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 |
|---|---|---|---|---|
29,823 | <p>Just a random thought here: Can cohomology theories (e.g. sheaf cohomology) on the Stone space $S_n(T)$ (the space of complete n-types) of a first-order theory $T$ tell us anything interesting (e.g. the classification of theories)? Is there any result in model theory that is obtained (probably most easily) by this kind of application of cohomology theories? Thanks!</p>
| Antongiulio Fornasiero | 7,986 | <p>You "only" need to change the topology of the stone space to make it interesting.
In o-minimality, the "spectral" topology is often used: see e.g. many papers by Edmundo.
A similar approach can be used in other topological structures, as long as the structure is definably connected; for structures that are (totally) definably disconnected (like the p-adics) you would need to come out with something different.</p>
|
1,053,683 | <p>How to show that
$$\sum_{n=1}^\infty\frac{1}{n^2+3n+1}=\frac{\pi\sqrt{5}}{5}\tan\frac{\pi\sqrt{5}}{2}$$
?</p>
<p><strong>My try:</strong></p>
<p>We have
$$n+3n+1=\left(n+\frac{3+\sqrt{5}}{2}\right)\left(n+\frac{3-\sqrt{5}}{2}\right),$$
so
$$\frac{1}{n^2+3n+1}=\frac{2}{\sqrt{5}}\left(\frac{1}{2n+3-\sqrt{5}}-\frac{1}{2n+3+\sqrt{5}}\right).$$
Then, I don't know how to proceed.</p>
| xpaul | 66,420 | <p>Note
$$ n^2+3n+1=(n+\frac{3}{2})^2+\left(\frac{\sqrt 5i}{2}\right)^2 $$
and
and hence
\begin{eqnarray}
\sum_{n=0}^\infty\frac{1}{n^2+3n+1}&=&\sum_{n=0}^\infty\frac1{(n+\frac{3}{2})^2+\left(\frac{\sqrt 5i}{2}\right)^2}\\
&=&\frac12\sum_{n=-\infty}^\infty\frac1{(n+\frac{3}{2})^2+\left(\frac{\sqrt 5i}{2}\right)^2}.
\end{eqnarray}
Then using the result from <a href="https://math.stackexchange.com/questions/386667/closed-form-for-sum-n-infty-infty-frac1n-a2b2">this</a>, you will get the answer.</p>
|
3,905,331 | <p>I need to prove that <span class="math-container">$\lim_\limits{n\to \infty}$$\sqrt{\frac{n^2+3}{2n+1}} = \infty$</span> (series) by using the definition:</p>
<p>"A sequence <span class="math-container">$a_n$</span> converges to <span class="math-container">$\infty$</span> if, for every number <span class="math-container">$M$</span>, there exists <span class="math-container">$N∈N$</span> such that whenever <span class="math-container">$n≥N$</span> it follows that <span class="math-container">$a_n>M$</span>."</p>
<p>I came with a proof which I'm not sure is valid, because I just learned the definition and maybe I didn't grasp it yet.</p>
<p>Please let me know what do you think of the following proof:</p>
<p>I'll find an <span class="math-container">$N$</span> such that whenever <span class="math-container">$n≥N$</span> it follows that <span class="math-container">$a_n>M$</span> for every <span class="math-container">$M$</span>:</p>
<p><span class="math-container">$M < \sqrt{\frac{n^2+3}{2n+1}} \Rightarrow M^2 < \frac{n^2+3}{2n+1} < \frac{n^2+n^2}{2n} = \frac{2n^2}{2n} = n$</span></p>
<p>Hence, we got <span class="math-container">$n > M^2$</span>, therefore we can say that when <span class="math-container">$N = M^2$</span> then <span class="math-container">$a_n>M$</span> for every <span class="math-container">$n > M$</span>.
End of proof.</p>
<p>Do you think it is valid or am I missing something? Any suggestion for a better proof using the definition?</p>
<p>Thanks!</p>
| fleablood | 280,126 | <p>Your proof is mostly okay but you do three things wrong. One is linguistically wrong. You say, and quote,</p>
<p>"I'll find an <span class="math-container">$N$</span> such that whenever <span class="math-container">$n≥N$</span> it follows that <span class="math-container">$a_n>M$</span> for every <span class="math-container">$M$</span>"</p>
<p>As stated, no such <span class="math-container">$N$</span> exists or can exist. <span class="math-container">$a_n >M$</span> for <em>some</em> <span class="math-container">$M$</span> but not for <em>every</em> <span class="math-container">$M$</span>. For any <span class="math-container">$a_n$</span> we can always have an <span class="math-container">$M = a_n + 1 > a_n$</span> so there can't be any <span class="math-container">$a_n> M$</span> for <em>all</em> <span class="math-container">$M$</span>.</p>
<p>The actually thing you wish to prove is that for any <span class="math-container">$M$</span> that we can choose as an arbitrary test limit we can find a point <span class="math-container">$N$</span> where suddenly, once we look at the values <span class="math-container">$a_n$</span> after <span class="math-container">$n=N$</span> we find they are all bigger than <em>that</em> <span class="math-container">$M$</span> we choose.</p>
<p>For example if we say <span class="math-container">$\lim_{n\to \infty} 2n = \infty$</span> what does that even <em>mean</em>? It means if we want to find a point where are the <span class="math-container">$a_n = 2n$</span> are bigger than <span class="math-container">$1000$</span> we can by looking at all <span class="math-container">$a_n$</span> where <span class="math-container">$n > 500$</span>. In those cases <span class="math-container">$a_n > 1000$</span>. But <span class="math-container">$a_{735} > 1000$</span> but we don't have <span class="math-container">$a_{735} > 1,000,000$</span>. But if we want to find a point where all <span class="math-container">$a_n > 1,000,000$</span> we can do that by looking at all <span class="math-container">$a_n$</span> where <span class="math-container">$n > 500,000$</span>. In those cases <span class="math-container">$a_n > 1,000,000$</span>. But we don't have <span class="math-container">$a_{678,923} > 10^{100}$</span>. Bue if we want to find a point where all <span class="math-container">$a_n > 10^{100}$</span> we just need <span class="math-container">$n > M = 5\cdot 10^{99}$</span>. Then if <span class="math-container">$n > N$</span> then <span class="math-container">$a_n > 10^{100}$</span>.</p>
<p>But we will <em>never</em> have a <span class="math-container">$N$</span> where <span class="math-container">$n>N$</span> means <span class="math-container">$a_n > M$</span> for <em>all</em> <span class="math-container">$M$</span>. The best we can do, and it is a heck of a <em>lot</em> and it is all we actually want. Is that for any target we want, we can always find an <span class="math-container">$N$</span> for <em><strong>that</strong></em> target.</p>
<p>If <span class="math-container">$M$</span> is an number as big as we want, it could be <span class="math-container">$10^{(10^{100000}}$</span>... or it could be <span class="math-container">$39$</span>, but we can't pick an <span class="math-container">$M$</span> that is actually "infinite". But if <span class="math-container">$M$</span> is "as big as we like", then if <span class="math-container">$N = \frac M2$</span> thenfor every <span class="math-container">$n>\frac M2$</span> we will find that all <span class="math-container">$a_{n; n>N} = 2n > M$</span>. That means whatever limit we want, we can <em>always</em> find a point where all <span class="math-container">$a_n$</span> past that point will all be bigger than <em>that</em> limit.</p>
<p>I'm picky that you understand what that means.</p>
<p>... okay... of my high horse</p>
<p>Secondly, you have you implication backwards.</p>
<p>You prove that if <span class="math-container">$a_n > M$</span> then <span class="math-container">$n > M^2$</span> but you didn't prove they other way around, if <span class="math-container">$n > M^2$</span> then <span class="math-container">$a_n > M$</span>.</p>
<p>Thirdly you found an upper limit rather than a lower limit.</p>
<p>We <em>want</em></p>
<p><span class="math-container">$\sqrt{\frac {n^2 + 3}{2n + 1}} > M$</span>.</p>
<p>And if we find anything <span class="math-container">$K > M$</span> we have to have <span class="math-container">$\sqrt{\frac {n^2 + 3}{2n + 1}}\ge K > M$</span>. If we have <span class="math-container">$K > \sqrt{\frac {n^2 + 3}{2n + 1}}$</span> and <span class="math-container">$K > M$</span> that won't help us at all! (Consider we want to prove that <span class="math-container">$5 > 7$</span> and we get that <span class="math-container">$2*4 > 5$</span> and <span class="math-container">$2*4 > 7$</span> and therefore <span class="math-container">$5 > 7$</span> .... well, that's just wrong.)</p>
<p><span class="math-container">$\sqrt {\frac {n^2 +3}{2n+1}} > M$</span> means</p>
<p><span class="math-container">$\frac {n^2 + 3}{2n+1} > M^2$</span> (assuming <span class="math-container">$M > 0$</span>). Now we must keep in mind this is what we <em>WANT</em> to show. THis is <em>not</em> what we actually know. We <em>CAN'T</em> conclude <span class="math-container">$\frac {n^2 + 3}{2n+1} > M^2\implies n >\frac {n^2 + 3}{2n+1} > M^2$</span> because we don't <em>KNOW</em> that <span class="math-container">$\frac {n^2 + 3}{2n+1} > M^2$</span>.</p>
<p>we need to work the other way around <span class="math-container">$\frac {n^2 + 3}{2n+1} > M^2\Leftarrow ... something ...$</span>.</p>
<p>And we have <span class="math-container">$\frac {n^2 + 3}{2n+1} > \frac {n^2}{2n+1} > \frac {n^2}{2n+n} = \frac n3$</span> (assuming <span class="math-container">$n\ge 1$</span>)</p>
<p>And <span class="math-container">$\frac n3 > M^2 \Leftarrow$</span> (is implied <em>BY</em>) <span class="math-container">$n > 3M^2$</span>.</p>
<p>So <em>if</em> <span class="math-container">$n > \max(1,3M^2)$</span> then <span class="math-container">$\frac {n^2 + 3}{2n+1} > \frac {n^2}{3n} = \frac n3 > M^2$</span> and so <span class="math-container">$a_n =\sqrt{\frac {n^2+3}{2n+1}} > |M| \ge M$</span>.</p>
<p>And we are done.</p>
|
3,461,531 | <p>I have to determine differentiability at <span class="math-container">$(0,1)$</span> of the following function:
<span class="math-container">$$f(x,y)=\frac{|x| y \sin(\frac{\pi x}{2})}{x^2+y^2}$$</span>
The partial derivatives both have value <span class="math-container">$0$</span> at <span class="math-container">$(0,1),$</span> and both are continuous on that point (I think I've got this part right), so the function must be differentiable at <span class="math-container">$(0,1).$</span> But when I checked for differentiability using the definition, the limit that should be <span class="math-container">$0$</span> doesn't exist, so I assume I'm doing something wrong when computing the limit. The following limit has to be <span class="math-container">$0$</span> if the function is differentiable at that point
<span class="math-container">$$\lim_{x,y\to(0,1)} \frac{|f(x,y)|}{\|(x,y)-(0,1)\|}$$</span></p>
<p>Doing the change <span class="math-container">$w=y-1$</span> we have:
<span class="math-container">$$\lim_{x,y\to(0,0)} \frac{|x (w+1)\sin(\frac{\pi x}{2})|}{(x^2+(w+1)^2)\sqrt{x^2+w^2}}$$</span> and then computing the limit along the line <span class="math-container">$x=w,$</span> it has the value <span class="math-container">$\pi /2\sqrt{2}$</span>, which contradicts that the limit is <span class="math-container">$0.$</span></p>
<p>What am I doing wrong?</p>
| Randall | 464,495 | <p>Take elements of <span class="math-container">$\mathbb{Z}$</span>, then remove elements of <span class="math-container">$\mathbb{Q}$</span>. What's left? Nothing. And for sure, <span class="math-container">$\varnothing$</span> is a subset of <span class="math-container">$\mathbb{N}$</span>. Along the same lines, <span class="math-container">$\mathbb{Z} \setminus \mathbb{Q}$</span> is a subset of the set of all pieces of furniture in my house. (For this, you must be okay with the fact that the empty set is a subset of every set.)</p>
|
3,516,776 | <p>I've been trying to solve this for limit comparison test with <span class="math-container">$a_n=a^\frac{1}{n}+a^{-\frac{1}{n}}-2 , b_n= \frac{1}{n}$</span>,
but
<span class="math-container">$\frac{a_n}{b_n}\rightarrow\ln{a}(a^{\frac{1}{x}}-a^{-\frac{1}{x}})\rightarrow 0$</span>.
Any help appreciated.</p>
| Luca Goldoni Ph.D. | 264,269 | <p>Let be <span class="math-container">$a>0$</span>. Since
<span class="math-container">$$
a^x + a^{ - x} - 2 = 2\left[ {\cosh (x\ln a) - 1} \right]
$$</span>
and since
<span class="math-container">$$
\mathop {\lim }\limits_{t \to 0} \left[ {\frac{{\cosh (t) - 1}}
{{t^2 }}} \right] = \frac{(\ln a)^2}
{2}
$$</span>
by setting <span class="math-container">$
t = \frac{{\ln a}}
{n}
$</span>
you get
<span class="math-container">$$
\begin{gathered}
\mathop {\lim }\limits_{n \to + \infty } n^{ - 2} \left( {a^{\frac{{\text{1}}}
{n}} + a^{ - \frac{{\text{1}}}
{n}} - 2} \right) = \hfill \\
= \mathop {\lim }\limits_{n \to + \infty } 2n^{ - 2} \left[ {\cosh \left( {\frac{{\ln a}}
{n}} \right) - 1} \right] = \frac{(\ln a)^2}
{2} \hfill \\
\hfill \\
\end{gathered}
$$</span>
and your series is convergent by asymptotic comparison test.</p>
|
3,135,386 | <p>Our teacher tells us to convert it this way <span class="math-container">$ 3^x = e^{\ln 3^x}= e^{x\cdot\ln 3}$</span> and then use the rule <span class="math-container">$e^u\cdot u'$</span> but I can't understand where <span class="math-container">$\ln$</span> comes from and how <span class="math-container">$\ln 3^x$</span> = <span class="math-container">$x\cdot \ln 3$</span>.</p>
| Michael Rybkin | 350,247 | <p>And are you familiar with this basic property of logarithms?
<span class="math-container">$$\log_{b}{x^y}=y\log_{b}{x}$$</span>
You can bring the power out front.</p>
<p>How about this fact?</p>
<p><span class="math-container">$$
a=e^{\ln{a}}, a >0
$$</span></p>
<p>Do you know what the derivative of an exponential function is?</p>
<p><span class="math-container">$$
(a^x)'=a^x\ln{a}
$$</span></p>
<p>This can be proven a number of ways. You can go back to the definition of the limit and prove it that way or you can use logarithmic differentiation: </p>
<p><span class="math-container">$$
y=a^x\\
\ln{y}=\ln{a^x}\\
\ln{y}=x\ln{a}\\
(\ln{y})'=(x\ln{a})'\\
\frac{1}{y}y'=\ln{a}\\
y'=y\ln{a}\\
y'=a^x\ln{a}
$$</span></p>
|
3,016,386 | <p>Hi I am struggling with this exercise, which may be perceived as simple. so I was trying to write tangents as follows:</p>
<p><span class="math-container">$$\tan(z)=-i\frac{e^{iz}-e^{-iz}}{e^{iz}+e^{-iz}}$$</span> and then <span class="math-container">$$z=a+bi$$</span>, which led me to <span class="math-container">$$ \tan z=-i\frac{\cos a(e^{-b}-e^{b})+i\sin a(e^{-b}+e^{b})}{\cos a(e^{-b}+e^{b})+i\sin a(e^{-b}-e^{b})}$$</span>, so I guess here I can multiply denominator by conjunction, but this is really a complicated computation on an exam... help appreciated</p>
| user | 505,767 | <p><strong>HINT</strong></p>
<p>We have that by the standard trick</p>
<p><span class="math-container">$$\tan(z)=-i\frac{e^{iz}-e^{-iz}}{e^{iz}+e^{-iz}}\cdot\frac{e^{i\bar z}+e^{-i\bar z}}{e^{i\bar z}+e^{-i\bar z}}=\ldots$$</span></p>
<p>then use the well know identities for <span class="math-container">$\cos$</span>, <span class="math-container">$\sin$</span>, <span class="math-container">$\cosh$</span> and <span class="math-container">$\sinh$</span>.</p>
|
3,016,386 | <p>Hi I am struggling with this exercise, which may be perceived as simple. so I was trying to write tangents as follows:</p>
<p><span class="math-container">$$\tan(z)=-i\frac{e^{iz}-e^{-iz}}{e^{iz}+e^{-iz}}$$</span> and then <span class="math-container">$$z=a+bi$$</span>, which led me to <span class="math-container">$$ \tan z=-i\frac{\cos a(e^{-b}-e^{b})+i\sin a(e^{-b}+e^{b})}{\cos a(e^{-b}+e^{b})+i\sin a(e^{-b}-e^{b})}$$</span>, so I guess here I can multiply denominator by conjunction, but this is really a complicated computation on an exam... help appreciated</p>
| MPW | 113,214 | <p>From the title, you're looking for values <span class="math-container">$w$</span> for which <span class="math-container">$\tan^{-1} w$</span> exists. So you want to solve the following equation for <span class="math-container">$z$</span> :</p>
<p><span class="math-container">$$w=\tan z\tag{requires $z\neq\tfrac{(2k+1)\pi}2$}$$</span>
<span class="math-container">$$w = \frac{(e^{iz}-e^{-iz})/2i}{(e^{iz}+e^{-iz})/2}$$</span>
<span class="math-container">$$iw=\frac{e^{2iz}-1}{e^{2iz}+1}$$</span>
<span class="math-container">$$iwe^{2iz}+iw =e^{2iz}-1$$</span>
<span class="math-container">$$(1-iw)(e^{iz})^2=1+iw$$</span>
<span class="math-container">$$(e^{iw})^2 = \frac{1+iw}{1-iw}\tag{requires $w\neq-i$}$$</span>
<span class="math-container">$$e^{iz} = \pm\sqrt{\frac{1+iw}{1-iw}}\tag{both roots included, unambiguous}$$</span>
<span class="math-container">$$z = -i\log\left( \pm\sqrt{\frac{1+iw}{1-iw}}\right) \tag{requires $w\neq i$; multi-valued}$$</span></p>
<p>This last is a formula for <span class="math-container">$\tan^{-1} w$</span>. It is evidently multi-valued and defined except for <span class="math-container">$w=\pm i$</span>.</p>
|
779,509 | <p>I know there is a nice way of getting the continued fraction expansion of quadratic irrationals mainly because they recur after a point, and if they recur after a point they are quadratic irrationals. When constructing the expansion you can multiply by conjugates (kind of), e.g. </p>
<p>$\sqrt 3 =1+\sqrt 3 -1 = 1+\frac {1}{\frac {\sqrt 3 +1}{2}} $</p>
<p>Where you use $(\sqrt 3 - 1)(\sqrt 3 +1)=2$.</p>
<p>Are there identities that would help with the construction for $ \sqrt[3]{2} $?</p>
<p>One I thought was useful in the first step to get [1; 3,...] was </p>
<p>$ (\sqrt[3]{2}-1)( \sqrt[3]{4} + \sqrt[3]{2}+1 )=1$,</p>
<p>So you get:</p>
<p>$ \sqrt[3]{2}=1+( \sqrt[3]{2}-1 )=1+\frac {1}{ \sqrt[3]{4} + \sqrt[3]{2}+1 }= 1+\frac {1}{3+ (\sqrt[3]{4} + \sqrt[3]{2}-2)} $</p>
<p>Thanks for the help.</p>
| Redu | 735,404 | <p>The accepted answer looks like based on Vincent's continued fractions method (1836). Downside is it's inefficiency. Say, the root is at <span class="math-container">$0.000001$</span> so <span class="math-container">$a_0=0$</span>. In order to calculate the next term <span class="math-container">$a_1$</span> you have to invert the polynomial and the root of the inverted polynomial appears at <span class="math-container">$1000000$</span>. This means at the next stage you have to perform <span class="math-container">$p(x)↦p(x+1)$</span> translation 1000000 times.</p>
<p>Instead one should find the lowest bound <span class="math-container">$b_0$</span> of the positive roots and perform a <span class="math-container">$p_0(x)↦p_0(x+b_0+1) = p_1(x)$</span> translation. We should also know that the lowest bound is just a bound and most probably doesn't yield an exact figure that pinpoints the smallest positive root. This means, we may have to perform subsequent lowest bound attempts on the translated polynomials up until there is no sign change remains in the coefficients of the last obtained polynomial, a.k.a. <span class="math-container">$var(p_n(x)) = 0$</span> which means there is no positive root according to <em>Descartes Rule of Sign Change</em>. Then the <span class="math-container">$i^{th}$</span>continued fraction coefficient <span class="math-container">$a_i$</span> would be the sum of all horizontal translation amounts less <span class="math-container">$1$</span> such as <span class="math-container">$a_i = b_{0,i}+b_{1,i}+...+b_{n,i}+n$</span>. Obviously if the subsequent translations yield a no sign change at the polynomial coefficients and at the same time <span class="math-container">$p(0) = 0$</span> case occurs then the continued fraction ends here. If not then we need to perform <span class="math-container">$p(\frac{1}{x-1})$</span> translation (one step to the right and invert) which is two translations one after the other like <span class="math-container">$p(x-1)$</span> and <span class="math-container">$p(\frac{1}{x})$</span>. Keep in mind that inverting a polynomial like <span class="math-container">$p(\frac{1}{x})$</span> is as simple as reversing the coefficients of the polynomial. Such as if <span class="math-container">$p(x) = x^4+7x^2+2x-5$</span> then <span class="math-container">$p(\frac{1}{x}) = -5x^4+2x^3+7x+1$</span> but we are only interested in the roots and should make the initial coefficient 1. Accordingly dividing all coefficients by <span class="math-container">$-5$</span> we can safely say that the <span class="math-container">$p(\frac{1}{x})$</span> of our interest would actually be <span class="math-container">$p(\frac{1}{x}) = x^4-\frac{2}{5}x^3-\frac{7}{5}x-\frac{1}{5}$</span>.</p>
<p>Now there are two crucial parts to this problem.</p>
<ol>
<li>Efficiently and most precisely finding the lowest bound.</li>
<li>Translating the polynomial by an arbitrary amount (<span class="math-container">$b_{lower}+1$</span>) in it's extended form.</li>
</ol>
<h3>1. Lowest Positive Bound</h3>
<p>There are several linear and quadratic algorithms to find the lower positive bound. Linear ones resolve faster but yielding a coarse lower bound while quadratic ones yield <em>much</em> finer bounds. Keep in mind that a sharper lower positive bound would minimize the translation count dramatically. According to my calculations on overall performance, I prefer a quadratic one. If you need to know more about them please drop a comment and i will extend the answer.</p>
<h3>2. Translation of a Polynomial by an Arbitrary Amount</h3>
<p>This is as simple as performing subsequent synthetic divisions, the degree of the polynomial many times. It's known as the Ruffini-Horner Method. A description of how it works can be found in <a href="https://stackoverflow.com/a/71992504/4543207">one of my SO answers</a>.</p>
|
3,154,332 | <p>I have a calculus question which i will display here as an image:
<a href="https://i.stack.imgur.com/8xN3P.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8xN3P.png" alt="enter image description here"></a></p>
<p>I am interested to understand part (b) of this question.
I actually got the answer, but i feel i need more to understand how to determine the maximum just via second derivative.
If one takes the function P(t) and graphs it, you can see the largest value of the derivative happen to be on the Maxima of the graph of the derivative.
Now if one was to take the derivative of this graph then you would get that the maximum rate of change happens when the second derivative is zero, which means its at the Points of Inflection.
So it seems that for Trig functions the maximum rate of change happens at the points of inflection.
So this is how i analyzed it, but i feel there is a better way to explain this.</p>
<p>Hope to get further insight from others here on the forum.</p>
| SNEHIL SANYAL | 636,469 | <p>The first derivative of any function <span class="math-container">$y=f(x)$</span> at point <span class="math-container">$x$</span> gives you the slope of the tangent at that point or the rate of change of the function <span class="math-container">$y=f(x)$</span> at the point <span class="math-container">$x$</span>. If the derivative is positive or negative slope is positive or negative accordingly. </p>
<p>Now, the second derivative of the function <span class="math-container">$y=f(x)$</span> is the rate of change of the first derivative of the function <span class="math-container">$y=f(x)$</span>, so if the second derivative is positive or negative then we can say the first derivative is increasing or decreasing continuously. </p>
<p>To get the maximum value of a function <span class="math-container">$y=f(x)$</span> observe that the slope of the tangent at that point is <span class="math-container">$0$</span> or the rate of change of the function itself is <span class="math-container">$0$</span>.
To get the maximum value of the rate of change <span class="math-container">$y'=f'(x)$</span>, you need to have the slope of tangent at the point as <span class="math-container">$0$</span>, for the function <span class="math-container">$y'=f'(x)$</span> which means <span class="math-container">$f''(x)=0$</span>. </p>
<p>Hope this helps...</p>
|
3,939,620 | <p>Given a polynomial of the form <span class="math-container">$R(z):=\frac{P(z)}{Q(z)}$</span> such that <span class="math-container">$R(z)$</span> has no real roots and <span class="math-container">$deg(Q) \geq deg(P) + 2$</span>, then the integral can be expressed as</p>
<p><span class="math-container">$$\int_{-\infty}^{+\infty} R(z)dz=2\pi i\sum_{a \in \Bbb H}{\mathrm{Res}\left ( f;a \right )}$$</span>
where <span class="math-container">$\Bbb H:=\{z \in \Bbb C: Im(z)>0\}$</span>.</p>
<p>Now for proving this statement, we have first to show that the following limit exists:
<span class="math-container">$$\int_{-\infty}^{+\infty} R(z)dz=\lim_{r \to -\infty}\int_{r}^{+\infty} R(z)dz+\lim_{r \to +\infty}\int_{-\infty}^{r} R(z)dz$$</span>
The argument they used there, is that <span class="math-container">$\exists M >0$</span> s.d. <span class="math-container">$ \vert R(z)\vert \leq \frac{M}{x^2}$</span>. But I don't see, why it exists such an <span class="math-container">$M$</span>. Many thanks for some help!</p>
| lab bhattacharjee | 33,337 | <p>What if <span class="math-container">$\sin x(\sin x+1)=0?$</span></p>
<p>Else we need <span class="math-container">$\sin x(\sin x+1)>0$</span></p>
<p>If <span class="math-container">$\sin x>0,\sin x+1>0\iff\sin x>-1\implies \sin x>$</span> max<span class="math-container">$(0,-1)$</span></p>
<p><span class="math-container">$\implies2n\pi\le x\le2n\pi+\pi$</span></p>
<p>What if <span class="math-container">$\sin x<0?$</span></p>
|
3,059,571 | <p><span class="math-container">$$\lim_{x\to \frac\pi2} \frac{(1-\tan(\frac x2))(1-\sin(x))}{(1+\tan(\frac x2))(\pi-2x)^3}$$</span></p>
<p>I only know of L'hopital method but that is very long. Is there a shorter method to solve this?</p>
| zipirovich | 127,842 | <p>Here's yet another approach. First of all, notice that the factor of <span class="math-container">$\left(1+\tan\dfrac{x}{2}\right)$</span> in the denominator is the only one that is not equal to zero. So it has nothing to do with the <span class="math-container">$\dfrac{0}{0}$</span> indeterminate form in this limit; it's only there (informally speaking) as an extra detractor. And we might as well take it out:</p>
<p><span class="math-container">$$\begin{split}
\lim_{x\to\frac{\pi}{2}}\frac{(1-\tan\frac{x}{2})(1-\sin x)}{(1+\tan\frac{x}{2})(\pi-2x)^3}&=\lim_{x\to\frac{\pi}{2}}\frac{1}{1+\tan\frac{x}{2}}\times\lim_{x\to\frac{\pi}{2}}\frac{(1-\tan\frac{x}{2})(1-\sin x)}{(\pi-2x)^3}\\
&=\frac{1}{2}\times\lim_{x\to\frac{\pi}{2}}\frac{(1-\tan\frac{x}{2})(1-\sin x)}{(\pi-2x)^3}.\end{split}$$</span></p>
<p>We could apply L'Hôpital's Rule now, but… Although the denominator looks much more manageable now, the numerator is still a product, and taking its derivatives will be difficult. Instead, let's break this limit further apart:</p>
<p><span class="math-container">$$\frac{1}{2}\times\lim_{x\to\frac{\pi}{2}}\frac{(1-\tan\frac{x}{2})(1-\sin x)}{(\pi-2x)^3}=\frac{1}{2}\times\lim_{x\to\frac{\pi}{2}}\frac{1-\tan\frac{x}{2}}{\pi-2x}\times\lim_{x\to\frac{\pi}{2}}\frac{1-\sin x}{(\pi-2x)^2}.$$</span></p>
<p>Of course, this is only valid when both limits in the product exist. But as we will see in a moment, they do, and so the desired answer is their product.</p>
<p>In fact, now both limits can be found by applying L'Hôpital's Rule — once for the first limit and twice for the second. Also, for the second limit we can use the Taylor series expansion of the numerator to find this limit — either expand in powers of <span class="math-container">$\left(x-\dfrac{\pi}{2}\right)$</span>, which is quite doable, albeit a bit ugly; or make the change of variables <span class="math-container">$y=x-\dfrac{\pi}{2}$</span> first. (Note: of course, we could use the Taylor series for <span class="math-container">$\tan$</span> for the first limit too; but it's a much lesser known and more advanced expansion involving Bernulli's numbers, and it doesn't appear in standard undergraduate Calculus courses.)</p>
|
2,699,942 | <p>I am confused about one thing during the lecture. </p>
<p>Let $x_n = n$ and $A_n = \{x_k | k \ge n\} = \{n, n+1, n+2, ...\}$.</p>
<p>Then, $\inf A_n = n $, and $\sup A_n = \infty$. </p>
<p>My lecturer also said that $\lim\inf x_n = \lim\inf A _n=\lim n$. </p>
<p>My thinking is that $\{x_n\}_{n=1}^{\infty}=\{1, 2, 3, .....\}$. Shouldn't $\inf x_n = 1$?? Then, $\lim \inf x_n =1$, which is not equal to $\lim n$.</p>
<p>Could you tell me if I am wrong? </p>
| mechanodroid | 144,766 | <p>Indeed strange since we usually define$$\liminf_{n\to\infty} A_n = \bigcup_{n=1}^\infty \left(\bigcap_{k=n}^\infty A_n\right)$$</p>
<p>which is clearly a set.</p>
<p>It can be shown that </p>
<p>$$\liminf_{n\to\infty} A_n = \{x \in \mathbb{R} : x \in A_n \text{ for all except finitely many } n \in \mathbb{N}\}$$</p>
<p>so in this case we have $$\liminf_{n\to\infty} A_n = \emptyset$$</p>
<hr/>
<p>Another interpretation is $$\lim_{n\to\infty}( \inf A_n) = \lim_{n\to\infty} n = \lim_{n\to\infty} x_n = \liminf_{n\to\infty} x_n = +\infty$$</p>
<p>since $\inf A_n = n = x_n$, $\forall n \in \mathbb{N}$.</p>
|
2,255,617 | <p>I am trying to learn how to do proofs by contradiction. The proof is,</p>
<p>"Prove by Contradiction that there are no positive real roots of $x^6 + 2x^3 +4x + 5$"</p>
<p>I understand that now I am attempting to prove that there is a positive real root of this equation, so I am able to contradict myself within the proof. I just don't even know where to start.</p>
| Angina Seng | 436,618 | <p>Certainly not. Take for instance $\Bbb Z$ and $\Bbb Q$.
Or $\Bbb R$ and $\Bbb C$.</p>
|
21,372 | <blockquote>
<p>Let $ y = \min \{ (x + 6), (4 – x) \}$, then find $y$.</p>
</blockquote>
<p>How to solve this problem?</p>
| Ross Millikan | 1,827 | <p>Hint: if $x$ is quite large and positive, which of the two choices will be smaller? If $x$ is large and negative, which will be smaller? Can you find the crossover point?</p>
|
1,989,259 | <p><strong>Can modus tollens be statement of proof by contradiction or is it just a specific case of contradiction?</strong></p>
<p>i.e we know that in general, proof by contradiction stated as follows</p>
<p><span class="math-container">$[P' \implies (q \land q')] \implies P$</span></p>
<p>And by modus tollens, we have</p>
<p><span class="math-container">$[(P' \implies q) \land q'] \implies P$</span></p>
<p>Here we assume <span class="math-container">$P'$</span> true and show q' happens, which should not happen: a contradiction. I tend to think modus tollens is foundation of proof by contradiction, but it seems just a specific case of contradictions...</p>
| Mauro ALLEGRANZA | 108,274 | <p>The two are equivalent.</p>
<p>For example, having <em>proof by contradiction</em> we get :</p>
<p>1) $(\lnot P \to Q) \land \lnot Q$ --- premise</p>
<p>2) $\lnot P \to Q$ --- from 1) by conjunction-elimination</p>
<p>3) $\lnot Q$ --- from 1) by conjunction-elimination</p>
<p>4) $\lnot P$ --- assumed [a]</p>
<p>5) $Q$ --- from 2) and 4) by <em>modus ponens</em></p>
<p>6) $Q \land \lnot Q$ --- from 3) and 5) by conjunction-introduction</p>
<p>7) $\lnot P \to (Q \land \lnot Q)$ --- from 4) and 6) by conditional proof, discharging [a]</p>
<p>8) $P$ --- from 7) and proof by contradiction and <em>modus ponens</em></p>
<blockquote>
<p>9) $(\lnot P \to Q) \land \lnot Q \vdash P$ --- from 1) and 8).</p>
</blockquote>
|
50,736 | <p>Hi guys,</p>
<p>I have recently started looking at polynomials $q_n$ generated by initial choices $q_0=1$, $q_1=x$ with, for $n\geq 0$, some recurrence formula</p>
<p>$$q_{n+2}=xq_{n+1}+c_n q_n$$</p>
<p>where $c_n$ is some function in $n$. The first few of these are</p>
<p>$$q_2=x^2+c_0$$
$$q_3=x^3+(c_0+c_1)x$$
$$q_4=x^4+(c_0+c_1+c_2)x^2+c_2c_0$$
$$q_5=x^5+(c_0+c_1+c_2+c_3)x^3+(c_0c_2+c_0c_3+c_1c_3)x$$
$$q_6=x^6+(c_0+c_1+c_2+c_3+c_4)x^4+(c_0c_2+c_0c_3+c_0c_4+c_1c_3+c_1c_4+c_2c_4)x^2$$$$+c_0c_2c_4$$</p>
<p>My question is whether there is a name for the coefficients of the powers of $x$. I realise that they can be written as certain formulations of elementary symmetric polynomials but I am ideally looking for a reference where the specific expressions are studied </p>
<p>Any help would be great :)</p>
| Ira Gessel | 10,744 | <p>These polynomials are closely related to continuants, which arise in studying continuing fractions. The $n$th continuant of a sequence $a_0$, $a_1$, $\ldots$ is defined by $K(0)=1$, $K(1)=a_1$, $K(n)=a_n K(n-1) + K(n-2)$. They are sums of products of $a_1,\dots, a_n$ in which
consecutive pairs are deleted. (See, for example, <a href="http://en.wikipedia.org/wiki/Continuant_(mathematics)" rel="nofollow">http://en.wikipedia.org/wiki/Continuant_(mathematics)</a>.)</p>
<p>Neither continuants nor the coefficients of the $c_n$ are symmetric polynomials; they cannot be expressed in terms of (or as "combinations of") elementary symmetric functions.</p>
|
396,794 | <p>Let's say that a (right) module <span class="math-container">$M$</span> is <em>well complemented</em> if every non-zero submodule of <span class="math-container">$M$</span> has an indecomposable direct summand (by the way, is there a better or more standard name for this property?). For instance, every module of finite <a href="https://en.wikipedia.org/wiki/Uniform_module#Uniform_dimension_of_a_module" rel="nofollow noreferrer">uniform dimension</a> is well complemented.</p>
<blockquote>
<p><strong>Question.</strong> Is the regular right module <span class="math-container">$R_R$</span> of a von Neumann regular ring <span class="math-container">$R$</span> well complemented?</p>
</blockquote>
<p>As a recall, a ring <span class="math-container">$R$</span> is <em>von Neumann regular</em> if, for every <span class="math-container">$x \in R$</span>, there exists <span class="math-container">$y \in R$</span> such that <span class="math-container">$x = xyx$</span>.</p>
| Benjamin Steinberg | 15,934 | <p>The answer is no. Take a compact totally disconnected space <span class="math-container">$X$</span> with no isolated points, like the Cantor set. Let <span class="math-container">$K$</span> be any field and let <span class="math-container">$R$</span> be the ring of locally constant functions <span class="math-container">$f\colon X\to K$</span> with pointwise operations. This is a commutative von Neumann regular ring. The idempotents of <span class="math-container">$R$</span> are precisely the characteristic functions <span class="math-container">$1_K$</span> of clopen sets <span class="math-container">$K$</span>. An orthogonal decomposition of <span class="math-container">$1_K$</span> into idempotents corresponds to writing <span class="math-container">$K$</span> as a disjoint union of clopen sets. Since <span class="math-container">$X$</span> has no isolated points, if <span class="math-container">$K$</span> is a nonempty clopen set, there are <span class="math-container">$x\neq y\in K$</span>. Then we can find a clopen subset <span class="math-container">$K'$</span> of <span class="math-container">$K$</span> with <span class="math-container">$x\in K'$</span> and <span class="math-container">$y\notin K'$</span>. Thus <span class="math-container">$1_K = 1_{K'}+1_{K\setminus K'}$</span> is a decomposition into orthogonal idempotents. If follows that <span class="math-container">$R$</span> has no primitive idempotents and hence no indecomposable summands (as an indecomposable summand is of the form <span class="math-container">$eR$</span> with <span class="math-container">$e$</span> primitive).</p>
|
2,551,233 | <p>There are 4 fair coins and 1 unfair coin that has only heads. We choose a coin and flip it three times. The result is HHH. What is the probability that the fourth flip is H? </p>
| Mario | 508,026 | <p>$ p(H|HHH) = p(H|unfair) + 4 \cdot p(H|fair)= p(unfair) + p(fair) \cdot p(H) = \dfrac{1}{5} + \dfrac{4}{5} \dfrac{1}{2} = \dfrac{6}{10}$</p>
|
2,551,233 | <p>There are 4 fair coins and 1 unfair coin that has only heads. We choose a coin and flip it three times. The result is HHH. What is the probability that the fourth flip is H? </p>
| Remy | 325,426 | <p>We would expect the probability to be greater than $\frac{1}{2}$. We must solve for $P(\text{Heads})$. We have</p>
<p>$$P(\text{Heads})=P(\text{Heads}|\text{fair})\cdot P(\text{fair})+P(\text{Heads}|\text{unfair})\cdot P(\text{unfair})$$</p>
<p>However, we cannot just say $P(\text{fair})=\frac{4}{5}$ because we are given that the first $3$ tosses are heads.</p>
<p>$$P(\text{fair|}HHH)=\frac{P(\text{fair} \cap HHH)}{P(HHH)}=\frac{\frac{4}{5}\cdot\frac{1}{2}^3}{\frac{4}{5}\cdot\frac{1}{2}^3+\frac{1}{5}\cdot\left(1\right)^3}=\frac{1}{3}$$</p>
<p>So we have that the probability that the coin was fair when we rolled those $3$ heads was $\frac{1}{3}$ and the probability that it was unfair is $\frac{2}{3}$.</p>
<p>Thus we have </p>
<p>$$\begin{align*}
P(\text{Heads})
&=P(\text{Heads|fair})\cdot P(\text{fair})+P(\text{Heads|unfair})\cdot P(\text{unfair})\\\\
&= \frac{1}{2}\cdot\frac{1}{3}+1\cdot\frac{2}{3}\\\\
&=\frac{5}{6}
\end{align*}$$</p>
<p>Which agrees with Nicola's answer.</p>
|
1,533,362 | <p>I need to prove this identity:</p>
<p>$\sum_{k=0}^n \frac{1}{k+1}{2k \choose k}{2n-2k \choose n-k}={2n+1 \choose n}$</p>
<p>without using the identity:</p>
<p>$C_{n+1}=\sum_{k=0}^n C_kC_{n-k}$.</p>
<p>Can't figure out how to.</p>
| Marko Riedel | 44,883 | <p>Suppose we seek to prove that</p>
<p><span class="math-container">$$\sum_{k=0}^n \frac{1}{k+1} {2k\choose k}
{2n-2k\choose n-k} = {2n+1\choose n}.$$</span></p>
<p>We get for the LHS</p>
<p><span class="math-container">$$[z^n] (1+z)^{2n} \sum_{k=0}^n \frac{1}{k+1} {2k\choose k}
\frac{z^k}{(1+z)^{2k}}.$$</span></p>
<p>Here the coefficient extractor <span class="math-container">$[z^n]$</span> combined with the factor <span class="math-container">$z^k$</span>
enforces the upper limit of the sum which we may thus extend to
infinity:</p>
<p><span class="math-container">$$[z^n] (1+z)^{2n} \sum_{k\ge 0} \frac{1}{k+1} {2k\choose k}
\frac{z^k}{(1+z)^{2k}}.$$</span></p>
<p>Now the generating function of the Catalan numbers is
<span class="math-container">$$\frac{1-\sqrt{1-4z}}{2z}$$</span> so this simplifies to</p>
<p><span class="math-container">$$\begin{align*}
& [z^n] (1+z)^{2n} \frac{1-\sqrt{1-4z/(1+z)^2}}{2z/(1+z)^2}
\\ = & [z^{n}] (1+z)^{2n+1}
\frac{1+z-\sqrt{(1+z)^2-4z}}{2z}
\\ = & [z^{n}] (1+z)^{2n+1}
\frac{1+z-\sqrt{(1-z)^2}}{2z}
\\ = & [z^n] (1+z)^{2n+1}
= {2n+1\choose n}
\end{align*}$$</span></p>
<p>as claimed.</p>
|
1,141,074 | <p>I need help with this integral: $$\int\frac{\sqrt{\tan x}}{\cos^2x}dx$$ I tried substitution and other methods, but all have lead me to this expression: $$2\int\sqrt{\tan x}(1+\tan^2 x)dx$$ where I can't calculate anything... Any suggestions? Thanks!</p>
| clathratus | 583,016 | <p>As you have noted, your integral simplifies to
<span class="math-container">$$2\int\sqrt{\tan x}\ \sec^2x\ dx$$</span>
If one makes the substitution <span class="math-container">$u=\tan x$</span>, one gets <span class="math-container">$du=\sec^2x dx$</span>, which reduces our integral to
<span class="math-container">$$2\int u^{1/2}du$$</span>
<span class="math-container">$$=2\frac{u^{3/2}}{3/2}+C$$</span>
<span class="math-container">$$=\frac{4u^{3/2}}{3}+C$$</span>
<span class="math-container">$$=\frac{4\tan^{3/2}x}{3}+C$$</span></p>
|
62,790 | <p>Among several possible definitions of ordered pairs - see below - I find Kuratowski's the least compelling: its membership graph (2) has one node more than necessary (compared to (1)), it is not as "symmetric" as possible (compared to (3) and (4)), and it is not as "intuitive" as (4) - which captures the intuition, that an ordered pair has a first and a second element.</p>
<p><img src="https://i.stack.imgur.com/RVRjq.png" alt="alt text"><a href="http://epublius.de/mathoverflow/orderedpairs.png" rel="noreferrer">(source)</a></p>
<p><sup><em>Membership graphs for possible definitions of ordered pairs (≙ top node, arrow heads omitted)</em></sup></p>
<pre><code>1: (x,y) := { x , { x , y } }
2: (x,y) := { { x } , { x , y } } (Kuratowski's definition)
3: (x,y) := { { x } , { { x } , y } }
4: (x,y) := { { x , 0 } , { 1 , y } } (Hausdorff's definition)
</code></pre>
<p>So my question is: </p>
<blockquote>
<p>Are there good reasons to choose
Kuratowski's definition (or did
Kuratowski himself give any) instead of
one of the more "elegant" - sparing,
symmetric, or intuitive -
alternatives?</p>
</blockquote>
| Community | -1 | <p>How one models ordered pairs is not particularly important; what matters is the existence of a pairing function $(-,-)$ along with functions $\text{first}(-)$ and $\text{second}(-)$ satisfying the requisite properties.</p>
<p>Which definition you use only matters for a brief period between the definition and the point where its properties have been proven, at which point you promptly forget the details of the definition -- so the only real point of deliberating definitions is to make this period as painless as possible for others.</p>
<p>I haven't thought through all of the possibilities, but I will offer an example that $\text{first}(-)$ is tricky to define for your definition 1. The 'obvious' choice seems to be
$$ z = \text{first}(P) \equiv z \in P \wedge \exists a: z \in a \in P $$
which turns out to depend on the axiom of foundation to be well-defined! (consider a $y$ satisfying $y = \{ x, y \}$)
I, personally, would have more confidence in being correct if I was trying to develop the properties of ordered pairs from definitions 2 or 4, rather than from 1 or 3.</p>
|
1,144,141 | <p>I have a question for my exam and I find it hard to understand.</p>
<p>I have to prove that the following formula is logically valid:</p>
<p><img src="https://i.stack.imgur.com/kdELq.jpg" alt="Example"></p>
<p>The professor told me to "push" all the symbols inside the brackets, and use the deduction theorem.</p>
<p>But I don't know how to do it, because I can't find the identities to push the "exist" symbol inside the brackets.</p>
<p>Your help is appriciated, thank you.</p>
<p>Alan</p>
| dtldarek | 26,306 | <p>Perhaps he meant something like this:</p>
<p>\begin{align}
\exists x\ \big(p(x) &\to \forall y\ p(y)\big) \\
\exists x\ \big(\neg p(x) &\lor \forall y\ p(y)\big) \\
\big(\exists x\ \neg p(x)\big) &\lor \big(\forall y\ p(y)\big) \\
\neg\big(\forall x\ p(x)\big) &\lor \big(\forall y\ p(y)\big) \\
\big(\forall x\ p(x)\big) &\to \big(\forall y\ p(y)\big) \\
\end{align}</p>
<p>Please note that the step from line 2 to 3 is not an equivalence,
if the universe is empty, then line 3 is true, but line 2 is false.</p>
<p>I hope this helps $\ddot\smile$</p>
|
257,623 | <p>Consider the following ellipse, generated by the bounding region of the following points</p>
<pre><code>ps = {{-11, 5}, {-12, 4}, {-10, 4}, {-9, 5}, {-10, 6}};
rec = N@BoundingRegion[ps, "FastEllipse"];
Graphics[{rec, Red, Point@ps}]
</code></pre>
<p><a href="https://i.stack.imgur.com/gvtUB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gvtUB.png" alt="enter image description here" /></a></p>
<p>We have that the ellipse 'rec' is given in the form</p>
<pre><code>Ellipsoid[{-10.4, 4.8}, {{2.77333, 0.853333}, {0.853333, 1.49333}}]
</code></pre>
<p>How can I retrieve the lengths of the two main axes of such ellipsoid? Following <a href="https://en.wikipedia.org/wiki/Ellipsoid#As_a_quadric" rel="nofollow noreferrer">this representation</a> and Mathematica's general definition of <code>Ellipsoid</code></p>
<p><a href="https://i.stack.imgur.com/vaCtl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vaCtl.png" alt="enter image description here" /></a></p>
<p>I tried the following, using the eigenvalues of <code>rec[[2]]</code></p>
<pre><code>eigs = Eigenvectors[Inverse[rec[[2]]]]
eigv = Eigenvalues[Inverse[rec[[2]]]];
lens = 2/Sqrt[eigv]
Out[]= {2.06559, 3.57771}
</code></pre>
<p>where the <code>2</code> factor comes from the fact what what I retrieve from the eigenvalues is actually half the length of the main axis. Indeed we get</p>
<pre><code>Graphics[{rec, Red, Point@ps,
Blue, Line[
RegionCentroid@rec + # & /@ {-(lens[[1]] eigs[[1]])/
2, (lens[[1]] eigs[[1]])/2}],
Line[RegionCentroid@rec + # & /@ {-(lens[[2]] eigs[[2]])/
2, (lens[[2]] eigs[[2]])/2}]}]
</code></pre>
<p><a href="https://i.stack.imgur.com/1N5gL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1N5gL.png" alt="enter image description here" /></a></p>
<p>Is this correct? Is there a quicker way of doing this?</p>
| Michael E2 | 4,999 | <p>Stolen from @J.M.'s answer, <a href="https://mathematica.stackexchange.com/a/239797/4999">https://mathematica.stackexchange.com/a/239797/4999</a>, with one correction (is that enough to make it not a duplicate?):</p>
<pre><code>Nodes = ps;
ellipsoidBR = BoundingRegion[Nodes, "FastEllipse"]; (* not "FastEllipsoid" *)
center = ellipsoidBR[[1]];
{vals, vecs} = Eigensystem[ellipsoidBR[[2]]];
{a, b} = Sqrt[vals];
major = N@{center - a vecs[[1]], center + a vecs[[1]]}
minor = N@{center - b vecs[[2]], center + b vecs[[2]]}
Graphics[{ellipsoidBR, Red, Point@ps, Green, Point@center,
Line@{major, minor}}, Frame -> True]
</code></pre>
<img src="https://i.stack.imgur.com/efGSn.png" width="400">
|
2,107,854 | <p>What is the limit when $n \to \infty$?</p>
<p>$$\lim_{n \to \infty} \frac{1}{n^4} \sum_{J=0}^{2n-1} J^3=?$$</p>
| Community | -1 | <p><strong>Hint:</strong> $$1^3+2^3+3^3+\cdots+k^3=\left(\frac{k(k+1)}{2}\right)^2.$$</p>
|
9,437 | <p>I love the way that Mathematica allows me to type in of formulas. It is really easy to type complicated expressions with shortcuts on the keyboard. It would be great if I could use Mathematica completely to publish my articles. The biggest reason I don't already do this is:</p>
<p>I can't find the proper tutorial for styling notebooks for PDF export. How is it possible to deal with page numbering, headers, footers, page breaks, or placing graphs in specific positions, instead of being limited to line by line text? Is this possible? I believe it is -- I'm fascinated with Mathematica's capabilities, but I don't yet have the skills to take advantage of all the features it offers. </p>
<p>If someone would write tutorial on styling notebooks, formatting, and exporting to PDF, I believe that it would be appreciated by many others. I have seen some texts where it's emphasized that they are written totally in Mathematica (and they are really styled well). </p>
<p>Thank you for any tips, ways of accomplishing these tasks and for sharing your experience. </p>
| Silvia | 17 | <p>You can adjust page size, page number style, headers, footers, etc from items under <strong>File</strong> -> <strong>Printing Settings</strong> menu. Or you can programmatically modify them by manipulating <code>Notebook</code>'s options: <code>PrintingCopies</code>, <code>PrintingStartingPageNumber</code>, <code>PrintingPageRange</code>, <code>PageHeaderLines</code>, <code>PageFooterLines</code>, <code>PrintingOptions</code>.</p>
<p><img src="https://i.stack.imgur.com/wcV4v.png" alt="Mathematica graphics" /></p>
<p><img src="https://i.stack.imgur.com/sED5g.png" alt="Mathematica graphics" /></p>
<p>Note: It seems <code>"PaperSize"</code> and <code>"PrintingMargins"</code> are calculated using a DPI value of 72, which I guess is not the DPI of the monitor but that of the default printer.</p>
<p><code>Graphics</code>, more generally, nearly any <code>Cell</code> expression, can be used in header and footer. One way is copy the <em><strong>entire</strong></em> <code>Cell</code>:</p>
<p><img src="https://i.stack.imgur.com/HaeIU.png" alt="Mathematica graphics" /></p>
<p>then paste into the <em>Headers and Footers</em> palette:</p>
<p><img src="https://i.stack.imgur.com/BwDsO.png" alt="Mathematica graphics" /></p>
<h1>A Simple Program Example</h1>
<p>A sample <code>Notebook</code>(Sorry I didn't find a convenient place to store large texts..):</p>
<pre><code>notebookStr = Uncompress["1:eJztWI1u2zYQ7jvsBTgBA5LN9fRrWd1QoE3qNkOcBLbRYbCMmZYYm4hCGhKVJf\
Xyjnuk3ZGyLdlp063dMAwlAkvi/fB4P9+R+XomB8M/vnry5EwqNpPyarw6Ylk2tkYLRkZUErWgiiRUkBkjKpdpygTh\
BRFSAYkRJtIy52JOqEhJKZIFFXP8BNE2QR2CXrOGEpxIP0YFMratFrFG7FZZkxYxhh0cSZEwfgNK5CWhxSFZ0BvNL7\
VIi3CF6lH1ec5BFVUyR94FozdgPS7DaK4WP5CD5D266J6qvoRfrYZmGcwAV7Fv3Ut5e0wVHcfxwSinKVdcCpr1ZH4d\
x1OYxL9nruu+jONfaRx/M4vjQ2LmLw9uD+Erjr+HD4d8h9Pz9ZxhetYL/SNya2bMHyxsHfNimdE7XKXM6Fl5PWM5S/\
dsep3T5YInBXyPT3mhxq/SOUMh8/WmZGO77dpBc0xbxGnDj90OupHvRB2v0/G8oNPpTmGBESi8EqwoQNS2O7YbBFHQ\
7dih7wWhjxzHtEBfmTX6LOXldYuY5wQtpEttgjWQpdA2/yS52Fpl9bliOfjZbQeBXR/edDJBDT2a7O3CifzGcLe7iP\
xuJ3RDJwg6vhd62sbzJU24ugNBT2ttkQuZ3c2l2LhK/9jtsOmdjta4Mxmhxorfs7tebfhGABzY3Y7IrQm4TnP4WsD3\
GiMIaxJBU8DVAl4zht2gJhBGoVsbZgWvqQXdov0wKDM2flEsWaIGFJJZ79cDDRDq9XO65oPCVEyoIcuAnc4yqJ9RXr\
I1+eSaztkFTVNIh8qayiRtQs3E6nvSkBzyd6zicJ0IeZyOi1wV00UmwUYxZ03dOwkdat1dpxaByK67x214x4lMiOtz\
rlczbbPqZmNnUjBNt07EsgRouG+Rn7lI5W+4A/L0OVmFod8iXTvcUPo0B5wqNHHldFvkRankNTg8AZbV5qNFwnuYuA\
C0VLDWkVxypoWc7eRQAbTB8wJcZqBghwEJ2mK9mudCKbWIfqBqIL5hNGV5ZYxGEIQ4DSGrobrLGFbFEZQrFCa+Wihk\
GX8cMwAilq7JPYBzhECo0upt3HsFhtxQxaZDhaiP5T624ng84lnKJqBnyzGShmc8BC+PHe3ULafJUGu7UUQPiyAm/2\
5Vb3ugDDupo/G3x3cA9DzBfYwk/LKiZuGgFAKWR6GxwfkWCUDzjwB7BTaGQmJzAwK5hBajWIEOJpUYQYnn2pJ2y6oS\
bGJqApIK3iBWIqV5ihbt5foRTRbrXHYcXRT63cO3SJfHpOoBk3vcuYlbtelGRkGATwR0L4DSFBNq81Gfv1/3i020P6\
mj/Y12ZuJZbePD1nzpZV962f+jl3VCZPGj/24nm3wYXGpgYn8AXJD0CLx8Mja/yDIwdZxX+LsBZrgFLBRZguMn/woc\
fyY0Rv/vouC2A7+lWalfrB7P2BlcV6wmfO60wve37t0u+kg3+diAV8eJnoSb5cPHiY1JJt7bHQ0XMgeWu2pXcUwe5O\
hDkS4Mj2ap0UEYJw9bu4Rf4OpnKCbAIzrH2ya+F1VIQVjhgWad/+ANs4nP2FsfC2O1YH3xKiNwkYcy48H4ntJCfczx\
7ASu9QraxuT9a+8U7/75b0cpefac9HKI0CuRTgdwgBWQXyyn2SMO/YvpZfL0lAtzEl71aFZUYN1IwC2HKXjNVztMny\
/RaMNhQR+uTsqFhTNGBCKUG3din9aEU3apGgSzrYdkqoKqkfp0LvglT7AZCU1w2roUlyw/zzlEZEuxLiDfc8oR5gwH\
gpImreCYAqeE2oDLhe+0I/ceeXF/6OSXOU2umGpsSBP7Zab4MmNvZM7fQbxotrdxzTdgc6iOXNsEEbva5wCfVaE0hq\
18OOnAMasb+JVlcO9xwLTQc91ugNHtBm3bi7zQrtm+O3WPsV6n0lvAEswu9Em3bWt87/Mkl4W8VNV1qiAHHf/pjKtD\
cnCeKImXoA4cu2zHOVwj4TG75IJvYr7z7y/DsimrHWYLvil4rS1mUCVg3FtecDgjIK3Kv3/YXKtqg5pWLBjTnZMqDI\
Gx6k/bOjzu"];
nbcontent = notebookStr // ToExpression // InputForm;
</code></pre>
<p>Define a function for display the headers and footers setting:</p>
<pre><code>Clear[HeaderFooterSettingView]
HeaderFooterSettingView[nbcontent_] :=
Function[hf,
Cases[nbcontent, (hf -> expr_) :> expr, \[Infinity]] //
If[# === {},
{{None, None, None}, {None, None, None}},
#[[1]]] & //
Column[{
Style[ToString[hf] <> " Settings:", 20],
Grid[Prepend[
Map[If[# === None,
Item[Spacer[20], Background -> LightBlue],
Style[InputForm@#, 8]] &, #, {2}]\[Transpose],
Item[#, Background -> LightYellow] & /@
{"Right page", "Left Page"}
]\[Transpose],
Dividers -> {
{False, Black, GrayLevel[.8], GrayLevel[.8]},
2 -> Directive[Black, Thick]}]
}] &] /@ {PageHeaders, PageFooters} //
Column[#, Frame -> All, FrameStyle -> GrayLevel[.8]] &
HeaderFooterSettingView@nbcontent
</code></pre>
<p><img src="https://i.stack.imgur.com/HpS4L.png" alt="enter image description here" /></p>
<p>Here the light-blue cells indicate empty slots for headers/footers.</p>
<p>Now we insert a <code>Graphics</code> at the right corner footer of right pages:</p>
<pre><code>nbcontentNew = nbcontent /.
(PageFooters -> expr_) :>
(PageFooters -> ReplacePart[expr,
{1, 3} ->
Cell[BoxData[ToBoxes[
Graphics[{Circle[], Inset[x^2 + y^2 == r^2, {0, 0}]},
Frame -> True, ImageSize -> 100]
]]]
]);
nbNew = nbcontentNew[[1]] // NotebookPut
</code></pre>
<p><code>NotebookPrint</code> gave a terrible result on my computer, so I manually selected the virtual pdf printer from <strong>Print</strong> dialog in the <strong>File</strong> menu to print the generated <code>Notebook</code>:</p>
<p><img src="https://i.stack.imgur.com/tprOg.jpg" alt="enter image description here" /></p>
|
2,040,041 | <p>I was able to think that the numerator will always be positive and will overpower the denominator as well. But couldn't proceed from there.</p>
| Black-horse | 170,518 | <p>Hint:</p>
<p>Let $f(x)=x^2e^x-2(e^x-(1+x)),x>0.$ Then
$$f^{'}(x)=x^2e^x+2xe^x-2(e^x-1)$$
and
$$f^{''}(x)=x^2e^x+2xe^x+2xe^x+2e^x-2e^x=x^2e^x+4xe^x>0.$$</p>
|
3,086,758 | <p>I know that if <span class="math-container">$\mathbb{E}[X]=\mathbb{E}[X|Y] , \mathbb{E}[Y]=\mathbb{E}[Y|X]$</span>, <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> can be dependent, for example a ‘uniform’ distribution in a unit circle.
Now we add the variance, if
<span class="math-container">$$\mathbb{E}[X]=\mathbb{E}[X|Y], \mathbb{E}[Y]=\mathbb{E}[Y|X], $$</span><span class="math-container">$$Var(X)=Var(X|Y), Var(Y)=Var(Y|X).$$</span>
Say the expectation and variance of <span class="math-container">$X$</span> are both not affected by <span class="math-container">$Y$</span>, and vice versa, then must <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> be independent?
In this case I can not find a counterexample just like the uniform circle.</p>
<p>If they are independent, how to prove it? If not, is there a counterexample?</p>
<p>Thanks!</p>
| jmerry | 619,637 | <p>Here's a discrete example:
<span class="math-container">$$100P(x,y)=\begin{array}{c|ccccc}x\backslash y&-2&-1&0&1&2\\ \hline -2&1&0&6&0&1\\-1&0&9&0&9&0\\0&6&0&36&0&6\\1&0&9&0&9&0\\2&1&0&6&0&1\end{array}$$</span>
Condition on any particular <span class="math-container">$x$</span>, and <span class="math-container">$y$</span> has mean zero and variance <span class="math-container">$1$</span>. Condition on any particular <span class="math-container">$y$</span>, and <span class="math-container">$x$</span> has mean zero and variance <span class="math-container">$1$</span>. But, of course, they're not independent, as the distribution's support isn't a Cartesian product.</p>
|
9,508 | <p>I need to write a coupon code system but I do not want to save each coupon code in the database. (For performance and design reasons.) Rather I would like to generate codes subsequent that are watermarked with another code.</p>
<p>They should like kind of fancy and random. Currently they look like this:</p>
<p>1: AKFCU2, 2: AKFDU2, 3: AKFDW2, 4: AKHCU2, 5: AKHCW2, ..., 200: CLFSU2, 201: CLFSW2, ...</p>
<p>It's obvious that subsequent codes look very similar as I just converted my code ingredients (the watermark and the integer in the front) to binary base and permutated the order by a fixed scheme. But to prevent people to easily guess another valid code or even accidently enter another valid code (thus making another code invalid by using it) I would prefer something more chaotic, like this:</p>
<p>1: FIOJ32, 2: X9NIU2, 3: SIUUN7, 4: XTVV4S, ...</p>
<p>In the end the problem is to find a bijective, discrete function on the domain {0,1}^27 (or alternatively {0,1,2,3,4, ..., [10^(8.5)]}) that is far away from being continuous. Also it should be as simple as possible to implement. (EDIT: I also need to implement the reverse function.)</p>
<p>Any suggestions for such a function?</p>
| Yuval Filmus | 1,277 | <p>Take any odd $a$ and calculate $x \mapsto ax + b \pmod{2^{27}}$.</p>
<p>EDIT: Here's a more sophisticated suggestion. The following functions are all invertible and easy to implement:</p>
<ul>
<li> Multiplication by an odd number modulo $2^{27}$.
<li> Addition of an arbitrary number modulo $2^{27}$.
<li> XOR of an arbitrary 27-bit number.
<li> Rotation of the bits (can be implemented using two shifts).
</ul>
<p>If you compose them repeatedly they become harder to break (but probably still easy for someone who already knows the general form of your cipher).</p>
<p>By composing, I mean you apply several of them in succession (you can apply a given mapping more than once with same or different parameters).</p>
<hr>
<p>Another suggestion is to use a random permutation. It's not difficult to generate a random permutation, and given a permutation to calculate its inverse. You can store both permutations in a file and load them to memory (if you have enough - it's 1GB for both) each time your main program starts.</p>
|
4,127,149 | <p>I understand that the addition and subtraction of complex number is the same as vector addition and subtraction. But what is the vector equivalent of multiplication and division of complex numbers?</p>
| Somos | 438,089 | <p>You asked</p>
<blockquote>
<p>But what is the vector equivalent of multiplication and division of complex numbers?</p>
</blockquote>
<p>The Wikipedia article on <a href="https://en.wikipedia.org/wiki/William_Rowan_Hamilton" rel="nofollow noreferrer">William Rowan Hamilton</a> states</p>
<blockquote>
<p>Hamilton was looking for ways of extending complex numbers (which can be viewed as points on a 2-dimensional plane) to higher spatial dimensions. He failed to find a useful 3-dimensional system (in modern terminology, he failed to find a real, three-dimensional skew-field), but in working with four dimensions he created quaternions.</p>
</blockquote>
<p>It also states</p>
<blockquote>
<p>In pure mathematics, quaternions show up significantly as one of the four finite-dimensional normed division algebras over the real numbers, with applications throughout algebra and geometry.</p>
</blockquote>
<p>The answer is that the vector equivalent is given by quaternions which are
an extension of three dimensional vectors, but with significant limitations.
For example, multiplication and division of
quaternions is not commutative. Also, vector quaternions are <strong>not</strong>
closed under products and quotients. The Wikipedia article <a href="https://en.wikipedia.org/wiki/History_of_quaternions" rel="nofollow noreferrer">History of quaternions</a> states</p>
<blockquote>
<p>In 1843, Hamilton knew that the complex numbers could be viewed as points in a plane and that they could be added and multiplied together using certain geometric operations. Hamilton sought to find a way to do the same for points in space. Points in space can be represented by their coordinates, which are triples of numbers and have an obvious addition, but Hamilton had difficulty defining the appropriate multiplication.</p>
</blockquote>
<p>and further on</p>
<blockquote>
<p>While he could not "multiply triples", he saw a way to do so for quadruples. By using three of the numbers in the quadruple as the points of a coordinate in space, Hamilton could represent points in space by his new system of numbers.</p>
</blockquote>
<p>So, in general, there is no vector equivalent of multiplication and division,
however, you may find <a href="https://en.wikipedia.org/wiki/Geometric_algebra" rel="nofollow noreferrer">geometric algebra</a>
of interest where vectors and scalars are generalized to algebraic systems
where multiplication is defined, but division is not.</p>
|
247,553 | <p>Let $f(x)=\binom{x}{2}+\binom{x}{4}+\cdots+\binom{x}{2u}$, where $u\in\mathbb{Z}^+$ and $\binom{x}{l}=\frac{x(x-1)\dots(x-l+1)}{l!}$ for all $l\in\mathbb{Z}^+$.
Then can we prove $f(x)$ is a convex function on $[0,+\infty)$?</p>
<p>Updates:</p>
<p>1) It was pointed out by @user44191 that, observing $\binom{x}{i}=\binom{x-1}{i}+\binom{x-1}{i-1}$, the question is equivalent to $\binom{x-1}{1}+\binom{x-1}{2}+\dots+\binom{x-1}{2u}$ is convex on $[0,+\infty)$.</p>
<p>2) Pointed out by @FedorPetrov @GeraldEdgar @H.H.Rugh:<br>
For $x<0$ each summand $\binom{x}{2i}$ is obviously convex, thus the question is equivalent to $f(x)$ is convex on $\mathbb{R}$.</p>
<p>3) Pointed out by @WłodzimierzHolsztyński:<br>
It has $(\Delta^2 f_u)(x) = 1+ f_{u-1}(x-2)$, where $(\Delta f_u)(x)=f_u(x)-f_u(x-1)$. Then we can conclude that $f(x)$ is discrete convex.</p>
| Liviu Nicolaescu | 20,302 | <p><em>This is not an answer to your question, is only an equivalent reformulation that seems promising. I write it as an answer only because of space constraints.</em></p>
<p>For any nonnegative integer $n$ and any $\newcommand{\bR}{\mathbb{R}}$ $x\in\bR$ we define</p>
<p>$$ a_n(x)=\sum_{k=0}^n \binom{x}{2k}, \;\;b_n(x)=\sum_{k=0}^n \binom{x}{2k+1},$$</p>
<p>where for any nonnegative integer $m$ we set</p>
<p>$$
\binom{x}{m}:=\frac{x(x-1)\cdots (x-m+1)}{m!}.
$$</p>
<p>For $t\in (-1,1)$ and $x\in\bR$ the series</p>
<p>$$ F_x(t):=\sum_{k=0}^\infty \binom{x}{k}t^k $$</p>
<p>is the Taylor series at $t=0$ of the function $t\mapsto (1+t)^x$ and thus</p>
<p>$$ F_x(t)=(1+t)^x,\;\;\forall |t|<1. $$</p>
<p>The generating series of the even binomial coefficients $\binom{x}{2m}$ is then</p>
<p>$$ F^0_x(t)=\sum_{m\geq 0}\binom{x}{2m}t^{2m}= \frac{1}{2}\Bigl(\, F_x(t)+F_x(-t)\,\Bigr)=\frac{(1+t)^x+(1-t)^x}{2}. $$</p>
<p>The generating series of the odd binomial coefficients $\binom{x}{2m+1}$ is then</p>
<p>$$ F^1_x(t)=\sum_{m\geq 0}\binom{x}{2m+1}t^{2m}= \frac{1}{2}\Bigl(\, F_x(t)-F_x(-t)\,\Bigr)=\frac{(1+t)^x-(1-t)^x}{2}. $$</p>
<p>The generating series of the sequence</p>
<p>$$a_n(x)=\sum_{k=0}^n \binom{x}{2k} $$</p>
<p>is</p>
<p>$$ A_x(t)=\sum_{n\geq 0} a_n(x)t^{2n}=\frac{1}{1-t^2} F_x^0(t)=\frac{(1+t)^x+(1-t)^{x}}{2(1-t^2)}. $$</p>
<p>The generating series of $b_n(x)$ is
$$
B_x(t)=\sum_{n\geq 0} b_n(x)t^{2n+1}=\frac{1}{1-t^2} F_x^1(t)=\frac{(1+t)^x-(1-t)^{x}}{2(1-t^2)}. $$</p>
<p>We have $\newcommand{\pa}{\partial}$</p>
<p>$$\pa^2_x A_x(t)=\sum_{n\geq 0} a_n''(x) t^{2n}. $$</p>
<p>The problem is equivalent to showing that, for any $x\in \bR$, the Taylor coefficients at $t=0$ of the function</p>
<p>$$[0,1)\ni t\mapsto \pa^2_xA_x(t) $$</p>
<p>are nonnegative, i.e. for any $x$, the function $t\mapsto \pa^2_x A_x(t)$ is absolutely monotonic on the $t$-interval $[0,1)$; for definition and properties of absolutely monotonic functions see Chap. IV of Widder's classical monograph <em>The Laplace Transform</em>.</p>
<p>Now observe that</p>
<p>$$ \pa^2_xA_x(t)=\frac{(1+t)^x\bigl(\,\log(1+t)\,\bigr)^2+(1-t)^x\bigl(\,\log(1-t)\,\bigr)^2}{2(1-t^2)}. $$</p>
<p>Actually we only need to prove that the even degree Taylor coefficients at $t=0$ of the function</p>
<p>$$ t\mapsto G_x(t)=(1+t)^x\frac{\log^2(1+t)}{1-t^2} $$</p>
<p>are positive for any $x\in\bR$.</p>
<p><strong>Remark.</strong> As observed in comments to the question, we can instead study the convexity of the function</p>
<p>$$ c_n(x)=\sum_{j=0}^{2n}\binom{x}{j}.$$</p>
<p>The generating series
$$ C_x(t) =\sum_{n\geq 0} c_n(x) t^{2n}, $$</p>
<p>is the even part of</p>
<p>$$ G_x(t)=\frac{(1+t)^x}{1-t}, $$</p>
<p>i.e.,</p>
<p>$$ C_x(t)=\frac{1}{2}\left( \frac{(1+t)^x}{1-t}+ \frac{(1-t)^x}{1+t}\right)=\frac{(1+t)^{x+1}+(1-t)^{x+1}}{2(1-t^2)} =A_{x+1}(t). $$</p>
|
247,553 | <p>Let $f(x)=\binom{x}{2}+\binom{x}{4}+\cdots+\binom{x}{2u}$, where $u\in\mathbb{Z}^+$ and $\binom{x}{l}=\frac{x(x-1)\dots(x-l+1)}{l!}$ for all $l\in\mathbb{Z}^+$.
Then can we prove $f(x)$ is a convex function on $[0,+\infty)$?</p>
<p>Updates:</p>
<p>1) It was pointed out by @user44191 that, observing $\binom{x}{i}=\binom{x-1}{i}+\binom{x-1}{i-1}$, the question is equivalent to $\binom{x-1}{1}+\binom{x-1}{2}+\dots+\binom{x-1}{2u}$ is convex on $[0,+\infty)$.</p>
<p>2) Pointed out by @FedorPetrov @GeraldEdgar @H.H.Rugh:<br>
For $x<0$ each summand $\binom{x}{2i}$ is obviously convex, thus the question is equivalent to $f(x)$ is convex on $\mathbb{R}$.</p>
<p>3) Pointed out by @WłodzimierzHolsztyński:<br>
It has $(\Delta^2 f_u)(x) = 1+ f_{u-1}(x-2)$, where $(\Delta f_u)(x)=f_u(x)-f_u(x-1)$. Then we can conclude that $f(x)$ is discrete convex.</p>
| fedja | 1,131 | <p>Since it looks like the best we have at the moment is a casework with some brute force estimates, I'll post it just to set the upper bound for the proof clumsiness. </p>
<p><strong><em>Case 1: $0\le x\le 1$.</em></strong></p>
<p>We have to show that the even coefficients of $F(t)=\frac{(1+t)^x\log^2(1+t)}{1-t^2}$ are non-negative. To this end, write $\log(1+t)=t\int_0^1\frac{du}{1+ut}$. Then it suffices to show that the even coefficients of $F_{u,v}(t)=\frac{(1+t)^x}{(1-t^2)(1+ut)(1+vt)}$ are non-negative. WLOG $u\le v$. Now write
$$
\frac{1}{(1+ut)(1+vt)}=\frac{1}{(v-u)t}\left[\frac{1}{1+ut}-\frac{1}{1+vt}\right]=\frac{1}{(v-u)}\int_u^v\frac{dw}{(1+wt)^2}\,.
$$
Thus, it is enough to consider $F_{w,w}$. Also, $\frac{1}{(1+wt)^2}=\frac{(1-wt)^2}{(1-w^2t^2)^2}$ and $\frac{(1+t)^x}{1-t^2}=\frac{1}{(1-t)^{x}(1-t^2)^{1-x}}$, so it is enough to show that the even coefficients of
$(1-wt)^2(1-t)^{-x}$ are non-negative. However, the sequence $x, \frac{x+1}2, \frac{x+2}3,\dots$ is non-decreasing for
$0\le x\le 1$, so the sequence of its partial products (which is exactly the sequence of the coefficients of $(1-t)^{-x}$) is log-convex and the result follows. Notice that in the last function only the first coefficient can be negative. Nevertheless, because of various other factors, this negativity can easily spread to other odd coefficients.</p>
<p><strong><em>Case 2: $x\ge 2$.</em></strong></p>
<p>It will suffice to show that <em>all</em> coefficients of
$$
G(t)=\frac{(1+t)^y}{1-t^2}[(1+t)\log(1+t)]^2=\frac{1}{(1-t)^{y}(1-t^2)^{1-y}}[(1+t)\log(1+t)]^2=U(t)V(t)^2
$$
are non-negative. To this end, we will investigate first the coefficients of $V(t)$. We have
$$
t^{-1}V(t)=1+\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k(k+1)}t^k\,.
$$
Thus, the sum of absolute values of all coefficients of $t^{-1}V(t)$ equals $2$ and the sum of negative coefficients is at most $a=\sum_{k\text{ even}}\frac 1{k(k+1)}\le \frac 16+\frac 12\frac 13=\frac 13$. Here we used a simple inequality
$$
\frac{1}{n(n+1)}+\frac{1}{(n+2)(n+3)}+\frac{1}{(n+4)(n+5)}+\dots
\\
\le
\frac{1}{n(n+1)}+\frac 12\left[\frac{1}{(n+1)(n+2)}+\frac{1}{(n+2)(n+3)}+\dots\right]=\frac{n+2}{2n(n+1)} \tag{$*$}
$$
Hence, the sum of absolute values of all negative coefficients of $t^{-2}V(t)^2$ is at most $2a(2-a)\le \frac 23\frac 53=\frac{10}9$ and that bound would be attained only if there were no cancellations. However, the coefficient at $t^2$ is $2\times 1\times(-\frac 16)+(\frac 12)^2$, which already shaves off $1/4$ from the trivial bound mentioned. Hence, the sum of absolute values of all negative coefficients of $t^{-2}V(t)^2$ is below $1$. Also, $t^{-2}V(t)^2=1+t+\dots$. </p>
<p>Thus, it will suffice to show that the coefficients of $U(t)$ are non-negative, bounded by $1$, and have the property that the sum of any two adjacent coefficients is at least $1$. </p>
<p>Note now that the second expression for $U(t)$ implies the non-negativity property immediately. To establish the rest, just observe that $(1+t)^y=1+a_1 t-a_2 t^2+a_3 t^3-\dots$ with $1\ge a_1\ge a_2\ge\dots\ge 0$.
Thus, the coefficients of $U(t)$ are $1,a_1,1-a_2, a_1+a_3, 1-a_2-a_4, a_1+a_3+a_3,\dots$ from which everything would follow either for trivial reasons, or from the properties of alternating series with diminishing terms, if we show that $a_1+a_3+a_5+\dots\le 1$. However, this sum is just
$\frac 12[(1+t)^y-(1-t)^y]$ evaluated at $t=1$, i.e. $2^{y-1}$.</p>
<p><strong><em>Case 3: $1\le x\le 2$</em></strong>.</p>
<p>We shall estimate the second derivative of each $f_n(x)=x(x-1)\dots(x-n+1)/n!$ from below separately, to which end we will note that $f_2''(x)=1$, $f_4(x)=\frac 1{24}g(x)$ with $g(x)=x(x-1)(x-2)(x-3)$, and $f_n(x)=g(x)h_n(x)$ with $h_n(x)=(x-4)(x-5)\dots(x-n+1)/n!$ for $n=6,8,10,\dots$. To take care of $g(x)$, use the change of variable $x=\frac 32+z$, $-\frac 12\le z\le\frac 12$, and write $g(x)=(z^2-\frac 14)(z^2-\frac 94)=z^4-\frac 52z^2+\frac{9}{16}$. Note that $g(x)\ge 0$ on $[1,2]$, $g''(x)=-5+12z^2\ge -5$ on $[1,2]$ and $|g'(x)|=|z|\cdot|5-4z^2|\le \frac 52$ on $[1,2]$. Also, $h_n''(x)\ge 0$,
$h(x)\le h(1)=\frac 1{2n(n-1)}$ and $|h'(x)|\le |h'(1)|=\frac 1{2n(n-1)}[\frac 13+\frac 14+\dots+\frac 1{n-2}]$. Using the Leibnitz rule
$$
f_n''=gh_n''+2g'h_n'+g''h_n
$$
and taking into account that the first term is non-negative, we get
$$
f_n''\ge -5\frac 1{2n(n-1)}[1+(\frac 13+\frac 14+\dots+\frac 1{n-2})]\,.
$$
Now it remains to show that the sum of the bounds on the right over even $n\ge 4$ is above $-1$. Rewrite this sum as
$$
\frac 52\left[S_4+(\frac 13+\frac 14)S_6+(\frac 15+\frac 16)S_8+\dots\right]
$$
where $S_n=\frac 1{(n-1)n}+\frac 1{(n+1)(n+2)}+\dots\le \frac{n+1}{2(n-1)n}$ by $(*)$. Thus, the whole sum is at most $\frac 52$ times
$$
\frac 5{24}+\sum_{k\text{ odd}, k\ge 3}\frac {2k+1}{k(k+1)}\frac{k+4}{2(k+2)(k+3)}\\
\le \frac 5{24}+\max_{k\ge 3}\frac{(2k+1)(k+4)}{(2k+2)(k+3)}\sum_{k\text{ odd}, k\ge 3}\frac 1{k(k+2)}\le\frac 5{24}+\frac{13}{12}\frac 16
$$
because
$$
\frac{(2k+1)(k+4)}{(2k+2)(k+3)}=1+\frac{k-2}{2(k+1)(k+3)}\le
1+\frac{1}{2(k+3)}\le\frac{13}{12}
$$
for $k\ge 3$. It remains to note that
$$
\frac 52\left[\frac 5{24}+\frac{13}{72}\right]=\frac{70}{72}<1\,.
$$</p>
|
853,659 | <p>Evaluate the integral:</p>
<p>$$\int \frac{x^6}{x^4-1} \, \mathrm{d}x$$</p>
<p>After a lot of help I have reached this point:</p>
<p>$x^2 = Ax^3 - Ax + Bx^2 - B + Cx^3 + Cx^2 + Cx + C + Dx^3 - Dx^2 + Dx - D$</p>
<p>But now I don't really know how to solve for $A, B, C$, and $D$. Please help!</p>
| amWhy | 9,003 | <p>First: Divide! Use polynomial division to get $$\frac{x^6}{x^4 - 1} = 1 + \frac{x^2}{x^4 - 1}$$ $\int 1\,dx = x + C$</p>
<p>For the second term:</p>
<p>Now factor the denominator, and decompose: $$x^4 - 1 = (x^2 + 1)(x^2-1) = (x^2 + 1)(x-1)(x+1)$$</p>
<p>So the set up we want for the second term is:</p>
<p>$$\frac{x^2}{x^4 - 1} = \dfrac{Ax + B}{x^2 + 1} +\frac{C}{x-1} + \frac D{x+1}$$</p>
<p>Now solve for A, B, C, D.</p>
|
4,473,543 | <p>For example, given <span class="math-container">$\color{green}{l_1:5x-2y-8=0}$</span> and <span class="math-container">$\color{blue}{l_2:3x+8y-8=0}$</span>,</p>
<p><a href="https://i.stack.imgur.com/sxZJ1m.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sxZJ1m.jpg" alt="img1" /></a></p>
<p>We can compute the set of lines that pass through the intersection of <span class="math-container">$l_1$</span> and <span class="math-container">$l_2$</span></p>
<p><span class="math-container">$$5x-2y-8+\lambda(3x+8y-8)=0$$</span>
<span class="math-container">$$(5+3\lambda)x+(-2+8\lambda)y+(-8-8\lambda)=0$$</span> for any <span class="math-container">$\lambda\in\Bbb{R}$</span></p>
<div class="s-table-container">
<table class="s-table">
<thead>
<tr>
<th><span class="math-container">$$\lambda$$</span></th>
<th><span class="math-container">$$(5+3\lambda)x+(-2+8\lambda)y+(-8-8\lambda)=0$$</span></th>
</tr>
</thead>
<tbody>
<tr>
<td><span class="math-container">$$-1$$</span></td>
<td><span class="math-container">$$x-5y=0$$</span></td>
</tr>
<tr>
<td><span class="math-container">$$1$$</span></td>
<td><span class="math-container">$$4x+3y-8=0$$</span></td>
</tr>
<tr>
<td><span class="math-container">$$2$$</span></td>
<td><span class="math-container">$$11x+14y-24=0$$</span></td>
</tr>
</tbody>
</table>
</div>
<p><a href="https://i.stack.imgur.com/5w1Kom.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5w1Kom.jpg" alt="img2" /></a></p>
<p>Why does this method work?</p>
<hr />
<p>The closest thread I could find is, <a href="https://math.stackexchange.com/questions/3743591/a-general-circle-through-the-intersection-points-of-line-l-and-circle-s-1-ha">A general circle through the intersection points of line <span class="math-container">$l$</span> and circle <span class="math-container">$S_1$</span> has the form <span class="math-container">$S_1+\lambda L$</span>. What is the significance of <span class="math-container">$\lambda$</span>?</a>.</p>
<blockquote>
<p>For example if we want to find lines through the point of intersection of 3x+4y+5=0 and 2x+y+4=0 . The required lines would be obtained by substituting different values of λ in 3x+4y+5+λ(2x+y+4)=0</p>
</blockquote>
<p>The accepted answer is,</p>
<blockquote>
<p>Let us take up the case of lines first.
Let <span class="math-container">$L_1(x,y)$</span> and <span class="math-container">$L_2(x,y)$</span> be two lines which intersect at <span class="math-container">$(a,b)\\$</span>.
Thus <span class="math-container">$$L_1(a,b)=0\\L_2(a,b)=0$$</span>Now let <span class="math-container">$L_3(x,y)$</span> be another line such that
<span class="math-container">$$L_3(x,y)=L_1(x,y)+\lambda L_2(x,y)$$</span>Now, if we are able to show that <span class="math-container">$L_3$</span> passes through <span class="math-container">$(a,b)$</span>,i.e. the intersection point of <span class="math-container">$L_1$</span> and <span class="math-container">$L_2$</span>, our job will be complete.To do this we put <span class="math-container">$(a,b)$</span> in our expression for <span class="math-container">$L_3$</span>
<span class="math-container">$$L_3(a,b)=L_1(a,b)+\lambda L_2(a,b)$$</span>
<span class="math-container">$$\Rightarrow L_3(x,y)=0+\lambda .0$$</span>
<span class="math-container">$$\Rightarrow L_3(x,y)=0$$</span>So as you can see, for any value of <span class="math-container">$\lambda$</span>, our line <span class="math-container">$L_3$</span> always passes through the intersection of lines <span class="math-container">$L_1$</span> and <span class="math-container">$L_2\\$</span>.
You can the same with any two curves(e.g. two circles) .</p>
</blockquote>
<p>What I understood from this answer is,</p>
<ul>
<li>Consider the lines <span class="math-container">$L_1(x, y)$</span> and <span class="math-container">$L_2(x, y)$</span> which intersept at <span class="math-container">$(a, b)$</span> such that <span class="math-container">$L_1(a, b)=0$</span> and <span class="math-container">$L_2(a, b)=0$</span></li>
<li>Assume that <span class="math-container">$L_3(x,y)=L_1(x,y)+\lambda L_2(x,y)$</span></li>
<li>Then this means that <span class="math-container">$L_3$</span> passes through <span class="math-container">$(a, b)$</span></li>
</ul>
<p>I do not understand how this proves that <span class="math-container">$L_3(x,y)=L_1(x,y)+\lambda L_2(x,y)$</span> spans a set of <em>distinct</em> lines that pass through <span class="math-container">$(a, b)$</span>.</p>
<p>In the case of circles of the form <span class="math-container">$x^2+y^2+Dx+Ey+F=0$</span> and <span class="math-container">$x^2+y^2+D'x+E'y+F'=0$</span> that intersect and are not concentric, we can't have <span class="math-container">$\lambda=-1$</span> because we would get their radical axis, not another circle.</p>
<p>I'm looking for a proof by deduction if possible (as opposed to assumption).</p>
| AlexSp3 | 963,724 | <p>Let's say that
<span class="math-container">$$L_1 : \{Ax+By+C=0\} \Longrightarrow \vec{L_1} =(A, B, C)$$</span>
<span class="math-container">$$L_2 : \{A'x+B'y+C'=0\} \Longrightarrow \vec{L_2} =(A', B', C')$$</span>
The intersection point can be found by solving the system by Cramer rule:
<span class="math-container">$$\begin{pmatrix}L_1 \\ L_2 \end{pmatrix} = \begin{pmatrix}A&B&C \\ A'&B'&C'\end{pmatrix}$$</span></p>
<p><span class="math-container">$$x=\frac{CB'-BC'}{AB'-BA'} \qquad y=\frac{AC'-CA'}{AB'-BA'}$$</span></p>
<p>Doing <span class="math-container">$R_2 = L_1 + \lambda(L_2) = L_3$</span>
<span class="math-container">$$\begin{pmatrix}L_1 \\ L_3 \end{pmatrix} = \begin{pmatrix}A&B&C \\ A+\lambda A'&B+\lambda B'&C+\lambda C'\end{pmatrix}$$</span></p>
<p><span class="math-container">$$x=\frac{C(B+\lambda B')-B(C+\lambda C')}{A(B+\lambda B')-B(A+\lambda A')}
\qquad y=\frac{A(C+\lambda C')-C(A+\lambda A')}{A(B+\lambda B')-B(A+\lambda A')}$$</span></p>
<p><span class="math-container">$$x=\frac{\lambda CB'-\lambda BC'}{\lambda AB'-\lambda BA'}
\qquad y=\frac{\lambda AC'-\lambda CA'}{\lambda AB'-\lambda BA'}$$</span>
<span class="math-container">$$x=\frac{CB'-BC'}{AB'-BA'}
\qquad y=\frac{AC'-CA'}{AB'-BA'}$$</span></p>
<p>Which means that <span class="math-container">$L_1 \cap L_2 = L_1 \cap L_3 $</span></p>
|
168,819 | <p>I was looking for a free PDF from which I can review MV calculus.</p>
<p>Specifically:</p>
<ol>
<li>MV Limits, Continuity, Differentiation.</li>
<li>Differentiation of vector and scalar fields</li>
<li>Surface/Multiple Integrals</li>
</ol>
<p>A succinct book would be great, (coherent) course notes and presentations would do as well.</p>
<p>I ran google searches with <code>filetype:pdf</code> but I couldn't find one which fits all my requirements.</p>
| Brian M. Scott | 12,042 | <p>You might try Paul Dawkins’ on-line <a href="http://tutorial.math.lamar.edu/Classes/CalcIII/CalcIII.aspx" rel="nofollow">Calculus III notes</a>, which can be downloaded in PDF format. I’ve not looked at them, but I’ve taught Calculus I and II from his notes for those courses and found them quite usable, though there are certainly books that are better. </p>
|
2,137,332 | <p>On the MathWorld page: </p>
<p><a href="http://mathworld.wolfram.com/FermatPseudoprime.html" rel="nofollow noreferrer">http://mathworld.wolfram.com/FermatPseudoprime.html</a></p>
<p>in the first table, I expect to see $561$ on every line, but it is not on the line for base $3$.</p>
<p>When you click on the link to the OEIS page, it also is missing from the list. Since $561$ is a Carmichael number, I expected it to be there. Is this a typo (and if so, how do I report it)? If not, what am I missing? Certainly $3^{561} \equiv 3 \pmod{561}$; is there a different definition of "Fermat pseudoprime" that leaves $561$ out?</p>
| Matematleta | 138,929 | <p>For $z=x+iy\ $ write $z=|z|e^{it}, $ where $|z|=\sqrt{x^{2}+y^{2}}\ $ and $\tan t=y/x.\ $Then, $z\mapsto |z|^2e^{2it},\ $ so $|z|\mapsto |z|^2$ and $t\mapsto 2t,\ $ which means that each point in the complex plane doubles its angle, and squares its modulus. </p>
<p>Therefore, the transformed triangle will be the region with vertices $(1,0),(0,i/2),(-1,0),(0,-i/2).$ </p>
<p>To find the boundary, consider each line segment joining the vertices of the triangle. For example, the line segment joining $(1,0)$ to $(0,i)$ has equation $y=1-x.\ $ Therefore for $0\le t\le \pi/4,\ $ we have $|z^2|=2x^2-2x+1,\ $ and $0\le 2t\le \pi/2,\ $ which means that the segment from $(1,0)$ to $(1/2,i/2)$ is transformed into a parabolic arc from $(1,0)$ to $(0,i/2).$ </p>
<p>Now, consider $\pi/4\le t\le \pi/2$ and repeat the argument, and similarly for the other edges of the triangle. </p>
|
157,876 | <p>Can anyone tell me how to find all normal subgroups of the symmetric group $S_4$?</p>
<p>In particular are $H=\{e,(1 2)(3 4)\}$ and $K=\{e,(1 2)(3 4), (1 3)(2 4),(1 4)(2 3)\}$ normal subgroups?</p>
| Arturo Magidin | 742 | <p>In any group, a subgroup is normal if and only if it is a union of conjugacy classes. </p>
<p>In $S_n$, the conjugacy classes are very easy: a conjugacy class consists exactly of all permutations of a given cycle structure. These corresponds to all possible partitions of $n$.</p>
<p>So, consider $S_4$. The conjugacy classes in $S_4$ are:</p>
<ol>
<li>The class of the $4$-cycles, cycle structure $(abcd)$, corresponding to the partition $4$. There are $6$ elements in this class.</li>
<li>The class of the $3$-cycles, cycle structure $(abc)(d)$, corresponding to the partition $3+1$. There are $8$ elements in this class.</li>
<li>The class of the product of two transpositions, cycle structure $(ab)(cd)$, corresponding to the partition $2+2$. There are $3$ elements in this class.</li>
<li>The class the transpositions, cycle structure $(ab)(c)(d)$, corresponding to the partition $2+1+1$. There are $6$ elements in this class.</li>
<li>The class of the identity, cycle structure $(a)(b)(c)(d)$, corresponding to the partition $1+1+1+1$. There is a single element in this class. </li>
</ol>
<p>Now, any subgroup that contains all transpositions is the whole group. </p>
<p>So we can consider only subgroups that don't contain the transpositions. Their order must be a divisor of $24$, and since it does not have the transpositions, it is at most $12$. So the order must be $1$, $2$, $3$, $4$, $6$, or $12$. Moreover, the order must the be sum of the sizes of some conjugacy classes, so it must be a sum of some of the numbers $1$, $3$, $8$, and $6$, and <em>must</em> include $1$.</p>
<p>One possibility is the trivial group, order $1$. We cannot get a normal subgroup of orders $2$ or $3$ (in particular, you $H$ cannot possibly be normal). The only way to get a subgroup of order $4$ is to take the class of the identity and the class of the product of two transpositions. This is your $K$; if it is a subgroup, then being a union of conjugacy classes shows that it is normal. So just check if it is a subgroup.</p>
<p>We cannot get a normal subgroup of order $6$, because we can't just take the conjugacy class of $4$-cycles (we need the identity). As for a subgroup of order $12$, we would need to take the identity ($1$ element), the class of products of two transpositions ($3$ elements), and the class of $3$-cycles ($8$ elements). This collection has a very familiar name...</p>
<p>And that's it! You cannot have any other normal subgroups. So, in summary: the trivial group, the whole group, and possibly $K$ (if it is a subgroup), and possibly this last collection (if it happens to be a subgroup). At most $4$, at least $2$. </p>
|
3,458,154 | <p>I came across this exercise in my Measure Theory workbook and I've been stuck on it. This is the question :</p>
<p>Let F be the set of all non-decreasing right-continuous functions <span class="math-container">$f : \mathbb{ R} \rightarrow \mathbb{R}$</span> with
<span class="math-container">$f(0) = 0$</span>.
Let <span class="math-container">$\mathcal M$</span> be the set of all measures <span class="math-container">$\mu$</span> on <span class="math-container">$(\mathbb R, \mathcal B)$</span> that are finite on every bounded interval
(that is, <span class="math-container">$\mu( [a, b] ) < \infty$</span> for any <span class="math-container">$a,b \in \mathbb R$</span> with <span class="math-container">$a < b$</span>). </p>
<p>Show that the map <span class="math-container">$\Lambda : \mathcal F \rightarrow \mathcal M$</span>, which sends
<span class="math-container">$f \in \mathcal F$</span> to its Lebesgue-Stieltjes measure <span class="math-container">$\lambda_f \in \mathcal M$</span>, is a bijection and describe how to construct
its two-sided inverse <span class="math-container">$F : \mathcal M \rightarrow \mathcal F$</span>.</p>
<p>(You can assume without proof that, for every <span class="math-container">$f \in \mathcal F$</span> and reals <span class="math-container">$a < b$</span>, we have <span class="math-container">$\lambda_f \in \mathcal M$</span> and
<span class="math-container">$\lambda_f ( (a, b] ) = f(b) − f(a)$</span>.)</p>
<p>Thanks</p>
| kam | 514,050 | <p>I am also trying to show the same problem. It is straightforward to show injectivity:</p>
<p>firstly take two functions <span class="math-container">$f_1, f_2$</span> that are not equal, that is to say they differ at some point <span class="math-container">$x\in{\mathbb{R}}$</span> s.t. <span class="math-container">$f_1(x)\neq{f_2(x})$</span>. Now consider the interval <span class="math-container">$(0,x]$</span>. Observe how <span class="math-container">$\lambda_{f_1}((0,x])=f_1(x)-f_1(0)=f_1(x)-0=f_1(x)$</span>. Similarly <span class="math-container">$\lambda_{f_2}((0,x])=f_2(x)-f_2(0)=f_2(x)-0=f_2(x)$</span>. And as these functions differ at one point <span class="math-container">$x$</span>. <span class="math-container">$\lambda_{f_1}\neq{\lambda_{f_2}}$</span>.</p>
<p>I am not sure how to show surjectivity though.</p>
|
3,458,154 | <p>I came across this exercise in my Measure Theory workbook and I've been stuck on it. This is the question :</p>
<p>Let F be the set of all non-decreasing right-continuous functions <span class="math-container">$f : \mathbb{ R} \rightarrow \mathbb{R}$</span> with
<span class="math-container">$f(0) = 0$</span>.
Let <span class="math-container">$\mathcal M$</span> be the set of all measures <span class="math-container">$\mu$</span> on <span class="math-container">$(\mathbb R, \mathcal B)$</span> that are finite on every bounded interval
(that is, <span class="math-container">$\mu( [a, b] ) < \infty$</span> for any <span class="math-container">$a,b \in \mathbb R$</span> with <span class="math-container">$a < b$</span>). </p>
<p>Show that the map <span class="math-container">$\Lambda : \mathcal F \rightarrow \mathcal M$</span>, which sends
<span class="math-container">$f \in \mathcal F$</span> to its Lebesgue-Stieltjes measure <span class="math-container">$\lambda_f \in \mathcal M$</span>, is a bijection and describe how to construct
its two-sided inverse <span class="math-container">$F : \mathcal M \rightarrow \mathcal F$</span>.</p>
<p>(You can assume without proof that, for every <span class="math-container">$f \in \mathcal F$</span> and reals <span class="math-container">$a < b$</span>, we have <span class="math-container">$\lambda_f \in \mathcal M$</span> and
<span class="math-container">$\lambda_f ( (a, b] ) = f(b) − f(a)$</span>.)</p>
<p>Thanks</p>
| Henry Garrett | 506,609 | <p>for surjectivity I think try:</p>
<p>Let <span class="math-container">$\mu \in M$</span> then set <span class="math-container">$f(x)= \mu ((0,x])$</span> for positive <span class="math-container">$x$</span> similar for negative. then show f must be increasing and right cts. </p>
|
683,513 | <p>There is much discussion both in the education community and the mathematics community concerning the challenge of (epsilon, delta) type definitions in real analysis and the student reception of it. My impression has been that the mathematical community often holds an upbeat opinion on the success of student reception of this, whereas the education community often stresses difficulties and their "baffling" and "inhibitive" effect (see below). A typical educational perspective on this was recently expressed by Paul Dawkins in the following terms: </p>
<p><em>2.3. Student difficulties with real analysis definitions. The concepts of limit and continuity have posed well-documented difficulties for students both at the calculus and analysis level of instructions (e.g. Cornu, 1991; Cottrill et al., 1996; Ferrini-Mundy & Graham, 1994; Tall & Vinner, 1981; Williams, 1991). Researchers identified difficulties stemming from a number of issues: the language of limits (Cornu, 1991; Williams, 1991), multiple quantification in the formal definition (Dubinsky, Elderman, & Gong, 1988; Dubinsky & Yiparaki, 2000; Swinyard & Lockwood, 2007), implicit dependencies among quantities in the definition (Roh & Lee, 2011a, 2011b), and persistent notions pertaining to the existence of infinitesimal quantities (Ely, 2010). Limits and continuity are often couched as formalizations of approaching and connectedness respectively. However, the standard, formal definitions display much more subtlety and complexity. That complexity often baffles students who cannot perceive the necessity for so many moving parts. Thus learning the concepts and formal definitions in real analysis are fraught both with need to acquire proficiency with conceptual tools such as quantification and to help students perceive conceptual necessity for these tools. This means students often cannot coordinate their concept image with the concept definition, inhibiting their acculturation to advanced mathematical practice, which emphasizes concept definitions.</em> </p>
<p>See <a href="http://dx.doi.org/10.1016/j.jmathb.2013.10.002" rel="nofollow noreferrer">http://dx.doi.org/10.1016/j.jmathb.2013.10.002</a> for the entire article (note that the online article provides links to the papers cited above).</p>
<p>To summarize, in the field of education, researchers decidedly have <em>not</em> come to the conclusion that epsilon, delta definitions are either "simple", "clear", or "common sense". Meanwhile, mathematicians often express contrary sentiments. Two examples are given below. </p>
<p><em>...one cannot teach the concept of limit without using the epsilon-delta definition. Teaching such ideas intuitively does not make it easier for the student it makes it harder to understand. Bertrand Russell has called the rigorous definition of limit and convergence the greatest achievement of the human intellect in 2000 years! The Greeks were puzzled by paradoxes involving motion; now they all become clear, because we have complete understanding of limits and convergence. Without the proper definition, things are difficult. With the definition, they are simple and clear.</em> (see Kleinfeld, Margaret; Calculus: Reformed or Deformed? Amer. Math. Monthly 103 (1996), no. 3, 230-232.) </p>
<p><em>I always tell my calculus students that mathematics is not esoteric: It is common sense. (Even the notorious epsilon, delta definition of limit is common sense, and moreover is central to the important practical problems of approximation and estimation.)</em> (see Bishop, Errett; Book Review: Elementary calculus. Bull. Amer. Math. Soc. 83 (1977), no. 2, 205--208.)</p>
<p>When one compares the upbeat assessment common in the mathematics community and the somber assessments common in the education community, sometimes one wonders whether they are talking about the same thing. How does one bridge the gap between the two assessments? Are they perhaps dealing with distinct student populations? Are there perhaps education studies providing more upbeat assessments than Dawkins' article would suggest? </p>
<p>Note 1. See also <a href="https://mathoverflow.net/questions/158145/assessing-effectiveness-of-epsilon-delta-definitions">https://mathoverflow.net/questions/158145/assessing-effectiveness-of-epsilon-delta-definitions</a></p>
<p>Note 2. Two approaches have been proposed to account for this difference of perception between the education community and the math community: (a) sample bias: mathematicians tend to base their appraisal of the effectiveness of these definitions in terms of the most active students in their classes, which are often the best students; (b) student/professor gap: mathematicians base their appraisal on their own scientific appreciation of these definitions as the "right" ones, arrived at after a considerable investment of time and removed from the original experience of actually learning those definitions. Both of these sound plausible, but it would be instructive to have field research in support of these approaches.</p>
<p>We recently published <a href="http://dx.doi.org/10.5642/jhummath.201701.07" rel="nofollow noreferrer">an article</a> reporting the result of student polling concerning the comparative educational merits of epsilon-delta definitions and infinitesimal definitions of key concepts like continuity and convergence, with students favoring the infinitesimal definitions by large margins.</p>
| StasK | 97,144 | <p>This is most likely a non-answer, but my (personal, strong, heavily biased, another math culture infused) opinion is that the confusion stems from things being presented in calculus classes in a bizarre illogical order, starting with complicated things (limits of change in functions, i.e., derivatives) and then, as a hindsight, dropping back to simper things (limits of sequences). </p>
<p>In teaching math to the elementary school students the basic arithmetics, we don't throw $\pi$ and $\sqrt{2}$ and ${\rm e}^\pi$ at them. Instead, we talk about 1, 2, 3, then 1+2=3, then $3 \times 4=12$, then introduce division... well, you all know. Natural numbers is a simpler set to digest than rational numbers, which are in turn easier to digest than real numbers. Now, think about limits: are limits on natural numbers easier to digest than limits on real numbers? </p>
<p>While you are thinking about it, take a look at <a href="http://rads.stackoverflow.com/amzn/click/007054235X" rel="nofollow">Rudin's book</a>:</p>
<ol>
<li>Real and complex numbers</li>
<li>Elements of set theory</li>
<li>Sequences and series</li>
<li>Continuity</li>
<li>Differentiation</li>
<li>The Riemann-Stieltjes integral</li>
</ol>
<p>etc. He does go in a logical order, from simpler objects to more complex: real line first, then mappings from natural numbers to real line (sequences), and limits for these; then mappings from reals to reals (functions) and limits on these (continuity). All of the calculus books I have been exposed to in my... uhm... childhood (this was in Soviet Union, so the books were in Russian) went in this order. No author tried to jump ahead of the train engine, and try to excite students with derivatives. Studying sequences first help establishing the concept of a limit. Fewer pathologies are possible with these: you cannot have jumps at infinity, unlike say what $\mathop{\rm sign} x$ does at zero. Once students learn to operate with limits on sequences (what should $N$ be so that $1/n^2$ is less than $10^{-6} \, \forall n\ge N$?), and understand quantifiers, you can start pushing into the world of functions.</p>
<p>To convert my non-answer to semi-answer, I'd be curious to see whether there are differences in reception of the Rudin's sequence of material with the <a href="http://www.stewartcalculus.com/media/10_home.php" rel="nofollow">Stewart's</a> sequence.</p>
|
2,355,852 | <p>Given are $m$ bins with equal probability of choosing one of them. Unknown number of balls $n$ is placed into the bins, and, at the end of placement, we observe number of empty bins $m_e$ and non-empty bins $m_{n}$.</p>
<p>Given $m$, $m_e$, $m_n$, what is the most likely number of balls $n$, which have been placed into bins?</p>
<p>(UPD) possible additional information: number of bins with <em>exactly</em> one ball can be also known.</p>
| G Cab | 317,234 | <p>1) <em>Premise</em> </p>
<p><em>Randomly throw (put) $n$ balls into $m$ bins (of unlimited capacity)</em><br>
means that you consider as equiprobable and indipendent events the<br>
<em>launch the $k$-th ball into the $j$-th bin</em><br>
i.e., a sequence of $n$ independent events, each having $m$ equiprobable results.<br>
Thus the space of elementary events is the $n$D (hyper)cube of side $1\dots m$, containing $m^n$ points
(i.e. sequences). </p>
<p>2) <em>the Problem</em></p>
<p>Let's call $m_e$ as $q$ to simplify notation.
Let's then call $N(n,m,q)$ the number of configurations with exactly $q$ bins empty, and clearly with the other $m-q$ containing at least one ball.<br>
The probability of having a configuration with exactly $q$ bins empty, out of $m$ bins in total and after having placed the $n$th ball is
$$ \bbox[lightyellow] {
P(n,m,q) = {1 \over {m^{\,n} }}N(n,m,q)
}$$
Keeping $m$ fixed, with some "abuse" of notation, but helping to make the development clear, allow to write the above as
$$ \bbox[lightyellow] {
P(n,m,q) = \wp (q|n) = {{\wp (q \wedge n)} \over {\wp (n)}}
}$$</p>
<p>Then the question is to find
$$ \bbox[lightyellow] {
\wp (n|q) = {{\wp (q \wedge n)} \over {\wp (q)}} = {{\wp (q \wedge n)} \over {\sum\limits_n {\wp (q|n)\;\wp (n)} }} = {{P(n,m,q)\wp (n)} \over {\sum\limits_n {\wp (q|n)\;\wp (n)} }}
} \tag {1}$$</p>
<p>This clearly show that we are missing an information in order to solve the problem: the P(n).<br>
That is we have to know the probability that the balls launched are $(0),\, 1,\, 2,\, \cdots$. </p>
<p><strong>Supposing it to be a uniform probability</strong>, between $0$ and, say, $u$, then it is constant and simplifies out, leaving
$$ \bbox[lightyellow] {
\wp (n|q)\left| {\;n\,{\rm uniform}\;{\rm distr}.} \right. = Q(n,m,q) = {{P(n,m,q)} \over {\sum\limits_{n\, \in \;{\rm range}} {P(n,m,q)} }}
} \tag {1.a}$$</p>
<p>3) <em>P(n,m,q) and Q(n,m,q) formula</em></p>
<p>In <a href="https://math.stackexchange.com/questions/2322858">this other post</a> it is explained how we arrive to
demonstrate that
$$ \bbox[lightyellow] {
\eqalign{
& N(n,m,q) = \left( \matrix{
m \cr
q \cr} \right)N(n,m - q,0) = {{m!} \over {q!}}\left\{ \matrix{
n \cr
m - q \cr} \right\}\quad \Rightarrow \cr
& \Rightarrow \quad P(n,m,q) = \left[ {0 = n} \right]\left[ {0 = m} \right]\left[ {0 = q} \right] + \left[ {1 \le m} \right]{{m!} \over {m^{\,n} \;q!}}\left\{ \matrix{
n \cr
m - q \cr} \right\} \cr}
} \tag {2}$$
where $[P]$ is the <a href="https://en.wikipedia.org/wiki/Iverson_bracket" rel="nofollow noreferrer"><em>Iverson bracket</em></a>
$$
\left[ P \right] = \left\{ {\begin{array}{*{20}c}
1 & {P = TRUE} \\
0 & {P = FALSE} \\
\end{array} } \right.
$$</p>
<p>Now it is
$$ \bbox[lightyellow] {
\eqalign{
& \sum\limits_{0\, \le \,k} {\left\{ \matrix{
k \hfill \cr
n \hfill \cr} \right\}\;\left( {{1 \over z}} \right)^{\,k} } \;\left| {\;0 \le {\rm integer}\;n} \right.\quad = \cr
& = {1 \over {\left( {z - 1} \right)\left( {z - 2} \right) \cdots \left( {z - n} \right)}} = {1 \over {\left( {z - 1} \right)^{\,\underline {\,n\,} } }} = z^{\,\overline {\, - n\,} } = {{\Gamma \left( {z - n} \right)} \over {\Gamma \left( z \right)}} \cr}
} $$
where $z^{\,\underline {\,n\,} }$ and $z^{\,\overline {\, n\,}} $ denote the falling and rising factorial respectively.</p>
<p>So that, for the range of $n$ going in the limit to $\infty$ we get (leaving apart the case $0=m$):
$$ \bbox[lightyellow] {
\sum\limits_{0\, \le \,n} {P(n,m,q)} = \sum\limits_{0\, \le \,n} {{{m!} \over {\;q!}}\left\{ \matrix{
n \cr
m - q \cr} \right\}\left( {{1 \over m}} \right)^{\,n} } = {{m!} \over {\;q!}}{{\Gamma \left( q \right)} \over {\Gamma \left( m \right)}}
} $$
and
$$ \bbox[lightyellow] {
\wp (n|q) = Q(n,m,q) = {1 \over {m^{\,n} \;}}{{\Gamma \left( m \right)} \over {\Gamma \left( q \right)}}\left\{ \matrix{
n \cr
m - q \cr} \right\}
} \tag {3}$$</p>
<p>It is worthy to note that, for $q=0$ and $1 \le m$, $ Q(n,m,q)$ is null. That is because the information <em>"there is no empty bin"</em>
just implies that $m \le n$, and the range assumed for $n$ extends to infinity.</p>
<p>4) <em>expected value for $n$</em></p>
<p>Since
$$ \bbox[lightyellow] {
\sum\limits_{0\, \le \,k} {k\left\{ \matrix{
k \hfill \cr
n \hfill \cr} \right\}\;z^{\,k} } \; = z{d \over {dz}}\sum\limits_{0\, \le \,k} {\left\{ \matrix{
k \hfill \cr
n \hfill \cr} \right\}\;z^{\,k} } = z{d \over {dz}}{{\Gamma \left( {1/z - n} \right)} \over {\Gamma \left( {1/z} \right)}} = {{\Gamma \left( {1/z - n} \right)} \over {z\;\Gamma \left( {1/z} \right)}}\left( {\psi (1/z) - \psi (1/z - n)} \right)
} $$</p>
<p>Then, always in the case of a uniform distribution for $n$, with a range estending in the limit to $\infty$, we get
$$ \bbox[lightyellow] {
\eqalign{
& E(n)\quad \left| {\,1 \le q \le m} \right.\quad = \cr
& = \sum\limits_{0\, \le \,n} {n\;Q(n,m,q)} = {{\Gamma \left( m \right)} \over {\Gamma \left( q \right)}}\sum\limits_{0\, \le \,n} {n\;\left\{ \matrix{
n \cr
m - q \cr} \right\}\left( {{1 \over {m\;}}} \right)^{\,n} } = \cr
& = m\left( {\psi (m) - \psi (q)} \right) \cr}
} \tag {4}$$
which is plotted below</p>
<p><a href="https://i.stack.imgur.com/iAhbA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/iAhbA.png" alt="empty_bins_1"></a></p>
<p>Again note that , for $q=0$ and $1 \le m$, the expected value of $n$ climbs to infinite, for the reason said above.</p>
<p>5) <em>max probable value for $n$</em></p>
<p>You are requesting for the mode of $Q(n,m,q)$.
Unfortunately its formula does not allow to find the mode analytically.<br>
However, the plot of $Q$ vs. $n$, at fixed $m$ and $q$, is clearly positive skewed.<br>
Therefore the mode will lay a little below the average, and thus the knowledge of
$E[n]$ is of help in finding the mode numerically.</p>
|
2,330,196 | <p>The question asks me to draw a Hasse diagram for the given set of rules.
$$ (\{n\in \mathbb N: n\mid 100\ \lor\ n = 75 \}, {}\mid{} ) $$</p>
<p>My approach is to write down the set satisfying for $n\mid 100$, but I dont get what's with "or" $n =75.$</p>
<p>Could someone help me figure out what that means? is it set of all $n\mid100$ or $n\mid75$?</p>
<p>I'm new to discrete any solution is much appreciated. </p>
| Francesco Polizzi | 456,212 | <p>Riemann-Roch implies that the degree of a canonical divisor on a compact Riemann surface of genus $g$ is $2g-2$. </p>
<p>On the other hand, a direct computation using differential forms shows that any canonical divisor on the Riemann sphere has degree $-2$, hence $g=0$. </p>
|
2,897,785 | <blockquote>
<p>Fix a $2\times 2$ real matrix $A$. Let $V$ be the set of all $2\times 2$ real matrices $X$ such that $AX=XA$. Show that $V$ is a vector space of dimension of at least 2.</p>
</blockquote>
<p>I'm struggling to see a good way to approach this problem. There's the brute force style method of algebraically manipulating 4 equations in 8 unknowns to show that there are (at least) two matrices $X$ that satisfy $AX=XA$ for any given $A$, but it seems like there should be a more insightful approach. Certainly the identity is in $V$, so there's one element in a basis. And the zero matrix is also in $V$, but this doesn't contribute to a basis as the columns are linearly dependent. And since we don't know that $A$ is invertible, we can't simply take $X=A^{-1}$. </p>
| user | 505,767 | <p>We can consider the following cases</p>
<ul>
<li>$A=0$ then X can be any matrix</li>
<li>$A\neq 0$ singular then $X$ can be $kI$ and $kA$</li>
<li>$A=kI$ then X can be any matrix</li>
<li>$A\neq kI$ not singular then $X$ can be $kI$ and $kA$</li>
</ul>
|
2,778,575 | <p>Given the equation: $\sin^2{x}+\cos{x}=0$</p>
<p>How is it solved?</p>
<p>I think: $\sin^2{x}=1-\cos^2{x}$, but even if I get a quadratic equation with one function (cos), how can I solve it?</p>
| Mohammad Riazi-Kermani | 514,496 | <p>$$\sin^2{x}+\cos{x}=0$$</p>
<p>$$1-\cos ^2 x +\cos x =0$$</p>
<p>$$\cos ^2 x -\cos x -1 =0$$</p>
<p>$$ \cos x = \frac {1-\sqrt 5 }{2}$$</p>
<p>$$x= \cos ^{-1} (\frac {1-\sqrt 5 }{2}) \approx 128.17 \text { degrees.}$$</p>
|
818,169 | <p>The variational distance is defined by,
$$
V(P,Q)=\sum _{i}|p_{i} -q_{i} |
$$
where $P=(p_{1} ,...,p_{n})$ and $Q=(q_{1} ,...,q_{n} )$ are discrete distributions.</p>
<p>It is fairly easy to see that $V$ is a metric, and in particular that it satisfies the triangle inequality. I now introduce class prior, $0<\alpha<1$, which represents the relative 'size' of the distributions,
$$
V(\alpha ,P,Q)=\sum _{i}|\alpha p_{i} -(1-\alpha )q_{i} |
$$
and would like to prove that it again satisfies the triangle inequality,</p>
<p>$$V(\alpha ,P,Q)+V(\beta ,Q,R)\ge V(\gamma ,P,R) $$
where $R$ is another distribution, $\beta$ and $\gamma$ are respective priors, and $\gamma$ is naturally,
$$\gamma =\frac{\alpha \beta }{\alpha \beta +(1-\alpha )(1-\beta )}
$$</p>
<p>I have confidence (numerically) that this triangle inequality indeed holds, but would like a formal proof. Any ideas?</p>
| Omri | 21,373 | <p>Actually, I've recently realized that the inequality is false. It is true only for equal class priors. Sorry about that. </p>
|
1,853,464 | <p>I am using the Lorentz Force Equation and the electric-cross-magnetic field velocity equation] to solve for the $E$ and $B$ fields given the known path of a particle moving in 3D. </p>
<p>So with that I have the following equations where a and v are known:
<a href="https://i.stack.imgur.com/oL7fg.gif" rel="nofollow noreferrer">Lorentz Form</a> and the
<a href="https://i.stack.imgur.com/xuxHP.gif" rel="nofollow noreferrer">E-cross-B Form</a></p>
<p>My question: Are these equations enough to solve for the $x, y, z$ components of $B$ and $E$?</p>
<p>----------Edit---------------</p>
<p>So this is actually being used as an analogy for the propagation of nano-scale self replicating cracks in 3D. In this analogy, the incoming tensile force is represented by the electric force, and the delamination is represented by the magnetic force. </p>
<p>So I have a parabaloid spiral shaped crack which will represent the motion of a charged particle. Since I know the shape/path I can directly get the position, velocity, and acceleration functions in each direction.</p>
<p>With that said, is there a way to use the two equations linked to find all components of the electric and magnetic fields?</p>
| Eugenio | 305,569 | <p>Well, I'm not sure if they are sufficient, but you could add <a href="https://en.wikipedia.org/wiki/Magnetostatics#Magnetostatics_as_a_special_case_of_Maxwell.27s_equations" rel="nofollow">magnetostatic equations</a> if your magnetic field is constant over time, or instead <a href="https://en.wikipedia.org/wiki/Faraday%27s_law_of_induction#Maxwell.E2.80.93Faraday_equation" rel="nofollow">the general equation</a>
$$\nabla \times \overrightarrow{E} = -\frac{\partial \overrightarrow{B}}{\partial t}$$</p>
|
1,369,641 | <p>I do not know how to set this problem up. Any insight as to how to get the equation would be great. </p>
<p>It is John's birthday and his parents want to make him a cake in the shape of a rectangular box. The height of the cake will be $15$ centimeters, and $2$ times the width plus $2$ times the length will be $180$ centimeters. Find the largest possible volume of cake that John can receive. Round your answer to $5$ decimal places.</p>
| Community | -1 | <p>Only the base are matters, as the height is fixed.</p>
<p>Then, $w+l=90$. You need to maximize $wl=w(90-w)$.</p>
<p>By canceling the derivative, $90-2w=0$, or $w=45$, and $l=45$. The largest area (volume) is given by a square (square prism).</p>
|
129 | <p>Is there some criterion for whether a space has the homotopy type of a closed manifold (smooth or topological)? Poincare duality is an obvious necessary condition, but it's almost certainly not sufficient. Are there any other special homotopical properties of manifolds?</p>
| Mike | 1,579 | <p>I'm relying on memory here. A good example, which is discussed in Madsen and Milgram's book on surgery and classifying spaces for topological, $PL$ and smooth manifolds is the set of $1$-connected Poincare duality spaces of dimension 5 with 4-skeleton h. e. to the $4$-skeleton of $S^2\times S^3$, which is $S^2\vee S^3$. </p>
<p><strong>a.</strong> $S^2\times S^3$, which is obviously a manifold</p>
<p><strong>b.</strong> a manifold homeomorphic to $SU(3)/SO(3)$ (though this fact isn't mentioned in Madsen & Milgram).</p>
<p><strong>c.</strong> a Poincaré duality space whose Spivak fibration cannot be reduced to a smooth vector bundle. This is proved using secondary cohomology operations based on Steenrod squares.
This was first proven by Gitler and Spivak(?).</p>
|
202,699 | <p><a href="https://i.stack.imgur.com/UqPw4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UqPw4.png" alt="enter image description here"></a></p>
<p>I try to solve for "t" at the various "x" from the function of </p>
<pre><code>f[t_, x_] =
0.5 Erfc[(x - 0.0236454911650369 t)/Sqrt[4*0.0108274976811351*t]] +
0.5 Exp[0.0236454911650369 x/0.0108274976811351]*
Erfc[(x + 0.0236454911650369 t)/Sqrt[4*0.0108274976811351*t]];
Table[{x, t /. NSolve[{f[t, x] == 0.05}, t, Reals ][[1]]}, {x, 0.5,
10, 0.5}]
</code></pre>
<p>and get coordinates as (x,t). But it does not work as shown in the picture. Please give me any advise on how to solve this problem. Thank you </p>
| Roman | 26,598 | <p>Here's a code snippet that I wrote a long time ago, specifically for the Hammer projection.</p>
<pre><code>HammerPlot::usage=
"HammerPlot[f] shows a Hammer projection (see http://en.wikipedia.org/wiki/Hammer_projection) of a function f[θ,φ]. "~~
"The option ViewPoint->{Θ,Φ,χ} places the viewer over the point (Θ,Φ) and "~~
"rotates the sphere by an angle χ around the line connecting the viewer and the center. "~~
"With ViewPoint->{0,0,0} the observer looks at the north pole, with the Gulf of Guinea to the right and the Bay of Bengal up. "~~
"With ViewPoint->{π/2,0,0} the observer looks at the Gulf of Guinea, with the Bay of Bengal up, the north pole left, and the south pole right. "~~
"With ViewPoint->{π/2,π/2,0} the observer looks at the Bay of Bengal, with the Gulf of Guinea down, the Pacific ocean up, the north pole left and the south pole right. "~~
"With the default setting ViewPoint->{π/2,0,-π/2} the observer looks at the Gulf of Guinea, with the equator going left-to-right, the north pole up, and the south pole down.";
Options[HammerPlot] = Join[Options[DensityPlot], {ViewPoint -> Automatic}];
SetOptions[HammerPlot, AspectRatio -> Automatic, PlotRange -> All,
Frame -> False, ColorFunction -> "SunsetColors"];
HammerPlot[f_, opts : OptionsPattern[]] := Module[{vp, θ, φ, χ},
(* read view point *)
vp = OptionValue[ViewPoint];
{θ, φ, χ} = If[VectorQ[vp, NumericQ] && Length[vp] == 3,
vp,
{π/2, 0, -π/2}];
DensityPlot[
f[ArcCos[1/(4 Sqrt[4 + (-8 + x^2) y^2 + 4 y^4]) ((x^4 + 8 (-1 + y^2)^2 + x^2 (-8 + 6 y^2)) Cos[θ] + Sin[θ] (x Sqrt[8 - x^2 - 4 y^2] (-4 + x^2 + 4 y^2) Cos[χ] - 2 y Sqrt[-(-8 + x^2 + 4 y^2) (4 + (-8 + x^2) y^2 + 4 y^4)] Sin[χ]))],
ArcTan[(x^4 + 8 (-1 + y^2)^2 + x^2 (-8 + 6 y^2)) Cos[φ] Sin[θ] - Sin[φ] (2 y Sqrt[(8 - x^2 - 4 y^2) (4 + (-8 + x^2) y^2 + 4 y^4)] Cos[χ] + x Sqrt[8 - x^2 - 4 y^2] (-4 + x^2 + 4 y^2) Sin[χ]) + Cos[θ] Cos[φ] (-x Sqrt[8 - x^2 - 4 y^2] (-4 + x^2 + 4 y^2) Cos[χ] + 2 y Sqrt[(8 - x^2 - 4 y^2) (4 + (-8 + x^2) y^2 + 4 y^4)] Sin[χ]), Cos[φ] (2 y Sqrt[(8 - x^2 - 4 y^2) (4 + (-8 + x^2) y^2 + 4 y^4)] Cos[χ] + x Sqrt[8 - x^2 - 4 y^2] (-4 + x^2 + 4 y^2) Sin[χ]) + Sin[φ] (8 Sin[θ] + (x^2 + 2 y^2) (-8 + x^2 + 4 y^2) Sin[θ] + Cos[θ] (-x Sqrt[8 - x^2 - 4 y^2] (-4 + x^2 + 4 y^2) Cos[χ] + 2 y Sqrt[(8 - x^2 - 4 y^2) (4 + (-8 + x^2) y^2 + 4 y^4)] Sin[χ]))]],
{x, -2, 2}, {y, -1, 1},
RegionFunction -> ((#1/2)^2 + #2^2 < 1 &),
Evaluate[FilterRules[{opts, Options[HammerPlot]}, Options[DensityPlot]]]]]
</code></pre>
<p>Try it out: Plot the <span class="math-container">$x=\sin(\theta)\cos(\phi)$</span> coordinate with</p>
<pre><code>HammerPlot[Sin[#1] Cos[#2] &]
</code></pre>
<p><a href="https://i.stack.imgur.com/DERHL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DERHL.png" alt="enter image description here"></a></p>
<p>Plot the <span class="math-container">$y=\sin(\theta)\sin(\phi)$</span> coordinate with</p>
<pre><code>HammerPlot[Sin[#1] Sin[#2] &]
</code></pre>
<p><a href="https://i.stack.imgur.com/mKBvj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mKBvj.png" alt="enter image description here"></a></p>
<p>Plot the <span class="math-container">$z=\cos(\theta)$</span> coordinate with</p>
<pre><code>HammerPlot[Cos[#1] &]
</code></pre>
<p><a href="https://i.stack.imgur.com/ivc8Q.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ivc8Q.png" alt="enter image description here"></a></p>
<p>These plots can be modified with all the options available for <code>DensityPlot</code>, for example <code>HammerPlot[Cos[#1] &, ColorFunction -> Hue, PlotLegends -> Automatic]</code>. Particularly the <code>ViewPoint</code> option is useful to center the viewer at arbitrary points over the sphere.</p>
|
456,892 | <p>Find all solutions of $4\cos^2(x)-4\sin(x)-5=0$ in the interval $(6\pi, 8\pi)$.</p>
<p>I tried to work it out and got: $4y^2-4y -9 = 0$, but I can't figure out what $\cos x = $from there to finish the problem.</p>
| Sachin | 68,597 | <p>$$
\begin{align}
4\cos^2x -4\sin x -5 & = 0 \\
\Rightarrow 4(1-\sin^2x)-4\sin x -5 &=0 \\
\Rightarrow 4\sin^2x+4\sin x +1 &=0
\end{align}
$$</p>
<p>Let $\sin x =t$ </p>
<p>$$\Rightarrow 4t^2+4t+1=0$$ </p>
<p>Solving this equation you get </p>
<p>$$\begin{align}
t & =\frac{-1}{2} \\
\Rightarrow \sin x & = \frac{-1}{2} \\
\Rightarrow \sin x & = 240^\circ ; 330^\circ
\end{align}
$$</p>
<p>[ As $\sin x$ is negative therefore it lies in third or fourth quadrant]</p>
|
204,365 | <p>Consider a positive matrix <code>M</code> and a positive vector <code>b</code>, e.g.</p>
<pre><code>nn = 1000;
M = Table[RandomReal[{0, 100}], {i, 1, nn}, {j, 1, nn}];
b = Table[RandomReal[{0, 100}], {i, 1, nn}];
</code></pre>
<p>I would like to find a positive vector <code>X</code></p>
<pre><code>X = Array[x,nn];
</code></pre>
<p>(each <code>x[i]>0</code>) such that given</p>
<pre><code>expr = M.X-b;
</code></pre>
<p>the quantity <code>expr.expr</code> is minimized. Is it possible to do that in Mathematica efficiently (so that it finishes within a few seconds/minutes)?</p>
| Carl Woll | 45,431 | <p>You can use the new in M12 function <a href="http://reference.wolfram.com/language/ref/QuadraticOptimization" rel="noreferrer"><code>QuadraticOptimization</code></a>, which minimizes functions of the form:</p>
<p><span class="math-container">$$\frac{1}{2} x . q. x + c . x$$</span></p>
<p>subject to linear constraints on <span class="math-container">$x$</span>. So, the first step is to figure out what <span class="math-container">$q$</span> and <span class="math-container">$c$</span> are for your example. To do this we expand your <code>expr</code>, but before doing so, note that for a vector <code>u</code> and a matrix <code>M</code> we have:</p>
<p><span class="math-container">$$u.M=M^T.u$$</span></p>
<p>Then, we can expand <code>expr</code> yielding:</p>
<p><span class="math-container">$$(M.X-b).(M.X-b) = X.M^T.M.X-2 b.M.X+b.b$$</span></p>
<p>Hence we have:</p>
<pre><code>q := 2 Transpose[M] . M
c := -2 b . M
</code></pre>
<p>Let's do a small example and compare to @Roman's answer. Setup:</p>
<pre><code>nn = 10;
SeedRandom[10];
M = Table[RandomReal[{0, 100}], {i, 1, nn}, {j, 1, nn}];
b = Table[RandomReal[{0, 100}], {i, 1, nn}];
X = Array[x, nn];
expr = M . X - b;
</code></pre>
<p>Comparison:</p>
<pre><code>r1 = X /. Last @ Minimize[{expr.expr, X > 0}, X]
r2 = QuadraticOptimization[{q, c}, {IdentityMatrix[nn], ConstantArray[0, nn]}]
</code></pre>
<blockquote>
<p>{0.221992, 0.188374, 0.131969, 0., 0., 0., 0., 0., 0.0849646, 0.028771}</p>
<p>{0.221992, 0.188374, 0.131969, 0., 0., 0., 0., 0., 0.0849646, 0.028771}</p>
</blockquote>
<p>where I used constraints that ensured that each vector element is nonnegative. For your larger example:</p>
<pre><code>nn = 1000;
SeedRandom[1];
M = Table[RandomReal[{0, 100}], {i, 1, nn}, {j, 1, nn}];
b = Table[RandomReal[{0, 100}], {i, 1, nn}];
res = QuadraticOptimization[{q, c}, {IdentityMatrix[nn], ConstantArray[0, nn]}]; //AbsoluteTiming
res[[;;20]]
</code></pre>
<blockquote>
<p>{1.48153, Null}</p>
<p>{0., 0., 0., 0.015694, 0., 0.00561439, 0., 0.0157487, 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.}</p>
</blockquote>
<p>So, it finishes on the order of a couple seconds, as requested.</p>
|
2,304,379 | <p>My textbook give the following definition.</p>
<blockquote>
<p>Let $G$ be any topological group. A representation of $G$ on a nonzero complex Hilbert space $V$ is a group homomorphism $\phi$ of $G$ into the group of bounded linear operators on $V$ with bounded inverses, such that the resulting map $ G\times V\to V$ is continuous. </p>
</blockquote>
<p>And it says that to have the continuity property it is enough to have that the map $g\mapsto \phi(g)v$ from $G$ to $V$ is continuous at $g=1$ for all $v\in V$ and a uniform bound for $\|\phi(g)\|$ in some neighbourhood of $1$.</p>
<p>I want to show that the map $G\times V\to V$ given by $(g,v)\mapsto \phi(g)v$ is continuous under these assumptions.</p>
<p><strong>My attempt</strong></p>
<p>Let $(g_0,v_0)\in G\times V$. Let $\epsilon>0$. We need to find an open neighbourhood $U\subseteq G\times V$ of $(g_0,v_0)$ such that $\|\phi(g)v-\phi(g_0)v_0\|<\epsilon$ for all $(g,v)\in U$. By the continuity of the map $G\to V$ by $g\mapsto \phi(g)v_0$ at $g=1$, there is an open neighbourhood $N$ of $1$ such that $\|\phi(g)v_0-v_0\|<\epsilon$ for all $g\in N$. On the other hand, for any $g\in N$ we have
\begin{align*}
\|\phi(g)v-\phi(g_0)v_0\|&=\|\phi(g_0^{-1}g)v-v_0\| \qquad \text{by the unitarity} \\&\leq \|\phi(g_0^{-1}g)v-\phi(g_0^{-1}g)v_0\|+\|\phi(g_0^{-1}g)v_0-v_0\| \\& \leq \|\phi(g_0^{-1}g)\|_{op} \|v-v_0\|+\underbrace{\|\phi(g_0^{-1}g)v_0-v_0\|}_\text{$(*)$}
\end{align*}</p>
<p>Here I could not find an upper bound for the term $(*)$. Could anyone help me? Thanks.</p>
<p>Update: I have realized that we have no unitarity assumption. So I will think on this. </p>
| Alex | 293,781 | <p>Here is an example of such a set with infinitely many elements: </p>
<p>The <a href="https://math.stackexchange.com/questions/1270822/">set of integers in $\mathbb R$</a> is closed, and any nonempty subset of integers is closed as well.</p>
|
1,102,638 | <p>Let $n\in \mathbb{N}$. Can someone help me prove this by induction:</p>
<p>$$\sum _{i=0}^{n}{i} =\frac { n\left( n+1 \right) }{ 2 } .$$</p>
| Alex Silva | 172,564 | <p><strong>Hint:</strong></p>
<p>Multiply the numerator and the denominator by $$\sqrt{1+x+x^2}+1.$$</p>
|
3,065,818 | <blockquote>
<p>If <span class="math-container">$$z=\dfrac{\sqrt{3}-i}{2}$$</span> then <span class="math-container">$$(z^{95}+i^{67})^{94}=z^n$$</span> then, <span class="math-container">$\text{find the smallest positive integral value of}$</span> <span class="math-container">$n$</span> <span class="math-container">$\text{where}$</span> <span class="math-container">$i=\sqrt{-1}$</span></p>
</blockquote>
<p><span class="math-container">$\text{My Attempt:}$</span> First of all I tried to convert <span class="math-container">$z$</span> into <span class="math-container">$\text{Euler's Form}$</span> so, <span class="math-container">$z=e^{-i(\frac{π}{6})}$</span>
Then, I raised <span class="math-container">$z$</span> to the <span class="math-container">$\text{95th}$</span> power. Then I'm getting stuck. And, not being able to proceed. Help. </p>
| Bill Dubuque | 242 | <p>Mimic <span class="math-container">$\rm\color{#c00}{subtractive}$</span> Euclidean algorithm on <span class="math-container">$\color{#c00}{(a,b)}.\,$</span> Clear if <span class="math-container">$\,a\!=\!b\,$</span> by <span class="math-container">$\,a,b\,$</span> coprime <span class="math-container">$\Rightarrow\, a\!=\!b\!=\!1$</span></p>
<p>Else wlog <span class="math-container">$\,a > b\,$</span> therefore <span class="math-container">$\ x^{\large\color{#c00}{a-b}} = (y/x)^{\large \color{#c00}b}\,$</span> and <span class="math-container">$\,y/x\in\Bbb Z\,$</span> via <a href="https://math.stackexchange.com/a/658058/242">Rational Root Test.</a></p>
<p>By induction we conclude: <span class="math-container">$\, x = n^{\large b},\ y/x = n^{\large a-b} \Rightarrow\, y = n^{\large a}\ \ \ {\rm\small QED}$</span></p>
|
638,875 | <p>Let $P$ be a $p$-group and let $A$ be maximal among abelian normal subgroups of $P$. Show that $A=C_P(A)$.</p>
<p>This is the second part of a problem in which I successfully proved the following:
Let $P$ be a finite $p$-group and let $U<V$ be normal subgroups of $P$. Show that there exists $W \triangleleft P$ with $U<W \le V$ and $|W:U|=p$.</p>
<p>I did this by observing that since $U<V$ are normal in $P$, $(V/U) \triangleleft P/U$ and so $(V/U) \cap Z(P/U)$ is nontrivial. Now suppose that $|V/U|=p$. Then it easily follows that $U \triangleleft V \triangleleft P$ and $|V:U|=p$. Now suppose that $|V/U|>p$. Then choose a subgroup of $(V/U) \cap Z(P/U)$ of order $p$, which is normal (since it is central) in $P/U$. This subgroup is of the form $W/U$ for some $W<P$. Then by the Correspondence Theorem, we have $U \triangleleft W \triangleleft P$ and $|W:U|=p$.</p>
<p>I have been told to apply the first part with $U=A$ and $V=C_P(A)$ and show that $W$ is abelian. I tried using the same strategy as above, i.e. choosing $W$ from $Z(P/U)$. However, abelian-ness isn't necessarily preserved under the canonical homomorphism from $P$ to $P/U$. Even if I could obtain such a $W$ I don't see how $W$ abelian implies that $A=C_P(A)$.</p>
<p>I would appreciate a hint to point me in the right direction with this. Thanks.</p>
| DonAntonio | 31,254 | <p>It is an easy-to-prove fact that if $\;|P|=p^n\;,\;\;p\;$ a prime, then for any $\;0\le k\le n\;$ there exists a normal subgroup of $\;P\;$ of order $\;p^k\;$ (induction + the center of finite $\;p-$groups is non-trivial).</p>
<p>The above proves your part one without problem.</p>
<p>Let's now try the following (BTW, the claim is true for any <em>nilpotent</em> group $\;P\;$ , not only finite $\;p-$ groups): We have the following</p>
<p>$$A\lhd P\implies C_P(A)\lhd N_P(A)=P$$</p>
<p>Suppose $\;A\lneqq C_P(A)\;$ , so by part one there exists a normal subgroup $\;W\;$ s.t. $\;A<W\le C_P(A)\;,\;\;[W:A]\;$ = p .</p>
<p>But in fact $\;A\;$ is central in $\;W\;$ : $\;A\le Z(W)\;$ , so that $\; W/Z(W)\;$ is cyclic and thus $\;W\;$ is abelian, and here you have your contradiction...</p>
|
4,326,073 | <p><a href="https://i.stack.imgur.com/RTyOy.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RTyOy.jpg" alt="enter image description here" /></a>
I came across questions in the free module section of my abstract algebra text. In the text, the notation <span class="math-container">$End_{R}(V)$</span> denotes the set of all <span class="math-container">$R$</span>-module endomorphisms of <span class="math-container">$M$</span>. <a href="http://homepage.math.uiowa.edu/%7Egoodman/algebrabook.dir/book.2.6.pdf" rel="nofollow noreferrer">Algebra: Abstract and Concrete, exercise 8.1.9 on p358 in the attached picture of the linked text</a> Onto the question:</p>
<blockquote>
<p>Let <span class="math-container">$V$</span> be a finite dimensional vector space over a field <span class="math-container">$K$</span>. Let <span class="math-container">$T\in End_{K}(V)$</span>. Give <span class="math-container">$V$</span> the corresponding <span class="math-container">$K[x]$</span>-module structure defined by <span class="math-container">$\sum_{i} \alpha_{i}x^{i}v=\sum_{i}\alpha_{i}T^{i}(v).$</span> Show that <span class="math-container">$V$</span> is not free as a <span class="math-container">$K[x]$</span>-module.</p>
</blockquote>
<p>In this question, I don't understand how in <span class="math-container">$\sum_{i} \alpha_{i}x^{i}v=\sum_{i}\alpha_{i}T^{i}(v).$</span> affects whether <span class="math-container">$V$</span> can be a free <span class="math-container">$K[x]$</span>-module. If I take a finite basis <span class="math-container">$B$</span> for <span class="math-container">$V$</span>, where <span class="math-container">$B=\{v_1, v_2,...v_n\}$</span>, with coefficients <span class="math-container">$\alpha_i \in K$</span>, then would it be that each of the <span class="math-container">$\alpha_i x^i v_{i}$</span> term in the identity <span class="math-container">$\alpha_i x^i v_{i} = \alpha_{i}T^{i}(v_i)$</span>, the coefficients <span class="math-container">$\alpha_{i}$</span> or <span class="math-container">$\alpha_{i}x^{i}$</span> might not equal to zero? Actually, I am assuming the <span class="math-container">$v$</span> in the definition <span class="math-container">$\alpha_i x^i v = \alpha_{i}T^{i}(v)$</span> refers to basis elements from <span class="math-container">$B$</span>, but I am not sure where the <span class="math-container">$x^{i}$</span> is suppose to come from. Is it from the vector space <span class="math-container">$V$</span> or from the field <span class="math-container">$K$</span> along with the <span class="math-container">$\alpha_i$</span>.</p>
<p>Thank you in advance.</p>
| Zieac | 991,422 | <p>For any <span class="math-container">$\varepsilon > 0$</span> we have <span class="math-container">$N$</span> such that any <span class="math-container">$n \ge N$</span>, <span class="math-container">$a - \varepsilon \le a_n \le a + \varepsilon$</span> holds.
Let <span class="math-container">$M_1 = \sum_{i = 1}^{N}{i(a - \varepsilon)}$</span> and <span class="math-container">$M_2 = \sum_{i = 1}^{N}{i(a + \varepsilon)}$</span> which is obviously finite and fixed.
Notice that
<span class="math-container">$$ \frac{M_1}{n^2} + \frac{1}{n^2}\sum_{i = N+1}^{n}{i(a - \varepsilon)} \le b_n \le \frac{M_2}{n^2} + \frac{1}{n^2}\sum_{i = N+1}^{n}{i(a + \varepsilon)}$$</span>
Simplify this inequality and we get
<span class="math-container">$$ \frac{M_1}{n^2} + \frac{a - \varepsilon}{2}\left(1 - \frac{N}{n}\right)\left(1 + \frac{N + 1}{n}\right) \le b_n \le \frac{M_2}{n^2} + \frac{a + \varepsilon}{2}\left(1 - \frac{N}{n}\right)\left(1 + \frac{N + 1}{n}\right)$$</span>
Take <span class="math-container">$n\to\infty$</span> then <span class="math-container">$(a - \varepsilon)/2 \le \lim_{n\to\infty}{b_n} \le (a + \varepsilon)/2$</span>.
Note that this holds for any <span class="math-container">$\varepsilon > 0$</span> so this proves what you want.</p>
|
4,326,073 | <p><a href="https://i.stack.imgur.com/RTyOy.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RTyOy.jpg" alt="enter image description here" /></a>
I came across questions in the free module section of my abstract algebra text. In the text, the notation <span class="math-container">$End_{R}(V)$</span> denotes the set of all <span class="math-container">$R$</span>-module endomorphisms of <span class="math-container">$M$</span>. <a href="http://homepage.math.uiowa.edu/%7Egoodman/algebrabook.dir/book.2.6.pdf" rel="nofollow noreferrer">Algebra: Abstract and Concrete, exercise 8.1.9 on p358 in the attached picture of the linked text</a> Onto the question:</p>
<blockquote>
<p>Let <span class="math-container">$V$</span> be a finite dimensional vector space over a field <span class="math-container">$K$</span>. Let <span class="math-container">$T\in End_{K}(V)$</span>. Give <span class="math-container">$V$</span> the corresponding <span class="math-container">$K[x]$</span>-module structure defined by <span class="math-container">$\sum_{i} \alpha_{i}x^{i}v=\sum_{i}\alpha_{i}T^{i}(v).$</span> Show that <span class="math-container">$V$</span> is not free as a <span class="math-container">$K[x]$</span>-module.</p>
</blockquote>
<p>In this question, I don't understand how in <span class="math-container">$\sum_{i} \alpha_{i}x^{i}v=\sum_{i}\alpha_{i}T^{i}(v).$</span> affects whether <span class="math-container">$V$</span> can be a free <span class="math-container">$K[x]$</span>-module. If I take a finite basis <span class="math-container">$B$</span> for <span class="math-container">$V$</span>, where <span class="math-container">$B=\{v_1, v_2,...v_n\}$</span>, with coefficients <span class="math-container">$\alpha_i \in K$</span>, then would it be that each of the <span class="math-container">$\alpha_i x^i v_{i}$</span> term in the identity <span class="math-container">$\alpha_i x^i v_{i} = \alpha_{i}T^{i}(v_i)$</span>, the coefficients <span class="math-container">$\alpha_{i}$</span> or <span class="math-container">$\alpha_{i}x^{i}$</span> might not equal to zero? Actually, I am assuming the <span class="math-container">$v$</span> in the definition <span class="math-container">$\alpha_i x^i v = \alpha_{i}T^{i}(v)$</span> refers to basis elements from <span class="math-container">$B$</span>, but I am not sure where the <span class="math-container">$x^{i}$</span> is suppose to come from. Is it from the vector space <span class="math-container">$V$</span> or from the field <span class="math-container">$K$</span> along with the <span class="math-container">$\alpha_i$</span>.</p>
<p>Thank you in advance.</p>
| Theo Bendit | 248,286 | <p>An application of <a href="https://en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem" rel="noreferrer">Stolz-Cesaro</a> works:
<span class="math-container">\begin{align}
\lim_{n \to \infty} \frac{\sum_{i=1}^n ia_i}{n^2} = \frac{a}{2} &\color{red}\impliedby \lim_{n \to \infty} \frac{\sum_{i=1}^{n+1} ia_i - \sum_{i=1}^n ia_i}{(n+1)^2 - n^2} = \frac{a}{2} \\
&\iff \lim_{n \to \infty} \frac{(n+1)a_n}{2n + 1} = \frac{a}{2} \\
&\impliedby \lim_{n \to \infty} \frac{n+1}{2n + 1} \cdot \lim_{n \to \infty} a_n = \frac{a}{2} \\
&\iff \frac{1}{2} \cdot \lim_{n \to \infty} a_n = \frac{a}{2},
\end{align}</span>
which is true. The red <span class="math-container">$\color{red}\impliedby$</span> marks where Stolz-Cesaro was used. Importantly, note that <span class="math-container">$n^2$</span> diverges to <span class="math-container">$\infty$</span> strictly monotonically, so the theorem applies.</p>
|
352,849 | <p>I have to show that $\lim \limits_{n\rightarrow\infty}\frac{n!}{(2n)!}=0$ </p>
<hr>
<p>I am not sure if correct but i did it like this :
$(2n)!=(2n)\cdot(2n-1)\cdot(2n-2)\cdot ...\cdot(2n-(n-1))\cdot (n!)$ so I have $$\displaystyle \frac{1}{(2n)\cdot(2n-1)\cdot(2n-2)\cdot ...\cdot(2n-(n-1))}$$ and $$\lim \limits_{n\rightarrow \infty}\frac{1}{(2n)\cdot(2n-1)\cdot(2n-2)\cdot ...\cdot(2n-(n-1))}=0$$ is this correct ? If not why ?</p>
| dtldarek | 26,306 | <p><strong>Hint:</strong></p>
<p>$$ 0 \leq \lim_{n\to \infty}\frac{n!}{(2n)!} \leq \lim_{n\to \infty} \frac{n!}{(n!)^2} = \lim_{k \to \infty, k = n!}\frac{k}{k^2} = \lim_{k \to \infty}\frac{1}{k} = 0.$$</p>
|
1,380,402 | <p>I'm developing a C++ program and I need to find a formula that given a number to reduce and a limit number, get a value between 0 and this limit number.</p>
<p>I don't know if it is allow to put C++ code here, but I want to show you my function:</p>
<pre><code>double Utils::reduceNumber(double numberToReduce, double limitNumber)
{
double factor = 0.0;
double result = 0.0;
factor = std::abs(numberToReduce / limitNumber);
if (factor != (int)factor)
factor = (int)factor + 1;
if (numberToReduce > 0)
result = numberToReduce - (byNumber * factor);
else
result = numberToReduce + (byNumber * factor);
return result;
}
</code></pre>
<p>For example, If I want to reduce −465.986246 in a limit between 0 and 24, I have to do this:</p>
<pre><code>−465.986246 + (24 x 20) = 14.013754
</code></pre>
<p>What is the formula to obtain that 20?</p>
| mathlove | 78,967 | <p>Let $a$ be a given number. Also, suppose that the limit is between $0$ and $N$.</p>
<p>If you want an integer $b$ such that
$$0\le a+Nb\le N\iff -\frac{a}{N}\le b\le \frac{N-a}{N},$$
then $$b=\left\lfloor\frac{N-a}{N}\right\rfloor$$
works where $\lfloor x\rfloor$ is the largest integer not greater than $x$.</p>
<p>For $a=−465.986246$ and $N=24$, we have $b=\left\lfloor\frac{24-(−465.986246)}{24}\right\rfloor=20$.</p>
|
1,380,402 | <p>I'm developing a C++ program and I need to find a formula that given a number to reduce and a limit number, get a value between 0 and this limit number.</p>
<p>I don't know if it is allow to put C++ code here, but I want to show you my function:</p>
<pre><code>double Utils::reduceNumber(double numberToReduce, double limitNumber)
{
double factor = 0.0;
double result = 0.0;
factor = std::abs(numberToReduce / limitNumber);
if (factor != (int)factor)
factor = (int)factor + 1;
if (numberToReduce > 0)
result = numberToReduce - (byNumber * factor);
else
result = numberToReduce + (byNumber * factor);
return result;
}
</code></pre>
<p>For example, If I want to reduce −465.986246 in a limit between 0 and 24, I have to do this:</p>
<pre><code>−465.986246 + (24 x 20) = 14.013754
</code></pre>
<p>What is the formula to obtain that 20?</p>
| anak | 133,414 | <p>Just setting up some notation:</p>
<p>For your limit number $l > 0$ and the number you want to "reduce", $n$, you want to find another number $x$ such that it satisfies: $$(n + l\cdot x) \in (0,l).$$</p>
<p>So for example, we know that $\frac{l}{2} \in (0,l)$ obviously, so we will say you want to satisfy the equation: $$n + l\cdot x = \frac{l}{2}.$$</p>
<p>Solving this equation for $x$ results in the following: $$x = \frac{l-2n}{2l}.$$</p>
<hr>
<p>Note this will always put the number in the centre of the interval $(0,l)$. You can also do this for other specific points in the interval $(0,l)$.</p>
<p>If you want some sort of random noise to not make it the exact centre of the interval each time, you can just let $\epsilon$ be your randomly generated number in the interval $(-1,1)$ and then set your $x$ as follows: $$x = \frac{l(1+\epsilon) - 2n}{2l}.$$</p>
<p>EDIT: I see that mathlove assumed you wanted an integer, so this solution likely isn't what you wanted (however you specify the 'factor' is a double in your code, so you might want to optimize that if that is the case...). </p>
|
51,752 | <p>Can someone give an argument, if possible using only the axioms of set theory, because I'm <strong>very</strong> weak there and have virtually no background, except the usual knowledge of the operation with sets one has to have when doing non-set theoretic non-research mathematics, why $\emptyset \in \emptyset$ or $\emptyset \subseteq \emptyset$ should or should not hold?</p>
| Zev Chonoles | 264 | <p>An axiomatic argument (as ccc points out, we must assume that the ZF axioms are in fact consistent) would proceed as follows: By the <a href="http://en.wikipedia.org/wiki/Axiom_of_empty_set">axiom of the empty set</a>, $\forall x(\neg x\in\emptyset)$. So in particular, it is false that $\emptyset\in\emptyset$. </p>
<p>Here is a more intuitive explanation. $\emptyset\in\emptyset$ is false because $\emptyset$ has no elements (by definition). In other words,
$$\emptyset=\{\}.$$
A set <em>containing</em> the empty set is a perfectly fine set, for example,
$$\{\emptyset\};$$
but it is evident that $\emptyset$ itself is not a set containing $\emptyset$ as an element.</p>
<p>However, it <em>is</em> true that $\emptyset\subseteq\emptyset$. For any sets $A$ and $B$, we say that $A\subseteq B$ precisely when $\forall x(x\in A\Rightarrow x\in B)$. There are two ways this is clear for the case $A=B=\emptyset$; firstly, it is true that $P\Rightarrow P$ for any statement $P$ (here the statement $P$ is "$x\in \emptyset$"), so $A\subseteq A$ for any set $A$, and in particular $\emptyset\subseteq\emptyset$. Secondly, for all $x$ it is false that $x\in\emptyset$, so the implication $x\in\emptyset\Rightarrow P$ is true for any statement $P$ (see <a href="http://en.wikipedia.org/wiki/Truth_table#Logical_implication">here</a>).</p>
|
1,019,078 | <p>Let $\alpha_1=[ 2,1,3,0] $
$\alpha_2=[ 1,1,1,-1] $, $\alpha_3=[ 2,-1,5,4] $, $\alpha_4=[ 1,2,0,-3] $, $\alpha_5=[ 3,1,6,1] $
be vectors from $\mathbb{R}^4$ . From vectors system ($\alpha_1,\alpha_2, \alpha_3, \alpha_4, \alpha_5 $) choose basis of vector space $V=lin(\alpha_1,\alpha_2, \alpha_3, \alpha_4, \alpha_5)\subset\mathbb{R}^4$ spanned by those vectors.
I need help with creating a proper matrix for this question.</p>
| egreg | 62,967 | <p>Consider the matrix
\begin{bmatrix}
2 & 1 & 2 & 1 & 3\\
1 & 1 & −1 & 2 & 1\\
3 & 1 & 5 & 0 & 6\\
0 & −1 & 4 & −3 & 1
\end{bmatrix}
and perform Gaussian elimination on it:
\begin{align}
\begin{bmatrix}
2 & 1 & 2 & 1 & 3\\
1 & 1 & −1 & 2 & 1\\
3 & 1 & 5 & 0 & 6\\
0 & −1 & 4 & −3 & 1
\end{bmatrix}
&\to
\begin{bmatrix}
1 & 1/2 & 1 & 1/2 & 3/2\\
1 & 1 & −1 & 2 & 1\\
3 & 1 & 5 & 0 & 6\\
0 & −1 & 4 & −3 & 1
\end{bmatrix}
\\&\to
\begin{bmatrix}
1 & 1/2 & 1 & 1/2 & 3/2\\
0 & 1/2 & −2 & 3/2 & -1/2\\
0 & -1/2 & 2 & -3/2 & 3/2\\
0 & −1 & 4 & −3 & 1
\end{bmatrix}
\\&\to
\begin{bmatrix}
1 & 1/2 & 1 & 1/2 & 3/2\\
0 & 1 & −4 & 3 & -1\\
0 & 0 & 0 & 0 & 1\\
0 & 0 & 0 & 0 & 0
\end{bmatrix}
\end{align}
The columns where a pivot is found are linearly independent and the others can be written as linear combination of them. So a basis is
$$
\{\alpha_1,\alpha_2,\alpha_5\}
$$</p>
<p>It's not the only solution, of course, but you can see that $\alpha_5$ will be in any basis extracted from that set, because it is not a linear combination of the other four vectors. If you go on with backwards elimination,
\begin{align}
\begin{bmatrix}
1 & 1/2 & 1 & 1/2 & 3/2\\
0 & 1 & −4 & 3 & -1\\
0 & 0 & 0 & 0 & 1\\
0 & 0 & 0 & 0 & 0
\end{bmatrix}
&\to
\begin{bmatrix}
1 & 1/2 & 1 & 1/2 & 0\\
0 & 1 & −4 & 3 & 0\\
0 & 0 & 0 & 0 & 1\\
0 & 0 & 0 & 0 & 0
\end{bmatrix}
\\&\to
\begin{bmatrix}
1 & 0 & 3 & -1 & 0\\
0 & 1 & −4 & 3 & 0\\
0 & 0 & 0 & 0 & 1\\
0 & 0 & 0 & 0 & 0
\end{bmatrix}
\end{align}
you see that
\begin{align}
\alpha_3&=3\alpha_1-4\alpha_2\\
\alpha_4&=-\alpha_1+3\alpha_2
\end{align}
because elementary row operation don't change linear relations between columns.</p>
|
184,824 | <p>I have two piecewise function</p>
<pre><code>equ1 = Piecewise[{{0.524324 + 0.0376478x, 0.639464 <= x <= 0.839322}}]
equ2 = Piecewise[{{-0.506432 + 1.48068x, 0.658914 <= x <= 0.77085}}]
</code></pre>
<p>Now, I am trying to solve <code>equ1 = equ2</code>.</p>
<p>Firstly I tried <code>FindRoot</code>: </p>
<pre><code>FindRoot[equ1 == equ2, x]
</code></pre>
<p>But the output is <code>x = 0</code>. I can only get the correct answer by set search starting point <code>0.7</code>. How can I direct get the answer without set starting point? </p>
<p>Secondly, I tried code Reduce: </p>
<pre><code>Reduce[equ1 == equ2, x]
</code></pre>
<p>However, the error appear. The good news is <code>Reduce</code> do provide the correct answer for my equation. The error is: </p>
<pre><code>Reduce::ratnz: Reduce was unable to solve the system with inexact coefficients. The answer was obtained by solving a corresponding exact system and numericizing the result.
</code></pre>
<p>Do I have other way to solve those two piecewise function? </p>
| Michael E2 | 4,999 | <p>Either:</p>
<pre><code>FindRoot[equ1 == equ2, {x, 0.7, 0.639464, 0.839322}]
(* {x -> 0.714299} *)
</code></pre>
<p>Or:</p>
<pre><code>NSolve[equ1 == equ2 && 0.639464 <= x <= 0.839322, x]
</code></pre>
<blockquote>
<p><code>NSolve::ratnz</code>:.... <em>[Unimportant warning.]</em></p>
</blockquote>
<pre><code>(* {{x -> 0.714299}} *)
</code></pre>
<p><em>Update:</em>
Here's a programmatic way to get the interval of support and a starting point for <code>FindRoot</code>, assuming that by <a href="https://mathematica.stackexchange.com/questions/184824/solve-piecewise-function/184831#comment481956_184824">a comment</a>, the OP means that they have to do this a hundred times:</p>
<pre><code>equAll = {equ1, equ2};
{funAll, domAll} = Transpose@ equAll[[All, 1, 1]];
support = Simplify[And @@ domAll];
endpoints = First@ RegionBounds@ ImplicitRegion[support, x];
midpoint = Mean@ endpoints; (* can be used as a starting point *)
</code></pre>
<p><strong>Alternatively,</strong> if you can define your piecewise functions to be undefined outside the interval of support, then <code>Solve</code> can be used directly:</p>
<pre><code>pw1 = Piecewise[{{0.524324 + 0.0376478 x, 0.639464 <= x <= 0.839322}}, Undefined];
pw2 = Piecewise[{{-0.506432 + 1.48068 x, 0.658914 <= x <= 0.77085}}, Undefined];
Solve[pw1 == pw2, x]
(* {{x -> 0.714299}} *)
</code></pre>
|
1,113,760 | <blockquote>
<p>$\frac{4}{3} e^{3x} + 2 e^{2x} - 8 e^x$</p>
</blockquote>
<p>I have some confusion especially because of the e </p>
<p>how can I approach the solution?</p>
<p>The solution of the x-intercept is 0.838</p>
<p>Many thanks</p>
| Eff | 112,061 | <p><strong>Hint:</strong> The $x$-intercept is when $\frac43 e^{3x}+2e^{2x}-8e^{x} = 0$. Now set $y = e^{x}$, so your equation is </p>
<p>$$\frac{4}{3}y^3+2y^2-8y = 0 $$</p>
<p>which means</p>
<p>$$y\left(\frac{4}{3}y^2+2y-8\right) = 0. $$</p>
|
4,530,792 | <p>I have the following sequence <span class="math-container">$\left \{k \sin \left(\frac{1}{k}\right) \right\}^{\infty}_{1}$</span>. I don't know how to show that this is monotonically increasing.</p>
<p>I tried taking the derivative of the corresponding function <span class="math-container">$f(x) = x \sin \left(\frac{1}{x}\right)$</span>, and show that <span class="math-container">$f^{\prime} \geq 0$</span> for <span class="math-container">$x \geq 1$</span>, but the derivative is kind of messy. The problem falls down to showing that <span class="math-container">$$\sin(\frac{1}{x}) - \frac{\cos{\frac{1}{x}}}{x} \geq 0.$$</span></p>
<p>I am open to other approaches too, maybe some outside the box thinking way that doesn't even need the first derivative. But it feels like one should be able to show that the inequality holds.
Thank you!</p>
| DonAntonio | 31,254 | <p>Observe that</p>
<p><span class="math-container">$$\sin\frac1x-\frac1x\cos\frac1x\ge0\iff\tan\frac1x\ge\frac1x$$</span></p>
<p>Put now <span class="math-container">$\;t:=\frac1x\;$</span>, so that you want to find about</p>
<p><span class="math-container">$$f(t):=\tan t-t\;,\;\;f'(t)=\frac1{\cos^2t}-1\ge0\;\;\forall t\;\Bbb R-\left\{\frac\pi2+k\pi\;,\;\;k\in\Bbb Z\right\}$$</span></p>
|
4,394,247 | <p>I know how to represent the sentence “there is exactly one person that is happy”,</p>
<p>∀y∀x((Happy(x)∧Happy(y))→(x=y))</p>
<p>Edit: ∃x∀y(y=x↔Happy(y))
(NOW, I actually know how to represent it)</p>
<p>Where x and y represent a person.</p>
<p>However, my problem is that I can’t figure out how to say “there are exactly 3 people that are happy” in predicate logic.</p>
| Tom Sharpe | 342,007 | <p>You haven't actually asserted that anyone <em>is</em> happy. So, You need to introduce a happy person, and then assert that anyone else who is happy is actually that person:
<span class="math-container">$$\exists x\left(\mathrm{Happy}(x)\wedge\left(\forall y\left(\mathrm{Happy}(y)\to (y=x)\right)\right)\right).$$</span></p>
|
71,184 | <p>I'm solving some problems for practice, and I've come across a something I don't quite understand... So here's the deal:</p>
<blockquote>
<p>$A = \{x \in \mathbb{N}: -1 \leq x < 2\}$</p>
<p>$B = \{x \in \mathbb{Z}: -10 < x \leq 0\}$</p>
<p>$C = \{n \in \mathbb{Z}: n = 2k + 1, k \in \mathbb{Z}\}$</p>
<p>a) $C \setminus(A\cap B)$</p>
<p>b) $(B\cup C)\setminus A$</p>
</blockquote>
<p>How am I supposed to solve this when $C$ has infinite members?</p>
| Zev Chonoles | 264 | <p>The two examples you gave are both <a href="http://en.wikipedia.org/wiki/Algebraic_functions" rel="nofollow">algebraic functions</a>, i.e. functions that satisfy a polynomial equation whose coefficients are rational functions. For example,
$$f=9-\sqrt{x}$$
satisfies the polynomial equation
$$(y-9)^2-x=y^2-18y+(81-x)=0,$$
i.e. $y=f$ is a root of this polynomial, and the coefficients of the polynomial are the rational functions $1$, $-18$, and $81-x$. However, as has been pointed out in the comments,
$$g=2x^2-5x+\tfrac{3}{x}$$
<strong>is itself a rational function</strong>, so there is a particularly simple polynomial it satisfies (specifically, a linear polynomial)
$$y-(2x^2-5x+\tfrac{3}{x})=0.$$</p>
|
71,184 | <p>I'm solving some problems for practice, and I've come across a something I don't quite understand... So here's the deal:</p>
<blockquote>
<p>$A = \{x \in \mathbb{N}: -1 \leq x < 2\}$</p>
<p>$B = \{x \in \mathbb{Z}: -10 < x \leq 0\}$</p>
<p>$C = \{n \in \mathbb{Z}: n = 2k + 1, k \in \mathbb{Z}\}$</p>
<p>a) $C \setminus(A\cap B)$</p>
<p>b) $(B\cup C)\setminus A$</p>
</blockquote>
<p>How am I supposed to solve this when $C$ has infinite members?</p>
| Gerry Myerson | 8,269 | <p>I suppose you could call them Puiseux polynomials by analogy with <a href="http://en.wikipedia.org/wiki/Puiseux_series" rel="nofollow">Puiseux series,</a> though I'm not sure anyone has ever done so. </p>
|
3,794,101 | <blockquote>
<p>Show that <span class="math-container">$f: \mathbb{R^3} \to \mathbb{R}$</span> <span class="math-container">$$f(x, y, z) = xy + z^2$$</span> is continuous.</p>
</blockquote>
<p>One could just deduce that since it's a polynomial it's continuous, but how would I show this using <span class="math-container">$(\varepsilon, \delta)$</span>? I'm not familiar on using the method with multivariate functions.</p>
| José Carlos Santos | 446,262 | <p>Note that if <span class="math-container">$(x,y,z),(x_0,y_0,z_0)\in\Bbb R^3$</span>, then<span class="math-container">\begin{align}\bigl|f(x,y,z)-f(x_0,y_0,z_0)\bigr|&=|xy-x_0y_0+z^2-z_0^{\,2}|\\&\leqslant|xy-x_0y_0|+|z^2-z_0^{\,2}|\\&=\bigl|(x-x_0)y_0+(y-y_0)x_0+(x-x_0)(y-y_0)\bigr|+\\&\phantom{=}+\bigl|(z-z_0)(z+z_0)\bigr|\\&\leqslant|x-x_0||y_0|+|y-y_0||x_0|+|x-x_0||y-y_0|+\\&\phantom{=}+|z-z_0|\bigl(|z-z_0|+2|z_0|\bigr).\end{align}</span>So, given <span class="math-container">$\varepsilon>0$</span>, take <span class="math-container">$\delta>0$</span> such that:</p>
<ul>
<li><span class="math-container">$\delta|y_0|<\dfrac\varepsilon5$</span>;</li>
<li><span class="math-container">$\delta|x_0|<\dfrac\varepsilon5$</span>;</li>
<li><span class="math-container">$\displaystyle\delta<\frac{\sqrt\varepsilon}{\sqrt5}$</span>;</li>
<li><span class="math-container">$2\delta|z_0|<\dfrac\varepsilon5$</span>.</li>
</ul>
<p>Then, assuming that <span class="math-container">$|x-x_0|,|y-y_0|,|z-z_0|<\delta$</span>, since <span class="math-container">$\bigl|f(x,y,z)-f(x_0,y_0,z_0)\bigr|$</span> is smaller than or equal to<span class="math-container">$$\overbrace{|x-x_0||y_0|}^{\phantom{\varepsilon/5}<\varepsilon/5}+\overbrace{|y-y_0||x_0|}^{\phantom{\varepsilon/5}<\varepsilon/5}+\overbrace{|x-x_0||y-y_0|}^{\phantom{\varepsilon/5}<\varepsilon/5}+\overbrace{|z-z_0|^2}^{\phantom{\varepsilon/5}<\varepsilon/5}+\overbrace{2|z-z_0||z_0|}^{\phantom{\varepsilon/5}<\varepsilon/5},$$</span>you have<span class="math-container">$$\bigl|f(x,y,z)-f(x_0,y_0,z_0)\bigr|<\varepsilon.$$</span></p>
|
979,432 | <p>i was recently watching a single variable calculus video of mit 18.01, lecture 23. in that it is said that average height of a point on semicircle with respect to arc length is 2/pi.I have a hard time to understand that point. i understand why average height of point on semi circle with respect to x is pi/4. but i dont understand with respect to arc length. plz can somebody help me. </p>
| aschepler | 2,236 | <p>The "average value" of any formula $\varphi$ with respect to any increasing variable $\xi$ is defined as</p>
<p>$$ \frac{\int \varphi \, d\xi}{\int d\xi} .$$</p>
<p>For a unit semicircle, arc length is equal to the angle $\theta$, so we can write the average of height $y$ with respect to arc length as</p>
<p>$$ \frac{\int_0^\pi y \, d\theta}{\int_0^\pi d\theta} = \frac{\int_0^\pi \sin\theta\, d\theta}{\int_0^\pi d\theta} .$$</p>
<p>From there it's simple to find the answer you quoted.</p>
|
3,379,837 | <p>I know that the two semigroups <span class="math-container">$(\{0,1,2,\dots \},\times)$</span> and <span class="math-container">$(\{0,1,2,\dots \},+)$</span> are not isomorphic because if we want to map identity elements together then it can be see that we can't have injective function between them,but what can we say about <span class="math-container">$(\{1,2,\dots \},\times)$</span> and <span class="math-container">$(\{0,1,2,\dots \},+)$</span>?</p>
| J.-E. Pin | 89,374 | <p>Suppose there is an isomorphism <span class="math-container">$f:(\Bbb{N},+) \to ((\Bbb{N}-\{0\}, \times)$</span>. Then since <span class="math-container">$f$</span> preserves idempotents, one has <span class="math-container">$f(0) = 1$</span>. Let <span class="math-container">$a = f(1)$</span>. Then for every <span class="math-container">$n >0$</span>, <span class="math-container">$f(n) = a^n$</span>. Thus <span class="math-container">$f(\Bbb{N}) = \{a^n \mid n \geqslant 0\}$</span> and hence <span class="math-container">$f$</span> is not a bijection, a contradiction.</p>
|
617,927 | <p>Find the taylor expansion of $\sin(x+1)\sin(x+2)$ at $x_0=-1$, up to order $5$.</p>
<p><strong>Taylor Series</strong></p>
<p>$$f(x)=f(a)+(x-a)f'(a)+\frac{(x-a)^2}{2!}f''(a)+...+\frac{(x-a)^r}{r!}f^{(r)}(a)+...$$</p>
<p>I've got my first term...</p>
<p>$f(a) = \sin(-1+1)\sin(-1+2)=\sin(0)\sin(1)=0$</p>
<p>Now, I've calculated $f'(x)=\sin(x+1)\cos(x+2)+\sin(x+2)\cos(x+1)$</p>
<p>So that $f'(-1) = \sin(1) = 0.8414709848$</p>
<p>This means my second term would be $(x+1)(0.8414709848).$</p>
<p>But, this doesn't seem to be nice and neat like the other expansions I have done and I can't figure out what I've done wrong.</p>
<p>Merry Christmas and thanks in advance.</p>
| Thomas Belulovich | 831 | <p>Hint: </p>
<p>$f′(x)=\sin(x+1)\cos(x+2)+\sin(x+2)\cos(x+1) = \sin(2x+3).$</p>
|
3,646,911 | <p>Exercise 14.7.4 from Dummit and Foote</p>
<blockquote>
<p>Let <span class="math-container">$K=\mathbb{Q}(\sqrt[n]{a})$</span>, where <span class="math-container">$a\in \mathbb{Q}$</span>, <span class="math-container">$a>0$</span> and suppose <span class="math-container">$[K:\mathbb{Q}]=n$</span>(i.e., <span class="math-container">$x^n-a$</span> is irreducible). Let <span class="math-container">$E$</span> be any subfield of <span class="math-container">$K$</span> and let <span class="math-container">$[E:\mathbb{Q}]=d$</span>. Prove that <span class="math-container">$E=\mathbb{Q}(\sqrt[d]{a})$</span>. [Consider <span class="math-container">$ N_{K/E}(\sqrt[n]a)\in E$</span>]</p>
</blockquote>
<p>Here is a solution in MSE. </p>
<p><a href="https://math.stackexchange.com/q/1785837/517603">Subfield of $\mathbb{Q}(\sqrt[n]{a})$</a></p>
<p>I rewrite the solution in that answer here. </p>
<p>Let <span class="math-container">$\alpha=\sqrt [n]a\in \mathbb R_+$</span> be the real positive <span class="math-container">$n$</span>-th root of <span class="math-container">$a$</span>, so that <span class="math-container">$K=\mathbb Q(\alpha)$</span>.<br>
Consider some intermediate field <span class="math-container">$\mathbb Q\subset E\subset K$</span> (with <span class="math-container">$d:=[E:\mathbb Q]$</span>) and define <span class="math-container">$\beta=N_{K/E}(\alpha)\in E$</span>.<br>
We know that <span class="math-container">$\beta=\Pi _\sigma \sigma (\alpha) $</span> where <span class="math-container">$\sigma$</span> runs through the <span class="math-container">$E$</span>-algebra morphisms <span class="math-container">$K\to \mathbb C$</span>.<br>
Now, <span class="math-container">$\sigma (\alpha)=w_\sigma \cdot\alpha$</span> for some suitable complex roots <span class="math-container">$w_\sigma$</span> of <span class="math-container">$1$</span> so that <span class="math-container">$$\beta=\Pi _\sigma \sigma (\alpha)=(\Pi _\sigma w_\sigma) \cdot\alpha^e\quad (\operatorname { where } e=[K:E]=\frac nd)$$</span> Remembering that <span class="math-container">$\beta\in E\subset K\subset \mathbb R$</span> is real and that the only real roots of unity are <span class="math-container">$\pm 1$</span> we obtain <span class="math-container">$\Pi _\sigma w_\sigma=\frac {\beta}{\alpha^e}=\pm 1$</span> and <span class="math-container">$\beta=\pm \alpha^e=\pm \sqrt [d]a$</span>.<br>
Thus we have <span class="math-container">$\mathbb Q(\beta)=\mathbb Q(\sqrt [d]a)\subset E$</span> with <span class="math-container">$ \sqrt [d]a$</span> of degree <span class="math-container">$d$</span> over <span class="math-container">$\mathbb Q$</span>.<br>
Since <span class="math-container">$[E:\mathbb Q]=d$</span> too we obtain <span class="math-container">$E=\mathbb Q(\sqrt [d]a)$</span>, just as claimed in the exercise.</p>
<p>Question:
In this line,
<span class="math-container">$$\beta=\Pi _\sigma \sigma (\alpha)=(\Pi _\sigma w_\sigma) \cdot\alpha^e\quad (\operatorname { where } e=[K:E]=\frac nd)$$</span>
<span class="math-container">$K$</span> is not necessarily galois extension.<span class="math-container">$|Aut(K/E)|$</span> need not be equal to <span class="math-container">$[K:E]=\frac nd$</span>.</p>
<p>It's there in all the solution I came across. I wonder why it has to be true. Please explain why <span class="math-container">$e$</span> in above equation has to be <span class="math-container">$[K:E]$</span></p>
| Gareth Ma | 623,901 | <p>It can be written as
<span class="math-container">$$x^{-5}\sum_{n=0}^5 \binom{5}{n}(-1)^n x^{10-2n}$$</span>
<span class="math-container">$$x^{-5}\sum_{n=0}^5 \binom{5}{n}(-1)^n (x^2)^{5-n}$$</span>
<span class="math-container">$$=x^{-5}(x^2-1)^5$$</span>
<span class="math-container">$$=(\frac{x^2-1}{x})^5=32$$</span></p>
<p>which then can be easily solved.</p>
|
3,646,911 | <p>Exercise 14.7.4 from Dummit and Foote</p>
<blockquote>
<p>Let <span class="math-container">$K=\mathbb{Q}(\sqrt[n]{a})$</span>, where <span class="math-container">$a\in \mathbb{Q}$</span>, <span class="math-container">$a>0$</span> and suppose <span class="math-container">$[K:\mathbb{Q}]=n$</span>(i.e., <span class="math-container">$x^n-a$</span> is irreducible). Let <span class="math-container">$E$</span> be any subfield of <span class="math-container">$K$</span> and let <span class="math-container">$[E:\mathbb{Q}]=d$</span>. Prove that <span class="math-container">$E=\mathbb{Q}(\sqrt[d]{a})$</span>. [Consider <span class="math-container">$ N_{K/E}(\sqrt[n]a)\in E$</span>]</p>
</blockquote>
<p>Here is a solution in MSE. </p>
<p><a href="https://math.stackexchange.com/q/1785837/517603">Subfield of $\mathbb{Q}(\sqrt[n]{a})$</a></p>
<p>I rewrite the solution in that answer here. </p>
<p>Let <span class="math-container">$\alpha=\sqrt [n]a\in \mathbb R_+$</span> be the real positive <span class="math-container">$n$</span>-th root of <span class="math-container">$a$</span>, so that <span class="math-container">$K=\mathbb Q(\alpha)$</span>.<br>
Consider some intermediate field <span class="math-container">$\mathbb Q\subset E\subset K$</span> (with <span class="math-container">$d:=[E:\mathbb Q]$</span>) and define <span class="math-container">$\beta=N_{K/E}(\alpha)\in E$</span>.<br>
We know that <span class="math-container">$\beta=\Pi _\sigma \sigma (\alpha) $</span> where <span class="math-container">$\sigma$</span> runs through the <span class="math-container">$E$</span>-algebra morphisms <span class="math-container">$K\to \mathbb C$</span>.<br>
Now, <span class="math-container">$\sigma (\alpha)=w_\sigma \cdot\alpha$</span> for some suitable complex roots <span class="math-container">$w_\sigma$</span> of <span class="math-container">$1$</span> so that <span class="math-container">$$\beta=\Pi _\sigma \sigma (\alpha)=(\Pi _\sigma w_\sigma) \cdot\alpha^e\quad (\operatorname { where } e=[K:E]=\frac nd)$$</span> Remembering that <span class="math-container">$\beta\in E\subset K\subset \mathbb R$</span> is real and that the only real roots of unity are <span class="math-container">$\pm 1$</span> we obtain <span class="math-container">$\Pi _\sigma w_\sigma=\frac {\beta}{\alpha^e}=\pm 1$</span> and <span class="math-container">$\beta=\pm \alpha^e=\pm \sqrt [d]a$</span>.<br>
Thus we have <span class="math-container">$\mathbb Q(\beta)=\mathbb Q(\sqrt [d]a)\subset E$</span> with <span class="math-container">$ \sqrt [d]a$</span> of degree <span class="math-container">$d$</span> over <span class="math-container">$\mathbb Q$</span>.<br>
Since <span class="math-container">$[E:\mathbb Q]=d$</span> too we obtain <span class="math-container">$E=\mathbb Q(\sqrt [d]a)$</span>, just as claimed in the exercise.</p>
<p>Question:
In this line,
<span class="math-container">$$\beta=\Pi _\sigma \sigma (\alpha)=(\Pi _\sigma w_\sigma) \cdot\alpha^e\quad (\operatorname { where } e=[K:E]=\frac nd)$$</span>
<span class="math-container">$K$</span> is not necessarily galois extension.<span class="math-container">$|Aut(K/E)|$</span> need not be equal to <span class="math-container">$[K:E]=\frac nd$</span>.</p>
<p>It's there in all the solution I came across. I wonder why it has to be true. Please explain why <span class="math-container">$e$</span> in above equation has to be <span class="math-container">$[K:E]$</span></p>
| Condo | 409,795 | <p>Factoring out the <span class="math-container">$x^5$</span> (because it doesn't depend on <span class="math-container">$n$</span>) we obtain <span class="math-container">$$x^5\sum_{n=0}^5(-1)^n{5 \choose n}x^{-2n}=32.$$</span> Now remark that <span class="math-container">$x^{-2n}=(\tfrac{1}{x^2})^n$</span>. So by the binomial theorem we have <span class="math-container">$$x^5(1-\tfrac{1}{x^2})^5=32.$$</span> Recalling that <span class="math-container">$2^5=32$</span> we can take the 5th root of both sides to obtain <span class="math-container">$x(1-\tfrac{1}{x^2})=2$</span>. Finally we multiply both sides by <span class="math-container">$x$</span> and rearrange to obtain the equation <span class="math-container">$$x^2-2x-1=0.$$</span> I'll leave the rest to you.</p>
|
833,827 | <p>I am trying to refresh on algorithm analysis. I am looking for a refresher on summation formulas.<br>
E.g.<br>
I can derive the $$\sum_{i = 0}^{N-1}i$$ to be N(N-1)/2 but I am rusty on the and more complex e.g. something like $$\sum_{i = 0}^{N-1}{\sum_{j = i+1}^{N-1}\sum_{k=j+1}^{N-1}}$$<br>
Is there a good refresher material for this?<br>
In my example my result of the inner most loop is:<br>
$$N(N-1)(N-2)/2$$</p>
<p>which is wrong though </p>
<p><strong>UPDATE</strong><br>
The sums I am describing are basically representing the following algorithm: </p>
<pre><code>for (i = 0; i < n; i++) {
for( j = i+1; j < n; j++) {
for (k = j +1; j < n; j++) {
//code
}
}
}
</code></pre>
<p>This algorithm is <code>O(N^3)</code> according to all textbooks by definition of its structure. I am not sure why the answers are giving me an <code>O(N^4)</code></p>
| vonbrand | 43,946 | <p>The other answers are right, but they assume the innermost loop does work that is proportional to $k$, while I believe you intend it to be constant. You are right, the total work done is $O(N^3)$. You can use the sums-of-powers formulas mentioned to get the precise value if needed.</p>
|
988,628 | <p>Problem : </p>
<p>For the series $$S = 1+ \frac{1}{(1+3)}(1+2)^2+\frac{1}{(1+3+5)}(1+2+3)^2+\frac{1}{(1+3+5+7)}(1+2+3+4)^2+\cdots $$ Find the nth term of the series. </p>
<p>We know that nth can term of the series can be find by using $T_n = S_n -S_{n-1}$ </p>
<p>$$S_n =1+ \sum \frac{(\frac{n(n+1)}{2})^2}{(2n-1)^2}$$ </p>
<p>$$\Rightarrow S_n =\frac{n^4+5n^2+2n^3-4n+1}{(2n-1)^2}$$ </p>
<p>But I think this is wrong, please suggest how to proceed thanks..</p>
| Deepak | 151,732 | <p>You don't need to bother with calculating $S_n - S_{n-1}$ here as the terms are explicitly given in the summation.</p>
<p>Here $T_n = \frac{(1+2+...+n)^2}{(1+3+5+...+(2n-1))}$</p>
<p>The numerator is the square of the first $n$ integers, so is equal to $(\frac{1}{2}n(n+1))^2$.</p>
<p>The denominator is the sum of the first $n$ odd integers. You can calculate the sum in a couple of ways, by direct application of the arithmetic series sum formula, or by subtracting the sum of even numbers from the sum of the first $2n$ integers. Either way, you'll figure out the denominator is equal to $n^2$. Now just do the division.</p>
|
988,628 | <p>Problem : </p>
<p>For the series $$S = 1+ \frac{1}{(1+3)}(1+2)^2+\frac{1}{(1+3+5)}(1+2+3)^2+\frac{1}{(1+3+5+7)}(1+2+3+4)^2+\cdots $$ Find the nth term of the series. </p>
<p>We know that nth can term of the series can be find by using $T_n = S_n -S_{n-1}$ </p>
<p>$$S_n =1+ \sum \frac{(\frac{n(n+1)}{2})^2}{(2n-1)^2}$$ </p>
<p>$$\Rightarrow S_n =\frac{n^4+5n^2+2n^3-4n+1}{(2n-1)^2}$$ </p>
<p>But I think this is wrong, please suggest how to proceed thanks..</p>
| Leucippus | 148,155 | <p>Consider the series
\begin{align}
S_{n} = 1 + \frac{(1+2)^{2}}{1+3} + \frac{(1+2+3)^{2}}{1+3+5} + \cdots + \frac{(1+2+\cdots+n)^{2}}{1+3+\cdots+(2n-1)}.
\end{align}
This series is seen as
\begin{align}
S_{n} &= 1 + \frac{1}{2^2}\binom{3}{2}^{2}+ \frac{1}{3^{2}} \binom{4}{2}^{2}+ \cdots + \frac{1}{n^{2}} \binom{n+1}{2}^{2} \\
&= \sum_{r=1}^{n} \frac{1}{r^{2}} \binom{r+1}{2}^{2} \\
&= \frac{1}{4} \sum_{r=1}^{n} (r+1)^{2} = \frac{1}{4} \sum_{r=2}^{n+1} r^{2} \\
&= \frac{1}{4} \left[ -1 + \frac{(n+1)(n+2)(2n+3)}{6} \right] \\
&= \frac{n}{24} \left(2 n^{2} + 9 n + 13 \right)
\end{align}</p>
<p>Making use of this formula it is quickly seen that
\begin{align}
S_{1} &= 1 \\
S_{2} &= 1 + \frac{3^{2}}{2^{2}} = 1 + \frac{(1+2)^{2}}{(1+3)}
\end{align}</p>
|
3,735,798 | <blockquote>
<p><strong>QUESTION:</strong> Given a square <span class="math-container">$ABCD$</span> with two consecutive vertices, say <span class="math-container">$A$</span> and <span class="math-container">$B$</span> on the positive <span class="math-container">$x$</span>-axis and positive <span class="math-container">$y$</span>-axis respectively. Suppose the other vertex <span class="math-container">$C$</span> lying in the first quadrant has coordinates <span class="math-container">$(u , v)$</span>. Then find the area of the square <span class="math-container">$ABCD$</span> in terms of <span class="math-container">$u$</span> and <span class="math-container">$v$</span>.</p>
</blockquote>
<hr>
<p><strong>MY APPROACH:</strong> I was trying to solve it out using complex numbers, but I need a minor help. I have assumed <span class="math-container">$A$</span> to be <span class="math-container">$(x_1+0i)$</span>, <span class="math-container">$B$</span> to be <span class="math-container">$(0+y_2i)$</span> and <span class="math-container">$C$</span> is <span class="math-container">$(u+vi)$</span>. We know that multiplying a point by <span class="math-container">$i$</span> basically rotates it by <span class="math-container">$90°$</span>, <strong>about the origin</strong>. Here, <span class="math-container">$C$</span> is nothing but the reflection of <span class="math-container">$A$</span> about the line <span class="math-container">$BD$</span>. So if I can somehow rotate <span class="math-container">$A$</span> about <span class="math-container">$B$</span> by <span class="math-container">$90°$</span> then we will get <span class="math-container">$x_1$</span> and <span class="math-container">$y_2$</span> in terms of <span class="math-container">$u$</span> and <span class="math-container">$v$</span>.
This is where I am stuck. How to rotate a point with respect to another?</p>
<blockquote>
<p>Note that this question has been asked before. But I want to know how to solve it using complex numbers..</p>
</blockquote>
<p>Any answers, possibly with a diagram will be much helpful..</p>
<p>Thank you so much..</p>
| toronto hrb | 802,748 | <p><a href="https://i.stack.imgur.com/qZVTM.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qZVTM.jpg" alt="enter image description here" /></a></p>
<p>The area you want is the difference between the large square and 4 triangles. If <span class="math-container">$u\ge v$</span>, you have another case.</p>
|
3,735,798 | <blockquote>
<p><strong>QUESTION:</strong> Given a square <span class="math-container">$ABCD$</span> with two consecutive vertices, say <span class="math-container">$A$</span> and <span class="math-container">$B$</span> on the positive <span class="math-container">$x$</span>-axis and positive <span class="math-container">$y$</span>-axis respectively. Suppose the other vertex <span class="math-container">$C$</span> lying in the first quadrant has coordinates <span class="math-container">$(u , v)$</span>. Then find the area of the square <span class="math-container">$ABCD$</span> in terms of <span class="math-container">$u$</span> and <span class="math-container">$v$</span>.</p>
</blockquote>
<hr>
<p><strong>MY APPROACH:</strong> I was trying to solve it out using complex numbers, but I need a minor help. I have assumed <span class="math-container">$A$</span> to be <span class="math-container">$(x_1+0i)$</span>, <span class="math-container">$B$</span> to be <span class="math-container">$(0+y_2i)$</span> and <span class="math-container">$C$</span> is <span class="math-container">$(u+vi)$</span>. We know that multiplying a point by <span class="math-container">$i$</span> basically rotates it by <span class="math-container">$90°$</span>, <strong>about the origin</strong>. Here, <span class="math-container">$C$</span> is nothing but the reflection of <span class="math-container">$A$</span> about the line <span class="math-container">$BD$</span>. So if I can somehow rotate <span class="math-container">$A$</span> about <span class="math-container">$B$</span> by <span class="math-container">$90°$</span> then we will get <span class="math-container">$x_1$</span> and <span class="math-container">$y_2$</span> in terms of <span class="math-container">$u$</span> and <span class="math-container">$v$</span>.
This is where I am stuck. How to rotate a point with respect to another?</p>
<blockquote>
<p>Note that this question has been asked before. But I want to know how to solve it using complex numbers..</p>
</blockquote>
<p>Any answers, possibly with a diagram will be much helpful..</p>
<p>Thank you so much..</p>
| Narasimham | 95,860 | <p>Let <span class="math-container">$a,b $</span> be x and y intercepts. Draw lines parallel to x-axis and y-axis.</p>
<p><span class="math-container">$$ a= v- u,\; b=u $$</span></p>
<p>The diagram will be helpful.</p>
<p><a href="https://i.stack.imgur.com/8lq5c.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8lq5c.png" alt="enter image description here" /></a></p>
<p><span class="math-container">$$ Area = a^2+b^2= 2 u^2 +v^2 - 2 u v $$</span></p>
|
1,675 | <p>This is a follow-up to <a href="https://mathoverflow.net/questions/1039/explicit-direct-summands-in-the-decomposition-theorem">this post</a> on the Decomposition Theorem. Hopefully, this will also invite some discussion about the theorem and perverse sheaves in general.</p>
<p>My question is how does one use the Decomposition Theorem in practice? Is there any way to pin down the subvarieties and local systems that appear in the decomposition. For example, how do you compute intesection homology complexes using this theorem? Does anyone have a link to a source with worked out examples?</p>
<p>Another related question: What is the deep part of the theorem? Is it the fact that the pushforward of a perverse sheaf is isomorphic to its perverse hypercohomology? Is it the fact that these pieces are semisimple? Or are these both hard statements? And what is so special about algebraic varieties?</p>
| Ben Webster | 66 | <p>The short answer is that in general its very hard. For special classes of maps like semi-small ones, it's not so bad (see the book of Chriss and Ginzburg), but for an arbitrary projective map, I don't know any reliable way of dealing with it. If you know the IC sheaves downstairs well, you can use point counting (there's an example of a computation like this in <a href="https://arxiv.org/abs/0905.0486" rel="nofollow noreferrer">my paper with Geordie Williamson</a>; it's in Section 4, I think).</p>
<p>To answer your last point:</p>
<p>The hard part of the theorem is developing the theory of weights, and proving that IC sheaves are pure. Since you can't have any extensions between pure sheaves of the same weight (this is roughly because it would require the Frobenius to act trivially on Ext^1), this proves that the pushforward (which is pure because proper pushforward preserves purity) is a direct sum of shifts of perverse sheaves, so really one proves both of those pieces simultaneously.</p>
|
2,793,384 | <p>I am taking a course in Algebraic Topology next semester, so I thought of starting to read about it on my own.</p>
<p><strong>My concern:</strong> I have read <em>Topology</em> by Munkres, and read the first chapter on algebraic topology from that book, so I have an idea about the fundamental group and covering spaces. Later I read the book <em>Topology</em> by Klaus Janich till the chapter on CW complexes. I really liked the book: it was not rigorous (somewhat unclear at some places) but enough to draw my interest in the subject. I was planning to start reading algebraic topology more seriously, so a senior suggested me the book <em>Algebraic Topology: A First Course</em> by William Fulton. I have read the first chapter, but I am having some problems. I actually don't have a good background in multivariable calculus. The author gives an overview of calculus in the plane and he introduces some basic concepts about multivariable calculus. I really like algebra; I have a good background in fields and Galois theory, ring and group theory (don't know free abelian groups yet). I am reading commutative algebra side by side. So I would really like to learn algebraic topology in an 'algebraic way', and I am concerned that Fulton in this book does this by another way (a differential approach or something). But at the same time I want to learn algebraic topology in an intuitive way (with pictures) and I've been told that this is the most expository book available on algebraic topology. Am I right about this book, I mean does it approach algebraic topology using category theory and other algebraic tools or in some other way? If so please tell me is it more important for intuition, so that I can read more on multivariable calculus and then start again?</p>
<p>I know this kind of question is not supported in this community, but I do need help about this. My course will be starting from next month, and I want to get a good concept about this subject. And I want to take up a course about algebraic geometry too. That's why I don't want to spend time experimenting. Thanks in advance.</p>
| Community | -1 | <p>I have a few suggestions:</p>
<p>1) Algebraic Topology by Hatcher is a very readable book that explains things moderately well in more of an informal manner - lots of diagrams for low dimensional things. If you already know about covering spaces and fundamental groups this book will be easily accessible. </p>
<p>2) My personal favourite is A Concise Course in Algebraic Topology by May. This is quite a short book - gets straight to the point and is quite algebraic. It's also quite easy to read. When you finish that you can even move onto its sequel, More Concise Algebraic Topology by May and Ponto. </p>
|
4,236,878 | <p>Given a symmetric matrix <span class="math-container">$S$</span> and positive definite matrix <span class="math-container">$B$</span>, with <span class="math-container">$S,B \in \mathbb{R}^{n \times n}$</span> can one prove that</p>
<p><span class="math-container">\begin{align*}
\text{tr}((S-B)B) \le -\mu(S) \text{tr}(B)
\end{align*}</span></p>
<p>where <span class="math-container">$\mu(S) < 0$</span> is the largest eigenvalue of <span class="math-container">$S$</span>? And does this hold if <span class="math-container">$\mu(S) > 0$</span>?</p>
| march | 852,914 | <p>Using the additive and cyclic properties of the trace, we can write
<span class="math-container">$$
\operatorname{Tr}((S-B)B) = \operatorname{Tr}(SB) -\operatorname{Tr}(B^2) = \operatorname{Tr}(BS) - \operatorname{Tr}(B^2).
$$</span>
Provided the matrix <span class="math-container">$B$</span> has all real eigenvalues<span class="math-container">$-$</span>which is the case here since <span class="math-container">$B$</span> is positive definite<span class="math-container">$-$</span>the trace of <span class="math-container">$B^2$</span> is positive. Therefore, we can drop it to get the inequality
<span class="math-container">$$
\operatorname{Tr}((S-B)B) \leq \operatorname{Tr}(BS).
$$</span>
Now, since <span class="math-container">$S$</span> is real symmetric, it has a basis of eigenvectors <span class="math-container">$\{v_n\}$</span> with real eigenvalues <span class="math-container">$\{s_n\}$</span>, and we can compute the trace in this basis as
<span class="math-container">$$
\operatorname{Tr}(BS) = \sum_n\langle BSv_n,v_n\rangle
=\sum_n\langle Bs_nv_n,v_n\rangle
=\sum_ns_n\langle Bv_n,v_n\rangle.
$$</span>
Letting <span class="math-container">$\mu(S)$</span> be the largest eigenvalue <span class="math-container">$s_n$</span> of <span class="math-container">$S$</span>. Then,
<span class="math-container">$$
\operatorname{Tr}(BS)
=\sum_ns_n\langle Bv_n,v_n\rangle
\leq \sum_n\mu(s)\langle Bv_n,v_n\rangle
=\mu(s)\sum_n\langle Bv_n,v_n\rangle
=\mu(s)\operatorname{Tr}(B),
$$</span>
and thus,
<span class="math-container">$$
\operatorname{Tr}((S-B)B) \leq \mu(s)\operatorname{Tr}(B).
$$</span>
If it happens that <span class="math-container">$\mu(S)<0$</span>, then, using the fact that <span class="math-container">$\operatorname{Tr}(B)>0$</span> (since <span class="math-container">$B$</span> is positive definite), we have
<span class="math-container">$$
\operatorname{Tr}((S-B)B) \leq \mu(s)\operatorname{Tr}(B)
=-|\mu(s)|\operatorname{Tr}(B) \leq |\mu(s)|\operatorname{Tr}(B)
=-\mu(s)\operatorname{Tr}(B),
$$</span>
but this trick won't work for <span class="math-container">$\mu(S)>0$</span>.</p>
|
2,107,685 | <p><a href="https://i.stack.imgur.com/GtU6e.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GtU6e.png" alt="laaa"></a></p>
<p>I have to represent the function on the left as a power series, and this is the solution to it but I don't know how to calculate this for example when n=1?</p>
| Disintegrating By Parts | 112,478 | <p>Whenever you apply a spectral projection $E(S)\ne I$ to $A$, you end up with $0$ in the point spectrum of $E(S)A=AE(S)$ because $\{AE(S)\}E(\sigma\setminus S)=0$ and $E(\sigma\setminus S) \ne 0$. So that special case always requires special attention.</p>
<p>If $(a,b)\subseteq\sigma(A)$, then $E(a,b) \ne 0$; otherwise $\mu \in (a,b)$ would lead to a bounded operator,
$$
R(\mu)=\int_{\sigma}\frac{1}{\lambda-\mu}dE(\mu),
$$
which would have to be the resolvent $R(\mu)=(A-\mu I)^{-1}$. And that would force $(a,b)\in\rho(A)$, contrary to assumption.</p>
<p>Suppose $[0,1]\subseteq\sigma(A)$. Then $E(a,b) \ne 0$ for $(a,b)\subseteq[0,1]$. I'll first consider your first question where you ask about the spectrum of $AE[1/4,1/2)$. Automatically $0\in\sigma(AE[1/4,1/2))$ because $\{AE[1/4,1/2)\}E(1/2,1)=0$ and $E(1/2,1)\ne 0$. For $\mu\ne 0$ and $\mu\notin [1/4,1/2]$,
$$
AE[1/4,1/2)-\mu I=(A-\mu I)E[1/4,1/2)-\mu E(\sigma\setminus[1/4,1/2))
$$
has a bounded inverse given by
$$
(AE[1/4,1/2)-\mu I)^{-1}=\int_{[1/4,1/2)}\frac{1}{\lambda-\mu}dE(\lambda)-\frac{1}{\mu}E(\sigma\setminus[1/4,1/2)).
$$
Therefore, $\sigma(AE[1/4,1/2))\subseteq [1/4,1/2]\cup\{0\}$. Conversely $0\in\sigma(AE[1/4,1/2))$ was noted above, and, for any $\mu\in(1/4,1/2)$, the projections $E(\mu-\delta,\mu+\delta)\ne 0$ for all $\delta > 0$, which gives the existence of a non-zero vector $x_{\delta}$ such that $E(\mu-\delta,\mu+\delta)x_{\delta}=x_{\delta}$ and, hence,
\begin{align}
\|AE[1/4,1/2)x_{\delta}-\mu x_{\delta}\|
& = \|(A-\mu I)E(\mu-\delta,\mu+\delta)x_{\delta}\| \\
& \le \delta \|E(\mu-\delta,\mu+\delta)x_{\delta}\| \\
& = \delta\|x_{\delta}\|.
\end{align}
So $AE[1/4,1/2)-\mu I$ cannot be continuously invertible, which proves
$$
\{0\}\cup (1/4,1/2) \subseteq \sigma(AE[1/4,1/2))
$$
Because the spectrum is closed,
$$
\{0\} \cup [1/4,1/2]\subseteq \sigma(AE[1/4,1/2)).
$$
The opposite inclusion was previously shown. So
$$
\sigma(AE[1/4,1/2))=\{0\}\cup[1/4,1/2].
$$
The operator $AE[1/4,1/2]$ is selfadjoint. So its norm is its spectral radius, which gives $\|AE[1/4,1/2)\|=1/2$.</p>
<p>I'll let you consider the other cases. Note for example that $S=[1/3,1/2]\cap\mathbb{Q}$ could be such that $E(S)=0$, or it could give $E(S)=E[1/3,1/2]$, or $E(S)=E(T)$ could hold for a lot of closed subsets $T$ of $[1/3,1/2]$ because the spectrum is closed, and every subset of $[1/3,1/2]\cap\mathbb{Q}$ could consist of eigenvalues.</p>
|
432,964 | <p>Let $X\in \mathbb{R}^{n \times n}$.
Then, is the function</p>
<p>$$ \text{Tr}\left( (X^T X )^{-1} \right)$$ </p>
<p>convex in $X$? ($\text{Tr}$ denotes the trace operator)</p>
| user1551 | 1,551 | <p>As pointed out in the above by user "1015" (who keeps changing his username ^_^ ), the set of all invertible matrices is not convex. Therefore your question does not make sense. However, $\operatorname{tr}\left((X^TX)\right)^{-1}$ is locally convex at every invertible matrix $X$ and this can be proved using the trick demonstrated in <a href="https://math.stackexchange.com/a/297649/1551">another thread</a> by Robert Israel.</p>
<p>Let $S$ be the unique positive definite square root of $(X^TX)^{-1}$. Let $H\in M_n(\mathbb{R})$. We want to show that the function $f(t)=\operatorname{tr}\left\{\left[(X+tH)^T(X+tH)\right]^{-1}\right\}$ is locally convex at $t=0$. Now,
\begin{align*}
&\left[ (X+tH)^T (X+tH) \right]^{-1}\\
=&\left[ X^TX + t(H^TX+X^TH) + t^2H^TH \right]^{-1}\\
=&\left[ S^{-2} + t(H^TX+X^TH) + t^2H^TH \right]^{-1}\\
=&S \left[ I + t S(H^TX+X^TH)S + t^2 SH^THS \right]^{-1} S\\
=&S \left[ I + t \left(S(H^TX+X^TH)S + t SH^THS\right) \right]^{-1} S\\
=&S \left[ I + t \left(S(H^TX+X^TH)S + t SH^THS\right) + t^2 \left(S(H^TX+X^TH)S + t SH^THS\right)^2 \right] S + O(t^3).
\end{align*}
Therefore, the second order term is equal to
$$
S \left[ SH^THS + \left(S(H^TX+X^TH)S\right)^2 \right] S
$$
which is positive semidefinite. Hence its trace (i.e. $f''(0)$) is nonnegative and $f$ is locally convex at $t=0$.</p>
|
426,499 | <p>Let <span class="math-container">$X$</span> be a separable metric space which is <em>homogeneous</em>, i.e. for every two points <span class="math-container">$x,y\in X$</span> there is a homeomorphism <span class="math-container">$h$</span> of <span class="math-container">$X$</span> onto itself such that <span class="math-container">$h(x)=y$</span>.</p>
<p>A compactification of <span class="math-container">$X$</span> is a compact metric space which contains a dense homeomorphic copy of <span class="math-container">$X$</span>.</p>
<p>Does <span class="math-container">$X$</span> have a homogeneous compactification?</p>
<p>Examples of homogeneous compactifications include the circle for the real line, the torus for the plane etc.</p>
| YCor | 14,094 | <p>Since you want a connected example:</p>
<p>A surface of infinite genus has no homogeneous compactification.</p>
<p>Indeed first observe a dense locally compact subset has to be open.</p>
<p>So the surface has to be open, and by homogeneity the compactification is a closed surface. But an open subset of a closed surface has (each component of) finite genus.</p>
|
4,545,300 | <p>Find the number of all <span class="math-container">$n$</span>, <span class="math-container">$1 \leq n \leq 25$</span> such that <span class="math-container">$n^2+15n+122$</span> is divisible by 6.</p>
<p><strong>My attempt</strong>. We know that:
<span class="math-container">\begin{align*}
n^2+15n+122 & \equiv n^2+3n+2 \pmod{6}
\end{align*}</span>
But <span class="math-container">$n^2+3n+2=(n+1)(n+2)$</span>, then <span class="math-container">$n^2+15n+122 \equiv (n+1)(n+2) \pmod{6}$</span>, now we have</p>
<p><span class="math-container">\begin{align*}
n(n^2+15n+122) & \equiv n(n+1)(n+2)\pmod{6} \\
n^3+15n^2+122n & \equiv 0 \pmod{6}
\end{align*}</span>
I have done this and I think I have complicated the problem even more.</p>
| Mike | 544,150 | <p><strong>If <span class="math-container">$n$</span> must be an integer</strong>: HINT: First, <span class="math-container">$n^2+15n+122$</span> is even for all integers <span class="math-container">$n$</span>. Then <span class="math-container">$n^2+15n+122$</span> will be divisble by <span class="math-container">$6$</span> iff <span class="math-container">$n^2+15n+122$</span> is a muliple of <span class="math-container">$3$</span>. So what integers <span class="math-container">$n \pmod 3$</span> is <span class="math-container">$n^2+15n+122$</span> a multiple of <span class="math-container">$3$</span>?</p>
<p>You can tell working <span class="math-container">$\pmod 3$</span>, that <span class="math-container">$n^2+15n+122$</span> is divisible by <span class="math-container">$3$</span> for every integer <span class="math-container">$n$</span> <em>that is not a multiple of 3</em>. [Indeed, <span class="math-container">$$n^2+15n+122 \pmod 3 = n^2+2 $$</span> <span class="math-container">$$= 0 \ \text{ if $n \pmod 3 \not = 0$}.] $$</span> Thus, <span class="math-container">$n^2+15n+122$</span> is divisible by <span class="math-container">$3$</span> iff <span class="math-container">$n \pmod 3$</span> is in <span class="math-container">$\{1,2\}$</span>.</p>
<p>Thus, from this and the top paragraph, <span class="math-container">$n^2+15n+122$</span> is divisible by <span class="math-container">$6$</span> iff <span class="math-container">$n \pmod 3$</span> is in <span class="math-container">$\{1,2\}$</span>.</p>
<p><strong>If <span class="math-container">$n$</span> is allowed to be any real number and not just integral:</strong> Note that, as <span class="math-container">$n^2+15n+122$</span> is continuous on <span class="math-container">$[0,25]$</span>, the function <span class="math-container">$n^2+15n+122$</span> in <span class="math-container">$[0,25]$</span> takes all values in <span class="math-container">$[122,1122]$</span>. There are precisely <span class="math-container">$166$</span> multiples of <span class="math-container">$6$</span> in <span class="math-container">$[122,1122]$</span>. As <span class="math-container">$n^2+15n+122$</span> is strictly increasing on <span class="math-container">$[0,25]$</span>, it follows that there is at most one real number <span class="math-container">$n \in [0,25]$</span> such that <span class="math-container">$n^2+15n+122 = y$</span>.</p>
<p>So for each of these <span class="math-container">$166$</span> multiples <span class="math-container">$y$</span> of <span class="math-container">$6$</span> in <span class="math-container">$[122,1122]$</span> there is exactly one real number <span class="math-container">$n \in [0,25]$</span> such that <span class="math-container">$n^2+15n+122 = y$</span>.</p>
|
4,545,300 | <p>Find the number of all <span class="math-container">$n$</span>, <span class="math-container">$1 \leq n \leq 25$</span> such that <span class="math-container">$n^2+15n+122$</span> is divisible by 6.</p>
<p><strong>My attempt</strong>. We know that:
<span class="math-container">\begin{align*}
n^2+15n+122 & \equiv n^2+3n+2 \pmod{6}
\end{align*}</span>
But <span class="math-container">$n^2+3n+2=(n+1)(n+2)$</span>, then <span class="math-container">$n^2+15n+122 \equiv (n+1)(n+2) \pmod{6}$</span>, now we have</p>
<p><span class="math-container">\begin{align*}
n(n^2+15n+122) & \equiv n(n+1)(n+2)\pmod{6} \\
n^3+15n^2+122n & \equiv 0 \pmod{6}
\end{align*}</span>
I have done this and I think I have complicated the problem even more.</p>
| B. Goddard | 362,009 | <p>If <span class="math-container">$n$</span> is even then <span class="math-container">$n^2+15n$</span> is even. If <span class="math-container">$n$</span> is odd, then <span class="math-container">$n^2+15n$</span> is still even. So <span class="math-container">$n^2+15n+122$</span> is even for every <span class="math-container">$n$</span>.</p>
<p>So we just need to determine when the expression is divisible by <span class="math-container">$3$</span>.</p>
<p><span class="math-container">$$n^2+15n+122 \equiv n^2 +2 \pmod{3}.$$</span></p>
<p>It's easy to check that <span class="math-container">$n=1$</span> and <span class="math-container">$n=2$</span> are the only solutions. So the answer is "all non multiples of <span class="math-container">$3$</span>". There are <span class="math-container">$8$</span> multiples of <span class="math-container">$3$</span> between <span class="math-container">$1$</span> and <span class="math-container">$25$</span>, so the final number is <span class="math-container">$25-8 = 17.$</span></p>
|
514 | <p>I just came back from my Number Theory course, and during the lecture there was mention of the Collatz Conjecture.</p>
<p>I'm sure that everyone here is familiar with it; it describes an operation on a natural number – <span class="math-container">$n/2$</span> if it is even, <span class="math-container">$3n+1$</span> if it is odd.</p>
<p>The conjecture states that if this operation is repeated, all numbers will eventually wind up at <span class="math-container">$1$</span> (or rather, in an infinite loop of <span class="math-container">$1-4-2-1-4-2-1$</span>).</p>
<p>I fired up Python and ran a quick test on this for all numbers up to <span class="math-container">$5.76 \times 10^{18}$</span> (using the powers of cloud computing and dynamic programming magic). Which is millions of millions of millions. And all of them eventually ended up at <span class="math-container">$1$</span>.</p>
<p>Surely I am close to testing every natural number? How many natural numbers could there be? Surely not much more than millions of millions of millions. (I kid.)</p>
<p>I explained this to my friend, who told me, "Why would numbers suddenly get different at a certain point? Wouldn't they all be expected to behave the same?"</p>
<p>To which I said, "No, you are wrong! In fact, I am sure there are many conjectures which have been disproved by counterexamples that are extremely large!"</p>
<p>And he said, "It is my conjecture that there are none! (and if any, they are rare)".</p>
<p>Please help me, smart math people. Can you provide a counterexample to his conjecture? Perhaps, more convincingly, several? I've only managed to find one! (Polya's conjecture). One, out of the many thousands (I presume) of conjectures. It's also one that is hard to explain the finer points to the layman. Are there any more famous or accessible examples?</p>
| mau | 89 | <p>The first example which came to my mind is the <a href="http://en.wikipedia.org/wiki/Skewes%27_number">Skewes' number</a>, that is the smallest natural number n for which π(n) > li(n). Wikipedia states that now the limit is near e<sup>727.952</sup>, but the first estimation was much higher.</p>
|
514 | <p>I just came back from my Number Theory course, and during the lecture there was mention of the Collatz Conjecture.</p>
<p>I'm sure that everyone here is familiar with it; it describes an operation on a natural number – <span class="math-container">$n/2$</span> if it is even, <span class="math-container">$3n+1$</span> if it is odd.</p>
<p>The conjecture states that if this operation is repeated, all numbers will eventually wind up at <span class="math-container">$1$</span> (or rather, in an infinite loop of <span class="math-container">$1-4-2-1-4-2-1$</span>).</p>
<p>I fired up Python and ran a quick test on this for all numbers up to <span class="math-container">$5.76 \times 10^{18}$</span> (using the powers of cloud computing and dynamic programming magic). Which is millions of millions of millions. And all of them eventually ended up at <span class="math-container">$1$</span>.</p>
<p>Surely I am close to testing every natural number? How many natural numbers could there be? Surely not much more than millions of millions of millions. (I kid.)</p>
<p>I explained this to my friend, who told me, "Why would numbers suddenly get different at a certain point? Wouldn't they all be expected to behave the same?"</p>
<p>To which I said, "No, you are wrong! In fact, I am sure there are many conjectures which have been disproved by counterexamples that are extremely large!"</p>
<p>And he said, "It is my conjecture that there are none! (and if any, they are rare)".</p>
<p>Please help me, smart math people. Can you provide a counterexample to his conjecture? Perhaps, more convincingly, several? I've only managed to find one! (Polya's conjecture). One, out of the many thousands (I presume) of conjectures. It's also one that is hard to explain the finer points to the layman. Are there any more famous or accessible examples?</p>
| J. W. Tanner | 615,567 | <p>Fermat conjectured that <span class="math-container">$F_n=2^{2^n}+1$</span> is prime for all <span class="math-container">$n$</span>, </p>
<p>but Euler showed that <span class="math-container">$ F_{5}=2^{2^{5}}+1=2^{32}+1=4294967297=641\times 6700417.$</span></p>
|
1,246,522 | <p>What are the prime ideals of $\mathbb F_p[x]/(x^2)$? I have been told that the only one is $(x)$, but I would like a proof of this. I want to say that a prime ideal of $\mathbb F_p[x]/(x^2)$ corresponds to a <strong>prime</strong> ideal $P$ of $\mathbb F_p[x]$ containing $(x^2)$. And then $P$ contains $(x)$ since it is prime. But I don't know if prime ideals correspond to prime ideals under the correspondence theorem, and I still can't seem to prove that if they do, $P$ can't be some non-principal ideal properly larger than $(x)$.</p>
<p>Some context: I'm considering why the prime ideals $\mathfrak p$ of $\mathcal O_K$, (with $K=\mathbb Q(\sqrt d)$ and $\textrm{Norm}(\mathfrak p)=p$, a ramified prime) are unique. My definition of a ramified prime is that $\mathcal O_K/(p) \cong \mathbb F_p[x]/(x^2)$ and I know nothing else about these primes.</p>
| rschwieb | 29,335 | <blockquote>
<p>and I still can't seem to prove that if they do [correspond], $P$ can't be some non-principal ideal properly larger than $(x)$.</p>
</blockquote>
<p>Assuming you convince yourself of the correspondence of prime ideals (which is just fine) here's an elementary way to see that there can only be one prime ideal in this ring.</p>
<p>Now $(x)$ is a maximal ideal of $\Bbb F[x]$ for any field $F$. This is easily seen since $F[x]/(x)\cong F$. The ideal $(x^2)=(x)^2$ therefore is a power of a maximal ideal.</p>
<p>We can say a little by abstracting: </p>
<p><strong>Proposition:</strong> If $R$ is any ring and $M$ is a maximal ideal, $R/M^n$ has exactly one prime ideal, namely $M/M^n$.</p>
<p><em>Proof</em>: A prime ideal of $R/M^n$ looks like $P/M^n$ where $P$ is a prime ideal of $R$ containing $M^n$. Plainly $M^n\subseteq P$, and by primeness $M\subseteq P$. But $M$ is maximal, so $M=P$. Thus there is only one prime ideal containing $M^n$, and only one prime ideal of $R/M^n$.</p>
|
26,451 | <p>I am trying to solve the following:</p>
<p>$\begin{align*}
&X \sim N(1,1)\\
&\mathrm{cov}(X, X^3) = \text{?}
\end{align*}$</p>
<p>where $\mathrm{cov}$ is the covariance.</p>
<p>How would you do this in <em>Mathematica</em>?</p>
<p>I have tried</p>
<pre><code>X = NormalDistribution[1, 1]
cov[x_, y_] := Mean[TransformedDistribution[a*b,
{a \[Distributed] x, b \[Distributed] y}]] - Mean[x] Mean[y]
cov[X, TransformedDistribution[a^3, a \[Distributed] X]]
</code></pre>
<p>But this doesn't seem to work.</p>
| wolfies | 898 | <p>Given $X$ ~ $N(\mu, \sigma^2)$ with pdf $f(x)$:</p>
<p>$$f=\frac{1}{\sqrt{2 \pi } \sigma } {\text{Exp} \left[-\frac{(x-\mu )^2}{2 \sigma ^2}\right]}; \text{ domain}[f]=\{x,-\infty ,\infty \}\land \{\mu \in \text{Reals},\sigma >0\};$$</p>
<p>Then, using the <code>mathStatica</code> package for <em>Mathematica</em>, the solution is simply:</p>
<pre><code> Cov[{x, x^3}, f]
</code></pre>
<blockquote>
<p>$3 \sigma ^2 \left(\mu ^2+\sigma ^2\right)$</p>
</blockquote>
<p>In your specific case, with $\mu =1$ and $\sigma^2=1$, the answer is thus 6. </p>
|
3,595,622 | <p><strong>Problem: Give an example of a linear continuum which is not the real line <span class="math-container">$\mathbb{R}$</span>, nor
topologically equivalent to a subspace of <span class="math-container">$\mathbb{R}$</span>.</strong></p>
<p><strong>Definition of Linear Continuum:</strong> Let X be a linearly ordered set with order <. We say that X is a linear continuum iff it satisfies the following two axioms:</p>
<p>(1) LUB: X has the least upper bound property.
(2) Betweenness: <span class="math-container">$\forall x$</span> < <span class="math-container">$y$</span> <span class="math-container">$\in X$</span>, <span class="math-container">$\exists z$</span> <span class="math-container">$\in X$</span>, such that <span class="math-container">$x < z < y$</span>.</p>
<p>This is how I did it, not sure whether its accurate or not. </p>
<p>I tried to prove that <span class="math-container">$I$</span> x <span class="math-container">$I$</span> is not connected under subspace topology of <span class="math-container">$\mathbb{R^{2}}$</span> under dictionary order. Since, <span class="math-container">${x}$</span> X <span class="math-container">$I$</span> is open in <span class="math-container">$I$</span> X <span class="math-container">$I$</span>, therefore, for each <span class="math-container">$x$</span> <span class="math-container">$\in$</span> <span class="math-container">$I$</span>, say (<span class="math-container">${a}$</span> X <span class="math-container">$[a,b]$</span>) <span class="math-container">$\cup$</span> <span class="math-container">${y}$</span> X <span class="math-container">$I$</span>, where <span class="math-container">$y \in [a,b]$</span>, we can clearly say that their intersection will be <span class="math-container">$\emptyset$</span>, i.e. (<span class="math-container">${a}$</span> X <span class="math-container">$[a,b]$</span>) <span class="math-container">$\cup$</span> {<span class="math-container">${y}$</span> X <span class="math-container">$I$</span>} = <span class="math-container">$I X I$</span>, & (<span class="math-container">${a}$</span> X <span class="math-container">$[a,b]$</span>) <span class="math-container">$\cap$</span> (<span class="math-container">${y}$</span> X <span class="math-container">$I$</span>) = <span class="math-container">$\emptyset$</span>. Hence, its not connected under subspace topology of <span class="math-container">$\mathbb{R}^{2}$</span> but its still a linear continuum. </p>
<p>I was trying to come up with something else but unfortunately I couldn't get a better example. Need help from someone on this. Appreciate your time and patience. </p>
| Henno Brandsma | 4,280 | <p>In fact, the lexicographically ordered square <span class="math-container">$I \times I$</span> is a classic example of a linear continuum, even compact, that is not a subspace of the reals, e.g. because it is not separable (and all subspaces of <span class="math-container">$\Bbb R$</span> are second countable hence separable). This can be seen as all subsets of the form <span class="math-container">$\{x\} \times (0,1)$</span> are open, and pairwise disjoint so any dense set must intersect all of them, and each time in a new point, so all dense sets are at least size continuum too. If <span class="math-container">$X$</span> has the property that all pairwise disjoint families of non-empty open sets are countable, it’s said to be a ccc space (ccc stands for countable chain condition, which is an historically grown term). So this square is not ccc, hence non-separable.</p>
<p>Another classic example is based on <span class="math-container">$\omega_1$</span> (or <span class="math-container">$\Omega$</span>, as Munkres calls it) the smallest uncountable well-ordered set. If we order <span class="math-container">$\omega_1 \times [0,1)$</span> lexicographically we get another linear continuum (the <em>long line</em>)( being a linear continuum already follows from the well-orderedness of the first space) that is not a subspace of the reals for the same reason (non-separable), even though it is even path-connected, unlike the lexicographically ordered square. It also is non-ccc. </p>
<p>In fact all separable linear continua can be embedded as subspaces of <span class="math-container">$\Bbb R$</span> and all known other examples of non-separable ones are also not ccc. It has long been an open problem whether a non-separable but ccc linear continuum exists and it turned out that in <strong>some</strong> models of set theory such so-called <em>Suslin lines</em> do exist and in <strong>other</strong> models they do not. So you cannot just construct an example of one of those (although they are intriguing objects when they do exist).</p>
|
3,805,989 | <p>I'm doing Exercise 4 in textbook Algebra by Saunders MacLane and Garrett Birkhoff.</p>
<p><a href="https://i.stack.imgur.com/JQww8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JQww8.png" alt="enter image description here" /></a></p>
<blockquote>
<p>Show that, if <span class="math-container">$F$</span> is a field, the group of all those automorphisms of <span class="math-container">$F[x]$</span> which leave all elements of <span class="math-container">$F$</span> fixed, consists of substitutions given by <span class="math-container">$x \mapsto a x+b, a \neq 0$</span> and <span class="math-container">$b$</span> in <span class="math-container">$F$</span>.</p>
</blockquote>
<p>Could you please verify if my understanding is correct? Thank you so much for your help!</p>
<hr />
<p><strong>My attempt:</strong></p>
<p>Consider a map <span class="math-container">$f: \sum a_n x^n \mapsto \sum a_n (ax+b)^n$</span>. It suffices to show that <span class="math-container">$f$</span> is an automorphism. It's trivial to show that it is a homomorphism. Hence it remains to show that it is bijective.</p>
<p>Let <span class="math-container">$p = \sum a_n x^n\in F[x]$</span>. By polynomial division, there are unique polynomials <span class="math-container">$q_1,r_1$</span> such that <span class="math-container">$p = (ax+b)q_1+r_1$</span> and <span class="math-container">$\deg r_1 < \deg q_1$</span>. Inductively, <span class="math-container">$p = \sum b_n (ax+b)^n$</span> for some <span class="math-container">$b_n$</span>'s. The surjectivity then follows. Because such <span class="math-container">$b_n$</span>'s are unique, the injectivity then follows.</p>
<hr />
<p><strong>Update:</strong> I add the proof for "If <span class="math-container">$f$</span> is an automorphism on <span class="math-container">$F[x]$</span> such that <span class="math-container">$f(c)=c$</span> for all <span class="math-container">$c \in F$</span>, then <span class="math-container">$f(x)=ax+b$</span> for some <span class="math-container">$a \neq 0$</span> and <span class="math-container">$b$</span> in <span class="math-container">$F$</span>" here.</p>
<p>If <span class="math-container">$\deg f(x) < 1$</span>, then <span class="math-container">$\operatorname{im} f \subseteq F$</span>. If <span class="math-container">$\deg f(x) > 1$</span>, then <span class="math-container">$\operatorname{im} f$</span> does not contain such polynomials whose degrees are <span class="math-container">$1$</span>. In both cases, <span class="math-container">$f$</span> is not surjective. As such, <span class="math-container">$\deg f(x) = 1$</span>.</p>
| Wuestenfux | 417,848 | <p>Hint: If <span class="math-container">$\sigma$</span> is such an automorphism, then <span class="math-container">$f = \sum_i a_ix^i$</span> maps to <span class="math-container">$f^\sigma = \sum_i a_i^\sigma (x^i)^\sigma = \sum_i a_i (x^\sigma)^i =\sum_i a_i g^i$</span>, where <span class="math-container">$x^\sigma = g$</span> is a polynomial in <span class="math-container">$x$</span>.</p>
<p>Since <span class="math-container">$f$</span> is an automorphism, <span class="math-container">$g$</span> must be a linear polynomial which you can show by comparison of degrees.</p>
|
3,506,659 | <p>Let <span class="math-container">$\sigma_i$</span> denote the <a href="https://en.wikipedia.org/wiki/Pauli_matrices" rel="nofollow noreferrer">Pauli matrices</a>:
<span class="math-container">$$
\sigma_1\equiv \begin{pmatrix}0&1\\1&0\end{pmatrix}, \quad
\sigma_2\equiv \begin{pmatrix}0&-i\\i&0\end{pmatrix}, \quad
\sigma_3\equiv \begin{pmatrix}1&0\\0&-1\end{pmatrix}.
$$</span>
It isn't hard to see that any <span class="math-container">$2\times 2$</span> unitary <span class="math-container">$U$</span> can be written in terms of these matrices as
<span class="math-container">$$ U = c_0 I + \sum_{k=1}^3 ic_k \sigma_k, $$</span>
for some real coefficients <span class="math-container">$c_j$</span> normalised to one: <span class="math-container">$\mathbf c\equiv(c_0,c_1,c_2,c_3)\in S^3$</span>.</p>
<p>It turns out to be the case that
<span class="math-container">$$ U\sigma_i U^\dagger = \sum_{j=1}^3 B_{ij} \sigma_j, \tag A$$</span>
for any <span class="math-container">$i\in\{1,2,3\}$</span>, with <span class="math-container">$B$</span> a unitary matrix.
I can see why this must be the case by direct analysis on <span class="math-container">$U\sigma_i U^\dagger$</span>: expanding <span class="math-container">$U$</span> in terms of Pauli matrices and using the known expressions for products of Pauli matrices to get to a final expression for <span class="math-container">$B_{ij}$</span>. My problem with this is that it's a somewhat tedious procedure, and the final expression doesn't make it particularly obvious that <span class="math-container">$B$</span> is always unitary.</p>
<p>I am looking for a better way to prove (A), especially because the expression seems to lend itself to be understood on more abstract grounds (I don't know much about Lie theory, but it seems to be saying something on the lines of <span class="math-container">$U(2)$</span> acting on its Lie algebra unitarily via the adjoint representation... if that makes sense).</p>
| glS | 173,147 | <p>While <a href="https://math.stackexchange.com/a/3506882/173147">the other answer</a> is definitely what I was looking for, I will also add how to find the explicit form of <span class="math-container">$B$</span>, for future reference.</p>
<p>The idea is to find what <span class="math-container">$U\sigma_i U^\dagger$</span> looks like, for <span class="math-container">$U=c_0 I+ ic_k\sigma_k$</span> (summing on repeated indices), using the following identities to handle products of Pauli matrices:
<span class="math-container">$$
\sigma_i \sigma_j = i\epsilon_{ijk}\sigma_k + \delta_{ij} I, \\
\sigma_i \sigma_j \sigma_k = i\epsilon_{ijk} I + (\delta_{ij}\sigma_k+\delta_{jk}\sigma_i-\delta_{ik}\sigma_j).
$$</span>
We then have
<span class="math-container">$$
U\sigma_i U^\dagger = (c_0 I +ic_j\sigma_j)\sigma_i(c_0 I -i c_k\sigma_k)
= c_0^2 \sigma_i - ic_0c_k\sigma_i \sigma_k + ic_0 c_j\sigma_j\sigma_i + c_j c_k\sigma_j\sigma_i\sigma_k.
$$</span>
Using <span class="math-container">$\sigma_i\sigma_k=-\sigma_k\sigma_i+2\delta_{ik}$</span> and the other given properties we have
<span class="math-container">$$
U\sigma_i U^\dagger = c_0^2 \sigma_i + \color{blue}{2ic_0 c_j \sigma_j\sigma_i} -2ic_0 c_i
+ ic_j c_k \epsilon_{jik} + (c_i c_k \sigma_k + c_j c_i \sigma_j - c_j c_j \sigma_i) \\
= c_0^2 \sigma_i + \color{blue}{(-2c_0 c_j \epsilon_{jik}\sigma_k +}\underbrace{\color{blue}{ 2ic_0 c_i)} -2ic_0 c_i}_{=0}
+ \underbrace{ic_j c_k \epsilon_{jik}}_{=0} + (c_i c_k \sigma_k + c_j c_i \sigma_j - c_j c_j \sigma_i) \\
= (c_0^2 - \|\mathbf c\|^2) \sigma_i + 2c_0 c_j \epsilon_{ijk}\sigma_k + 2c_i (\mathbf c\cdot \boldsymbol\sigma)
= (2c_0^2 - 1) \sigma_i + 2c_0 c_j \epsilon_{ijk}\sigma_k + 2c_i (\mathbf c\cdot \boldsymbol\sigma).
$$</span>
In other words, the <span class="math-container">$B$</span> in <span class="math-container">$U\sigma_i U^\dagger = B_{ij}\sigma_j$</span> is then given by
<span class="math-container">$$
B_{ij} = (2c_0^2-1)\delta_{ij} + 2c_0 c_k \epsilon_{ikj} + 2c_i c_j.
$$</span>
I don't know of an easy way to see that this is unitary.</p>
|
4,235,607 | <p>Let <span class="math-container">$x_1,\dots, x_n \geq 0$</span> be a sequence of numbers such that <span class="math-container">$\sum_{i=1}^n x_i = 1$</span>. For every <span class="math-container">$k \geq 1$</span>, I conjecture (and need to prove) that
<span class="math-container">$$
\frac{\sum_{1\leq i\neq j\leq n} x_i x_j \left( (1-x_i-x_j)^{k-1}- (1-x_i)^k(1-x_j)^k\right)}{\sum_{i=1}^n x_i \left(1-(1-x_i)^k\right)} \leq \frac{C}{k}
$$</span>
where <span class="math-container">$C>0$</span> is some absolute constant, which I suspect can be taken to be <span class="math-container">$C=2$</span>.</p>
<p>I don't really know how to handle this elegantly, or even how to handle it at all. I'd be happy with any proof, but a short one would be appreciated. It's one of these statements which <em>seems</em> to scream for a nice "convexity" or "symmetry" or other "beautiful rabbit out of hat" argument, but any proof at all (or counterexample -- though that'd be quite annoying) would be appreciated.</p>
<p>The "natural" case where all <span class="math-container">$x_i$</span> are either 0 or some value <span class="math-container">$1/m$</span> (i.e., <span class="math-container">$x_1=\dots=x_m=1/m$</span>, and <span class="math-container">$x_{m+1}=\dots=x_n=0$</span>) has an upper bound with a closed form
<span class="math-container">$$
k\frac{(1-\frac{2}{m})^{k-1}-(1-\frac{1}{m})^{2k}}{1-(1-\frac{1}{m})^k} \leq 2
$$</span>
which supports the conjecture. I suspect this is the worst case, but am not sure how to formally argue it.</p>
| Clement C. | 75,808 | <ul>
<li><strong>A failed attempt, which gives the <span class="math-container">$2/k$</span> dependence but "loses" the denominator.</strong></li>
</ul>
<p>We add back the diagonal terms of the double sum, and bound the numerator <span class="math-container">$N_k(x)$</span> as
<span class="math-container">\begin{align}
N_k(x) &= \sum_{1\leq i\neq j\leq n} x_i x_j \left( (1-x_i-x_j)^{k-1}- (1-x_i)^k(1-x_j)^k\right) \\
&\leq \sum_{1\leq i, j\leq n} x_i x_j \left( (1-x_i-x_j)^{k-1}- (1-x_i)^k(1-x_j)^k\right) \\
&\leq \sum_{1\leq i, j\leq n} x_i x_j \left( (1-x_i)^{k-1}(1-x_j)^{k-1}- (1-x_i)^k(1-x_j)^k\right) \\
&= \sum_{1\leq i, j\leq n} x_i x_j (1-x_i)^{k-1}(1-x_j)^{k-1} \left( x_i+x_j - x_ix_j\right) \\
&= 2\sum_{i=1}^n x_i^2 (1-x_i)^{k-1} \sum_{j=1}^n x_j(1-x_j)^{k-1} - \left(\sum_{i=1}^n x_i^2 (1-x_i)^{k-1}\right)^2 \\
&= \left(\sum_{i=1}^n x_i^2 (1-x_i)^{k-1}\right)\left(\sum_{i=1}^n x_i(2-x_i) (1-x_i)^{k-1}\right)
\end{align}</span></p>
<p>We can bound the first factor as
<span class="math-container">$$
\sum_{i=1}^n x_i^2 (1-x_i)^{k-1}
\leq \left(\sum_{i=1}^n x_i\right) \sup_{y\in[0,1]} y (1-y)^{k-1}
= 1\cdot \frac{1}{k}\left(1-\frac{1}{k}\right)^{k-1}
\leq \frac{1}{k}
$$</span></p>
<p>For the second, we can write
<span class="math-container">$
\sum_{i=1}^n x_i(2-x_i) (1-x_i)^{k-1} \leq
2.
$</span></p>
<p>Ufortunately, this does not seem to lead to a better bound, since that second term is <em>roughly</em> of the form <span class="math-container">$cst\cdot\sum_{i=1}^n x_i(1-x_i)^{k}$</span>, while we would like to compare it to <span class="math-container">$cst(1-\sum_{i=1}^n x_i(1-x_i)^{k})$</span> (which can be much, more smaller, as we can have <span class="math-container">$\sum_{i=1}^n x_i(1-x_i)^{k} \approx 1$</span>).</p>
<ul>
<li><strong>Some evidence, based on Taylor expansions.</strong>
If we just take the linear part of the Taylor expansions (which we cannot in general, of course) and ignore the rest, we get:
<span class="math-container">$$\begin{align}
&\frac{\sum_{i\neq j} x_i x_j \left( (1-x_i-x_j)^{k-1}- (1-x_i)^k(1-x_j)^k\right)}{\sum_{i=1}^n x_i \left(1-(1-x_i)^k\right)} \\&\stackrel{\color{red}⚠️}{\approx}
\frac{\sum_{i\neq j} x_i x_j \left( (1-(k-1)(x_i+x_j)- (1-k (x_i +x_j)\right)}{\sum_{i=1}^n x_i \left(1-(1-k x_i)\right)} \\
&=
\frac{\sum_{i\neq j} x_i x_j \left(x_i+x_j\right)}{k\sum_{i=1}^n x_i^2} \\
&\leq
\frac{\sum_{i, j} x_i x_j \left(x_i+x_j\right)}{k\sum_{i=1}^n x_i^2}
= \frac{2\sum_{i=1}^n x_i \sum_{j=1}^n x_j^2}{k\sum_{i=1}^n x_i^2} \\
&= \frac{2}{k}
\end{align}$$</span>
using <span class="math-container">$\sum_{i=1}^n x_i=1$</span> in the last equality.</li>
</ul>
|
4,235,607 | <p>Let <span class="math-container">$x_1,\dots, x_n \geq 0$</span> be a sequence of numbers such that <span class="math-container">$\sum_{i=1}^n x_i = 1$</span>. For every <span class="math-container">$k \geq 1$</span>, I conjecture (and need to prove) that
<span class="math-container">$$
\frac{\sum_{1\leq i\neq j\leq n} x_i x_j \left( (1-x_i-x_j)^{k-1}- (1-x_i)^k(1-x_j)^k\right)}{\sum_{i=1}^n x_i \left(1-(1-x_i)^k\right)} \leq \frac{C}{k}
$$</span>
where <span class="math-container">$C>0$</span> is some absolute constant, which I suspect can be taken to be <span class="math-container">$C=2$</span>.</p>
<p>I don't really know how to handle this elegantly, or even how to handle it at all. I'd be happy with any proof, but a short one would be appreciated. It's one of these statements which <em>seems</em> to scream for a nice "convexity" or "symmetry" or other "beautiful rabbit out of hat" argument, but any proof at all (or counterexample -- though that'd be quite annoying) would be appreciated.</p>
<p>The "natural" case where all <span class="math-container">$x_i$</span> are either 0 or some value <span class="math-container">$1/m$</span> (i.e., <span class="math-container">$x_1=\dots=x_m=1/m$</span>, and <span class="math-container">$x_{m+1}=\dots=x_n=0$</span>) has an upper bound with a closed form
<span class="math-container">$$
k\frac{(1-\frac{2}{m})^{k-1}-(1-\frac{1}{m})^{2k}}{1-(1-\frac{1}{m})^k} \leq 2
$$</span>
which supports the conjecture. I suspect this is the worst case, but am not sure how to formally argue it.</p>
| fedja | 12,992 | <p>If you don't care about the constant, the inequality is rather simple. For all practical purposes (i.e., up to an absolute constant factor) <span class="math-container">$1-(1-x_i)^k\asymp\min(kx_i,1)=y_i$</span>. Also, we trivially have <span class="math-container">$\sum_i y_i\le k\sum_i x_i=k$</span>. Now we note that <span class="math-container">$(1-x_i-x_j)^{k-1}-(1-x_i)^{k-1}(1-x_j)^{k-1}\le 0$</span> (just because <span class="math-container">$1-x_i-x_j\le(1-x_i)(1-x_j)$</span>). Thus it suffices to prove that
<span class="math-container">$$
\sum_{i,j}x_ix_j[1-(1-x_i)(1-x_j)](1-x_i)^{k-1}(1-x_j)^{k-1}\le \frac Ck\sum_i x_iy_i\,.
$$</span>
Now,<span class="math-container">$1-(1-x_i)(1-x_j)\le x_i+x_j$</span> and <span class="math-container">$x_i(1-x_i)^{k-1}\le\min(x_i,\frac 1k)=\frac 1ky_i$</span>, so the LHS is bounded by
<span class="math-container">$$
\frac 1{k^2}\sum_{i,j}(x_i+x_j)y_iy_j=\frac 2{k^2}\sum_{i,j}x_iy_iy_j
\\
=\frac 2{k^2}\left[\sum_{i}x_iy_i\right]\left[\sum_{j}y_j\right]\le
\frac 2{k^2}\left[\sum_{i}x_iy_i\right]k= \frac 2k\sum_{i}x_iy_i
$$</span>
and we are done.</p>
<p>This crude computation doesn't yield <span class="math-container">$C=2$</span> though. To get <span class="math-container">$C=2$</span> in this way, you just want to define <span class="math-container">$y_i=1-(1-x_i)^k$</span> (which is always <span class="math-container">$\le\min(kx_i,1)$</span>, so the final computation is fine) and show that the inequality <span class="math-container">$kx_i(1-x_i)^{k-1}\le y_i$</span> still holds. This is also easy once you realize that <span class="math-container">$1-(1-x)^k=\int_0^x k(1-t)^{k-1}\,dt$</span> and you can now rewrite the above argument with this slick definition in the same way, but, of course, that is not how one would guess it in the first place, so I included the crude computation based on the simple idea to replace hard functions with equivalent simple ones :-)</p>
|
374,105 | <p>Does $\exists$ on the hyperbolic plane, a convex quadrilateral $Q$ and a convex pentagon $P$ with the same angle sum? I found this question to be rather interesting.</p>
| Incnis Mrsi | 168,952 | <p>Hint: a regular quadrilateral can have <em>any</em> value of interior angles between 0 and π/2, and a regular pentagon can have <em>any</em> value of interior angles between 0 and 3π/5.</p>
|
16,802 | <p>In an attempt to squeeze more plots and controls into the limited space for a demo UI, I am trying to remove any extra white spaces I see.</p>
<p>I am not sure what options to use to reduce the amount of space between the ticks labels and the actual text that represent the labels on the axes.</p>
<p>Here is a small <code>Plot</code> example using <code>Frame->True</code> (I put an outside <code>Frame</code> as well, just for illustration, it is not part of the problem here)</p>
<pre><code>Framed[
Plot[Sin[x], {x, -Pi, Pi},
Frame -> True,
FrameLabel -> {{Sin[x], None}, {x,
Row[{"This is a plot of ", Sin[x]}]}},
ImagePadding -> {{55, 10}, {35, 20}},
ImageMargins -> 0,
FrameTicksStyle -> 10,
RotateLabel -> False
],
FrameMargins -> 0
]
</code></pre>
<p><img src="https://i.stack.imgur.com/JRzf3.png" alt="Mathematica graphics"></p>
<p>Is there an option or method to control this distance?</p>
<p>Notice that <code>ImagePadding</code> affects distance below the frame label, and not between the frame label and the ticks. Hence changing <code>ImagePadding</code> will not help here.</p>
<p>Depending on the plot and other things, this space can be more than it should be. The above is just a small example I made up. Here is a small part of a UI, and I think the space between the <code>t(sec)</code> and the ticks is too large. I'd like to reduce it by few pixels. I also might like to push the top label down closer to the plot by few pixels also.</p>
<p><img src="https://i.stack.imgur.com/jruyf.png" alt="Mathematica graphics"></p>
<p>I am Using V9 on windows.</p>
<p><strong>update 12/22/12</strong></p>
<p>Using Labeld solution by @kguler below is a good solution, one just need to be little careful with the type-sitting for the labels. <code>Plot</code> automatically typeset things as <code>Text</code> in <code>TraditionalFormat</code>, which is a nice feature. To do the same when using <code>Labeled</code> one must do this manually using <code>TraditionalForm</code> and <code>Text</code> as well. </p>
<p>Here is example to show the difference</p>
<p><strong>1)</strong> Labeled used just with <code>TraditionalForm</code>. The left one uses <code>Plot</code> and the right one uses <code>Labeled</code> with <code>TraditionalForm</code>. Notice the difference in how labels look.</p>
<pre><code>Grid[{
{
Plot[Sin[x], {x, -Pi, Pi}, Frame -> True,
FrameLabel -> {{Sin[x], None}, {x, E Tan[x] Sin[x]}},
ImageSize -> 300, FrameTicksStyle -> 10, FrameStyle -> 16,RotateLabel -> False],
Labeled[
Plot[Sin[x], {x, -Pi, Pi}, Frame -> True, ImageSize -> 300],
TraditionalForm /@ {Sin[x], x, E Tan[x] Sin[x]}, {Left, Bottom, Top},
Spacings -> {0, 0, 0}, LabelStyle -> "Panel"]
}
}, Frame -> All]
</code></pre>
<p><img src="https://i.stack.imgur.com/MXRlR.png" alt="Mathematica graphics"></p>
<p><strong>2)</strong> Now we do the same, just need to add <code>Text</code> to get the same result as <code>Plot</code>.</p>
<pre><code>Grid[{
{
Plot[Sin[x], {x, -Pi, Pi}, Frame -> True, FrameTicksStyle -> 10,
FrameStyle -> 16,
FrameLabel -> {{Sin[x], None}, {x, E Tan[x] Sin[x]}},
ImageSize -> 300, RotateLabel -> False],
Labeled[
Plot[Sin[x], {x, -Pi, Pi}, Frame -> True, ImageSize -> 300],
Text /@ TraditionalForm /@ {Sin[x], x, E Tan[x] Sin[x]}, {Left,
Bottom, Top}, Spacings -> {0, 0, 0}, LabelStyle -> "Panel"]
}
}, Frame -> All]
</code></pre>
<p><img src="https://i.stack.imgur.com/CHhZF.png" alt="Mathematica graphics"></p>
<p><strong>Update 12/22/12 (2)</strong></p>
<p>There is a big problem with controlling the spacing. </p>
<p><code>Labeled</code> spacing only seem to work for horizontal and vertical spacing, taken togother. </p>
<p>i.e. One can't control spacing on each side of the plot separately? Here is an example, where I tried to move the bottom axes label up, this ended with moving the top label down as well. Which is not what I want.</p>
<pre><code>Labeled[Plot[Sin[x], {x, -Pi, Pi}, Frame -> True, ImageSize -> 300],
Text /@ TraditionalForm /@ {Sin[x], x, E Tan[x] Sin[x]}, {Left,
Bottom, Top}, Spacings -> {-.2, -0.7}]
</code></pre>
<p><img src="https://i.stack.imgur.com/q0Wdo.png" alt="Mathematica graphics"></p>
<p>Will see if there is a way to control each side spacing on its own. Trying <code>Spacing->{-0.2,{-0.7,0}}</code> does not work, it seems to take the zero in this case and ignored the <code>-0.7</code></p>
<p>This gives the same result as above:</p>
<pre><code>Labeled[Plot[Sin[x], {x, -Pi, Pi}, Frame -> True, ImageSize -> 300],
Text /@ TraditionalForm /@ {Sin[x], x, E Tan[x] Sin[x]}, {Left,
Bottom, Top}, Spacings -> {-.2, -0.7, .0}]
</code></pre>
<p><img src="https://i.stack.imgur.com/6JCgC.png" alt="Mathematica graphics"></p>
<p>ps. there might be a way to specify the spacing for each side with some tricky syntax. I have not figured it out yet. Still trying thing....
<a href="http://reference.wolfram.com/mathematica/ref/Spacings.html" rel="noreferrer">http://reference.wolfram.com/mathematica/ref/Spacings.html</a></p>
<p><strong>update 12/22/12 (3)</strong>
Using combination of <code>ImagePadding</code> and <code>Spacing</code> should have worked, but for some reason, the top label now is cut off. Please see screen shot. Using V9 on windows</p>
<p><img src="https://i.stack.imgur.com/kbT3G.png" alt="enter image description here"></p>
<p>Note: The above seems to be related to the issue reported here:
<a href="https://mathematica.stackexchange.com/questions/16248/some-graphics-output-do-not-fully-render-on-the-screen-until-an-extra-click-is-m">some Graphics output do not fully render on the screen until an extra click is made into the notebook</a></p>
<p>Need an extra click inside the notebook. Then label become un-chopped !</p>
| kglr | 125 | <p>The most convenient way I found is to wrap <code>Plot</code> (without <code>FrameLabels</code> and <code>PlotLabel</code> and with appropriate <code>ImagePadding</code> and <code>ImageMargins</code>) inside <code>Labeled</code> and use the <code>Spacings</code> option to position the labels:</p>
<pre><code> Labeled[Plot[Sin[x], {x, -Pi, Pi}, Frame -> True,
ImagePadding -> {{20, 1}, {15, 2}}, ImageMargins -> 0,
FrameTicksStyle -> 10],
TraditionalForm /@ {Sin[x], x, Row[{"This is a plot of ", Sin[x]}]},
{Left, Bottom, Top},
Spacings -> {0, 0}, LabelStyle -> "Panel"]
</code></pre>
<p><img src="https://i.stack.imgur.com/iJKau.png" alt="enter image description here" /></p>
|
16,802 | <p>In an attempt to squeeze more plots and controls into the limited space for a demo UI, I am trying to remove any extra white spaces I see.</p>
<p>I am not sure what options to use to reduce the amount of space between the ticks labels and the actual text that represent the labels on the axes.</p>
<p>Here is a small <code>Plot</code> example using <code>Frame->True</code> (I put an outside <code>Frame</code> as well, just for illustration, it is not part of the problem here)</p>
<pre><code>Framed[
Plot[Sin[x], {x, -Pi, Pi},
Frame -> True,
FrameLabel -> {{Sin[x], None}, {x,
Row[{"This is a plot of ", Sin[x]}]}},
ImagePadding -> {{55, 10}, {35, 20}},
ImageMargins -> 0,
FrameTicksStyle -> 10,
RotateLabel -> False
],
FrameMargins -> 0
]
</code></pre>
<p><img src="https://i.stack.imgur.com/JRzf3.png" alt="Mathematica graphics"></p>
<p>Is there an option or method to control this distance?</p>
<p>Notice that <code>ImagePadding</code> affects distance below the frame label, and not between the frame label and the ticks. Hence changing <code>ImagePadding</code> will not help here.</p>
<p>Depending on the plot and other things, this space can be more than it should be. The above is just a small example I made up. Here is a small part of a UI, and I think the space between the <code>t(sec)</code> and the ticks is too large. I'd like to reduce it by few pixels. I also might like to push the top label down closer to the plot by few pixels also.</p>
<p><img src="https://i.stack.imgur.com/jruyf.png" alt="Mathematica graphics"></p>
<p>I am Using V9 on windows.</p>
<p><strong>update 12/22/12</strong></p>
<p>Using Labeld solution by @kguler below is a good solution, one just need to be little careful with the type-sitting for the labels. <code>Plot</code> automatically typeset things as <code>Text</code> in <code>TraditionalFormat</code>, which is a nice feature. To do the same when using <code>Labeled</code> one must do this manually using <code>TraditionalForm</code> and <code>Text</code> as well. </p>
<p>Here is example to show the difference</p>
<p><strong>1)</strong> Labeled used just with <code>TraditionalForm</code>. The left one uses <code>Plot</code> and the right one uses <code>Labeled</code> with <code>TraditionalForm</code>. Notice the difference in how labels look.</p>
<pre><code>Grid[{
{
Plot[Sin[x], {x, -Pi, Pi}, Frame -> True,
FrameLabel -> {{Sin[x], None}, {x, E Tan[x] Sin[x]}},
ImageSize -> 300, FrameTicksStyle -> 10, FrameStyle -> 16,RotateLabel -> False],
Labeled[
Plot[Sin[x], {x, -Pi, Pi}, Frame -> True, ImageSize -> 300],
TraditionalForm /@ {Sin[x], x, E Tan[x] Sin[x]}, {Left, Bottom, Top},
Spacings -> {0, 0, 0}, LabelStyle -> "Panel"]
}
}, Frame -> All]
</code></pre>
<p><img src="https://i.stack.imgur.com/MXRlR.png" alt="Mathematica graphics"></p>
<p><strong>2)</strong> Now we do the same, just need to add <code>Text</code> to get the same result as <code>Plot</code>.</p>
<pre><code>Grid[{
{
Plot[Sin[x], {x, -Pi, Pi}, Frame -> True, FrameTicksStyle -> 10,
FrameStyle -> 16,
FrameLabel -> {{Sin[x], None}, {x, E Tan[x] Sin[x]}},
ImageSize -> 300, RotateLabel -> False],
Labeled[
Plot[Sin[x], {x, -Pi, Pi}, Frame -> True, ImageSize -> 300],
Text /@ TraditionalForm /@ {Sin[x], x, E Tan[x] Sin[x]}, {Left,
Bottom, Top}, Spacings -> {0, 0, 0}, LabelStyle -> "Panel"]
}
}, Frame -> All]
</code></pre>
<p><img src="https://i.stack.imgur.com/CHhZF.png" alt="Mathematica graphics"></p>
<p><strong>Update 12/22/12 (2)</strong></p>
<p>There is a big problem with controlling the spacing. </p>
<p><code>Labeled</code> spacing only seem to work for horizontal and vertical spacing, taken togother. </p>
<p>i.e. One can't control spacing on each side of the plot separately? Here is an example, where I tried to move the bottom axes label up, this ended with moving the top label down as well. Which is not what I want.</p>
<pre><code>Labeled[Plot[Sin[x], {x, -Pi, Pi}, Frame -> True, ImageSize -> 300],
Text /@ TraditionalForm /@ {Sin[x], x, E Tan[x] Sin[x]}, {Left,
Bottom, Top}, Spacings -> {-.2, -0.7}]
</code></pre>
<p><img src="https://i.stack.imgur.com/q0Wdo.png" alt="Mathematica graphics"></p>
<p>Will see if there is a way to control each side spacing on its own. Trying <code>Spacing->{-0.2,{-0.7,0}}</code> does not work, it seems to take the zero in this case and ignored the <code>-0.7</code></p>
<p>This gives the same result as above:</p>
<pre><code>Labeled[Plot[Sin[x], {x, -Pi, Pi}, Frame -> True, ImageSize -> 300],
Text /@ TraditionalForm /@ {Sin[x], x, E Tan[x] Sin[x]}, {Left,
Bottom, Top}, Spacings -> {-.2, -0.7, .0}]
</code></pre>
<p><img src="https://i.stack.imgur.com/6JCgC.png" alt="Mathematica graphics"></p>
<p>ps. there might be a way to specify the spacing for each side with some tricky syntax. I have not figured it out yet. Still trying thing....
<a href="http://reference.wolfram.com/mathematica/ref/Spacings.html" rel="noreferrer">http://reference.wolfram.com/mathematica/ref/Spacings.html</a></p>
<p><strong>update 12/22/12 (3)</strong>
Using combination of <code>ImagePadding</code> and <code>Spacing</code> should have worked, but for some reason, the top label now is cut off. Please see screen shot. Using V9 on windows</p>
<p><img src="https://i.stack.imgur.com/kbT3G.png" alt="enter image description here"></p>
<p>Note: The above seems to be related to the issue reported here:
<a href="https://mathematica.stackexchange.com/questions/16248/some-graphics-output-do-not-fully-render-on-the-screen-until-an-extra-click-is-m">some Graphics output do not fully render on the screen until an extra click is made into the notebook</a></p>
<p>Need an extra click inside the notebook. Then label become un-chopped !</p>
| Szabolcs | 12 | <p>When using <a href="http://scidraw.nd.edu/" rel="noreferrer">SciDraw</a>, we have full control over all label positions. The key option names are variation on <code>TextNudge</code>. The value specified in this option will be added to the label position.</p>
<p>Example with frame labels:</p>
<pre><code>Figure[
{
FigurePanel[{}, FrameLabel -> {x, y},
XFrameTextNudge -> 10, (* shift up by 10 pt *)
YFrameTextNudge -> {10, 0} (* shift right by 10 pt *)
]
}
]
</code></pre>
<p><a href="https://i.stack.imgur.com/FY65b.png" rel="noreferrer"><img src="https://i.stack.imgur.com/FY65b.png" alt="enter image description here"></a></p>
<p>SciDraw of course draws its own frames and requires working in (and learning) its own framework. It will not help in changing the position of <em>standard</em> frame labels.</p>
<p>The ability of SciDraw to move labels is one of the several reasons why I usually use it for publication figures. I you have the time and are willing to learn it, it is a good option to deal with the label positioning problem.</p>
|
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