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 |
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1,520,643 | <p>If A² = I, prove that the matrix A is diagonalizable.</p>
<p>I have computed the eigenvalues to be 1 or -1 but I'm not sure how to proceed from here. </p>
<p>I'm thinking along the lines of "since rank(A + I) + rank(A - I) = n, therefore there exists n linearly independent vectors which corresponds to n eigenvectors. Hence, A is diagonalizable". Is that correct?</p>
<p>Thanks in advance!</p>
| Hirshy | 247,843 | <p>Hint: Let $\mu_A$ be the minimal polynomial of $A$. $A$ is diagonalizable iff $$\mu_A(x)=\prod\limits_{k=1}^m (x-x_k),m\leq n$$ with $x_i\neq x_j$ for $i\neq j$ being the eigenvalues of $A$.</p>
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2,484,004 | <p><span class="math-container">$M^*(a,b)=b-a$</span> we know that this fact but how we can prove closed intervals are Lebesgue measurable. I tried to prove by using <span class="math-container">$\cap ((a-\frac1n),(b+\frac1n))$</span> But ı totaly stucked :( please help me guys</p>
| jonsno | 310,635 | <p>We have the function:
$$f(x,y) = \left(x+\frac{3}{4}\right)^2 + y^2 + \frac{7}{16}$$
This is a (imaginary?) <em>circle</em> with centre $C(-\frac{3}{4}, 0)$. The value of $f(x,y)$ at $P(x,y)$ depends on distance of $P$ from $C$ ($d^2 + 7/16$, where $d$ is distance of $P$ from $C$).</p>
<p>The other equation is an <em>ellipse</em>:</p>
<p>$$\frac{x^2}{1/4} + y^2 = 1$$</p>
<p>Thus you have to find such point $P$ on ellipse whose distance from $C$ is maximum.</p>
<p><strong>For solving</strong>: Using your method, we have:</p>
<p>$$y^2 = 1-4x^2$$</p>
<p>Substitute in original equation and call this $g(x)$:</p>
<p>$$g(x) = 2-3x^2 + \frac{3}{2}x$$</p>
<p>$g(x)$ is maximum at $x = \frac{1}{4}$. This is easily verified from $g'(x)$.</p>
<p>Thus you get two points where $f(x,y)$ is maximum : $(\frac{1}{4}, \pm \frac{\sqrt{3}}{2})$</p>
|
3,134,991 | <p>If nine coins are tossed, what is the probability that the number of heads is even?</p>
<p>So there can either be 0 heads, 2 heads, 4 heads, 6 heads, or 8 heads.</p>
<p>We have <span class="math-container">$n = 9$</span> trials, find the probability of each <span class="math-container">$k$</span> for <span class="math-container">$k = 0, 2, 4, 6, 8$</span></p>
<p><span class="math-container">$n = 9, k = 0$</span></p>
<p><span class="math-container">$$\binom{9}{0}\bigg(\frac{1}{2}\bigg)^0\bigg(\frac{1}{2}\bigg)^{9}$$</span> </p>
<p><span class="math-container">$n = 9, k = 2$</span></p>
<p><span class="math-container">$$\binom{9}{2}\bigg(\frac{1}{2}\bigg)^2\bigg(\frac{1}{2}\bigg)^{7}$$</span> </p>
<p><span class="math-container">$n = 9, k = 4$</span>
<span class="math-container">$$\binom{9}{4}\bigg(\frac{1}{2}\bigg)^4\bigg(\frac{1}{2}\bigg)^{5}$$</span></p>
<p><span class="math-container">$n = 9, k = 6$</span></p>
<p><span class="math-container">$$\binom{9}{6}\bigg(\frac{1}{2}\bigg)^6\bigg(\frac{1}{2}\bigg)^{3}$$</span></p>
<p><span class="math-container">$n = 9, k = 8$</span></p>
<p><span class="math-container">$$\binom{9}{8}\bigg(\frac{1}{2}\bigg)^8\bigg(\frac{1}{2}\bigg)^{1}$$</span></p>
<p>Add all of these up: </p>
<p><span class="math-container">$$=.64$$</span> so there's a 64% chance of probability?</p>
| MCCCS | 357,924 | <p><span class="math-container">$$=\frac{\color{red}{\binom{9}{0}}+\color{blue}{\binom{9}{2}}+\color{orange}{\binom{9}{4}}+\color{green}{\binom{9}{6}}+\color{purple}{\binom{9}{8}}}{\color{red}{\binom{9}{0}}+\color{purple}{\binom{9}{1}}+\color{blue}{\binom{9}{2}}+\color{green}{\binom{9}{3}}+\color{orange}{\binom{9}{4}}+\color{orange}{\binom{9}{5}}+\color{green}{\binom{9}{6}}+\color{blue}{\binom{9}{7}}+\color{purple}{\binom{9}{8}}+\color{red}{\binom{9}{9}}}$$</span></p>
<p><span class="math-container">$$=\frac{\color{red}{\binom{9}{0}}+\color{blue}{\binom{9}{2}}+\color{orange}{\binom{9}{4}}+\color{green}{\binom{9}{3}}+\color{purple}{\binom{9}{1}}}{\color{red}{\binom{9}{0}}+\color{purple}{\binom{9}{1}}+\color{blue}{\binom{9}{2}}+\color{green}{\binom{9}{3}}+\color{orange}{\binom{9}{4}}+\color{orange}{\binom{9}{5}}+\color{green}{\binom{9}{6}}+\color{blue}{\binom{9}{7}}+\color{purple}{\binom{9}{8}}+\color{red}{\binom{9}{9}}}$$</span></p>
<p><span class="math-container">$$=\frac{\color{red}{\binom{9}{0}}+\color{purple}{\binom{9}{1}}+\color{blue}{\binom{9}{2}}+\color{green}{\binom{9}{3}}+\color{orange}{\binom{9}{4}}}{\color{red}{\binom{9}{0}}+\color{purple}{\binom{9}{1}}+\color{blue}{\binom{9}{2}}+\color{green}{\binom{9}{3}}+\color{orange}{\binom{9}{4}}+\color{orange}{\binom{9}{5}}+\color{green}{\binom{9}{6}}+\color{blue}{\binom{9}{7}}+\color{purple}{\binom{9}{8}}+\color{red}{\binom{9}{9}}}$$</span></p>
<p><span class="math-container">$$=\frac{\color{red}{\binom{9}{0}}+\color{purple}{\binom{9}{1}}+\color{blue}{\binom{9}{2}}+\color{green}{\binom{9}{3}}+\color{orange}{\binom{9}{4}}}{\color{red}{\binom{9}{0}}+\color{purple}{\binom{9}{1}}+\color{blue}{\binom{9}{2}}+\color{green}{\binom{9}{3}}+\color{orange}{\binom{9}{4}}+\color{orange}{\binom{9}{4}}+\color{green}{\binom{9}{3}}+\color{blue}{\binom{9}{2}}+\color{purple}{\binom{9}{1}}+\color{red}{\binom{9}{0}}}$$</span></p>
<p><span class="math-container">$$=\frac{a}{a+a}$$</span></p>
<p><span class="math-container">$$=\frac{1}{2}$$</span></p>
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3,134,991 | <p>If nine coins are tossed, what is the probability that the number of heads is even?</p>
<p>So there can either be 0 heads, 2 heads, 4 heads, 6 heads, or 8 heads.</p>
<p>We have <span class="math-container">$n = 9$</span> trials, find the probability of each <span class="math-container">$k$</span> for <span class="math-container">$k = 0, 2, 4, 6, 8$</span></p>
<p><span class="math-container">$n = 9, k = 0$</span></p>
<p><span class="math-container">$$\binom{9}{0}\bigg(\frac{1}{2}\bigg)^0\bigg(\frac{1}{2}\bigg)^{9}$$</span> </p>
<p><span class="math-container">$n = 9, k = 2$</span></p>
<p><span class="math-container">$$\binom{9}{2}\bigg(\frac{1}{2}\bigg)^2\bigg(\frac{1}{2}\bigg)^{7}$$</span> </p>
<p><span class="math-container">$n = 9, k = 4$</span>
<span class="math-container">$$\binom{9}{4}\bigg(\frac{1}{2}\bigg)^4\bigg(\frac{1}{2}\bigg)^{5}$$</span></p>
<p><span class="math-container">$n = 9, k = 6$</span></p>
<p><span class="math-container">$$\binom{9}{6}\bigg(\frac{1}{2}\bigg)^6\bigg(\frac{1}{2}\bigg)^{3}$$</span></p>
<p><span class="math-container">$n = 9, k = 8$</span></p>
<p><span class="math-container">$$\binom{9}{8}\bigg(\frac{1}{2}\bigg)^8\bigg(\frac{1}{2}\bigg)^{1}$$</span></p>
<p>Add all of these up: </p>
<p><span class="math-container">$$=.64$$</span> so there's a 64% chance of probability?</p>
| ThisIsNotAnId | 24,567 | <p>Here's an analytical answer with greater emphasis on reasoning than anything specific to probability which might lend greater insight into the problem.</p>
<p>Consider if there was only one coin. The probability to have an even number of heads is <span class="math-container">$1\over2$</span>, since there are two possible outcomes and we are only interested in one of them.</p>
<p>Now, let there be <span class="math-container">$N \gt 1$</span> coin tosses. The <span class="math-container">$N^{\text{th}}$</span> coin toss will either give a heads or tails. If the <span class="math-container">$N-1$</span> tosses resulted in an even number of heads the probability of the <span class="math-container">$N$</span> coin tosses resulting in an even number of heads is <span class="math-container">$1\over2$</span> since the <span class="math-container">$N^{\text{th}}$</span> coin toss will add either <span class="math-container">$0$</span> or <span class="math-container">$1$</span> to the number of heads from the <span class="math-container">$N-1$</span> tosses, and we are only concerned with the parity of the count. </p>
<p>The conclusion is the same for the <span class="math-container">$N-1$</span> coin tosses resulting in an odd number of heads.</p>
<p>This approach is valid as this reasoning applies to all possible values of <span class="math-container">$N$</span> in our given domain.</p>
<p><span class="math-container">$\therefore$</span> The probability of <span class="math-container">$N$</span> coin tosses resulting in an even number of heads is <span class="math-container">$1\over2$</span>, with <span class="math-container">$N \in \mathbb{N}$</span>.</p>
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2,406,061 | <p>I am also confused about whether these are symbols or have some meaning of their own.
PS- I know that <span class="math-container">$\operatorname{d}y\over\operatorname{d}x$</span> geometrically represents the slope. But, I've come across <span class="math-container">$\operatorname{d}x\over\operatorname{d}y$</span> to make problems easier. What does <span class="math-container">$\operatorname{d}x\over\operatorname{d}y$</span> mean?</p>
| Ravenex | 442,239 | <p>In the terms dx and dy, the d is for delta or "change in". So they represent the change in y and the change in x as a function, usually in terms of each other but sometimes another parameter. So dy/dx as you said is the slope, or change in x divided by the change in y, dy/dx is simply the inverse slope.</p>
<p>The different between dy and ∆y or dx and ∆x is that dy is a function that can be solved at any point to give the change in y at that point in relation to another variable, where as ∆y is a numerical value representing the difference in y between two points.</p>
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2,406,061 | <p>I am also confused about whether these are symbols or have some meaning of their own.
PS- I know that <span class="math-container">$\operatorname{d}y\over\operatorname{d}x$</span> geometrically represents the slope. But, I've come across <span class="math-container">$\operatorname{d}x\over\operatorname{d}y$</span> to make problems easier. What does <span class="math-container">$\operatorname{d}x\over\operatorname{d}y$</span> mean?</p>
| P. Siehr | 457,090 | <p>Let's first start with $Δx$: </p>
<p>If you have two real numbers $x_0, x_1$, lets say $x_1>x_0$ you can calculate the different $x_1-x_0$. We define that difference as $Δx$:
$$Δx:=x_1-x_0$$
With that difference we can also write:
$$x_1 = x_0 + Δx.$$
or in words: If we add some change $Δx$ to $x_0$ we get $x_0+Δx$. That sounds trivial, but well it's just the definition of "change".</p>
<hr>
<p>Well let us now look at functions. At first, let $f$ be a very simple function - a linear mapping:
$$f:ℝ→ℝ \qquad x↦f(x)=mx + c.$$
Sometimes I will use $y=f(x)$, to simplify the notation. In that definition $m$ is the slope of the linear function, and $c$ the shift on the $y$-axis, since $f(0)=c$. </p>
<p>Since we learned how to write the change in $x$, we can try to write the change in $f(x)$, too. So we have:
$$f(x+Δx) = m(x+Δx)+c = mx+mΔx+c.$$
Ok, that seems to be very boring.</p>
<p>But remember what we said above: Δx represents the change from $x_0$ to $x_0+Δx$. So if we now divide the change on the $y$-axis $Δy$ with the change on the $x$-axis we should get the something like the "rate of change" or the rate how "fast" $y$ changes relative to $x$. So let's do that:
\begin{align*}
\frac{Δy}{Δx}&=\frac{y_1-y_0}{Δx}=\frac{f(x_1) - f(x_0)}{Δx} = \frac{f(x_0+Δx) - f(x_0)}{Δx}\\ &=\frac{(mx+mΔx+c) - (mx+c)}{Δx}=\frac{mΔx}{Δx} = m.
\end{align*}
Pretty cool that we got the slope, as rate of change. Now we also see why that is the rate of change, if we divide by $Δx$:
$$Δy = mΔx$$
So $m$ is the factor that tells you how large the change $Δ$ of $y$ is, if $x$ changes by $Δx$.</p>
<p>This quotient is called difference quotient or ratio of the differences.</p>
<hr>
<p>Enough with these boring linear functions - we now look at arbitrary, but nice¹ functions:
$$f:ℝ→ℝ,\qquad x↦f(x).$$</p>
<p>We can still look at the difference quotient - why the heck not?
$$\frac{Δy}{Δx} = \frac{y_1-y_0}{(x_0+Δx) - x_0} = \frac{f(x_0+Δx) - f(x_0)}{(x_0+Δx) - x_0}$$
Uhm, well, now we are kind of stuck. But that is actually not that bad. Let's have a look at this picture from the German Wikipedia. Yes, I know it's German, but I think you will understand. Sekante = secant, Tangente = tanget, Funktionsgraph = graph, $x=x$ ...
<a href="https://i.stack.imgur.com/bUj7F.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bUj7F.png" alt="enter image description here"></a></p>
<p>You can see the two crosses marking $(x_0,f(x_0))$ and $(x_0+Δx, f(x_0+Δx))$. So, our difference quotient represents the secant of the function $f$. Cool, but what does that mean? </p>
<p>Here is an example: <br>
Today was really nice weather, and I made a trip by bike. Sometimes I was very slow, when there was a hill in my way. But on the other side I could ride very fast. <br>
Since I own a smartphone, I was able to track the distance I travelled, and omniscient Google could also show me the graph of the distance I travelled as a function $f(x)$ at time $x$. </p>
<p>I started in the morning at $x_0=10:30$ and arrived at $x_1=17:00 = x_0+6.5\text{h}$ at my destination. Google says I travelled 130km. So I can calculate:
$$\frac{Δy}{Δx} = \frac{130\text{km}}{6.5\text{h}} = 20 \frac{\text{km}}{\text{h}},$$
which is the <em>average travelling velocity</em>. And that is exactly what the slope of the secant represents. Just look at the picture above. (Well, I have to admit. If that red curve would represent the bike trip, it would be a very lame bike trip ...)</p>
<p>Now the average travelling speed is nice to know. But it would be very cool to know how fast my top speed was, don't you think? So now we are not interested in the average speed, but the speed in one point / current speed. Let's name it $f'(x_0)$. </p>
<p>Well, I don't know the definition of $f'$, yet. But if I calculate the average speed between two points $x_0$ and $x_1+Δx$ with a tiny, change $Δx$, doesn't that sound like a good approximation for "current speed"? </p>
<p>And since the average speed is the only thing that I know how to calculate, why not make $Δx$ smaller, and smaller, and smaller? <br> When we look at this process of making things "smaller and smaller" we call it a limit process, were we hope that at the end there is one value - the limit. If there is this limit, then it we use it to define "current speed":
$$f'(x_0):= \lim_{Δx→0}\frac{Δy}{Δx} = \lim_{Δx→0}\frac{f(x_0+Δx)-f(x_0)}{Δx}$$</p>
<p>For a second let's take a step back, and look at the picture above. Imagine how the secant looks, if we make $Δx$ smaller and smaller. It will eventually hit the tangent. (<a href="https://en.wikipedia.org/wiki/Derivative#/media/File:Derivative_GIF.gif" rel="nofollow noreferrer">Here</a> you can look at an animation of that limit process, with a different function [$h=Δx$].)</p>
<p>So we learned that the secant represents the "average speed" and the tangent represents the "current speed". Pretty cool, don't you think?</p>
<hr>
<p>Now what about this $\mathrm{d}x$?</p>
<p>Since mathematicians are usually very lazy² - I think everyone on this site will agree ;) - we like to invent new notations to write less.</p>
<p>So we write:
$$f'(x_0)=\frac{\mathrm{d}f(x)}{\mathrm{d}x}\big|_{x=x_0} = \frac{\mathrm{d}f(x_0)}{\mathrm{d}x} =\frac{\mathrm{d}y}{\mathrm{d}x}.$$</p>
<p>In that notation we hide that actually there is this limit process above. These $\mathrm{dx}$'s or $\mathrm{dy}$'s are also called infinitesimals (=infinitely small increments). And the guy who invented this notation is <a href="https://en.wikipedia.org/wiki/Leibniz%27s_notation" rel="nofollow noreferrer">Gottfried Leibniz</a> (not to be confused with the <a href="https://upload.wikimedia.org/wikipedia/commons/thumb/4/4c/Butterkeks.jpg/800px-Butterkeks.jpg" rel="nofollow noreferrer">cookie</a>). The invention of this notation is a really cool thing, that helped a lot in modern calculus, after it was improved by Weierstraß.</p>
<hr>
<p>Well almost done, but there is one part missing, that I did not answer so far: </p>
<blockquote>
<p>But, I've come across dx/dy to make problems easier. </p>
</blockquote>
<p>If you explain what you mean with that, I will gladly edit my answer. I just don't understand what you mean. Can you give an example of a problem that get's easier?</p>
<hr>
<p>¹Nice meaning, that you don't have any jumps or "corners", as in $f(x)=|x|$, in the function values.
²Actually physicists are even lazier when it comes to writing ...</p>
|
1,291,107 | <p>Let $X$ be random variable and $f$ it's density. How can one calculate $E(X\vert X<a)$?</p>
<p>From definition we have:</p>
<p>$$E(X\vert X<a)=\frac{E\left(X \mathbb{1}_{\{X<a\}}\right)}{P(X<a)}$$</p>
<p>Is this equal to:</p>
<p>$$\frac{\int_{\{X<a\}}xf(x)dx}{P(X<a)}$$</p>
<p>? If yes, then how one justify it? Thanks. I'm conditional expectation noob.</p>
<p>Also, what is $E(X|X=x_0)$? In discrete case it is $x_0$...</p>
| Renato Faraone | 217,700 | <p>I know that the problem has already been answered but I want to show you a more general method, let's suppose that you don't se how to rewrite the equation:</p>
<p>$-x^4+16x^3-90x^2+199x-124=0$</p>
<p>Or I prefer to write:</p>
<p>$x^4-16x^3+90x^2-199x+124=0$</p>
<p>You can use something called the Ruffini rule: search for integers divisor (both positive and negative) of the constant term and then set $x$ equals to the them and see if one of them is a solution. Starting from one we have:</p>
<p>$1-16+90-199+124=0$</p>
<p>So $x=1$ is a solution, now via Ruffini's rule, which can be seen <a href="http://en.wikipedia.org/wiki/Ruffini's_rule" rel="nofollow noreferrer">here</a>, we can rewrite the equation as:</p>
<p>$(x-1)(x^3-15x^2+75x-124)=0$</p>
<p>Now you have $3$ options to end this exercise:</p>
<blockquote>
<blockquote>
<p>$1.)$ Note that the second factor is a perfect cube;</p>
<p>$2.)$ Use Ruffini's rule again;</p>
<p>$3.)$ Use the general formula for third degree equation (which I'd not advise you to if your interested only in real solution).</p>
</blockquote>
</blockquote>
<p>This is a more general method so I hope this will help you in the future!</p>
|
4,437,921 | <p>Given <span class="math-container">$X$</span>~<span class="math-container">$N(0,\sigma^2_X)$</span> and <span class="math-container">$Y$</span>~<span class="math-container">$N(0,\sigma^2_Y)$</span> (<span class="math-container">$X$</span> and <span class="math-container">$Y$</span> are normally distributed random variables with expectations given by <span class="math-container">$\mu_X=\mu_Y=0$</span> and variances given by <span class="math-container">$\sigma^2_X$</span> and <span class="math-container">$\sigma^2_Y)$</span>, can anyone help me figure out the expectation and variance of <span class="math-container">$Z=\sin(X)\cos(Y)$</span>?</p>
<p>NOTE: <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> are independent. I have found some threads that purport the following:</p>
<p><span class="math-container">$E[\cos(X)]=e^{-\sigma^2_X/2}$</span></p>
<p><span class="math-container">$E[\sin(X)]=0$</span></p>
<p>but i still have trouble using this info to deduce <span class="math-container">$Z=\sin(X)\cos(Y)$</span></p>
<p>I just realized the answer is zero. Ayyyyye</p>
<h2><em><strong>EDIT:</strong></em></h2>
<p>I am also curious to know...</p>
<p><span class="math-container">$E[\sin X \cos X]$</span></p>
<p>where <span class="math-container">$X$</span>~<span class="math-container">$N(0,\sigma)$</span>. Here, the domain of <span class="math-container">$X$</span> is any real number, but any such number actually corresponds to an angle in radians.</p>
| RyRy the Fly Guy | 412,727 | <p>In response to the edit, note that <span class="math-container">$Z = \sin X \cos X = g(X)$</span> is a function <span class="math-container">$g$</span> of the normal random variable <span class="math-container">$X$</span>~<span class="math-container">$N(0,\sigma)$</span> whose pdf is given by <span class="math-container">$f(x)=\frac{1}{\sigma\sqrt{2\pi}}e^{\frac{-1}{2} x^2\sigma^{-2}}$</span>. By definition of the expectation of a continuous random variable, we have</p>
<p><span class="math-container">$\mathbb{E} \big[ Z ] = \mathbb{E} \big[ \sin X \cos X ] = \mathbb{E} \big[ g(X) ] = {\huge\int}_0^{2\pi} g(x) f(x) dx = {\huge\int}_0^{2\pi} \sin x \cos x f(x) dx$</span></p>
<p>Let us parse the integral as follows</p>
<p><span class="math-container">$= {\huge\int}_0^{\pi} \sin x \cos x f(x) dx + {\huge\int}_{\pi}^{2\pi} \sin x \cos x f(x) dx $</span></p>
<p>Now consider that <span class="math-container">$\sin x = \sin (x-2\pi)$</span> and <span class="math-container">$\cos x = \cos (x-2\pi)$</span>. Hence, we may rewrite the second integral as follows</p>
<p><span class="math-container">$= {\huge\int}_0^{\pi} \sin x \cos x f(x) dx + {\huge\int}_{\pi}^{2\pi} \sin (x-2\pi) \cos (x-2\pi) f(x) dx $</span></p>
<p>Now, we shift the limits of the second integral using this helpul identity <span class="math-container">$\int_a^b h(x) dx = \int_{a+c}^{b+c} h(x-c) dx = \int_{a-c}^{b-c} h(x+c) dx$</span> so that we have</p>
<p><span class="math-container">$= {\huge\int}_0^{\pi} \sin x \cos x f(x) dx + {\huge\int}_{\pi-2\pi}^{2\pi-2\pi} \sin (x-2\pi+2\pi) \cos (x-2\pi+2\pi) f(x+2\pi) dx $</span></p>
<p><span class="math-container">$= {\huge\int}_0^{\pi} \sin x \cos x f(x) dx + {\huge\int}_{-\pi}^{0} \sin x \cos x f(x+2\pi) dx $</span></p>
<p><span class="math-container">$= {\huge\int}_0^{\pi} \sin x \cos x f(x) dx + {\huge\int}_{0}^{\pi} \sin (-x) \cos (-x) f(-x-2\pi) dx $</span></p>
<p><span class="math-container">$= {\huge\int}_0^{\pi} \sin x \cos x f(x) dx + {\huge\int}_{0}^{\pi} \sin (-x) \cos x f(-x-2\pi) dx $</span></p>
<p><span class="math-container">$= {\huge\int}_0^{\pi} \sin x \cos x f(x) dx - {\huge\int}_{0}^{\pi} \sin x \cos x f(-x-2\pi) dx $</span></p>
<p>Note that <span class="math-container">$f(-x-2\pi) = \frac{1}{\sigma\sqrt{2\pi}} e^{\frac{-1}{2} (-x-2\pi)^2\sigma^{-2}} = \frac{1}{\sigma\sqrt{2\pi}} e^{\frac{-1}{2} \big[-1(x+2\pi)\big]^2\sigma^{-2}} = \frac{1}{\sigma\sqrt{2\pi}} e^{\frac{-1}{2} (x+2\pi)^2\sigma^{-2}} = f(x + 2\pi)$</span>, so we have</p>
<p><span class="math-container">$= {\huge\int}_0^{\pi} \sin x \cos x f(x) dx - {\huge\int}_{0}^{\pi} \sin x \cos x f(x+2\pi) dx $</span></p>
<p>In applications where the pdf <span class="math-container">$f$</span> describes a <a href="https://en.wikipedia.org/wiki/Wrapped_distribution" rel="nofollow noreferrer">wrapped probability distribution</a>, that is, a distribution corresponding to data values that lie along a unit circle, then we have <span class="math-container">$f(y) = f(y \mod 2\pi)$</span>. If we let <span class="math-container">$y = x + 2\pi$</span>, then <span class="math-container">$y \mod 2\pi = (x + 2\pi) \mod 2\pi = x$</span> where <span class="math-container">$x \in [0,\pi]$</span>. Thus, <span class="math-container">$f(y) = f(x + 2\pi) = f(y \mod 2\pi) = f(x)$</span> and the result is</p>
<p><span class="math-container">$= {\huge\int}_0^{\pi} \sin x \cos x f(x) dx - {\huge\int}_{0}^{\pi} \sin x \cos x f(x) dx = 0$</span></p>
<p>So in summary, <span class="math-container">$\mathbb{E} \big[ \sin X \cos X ] = 0$</span></p>
|
135,675 | <p>Let $D$ be the Dirac-Operator on $\mathbb{R}^n$ or more generally the Dirac spinor bundle $\mathcal{S}\to M$ of a (semi-)Riemannian spin manifold $M$. Then we consider $D$ as an unbouded Operator on $\mathcal{H}=L^2(\mathbb{R}^n)$ with domain $C^\infty_c(\mathbb{R}^n,\mathbb{C}^N)$. Then it is said that the operator $f\langle D\rangle^{-n}$ is compact, where $f\in C^\infty_c(\mathbb{R}^n,\mathbb{C})$ is considered as a multiplication operator on $\mathcal{H}$ and $\langle D\rangle:=\sqrt{D^\dagger D+ DD^\dagger}$.</p>
<p>Since I am not really a crack in functional analysis, it is not even obvious for my how exactly $\langle D\rangle$ works. I suspect that the Operator $D^\dagger D+DD^\dagger$ is (essentially) self-adjoint and then the spectral theorem is used for defining $\langle D\rangle$ and its powers $\langle D\rangle^{-n}$.</p>
<p>But what is even more mysterious to me is the claim that $f\langle D\rangle^{-n}$ is actually compact (note that $f$ has compact support, however). Why is this true?</p>
| paul garrett | 15,629 | <p>I think it is useful to ask the simpler question, why $f\cdot (1-\Delta)^{-1}$ is compact, on $\mathbb R^n$, when $f$ is a test function. Part of the point is that $\Delta$ itself (nevermind the Dirac operator) does <em>not</em> have compact resolvent on $\mathbb R^n$, essentially because Fourier inversion shows that the spectrum is purely continuous. An even nicer case of positive outcome is the Schrodinger Hamiltonian $-\Delta+|x|^2$ on $\mathbb R^n$, which provably has discrete spectrum (without looking at specific formulaic aspects), for general geometric reasons. </p>
|
70,803 | <p>Let $S$ be the sphere in $\mathbb{R}^3$ and $C:[0,1]\to S$ a continuously differentiable curve on $S$. Let $T:[0,1]\to\mathbb{R}^3$ denote the tangent vector of $C$. Let $P(t)$ be the plane containing $C(t)$ and having normal vector $T(t)$.</p>
<p>Given a size $d$ of the "paint brush" we define the "brush" $b:[0,1]\to \mathcal{P}(S)$ by letting $b(t)$ be the points of $S$ that are at most a distance $d$ (metric on the sphere) from $C(t)$ that are contained in $P(t)$.</p>
<p>We can think of this as saying the "brush" $b(t)$ is an arc on the sphere that is "orthogonal" to the motion $C(t)$ of the "paint brush".</p>
<p>Given $d$ what is the arclength of the shortest curves such that $\cup_{t\in[0,1]} b(t) = S$. This says that the "paint brush" covered the sphere.</p>
| Joseph O'Rourke | 6,094 | <p>The model is that used by Henryk Gerlach and Heiko von der Mosel in their 2010 paper "On sphere-filling ropes" <a href="http://arxiv.org/abs/1005.4609" rel="noreferrer">arXiv:1005.4609v1 (math.GT)</a> may be relevant.
Their question is different: What is the longest rope of a given thickness on a sphere?
But their explicit solutions are packings, and it seems they could be converted to
painting paths.
Here is a piece of their Fig. 6:
<br />
<img src="https://i.stack.imgur.com/Fem4H.jpg" alt="Rope on sphere"></p>
|
991,878 | <p>How can it be proven that a cycle of length k is an even permutation if and only if k is odd?
I know it can be done using the fact that a permutation which exchanges two elements but leaves the rest unchanged is an odd permutation.</p>
| Paul | 17,980 | <p>Hints:</p>
<p>You should discuss that $\log a> 0$, $\log a<0$ and $\log a=0$ .</p>
|
991,878 | <p>How can it be proven that a cycle of length k is an even permutation if and only if k is odd?
I know it can be done using the fact that a permutation which exchanges two elements but leaves the rest unchanged is an odd permutation.</p>
| David K | 139,123 | <p>To follow up on Macavity's excellent answer, here's a proof that $x = 1$ is
the <em>only</em> value of $x$ that satisfies
$x \log a \leq a - 1$ for all positive real values of $a.$</p>
<p>Let $x \log a \leq a - 1.$
For $a > 1,$ we have $\log a > 0,$ and therefore
$$x \leq \frac{a - 1}{\log a}.$$
Now consider what happens for values of $a$ greater than but very close to $1.$
We have
$$\lim_{a\to 1^+} \frac{a - 1}{\log a} = \lim_{a\to 1^+} \frac{1}{1/a} = 1.$$
If we set $x$ to any value greater than $1,$ we will then have
$x > \frac{a - 1}{\log a}$ and therefore $x \log a > a - 1$ for some values of $a.$</p>
<p>For $a < 1,$ we have $\log a < 0,$ and therefore
$$x \geq \frac{a - 1}{\log a}.$$
We have
$$\lim_{a\to 1^+} \frac{a - 1}{\log a} = \lim_{a\to 1^+} \frac{1}{1/a} = 1.$$
So if $x < 1$ we will then have
$x < \frac{a - 1}{\log a}$ and therefore $x \log a > a - 1$ for some values of $a.$
The value $x = 1$ is the only remaining value that satisfies
$x \log a \leq a - 1$ for all positive real values of $a.$</p>
|
2,002,201 | <p>simplify <span class="math-container">$\sqrt{(45+4\sqrt{41})^3}-\sqrt{(45-4\sqrt{41})^3}$</span>.</p>
<blockquote>
<p>1.<span class="math-container">$90^{\frac{3}{2}}$</span></p>
<p>2.<span class="math-container">$106\sqrt{41}$</span></p>
<p>3.<span class="math-container">$4\sqrt{41}$</span></p>
<p>4.<span class="math-container">$504$</span></p>
<p>5.<span class="math-container">$508$</span></p>
</blockquote>
<p><strong>My attempt</strong>:I do like this but I didn't get any of those five.</p>
<p><span class="math-container">$\sqrt{(45+4\sqrt{41})^3}-\sqrt{(45-4\sqrt{41})^3}={\sqrt{45+4\sqrt{41}}}^3-\sqrt{45-4\sqrt{41}}^3$</span></p>
<p><span class="math-container">$=(\sqrt{45+4\sqrt{41}}-\sqrt{45-4\sqrt{41}})(45+4\sqrt{41}+45-4\sqrt{41}+(\sqrt{45+4\sqrt{41}})(\sqrt{45-4\sqrt{41}})$</span></p>
<p>Now I do the nested radicals formula and I get <span class="math-container">$254\sqrt{41}$</span> which is none of those where did I mistaked?</p>
| fleablood | 280,126 | <p>Is this the nested radical formula?</p>
<p>$\sqrt{45 \pm 4\sqrt{41}} = a \pm b\sqrt{41}$</p>
<p>$45 \pm 4\sqrt{41} = (a^2 + 41b^2) \pm 2ab\sqrt{41}$</p>
<p>$a^2 + 41b^ = 45; 2ab = 4 \implies a=2;b = 1$</p>
<p>So $\sqrt{45 \pm 4\sqrt{41}} = |2 \pm \sqrt{41}|= \pm 2 + \sqrt{41}$</p>
<p>Plugging that into: $=(\sqrt{45+4\sqrt{41}}-\sqrt{45-4\sqrt{41}})(45+4\sqrt{41}+45-4\sqrt{41}+(\sqrt{45+4\sqrt{41}})(\sqrt{45-4\sqrt{41}}))$</p>
<p>We get $(2+\sqrt{41} - (-2+\sqrt{41}))(90+(2 + \sqrt{41})(-2+\sqrt{41}))=$</p>
<p>$4(90 + (-4 + 41)) = 508$</p>
<p>===</p>
<p>It would have been easier not to do all the expanding:</p>
<p>$(2 + \sqrt{41})^3 - (-2+\sqrt{41})^3=$</p>
<p>$(2^3 + 3*2^2*\sqrt{41} + 3*2*\sqrt{41}^2+\sqrt{41}^3)- (-2^3 + 3*2^2*\sqrt{41} - 3*2*\sqrt{41}^2+\sqrt{41}^3)=$</p>
<p>$2(8 + 6*41) = 508$.</p>
|
4,510,384 | <p>In exercise 2.13 of page 43 of the book <a href="https://rads.stackoverflow.com/amzn/click/com/0134746759" rel="nofollow noreferrer" rel="nofollow noreferrer">Mathematical Proofs: A Transition to Advanced Mathematics</a> the reader is asked to state the logical negation of some statements. Of these, I find the authors' answer to one of them baffling.</p>
<p>The statement to negate is:</p>
<blockquote>
<p>"Two sides of the triangle have the same length."</p>
</blockquote>
<p>The authors' negation of the statement is:</p>
<blockquote>
<p>"The sides of the triangle have different lengths".</p>
</blockquote>
<p>Am I mistaken in assuming that when negating a statement, one is supposed to state what previously presumed false as true and vice versa? If one assumes the proposition "Two sides of the triangle have the same length." to be true, is it erroneous to conclude that the negation would be "The sides of the triangle have different lengths" or (exclusively) 'Three sides of the triangle have the same length'? I thank your aid in advance.</p>
| Kuhlambo | 220,642 | <p>I hope it's possible to be logically consistent and easy to understand, I'll try to be both.</p>
<p>I think both versions of the sentence are equivalent, and I'll try to prove it, or at least to convince you.</p>
<p><strong>Hypothesis:</strong>
The sides are of different length.
<span class="math-container">$ \Leftrightarrow$</span> No two sides are of equal length.</p>
<p><strong>Proof:</strong> let <span class="math-container">$a,b,c$</span> be the lengths of the sides.</p>
<p><em>Now let all three numbers be different:</em>
<span class="math-container">$\Leftrightarrow a\neq b \land a\neq c \land c \neq b $</span> that is obvious right? This obviously implies that no two sides should be equal.</p>
<p><em>Now we look at the ops idea and say</em>: Let no two of the numbers be equal.</p>
<p>This is not as intuitive I think, but I hope that it will become clear.</p>
<p>We start by saying <span class="math-container">$ a\neq b $</span> so far so good but what about <span class="math-container">$c$</span>? It can't be equal to <span class="math-container">$a$</span> because then two numbers would be equal, so we need to add that with a logical and because the inequalies must both be fulfilled, one or the other would not be enough: <span class="math-container">$\ a\neq b \land a\neq c$</span>.
Now it is getting more obvious right? <span class="math-container">$c$</span> also can't equal <span class="math-container">$b$</span> and again we need that to be true at the same time.</p>
<p>So the result is: let no two of them be equal <span class="math-container">$\Rightarrow a\neq b \land a\neq c \land c \neq b $</span>.</p>
<p><strong>Concluding, that no two sides being equal implies all sides being different and vice versa.</strong></p>
<p>We could have started with any two of the sides and would have had to add the other inequalies.</p>
|
4,510,384 | <p>In exercise 2.13 of page 43 of the book <a href="https://rads.stackoverflow.com/amzn/click/com/0134746759" rel="nofollow noreferrer" rel="nofollow noreferrer">Mathematical Proofs: A Transition to Advanced Mathematics</a> the reader is asked to state the logical negation of some statements. Of these, I find the authors' answer to one of them baffling.</p>
<p>The statement to negate is:</p>
<blockquote>
<p>"Two sides of the triangle have the same length."</p>
</blockquote>
<p>The authors' negation of the statement is:</p>
<blockquote>
<p>"The sides of the triangle have different lengths".</p>
</blockquote>
<p>Am I mistaken in assuming that when negating a statement, one is supposed to state what previously presumed false as true and vice versa? If one assumes the proposition "Two sides of the triangle have the same length." to be true, is it erroneous to conclude that the negation would be "The sides of the triangle have different lengths" or (exclusively) 'Three sides of the triangle have the same length'? I thank your aid in advance.</p>
| ryang | 21,813 | <blockquote>
<p>"Two sides of the triangle have the same length."</p>
</blockquote>
<p>In other words, the triangle is isosceles (possibly equilateral).</p>
<p>So, the statement's <strong>negation</strong> is</p>
<ul>
<li><strong>Each side of the triangle has a distinct length.</strong></li>
<li><strong>Each side of the triangle has a different length</strong> (from the other two)<strong>.</strong></li>
<li><strong>The triangle's sides have different lengths from one another.</strong></li>
</ul>
<p>A formalisation of the given sentence: <span class="math-container">$$\exists(x,y)\;(Lxy\land x\ne y);$$</span> its negation: <span class="math-container">$$\forall (x,y)\;(\lnot Lxy\lor x=y).$$</span></p>
<blockquote>
<p>the negation would be "The sides of the triangle have different lengths"</p>
</blockquote>
<p>Correct.</p>
<blockquote>
<p>or (exclusively) 'Three sides of the triangle have the same length'?</p>
</blockquote>
<p>Incorrect. Every equilateral triangle is automatically isosceles; that is, if a triangle is equilateral, then it <em><strong>does</strong></em> have two sides with a common length, which is <strong>not</strong> saying that it has <em><strong>exactly two</strong></em> sides with a common length.</p>
<p>After all, if I I have exactly 10 fingers, then</p>
<ul>
<li>“I have 2 fingers” is absolutely true (albeit misleading);</li>
<li>“I have at least 2 fingers” is also true;</li>
<li>“I have exactly 2 fingers” is false.</li>
</ul>
<hr />
<p><strong>Addendum</strong></p>
<p>Rosie F:</p>
<blockquote>
<p>@fleablood "we should... teach the students that statements only imply what they explicitly state." That principle seems reasonable. So how is it consistent with your contention that the statement means "at least two", and that to interpret it as "exactly two" is wrong? The statement explicitly stated "two", and did not state "at least" (or any other modifier). It seems that students need to learn both to make and to understand the difference between the statements.</p>
</blockquote>
<p>I'm not fleablood, but the point is that ‘at least two’ is the most conservative interpretation: notice that ‘exactly two’ means ‘at least two <em>and</em> at most two’, which contains an unwarranted assumption that restricts the possibilities.</p>
|
3,357,697 | <p>I'm doing exercise II.4.5 in textbook Analysis I by Amann.</p>
<p>Could you please verify if my attempt contains logical mistakes/gaps! Thank you so much!</p>
<blockquote>
<p><a href="https://i.stack.imgur.com/usrr7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/usrr7.png" alt="enter image description here"></a></p>
</blockquote>
<p><strong>My attempt:</strong></p>
<blockquote>
<p><strong>Lemma:</strong> <span class="math-container">$f:X \to Y$</span> is continuous iff <span class="math-container">$f[\overline{A}] \subseteq \overline{f[A]}$</span> for all <span class="math-container">$A \subseteq X$</span>.</p>
</blockquote>
<p>Assume that <span class="math-container">$f : \overline{A} \rightarrow\{0,1\}$</span> is continuous, then the restriction <span class="math-container">$f \restriction A : A \rightarrow\{0,1\}$</span> is continuous because <span class="math-container">$A \subseteq \overline{A}$</span>. Because <span class="math-container">$A$</span> is connected, then <span class="math-container">$f \restriction A$</span> is not surjective. WLOG, we assume <span class="math-container">$f[A] = \{0\}$</span>. Since <span class="math-container">$\{0\}$</span> is both closed and open in <span class="math-container">$\{0,1\}$</span>, <span class="math-container">$\overline{f[A]} = \overline{\{0\}} = \{0\}$</span>. By <strong>Lemma</strong>, <span class="math-container">$f[\overline{A}] \subseteq \overline{f[A]}= \{0\}$</span>, so <span class="math-container">$f$</span> is not surjective. Hence <span class="math-container">$\overline{A}$</span> is connected.</p>
| Badam Baplan | 164,860 | <p>McCoy's theorem extends as corollary to multivariate polynomials with very little work.</p>
<p>In <span class="math-container">$R[x, y] = R[x][y]$</span> a zero divisor <span class="math-container">$f$</span> is annihilated by an element <span class="math-container">$g\in R[x]$</span>, using McCoy's theorem.</p>
<p>Writing <span class="math-container">$f = \sum f_i y^i$</span> with <span class="math-container">$f_i \in R[x]$</span>, we see by comparing coefficients that <span class="math-container">$f_i g = 0$</span>.</p>
<p>Then let <span class="math-container">$f' = \sum f_i x^{n_i} \in R[x]$</span> be a polynomial where the <span class="math-container">$n_i \in \mathbb{N}$</span> are chosen large enough so that the <span class="math-container">$f_i$</span> do not 'interact.'</p>
<p>Clearly <span class="math-container">$gf' = 0$</span>, and McCoy's theorem now shows that <span class="math-container">$r f' = 0$</span> for some <span class="math-container">$r \in R$</span>. We deduce that <span class="math-container">$r f_i = 0$</span> for all <span class="math-container">$i$</span>, so also <span class="math-container">$rf = 0$</span>.</p>
|
3,357,697 | <p>I'm doing exercise II.4.5 in textbook Analysis I by Amann.</p>
<p>Could you please verify if my attempt contains logical mistakes/gaps! Thank you so much!</p>
<blockquote>
<p><a href="https://i.stack.imgur.com/usrr7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/usrr7.png" alt="enter image description here"></a></p>
</blockquote>
<p><strong>My attempt:</strong></p>
<blockquote>
<p><strong>Lemma:</strong> <span class="math-container">$f:X \to Y$</span> is continuous iff <span class="math-container">$f[\overline{A}] \subseteq \overline{f[A]}$</span> for all <span class="math-container">$A \subseteq X$</span>.</p>
</blockquote>
<p>Assume that <span class="math-container">$f : \overline{A} \rightarrow\{0,1\}$</span> is continuous, then the restriction <span class="math-container">$f \restriction A : A \rightarrow\{0,1\}$</span> is continuous because <span class="math-container">$A \subseteq \overline{A}$</span>. Because <span class="math-container">$A$</span> is connected, then <span class="math-container">$f \restriction A$</span> is not surjective. WLOG, we assume <span class="math-container">$f[A] = \{0\}$</span>. Since <span class="math-container">$\{0\}$</span> is both closed and open in <span class="math-container">$\{0,1\}$</span>, <span class="math-container">$\overline{f[A]} = \overline{\{0\}} = \{0\}$</span>. By <strong>Lemma</strong>, <span class="math-container">$f[\overline{A}] \subseteq \overline{f[A]}= \{0\}$</span>, so <span class="math-container">$f$</span> is not surjective. Hence <span class="math-container">$\overline{A}$</span> is connected.</p>
| Wuestenfux | 417,848 | <p>Well, use the lexicographic ordering on the polynomial ring <span class="math-container">$R=K[x_1,\ldots,x_n]$</span>.</p>
<p>Given two polynomials <span class="math-container">$f,g\ne 0$</span> in <span class="math-container">$R$</span>.
Suppose <span class="math-container">$x^\alpha$</span> and <span class="math-container">$x^\beta$</span> are the largest monomials involved in <span class="math-container">$f$</span> and <span class="math-container">$g$</span> w.r.t. the lex ordering, respectively.</p>
<p>Then the largest monomial involved in <span class="math-container">$fg$</span> w.r.t. the lex ordering is <span class="math-container">$x^{\alpha+\beta}$</span>, which shows that <span class="math-container">$fg\ne 0$</span>. </p>
<p>This is a direct extension of the case of one variable.</p>
|
451,722 | <p>I want to find the projection of the point $M(10,-12,12)$ on the plane $2x-3y+4z-17=0$. The normal of the plane is $N(2,-3,4)$.</p>
<p>Do I need to use Gram–Schmidt process? If yes, is this the right formula?</p>
<p>$$\frac{N\cdot M}{|N\cdot N|} \cdot N$$</p>
<p>What will the result be, vector or scalar?</p>
<p>Thanks!</p>
| Tony Piccolo | 71,180 | <p>You can use calculus to minimize the distance (easier: the square of the distance) of M from the generic point of the plane.<br>
Use the equation of the plane to drop a variable, obtaining a function of two independent variables: compute the partial derivatives and find the stationary point. That's all.</p>
|
75,880 | <p>Say $f:X\rightarrow Y$ and $g:Y\rightarrow X$ are functions where $g\circ f:X\rightarrow X$ is the identity. Which of $f$ and $g$ is onto, and which is one-to-one?</p>
| Martin Sleziak | 8,297 | <p>You already have several answers which can help you remember the theorem. If you're looking for a proof (and have problems with showing it yourself), you might try to have a look at these links:</p>
<ul>
<li><a href="http://www.proofwiki.org/wiki/Injection_if_Composite_is_an_Injection" rel="nofollow noreferrer">http://www.proofwiki.org/wiki/Injection_if_Composite_is_an_Injection</a> </li>
<li><a href="http://www.proofwiki.org/wiki/Surjection_if_Composite_is_a_Surjection" rel="nofollow noreferrer">http://www.proofwiki.org/wiki/Surjection_if_Composite_is_a_Surjection</a></li>
</ul>
<p>or these questions/answers:</p>
<ul>
<li><a href="https://math.stackexchange.com/questions/28123/if-gfx-is-one-to-one-injective-show-fx-is-also-one-to-one-given-that">If g(f(x)) is one-to-one (injective) show f(x) is also one-to-one (given that...)</a></li>
<li><a href="https://math.stackexchange.com/questions/5627/surjection-on-composed-function">Surjection on composed function?</a></li>
<li><a href="https://math.stackexchange.com/questions/22572/injective-and-surjective-functions">Injective and Surjective Functions</a></li>
</ul>
<p>More general results are here:</p>
<ul>
<li><a href="http://www.proofwiki.org/wiki/Injection_iff_Left_Inverse" rel="nofollow noreferrer">http://www.proofwiki.org/wiki/Injection_iff_Left_Inverse</a></li>
<li><a href="http://www.proofwiki.org/wiki/Surjection_iff_Right_Inverse" rel="nofollow noreferrer">http://www.proofwiki.org/wiki/Surjection_iff_Right_Inverse</a></li>
</ul>
|
1,146,759 | <p>A covering of a group $G$ a family $\{S_i\}_{i \in I}$ of subsets of $G$ such that $G = \displaystyle \bigcup _{i \in I} S_i$.</p>
<p>Why is true that: A group covered by finitely many cyclic subgroups is either cyclic or finite?</p>
<p>Remark:
Is true that by Baer (see D. Robinson, Finiteness Conditions and Generalized Soluble Groups, p.105) that:
Theorem 4.16: A group is central-by-finite if and only if it has a finite covering consisting of abelian subgroups?</p>
| Hagen von Eitzen | 39,174 | <p>Some partial elementary results:</p>
<p>Assume $G$ is infinite and has a finite covering by finitely many cyclic subgroups $S_i=\langle s_i\rangle$, $i\in I$. We have to show that $G$ is cyclic. </p>
<p>If $S_i\subseteq S_j$ for some $i\ne j$, we may drop $S_i$. Hence we may assume wlog. that $S_i\not\subseteq S_j$ for $i\ne j$.
Consequently, none of the $S_i$ is properly contained in a larger cyclic subgroup.
Any conjugate of an $S_i$ is contained in some $S_{i'}$ and by the previous remark must in fact equal $S_{i'}$.
Thus $G$ acts on the set $\{\,S_i:i\in I\,\}$ by conjugation.</p>
<p>If all $S_i$ are finite then so is their union, hence some of the $S_i$ must be
infinite.
Hence there exist subsets $J\subseteq I$ such that $\bigcap_{j\in J}S_j$ is nontrivial and each $S_j$, $j\in J$, is infinite. Let $J$ be a maximal such subset and let the corresponding intersection be $\langle a\rangle$.</p>
<p><strong>Claim.</strong> $\bigcup_{j\in J}S_j=G$.</p>
<p><em>Proof.</em>
Assume $g\notin \bigcup_{j\in J}S_j$. Then the infinitely many $a^kg$ are not in any $S_j$ with $j\in J$. By pigeon-hole we find distinct $a^kg, a^lg$ in the same $S_i$, $i\notin J$, and then also $a^{k-l}\in S_i$. This contradicts maximality of $J$. $_\square$</p>
<p>So wlog. $I=J$.
Since $a$ commutes with each $s_i$, the center has finite index in $G$, and each $S_I$ has finite index.</p>
|
1,598,545 | <p>Maybe I am not well versed with the actual definition of mean, but I have a doubt. On most resources, people say that arithmetic mean is the sum of $n$ observations divided by n. So my first question: </p>
<blockquote>
<p>How does this formula work? Is there any derivation to it? If not,
then while creating this definition, what was the creator thinking?</p>
</blockquote>
<p>Okay, so using my intuition, I thought that it is the value that lies in the centre. And it worked for some cases, like the mean of $1$ , $2$ and $3$ is $2$ , which is the central value. But, lets imagine a number line from numbers $0$ to $9$. Now, I choose $3$ numbers, say $1$, $8$ and $9$. By the formula, I get the mean is equal to $6$. But, if mean really is a central value, shouldn't it be $5$(I know we call $5$ the median in this case)? But it seems like the mean is getting closer to $8$ and $9$, which means it is not central? So my final question?</p>
<blockquote>
<p>Have I imagined mean incorrectly? What kind of central value really
mean is?</p>
</blockquote>
| Kamil Jarosz | 183,840 | <p>You are talking about median, given some sequence of <strong>ordered</strong> numbers </p>
<p>$$1,1,2,2,3,3,\color{red}{3},6,6,7,8,9,9\tag{1}$$</p>
<p>The number at the center is called a median. When there are even number of elements,</p>
<p>$$1,1,2,2,3,\color{red}{3},\color{red}{6},6,7,8,9,9$$</p>
<p>Median is defined as a mean of those two numbers, namely $4.5$.</p>
<p>Different thing goes on with arithmetic mean. This is the sum of elements divided by number of them. Arithmetic mean of $(1)$ is
$$\frac{1+1+2+2+3+3+3+6+6+7+8+9+9}{13}=\frac{60}{13}\approx 4.6153$$</p>
<p>which clearly differ from its median. The median cares only about numbers that are in half of the sequence, arithmetic mean cares about all numbers.</p>
|
3,845,475 | <p>Here's what I'm tasked with showing:</p>
<p>Let <span class="math-container">$(a_n)$</span> be a convergent sequence with <span class="math-container">$a_n\rightarrow a$</span> as <span class="math-container">$n\rightarrow\infty$</span>. By the Algebraic Limit Theorem, we know that <span class="math-container">$(a_n^2)\rightarrow a^2$</span>. Now prove this using the definition of convergence.</p>
<p>In doing so, I have the following:</p>
<p>Let <span class="math-container">$\epsilon>0$</span> be given. Choose some <span class="math-container">$N\in\mathbb{N}$</span> so that for all <span class="math-container">$n\geq N$</span>, <span class="math-container">$|a_n^2-a^2|<\epsilon$</span>. By algebra, <span class="math-container">$|a_n^2-a^2|=|a_n-a||a_n+a|$</span>. Consider <span class="math-container">$|a_n+a|$</span>. By the triangle inequality, <span class="math-container">$|a_n+a|\leq|a_n|+|a|$</span>, thus <span class="math-container">$|a_n|$</span> is bounded by some <span class="math-container">$M\in\mathbb{N}$</span>.</p>
<p>I know I'm trying to choose <span class="math-container">$M$</span> so that <span class="math-container">$|a_n-a|<\epsilon+M+|a|$</span>. Where should I take it from here?</p>
| Dixon | 1,072,039 | <p>Given that the first ace is the 20th card to appear:
This suggests no Ace was chosen in the first 19 card flips. No other information was given, so every other card is subject to random choice in the first 19 card flips. On the 20th flip we are told an Ace is flipped. This gives the equation for P(A)</p>
<p><span class="math-container">$P(A) = \binom{48}{19} * \binom{4}{1}$</span></p>
<p>On the 21st flip we are told an Ace of Spades is flipped. This means there are only 3 Aces to be randomly selected on the 20th flip. On the 21st flip there are 32 cards to chose from and we need the probability that the Ace of Spades is chosen.</p>
<p><span class="math-container">$P(BA) = \binom{48}{19} * \binom{3}{1} * \dfrac1{32}$</span></p>
<p>So the final equation is</p>
<p><span class="math-container">$P(B \mid A) = \dfrac{\binom{48}{19} * \binom{3}{1} * \dfrac1{32}}{\binom{48}{19} * \binom{4}{1}}$</span></p>
<p>On the 21st flip we are told the Two of Clubs is flipped. This means in addition to the 4 Aces, the Two of Clubs is not chosen in the first 19 flips. This leaves 47 cards to be chosen at random in the first 19 flips. On the 20th flip one of four Aces are chosen. On the 21st flip the Two of Clubs is chosen.</p>
<p><span class="math-container">$P(CA) = \binom{47}{19} * \binom{4}{1} * \dfrac1{32}$</span></p>
<p>So</p>
<p><span class="math-container">$P(C \mid A) = \dfrac{\binom{47}{19} * \binom{4}{1} * \dfrac1{32}}{\binom{48}{19} * \binom{4}{1}}$</span></p>
|
1,477,058 | <p>Let $\textbf{$\gamma$}: \mathbb{R} \rightarrow \mathbb{R}^n$ be a smooth curve and let $T \in \mathbb{R}$. We say that $\textbf{$\gamma$}$ is $T$-periodic if $$\textbf{$\gamma$}(t+T)=\textbf{$\gamma$}(t) \text{ for all } t \in \mathbb{R}.$$
If $\textbf{$\gamma$}$ is not constant and is $T$-periodic for some $T\neq 0$, then $\textbf{$\gamma$}$ is said to be closed. </p>
<p>The period of a closed curve $\textbf{$\gamma$}$ is the smallest positive number $T$ such that $\textbf{$\gamma$}$ is $T$-periodic. </p>
<p>A curve $\textbf{$\gamma$}$ is said to have a self-intersection at a point $p$ of the curve if there exist parameter values $a\neq b$ such that </p>
<ul>
<li>$\textbf{$\gamma$}(a) = \textbf{$\gamma$}(b) = p$, and </li>
<li>if $\textbf{$\gamma$}$ is closed with period $T$, then $a-b$ is not an integer multiple of $T$. </li>
</ul>
<p>$$$$ </p>
<hr>
<p>$$$$ </p>
<p>I am looking at the following exercise: </p>
<p>Show that the Cayley sextic
$$\textbf{$\gamma$}(t) = \left (\cos^3t \cos {3t}, \cos^3t \sin {3t}\right ), t \in \mathbb{R}$$
is a closed curve which has exactly one self-intersection. What is its period? </p>
<p>$$$$ </p>
<p>I have done the following: </p>
<p>Since $\textbf{$\gamma$} (0)=(1,0)$ and $\textbf{$\gamma$}\left (\frac{\pi}{6}\right )=\left (0, \frac{\sqrt{3}}{2}\right )$, i.e., $\textbf{$\gamma$} (0)\neq \textbf{$\gamma$} \left (\frac{\pi}{6}\right )$ we have that $\textbf{$\gamma$}$ is not constant. </p>
<p>We have that $$\cos (t+2\pi )=\cos t , \ \ \cos (3(t+2\pi ))=\cos (3t) , \ \ \sin (3(t+2\pi ))=\sin (3t)$$
so then $$\textbf{$\gamma$}(t+2\pi ) = \left ((\cos (t+2\pi ))^3 \cos {[3(t+2\pi )]}, (\cos (t+2\pi ))^3 \sin {[3(t+2\pi )]}\right )=\left (\cos^3t \cos {3t}, \cos^3t \sin {3t}\right )=\textbf{$\gamma$}(t), \forall t$$ </p>
<p>That means that the non-constant curve $\textbf{$\gamma$}$ is $2\pi$-periodic. </p>
<p>So, $\textbf{$\gamma$}$ is a closed curve. </p>
<p>Is this correct? </p>
<p>Is $2\pi$ the period? Is this the smallest positive number $T$ such that $\textbf{$\gamma$}$ is $T$-periodic? Or does the period change when we have $\cos 3t$ and $\sin 3t$ instead of $\cos t$ and $\sin t$ ? </p>
<p>$$$$ </p>
<p><strong>EDIT:</strong> </p>
<p>We have that $$\textbf{$\gamma$}\left (\frac{\pi}{3}\right )=\left (-\frac{1}{8}, 0 \right ) \\ \textbf{$\gamma$}\left (\frac{2\pi}{3}\right )=\left (-\frac{1}{8}, 0\right )$$ </p>
<p>So, when we take $p=\left (-\frac{1}{8}, 0\right )$ we have that $\textbf{$\gamma$}\left (\frac{\pi}{3}\right )=\textbf{$\gamma$}\left (\frac{2\pi}{3}\right )=\left (-\frac{1}{8}, 0\right )$ </p>
<p>Since the period is $\pi$, we have that $\frac{2\pi}{3}-\frac{\pi}{3}=\frac{\pi}{3}=\frac{1}{3}(\pi)$, i.e., $\frac{2\pi}{3}-\frac{\pi}{3}$ is not an integer multiple of the period. </p>
<p>That means that $\gamma$ has a self-intersection at $p=\left (-\frac{1}{8}, 0\right )$. </p>
<p>Now we want to show that this self-intersection is unique. </p>
<p>To do that we want to find all points $x,y$ within one period of $[0,\pi]$ such that $\gamma(x)=\gamma(y)$. That is, find all solutions of the simultaneous system
$$
\begin{align*}
\cos^3(x) \cos(3x)&=\cos^3(y) \cos(3y) \\
\cos^3(x) \sin(3x)&=\cos^3(y) \sin(3y).
\end{align*}
$$</p>
<p>$$\frac{\cos^3(x) \cos(3x)}{\cos^3(x) \sin(3x)}=\frac{\cos^3(y) \cos(3y)}{\cos^3(y) \sin(3y)} \Rightarrow \frac{ \cos(3x)}{\sin(3x)}=\frac{ \cos(3y)}{ \sin(3y)} \\ \Rightarrow \cot (3x)=\cot (3y) \Rightarrow 3x=3y+k\pi \Rightarrow x=y+\frac{k}{3}\pi$$ </p>
<p>Since the period is $\pi$, the only possible values for $k$ are $k=0$, $k=1$ and $k=2$. </p>
<p>For $k=0$ we have $3x=3y \Rightarrow x=y$. But for $x=y$ the second condition of the definition of a self-intersection is not satisfied. </p>
<p>So, for $k=1,2$ we have that $3x=3y+k\pi \Rightarrow x=y+\frac{k\pi}{3}$. </p>
<p>The first condition of the definition is satisfied with each of these values. </p>
<p>As for the second definition we want that $x-y=\frac{k\pi}{3}$ is not an integer multiple of $\pi$, which is also satisfied with each of these values for $k$. </p>
<p>So we want somehow to show that the points that we get this relation are $\frac{\pi}{3}$ and $\frac{2\pi}{3}$, or not? </p>
<p>But how? I don't have any idea...</p>
| A.Γ. | 253,273 | <p>The period $2\pi$ is not right. Let's complexify $\gamma(t)=(x(t),y(t))$ as $x(t)+iy(t)$:
$$
\gamma=\cos^3(t)e^{i3t}=(\cos(t)e^{it})^3=\Bigl(\frac{e^{it}+e^{-it}}{2}e^{it}\Bigr)^3=\Bigl(\frac{z^2+1}{2}\Bigr)^3
$$
where $z=e^{it}$. The curve depends only on $z^2$ which is $\pi$ periodic, hence, the period is at most $\pi$. It is easy to see that it is actually $\pi$ since $\zeta=\frac{z^2+1}{2}$ runs along the small circle with the center at $1/2$ and radius $1/2$, so to get the initial value $\zeta^3=\gamma=1$ again it needs to make the whole loop which corresponds to $t=\pi$.</p>
<p>Self-intersection is guessed right. It can also be concluded without finding it explicitly as the function $\zeta^3$ is not univalent function inside the small circle mentioned above.</p>
|
1,477,058 | <p>Let $\textbf{$\gamma$}: \mathbb{R} \rightarrow \mathbb{R}^n$ be a smooth curve and let $T \in \mathbb{R}$. We say that $\textbf{$\gamma$}$ is $T$-periodic if $$\textbf{$\gamma$}(t+T)=\textbf{$\gamma$}(t) \text{ for all } t \in \mathbb{R}.$$
If $\textbf{$\gamma$}$ is not constant and is $T$-periodic for some $T\neq 0$, then $\textbf{$\gamma$}$ is said to be closed. </p>
<p>The period of a closed curve $\textbf{$\gamma$}$ is the smallest positive number $T$ such that $\textbf{$\gamma$}$ is $T$-periodic. </p>
<p>A curve $\textbf{$\gamma$}$ is said to have a self-intersection at a point $p$ of the curve if there exist parameter values $a\neq b$ such that </p>
<ul>
<li>$\textbf{$\gamma$}(a) = \textbf{$\gamma$}(b) = p$, and </li>
<li>if $\textbf{$\gamma$}$ is closed with period $T$, then $a-b$ is not an integer multiple of $T$. </li>
</ul>
<p>$$$$ </p>
<hr>
<p>$$$$ </p>
<p>I am looking at the following exercise: </p>
<p>Show that the Cayley sextic
$$\textbf{$\gamma$}(t) = \left (\cos^3t \cos {3t}, \cos^3t \sin {3t}\right ), t \in \mathbb{R}$$
is a closed curve which has exactly one self-intersection. What is its period? </p>
<p>$$$$ </p>
<p>I have done the following: </p>
<p>Since $\textbf{$\gamma$} (0)=(1,0)$ and $\textbf{$\gamma$}\left (\frac{\pi}{6}\right )=\left (0, \frac{\sqrt{3}}{2}\right )$, i.e., $\textbf{$\gamma$} (0)\neq \textbf{$\gamma$} \left (\frac{\pi}{6}\right )$ we have that $\textbf{$\gamma$}$ is not constant. </p>
<p>We have that $$\cos (t+2\pi )=\cos t , \ \ \cos (3(t+2\pi ))=\cos (3t) , \ \ \sin (3(t+2\pi ))=\sin (3t)$$
so then $$\textbf{$\gamma$}(t+2\pi ) = \left ((\cos (t+2\pi ))^3 \cos {[3(t+2\pi )]}, (\cos (t+2\pi ))^3 \sin {[3(t+2\pi )]}\right )=\left (\cos^3t \cos {3t}, \cos^3t \sin {3t}\right )=\textbf{$\gamma$}(t), \forall t$$ </p>
<p>That means that the non-constant curve $\textbf{$\gamma$}$ is $2\pi$-periodic. </p>
<p>So, $\textbf{$\gamma$}$ is a closed curve. </p>
<p>Is this correct? </p>
<p>Is $2\pi$ the period? Is this the smallest positive number $T$ such that $\textbf{$\gamma$}$ is $T$-periodic? Or does the period change when we have $\cos 3t$ and $\sin 3t$ instead of $\cos t$ and $\sin t$ ? </p>
<p>$$$$ </p>
<p><strong>EDIT:</strong> </p>
<p>We have that $$\textbf{$\gamma$}\left (\frac{\pi}{3}\right )=\left (-\frac{1}{8}, 0 \right ) \\ \textbf{$\gamma$}\left (\frac{2\pi}{3}\right )=\left (-\frac{1}{8}, 0\right )$$ </p>
<p>So, when we take $p=\left (-\frac{1}{8}, 0\right )$ we have that $\textbf{$\gamma$}\left (\frac{\pi}{3}\right )=\textbf{$\gamma$}\left (\frac{2\pi}{3}\right )=\left (-\frac{1}{8}, 0\right )$ </p>
<p>Since the period is $\pi$, we have that $\frac{2\pi}{3}-\frac{\pi}{3}=\frac{\pi}{3}=\frac{1}{3}(\pi)$, i.e., $\frac{2\pi}{3}-\frac{\pi}{3}$ is not an integer multiple of the period. </p>
<p>That means that $\gamma$ has a self-intersection at $p=\left (-\frac{1}{8}, 0\right )$. </p>
<p>Now we want to show that this self-intersection is unique. </p>
<p>To do that we want to find all points $x,y$ within one period of $[0,\pi]$ such that $\gamma(x)=\gamma(y)$. That is, find all solutions of the simultaneous system
$$
\begin{align*}
\cos^3(x) \cos(3x)&=\cos^3(y) \cos(3y) \\
\cos^3(x) \sin(3x)&=\cos^3(y) \sin(3y).
\end{align*}
$$</p>
<p>$$\frac{\cos^3(x) \cos(3x)}{\cos^3(x) \sin(3x)}=\frac{\cos^3(y) \cos(3y)}{\cos^3(y) \sin(3y)} \Rightarrow \frac{ \cos(3x)}{\sin(3x)}=\frac{ \cos(3y)}{ \sin(3y)} \\ \Rightarrow \cot (3x)=\cot (3y) \Rightarrow 3x=3y+k\pi \Rightarrow x=y+\frac{k}{3}\pi$$ </p>
<p>Since the period is $\pi$, the only possible values for $k$ are $k=0$, $k=1$ and $k=2$. </p>
<p>For $k=0$ we have $3x=3y \Rightarrow x=y$. But for $x=y$ the second condition of the definition of a self-intersection is not satisfied. </p>
<p>So, for $k=1,2$ we have that $3x=3y+k\pi \Rightarrow x=y+\frac{k\pi}{3}$. </p>
<p>The first condition of the definition is satisfied with each of these values. </p>
<p>As for the second definition we want that $x-y=\frac{k\pi}{3}$ is not an integer multiple of $\pi$, which is also satisfied with each of these values for $k$. </p>
<p>So we want somehow to show that the points that we get this relation are $\frac{\pi}{3}$ and $\frac{2\pi}{3}$, or not? </p>
<p>But how? I don't have any idea...</p>
| Christian Blatter | 1,303 | <p>We had the same question a few days ago (<a href="https://math.stackexchange.com/questions/1478030/find-the-self-interection-differential-geometry/1478134#1478134">Find the self-interection.Differential Geometry</a>). I'm extending my accepted answer there to a full solution.</p>
<p>Write your curve in the form
$$\gamma: \quad t\mapsto z(t):=x(t)+iy(t)={1\over8}(e^{it}+e^{-it})^3\>e^{3it}\ .$$
Substituting $t:={u\over2}$ we obtain, after a simple calculation, the parametrization
$$\gamma:\quad s\mapsto{1\over8}\bigl(1+e^{iu}\bigr)^3 \qquad(u\in{\mathbb R})\ .$$
This already shows that in terms of the original variable $t$ the period is at most $\pi$. </p>
<p>In order to find possible double points (or even a smaller period) we have to investigate in detail for which pairs $(u,v)$ with $$0\leq u<v<2\pi\tag{1}$$ the equation
$$\bigl(1+e^{iu}\bigr)^3=\bigl(1+e^{iv}\bigr)^3$$
is true. This equation is satisfied iff $1+e^{iu}$ and $1+e^{iv}$ coincide up to a third root of unity. The case $1+e^{iu}=1+e^{iv}$ gives no solution within the bounds $(1)$. Therefore it remains to discuss the cases
$${\rm(a)}\quad 1+e^{iu}=\omega\bigl(1+e^{iv}\bigr),\qquad{\rm(b)}\quad 1+e^{iu}=\bar\omega\bigl(1+e^{iv}\bigr)\ ,$$
where $\omega:=e^{2\pi i/3}$, which is the same as
$${\rm(a)}\quad \cos{u\over2}=\omega e^{i(v-u)/2}\>\cos{v\over2},\qquad {\rm(b)}\quad \cos{u\over2}=\bar\omega e^{i(v-u)/2}\>\cos{v\over2}\ .\tag{2}$$
In case (a) we want $\omega e^{i(v-u)/2}\in{\mathbb R}$ within the bounds $(1)$. This can only be done with ${v-u\over2}={\pi\over3}$, and $(2)$(a) then implies $\cos{u\over2}=-\cos{v\over2}$. The pair $u={2\pi\over3}$, $v={4\pi\over3}$ is the only one that satisfies these conditions, and gives rise to a double point $z_*$ of our curve, namely
$$z_*={1\over8}\bigl(1+e^{2\pi i/3}\bigr)^3=-{1\over8}\ .$$</p>
<p>In case (b) we want $\bar \omega e^{i(v-u)/2}\in{\mathbb R}$ within the bounds $(1)$. This can only be done with ${v-u\over2}={2\pi\over3}$, and $(2)$(b) then implies $\cos{u\over2}=\cos{v\over2}$. The latter equation is not satisfiable within the bounds $(1)$.</p>
|
1,477,058 | <p>Let $\textbf{$\gamma$}: \mathbb{R} \rightarrow \mathbb{R}^n$ be a smooth curve and let $T \in \mathbb{R}$. We say that $\textbf{$\gamma$}$ is $T$-periodic if $$\textbf{$\gamma$}(t+T)=\textbf{$\gamma$}(t) \text{ for all } t \in \mathbb{R}.$$
If $\textbf{$\gamma$}$ is not constant and is $T$-periodic for some $T\neq 0$, then $\textbf{$\gamma$}$ is said to be closed. </p>
<p>The period of a closed curve $\textbf{$\gamma$}$ is the smallest positive number $T$ such that $\textbf{$\gamma$}$ is $T$-periodic. </p>
<p>A curve $\textbf{$\gamma$}$ is said to have a self-intersection at a point $p$ of the curve if there exist parameter values $a\neq b$ such that </p>
<ul>
<li>$\textbf{$\gamma$}(a) = \textbf{$\gamma$}(b) = p$, and </li>
<li>if $\textbf{$\gamma$}$ is closed with period $T$, then $a-b$ is not an integer multiple of $T$. </li>
</ul>
<p>$$$$ </p>
<hr>
<p>$$$$ </p>
<p>I am looking at the following exercise: </p>
<p>Show that the Cayley sextic
$$\textbf{$\gamma$}(t) = \left (\cos^3t \cos {3t}, \cos^3t \sin {3t}\right ), t \in \mathbb{R}$$
is a closed curve which has exactly one self-intersection. What is its period? </p>
<p>$$$$ </p>
<p>I have done the following: </p>
<p>Since $\textbf{$\gamma$} (0)=(1,0)$ and $\textbf{$\gamma$}\left (\frac{\pi}{6}\right )=\left (0, \frac{\sqrt{3}}{2}\right )$, i.e., $\textbf{$\gamma$} (0)\neq \textbf{$\gamma$} \left (\frac{\pi}{6}\right )$ we have that $\textbf{$\gamma$}$ is not constant. </p>
<p>We have that $$\cos (t+2\pi )=\cos t , \ \ \cos (3(t+2\pi ))=\cos (3t) , \ \ \sin (3(t+2\pi ))=\sin (3t)$$
so then $$\textbf{$\gamma$}(t+2\pi ) = \left ((\cos (t+2\pi ))^3 \cos {[3(t+2\pi )]}, (\cos (t+2\pi ))^3 \sin {[3(t+2\pi )]}\right )=\left (\cos^3t \cos {3t}, \cos^3t \sin {3t}\right )=\textbf{$\gamma$}(t), \forall t$$ </p>
<p>That means that the non-constant curve $\textbf{$\gamma$}$ is $2\pi$-periodic. </p>
<p>So, $\textbf{$\gamma$}$ is a closed curve. </p>
<p>Is this correct? </p>
<p>Is $2\pi$ the period? Is this the smallest positive number $T$ such that $\textbf{$\gamma$}$ is $T$-periodic? Or does the period change when we have $\cos 3t$ and $\sin 3t$ instead of $\cos t$ and $\sin t$ ? </p>
<p>$$$$ </p>
<p><strong>EDIT:</strong> </p>
<p>We have that $$\textbf{$\gamma$}\left (\frac{\pi}{3}\right )=\left (-\frac{1}{8}, 0 \right ) \\ \textbf{$\gamma$}\left (\frac{2\pi}{3}\right )=\left (-\frac{1}{8}, 0\right )$$ </p>
<p>So, when we take $p=\left (-\frac{1}{8}, 0\right )$ we have that $\textbf{$\gamma$}\left (\frac{\pi}{3}\right )=\textbf{$\gamma$}\left (\frac{2\pi}{3}\right )=\left (-\frac{1}{8}, 0\right )$ </p>
<p>Since the period is $\pi$, we have that $\frac{2\pi}{3}-\frac{\pi}{3}=\frac{\pi}{3}=\frac{1}{3}(\pi)$, i.e., $\frac{2\pi}{3}-\frac{\pi}{3}$ is not an integer multiple of the period. </p>
<p>That means that $\gamma$ has a self-intersection at $p=\left (-\frac{1}{8}, 0\right )$. </p>
<p>Now we want to show that this self-intersection is unique. </p>
<p>To do that we want to find all points $x,y$ within one period of $[0,\pi]$ such that $\gamma(x)=\gamma(y)$. That is, find all solutions of the simultaneous system
$$
\begin{align*}
\cos^3(x) \cos(3x)&=\cos^3(y) \cos(3y) \\
\cos^3(x) \sin(3x)&=\cos^3(y) \sin(3y).
\end{align*}
$$</p>
<p>$$\frac{\cos^3(x) \cos(3x)}{\cos^3(x) \sin(3x)}=\frac{\cos^3(y) \cos(3y)}{\cos^3(y) \sin(3y)} \Rightarrow \frac{ \cos(3x)}{\sin(3x)}=\frac{ \cos(3y)}{ \sin(3y)} \\ \Rightarrow \cot (3x)=\cot (3y) \Rightarrow 3x=3y+k\pi \Rightarrow x=y+\frac{k}{3}\pi$$ </p>
<p>Since the period is $\pi$, the only possible values for $k$ are $k=0$, $k=1$ and $k=2$. </p>
<p>For $k=0$ we have $3x=3y \Rightarrow x=y$. But for $x=y$ the second condition of the definition of a self-intersection is not satisfied. </p>
<p>So, for $k=1,2$ we have that $3x=3y+k\pi \Rightarrow x=y+\frac{k\pi}{3}$. </p>
<p>The first condition of the definition is satisfied with each of these values. </p>
<p>As for the second definition we want that $x-y=\frac{k\pi}{3}$ is not an integer multiple of $\pi$, which is also satisfied with each of these values for $k$. </p>
<p>So we want somehow to show that the points that we get this relation are $\frac{\pi}{3}$ and $\frac{2\pi}{3}$, or not? </p>
<p>But how? I don't have any idea...</p>
| amcalde | 168,694 | <p>Just using trig:</p>
<p>$T = \pi$</p>
<p>see this by computing
$\gamma(0) = (1,0) = \gamma(T)$</p>
<p>Solve this using the second component first so that $$0 = \textrm{sin}(3T)\textrm{cos}^3(T)$$
Then $T = n \pi/3$, Do the same thing for the first component and you must have $n$ is a multiple of 3. In fact this is sufficient for all $t$, $\gamma(t+T) = \gamma(t)$.</p>
<p>The norm of the vector is $|\gamma(t)| = \cos^6(t)$ so to have an intersection we must have $\cos^6(t) = \cos^6(s)$ or in other words $s = \pi - t$ in this interval of $[0,\pi]$ This condition for the intersection is satisfied by the first component easily but leads to the following for the second component:
$$\cos^3(t) \sin (3t) = -\cos^3(t) \sin (3t)$$ or in other words $$0 =\cos^3(t) \sin (3t) $$
The only possible intersections are then at $t = {0, \pi/3, \pi/2, 2\pi/3}$
Let $t<s$ from above then you must have either $t =0$ or $t = \pi/3$. But $t=0$ leads to a degenerate case so $t = \pi/3$ and $s = 2\pi/3$. Which you can check with substitution directly.</p>
<p>So
$\gamma(\pi/3) = \gamma(2\pi/3)$ is the unique intersection.</p>
|
320,557 | <p>let $S$ be the set of pairs $(x,y)$ where x,y are orthogonal unit vectors in $\mathbb R^3$. i am trying to show this is a topological manifold. for starters one needs to define a suitable topology on it. i was thinking let a set $U$ be open in $S$ iff $U \cap S^2$ (intersection with sphere) is open in $S^2$ in the subspace topology? am i going in the right direction? I'd really appreicate some help. thank you </p>
<p>this is really about topological manifold, sorry for not stating it earlier. </p>
<p>here is the definition:</p>
<blockquote>
<p>a manifold is a second countable Hausdorff space that is locally
Euclidean.</p>
</blockquote>
| Albert | 19,331 | <p>It is a smooth manifold of dimension 3.
Indeed : let $F : S^2 \times S^2 \rightarrow \mathbb{R}$ be defined by $F(x,y)=\sum_i x_i y_i$ (the standard inner product). $F$ is a smooth map on the smooth manifold $S^2 \times S^2$ and its derivative does not vanish so it is a submersion. Bu the submersion property, the set of its zeroes is a manifold of dimension $2*2-1=3$.</p>
|
18,960 | <p>I am making this post in regards to the ongoing delete/undelete skirmish (let's at least change the monotonicity of the use of "war"). The old version of the question is <a href="https://math.stackexchange.com/revisions/172652/3">here</a>, the current version (after edits today) <a href="https://math.stackexchange.com/q/172652/11619">here</a>, and the answer <a href="https://math.stackexchange.com/questions/172652/conics-passing-through-integer-lattice-points/172683#172683">here</a> </p>
<p>There are two facts:</p>
<ol>
<li>The original post is of poor quality: it is a problem statement of various simultaneous problems, which shows no effort from the OP.</li>
<li>Nonetheless, the post has received a very good answer, which is worth keeping (in my personal opinion) for the sake of future users.</li>
</ol>
<p>As a consequence, the moderator team decided to undelete the question some days ago. The question was again deleted by two of the three users that had deleted it before. Since the question is already closed, I find no reason to delete it, given it signifies the deletion of a useful answer. </p>
<p>The mod team (more precisely, Pedro and Jyrki) have undeleted, locked, edited, unlocked and reopened the question today.</p>
| Thomas | 26,188 | <p>I might take a controversial stand on this one. This is just my opinion.</p>
<p>The original question was a statement only question and should have been closed and deleted despite the existence of a good answer. As a principle I think it is wrong for anyone (even a moderator) to radically change the question without the approval of the OP. The only exception is if the OP included further detail in comments.</p>
<p>I think the better way to have dealt with this one would have been for a moderator to have posted the new question and to have strongly encouraged André Nicolas to post his answer on the new question. If not this, then I would at least seek approval from André Nicolas to move his answer to the new question (if this is even possible).</p>
<p>I don't believe it is up to the moderators to improve questions as was done in this case. </p>
<p>Granted, I don't think that the OP has been around for some time, and it might be unlikely that they would see communication from a moderator seeking approval to change the question. But, as a principle, I do think that it is important not to radically change someone's post without approval. Now the OP's name is attached to a post that he didn't make.</p>
|
4,580,470 | <p>Suppose <span class="math-container">$X$</span> is a Geometric random variable (with parameter <span class="math-container">$p$</span> and range <span class="math-container">$\{k\geq 1\}$</span>).</p>
<p>Let <span class="math-container">$M$</span> be a positive integer.</p>
<p>Let <span class="math-container">$Z:=\min\{X,M\}$</span>. We want to calculate the expectation <span class="math-container">$\mathbb E[Z]$</span>.</p>
<hr />
<p>My professor solved the problem by starting with <span class="math-container">$$\mathbb E[Z]=\sum_{k=1}^\infty \mathbb P[Z\geq k].$$</span></p>
<p>I can understand all other steps in his method except this starting step. Why shouldn't we have <span class="math-container">$\mathbb E[Z]=\sum_{k=1}^\infty k\mathbb P[Z= k]$</span> by the definition of expectation instead?</p>
<p>I am really confused by this starting step. Any help with understanding the step <span class="math-container">$\mathbb E[Z]=\sum_{k=1}^\infty \mathbb P[Z\geq k]$</span> will be really appreciated. Thanks!</p>
| Siong Thye Goh | 306,553 | <p>The equation holds for nonnegative random variable.</p>
<p><span class="math-container">\begin{align}
E[Z]&=\sum_{k=1}^\infty kP[Z=k]\\
&=P(Z=1)\\
&+P(Z=2) + P(Z=2)\\
&+P(Z=3) + P(Z=3) + P(Z=3)+\ldots\\
\vdots\\
&=P(Z\ge 1) + P(Z\ge 2)+P(Z\ge 3)+\ldots\\
&=\sum_{k=1}^\infty P(Z \ge k)
\end{align}</span></p>
<p>where the third equation is due to we sum the terms columnwise.</p>
|
2,547,933 | <p>Consider the integral $I=\displaystyle\int_{R}\int f(x,y)dx dy$ over the region $R$, given by the triangle with vertices $(0,0),(1,1)$ and $(2,0)$. </p>
<p>This is an isosceles triangle with one side lying along the $x-$axis. So, our domain is not "nice" to find the bounds for integral I assume, since even if we write $0\le x \le 2$, we can not give bounds for $y$ easily. To find this integral, the book I am reading makes the following transformation: $u = y-x$, $v=y+x$. After that, our new domain becomes a right angled triangle with the perpendicular edges lying on the $u$ and $v$ axis.</p>
<p>Finally, my question is how can we conclude this transformations? In general, setting up $u = x+y, v= x-y$ works quite nice for triangular/rectangular domains but is there a rule for this?</p>
<p>Thank you. </p>
| Mark Bennet | 2,906 | <p>Expand both using the binomial theorem</p>
<p>$$\left(1+\frac 1n\right)^n=1^n+n\cdot\frac 1n\cdot1^{n-1}+\binom n2\left(\frac 1n\right)^21^{n-2}+\binom n3\left(\frac 1n\right)^31^{n-3}+\dots =$$$$=1+1+\frac {n(n-1)}{2n^2}+\frac {n(n-1)(n-2)}{6n^3}+\dots$$while $$\left(1-\frac 1n\right)^n=1-1+\frac {n(n-1)}{2n^2}-\frac {n(n-1)(n-2)}{6n^3}+\dots$$</p>
<p>Subtract the second from the first and you get $$2+\frac {n(n-1)(n-2)}{3n^3}+\dots$$ a sum of positive terms. So the difference between the two expressions is at least $2$. If the limits exist (they do), the difference between the limits must be at least $2$.</p>
<p>If you analyse the difference you will note that it is increasing with $n$ (but bounded) so even though the $(1\pm \frac 1n)$ part tends to zero, raising to the $n^{th}$ power in this case gives expressions which get further apart rather than closer together.</p>
|
920,782 | <p>How do I find the number of integral solutions to the equation - </p>
<p>$$2x_1 + 2x_2 + \cdots + 2x_6 + x_7 = N$$</p>
<p>$$x_1,x_2,\ldots,x_7 \ge 1$$</p>
<p>I just thought that I should reduce this a bit more, so I replace $x_i$ with $(y_i+1)$, so we have:</p>
<p>$$y_1 + y_2 + \cdots + y_6 = \tfrac{1}{2}(N + 13 - y_7)$$</p>
<p>$$y_1,y_2,\ldots,y_7 \ge 0$$</p>
<p>I will be solving this as a programming problem by looping over $y_7$ from $[0, N+13]$. How do I find the number of solutions to this equation in each looping step?</p>
| Felix Marin | 85,343 | <p>$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
\newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
\newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
\newcommand{\dd}{{\rm d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
\newcommand{\fermi}{\,{\rm f}}
\newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
\newcommand{\half}{{1 \over 2}}
\newcommand{\ic}{{\rm i}}
\newcommand{\iff}{\Longleftrightarrow}
\newcommand{\imp}{\Longrightarrow}
\newcommand{\pars}[1]{\left(\, #1 \,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\pp}{{\cal P}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
\newcommand{\sech}{\,{\rm sech}}
\newcommand{\sgn}{\,{\rm sgn}}
\newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
With $\ds{0 < a < 1}$:</p>
<blockquote>
<p>\begin{align}
&\color{#66f}{\large\sum_{x_{1} = 0}^{\infty}\ldots\sum_{x_{7} = 0}^{\infty}
\delta_{2x_{1} + \cdots + 2x_{6} + x_{7},N}}
=\sum_{x_{1} = 0}^{\infty}\ldots\sum_{x_{7} = 0}^{\infty}
\oint_{\verts{z}\ =\ a}
{1 \over z^{-2x_{1}\ -\ \cdots\ -\ 2x_{6}\ -\ x_{7}\ +\ N\ +\ 1}}
\,{\dd z \over 2\pi\ic}
\\[3mm]&=\oint_{\verts{z}\ =\ a}{1 \over z^{N + 1}}
\pars{\sum_{x = 0}^{\infty}z^{2x}}^{6}\sum_{y = 0}^{\infty}z^{y}
\,{\dd z \over 2\pi\ic}
=\oint_{\verts{z}\ =\ a}{1 \over z^{N + 1}}\,{1 \over \pars{1 - z^{2}}^{6}}\,
{1 \over 1 - z}\,{\dd z \over 2\pi\ic}
\\[3mm]&=\oint_{\verts{z}\ =\ a}{1 \over z^{N + 1}}
\sum_{n = 0}^{\infty}{-6 \choose n}\pars{-1}^{n}z^{2n}
\sum_{k = 0}^{\infty}z^{k}\,{\dd z \over 2\pi\ic}
\\[3mm]&=\sum_{n = 0}^{\infty}\sum_{k = 0}^{\infty}\pars{-1}^{n}
{6 + n - 1 \choose n}\pars{-1}^{n}\oint_{\verts{z}\ =\ a}
{1 \over z^{N - 2n - k+ 1}}\,{\dd z \over 2\pi\ic}
\\[3mm]&=\sum_{n = 0}^{\infty}\sum_{k = 0}^{\infty}
{n + 5 \choose n}\delta_{k,N - 2n}
=\left.\sum_{n = 0}^{\infty}{n + 5 \choose 5}\right\vert_{\,N\ -\ 2n\ \geq\ 0}
=\sum_{n = 0}^{\floor{N/2}}{n + 5 \choose 5}
\\[3mm]&=\sum_{n = 0}^{\floor{N/2}}\oint_{\verts{z}\ =\ 1}
{\pars{1 + z}^{n + 5} \over z^{6}}\,{\dd z \over 2\pi\ic}
=\oint_{\verts{z}\ =\ 1}
{\pars{1 + z}^{5} \over z^{6}}\sum_{n = 0}^{\floor{N/2}}\pars{1 + z}^{n}
\,{\dd z \over 2\pi\ic}
\\[3mm]&=\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{5} \over z^{6}}
{\pars{1 + z}^{\floor{N/2} + 1} - 1 \over \pars{1 + z} - 1}\,{\dd z \over 2\pi\ic}
\\[3mm]&=\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{\floor{N/2} + 6} - 1 \over z^{7}}
\,{\dd z \over 2\pi\ic}
-\oint_{\verts{z}\ =\ 1}{\pars{1 + z}^{5} \over z^{7}}
\,{\dd z \over 2\pi\ic}
=\color{#00f}{\large{\floor{N/2} + 6 \choose 6}} - {5 \choose 6}
\\[3mm]&=\color{#00f}{{\pars{\floor{N/2} + 6}\pars{\floor{N/2} + 5}
\pars{\floor{N/2} + 4}\pars{\floor{N/2} + 3}\pars{\floor{N/2} + 2}
\pars{\floor{N/2} + 1} \over 720}}
\end{align}</p>
</blockquote>
|
504,524 | <p>I'm trying to learn probability and statistics but I can't really get my head around this one. I realize after drawing the first card there will only be 51 cards in the deck but I'm having trouble calculating the chance that the second one is an Ace if I don't know what the first card is?</p>
<p>Assuming that the initial card was an Ace, I could say the probability of the second being an Ace is 3/51... but how do I approach this with an unknown card?</p>
| Community | -1 | <p>Hint: If the first one you draw is an Ace then the probability for the second one to be an Ace is $\frac3{51}$;</p>
<p>however if the first one drawn is <strong>not</strong> an Ace then the probability is $\frac4{51}$.</p>
|
504,524 | <p>I'm trying to learn probability and statistics but I can't really get my head around this one. I realize after drawing the first card there will only be 51 cards in the deck but I'm having trouble calculating the chance that the second one is an Ace if I don't know what the first card is?</p>
<p>Assuming that the initial card was an Ace, I could say the probability of the second being an Ace is 3/51... but how do I approach this with an unknown card?</p>
| kiss my armpit | 26,975 | <p>There are $52\times 51$ outcomes when you draw 2 cards one after another without returning the first drawn card to the deck.</p>
<p>There are $48\times 4$ outcomes when the first card is not Ace and the second card is Ace.
There are $4\times 3$ outcomes when both cards are Ace. The total outcomes are $48\times 4 + 4 \times 3 = 51\times 4$.</p>
<p>The probability is $\frac{51\times 4}{52\times 51}=\frac{51}{52}\times\frac{4}{51}=\frac{4}{52}=\frac{1}{13}$.</p>
|
3,762,624 | <p>This is a pretty common question in probability and there are already a few answers on the site. HOWEVER, my Professor has put a little twist on it and I can't piece it together anymore.</p>
<p><strong>QUESTION:</strong></p>
<p>Let <span class="math-container">$(X_n)$</span> be a sequence of geometric random variables, each with parameter <span class="math-container">$p_n$</span> respectively. Suppose <span class="math-container">$p_n \rightarrow 0$</span>. Let <span class="math-container">$(\theta_n)$</span> be a sequence of positive real numbers with <span class="math-container">$\frac{p_n}{\theta_n} \rightarrow \lambda$</span> for some <span class="math-container">$0<\lambda < \infty$</span>. Let <span class="math-container">$Y_n = \theta_n X_n$</span>. Show that <span class="math-container">$Y_n$</span> converges in distribution to an exponential distribution with parameter <span class="math-container">$\lambda$</span>.</p>
<p><strong>MY ATTEMPT:</strong></p>
<p>Fix <span class="math-container">$x \in \mathbb{R}^+$</span>. Let <span class="math-container">$Z$</span> be an exponential random variable with parameter <span class="math-container">$\lambda$</span>. Then we have
<span class="math-container">\begin{align*}
P(Y_n \leq x) &=P\bigg(X_n \leq \frac{x}{\theta_n}\bigg)\\
&= 1-(1-p_n)^{\big\lfloor \frac{x}{\theta_n}\big\rfloor} \\
& \approx 1-e^{-\frac{p_n}{\theta_n} x} \tag{large n}\\
& \rightarrow 1- e^{-\lambda x} \\
&= P(Z \leq x),
\end{align*}</span>
as desired.</p>
<p><strong>PROBLEMS WITH MY SOLUTION</strong></p>
<p>In my third line, I am just using my intuition. I am unable to prove this, nor can I find any theorems which might help.</p>
<p>Any input is appreciated! Thanks :)</p>
| Ben Grossmann | 81,360 | <p><strong>Counterexample:</strong> Take <span class="math-container">$t = 1$</span> and define
<span class="math-container">$$
g(x) = x^2, \quad f(x) = 2x^2 - \frac 13 x^3.
$$</span>
In particular, the ratios come out to
<span class="math-container">$$
\frac{tg(t)}{\int_0^tg(s)\,ds} = 3, \quad
\frac{tf(t)}{\int_0^tf(s)\,ds} = \frac{2 - \frac 13}{\frac 23 - \frac 1{12}} = \frac{20}{7} < 3.
$$</span></p>
|
337,930 | <p>Given two polynomials</p>
<p>$$
p(x) = a_0 + a_1 x + a_2 x^2 + \ldots + a_{n-1}x^{n-1} \\
q(x) = b_0 + b_1 x + b_2 x^2 + \ldots + b_{n}x^{n}
$$</p>
<p>And the series expansion from their rational polynomial</p>
<p>$$
\frac{p(x)}{q(x)} = c_0 + c_1 x + c_2 x^2 + \ldots
$$</p>
<p>is it possible to recover the the original polynomials $a_n$, $b_n$ from only the series $c_n$ via the solution of a linear system? </p>
| Rohan Shinde | 463,895 | <p>It can be proven combinatorially by noting that any combination of $r$ objects from a group of $m+n$ objects must have some $0\le k\le r$ objects from group $m$ and the remaining from group $n$.</p>
|
622,076 | <p>Continuity $\Rightarrow$ Intermediate Value Property. Why is the opposite not true? </p>
<p>It seems to me like they are equal definitions in a way. </p>
<p>Can you give me a counter-example? </p>
<p>Thanks</p>
| DanZimm | 37,872 | <p>To begin I want to state the IVP considering I messed up on the definition:</p>
<blockquote>
<p>Let $I$ be an open interval and $f : I \to \mathbb{R}$ then $f$ has the IVP iff Given $a,b \in I : a \le b$
$$
\forall \; y \text{ between } f(a),f(b) \; \exists \; x \in [a,b] : f(x) = y
$$</p>
</blockquote>
<p>The following function <s>has IVP</s> on $\mathbb{R}$ but it is not continuous on all of $\mathbb{R}$ EDIT: this a discontinuous surjective function ;)
$$
f: \mathbb{R} \to \mathbb{R} ,\; f(x) = \left\{ \begin{array}{c} \frac{1}{x} : x \neq 0 \\ 0 : x = 0 \end{array}\right.
$$
Like everyone else has said this guy named Darboux (pretty cool guy) came up with the following theorem:</p>
<blockquote>
<p>Given an open interval $I$ and $f$ a differentiable function (NOTE: $f$ doesn't necessarily have to be $C^1$!) s.t. $f : I \to \mathbb{R}$, $\frac{\mathrm{d} f}{\mathrm{d} x} = f'$ has IVP on $I$.</p>
</blockquote>
<p>So a pretty common example is putting
$$
f : (-1,1) \to \mathbb{R}, \; f(x) = \left\{ \begin{array}{rl} x^2 \sin \left( \frac{1}{x} \right) : & x \neq 0 \\ 0 : & x = 0 \end{array} \right.
$$
and then seeing that $f'(x) = 2x \sin \left( \frac{1}{x} \right) - \cos \left( \frac{1}{x} \right) : x \neq 0$ (chain rule, in case you brain fart often like I do) and thus $f'$ is clearly discontinuous at $x=0$ but by Darboux's Theorem it has IVP!</p>
|
622,076 | <p>Continuity $\Rightarrow$ Intermediate Value Property. Why is the opposite not true? </p>
<p>It seems to me like they are equal definitions in a way. </p>
<p>Can you give me a counter-example? </p>
<p>Thanks</p>
| N. S. | 9,176 | <p>The next theorem might be of interest to you, it really shows that the class of functions with the IVP is very big.</p>
<p><strong>Theorem (Sierpinski)</strong> Let $f : \mathbb R \to \mathbb R$ be any function. Then there exists $f_1,f_2 : \mathbb R \to \mathbb R$ such that $f=f_1+f_2$ and $f_1,f_2$ satisfy the Intermediate Value Property.</p>
<p>Moreover, in the above Theorem, $f_1,f_2$ can be chosen to be discontinuous at all points.</p>
|
442,950 | <p>I would like to show <span class="math-container">$\lim\limits_{r\to\infty}\int_{0}^{\pi/2}e^{-r\sin \theta}\text d\theta=0$</span>.</p>
<p>Now, of course, the integrand does not converge uniformly to <span class="math-container">$0$</span> on <span class="math-container">$\theta\in [0, \pi/2]$</span>, since it has value <span class="math-container">$1$</span> at <span class="math-container">$\theta =0$</span> for all <span class="math-container">$r\in \mathbb{R}$</span>. </p>
<p>If <span class="math-container">$F(r) = \int_{0}^{\pi/2}e^{-r\sin \theta}\text d\theta$</span>, we can find the <span class="math-container">$j$</span>th derivative <span class="math-container">$F^{(j)}(r) = (-1)^j\int_{0}^{\pi/2}\sin^{j}(\theta)e^{-r\sin\theta}\text d\theta$</span>, but I don't see how this is helping.</p>
<p>The function is strictly decreasing on <span class="math-container">$[0,\pi/2]$</span>, since <span class="math-container">$\partial_{\theta}(e^{-r\sin\theta})=-r\cos\theta e^{-r\sin \theta}$</span>, which is strictly negative on <span class="math-container">$(0,\pi/2)$</span>.</p>
<p>Any ideas?</p>
| Potato | 18,240 | <p>Split it into two pieces: one integral over $[0,\epsilon]$ and another over $[\epsilon, \pi/2]$. Since the integrand is bounded on the first piece, you can make it arbitrarily small by choosing $\epsilon$ small. On the other piece, it converges uniformly to zero. I think you can take it from there. </p>
|
2,836,539 | <p>Recently I run into this integral</p>
<p>$$\mathcal{J} = \int_{0}^{\pi/2} \frac{x \log \left ( 1-\sin x \right )}{\sin x} \, \mathrm{d}x$$</p>
<p>I don't know to what it evaluates. I tried several approaches.</p>
<p><strong>1st:</strong> Differentiation under the integral sign</p>
<p>Consider the function $\displaystyle f(\alpha)= \int_{0}^{\pi/2} \frac{x \log \left ( 1-\alpha\sin x \right )}{\sin x} \, \mathrm{d}x$. Hence</p>
<p>\begin{align*}
\frac{\mathrm{d} }{\mathrm{d} \alpha} f(\alpha) &= \frac{\mathrm{d} }{\mathrm{d} \alpha} \int_{0}^{\pi/2} \frac{x \log \left ( 1-\alpha\sin x \right )}{\sin x} \, \mathrm{d}x \\
&= \int_{0}^{\pi/2} \frac{\partial }{\partial \alpha} \frac{x \log \left ( 1-\alpha\sin x \right )}{\sin x} \, \mathrm{d}x \\
&= -\int_{0}^{\pi/2} \frac{x \sin x}{\sin x \left ( 1- \alpha \sin x \right )} \, \mathrm{d}x\\
&=- \int_{0}^{\pi/2} \frac{x}{1- \alpha \sin x} \, \mathrm{d}x
\end{align*}</p>
<p>And the last integral equals? </p>
<p><strong>2nd:</strong> Taylor series expansion</p>
<p><em>Lemma:</em> It holds that</p>
<p>$$x \sin^n x = \left\{\begin{matrix}
2^{1-n}\displaystyle\mathop{\sum}\limits_{k=0}^{\frac{n-1}{2}}(-1)^{\frac{n-1}{2}-k}\binom{n}{k}\,x\sin\big((n-2k)x\big) & , & n \;\; \text{odd} \\\\
2^{-n}\displaystyle\binom{n}{\frac{n}{2}}\,x+2^{1-n}\mathop{\sum}\limits_{k=0}^{\frac{n}{2}-1}(-1)^{\frac{n}{2}-k}\binom{n}{k}\,x\cos\big((n-2k)x\big) & , & n \;\; \text{even}
\end{matrix}\right.$$</p>
<p>Hence,</p>
<p>\begin{align*}
\int_{0}^{\pi/2} \frac{x \log \left ( 1-\sin x \right )}{\sin x} \, \mathrm{d}x &= -\int_{0}^{\pi/2} \frac{x}{\sin x} \sum_{n=1}^{\infty} \frac{\sin^n x}{n} \, \mathrm{d}x \\
&=-\sum_{n=1}^{\infty} \frac{1}{n} \int_{0}^{\pi/2} x \sin^{n-1} x \, \mathrm{d}x
\end{align*}</p>
<p>However the lemma does not help at all. In fact, if someone substitutes the RHS what it seems to be in there is an $\arcsin $ Taylor expansion. The series that remains to be evaluated is very daunting.</p>
<p>To sum up, I don't know to what this integral evaluates. I don't even know if a nice closed form exists neither do I expect one. But , I still hope. </p>
| Quanto | 686,284 | <p>After the substitution <span class="math-container">$t=\tan\frac x2$</span>, it can be shown that the integral reduces to</p>
<p><span class="math-container">\begin{align}
& \int_{0}^{\pi/2} \frac{x \log \left ( 1-\sin x \right )}{\sin x} \, {d}x\\
=&\> 4\int_0^1 \frac{\tan^{-1}t\ln t}tdt
= -2\int_0^1 \frac{\ln^2t}{1+t^2}dt = -\frac{\pi^3}8
\end{align}</span></p>
|
3,832,383 | <p>Below is a problem I found, however, after many attemps I can not seem to get a solution.</p>
<p><strong>Problem:</strong> Let <span class="math-container">$E \subset [0,1]$</span>. Show that if <span class="math-container">$m^*(E) + m^*([0,1] \setminus E) = 1$</span>, then <span class="math-container">$E$</span> is measurable.</p>
<p>(my attempted) <strong>Solution:</strong> Notice, the following is true <span class="math-container">$[0,1] \setminus E = [0,1] \cap E^c$</span>. Therefore, we can rewrite the given equations as</p>
<p><span class="math-container">$$m^*(E) + m^*([0,1] \cap E^c) = 1. \quad (i)$$</span></p>
<p>Also, notice that <span class="math-container">$E = [0,1] \cap E$</span>, hence, we can rewrite <span class="math-container">$(i)$</span> as the following</p>
<p><span class="math-container">$$m^*([0,1] \cap E) + m^*([0,1] \cap E^c) = 1. \quad (ii)$$</span></p>
<p>Since <span class="math-container">$m^*([0,1]) = 1$</span> we can, again, rewrite <span class="math-container">$(ii)$</span> as follows</p>
<p><span class="math-container">$$m^*([0,1] \cap E) + m^*([0,1] \cap E^c) = m^*([0,1]). \quad (iii)$$</span></p>
<p>For this path, this is where I get stuck. In other words, obviously if <span class="math-container">$[0,1]$</span> could be replaced by any set <span class="math-container">$A \subseteq \mathbb{R}$</span> then, sure, <span class="math-container">$E$</span> is measurable. Therefore, I do not think this is the "correct path" to take.</p>
<p>I feel like one way is, possibly, to show that for any <span class="math-container">$\epsilon > 0$</span>, there exists a closed set F, such that, <span class="math-container">$E \subseteq F$</span> and <span class="math-container">$m^*(F \setminus E) < \epsilon$</span>. Taking <span class="math-container">$F = [0,1]$</span>, we have <span class="math-container">$E \subset F$</span> and
<span class="math-container">$m^*(F \setminus E) < 1 - m^*(E)$</span>. Hence, if I take <span class="math-container">$\epsilon = 1 - m^*(E)$</span> can I conclude the proof?</p>
| Angelo | 771,461 | <p><span class="math-container">$(2)\implies(1)$</span></p>
<p><strong>Proof:</strong></p>
<p>Assume that <span class="math-container">$\;E\;$</span> is a subset of <span class="math-container">$\;[0,1]\;$</span> such that (2) holds.</p>
<p>Let <span class="math-container">$\;Y\;$</span> be a <span class="math-container">$\;G_\delta\;$</span> set (<span class="math-container">$Y$</span> is Lebesgue measurable) such that <span class="math-container">$\;E\subseteq Y\subseteq [0,1]\;$</span> and <span class="math-container">$\;m(Y)=m^*(E)\;$</span>.</p>
<p>From <span class="math-container">$\;(2)\;$</span> it follows that</p>
<p><span class="math-container">$m^*(E)=m(Y)=m^*(Y)\ge m^*(Y\cap E)+m^*(Y\cap E^c)=$</span></p>
<p><span class="math-container">$=m^*(E)+m^*(Y\cap E^c)\;,\;$</span> hence</p>
<p><span class="math-container">$m^*(Y\cap E^c)=0\;$</span> and so <span class="math-container">$\;Z=Y\cap E^c\;$</span> is Lebesgue measurable.</p>
<p>Since <span class="math-container">$\;E=Y-Z\;,\;$</span> it is Lebesgue measurable as well.</p>
<p><strong>Addendum:</strong></p>
<p>Now we are going to prove that there exists a <span class="math-container">$\;G_\delta\;$</span> set <span class="math-container">$\;Y\;$</span> such that <span class="math-container">$\;E\subseteq Y\subseteq [0,1]\;$</span> and <span class="math-container">$\;m(Y)=m^*(E)\;$</span>.</p>
<p><span class="math-container">$m^*(E)=\inf\left\{\sum\limits_{n=1}^\infty m(I_n): \left(I_n\right)_{n\in\mathbb{N}}\text{ is a sequence of open}\\\text{ intervals such that } E\subseteq\bigcup_\limits{n=1}^\infty I_n\right\}.$</span></p>
<p>Let <span class="math-container">$\;A=\bigcup_\limits{n=1}^\infty I_n\implies A\;$</span> is a open set, <span class="math-container">$\;A\supseteq E\;$</span> and <span class="math-container">$\;m(A)\le\sum\limits_{n=1}^\infty m(I_n)\;.$</span></p>
<p>There exists a sequence <span class="math-container">$\;\left(A_k\right)_{k\in\mathbb{N}}\;$</span> of open sets such that <span class="math-container">$\;A_k\supseteq E\;$</span> and <span class="math-container">$\;m(A_k)<m^*(E)+\cfrac{1}{k}\;,\;$</span> for all <span class="math-container">$\;k\in\mathbb{N}\;.$</span></p>
<p>Let <span class="math-container">$\;B_k=A_k\cap\left]-\frac{1}{k},1+\frac{1}{k}\right[\;$</span> for all <span class="math-container">$\;k\in\mathbb{N}\;.$</span></p>
<p>Hence,</p>
<p><span class="math-container">$B_k\;$</span> is a open set, <span class="math-container">$\;E\subseteq B_k\subseteq\left]-\frac{1}{k},1+\frac{1}{k}\right[\;$</span> and <span class="math-container">$\;m(B_k)<m^*(E)+\cfrac{1}{k}\;,\;$</span> for all <span class="math-container">$\;k\in\mathbb{N}\;.$</span></p>
<p>Let <span class="math-container">$\;Y=\bigcap\limits_{k=1}^\infty B_k\implies Y\;$</span> is a <span class="math-container">$\;G_\delta\;$</span> set.</p>
<p>It results that <span class="math-container">$\;E\subseteq Y\subseteq\left]-\frac{1}{k},1+\frac{1}{k}\right[\;$</span> for all <span class="math-container">$\;k\in\mathbb{N}\;,$</span></p>
<p>so <span class="math-container">$\;E\subseteq Y\subseteq[0,1]\;,\;$</span> moreover,</p>
<p><span class="math-container">$m(Y)<m^*(E)+\cfrac{1}{k}\;$</span> for all <span class="math-container">$\;k\in\mathbb{N}\;,\;$</span> hence</p>
<p><span class="math-container">$\;m(Y)\le m^*(E)\;.$</span></p>
<p>Since <span class="math-container">$\;E\subseteq Y\;,\;$</span> it follows that <span class="math-container">$\;m^*(E)\le m(Y)\;,\;$</span> hence <span class="math-container">$\;m(Y)=m^*(E)\;.$</span></p>
|
871,542 | <p>I have the following theorem:</p>
<blockquote>
<p>Let <span class="math-container">$\rho$</span> be the traffic intensity.</p>
<p>a) If <span class="math-container">$\rho<1$</span>, then <span class="math-container">$X$</span> is positive recurrent.</p>
<p>b) If <span class="math-container">$\rho>1$</span>, then <span class="math-container">$X$</span> is transient.</p>
<p>c) If <span class="math-container">$\rho=1$</span>, then <span class="math-container">$X$</span> is null-reсurrent.</p>
</blockquote>
<p>It is not important what exactly <span class="math-container">$\rho$</span> and <span class="math-container">$X$</span> are. Let's assume I have proved a) and b). Then it remains to prove the case <span class="math-container">$\rho=1$</span>:</p>
<p>Can I say the following: <span class="math-container">$X$</span> is transient if and only if <span class="math-container">$\rho>1$</span>, and is positive recurrent if and only if <span class="math-container">$\rho<1$</span>. If <span class="math-container">$\rho=1$</span> then <span class="math-container">$X$</span> has no choice but null-recurrent.</p>
<p>Is that correct?</p>
| Mark Fantini | 88,052 | <p>He mentions that $Z(t) = e^{it}.$ You have $$|Z(t)| = |e^{it}| = \sqrt{\cos^2(t) + \sin^2(t)} = 1.$$</p>
|
2,780,731 | <p>In school, I have recently been learning about simple differential equations. We know that the solution of $y'=y$ is $y=Ae^x$, where $A$ is a constant. But how can we know that it is the <strong>only</strong> solution? The only thing I can figure out is that $y$ is continuously differentiable. Help me, please.</p>
| Mohammad Riazi-Kermani | 514,496 | <p>Suppose $y$ is a solution to $y'=y$</p>
<p>Multiply both sides by $e^{-x}$ to get $$ y'e^{-x} = ye^{-x}$$</p>
<p>$$y'e^{-x}-ye^{-x}=0$$</p>
<p>$$ \frac {d}{dx} (ye^{-x}) =0$$
$$ye^{-x} =A$$
$$ y=Ae^{x} $$</p>
|
1,598,006 | <p>(Here, $B$ is relatively compact means the closure of $B$ is compact.)</p>
<blockquote>
<ol>
<li><p>$\hat A$ is compact.</p></li>
<li><p>$\hat A=\hat {\hat A}$.</p></li>
<li><p>$\hat A$ is connected.</p></li>
<li><p>$\hat A=X$.</p></li>
</ol>
</blockquote>
<p>I try to eliminate the options by using an example.</p>
<p>Consider $X=\Bbb R - \{1,2,3\}$ with metric topology and let $A=(-\infty,1)$.</p>
<p>Then $\hat A=(-\infty,1) \cup (1,2) \cup (2,3)$.</p>
<p>Hence options 1,3,4 are false.</p>
<p>So I select option 2 as an answer.</p>
<p>Is my method correct?</p>
| DanielWainfleet | 254,665 | <p>For (1) let $X=R$ and $A=(0,1)$. The connected components of $X-A$ are $(-\infty,0]$ and $[1,\infty)$ which are closed in $X$ but not compact.So $\hat A=A$, which is not compact.For (3) let $X=R^2$ and let $A$ contain exactly 2 points.Then $X-A$ is connected but $\overline {X-A}=X$ is not compact. So $\hat A=A$, which is not connected. Either of these will take care of (4)..... I have also seen the term "pre-compact" as a synonym for "relatively compact."</p>
|
1,600,063 | <p>I am trying to prove the following statement:</p>
<p>Given any two real numbers $x,y$ with $x<y$, there exists a rational number $q$ that satisfies $x<q<y$.</p>
<p>I got stuck at one point of the proof, so this is what I thought of:</p>
<p>I want to find a rational number $q$, which can be expressed as $q=\dfrac{a}{b}$, with $a$ integer and $b$ a positive integer, such that $$x<\dfrac{a}{b}<y$$</p>
<p>This is equivalent to $$(*) \space \space bx<a<by,$$</p>
<p>so if I could find an integer $a$ and a positive integer $b$ that satisfy $(*)$, then I would be done. </p>
<p>This is an exercise from Tao's Analysis I which follows right after the archimidean property, so it occurred to me to use this property:</p>
<p>Pick $x,y$ two real numbers with $x<y$, then we have $y-x$ is positive. By the archimedean property, there exists a positive integer $b$ such that $b(y-x)>1$. From here I don't know how to deduce that the integer $a$ satisfying $(*)$ exists. Any help would be appreciated.</p>
| Hagen von Eitzen | 39,174 | <p>Let $a$ be the least integer $>bx$ and show that then $a<by$, for otherwise $a-1$ would also be $>bx$.</p>
|
3,752,771 | <p>I wanted to get the full probability of 2 attempts made at 60% chance of success.</p>
<p>I was looking at a different chain of math and found my probability to hit an enemy is 60% per each attack but I was wondering how it would look at all the outcomes and the probability of it.</p>
<blockquote>
<p>6/10 * 6/10 = 36%</p>
</blockquote>
<p>of both attempts failing and both attempts succeeding with a 64% chance of at least 1 attempt succeeding?</p>
<p>I didn't understand how it would look as a failure adding up to 136%</p>
| Alex | 38,873 | <p>You got 136% by adding <span class="math-container">$60\%+40\%+36\%$</span>, i.e. <span class="math-container">$P(S) + P(F) + P(SS)$</span>. This is incorrect because these outcomes do not partition sample space (full set of outcomes).</p>
<p>Your sample space are the following outcomes:
<span class="math-container">$$
\Omega = \{SS, SF, FS, FF\}
$$</span>
The probability of these 4 events must sum to 1. Here, <span class="math-container">$P(SS)=0.36$</span>, assuming outcomes are independent and two failues, <span class="math-container">$P(FF) = 0.24$</span>. You can compute the rest.</p>
|
2,481,046 | <p>I have a question that asks to show that $S^2 = \{(x,y,z) \in \mathbb{R}^3|x^2+y^2+z^2=1\}$ is a differentiable manifold. My professor says that one way to do this is to define the following 6 parametrizations of the sphere, which cover the entire sphere.</p>
<p>$\vec{\phi_{i}}:V \to \mathbb{R}^3$ where $V = \{(u,v) \in \mathbb{R}^2|u^2+v^2<1\}$</p>
<p>$\vec{\phi_{1}}(u,v) = (u,v,\sqrt{1-u^2-v^2}) \qquad (z>0)$</p>
<p>$\vec{\phi_{2}}(u,v) = (u,v,-\sqrt{1-u^2-v^2}) \qquad (z<0)$</p>
<p>$\vec{\phi_{3}}(u,v) = (u,\sqrt{1-u^2-v^2},v) \qquad (y>0)$</p>
<p>$\vec{\phi_{4}}(u,v) = (u,-\sqrt{1-u^2-v^2},v) \qquad (y<0)$</p>
<p>$\vec{\phi_{5}}(u,v) = (\sqrt{1-u^2-v^2},u,v) \qquad (x>0)$</p>
<p>$\vec{\phi_{6}}(u,v) = (-\sqrt{1-u^2-v^2},u,v) \qquad (x<0)$</p>
<p>I don't understand what these parameterizations mean at all and I don't understand what a parameterization is. From what I can read online, it's some function but I'm not sure why this specific function with $u$ and $v$ is what we're using to cover the entire sphere. Can someone explain this to me please?</p>
| Jack D'Aurizio | 44,121 | <p>The function $\sin\left(x+\frac{1}{x}\right)$ is continuous and bounded between $-1$ and $1$ on the interval $(0,1)$, hence the given integral is trivially convergent</p>
<p>By enforcing the substitution $x+\frac{1}{x}\mapsto z$ we have</p>
<p>$$\begin{eqnarray*} \int_{0}^{1}\sin\left(x+\tfrac{1}{x}\right)\,dx &=& \frac{1}{2}\int_{2}^{+\infty}\sin(z)\left[\tfrac{z}{\sqrt{z^2-4}}-1\right]\,dz\\&=&\int_{1}^{+\infty}\sin(2u)\left[\tfrac{u}{\sqrt{u^2-1}}-1\right]\,du\\&=&\int_{1}^{+\infty}\frac{\sin(2u)}{(u+\sqrt{u^2-1})\sqrt{u^2-1}}\,du\\&=&\int_{0}^{+\infty}\frac{\sin(2v+2)}{(v+1+\sqrt{v^2+2v})\sqrt{v^2+2v}}\,dv\end{eqnarray*}$$
where the last integral is clearly convergent by the Dirichlet test, since $\sin(2v+2)$ has a bounded primitive and $(v+1+\sqrt{v^2+2v})\sqrt{v^2+2v}$ is an increasing and unbounded function behaving like $\sqrt{2v}$ in a right neighbourhood of the origin. To apply the (inverse) Laplace transform to the last integral is a god way for deriving accurate numerical approximations, like $\color{blue}{0.37619956866}$.</p>
|
2,456,976 | <p>$ f(x,y) = \begin{cases} \dfrac{x^3+y^3}{x^2+y^2} &\quad\text{if} [x,y] \neq [0,0]\\[2ex] 0 &\quad\text{if}[x,y] = [0,0]\\ \end{cases} $</p>
<p>The only point it could be discontinuous in is <code>[0,0]</code>. How do I find the limit of the function for $(x,y) \rightarrow (0,0)$? $ \lim_{(x,y) \rightarrow (0,0)} \frac{x^3+y^3}{x^2+y^2} $ seems pretty hard to analyse.</p>
| Yaddle | 333,729 | <p>For $x,y \in \mathbb R \setminus \{0\}$ we have</p>
<p>$$\vert f(x,y)\vert = \left\vert\frac{x^3 + y^3}{x^2+ y^2}\right\vert \leq \left\vert \frac{x^3}{x^2+ y^2} \right\vert + \left\vert \frac{y^3}{x^2+ y^2}\right\vert \leq \left\vert \frac{x^3}{x^2} \right\vert + \left\vert \frac{y^3}{y^2}\right\vert = \vert x \vert + \vert y \vert \to 0$$
for $(x, y) \to (0, 0)$. If $x = y = 0$ we have $f(x,y) = 0$. Thus it follows that $\lim_{(x,y) \to (0,0)} f(x,y) = 0$. Therefore we can deduce that $f$ is continuous at $(0,0)$. </p>
|
1,105,454 | <p>We have $f: \Bbb{R} \rightarrow \Bbb{R}$ defined as follows:</p>
<p>$$f(x) = \begin{cases} a, & \mbox{if } x=0 \\ \sin\frac{b}{|x|}, & \mbox{if } x\neq 0 \end{cases}$$</p>
<p>The problem asks us to tell for which $a,b \in \Bbb{R}$, $f$ is continuous.</p>
<p>Intuitively I should find $\lim_{x\rightarrow 0} \sin \frac{b}{|x|}$ and that would be somehow dependent on $b$. Then we can set $a$ equal to that limit, and the function should be continuous. But how to find $\lim_{x\rightarrow 0} \sin \frac{b}{|x|}$ , I don't have a clue - here we have parametrized $\sin$ fuction with an arguement that seems to be approaching $+\infty$ or $-\infty$ depending on $b$. But it doesn't appear that $\sin \frac{b}{|x|}$ would have any limit at all because as $y\rightarrow \infty$, $\sin y$ will always "alternate" between $1$ and $-1$. So maybe the answer is that such $a,b$ don't exist then? So - what should I do with this problem?</p>
| hmakholm left over Monica | 14,366 | <p>Speaking of the "value" or "result" of the operation would by far be the most understandable.</p>
|
1,105,454 | <p>We have $f: \Bbb{R} \rightarrow \Bbb{R}$ defined as follows:</p>
<p>$$f(x) = \begin{cases} a, & \mbox{if } x=0 \\ \sin\frac{b}{|x|}, & \mbox{if } x\neq 0 \end{cases}$$</p>
<p>The problem asks us to tell for which $a,b \in \Bbb{R}$, $f$ is continuous.</p>
<p>Intuitively I should find $\lim_{x\rightarrow 0} \sin \frac{b}{|x|}$ and that would be somehow dependent on $b$. Then we can set $a$ equal to that limit, and the function should be continuous. But how to find $\lim_{x\rightarrow 0} \sin \frac{b}{|x|}$ , I don't have a clue - here we have parametrized $\sin$ fuction with an arguement that seems to be approaching $+\infty$ or $-\infty$ depending on $b$. But it doesn't appear that $\sin \frac{b}{|x|}$ would have any limit at all because as $y\rightarrow \infty$, $\sin y$ will always "alternate" between $1$ and $-1$. So maybe the answer is that such $a,b$ don't exist then? So - what should I do with this problem?</p>
| MphLee | 67,861 | <p>Generalizing the question to functions, note that $\circ$ is just a <em>binary function</em> $\circ(x,y)=z$ but expressed with <a href="http://en.wikipedia.org/wiki/Infix_notation" rel="nofollow"><em>infixed notation</em></a> $x\circ y$, in my opinion the common terminology is: <em>given a</em> $n$*-ary function* $f:X_0\times X_1\times...\times X_n\rightarrow Y$</p>
<p>$$f(x_0,x_1,...x_n)=y$$</p>
<blockquote>
<p><strong>arguments</strong> <strong>:</strong> $x_0,x_1,...x_n$ <em>are the elements af the domains</em> $X_0, X_1,...X_n$</p>
<p><strong>value</strong> <strong>:</strong> $y$ <em>is the image of the</em> $n$*-uple* $(x_0,x_1,...x_n)$ <em>via the function</em> $f$ <em>and belongs to the codomain</em> $Y$</p>
</blockquote>
<p><strong>Note:</strong> from <a href="http://en.wikipedia.org/wiki/Value_(mathematics)" rel="nofollow">wikipedia</a> we get this definition</p>
<blockquote>
<p>A <strong>value</strong> of a function is the result associated to a value of its <a href="http://en.wikipedia.org/wiki/Argument_of_a_function" rel="nofollow"><strong>argument</strong></a> (also called variable of the function)</p>
</blockquote>
|
2,661,468 | <p>the number of not identically zero functions $f:\mathbb{R} \to \mathbb{R}$ satisfying the equation $f(xy)=f(x)f(y)$ and $f(x+z)=f(x)+f(z)$ for some $z$ not equal to zero</p>
<ol>
<li>one</li>
<li>finite</li>
<li>countable</li>
<li>uncountable</li>
</ol>
<p>it seems like the question asks about the number of homomorphisams from R to R.but I think there are uncountable such homomorphisams. how ever the answer give is one..whats my mistake in thinking.how should I proceed..please help </p>
| lhf | 589 | <p>Since $f$ is not identically zero, take $y$ such that $f(y)\ne0$. Then $f(y)=f(1y)=f(1)f(y)$ implies $f(1)=1$.</p>
<p>Take $x=0$ in $f(x+z)=f(x)+f(z)$ and get $f(z)=f(0)+f(z)$, which implies $f(0)=0$.</p>
<p>Therefore, $f$ is a ring homomorphism $\mathbb{R} \to \mathbb{R}$.</p>
<p>Now see <a href="https://math.stackexchange.com/questions/449404/is-an-automorphism-of-the-field-of-real-numbers-the-identity-map">this question</a>.</p>
|
891,575 | <p>The circumference of a circle has length 90 centimeters, Three points on the circle divide the circle into three equal lengths. Three ants A, B, and C start to crawl clockwise on the circle, with starting from one of the three points. Initially A is ahead of B and B is ahead of C. Ant A crawls 3 centimeters per second, ant V 5 centimeters, and and C 10 centimeters. How long does it take for the three ants to arrive at the same spot for the first time?</p>
<p>I tried making a list and writing down the numbers, but they seem to never be the same. I know the distance formula is <em>d=rt</em>, but I don't know how to use it to solve this problem. Any help? Thanks!</p>
| DJohnM | 58,220 | <p>Another way of simplifying the problem:</p>
<p>A is at 4 o'clock, B is at 12 o'clock, and C is at 8 o'clock.</p>
<p>Let's subtract $5$ cm/s from each ant's speed. So now C is going clockwise at only $5$ cm/s, B is stationary, and A is going <strong>counter-clockwise</strong> at $2$ cm/s. And they're all still $30$ cm apart. Subtracting the same amount from each ant's speed doesn't change the solution to the problem: in physics terms, it's equivalent to working in ant B's frame of reference, instead of the circle's.</p>
<p>C will reach B after $\frac{30}{5}$ sec, and thereafter every $\frac{90}{5}$ sec, or after $6, 24, 42, 60, 78$, ... seconds</p>
<p>A will reach B after $\frac{30}{2}$ sec, and thereafter every $\frac{90}{2}$ sec, or after $15, 60, 105$, ... seconds</p>
<p>The first double hit is after 60 seconds...</p>
|
630,966 | <p>Most universities have a 3rd year undergraduate analysis course in which metric spaces are studied in depth (compactness, completeness, connectedness, etc...). However, in practice it seems that most of these metric spaces are normed vector spaces. Why not just cover normed vector spaces instead of metric spaces? </p>
<p>Even if we lose some generality, normed vector spaces feel more natural and interesting, in my opinion, at least. </p>
| Ittay Weiss | 30,953 | <p>Metric spaces are far more general than normed spaces. The metric structure in a normed space is very special and possesses many properties that general metric spaces do not necessarily have. </p>
<p>Metric spaces are also a kind of a bridge between real analysis and general topology. With every metric space there is associated a topology that precisely captures the notion of continuity for the given metric. That means that many topological properties can be understood in the context of metric spaces, encompassing many many examples of varying degrees of complexity, while still having a notion of distance. The notion of distance is typically seen as much less abstract than that of a topology, with many of the proofs just being a re-formulation of the standard proofs from real analysis. In fact, somewhat less known is that a naturally occurring generalization of metric spaces (i.e., allowing the metric to attain values not just in the range $[0,\infty]$ but rather in what is known as a value quantale) is as general as topological spaces. That is, every topological space arises as the associated topology for some metric structure in that sense. </p>
<p>Also, many universities are somewhat reluctant to introduce abstract notions early on and so even though the limits in real analysis can be taught just as special cases of metric analysis, and thus introduce metric spaces in the first year, universities often choose to postpone metric spaces. However, this is not always the case as I had the opportunity to teach metric spaces in a first year course (at Utrecht University). </p>
<p>So, this is largely a question of tradition and certainly varies between universities. As is commonly the case, university curricula tend to follow historical developments and tend to adapt and change rather slowly. Metric spaces were introduced in 1906 by Frechet and thus are quite newer than the traditional primary objects of study such as $\mathbb R$ and various function spaces. This may explain why the latter are often taught first. </p>
|
2,694,740 | <p>$$\frac{2.10^{-7} - 0,4.10^{-6}}{10^{-8}} = ? $$</p>
<p>These questions are making me confused because we're dealing with the terms like $10^x$. What are your professional tips? </p>
<p><strong>My attempt:</strong></p>
<p>$$\frac{2.10^{-7} - 4.10^{-7}}{10^{-8}} \tag{1} $$
$$\frac{ -8.10^{-7}}{10^{-8}} \tag{2} $$</p>
<p>And that's where I'm stuck. </p>
| Dr. Sonnhard Graubner | 175,066 | <p>multplying numerator and denominator by $10^{8}$ we get
$$2\cdot 10^{-7}\cdot 10^{8}-0.4\cdot 10^{-6}\cdot 10^{8}$$ and we get $$2\cdot 10-0.4\cdot 10^2=...$$</p>
|
2,867,404 | <p>I have a problem how to get the area from the picture.
Some ideas I got are not good enough to get the correct value of the whole element.</p>
<p><a href="https://i.stack.imgur.com/PeQ5O.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PeQ5O.jpg" alt="enter image description here"></a></p>
| Mladen Uzelac | 580,641 | <p>I asked someone and this answer will satisfy me</p>
<p><a href="https://i.stack.imgur.com/Qsq8g.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Qsq8g.jpg" alt="enter image description here"></a></p>
|
87,466 | <p>I have some code which involves tiny numbers being put to the power of very large numbers. The function I'm looking at is</p>
<p>$\varphi = \omega(T) \left(1 - (1 - \epsilon)^{n_{e}(T)} \right)$</p>
<p>when $\epsilon $ is very small (~$10^{-16}$) and $n_{e}$ is large ($> 10^{10}$). Both $n_{e}$ and $\omega$ are known functions of $T$. At a known value ($T = 37$) the value of $\varphi$ is known, and we can in theory estimate $\epsilon$ by</p>
<p>$\epsilon = 1 - \left(1 - \frac{\varphi}{\omega} \right)^{\frac{1}{n_{e}}}$</p>
<p>However, when I put this into Mathematica I'm getting very strange behaviour; here's a MWE when $\varphi(37) = 0.2506$; </p>
<pre><code>gammaex = 0.2506;
omega[t_] := 2.43163218375*10^7*Exp[1700*(1/298.15 - 1/(273.15 + t))];
w[t_] := (3.414105049212413*10^12)/(omega[t]);
v[t_] := Sqrt[661.6469313477045*(t + 273.15)];
ne[t_] := (v[t]*5.104757516005496*(10^7));
epsilon = SetAccuracy[ 1 - ( 1 - gammaex/w[37])^(1/ne[37]), 30];
test = w[37]*(1 - (1 - epsilon)^(ne[37]))
</code></pre>
<p>Now when I evaluate $\varphi(37)$ via the last line I should get the value 0.2506 but I do not; instead I get $\varphi = 0.289104$, over 15% off the true value. I thought maybe this was a precision problem, so I tried some other commands (Surd, Power...) and got the same wrong value. The output value for the epsilon is $\epsilon \approx 1.11 \times 10^{-16} $ from mathematica.</p>
<p>However I've evaluated this with WolframAlpha and got a markedly different value of $\epsilon \approx 9.61 \times 10^{-17}$, and as <a href="http://www.wolframalpha.com/input/?i=%28112608%29*%281%20-%20%20%28%201%20-9.61369%20%C3%97%2010%5E-17%29%5E%282.31246*10%5E10%29%29" rel="nofollow">this link demonstrates</a> the calculation seems to work with WolframAlpha, returning close to the expected value around 0.25*. Any ideas why the calculations are different, and how I can make Mathematica behave with such extreme values? </p>
<p>Incidentally, if I dump the WolframAlpha equation ( (112608)*(1 - (1 - 9.61369*10^-17)^(2.31246*10^10)) ) directly into Mathematica, it still comes up with the wrong value, so I assume it's a radical issue?</p>
| Fred Simons | 20,253 | <p>As J. M. already suggested, this is a machine precision issue. So let us do your computation with arbitrary precision numbers. For doing so, all machine numbers have to be replaced with arbitrary precision numbers, otherwise the computation falls back to machine numbers. In the following command I have done this by placing `30 after each machine number.</p>
<pre><code>gammaex=0.2506`30;
omega[t_]:=2.43163218375`30*10^7*Exp[1700*(1/298.15`30-1/(273.15`30+t))];
w[t_]:=(3.414105049212413`30*10^12)/(omega[t]);
v[t_]:=Sqrt[661.6469313477045`30*(t+273.15`30)];
ne[t_]:=(v[t]*5.104757516005496`30*(10^7));
epsilon=1-(1-gammaex/w[37])^(1/ne[37]);
test=w[37]*(1-(1-epsilon)^(ne[37]))
FullForm[test]
</code></pre>
<p>Now the result is what you expected:</p>
<pre><code>(* 0.25060000000000000000000000000 *)
(* 0.2506`28.52376207789293 *)
</code></pre>
|
3,995,728 | <p>For the integral
<span class="math-container">$$
\int_1^2 \frac {3^x + 2}{3^{2x} + 3^x} dx
$$</span>
choose the interval of its result: <span class="math-container">$(-\infty, 0], (0, \frac 12], (\frac 12, 1], (1, 3], (3, \infty)$</span>. According to the author of this task you do not have to compute the integral itself to determine the interval. I have managed to compute the integral but I am unsure how to determine the interval without computing it. How can I do that?</p>
| Ethan Bolker | 72,858 | <p>Hint. In the upper circle you have the information you need to find the <span class="math-container">$\sin$</span> of half the central angle.</p>
|
441,792 | <p>There are many objects in mathematics that have the term "chiral" in their name, for instance, chiral algebra by Beilinson and Drinfeld, chiral de Rham complex, chiral Koszul duality etc. Some people told me that chiral algebras are <span class="math-container">$2$</span>-dimensional analogue of associative algebras, which are considered to be <span class="math-container">$1$</span>-dimensional. However, I don't understand its precise meaning since the definition of a vertex operator algebra is so complicated. Does the term chiral has something to do with this <span class="math-container">$2$</span>-dimensionality?</p>
<p>For a vertex operator algebra <span class="math-container">$V$</span>, Yongchang Zhu constructed an associative algebra <span class="math-container">$A(V)$</span> out of <span class="math-container">$V$</span>, such that there is a bijection between the set of isomorphism classes of irreducible positive energy representations of <span class="math-container">$V$</span> and that of simple <span class="math-container">$A(V)$</span>-modules. For an associative algebra <span class="math-container">$A$</span>, Tomoyuki Arakawa calls <span class="math-container">$V$</span> to be the <em>chiralization</em> of <span class="math-container">$A$</span> if <span class="math-container">$A\simeq A(V)$</span> as associative algebras. What's the meaning of chiralization here?</p>
<p>There are some other explanations for the term chiral that I have ever heard. For example, in electromagnetism, chirality means the handedness of electromagnetic waves associated with their polarization. Some others also told me that in the <span class="math-container">$2$</span>-dimensional setting, chiral means holomorphic.</p>
<p>I want to know the geometry/physics behind the term chiral. A philosophical answer is welcome, but a mathematical/physical answer is better.</p>
| Carlo Beenakker | 11,260 | <p><strong>Q:</strong> <em>What is the geometry/physics behind the term chiral?</em></p>
<p>In the physics context, a Hamiltonian <span class="math-container">$H$</span> is said to possess chiral symmetry if it anticommutes with a unitary involution <span class="math-container">$C$</span>. Eigenstates <span class="math-container">$\Psi_\pm$</span> of <span class="math-container">$C$</span> are said to be of left-handed or right-handed chirality, depending on the sign of the eigenvalue <span class="math-container">$\pm 1$</span>. Chiral symmetry ensures that the dynamics generated by the Hamiltonian conserves the chirality.</p>
<p>In specific contexts the chirality is related to a spin degree of freedom, such that the particle moves parallel or antiparallel to the direction of the spin, depending on its chirality.</p>
<p><strong>Q:</strong> <em>What is the meaning of chiral in the context of vertex algebras?</em></p>
<p>A conformal field theory is called chiral if it contains only particles of one single chirality. It is my understanding that vertex algebras were introduced as an algebraic description of a chiral CFT. The connection with the physics concept of chirality seems to have been lost.</p>
|
1,419,315 | <p>I have a particular scenario.</p>
<p>In this scenario, we have the standard cubic equation,</p>
<pre><code>ax^3 + bx^2 + cx + d = y
</code></pre>
<p>as well as 3 points that are graphed, <a href="https://i.imgur.com/VCZKuGW.png" rel="nofollow noreferrer">as can be seen in this graph</a>. (The line is irrelevant for now)</p>
<p>Assume that </p>
<pre><code>a, b, and c
</code></pre>
<p>will all adapt themselves so that they can fit all three dots as well as </p>
<pre><code>d
</code></pre>
<p>and create a cubic line.</p>
<p>The problem at hand, now, is to find some sort of general equation that would find the value of d that would lead to having the smallest sum of all coefficients, those being </p>
<pre><code>a, b, c and d
</code></pre>
<p>The ending result doesn't have to be a specific number, though. A range of values a reasonable amount of numbers wide or so that is guaranteed to contain the number is acceptable as well.</p>
<p>If any clarifications or more details are needed, then feel free to comment and I'll add anything necessary.</p>
<p><strong>EDIT</strong></p>
<p>I should have mentioned that by minimum, I mean the smallest absolute value of the equation, so in other words</p>
<pre><code>|a| + |b| + |c| + |d|
</code></pre>
<p><a href="https://jsfiddle.net/gamea12/crkg1f6c/" rel="nofollow noreferrer">Here's a program that simulates the scenario</a></p>
| Eugene Zhang | 215,082 | <p>A simple way is to calculate all principle minors of <span class="math-container">$A$</span>. If they are all positive, then <span class="math-container">$A$</span> is positive definite.</p>
<p>For example, <span class="math-container">$|A|_1=2>0$</span></p>
<p><span class="math-container">$$
|A|_2=\left|\begin{array}{}{\quad2 \quad-1\\ -1\quad 2}
\end{array}\right|=3>0
$$</span>
Then calculate <span class="math-container">$|A|_3=|A|$</span>.</p>
<p>If <span class="math-container">$|A|_i\geqslant0,1\leqslant i\leqslant n$</span>, then <span class="math-container">$A$</span> is semi-positive definite.</p>
<p>If <span class="math-container">$|A|_i<0$</span> for <span class="math-container">$i$</span> is odd and <span class="math-container">$|A|_i>0$</span> for <span class="math-container">$i$</span> is even, then <span class="math-container">$A$</span> is negative definite.</p>
<p>If <span class="math-container">$|A|_i\leqslant 0$</span> for <span class="math-container">$i$</span> is odd and <span class="math-container">$|A|_i\geqslant 0$</span> for <span class="math-container">$i$</span> is even, then <span class="math-container">$A$</span> is semi-negative definite.</p>
|
509,635 | <p>If every chain in a lattice is complete (we take the empty set to be a chain), does that mean that the lattice is complete? If yes, why? </p>
<p>My intuition says yes, and the reasoning is that we should somehow be able to define a supremum of any subset of the lattice to be the same as the supremum of some chain related to that lattice, but I've not abled to make more progress on the same. ANy suggestions? </p>
| Doug | 725,720 | <p>By completeness, I mean every nonempty subset has a sup and inf. I am not sure if a chain-complete poset is complete and my personal guess is negative. Here is a proof showing a nonempty chain-complete lattice <span class="math-container">$L$</span> is complete, with the lattice property emphasised.</p>
<p>By Zorn's lemma, <span class="math-container">$L$</span> has a maximal element <span class="math-container">$M$</span>. It is the maximum: For every <span class="math-container">$x\in L$</span>, <span class="math-container">$x\vee M$</span> exists because <span class="math-container">$L$</span> is a lattice. Now <span class="math-container">$x\vee M\ge M$</span>, so <span class="math-container">$M=x\vee M$</span> by maximality of <span class="math-container">$M$</span>. Then <span class="math-container">$M\ge x$</span>. So, <span class="math-container">$M=\max(L)$</span>. Idem, <span class="math-container">$\min(L)$</span> exists.</p>
<p>For every nonempty subset <span class="math-container">$A\subset L$</span>, let <span class="math-container">$B:=\{x\in L:x\le a,\forall a\in A\}$</span>. Then <span class="math-container">$\min(L)\in B$</span>, so <span class="math-container">$B$</span> is nonempty. For every chain <span class="math-container">$C\subset B$</span>, <span class="math-container">$\sup_L(C)$</span> exists by chain-completeness. For every <span class="math-container">$a\in A$</span>, <span class="math-container">$c\in C$</span>, we have <span class="math-container">$c\le a$</span>, so <span class="math-container">$\sup_L(C)\le a$</span> and then <span class="math-container">$\sup_L(C)\in B$</span>. By Zorn's lemma again, <span class="math-container">$B$</span> has a maximal element <span class="math-container">$b$</span>. Again, it is <span class="math-container">$\max(B)$</span>: For every <span class="math-container">$b'\in B$</span>, <span class="math-container">$b\vee b'$</span> exists because <span class="math-container">$L$</span> is a lattice. For every <span class="math-container">$a\in A$</span>, <span class="math-container">$b\le a$</span> and <span class="math-container">$b'\le a$</span> give <span class="math-container">$b\vee b'\le a$</span>. So, <span class="math-container">$b\vee b'\in B$</span>. As <span class="math-container">$b$</span> is maximal in <span class="math-container">$B$</span> and <span class="math-container">$b\vee b'\ge b$</span>, we get <span class="math-container">$b=b\vee b'\ge b'$</span>. So, <span class="math-container">$b=\max(B)=\inf_L(A)$</span>. Idem, <span class="math-container">$\sup_L(A)$</span> exists. The proof is completed.</p>
|
3,537,654 | <p><span class="math-container">$$\lim_{x\to 0^{+}} (\tan x)^x$$</span></p>
<p><span class="math-container">$$\lim_{x\to 0^{+}} e^{\ln((\tan x)^x)}=\lim_{x\to 0^{+}} e^{x\ln(\tan x)}=\lim_{x\to 0^{+}} e^{x[\ln(\sin x)-\ln(\cos x)]}$$</span></p>
<p>We can continue to create an expression that may help us use L'Hospital but it does not seem to be correct </p>
<p>P.S can we write:</p>
<p><span class="math-container">$$1=(\frac{-1}{-1})^x\leq \lim_{x\to 0^+}\Bigl(\frac{\sin x}{\cos x}\Bigr)^x\leq \Bigl(\frac{1}{1}\Bigr)^x=1$$</span>?</p>
| Bernard | 202,857 | <p>Compute first the limit of the logarithm, using <em>equivalence</em>:</p>
<p><span class="math-container">$\tan x\sim_0 x,\:$</span> so
<span class="math-container">$\quad \ln\bigl((\tan x)^x\bigr)=x\ln(\tan x)\sim_0x\ln x \xrightarrow[x\to 0]{} 0.$</span></p>
|
3,537,654 | <p><span class="math-container">$$\lim_{x\to 0^{+}} (\tan x)^x$$</span></p>
<p><span class="math-container">$$\lim_{x\to 0^{+}} e^{\ln((\tan x)^x)}=\lim_{x\to 0^{+}} e^{x\ln(\tan x)}=\lim_{x\to 0^{+}} e^{x[\ln(\sin x)-\ln(\cos x)]}$$</span></p>
<p>We can continue to create an expression that may help us use L'Hospital but it does not seem to be correct </p>
<p>P.S can we write:</p>
<p><span class="math-container">$$1=(\frac{-1}{-1})^x\leq \lim_{x\to 0^+}\Bigl(\frac{\sin x}{\cos x}\Bigr)^x\leq \Bigl(\frac{1}{1}\Bigr)^x=1$$</span>?</p>
| trancelocation | 467,003 | <p>You surely know</p>
<ul>
<li><span class="math-container">$\lim_{x\to 0}\frac{\sin x}{x} = 1 \Rightarrow \lim_{x\to 0}\frac{\tan x}{x} = \lim_{x\to 0}\left(\frac{\sin x}{x}\cdot \frac 1{\cos x} \right) = 1$</span>.</li>
<li>Besides this, it is easy to show that <span class="math-container">$\lim_{x\to 0^+}x^x = 1$</span>.</li>
</ul>
<p>Hence,</p>
<p><span class="math-container">$$(\tan x)^x = \underbrace{\left(\frac{\tan x}{x}\right)^x}_{\stackrel{x\to 0^+}{\longrightarrow}1^0=1}\cdot \underbrace{x^x}_{\stackrel{x\to 0^+}{\longrightarrow}1} \stackrel{x\to 0^+}{\longrightarrow}1$$</span></p>
|
3,371,964 | <p>Let be <span class="math-container">$O_{2}$</span> the orthogonal group, that is, the group of reflections and rotations of <span class="math-container">$\mathbb{R}^{2}$</span>. His center is <span class="math-container">$\{ \pm I\} \simeq \mathbb{Z}_{2}$</span>. I'm having problems to study the center of the quotient <span class="math-container">$\frac{O_{2}}{\{ \pm I\}}$</span>. Someone could clarify?</p>
<p>Thanks in advance.</p>
| Inácio | 400,062 | <p>I was ignoring the part "anticommutes with every orthogonal matrix". Indeed, we can look to commutators:</p>
<p><span class="math-container">\begin{align}
[Rot(\phi), Ref(\psi)] &= Rot(-2\phi)\\
[Rot(\phi), Rot(\psi)] &= I_{2}\\
[Ref(\phi), Ref(\psi)] &= Rot(4(\phi - \psi))
\end{align}</span></p>
<p>If <span class="math-container">$AZ(O_{2})$</span> <span class="math-container">$\in$</span> <span class="math-container">$Z\Big(\frac{O_{2}}{Z(O_{2})}\Big)$</span>:</p>
<p><span class="math-container">$$ (AB)Z(O_{2}) = (BA)Z(O_{2}) \Rightarrow [A,B] = \pm I_{2}$$</span></p>
<p>Looking to the commutators above, because <span class="math-container">$B$</span> is arbitrary, <span class="math-container">$A$</span> must be a rotation, say <span class="math-container">$Rot(\phi)$</span>, and <span class="math-container">$Rot(-2\phi) = \pm I$</span>, which implies <span class="math-container">$\phi = 0, \pi$</span> or <span class="math-container">$\pm \frac{\pi}{2}$</span>. Hence:
<span class="math-container">$$Z\Big(\frac{O_{2}}{Z(O_{2})}\Big) \simeq \mathbb{Z}_{2}$$</span></p>
<p>So, the center of the quotient isn't trivial.</p>
|
1,263,865 | <p>So I have that $700=7\cdot2^2\cdot5^2$ and I got that $3^2\equiv1\pmod2$ so then $3^{1442}\equiv1\pmod2$ also $3^2\equiv1\pmod{2^2}$ so $3^{1442}\equiv1\pmod{2^2}$ which covers one of the divisors of $700$. Im not sure if I'm supposed to use $2$ or $2^2$ and I was able to find that $3^2\equiv-1\pmod5$ so $3^{1442}\equiv-1\pmod5$, For mod $7$ I wasn't able to come up with an answer in a way like the other two, and I'm not really sure how to do this to find the least non negative residue </p>
| Adhvaitha | 228,265 | <p>Go $\pmod4$, $\pmod7$ and $\pmod{25}$. We have
\begin{align}
3^2 \equiv 1\pmod4\\
3^6 \equiv 1\pmod7\\
3^{20} \equiv 1\pmod{25}
\end{align}
This gives us that
\begin{align}
3^{60} \equiv 1\pmod4\\
3^{60} \equiv 1\pmod7\\
3^{60} \equiv 1\pmod{25}
\end{align}
This means
$$3^{60} \equiv 1\pmod{700}$$
Note that $3^{1442} = 3^{24\cdot60+2} = \left(3^{60}\right)^{24} \cdot 3^2$. Hence, we obtain
$$3^{1442} \equiv 3^2\pmod{700} \equiv9\pmod{700}$$</p>
|
225,128 | <p>I'm trying to define a function accepting only real values like this:</p>
<pre><code>f[x_Real] := x^2
f[0]
</code></pre>
<p>But it outputs</p>
<blockquote>
<p><code>f[0] </code></p>
</blockquote>
<p>and doesn't output 0.</p>
<p>Is there any reason why <code>f[x_Real]</code> doesn't work? I tested <code>f[x_Integer]</code> and <code>f[x_Complex]</code> and they both seem to work.</p>
| Natas | 67,431 | <p>Let me explain what went wrong.</p>
<p>When you define a function <code>f[pattern] := ...</code> then this will check at every occurrence of <code>f[x]</code> if <code>x</code> matches <code>pattern</code>.</p>
<p>Your pattern was <code>x_Real</code> which means "match any expression whose <code>Head</code> is <code>Real</code>". In Mathematica floating point numbers have the <code>Head</code> <code>Real</code>.</p>
<pre><code>In[1]:= Head[0]
Out[1]= Integer
In[2]:= Head[0.]
Out[2]= Real
</code></pre>
<p>However, you do not care whether or not the argument has had <code>Real</code>, rather you want it to be an element of the set <code>Reals</code>. This is what Artes suggested in the comments by using a <code>Condition</code></p>
<pre><code>f[x_] /; Element[x, Reals] := x^2
</code></pre>
<p>Alternatively you could also use a <code>PatternTest</code></p>
<pre><code>f[x_?Element[#, Reals]&] := x^2
</code></pre>
|
2,204,944 | <p>A line is a collection of infinitely many points. By definition, a point has no dimensions. But, how can infinitely many dimensionless points give rise to a line with a dimension. This is the same case with planes, solids and higher dimensions too...</p>
<p>Thanks in advance for any help..!!</p>
| Graham Kemp | 135,106 | <p><strong>Long story short:</strong> The random variables have identical expectations when they follow <em>identical distributions</em>.</p>
<hr>
<p>I have a deck of these four cards, with values 1,2,3,4. I shuffle the deck and place two cards face down, one on your left and one on your right. </p>
<p>What is your expectation for the value of the card on your left?</p>
<p>What is your expectation for the value of the card on your right?</p>
<hr>
<p><em>Hint:</em> I have not told you which I placed down first. Does it matter?</p>
<p>Indeed, let us go ahead an place the other two cards face down. What is your expectation for the values of each of these as well?</p>
<hr>
<p>You see, the dependency of the cards does means that the <em>conditional expectation</em>, $\mathsf E(X_2\mid X_1)$ will indeed not be constant but rather a function of $X_1$; however we are <em>not discussing</em> the conditional expectations, just the expectation.</p>
<p>Still, let us examine the conditional expectation and use the law of iterated expectation:</p>
<p>$$\begin{align}\mathsf E(X_2\mid X_1=k) ~&=~\begin{cases} \tfrac 13(~~~~~~~2+3+4)&:&k=1\\\tfrac 13(1+~~~~~~~3+4)&:&k=2\\\tfrac 13(1+2+~~~~~~~4)&:&k=2\\\tfrac 13(1+2+3~~~~~~~)&:&k=4 \end{cases}\\[2ex] \mathsf E(X_2) ~&=~ \mathsf E(\mathsf E(X_2\mid X_1)) \\[1ex] &=~ \tfrac 14\big(\mathsf E(X_2\mid X_1=1)+\mathsf E(X_2\mid X_1=2)+\mathsf E(X_2\mid X_1=3)+\mathsf E(X_2\mid X_1=4)\big)\\[1ex] &=~ \tfrac 14(1+2+3+4) \\[1ex] &=~ \mathsf E(X_1)\end{align}$$</p>
<hr>
<p><strong>Contrast with Coupon Collection. Why is it not the same?</strong></p>
<p>Where as the random variables for the cards are the values of the cards drawn, the random variables used in the coupon collection are: the count of draws (with replacement) after the $k-1$-th new value until the $k$ new value is draw ("new value" meaning a value not previously drawn).</p>
<p>That is, with the same deck of cards, let $Y_1$ be the count of draws until the first new value is drawn, $Y_2$ be the count of draws <em>after</em> that until the second new value is drawn, $Y_3$ be the count of draws <em>after</em> that until the third new value is drawn, and $Y_4$ be the count of draws <em>after</em> that until the last new value is drawn. </p>
<p>$$\mathsf E(Y_1)=1, \mathsf E(Y_2)=4/3, \mathsf E(Y_3)=4/2, \mathsf E(Y_4)=4$$</p>
<p>These <em>random variables</em> have different expectations <em>not</em> because they are dependent or independent,† but rather <em>it is</em> because the random variables have <em>different distributions</em>.‡</p>
<p>(† They are in fact <em>independent</em>.)</p>
<p>(‡ They follow Geometric Distributions with <em>different rates</em>.)</p>
|
834,228 | <p>$$u_{1}=2, \quad u_{n+1}=\frac{1}{3-u_n}$$
Prove it is decreasing and convergent and calculate its limit.
Is it possible to define $u_{n}$ in terms of $n$?</p>
<p>In order to prove it is decreasing, I calculated some terms but I would like to know how to do it in a more "elaborated" way.</p>
| Claude Leibovici | 82,404 | <p>If there is a limit, it will be defined by $$L=\frac{1}{3-L}$$ which reduces to $L^2-3L+1=0$. You need to solve this quadratic and discard any root greater than $2$ since this is the starting value and that you proved that the terms are decreasing.</p>
|
834,228 | <p>$$u_{1}=2, \quad u_{n+1}=\frac{1}{3-u_n}$$
Prove it is decreasing and convergent and calculate its limit.
Is it possible to define $u_{n}$ in terms of $n$?</p>
<p>In order to prove it is decreasing, I calculated some terms but I would like to know how to do it in a more "elaborated" way.</p>
| JimmyK4542 | 155,509 | <p>As others have said, you don't need a formula for $u_n$ to show that it is strictly decreasing and converges to $\dfrac{3-\sqrt{5}}{2}$. </p>
<p>However, You can get an exact formula for $u_n$. Lets assume $u_n = \dfrac{a_n}{b_n}$ for integers $a_n, b_n$. </p>
<p>Then, $\dfrac{a_{n+1}}{b_{n+1}} = u_{n+1} = \dfrac{1}{3-u_n} = \dfrac{1}{3-\dfrac{a_n}{b_n}} = \dfrac{b_n}{3b_n-a_n}$. </p>
<p>So, we can let $a_{n+1} = b_n$, and $b_{n+1} = 3b_n - a_n$. </p>
<p>Thus, $a_{n+1} = b_n = 3b_{n-1} - a_{n-1} = 3a_n - a_{n-1}$</p>
<p>The solution to this recurrence is $a_n = C_1\left(\dfrac{3+\sqrt{5}}{2}\right)^n + C_2\left(\dfrac{3-\sqrt{5}}{2}\right)^n$. </p>
<p>Since $u_1 = 2$ and $u_2 = 1$, we have $a_1 = 2$ and $a_2 = 1$. Solve for $C_1$ and $C_2$ to get: $a_n = \left(1-\dfrac{2}{\sqrt{5}}\right)\left(\dfrac{3+\sqrt{5}}{2}\right)^{n-1} + \left(1+\dfrac{2}{\sqrt{5}}\right)\left(\dfrac{3-\sqrt{5}}{2}\right)^{n-1}$. </p>
<p>Then, $u_n = \dfrac{a_n}{b_n} = \dfrac{a_n}{a_{n+1}} = $ (really ugly expression). </p>
<p>Needless to say, the problem doesn't intend for you to find an explicit formula for $u_n$, although a formula does exist. </p>
|
3,964,429 | <p>Zeckendorf : <em>Every positive integer N can be expressed uniquely as a sum of distinct non-consecutive Fibonacci numbers</em></p>
<p>I was wondering if this theorem can be applied with the extended Fibonacci numbers, and especially I am looking for a way to <strong>find the Zeckendorf-like representation of <span class="math-container">$ N $</span>, with Fibonacci numbers <span class="math-container">$ F_n $</span> having only negative indexes</strong>. (First for <span class="math-container">$ N \in \mathbb{N} $</span> then maybe <span class="math-container">$ N \in \mathbb{Z} $</span>)</p>
<p><a href="https://mathworld.wolfram.com/ZeckendorfRepresentation.html" rel="nofollow noreferrer">Mathworld</a> states that this theorem only applies on positive numbers but do not says if Fibonacci numbers can be negative.
Online <a href="https://www.dcode.fr/zeckendorf-representation" rel="nofollow noreferrer">Zeckendorf Representation</a> pages show positive indexes only and mostly use Binet's formula.</p>
<p>I've read that Binet's formula can be applied to generalized Fibonacci numbers (<a href="https://proofwiki.org/wiki/Euler-Binet_Formula/Negative_Index" rel="nofollow noreferrer">Proof</a>), but it does not prove that Zeckendorf theorem can be used as well.</p>
<p><strong>EDIT :</strong> I missed a paragraph in <a href="https://en.wikipedia.org/wiki/Zeckendorf%27s_theorem#Representation_with_negafibonacci_numbers" rel="nofollow noreferrer">this Wikipedia page</a> called Representation with Negafibonacci numbers that states this Zeckendorf-like representation exists and is unique. I probably can use <a href="https://en.wikipedia.org/wiki/NegaFibonacci_coding" rel="nofollow noreferrer">NegaFibonacci</a> coding method to find the indexes.</p>
| Crypto | 234,832 | <p><a href="https://www.fq.math.ca/Scanned/30-2/bunder.pdf" rel="nofollow noreferrer">This paper</a> that proves the existence and shows a method</p>
<p><em>(we prove that) every integer can be represented uniquely as a sum of nonconsecutive Fibonacci numbers Fi where i < 0 and we specify an algorithm that leads to this representation</em></p>
|
4,206,039 | <p>Find the radius of convergence of the following power series <span class="math-container">$$\sum_{n=1}^\infty \frac{(-1)^n z^{n(n+1)}}{n}$$</span></p>
<p>Here's my working
<span class="math-container">$$\lim_{n\to \infty}| \frac{(-1)^{n+1} z^{(n+1)(n+2)}}{n+1} \frac{n}{(-1)^nz^{n(n+1)}}|$$</span>
<span class="math-container">$$=\lim_{n\to \infty}\big| \frac{(-1)^n(-1) z^{(n^2+3n+2)}}{n+1} \frac{n}{(-1)^nz^{n^2}z^n}\big|$$</span>
<span class="math-container">$$= \lim_{n\to \infty}\big| \frac{ -z^{3n}z^2}{n+1} \frac{n}{{}z^n}\big|$$</span>
<span class="math-container">$$=\lim_{n\to \infty}\big| \frac{ -nz^{2n}z^2}{n+1}\big|$$</span>
I am stuck after this. Is the limit greater than 1 since anything raised to the power infinity is very huge? How do I do this? Also what happens if z=i?</p>
| DonAntonio | 31,254 | <p>You could try to do the following to see your series as a "usual" power series:</p>
<p><span class="math-container">$$\frac{(-1)^n z^{n(n+1)}}n=a_mz^m\;,\;\;\text{when}\;\;a_m=\begin{cases}0,&m\neq2,6,12,...,k(k+1)\\{}\\\cfrac{(-1)^m}m,&m=k(k+1)\end{cases}\;,\;\;k\in\Bbb N$$</span></p>
<p>From here, using Cauchy-Hadamard theorem:</p>
<p><span class="math-container">$$\frac1R=\limsup_{n\to\infty}\sqrt[n]{|a_n|}=\lim_{n\to\infty}\frac1{\sqrt[n]n}=1$$</span></p>
<p>so <span class="math-container">$\;R=1\;$</span> .</p>
<p>It's important to note that the series could be <em>not</em> considered a power series per dry definition (depending on the author), but it is possible to turn it into one as above.</p>
|
2,073,923 | <p>It seems the number of nonnegative integer solutions to the equation $xyz=n$ is given by
$$\sum\limits_{d \mid n} \tau(d)$$</p>
<p>$\tau$ is the number of divisors function. I'm wondering if there is a way to simplify this sum. Really appreciate any kind of help. Thank you.</p>
<hr>
<p>Here is my attempt so far
$$xyz = n$$</p>
<p>$x$ can be any of the factors of $n$ and the product $yz$ will be $n/x$.
Since $yz$ sees all the factors of $n$, the number of nonnegative integer solutions to $xyz=n$ is simply the sum of divisors of the product $yz$.</p>
<p>Edit : Special thanks to @Tryss for identifying an error in the formula. I've fixed it now..</p>
| Jack D'Aurizio | 44,121 | <p>Factor $n$ as $p_1^{\alpha_1}\cdot\ldots\cdot p_k^{\alpha_k}$. Then every solution is associated with three vectors (the exponents in the factorizations of $x,y,z$) with non-negative integer components and sum given by $(\alpha_1,\ldots,\alpha_k)$. By <a href="https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)" rel="nofollow noreferrer">stars and bars</a>, it follows that the number of solutions is given by
$$ \prod_{h=1}^{k}\frac{(\alpha_h+2)(\alpha_h+1)}{2}.$$</p>
|
204,612 | <blockquote>
<p>Is it possible to verify the following <code>lhs,rhs</code> involving the sums
are equal, with Mathematica?</p>
</blockquote>
<p>I can verify it for individual values of <span class="math-container">$d$</span> variable:</p>
<pre><code>ClearAll[d, q, h, eq1, eq2, x, lhs, rhs];
eq1[d_: d, q_: q, h_: h] := Sum[
((-1)^(d + 1 - k))*(1/((d + 1) Factorial[(k - 1)] Factorial[(d + 1 - k)]))*
Product[(q/h + (k - i)), {i, 1, d + 1}] *
Product[(1 - j h), {j, k, d}] *
Product[(1 + l h), {l, d - k + 2, d}]
, {k, 1, d + 1}]
eq2[d_: d, q_: q, h_: h] := Sum[
((-1)^(d - k))*(1/((d) Factorial[(k - 1)] Factorial[(d - k)]))*
Product[(q/h + (k - i)), {i, 1, d}] *
Product[(1 + (k - j) h), {j, 1, d}]
, {k, 1, d}]
lhs[d_: d, x_: x, h_: h] := (x - 1)^d eq1[d, 1/(x - 1), h]
rhs[d_: d, x_: x, h_: h] := (x)^d eq2[d, 1/x, (((x - 1) h)/x)]
t = 0;
Do[r = PossibleZeroQ[lhs[d] - rhs[d]]; Print[d, " ", r], {d, 1, 100}];
</code></pre>
<p>And it is true for first <span class="math-container">$100$</span> values of <span class="math-container">$d$</span>, for example.</p>
<p>But <code>PossibleZeroQ[lhs[d] - rhs[d]]</code> returns <code>false</code> for general <span class="math-container">$d$</span>. </p>
<p>I've tried simplifying and expanding, with no success.</p>
<p>I've added method <code>Method -> "ParallelBestQuality"</code> to both sums to obtain expressions in terms of <code>Gamma</code> and <code>HypergeometricPFQRegularized</code>, as following:</p>
<p><br>
lhs: </p>
<pre><code>((-1)^d (-h)^d (-1 + x)^d Gamma[1 + d - 1/h] Gamma[1 + 1/(h (-1 + x))] HypergeometricPFQRegularized[{-d, -((1 + d h)/ h), 1 + 1/(h (-1 + x))},{(-1 + h)/h, -d + 1/(h (-1 + x))}, 1])/Gamma[2 + d]
</code></pre>
<p>rhs:</p>
<pre><code>((-1)^(1 + d) (h (-1 + 1/x))^d x^d Gamma[1 + 1/(h (-1 + x))] Gamma[(x + h (-1 + d + x - d x))/(h (-1 + x))] Gamma[d + x/(h - h x)] HypergeometricPFQRegularized[{1 - d, 1 + 1/(h (-1 + x)), 1 + x/(h (-1 + x))}, {1 - d + 1/(h (-1 + x)), (x + h (-1 + d + x - d x))/(h (-1 + x))},1])/(Gamma[1 + d] Gamma[x/(h - h x)])
</code></pre>
<blockquote>
<p>Is it possible for Mathematica to simplify one of these two into the
other? (Show difference is <span class="math-container">$0$</span>?)</p>
</blockquote>
<hr>
<p>Solving this will also answer the following question from MSE: <a href="https://math.stackexchange.com/q/3330637/318073">Equality like Pascal triangle</a>.</p>
<p>(The source of sums)</p>
| JimB | 19,758 | <p>I know you want the expectation for any positive value of <span class="math-container">$\kappa$</span> but here is the resulting expectation for integer values of <span class="math-container">$\kappa$</span>:</p>
<pre><code>mean[κ_, b_, μ_, σ_, ρ_, ω_] := b Exp[μ + σ^2/2] *
Sum[Binomial[κ, i] ω^i Exp[i ρ (μ + (1 + i ρ/2) σ^2)],
{i, 0, κ}]
</code></pre>
<p>For example, </p>
<pre><code>mean[4, b, μ, σ, ρ, ω]
</code></pre>
<p><span class="math-container">$$b e^{\mu +\frac{\sigma ^2}{2}} \left(\omega ^4 e^{4 \rho \left(\mu +(2 \rho +1) \sigma ^2\right)}+4 \omega ^3 e^{3 \rho \left(\mu +\left(\frac{3 \rho }{2}+1\right) \sigma ^2\right)}+6 \omega ^2 e^{2 \rho \left(\mu +(\rho +1) \sigma ^2\right)}+4 \omega e^{\rho \left(\mu +\left(\frac{\rho }{2}+1\right) \sigma ^2\right)}+1\right)$$</span></p>
<p>As a partial check on that formula:</p>
<pre><code>d = TransformedDistribution[b*x*(1 + ω*x^ρ)^κ, x \[Distributed] LogNormalDistribution[μ, σ],
Assumptions -> {b > 0, w ∈ Reals, ρ > 0, κ >= 1}];
ExpandAll[mean[#, b, μ, σ, ρ, ω] - Mean[d /. κ -> #]] & /@ Range[6]
(* {0, 0, 0, 0, 0, 0} *)
</code></pre>
|
235,945 | <p>Hello please help me with these trig identities and double angles as I am not sure where I am going wrong but I keep getting the wrong answer </p>
<p>This is the problem
$$
\sin(\theta+30) = 2\cos(\theta)
$$
This is my one of my incorrect solutions</p>
<p>$$\sin(\theta +30) = 2\cos(\theta)$$
$$\sin(\theta)\cos(30) + \sin(30)\cos(\theta)=2(1 - \sin(\theta))$$
$$\sin(\theta)(\frac{\sqrt3}{2})+(\frac{1}{2})(1 - \sin(\theta))=2-2\sin(\theta)$$
$$\frac{\sin(\theta)\sqrt3+1-\sin(\theta)}{2}+2 \sin(\theta)=2$$</p>
<p>I get stuck and I am not sure what to do with this problem.</p>
<p>Please help as I am trying to self teach my A -level maths.</p>
<p>Thanks in advance</p>
| Will Jagy | 10,400 | <p>As $\cos^2 \theta + \sin^2 \theta = 1,$ I'm afraid
$$ \cos \theta = \pm \sqrt{1 - \sin^2 \theta} $$
which is useless for your purposes. So leave the cosine on you right-hand side as it is, your substitution is just wrong. </p>
<p>Alright, if you do it properly, you get a relationship between $\sin \theta$ and $\cos \theta$ which can be rewritten as specifying a value for $\tan \theta.$</p>
<p>Meanwhile, I am a big fan of drawing graphs. I recommend you get some graph paper or quadrille paper and draw $\theta,y$ axes with, say, $0 \leq \theta \leq 360^\circ$ and then draw your two curves, $y = \sin (\theta + 30^\circ)$ and $y = 2 \cos \theta.$ It will be fairly clear when they cross.</p>
<p>Here is a jpeg of empty axes I just made. Since one axis is degrees and one real numbers, I did not worry about relative scale. I also used negative degrees, that aspect does not really matter, but I can understand if degrees up to 360 would be preferred. Anyway, anyone can email me for a jpeg. Believe me, the practice in drawing these things is invaluable. I guess I will draw this specific graph next.</p>
<p>==================</p>
<p><img src="https://i.stack.imgur.com/24Qxv.jpg" alt="enter image description here"></p>
<p>==================</p>
<p>Alright, I did the graphs. As you can see, (and confirm yourself once you see the likely candidate points), the graphs cross at $60^\circ$ and $-120^\circ.$ By adding $360^\circ,$ we find that $-120^\circ$ is equivalent to $240^\circ.$ And these are the places where $\tan \theta = \sqrt 3.$</p>
<p>==================</p>
<p><img src="https://i.stack.imgur.com/gHXUT.jpg" alt="enter image description here"></p>
<p>================== </p>
<p>Doing these graphs yourself really does help. It's true.</p>
|
235,945 | <p>Hello please help me with these trig identities and double angles as I am not sure where I am going wrong but I keep getting the wrong answer </p>
<p>This is the problem
$$
\sin(\theta+30) = 2\cos(\theta)
$$
This is my one of my incorrect solutions</p>
<p>$$\sin(\theta +30) = 2\cos(\theta)$$
$$\sin(\theta)\cos(30) + \sin(30)\cos(\theta)=2(1 - \sin(\theta))$$
$$\sin(\theta)(\frac{\sqrt3}{2})+(\frac{1}{2})(1 - \sin(\theta))=2-2\sin(\theta)$$
$$\frac{\sin(\theta)\sqrt3+1-\sin(\theta)}{2}+2 \sin(\theta)=2$$</p>
<p>I get stuck and I am not sure what to do with this problem.</p>
<p>Please help as I am trying to self teach my A -level maths.</p>
<p>Thanks in advance</p>
| josh | 11,815 | <p>Your addition identity is almost correct. You can't say $\cos(\theta) = 1 - \sin(\theta)$. However, you can say $\cos^2(\theta) = 1 - \sin^2(\theta)$ from the Pythagorean Identity. Just be aware that you CANNOT drop the squares in this equation by taking the square root of both sides. Now,
\begin{array}{ccc}
\sin(\theta + 30) & = & \sin(\theta)\cos(30) + \cos(\theta)\sin(30) \\
& = & \sin(\theta)\frac{\sqrt{3}}{2} + \cos(\theta)\frac{1}{2} \\
& = & \frac{1}{2}\left[\sqrt{3}\sin(\theta) + \cos(\theta)\right]
\end{array}</p>
<p>If this expression was equal to $2\cos(\theta)$, we would have:</p>
<p>\begin{array}{ccc}
\frac{1}{2}\left[\sqrt{3}\sin(\theta) + \cos(\theta)\right] & = &2\cos(\theta) \\
\sqrt{3}\sin(\theta) + \cos(\theta)& = & 4\cos(\theta) \\
\sqrt{3}\sin(\theta) & = & 3\cos(\theta) \\
\sin(\theta) & = & \sqrt{3}\cos(\theta)\\
\end{array}</p>
<p>Since this doesn't hold for all $\theta$, this is not an identity. As mentioned in the comments, you can get to $\tan(\theta)$ from what I ended up with. Then, you can solve for $\theta$.</p>
|
31,099 | <p>I was wondering what "anti-optimization" is about? Is it related to optimization? What topics does it cover? </p>
<p>All I can find out from Google is <a href="http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6TJ4-42WP6GH-M&_user=10&_coverDate=07/31/2001&_rdoc=1&_fmt=high&_orig=gateway&_origin=gateway&_sort=d&_docanchor=&view=c&_rerunOrigin=google&_acct=C000050221&_version=1&_urlVersion=0&_userid=10&md5=abd081ecba664ee663f2157095a2bb4a&searchtype=a" rel="nofollow">this paper</a>. It looks like having some relation with probability and optimization?</p>
<p>Thanks and regards!</p>
| John | 99,531 | <p>Anti-optimization is a buzzword for "worst case" and the "hybrid optimization/anti-optimization" is a buzzword for <a href="http://en.wikipedia.org/wiki/Maximin_%28decision_theory%29%22maximin%22" rel="nofollow">Maximin</a>.</p>
<p>It should be pointed out that the invention of such buzzwords does not represent any scientific progress. In fact, it is counter-productive distraction. This is so especially in cases where the use of such buzzwords gives the (false) illusion that the theory/method described is new, while in fact the only thing new is the use of new buzzwords. </p>
|
1,984,178 | <p>I have a problem with the following exercise:</p>
<p>We have the operator $T: l^1 \to l^1$ given by</p>
<p>$$T(x_1,x_2,x_3,\dots)=\left(\left(1-\frac11\right)x_1, \left(1-\frac12\right)x_2, \dots\right)$$ for $(x_1,x_2,x_3,\dots)$ in $l^1$. Showing that this operator is bounded is easy, but I am really desperate with showing that the norm $\|T\| = 1$.</p>
<p>I know that for bounded operators the norm is defined as $\|T\|=\sup{\left\{\|T(x)\|: \|x\| \le 1\right\}}$.</p>
<p>I am also wondering if there exists a x in $ l^1$ such that $\|x\|=1 $</p>
<p>and $\|T(x)\|= \|T\|$</p>
<p>Thank you! :)</p>
| Hermès | 127,149 | <p><strong>Hint:</strong> Define $X^n = (\underbrace{\frac{1}{n}, \dots ,\frac{1}{n}}_{n}, 0, 0, ...)$. Show that for every $\epsilon >0$, there exists $N$ such that $||X^N||_{l^1} > 1-\epsilon$, thus proving the statement.</p>
|
1,662,226 | <p>Find a sufficient statistic for $σ^2$ with $μ$ known, where $X_i$ is a random sample from $N(μ,σ^2)$</p>
<p>I was able to find a sufficient statistic for $μ$ with $σ^2$ known, but I'm stuck on finding one for $σ^2$ when $μ$ is known. Can anyone give me some help? </p>
<p>I was using the factorization method before, is this the best way?</p>
| egreg | 62,967 | <p>Just square:</p>
<p>$0.4^2=0.16$</p>
<p>$0.04^2=0.0016$</p>
<p>$0.2^2=0.04$</p>
<p>$0.02^2=0.0004$</p>
<p>$0.13^2=0.0169$</p>
<p>Can you choose?</p>
|
1,662,226 | <p>Find a sufficient statistic for $σ^2$ with $μ$ known, where $X_i$ is a random sample from $N(μ,σ^2)$</p>
<p>I was able to find a sufficient statistic for $μ$ with $σ^2$ known, but I'm stuck on finding one for $σ^2$ when $μ$ is known. Can anyone give me some help? </p>
<p>I was using the factorization method before, is this the best way?</p>
| colormegone | 71,645 | <p>If you rationalize the denominator of your ratio to get $ \ \frac{ \sqrt{10} }{25} \ $ , you can use the fact that $ \ \sqrt{10} \ $ is a little bigger than 3 to estimate that the number in question is a bit larger than $ \ \frac{3}{25} \ = \ 0.12 \ $ . No other choice but (E) is close to that.</p>
|
1,200,919 | <p>Let $x$ be the solution of the equation $x^x=2$. Is $x$ irrational? How to prove this?</p>
| Vincenzo Oliva | 170,489 | <p>Suppose $x$ is rational. Then there exist two integers $a,b$ such that $$\left(\frac{a}{b}\right)^{a/b}=2 \\ \frac{a}{b} = 2^{b/a}.$$ But that's impossible because the RHS is rational only for $a=1$, which actually makes it also integer, while with $a=1$ the LHS is non-integer for all $b>1$. Checking that $(a,b)=(1,1)$ yields the false identity $1=2$ concludes the proof.</p>
|
1,821,927 | <p>Let $V = \big\{z: |z|<5,\text{Im}(z)>0 \big\}$. Let $f$ analytic in $V$, continuous in $\overline{V}$ and suppose $$\forall x \in \left[ -5,5\right]:\ f\left( x\right) \in \mathbb{R}$$
Show that $$\limsup_{n \rightarrow \infty} \root{n}\of{\frac{f^{(n)}(1)}{n!}} \le \frac{1}{4}$$
<br><br><br>
I tried expanding to power series near $z=1$ but it's impossible since $1\in \partial V$, then I tried device an analytic expansion of $f$ to the disc $|z| \leq 5$ by defining
$$g(z) = \cases{f(z), \qquad &\text{Im}(z) \geq0 \\ -f( \overline{z}), \qquad &\text{Im}(z) < 0}$$
But I cant show that $g$ is analytic in the whole disc.
How can I show that $f$ is holomorphicaly extendible to the whole disk of radius 5?
<br>Help<br><br><br></p>
| mercio | 17,445 | <p>Let $V'$ be the complex conjugate of $V$, $\gamma$ the path following the border of $V$ in a direct orientation, $\gamma'$ the path following the border of the $V'$ also in a direct orientation, and $W$ the open ball of radius $5$ centered at $0$ (so that $\overline W = \overline V \cup \overline {V'}$)</p>
<p>Then we can define an continuous extension $g$ of $f$ to $\overline W$ by $g(z) = \overline {f(\overline z)}$ for $z \in \overline {V'}$.</p>
<p>Since $g$ is continuous on $\overline W$ and analytic on $V$ and $V'$, for any $z \in V$ you have by the residue theorem the equalities $2i\pi g(z) = \int_\gamma g(w)dw/(w-z)$ and $0 = \int_{\gamma'} g(w)dw/(w-z) $.</p>
<p>Adding the two integrals you get that $2i\pi g(z) = \int_C g(w)dw/(w-z)$ where $C$ is the circle of radius $5$.</p>
<p>The same result is true if $z \in V'$, and because both sides extend continuously to $W$, the result is also true for $z \in W$, and this shows that $g$ is analytic on $W$.</p>
|
3,450,581 | <p>I know convergence-preserving functions have been discussed a fair amount in the past; however, I was a looking at <a href="https://math.stackexchange.com/questions/1337042/sum-a-n-converges-iff-sum-fa-n-converges/1337057#1337057">another post</a>, and I saw the following result: if <span class="math-container">$f$</span> is Lipschitz and <span class="math-container">$f(0)=0$</span>, then <span class="math-container">$\sum |a_n| <\infty \Rightarrow \sum |f(a_n)|<\infty$</span>. How exactly would I go about proving this statement? I'm also guessing from other posts that this is a sufficient but not necessary condition?</p>
<p>Edit: made it absolute convergence instead.</p>
| Cye Waldman | 424,641 | <p>According to <em>A Catalog of Special Plane Curves</em>, J. Dennis Lawrence, Dover, 1972, the equations for the epicycloid are</p>
<p><span class="math-container">$$x=m\cos t-b\cos mt/b\\
y=m\sin t-b\sin mt/b
$$</span></p>
<p>where <span class="math-container">$t\in[-\pi,\pi]$</span> and <span class="math-container">$m=a+b$</span> (i.e., the radii).</p>
<p>However, it states that <span class="math-container">$L=8m$</span> (<strong>if <span class="math-container">$m/b$</span> is an integer</strong>). Therein lies your problem.</p>
|
206,318 | <p>For plots like the one shown below, what is the syntax for adding filling between particular lines and the axis, but only in the negative region:</p>
<p><a href="https://i.stack.imgur.com/X4TME.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/X4TME.png" alt="enter image description here"></a></p>
| Kuba | 5,478 | <pre><code>Want // ClearAll;
x = 111;
Want /: SetDelayed[lhs_, Want[s_String]] := ToExpression[
s,
InputForm,
Function[rhs, SetDelayed @@ Hold[lhs, rhs], HoldAll]
]
H[x_, y_] := Want["x^2+y^2"]
H[3, 4]
</code></pre>
<blockquote>
<p>25</p>
</blockquote>
|
4,347,308 | <p><em><strong>Definition:</strong></em></p>
<p>Let <span class="math-container">$(X,\mathscr{A},\mu)$</span> be a measurable space, an atom of the measure <span class="math-container">$\mu$</span> is a set <span class="math-container">$A \in\mathscr{A}$</span> with the property that
<span class="math-container">$\mu(A) > 0$</span> and for any <span class="math-container">$B\in \sigma (A)$</span> either <span class="math-container">$\mu(B) = 0$</span>, or <span class="math-container">$\mu(A \setminus B) = 0$</span>. If a measure has
atoms it is called atomic; in the opposite case, the measure is called non-atomic (or
atomeless). A measure is called purely atomic if <span class="math-container">$X$</span> can be written as the union of a
finite or countable number of atoms.</p>
<p>From the definition of atoms, we get the following corollary:</p>
<p><em><strong>Corollary:</strong></em></p>
<p><em>Every purely atomic measure is an atomic measure.</em></p>
<p>I am trying to find an example of an atomic measure that is not purely atomic, can anyone help me?</p>
| heropup | 118,193 | <p>Don't do more work than you need to. The better approach is to perform the polynomial long division first:</p>
<p><span class="math-container">$$\frac{1+x^3}{1+x^2} = x + \frac{1-x}{1+x^2}.$$</span> Now apply your series formula:</p>
<p><span class="math-container">$$\begin{align}
x + \frac{1-x}{1+x^2}
&= x + \sum_{n=0}^\infty (1-x)(-x^2)^n \\
&= x + \sum_{n=0}^\infty (-1)^n (x^{2n} - x^{2n+1}).\\
\end{align}$$</span>
You could leave this as is, or you could observe that the <span class="math-container">$x$</span> outside the sum cancels with a term in the sum: when <span class="math-container">$n = 0$</span>, the summand is <span class="math-container">$1 - x$</span>, thus an alternative form is
<span class="math-container">$$\frac{1+x^3}{1+x^2} = 1 + \sum_{n=1}^\infty (-1)^n (x^{2n} - x^{2n+1}). \tag{1}$$</span>
But this is still somewhat unsatisfactory, since what would be ideal is to compute <span class="math-container">$f^{(n)}(0)$</span> in terms of <span class="math-container">$n$</span>; i.e., the Maclaurin expansion has the form <span class="math-container">$$f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} x^n.$$</span> Using <span class="math-container">$(1)$</span> to write out the first few terms,
<span class="math-container">$$\frac{1+x^3}{1+x^2} = 1 - x^2 + x^3 + x^4 - x^5 - x^6 + x^7 + x^8 - \cdots,$$</span> so this suggests <span class="math-container">$|f^{(n)}(0)| = n!$</span> except when <span class="math-container">$n = 1$</span>, and the sign is positive if <span class="math-container">$n$</span> is congruent to <span class="math-container">$0$</span> or <span class="math-container">$3$</span> modulo <span class="math-container">$4$</span>; i.e., <span class="math-container">$$f^{(n)}(0) = \begin{cases} 0 & n = 1, \\ n! & n \in \{4m, 4m+3\}, \\ -n! & n \in \{4m+1, 4m+2\} \cap n \ne 1. \end{cases}$$</span></p>
|
153,409 | <p>Would you please tell me whether there is any wrong on this problem? given that $g$ is continuous on $[0,\infty)\rightarrow \mathbb{R}$ satisfying $\int_{0}^{x^2(1+x)}g(t)dt=x \forall x\in [0,\infty)$ then I need to find what is $g(2)$?</p>
| JLA | 30,952 | <p>Note that
$$
\frac{d}{dx}\int_0^{x^2(1+x)}g(t)\,dt=g(x^2(1+x))(2x+3x^2)=\frac{d}{dx}x=1.
$$
So now solve for $g(2)$.</p>
|
716,498 | <p>My Algebraic Topology book says </p>
<blockquote>
<p>Let $\Bbb{R}^n$ denote Euclidean n-space. Then $\pi_1(\Bbb{R}^n,x_0)$ is the trivial subgroup (the group consisting of the identity alone).</p>
</blockquote>
<p>I wonder why that is. I can imagine infinite continuous "loops" in $\Bbb{R}^3$ that start and end at $x_0$.</p>
<p>Thanks in advance!</p>
| Bruno Stonek | 2,614 | <p>The problem with your last sentence is that $\pi_1(X,x_0)$ is not the set of loops based on $x_0$, but of <em>homotopy classes</em> of loops based on $x_0$.</p>
<p>Can you see why every loop in $\mathbb R^n$ based on $x_0\in \mathbb R^n$ is homotopic to the constant map based in $x_0$?</p>
|
323,109 | <p>Could someone help with the following integration:
$$\int^0_\pi \frac{x \sin x}{1+\cos^2 x}$$</p>
<p>So far I have done the following, but I am stuck:</p>
<p>I denoted $ y=-\cos x $ then:
$$\begin{align*}&\int^{1}_{-1} \frac{\arccos(-y) \sin x}{1+y^2}\frac{\mathrm dy}{\sin x}\\&= \arccos(-1) \arctan 1+\arccos 1 \arctan(-1) - \int^1_{-1}\frac{1}{\sqrt{1-y^2}}\frac{1}{1+y^2} \mathrm dy\\&=\frac{\pi^2}{4}-\int^{1}_{-1}\frac{1}{\sqrt{1-y^2}}\frac{1}{1+y^2} \mathrm dy\end{align*}$$</p>
<p>Then I am really stuck. Could someone help me?</p>
| Community | -1 | <p>Let $$I = \int_0^{\pi} \dfrac{x \sin(x)}{1+\cos^2(x)} dx = \int_{-\pi/2}^{\pi/2} \dfrac{(x+\pi/2) \sin(x+\pi/2)}{1 + \cos^2(x+\pi/2)} dx = \int_{-\pi/2}^{\pi/2} \dfrac{(x+\pi/2) \cos(x)}{1 + \sin^2(x)} dx $$
Now
$$\int_{-\pi/2}^{\pi/2} \dfrac{(x+\pi/2) \cos(x)}{1 + \sin^2(x)} dx = \int_{-\pi/2}^{\pi/2} \underbrace{\dfrac{x \cos(x)}{1 + \sin^2(x)}}_{\text{Odd function}} dx + \dfrac{\pi}2 \cdot \int_{-\pi/2}^{\pi/2} \dfrac{\cos(x)}{1 + \sin^2(x)} dx $$
Hence, we get that
$$I = \dfrac{\pi}2 \cdot \int_{-\pi/2}^{\pi/2} \dfrac{\cos(x)}{1 + \sin^2(x)} dx = \dfrac{\pi}2 \cdot \int_{-1}^1 \dfrac{dt}{1+t^2} = \dfrac{\pi}2 \cdot \left( \dfrac{\pi}4 - \dfrac{-\pi}4\right) = \dfrac{\pi^2}4$$
The integral you are after is $-I$ and hence the answer is $-\dfrac{\pi^2}4$.</p>
|
323,109 | <p>Could someone help with the following integration:
$$\int^0_\pi \frac{x \sin x}{1+\cos^2 x}$$</p>
<p>So far I have done the following, but I am stuck:</p>
<p>I denoted $ y=-\cos x $ then:
$$\begin{align*}&\int^{1}_{-1} \frac{\arccos(-y) \sin x}{1+y^2}\frac{\mathrm dy}{\sin x}\\&= \arccos(-1) \arctan 1+\arccos 1 \arctan(-1) - \int^1_{-1}\frac{1}{\sqrt{1-y^2}}\frac{1}{1+y^2} \mathrm dy\\&=\frac{\pi^2}{4}-\int^{1}_{-1}\frac{1}{\sqrt{1-y^2}}\frac{1}{1+y^2} \mathrm dy\end{align*}$$</p>
<p>Then I am really stuck. Could someone help me?</p>
| L. F. | 56,837 | <p>$$I=\int_0^{\pi} \frac{-x\sin x}{1+\cos^2 x}\,dx=\int_0^{\pi} \frac{(x-\pi)\sin x}{1+\cos^2 x}dx\quad(x\to \pi-x)$$</p>
<p>$$\Rightarrow I=\frac{\pi}{2}\int_0^{\pi}\frac{-\sin x}{1+\cos^2 x}\,dx$$</p>
<p>Let $t=\cos x:$</p>
<p>$$I=\frac{\pi}{2}\int_{-1}^{1}-\frac{1}{1+t^2}\,dt=-\frac{\pi^2}{4}$$</p>
|
446,148 | <blockquote>
<p>Let $z_1, . . . , z_n$ and $w_1, . . . , w_n$ be complex numbers. Show
that $$|z_1w_1 + ··· + z_n w_n|^2 ≤ \sum ^n _{j=1} |z_j|^2 \sum ^n
_{j=1}|w_j|^2$$</p>
</blockquote>
<p>I basically tried to use the proof given for real numbers but I feel that something must be wrong. Could somebody look over the proof? </p>
<p>Let $z=(z_1,...,z_n)$ and $w=(w_1,...,w_n)$. Then:</p>
<p>$$0 \leq \sum_j^n\bigg(||z|||w_j|-||w|||z_j|\bigg)^2=||z||^2||w||^2-2||z||||w||\sum_j^n|z_jw_j|+||z||^2||w||^2=2||z||||w||\bigg(||z||||w||-\sum_j^n|z_jw_j|\bigg)$$</p>
<p>But then</p>
<p>$$|\sum_j^nz_jw_j|\leq \sum_j^n|z_jw_j|\leq ||z||||w||$$</p>
<p>And just squaring is left.</p>
<p>Thanks!</p>
| Emily | 31,475 | <p>Let $\omega = z_1w_1 + z_2w_2 + \cdots + z_nw_n$. In general, $|\omega|^2 = \omega \overline{\omega}$.</p>
<p>$$\omega \overline{\omega} = (z_1w_1+z_2w_2 + \cdots + z_nw_n)(\overline{z_1}\overline{w_1} + \cdots + \overline{z_n}\overline{w_n}) \\
= z_1\overline{z_1}w_1\overline{w_1} + z_1\overline{z_2}w_1\overline{w_2} \cdots + z_n\overline{z_n}w_n\overline{w_n} \\
= |z_1|^2|w_1|^2 + \cdots + |z_n|^2 |w_1|^2 + \textrm{mixed terms}$$</p>
<p>now It should be clear from here.</p>
|
446,148 | <blockquote>
<p>Let $z_1, . . . , z_n$ and $w_1, . . . , w_n$ be complex numbers. Show
that $$|z_1w_1 + ··· + z_n w_n|^2 ≤ \sum ^n _{j=1} |z_j|^2 \sum ^n
_{j=1}|w_j|^2$$</p>
</blockquote>
<p>I basically tried to use the proof given for real numbers but I feel that something must be wrong. Could somebody look over the proof? </p>
<p>Let $z=(z_1,...,z_n)$ and $w=(w_1,...,w_n)$. Then:</p>
<p>$$0 \leq \sum_j^n\bigg(||z|||w_j|-||w|||z_j|\bigg)^2=||z||^2||w||^2-2||z||||w||\sum_j^n|z_jw_j|+||z||^2||w||^2=2||z||||w||\bigg(||z||||w||-\sum_j^n|z_jw_j|\bigg)$$</p>
<p>But then</p>
<p>$$|\sum_j^nz_jw_j|\leq \sum_j^n|z_jw_j|\leq ||z||||w||$$</p>
<p>And just squaring is left.</p>
<p>Thanks!</p>
| Pedro | 23,350 | <p>You have a canonical inner product in $\Bbb C^n$ given by $x\cdot y=\displaystyle \sum_{i=1}^n x_i\overline{y_i}$. Note $x\cdot x=\displaystyle \sum_{i=1}^n x_i\overline{x_i}=\displaystyle \sum_{i=1}^n |x_i|^2$</p>
<p><em>Claim</em> Let $V$ be a $\Bbb C$-vector space, and $\langle \cdot,\cdot\rangle$ an inner product. If we define a norm $\lVert x\rVert :=\langle x,x\rangle^{1/2}$, then we always have $$|\langle x,y\rangle|^2\leqslant \lVert x\rVert ^2\lVert y\rVert^2$$</p>
<p><em>Proof</em> If $x$ or $y$ are zero, the inequality is true, since $\langle x,0\rangle=0$ and $\lVert 0\rVert =0$. We may assume then that both norms are nonzero. Since $\langle z,z\rangle\geqslant 0$ for any choice of $z$, we may take $$z=x-\frac{\langle x,y\rangle y}{\lVert y\rVert ^2} $$</p>
<p>Using the inner product is (sesqui)linear we get </p>
<p>$$\begin{align}
\langle z,z\rangle
&= \langle {x,x} \rangle - \frac{1}{\lVert y\rVert^2}\langle {x,\langle x,y \rangle y} \rangle - \frac{1}{\lVert y\rVert^2}\langle {\langle x,y \rangle y,x} \rangle + \frac{{\langle x,y \rangle \overline {\langle x,y \rangle} }}{{\lVert y\rVert^2\lVert y\rVert^2}}\langle {y,y} \rangle \\
&= {\| x \|^2} - \frac{{| \langle x,y \rangle |^2}}{\lVert y\rVert^2} - \frac{{| \langle x,y \rangle |^2}}{\lVert y\rVert^2} + \frac{{| \langle x,y \rangle |^2}}{\lVert y\rVert^2} \\
&= {\| x \|^2} - \frac{{| \langle x,y \rangle |^2}}{\| y \|^2} \geqslant 0\end{align} $$
as we wanted. We made repeated use of $$\langle a+b,c\rangle=\langle a,c\rangle+\langle b,c\rangle\\ \langle a,b+c\rangle=\langle a,b\rangle+\langle a,c\rangle\\\langle \alpha a,b\rangle=\alpha\langle a,b\rangle\\\langle a,\beta b\rangle=\overline \beta\langle a,b\rangle \\z\overline z=|z|^2$$</p>
|
3,845,570 | <p>Premises: <span class="math-container">$\neg(A \to B)\ ,\ \neg B \to C$</span> .</p>
<p>Conclusion: <span class="math-container">$C$</span></p>
<p>My intuition is that I should do a sub-derivation where I prove <span class="math-container">$\neg C$</span> is an absurdity. However, I soon run into issues. If I could prove that <span class="math-container">$B$</span> is an absurdity, that would work also, but I'm not sure how to do so using the first premise.</p>
| Tortar | 704,856 | <p>Write the first premise as <span class="math-container">$\neg\neg(A \land \neg B) \equiv A \land \neg B $</span> , so <span class="math-container">$\neg B$</span> is true. Therefore, from the second premise it follows <span class="math-container">$C$</span>.</p>
|
3,845,570 | <p>Premises: <span class="math-container">$\neg(A \to B)\ ,\ \neg B \to C$</span> .</p>
<p>Conclusion: <span class="math-container">$C$</span></p>
<p>My intuition is that I should do a sub-derivation where I prove <span class="math-container">$\neg C$</span> is an absurdity. However, I soon run into issues. If I could prove that <span class="math-container">$B$</span> is an absurdity, that would work also, but I'm not sure how to do so using the first premise.</p>
| Graham Kemp | 135,106 | <blockquote>
<p>My intuition is that I should do a sub-derivation where I prove <span class="math-container">$\neg C$</span> is an absurdity. However, I soon run into issues. If I could prove that <span class="math-container">$B$</span> is an absurdity, that would work also, but I'm not sure how to do so using the first premise.</p>
</blockquote>
<p>The contradiction of the first premise requires but a conditional proof: derive <span class="math-container">$B$</span> under the assumption of <span class="math-container">$A$</span>.</p>
<p>Since your instinct was to derive that contradiction under the assumption of <span class="math-container">$B$</span>, go with that.</p>
<ul>
<li><span class="math-container">$\neg(A\to B)$</span> by premise</li>
<li><span class="math-container">$\neg B\to C$</span> by premise
<ul>
<li><span class="math-container">$\neg C$</span> by supposition
<ul>
<li><span class="math-container">$B$</span> by supposition
<ul>
<li><span class="math-container">$A$</span> by supposition</li>
<li><span class="math-container">$B$</span> by reiteration</li>
</ul>
</li>
<li><span class="math-container">$A\to B$</span> by deduction (conditional introduction)</li>
<li>Contradiction!</li>
</ul>
</li>
<li><span class="math-container">$\neg B$</span> by denial (negation introduction)</li>
<li><span class="math-container">$C$</span> by affirmation (modus ponens, or conditional elimination)</li>
<li>Contradiction!</li>
</ul>
</li>
<li><span class="math-container">$\neg\neg C$</span> by denial</li>
<li><span class="math-container">$C$</span> by double negation elimination.</li>
</ul>
<p>Which is a valid proof, but a little inspection reveals some redundancy which we may prune away.</p>
<ul>
<li><span class="math-container">$\neg(A\to B)$</span> by premise</li>
<li><span class="math-container">$\neg B\to C$</span> by premise
<ul>
<li><span class="math-container">$B$</span> by supposition
<ul>
<li><span class="math-container">$A$</span> by supposition</li>
<li><span class="math-container">$B$</span> by reiteration</li>
</ul>
</li>
<li><span class="math-container">$A\to B$</span> by deduction (conditional introduction)</li>
<li>Contradiction!</li>
</ul>
</li>
<li><span class="math-container">$\neg B$</span> by denial (negation introduction)</li>
<li><span class="math-container">$C$</span> by affirmation (modus ponens, or conditional elimination)</li>
</ul>
|
2,352,721 | <h2>Question</h2>
<blockquote>
<p>Four fair six-sided dice are rolled. The probability that the sum of the results being <span class="math-container">$22$</span> is <span class="math-container">$$\frac{X}{1296}.$$</span> What is the value of <span class="math-container">$X$</span>?</p>
</blockquote>
<h2>My Approach</h2>
<p>I simplified it to the equation of the form:</p>
<blockquote>
<p><span class="math-container">$x_{1}+x_{2}+x_{3}+x_{4}=22, 1\,\,\leq x_{i} \,\,\leq 6,\,\,1\,\,\leq i \,\,\leq 4 $</span></p>
</blockquote>
<p>Solving this equation results in:</p>
<p><span class="math-container">$x_{1}+x_{2}+x_{3}+x_{4}=22$</span></p>
<p>I removed restriction of <span class="math-container">$x_{i} \geq 1$</span> first as follows-:</p>
<p><span class="math-container">$\Rightarrow x_{1}^{'}+1+x_{2}^{'}+1+x_{3}^{'}+1+x_{4}^{'}+1=22$</span></p>
<p><span class="math-container">$\Rightarrow x_{1}^{'}+x_{2}^{'}+x_{3}^{'}+x_{4}^{'}=18$</span></p>
<p><span class="math-container">$\Rightarrow \binom{18+4-1}{18}=1330$</span></p>
<p>Now i removed restriction for <span class="math-container">$x_{i} \leq 6$</span> , by calculating the number of <strong>bad cases</strong> and then subtracting it from <span class="math-container">$1330$</span>:</p>
<p>calculating <strong>bad combination</strong> i.e <span class="math-container">$x_{i} \geq 7$</span></p>
<p><span class="math-container">$\Rightarrow x_{1}^{'}+x_{2}^{'}+x_{3}^{'}+x_{4}^{'}=18$</span></p>
<p>We can distribute <span class="math-container">$7$</span> to <span class="math-container">$2$</span> of <span class="math-container">$x_{1}^{'},x_{2}^{'},x_{3}^{'},x_{4}^{'}$</span> i.e<span class="math-container">$\binom{4}{2}$</span></p>
<p>We can distribute <span class="math-container">$7$</span> to <span class="math-container">$1$</span> of <span class="math-container">$x_{1}^{'},x_{2}^{'},x_{3}^{'},x_{4}^{'}$</span> i.e<span class="math-container">$\binom{4}{1}$</span> and then among all others .</p>
<p>i.e</p>
<p><span class="math-container">$$\binom{4}{1} \binom{14}{11}$$</span></p>
<p>Therefore, the number of bad combinations equals <span class="math-container">$$\binom{4}{1} \binom{14}{11} - \binom{4}{2}$$</span></p>
<p>Therefore, the solution should be:</p>
<p><span class="math-container">$$1330-\left( \binom{4}{1} \binom{14}{11} - \binom{4}{2}\right)$$</span></p>
<p>However, I am getting a negative value. What am I doing wrong?</p>
<p><strong>EDIT</strong></p>
<p>I am asking for my approach, because if the question is for a larger number of dice and if the sum is higher, then predicting the value of dice will not work.</p>
| Felix Marin | 85,343 | <p>$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$</p>
<blockquote>
<p>The number of configurations that satisfies "the sum is $\ds{22}$" is given by: </p>
</blockquote>
<p>\begin{align}
X & =
\sum_{d_{1} = 1}^{6}\sum_{d_{2} = 1}^{6}\sum_{d_{3} = 1}^{6}\sum_{d_{4} = 1}^{6}
\bracks{z^{22}}z^{d_{1} + d_{2} + d_{3} + d_{4}} =
\bracks{z^{22}}\pars{\sum_{d = 1}^{6}z^{d}}^{4} =
\bracks{z^{22}}\pars{z\,{z^{6} - 1 \over z - 1}}^{4}
\\[5mm] & =
\bracks{z^{18}}{1 - 4z^{6} + 6z^{12} - 4z^{18} + z^{24}\over \pars{1 - z}^{4}}
\\[5mm] & =
\bracks{z^{18}}\pars{1 - z}^{-4} - 4\bracks{z^{12}}\pars{1 - z}^{-4} + 6\bracks{z^{6}}\pars{1 - z}^{-4} - 4\bracks{z^{0}}\pars{1 - z}^{-4}
\\[5mm] & =
{-4 \choose 18}\pars{-1}^{18} - 4{-4 \choose 12}\pars{-1}^{12} +
6{-4 \choose 6}\pars{-1}^{6} - 4 =
{21 \choose 18} - 4{15 \choose 12} + 6{9 \choose 6} - 4
\\[5mm] & =
1330 -4 \times 455 + 6 \times 84 - 4 = \bbx{10}
\end{align}</p>
|
2,352,721 | <h2>Question</h2>
<blockquote>
<p>Four fair six-sided dice are rolled. The probability that the sum of the results being <span class="math-container">$22$</span> is <span class="math-container">$$\frac{X}{1296}.$$</span> What is the value of <span class="math-container">$X$</span>?</p>
</blockquote>
<h2>My Approach</h2>
<p>I simplified it to the equation of the form:</p>
<blockquote>
<p><span class="math-container">$x_{1}+x_{2}+x_{3}+x_{4}=22, 1\,\,\leq x_{i} \,\,\leq 6,\,\,1\,\,\leq i \,\,\leq 4 $</span></p>
</blockquote>
<p>Solving this equation results in:</p>
<p><span class="math-container">$x_{1}+x_{2}+x_{3}+x_{4}=22$</span></p>
<p>I removed restriction of <span class="math-container">$x_{i} \geq 1$</span> first as follows-:</p>
<p><span class="math-container">$\Rightarrow x_{1}^{'}+1+x_{2}^{'}+1+x_{3}^{'}+1+x_{4}^{'}+1=22$</span></p>
<p><span class="math-container">$\Rightarrow x_{1}^{'}+x_{2}^{'}+x_{3}^{'}+x_{4}^{'}=18$</span></p>
<p><span class="math-container">$\Rightarrow \binom{18+4-1}{18}=1330$</span></p>
<p>Now i removed restriction for <span class="math-container">$x_{i} \leq 6$</span> , by calculating the number of <strong>bad cases</strong> and then subtracting it from <span class="math-container">$1330$</span>:</p>
<p>calculating <strong>bad combination</strong> i.e <span class="math-container">$x_{i} \geq 7$</span></p>
<p><span class="math-container">$\Rightarrow x_{1}^{'}+x_{2}^{'}+x_{3}^{'}+x_{4}^{'}=18$</span></p>
<p>We can distribute <span class="math-container">$7$</span> to <span class="math-container">$2$</span> of <span class="math-container">$x_{1}^{'},x_{2}^{'},x_{3}^{'},x_{4}^{'}$</span> i.e<span class="math-container">$\binom{4}{2}$</span></p>
<p>We can distribute <span class="math-container">$7$</span> to <span class="math-container">$1$</span> of <span class="math-container">$x_{1}^{'},x_{2}^{'},x_{3}^{'},x_{4}^{'}$</span> i.e<span class="math-container">$\binom{4}{1}$</span> and then among all others .</p>
<p>i.e</p>
<p><span class="math-container">$$\binom{4}{1} \binom{14}{11}$$</span></p>
<p>Therefore, the number of bad combinations equals <span class="math-container">$$\binom{4}{1} \binom{14}{11} - \binom{4}{2}$$</span></p>
<p>Therefore, the solution should be:</p>
<p><span class="math-container">$$1330-\left( \binom{4}{1} \binom{14}{11} - \binom{4}{2}\right)$$</span></p>
<p>However, I am getting a negative value. What am I doing wrong?</p>
<p><strong>EDIT</strong></p>
<p>I am asking for my approach, because if the question is for a larger number of dice and if the sum is higher, then predicting the value of dice will not work.</p>
| N. F. Taussig | 173,070 | <p>@expiTTp1z0 has addressed where you made your error. </p>
<p>I am going to show you how you can reduce the given problem a simpler one by using symmetry.</p>
<p>You wish to find the number of solutions of the equation
$$x_1 + x_2 + x_3 + x_4 = 22 \tag{1}$$
in positive integers not exceeding $6$. Since $x_k$, $1 \leq k \leq 4$, is a positive integer satisfying $x_k \leq 6$, then $y_k = 7 - x_k$ is also a positive integer not exceeding $6$. Substituting $7 - y_k$ for $x_k$, $1 \leq k \leq 4$, in equation 1 yields
\begin{align*}
7 - y_1 + 7 - y_2 + 7 - y_3 + 7 - y_4 & = 22\\
-y_1 - y_2 - y_3 - y_4 & = -6\\
y_1 + y_2 + y_3 + y_4 & = 6 \tag{2}
\end{align*}
Equation 2 is an equation in the positive integers when the given restrictions are imposed. A particular solution of equation corresponds to placing an addition sign in three of the five spaces between successive ones in a row of six ones. For instance,
$$1 + 1 1 + 1 + 1 1$$
corresponds to the solution $y_1 = 1$, $y_2 = 2$, $y_3 = 1$, and $y_4 = 2$ (or $x_1 = 6$, $x_2 = 5$, $x_3 = 6$, $x_4 = 5$) while
$$1 1 1 + 1 + 1 + 1$$
corresponds to the solution $y_1 = 3$, $y_2 = y_3 = y_4 = 1$ (or $x_1 = 4$, $x_2 = x_3 = x_4 = 6$). Thus, the number of solutions of equation 2 in the positive integers is the number of ways of selecting which three of the five spaces between successive ones in a row of six ones will be filled by addition signs, which is
$$\binom{5}{3} = 10$$ </p>
<p><em>Note</em>: If you prefer to work in the nonnegative integers, then we wish to find the number of solutions of the equation
$$x_1' + x_2' + x_3' + x_4' = 18 \tag{$1'$}$$
subject to the restrictions that $x_k' = x_k - 1 \leq 6 - 1 = 5$ for $1 \leq k \leq 4$. Since $x_k'$, $1 \leq k \leq 5$, is a nonnegative integer not exceeding $5$, so is $y_k' = 5 - x_k'$. Substituting $5 - y_k'$ for $x_k'$ in equation 1' and simplifying yields
$$y_1' + y_2' + y_3' + y_4' = 2 \tag{$2'$}$$
which is an equation in the nonnegative integers. A particular solution corresponds to the insertion of three addition signs in a row of two ones. For instance,
$$+ + + 1 1$$
corresponds to the solution $y_1 = y_2 = y_3 = 0$, $y_4 = 2$ (or $x_1 = x_2 = x_3 = 6$, $x_4 = 4$), while
$$1 + 1 + +$$
corresponds to the solution $y_1 = y_2 = 1$, $y_3 = y_4 = 0$ (or $x_1 = x_2 = 5$, $x_3 = x_4 = 6$). Thus, the number of solutions of equation 2 in the nonnegative integers is
$$\binom{2 + 3}{3} = \binom{5}{3}$$
since we must choose which three of the five positions (two ones and three addition signs) will be filled with addition signs. </p>
|
271,824 | <p>I have a list= {4, 8, 10, 11, 12, 14, 16, 7, 9}</p>
<p>How can i partition the list by group of Arithmetic Progression with common difference 1 :</p>
<p>{{4}, {8}, { 10, 11, 12}, {14}, {16}, {7}, {9}}</p>
| bmf | 85,558 | <pre><code>(li1 /. Thread[
Rule[Complement[Union[li1, li2], Intersection[li1, li2]], x]])
</code></pre>
<blockquote>
<p><code>{x, {2, 9}, x, {4, 7}, {5, 6}, x, x, x, {9, 2}, x}</code></p>
</blockquote>
|
946,973 | <p>After completing the square, what are the solutions to the quadratic equation below?
<span class="math-container">$$x^2 + 2x = 25$$</span></p>
<p><img src="https://i.stack.imgur.com/AoFhV.png" alt="enter image description here" /></p>
<p>Honstely I think it's B. But I'm not sure.</p>
| Jasser | 170,011 | <p>You can verify your answrt using this.
The solution to general quadratic equation $ax^2+bx+c=0$ is given by the formula $\frac {-b+\sqrt {b^2-4ac}}{2a}, \frac {-b-\sqrt {b^2-4ac}}{2a}$
The given equation is $x^2-2x-25=0$
There fore after applying this formula you will get option c as your answer.</p>
|
3,154,244 | <p>I tried to ask this in a different way and did not correctly explain myself.</p>
<p>I am ok integrating the line <span class="math-container">$y = x$</span> , let us say from <span class="math-container">$0$</span> to <span class="math-container">$2$</span> using calculus.
If I want to get the square I can easily multiply by two and using calculus the dimensions work. The area is a square and when we integrate we have a square.</p>
<p>Here is my question. If I use a straight line above the x axis my equation becomes <span class="math-container">$y$</span> = ( some constant) . Now I use calculus and integrate from 0 to 2 , I also get the correct answer BUT I have to imagine it is a rectangle because when you ingtegrate you have <em>one</em> <span class="math-container">$x$</span> term and it is not a <em>square.</em> The answers match OK it's the dimensions that bother me. </p>
<p>I did not do a good job explaining this on my previous question. Sorry </p>
| SNEHIL SANYAL | 636,469 | <p>Consider a rectangle formed by the equation <span class="math-container">$y=K$</span> extending from <span class="math-container">$x=a$</span> to <span class="math-container">$x=b$</span>.
<a href="https://i.stack.imgur.com/OUM4D.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OUM4D.png" alt="enter image description here"></a>
<a href="https://i.stack.imgur.com/4aEJS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4aEJS.png" alt="enter image description here"></a></p>
<p>Calculate the area of a small rectangular strip formed by the coordinates <span class="math-container">$(x,0)$</span>, <span class="math-container">$(x+dx,0)$</span>,<span class="math-container">$(x+dx,y)$</span> and <span class="math-container">$(x,y)$</span>.</p>
<p>The area of this rectangle will be, <span class="math-container">$ydx$</span>. Convert the whole rectangular area into smaller rectangles of area <span class="math-container">$ydx$</span> and sum all of them from <span class="math-container">$x=a$</span> to <span class="math-container">$x=b$</span> using integration.</p>
<p>The area under the curve <span class="math-container">$y=f(x)$</span> bounded by the X Axis and the lines <span class="math-container">$x=a$</span> and <span class="math-container">$x=b$</span> is given by,</p>
<p><span class="math-container">$$\int_{x=a}^{x=b}ydx=\int_{x=a}^{x=b}f(x)dx=\int_{x=a}^{x=b}Kdx=K(b-a)$$</span>
Hope this helps...</p>
|
1,158,956 | <p>To show that orthogonal complement of a set A is closed.</p>
<p>My try: I first show that the inner product is a continuous map. Let $X$ be an inner product space. For all $x_1,x_2,y_1,y_2 \in X$, by Cauchy-Schwarz inequality we get,
$$|\langle x_1,y_1\rangle - \langle x_2,y_2\rangle| = |\langle x_1- x_2,y_1\rangle + \langle x_2, y_1-y_2\rangle| $$
$$\leq \|x_1- x_2\|\cdot\|y_1\| +\|x_2\|\cdot\| y_1-y_2\|$$</p>
<p>This implies continuity of inner products.</p>
<p>Let $A \subset X$ and $y \in A^\perp$. To show that $ A^\perp$ is closed, we have to show that if $(y_n)$ is convergent sequence in $ A^\perp$, then the limit $y$ also belong to $ A^\perp$.</p>
<p>Let $x \in A$, then using that inner product is a continuous map,
$$\langle x,y\rangle = \langle x, \lim_{n\to \infty} (y_n)\rangle = \lim_{n\to \infty} \langle x, y_n\rangle = 0.$$</p>
<p>Since $\langle x, y_n\rangle = 0$ for all $x \in A$ and $y_n \in A^\perp$. Hence $y \in A^\perp$.</p>
<p>Is the approach\the proof correct??</p>
<p>Thank You!!</p>
| Sam Wong | 507,382 | <p>Let <span class="math-container">$\{y_n\}_{n=1}^\infty \in A^\perp$</span> s.t. <span class="math-container">$y_n \to y.$</span></p>
<p>Then <span class="math-container">$\forall x\in A,$</span> we have <span class="math-container">$$\langle y , x\rangle=\lim_{n\to\infty} \langle y_n , x\rangle=\lim_{n\to\infty}0=0,$$</span></p>
<p>where the first equality holds because of the (norm) convergence of <span class="math-container">$y_n$</span> to <span class="math-container">$y$</span> and of the Cauchy Schwartz inequality.</p>
<p>So <span class="math-container">$y\in A^\perp$</span> as desired.</p>
|
2,166,075 | <blockquote>
<p>Prove $a_1+\cdots+a_n=\dfrac{(a_1+a_n)n}{2}$ inductively.</p>
</blockquote>
<p>Where $a_i=a_{i+1}-r$.</p>
<p>I tried to start proving it inductively, but any try lead to a bad conclusion, so I ended up proving it by making $a_n$ depend on $a_i$.</p>
<p>But I didn't know how to prove it inductively, so there is the problem.</p>
<p><strong>EDIT:</strong></p>
<p>I'm looking for a valid induction steps to reach the conclusion.</p>
| Tsemo Aristide | 280,301 | <p>$a_1+a_2={2{a_1+a_2\over2}}$. You have $a_2=a_1+r$ and recursively $a_n=a_1+(n-1)r$, $a_1+...+a_n=a_1+r+a_1+2r+...+a_1+(n-1)r=na_1+r{n(n-1)\over 2}$ =${{na_1+na_1+rn(n-1)}\over 2}$ =$n{{a_1+a_1r(n-1)}\over 2}=n{{a_1+a_n}\over 2}.$</p>
|
418,647 | <p>Sorry if the question is dumb. I am trying to learn representation theory of finite groups from J.P.Serre's book by myself. In section 2.6 on canonical decomposition, he says that let V be a representation of a finite group G, $W_1,...,W_h$ be the distinct irreducible representations of G, and let V = $U_1 \oplus ... \oplus U_m$ be some decomposition of V into irreducible subrepresentations. Then we can write V = $V_1\oplus ...\oplus V_h$, where $V_i$ is the direct sum of irreducible subrepresentations among $U_i$'s which are $isomorphic$ to $W_i$. This much is clear. But then he says that :</p>
<blockquote>
<p>Next, if needed, one chooses a decomposition of $V_i$ into a direct sum of irreducible representations each isomorphic to $W_i$: $$V_i = W_i \oplus ...\oplus W_i$$
The last decomposition can be done in infinitely many ways; it is just as arbitrary as the choice of a basis in a vector space.</p>
</blockquote>
<p>I am confused with this part. I understand $external$ direct sums of same spaces, but how is the $internal$ direct sum of same spaces $W_i$ defined in general? I think I might be facing some notational difficulty.
Thanks in advance.</p>
| Jared | 65,034 | <p>If you allow loops and/or multiple edges between vertices, then such a graph exists. Take $1$ vertex with $17$ loops, or two vertices with $17$ edges between them, and let the other vertices be isolated.</p>
<p>Now assuming we are working with a simple graph (no loops, and no multiple edges), then no such graph exists. This is because the maximum number of edges that can exist on a simple graph of $6$ vertices is $15$. Can you see why?</p>
|
2,965,989 | <p>Why <span class="math-container">$$p(y)=\int_0^\infty x\delta (y-x)dx=y\ \ ?$$</span></p>
<p>For me, <span class="math-container">$$p(y)=\int_0^\infty x\delta (y-x)dx=\int_{\{y\}}xdx=0.$$</span></p>
<p>If it would be written <span class="math-container">$\int_0^\infty xd\delta _y$</span>, then I would be agree with the answer. But here it's written <span class="math-container">$$\int_0^\infty x\boldsymbol 1_{x=y}(x)dx$$</span>
what I interpret as <span class="math-container">$\int_0^\infty x\boldsymbol 1_{x=y}(x)dx$</span>.</p>
| GEdgar | 442 | <p>Mathematical approach. The <a href="https://en.wikipedia.org/wiki/Distribution_(mathematics)" rel="nofollow noreferrer">Schwartz distribution</a> <span class="math-container">$\delta(y-x)\;dx$</span> is a <a href="https://en.wikipedia.org/wiki/Measure_(mathematics)" rel="nofollow noreferrer">measure</a>. It is the unit point mass at the point <span class="math-container">$y$</span>. Let's write <span class="math-container">$\epsilon_y$</span> for that unit mass. So we have
<span class="math-container">$$
\int_0^\infty x\delta(y-x)\;dx \qquad\text{is notation meaning}\qquad \int_0^\infty x\;\epsilon_y(dx)
$$</span>
If <span class="math-container">$y>0$</span>, we get answer <span class="math-container">$y$</span>. If <span class="math-container">$y<0$</span> we get the answer <span class="math-container">$0$</span>. If <span class="math-container">$y=0$</span>, then this is ambiguous. For the integral from <span class="math-container">$0$</span> to <span class="math-container">$\infty$</span> do we mean the set <span class="math-container">$[0,\infty)$</span> or the set <span class="math-container">$(0,\infty)$</span> ?? A mathematican would write either
<span class="math-container">$$
\int_{[0,\infty)} x\;\epsilon_y(dx)\qquad\text{or}\qquad \int_{(0,\infty)} x\;\epsilon_y(dx)
$$</span>
instead of something ambiguous.</p>
|
3,700,299 | <p>I want to show that <span class="math-container">$\int\limits_{-\infty}^\infty e^{-\pi x^2}dx = 1$</span>.</p>
<p>By definition
<span class="math-container">$$\int\limits_{-\infty}^\infty e^{-\pi x^2}dx = \lim\limits_{t\to\infty}\int\limits_{-t}^t e^{-\pi x^2}dx$$</span>
and since the integrand <span class="math-container">$e^{-\pi x^2}$</span> is an even function
<span class="math-container">$$\int\limits_{-\infty}^\infty e^{-\pi x^2}dx = \lim\limits_{t\to\infty}\int\limits_{-t}^t e^{-\pi x^2}dx = 2\lim\limits_{t\to\infty}\int\limits_0^t e^{-\pi x^2}dx$$</span>
i.e. we can equivalently show that <span class="math-container">$\lim\limits_{t\to\infty}\int\limits_0^t e^{-\pi x^2}dx=\frac{1}{2}$</span>.</p>
<p>Since the antiderivative of <span class="math-container">$e^{-x^2}$</span> is given by the <a href="https://en.wikipedia.org/wiki/Error_function" rel="nofollow noreferrer">error function</a> we can't straightforwardly evaluate the integral, so I tried to use the power series expansion, hoping to be able to see that the resulting series will converge to <span class="math-container">$\frac{1}{2}$</span>:</p>
<p><span class="math-container">$$|\int\limits_0^t e^{-\pi x^2}dx-\frac{1}{2}| = |\int\limits_0^t\sum\limits_{n=0}^\infty\frac{\pi^n\cdot x^{2n}}{n!}dx - \frac{1}{2}| = |\sum\limits_{n=0}^\infty\frac{\pi^n\cdot t^{2n+1}}{n!\cdot(n+1)}-\frac{1}{2}|$$</span></p>
<p>However, I'm in a doubt that it converges and a quick check in Wolfram Mathematica shows indeed that with <span class="math-container">$t\to\infty$</span> the resulting series will diverge.</p>
<p>What am I doing wrong? Can anybody help me with a proof for this problem? Any help will be really appreciated.</p>
| Harish Chandra Rajpoot | 210,295 | <p>Let <span class="math-container">$\pi x^2=t\implies dx=\dfrac{t^{-1/2}dt}{2\sqrt{\pi}}$</span>
<span class="math-container">$$\int_{-\infty}^{\infty}e^{-\pi x^2} dx=2\int_{0}^{\infty}e^{-\pi x^2} dx$$</span>
<span class="math-container">$$=2\int_{0}^{\infty}e^{-t}\dfrac{t^{-1/2}dt}{2\sqrt{\pi}}$$</span>
<span class="math-container">$$=\frac{1}{\sqrt{\pi}}\int_{0}^{\infty}e^{-t}t^{-1/2} dt$$</span>
Using Laplace transform: <span class="math-container">$L[t^n]=\int_0^{\infty}e^{-st}t^{n}dt=\dfrac{\Gamma(n+1)}{s^{n+1}}$</span>,
<span class="math-container">$$=\frac{1}{\sqrt{\pi}}L[t^{-1/2}]_{s=1}$$</span>
<span class="math-container">$$=\frac{1}{\sqrt{\pi}}\left[\frac{\Gamma(-\frac12+1)}{s^{-\frac12+1}}\right]_{s=1}$$</span>
<span class="math-container">$$=\frac{1}{\sqrt{\pi}}\left[\frac{\Gamma(\frac12)}{s^{\frac12}}\right]_{s=1}$$</span>
<span class="math-container">$$=\frac{1}{\sqrt{\pi}}\left[\frac{\sqrt{\pi}}{1}\right]\quad (\because \Gamma(1/2)=\sqrt{\pi})$$</span>
<span class="math-container">$$=1$$</span></p>
|
3,679,386 | <p>defining matrix exponentiation for natural numbers by repeated multiplication and defining it for <span class="math-container">$\frac{1}{n}$</span> by: <span class="math-container">$A^{\frac{1}{n}}$</span> is the matrix s.t. <span class="math-container">$(A^{\frac{1}{n}})^n=A$</span>.
for a rational number <span class="math-container">$a=\frac{p}{q}$</span> <span class="math-container">$A^{a}=(A^{p})^{\frac{1}{q}}$</span>
how do i prove that <span class="math-container">$A^a=(A^{\frac{1}{q}})^p$</span>?
I'm really unsure how to approach this, can i get some hints to the right direction?</p>
| M. Wang | 549,229 | <p>If <span class="math-container">$\sum_{n=1}^\infty z_n$</span> converges then the sequence <span class="math-container">$a_m = \sum_{n=1}^m z_n$</span> is convergent. So in particular, it is a Cauchy sequence, which means that for all <span class="math-container">$\epsilon > 0$</span> there exists some <span class="math-container">$M$</span> such that for all <span class="math-container">$m_1 \geq m_2 \geq M$</span>:</p>
<p><span class="math-container">$$|a_{m_1} - a_{m_2}| < \epsilon.$$</span></p>
<p>In particular this holds for <span class="math-container">$m_1 = m+1$</span> and <span class="math-container">$m_2 = m$</span> whenever <span class="math-container">$m \geq M$</span>. Then</p>
<p><span class="math-container">$$|a_{m_1} - a_{m_2}| = |\sum_{n=1}^{m+1} z_n -\sum_{n=1}^m z_n| = |z_{m+1}| < \epsilon.$$</span></p>
<p>So for all <span class="math-container">$m \geq M+1$</span> we have <span class="math-container">$|z_m| < \epsilon$</span>. Since <span class="math-container">$\epsilon$</span> was arbitrary we have <span class="math-container">$\lim_{n \rightarrow \infty} z_n = 0$</span>. </p>
|
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