n+\frac{1}{x^n}$ is an integer.</p>
<p><em>Case-2:</em> If $n$ is odd then for some $l\in\mathbf{Z^+}$, $n = 2l+1$ thus the <em>Binomial-Theorem</em> implies that
$$(x+\frac{1}{x})^n - \sum_{j=1}^{l}\binom{n}{2j-1}\left(x^{2j-1}+\frac{1}{x^{2j-1}}\right) = x^n+\frac{1}{x^n}$$
and by using the same reasoning as in the previous case we can deduce that $x^n+\frac{1}{x^n}$ is an integer.</p>
| Atmos | 516,446 | <p>I think it is easier to see that</p>
<blockquote>
<p>$$
x^{n+2}+\frac{1}{x^{n+2}}=\left(x+\frac{1}{x}\right)\left(x^{n+1}+\frac{1}{x^{n+1}}\right)-\left(x^n+\frac{1}{x^n}\right)
$$</p>
</blockquote>
|
3,497,420 | <p>Consider the function <span class="math-container">$$f(x,y)=x^6-2x^2y-x^4y+2y^2.$$</span> The point <span class="math-container">$(0,0)$</span> is a critical point. Observe,
<span class="math-container">\begin{align*}
f_x & = 6x^5-4xy-4x^3y, f_x(0,0)=0\\
f_y & = 2x^2-x^4+4y. f_y(0,0)=0\\
f_{xx} & = 30x^4-4y-12x^2y, f_{xx}(0,0)=0\\
f_{xy} & = 4x-4x^3, f_{xy}(0,0)=0\\
f_{yy} & = 4, f_{yy}=4
\end{align*}</span></p>
<p>So, in order to determine the nature of the above critical point, we need to check the Hessian at <span class="math-container">$(0,0)$</span> which is <span class="math-container">$0$</span> and hence the test is inconclusive. <span class="math-container">$$ H(x,y)= \det \begin{pmatrix} f_{xx} & f_{xy}\\ f_{yx} & f_{yy} \end{pmatrix}=\det \begin{pmatrix} 0 & 0 \\ 0 & 4 \end{pmatrix}=0$$</span>So, I tried to see the function on slices like <span class="math-container">$y=0$</span> and <span class="math-container">$y=x$</span> but nothing worked. So please suggest me how do I find the nature of the critical point in this case?</p>
| Cesareo | 397,348 | <p>With <span class="math-container">$g(x) = x^2$</span> we have</p>
<p><span class="math-container">$$
f(g(x),y) = g(x)^3-2g(x)y-g(x)^2y +2y^2
$$</span></p>
<p>or</p>
<p><span class="math-container">$$
f(g,y)=g^3-2g y-g^2y+2y^2
$$</span></p>
<p>and the hessian of <span class="math-container">$f(g,y)$</span> is</p>
<p><span class="math-container">$$
H(g,y) = \left(
\begin{array}{cc}
6 g-2 y & -2 g-2 \\
-2 g-2 & 4 \\
\end{array}
\right)
$$</span></p>
<p>and also</p>
<p><span class="math-container">$$
H(0,0) = \left(
\begin{array}{cc}
0 & -2 \\
-2 & 4 \\
\end{array}
\right)
$$</span></p>
<p>with eigenvalues </p>
<p><span class="math-container">$$
\left\{2 \left(1+\sqrt{2}\right),2 \left(1-\sqrt{2}\right)\right\}
$$</span></p>
<p>characterizing a saddle point.</p>
|
3,826,994 | <p>I would like to find <span class="math-container">$z$</span> which minimizes the below, when <span class="math-container">$x$</span> is held at a specific value.</p>
<p><span class="math-container">$f(x,z) =\sqrt{\sqrt{x^2 + z^2} - 0.25}$</span></p>
<p>For example; I would like to find the value of <span class="math-container">$z$</span> which minimizes the function when <span class="math-container">$x = 0.5$</span></p>
| Soumyadwip Chanda | 823,370 | <p>The expression means that z subtends an angle <span class="math-container">$\frac{\pi}{2}$</span> at the points <span class="math-container">$2i$</span> and <span class="math-container">$6$</span></p>
<p>Ponder upon the following visual</p>
<p><a href="https://i.stack.imgur.com/BaDQy.png" rel="noreferrer"><img src="https://i.stack.imgur.com/BaDQy.png" alt="enter image description here" /></a></p>
<p>The green arrows show some more positions which are possible for <span class="math-container">$z$</span>. Clearly, as it subtends <span class="math-container">$90^o$</span> at those points, so the locus of <span class="math-container">$z$</span> is a semicircle, the end point of hose diameter is <span class="math-container">$(0,2)$</span> and <span class="math-container">$(6,0)$</span></p>
|
1,572,045 | <p>This is maybe a stupid question, but I want to find the roots of:</p>
<blockquote>
<p>$$2(x+2)(x-1)^3-3(x-1)^2(x+2)^2=0$$</p>
</blockquote>
<p>What that I did:</p>
<p>$$\underbrace{2(x+2)(x-1)(x-1)(x-1)}_{A}-\underbrace{3(x-1)(x-1)(x+2)(x+2)}_{B}=0$$</p>
<p>So the roots are when $A$ and $B$ are both zeros when $x=1$ and $x=-2$ </p>
<p>My questions:</p>
<p>$1)$ Is there an easy way to see that $x=-8$ is a root too?</p>
<p>$2)$ The degree of this polynomial is $4$, so I should have $4$ roots, and here I have only $3$</p>
| Leox | 97,339 | <p><strong>Ніnt:</strong>
$$2(x+2)(x-1)^3-3(x-1)^2(x+2)^2=- \left( x+8 \right) \left( x+2 \right) \left( x-1 \right) ^{2}.$$</p>
|
1,572,045 | <p>This is maybe a stupid question, but I want to find the roots of:</p>
<blockquote>
<p>$$2(x+2)(x-1)^3-3(x-1)^2(x+2)^2=0$$</p>
</blockquote>
<p>What that I did:</p>
<p>$$\underbrace{2(x+2)(x-1)(x-1)(x-1)}_{A}-\underbrace{3(x-1)(x-1)(x+2)(x+2)}_{B}=0$$</p>
<p>So the roots are when $A$ and $B$ are both zeros when $x=1$ and $x=-2$ </p>
<p>My questions:</p>
<p>$1)$ Is there an easy way to see that $x=-8$ is a root too?</p>
<p>$2)$ The degree of this polynomial is $4$, so I should have $4$ roots, and here I have only $3$</p>
| SchrodingersCat | 278,967 | <p>$$2(x+2)(x-1)^3-3(x-1)^2(x+2)^2=0$$
$$(x-1)^2(x+2)\left[2(x-1)-3(x+2)\right]=0$$
$$(x-1)^2(x+2)(-x-8)=0$$
$$(x-1)^2(x+2)(x+8)=0$$</p>
<p>So this polynomial of degree $4$ has $4$ real roots i.e. $1,1,-2,-8$.</p>
|
1,044,507 | <p>Sample:
$$∀x ∈ R+,∃y ∈ R+, x < y ⇒ x > y$$</p>
<p>Say I tried <code>y = 5</code>. Do I need to check if the consequent is true for just the x values less than 5?</p>
<p>Secondly, Since there is no value y that makes the antecedent true, is this statement true since there are no counter examples? The implication is never used.</p>
<p>I'm looking for a small explanation, rather than a solution (don't treat it like a homework question looking for an answer).</p>
| Dan Christensen | 3,515 | <p>I'm not sure what you mean by "check if the consequent is true," but we can prove </p>
<p>$∀x ∈ R+:∃y ∈ R+: [x < y \implies P(x,y)]$ </p>
<p>for <em>any</em> binary predicate $P$. As Henning pointed out, the required $y$ could be just $x$ itself.</p>
<p>The proof makes use of the following facts:</p>
<p>$\forall x: x=x$</p>
<p>$\forall x\in R+: \neg x\lt x$</p>
<p>Proof:</p>
<ol>
<li><p>Suppose $a\in R+$</p></li>
<li><p>$a=a$</p></li>
<li><p>$\exists x:x=a$ (from 2)</p></li>
<li><p>$b=a$ (from 3, introducing $b$)</p></li>
<li><p>$b\in R+$ (from 1 and 4)</p></li>
<li><p>$\neg b\lt b$ (from 5)</p></li>
<li><p>Suppose $a\lt b$</p></li>
<li><p>$b\lt b$ (from 4 and 7)</p></li>
<li><p>$\neg b\lt b \implies P(a,b)$ (from 8)</p></li>
<li><p>$P$ (from 6 and 9)</p></li>
<li><p>$a\lt b \implies P(a,b)$ (from 7 and 10)</p></li>
<li><p>$\forall x:[x\in R+ \implies \exists y:[y\in R+ [\land \implies P(x,y)]]]$ (from 1 and 11)</p></li>
</ol>
<p>or equivalently</p>
<p>$∀x ∈ R+:∃y ∈ R+: [x \lt y \implies P(x,y)]$ </p>
|
11,178 | <p>As far as I know, one way to take a homotopy colimit in a model category is to replace (up to acyclic fibration) all arrows in the diagram with cofibrations, and take the strict colimit of the resulting diagram.</p>
<p>In Top with the model structure given by Serre fibrations, cofibrations, and weak equivalences, if one wants to obtain a homotopy pushout of the diagram <span class="math-container">$X \leftarrow A \rightarrow Y$</span>, it is "enough" to replace only one of these arrows with a cofibration: that is, there is a natural map (by the universal property of the pushout) <span class="math-container">$Cyl(X)\cup_A Cyl(Y) \to Cyl(X) \cup_A Y$</span> that is a homotopy equivalence of spaces.</p>
<blockquote>
<p>Question 1: What conditions on the model category <span class="math-container">$\mathcal{C}$</span> (or objects <span class="math-container">$X,Y,A$</span>) will guarantee that the natural map <span class="math-container">$Cyl(X) \cup_A Cyl(Y) \to Cyl(X) \cup_A Y$</span> is a weak equivalence?</p>
<p>Question 2: This question is less precise, but if the map above is a weak equivalence, does that mean <span class="math-container">$Cyl(X) \cup_A Y$</span> is a good model for the homotopy pushout?</p>
</blockquote>
| Charles Rezk | 437 | <p>(I'll assume that in a general model category $\mathcal{C}$, $\mathrm{Cyl}(X)$ really means: a factorization of $A\to X$ into a cofibration $A\to \mathrm{Cyl}(X)$ followed by a trivial fibration $\mathrm{Cyl}(X)\to X$.)</p>
<p>A sufficient condition on objects for the map in question 1 to be a weak equivalence, is that the objects $X,Y,A$ be <em>cofibrant</em>. (The fact you want to use is the statement due to Reedy (which you can find at the start of the chapter on Proper Model Categories in Hirschhorn's book), that a pushout of a weak equivalence between cofibrant objects along a cofibration is a weak equivalence.)</p>
<p>A sufficient condition on $\mathcal{C}$ for the map in question 1 to be a weak equivalence, is that the model category be <em>left proper</em>.</p>
<p>It's an interesting fact that in Top, the this also works if $\mathrm{Cyl}(X)$ denotes the "classical" mapping cylinder construction ($X$ union a cylinder on $A$), which isn't necessarily a cofibration in the Quillen model structure. The slickest proof is to use the "excisive triad theorem", as proved by May in <em>A Consise Course in Algebraic Topology</em>, p.79. This also leads to a proof that Top is left proper.</p>
|
1,354,490 | <p>Prove this function is injective $f(x)=x+\mod(x,7)$.</p>
<p><strong>Attempt:</strong></p>
<p>I tried separating in two cases: $x \equiv y \pmod 7$ and $x \not \equiv y \pmod 7 $:</p>
<p>First case:
$$f(x)=f(y) \iff x+\mod(x,7)=y+ \mod (y,7)\implies x= y
$$</p>
<p>But I couldn't prove the second case.</p>
| Steven Alexis Gregory | 75,410 | <p>We have $f:\mathbb Z \to \mathbb Z$ defined by
$f(x) = x + \operatorname{mod}(x,7)$</p>
<p>Define $g:\mathbb Z \to \mathbb Z$ by $g(y) = y - \operatorname{mod}(4y,7)$</p>
<p>Suppose $x = 7a + b$ where $0 \le b \lt 7$. Then $f(x) = 7a + b + b = 7a + 2b$.</p>
<p>\begin{align}
g(f(x))
&= g(7a + 2b)\\
&= 7a + 2b - \operatorname{mod}(4(7a+2b),7)\\
&= 7a + 2b - \operatorname{mod}(28a+8b,7)\\
&= 7a + 2b - b\\
&= x
\end{align}</p>
<p>Now suppose have $f(x_1) = f(x_2)$.</p>
<p>Then $g(f(x_1)) = g(f(x_2))$.</p>
<p>Hence $x_1 = x_2$.</p>
<p>So $f$ is injective.</p>
|
4,227,800 | <p>Let <span class="math-container">$p=10007$</span> which is prime. I want to find the number of matrices X of <span class="math-container">$2\times2$</span> dimension with elements from <span class="math-container">$\mathbb{Z}_p$</span> for which <span class="math-container">$X^2\equiv I$</span> (mod <span class="math-container">$p$</span>). Where <span class="math-container">$I$</span> - identity matrix. How can I solve this problem? I tried to write down the square of matrix so that I get the system of congruences, but I can't figure out what to do after that, and I don't think this is the right way to solve this.</p>
| Alan Abraham | 823,763 | <p>Let's say <span class="math-container">$X=\begin{bmatrix} a&b\\c&d\end{bmatrix}$</span>, after computing <span class="math-container">$X^2$</span>, we get the following system of congruences
<span class="math-container">$$\begin{cases} a^2+bc\equiv 1\mod p\\b(a+d)\equiv 0\mod p\\c(a+d)\equiv 0\mod p\\d^2+bc\equiv 1\mod p\end{cases}$$</span></p>
<p>The <span class="math-container">$2$</span>nd and <span class="math-container">$3$</span>rd equations are simple to solve, so we will do casework on the those. We either have <span class="math-container">$b,c\equiv 0\mod p$</span> or <span class="math-container">$a+d\equiv 0\mod p$</span>. There is also some overlap between these two cases, so we have to watch out for overcounting</p>
<p><strong>Case 1:</strong> <span class="math-container">$b,c\equiv 0\mod p$</span>
Then we have
<span class="math-container">$$\begin{cases} a^2\equiv 1\mod p\\d^2\equiv 1\mod p\end{cases}$$</span>
There are clearly <span class="math-container">$2$</span> solutions for each of the congruences (<span class="math-container">$a,d\equiv \pm 1\mod p$</span>), for a total of <span class="math-container">$4$</span> solutions. Since the <span class="math-container">$2$</span> solutions when <span class="math-container">$a\equiv -d\mod p$</span> overlap with the second case, we will exclude those for a total of <span class="math-container">$\boxed{2}$</span>.</p>
<p><strong>Case 2:</strong> <span class="math-container">$a+d\equiv 0\mod p$</span>
Then we have
<span class="math-container">$$bc\equiv 1-a^2\equiv 1-d^2\mod p$$</span>
We will split this case into sub cases depending on the values of <span class="math-container">$a,d$</span>.</p>
<p><strong>Case 2a:</strong> <span class="math-container">$a^2,d^2\equiv 1\mod p$</span>
The solutions for <span class="math-container">$a,d$</span> are <span class="math-container">$a\equiv \pm 1\mod p,d\equiv \mp 1\mod p$</span>. Hence there are <span class="math-container">$2$</span> solutions for <span class="math-container">$a,d$</span>.
To find the number of solutions for <span class="math-container">$b,c$</span>, we need to find the number of solutions to
<span class="math-container">$$bc\equiv 0\mod p$$</span>
This only happens if at least <span class="math-container">$1$</span> of <span class="math-container">$b,c$</span> is equivalent to <span class="math-container">$0\mod p$</span>. We must be careful to not double count the solutions of <span class="math-container">$b,c\equiv 0\mod p$</span>. We get a total of <span class="math-container">$p+p-1=2p-1$</span> solutions for <span class="math-container">$b,c$</span>.</p>
<p>Hence, there are <span class="math-container">$2(2p-1)=\boxed{4p-2}$</span> solutions for this subcase.</p>
<p><strong>Case 2b:</strong> <span class="math-container">$a^2,d^2\equiv k\not\equiv 1\mod p$</span>
Since <span class="math-container">$a,d\not\equiv -1,1\mod p$</span>, there are <span class="math-container">$p-2$</span> ordered pairs to choose for <span class="math-container">$(a,d)$</span>. Each of these generate a <span class="math-container">$k$</span> that satisfies <span class="math-container">$1-k\not\equiv 0\mod p$</span>.</p>
<p>To find the solutions for <span class="math-container">$b,c$</span>, these satisfy
<span class="math-container">$$bc\equiv 1-k\mod p$$</span>
Since <span class="math-container">$1-k\not\equiv 0\mod p$</span>, we have that
<span class="math-container">$$gcd(b,p)=1\implies c\equiv (1-k)b^{-1}\mod p$$</span>
Hence, there is a unique solution for all <span class="math-container">$b\in[1,p-1]$</span>.</p>
<p>The total in this subcase is <span class="math-container">$(p-1)(p-2)=\boxed{p^2-3p+2}$</span></p>
<hr />
<p>Totaling all of these up, the number of solutions for <span class="math-container">$X$</span> is
<span class="math-container">$$2+(4p-2)+(p^2-3p+2)$$</span>
<span class="math-container">$$\boxed{\boxed{p^2+p+2}}$$</span></p>
<p>This formula should only work when <span class="math-container">$p>2$</span>. This is because throughout the process we assumed <span class="math-container">$1\not\equiv -1\mod p$</span>. Upon substituting <span class="math-container">$p=10007$</span>, we get the total number of solutions is <span class="math-container">$\boxed{100150058}$</span>.</p>
|
4,227,800 | <p>Let <span class="math-container">$p=10007$</span> which is prime. I want to find the number of matrices X of <span class="math-container">$2\times2$</span> dimension with elements from <span class="math-container">$\mathbb{Z}_p$</span> for which <span class="math-container">$X^2\equiv I$</span> (mod <span class="math-container">$p$</span>). Where <span class="math-container">$I$</span> - identity matrix. How can I solve this problem? I tried to write down the square of matrix so that I get the system of congruences, but I can't figure out what to do after that, and I don't think this is the right way to solve this.</p>
| paul garrett | 12,291 | <p>As in @ThomasAndrews' comment, there is indeed a systematic (not-so-computational) way to approach this, maybe less lending itself to computational errors and such.</p>
<p>So, for a two-by-two matrix with <span class="math-container">$x^2=1$</span>, certainly <span class="math-container">$x^2-1=0$</span>, so either <span class="math-container">$x-1=0$</span>, or <span class="math-container">$x+1=0$</span>, or neither holds, but <span class="math-container">$x^2-1=0$</span>.</p>
<p>In the first two cases, <span class="math-container">$x$</span> is <span class="math-container">$\pm$</span> the identity matrix, that is, has either eigenvalue <span class="math-container">$1$</span> (with multiplicity <span class="math-container">$2$</span>) or eigenvalue <span class="math-container">$-1$</span> (with multiplicity <span class="math-container">$2$</span>).</p>
<p>In the third case, necessarily <span class="math-container">$x$</span> has eigenvalues <span class="math-container">$1$</span> and <span class="math-container">$-1$</span>. By standard linear algebra (!!!), this means that there is an invertible two-by-two matrix <span class="math-container">$A$</span> such that
<span class="math-container">$$
x \;=\; Ax_oA^{-1}
$$</span>
with
<span class="math-container">$$
x_o \;=\; \pmatrix{1 & 0 \cr 0 & -1}
$$</span>
Two matrices <span class="math-container">$A$</span> and <span class="math-container">$B$</span> produce the same <span class="math-container">$x$</span> if and only if
<span class="math-container">$$
x_o \;=\; (B^{-1}A)x_o(B^{-1}A)^{-1}
$$</span>
That is, <span class="math-container">$B^{-1}A$</span> is in the <em>normalizer</em> of <span class="math-container">$x_o$</span>. We can check that this normalizer is a <em>subgroup</em> of the two-by-two invertible matrices. For <span class="math-container">$g=\pmatrix{a & b\cr c & d}$</span>, the condition <span class="math-container">$gx_og^{-1}=x_o$</span> (multiplied out) is equivalent to <span class="math-container">$b=c=0$</span>, <em>if</em> <span class="math-container">$-1\not=+1$</span>, which does hold for <span class="math-container">$p\not=2$</span>. That is, for <span class="math-container">$p\not=2$</span>, the normalizer consists of diagonal matrices.</p>
<p>Thus, the collection of matrices <span class="math-container">$x$</span> with eigenvalues <span class="math-container">$1$</span> and <span class="math-container">$-1$</span> is in bijection with the coset space of two-by-two invertible matrices by the subgroup of diagonal invertible matrices.</p>
<p>The number of diagonal matrices is <span class="math-container">$(q-1)^2$</span>, while the cardinality of two-by-two invertible matrices is <span class="math-container">$(q^2-1)(q^2-q)$</span>. Thus, the number of such matrices <span class="math-container">$x$</span> is the quotient, namely, <span class="math-container">$(q+1)q$</span>.</p>
<p>Thus, adding the identity and negative identity, we get <span class="math-container">$(q+1)q+1+1=q^2+q+2$</span>. (Matching Alan Abrahams more concrete computation.)</p>
|
177,144 | <p>If G is a finite group, I understand that the category of RO(G)-graded spectra, when rationalized, becomes Quillen equivalent to the category of Mackey functors valued in chain complexes of rational vector spaces.</p>
<p>How does RO(G) act on the category of Mackey functors? For instance, if F = F(G/H) is a Mackey functor and V is a real representation of G, what is (S^V F)(G/H)?</p>
| Peter May | 14,447 | <p>The term ``RO(G)-graded spectrum'' is a misnomer, not to be used.
It is never used in the literature I'm familiar with, and it never
should be. There are several Quillen equivalent models for the
category of genuine $G$-spectra, and such animals represent
$RO(G)$-graded cohomology theories, but it is meaningless to
think of them as $RO(G)$-graded themselves. They are not. </p>
<p>The zeroth stable homotopy group system of the sphere $G$-spectrum
(take $H$-fixed point spectra and then take $\pi_0$ of those) is the
Burnside ring Mackey functor $\mathbf{A}(G)$. It is a Green functor
(think of it as a ring in the category of Mackey functors), and any Mackey functor
is a module over it. That is where the action takes place. The category
of rational $G$-spectra, $G$-finite, is equivalent to the category of
chain complexes in the category of rational Mackey functors. </p>
<p>Via permutation
representations, we get a map of Green functors from $\mathbf{A}(G)$ to the
representation ring Mackey functor $\mathbf{RO}(G)$ that sends $G/H$ to $RO(H)$.
That is one way that $RO(G)$ enters the picture.</p>
<p>In equivariant cohomology theory, the use of $RO(G)$ is a great convenience, but it is not the thing most
intrinsic to the mathematics. Ignore the multiplication on $RO(G)$,
which is irrelevant to its use for grading theories,
and think of it just as an abelian group. Send a representation $V$
to the isomorphism class of the suspension $G$-spectrum of the one-point
compactification $S^V$. This induces a homomorphism from $RO(G)$ into
the Picard group $Pic(Ho G\mathcal S)$ of the stable homotopy category of
$G$-spectra, namely the abelian group of equivalence classes of
$G$-spectra that are invertible under the smash product. That homomorphism
is neither a monomorphism nor an epimorphism. See <a href="http://www.math.uchicago.edu/~may/PAPERS/FLMJan01.pdf">http://www.math.uchicago.edu/~may/PAPERS/FLMJan01.pdf</a>
for a discussion of that Picard group. Logically, equivariant cohomology
theories really should be graded on $Pic(Ho G\mathcal S)$, but that group is much less
convenient than $RO(G)$. </p>
|
2,353,142 | <p>Solve:
$$(\cot^{-1} (x))^2 - 4\cot^{-1} (x) + 3 \geq 0$$.</p>
<p>My Attempt:
$$(\cot^{-1} (x))^2 - 4\cot^{-1} (x) + 3 \geq 0$$.
Let $\cot^{-1} (x)=t$. then</p>
<p>$$t^2-4t+3\geq 0$$
$$(t-3)(t-1)\geq 0$$
Either, $\cot^{-1} (x) \leq 1$</p>
<p>Or, $\cot^{-1} (x) \geq 3$</p>
<p>I solved till here, but couldn't get the answer given in book. The answer in book is $x \in (-\infty, \cot (3)] \cup [\cot (3), \infty )$.</p>
| Davide Giraudo | 9,849 | <p>Assume that $p\gt 1$ is such that inequality $2|x||y| \leqslant x^p + y^p$ holds for any $x,y\in\geqslant 0$. In particular, it holds for $x=y\geqslant 0$ hence we have for any positive $x$ the following inequality:
$$\tag{*} x^2 \leqslant x^p .$$
If $p\gt 2$, then divide in (*) by $x^2$ and let $x$ go to zero (or small enough) to get a contradiction. </p>
<p>If $p\lt 2$, then divide in (*) by $x^p$ and let $x$ go to infinity (or large enough) to get a contradiction. </p>
<p>Finally, the equalily holds for $p=2$ since $\left(\left|x\right|-\left|y\right|\right)^2\geqslant 0$. </p>
|
211,427 | <p>Can't seem to figure this one out. Could anyone help me out and explain it to me?<br>
Thank you.</p>
<p>Let $P$ and $Q$ be relations on $Z$ by x$P$y iff x + 1 <= y and a$Q$b iff a + 2 <= b. Prove that P $\circ$ Q = {(p,q) belonging to ZxZ | p + 3 <= q} </p>
| Yury | 35,791 | <p>Every bi-linear form $\sigma:V\times V \to F$ defines a linear operator $A$ from $V$ to $V^*$; $A$ maps vector $v\in V$ to linear functional $l_v$ such that $l_v(u) = \sigma(u,v)$. The correspondence between $\sigma$ and $A$ is “canonical” — it doesn't depend on the basis. </p>
<p>Now suppose that we fix a basis $e_1, \dots, e_n$ in $V$. Define a basis $e_1^*, \dots, e_k^*$ in $V^*$ as follows: $e_i^*(e_i) = 1$ and $e_i^*(e_j) = 0$ if $j\neq i$. Let $\varphi$ be the map from $V^*$ to $V$ that sends $e_i^*$ to $e_i$. Then the operator $[L]_{B}$ is equal to $\varphi A$. Note that unlike the definition of $A$, the definition of $\varphi$ depends on the choice of the basis in $V$.</p>
|
3,411,081 | <p>A group of 12 people are going out to a concert on Saturday night. The group will take three cars with four people in each car. If they distribute themselves at random, what is the probability that A and B will be in the same car?</p>
<p>I tried (12C2*10C2*8C4*4C4)/(12C4*8C4*4C4) because you're choosing two first and then sorting the rest. This gave me a more than 100% probability.</p>
<p>The answer at the back is 0.273 assuming the cars are non-distinct.
The closest I came to it was 4C2/12C3 but I'm not sure why this works.</p>
| amd | 265,466 | <p>Write down <span class="math-container">$1+1+\cdots+1=X$</span>. Now group runs of adjacent ones by going from left to right and for each <span class="math-container">$+$</span> sign you encounter deciding whether or not a group boundary occurs there. Parenthesize each group to get a partition of <span class="math-container">$X$</span> into a sum of nonnegative integers. Each possible choice of group boundaries corresponds to a distinct partition if order matters (prove this!) and there are <span class="math-container">$X-1$</span> decision points, therefore there are <span class="math-container">$2^{X-1}$</span> such partitions of <span class="math-container">$X$</span>.</p>
|
2,592,007 | <blockquote>
<p>Let $V$ be a vector space and $W,U\subseteq V$ subspaces s.t $W\not \subseteq U$
$\dim(V)=5, \dim(W)=2, \dim(U)=4$ </p>
<p>Prove\Disprove: $\dim(U\cap W)=1$</p>
</blockquote>
<p>So I started with \begin{align} & \dim(W+U)=\dim(U)+\dim(W)-\dim(U\cap W) \\[10pt]
\iff & \dim(W+U)=4+2-\dim(U\cap W)\end{align}</p>
<p>Now in the following steps I am not sure</p>
<p>Because $W,U\subseteq V$ subspaces $\dim(W+U)\leq \dim(V)=5$ So $1\leq \dim(U\cap W)$</p>
<p>What can I conclude from $W\not \subseteq U$ ? How should I continue?</p>
| angryavian | 43,949 | <p>It remains to prove $\dim(U \cap W) \le 1$.
You already know $\dim(U \cap W) \le \dim(W) = 2$.
If you can show that $W \not\subseteq U$ implies $\dim(U \cap W) \ne 2$, then you are finished.</p>
<hr>
<p>Proving the unproved claim above (via contrapositive): if $\dim(U \cap W) = 2$, then $U \cap W = W$ because $\dim(W)=2$. This implies $W = U \cap W \subseteq U$.</p>
|
1,270,042 | <p>$$(a+5)(b-1)=ab-a+5b-5=20-5=15.$$</p>
<p>So, both $a + 5$ and $b-1$ divide $15$. </p>
<p>Then, $a + 5$ is one of $15, -15, 3, -3, 5, -5, 1, -1$, so $a$ is one of $10, -20, -2, -8, 0, -10, -4, -6$ and $b – 1$ is one of $15, -15, 3, -3, 5, -5, 1, -1$, so $b = 14, -14, 4, -2, 6, -4, 2, 0$.</p>
<p>Could all possibilities for $a, b$ found by considering $(a+5)(b-1)$ be just random(not in a probability sense) and not connected to $ab = a - 5b + 20$ at all? In other words, could it be that if some of the possible $a, b$ found this way happen to satisfy $ab = a - 5b + 20$, then it's just a coincidence? </p>
| Edwin Gray | 231,072 | <p>From ab = a -5b +20, we have b(a + 5) = a + 5 + 15.
Therefore a + 5 divides 15 and a must be positive.Hence a = 10.
Then 15b = 15 + 15, and b = 2. Edwin Gray</p>
|
872,889 | <p>What determinant is zero? What equation does this give for the plane?</p>
<p>I need some help here, am pretty stuck</p>
| GEdgar | 442 | <p>Five years ago or so I worked on some "tree-type" continued fractions...
$$
T =
1+\frac{\displaystyle
1+\frac{\displaystyle
1+\frac{\displaystyle
1+\frac{1+\cdots\;}{2+\cdots\;}
}{\displaystyle
2+\frac{2+\cdots\;}{4+\cdots\;}
}
}{\displaystyle
2+\frac{\displaystyle
2+\frac{2+\cdots\;}{4+\cdots\;}
}{\displaystyle
4+\frac{4+\cdots\;}{8+\cdots\;}
}
}
}{\displaystyle
2+\frac{\displaystyle
2+\frac{\displaystyle
2+\frac{2+\cdots\;}{4+\cdots\;}
}{\displaystyle
4+\frac{4+\cdots\;}{8+\cdots\;}
}
}{\displaystyle
4+\frac{\displaystyle
4+\frac{4+\cdots\;}{8+\cdots\;}
}{\displaystyle
8+\frac{8+\cdots\;}{16+\cdots\;}
}
}
} \;.
$$</p>
<p>But it never went very far. I did submit one of these as a problem...
<em>Math. Mag.</em> <strong>79</strong> (2006) p.151</p>
|
2,512,363 | <h2>Defining the barycentre and finding its variance</h2>
<p>I have a set of $N$ points at the locations $x_i$ which has weights $W_i$, $i=1,\ldots, N$ and want to find the barycenter (or center of gravity) </p>
<p>$$ B = {\sum_{i=1}^N W_i x_i\over \sum_{i=1}^N W_i}.$$</p>
<p>I want to find the variance of $B$.
I assume that $W_i$ are i.i.d. and positive with expected value $Ew_i=\mu$ and variance $Var[W_i]=E[w_i-\mu]^2=\sigma^2 $.
I also assume that $N$ is large, so $\sum_{i=1}^N w_i$ is well approximated by $N\mu$ (justified below).</p>
<p>When I do the calculations (see one example below) I end up with
$$ Var [B] = {\sigma^2 \over N^2 \mu^2} \sum_{i=1}^N x_i^2. $$</p>
<h2>The contradiction</h2>
<p>As explained below, I think that the barycentre variance should be independent of a shift of all the $x_i$, that is $Var[B]$ should not change if all $x_i$ is shifted to $x_i+c$. But $\sum_{i=1}^N x_i^2$, and therefore my estimated $Var[B]$, will change a shift of the $x_i$. </p>
<p>Where is the error: In my calculations, or the expectation that the variance should be independent of shift in $x$.</p>
<hr>
<h2>One derivation of the variance</h2>
<p>$$Var [B ]
= Var \left[{\sum_{i=1}^N W_i x_i \over \sum_{i=1}^N w_i}\right]
= {1\over N^2 \mu^2} \sum_{i=1}^N x_i^2 Var [W_i]
={\sigma^2\over N^2 \mu^2} \sum_{i=1}^N x_i^2$$</p>
<p>Here I used that $Var [\sum_i X_i] = \sum_i Var[ X_i]$ when the $X_i$ are independent, that $Var [cX]=c^2 Var[X]$, and that $Var[ w_i] =\sigma^2$.</p>
<p>I have also done the same in a more direct fashion, with the same result. It is kind of messy, so I have omitted it.</p>
<h2>Why do I expect $Var [B]$ to be invariant to a shift of $x_i$?</h2>
<p>The expected value of $B$ is the mean of the $x_i$:</p>
<p>$$E[B]=E\left[{\sum_{i=1}^N E[W_i] x_i\over \sum_{i=1}^N W_i}\right]
={\sum_{i=1}^N E[W_i] x_i\over N\mu}
={\mu\sum_{i=1}^N x_i\over N\mu}={\sum_{i=1}^N x_i\over N}=\bar x_i.$$ </p>
<p>Moving all $x_i$ by $c$ will move every barycentre estimate by $c$:
$$ \tilde B = {\sum_{i=1}^N W_i (x_i+c)\over \sum_{i=1}^N W_i}
= {\sum_{i=1}^N W_i x_i\over \sum_{i=1}^N W_i} +{\sum_{i=1}^N W_i c\over \sum_{i=1}^N W_i}=
{\sum_{i=1}^N W_i x_i\over \sum_{i=1}^N W_i} +c=B+c.$$</p>
<p>The variance of $B$ is $Var[B]=E[(B-E[B])^2]$.
Moving all the $x_i$ by $c$ adds $c$ to $B$ and $E[B]$, so $B-E[B]$ and the variance should stay the same.</p>
<p>To check this I did a <a href="https://gist.github.com/anonymous/30f09acb70b920b63f110bd1de32ee79" rel="nofollow noreferrer">MATLAB simulation</a>. In that simulation the estimated variances and barycentres (after i remove the displacement $c$) are identical.</p>
<h2>Why do I think $\sum_{i=1}^N w_i =N\mu$ is a good approximation</h2>
<p>Although the variance of $\sum_{i=1}^N w_i$ grow with $N$, this means that the standard deviation grow with $\sqrt N$.
And since the expected value grow with $N$, the relative spread will decrease. </p>
<p>It is the relative spread that is important for this case. Say that for one $N$ the expected value is 100 and the standard deviation is 1. Then, if $\sum_{i=1}^N w_i $ miss $N\mu$ with one standard deviation my estimate of $B$ is off by 1%.
If I increase N by a factor of a 100, the expected value will be 10 000 and the standard deviation will be 10.
So if $\sum_{i=1}^N w_i $ miss $N\mu$ with one standard deviation my estimate of $B$ is off by 0.1%.</p>
<p>Which mean that for large enough N the error will be negligible.</p>
<hr>
<h2>Context</h2>
<p>I'm trying to repeat a derivation for the variance of an estimated arrival time from <a href="http://ieeexplore.ieee.org/document/5607320/" rel="nofollow noreferrer">this paper</a>. The barycentre is used to find the time when the echo from the seafloor is received for seabed mapping echo sounder. </p>
<p>As a simplified model, the part of the time sequence used in the barycentre calculation can be modelled as a sequence of independent, identically (Rayleigh) distributed, amplitude values. These amplitude values are equally spaced in time.</p>
<p>In this question the $W_i$ represent the amplitudes, the $x_i$ it the sample times and $B$ is the estimated time of arrival.
The paper states that it can be show analytically that </p>
<p>$$Var[B]= \frac{\Delta x^2}{12(N+1)}\left(\frac{4}{\pi}-1\right)N(N+2),$$</p>
<p>where $\Delta x$ is the distance between the $x_i$ ($x_i=x_0+i \Delta x$).
I have translated the equation to the notation used in this question.
No reference is given, although I am searching in related papers.</p>
<p>The middle factor in that equation $\left(4/\pi-1\right)$ is equal to the ratio the variance and the mean value squared of a <a href="https://en.wikipedia.org/wiki/Rayleigh_distribution" rel="nofollow noreferrer">Rayleigh</a> distributed variable.</p>
| Mixopteryx | 187,956 | <p>Through <a href="https://tel.archives-ouvertes.fr/tel-00783671" rel="nofollow noreferrer" title="Ladroit PhD thesis">this PhD thesis</a> (chapter 3.2.3) i found the following solution.</p>
<p>For simplicity, define
$$F_i = {W_i\over\sum W_i}, \text{ so that } B=\sum_{i=1}^N F_i x_i.$$
(As Deans answer suggest)</p>
<p>Then
$$E[(B-E[B])^2]={\sum_{i=1}^N \sum_{j=1}^N (x_i-\bar x) (x_j-\bar x) E[(F_i-E[F_i]) (F_j-E[F_j])} $$
$${=\sum_{i=1}^N (x_i-\bar x)^2 E[(F_i-E[F_i])^2]}
{=\sum_{i=1}^N (x_i-\bar x)^2 Var[F_i]}$$
if we assume that the $F_i$ are uncorrelated. $\bar x$ is the mean of the $x_i$.</p>
<p>The first step holds since
$${\sum_{i=1}^N \sum_{j=1}^N (x_i-\bar x) (x_j-\bar x) E[(F_i-E[F_i]) (F_j-E[F_j])}=E\left[\left(\sum_{i=1}^N (x_i-\bar x) (F_i-E[F_i])\right)^2\right],$$
and
$$\sum_{i=1}^N (x_i-\bar x) (F_i-E[F_i])=\sum_{i=1}^N( x_iF_i-x_iE[F_i]) + \bar x \sum_{i=1}^N(E[F_i]-F_i)=B-E[B].$$
Where I used that $\sum_{i=1}^NE[F_i]=1$ and $\sum_{i=1}^NF_i=1$. </p>
|
1,919,912 | <blockquote>
<p>Let $D$ be the Integral Domain with characteristic $m>0$. Prove that $m$ is prime.</p>
</blockquote>
<p>My Proof: </p>
<p>Since the characteristic of $D$ is $m$, $m\cdot b=0$ for all $b\in D$ and if $n\cdot b=0$ for all $b\in D$, then $m\leq n$. </p>
<p>Assume that $m$ is composite number. Then $m=n_1n_2$ where $n_1,n_2>1$.</p>
<p>Let $a\in D$. Then $m\cdot a=0$ i.e., $n_1(n_2\cdot a)=0$. Take $c=n_2\cdot a$, then $n_1\cdot c=0$. Since $a$ is an arbitrary element of $D$, $c$ is also an arbitrary element of $D$. So, we have $n_1\cdot c=0$ for all $c\in D$ and $m> n_1$. This is a contradiction. So, our assumption that $m$ is composite number, is false. Hence $m$ is a prime number.</p>
| egreg | 62,967 | <p>If your definition of integral domain requires an identity, it's easy: the unique ring homomorphism (preserving identities)
$$
\chi\colon\mathbb{Z}\to D
$$
has kernel $m\mathbb{Z}$, so $\mathbb{Z}/m\mathbb{Z}$ is isomorphic to a subring of $D$ and therefore is a domain. Hence either $m=0$ or $m$ is prime.</p>
<p>If $D$ is not required to have an identity, you can still consider the field of quotients $Q$ of $D$, which has the same characteristic as $D$ and so the previous argument works.</p>
|
1,919,912 | <blockquote>
<p>Let $D$ be the Integral Domain with characteristic $m>0$. Prove that $m$ is prime.</p>
</blockquote>
<p>My Proof: </p>
<p>Since the characteristic of $D$ is $m$, $m\cdot b=0$ for all $b\in D$ and if $n\cdot b=0$ for all $b\in D$, then $m\leq n$. </p>
<p>Assume that $m$ is composite number. Then $m=n_1n_2$ where $n_1,n_2>1$.</p>
<p>Let $a\in D$. Then $m\cdot a=0$ i.e., $n_1(n_2\cdot a)=0$. Take $c=n_2\cdot a$, then $n_1\cdot c=0$. Since $a$ is an arbitrary element of $D$, $c$ is also an arbitrary element of $D$. So, we have $n_1\cdot c=0$ for all $c\in D$ and $m> n_1$. This is a contradiction. So, our assumption that $m$ is composite number, is false. Hence $m$ is a prime number.</p>
| QED | 91,884 | <p>As an integral domain has no zero divisors, if it has a finite characteristic it is itself finite. So it is a field and contains an unit element, $1$. Let $m=a.b$ if possible, for some integers $a,b>1$. Then $a.1=a\in D$. Let all the elements of $D$ be $x_1,\cdots,x_n$. If $a.x_i=a.x_j$ for $i\neq j$ then as $D$ has no zero divisors $x_i-x_j=0$, contradicting $i\neq j$. Hence $a.x_1,\cdots,a.x_n$ constitute all elements of $D$. But $b.(a.x_i)=0$ for all $i$. But then $b<m$ satisfies $b.x=0$ for all $x\in D$, contradicting the fact that the characteristic is $m$.</p>
|
817,386 | <p>For a function $f$ I know that: $$\int{f'(r)dr}=f(r)$$ where $f(r)$ is known. knowing the result of this integral how can i calculate $$\int{(f'(r))^2dr}$$ Is there any relation between these integrals?</p>
| Robert Bryant | 84,371 | <p>While there is no explicit formula of the exact kind you desire, if one is willing to reparametrize, there is an integral-free parametrization of the curves of the form $\bigl(x,f(x),g(x)\bigr)$ for which $g'(x) = (f'(x))^2$: As long as $f''(x)$ is non-vanishing, one can reparametrize such a curve in the form
$$
\bigl(x,f(x),g(x)\bigr)
= \bigl(u''(t),\ t\,u''(t){-}u'(t),\ t^2u''(t){-}2tu'(t){+}2u(t)\,\bigr)
$$
where $u$ is a function of $t = f'(x)$. Conversely, if $u$ is an arbitrary function of $t$ for which $u'''(t)$ is nonvanishing, the curve on the right hand side of the equation is always of the form $\bigl(x,f(x),g(x)\bigr)$ for some functions $f$ and $g$ satisfying $g'(x) = (f'(x))^2$.</p>
<p>Perhaps this is the sort of thing that one can use when one is trying to get an integral-free description of the relationship between the two functions $f$ and $g$ (and $x$, of course).</p>
|
1,800,233 | <blockquote>
<p>If $n$ is composite and $\phi{(n)} | (n - 1)$ then prove that $n$ has at least four distinct prime factors.</p>
</blockquote>
<p><strong>Attempt:</strong></p>
<p>Since $n$ is not a prime, let's first take the case that $n$ is squarefree. Then $n = a_1 \cdot a_2 \cdots a_r$ where $a_i$ are the prime factors of $n$ listed in ascending order. Thus, $\dfrac{n-1}{\phi(n)} = \dfrac{a_1a_2\cdots a_r-1}{(a_1-1)(a_2-1)\cdots (a_n-1)}$. The denominator has a factor of $2^n$.</p>
<p>I am not sure how to continue from here.</p>
| alphacapture | 334,625 | <p>This solves the 2 prime factor case, which together with user5713492's answer, solves the problem:</p>
<p>Suppose we have primes $p$ and $q$ such that $\phi(pq)=(p-1)(q-1)|pq-1$. Then we have:</p>
<p>$$pq\equiv1\pmod{(p-1)(q-1)}$$</p>
<p>so</p>
<p>$$pq\equiv1\pmod{p-1}$$</p>
<p>and also</p>
<p>$$p\equiv1\pmod{p-1}$$</p>
<p>so</p>
<p>$$pq\equiv q\pmod{p-1}$$</p>
<p>hence</p>
<p>$$q\equiv1\pmod{p-1}$$</p>
<p>and so</p>
<p>$$q\geq p.$$</p>
<p>Similarly,</p>
<p>$$p\geq q,$$</p>
<p>so $p=q$, which contradicts that it must be squarefree, as noted in the comments.</p>
|
3,310,038 | <p>If <span class="math-container">$U$</span> is an open set of <span class="math-container">$\mathbb{R}^{m}$</span>, do we have that <span class="math-container">$U\times \mathbb{R}^{n-m}$</span> is an open set of <span class="math-container">$\mathbb{R}^{n}$</span>? </p>
<p>Here <span class="math-container">$\mathbb{R}^{n},\mathbb{R}^{m}$</span> both equip with the standard topology.</p>
| Community | -1 | <p>Yes, we do. Because <span class="math-container">$U×\Bbb R^{n-m}$</span> is the product of open sets. Those sets are by definition open in <span class="math-container">$\Bbb R^n$</span>.</p>
<p>You may wish to look up <a href="https://en.m.wikipedia.org/wiki/Product_topology" rel="nofollow noreferrer">product topology </a>, or "box topology", which agrees with the product topology on finite products, or "standard topology on <span class="math-container">$\Bbb R^n$</span>".</p>
<p>You could use the projection, <span class="math-container">$\pi$</span>, onto the first <span class="math-container">$m$</span> coordinates. The standard topology is defined to be the coarsest topology which makes this map continuous. But <span class="math-container">$U$</span> is open. Thus, by definition, <span class="math-container">$U×\Bbb R^{n-m}=\pi^{-1}(U)$</span> is open. </p>
|
3,006,952 | <p><strong>Find coordinate in first quadrant which tangent line to <span class="math-container">$x^3-xy+y^3=0$</span> has slope 0</strong></p>
<p>First, I do implicit differentiation:</p>
<p><span class="math-container">$\frac{3x^2-y}{x-3y^2}=y'$</span></p>
<p>so I look at the numerator and go hmmm if i put in (1,3) that makes the slope 0.</p>
<p>But then I graph it on a software and i get the following image-</p>
<p><a href="https://i.stack.imgur.com/IVJ0D.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IVJ0D.png" alt="enter image description here"></a></p>
<p>Clearly, this is an incorrect point. I did double check that the eqn i typed in was correct and that i did the implicit differentiation right</p>
| user | 505,767 | <p>We have that</p>
<p><span class="math-container">$$x^3-xy+y^3=0\implies 3x^2dx-dx-xdy+3y^2dy=0 \implies \frac{dy}{dx}=\frac{3x^2-y}{x-3y^2}=0$$</span></p>
<p>that is</p>
<p><span class="math-container">$$3x^2=y \implies (x,y)=(t,3t^2)$$</span></p>
<p><span class="math-container">$$x^3-xy+y^3=0 \iff t^3-3t^3+27t^6=0\iff t^3(27t^3-2)=0$$</span></p>
<p>that is</p>
<p><span class="math-container">$$(x,y)=\left(\frac{\sqrt[3]2}{3},\sqrt[3]4\right)$$</span></p>
|
3,006,952 | <p><strong>Find coordinate in first quadrant which tangent line to <span class="math-container">$x^3-xy+y^3=0$</span> has slope 0</strong></p>
<p>First, I do implicit differentiation:</p>
<p><span class="math-container">$\frac{3x^2-y}{x-3y^2}=y'$</span></p>
<p>so I look at the numerator and go hmmm if i put in (1,3) that makes the slope 0.</p>
<p>But then I graph it on a software and i get the following image-</p>
<p><a href="https://i.stack.imgur.com/IVJ0D.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/IVJ0D.png" alt="enter image description here"></a></p>
<p>Clearly, this is an incorrect point. I did double check that the eqn i typed in was correct and that i did the implicit differentiation right</p>
| Andrei | 331,661 | <p>You solved only half of the problem. You have that the derivative is <span class="math-container">$0$</span>, but you also need to use the fact that the point is on the graph of your line. You have two equations with two unknowns. Since they are not linear equations, you might have multiple solutions.</p>
<p>Just plug in <span class="math-container">$y=3x^2$</span> into your original equation, and solve for <span class="math-container">$x$</span></p>
|
1,999,834 | <p>Let $\varphi : G \rightarrow H$ be a group homomorphism with kernel $K$ and let $a,b \in \varphi(G)$. Let $X = \varphi^{-1}(a)$ and $Y = \varphi^{-1}(b)$. Fix $u \in X$. Let $Z=XY$. Prove that for every $w \in Z$ that there exists $v \in Y$ such that $uv=w$. This is Dummit and Foote exercise 3.1.2.</p>
<p>My attempt:</p>
<p>Suppose we assume that $v = u^{-1}w$</p>
<p>I try to show that $v \in Y$</p>
<p>$\varphi(v) = \varphi(u^{-1})\varphi(w) = a^{-1}\varphi(w)$</p>
<p>If I could somehow show $\varphi(w) = ab$, then $\varphi(v) = b$ so that $v \in Y$ but I think I am going in circles.</p>
| Math Helper | 374,223 | <p>You are also supposing that $w\in Z$ correct? That is, $w\in XY$ so that $w=mn$ where $m\in X$ and $n\in Y$ and so $\phi(w)=\phi(mn)=\phi(m)\phi(n)=ab$.</p>
|
3,371,922 | <p>The definition of the limit states that limit of <span class="math-container">$f(x)$</span> when <span class="math-container">$x$</span> approaches <span class="math-container">$c$</span> is <span class="math-container">$L$</span> iff for every <span class="math-container">$\epsilon > 0$</span> there exists <span class="math-container">$\delta > 0$</span> such that <span class="math-container">$|f(x) - L | < \epsilon$</span> and <span class="math-container">$0 < |x - c| < δ )$</span>.</p>
<p>This states that <span class="math-container">$f(x)$</span> can reach <span class="math-container">$L ( L- ε < f(x) < L + \epsilon)$</span>, while <span class="math-container">$x$</span> cannot reach <span class="math-container">$c ( 0 < |x-c|)$</span>.</p>
<p>The informal definition says that limit means the value the function approaches as the input approaches some value. (They use the same word.) Why can one reach its corespondent value (<span class="math-container">$L$</span>) while the other can't (<span class="math-container">$x$</span> to equal <span class="math-container">$c$</span>)? Why f(x) can equal L , and x can't equal c. What is the intuitive answer to this question?</p>
| José Carlos Santos | 446,262 | <p>The idea is that if, for instance,<span class="math-container">$$f(x)=\begin{cases}1&\text{ if }x=0\\2&\text{ otherwise,}\end{cases}$$</span>then <span class="math-container">$\lim_{x\to0}f(x)=2$</span>. The fact that <span class="math-container">$f(0)=1$</span> is not relevant here. What matters is that when <span class="math-container">$x$</span> is close to <span class="math-container">$0$</span> <em>and distinct from</em> <span class="math-container">$0$</span>, then <span class="math-container">$f(x)$</span> is close to <span class="math-container">$2$</span>. And in this case it is actually equal to <span class="math-container">$2$</span>. So, if we had imposed that <span class="math-container">$0<\bigl\lvert f(x)-L\bigr\rvert$</span>, the limit would not exist.</p>
|
2,578,444 | <blockquote>
<p><span class="math-container">$\tan x> -\sqrt 3$</span></p>
</blockquote>
<p>How do I solve this inequality?</p>
<p>From the <a href="https://www.desmos.com/calculator/qb8bg1vbsf" rel="nofollow noreferrer">graph</a> it is evident that <span class="math-container">$\tan x>-\sqrt 3$</span> for <span class="math-container">$\left(\dfrac{2\pi}3 , \dfrac{5\pi} 3\right)$</span> <span class="math-container">$\forall x\in (0, 2\pi)$</span>.</p>
<p>Generalising this solution we get <span class="math-container">$\left(2n\pi +\dfrac{2\pi}3 , 2n\pi+\dfrac{5\pi} 3\right) \forall n \in \mathbb{Z}$</span> as the answer.</p>
<p>But the answer given is: <span class="math-container">$\left(n\pi - \dfrac \pi 3, n\pi + \dfrac \pi 2\right)$</span></p>
<p>Where have I gone wrong?</p>
| Roger Figueroa Quintero | 253,300 | <p>There is an error when analyzing the function at $\pi/2$, since at that point there is a discontinuity. The correct way to proceed is analyzing the graph between $-\pi/2$ and $\pi/2$, in this way you get that the solution of $\tan(x)>-\sqrt{3}$ is $(-\pi/3,\pi/2)$ and adding the period $\pi$ you get $(n\pi-\pi/3,n\pi+\pi/2)$ $\forall n \in \mathbb{Z}$. </p>
|
3,691,255 | <p>Pierre runs a game at a fair, where each player is guaranteed to win $10. </p>
<p>Players pay a certain amount each time they roll an unbiased die, and must keep rolling until a ‘6’ occurs. </p>
<p>When a ‘6’ occurs, Pierre gives the player $10 and the game concludes. </p>
<p>On average, Pierre wishes to make a profit of $2 per game. How much does he need to charge for each roll of the die?</p>
| Alexey Burdin | 233,398 | <p>The probability of not rolling <span class="math-container">$6$</span> for <span class="math-container">$k$</span> times is <span class="math-container">$\left(\frac{5}{6}\right)^k$</span>.<br>
The probability of not rolling <span class="math-container">$6$</span> for <span class="math-container">$k-1$</span> first times and then rolling <span class="math-container">$6$</span> on the <span class="math-container">$k$</span>th roll is <span class="math-container">$\frac16\left(\frac{5}{6}\right)^{k-1}$</span>.<br>
So to find the expected number of rolls we are just to find the sum
<span class="math-container">$\sum\limits_{k=1}^\infty k\frac16\left(\frac{5}{6}\right)^{k-1}$</span>.<br>
Let <span class="math-container">$S_n=\sum\limits_{k=0}^{n} (k+1)\left(\frac{5}{6}\right)^{k}$</span>,
<span class="math-container">$$\begin{align*}
\frac{5}{6}S_n
&=\sum\limits_{k=0}^{n} (k+1)\left(\frac{5}{6}\right)^{k+1}\\
%&=\frac56+\sum\limits_{k=1}^{n} (k+1)\left(\frac{5}{6}\right)^{k+1}\\
&=\sum\limits_{k=1}^{n+1} k\left(\frac{5}{6}\right)^{k}\\
&=\sum\limits_{k=0}^{n+1} k\left(\frac{5}{6}\right)^{k}\\
&=\sum\limits_{k=0}^{n+1} (k+1)\left(\frac{5}{6}\right)^{k}-
\sum\limits_{k=0}^{n+1} \left(\frac{5}{6}\right)^{k}\\
&=\sum\limits_{k=0}^{n+1} (k+1)\left(\frac{5}{6}\right)^{k}-
\frac{(5/6)^{n+2} - 1}{5/6-1}\\
&=\sum\limits_{k=0}^{n} (k+1)\left(\frac{5}{6}\right)^{k}
+(n+2)\left(\frac{5}{6}\right)^{n+1}
-\frac{(5/6)^n - 1}{5/6-1}\\
&=S_n
+(n+2)\left(\frac{5}{6}\right)^{n+1}
+6\left((5/6)^n - 1\right),
\end{align*}$$</span>
<span class="math-container">$$\begin{align*}
S_n&=-6\left((n+2)\left(\frac{5}{6}\right)^{n+1}
+6\left(\left(\frac{5}{6}\right)^{n} - 1\right)\right)
\end{align*}$$</span>
<span class="math-container">$$\begin{align*}
\lim\limits_{n\to\infty}S_n&=-6\left(0
+6\left(0 - 1\right)\right)=36
\end{align*}$$</span>
So the expected number of rolls is <span class="math-container">$\frac{1}{6}\cdot 36=6$</span> and a roll cost to have income <span class="math-container">$2$</span> per game on average is <span class="math-container">$\frac{10+2}{6}=2$</span></p>
|
661,771 | <p>I am stuck on the following problem that says : </p>
<blockquote>
<p>Which of the following is a solution to the differential equation $y'=|y|^{\frac12},y(0)=0\,$ where square root means the
positive square root ? </p>
<ol>
<li><p>$y(t)=\frac{t^2}{4}$ </p></li>
<li><p>$y(t)=-\frac{t^2}{4}$ </p></li>
<li><p>$y(t)=\frac{t|t|}{4}$ </p></li>
<li><p>$y(t)=-\frac{t|t|}{4}$ </p></li>
</ol>
</blockquote>
<p>MY ATTEMPT: Taking $y>0,$ I get from the differential equation $y'=|y|^{\frac12},y(0)=0 \implies 2\sqrt y=x $. Now, after
looking at the options ,I am not sure which of the aforementioned options is correct . Can someone help? </p>
| Community | -1 | <p>No I do not think this is correct. The idea seems correct, but the execution was poor. You should specify that $y\in X\backslash D$. I am also not sure how you justify your last inequality. If $t$ is arbitrary in $X\backslash D$ we cannot conclude $d(z,t)<r_1$ and $-d(z,t)>-r_1.$ </p>
<p>Here is how I would solve the problem:</p>
<p>Let $X$ be a metric space, $p\in X$, $r>0$. </p>
<p>Let $A=\{q\in X :d(p,q)\leq r\}$. Let $\{q_k\}\in A$ with $d(q_k,q)\rightarrow 0$. </p>
<p>We want to show $q\in A$. </p>
<p>By the triangle inequality we have $$d(q,q_k)+d(q_k,p)\ge d(q,p)$$. $\Rightarrow$ $$d(q,q_k)+r\ge d(q,p)$$ (Since $\{q_k\}\subseteq A)$</p>
<p>Now take the limit as $k\rightarrow \infty$ of both sides and we have $$0+r\ge d(p,q)$$
$$\Rightarrow q\in A$$ $\Rightarrow A$ is closed $$QED$$</p>
|
2,067,097 | <p>Given are two points on a line with coordinates. How do we calculate the third forming a perfect 60 degree triangle? So we have X,Y, but need Z...</p>
<p>X: 0,0    ( 0,0 i.e. horizontal, vertical )<br>
Y: 50, 0 <br>
Z: 25, ??</p>
<p>How to calculate the missing horizontal coordinate for Z? Forming a perfect 60 degree triangle?</p>
<p>EDIT:
I'm trying to figure out a formula that by providing only 50 as the length of one side of the triangle all the point coordinates can be figured out. The first 2 and half of them are easy (0,0 50, 0, ??, 25)... but how to calculate the number at '??' is what I'm trying to figure out.</p>
<p>EDIT2 (@Dennis):
It would probably help if I explained that my coordinates are not of the actual point but a 100px wide circle that the point is in the middle of. I made a small video to show you guys exactly what I'm doing and will post is just as soon as it's done uploading.</p>
<p>EDIT3:
Here is a video showing exactly what I mean and why I came up with the question in the first place: <a href="http://archebian.org/videos/math/triangle-question.mp4" rel="nofollow noreferrer">http://archebian.org/videos/math/triangle-question.mp4</a></p>
| Zin | 400,692 | <h2>With the video explanation</h2>
<p>You are asking for the intersection of two circles (this is <strong>not</strong> creating an <em>equilateral</em> triangle). You can take the equations of your circles $c_1: x^2+y^2=50^2$ and $c_2: (x-50)^2+y^2=50^2$.</p>
<p>$$x = \frac{d^2-r^2+R^2}{2d}$$
For your question the distance between the center points is $d=50$ and $r=R=50$. And surprisingly you'll find it at 25. This you can put into the circle equation again and find $$y=\sqrt{50^2 - 25^2}\approx 43,30127$$ Just as in your picture.</p>
<p>You find the solution here: <a href="http://mathworld.wolfram.com/Circle-CircleIntersection.html" rel="nofollow noreferrer">http://mathworld.wolfram.com/Circle-CircleIntersection.html</a>.</p>
<h2>For the equilateral part</h2>
<p>What about the height of an equilateral triangle being $\frac{\sqrt{3}}{2}a$ where $a$ is the length of your side. With this you should see that 25 isn't suitable for your chosen height of 25.</p>
<p>e.g. (0,0); (50,0); (25,21.65) is close to perfect. In your post I see the 25 in the y coordinate, which doesn't make sense to me.</p>
|
427,564 | <p>I'm supposed to give a 30 minutes math lecture tomorrow at my 3-grade daughter's class. Can you give me some ideas of mathemathical puzzles, riddles, facts etc. that would interest kids at this age?</p>
<p>I'll go first - Gauss' formula for the sum of an arithmetic sequence.</p>
| AD - Stop Putin - | 1,154 | <p>Let them try to create maps that need as few colours as possible. The rule is that two counties are neighbours if their borders meet in more than a finite number of points and neighbours should not have the same colour. Hopefully, this might lead to an interesting discussion about the 4-colour theorem. </p>
<p><a href="https://www.google.se/search?q=four%20colour%20theorem&client=firefox-a&hs=6Ln&rls=org.mozilla%3asv-SE%3aofficial&tbm=isch&tbo=u&source=univ&sa=X&ei=Dg_HUersAbDc4QTf6IHYCg&ved=0CDoQsAQ&biw=1252&bih=557">Here are example "maps".</a></p>
<p><a href="http://en.wikipedia.org/wiki/Four_color_theorem">Here is the Wikipedia page.</a> </p>
|
427,564 | <p>I'm supposed to give a 30 minutes math lecture tomorrow at my 3-grade daughter's class. Can you give me some ideas of mathemathical puzzles, riddles, facts etc. that would interest kids at this age?</p>
<p>I'll go first - Gauss' formula for the sum of an arithmetic sequence.</p>
| ZeerakW | 83,653 | <p>I think Map Colouring and Probability are a great ideas, as mentioned above.</p>
<p>I would also suggest finding the highest prime, it requires multiplication/division, which is always good to practice.</p>
<p>Otherwise you could consider having them enact Bubble Sort[1], based on their height or birthday.</p>
<p>[1] <a href="https://en.wikipedia.org/wiki/Bubble_sort" rel="nofollow">https://en.wikipedia.org/wiki/Bubble_sort</a></p>
|
427,564 | <p>I'm supposed to give a 30 minutes math lecture tomorrow at my 3-grade daughter's class. Can you give me some ideas of mathemathical puzzles, riddles, facts etc. that would interest kids at this age?</p>
<p>I'll go first - Gauss' formula for the sum of an arithmetic sequence.</p>
| Yuri | 83,680 | <p>I think most posters overestimate what you can learn a bunch of nine year olds in 30 minutes. Remember you are not 1-on-1 with the brightest student, but you need to have the full class being able to follow you. They maybe have had division and such in class but I imagine it isn't their second nature yet, so you can't depend on it for such a short lecture.</p>
<p>Personally I would try to learn them to count in base-9 and conversions to and from base 10.</p>
|
2,196,936 | <p>How to prove that $7p + 3^p -4$ is not a perfect square? </p>
<p>I calculated: $\left(\frac{7p+3^p-4}{p}\right) = \left(\frac{-1}{p}\right)$. So if $p \equiv 3 \mod 4$, the result is $-1$. So in that case, $7p+3^p -4$ can't be a square. But what about the case $p \equiv 1 \mod 4$? Any hints? Thanks in advance.</p>
| Alex R. | 22,064 | <p>If $p=4k+1$, then:</p>
<p>$$7p+3^p-4\equiv -1+(-1)\pmod{4}=-2\pmod{4}.$$</p>
<p>But perfect squares must be 0,1 mod 4. </p>
|
3,779,589 | <p>Let the metric <span class="math-container">$d$</span> be defined as
<span class="math-container">$$
d(f,g) =\sup_{x\in[0,1]}|f(x)-g(x)|,
$$</span>
and let<br />
<span class="math-container">$$
H(x) = \begin{cases} 0 \text{ if } x \leq \frac{1}{2}\\ 1 \text { if } x > \frac{1}{2} \end{cases}.
$$</span>
Is <span class="math-container">$f(x) = x$</span> in <span class="math-container">$B_\frac{1}{2}(H)$</span> ?</p>
<p><strong>My answer</strong>. No, because
<span class="math-container">$$
d(H(x),f(x)) = \sup_{x\in[0,1]}|f(x)-H(x)| = \frac{1}{2}.
$$</span>
Therefore, <span class="math-container">$f(x) \not \in B_\frac{1}{2}(H)$</span></p>
<p>I am not sure if my answer is correct, and I found that it is hard to visualize this metric. Can someone helps me on this?</p>
| PrincessEev | 597,568 | <p>I wouldn't call this approach "by contradiction;" rather, if anything, it is typically referred to as the complementary approach. Want to find a quantity satisfying a condition? The complementary approach is to find the overall quantity disregarding such a condition, and then subtract off the ones <em>not</em> satisfying that condition.</p>
<p>Your reasoning is also okay, but it could be a bit more succinct. Here's how I would do it ...</p>
<hr />
<p>Define</p>
<p><span class="math-container">$$A_n := \{ k \in \Bbb Z \mid k \in [1,200] \text{ and } n \text{ divides } k \}$$</span></p>
<p>Thus, for instance, <span class="math-container">$A_2 = \{2,4,6,\cdots,200\}$</span>.</p>
<p>In your problem, you want to find those integers <span class="math-container">$n \in [1,200]$</span> where neither <span class="math-container">$2$</span> nor <span class="math-container">$5$</span> divide <span class="math-container">$n$</span>. The complementary approach, then, is to note that you have <span class="math-container">$200$</span> possible numbers, and then subtract off the invalid ones. Thus, you would subtract off <span class="math-container">$|A_2|$</span> and <span class="math-container">$|A_5|$</span>. But beware! This "double subtracts" the members of <span class="math-container">$|A_{10}|$</span> since all multiples of ten are in both sets. (This is a particular instance of the principle of inclusion and exclusion.) Thus you have to add back <span class="math-container">$|A_{10}|$</span>.</p>
<p>Thus, the quantity you seek is</p>
<p><span class="math-container">$$200 - |A_2| - |A_5| + |A_{10}|$$</span></p>
<p>The benefit of this framing is that, if <span class="math-container">$n$</span> divides <span class="math-container">$200$</span>, then <span class="math-container">$|A_n| = 200/n$</span>, making this calculation satisfyingly easily. We see that it is, for you,</p>
<p><span class="math-container">$$200 - 100 - 40 + 20 = 80$$</span></p>
<p>as you proposed!</p>
<hr />
<p>Your method is not invalid, but I feel like this framing of the problem makes everything a lot easier to follow, personally. For instance, you don't have to go through every series of inequalities; the definition of the <span class="math-container">$A_n$</span> makes those results almost obvious, and generalizes all three cases you check. But on the other hand, if you prefer your method and framing, feel free to use it - probably a matter of preference in the end.</p>
|
3,041,907 | <p>I am unable to isolate the variable <span class="math-container">$x$</span> of this inequality <span class="math-container">$y \leq \sqrt{2x-x^2}$</span> ( where <span class="math-container">$0 \leq y \leq 1 $</span>)</p>
<p>Is it correct doing this: <span class="math-container">$y^2 \leq 2x-x^2$</span>?
I found that <span class="math-container">$y^2 \leq x \leq 2-y^2$</span> and <span class="math-container">$0 \leq x \leq 2$</span>. Is it correct?</p>
<p>From here I am not sure how to proceed.
Thanks in advance for any help.</p>
| hamam_Abdallah | 369,188 | <p>Let <span class="math-container">$f(x)=\sqrt{2x-x^2}$</span></p>
<p>its domain is <span class="math-container">$[0,2]$</span>.</p>
<p><span class="math-container">$$f'(x)=\frac{1-x}{f(x)}$$</span></p>
<p>the maximum is <span class="math-container">$f(1)=1$</span>.</p>
<p>Let <span class="math-container">$y\in[0,1]$</span></p>
<p>we look for <span class="math-container">$x$</span> such that
<span class="math-container">$$x^2-2x+y^2=0$$</span></p>
<p>which gives two solutions, one in <span class="math-container">$[0,1]$</span> and the other in <span class="math-container">$[1,2]$</span> :
<span class="math-container">$$x=1\pm\sqrt{1-y^2}$$</span></p>
|
3,041,907 | <p>I am unable to isolate the variable <span class="math-container">$x$</span> of this inequality <span class="math-container">$y \leq \sqrt{2x-x^2}$</span> ( where <span class="math-container">$0 \leq y \leq 1 $</span>)</p>
<p>Is it correct doing this: <span class="math-container">$y^2 \leq 2x-x^2$</span>?
I found that <span class="math-container">$y^2 \leq x \leq 2-y^2$</span> and <span class="math-container">$0 \leq x \leq 2$</span>. Is it correct?</p>
<p>From here I am not sure how to proceed.
Thanks in advance for any help.</p>
| egreg | 62,967 | <p>Since you're assuming <span class="math-container">$y\ge0$</span>, the given inequality is equivalent to <span class="math-container">$y^2\le2x-x^2$</span>, that is, <span class="math-container">$x^2-2x+y^2\le0$</span>.</p>
<p>The discriminant of <span class="math-container">$x^2-2x+y^2$</span> is <span class="math-container">$4(1-y^2)$</span>. If <span class="math-container">$y>1$</span>, the inequality <span class="math-container">$x^2-2x+y^2\le0$</span> is satisfied for no <span class="math-container">$x$</span>.</p>
<p>If <span class="math-container">$y\le1$</span>, the inequality is satisfied for <span class="math-container">$1-\sqrt{1-y^2}\le x\le 1+\sqrt{1-y^2}$</span>. However we still need <span class="math-container">$x\ge y$</span>; note that <span class="math-container">$1-\sqrt{1-y^2}\ge y$</span> (prove it). Thus the solutions are
<span class="math-container">$$
y\le x\le 1+\sqrt{1-y^2}\qquad (\text{for $y\le1$})
$$</span></p>
<p>The following graph confirms this.</p>
<p><a href="https://i.stack.imgur.com/UtFuC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UtFuC.png" alt="enter image description here"></a></p>
|
240,461 | <p>What's the mathematica command to get the <strong>numerical value</strong> of :</p>
<p><span class="math-container">$$PV\int_0^\infty \frac{\tan x}{x}\text{d}x?$$</span></p>
<p>where <span class="math-container">$PV$</span> is the principal value.</p>
| Michael E2 | 4,999 | <p>The method <code>"PrincipalValue"</code> does not work on an integral with infinitely many poles, since each pole must be specified. Confining to a finite region of integration is problematic on integrals like <span class="math-container">$\int_0^\infty (\tan x/x)\,dx$</span> that converge slowly. <code>NIntegrate</code> has built in the <code>"ExtrapolatingOscillatory"</code> strategy, but even if it would work together with the <code>"PrincipalValue"</code> strategy, you would still have the problem of specifying infinitely many poles. Instead, we can manually call <code>NSum</code> on the principal value integral over each period of <code>Tan[x]</code>. <code>NSum</code> will do the extrapolation for us. (Something similar was done <a href="https://mathematica.stackexchange.com/a/147959/4999">here</a> and is shown <a href="https://reference.wolfram.com/language/tutorial/NIntegrateIntegrationStrategies.html#50593814" rel="nofollow noreferrer">in the documenation</a>.) The series still converges slowly, and the results are not as good as my other answer.</p>
<pre><code>ClearAll[cc];
cc[n_Integer] := cc[n] =
NIntegrate[Tan[x]/x, {x, 0 + n Pi, Pi/2 + n Pi, Pi + n Pi},
WorkingPrecision -> 50,
Method -> {"PrincipalValue", Method -> "GaussKronrodRule"}];
NSum[cc[n], {n, 0, Infinity}, NSumTerms -> 50,
WorkingPrecision -> 50] // AbsoluteTiming
(* {1.76767, 1.57067} *)
</code></pre>
<hr />
<p><strong>Alternative</strong></p>
<p>We can speed up the principal value integral by subtracting a null integral that smashes the pole (something similar was done <a href="https://mathematica.stackexchange.com/a/214701/4999">here</a>). The slow convergence persists, but the following is twice as fast as with the <code>"PrincipalValue"</code> method.</p>
<pre><code>ClearAll[cc];
cc[n_Integer] := cc[n] = NIntegrate[
Tan[x]/x - Tan[x]/(Pi/2 + n Pi),
{x, 0 + n Pi, Pi + n Pi},
WorkingPrecision -> 50, Method -> "GaussKronrodRule"];
NSum[cc[n], {n, 0, Infinity}, NSumTerms -> 1000,
WorkingPrecision -> 50] // AbsoluteTiming
(* {10.2142, 1.57078947} *)
Last@% - Pi/2 (* error *)
(* -6.85*10^-6 *)
</code></pre>
|
1,319,476 | <p>This is a question related to another posted question:</p>
<p>The answer to the following question "Find all solutions to: $e^{ix}=i$" is as follows: </p>
<p>"Euler's formula: $e^{ix}=\cos(x)+i\sin(x)$,</p>
<p>so: $ \cos x+i\sin x=0+1⋅i$</p>
<p>compare real and imaginary parts
$\sin(x)=1$
and
$\cos(x)=0$</p>
<p>$x=\frac{(4n+1)π}2$, $n∈$
(W stands for set of whole number W={0,1,2,3,.......,n})."</p>
<p>My question: Where does $x=\frac{(4n+1)π}2$, $n∈$ come from? </p>
<p>My steps: </p>
<ol>
<li><p>$\cos(x) + i\sin(x) = 0 + i(1)$</p></li>
<li><p>$\cos(x) = i(1 - \sin(x))$</p></li>
<li><p>... </p></li>
<li><p>how does $x=\frac{(4n+1)π}2$ follow? </p></li>
</ol>
| Lucian | 93,448 | <p><strong>Hint:</strong></p>
<ul>
<li><p>$i$ is <strong>a point on the unit circle</strong>.</p></li>
<li><p>$e^{ix}$ is <em>also</em> <strong>a point on the unit circle</strong>, lying <em>x</em> radians away from $(1,0)$, in trigonometric or counterclockwise direction.</p></li>
</ul>
<p>So, to answer a question with a question, What is the position of $i$ on the unit circle, and How many radians away from $(1,0)$ does it lie, in trigonometric or counterclockwise direction ?</p>
|
2,507,613 | <p>I am trying to teach myself group theory and I recently came across the topic of Isomorphisms.
I know that 2 groups are isomorphic if there is a one-on-one correspondence between their elements.
So if the groups have a different order, does that mean they are not isomorphic? Such as a group $S$ and its permutation group $S_{n}$ for $n>1$.</p>
| José Carlos Santos | 446,262 | <p>The answer is “yes”, but your definition of isomorphism is not correct. An isomorphism between groups is a bijection <span class="math-container">$\varphi$</span> which preserves the product, that is, such that <span class="math-container">$\varphi(x.y)=\varphi(x).\varphi(y)$</span> for every <span class="math-container">$x$</span> and every <span class="math-container">$y$</span> in the domain. But since it must be a bijection then, yes, groups with distinct orders cannot be isomorphic.</p>
|
2,445,655 | <p>Challenge: A Good Deal</p>
<p>You are currently learning some important aspects of collusion and cartels. This challenge puts you in the position of a bad guy, namely a price-fixing sales manager. Suppose that you find yourself in a so-called “smoke-filled room” to fix prices for the upcoming year with the sales manager of a competing firm, Snitch Inc.. Suddenly the door opens and one of the employees of Snitch Inc. enters the room. The employee knows that price-fixing is illegal and immediately grabs his cellphone to inform the competition authority. You are fully aware that you are now facing a serious risk of getting a fine or even a jail sentence. Whereas your fellow sales manager is simply in shock, not knowing what to do, you as a University School of Business and Economics alumni are quick to
react and you try to save the situation. Your idea is to bribe the employee. You expect that the employee requires a bribe of at least €100 to remain silent. In other words, the reservation price of the employee is €100. For simplicity, assume in the following that this expectation is correct. At the same time, suppose that you are not willing to offer more than €200 for otherwise you would rather save your money and spend it on a good lawyer instead. In other words, your reservation price is €200. You consider your chances and are thinking about making an offer. For simplicity, assume in the following that all parties aim to maximize the gains from trade and that offers can be any positive real number (all values weakly above zero, that is).</p>
<p>Try to provide a clear and concise answer to the following four questions.
For the first two questions suppose that the employee is still slightly in shock and therefore can only respond by either accepting or rejecting your offer.</p>
<ol>
<li>How many Nash equilibria does this game have, if any? Explain your answer.</li>
<li>How many subgame perfect Nash equilibria does this game have, if any? Explain your answer.</li>
</ol>
<p>Now suppose that you are dealing with an, somewhat cocky, employee who is brave
enough to start negotiating with you. That is, in questions 3 and 4 below, the
employee, instead of simply accepting or rejecting the offer, may now make a
counteroffer instead. For simplicity, assume that both parties get a zero payoff in case of “no deal”.</p>
<ol start="3">
<li>Suppose that the employee indeed does make a counteroffer and that you will either accept or reject (in other words, the game ends after your response to the counteroffer). What is the subgame perfect Nash equilibrium in this case? </li>
</ol>
<p>You fear that these negotiations may take quite some time and as a business man you know that time is money. Suppose that your impatience as well as that of the
employee is given by a common discount factor 0 < δ < 1 (which is known to you and the employee). The interpretation is that the utility of a money amount K “tomorrow” is equal to the utility of an amount of δK “today”. Your goal is to make an acceptable offer the first round while at the same time saving as much money as possible.</p>
<ol start="4">
<li>What is the subgame perfect Nash equilibrium outcome in this case? Do you (i.e., the sales manager) or the employee benefit from your impatience? Explain your answer.</li>
</ol>
| sharpe | 217,520 | <p>Define $t:=-x$. Then,
\begin{align*}
&\lim_{x \to -\infty}2x+\sqrt{4x^2+x}=\lim_{t \to \infty} \left( -2t+\sqrt{4t^2-t} \right)\\
&=\lim_{t \to \infty} \left( -2t+\sqrt{4t^2-t} \right)\cdot\frac{2t+\sqrt{4t^2-t}}{2t+\sqrt{4t^2-t}} =\lim_{t \to \infty}\frac{-t}{2t+\sqrt{4t^2-t}} \\
&=\lim_{t \to \infty}\frac{-1}{2+\frac{\sqrt{4t^2-t}}{t}}=\lim_{t \to \infty}\frac{-1}{2+\sqrt{\frac{4t^2-t}{t^2}}}=\lim_{t \to \infty}\frac{-1}{2+\sqrt{\frac{4t^2-t}{t^2}}}\\
&=\lim_{t \to \infty}\frac{-1}{2+\sqrt{4-(1/t)}}=\frac{-1}{2+\sqrt{4}}=-1/4.
\end{align*}</p>
|
2,445,655 | <p>Challenge: A Good Deal</p>
<p>You are currently learning some important aspects of collusion and cartels. This challenge puts you in the position of a bad guy, namely a price-fixing sales manager. Suppose that you find yourself in a so-called “smoke-filled room” to fix prices for the upcoming year with the sales manager of a competing firm, Snitch Inc.. Suddenly the door opens and one of the employees of Snitch Inc. enters the room. The employee knows that price-fixing is illegal and immediately grabs his cellphone to inform the competition authority. You are fully aware that you are now facing a serious risk of getting a fine or even a jail sentence. Whereas your fellow sales manager is simply in shock, not knowing what to do, you as a University School of Business and Economics alumni are quick to
react and you try to save the situation. Your idea is to bribe the employee. You expect that the employee requires a bribe of at least €100 to remain silent. In other words, the reservation price of the employee is €100. For simplicity, assume in the following that this expectation is correct. At the same time, suppose that you are not willing to offer more than €200 for otherwise you would rather save your money and spend it on a good lawyer instead. In other words, your reservation price is €200. You consider your chances and are thinking about making an offer. For simplicity, assume in the following that all parties aim to maximize the gains from trade and that offers can be any positive real number (all values weakly above zero, that is).</p>
<p>Try to provide a clear and concise answer to the following four questions.
For the first two questions suppose that the employee is still slightly in shock and therefore can only respond by either accepting or rejecting your offer.</p>
<ol>
<li>How many Nash equilibria does this game have, if any? Explain your answer.</li>
<li>How many subgame perfect Nash equilibria does this game have, if any? Explain your answer.</li>
</ol>
<p>Now suppose that you are dealing with an, somewhat cocky, employee who is brave
enough to start negotiating with you. That is, in questions 3 and 4 below, the
employee, instead of simply accepting or rejecting the offer, may now make a
counteroffer instead. For simplicity, assume that both parties get a zero payoff in case of “no deal”.</p>
<ol start="3">
<li>Suppose that the employee indeed does make a counteroffer and that you will either accept or reject (in other words, the game ends after your response to the counteroffer). What is the subgame perfect Nash equilibrium in this case? </li>
</ol>
<p>You fear that these negotiations may take quite some time and as a business man you know that time is money. Suppose that your impatience as well as that of the
employee is given by a common discount factor 0 < δ < 1 (which is known to you and the employee). The interpretation is that the utility of a money amount K “tomorrow” is equal to the utility of an amount of δK “today”. Your goal is to make an acceptable offer the first round while at the same time saving as much money as possible.</p>
<ol start="4">
<li>What is the subgame perfect Nash equilibrium outcome in this case? Do you (i.e., the sales manager) or the employee benefit from your impatience? Explain your answer.</li>
</ol>
| farruhota | 425,072 | <p>Note that for $x\to -\infty$, $4x^2+x=(2x+\frac14)^2-\frac{1}{16} \sim (2x+\frac14)^2.$ Hence:
$$\lim_{x\to-\infty} \left(2x+\sqrt{4x^2+x}\right)=\lim_{x\to-\infty}\left(2x-\left(2x+\frac14\right)\right)=-\frac14.$$</p>
|
4,135,472 | <p><strong>What is the clearest and simplest way of proving that <span class="math-container">$[x]+[x+1/2]=[2x]$</span>? (Where <span class="math-container">$[x]$</span> is the greatest integer function)</strong></p>
<p>According to Bartleby, if <span class="math-container">$x=m$</span> for <span class="math-container">$m\in \mathbb{Z}$</span>, then <span class="math-container">$[x]=m$</span> and <span class="math-container">$[x+1/2]=m$</span>. Hence, <span class="math-container">$[x]+[x+1/2]=2m=[2x]$</span>. Hence when <span class="math-container">$m\in\mathbb{Z}$</span>, <span class="math-container">$[x]+[x+1/2]=[2x]$</span>.</p>
<p>When <span class="math-container">$m<x<m+1$</span>, <span class="math-container">$[x]=[m+\{x\}]=m+[\{x\}]$</span> and <span class="math-container">$[x+1/2]=[m+\{x\}+1/2]=m+[\{x\}+1/2]$</span>. Hence <span class="math-container">$[x]+[x+1/2]=2m+[\{x\}]+[\{x\}+1/2]$</span>. Since <span class="math-container">$0 \le \{x\} < 1$</span> and <span class="math-container">$1/2 \le \{x\}+1/2 < 3/2$</span>; we have <span class="math-container">$0 \le [\{x\}] < 1$</span> and <span class="math-container">$0\le[\{x\}+1/2]< 1$</span>. Hence <span class="math-container">$2m < 2m+[\{x\}]+[\{x\}+1/2]\le 2m+1$</span> which implies <span class="math-container">$[2x]$</span>. Hence <span class="math-container">$[x]+[x+1/2]=[2x]$</span>.</p>
<p>Hence <span class="math-container">$[x]+[x+1/2]=[2x]$</span></p>
<p>Is my solution clear or is there a better one?</p>
| WhatsUp | 256,378 | <p>For <span class="math-container">$x > 0$</span>:</p>
<p><span class="math-container">$[2x]$</span> is the number of positive integers <span class="math-container">$\leq 2x$</span>. Count them by separating the even and odd numbers.</p>
<p>For general <span class="math-container">$x$</span>: add a sufficiently large integer to it.</p>
|
3,145,896 | <h1>Solve for <span class="math-container">$x$</span></h1>
<p>I have an equation that I have been working on solving; I know the solution, but I cannot get to it myself. Almost every simplification I do reverts back to a previous step. Can anyone show me how to solve for <span class="math-container">$x$</span> in this equation?</p>
<h3>Equation:</h3>
<p><span class="math-container">$$\log_6(2x-3)+\log_6(x+5)=\log_3x$$</span></p>
<h3>Solution:</h3>
<p><span class="math-container">$$x ≅ \frac{3347}{2000} ≅ 1.6735$$</span>
<br>
<strong>Note:</strong> upon further analysis of the answer, while close, it does not seem to be the <em>exact</em> solution.</p>
<hr>
<h3>What I Have Tried So Far</h3>
<p><span class="math-container">$$\log_6(2x-3) + \log_6(x + 5) = \log_3x$$</span>
<span class="math-container">$$\frac{\log(2x-3)}{\log6} + \frac{\log(x + 5)}{\log6} = \frac{\log x}{\log3}$$</span>
<span class="math-container">$$\log3 \cdot \log(2x-3) + \log3 \cdot \log(x + 5) = \log6 \cdot \log x$$</span>
<span class="math-container">$$\log3 \cdot \log \left[(2x - 3)(x + 5)\right] = \log6 \cdot \log x$$</span>
<span class="math-container">$$\frac{\log \left[(2x - 3)(x + 5)\right]}{\log_3 10} = \frac{\log6}{\log_x10}$$</span>
<span class="math-container">$$\log_x10 \cdot \log \left[(2x - 3)(x + 5)\right] = \log_3 10 \cdot \log6$$</span>
<span class="math-container">$$\log_x \left[(2x - 3)(x + 5)\right] = \log_3 6$$</span>
<span class="math-container">$$\log_x3 \cdot \log_x \left[(2x - 3)(x + 5)\right] = \frac{\log_3 6}{\log_3 x}$$</span>
<span class="math-container">$$\log_x \left[(2x - 3)(x + 5)\right]^{\ \log_x3} = \log_x 6$$</span>
<span class="math-container">$$\left[(2x - 3)(x + 5)\right]^{\ \log_x3} = 6$$</span>
<span class="math-container">$$(2x - 3)(x + 5) = x^{\log_3 6}$$</span></p>
<p><br>
I know these steps aren't really working towards the solution at points; I was sort of just playing around with the equation. Regardless, I really don't know how to go about moving forward from here.</p>
<hr>
| Community | -1 | <p>Taking the base-<span class="math-container">$6$</span> antilogarithm,</p>
<p><span class="math-container">$$(2x-3)(x+5)=6^{\log_3x}=e^{\ln6\ln x/\ln3}=x^{\ln6/\ln3}.$$</span></p>
<p>Because of the irrational exponent, there is no closed-form solution and you need to use a numerical method.</p>
|
3,242,553 | <p>I got two sequences of stochastic process <span class="math-container">$(X_n(t))_{t \in [0,1]}$</span> and <span class="math-container">$(Y_{n}(t))_{t \in [0,1]}$</span>, defined on a probability space <span class="math-container">$(\Omega, \mathcal{F},P)$</span>, and know that their distance in the sup-norm on <span class="math-container">$[0,1]$</span> converges to <span class="math-container">$0$</span> almost surely, i.e.</p>
<p><span class="math-container">$\sup \limits_{t \in [0,1]} \vert X_n(t) - Y_n(t) \vert \to 0 \quad P-a.s., \quad n \to \infty$</span>, </p>
<p>or equivalently</p>
<p><span class="math-container">$P \left (\lim \limits_{n \to \infty} \sup \limits_{t \in [0,1]} \vert X_n(t) - Y_n(t) \vert = 0 \right ) = 1$</span>.</p>
<p>Now I'm wondering if this also implies the (pointwise) convergence in distribution of <span class="math-container">$X_n$</span> to <span class="math-container">$Y_n$</span>. This result seems very intuitive, but how does one formally show this?</p>
<p>Thanks!</p>
| kimchi lover | 457,779 | <p>What you want is <a href="https://en.wikipedia.org/wiki/Slutsky%27s_theorem" rel="nofollow noreferrer">Slutsky's theorem</a>. If, for some <span class="math-container">$t$</span>, the sequence <span class="math-container">$X_n(t)$</span> converges in distribution, and if <span class="math-container">$Y_n(t)-X_n(t)$</span> converges to <span class="math-container">$0$</span> in probability, then <span class="math-container">$Y_n(t)$</span> converges to the same limit law as <span class="math-container">$X_n(t)$</span>. In your case you have almost sure convergence of <span class="math-container">$Y_n(t)-X_n(t)$</span> to <span class="math-container">$0$</span>, which is stronger than what you need.</p>
|
3,917,255 | <p>Why does Chebyshev's inequality demand that <span class="math-container">$\mathbb{E(}X^2) < \infty$</span>?</p>
| Emmanuel C. | 619,398 | <p>Indeed, <span class="math-container">$f$</span> is a closed map. Take a closed subset <span class="math-container">$F\subseteq \mathbb S^1\times\mathbb S^1$</span>. Since <span class="math-container">$\mathbb S^1\times \mathbb S^1$</span> is compact, it follows that <span class="math-container">$F$</span> is compact. Then, by continuity, <span class="math-container">$f(F)$</span> is compact. Finally, since <span class="math-container">$f(F)$</span> is compact in a Hausdorff space (because <span class="math-container">$\mathbb S^1\times\mathbb S^1$</span> is a Hausdorff space), <span class="math-container">$f(F)$</span> is closed.</p>
|
2,903,163 | <p>I don't really know whether to put this in Physics forums since it is relating to Mechanics, or Math since the question is actually about the math being done. Don't criticize me over it.</p>
<p>So for the question: I was doing some review problems on Lagrange's equations, KE+PE, and I found <a href="http://wwwf.imperial.ac.uk/%7Epavl/ASHEET2.PDF" rel="nofollow noreferrer">this document</a>.</p>
<p>In the first question's solution, the writer differentiates without explaining the step. They have these:</p>
<p><span class="math-container">$$\begin{cases}
x = r \sin(\theta) \cos(\phi)\\[5 pt]
y = r \sin(\theta) \sin(\phi)\\[5 pt]
z = r \cos(\theta)
\end{cases}
$$</span></p>
<p>and this:</p>
<p><span class="math-container">$$T = {m\over 2}(\dot x^2 +\dot y^2 +\dot z^2)$$</span></p>
<p>I never really studied the spherical coordinate system much, and obviously never thought about the derivatives of the conversion into Cartesian. Can someone find or explain the process of taking the derivatives of the first three equations, plugging into the equation for Kinetic Energy, and simplifying? There is a probably a different calculus method for the coordinate system, which I don't know. Thanks!</p>
<p>EDIT: While doing taking the derivatives, was the method used actually a separate form of calculus beyond I and II, or was it normal first-order differentiation? If so, how? Here is the part I am speaking of:</p>
<blockquote>
<p><strong>Solution:</strong> The kinetic energy is <span class="math-container">$T=\frac m2(\dot x^2+\dot y^2+\dot z^2)$</span>. We substitute
<span class="math-container">$$\begin{cases}
x = r \sin(\theta) \cos(\phi)\\[5 pt]
y = r \sin(\theta) \sin(\phi)\\[5 pt]
z = r \cos(\theta)
\end{cases}
$$</span>
Differentiating these, substituting into <span class="math-container">$T$</span>, and simplifying, we find
<span class="math-container">$$T=\frac m2 (\dot r^2 +r^2\dot\theta^2+r^2\sin^2\theta\dot\phi^2).$$</span></p>
</blockquote>
| Robert Lewis | 67,071 | <p>To convert the cartesian expression for kinetic energy,</p>
<p>$T = \dfrac{m}{2}(\dot x^2 + \dot y^2 + \dot z^2) \tag 1$</p>
<p>into sperical coordinates $r,\phi, \theta$ such that</p>
<p>$x = r \sin \theta \cos \phi, \tag 2$</p>
<p>$y = r\sin \theta \sin \phi, \tag 3$</p>
<p>$z = r\cos \theta, \tag{4}$</p>
<p>we merely need employ two standard results from elementary calculus, namely, the Leibniz product rule and the chain rule; the calculations are all in the realm of basic first-order differentiation using these two principles. I will start by illustrating how these concepts apply to $z$ (4), since it is the simplest of the three expressions (2)-(4); from (4), by the product rule, where I use $\dot{}$ and ${}´$ both to represent the $t$-derivative,</p>
<p>$\dot z = \dot r \cos \theta + r (\cos \theta)'; \tag 5$</p>
<p>we apply the chain rule to (5):</p>
<p>$(\cos \theta)' = \left (\dfrac{d\cos \theta}{d\theta} \right ) \dot \theta = -\dot \theta \sin \theta; \tag 6$</p>
<p>thus (5) becomes</p>
<p>$\dot z = \dot r \cos \theta - r \dot \theta \sin \theta; \tag 7$</p>
<p>we similarly handle $x$ as in (2): again, the product rule yields</p>
<p>$\dot x = \dot r \sin \theta \cos \phi + r(\sin \theta)'\cos \phi + r\sin \theta (\cos \phi)', \tag 8$</p>
<p>and again we apply the chain rule, this time twice:</p>
<p>$(\sin \theta)' = \dfrac{d\sin \theta}{d\theta} \dot \theta = \dot \theta \cos \theta, \tag 9$</p>
<p>$(\cos \phi)' = \dfrac{d\cos \phi}{d \phi} \dot \phi = -\dot \phi \sin \phi; \tag{10}$</p>
<p>assembling (8)-(10) together:</p>
<p>$\dot x = \dot r \sin \theta \cos \phi + r\dot \theta \cos \theta \sin \phi - r\dot \phi \sin \theta \sin \phi; \tag{11}$</p>
<p>y a parellel procedure, first using the Leibniz and the chain rule, we also have</p>
<p>$\dot y = \dot r \sin \theta \sin \phi + r \dot \theta \cos \theta \sin \phi + r \dot \phi \sin \theta \cos \phi; \tag{12}$</p>
<p>with (7), (11)-(12) at hand, calculating $\dot x^2 + \dot y^2 + \dot z^2$ in sphericals involves no more than a good slug o' algebra; but there is really nothing to see in it that hasn't been very nicely and more than adequately presented by our colleagues David Morgante and
Ahmed S. Ataalla.</p>
<p>So I think I'll leave off now. My main point and interest here has been to point out how the Leibniz and chain rules, both results of basic calculus, are used to effect the transformation of the velocities, which then leads to the expression for $T$ in spherical coordinates, as others have shown.</p>
|
477,477 | <p>Prove that $e^x=-x^2+2x+5 $ have exactly two solutions.</p>
<p>Is it enoguht that Vertex of the parabola is over $y=e^x$ and arms of it looks down</p>
| Ross Millikan | 1,827 | <p>The basic idea is that quadrilaterals average $90^\circ$ angles. If four of them meet at every corner, you wont have the required $720^\circ$ deficit to make a closed sphere. You need eight three-way vertices to get it to close. This is like the requirement that using hexagons and pentagons you need 12 pentagons for a full sphere. As one approximation, you can take a spherical cap and choose four points on the bottom circle to identify as "vertices" of your "quadrilateral", but maybe this is too trivial. You can also inscribe a cube in a sphere and project the edges outward to make a tessellation. The edges will not be geodesics. You can then subdivide the squares with quadrilaterals to make a tessellation with more elements.</p>
|
2,868,595 | <p>A Vitali set is a subset $V$ of $[0,1]$ such that for every $r\in \mathbb R$ there exists one and only one $v\in V$ for which $v-r \in \mathbb Q$. Equivalently, $V$ contains a single representative of every element of $\mathbb R / \mathbb Q$.</p>
<p>The proof I read is in this short article on Wikipedia: <a href="https://en.wikipedia.org/wiki/Vitali_set" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Vitali_set</a></p>
<p>Under "proof", the second to last inequality $1 \leq \sum \lambda (V_k) \leq 3$ is claimed to result from the previous inequality $[0,1] \subset \bigcup V_k \subset [-1,2]$ simply using sigma-additivity. There must be some missing argument to claim that the sum of the measures, although greater than the measure of the union, is still less than the measure of $[-1,2]$.</p>
<p>What is the missing argument ?</p>
| DanielWainfleet | 254,665 | <p>Notation: $x+S=S+x=\{x+s:s\in S\}$ when $x\in \Bbb R$ and $S\subset \Bbb R.$</p>
<p>Let $q_0$ be the unique member of $V\cap \Bbb Q.$ Let $V_0=V+(-q_0).$ Let $W_0=\cup_{n\in \Bbb Z}(n+V_0).$ For $q\in (0,1)\cap \Bbb Q$ let $W_q=q+W_0.$</p>
<p>Observe that $(-q)+W_q=W_0$ when $q\in \Bbb Q\cap [0,1)$ and that if $q,q'$ are distinct members of $[0,1)\cap \Bbb Q$ then $W_q$ and $W_{q'}$ are disjoint.</p>
<p>Observe that $\{W_q\cap [0,1):q\in \Bbb Q \cap [0,1)\}$ is a partition of $[0,1).$ </p>
<p>Observe that if $V$ is Lebesgue-measurable then so is $W_q\cap [0,1)$ for each $q\in \Bbb Q \cap [0,1).$ </p>
<p>Let $m'(S)$ denote the Lebesgue outer measure of any $S\subset \Bbb R.$ </p>
<p>For any $q\in \Bbb Q \cap [0,1)$ we have $$m'(W_q\cap [0,1))=m'((\;(-q)+W_q)\;)\cap [-q,1-q))=m'(W_0\cap [-q,1-q))=$$ $$=m'(W_0\cap [-q,0))+m'(W_0\cap [0,1-q))=$$ $$=m'(1+(W_0\cap [-q,0)\;)+m'(W_0\cap [0,1-q))=$$ $$=m'(W_0\cap [1-q,1))+m'(W_0\cap [0,1-q))=$$ $$=m'(W_0\cap [0,1)).$$</p>
<p>Suppose $V$ is Lebesgue-measurable. Then $\{W_q\cap [0,1): q\in \Bbb Q\cap [0,1)\}$ is a partition of $[0,1)$ into a countably infinite family of pair-wise disjoint measurable subsets, each with the same measure $r.$ But if $r=0$ this implies that $[0,1)$ is a measure-$0$ set, and if $r>0$ this implies the measure of $[0,1)$ is $\infty.$</p>
<p>The sets $W_q$ can be described as follows: Consider $\Bbb R$ as a vector space over the field $\Bbb Q.$ Let $B$ be a Hamel (vector-space ) basis for $\Bbb R$ over $\Bbb Q,$ with $1=b_1\in B.$ Let $p:\Bbb R\to \Bbb Q$ be the projection of $\Bbb R$ onto the "$b_1$" co-ordinate. Let $W_q=\{x\in \Bbb R: p(x)-q\in \Bbb Z\}.$ The Axiom of Choice is needed to prove that $B$ exists.</p>
|
2,476,181 | <p>Where Ω = {1,2,...,p}, all Ω are equally likely and p is prime how would I show that if A and B are independent events then at least one of A and B is either ∅ or Ω?</p>
| Bram28 | 256,001 | <p>What you calculated is all the different arrangements of $9$ consonants and $4$ vowels, and you did this correctly, but you forgot to arrange the vowels <em>among</em> the consonants, so you need to multiply all of this by ${13 \choose 4}$</p>
|
2,197,790 | <h3>Question</h3>
<blockquote>
<p>A sequence $\{a_n\}$ of real numbers is said to be a Cauchy sequence of for
each $\epsilon$ > 0 there exists a number $N > 0$ such that m, $n > N$ implies
that $|a_n − a_m| <\epsilon$.</p>
<p>Prove that every convergent sequence is a Cauchy sequence</p>
</blockquote>
<hr>
<h3>Attempt</h3>
<p>This is my first time hearing what a cauchy sequence is. I have no idea how to even start this. I googled cauchy sequence and I think its when $a_n$ converges to $a_{n+1}$? </p>
<p>Attempt:</p>
<p>WTS: $\exists a_m \in \mathbb R, \forall \epsilon > 0, \exists N > 0$, such that for all $n \in \mathbb N$, if $n > N$, then $|a_n - a_m| < \epsilon$</p>
<p>Let $\epsilon > 0$ be arbitrary</p>
<p>Choose N such that for $n > N$ we have $|a_n - a_m| < \epsilon$</p>
<p>Suppose $n > N$, then </p>
<p>??</p>
<p>Could someone point me to the right direction? Thx.</p>
| Jacob Wakem | 117,290 | <p>If it is not Cauchy, it is obviously not convergent: for |a_n -a_m| is greater than some epsilon for arbitrarily large n,m. This assures a_n and a_m do not both get within epsilon/2 of some a_infinity. </p>
|
25,260 | <h2>TL;DR:</h2>
<hr />
<p>Tell me which topics should i study the most, based on this three tests:</p>
<p>Mathematics (A):
<a href="https://www.studyinjapan.go.jp/ja/_mt/2021/06/2020_ga_math_a.pdf" rel="nofollow noreferrer">2020</a>
<a href="https://www.studyinjapan.go.jp/ja/_mt/2021/06/2019_ga_math_a.pdf" rel="nofollow noreferrer">2019</a>
<a href="https://www.studyinjapan.go.jp/ja/_mt/2021/06/2018_ga_math_a.pdf" rel="nofollow noreferrer">2018</a></p>
<p>This question may sound a bit weird, since the natural answer would be "study whats already is on the test" but i'm wanna share a bit of context for this.</p>
<h4>Context:</h4>
<hr />
<p>So i'm currently studying for the Japanese Scholarship known as MEXT in which every year the japanese embassy of each country gave the opportunity to enroll on a japanese university and go to study in that country. As an undergraduate student this is a big deal specially if there's a lack of opportunities in your current country.</p>
<p>I also want you to know that here in my country, there is a lack of proper mathematical education when it comes to high school, pretty much anyone who obtained a bachelor degree's are not well prepared to face one of this japanese test designed for undergraduate students since the test, present to you calculus, trigonometry, modular-arithmetic, number-theory, etc. problems which you don't see until you reach third semester of any university engineer-based career (at least here where i live)</p>
<p>Finally i want to tell that i've been studying calculus for a bit and i do know the basics and i don't have any problem with arithmetic or algebra.</p>
<h2>So here is the question:</h2>
<hr />
<p>Could somebody look at those test and tell me which topics should i study the most? Below (in the comment section) you will find the three test that were used in past examinations and contains all the questions, i also will leave a link to the answers in case someone want to deep onto this:</p>
<h4>Current approach and status:</h4>
<hr />
<p>In my current approach i tried solving most of the exercises presented and always fall in the same loop of no knowing, research, try solving, fail, research again, try solving again, failing, and finally making a question here. This of course may not be the best for a huge variety of topic since you have to make sure to <strong>understand</strong> what you are doing before going into any kind of examination but with so many themes its hard to pick one thing to study since you're pretty much dropping everything else in order to solve only one thing.</p>
<p><a href="https://i.stack.imgur.com/MhGqK.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MhGqK.png" alt="flow chart" /></a></p>
<h3>The ideal:</h3>
<hr />
<p>When making this post i'm hoping someone could orient me through a path of things i should know and preferably master in order to be well prepared for the test.</p>
| Alexander Woo | 4,991 | <p>I would say the assumption is that people heading to mathematics graduate school know about the Wronskian, but this assumption isn't universally true.</p>
<p>Certainly, anyone who has studied a semester of differential equations (and is heading to graduate school) should know it.</p>
<p>However, there is a substantial minority of people interested in pure mathematics of a generally algebraic bent who never study differential equations. (I was one of these people until I was assigned to teach it!)</p>
<p>There really isn't anything that's universally expected (rather than merely generally assumed) of people heading to mathematics graduate school other than an ability to read and write proofs, and some folks doing mathematical modelling might even disagree with that.</p>
|
74,271 | <p>Hello, all!</p>
<p>I have a big sum of log-normal (with location parameter $\mu$ and scale parameter $\sigma$) random variables $X_i$ $\sum_{i=1}^N X_i$ with $N \gg 1$.
How could I estimate convergence rate to a gaussian distribution relative to $\mu$ and $\sigma$?</p>
<p>Thank you.</p>
| Igor Rivin | 11,142 | <p>Log normal distribution has finite variance, so if you subtract the mean, the magic words are "Berry-Esseen theorem". If you don't subtract the mean, the sum diverges.</p>
|
393,378 | <p>Let <span class="math-container">$K=\mathbb{Q}(x)$</span> be the rational functions in one variable <span class="math-container">$x$</span> and let the automorphisms <span class="math-container">$\phi,\psi$</span> of <span class="math-container">$K$</span> be defined as <span class="math-container">$\phi(x)=-\frac{1}{x+1}$</span> and <span class="math-container">$\psi(x)=\frac{1}{x}$</span>.</p>
<p>Let <span class="math-container">$G$</span> be the group generated by <span class="math-container">$\phi,\psi$</span>, then <span class="math-container">$G=\langle \phi,\psi|\phi^3=\psi^2=1,\phi\psi=\psi\phi^2\rangle $</span>.</p>
<p>To be specific, <span class="math-container">$ G=\{1,\phi,\phi^2,\psi,\phi\psi,\psi\phi\} $</span> and
<span class="math-container">$$ \phi(x)=-\frac{1}{x+1}\\ \phi^2(x)=-\frac{x+1}{x}\\ \psi(x)=\frac{1}{x}\\ \phi\psi(x)=-x-1\\ \psi\phi(x)=-\frac{x}{x+1} $$</span>
let <span class="math-container">$K_0$</span> be the invariant subfield of <span class="math-container">$K$</span> under the <span class="math-container">$G$</span>-action.</p>
<p><strong>The point is to show that <span class="math-container">$K_0$</span> is a simple extension of <span class="math-container">$\mathbb{Q}$</span>.</strong></p>
<p>The threads I hold: Regard <span class="math-container">$K$</span> as a <span class="math-container">$G$</span>-module then <span class="math-container">$K_0$</span> is nothing but the 0-th cohomology of <span class="math-container">$G$</span> with coefficient <span class="math-container">$K$</span> and <span class="math-container">$\text{Gal}(K/K_0)=G$</span>. On the other hand set <span class="math-container">$N=\sum\limits_{g\in G}g$</span> to be the norm element of <span class="math-container">$G$</span>, then there is a cannonical map <span class="math-container">$\alpha$</span> form <span class="math-container">$K$</span> to <span class="math-container">$K_0$</span> sending every rational function <span class="math-container">$f\in K$</span> to <span class="math-container">$Nf$</span>. And I guess <span class="math-container">$\alpha$</span> is a surjection which I'm not sure. Notice that <span class="math-container">$Nx=-3$</span> and <span class="math-container">$N(x^2)=(t+1)^2+(\frac{1}{t+1})^2+t^2+\frac{1}{t^2}+(\frac{t+1}{t})^2+(\frac{t}{t+1})^2$</span> and I have a vague sense that <span class="math-container">$N(x^i)$</span> could be expressed in <span class="math-container">$N(x^2)$</span> for any <span class="math-container">$i\in\mathbb{Z}$</span>. So I guess <span class="math-container">$K_0=\mathbb{Q}(N(x^2))$</span>. Again, I'm not sure about this.</p>
| GNiklasch | 49,003 | <p>As hinted in a comment, this is a special case of <a href="https://en.wikipedia.org/wiki/L%C3%BCroth%27s_theorem" rel="nofollow noreferrer">Lüroth's theorem</a>, but it's not hard to find an explicit generator for <span class="math-container">$K_0$</span>.</p>
<p>Note that <span class="math-container">$$x+\phi(x)+\phi^2(x) = \frac{x^3-3x-1}{x(x+1)}$$</span> is invariant under <span class="math-container">$\phi$</span>, while <span class="math-container">$\psi$</span> interchanges it with <span class="math-container">$$\frac{1-3x^2-x^3}{x(x+1)}.$$</span> The sum of the two, as you've already noted, is a constant; but the product of the two or, for convenience, minus the product of the two is a non-constant element <span class="math-container">$$y=\frac{(x^3-3x-1)(x^3+3x^2-1)}{x^2(x+1)^2}$$</span>
of <span class="math-container">$K$</span> and invariant under the whole group, thus in <span class="math-container">$K_0$</span>. Rearranging this last displayed equation gives a monic equation for <span class="math-container">$x$</span> of degree 6 over <span class="math-container">$\mathbb{Q}(y)$</span>, thus <span class="math-container">$\mathbb{Q}(y)$</span> is already all of <span class="math-container">$K_0$</span>.</p>
<p>(One moral is: When life gives you a <em>tower</em> of field extensions, it's usually helpful to exploit this.)</p>
|
1,568,233 | <p>$$\sum_{n=1}^{\infty}n^210^{-n} = \frac{110}{3^6}$$
I noticed this while playing around on my calculator. Is it true and how come?</p>
| Bernard | 202,857 | <p>This is because:
\begin{align*}
\sum_n n^2x^n&=x^2\sum_n n(n-1)x^{n-2}+x\sum_n nx^{n-1}\\
&=x^2\Bigl(\frac 1{1-x}\Bigr)''+x\Bigl(\frac 1{1-x}\Bigr)'\\
&=\frac{2x^2}{(1-x)^3}+\frac x{(1-x)^2}=\frac{x^2+x}{(1-x)^3}=\color{red}{\frac{x(x+1)}{(1-x)^3}}.
\end{align*}
Then set $x=\dfrac1{10}$.</p>
|
4,638,170 | <p>I'm trying to write a proof to show that a tree structure of finite nodes terminate.</p>
<p>Suppose we can say that either a node is a parent of another node (<span class="math-container">$Pqp$</span>: <span class="math-container">$q$</span> is the parent of <span class="math-container">$p$</span>), or it is a terminal node (<span class="math-container">$Tp$</span>: <span class="math-container">$p$</span> does not have a parent).</p>
<ol>
<li><span class="math-container">$\forall_p \space Tp \otimes \exists_q Pqp$</span></li>
</ol>
<p>Every node that has a parent only has one parent.</p>
<ol start="2">
<li><span class="math-container">$\forall_{pqr} \space (Pqp \space \land \space Prp) \rightarrow q=r$</span></li>
</ol>
<p>We say that if node <span class="math-container">$q$</span> comes before node <span class="math-container">$p$</span> (<span class="math-container">$Bqp$</span>), then any node that comes before <span class="math-container">$q$</span> also comes before <span class="math-container">$p$</span>. Additionally, we deny that <span class="math-container">$p$</span> comes before <span class="math-container">$q$</span>. This avoids cyclical chains.</p>
<ol start="3">
<li><span class="math-container">$\forall_{pqr} \space Bqp \rightarrow ((Brq \rightarrow Brp) \space \land \space \lnot Bpq)$</span></li>
</ol>
<p>Any node that is the parent of a nother node must also come before it.</p>
<ol start="4">
<li><span class="math-container">$\forall_{pq} \space Ppq \rightarrow Bpq$</span></li>
</ol>
<p>Suppose we know that at least one node exists.</p>
<ol start="5">
<li><span class="math-container">$\exists p$</span></li>
</ol>
<p>I want to conclude that there must be a terminal node <span class="math-container">$\therefore \exists_p \space Tp$</span> however, one can always find a counter-example using premise 1 in that <span class="math-container">$p$</span> has a parent -- resulting in an infinite regress.</p>
<p>(tl;dr)</p>
<p>So my struggle is, I want to formalise the assumption that there are finite nodes. In all of the definitions of a finite set that I've seen, you need to specify some maximum N. In this case, I want to be intentionally vague about it -- only claiming that it's finite; not how it's finite.</p>
<p>An idea I had was to assume that the domain of discourse is positive integers with something like <span class="math-container">$N\in \mathbb{N_1}$</span> and <span class="math-container">$\forall_p \space p < N$</span> but something about this solution feels kind of hacky.</p>
<p>So is there a better way to do this and, if so what is it?</p>
| Samuel Adrian Antz | 1,045,826 | <p>As requested by ronno, here is my comment as a full answer with more details (like both directions) and references (as a more general view on the concepts might be helpful). For completeness, I have also included the backwards direction:</p>
<p><strong>Lemma</strong>: <span class="math-container">$X$</span> connected <span class="math-container">$\Leftarrow X$</span> path-connected</p>
<p>(The backwards direction does indeed not need the condition of <span class="math-container">$X$</span> being locally path-connected.)</p>
<p><em>Proof</em>: Assume <span class="math-container">$X$</span> is path-connected, but not connected, then there are non-empty, disjoint and open subsets <span class="math-container">$U,V\subset X$</span> with <span class="math-container">$U\cup V=X$</span>. Since they are non-empty, there are points <span class="math-container">$x\in U$</span> and <span class="math-container">$y\in V$</span> and since <span class="math-container">$X$</span> is path-connected, there is a path <span class="math-container">$\gamma\colon[0;1]\rightarrow X$</span> with <span class="math-container">$\gamma(0)=x$</span> and <span class="math-container">$\gamma(1)=y$</span>. Consider <span class="math-container">$\gamma^{-1}(U),\gamma^{-1}(V)$</span>:</p>
<ol>
<li>They are non-empty as <span class="math-container">$0\in\gamma^{-1}(x)\subseteq\gamma^{-1}(U)$</span> and <span class="math-container">$1\in\gamma^{-1}(y)\subseteq\gamma^{-1}(V)$</span>.</li>
<li>They are disjoint as <span class="math-container">$\gamma^{-1}(U)\cap\gamma^{-1}(V)=\gamma^{-1}(U\cap V)=\gamma^{-1}(\emptyset)=\emptyset$</span>.</li>
<li>They are open as preimages of open sets under a continuous map.</li>
<li>They fulfill <span class="math-container">$\gamma^{-1}(U)\cup\gamma^{-1}(V)=\gamma^{-1}(U\cup V)=\gamma^{-1}(X)=[0;1]$</span></li>
</ol>
<p>Hence <span class="math-container">$[0;1]$</span> would not be connected, which is not the case and hence yields a contradiction. <span class="math-container">$\square$</span></p>
<p>The last step has to be proven the hard way: If it could be parted into two non-empty disjoint and open subsets <span class="math-container">$U,V\subset[0;1]$</span> with <span class="math-container">$U\cup V=[0;1]$</span> (with <span class="math-container">$0\in U$</span> without loss of generality), then we would get the contradiction <span class="math-container">$\sup U\in U$</span> and <span class="math-container">$\sup U\in V$</span>, so <span class="math-container">$U\cap V\neq\emptyset$</span>.</p>
<p><strong>Be aware</strong>: It might be tempting at first, but <span class="math-container">$[0;1]$</span> being connected must not be proven using that it is obviously path-connected and therefore connected because of the upper lemma as this is a circular conclusion.</p>
<p><strong>Lemma</strong>: <span class="math-container">$X$</span> connected and locally path-connected <span class="math-container">$\Rightarrow X$</span> path-connected</p>
<p><em>Proof</em>: Let <span class="math-container">$x\in X$</span> be a point and denote by <span class="math-container">$U=[x]_\sim\subseteq X$</span> its path-connected component, where <span class="math-container">$x\sim y$</span> iff there is a path <span class="math-container">$\gamma\colon[0;1]\rightarrow X$</span> with <span class="math-container">$\gamma(0)=x$</span> and <span class="math-container">$\gamma(1)=y$</span>. (The relation is reflexive because of the constant path <span class="math-container">$\epsilon_x\colon[0;1]\rightarrow X,t\mapsto x$</span>, symmetric because of the inverse path <span class="math-container">$\overline\gamma\colon[0;1]\rightarrow X,t\mapsto\gamma(1-t)$</span> and transitive because of the <a href="https://en.wikipedia.org/wiki/Path_(topology)#Path_composition" rel="nofollow noreferrer">path composition</a>.)</p>
<p>Let <span class="math-container">$y\in U$</span>, so there is a path <span class="math-container">$\gamma$</span> from <span class="math-container">$x$</span> to <span class="math-container">$y$</span>. Since <span class="math-container">$X$</span> is locally path-connected, there is an open and path-connected neighborhood <span class="math-container">$V$</span> of <span class="math-container">$y$</span>, so for every point <span class="math-container">$z\in V$</span>, there is a path <span class="math-container">$\delta$</span> from <span class="math-container">$y$</span> to <span class="math-container">$z$</span>. With path composition, <span class="math-container">$\gamma*\delta$</span> is a path from <span class="math-container">$x$</span> to <span class="math-container">$z$</span>, hence <span class="math-container">$V\subset U$</span> and <span class="math-container">$U$</span> <strong>is open</strong>.</p>
<p>Let <span class="math-container">$z\in\overline U$</span>. Since <span class="math-container">$X$</span> is locally path-connected, there is an open and path-connected neighborhood <span class="math-container">$V$</span> of <span class="math-container">$z$</span> with <span class="math-container">$U\cap V\neq\emptyset$</span>. Choose <span class="math-container">$y\in U\cap V$</span>. Because of <span class="math-container">$y\in U$</span>, there is a path <span class="math-container">$\gamma$</span> from <span class="math-container">$x$</span> to <span class="math-container">$y$</span> and because of <span class="math-container">$y\in V$</span>, there is a path <span class="math-container">$\delta$</span> from <span class="math-container">$y$</span> to <span class="math-container">$z$</span>. With path composition, <span class="math-container">$\gamma*\delta$</span> is a path from <span class="math-container">$x$</span> to <span class="math-container">$z$</span>, hence <span class="math-container">$z\in U$</span> and <span class="math-container">$U$</span> <strong>is closed</strong>, meaning <span class="math-container">$U^\complement$</span> is open.</p>
<p>We get the partition <span class="math-container">$X=U\cup U^\complement$</span> into disjoint open subsets and since <span class="math-container">$X$</span> is connected, they cannot both be non-empty. Since <span class="math-container">$x\in U$</span> (as <span class="math-container">$\sim$</span> is reflexive) and <span class="math-container">$U$</span> is therefore non-empty, <span class="math-container">$U^\complement=\emptyset$</span> must be empty, so <span class="math-container">$X=U$</span>, which means, that <span class="math-container">$X$</span> is path-connected. <span class="math-container">$\square$</span></p>
<p><strong>Be aware</strong>: Since I sometimes encounter people believing the following myth, I want to add an important remark for dealing with local properties in topology: (Path-)connectedness does not imply local (path-)connectedness! The <a href="https://en.wikipedia.org/wiki/Topologist%27s_sine_curve" rel="nofollow noreferrer">Warsaw sine curve</a> is connected, but not locally connected (See "<em>Counterexamples in Topology</em>" by Steen and Seebach found <a href="https://xn--webducation-dbb.com/wp-content/uploads/2019/04/Lynn-Arthur-Steen-J.-Arthur-Seebach-Jr.-Counterexamples-in-topology-Dover-Publications-1995.pdf" rel="nofollow noreferrer">here</a>, Remark 6 on page 138.) The <a href="https://en.wikipedia.org/wiki/Comb_space" rel="nofollow noreferrer">Comb space</a> is path-connected, but not locally path-connected. (It is also contractible, but not locally contractible.)</p>
|
1,898,207 | <blockquote>
<p>A man speaks the truth $8$ out of $10$ times. A fair die is thrown. The man says that the number on the upper face is $5$. Find the probability that the original number on the upper face is $5$.</p>
</blockquote>
<p>While solving I find two ways (shown in the image). I think one of them is correct and other one is incorrect. Please tell me which is the correct one and why.</p>
<p>Any advice on solving tricky problems (on Bayes theorem) is welcome.<a href="https://i.stack.imgur.com/DRzEk.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DRzEk.jpg" alt="enter image description here"></a></p>
| Patrick Stevens | 259,262 | <p>Your first one is false. $P(X \mid T)$ is $1$, not $\frac{1}{6}$. Given that the man speaks the truth, the die definitely showed $5$.</p>
<p>Your second method is much more natural, and it is correct (assuming you plugged the numbers in correctly).</p>
|
255,761 | <p>Given a graph of n vertices, it is possible to plot a discrete signal (or function) of n samples on the vertices of the graph, so that one can visualize the features of the signal on the graph (see the attached image)?</p>
<p><a href="https://i.stack.imgur.com/4s3Qz.png" rel="noreferrer"><img src="https://i.stack.imgur.com/4s3Qz.png" alt="graph with signal" /></a></p>
| flinty | 72,682 | <pre><code>g = PetersenGraph[];
(* get the 3d coordinates of the graph vertices *)
coords = Append[#, 0] & /@ GraphEmbedding[g];
points = Point[coords];
(* create lines between the edges *)
connections = EdgeList[g];
lines = Line[coords[[#]]] & /@ (connections /. UndirectedEdge -> List);
(* generate some signals on each vertex as lines pointing upwards *)
SeedRandom[1];
signals = RandomReal[1, VertexCount@g];
signalsLines = MapThread[
Line[{#1, Append[Most@#1, #2]}] &,
{coords, signals}];
(* draw it all in ortho projection *)
Graphics3D[{
{Red, Dashed, lines},
{Red, PointSize[Large], points},
{Blue, signalsLines}
}, Boxed -> False, ViewProjection -> "Orthographic"]
</code></pre>
<p><a href="https://i.stack.imgur.com/nndHf.png" rel="noreferrer"><img src="https://i.stack.imgur.com/nndHf.png" alt="enter image description here" /></a></p>
|
255,761 | <p>Given a graph of n vertices, it is possible to plot a discrete signal (or function) of n samples on the vertices of the graph, so that one can visualize the features of the signal on the graph (see the attached image)?</p>
<p><a href="https://i.stack.imgur.com/4s3Qz.png" rel="noreferrer"><img src="https://i.stack.imgur.com/4s3Qz.png" alt="graph with signal" /></a></p>
| kglr | 125 | <pre><code>SeedRandom[1]
signals = RandomReal[1, 10];
</code></pre>
<p>To get a 3D graph that looks like the one in OP, we can use <code>signals</code> to specify a custom <code>VertexShapeFunction</code> and use it with <code>PetersenGraph</code>:</p>
<pre><code>PetersenGraph[
EdgeStyle -> Directive[Dashed, Red],
VertexCoordinates -> Append[0] /@ GraphEmbedding[PetersenGraph[]],
EdgeShapeFunction -> "Line",
VertexShapeFunction -> ({Thick, Blue,
Line[{#, # + {0, 0, signals[[#2]]}}], Red, Sphere[#, .03]} &),
ImageSize -> 500
]
</code></pre>
<p><a href="https://i.stack.imgur.com/vh1oo.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vh1oo.png" alt="enter image description here" /></a></p>
<p>More generally, we can use <code>signals</code> as the setting for the options <a href="https://reference.wolfram.com/language/ref/VertexSize.html" rel="nofollow noreferrer"><code>VertexSize</code></a> and/or <a href="https://reference.wolfram.com/language/ref/VertexStyle.html" rel="nofollow noreferrer"><code>VertexStyle</code></a></p>
<pre><code>g0 = PetersenGraph[ImageSize -> 300];
g1 = PetersenGraph[ImageSize -> 300,
VertexSize -> {v_ :> signals[[v]]},
VertexStyle -> {v_ :> ColorData["Rainbow"]@Rescale[signals][[v]]},
EdgeStyle -> Directive[Dashed, Red],
EdgeShapeFunction -> "Line",
BaseStyle -> FaceForm[Opacity[.5]]];
Row[{g0, g1}]
</code></pre>
<p><a href="https://i.stack.imgur.com/padeh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/padeh.png" alt="enter image description here" /></a></p>
<p>To get a 3D graph object simply wrap <code>g1</code> with <code>Graph3D</code>. Use a custom <code>VertexShapeFunction</code> if you need to depict vertices as lines:</p>
<pre><code>Row[{Graph3D @ g1,
Graph3D[g1,
VertexShapeFunction -> ({Thick, Line @ {#, # + {0, 0, signals[[#2]]}}} &)]}]
</code></pre>
<p><a href="https://i.stack.imgur.com/LSgCR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LSgCR.png" alt="enter image description here" /></a></p>
<p>We can also use a more flexible custom <code>VertexShapeFunction</code> which allows any 3D graphics primitive as vertex shape:</p>
<pre><code>ClearAll[vShapeF, tF]
tF[x_] := Translate[
Scale[x, If[FreeQ[_Sphere]@x, {.1, .1, #3[[3]]}, #3], {0, 0, 0}], #] &;
vShapeF[prim_: Automatic, d___] :=
Module[{pnt = {AbsolutePointSize[5], Black, Opacity[1], Point[{0, 0, 0}]},
dir = Sequence[Opacity[.5], AbsoluteThickness[2], CapForm[None], d]},
Switch[prim,
Automatic | "Automatic" | Sphere | "Sphere",
tF[{dir, Sphere[], pnt}],
Line | "Line", tF[{dir, Line[{{0, 0, 0}, {0, 0, 1}}], pnt}],
Tube | "Tube", tF[{dir, Tube[{{0, 0, 0}, {0, 0, 1}}, 1], pnt}],
Cylinder | "Cylinder",
tF[{dir, Cylinder[{{0, 0, 0}, {0, 0, 1}}, 1], pnt}],
Cuboid | "Cuboid", tF[{dir, Cuboid[{-1, -1, 0}, {1, 1, 1}], pnt}],
_, tF[{dir, prim, pnt}]]]
</code></pre>
<p><em><strong>Examples:</strong></em></p>
<pre><code>triangleWaveCube = ChartElementData["TriangleWaveCube",
"AngularFrequency" -> 5, "RadialAmplitude" -> 0.4][{{-1, 1}, {-1, 1}, {0, 1}}];
Multicolumn[
Graph3D[g1, VertexShapeFunction -> vShapeF[ToExpression @ #],
BoxRatios -> If[# == "Automatic", Automatic, {1, 1, 1/2}],
PlotLabel -> Style[#, 16, Black],
ImageSize -> 400] & /@
{"Automatic", "Line", "Tube", "Cylinder", "Cuboid", "triangleWaveCube"},
2, Appearance -> "Horizontal"]
</code></pre>
<p><a href="https://i.stack.imgur.com/4I0cP.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4I0cP.png" alt="enter image description here" /></a></p>
|
3,242,844 | <p><span class="math-container">$$\int_0^{\pi/6} \frac{x\cos x}{1+2\cos x}dx$$</span></p>
<p>Does it have a closed solution? <a href="https://www.wolframalpha.com/input/?i=int%20%5Cfrac%7Bxcos%20x%7D%7B1%2B2cos%20x%7Ddx%20from%200%20to%20%5Cpi%2F6" rel="nofollow noreferrer">WA</a> outputs this result.</p>
| clathratus | 583,016 | <p><strong>(Basically) Complete Answer</strong></p>
<p>We define
<span class="math-container">$$f(t)=\int_0^t \frac{\cos x}{1+2\cos x}dx$$</span>
Then from integration by parts,
<span class="math-container">$$J=\int_0^{\pi/6}\frac{x\cos x}{1+2\cos x}dx=\frac{\pi}{6}f\left(\frac\pi6\right)-\int_0^{\pi/6}f(t)dt\, .$$</span>
First, I will find a closed form for <span class="math-container">$f(t)$</span>, then I will proceed with <span class="math-container">$\int_0^{\pi/6}f(t)dt$</span>.</p>
<hr>
<p>We see that
<span class="math-container">$$f(t)=\frac12\int_0^{t}\frac{-1+1+2\cos x}{1+2\cos x}dx=\frac{t}{2}-\frac12\int_0^{t}\frac{dx}{1+2\cos x}$$</span>
In the remaining integral, we use <a href="https://en.wikipedia.org/wiki/Tangent_half-angle_substitution#The_substitution" rel="nofollow noreferrer"><span class="math-container">$u=\tan(x/2)$</span></a> to get
<span class="math-container">$$\begin{align}
\int_0^t\frac{dx}{1+2\cos x}&=2\int_0^{\tan(t/2)}\frac1{1+2\frac{1-u^2}{1+u^2}}\frac{du}{1+u^2}\\
&=2\int_0^{\tan(t/2)}\frac{du}{3-u^2}\\
&=\frac{2}{\sqrt3}\int_0^{\frac{1}{\sqrt3}\tan(t/2)}\frac{du}{1-u^2}\\
&=\frac{2}{\sqrt3}\tanh^{-1}\left[\frac1{\sqrt3}\tan\frac{t}2\right].
\end{align}$$</span>
So
<span class="math-container">$$f(t)=\frac{t}{2}-\frac{1}{\sqrt3}\tanh^{-1}\left[\frac1{\sqrt3}\tan\frac{t}2\right].$$</span>
This gives
<span class="math-container">$$\begin{align}
J&=\frac{\pi^2}{72}+\frac\pi{12\sqrt3}\ln(\sqrt{3}-1)-\int_0^{\pi/6}\left[\frac{t}{2}-\frac{1}{\sqrt3}\tanh^{-1}\left(\frac1{\sqrt3}\tan\frac{t}2\right)\right]dt\\
&=\frac{\pi^2}{144}+\frac\pi{12\sqrt3}\ln(\sqrt3-1)+\frac1{\sqrt3}\int_0^{\pi/6}\tanh^{-1}\left(\frac1{\sqrt3}\tan\frac{t}2\right)dt
\end{align}$$</span></p>
<hr>
<p>The next integral is
<span class="math-container">$$P=\int_0^{\pi/6}\tanh^{-1}\left(\frac1{\sqrt3}\tan\frac{t}2\right)dt.$$</span>
Set <span class="math-container">$x=\frac1{\sqrt3}\tan\frac{t}{2}$</span> to get
<span class="math-container">$$P=2\sqrt3\int_0^{\frac2{\sqrt3}-1}\frac{\tanh^{-1}(x)}{1+3x^2}dx$$</span>
Then recall that for <span class="math-container">$|z|<1$</span>,
<span class="math-container">$$\tanh^{-1}(z)=\sum_{n\geq0}\frac{x^{2n+1}}{2n+1}$$</span>
So
<span class="math-container">$$P=2\sqrt3\sum_{n\geq0}\frac1{2n+1}\int_0^{\frac2{\sqrt3}-1}\frac{x^{2n+1}}{3x^2+1}dx.$$</span>
So we define
<span class="math-container">$$\begin{align}
j_n&=2\int_0^{\frac2{\sqrt3}-1}\frac{x^{2n}}{3x^2+1}xdx\\
&=\int_0^{7/3-4/\sqrt3}\frac{x^n}{3x+1}dx\\
&=\frac1{3^{n+1}}\int_1^{8-4\sqrt3}\frac{(x-1)^n}{x}dx.
\end{align}$$</span>
Then we notice that <span class="math-container">$j_0=\frac13\ln(8-4\sqrt3)$</span> so that we can proceed for <span class="math-container">$n\geq1$</span> with the binomial theorem:
<span class="math-container">$$\begin{align}
j_n&=\frac1{3^{n+1}}\int_1^{8-4\sqrt3}\frac{(x-1)^n}{x}dx\\
&=\frac1{3^{n+1}}\int_1^{8-4\sqrt3}\frac{1}{x}\sum_{k=0}^{n}(-1)^{n-k}{n\choose k}x^kdx\\
&=\frac1{3^{n+1}}\int_1^{8-4\sqrt3}\left[\frac{(-1)^n}{x}+\sum_{k=1}^{n}(-1)^{n-k}{n\choose k}x^{k-1}\right]dx\\
&=\frac{(-1)^n}{3^{n+1}}\ln(8-4\sqrt3)+\frac{1}{3^{n+1}}\sum_{k=1}^{n}(-1)^{n-k}{n\choose k}\int_1^{8-4\sqrt3}x^{k-1}dx\\
&=\frac{(-1)^n}{3^{n+1}}\ln(8-4\sqrt3)+\frac{1}{3^{n+1}}\sum_{k=1}^{n}\frac{(-1)^{n-k}}{k}{n\choose k}\left[(8-4\sqrt3)^k-1\right].
\end{align}$$</span>
So
<span class="math-container">$$\begin{align}
P&=\sqrt{3}\left[j_0+\sum_{n\geq1}\frac{j_n}{2n+1}\right]\\
&=\frac1{\sqrt3}\ln a+\frac1{\sqrt3}\sum_{n\geq1}\frac{1}{3^n(2n+1)}\left[(-1)^n\ln a+\sum_{k=1}^{n}\frac{(-1)^{n-k}}{k}{n\choose k}(a^k-1)\right]\\
&=\frac{\ln a}{\sqrt3}\left(1+\sum_{n\geq1}\frac{(-1)^n}{3^n(2n+1)}\right)+\frac1{\sqrt3}\sum_{n\geq1}\frac{(-1)^n}{3^n(2n+1)}\sum_{k=1}^{n}\frac{(-1)^{k}}{k}{n\choose k}(a^k-1)\\
&=\frac\pi6\ln a+\frac{1}{\sqrt3}S(a)
\end{align}$$</span>
Where <span class="math-container">$a=8-4\sqrt3$</span> and <span class="math-container">$$S(x)=\sum_{n\geq1}\frac{(-1)^n}{3^n(2n+1)}\sum_{k=1}^{n}\frac{(-1)^{k}}{k}{n\choose k}(x^k-1).$$</span>
Note that we used <span class="math-container">$$1+\sum_{n\geq1}\frac{(-1)^n}{3^n(2n+1)}=\sum_{n\geq0}\frac{(-1)^n}{3^n(2n+1)}=\frac\pi{2\sqrt3}$$</span>
which comes from
<span class="math-container">$$\tan^{-1}(z)=\sum_{n\geq0}\frac{(-1)^n}{2n+1}z^{2n+1}\qquad |z|\leq1.$$</span></p>
<blockquote>
<p>Anyway, we combine results (and do a little algebra with the <span class="math-container">$\ln$</span> terms) to get
<span class="math-container">$$J=\frac{\pi^2}{144}+\frac{\pi}{6\sqrt3}\ln\left(4\sqrt{11\sqrt3-19}\right)+\frac13 S(8-4\sqrt3).$$</span>
I will update my answer once I figure out how to compute <span class="math-container">$S(x)$</span>.</p>
</blockquote>
<p><strong>Update:</strong> See <a href="https://math.stackexchange.com/a/3244970/583016">this answer</a> for a way to find a closed form for <span class="math-container">$S(q)$</span> involving <a href="https://en.wikipedia.org/wiki/Spence%27s_function" rel="nofollow noreferrer"><span class="math-container">$\mathrm{Li}_2(z)$</span></a>.</p>
|
42,301 | <p>everyone! I am sorry, but I am an abcolute novice of Mathematica (to be more precise this is my first day of using it) and even after surfing the web and all documents I am not able to solve the following system: </p>
<pre><code>Solve[{y*(((y*x)/(beta*b))^(1/(beta - 1)) - v) - c*alpha ==
0, ((x/alpha))*(((y*x)/(alpha*beta*b))^(1/(beta - 1)) -
v) + (((y*x)/alpha) -
2*alpha*((yx)/(2*beta*b))^(1/(beta - 1)))*(1/(beta -
1))*(x/(alpha*beta*
b))*((y*x)/(alpha*beta*b))^((2 - beta)/(1 - beta)) == 0}, {x,
y}]
</code></pre>
<p>What I need is to solve following systems, getting x and y expressed through all these symbols. Is it even possible? Thank you in advance. </p>
| sakra | 68 | <p>An alternate solution using <code>Fold</code>:</p>
<pre><code>Fold[If[Last[#2] < Last[#1], #2, #1] &, {0, 0, Infinity}, mya]
</code></pre>
<p>If the list is known to be non-empty, the following solution is faster:</p>
<pre><code>Fold[If[Last[#2] < Last[#1], #2, #1] &, First[mya], Rest[mya]]
</code></pre>
|
1,835,158 | <p>I wont to choose three random integer point in origin $|x|\leq r, |y|\leq r$ at plane as $(a_{1},b_{1}),(a_{2},b_{2}),(a_{3},b_{3})$.
What the probability that this three point create a right triangle ( it is depend to r? what about isosceles triangle?
I think that its zero but I cant proof. Thank you.</p>
| Jamal Farokhi | 244,152 | <p>My attempt:
If $r\in \mathbb{N}$ then there are $(2r+1)^{2}$ integer pair point and also there are $(2r+1)^{6}$ way to chose pair three integer number. So lets that $N(r)=(2r+1)^{6}$. We try to find all points that they create a right triangle. We denoted them by $N^{\prime}(r)$. For instance in this two figure, for $r=2,r=3$<a href="https://i.stack.imgur.com/sm4im.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sm4im.gif" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/42RvN.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/42RvN.gif" alt="enter image description here"></a></p>
<p>$$N(2)=5^{6}, \ \ N^{\prime}(2)=?$$
$$N(2)=7^{6}, \ \ N^{\prime}(3)=?$$
We guess that $N^{\prime}(r)$ is a recursive sequence such as $N^{\prime}(r)=N^{\prime}(r-1)+h(r)$. But How about $h(r)$?. I don't know!!!
So $$P_{r}(Right \ \ triangle)=\dfrac{N^{\prime}(r)}{N(r)}=\dfrac{N^{\prime}(r-1)+h(r)}{N(r)}$$
And so:
$$\lim_{r\rightarrow \infty}P_{r}( Right \ \ triangle )=\Huge{?!!?}$$
Thank you.</p>
|
14,458 | <p>I want to be able to plot several numerical solutions of an ODE, plus its analytical solution in one plot, in order to see how the numerical solutions converge towards the analytical one with respect to the number of steps. The method I'm using is Euler's method for the equation
$ y'(t) = 1-t +4y(t), y(0)=1$</p>
<p>The code I have so far is: </p>
<pre><code>y[0]=1;
Do[y[n+1]=y[n]+0.01(1-0.01n+4y[n]), {n,0,99}]
y[100]
</code></pre>
<p>Is this doable? Thanks in advance for any help :)</p>
| J. M.'s persistent exhaustion | 50 | <p>Sascha showed you how to use the built-in <code>"ExplicitEuler"</code> option. You mention</p>
<blockquote>
<p>I want to be able to plot several numerical solutions of an ODE, plus its analytical solution in one plot, in order to see how the numerical solutions converge...</p>
</blockquote>
<p>Here's one way to generate <a href="http://books.google.com/books?hl=en&id=F93u7VcSRyYC&pg=PA39" rel="nofollow noreferrer">"Lady Windermere's fan"</a>:</p>
<pre><code>yTrue = y /. First @ DSolve[{y'[t] == 1 - t + 4 y[t], y[0] == 1}, y, t];
pts = Table[
With[{ya = y /. First @
NDSolve[{y'[t] == 1 - t + 4 y[t], y[0] == 1}, y, {t, 0, 1},
Method -> "ExplicitEuler", StartingStepSize -> 2^-k]},
Transpose[Append[ya["Coordinates"], ya["ValuesOnGrid"]]]], {k, 1, 5}];
Show[Plot[yTrue[t], {t, 0, 1}, PlotStyle -> Directive[Thick, Dashed, Gray]],
ListPlot[pts, Joined -> True, Mesh -> All, PlotMarkers -> Automatic],
Axes -> None, Frame -> True]
</code></pre>
<p><img src="https://i.stack.imgur.com/ep07X.png" alt="Lady Windermere's fan"></p>
<p>Here, the gray dashed curve is the solution obtained from <code>DSolve[]</code>.</p>
|
4,174 | <p>I'm developing a course that focuses on the transistion from arithmetic to algebraic thinking, particularly in grades 5-8. We will do this through focus on the common core. I'm also putting together a collection of suggested readings from the math education literature. I would be interested to hear your suggestions for suggested readings.</p>
| JPBurke | 759 | <p>Early algebra research necessarily deals with the development of algebraic reasoning and questions like "what is algebra" and "what counts as algebraic thinking and reasoning?" And my own readings on early algebra have helped me to focus on what about algebraic thinking are students developing, apart from the manipulation of symbols. This is where, I think, some of the early algebra research may have more general interest, since some of the authors deal with what students are capable of that is connected to algebra well before they are doing what we are familiar with in an Algebra 1 course.</p>
<p>A book collecting some different research on Early Algebra is:</p>
<p><a href="http://rads.stackoverflow.com/amzn/click/0805854738">Kaput, J. J., Carraher, D. W., & Blanton, M. L. (2007). Algebra in the early grades.</a></p>
<p>There is a review of it in JRME <a href="http://www.jstor.org/discover/10.2307/20720131?uid=3739696&uid=377752083&uid=38070&uid=2&uid=3&uid=364742151&uid=67&uid=62&uid=38069&uid=3739256&uid=60&sid=21104579409983">(Chazan & Edwards, 2010)</a>, and <a href="https://www.education.umd.edu/MathEd/info/DChazanScans/2010JRME03-203.pdf">someone is sharing the review here as a PDF (in case you don't have access)</a>.</p>
<p>From the review:</p>
<blockquote>
<p>In addition to these direct assaults on the rhetorical challenge posed by early
algebra, one of the strengths of the volume is the richness of the presentations aimed
at helping readers appreciate how early algebra is not the sort of activity that a
reader might associate with a high school Algebra 1 course. In particular, the seven
chapters in Part II offer theoretical arguments for children’s capacity to learn
algebra early and empirical studies of children’s algebraic reasoning in a variety of
different elementary-level classroom and curricular contexts. These chapters
provide illustrations of what early algebra curriculum and instruction look like, as
well as existence proofs of children’s capacity to do this work. (p. 204)</p>
</blockquote>
<p>Even though you're not talking about algebrafying early mathematics, and depending on what sort of things you want students to consider and even argue about, <a href="http://rads.stackoverflow.com/amzn/click/0805854738">this book</a> (along with other early algebra research) may provide perspectives that spark thoughtful discussions and reflections.</p>
|
3,795,072 | <p><strong>QUESTION</strong></p>
<blockquote>
<p>What is the sign of <span class="math-container">$$\frac{d}{dx} \bigg(\frac{x+1}{x}+\bigg(\frac{2x}{x+1}\bigg)^{\dfrac{\ln(6/5)}{\ln(31/25)}}\bigg)$$</span> when <span class="math-container">$x > {10}^{500}$</span>?</p>
</blockquote>
<p><strong>MY ATTEMPT</strong></p>
<p><a href="https://www.wolframalpha.com/input/?i=d%2Fdx%20%28%28x%2B1%29%2Fx%20%2B%20%282x%2F%28x%2B1%29%29%5E%28ln%286%2F5%29%2Fln%2831%2F25%29%29%29" rel="nofollow noreferrer">WolframAlpha</a> gives
<span class="math-container">$$
\begin{split}
p(x)
&= \frac{d}{dx}
\left(\frac{x+1}{x}
+ \left(\frac{2x}{x+1}\right)^{\dfrac{\ln(6/5)}{\ln(31/25)}}\right) \\
&= -\frac{x+1}{x^2} + \frac{1}{x}
+ \frac{\left(\frac{1}{x+1} - \frac{x}{(x+1)^2}\right)
2^{\dfrac{\ln(6/5)}{\ln(31/25)}}
\ln(\frac{6}{5})
\left(\frac{x}{x+1}\right)^{\dfrac{\ln(6/5)}{\ln(31/25)} - 1}}
{\ln(\frac{31}{25})}.
\end{split}
$$</span></p>
<p>Solving for the root(s) of <span class="math-container">$p(x)=0$</span> (again, using <a href="https://www.wolframalpha.com/input/?i=Solve%20for%20x%20if%20%28d%2Fdx%20%28%28x%2B1%29%2Fx%20%2B%20%282x%2F%28x%2B1%29%29%5E%28ln%286%2F5%29%2Fln%2831%2F25%29%29%29%20%3D%200%29" rel="nofollow noreferrer">WolframAlpha</a>) gives the approximate solution
<span class="math-container">$$x \approx 3.89605$$</span>
and a message that <strong>Standard computation time (has been) exceeded</strong>.</p>
<p>Evaluating <span class="math-container">$p(x)$</span> at successively larger powers of <span class="math-container">$10$</span> (still using WolframAlpha), starting from <span class="math-container">$x=10$</span>, I obtain the following:
<span class="math-container">$$p(10) > 0$$</span>
<a href="https://www.wolframalpha.com/input/?i=Evaluate%20-%28%28x%2B1%29%2Fx%5E2%29%20%2B%20%281%2Fx%29%20%2B%20%28%28%281%2F%28x%2B1%29%20-%20%28x%2F%28x%2B1%29%5E2%29%29%282%5E%28ln%286%2F5%29%2Fln%2831%2F25%29%29%29%28ln%286%2F5%29%29%28x%2F%28x%2B1%29%29%5E%28%28ln%286%2F5%29%2Fln%2831%2F25%29%29-1%29%29%2F%28ln%2831%2F25%29%29%29%20at%20x%20%3D%2010" rel="nofollow noreferrer">WolframAlpha computation for <span class="math-container">$p(10)$</span></a>
<span class="math-container">$$p(100) > 0$$</span>
<a href="https://www.wolframalpha.com/input/?i=Evaluate%20-%28%28x%2B1%29%2Fx%5E2%29%20%2B%20%281%2Fx%29%20%2B%20%28%28%281%2F%28x%2B1%29%20-%20%28x%2F%28x%2B1%29%5E2%29%29%282%5E%28ln%286%2F5%29%2Fln%2831%2F25%29%29%29%28ln%286%2F5%29%29%28x%2F%28x%2B1%29%29%5E%28%28ln%286%2F5%29%2Fln%2831%2F25%29%29-1%29%29%2F%28ln%2831%2F25%29%29%29%20at%20x%20%3D%20100" rel="nofollow noreferrer">WolframAlpha computation for <span class="math-container">$p(100)$</span></a>
<span class="math-container">$$p(1000) > 0$$</span>
<a href="https://www.wolframalpha.com/input/?i=Evaluate%20-%28%28x%2B1%29%2Fx%5E2%29%20%2B%20%281%2Fx%29%20%2B%20%28%28%281%2F%28x%2B1%29%20-%20%28x%2F%28x%2B1%29%5E2%29%29%282%5E%28ln%286%2F5%29%2Fln%2831%2F25%29%29%29%28ln%286%2F5%29%29%28x%2F%28x%2B1%29%29%5E%28%28ln%286%2F5%29%2Fln%2831%2F25%29%29-1%29%29%2F%28ln%2831%2F25%29%29%29%20at%20x%20%3D%201000" rel="nofollow noreferrer">WolframAlpha computation for <span class="math-container">$p(1000)$</span></a>
<span class="math-container">$$p(10000) > 0$$</span>
<a href="https://www.wolframalpha.com/input/?i=Evaluate%20-%28%28x%2B1%29%2Fx%5E2%29%20%2B%20%281%2Fx%29%20%2B%20%28%28%281%2F%28x%2B1%29%20-%20%28x%2F%28x%2B1%29%5E2%29%29%282%5E%28ln%286%2F5%29%2Fln%2831%2F25%29%29%29%28ln%286%2F5%29%29%28x%2F%28x%2B1%29%29%5E%28%28ln%286%2F5%29%2Fln%2831%2F25%29%29-1%29%29%2F%28ln%2831%2F25%29%29%29%20at%20x%20%3D%2010000" rel="nofollow noreferrer">WolframAlpha computation for <span class="math-container">$p(10000)$</span></a></p>
<p>However, when I try to plug in, say, <span class="math-container">$x = {10}^{500} + 1$</span> for <span class="math-container">$p(x)$</span> in WolframAlpha, it returns a <strong>Standard computation time exceeded</strong> message.</p>
<p>I was wondering whether it would be possible to prove that <span class="math-container">$p(x) > 0, \forall x > {10}^{500}$</span>? I hope somebody with more computational power can help out with this one.</p>
| Jean Marie | 305,862 | <p>This is not a matter of computational power.</p>
<p>Consider your function as a composition :</p>
<p><span class="math-container">$$f(u)=u+\left(\frac{2}{u}\right)^{\alpha}$$</span></p>
<p><span class="math-container">$$\text{with} \ u:=\frac{x+1}{x}=1+\frac{1}{x}\tag{1}$$</span></p>
<p>where <span class="math-container">$\alpha\approx 0.84757$</span>.</p>
<p>Observe that the range of <span class="math-container">$u$</span> is interval <span class="math-container">$[1,2]$</span>.</p>
<p>then write the derivative in the following way :</p>
<p><span class="math-container">$$f'(u(x)).u'(x)=...$$</span></p>
<p>Computations (you have done) give</p>
<p><span class="math-container">$$u'(x)=-\frac{1}{x^2}<0 \ \text{and} \ f'(u)=1-K\tfrac{1}{u^{\alpha+1}}$$</span></p>
<p>(with <span class="math-container">$K=\alpha 2^{\alpha} \approx 1.52517$</span>)</p>
<p>As <span class="math-container">$x \to \infty, \ u \to 1_+$</span>, therefore <span class="math-container">$f'(u)$</span> becomes negative for all <span class="math-container">$u<u_0=1/K^{\alpha+1}\approx 0.458466$</span>.</p>
<p>The monotonicity of the inverse function <span class="math-container">$x=\frac{1}{u-1}$</span> (see (1)) gives the existence of <span class="math-container">$x_0:=\frac{1}{u_0-1}\approx 1.846607$</span> such that <span class="math-container">$x>x_0 \implies f'(u)<0$</span>.</p>
<p>Therefore, we have, for <span class="math-container">$x>x_0$</span>, a positive result (product of two negative quantities <span class="math-container">$f'(u)$</span> and <span class="math-container">$u'(x)$</span>).</p>
|
3,795,072 | <p><strong>QUESTION</strong></p>
<blockquote>
<p>What is the sign of <span class="math-container">$$\frac{d}{dx} \bigg(\frac{x+1}{x}+\bigg(\frac{2x}{x+1}\bigg)^{\dfrac{\ln(6/5)}{\ln(31/25)}}\bigg)$$</span> when <span class="math-container">$x > {10}^{500}$</span>?</p>
</blockquote>
<p><strong>MY ATTEMPT</strong></p>
<p><a href="https://www.wolframalpha.com/input/?i=d%2Fdx%20%28%28x%2B1%29%2Fx%20%2B%20%282x%2F%28x%2B1%29%29%5E%28ln%286%2F5%29%2Fln%2831%2F25%29%29%29" rel="nofollow noreferrer">WolframAlpha</a> gives
<span class="math-container">$$
\begin{split}
p(x)
&= \frac{d}{dx}
\left(\frac{x+1}{x}
+ \left(\frac{2x}{x+1}\right)^{\dfrac{\ln(6/5)}{\ln(31/25)}}\right) \\
&= -\frac{x+1}{x^2} + \frac{1}{x}
+ \frac{\left(\frac{1}{x+1} - \frac{x}{(x+1)^2}\right)
2^{\dfrac{\ln(6/5)}{\ln(31/25)}}
\ln(\frac{6}{5})
\left(\frac{x}{x+1}\right)^{\dfrac{\ln(6/5)}{\ln(31/25)} - 1}}
{\ln(\frac{31}{25})}.
\end{split}
$$</span></p>
<p>Solving for the root(s) of <span class="math-container">$p(x)=0$</span> (again, using <a href="https://www.wolframalpha.com/input/?i=Solve%20for%20x%20if%20%28d%2Fdx%20%28%28x%2B1%29%2Fx%20%2B%20%282x%2F%28x%2B1%29%29%5E%28ln%286%2F5%29%2Fln%2831%2F25%29%29%29%20%3D%200%29" rel="nofollow noreferrer">WolframAlpha</a>) gives the approximate solution
<span class="math-container">$$x \approx 3.89605$$</span>
and a message that <strong>Standard computation time (has been) exceeded</strong>.</p>
<p>Evaluating <span class="math-container">$p(x)$</span> at successively larger powers of <span class="math-container">$10$</span> (still using WolframAlpha), starting from <span class="math-container">$x=10$</span>, I obtain the following:
<span class="math-container">$$p(10) > 0$$</span>
<a href="https://www.wolframalpha.com/input/?i=Evaluate%20-%28%28x%2B1%29%2Fx%5E2%29%20%2B%20%281%2Fx%29%20%2B%20%28%28%281%2F%28x%2B1%29%20-%20%28x%2F%28x%2B1%29%5E2%29%29%282%5E%28ln%286%2F5%29%2Fln%2831%2F25%29%29%29%28ln%286%2F5%29%29%28x%2F%28x%2B1%29%29%5E%28%28ln%286%2F5%29%2Fln%2831%2F25%29%29-1%29%29%2F%28ln%2831%2F25%29%29%29%20at%20x%20%3D%2010" rel="nofollow noreferrer">WolframAlpha computation for <span class="math-container">$p(10)$</span></a>
<span class="math-container">$$p(100) > 0$$</span>
<a href="https://www.wolframalpha.com/input/?i=Evaluate%20-%28%28x%2B1%29%2Fx%5E2%29%20%2B%20%281%2Fx%29%20%2B%20%28%28%281%2F%28x%2B1%29%20-%20%28x%2F%28x%2B1%29%5E2%29%29%282%5E%28ln%286%2F5%29%2Fln%2831%2F25%29%29%29%28ln%286%2F5%29%29%28x%2F%28x%2B1%29%29%5E%28%28ln%286%2F5%29%2Fln%2831%2F25%29%29-1%29%29%2F%28ln%2831%2F25%29%29%29%20at%20x%20%3D%20100" rel="nofollow noreferrer">WolframAlpha computation for <span class="math-container">$p(100)$</span></a>
<span class="math-container">$$p(1000) > 0$$</span>
<a href="https://www.wolframalpha.com/input/?i=Evaluate%20-%28%28x%2B1%29%2Fx%5E2%29%20%2B%20%281%2Fx%29%20%2B%20%28%28%281%2F%28x%2B1%29%20-%20%28x%2F%28x%2B1%29%5E2%29%29%282%5E%28ln%286%2F5%29%2Fln%2831%2F25%29%29%29%28ln%286%2F5%29%29%28x%2F%28x%2B1%29%29%5E%28%28ln%286%2F5%29%2Fln%2831%2F25%29%29-1%29%29%2F%28ln%2831%2F25%29%29%29%20at%20x%20%3D%201000" rel="nofollow noreferrer">WolframAlpha computation for <span class="math-container">$p(1000)$</span></a>
<span class="math-container">$$p(10000) > 0$$</span>
<a href="https://www.wolframalpha.com/input/?i=Evaluate%20-%28%28x%2B1%29%2Fx%5E2%29%20%2B%20%281%2Fx%29%20%2B%20%28%28%281%2F%28x%2B1%29%20-%20%28x%2F%28x%2B1%29%5E2%29%29%282%5E%28ln%286%2F5%29%2Fln%2831%2F25%29%29%29%28ln%286%2F5%29%29%28x%2F%28x%2B1%29%29%5E%28%28ln%286%2F5%29%2Fln%2831%2F25%29%29-1%29%29%2F%28ln%2831%2F25%29%29%29%20at%20x%20%3D%2010000" rel="nofollow noreferrer">WolframAlpha computation for <span class="math-container">$p(10000)$</span></a></p>
<p>However, when I try to plug in, say, <span class="math-container">$x = {10}^{500} + 1$</span> for <span class="math-container">$p(x)$</span> in WolframAlpha, it returns a <strong>Standard computation time exceeded</strong> message.</p>
<p>I was wondering whether it would be possible to prove that <span class="math-container">$p(x) > 0, \forall x > {10}^{500}$</span>? I hope somebody with more computational power can help out with this one.</p>
| gt6989b | 16,192 | <p>Define <span class="math-container">$a = \frac{\ln(6/5)}{\ln(31/25)} \approx 0.847$</span> and let
<span class="math-container">$$
f(x)
= \frac{x+1}{x}+\left(\frac{2x}{x+1}\right)^a
= 1 + \frac1x + 2^a \left(1 - \frac{1}{x+1}\right)^a.
$$</span>
Then,
<span class="math-container">$$
\begin{split}
f'(x)
&= -\frac{1}{x^2} + a 2^a \left(1 - \frac{1}{x+1}\right)^{a-1}
\frac{d}{dx} \left[1 - \frac{1}{x+1}\right] \\
&= -\frac{1}{x^2} + a 2^a \left(1 - \frac{1}{x+1}\right)^{a-1} \frac{1}{(x+1)^2}
\end{split}
$$</span>
Can you finish?</p>
|
587,217 | <p>I know that the units of 2 by 2 matrices with integer entries must have a determinant of 1 or -1, and I have proved that if the determinant is zero then the matrix is not a unit, however I am wondering how you would go about proving that matrices with determinants other than 1 and -1 are not units?</p>
| DonAntonio | 31,254 | <p>An element $\;r\;$ in a ring is a unit if there exists another element $\;x\;$ there s.t. $\;rx=1\;$ .</p>
<p>If $\;A,B \;$ are square integer matrices ,then</p>
<p>$$AB=1\implies \det A=\frac1{\det B}\;$$</p>
<p>But the rightmost number is <em>not</em> an integer if $\;\det B\neq\pm 1\;$ ...</p>
|
207,185 | <p>How would I go about proving this without a truth table?</p>
<p>$[(p \lor q) \implies r ] \implies [ \neg r \implies (\neg p \land \neg q)]$</p>
| Blue | 409 | <p><strong>Fact.</strong> If $x$, $y$, $z$ are points on the complex plane, then the circumcenter of $\triangle xyz$ is the point</p>
<p>$$ i \frac{x(z\overline{z} - y\overline{y})+ y(x\overline{x}-z\overline{z})+z(y\overline{y}-x\overline{x})}{x(\overline{z} - \overline{y})+ y(\overline{x}-\overline{z})+z(\overline{y}-\overline{x})}$$</p>
<p>where "$\overline{x}$" indicates the complex conjugate of $x$.</p>
<hr>
<p>Let us consider $\triangle abc$, defining $d$, $e$, $f$ on sides $bc$, $ca$, $ab$, respectively:</p>
<p>$$d := b + \alpha(c-b) \qquad e := c+\beta(a-c) \qquad f := a+\gamma(b-a)$$</p>
<p>for real scalars $\alpha, \beta, \gamma$.</p>
<p>Let $p$, $q$, $r$ be the respective circumcenters of $\triangle aef$, $\triangle dbf$, $\triangle dec$; for simplicity, we take $a$, $b$, $c$ on the unit circle so that $\overline{a}=1/a$, etc. Then,</p>
<p>$$\begin{align}
p &= i\;\frac{c(b-a)+ \beta b ( a-c) + \gamma c( a - b )}{b-c} \\[6pt]
q &= i\;\frac{a(c-b)+ \gamma c ( b-a) + \alpha a( b - c )}{c-a} \\[6pt]
r &= i\;\frac{b(a-c)+ \alpha a ( c-b) + \beta b( c - a )}{a-b}
\end{align}$$</p>
<p>We deduce</p>
<p>$$\frac{p-q}{a-b} = \frac{q-r}{b-c} = \frac{r-p}{c-a} =: u$$</p>
<p>which, by taking the modulus, proves $\triangle abc \sim \triangle pqr$. A little algebra reveals that
$$u + \overline{u} = 1$$</p>
<p>Now ... A point, $m$, (which <em>may or may not be</em> the Miquel point) is the center of spiral similarity from $\triangle abc$ and $\triangle pqr$, if and only if</p>
<p>$$\frac{a-m}{p-m} = \frac{b-m}{q-m} = \frac{c-m}{r-m}$$</p>
<p>(This characterization is what inspired me to approach the problem in the complex plane.) Solving for $m$ via the first equality gives</p>
<p>$$m = \frac{aq-bp}{(a-b)-(p-q)} = \frac{r - c u}{1-u}$$</p>
<p>whence
$$m-r = \frac{u(r-c)}{1-u} = \frac{u}{\overline{u}}(r-c) \qquad\qquad
\overline{m}-\overline{r} = \frac{\overline{u}}{u}(\overline{r}-\overline{c})$$
so that
$$|m-r|^2 = (m-r)(\overline{m}-\overline{r}) = (r-c)(\overline{r}-\overline{c}) = |c-r|^2$$</p>
<p>indicating that $c$ and $m$ are equidistant from $r$: thus, $m$ is on the circumcircle of $\triangle dec$. By symmetry, it is also on the circumcircles of $\triangle aef$ and $\triangle dbf$; the center of spiral similarity is in fact the Miquel point.</p>
|
2,643,099 | <p>Can someone help me explain why it is true that</p>
<p>$$\sin(\pi/2-\theta)=\sqrt{1-\sin^2\theta}$$</p>
<p>When answering please explain the different relation which is used</p>
<p>Thanks</p>
| MrYouMath | 262,304 | <p>You should be asking why </p>
<p>$$\sin^2(\pi/2-\theta) = 1- \sin^2(\theta).$$</p>
<p>By the trigonometric Pythagorean theorem, we know that</p>
<p>$$\cos^2(\theta)+\sin^2(\theta)=1 \implies \cos^2(\theta)=1-\sin^2(\theta)$$</p>
<p>is valid.</p>
<p>If you additionally use the complementary formula for trigonometric functions $\cos(\theta)=\sin(\pi/2-\theta)$ then you can conclude what you wanted to show in the first place.</p>
<p>The reason why
$$\sin(\pi/2-\theta) = \sqrt{1- \sin^2(\theta)}$$</p>
<p><strong>is not always true</strong> is because you dropped the minus sign. The sign must be chosen according to the quadrant in which the angle is located.</p>
<p>EDIT: If $-\pi/2\leq \theta \leq \pi/2$ (first and fourth quadrant) then the sign is positive because the cosine function is positive for this interval. If $\pi/2 < \theta <3\pi/2$ (second and third quadrant) then the sign is negative because the cosine function is negative for this interval. </p>
|
3,161,371 | <p>For <span class="math-container">$p, q$</span> prime, if <span class="math-container">$q$</span> divides an integer <span class="math-container">$n$</span> but <span class="math-container">$p$</span> does not, show that <span class="math-container">$\text{gcd}(n, pq) = q$</span></p>
<p>This statement sort of reminds me of Euclid's Lemma, but I haven't been able to progress much. </p>
<p>I tried writing <span class="math-container">$n = kq$</span> for some integer <span class="math-container">$k$</span>. Then we have <span class="math-container">$\text{gcd}(kq, pq)$</span>, where <span class="math-container">$p$</span> and <span class="math-container">$q$</span> are prime. I don't really know how to progress from here.</p>
| Robert Israel | 8,508 | <p>Hint: there are only <span class="math-container">$4$</span> divisors of <span class="math-container">$p\cdot q$</span>. Which could be <span class="math-container">$\gcd(n,p\cdot q)$</span>?</p>
|
2,352,821 | <p>So, in order to obtain the required answer, I tried to apply some Taylor expansions, which led me to nowhere actually.
After a while I tried to use the summation theorem </p>
<p>$\sum_{n=-\infty}^{+\infty}{f\left(n\right)}=-\sum_{i=1}^{m}{Res_{z=z_i}{\pi\cot\left(\pi z\right)f\left(z\right)}}$ at $f\left(z\right)$'s poles.</p>
<p>Residue at $z=-\frac{1}{2}$ equals to residue at $z=\frac{1}{2}$, both of them are $-\frac{\pi ^2}{16}$.
What I got seems to be not the correct answer after all:</p>
<p>$-\sum_{i=1}^{m}{Res_{z=z_i}{\pi\cot\left(\pi z\right)f\left(z\right)}} = -\frac{\pi ^2}{8}$, so that $\sum_{n=-\infty}^{+\infty}{f\left(n\right)}=\frac{\pi ^2}{8}$
Since the $f\left(z\right)$ is even, I get $\sum_{n=0}^{+\infty}{f\left(n\right)}=\frac{\pi ^2}{16}$</p>
<p>However, the initial task was to find the sum of $\sum_{n=1}^{+\infty}{f\left(n\right)}$. I supposed that $\sum_{n=1}^{+\infty}{f\left(n\right)}=\sum_{n=0}^{+\infty}{f\left(n\right)}-f\left(0\right)=\frac{\pi ^2}{16}-1$, which is incorrect, as according to Mathematica, I should get $\frac{\pi ^2}{16}-\frac{1}{2}$. I don't know where is my mistake at this point. Perhaps, the whole approach is not well executed. </p>
| Jack D'Aurizio | 44,121 | <p>Since $\frac{1}{4n^2-1}=\frac{1}{2}\left(\frac{1}{(2n-1)}-\frac{1}{(2n+1)}\right)$ by squaring we get
$$\sum_{n\geq 1}\frac{1}{(4n^2-1)^2} = \frac{1}{4}\sum_{n\geq 1}\frac{1}{(2n-1)^2}+\frac{1}{4}\sum_{n\geq 1}\frac{1}{(2n+1)^2}-\frac{1}{4}\sum_{n\geq 1}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)$$
where the first two series in the RHS depend on $\zeta(2)-\frac{1}{4}\zeta(2)=\frac{\pi^2}{8}$ and the last one is telescopic:
$$ \sum_{n\geq 1}\frac{1}{(4n^2-1)^2}=\color{blue}{\frac{\pi^2-8}{16}}.$$</p>
|
4,613,982 | <p>Calculate the integral <span class="math-container">$$\int_{-\infty }^{\infty } \frac{\sin(\Omega x)}{x\,(x^2+1)} dx$$</span> given <span class="math-container">$$\Omega >>1 $$</span></p>
<p><a href="https://i.stack.imgur.com/mPVqq.jpg" rel="nofollow noreferrer">I tried but couldn't find C1</a></p>
| Lai | 732,917 | <p><span class="math-container">$$
\begin{aligned}
\int_{-\infty}^{\infty} \frac{\sin (\Omega x)}{x\left(x^2+1\right)} d x =&\int_{-\infty}^{\infty} \frac{x \sin (\Omega x)}{x^2\left(x^2+1\right)} d x \\
= & \underbrace{\int_{-\infty}^{\infty} \frac{\sin (\Omega x)}{x} d x}_{=\pi}- \underbrace{ \int_{-\infty}^{\infty} \frac{x \sin (\Omega x)}{x^2+1} d x}_{J} \\
\end{aligned}
$$</span>
Using contour integration along anti-clockwise direction of the path <span class="math-container">$$\gamma=\gamma_{1} \cup \gamma_{2} \textrm{ where } \gamma_{1}(t)=t+i 0(-R \leq t \leq R) \textrm{ and } \gamma_{2}(t)=R e^{i t} (0<t<\pi) ,$$</span>
we have
<span class="math-container">$$
\begin{aligned}
J & =\operatorname{Im}\left[\operatorname{Res}\left(\frac{z e^{\Omega z i}}{z^2+1}, z=i\right)\right]\\
& =\operatorname{Im}\left(2 \pi i \cdot \lim _{z \rightarrow i} \frac{z e^{\Omega z i}}{z+i}\right) \\
& =\operatorname{Im}\left(2 \pi i \cdot \frac{1}{2} e^{-\Omega}\right) \\
& =\pi e^{-\Omega}
\end{aligned}
$$</span>
We can now conclude that <span class="math-container">$$
\boxed{I=\pi\left(1-e^{-\Omega}\right)}
$$</span></p>
|
4,291,880 | <p>I don't understand how you would take the conjugate of a quadratic equation and how it would be useful to solve this question.</p>
<p>I would normally show it by saying if <span class="math-container">$b$</span> is real, then it is equal to <span class="math-container">$\alpha$</span> times <span class="math-container">$\beta$</span>, say <span class="math-container">$a$</span> is equal to <span class="math-container">$-(\alpha + \beta)$</span> and then just show it, but this doesn't involve conjugates as far as I can tell.</p>
| Mohammad Riazi-Kermani | 514,496 | <p>Note that the conjugate of <span class="math-container">$z^2$</span> is <span class="math-container">$\bar z^2$</span></p>
<p>For real numbers <span class="math-container">$a$</span> and <span class="math-container">$b$</span> the conjugate of <span class="math-container">$z^2 + az + b$</span> is <span class="math-container">$\bar z^2 + a\bar z + b$</span></p>
<p>Therefore if z is a solution of <span class="math-container">$z^2 + az + b=0$</span> so is <span class="math-container">$\bar z$</span>, which means the complex solutions appear in conjugate pairs.</p>
|
7,715 | <p>I am starting on a Phd program and am supposed to read Colliot Thelene and Sansuc's article
on R-equivalence for tori. I find it very difficult and although I have some knowledge over schemes , I am completely baffled by this scalar restriction business of having a field extension $K/k$ , a torus over $K$ and "restricting" it to $k$.
I would be very gratefull for a reference or even better by some explanation . I found nothing in my standard books (Hartshorne, Qing Liu, Mumford etc) so I hope this question is appropriate for the site. Thank you.</p>
| Pete L. Clark | 1,149 | <p>As said, the sought after concept is also known as Weil restriction. In a word, it is the algebraic analogue of the process of viewing an <span class="math-container">$n$</span>-dimensional complex variety as a <span class="math-container">$(2n)$</span>-dimensional real variety.</p>
<p>The setup is as follows: let <span class="math-container">$L/K$</span> be a finite degree field extension and let <span class="math-container">$X$</span> be a scheme over <span class="math-container">$L$</span>. Then the Weil restriction <span class="math-container">$W_{L/K} X$</span> is the <span class="math-container">$K$</span>-scheme representing the following functor on the category of K-algebras:</p>
<p><span class="math-container">$A\mapsto X(A \otimes_K L)$</span>.</p>
<p>In particular, one has <span class="math-container">$W_{L/K} X(K) = X(L)$</span>.</p>
<p>By abstract nonsense (Yoneda...), if such a scheme exists it is uniquely determined by the above functor. For existence, some hypotheses are necessary, but I believe that it exists whenever <span class="math-container">$X$</span> is reduced of finite type.</p>
<p>Now for a more concrete description. Suppose <span class="math-container">$X = \mathrm{Spec} L[y_1,...,y_n]/J$</span> is an affine scheme. Let <span class="math-container">$d = [L:K]$</span> and <span class="math-container">$a_1,...,a_d$</span> be a <span class="math-container">$K$</span>-basis of <span class="math-container">$L$</span>. Then we make the following "substitution":</p>
<p><span class="math-container">$$y_i = a_1 x_{i1} + ... + a_d x_{id},$$</span></p>
<p>thus replacing each <span class="math-container">$y_i$</span> by a linear expression in d new variables <span class="math-container">$x_{ij}$</span>. Moreover,
suppose <span class="math-container">$J = \langle g_1,...,g_m \rangle$</span>; then we substitute each of the above equations into <span class="math-container">$g_k(y_1,...,y_n)$</span> getting a polynomial in the <span class="math-container">$x$</span>-variables, however still with <span class="math-container">$L$</span>-coefficients. But now using our fixed basis of <span class="math-container">$L/K$</span>, we can regard a single polynomial with <span class="math-container">$L$</span>-coefficients as a vector of <span class="math-container">$d$</span> polynomials with <span class="math-container">$K$</span> coefficients. Thus we end up with <span class="math-container">$md$</span> generating polynomials in the <span class="math-container">$x$</span>-variables, say generating an ideal <span class="math-container">$I$</span> in <span class="math-container">$K[x_{ij}]$</span>, and we put <span class="math-container">$\mathrm Res_{L/K} X = \mathrm{Spec} K[x_{ij}]/I$</span>.</p>
<p>A great example to look at is the case <span class="math-container">$X = G_m$</span> (multiplicative group) over <span class="math-container">$L = \mathbb{C}$</span> (complex numbers) and <span class="math-container">$K = \mathbb{R}$</span>. Then <span class="math-container">$X$</span> is the spectrum of</p>
<p><span class="math-container">$$\mathbb{C}[y_1,y_2]/(y_1 y_2 - 1);$$</span></p>
<p>put <span class="math-container">$y_i = x_{i1} + \sqrt{-1} x_{i2}$</span> and do the algebra. You can really see that the
corresponding real affine variety is <span class="math-container">$\mathbb{R}\left[x,y\right]\left[(x^2+y^2)^{-1}\right]$</span>, as it should be: see e.g.
p. 2 of</p>
<p><a href="http://alpha.math.uga.edu/%7Epete/SC5-AlgebraicGroups.pdf" rel="nofollow noreferrer">http://alpha.math.uga.edu/~pete/SC5-AlgebraicGroups.pdf</a></p>
<p>for the calculations.</p>
<p>Note the important general property that for a variety <span class="math-container">$X/L$</span>, the dimension of the Weil restriction from <span class="math-container">$L$</span> down to <span class="math-container">$K$</span> is <span class="math-container">$[L:K]$</span> times the dimension of <span class="math-container">$X/L$</span>. This is good to keep in mind so as not to confuse it with another possible interpretation of "restriction of scalars", namely composition of the map <span class="math-container">$X \to \mathrm{Spec} L$</span> with the map <span class="math-container">$\mathrm{Spec} L \to Spec K$</span> to
give a map <span class="math-container">$X \to \mathrm{Spec} K$</span>. This is a much weirder functor, which preserves the dimension but screws up things like geometric integrality. (When I first heard about "restriction of
scalars", I guessed it was this latter thing and got very confused.)</p>
|
4,401,460 | <p>I'm wondering what other tools there are aside from radicals can be used to extend fields in the context of solving polynomials. Since <span class="math-container">$S_5$</span> isn't solvable, constructing a field with a Galois group of <span class="math-container">$S_5$</span> with respect to <span class="math-container">$\mathbb{Q}$</span> can't be a radical extension, but is there some other function or operation that could be used? In other words, a quintic formula with radicals doesn't exist, but is there some function that isn't a purpose-built "this function yields solutions to a polynomial" function that could be used to solve quintics or higher polynomials?</p>
| MangoPizza | 956,791 | <p>Let <span class="math-container">$f(x)$</span> denote the expected number of runs till we get remainder <span class="math-container">$0$</span> with <span class="math-container">$x$</span> denoting the current remainder mod <span class="math-container">$3$</span>.</p>
<p>Note that each remainder has <span class="math-container">$\frac{1}{3}$</span> chance of occurring.</p>
<p><span class="math-container">$$f(1) = \frac{1}{3}(1 + f(2)) + \frac{1}{3}(f(1) + 1) + \frac{1}{3}(1)$$</span>
<span class="math-container">$$f(2) = \frac{1}{3}(1 + f(1)) + \frac{1}{3}(f(2) + 1) + \frac{1}{3}(1)$$</span></p>
<p>Note that in case of <span class="math-container">$f(1)$</span>, if we get remainder <span class="math-container">$2$</span> on the roll (numbers <span class="math-container">$2$</span> and <span class="math-container">$5$</span>), we end the process, so we count only <span class="math-container">$1$</span> roll.
Similarly, for <span class="math-container">$f(2)$</span>.</p>
<p>We get <span class="math-container">$f(1) = f(2)$</span>. Solving, we get <span class="math-container">$f(1) = f(2) = 3$</span>.</p>
<p>Let us now iterate on the value of the first die roll. (Initial remainder is <span class="math-container">$0$</span>)</p>
<p><span class="math-container">$$ans = \frac{1}{3}(f(2) + 1) + \frac{1}{3}(f(1) + 1) + \frac{1}{3}(1)$$</span></p>
<p><span class="math-container">$$ans = 3$$</span></p>
|
4,401,460 | <p>I'm wondering what other tools there are aside from radicals can be used to extend fields in the context of solving polynomials. Since <span class="math-container">$S_5$</span> isn't solvable, constructing a field with a Galois group of <span class="math-container">$S_5$</span> with respect to <span class="math-container">$\mathbb{Q}$</span> can't be a radical extension, but is there some other function or operation that could be used? In other words, a quintic formula with radicals doesn't exist, but is there some function that isn't a purpose-built "this function yields solutions to a polynomial" function that could be used to solve quintics or higher polynomials?</p>
| JMP | 210,189 | <p>The probability that the game is a success in the <span class="math-container">$k^{th}$</span> roll is given by the coefficient of <span class="math-container">$x^k$</span> in</p>
<p><span class="math-container">$$\frac12\sum_{k=1}^\infty \left(\frac{2x}3\right)^k \tag{1}$$</span></p>
<p>The expected number of throws is</p>
<p><span class="math-container">$$\frac12\sum_{k=1}^\infty k\left(\frac{2x}3\right)^k \tag{2}$$</span></p>
<p>evalulated at <span class="math-container">$x=1$</span>.</p>
<p>The differential of <span class="math-container">$(1)$</span> is</p>
<p><span class="math-container">$$\frac12\frac23\sum_{k=1}^\infty k\left(\frac{2x}3\right)^{k-1}$$</span></p>
<p>which equals <span class="math-container">$(2)$</span> when evaluated at <span class="math-container">$x=1$</span>.</p>
<p><span class="math-container">$(1)$</span> can also be written as</p>
<p><span class="math-container">$$\frac12\left(-1+\left(1-\frac{2x}{3}\right)^{-1}\right)$$</span></p>
<p>This differentiates to</p>
<p><span class="math-container">$$\frac13\left(1-\frac{2x}{3}\right)^{-2}$$</span></p>
<p>which evaluates to <span class="math-container">$3$</span> at <span class="math-container">$x=1$</span>.</p>
<p>Therefore <span class="math-container">$E(X)=3$</span>.</p>
|
4,401,460 | <p>I'm wondering what other tools there are aside from radicals can be used to extend fields in the context of solving polynomials. Since <span class="math-container">$S_5$</span> isn't solvable, constructing a field with a Galois group of <span class="math-container">$S_5$</span> with respect to <span class="math-container">$\mathbb{Q}$</span> can't be a radical extension, but is there some other function or operation that could be used? In other words, a quintic formula with radicals doesn't exist, but is there some function that isn't a purpose-built "this function yields solutions to a polynomial" function that could be used to solve quintics or higher polynomials?</p>
| drhab | 75,923 | <p>If <span class="math-container">$n$</span> denotes an arbitrary integer and <span class="math-container">$D$</span> is the result of throwing a fair die then:<span class="math-container">$$\Pr(3\text{ divides }n+D)=\frac13$$</span></p>
<p>This because for <em>every</em> integer <span class="math-container">$n$</span> the set <span class="math-container">$\{n+1,n+2,n+3,n+4,n+5,n+6\}$</span> contains exactly <span class="math-container">$2$</span> numbers that are divisible by <span class="math-container">$3$</span>.</p>
<p>Applying that for <span class="math-container">$\mu:=\mathbb EX$</span> we find:<span class="math-container">$$\mu=\frac13\cdot1+\frac23(1+\mu)=1+\frac23\mu$$</span>
From this we conclude that:<span class="math-container">$$\mu=3$$</span></p>
|
918,689 | <p>of 5 be selected that contain /at least/ 1 of the broken bulbs?</p>
<p>So far, I have tried only 1 method, as it's the only one I've been taught, but I don't know if I am doing it right.
I tried doing C(100,1)/C(100,5) but it just doesn't seem right. Is it? If it isn't, what am I doing wrong?</p>
| user84413 | 84,413 | <p>Using subtraction (as described in another answer) is probably the best way to do this, but you could also break it up into two cases: </p>
<p>where you get exactly 1 defective bulb, and where you get both of the defective bulbs.</p>
<p>This approach gives an answer of $\displaystyle\binom{2}{1}\binom{98}{4}+\binom{2}{2}\binom{98}{3}$.</p>
|
2,114,446 | <p>But, just to get across the idea of a generating function, here is how a generatingfunctionologist might answer the question: the nth Fibonacci number, $F_{n}$, is the coefficient of $x^{n}$ in the expansion of the function $\frac{x}{(1 − x − x^2)}$ as a power series about the origin.</p>
<p>I am reading a book about generating function, however, I got a little rusted about power series. could anyone give me a quick review about what the statement above is saying?</p>
<p>namely, </p>
<p>$F_{n}$, is the coefficient of $x_{n}$ in the expansion of the function $\frac{x}{(1 − x − x^2)}$ as a power series about the origin</p>
| angryavian | 43,949 | <p>If you consider the power series
$$f(x) = F_0 + F_1 x + F_2 x^2 + \cdots$$
then the relation $F_{n+2} = F_n + F_{n+1}$ implies
$$x^2 f(x) + x f(x) - F_0 x = f(x) - F_0 - F_1 x.$$
(Write out each term as a power series, and combine terms.)
Rearranging and plugging in $F_0=0$ and $F_1=1$ gives
$$(1-x-x^2) f(x) = x.$$</p>
<hr>
<p>More detail:</p>
<p>\begin{align}
x^2 f(x) + x f(x)
&= (F_0 x^2 + F_1 x^3 + F_2 x^4 + \cdots) + (F_0 x + F_1 x^2 + F_2 x^3 + \cdots)
\\
&=F_0 x + (F_0+F_1)x^2 + (F_1+F_2)x^3 + (F_2+F_3)x^4+\cdots\\
&= F_0 x + F_2 x^2 + F_3 x^3 + F_4 x^4 + \cdots
\\
&= F_0 x - F_0 - F_1 x + (F_0 + F_1x + F_2x^2 + F_3 x^3 + \cdots)
\\
&= F_0 x - F_0 - F_1 x + f(x).
\end{align}</p>
|
1,579,371 | <p>I'm studying for my number theory test tomorrow, and these are the last questions in my study guide. I think I understand Fermat's factorization, however, I can't tell how my professor wants us to answer these questions. One of them is going to be on the exam.</p>
<ol>
<li><p>Set <span class="math-container">$n= 87463$</span> and <span class="math-container">$q(x) = x^2 - n$</span>. Explain how to use the congruences
<span class="math-container">\begin{eqnarray*}
q(265) &=& -2\times3\times13^2\times17,\\
q(278) &=& -3^3\times13\times29,\\
q(296) &=& 3^2\times17,\\
q(299) &=& 2\times3\times17\times19,\\
q(307) &=& 2\times3^2\times13\times29,\\
q(316) &=& 3^6\times17,
\end{eqnarray*}</span>
to factor <span class="math-container">$n$</span>.</p>
</li>
<li><p>Explain how the congruences below prove that <span class="math-container">$n = 2821$</span> is composite
<span class="math-container">\begin{eqnarray*}
2^{705} &=& 2605 \pmod n,\\
2^{1410} &=& 1520 \pmod n,\\
2^{2820} &=& 1 \pmod n.
\end{eqnarray*}</span>
I'm not so sure if these questions are related or not. In the second one, it is easy to see <span class="math-container">$705 = \frac{n-1}4$</span>, <span class="math-container">$1410 = \frac{n-1}{2}$</span> and <span class="math-container">$2820 = n-1$</span> however I'm not sure on which property to use here.</p>
</li>
</ol>
| Raymond Manzoni | 21,783 | <p>I'll try to prove Tito Piezas III's (+1) neat conjecture :
$$\tag{1}32\;\Re\operatorname{Li}_3\left(i\,\phi^k\right)\stackrel{\color{blue}?}=\operatorname{Li}_3\left(\phi^{-4k}\right)-4\operatorname{Li}_3\left(\phi^{-2k}\right)+10\,k\operatorname{Li}_3\left(\phi^{-2}\right)-\frac{4k^3+5k}{6}\,\ln^3(\phi^2)-8k\,\zeta(3)$$
The classical identity $\;\displaystyle\frac {\operatorname{Li}_3(x^2)}4=\operatorname{Li}_3(x)+\operatorname{Li}_3(-x)\;$ (c.f. Lewin $1981$ "Polylogaritms and associated functions" p.$154$) allows to rewrite the two first terms at the right as :
$$\tag{2}32\;\Re\operatorname{Li}_3\left(i\,\phi^k\right)\stackrel{\color{blue}?}=4\operatorname{Li}_3\left(-\phi^{-2k}\right)+10\,k\operatorname{Li}_3\left(\phi^{-2}\right)-\frac{4k^3+5k}{6}\,\ln^3(\phi^2)-8k\,\zeta(3)$$
and gives further :
$\;\dfrac {\operatorname{Li}_3\left(-\phi^{2k}\right)}4=\operatorname{Li}_3\left(i\,\phi^k\right)+\operatorname{Li}_3\left(-i\,\phi^k\right)=2\Re\operatorname{Li}_3\left(i\,\phi^k\right)\;$ that is :
$$\tag{3}4\operatorname{Li}_3\left(-\phi^{2k}\right)\stackrel{\color{blue}?}=4\operatorname{Li}_3\left(-\phi^{-2k}\right)+10\,k\operatorname{Li}_3\left(\phi^{-2}\right)-\frac{4k^3+5k}{6}\,\ln^3(\phi^2)-8k\,\zeta(3)$$</p>
<p>Another usual identity is $\;\displaystyle \operatorname{Li}_3(-x)-\operatorname{Li}_3(-1/x)=-\zeta(2)\log(x)-\frac{\ln^3(x)}6\;$ (same Lewin page)<br>
applied to $\,x:=\phi^{2k}\,$ $(3)$ becomes :
\begin{align}
-8k\zeta(2)\ln(\phi)-\frac{4k^3\ln^3(\phi^{2})}6\stackrel{\color{blue}?}=&10\,k\operatorname{Li}_3\left(\phi^{-2}\right)-\frac{4k^3+5k}{6}\,\ln^3(\phi^2)-8k\,\zeta(3)\\
-8k\zeta(2)\ln(\phi)+\frac{5k\ln^3(\phi^{2})}6+8k\,\zeta(3)\stackrel{\color{blue}?}=&10\,k\operatorname{Li}_3\left(\phi^{-2}\right)\\
\tag{4}\operatorname{Li}_3\left(\phi^{-2}\right)=&\frac 45(\zeta(3)-\zeta(2)\ln(\phi))+\frac 23{\ln^3(\phi)}\\
\end{align}
This last identity is proved in Lewin's $1991$ book (<a href="https://books.google.com/books?id=iJnyBwAAQBAJ&pg=PA2" rel="noreferrer">page $2$</a> with $\,\rho=\dfrac 1{\phi}\,$) and may be compared with <a href="http://www.wolframalpha.com/input/?i=polylog%283,%282%2F%281%2Bsqrt%285%29%29%29%5E2%29" rel="noreferrer">alpha's evaluation</a> knowing that $\,\operatorname{csch}^{-1}(2)=\ln(\phi)$.</p>
|
2,491,577 | <p>What is the meaning of phrase,<strong>"Compactness and Connectedness are intrinsic properties of a topological space"</strong>?</p>
| Pedro | 70,305 | <p>According to the <a href="https://books.google.com.br/books?id=ehmMCgAAQBAJ&pg=PA197&lpg=PA197&dq=%22intrinsic+properties%22+topology&source=bl&ots=B-udXETF1Z&sig=651ngQs6klMPghpzBFnF1PdyQog&hl=pt-BR&sa=X&ved=0ahUKEwiF7uzorY_XAhVIvZAKHaeOC74Q6AEITjAG#v=onepage&q=%22intrinsic%20properties%22%20topology&f=false" rel="nofollow noreferrer"><em>Handbook of mathematics</em></a> (see p. 197), it means that compactness and connectedness are <a href="https://www.encyclopediaofmath.org/index.php/Topological_invariant" rel="nofollow noreferrer"><em>Topological invariants</em></a>.</p>
|
413,778 | <p>Let <span class="math-container">$G$</span> be a finite abelian group, <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> be two non-empty subsets of <span class="math-container">$G$</span> of equal size. Suppose that for each irreducible character <span class="math-container">$\chi$</span> of <span class="math-container">$G$</span> we have <span class="math-container">$\sum_{x\in X}\chi(x)=\sum_{y\in Y}\chi(y)$</span>. Is it true that <span class="math-container">$X=Y$</span> in general?</p>
| Absol | 91,692 | <p><span class="math-container">$\DeclareMathOperator\Irr{Irr}$</span>You can see this from the fact that for abelian groups, irreducible characters form a <span class="math-container">$\mathbb{C}$</span>-basis of the space of functions from <span class="math-container">$G$</span> to <span class="math-container">$\mathbb{C}$</span>. These functions correspond bijectively to linear transformations from the group algebra <span class="math-container">$\mathbb{C}G$</span> to <span class="math-container">$\mathbb{C}$</span>, and <span class="math-container">$\Irr(G)$</span> is also a basis for the space of such linear transformations. Therefore, <span class="math-container">$\chi(\sum_{x\in X}x)=\sum_{x\in X}\chi(x)=\sum_{y\in Y}\chi(y)=\chi(\sum_{y\in Y}y)$</span> for each <span class="math-container">$\chi\in \Irr(G)$</span> implies that <span class="math-container">$f(\sum_{x\in X}x)=f(\sum_{y\in X}y)$</span> for every linear transformation <span class="math-container">$f:\mathbb{C}G\rightarrow \mathbb{C}$</span>. This happens if and only if <span class="math-container">$\sum_{x\in X}x=\sum_{y\in Y}y$</span>.</p>
<p>There is another way to see this. In general, for split finite dimensional algebra <span class="math-container">$A$</span> over a field <span class="math-container">$k$</span>, the common zero set of irreducible characters <span class="math-container">$\Irr_k(A)$</span> is <span class="math-container">$J(A)+[A,A]$</span>, where <span class="math-container">$J(A)$</span> is the Jacobson radical of <span class="math-container">$A$</span> and <span class="math-container">$[A,A]$</span> is the commutator subalgebra of <span class="math-container">$A$</span> generated by elements of the form <span class="math-container">$[a,b]=ab-ba$</span>.</p>
<p>If <span class="math-container">$G$</span> is a finite abelian group, then the group algebra <span class="math-container">$\mathbb{C}G$</span> is finite dimensional, split, semisimple and commutative. Hence, both <span class="math-container">$J(A)$</span> and <span class="math-container">$[A,A]$</span> are trivial, and the above theorem says that the common zero set of irreducible complex characters is trivial. In particular, if two elements <span class="math-container">$a$</span>, <span class="math-container">$b$</span> of the group algebra satisfy <span class="math-container">$\chi(a)=\chi(b)$</span> for each <span class="math-container">$\chi\in \Irr(G)$</span>, then <span class="math-container">$a-b$</span> is in the common zero set, which is <span class="math-container">$\{0\}$</span>.</p>
|
2,543,169 | <p>The question is pretty self explanatory, but I’ve encountered situations where, for the length of some vector $\vec{a}$, to denote the length (or magnitude, which ever you prefer) as either $\| \vec{a}\|= \sqrt{a_1^2+a_2^2+\ldots+a_n^2}$ or $|\vec{a}|= \sqrt{a_1^2+a_2^2+\ldots+a_n^2}$ and I was wondering which notation is more widely accepted, per say? I’ve tried researching this and different websites actually use different notation. </p>
<p>Any help is appreciated, thank you.</p>
| user | 505,767 | <p>The first one is the more correct, in the sense that is preferable when you are dealing both with vectors and numbers and it is necessary to avoid misunderstanding.</p>
<p>The notation $|\cdot|$ indicates the absolute value for numbers, but it is also frequently and widely used for vectors when it is clear from the context.</p>
|
2,543,169 | <p>The question is pretty self explanatory, but I’ve encountered situations where, for the length of some vector $\vec{a}$, to denote the length (or magnitude, which ever you prefer) as either $\| \vec{a}\|= \sqrt{a_1^2+a_2^2+\ldots+a_n^2}$ or $|\vec{a}|= \sqrt{a_1^2+a_2^2+\ldots+a_n^2}$ and I was wondering which notation is more widely accepted, per say? I’ve tried researching this and different websites actually use different notation. </p>
<p>Any help is appreciated, thank you.</p>
| Fly by Night | 38,495 | <p>You see both sets of notation. For vectors you see ${\bf v}$, $\vec{v}$ and $\underline{v}$, perhaps others. For the norm you see $|\cdot|$ and $\| \cdot \|$. It depends if you're in high-school or university, do physics or pure maths. </p>
<p>As a mathematician, I prefer $\|{\bf v}\|$ for the norm of a vector. By hand, that would be written as $\| \underline{v}\|$, but that's just because it's hard to write bold font by hand. </p>
<p>(Same reason we use $\mathbb R$ for what was traditionally ${\bf R}$)</p>
<p>I like to use $|\cdot|$ for the modulus function, a.k.a. absolute value, which applies to scalars. It might seem silly when both $|\cdot|$ and $\| \cdot \|$ measure "size" in some way, but it makes it easier for the reader to see what is a vector and what is a scalar. </p>
<p>(To be technical: A vector space has a set of vectors, and an accompanying scalar field. There is often an idea of "size" in the vector space, and an idea of "size" in the scalar field. I like to use $\| \cdot \|$ for the norm in the vector space and $| \cdot |$ for the norm in the scalar field.)</p>
<p>For example, given two intersecting lines with respective direction vectors ${\bf u}$ and ${\bf v}$, the acute angle of intersection $\theta$ satisfies $|{\bf u} \cdot {\bf v}| = \|{\bf u}\| \|{\bf v}\| \cos\theta$. The scalar/inner/dot product ${\bf u} \cdot {\bf v}$ is a scalar, and so $|{\bf u} \cdot {\bf v}|$ is the absolute value/modulus of that scalar. On the other hand ${\bf u}$ and ${\bf v}$ are vectors and $\|{\bf u}\|$ and $\|{\bf v}\|$ are the norms of those vectors.</p>
<p>I hope this makes sense.</p>
|
4,118 | <p>I've recently dipped my toes into the world of number theory; and I've bought a book that to me is quite unconventional: R. P. Burn, <em>A Pathway into Number Theory</em>. I've yet to put the book through its paces, but it seems agreeable enough to me. The book is unique in that it poses a sequence of questions to you in the hope that you'll be able to answer them and by thus doing so, begin to discover the fundamentals of number theory.</p>
<p>This is a style of learning that I find agreeable as the knowledge I gain this way is assimilated and retained better. I like being able to discover for myself however most times I don't have the necessary direction (I am self-studying) but that's where this textbook comes in. I feel that this book in the process of nudging you in the right direction also helps you think more like a mathematician (from my very limited experience with it). </p>
<p>I enjoy books that give you a "pathway", although I guess this is the aim of all textbooks. Is it possible for anyone to recommend texts that take a similar aided discovery/inquiry based approach? </p>
| Michael Joyce | 1,397 | <p><em><a href="https://math.dartmouth.edu/news-resources/electronic/kpbogart/" rel="nofollow noreferrer">Combinatorics Through Guided Discovery</a></em> by the late Kenneth Bogart is a great introduction to combinatorics through a guided set of problems and is freely available for download at the link given above.</p>
|
513,779 | <p>If $a,b\in\mathbb{N}$ are odd</p>
<p>then demonstrate:
$$ {\sqrt{a^2 + b^2}} \not\in \mathbb{Q}$$ </p>
<p>I try to guess that $$ {\sqrt{a^2 + b^2}} \in\mathbb{Q}.$$ Then i write $$ {\sqrt{a^2 + b^2}= m/n}.$$ After that: $$ {n\sqrt{a^2 + b^2}= m}$$ , I raised at squared and i have like $$ n^2(a^2+ b^2)=m^2 $$ and i thought that $$ (a^2 +b^2)$$ is even so m^2 is even. After this I write $$m^2= 4k^2.$$ In the and I have this ecuation $$a^2 + b^2 = 4k^2/ n^2$$ This fraction is irreducible..I think</p>
| André Nicolas | 6,312 | <p><strong>Hint:</strong> The square of an odd number is of the shape $4k+1$, indeed $8k+1$. </p>
|
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