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 |
|---|---|---|---|---|
2,461,820 | <p>I am a bit confused with the following question, I get that P(T|D) = 0.95 and P(D) = 0.0001 however because i'm unable to work out P(T|~D) i'm struggling to apply the theorem, am i missing something? Also i'm unsure about what to do with the information relating to testing negative when you don't have the disease correctly 95% of the time</p>
<p>You take a test T to tell whether you have a disease D. The test comes back positive. You know that test is
95% accurate (the probability of testing positive when you do have the disease is 0.95, and the probability of
testing negative when you don’t have the disease is also 0.95). You also know that the disease is rare, only 1
person is 10,000 gets the disease.
What is the probability that you have the disease?
How would this change if the disease was more common, say affecting 1 person in 100?</p>
| Alekos Robotis | 252,284 | <p>$\ln x:= n$ such that $e^n=x$. So, $\ln e=n$ such that $e^n=e.$ So, $\ln e=1$. Now, $\ln 1=n$ such that $e^n=1$, which implies that $\ln 1=0$.</p>
<p>$$ \ln(\ln e)=\ln(1)=0.$$</p>
|
54,541 | <p>Apparently, Mathematica has no real sprintf-equivalent (unlike any other high-level language known to man). <a href="https://mathematica.stackexchange.com/questions/970/sprintf-or-close-equivalent-or-re-implementation">This has been asked before</a>, but I'm wondering if the new <code>StringTemplate</code> function in Mathematica 10 can be extended to include such formatting capabilities.</p>
<p>What I have in mind is a function that takes a <code>TemplateObject</code>, and looks for "formatting specification strings" immediately after <code>TemplateSlot</code>'s and <code>TemplateExpression</code>'s and replaces them with <code>TemplateExpression</code>'s containing appropriate formatting code. So, for example, you could write:</p>
<pre><code>st = applyFormat@StringTemplate["Number: `1`%.2 some other text"]
</code></pre>
<p>and you would get something equivalent to: </p>
<pre><code>TemplateObject[{"Number: ",
TemplateExpression[ToString[NumberForm[TemplateSlot[1], {\[Infinity], 2}]]],
" some other text"}, InsertionFunction -> TextString,
CombinerFunction -> StringJoin]
</code></pre>
<p>I'm not particularly picky about the syntax (it doesn't have to mimic sprintf), as long as:</p>
<ul>
<li>it's easy to write and easy to read</li>
<li>it supports Mathematica's number formatting functions (<code>AccountingForm</code>, <code>ScientificForm</code>...)</li>
<li>it's extensible (e.g by delegating the formatting to a pattern that can be overwritten/extended)</li>
<li>it's compatible with existing <code>StringTemplate</code> templates</li>
</ul>
<p>I've started a function that does this, but I'm curious if you have better ideas (both implementation- and syntax-wise), so I'm posting it as an answer, not as part of the question.</p>
| Taliesin Beynon | 7,140 | <p>I do want to support something like this with some custom syntax, though I can't guarantee when that will actually happen.</p>
<p>In the meantime, you can do this:</p>
<pre><code>StringTemplate["Pi is <*NumberForm[N[#],3]*>!"][Pi]
</code></pre>
<p>which produces:</p>
<pre><code>"Pi is 3.14!"
</code></pre>
<p>Unfortunately TextString isn't that smart yet about NumberForm, and doesn't support EngineeringForm and ScientificForm at all, though it will do that in a coming point release. </p>
<p>At least TextString gets us away from the braindead default behavior of OutputForm in ToString. ToString is basically broken for producing human-readable strings.</p>
|
3,413,253 | <p>Can anyone give me some examples and non examples of Lindelöf or second countable space and spaces that is Lindelöf but not second countable? And I understand the definition but find it is hard to visualize and imagine.
I have tried google it but it turns out I only found some silly examples like finite set or empty set.</p>
<p>In general, how can one construct a topological space that is Lindelöf or second countable?</p>
<p>Someone in stack exchange said the real line with discrete topology is Lindelöf, but I do not think so. We can simply construct an open cover defined by the collection of all the singleton set. And this open cover is well defined since singleton set is open in discrete topology. Hence, by definition it is not Lindelöf.</p>
<p>Last question, is (0,1) in the real line equipped with usual topology Lindelöf? I think it is Lindelöf but I could not give any formal proof.
(0,1) fails to be a compact set since we can construct an open cover defined by (1/n,1-1/n) but this open cover does not work so well for arguing for Lindelöf property since quotient number is dense in (0,1). So intuitively I think it is Lindelöf.</p>
<p>I wrote a pretty long question. My mothertongue is not English. Hopefully, you guys can understand me.</p>
| Mirko | 188,367 | <p><a href="https://en.wikipedia.org/wiki/Lower_limit_topology" rel="noreferrer">The Sorgenfrey line</a>, also called the lower-limit topology on the real line. It has a basis of intervals <span class="math-container">$[a,b)$</span> (or some authors prefer <span class="math-container">$(a,b]$</span>, upper-limit topology). </p>
<p>It is hereditarily Lindelof (and hereditarily separable, i.e. every subspace has a countable dense subset), but it is not second countable. </p>
<p>Real line with discrete topology is not Lindelof. </p>
<p><span class="math-container">$(0,1)$</span> with the usual topology is Lindelof, it is homeomorphic to the real line.
The real line is <span class="math-container">$\sigma$</span>-compact, that is union of countably many compact subspaces. The real line is the union <span class="math-container">$\cup_n[-n,n]$</span>, clearly <span class="math-container">$\sigma$</span>-compact.
The interval <span class="math-container">$(0,1)=\cup_n [\frac1n,1-\frac1n]$</span> is <span class="math-container">$\sigma$</span>-compact. </p>
<p>Every <span class="math-container">$\sigma$</span>-compact space is Lindelof (easy to prove). </p>
<p>In particular, clearly every countable space is Lindelof. But not every countable space is second countable (even with nice separation axioms). For example, the countable sequential fan is not second countable. It is obtained by taking a disjoint family of countably many convergent sequences, and
"gluing" their limit points into one limit point, via a quotient map. The result is not first countable at that point. (I couldn't find a suitable online reference to the countable sequential fan, but it has similar properties to the quotient space <span class="math-container">$\Bbb R/\Bbb N$</span>, which is also not first countable, and likely discussed in most topology books.) </p>
<p>There is an online searchable database (called <span class="math-container">$\pi$</span>-base), you can make a query asking for Lindelof, not second countable spaces. For many more examples see </p>
<p><a href="https://topology.jdabbs.com/spaces?q=lindelof%20%2B%20~Second%20Countable" rel="noreferrer">https://topology.jdabbs.com/spaces?q=lindelof%20%2B%20~Second%20Countable</a></p>
<p>Every second countable space is Lindelof (you may need to assume some separation axioms, often included in the definitions). </p>
<p>One of the examples at <span class="math-container">$\pi$</span>-base is the one-point Lidelofication of uncountable discrete space. Take any uncountable set, and a point <span class="math-container">$p$</span>, and isolate all point but <span class="math-container">$p$</span>. The neighborhoods of <span class="math-container">$p$</span> are co-countable (that is, they have a countable complement). The definition easily implies that the result is a Lindelof space. But it is not first-countable at <span class="math-container">$p$</span>, and hence not second countable. Often (in the case when "uncountable" is taken to be the first uncountable cardinal) this space is described as the set of all countable ordinals <span class="math-container">$\omega_1=\{\alpha:\alpha<\omega_1\}$</span> together with the first uncountable ordinal <span class="math-container">$\omega_1$</span>, so <span class="math-container">$X=[0,\omega_1]=\{\alpha:\alpha\le\omega_1\}$</span>, with all countable ordinals <span class="math-container">$\alpha<\omega_1$</span> isolated, and with basic neighborhoods of <span class="math-container">$\omega_1$</span>
of the form <span class="math-container">$(\alpha,\omega_1]$</span>, with <span class="math-container">$\alpha<\omega_1$</span>. This is, in addition, an example of a Linearly Ordered Topological Space (LOTS), with only one non-isolated point. (It is a LOTS under a somewhat different order, one may insert a decreasing sequence in front of every limit ordinal.) </p>
<p>The last (and some of the previous examples) are not second countable, because they are not even first countable (and, for the one-point Lindelofication the point <span class="math-container">$p$</span> is not even a <span class="math-container">$G_\delta$</span> point, that is, it is not the intersection of any countable family of open sets, and the space is not <span class="math-container">$\sigma$</span>-compact). On the other hand the Sorgenfrey line is perfectly normal: Every closed set (and in particular every point) is the intersection of a countable family of open sets (the proof that the Sorgenfrey line is hereditarily Lindelof uses this). But it is not second countable, since for every basis, and for every <span class="math-container">$x$</span> there must be a basic element <span class="math-container">$B_x$</span> with <span class="math-container">$x\in B_x\subseteq[x,\infty)$</span> and clearly if <span class="math-container">$x\neq y$</span> then <span class="math-container">$B_x\neq B_y$</span>. </p>
<p>An example of a space that is separable but not second countable and not Lindelof is the <a href="https://en.wikipedia.org/wiki/Moore_plane" rel="noreferrer">Moore, or Niemytzki plane</a> (also called tangent-disk space, usually available in topology texts). </p>
<p>There are also compact spaces (which of course is stronger than Lindelof) that are not second countable. One such example closely related to the Sorgenfrey line is the <a href="https://en.wikipedia.org/wiki/Split_interval" rel="noreferrer">Alexandrov double arrow space</a>, also called split interval.
Another is the <a href="https://math.stackexchange.com/q/1331117">Alexandroff double circle</a>. Note that every second countable space in hereditarily Lindelof (since every subspace is second countable, and hence Lindelof). The Alexandroff double circle is compact (and hence Lindelof), but has an uncountable discrete subspace, which of course is not Lindelof. Hence the Alexandroff double circle is not second countable. </p>
|
72,854 | <p>Hi everybody,</p>
<p>Does there exist an explicit formula for the Stirling Numbers of the First Kind which are given by the formula
$$
x(x-1)\cdots (x-n+1) = \sum_{k=0}^n s(n,k)x^k.
$$</p>
<p>Otherwise, what is the computationally fastest formula one knows?</p>
| Gottfried Helms | 7,710 | <p>In Pari/GP; one could simplify for either readability, speed or memory organisation for big matrices:</p>
<pre>
{ makemat_St1(dim=n) = local(f, M);
M=matid(dim);
f=1;
for(r=2,dim, \\ comp diagonal and first column
M[r,1]=f;f*=(r)
);
for(c=2,dim, \\ compute core entries
for(r=c+1,dim,
M[r,c]=M[r-1,c-1]+(r-1)*M[r-1,c]
)
);
f1=1; \\ apply signs
for(r=2,dim,
f1*=-1;f2=-f1;
for(c=1,r-1,
f2*=-1;M[r,c]*=f2
)
);
return(M) }
</pre>
<p><hr>
A shorter form is this</p>
<pre>
{makemat_st1(dim=6) = local(m); \\ give it a default dimension of 6
m=matrix(dim,dim);
m[1,1]=1;
for(r = 2,dim,
m[r,1]= 0 - (r-1)*m[r-1,1] ; \\ first column has no up-left neighbour
for(c = 2,r,
m[r,c]= m[r-1,c-1] - (r-1)*m[r-1,c]
);
);
return(m);}
</pre>
|
635,301 | <p>I need some help with the following problem: </p>
<blockquote>
<p>Let $f:\Bbb C \to \Bbb C$ be continuous satisfying that $f(\Bbb C)$ is an open set and that $|f(z)| \to \infty$ as $z\to \infty$. Prove that $f(\Bbb C)=\Bbb C$. </p>
</blockquote>
<p>My idea on this one is to prove by contradiction and assume that $S=f(\Bbb C)\ne\Bbb C$ to get some contradiction with the given two properties of the function. But I have no idea on how to proceed next. </p>
<p>Thanks in advance. </p>
| Robert Israel | 8,508 | <p>Hint: Suppose $w$ is in the boundary of $f(\mathbb C)$. Take a sequence $z_n$ with $f(z_n) \to w$. There is a subsequence ...</p>
|
2,793,077 | <blockquote>
<p>Suppose $R$ is a Boolean ring. Prove that $a+a=0$ for all $a\in R$.
Also prove that $R$ is commutative. Give an example (with explanation)
of a Boolean ring.</p>
</blockquote>
<p>From what I know, a Boolean ring is a ring for which $a^2=a$ for all $a\in R$.</p>
<p>Under addition a ring is a commutative group. </p>
<p>$a + b = b + a$ (commutative)</p>
<p>$(a + b) + c = a + (b + c)$ (associative)</p>
<p>$a + (-a) = 0$ (inverse exists for every element)</p>
<p>$a + 0 = a$ (identity exists)</p>
<p>Where $a,b,c \in R$</p>
<p>But I'm not really sure how to proceed with the proof from here. Any idea?</p>
| Kempa | 825,616 | <p><span class="math-container">$a+a=(a+a)^2=a^2+a^2+a^2+a^2=a+a+a+a\Rightarrow 0=2a$</span></p>
|
2,793,077 | <blockquote>
<p>Suppose $R$ is a Boolean ring. Prove that $a+a=0$ for all $a\in R$.
Also prove that $R$ is commutative. Give an example (with explanation)
of a Boolean ring.</p>
</blockquote>
<p>From what I know, a Boolean ring is a ring for which $a^2=a$ for all $a\in R$.</p>
<p>Under addition a ring is a commutative group. </p>
<p>$a + b = b + a$ (commutative)</p>
<p>$(a + b) + c = a + (b + c)$ (associative)</p>
<p>$a + (-a) = 0$ (inverse exists for every element)</p>
<p>$a + 0 = a$ (identity exists)</p>
<p>Where $a,b,c \in R$</p>
<p>But I'm not really sure how to proceed with the proof from here. Any idea?</p>
| Gustavo Andres Pava Parra | 714,992 | <p>Take <span class="math-container">$x \in R$</span>, with <span class="math-container">$R$</span> boolean ring, in particular is a ring, therefore <span class="math-container">$(x+x) \in R$</span> and <span class="math-container">$(x+x)^{2}=(x+x)$</span>. Thus
<span class="math-container">\begin{align*}
(x+x)^{2} &= (x+x)(x+x) \\
&=(x+x)x + (x+x)x \\
&= x^2 + x^2 + x^2 + x^2
\end{align*}</span>
since <span class="math-container">$x^2 = x$</span> and <span class="math-container">$(x+x)^2 = (x+x)$</span> because <span class="math-container">$R$</span> is boolean, then <span class="math-container">\begin{align*}
(x+x)^{2} &= x+x+x +x\\
&= (x+x) + (x+x)\\
&= (x+x)
\end{align*}</span>
and finally we add both sides the right inverse of <span class="math-container">$-(x+x)$</span>.
<span class="math-container">\begin{align*}
(x+x) + (x+x) -(x+x) &= (x+x) - (x+x)
\end{align*}</span>
we conclude<br />
<span class="math-container">\begin{align*}
(x+x) = 2x = 0
\end{align*}</span></p>
|
387,096 | <blockquote>
<p>Suppose $B$ is an $m \times n$ matrix. Prove that $BB^T$ is positive semidefinite.</p>
</blockquote>
<p>Can someone give a fairly good proof?</p>
<p>Inputs are greatly appreciated. The question is listed above.</p>
| Berci | 41,488 | <p><strong>Hint:</strong> For column vectors $a,b$, their inner (or scalar or dot) product is
$$\langle a,b\rangle = a^Tb$$
using matrix product, and $a^Ta=\langle a,a\rangle=\|a\|^2$.</p>
|
387,096 | <blockquote>
<p>Suppose $B$ is an $m \times n$ matrix. Prove that $BB^T$ is positive semidefinite.</p>
</blockquote>
<p>Can someone give a fairly good proof?</p>
<p>Inputs are greatly appreciated. The question is listed above.</p>
| robjohn | 13,854 | <p>$BB^T$ is positive semidefinite if for any vector $x$
$$
xBB^Tx^T\ge0
$$
Let $u=xB$. Then,
$$
xBB^Tx^T=uu^T=|u|^2\ge0
$$</p>
|
1,878,806 | <p>I am a graduate school freshman.</p>
<p>I did not take a probability lecture.</p>
<p>So I don't have anything about Probability.</p>
<p>Could you suggest Probability book No matter What book level?</p>
| Community | -1 | <p>I think the book <em>Probability Theory</em> by Heinz Bauer is a very good text on probability theory. It contains an extensive discussion of all the basic parts of the theory and is very readable. The book requires, however, a modest background in measure theory. </p>
<p>The original version of the book from 1973, <em>Probability Theory and Elements of Measure Theory</em>, contains all the necessary background in measure theory. Later versions of the book are split into two books, the parts on measure and probability theory are published as <em>Measure and Integration Theory</em> and <em>Probability Theory</em>, respectively. </p>
|
1,878,806 | <p>I am a graduate school freshman.</p>
<p>I did not take a probability lecture.</p>
<p>So I don't have anything about Probability.</p>
<p>Could you suggest Probability book No matter What book level?</p>
| Jean Marie | 305,862 | <p>@EHH I completely agree for Grimmett ! A very valuable book also by Grimmett and a coauthor (Stirzacker): (<a href="http://www.oupcanada.com/catalog/9780198572220.html" rel="nofollow">http://www.oupcanada.com/catalog/9780198572220.html</a>). It has a somewhat larger scope (including stochastic processes). A very good point: a huge amount of exercices with an invaluable book of solutions.</p>
|
848,415 | <p>If the limit of one sequence $\{a_n\}$ is zero and the limit of another sequence $\{b_n\}$ is also zero does that mean that $\displaystyle\lim_{n\to\infty}(a_n/b_n) = 1$?</p>
| M. Vinay | 152,030 | <p>No. Let $a_n = \dfrac 1 n$ and $b_n = \dfrac 1 {n^2}$.</p>
|
3,537,226 | <p>A deck contains six cards, one pair labelled '1', another pair labelled '2' and the last labelled '3'. The deck is shuffled and you a pair of cards at a time until there are no cards left. A pair of cards <span class="math-container">$(i,j)$</span> is called acceptable if <span class="math-container">$|i-j|\leq1$</span>. What is the probability you have drawn only acceptable pairs? How does your answer change if there are <span class="math-container">$n$</span> pairs and the condition becomes <span class="math-container">$|i-j|\leq k$</span>?</p>
<p>I'm quite stuck on the last bit of the problem. Here's my approach to the first part:</p>
<p>My solution so far:
My idea is that as long as a pair <span class="math-container">$(1,3)$</span> or <span class="math-container">$(3,1)$</span> is drawn, then set contains an unacceptable pair. The probability <span class="math-container">$(1,3)$</span> or <span class="math-container">$(3,1)$</span> is drawn first is
<span class="math-container">$$2\left(\frac26\times\frac25\right)=\frac4{15},$$</span></p>
<p>the probability <span class="math-container">$(1,3)$</span> or <span class="math-container">$(3,1)$</span> is drawn second is
<span class="math-container">$$\left(1-\frac4{15}\right)\frac4{15}=\frac{44}{225},$$</span></p>
<p>and the probability <span class="math-container">$(1,3)$</span> or <span class="math-container">$(3,1)$</span> is drawn last is
<span class="math-container">$$\left(1-\frac4{15}-\frac{44}{225}\right)\frac4{15}=\frac{484}{3375},$$</span></p>
<p>so the probability of having only acceptable pairs is
<span class="math-container">$$1-\frac4{15}-\frac{44}{225}-\frac{484}{3375}=1-\frac{900}{3375}-\frac{660}{3375}-\frac{484}{3375}=\frac{1331}{3375}.$$</span></p>
<p>(Correct me if this is wrong please!)</p>
<p>However, I am unsure of how to extend this to a more general number of cards and relaxed constraint. Taking the complement seems to be less efficient compared to finding the actual probability, but I'm not sure if there's a closed form. Qualitatively, all I can see is the probability dropping to zero as the number of cards increases. Could someone provide better insight? Cheers!</p>
| Christopher Wells | 720,910 | <p>I'm afraid your answer to the first half of the question is incorrect.</p>
<p>For one thing, there are 720 ways to shuffle a six-card deck, so whatever the probability is it must be a fraction with 720 on the denominator. Or at least a factor of 720.</p>
<p>You're correct that <span class="math-container">$2 \times \frac{2}{6} \times \frac{2}{5}$</span> is the probability that the first pair is unacceptable. But it's not the probability that the second pair is given that the first was acceptable. That would be something different, as there are only 4 cards left in the deck and they're not an equal distribution of all three numbers.</p>
<hr>
<p>The 3s can only be acceptably paired with the other 3 or with a 2. In particular, if one 3 is paired with a 2, the other 3 must be paired with the other 2.</p>
<p>The probability that the 3s are paired with each other is <span class="math-container">$\frac{1}{5}$</span>. This leaves the 1s and 2s to be paired - any arrangement of these is acceptable.</p>
<p>If they are not paired with each other, they must both be paired with the 2s. The probability of this is <span class="math-container">$\frac{2}{15}$</span>. This leaves the 1s to be paired with each other, which is acceptable.</p>
<p>In total, the probability that all pairs are acceptable is:
<span class="math-container">$$\frac{1}{3}$$</span></p>
<hr>
<p>You can extend this approach to more cards. Let <span class="math-container">$n > 3$</span>. In a deck with <span class="math-container">$n$</span> pairs numbered <span class="math-container">$1$</span> to <span class="math-container">$n$</span>, the <span class="math-container">$n$</span>s can only be acceptably paired with each other or with the <span class="math-container">$n-1$</span>s.</p>
<p>If they are paired with each other, the rest of the cards must be an acceptable arrangement of a deck with pairs numbered <span class="math-container">$1$</span> to <span class="math-container">$n-1$</span>.</p>
<p>If they are paired with the <span class="math-container">$n-1$</span>s, the rest of the cards must be an acceptable arrangement of a deck with pairs numbered <span class="math-container">$1$</span> to <span class="math-container">$n-2$</span>.</p>
<p>By finding the probability of the <span class="math-container">$n$</span>s being paired with each other. and the probability of them being paired with the <span class="math-container">$n-1$</span>s, you can get a recursive formula for the probability that all pairs are acceptable.</p>
<hr>
<p>I don't know if this solution extends to weaker constraints, too.</p>
|
406,437 | <p>Calculus the extreme value of the $f(x,y)=x^{2}+y^{2}+xy+\dfrac{1}{x}+\dfrac{1}{y}$</p>
<p>pleasee help me.</p>
| Vishal Gupta | 60,810 | <p><strong>EDIT:</strong> As pointed out in the comments, the following example is not associative.</p>
<p>What about the ring of matrices? It is non-commutative and perhaps as easy to understand as the real numbers. You may also consider the subset of invertible matrices.</p>
<p>You can consider the abelian group of matrices and define the multiplication as</p>
<p>$ X*Y := XY - YX $.</p>
<p>Then you have a non-commutative ring. This is in fact the Lie algebra of all matrices.</p>
|
2,377,598 | <p>I am a Ph.D student in computer science, and I work on graph isomorphism. My research work requires some level of mathematics (mostly group theory ). I have done basic level abstract algebra course. I try to write down the theorems on peace of paper and try to understand them; I usually repeat this process four five times. Some time by doing this I understand the theorem and its proof, but there are times when I find it difficult. The biggest problem I have faced is that theorems related to abstract algebra are really abstract I mean there is no way to create an intuition (Is it true ?). </p>
<blockquote>
<p><strong>My question</strong> : How to create an intuition for abstract algebra theorems?</p>
</blockquote>
<p><strong>For Example</strong> </p>
<p><a href="https://i.stack.imgur.com/LNbjy.png" rel="noreferrer"><img src="https://i.stack.imgur.com/LNbjy.png" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/27Rzy.png" rel="noreferrer"><img src="https://i.stack.imgur.com/27Rzy.png" alt="enter image description here"></a></p>
| Ethan Bolker | 72,858 | <p>I think the best way to understand abstractions intuitively is to study lots of examples. To understand group actions, write some down. Then think about whether each is transitive, whether it's primitive, look at the stabilizers of elements. Verify that the theorem is true; try to understand why in each particular case. Look for examples where some hypothesis fails and see whether (and why) the conclusion fails.</p>
<p>That learning strategy reproduces (in part) the explorations that led to the useful abstractions of group theory. Mathematicians studying the symmetries of various geometrical objects realized that they could reason about the symmetry of just about anything by inventing an abstract language whose definitions captured the essence of the properties of symmetries.</p>
<p>Unfortunately, often the abstractions - the definitions and theorems - take center stage in teaching and learning. Students find it hard to understand what's going on without the examples. I would tell my students that for me the category of groups consisted of those groups with which I was personally familiar. I assigned homework that called for working out examples at least as much as proving theorems.</p>
<p>In your own discipline (computer science) there are analogous historical trends leading from examples to abstractions: the concepts of object oriented or functional programming languages, the development of abstract tools to reason about databases.</p>
|
2,377,598 | <p>I am a Ph.D student in computer science, and I work on graph isomorphism. My research work requires some level of mathematics (mostly group theory ). I have done basic level abstract algebra course. I try to write down the theorems on peace of paper and try to understand them; I usually repeat this process four five times. Some time by doing this I understand the theorem and its proof, but there are times when I find it difficult. The biggest problem I have faced is that theorems related to abstract algebra are really abstract I mean there is no way to create an intuition (Is it true ?). </p>
<blockquote>
<p><strong>My question</strong> : How to create an intuition for abstract algebra theorems?</p>
</blockquote>
<p><strong>For Example</strong> </p>
<p><a href="https://i.stack.imgur.com/LNbjy.png" rel="noreferrer"><img src="https://i.stack.imgur.com/LNbjy.png" alt="enter image description here"></a></p>
<p><a href="https://i.stack.imgur.com/27Rzy.png" rel="noreferrer"><img src="https://i.stack.imgur.com/27Rzy.png" alt="enter image description here"></a></p>
| Santana Afton | 274,352 | <p>One way to build intuition is to reframe results in terms of objects and relationships that you know of and are familiar with. Since it seems as though you're looking at group action in the context of graph theory, this might be a good route to go.</p>
<p>For example, consider the first theorem you state about the primitivity of a transitive group action. For a graph $X$, think of $\Omega = V(X)$ and $G=\operatorname{Aut}(X)$. Then a system of imprimitivity can be seen as a partition of $X$ into subgraphs that 'look the same.' Now, the theorem states</p>
<blockquote>
<p>Assume that $\operatorname{Aut}(X)$ acts transitivity on $V(X)$, and consider $v\in V(X)$. Then $\operatorname{Aut}(X)$ has a system of imprimitivity if and only if there is some nontrivial subgroup $H$ containing $\operatorname{stab}(v)$. </p>
</blockquote>
<p>This might start to seem more concrete. Since $\operatorname{Aut}(X)$ acts transitively on $X$, the subgroup that sends these subgraphs to themselves not only contains $\operatorname{stab}(v)$, but strictly contains it, since $v$ can also be mapped to any other vertex in its partition. </p>
|
217,244 | <p>I would like to know some of the most important definitions and theorems of definite and semidefinite matrices and their importance in linear algebra.
Thanks for your help</p>
| EuYu | 9,246 | <p>There are many uses for definite and semi-definite matrices. I can give just a few examples although undoubtedly I will be missing many.</p>
<ol>
<li><p>Positive-definite matrices are the matrix analogues to positive <em>numbers</em>. It is generally not possible to define a consistent notion of "positive" for matrices other than symmetric matrices. As a consequence, positive definite matrices are a special class of symmetric matrices (which themselves are another very important, special class of matrices). It turns out that many useful matrices fall under this class such the covariance matrix, overlap matrices used in quantum chemistry and dynamical matrices used in calculation of molecular vibrations (which is positive semi-definite).</p></li>
<li><p>Definiteness is a useful measure for optimization. Quadratic forms on positive definite matrices<br>
$$\mathbf{x}^\mathrm{T}A\mathbf{x}$$
are always positive for non-zero $\mathbf{x}$ and are convex. Analogous results hold for negative-definite matrices. This is a very desirable property for optimization since it guarantees the existences of maxima and minima. It is properties like these for example, that allow you to use the Hessian matrix to optimize multivariate functions.</p></li>
<li><p>Perhaps equally (or more) important, especially to a mathematician, is the fact that the theory of (semi)definite matrices is an incredibly rich and beautiful field. There are chains of elegant results concerning these matrices, especially for positive-definite matrices. That is motivation enough.</p></li>
</ol>
|
2,536,185 | <p><strong>Lemma 3.21</strong>. </p>
<p>Let $S$ be a subset of $R$. Then $\bar{S}$ is a closed set. $\bar{S}$ denotes the closure of $S$.</p>
<p>The following is a proof of Lemma 3.21:</p>
<p><em>Proof</em>. </p>
<blockquote>
<p>By Corollary 3.16 it is enough to show that $\bar{S^c}$
is open.</p>
</blockquote>
<p>I understand that if we show that $\bar{S^c}$ is open, then $\bar{S}$ is closed, since Corollary 3.16 states that "a nonempty set $E$ is closed iff $E^c$
is open."</p>
<blockquote>
<p>We will show that
$\bar{S^c}$
is open using Lemma 3.15 (c) with $S$ replaced by $\bar{S^c}$
. So let $x \in \bar{S^c}$, if we can
show that $x$ has a neighborhood $U$ which is contained in $\bar{S^c}$
then we will be done (by that Lemma).</p>
</blockquote>
<p>Lemma 3.15 (c) states that "for every point $x \in S$, there exists $\epsilon > 0$ s.t. $N(x,\epsilon) ⊂ S$." So, I see why we replace $S$ with $\bar{S^c}$. Lemma 3.15 (a) states that "$S$ is open" and Lemma 3.15 (a) and (c) are equivalent. So, essentially, we would show that $\bar{S^c}$ is open.</p>
<blockquote>
<p>By the contrapositive of Lemma 3.20, there is a neighborhood $N$ of $x$ which is contained in $\bar{S^c}$.</p>
</blockquote>
<p>I understand that this is the contrapositive of Lemma 3.20. Lemma 3.20 states that "if $S$ is a nonempty set of real numbers then $x \in \bar{S}$ iff every neighborhood of $x$ contains at least one point $\in S$."</p>
<blockquote>
<p>If $y \in U$, then since $N$ is open, by the contrapositive of the last assertion in Lemma 3.20 with $x$ replaced by $y$, we see that $y \notin \bar{S}$. </p>
</blockquote>
<p>I understand how we arrive at the contrapositive of the statement. But, why do we replace $x$ with $y$? And is $N$ referring to an arbitrary neighborhood from Lemma 3.20?</p>
<blockquote>
<p>So $N ⊂ \bar{S^c}$
. Hence by Lemma 3.15 (c), $U = \bar{S^c}$
is open.</p>
</blockquote>
<p>Since the contrapositive states that $y \notin \bar{S}$, I get why $y \in \bar{S^c}$. But, I don't follow why $N ⊂ \bar{S^c}$ and why $U = \bar{S^c}$.</p>
| rtybase | 22,583 | <p>From the set theory perspective, let's note
$$P \overset{\text{def}}{=}\{p \in \mathbb{N} \mid p \text{ - prime}\}$$
$$P_{\leq n} \overset{\text{def}}{=} \{p \in P \mid p \leq n \}$$
$$M_n \overset{\text{def}}{=} \{p \in P \mid p \mid n\}$$
$\color{red}{M_n \ne \varnothing, \forall n\geq2}$. Then, it's easy to show
$$P_n=P_{\leq n} \setminus M_n \tag{1}$$
$$M_n \subset P_{\leq n} \tag{2}$$
Now, let's assume $\exists n \ne m$, in fact we can assume $\color{red}{n>m}$, such that $\color{blue}{P_n = P_m}$ then
$$P_{\leq n} \setminus M_n = P_{\leq m} \setminus M_m \overset{(2)}{\Rightarrow} P_{\leq n}=\left(P_{\leq n} \setminus M_n \right) \bigcup M_n=\left(P_{\leq m} \setminus M_m \right) \bigcup M_n$$
or
$$P_{\leq n}=\left(P_{\leq m} \setminus M_m \right) \bigcup M_n \tag{3}$$
given $\color{red}{P_{\leq m} \subset P_{\leq n}}$ and $(2)$, it's easy to deduct that
$$M_m \subset M_n \tag{4}$$</p>
<p>There are 3 possible cases to exploit with $(4)$:</p>
<hr>
<p><strong>Case 1 (incomplete).</strong> $\forall k \in M_n, k\in M_m$ or $M_n=M_m$ (like $n=2^2 \cdot 3^2, m=2 \cdot 3$ for example), then $\color{green}{P_{\leq n} = P_{\leq m}}$. If $m=\prod\limits_{i} p_{k_i}^{\alpha_i}$ then $n=\prod\limits_{i} p_{k_i}^{\beta_i}$ with $\beta_i \geq \alpha_i$ and at least one $\beta_{i^{*}} > \alpha_{i^{*}}$. Since all $p_i \geq 2$ this means $$m < 2m \leq n$$ From <a href="https://en.wikipedia.org/wiki/Bertrand%27s_postulate" rel="nofollow noreferrer">Bertrand postulate</a> there is $p^{*}$-prime in between $m$ and $2m$ with $p^{*} \notin P_{\leq m}$ and $p^{*} \in P_{\leq n}$, thus $\color{green}{P_{\leq m} \ne P_{\leq n}}$ - <strong>contradiction</strong>.</p>
<p><em>Obviously $\beta_i \geq \alpha_i$ doesn't cover all the cases, e.g. $n=2 \cdot 3^2, m=2^2 \cdot 3$.</em></p>
<hr>
<p><strong>Case 2.</strong> $\exists k \in M_n, k\notin M_m$ and $k \in P_{\leq m}$. Since $k \in P_{\leq m}$ and $k\notin M_m$ then $k \in P_m=P_{\leq m} \setminus M_m$, but because $k \in M_n$ and $k \in P_{\leq n}$ then $k \notin P_n=P_{\leq n} \setminus M_n$. Thus $\color{blue}{P_n \ne P_m}$ - <strong>contradiction</strong>.</p>
<hr>
<p><strong>Case 3.</strong> $\exists k \in M_n, k\notin M_m$ and $k \notin P_{\leq m}$. Then $n=k\cdot t \geq k > m$</p>
<ul>
<li>if $t=1$ then (given $k$ is prime) $n$ is prime and $P_n$ will contain all the prime factors of $m$, but $P_m$ will not contain prime factors of $m$, thus $\color{blue}{P_n \ne P_m}$ - <strong>contradiction</strong>.</li>
<li>if $t\geq2$ then between $n$ and $\frac{n}{2}$ exist a prime $p^{*}$ (<strong>proposition 1</strong>, below). Because $p^{*}>\frac{n}{2}=\frac{kt}{2}
\geq k>m \Rightarrow p^{*} \notin P_m$. Also $p^{*} \nmid n$, otherwise $n=qp^{*}\geq 2p^{*} > n$. Thus $p^{*} \in P_n$ and as a result $\color{blue}{P_n \ne P_m}$ - <strong>contradiction</strong>.</li>
</ul>
<hr>
<p><strong>Proposition 1.</strong> $\forall n \geq 3$ $\exists p$ - prime s.t. $\frac{n}{2}<p<n$</p>
<p>Let's take $k=\left \lfloor \frac{n}{2} \right \rfloor+1 \geq \frac{n}{2}$. <a href="https://en.wikipedia.org/wiki/Bertrand%27s_postulate" rel="nofollow noreferrer">Bertrand postulate</a> says $\exists p$ - prime s.t. $k < p < 2k-2$ or
$$\color{red}{\frac{n}{2}\leq k} < p < 2k-2=\color{blue}{2 \left \lfloor \frac{n}{2} \right \rfloor \leq n} \tag{5}$$
simply because:</p>
<ul>
<li>for $n=2r$ (even) we have $\color{red}{\frac{n}{2}=r < r+1=\left \lfloor \frac{n}{2} \right \rfloor+1=k}$ and $\color{blue}{2 \left \lfloor \frac{n}{2} \right \rfloor=2r=n}$</li>
<li>for $n=2r+1$ (odd) we have $\color{red}{\frac{n}{2}=r+\frac{1}{2}<r+1=\left \lfloor r+\frac{1}{2} \right \rfloor+1=\left \lfloor \frac{n}{2} \right \rfloor+1=k}$ and $\color{blue}{2 \left \lfloor \frac{n}{2} \right \rfloor=2 \left \lfloor r+\frac{1}{2} \right \rfloor=2r<2r+1=n}$</li>
</ul>
<p><a href="https://en.wikipedia.org/wiki/Bertrand%27s_postulate" rel="nofollow noreferrer">Bertrand postulate</a> says $(5)$ is true for $k>3$ or $n >4$. For $n=4$ we have $\frac{4}{2}<3<4$ and for $n=3$ we have $\frac{3}{2}<2<3$.</p>
|
4,085,635 | <p>What is the volume of an n-dimension cube? Consider the length of each side to be <span class="math-container">$a$</span>. How to solve this problem?</p>
| User12345 | 133,683 | <p>This can be rigorously proved by induction (which may or may not be intuitive to you).</p>
<p>I assume the result is clear for <span class="math-container">$n=3$</span>, i.e. the triangle (let me know if it's not). For <span class="math-container">$n \geq 4$</span>, let <span class="math-container">$l_1 \leq \cdots \leq l_{n-1} \leq l_n$</span> denote the sequence of lengths. Now let
<span class="math-container">$$
\epsilon = \frac{1}{2}\min_j\left\{\sum_{i=1}^n l_i - 2l_j, l_1 \right\},
$$</span>
and note <span class="math-container">$\epsilon > 0$</span> as it is a minimum of positive numbers (where positivity comes from the assumption on the <span class="math-container">$l_i$</span>'s). Consider the sequence <span class="math-container">$a_1 = l_1 + l_2 - \epsilon, a_2=l_3, \cdots, a_{n-1}=l_n$</span>; note <span class="math-container">$a_1 = l_1 + l_2 - \epsilon > l_2 > 0$</span>. Thus we have a collection of <span class="math-container">$n-1$</span> lengths. Moreover, every sum of <span class="math-container">$n - 2$</span> of these is greater than the remaining length. Indeed, if <span class="math-container">$a_1$</span> is in the sum, we have
<span class="math-container">$$
\sum_{\substack{1 \leq i \leq n - 1 \\ i \neq j}}a_i
= \sum_{i = 1}^{n-1} a_i - a_i
= \sum_{i = 1}^n l_i - l_{j+1} - \epsilon
> l_{j+1} = a_{j}
$$</span>
because <span class="math-container">$\epsilon < \sum_{i=1}^n l_i - 2l_{j+1}$</span>. For the remaining sum, first note <span class="math-container">$l_1 + l_2 \leq \sum_{i=3}^{n} l_i$</span>; indeed, because <span class="math-container">$n \geq 4$</span> we have <span class="math-container">$l_1 \leq l_3$</span> and <span class="math-container">$l_2 \leq l_4$</span>. Now we may compute
<span class="math-container">$$
\sum_{i=2}^{n-1}a_i
= \sum_{i=3}^n l_i \geq l_1 + l_2 > l_1 + l_2 - \epsilon = a_1
$$</span>
Hence by induction we see that <span class="math-container">$a_1, a_2, \cdots, a_{n-1}$</span> are the side lengths of some polygon. Finally, note <span class="math-container">$l_1, l_2, l_1 + l_2 - \epsilon$</span> are the side lengths of a triangle (verify at your leisure). Thus we may attach the triangle with these side lengths (identifying the sides of length <span class="math-container">$l_1 + l_2 - \epsilon$</span>) to our newly-constructed polygon to obtain a polygon with side lengths <span class="math-container">$l_1, l_2, \cdots, l_n$</span> and complete the proof!</p>
|
1,582,275 | <p>Suppose that $B = S^{-1}AS$ for some $n \times n$ matrices $A$, $B$, and $S$.</p>
<ol>
<li>Show that if $x \in \ker(B)$ then $Sx \in \ker(A)$.</li>
</ol>
<p>Proof: $B = S^{-1}AS$ implies that $SB = AS$ which implies that $SBx = ASx = 0$, that is $Sx \in \ker(A)$.</p>
<ol start="2">
<li>Show that the linear transformation $T : \ker(B) \to ker(A), \, x \mapsto Sx$ is an isomorphism. </li>
</ol>
<p>I know how to prove part 1, but I am not sure what to do for part 2. </p>
| egreg | 62,967 | <p>The proof of 1 is good. Now, the map $T$ is well defined and obviously linear as it's the multiplication by a matrix.</p>
<p>If $x\in\ker T$, then $Sx=0$, so $x=0$ because $S$ is invertible. Therefore $T$ is injective. Since $A$ and $B$ have the same rank, their null spaces (kernels, in your terminology) have the same dimension (rank-nullity theorem) and injectivity implies surjectivity.</p>
<p>Alternatively, show that $U\colon \ker A\to\ker B$ is well defined by $y\mapsto S^{-1}y$. Obviously, $T$ and $U$ are inverse of each other.</p>
|
638,164 | <p>Let $\textbf{F}\left ( x, y \right )=\left ( -\frac y{x^2+y^2},\frac x{x^2+y^2} \right )$ be a vector field in $\mathbb{R}^2-\left \{ \textbf{0} \right \}$.</p>
<p>I know that the potential function of $\textbf{F}$ on $x>0$ is $\arctan \left ( \frac yx \right )$.</p>
<p>But I want to know the potential function of $\textbf{F}$ on $\mathbb{R}^2-\left \{ \textbf{0} \right \}$. Does it exist?</p>
| ShreevatsaR | 205 | <p>Here's a proof along the lines of your agument.</p>
<p>Consider your special number (in your notation)
$$S = [1, 2, 3, \dots, (n-3), (n-2), n]_{n+1}.$$
This number has $n$ in its units $(= (n+1)^0)$ place, $n-2$ in its $n+1$ $(= (n+1)^1)$ place, $n-3$ in its $(n+1)^2$ place, and so on, until $1$ in its $(n+1)^{n-2}$ place. So your number is
$$S = n + \sum_{i=1}^{n-2} (n-1-i)(n+1)^i = n + \sum_{k=1}^{n-2}k(n+1)^{n-1-k} = \frac{(n+1)^n-1}{n^2} = \frac{R}{n},$$
where $R$ is defined as $nS = \dfrac{(n+1)^n - 1}{n} = [1, 1, 1, \dots, 1, 1]_{n+1}$ with $n$ $1$s. (In other words, $R$ is the <a href="https://en.wikipedia.org/wiki/Repunit" rel="nofollow">repunit</a> $R_n^{(n+1)}$.)</p>
<p>For example, for base $10$ (n = $9$), we have
$$S = 12345679 = \dfrac{111111111}{9} = \dfrac{999999999}{9^2} = \dfrac{10^9 - 1}{9^2}.$$</p>
<hr>
<p>Now that we know what our special number $S$ is, let's prove facts about it.</p>
<p>Consider $kS$. As $S = R/n$, we can also calculate $kS = kR/n$ by dividing $kR$ by $n$. To talk about the digits in $kR/n$, let's formalize the division procedure.</p>
<p>The result of dividing a number $a = [a_1, a_2, \dots, a_m]_{n+1}$ by $b$ is the quotient $[q_1, q_2, \dots, q_m]_{n+1}$ and remainder $r_m$, given by:</p>
<ul>
<li>$r_0 = 0$, $q_0 = 0$, and</li>
<li>$q_i = \left\lfloor (r_{i-1} (n + 1) + a_i) / b \right\rfloor$ and $r_i = (r_{i-1}(n + 1) + a_i) - q_i b$ for $i = 1, 2, \dots, m$.</li>
</ul>
<p>(Basically, $[q_1, q_2, \dots, q_i]_{n+1}$ and $r_i$ are the quotient and remainder respectively, on dividing the number formed by the first $i$ digits of $a$, by $b$.)</p>
<p>Now for the special case where $a = kR$ and $k < n$ and $b = n$, we have $m = n$ and $a_i = k$ for each $i$ so </p>
<ul>
<li>($i=1$): $q_1 = \left\lfloor k / n \right\rfloor$ and $r_1 = k - nq_1 \equiv k \mod n$,</li>
<li>($i=2$): $q_2 = \left\lfloor (r_1(n+1) + k) / n \right\rfloor = r_1 + \left\lfloor (r_1 + k) / n \right\rfloor$, and $r_2 = r_1(n+1) + k -nq_2 \equiv r_1 + k \equiv 2k \mod n$,<br>
and in general, by induction</li>
<li>$q_i = \left\lfloor (r_{i-1}(n+1) + k) / n \right\rfloor = r_{i-1} + \left\lfloor (r_{i-1} + k) / n \right\rfloor$, and $r_i = r_{i-1}(n+1) + k - nq_i \equiv ik \mod n$.</li>
</ul>
<p>We'd like to prove that all $n$ $q_i$'s are distinct (when $0 < k < n$, and $k$ is relatively prime to $n$), which (as there are $n$ of them) will prove the claim that they consist of all digits $0$ to $n$ except one.</p>
<p>Note that the $r_i$s have already been determined: $r_i$ is the unique number in $[0, n-1)$ that is congruent to $ik$ modulo $n$, namely $r_i = ik - c_in$ where $c_i = \lfloor ik / n \rfloor$. Then,
$$\begin{align}
q_i
&= r_{i-1} + \left\lfloor (r_{i-1} + k) / n \right\rfloor \\
&= (i-1)k - c_{i-1}n + \left\lfloor ((i-1)k - c_{i-1}n + k) / n \right\rfloor \\
&= (i-1)k + \lfloor ik / n \rfloor - c_{i-1}(n+1)
\end{align}$$
(Alternatively, we can calculate $q_i$ as $q_i = (r_{i-1}(n+1) + k - r_i)/n$ and get the same expression.)</p>
<p>As $q_i$ is a digit in base $n+1$, we know that $0 \le q_i < n+1$, so let's look at $q_i$ modulo $n+1$: from the above,
$$q_i \equiv (i-1)k + \lfloor ik / n \rfloor \mod (n+1).$$</p>
<p>The distinctness of the $q_i$ thus boils down the following</p>
<blockquote>
<p><strong>Lemma.</strong> $a + \lfloor a/n \rfloor \equiv b + \lfloor b/n \rfloor \mod (n+1) \iff a \equiv b \mod n$</p>
</blockquote>
<p><strong>Proof:</strong> Suppose $a$ has more than one digit in base $(n+1)$, say $a = \alpha(n+1) + \beta$. Then, modulo $n+1$, we have $a + \lfloor a/n \rfloor \equiv \beta + \lfloor (\alpha(n+1) + \beta)/n \rfloor = (\alpha + \beta) + \lfloor (\alpha + \beta) / n \rfloor = a' + \lfloor a'/n \rfloor$ for $a' = \alpha + \beta$. Now if $a'$ has more than one digit in base $n+1$, i.e. if $a' \ge n + 1$, we can repeat the process. By induction, $a + \lfloor a/n\rfloor$ is the same as that of the (repeated) sum of its digits (also known as <a href="https://en.wikipedia.org/wiki/Digital_root" rel="nofollow">digital root</a>) which is precisely what $a \bmod n$ is (except possibly in the case when $a \bmod n = 0$, but note that $n + \lfloor n/n \rfloor = n + 1 \equiv 0 \mod (n+1)$, so that's covered as well.) $\Box$</p>
<p>So, if some two $q_i$ and $q_j$ are equal, we must have
$(i-1)k + \lfloor ik/n \rfloor \equiv (j-1)k + \lfloor jk/n \rfloor \mod (n+1)$, so (adding $k$ to both sides)
$ik + \lfloor ik/n \rfloor \equiv jk + \lfloor jk/n \rfloor \mod (n+1)$, which by our lemma means that $ik \equiv jk \mod n$, which in turn (when $k$ is relatively prime to $n$) means that $i \equiv j \mod n$.</p>
<p>This proves that the $n$ digits of $kS = kR/n$, namely $q_1$ to $q_n$, are all distinct. </p>
<p>(By the way, of the $n+1$ possible digits, the digit that is not attained among the $n$ $q_i$s is the one that can't appear as $(i-1)k + \lfloor ik/n \rfloor = ik + \lfloor ik/n \rfloor - k$. This is $n - k$, as $n$ is the one that can't be attained as $ik + \lfloor ik /n \rfloor$: by our lemma, if $ik \equiv c \mod n$ with $0 \le c < n$, then $ik + \lfloor ik/n \rfloor \equiv c + \lfloor c/n \rfloor \equiv c \mod (n+1)$.)</p>
<hr>
<p>That does it for $k < n$, and for $n < k < n^2$, just notice that if $k = an + b$, then $kS = kR/n = aR + bR/n$. As $aR$ has all its digits equal to $a$, we have that the $i$th digit of $kS$ is congruent, modulo $n+1$, to $a$ plus the $i$th digit of $bS$. This simply corresponds to shifting all the digits by $a$, and therefore doesn't affect distinctness.</p>
|
2,468,155 | <p>This problem is from Challenge and Thrill of Pre-College Mathematics:
Prove that $$ (a^3+b^3)^2\le (a^2+b^2)(a^4+b^4)$$</p>
<p>It would be really great if somebody could come up with a solution to this problem.</p>
| DeepSea | 101,504 | <p>To be honest, I do not look at the answers before mine and tried it myself with the most innovative thought I am able to have: $(a^3+b^3)^2 = (a^2\cdot a + b^2\cdot b)^2 \le (a^4+b^4)(a^2+b^2)$ by Cauchy-Schwarz inequality. As you can see how "beautiful" the CS inequality is....</p>
|
16,290 | <p>Hi I am new here and have a calculus question that came up at work.</p>
<p>Suppose you have a $4' \times 8'$ piece of plywood. You need 3 circular pieces all equal diameter. What is the maximum size of circles you can cut from this piece of material?
I would have expected I could write a function for the area of the 3 circles in terms of $x$ and $y$, then differentiate it, find a point of maxima/minima and go from there.</p>
<p>My coworker did cut three $33''$ circles and that solved the real-world problem. But my passion would be to find the mathematical answer to this. I hope that my new stackexchange.com friends have the same passion, and can help me find the answer to this in general terms. </p>
<p>What I mean by that is someone says I have a piece of material Q units by 2Q units, what are the three circles of maximum size?... I hope you understand what I am asking.
I am looking to be a friend and contributor
BD</p>
| Christian Blatter | 1,303 | <p>Let $Q:=[-2,2]\times[-1,1]$ be the given rectangle and let $r_0$ be the radius computed by Isaac for this rectangle. I shall prove that 3 circles of radius $r>r_0$ cannot be placed into $Q$ without overlap. The midpoints of these circles would have to lie in the smaller rectangle $Q':=[-2+r, 2-r]\times[-1+r,1-r]$. The left half $Q'_-$ of $Q'$ has a diameter $<2r_0<2r$ (by definition of $r_0$). It follows that $Q'_-$ can contain at most one center of non-overlapping circles of radius $r$, and the same is true for the right half $Q'_+$.</p>
|
2,725,831 | <blockquote>
<p>If we define $$f(x)=\left\lfloor \frac {x^{2x^4}}{x^{x^2}+3}\right\rfloor$$ and we have to find unit digit of $f(10)$</p>
</blockquote>
<p>I had tried approximation, factorization and substitutions like $x^2=u$ but it proved of no use. Moreover the sequential powers are feeling the hell out of me. Can someone please provide me with some hints</p>
| CY Aries | 268,334 | <p>When $x=10$, $\displaystyle \frac {x^{2x^4}}{x^{x^2}+3}=\frac {10^{20000}}{10^{100}+3}$.</p>
<p>Let $y=10^{100}$. $\displaystyle \frac {10^{20000}}{10^{100}+3}=\frac{y^{200}}{y+3}$.</p>
<p>When $y^{200}$ (as a polynomial) is divided by $y+3$, the remainder is $(-3)^{200}=3^{200}$.</p>
<p>So, $\displaystyle \frac{y^{200}}{y+3}=Q(y)+\frac{3^{200}}{y+3}$ for some polynomial $Q(y)$.</p>
<p>Put $y=0$, $Q(0)=-3^{199}=(-1)(81)^{49}(27)\equiv 3$ (mod $10$).</p>
<p>Note that $\displaystyle \frac{3^{200}}{10^{100}+3}=\frac{9^{100}}{10^{100}+3}<1$.</p>
<p>$$ \frac {10^{20000}}{10^{100}+3}=Q(10^{100})=10^{100}\times \textrm{some positive integer}+3+\frac{9^{100}}{10^{100}+3}$$</p>
<p>The unit digit of $\displaystyle \left\lfloor\frac {10^{20000}}{10^{100}+3}\right\rfloor$ is $3$.</p>
|
4,646,650 | <p>I believe there will be values of <span class="math-container">$x$</span> for which the inequality <span class="math-container">$x^3 - 2x + 2 \ge 3 - x^2$</span> is true and values for which it is not true, because:</p>
<ul>
<li><em>LHS asymptotically increases but RHS decreases for increasingly positive values of <span class="math-container">$x$</span></em></li>
<li><em>LHS asymptotically decreases faster than RHS for increasingly negative values of <span class="math-container">$x$</span></em></li>
</ul>
<p>But I don't know how to reason much further about <span class="math-container">$x^3 - 2x + 2 \ge 3 - x^2$</span>.<br></p>
<p>My idea is that if I get the inequality in a form I can reason about, I can determine what values of <span class="math-container">$x$</span> will satisfy the condition. I can reason about</p>
<p><span class="math-container">$$\frac{a}{b} > 0$$</span></p>
<p>since either <span class="math-container">$a,b > 0$</span> or <span class="math-container">$a,b < 0$</span> for <span class="math-container">$\frac{a}{b} > 0$</span>.</p>
<p>I can also reason about the signs of <span class="math-container">$a,b$</span> if they are expressed as a product of factors, since any even multiple of negative factors gives a positive (or zero) product. I think this will put me in a good spot to reason about what conditions must be met for the signs to satisfy the inequality <span class="math-container">$\frac{a}{b} > 0$</span> (although in my case, I am including zero).</p>
<p>Thus, my plan is to transform the inequality into the form LHS = <code>a quotient of factorised terms</code> and RHS = 0. But I'm not sure how.</p>
| David G. Stork | 210,401 | <p>Solve the cubic to find <span class="math-container">$x = -1.8, -.445, 1.25$</span>.</p>
<p><a href="https://i.stack.imgur.com/fkVsf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fkVsf.png" alt="enter image description here" /></a></p>
|
4,467,763 | <p>I have the equation</p>
<p><span class="math-container">$A\vec{x} = \vec{b} \tag{1}.$</span></p>
<p>where <span class="math-container">$A$</span> is an <span class="math-container">$m\times n$</span> matrix of rank <span class="math-container">$m$</span>, so that <span class="math-container">$m<n$</span> and the system is underdetermined. As I understand it, I can get the minimum L2 norm solution of the system by premultiplying <span class="math-container">$\vec{b}$</span> with the Moore-Penrose right inverse:</p>
<p><span class="math-container">$\vec{x} = A^T(AA^T)^{-1}\vec{b}\tag{2}.$</span></p>
<p>What I don't understand is how I get from (1) to (2), i.e. what are the algebraic steps? I'm specifically interested in knowing how to do it with just basic matrix manipulations, without calculus.</p>
| user1551 | 1,551 | <p>Suppose <span class="math-container">$x=u$</span> is a solution to the equation <span class="math-container">$Ax=b$</span>. Let <span class="math-container">$u=u_0+u_1$</span> where <span class="math-container">$u_0\in\ker A$</span> and <span class="math-container">$u_1\in(\ker A)^\perp$</span>. If <span class="math-container">$u_0\ne0$</span>, then <span class="math-container">$Au_1=A(u_0+u_1)=b$</span> but <span class="math-container">$\|u_1\|<\sqrt{\|u_0\|^2+\|u_1\|^2}=\|u\|$</span>, meaning that <span class="math-container">$u_1$</span> is also a solution to <span class="math-container">$Ax=b$</span> but with a smaller norm than <span class="math-container">$u$</span>.</p>
<p>Therefore, if <span class="math-container">$x=u$</span> is a least-norm solution to <span class="math-container">$Ax=b$</span>, its <span class="math-container">$u_0$</span> component must be zero. That is, <span class="math-container">$x$</span> must lie inside <span class="math-container">$(\ker A)^\perp$</span>.</p>
<p>But what is <span class="math-container">$(\ker A)^\perp$</span>? For any <span class="math-container">$x\in\ker A$</span>, we have <span class="math-container">$\langle x,A^Ty\rangle=\langle Ax,y\rangle=0$</span> for all <span class="math-container">$y\in\mathbb R^m$</span>. Therefore <span class="math-container">$\operatorname{range} A^T\subseteq(\ker A)^\perp$</span>. As <span class="math-container">$\operatorname{rank}(A^T)=\operatorname{rank}(A)=n-\dim(\ker A)=\dim(\ker A)^\perp$</span>, we conclude that <span class="math-container">$\operatorname{range} A^T=(\ker A)^\perp$</span>.</p>
<p>Thus any least-norm solution <span class="math-container">$x$</span> to the equation <span class="math-container">$Ax=b$</span> must be in the form of <span class="math-container">$x=A^Ty$</span> for some <span class="math-container">$y$</span>. Substitute the latter into the former, we obtain <span class="math-container">$AA^Ty=b$</span>. Therefore, when <span class="math-container">$(AA^T)^{-1}$</span> exists, we have <span class="math-container">$y=(AA^T)^{-1}b$</span> and the least-norm solution is uniquely determined as <span class="math-container">$x=A^Ty=A^T(AA^T)^{-1}b$</span>.</p>
|
2,933,753 | <p>Given two finite groups <span class="math-container">$G, H$</span>, we are going to say that <span class="math-container">$G<_oH$</span> if either</p>
<p>a. <span class="math-container">$|G|<|H|$</span></p>
<p>or </p>
<p>b. <span class="math-container">$|G|=|H|$</span> and <span class="math-container">$\displaystyle\sum_{g\in G} o(g)<\sum_{h\in H} o(h)$</span>,</p>
<p>where <span class="math-container">$o(g)$</span> denotes the order of the element <span class="math-container">$g$</span> (has this ordering a name?).</p>
<p>What is the smallest example (in this ordering) of a pair of nonisomorphic groups such that <span class="math-container">$G$</span> and <span class="math-container">$H$</span> are incomparable, i.e., such that they have same cardinal and same sum of orders of elements?</p>
| Hw Chu | 507,264 | <p>Feed the following code to Magma(<a href="http://magma.maths.usyd.edu.au/calc/" rel="nofollow noreferrer">http://magma.maths.usyd.edu.au/calc/</a>):</p>
<pre><code>for grouporder in [1..24] do
printf "Checking sum of order for groups or order %o:", grouporder;
numgroup := NumberOfSmallGroups(grouporder);
sset := [];
for i in [1..numgroup] do
sumorder := 0;
G := SmallGroup(grouporder,i);
for j in G do
sumorder := sumorder + Order(j);
end for;
Append(~sset, sumorder);
end for;
sset;
printf "\n";
end for;
</code></pre>
<p>You can see that there are two groups with "signature" <span class="math-container">$(16,47)$</span>. Much thanks to Travis for pointing it out. Two of them are non-abelian; they have GAP id <span class="math-container">$(16,3)$</span> and <span class="math-container">$(16,13)$</span>. Please refer the groupprop page for the description. The third is <span class="math-container">$\mathbb Z/4\mathbb Z \times (\mathbb Z/2\mathbb Z)^2$</span>. All have 1 element of order 1, 7 elements of order 2, and 8 elements of order 4.</p>
<p>Also, there are three groups with "signature" <span class="math-container">$(16,55)$</span>.</p>
<p>The groups are the following:</p>
<p><span class="math-container">$G_1 = (\mathbb Z/4\mathbb Z)^2$</span>; <span class="math-container">$G_2 = \langle a, b \ | \ a^4 = b^4 = 1, ba = ab^3\rangle$</span> (the semi-direct product of two copies of <span class="math-container">$\mathbb Z/4\mathbb Z$</span>), <span class="math-container">$G_3 = Q \times \mathbb Z/2\mathbb Z$</span>, where <span class="math-container">$Q$</span> is the quaternion group. All have 1 element of order 1, 3 elements of order 2, and 12 elements of order 4.</p>
<p><strong>Edit:</strong> This is copied from Travis' comment below. There are two groups of signature <span class="math-container">$(16,87)$</span> as well. One of them is the abelian group <span class="math-container">$\mathbb Z/8\mathbb Z \times \mathbb Z/2\mathbb Z$</span>, and the other is the non-abelian group with GAP ID <span class="math-container">$(16, 6)$</span> or is also called <span class="math-container">$M_4(2)$</span>. They both have 1, 3, 4, and 8 elements of order 1, 2, 4, and 8, respectively.</p>
|
114,487 | <p>I have a stack of images (usually ca 100) of the same sample. The images have intrinsic variation of the sample, which is my signal, and a lot of statistical noise. I did a principal components analysis (PCA) on the whole stack and found that components 2-5 are just random noise, whereas the rest is fine. How can I produce a new stack of images where the noise components are filtered out?</p>
<p>EDIT:</p>
<p>I am sorry I was not as active as you yesterday. I must admit I am bit overwhelm by the depth and yet simplicity of your answers. It is hard for me to choose one, since all of them work great and give what I actually wanted.</p>
<p>I feel that I need to elaborate a bit more the problem I am working on. Unfortunately, my supervisor does not allow me to upload nay data before we have published the final results, so I have to work in abstract terms. We have an atomic cloud cooled to a temperature of 10 µK. Due to inter-atomic and laser interaction, the atomic cloud (all of the atoms a whole) is excited and starts to oscillates in different vibrational modes. This dynamic behavior is of great interest to us, since it provides and insight to the inter-atomic physics. </p>
<p>The problem is that most of the relevant variations are obscured by noise due to the imaging process. The noise usually is greatly suppressed if you take two images one with Noise+Signal and Noise only and then subtract them. However, this does not work if the noise in the two images is not correlated, which sadly is our case. Therefore, we decided to use PCA, because there you can clearly see the oscillation modes and filter everything that is crap. If you are interested in using PCA to visualize dynamics, you can have a look at this paper by different group:</p>
<p><a href="http://iopscience.iop.org/article/10.1088/1367-2630/16/12/122001" rel="noreferrer">http://iopscience.iop.org/article/10.1088/1367-2630/16/12/122001</a></p>
<p>I deeply thank everybody who contributed.</p>
| Anton Antonov | 34,008 | <h2>Start data</h2>
<p>First let us get some images. I am going to use the MNIST dataset for clarity. (And because <a href="https://mathematicaforprediction.wordpress.com/2013/08/26/classification-of-handwritten-digits/" rel="nofollow noreferrer">I experimented with similar data some time ago</a>.)</p>
<pre><code>MNISTdigits = ExampleData[{"MachineLearning", "MNIST"}, "TestData"];
testImages = RandomSample[Cases[MNISTdigits, (im_ -> 0) :> im], 100]
</code></pre>
<p><a href="https://i.stack.imgur.com/EfDgn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EfDgn.png" alt="enter image description here"></a></p>
<p>Let us convince ourselves that all images have the same dimension:</p>
<pre><code>Tally[ImageDimensions /@ testImages]
dims = %[[1, 1]];
(* {{{28, 28}, 100}} *)
</code></pre>
<p>Let us create noisy images. Here are the options:</p>
<p><a href="https://i.stack.imgur.com/b5gcH.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/b5gcH.png" alt="enter image description here"></a></p>
<p>Below I will show results with 2 and 5.</p>
<p>The listing/selection of these options comes from asking ourselves what we know about the data. What does "of the same sample" in the question mean? In terms of digit writing:</p>
<ol>
<li><p>is it the same instance of digit writing but obtained by different channels that add noise, or </p></li>
<li><p>is it different instances of digit writing obtained by a channel that adds noise?</p></li>
</ol>
<h2>Noisy images</h2>
<pre><code>noisyTestImages =
Table[ImageEffect[
testImages[[12]], {"GaussianNoise", RandomReal[1]}], {100}];
</code></pre>
<p>RandomSample[Thread[{testImages[[12]], noisyTestImages}], 15]</p>
<p><a href="https://i.stack.imgur.com/1oW4R.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1oW4R.png" alt="enter image description here"></a></p>
<pre><code>noisyTestImages5 =
Table[ImageEffect[
testImages[[i]], {"GaussianNoise", RandomReal[1]}], {i, 100}];
RandomSample[Thread[{testImages, noisyTestImages5}], 15]
</code></pre>
<p><a href="https://i.stack.imgur.com/4bqIe.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4bqIe.png" alt="enter image description here"></a></p>
<h2>Linear vector space representation</h2>
<p>Unfold the images into vectors and stack together into a matrix:</p>
<pre><code>noisyTestImagesMat = (Flatten@*ImageData) /@ noisyTestImages;
Dimensions[noisyTestImagesMat]
(* {100, 784} *)
</code></pre>
<h2>No centralizing and standardizing</h2>
<p>The vector representations of the images I use have 0's as the important signal values. It is better to use the negated images as in <a href="https://mathematica.stackexchange.com/a/114565/34008">my second answer</a> (not done in this one).</p>
<p>Initially I thought that centralizing and normalizing is detrimental to the described process. Thanks to arguments in the comments and <a href="https://mathematica.stackexchange.com/a/114539/34008">Rahul's answer</a> now I think that centralizing is just useless for the data in my answer -- see <a href="https://i.imgur.com/S40OjMp.png" rel="nofollow noreferrer">this comparison</a> of centralized and non-centralized results using the code in Rahul's answer. </p>
<p>My other reason for not using centralization is that I would like to have interpretable basis when using matrix factorizations of data. (See the NNMF basis in <a href="https://mathematica.stackexchange.com/a/114565/34008">my second answer</a>.) We will have at least one such vector with SVD if we do not centralize. The question does not specify which of the results of PCA are going to be used: the de-noised images or the obtained components.</p>
<h2>De-noising</h2>
<p>Apply PCA (SVD):</p>
<pre><code>{U, S, V} = SingularValueDecomposition[noisyTestImagesMat, 12];
</code></pre>
<p>Let us look at the singular values:</p>
<pre><code>ListPlot[Diagonal[S], PlotRange -> All, PlotTheme -> "Detailed"]
</code></pre>
<p><a href="https://i.stack.imgur.com/dBmEs.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dBmEs.png" alt="enter image description here"></a></p>
<p>The main step of the denoising process -- zeroing the basis vectors we think correspond to noise:</p>
<pre><code>dS = S;
Do[dS[[i, i]] = 0, {i, Range[2, 12, 1]}]
</code></pre>
<p>Reconstruct the stack of images matrix and plot it:</p>
<pre><code>newMat = U.dS.Transpose[V];
MatrixPlot[newMat]
</code></pre>
<p><a href="https://i.stack.imgur.com/2TBAj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2TBAj.png" alt="enter image description here"></a></p>
<p>Convert to denoised images:</p>
<pre><code>denoisedImages = Map[Image[Partition[#, dims[[2]]]] &, newMat];
</code></pre>
<p>Compare noisy and denoised images:</p>
<pre><code>Transpose[{noisyTestImages, denoisedImages}]
</code></pre>
<p><a href="https://i.stack.imgur.com/zQ0RO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zQ0RO.png" alt="enter image description here"></a></p>
<h3>The harder set</h3>
<p>Repeating the above steps for option 5 (all images same noise type with different parameters) we get this:</p>
<p><a href="https://i.stack.imgur.com/ThOcT.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ThOcT.png" alt="enter image description here"></a></p>
<p>Not as de-noised as with the more regular set but still good.</p>
<h2>Post processing</h2>
<p>As it can be seen with the last result the procedure finds the "mean image" ("mean digit" in this case.) This can be exploited with some sort of image normalization:</p>
<pre><code>ImageMultiply[ImageAdd[denoisedImages], 1/Length[denoisedImages]]
Binarize[%, 0.7]
</code></pre>
<p><a href="https://i.stack.imgur.com/LKbVM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LKbVM.png" alt="enter image description here"></a></p>
|
575,232 | <p>Consider all strings whose letters belong to the set:</p>
<p>$A = \{ a, b, c, d, e\}$</p>
<p>How many strings of length $6$ are there that contain exactly one $a$?</p>
<p>Attempt:</p>
<p>Since we are only using $\frac{4}{5}$ letters for the rest of the string, </p>
<p>There are $1* 4^5$ strings that contain exactly one $a$.</p>
<p>Book answer: $6 * 4^5$</p>
<p>What am I doing wrong? Where does the six come from?</p>
| Will Orrick | 3,736 | <p>This answer is a small attempt to address your third question. You may find the article,</p>
<p>Jethro van Ekeren, <a href="http://webpages.math.luc.edu/%7Eptingley/oldseminars/QuantumGroupsSpring2011/lecture6b.pdf" rel="nofollow noreferrer"><em>The six-vertex model, <span class="math-container">$R$</span>-matrices, and quantum groups</em></a>,</p>
<p>to be helpful.</p>
<p>One of the main sources in the development of quantum groups was the field of <em>exactly solvable models in statistical mechanics.</em> A number of simple mathematical models were devised at the same time as quantum mechanics was being developed in order to understand phase transitions (changes of state) that occur in certain magnetic materials like iron, and that can also be used to understand properties of more familiar changes of state like the liquid-gas transition. One of these models was the Ising model, which is a classical (non-quantum) model. Similar models that came much later were the six-vertex and eight-vertex models. On the quantum mechanical side, there was the Heisenberg spin chain, among others.</p>
<p>The remarkable thing about these models is that, at least for certain values of their parameters, it is possible to compute certain physically interesting quantities, such as the free energy, exactly in the limit of infinite system size (the thermodynamic limit). In physics, symmetries of a system are associated with conservation laws. What makes these particular systems exactly solvable is that, in the thermodynamic limit, they have an infinite-dimensional group of symmetries, and therefore infinitely many conservation laws. (Without these conservation laws, the computation of the free energy become increasingly intractable as the system size grows.) These infinite-dimensional symmetry groups are mathematically interesting. In 1944 Lars Onsager computed the free energy of the two-dimensional Ising model with the external magnetic field parameter set equal to <span class="math-container">$0$</span> by introducing a certain infinite-dimensional algebra which was later discovered to be connected to Kac–Moody algebras.</p>
<p>Onsager's solution was considered by physicists to be difficult, and a number of alternative solution methods were discovered in subsequent decades. One of these—the method of commuting transfer matrices—was particularly fruitful in terms of generalizations. The Ising model is defined on a lattice—a two-dimensional square lattice—in Onsager's work. (This is intended as a simplified model of the crystalline lattice of real metals.) There is a binary (two-valued) variable (a "spin") associated with each lattice site, and the spins experience interactions with their nearest neighbors. The transfer matrix is an operator that corresponds to adding a row of sites to the lattice. In the Ising model, six-vertex model, and other exactly solvable two-dimensional models, the key to solvability is that transfer matrices with different values of a certain parameter commute with each other. The quantum mechanical models mentioned above, such as the Heisenberg spin chain, are one-dimensional models with Hamiltonian (energy operator) given by a matrix in the same commuting family. The members of this commuting family can be thought of as physical operators that represent conserved quantities of the spin chain (since they commute with the Hamiltonian, these quantities do not change with time).</p>
<p>Rodney Baxter discovered that commutativity of transfer matrices is implied by what is now known as the Yang–Baxter equation, which involves an operator called the <span class="math-container">$R$</span>-matrix. The <span class="math-container">$R$</span>-matrix can be regarded as the fundamental building block out of which transfer matrices are constructed, corresponding to the interactions of a single lattice site with its neighbors. The Yang–Baxter equation relates two different ways in which three sites can interact. It has a graphical representation closely connected with knot theory.</p>
<p>Quantum groups arise as algebraic structures in which <span class="math-container">$R$</span>-matrices satisfying the Yang–Baxter equation naturally arise. These solutions to the Yang–Baxter equation give rise to new families of commuting transfer matrices, and therefore to new exactly solvable models describing new types of phase transitions.</p>
<p>There are connections between these models and other parts of physics. These particular exactly solvable models have critical points, at which the physical system exhibits a form of scale invariance. Scale invariance actually implies the stronger conformal invariance, and these models at their critical points have continuum limits called conformal field theories. These are a key ingredient in string theory, and play a role in many interesting mathematical developments as well (for example in the proof of the Moonshine conjectures).</p>
|
3,336,742 | <p>Can we evaluate <span class="math-container">$\displaystyle\sum_{n=1}^\infty\frac{H_n^2}{n^32^n}$</span> ?</p>
<p>where <span class="math-container">$H_n=\sum_{k=1}^n\frac1n$</span> is the harmonic number.</p>
<p>A related integral is <span class="math-container">$\displaystyle\int_0^1\frac{\ln^2(1-x)\operatorname{Li}_2\left(\frac x2\right)}{x}dx$</span>.</p>
<p>where <span class="math-container">$\operatorname{Li}_2(x)=\sum_{n=1}^\infty\frac{x^n}{n^2}$</span> is the dilogarithmic function.</p>
<hr>
<p>Here is how the integral and the sum are related:</p>
<p>From <a href="https://math.stackexchange.com/questions/3263927/prove-frac-partial-partial-m-textbn-m-textbn-m-sum-k-0n-1-f/3271177#3271177">here</a> we have </p>
<p><span class="math-container">$$\int_0^1x^{n-1}\ln^2(1-x)\ dx=\frac{H_n^2+H_n^{(2)}}{n}$$</span></p>
<p>Divide both sides by <span class="math-container">$n^22^n$</span> then sum up we get</p>
<p><span class="math-container">$$\sum_{n=1}^\infty \frac{H_n^2+H_n^{(2)}}{n^32^n}=\int_0^1\frac{\ln^2(1-x)}{x}\sum_{n=1}^\infty \frac{x^n}{n^22^n}dx=\int_0^1\frac{\ln^2(1-x)\operatorname{Li}_2(x/2)}{x}dx$$</span></p>
| Myunghyun Song | 609,441 | <p>Using the generating function of <span class="math-container">$\displaystyle\{H_k^2\}_{k=1}^\infty$</span>:
<span class="math-container">$$
\frac{\text{Li}_2(x) +\ln^2(1-x)}{1-x} = \sum_{k=1}^\infty H_k^2 x^k
$$</span> we can observe that
<span class="math-container">\begin{align*}
S =& \sum_{k=1}^\infty {H_k^2 \over 2^k} \frac 1{k^2}\\
=& \sum_{k=1}^\infty {H_k^2 \over 2^k} \frac 1 2\int_0^1 x^{k-1}\ln^2 x\ dx\\
=&\frac 1 2 \int_0^1 \left(\sum_{k=1}^\infty H_k^2\left(\frac x 2\right)^k\right)\frac{\ln^2 x}{x} dx\\
=& \frac 1 2 \int_0^{\frac 1 2}\left(\sum_{k=1}^\infty H_k^2 x ^k\right) \frac{\ln^2 (2x)}{x} dx \\
=& \frac 1 2 \int_0^{\frac 1 2} \frac{\text{Li}_2(x) +\ln^2(1-x)}{1-x} \frac{\ln^2 x}{x} dx \\
&+ \ln 2\int_0^{\frac 1 2} \frac{\text{Li}_2(x) +\ln^2(1-x)}{1-x} \frac{\ln x}{x} dx \\
&+\frac{\ln^2 2}{2}\int_0^{\frac 1 2} \frac{\text{Li}_2(x) +\ln^2(1-x)}{1-x} \frac{dx}{x}\\
=&: I_1 + I_2 + I_3.
\end{align*}</span></p>
<hr>
<p>For <span class="math-container">$I_1$</span>, we have
<span class="math-container">\begin{align*}
I_1=&\frac 1 2 \int_0^{\frac 12 } \frac{\big[\text{Li}_2(x)+\ln^2(1-x)\big]\ln^2 x}{x(1-x)}dx \\
=& \frac 1 2 \int_0^{\frac 12 } \frac{\text{Li}_2(x)\ln^2 x}{x}dx+\frac 1 2 \int_0^{\frac 12 } \frac{\text{Li}_2(x)\ln^2 x}{1-x}dx +\frac 1 2 \int_0^{\frac 1 2} \frac{\ln^2 (1-x)\ln^2 x}{x(1-x)} dx\\
=&:I_1'+I_1''+I_1'''.
\end{align*}</span></p>
<hr>
<p>For <span class="math-container">$I_1'$</span>, we integrate by parts twice to get
<span class="math-container">\begin{align*}
I_1' \underset{\text{IBP}}{=}& \frac 1 2\left[ \text{Li}_3(x)\ln^2 x\right]^{1/2}_0 -\int_0^{\frac 1 2}\frac{\text{Li}_3(x) \ln x}{x} dx\\
\underset{\text{IBP}}{=}&\frac{\ln^2 2\ \text{Li}_3(1/2)}2 - \left[\text{Li}_4(x)\ln x\right]^{1/2}_0 +\int_0^{\frac 1 2 }\frac{\text{Li}_4(x)}x dx\\
=&\frac{\ln^2 2\ \text{Li}_3(1/2)}2 + \ln 2\ \text{Li}_4(1/2)+\text{Li}_5(1/2)\\
=&\boxed{\text{Li}_5(1/2)+\ln 2\ \text{Li}_4(1/2)+\frac{7\ln^2 2}{16}\zeta(3)-\frac {\pi^2\ln^3 2}{24}+\frac{\ln^5 2}{12}}
\end{align*}</span> where the well-known value of
<span class="math-container">$
\text{Li}_3(1/2) = \frac 78 \zeta(3) -\frac{\pi^2\ln 2}{12}+\frac{\ln^3 2}{6}
$</span> is used to simplify.</p>
<hr>
<p>For <span class="math-container">$I_1''$</span>, by integrating by parts,
<span class="math-container">\begin{align*}
I_1'' \underset{\text{IBP}}{=}& \frac 1 2 \int_0^{\frac 1 2} \ln(1-x)\left[\frac{2\ln x\text{Li}_2(x)}{x} - \frac{\ln(1-x)\ln^2 x}{x}\right]dx +\frac{\ln^3 2}{2}\text{Li}_2(1/2)\\
=& {\int_0^{\frac 1 2} \ln x\frac{\ln(1-x)\text{Li}_2(x)}{x} dx}-\underbrace{\frac 1 2 \int_0^{\frac 1 2} \frac{\ln^2(1-x)\ln^2 x}{x}dx}_{=:J} +\frac{\ln^3 2}{2}\text{Li}_2(1/2)\\
\underset{\text{IBP}}{=}&\frac{\ln 2}2\text{Li}_2^2(1/2) +{\frac 1 2\int_0^{\frac 1 2}\frac{\text{Li}_2^2(x)}{x} dx}-J +\frac{\ln^3 2}{2}\text{Li}_2(1/2)\\
=&\frac{\pi^4\ln 2}{288} -\frac{\ln^5 2}{8}+{\frac 1 2\int_0^{\frac 1 2}\frac{\text{Li}_2^2(x)}{x} dx}-J.
\end{align*}</span>
The well-known value of <span class="math-container">$\text{Li}_2(1/2) = \frac{\pi^2}{12} - \frac{\ln^2 2}{2}$</span> is used to simplify. In fact, the integral <span class="math-container">${\int_0^{1/2}\frac{\text{Li}_2^2(x)}{x} dx}$</span> was already evaluated in my previous answer <a href="https://math.stackexchange.com/questions/3305999/compute-int-01-2-frac-left-operatornameli-2x-right2x-dx/3325928#3325928">here</a>:
<span class="math-container">\begin{align*}
{\int_0^{\frac 1 2}\frac{\text{Li}_2^2(x)}{x} dx} = &-2\text{Li}_5(1/2) -2\ln 2\ \text{Li}_4(1/2)+\frac{27}{32}\zeta(5) +\frac{7\pi^2}{48}\zeta(3)-\frac{7\ln^2 2}{8}\zeta(3) \\
&-\frac{\pi^4\ln 2}{144} +\frac{\pi^2\ln^3 2}{12} - \frac{7\ln^5 2}{60}.
\end{align*}</span></p>
<hr>
<p>For <span class="math-container">$J$</span>, we make substitution <span class="math-container">$y= \frac{x}{1-x}$</span> to get
<span class="math-container">\begin{align*}
J=&\frac 1 2\int_0^1 \frac{\ln^2(1+y)\ln^2\left(\frac{y}{1+y}\right)}{y(1+y)}dy\\
=&\frac 1 2\int_0^1 \frac{\ln^2(1+y)\ln^2\left(\frac{y}{1+y}\right)}{y}dy-\frac 1 2\int_0^1 \frac{\ln^2(1+y)\ln^2\left(\frac{y}{1+y}\right)}{1+y}dy\\
=&:K-L.
\end{align*}</span></p>
<hr>
<p>For <span class="math-container">$K$</span>, expanding <span class="math-container">$\ln^2\left(\frac y {1+y}\right)=\big[\ln y -\ln(1+y)\big]^2$</span> and integrating by parts we get
<span class="math-container">\begin{align*}
K =&\frac 1 2\int_0^1 \frac{\ln^2(1+y)\ln^2\left(\frac{y}{1+y}\right)}{y}dy\\
=& \frac 12{ \int_0^1\frac{\ln^2 y\ln^2(1+y)}{y} dy}-{\int_0^1 \frac{\ln y\ln^3(1+y)}{y} dy}+\frac 12 \int_0^1 \frac{\ln^4(1+y)}{y}dy\\
\underset{\text{IBP}}{=}&-\frac 1 3 \int_0^1 \frac{\ln^3 y \ln(1+y)}{1+y}dy + \frac 3 2\int_0^1 \frac{\ln^2 y\ln^2(1+y)}{1+y}dy +\frac 12 \int_0^1 \frac{\ln^4(1+y)}{y}dy
\end{align*}</span> Doing the same thing for <span class="math-container">$L$</span>,
<span class="math-container">\begin{align*}
L = & \frac 1 2\int_0^1 \frac{\ln^2(1+y)\ln^2\left(\frac{y}{1+y}\right)}{1+y}dy\\
=& \frac 12 \int_0^1 \frac{\ln^2 y\ln^2(1+y)}{1+y}dy- {\int_0^1 \frac{\ln y\ln^3(1+y)}{1+y} dy} +\frac 12 \int_0^1 \frac{\ln^4(1+y)}{1+y} dy\\
\underset{\text{IBP}}{=}& \frac 12 \int_0^1 \frac{\ln^2 y\ln^2(1+y)}{1+y}dy+\frac 1 4 \int_0^1 \frac{\ln^4(1+y)}y dy +\frac{\ln^5 2}{10}.
\end{align*}</span> This gives that
<span class="math-container">\begin{align*}
J=&K-L\\
=&-\frac 1 3 \int_0^1 \frac{\ln^3 y \ln(1+y)}{1+y}dy + \int_0^1 \frac{\ln^2 y\ln^2(1+y)}{1+y}dy +\frac 14 \int_0^1 \frac{\ln^4(1+y)}{y}dy -\frac{\ln^5 2}{10}\\
=&:-V_1+V_2+V_3 -\frac{\ln^5 2}{10}.
\end{align*}</span></p>
<hr>
<p>For <span class="math-container">$V_1$</span>, we can use the Maclaurin series of <span class="math-container">$\frac{\ln (1+y)}{1+y} = \sum_{k=0}^\infty (-1)^{k-1} H_k y^k$</span> to get
<span class="math-container">\begin{align*}
V_1=&\frac 1 3\int_0^1 \frac{\ln^3 y\ln(1+y)}{1+y} dy \\
=& \frac 1 3\sum_{k=0}^\infty (-1)^{k-1}H_k {\int_0^1 y^k\ln^3 y\ dy} \\
=& \frac{-6}{3}\sum_{k=0}^\infty \frac{(-1)^{k-1}H_k}{(k+1)^4}\\
=&2\sum_{k=0}^\infty \frac{(-1)^k \left(H_{k+1}-\frac 1{k+1}\right)}{(k+1)^4} \\
=&2 \sum_{k=1}^\infty \frac{(-1)^{k-1}H_k}{k^4} -2 \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k^5}\tag{$k+1\mapsto k$}\\
=&2\left(\frac{59}{32}\zeta(5) - \frac{\pi^2\zeta(3)}{12}\right)-2\cdot \frac{15}{16}\zeta(5)\\
=&\frac{29}{16}\zeta(5) - \frac{\pi^2}{6}\zeta(3)
\end{align*}</span> where the known value of alternating Euler sum <a href="https://math.stackexchange.com/questions/1883098/alternating-euler-sums/1883731#1883731"><span class="math-container">$\sum_{k=1}^\infty \tfrac{(-1)^{k-1}H_k}{k^4}$</span></a> is used.</p>
<hr>
<p>For <span class="math-container">$V_2$</span>, we consider the algebraic identity
<span class="math-container">$$
6a^2b^2 = (a-b)^4 - a^4 +4a^3b +4ab^3 -b^4
$$</span> with <span class="math-container">$a=\ln y$</span> and <span class="math-container">$b = \ln(1+y)$</span> to get
<span class="math-container">\begin{align*}
V_2 =&\frac 1 6{\int_0^1 \frac{\ln^4\left(\frac y{1+y}\right)}{1+y} dy}-\frac 1 6\int_0^1 \frac{\ln^4 y}{1+y} dy +\underbrace{\frac 2 3\int_0^1 \frac{\ln^3 y\ln(1+y)}{1+y} dy}_{=2V_1}\\
&+\frac 2 3\underbrace{\int_0^1 \frac{\ln y\ln^3(1+y)}{1+y} dy}_{=- V_3\text{ by IBP}} -\frac 1 6\int_0^1 \frac{\ln^4(1+y)}{1+y} dy\\
=&\frac 1 6 \int_0^{\frac 1 2} \frac{\ln^4 x}{1-x} dx -\frac 1 6\int_0^1 \frac{\ln^4 y}{1+y} dy +2V_1-\frac 2 3 V_3 -\frac{\ln^5 2}{30}.\tag{$\tfrac y{1+y}= x$}
\end{align*}</span>
For the first integral, we have
<span class="math-container">\begin{align*}
W:=&\frac 1 6{\int_0^{\frac 1 2 }\frac{\ln^4 x}{1-x} dx}\tag{$2x\mapsto x$} \\
=& \frac 1 6\int_0^1 \frac{\ln^4(\tfrac x 2)}{2-x}dx\\
=&\frac 1 6\sum_{k=1}^\infty \frac{1}{2^k}\int_0^1 x^{k-1}\Big[\ln^4 x -4\ln 2 \ln^3 x + 6\ln^2 2\ln^2 x - 4\ln^3 2 \ln x + \ln^4 2\Big]dx\\
=&\frac 1 6\sum_{k=1}^\infty \frac{1}{2^k}\left[\frac{24}{k^5} + \frac{24\ln 2}{k^4} +\frac {12\ln^2 2}{k^3} +\frac{4\ln^3 2}{k^2} +\frac{\ln^4 2}{k}\right]\\
=&4\text{Li}_5(1/2) +4\ln 2\text{Li}_4(1/2) + 2\ln^2 2\text{Li}_3(1/2) + \frac{2\ln^3 2}{3}\text{Li}_2(1/2) + \frac{\ln^5 2}6\\
=&4\text{Li}_5(1/2) +4\ln 2\text{Li}_4(1/2) + \frac{7\ln^2 2}{4}\zeta(3)-\frac{\pi^2\ln^3 2}{9}+ \frac{\ln^5 2}6.
\end{align*}</span></p>
<p>For the second integral, we have
<span class="math-container">\begin{align*}
\frac 1 6\int_0^1 \frac{\ln^4 y}{1+y} dy
=& \frac 1 6\sum_{k=1}^\infty (-1)^{k-1} \int_0^1 y^{k-1}\ln^4 y \ dy \\
=& \frac 1 6 \sum_{k=1}^\infty(-1)^{k-1} \frac{24}{k^5}\\
=&\frac{15}{4}\zeta(5).
\end{align*}</span></p>
<p>This gives
<span class="math-container">$$
V_2 = W +2V_1-\frac 2 3 V_3-\frac{15}{4}\zeta(5)-\frac{\ln^5 2}{30}.
$$</span></p>
<hr>
<p>For <span class="math-container">$V_3$</span> we have
<span class="math-container">\begin{align*}
V_3=&\frac 14 \int_0^1 \frac{\ln^4(1+y)}{y}dy \tag{$y\mapsto y+1$}\\
=& \frac 1 {4}{ \int_1^2 \frac{\ln^4 y }{y-1} dy} \tag{$\tfrac 1 y\mapsto y$}\\
=&\frac 1 {4} \int_{\frac 1 2}^1 \frac{\ln^4 y}{y(1-y)} dy\\
=&\frac 1 {4} \int_{\frac 1 2}^1 \frac{\ln^4 y}{y} dy+\frac 1 {4} \int_{\frac 1 2}^1 \frac{\ln^4 y}{1-y} dy\\
=&\frac{\ln^5 2}{20} + \frac 1 {4} \int_0^1 \frac{\ln^4 y}{1-y} dy -\frac 1 {4} \underbrace{\int_0^{\frac 1 2}\frac{\ln^4 y}{1-y}dy}_{=6 W}\\
=&\frac{\ln^5 2}{20} +\frac 1 {4} \sum_{k=1}^\infty \int_0^1 y^{k-1}\ln^4 y\ dy- \frac 3 2 W\\
=&\frac{\ln^5 2}{20} +6\zeta(5)- \frac 3 2 W.
\end{align*}</span></p>
<hr>
<p>Combining <span class="math-container">$V_1$</span>, <span class="math-container">$V_2$</span> and <span class="math-container">$V_3$</span>, we get
<span class="math-container">\begin{align*}
J = & V_2 -V_1+V_3 -\frac{\ln^5 2}{10}\\
=& \left[W+2V_1 -\frac 2 3 V_3-\frac{15}{4}\zeta(5)-\frac{\ln^5 2}{30}\right]-V_1+V_3 -\frac{\ln^5 2}{10}\\
=& W+V_1+\frac 1 3 V_3-\frac{15}{4}\zeta(5)-\frac{2\ln^5 2}{15}\\
=&\frac 1 2 W+V_1 -\frac 7 4\zeta(5) -\frac{7\ln^5 2}{60}\\
=&2\text{Li}_5(1/2) +2\ln 2\ \text{Li}_4(1/2) +\frac 1 {16}\zeta(5) -\frac{\pi^2}6 \zeta(3) +\frac {7\ln^2 2}{8} \zeta(3) -\frac{\ln^2 2\pi^3}{18}-\frac{\ln^5 2}{30}.
\end{align*}</span>
This gives
<span class="math-container">\begin{align*}
I_1'' =&\frac{\pi^4\ln 2}{288} -\frac{\ln^5 2}{8}+\frac 1 2\int_0^{\frac 1 2}\frac{\text{Li}_2^2(x)}{x} dx-J\\
=&\boxed{\small -3\text{Li}_5(1/2) -3\ln 2\text{Li}_4(1/2) +\frac{23}{64}\zeta(5) +\frac {23\pi^2}{96}\zeta(3) -\frac {21\ln^2 2}{16}\zeta(3) +\frac{7\pi^2\ln^3 2}{72} - \frac{3\ln^5 2}{20}.}
\end{align*}</span></p>
<hr>
<p>For <span class="math-container">$I_1'''$</span>, we exploit the symmetric nature of the integrand to write
<span class="math-container">\begin{align*}
I_1''' :=& \frac 1 2\int_0^{\frac 1 2} \frac{\ln^2 x \ln^2 (1-x)}{x(1-x)} dx\\
=& \frac 1 4\int_0^1 \frac{\ln^2 x \ln^2 (1-x)}{x(1-x)} dx \\
=& \frac 1 4\underbrace{\int_0^1 \frac{\ln^2 x \ln^2 (1-x)}{x} dx}_{1-x\mapsto x}+\frac 1 4{\int_0^1 \frac{\ln^2 x \ln^2 (1-x)}{1-x} dx}\\
=&\frac 1 2\int_0^1 \frac{\ln^2 x \ln^2 (1-x)}{1-x} dx\\
=&\frac 1 2 \left[\frac{\partial^4}{\partial x^2 \partial y^2 } \text{B}(x,y)\right]_{x=1,y=0^+}
\end{align*}</span> where <span class="math-container">$\text{B}(x,y)=\tfrac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$</span> is Euler's Beta function. Now we can use the fact that
<span class="math-container">\begin{align*}
\lim_{y\to 0^+}\frac{\partial^3 }{\partial x\partial y^2 }\text{B}(x,y)
=&-\frac 1 3\psi'''(x)+\psi''(x)\Big[\psi(x) +\gamma\Big] + \psi'(x)\Big[\psi'(x)-\zeta(2) - \big[\psi(x) + \gamma\big]^2\Big]
\end{align*}</span> to obtain
<span class="math-container">\begin{align*}
I_1'''=& \frac 1 2\frac d{dx}\left[\frac{\partial^3 }{\partial x\partial y^2 }\text{B}(x,0^+)\right]_{x=1} \\
=& -\frac 1 6 \psi''''(1) +\psi'(1)\psi''(1) \\
=&\boxed{4\zeta(5) -\frac{\pi^2}3 \zeta(3)}
\end{align*}</span> where the values of <span class="math-container">$\psi(1) +\gamma = 0$</span>, <span class="math-container">$\psi'(1) =\zeta(2)$</span>, <span class="math-container">$\psi''(1) =-2\zeta(3)$</span> and <span class="math-container">$\psi''''(1) = -24\zeta(5)$</span> are used.</p>
<hr>
<p>Finally, from <span class="math-container">$I_1 = I_1'+I_1''+I_1'''$</span> we get</p>
<blockquote>
<p><span class="math-container">\begin{align*}
I_1 =& -2\text{Li}_5(1/2) - 2\ln 2\text{Li}_4(1/2) + \frac {279}{64}\zeta(5) -\frac {3\pi^2}{32}\zeta(3)-\frac {7\ln^2 2}{8} \zeta(3)+\frac {\pi^2\ln^3 2}{18}-\frac{\ln^5 2}{15}.
\end{align*}</span></p>
</blockquote>
<hr>
<p>For <span class="math-container">$I_2$</span>, we observe that
<span class="math-container">\begin{align*}\require{cancel}
I_2 =& \ln 2 \int_0^{\frac 12} \frac{\big[\text{Li}_2(x) + \ln^2(1-x)\big]\ln x}{x(1-x)} dx \\
=& \ln 2 {\int_0^{\frac 12} \frac{\text{Li}_2(x) \ln x}{1-x} dx}+\ln 2\int_0^{\frac 12} \frac{\ln^2(1-x)\ln x}{1-x} dx+ \ln 2 \int_0^{\frac 12} \frac{\big[\text{Li}_2(x) + \ln^2(1-x)\big]\ln x}{x} dx\\
\underset{\text{IBP}}{=}&-\ln^3 2\ \text{Li}_2(1/2)+\ln 2 \int_0^{\frac 12} \ln(1-x)\frac{-\cancel{\ln(1-x)\ln x}+\text{Li}_2(x)}{x} dx \\
&+\ln2 \int_0^{\frac 1 2} \frac{\ln^2(1-x)\ln x}{1-x} dx + \ln 2{ \int_0^{\frac 12} \frac{\big[\text{Li}_2(x)+\cancel{ \ln^2(1-x)}\big] \ln x}{x} dx}\\
\underset{\text{IBP}}{=}&\small-\ln^3 2 \text{Li}_2(1/2)-\tfrac{\ln 2}{2} \left[\text{Li}^2_2(x)\right]^{1/2}_0-\frac{\ln^5 2}3+{\frac{\ln 2}3{\int_0^{\frac 1 2} \frac{\ln^3(1-x)}{x} dx}} -\ln^2 2\ \text{Li}_3(1/2)-\ln 2\int_0^{\frac 1 2} \frac{\text{Li}_3(x)}{x} dx\normalsize\\
=&-\ln^3 2\ \text{Li}_2(1/2) -\tfrac{\ln 2}{2} \text{Li}^2_2(1/2)-\frac{\ln^5 2}3+ \small\underbrace{\frac{\ln 2}3{\int_{\frac 1 2}^1 \frac{\ln^3 x}{1-x} dx}}_{1-x\mapsto x, \ =:I_2'}\normalsize-\ln^2 2\ \text{Li}_3(1/2)-\ln 2\ \text{Li}_4(1/2)\\
=&-\ln 2\ \text{Li}_4(1/2) -\frac{7\ln^2 2}8 \zeta(3) -\frac{\pi^4 \ln 2}{288}+\frac {\pi^2\ln^3 2}{24} -\frac{\ln^5 2}{8} + I_2'.
\end{align*}</span></p>
<hr>
<p>For <span class="math-container">$I_2'$</span>, by integrating by parts, we have
<span class="math-container">\begin{align*}
I_2'
=& \frac{\ln 2}3\int_{\frac 1 2}^1 \frac{\ln^3 x}{1-x} dx\\
=&\frac{\ln 2}3\int_{0}^1 \frac{\ln^3 x}{1-x} dx -\frac{\ln 2}3{\int_{0}^{\frac 12} \frac{\ln^3 x}{1-x} dx}\tag{$x=\tfrac y 2$}\\
=&\frac{\ln 2}3\sum_{k=1}^\infty {\int_0^1 x^{k-1}\ln^3 x\ dx}-\underbrace{\frac{\ln 2}3\int_{0}^{1} \frac{\ln^3 (\tfrac y 2)}{2-y} dy}_{=:A}\\
=&-\frac{\pi^4\ln 2}{45}-A.
\end{align*}</span>
<span class="math-container">\begin{align*}
A=&\frac{\ln 2}3\sum_{k=1}^\infty \frac 1 {2^k}\int_{0}^{1} y^{k-1}\ln^3 (\tfrac y 2) dy\\
=&\frac{\ln 2}3\sum_{k=1}^\infty \frac 1 {2^k}\int_{0}^{1} y^{k-1}\left[\ln^3 y - 3\ln 2\ln^2 y +3\ln^2 2\ln y -\ln^3 2\right]dy\\
=&-\frac{\ln 2}3\sum_{k=1}^\infty \frac 1 {2^k}\left[\frac 6 {k^4}+\frac{6\ln 2}{k^3} +\frac{3\ln^2 2}{k^2} +\frac{\ln^3 2}{k}\right]\\
=& -2\ln 2\ \text{Li}_4(1/2) - 2\ln^2 2\ \text{Li}_3(1/2)-\ln^3 2\ \text{Li}_2(1/2)-\frac{\ln^5 2}3\\
=&-2\ln 2\ \text{Li}_4(1/2)-\frac{7\ln^2 2}{4}\zeta(3)+\frac{\pi^2\ln^3 2}{12}-\frac{\ln^5 2}{6}.
\end{align*}</span></p>
<hr>
<p>This gives
<span class="math-container">$$
I_2'= 2\ln 2\ \text{Li}_4(1/2)+\frac{7\ln^2 2}{4}\zeta(3)-\frac{\pi^4\ln 2}{45}-\frac{\pi^2\ln^3 2}{12}-\frac{\ln^5 2}{6}
$$</span> and</p>
<blockquote>
<p><span class="math-container">\begin{align*}
I_2=\ln 2\ \text{Li}_4(1/2)+\frac{7\ln^2 2}{8}\zeta(3)-\frac{37\pi^4\ln 2}{1440}-\frac{\pi^2\ln^3 2}{24}+\frac{\ln^5 2}{24}.
\end{align*}</span></p>
</blockquote>
<hr>
<p>For <span class="math-container">$I_3$</span>, we have
<span class="math-container">\begin{align*}
I_3=&\frac{\ln^2 2}{2}\int_0^{\frac 1 2} \frac{\text{Li}_2(x)+\ln^2(1-x)}{x(1-x)}dx\\
=&\frac{\ln^2 2}{2}\int_0^{\frac 1 2} \frac{\text{Li}_2(x)}{x}dx+\frac{\ln^2 2}{2}\int_0^{\frac 1 2} {\left[\frac{\ln^2(1-x)}{x}+\frac{\text{Li}_2(x)}{1-x}\right]}dx+\frac{\ln^2 2}{2}\int_0^{\frac 1 2} \frac{\ln^2(1-x)}{1-x}dx\\
=&\frac{\ln^2 2}{2}\text{Li}_3(1/2) +\frac{\ln^2 2}{2}\big[-\ln(1-x)\text{Li}_2(x)\big]^{1/2}_0+\frac{\ln^5 2}6.
\end{align*}</span> Using the well-known values of <span class="math-container">$\text{Li}_3(1/2)$</span> and <span class="math-container">$\text{Li}_2(1/2)$</span>, this simplifies to</p>
<blockquote>
<p><span class="math-container">$$
I_3 =\frac {7\ln^2 2}{16}\zeta(3).
$$</span></p>
</blockquote>
<hr>
<p>From <span class="math-container">$S = I_1+I_2 + I_3$</span>, we finally get</p>
<blockquote>
<p><span class="math-container">\begin{align*}
\sum_{k=1}^\infty \frac{H_k^2}{k^32^k} =& -2\text{Li}_5(1/2) -\ln 2\ \text{Li}_4(1/2) + \frac{279}{64}\zeta(5) - \frac{3\pi^2}{32}\zeta(3) + \frac{7\ln^2 2}{16}\zeta(3) - \frac{37\pi^4 \ln 2}{1440}\\
& + \frac{\pi^2 \ln^3 2}{72} - \frac{\ln^5 2}{40}.
\end{align*}</span></p>
</blockquote>
<p>We can observe that the values of <span class="math-container">$I_2$</span> and <span class="math-container">$I_3$</span> can be used to evaluate sums of lower order in a similar way:</p>
<blockquote>
<p><span class="math-container">\begin{eqnarray*}
&\sum_{k=1}^\infty \frac{H_k^2}{k^2 2^k} = -\frac 1{\ln 2} I_2 - \frac{2}{\ln 2} I_3=-\text{Li}_4(1/2) -\frac{7\ln 2}4\zeta(3) +\frac{37 \pi^4}{1440} +\frac{\pi^2\ln^2 2}{24} -\frac{\ln^4 2}{24},\\
&\sum_{k=1}^\infty \frac{H_k^2}{k 2^k} = \frac{2}{\ln^2 2} I_2 = \frac{7}{8}\zeta(3).
\end{eqnarray*}</span></p>
</blockquote>
|
422,084 | <p>$\mathbb{R}^2$ with different topologies on it are homeomorphic as a topological space?
for example with discrete topology and usual topology, what I need is a continous bijection with inverse is continous, from usual to discrete any continous map is finally constant map,so I think they are not homeomorphic.
thank you for help and discussion</p>
| Dylan Yott | 62,865 | <p>Suppose there exists a homeomorphism $f$ between $\mathbb R^{2}$ with the usual topology, and $\mathbb R^{2}$ with the discrete topology. Then $f(S^{1})$ is compact, as continuous functions preserve compactness, and it is also infinite, as $S^{1}$ has infinitely many points. However in a discrete space, only finite sets are compact. So there cannot exist a homeomorphism. In fact, there cannot even be an injective continuous such $f$, as that is all we used in the above. </p>
|
791,020 | <p>From my textbook. </p>
<p>$$\sum\limits_{k=0}^\infty (-\frac{1}{5})^k$$</p>
<p>My work:</p>
<p>So a constant greater than or equal to $1$ raised to ∞ is ∞.</p>
<p>A number $n$ for $0<n<1$ is $0$. So when taking the limit of this series you get 0 but when formatting the problem a different way $(-1)^k/(5^k)$ it seems like an alternating series. Can someone help me figure this out?</p>
| user84413 | 84,413 | <p>One way to do this is to rewrite the expression as $(1+(x-3))^{1/2}$, and then substitute $u=x-3$ and expand $(1+u)^{1/2}$ using the Binomial Series
$\displaystyle(1+u)^r=\sum_{n=0}^{\infty}\binom{r}{n}u^n$.</p>
|
1,332,420 | <p>Solution of this task is ${7\pi\over2}$ but I don't know how to get to this solution.I used the formula for double angle for $\sin{2x}$ and $\cos{2x}$ and moved everything on one side of the equation and made $2\cos{x}$ as common factor to get this :$$2\cos{x}(\sin{x} - {\sqrt{2}\over2} - \cos{x}) = 0$$
Then I can get that $\cos{x} = 0$ and that solution for this particular x are $x = {\pi\over2}$ and $x = {3\pi\over2}$. But I don't know what to do next inside these brackets to get the rest.</p>
<p>EDIT: So I get at the end that $\sin{2x} = {1\over2}$ but solutions I get then are:
$$\sin{2x} = {1\over 2}$$
$$2x = {\pi\over6} \quad \text{&}\quad 2x = {5\pi\over6}$$
$$x = {\pi\over12} \quad \text{&}\quad x = {5\pi\over12}$$
And when then I sum up those two with already found:
$${\pi\over2}+{3\pi\over2} + {\pi\over12}+{5\pi\over12}=$$
$$={6\pi\over12}+{18\pi\over12} + {\pi\over12}+{5\pi\over12}=$$
$$={30\pi\over12}={5\pi\over2}$$
Which is not the solution I was looking for</p>
| Mythomorphic | 152,277 | <p>$$\sin{x} - {\sqrt{2}\over2} - \cos{x} = 0$$
$$\sin x-\cos x=\frac{\sqrt2} 2$$
$$(\sin x-\cos x)^2=\frac12$$
$$\sin^2x+\cos^2x-2\sin x\cos x=\frac12$$
$$1-\sin 2x=\frac12$$
$$\sin 2x=\frac12$$</p>
|
1,985,653 | <p>I want to show that {$\sqrt 2$} $\cup$ {$\sqrt 2$ + $1/n$ : $n$ $∈$ $\mathbb{N}$} is closed. I'm having trouble. I've been trying to show that the complement is open, but the presence of {$\sqrt 2$} is confusing me.</p>
| Evan Aad | 37,058 | <p>A set is closed iff it contains all its limit points. The only limit point of the given set is $\sqrt{2}$.</p>
|
147,378 | <p>I have the following equation:</p>
<p>$$\frac{dx}{dt}+x=4\sin(t)$$</p>
<p>For solving, I find the homogenous part as:
$$f(h)=C*e^{-t}$$</p>
<p>Then finding $f(a)$ and $df(a)$:
$$f(a)=4A\sin(t)+4B\cos(t)$$
$$df(a)=4A\cos(t)-4B\sin(t)$$</p>
<p>Substituting in orginal equation:</p>
<p>$$4A\cos(t)-4B\sin(t)+4A\sin(t)+4B\cos(t)=4\sin(t)$$</p>
<p>I have to find numerical values of $A$ and $B$ but I absoloutly have no idea how can I solve this, I am also not sure if the steps I did are correct or not. Would somone please help with this equation?</p>
<p>The final answer should be substituted in:
$$x=f(h)+f(a)$$</p>
| Robert Mastragostino | 28,869 | <p>multiples of $\cos(t)$ can't add to make $\sin(t)$, and vice versa. So this can be split into two equations:
$$4A\cos(t)+4B\cos(t)=0$$
$$4A\sin(t)-4B\sin(t)=4\sin(t)$$
simplifying we get
$$A+B=0$$
$$A-B=1$$
and so
$$A=\frac{1}{2}, B=-\frac{1}{2}$$</p>
<p>So your final solution is $$x=f(h)+f(a)=Ce^{-t}+2\sin(t)-2\cos(t)$$</p>
|
147,378 | <p>I have the following equation:</p>
<p>$$\frac{dx}{dt}+x=4\sin(t)$$</p>
<p>For solving, I find the homogenous part as:
$$f(h)=C*e^{-t}$$</p>
<p>Then finding $f(a)$ and $df(a)$:
$$f(a)=4A\sin(t)+4B\cos(t)$$
$$df(a)=4A\cos(t)-4B\sin(t)$$</p>
<p>Substituting in orginal equation:</p>
<p>$$4A\cos(t)-4B\sin(t)+4A\sin(t)+4B\cos(t)=4\sin(t)$$</p>
<p>I have to find numerical values of $A$ and $B$ but I absoloutly have no idea how can I solve this, I am also not sure if the steps I did are correct or not. Would somone please help with this equation?</p>
<p>The final answer should be substituted in:
$$x=f(h)+f(a)$$</p>
| Milo Wielondek | 5,877 | <p>Another approach would be to consider the given equation as one of the form</p>
<p>$$
\frac{dx}{dt}+Px=Q
$$</p>
<p>where $P$ and $Q$ are functions of $t$. Such equation can be solved by multiplying both sides by the integrating factor $e^{\int P\,dt}$.</p>
<p>Applying this to your equation, we get
$$
\begin{align}
e^{t}\frac{dx}{dt}+e^t x=e^t 4 \sin{t}\\
\frac{d}{dt}\left(xe^t\right)=4\,e^t \sin{t}\\
xe^t=4\int e^t \sin{t}\ dx\\
\end{align}
$$</p>
<p>Solving the above by integration by parts we get</p>
<p>$$
\begin{align}
xe^t&=\frac{4e^t(\sin t -\cos t)}{2} + c\\
x&=2\sin t -2\cos t+ce^{-t}
\end{align}
$$</p>
|
4,274,085 | <p>A linear equation in one variable <span class="math-container">$x$</span>, <span class="math-container">$$ax+b=k$$</span> has only one non-negative integer solution. For example, <span class="math-container">$2x+3=85$</span> has a solution 41.</p>
<p>How to find the number of non-negative integer solutions of a linear equation in two variables <span class="math-container">$x$</span> and <span class="math-container">$y$</span>, for example, <span class="math-container">$5x+3y=100$</span> and in general, <span class="math-container">$$ax+by=k$$</span></p>
<p>By brute-force search, I found 7 non-negative integer solutions of <span class="math-container">$5x+3y=100$</span>. They are <span class="math-container">$(2, 30), (5, 25), (8, 20), (11, 15), (14, 10), (17, 5), \text{ and }(20, 0)$</span>.</p>
<p><em>P.S.</em> I came to know that a linear equation in two variables is also know as linear Diophantine equation. I also came across <a href="https://en.wikipedia.org/wiki/Ku%E1%B9%AD%E1%B9%ADaka" rel="nofollow noreferrer">Kuṭṭaka</a>, Aryabhata's algorithm for solving linear Diophantine equations in two variables.</p>
| Kuifje | 273,220 | <p>Or, you could solve a linear program with a solver (as the tag <code>linear-programming</code> is in your question), or more precisely an integer program (MIP), and add a no good cut to exclude the current solution, until there are no more left.</p>
<p>The variables are <span class="math-container">$(x,y)\in \mathbb{N}^2$</span>, there is no objective function (add a dummy one, e.g., <span class="math-container">$0$</span>), and the unique constraint is
<span class="math-container">$$
ax+by = k
$$</span>
The no good constraint should cut out a given solution <span class="math-container">$(\hat{x},\hat{y})$</span>. An option is to add:
<span class="math-container">$$
|x-\hat{x}|+|y-\hat{y}| \ge 1
$$</span></p>
<p>So for example, if the first iteration gives you <span class="math-container">$(x,y)=(2,30)$</span>, you can solve
<span class="math-container">$$
5x + 3y = 100 \\
|x-2|+|y-30| \ge 1
$$</span>
And it should give you a solution among the other ones you found by brute force, perhaps <span class="math-container">$(5,25)$</span>. So the third iteration would be:
<span class="math-container">$$
5x + 3y = 100 \\
|x-2|+|y-30| \ge 1 \\
|x-5|+|y-25| \ge 1
$$</span></p>
<p>Once the program becomes infeasible, you have your set of all feasible solutions.</p>
|
335,483 | <p>Let $N$ be a set of non-negative integers. Of course we know that $a+b=0$ implies that $a=b=0$ for $a, b \in N$.</p>
<p>How do (or can) we prove this fact if we don't know the subtraction or order?</p>
<p>In other words, we can only use the addition and multiplication.</p>
<p>Please give me advise.</p>
<p>EDIT</p>
<p>The addition law mean that for $a, b \in N$, there is an element $a+b$ in $N$ and this operation is associative.
The multiplication law means that for $a, b \in N$, there is an element $ab$ in $N$ and this operation is associative.
Also the distribution laws hold.</p>
<hr>
<p>EDIT2</p>
<p>Let me rephrase the question since I don't want arguments on orders.</p>
<p>Let $N$ be a set with operation $+$ and $\times$.</p>
<p>$N$ is a monoid with the operation $+$ and $\times$ respectively. There is an unit element $0\in N$.</p>
<p>The distribution laws hold as in the case of the set of integers.</p>
<p>Can we prove the fact above with this assumption?</p>
| Thomas Andrews | 7,933 | <p>There are three axioms that you need to prove this:</p>
<ol>
<li>For all non-negative integers $n$, $n+1\neq 0$</li>
<li>For all non-negative integers $a,b$, $a+(b+1)= (a+b)+1$</li>
<li>Induction</li>
</ol>
<p>Theorem: For all $b$, either $b=0$ or $a+b\neq 0$.</p>
<p>Proof by induction: If $b=0$ we are done.</p>
<p>Now assume it is true for $b$. Then $a+(b+1)=(a+b)+1$ by (2). But by $1$, $(a+b)+1\neq 0$. So we get our result.</p>
<p>Now, if $a+b=0$ then $b=0$ and then $a+b=a+0=a$ so $a=0$ as well.</p>
<p>You can't do it just from addition law and multiplication law because those laws are true for all the integers, and it is not true for all the integers.</p>
|
335,483 | <p>Let $N$ be a set of non-negative integers. Of course we know that $a+b=0$ implies that $a=b=0$ for $a, b \in N$.</p>
<p>How do (or can) we prove this fact if we don't know the subtraction or order?</p>
<p>In other words, we can only use the addition and multiplication.</p>
<p>Please give me advise.</p>
<p>EDIT</p>
<p>The addition law mean that for $a, b \in N$, there is an element $a+b$ in $N$ and this operation is associative.
The multiplication law means that for $a, b \in N$, there is an element $ab$ in $N$ and this operation is associative.
Also the distribution laws hold.</p>
<hr>
<p>EDIT2</p>
<p>Let me rephrase the question since I don't want arguments on orders.</p>
<p>Let $N$ be a set with operation $+$ and $\times$.</p>
<p>$N$ is a monoid with the operation $+$ and $\times$ respectively. There is an unit element $0\in N$.</p>
<p>The distribution laws hold as in the case of the set of integers.</p>
<p>Can we prove the fact above with this assumption?</p>
| Sean Eberhard | 23,805 | <p>Use the Mazur swindle! Namely, if $a+b=0$ then</p>
<p>\begin{align*}
0 &= 0 + 0 + 0 + \cdots\\
&= (a+b)+(a+b)+(a+b)+\cdots\\
&= a+(b+a)+(b+a)+\cdots \\
&= a + 0 + 0 + \cdots\\
&= a.
\end{align*}</p>
<p>Regrouping the infinite sum is justified because everything is nonnegative. I leave it as an exercise to identify exactly which axioms of arithmetic we've used.</p>
|
32,997 | <p>I have made a rather obvious yet peculiar observation while calculating with quadratic inequalities. Take a simple quadratic inequality like the one below</p>
<p>$\frac{x^2+1}{x}>1$</p>
<p>by multiplying both sides by $x$, then subtracting $x$ from both sides we get</p>
<p>$x^2-x+1>0$</p>
<p>Hence, both the above inequalities are one and the same in theory. However, the solution to the first inequality is $x>0$, while the second inequality is satisfied for all $x\in\mathbb R$.</p>
<p>It is not difficult to explain this simple case, seeing that the LHS of the first inequality will turn negative for negative $x$, but my question is more of a general nature, if you like. I wonder, why do two identical (but paraphrased/rewritten) inequalities have different solutions?</p>
| abel | 9,252 | <p>the method i describe below helps finding a complete solution by sticking to the simple principle that for polynomials and rational functions only candidates for the sign change is across the zeros of the numerator and denominator.</p>
<p>instead of looking at $\frac{x^2 + 1}{x} > 1$ let us look at $\frac{x^2 + 1}{x} -1$ which simplifies to the rational function $\frac{x^2 - x + 1}{x}$. the numerator has no real roots so it is of one sign which is positive because the value at 0 is 1. sign of rational function is now the sign of numerator. </p>
|
1,811,581 | <p>When discussing the order relation on $\mathbb{C}$, it is said that such a statement as $z_1 < z_2$ where $z_1, z_2 \in \mathbb{C}$ is meaningless, unless $z_1$ and $z_2$ are real.</p>
<p>My question is, when will a complex number $z$ be real? I know that if $\bar{z}$ is the conjugate of $z$, then</p>
<p>$$z + \bar{z} = 2a$$
$$z\bar{z} = a^2+b^2$$</p>
<p>produce real numbers, but it is easy to add $0i$ to either equation to produce a complex number.</p>
| AndreasS | 32,580 | <p>Yes, the (linear) span is a vector space. By definition it is the smallest vector space that contains all the elements in the set. In particular it will contain all linear combinations of those elements (and will in fact contain exactly all linear combinations that can be formed with those elements).</p>
<p>So in your example, if we consider $\{[1,0],[0,1]\} \subset \mathbb{R}^{2}$ we will get $$span\{[1,0],[0,1]\} = \mathbb{R}^{2}$$
because for every $[x,y] \in \mathbb{R}$ we can find a linear combination of $[1,0],[0,1]$ that represents $[x,y]$:
$$x\cdot[1,0] + y\cdot[0,1] = [x,y].$$</p>
<p>With similar arguments we get
get $$span\{[1,0]\} = \{[x,0] \in \mathbb{R}^{2}: x\in\mathbb{R}\}.$$</p>
<p><a href="https://en.wikipedia.org/wiki/Linear_span" rel="nofollow">Wikipedia</a> is also a good starting point for more on linear spans.</p>
|
367,885 | <p>Denote the set of all bounded sequences in $\mathbb{R}$ by $l^{\infty}$, endowed with the sup norm $\lVert \rVert _{\infty}$. Define a map $T:l^{\infty} \to l^{\infty}$ as follows:</p>
<p>$$(x_n)_n \mapsto \Big(\frac{x_n}{n}\Big)_n.$$</p>
<p>Then the question is:</p>
<blockquote>
<p>Show that the image of $l^{\infty}$ under $T$ is not a closed set in $l^{\infty}$.</p>
</blockquote>
<p><em><strong>My attempts:</em></strong></p>
<p>As it is enough to show the existence of a sequence $(t_n)$ of bounded sequences from $T(l^{\infty})$ which converges to a sequence outside of it, I first considered the constant sequence $$(1,1, \ldots, 1, \ldots) = t_0$$ $(\notin T(l^{\infty}))$ and tried to look for a sequence $(t_n)$ of sequences in $T(l^{\infty})$ which goes to it. The following are two (failed) attempts to construct such a sequence. </p>
<p>First I thought the sequences $$t_n = (1,1,\ldots,1,\frac{1}{(n+1)^2},\frac{1}{(n+2)^2},\ldots)$$ would work. But they didn't, because
$$\lVert t_n-t_0\rVert_{\infty} = \sup_{k\geq n}\Big|1-\frac{1}{k^2}\Big|=1 \nrightarrow 0$$ as $n\to \infty$ and thus $(t_n) \nrightarrow t_0$.</p>
<p>Also I thought about the sequences $$t_n = (1,1,\ldots,1,1+\frac{1}{n+1},1+\frac{1}{n+2},\ldots,1+\frac{1}{n+m},\ldots).$$ In this case $(t_n) \to t_0$ as $n \to \infty$, but none of the $t_n$'s is an element of $T(l^{\infty})$. So this sequence also is of no use.</p>
<p>Any help? Thanks a lot in advance.</p>
| vadim123 | 73,324 | <p>Consider $t_0=(\frac {1}{\sqrt{n}})_n$. This too is not in the image of $T$, since its preimage is $(\sqrt{n})_n$. Now define $t_m=(\frac{1}{\sqrt{1}},\frac{1}{\sqrt{2}},\ldots,\frac{1}{\sqrt{m}},\frac{\sqrt{m}}{m+1},\frac{\sqrt{m}}{m+2},\ldots)$, the image of $(\sqrt{1},\sqrt{2},\ldots,\sqrt{m},\sqrt{m},\sqrt{m},\ldots)$. But now $\|t_m-t_0\|_\infty=\sup_{n> m}\{|\frac{1}{\sqrt{n}}-\frac{\sqrt{m}}{n}|\}=\sup_{n> m}\{\frac{1}{\sqrt{n}}|\sqrt{\frac{m}{n}}-1|\}<\frac{1}{m}$, because $0<|\sqrt{\frac{m}{n}}-1|<1$</p>
|
367,885 | <p>Denote the set of all bounded sequences in $\mathbb{R}$ by $l^{\infty}$, endowed with the sup norm $\lVert \rVert _{\infty}$. Define a map $T:l^{\infty} \to l^{\infty}$ as follows:</p>
<p>$$(x_n)_n \mapsto \Big(\frac{x_n}{n}\Big)_n.$$</p>
<p>Then the question is:</p>
<blockquote>
<p>Show that the image of $l^{\infty}$ under $T$ is not a closed set in $l^{\infty}$.</p>
</blockquote>
<p><em><strong>My attempts:</em></strong></p>
<p>As it is enough to show the existence of a sequence $(t_n)$ of bounded sequences from $T(l^{\infty})$ which converges to a sequence outside of it, I first considered the constant sequence $$(1,1, \ldots, 1, \ldots) = t_0$$ $(\notin T(l^{\infty}))$ and tried to look for a sequence $(t_n)$ of sequences in $T(l^{\infty})$ which goes to it. The following are two (failed) attempts to construct such a sequence. </p>
<p>First I thought the sequences $$t_n = (1,1,\ldots,1,\frac{1}{(n+1)^2},\frac{1}{(n+2)^2},\ldots)$$ would work. But they didn't, because
$$\lVert t_n-t_0\rVert_{\infty} = \sup_{k\geq n}\Big|1-\frac{1}{k^2}\Big|=1 \nrightarrow 0$$ as $n\to \infty$ and thus $(t_n) \nrightarrow t_0$.</p>
<p>Also I thought about the sequences $$t_n = (1,1,\ldots,1,1+\frac{1}{n+1},1+\frac{1}{n+2},\ldots,1+\frac{1}{n+m},\ldots).$$ In this case $(t_n) \to t_0$ as $n \to \infty$, but none of the $t_n$'s is an element of $T(l^{\infty})$. So this sequence also is of no use.</p>
<p>Any help? Thanks a lot in advance.</p>
| Diego | 41,463 | <p>Let $t_0=(1,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{3}},\ldots)$ and $t_m=(1,\frac{1}{\sqrt{2}},\ldots,\frac{1}{\sqrt{m}},0,0,\ldots)$. Clearly $t_m\rightarrow t_0$, $t_m$ is in the image of $T$ and $t_0$ is not in the image of $T$.</p>
|
2,151,491 | <p>For which prime numbers $p$ does the decimal for $\frac{1}{p}$ have cycle length $p-1$? I started with some simple examples to find an idea to solve:</p>
<p>$\frac{1}{2}=0.5,\frac{1}{3}=\overline{3},\frac{1}{5}=0.2,\frac{1}{7}=0.\overline{142857},\frac{1}{11}=0.\overline{09},\frac{1}{13}=0.\overline{0769230}$</p>
<p>Here I only had $7$ Then I can't find idea for solution. Any hints?</p>
| Peter | 82,961 | <p>The primes upto $1000$ with period length $p-1$ are :</p>
<pre><code>? q=0;forprime(s=1,1000,if(gcd(s,10)==1,if(znorder(Mod(10,s))==s-1,print1(s," ")
;q=q+1;if(Mod(q,18)==0,print))))
7 17 19 23 29 47 59 61 97 109 113 131 149 167 179 181 193 223
229 233 257 263 269 313 337 367 379 383 389 419 433 461 487 491 499 503
509 541 571 577 593 619 647 659 701 709 727 743 811 821 823 857 863 887
937 941 953 971 977 983
?
</code></pre>
|
1,502,309 | <p>The initial notation is:</p>
<p>$$\sum_{n=5}^\infty \frac{8}{n^2 -1}$$</p>
<p>I get to about here then I get confused.</p>
<p>$$\left(1-\frac{3}{2}\right)+\left(\frac{4}{5}-\frac{4}{7}\right)+...+\left(\frac{4}{n-3}-\frac{4}{n-1}\right)+...$$</p>
<p>How do you figure out how to get the $\frac{1}{n-3}-\frac{1}{n-1}$ and so on? Like where does the $n-3$ come from or the $n-1$.</p>
| Michael Medvinsky | 269,041 | <p>\begin{align}
\sum_{n=5}^\infty \frac{8}{n^2 -1}
&=\sum_{n=5}^\infty \frac{8}{(n -1)(n+1)}\\
&=\sum_{n=5}^\infty \left(\frac{4}{n-1}-\frac{4}{n+1} \right)\\
&=\left(\frac 44 - \color{red}{\frac 46} \right)
+ \left(\frac 45 - \color{blue}{\frac 47} \right)
+ \left(\color{red}{\frac 46} - \color{green}{\frac 48} \right)
+ \left(\color{blue}{\frac 47} - \frac 49 \right)
+ \left(\color{green}{\frac 48} - \frac{4}{10} \right) \dots \\
& =\frac 44 +\frac{4}{5} \\
&=\frac{9}{5}
\end{align} </p>
|
1,502,309 | <p>The initial notation is:</p>
<p>$$\sum_{n=5}^\infty \frac{8}{n^2 -1}$$</p>
<p>I get to about here then I get confused.</p>
<p>$$\left(1-\frac{3}{2}\right)+\left(\frac{4}{5}-\frac{4}{7}\right)+...+\left(\frac{4}{n-3}-\frac{4}{n-1}\right)+...$$</p>
<p>How do you figure out how to get the $\frac{1}{n-3}-\frac{1}{n-1}$ and so on? Like where does the $n-3$ come from or the $n-1$.</p>
| ncmathsadist | 4,154 | <p>$$S = \sum_{n=5}^\infty \frac{8}{n^2 -1}=
\lim_{N\to\infty} \sum_{n=5}^N \frac{8}{n^2 -1}
= \lim_{N\to\infty} \sum_{n=5}^N \left\{ {4\over n-1} - {4\over n+1}\right\} = \lim_{N\to\infty} \sum_{n=5}^N {4\over n-1} - \sum_{n=5}^N {4\over n+1}$$</p>
<p>Now reindex to get </p>
<p>$$S = \lim_{N\to\infty} \sum_{n=4}^{N-1} {4\over n} - \sum_{n=6}^{N+1} {4\over n} = \lim_{N\to\infty} 1 + {4\over 5} - {4\over N}- {4\over N+1} = {9\over 5}$$</p>
|
3,256,881 | <p>How to prove that?
I try to use the comparasion test, but i don't know with that function compare.</p>
| XT Chen | 618,783 | <p>As Azif said this function is bounded. It's continuous and consequently integrable on <span class="math-container">$[δ,1]$</span> and |<span class="math-container">$\int_o^{\delta} \sin(x+\frac{1}{x}) \mathrm{d}x| \le \delta$</span> which converges to <span class="math-container">$0$</span> as <span class="math-container">$\delta \rightarrow 0$</span> so it is intégrable.</p>
|
3,501,052 | <p>I want to find the number of real roots of the polynomial <span class="math-container">$x^3+7x^2+6x+5$</span>.
Using Descartes rule, this polynomial has either only one real root or 3 real roots (all are negetive). How will we conclude one answer without doing some long process?</p>
| Claude Leibovici | 82,404 | <p>If you do not want to use the cubic discriminant, the easy solution has been proposed by Ross Millikan.</p>
<p>Consider that you lookf for the zero's of function
<span class="math-container">$$f(x)=x^3+7x^2+6x+5$$</span> for which
<span class="math-container">$$f'(x)=3x^2+14x+6 \qquad \text{and} \qquad f''(x)=6x+14$$</span> The first derivative cancels for
<span class="math-container">$$x_\pm=-\frac{1}{3} \left(7\pm\sqrt{31}\right)$$</span> and you have
<span class="math-container">$$f''(x_+)=2 \sqrt{31}>0\qquad \text{and} \qquad f''(x_-)=-2 \sqrt{31}<0$$</span> So, by the second derivative test, <span class="math-container">$x_+$</span> corresponds to a local minimum and <span class="math-container">$x_-$</span> to a local maximum.</p>
<p>Now
<span class="math-container">$$f(x_+)=\frac{443-62 \sqrt{31}}{27} >0\qquad \text{and} \qquad f(x_-)=\frac{443+62 \sqrt{31}}{27} >0$$</span> Since <span class="math-container">$f(x_-)>0$</span> there is only one real root which will be on the left of <span class="math-container">$x_-$</span>.</p>
<p>Based on this simply acquired information, we can even obtain an <em>estimate</em> of the solution writing
<span class="math-container">$$f(x)=f(x_-)+\frac 12 f''(x_-)(x-x_-)^2\implies x =x_--\sqrt{-2\frac {f(x_-)}{f''(x_-)}}$$</span>
<span class="math-container">$$x=-\frac{1}{3} \left(7+\sqrt{31}\right)-\sqrt{\frac{443+62 \sqrt{31}}{27 \sqrt{31}}}\approx -6.47905$$</span> while the exact solution is
<span class="math-container">$$x=-\frac{1}{3} \left(7+2 \sqrt{31} \cosh \left(\frac{1}{3} \cosh
^{-1}\left(\frac{443}{62 \sqrt{31}}\right)\right)\right)\approx -6.15745$$</span></p>
|
3,501,052 | <p>I want to find the number of real roots of the polynomial <span class="math-container">$x^3+7x^2+6x+5$</span>.
Using Descartes rule, this polynomial has either only one real root or 3 real roots (all are negetive). How will we conclude one answer without doing some long process?</p>
| 2'5 9'2 | 11,123 | <p>If you take <span class="math-container">$$f(x)=x^3+7x^2+6x+5$$</span> and subtract <span class="math-container">$5$</span>, you have <span class="math-container">$$g(x)=(x+6)(x+1)x$$</span> a cubic with roots at <span class="math-container">$-6$</span>, <span class="math-container">$-1$</span>, and <span class="math-container">$0$</span>. Note the graph goes down through the root at <span class="math-container">$-1$</span> and up through the root at <span class="math-container">$0$</span>.</p>
<p>To recover the original, you need to add <span class="math-container">$5$</span> to <span class="math-container">$g$</span>. It's hard to believe that adding <span class="math-container">$5$</span> will fail to pull the entire segment of the graph between <span class="math-container">$-1$</span> and <span class="math-container">$0$</span> up above the <span class="math-container">$x$</span>-axis. Indeed, if <span class="math-container">$x$</span> is in <span class="math-container">$[-1,0]$</span>, then <span class="math-container">$$\lvert g(x)\rvert=\lvert (x+6)(x+1)x\rvert >5\lvert(x+1)x\rvert\geq5\left(\frac14\right)$$</span> So <span class="math-container">$g$</span>'s local minimum is greater than <span class="math-container">$-\frac54$</span>. Adding <span class="math-container">$5$</span> definitely takes the local minimum up higher than the <span class="math-container">$x$</span>-axis. This leaves only one root for <span class="math-container">$f$</span>.</p>
|
646,010 | <p>So i kinda think i have figured this out, i'm not very good at math, and need a formula to figure out some stats for a game i'm playing.</p>
<p>I have a Weapon with a reload speed of X sec.. however, i also have a modifier attached, that will make the weapon reload faster by +Y%</p>
<p>i made this formula, mostly by guessing, as i got no clue what i am doing.</p>
<pre><code>100/(100+Y)*X
</code></pre>
<p>the results i am getting looks right to me, but is the formula ok?</p>
| TheOscillator | 95,893 | <p>Consider the Gamma function
$$\Gamma(s)=\int_{0}^{\infty}{t^{s-1}\cdot e^{-t}dt} ,s\in \mathbb{R^{+}}$$ now consider the substitution $t=iu^{2}$ which yields the following $$\Gamma(s)=2e^{i\frac{\pi}{2}s}\int_{i0}^{i\infty}{u^{2s-1}\cdot e^{-i u^{2}}du}=_{(1)} 2e^{i\frac{\pi}{2}s}\int_{0}^{\infty}{u^{2s-1}\cdot e^{-i u^{2}}du}$$
Primarly we are aiming for evaluating $$\frac{\partial^{2}}{\partial s^{2}}[\frac{\Gamma(s)}{2}\cdot e^{-i \frac{\pi}{2}s}]_{s=\frac{1}{2}}= 4\int_{0}^{\infty}{log^{2}(u) \cdot e^{-i u^{2}}du}$$ Simplifying the derivatives on the LHS followed by an identification of the imaginary part we have that $$\int_{0}^{\infty}{log^{2}(u) sin(u^{2})du}=\frac{1}{8}\sqrt{\frac{\pi}{2}}(\gamma +log4 -\frac{\pi}{2})^{2}$$</p>
<p>I would appreciate if anyone could help me justify the equality step (1)</p>
|
4,518,908 | <p>For sufficiently large integer <span class="math-container">$m$</span>, in order to prove</p>
<p><span class="math-container">$\frac{(m+1)}{m}<\log(m)$</span></p>
<p>is it sufficient to point out that</p>
<p><span class="math-container">$ \displaystyle\lim_{m \to \infty} \frac{(m+1)}{m}=1 $</span></p>
<p>while</p>
<p><span class="math-container">$ \displaystyle\lim_{m \to \infty} \log(m)=\infty $</span>?</p>
| Taladris | 70,123 | <p>Yes, <span class="math-container">$\forall x {\in} \mathbb{R} \left(P(x) \implies \left(\forall y {\in}\mathbb{R} \;P(x + y) \right)\right)$</span> implies that <span class="math-container">$\left(\forall x {\in}\mathbb{R}\; P(x)\right) \text{ or } \left(\forall x {\in}\mathbb{R}\;\lnot P(x)\right)$</span>.</p>
<p>To see why, we can remark that <span class="math-container">$\forall y {\in}\mathbb{R} \;P(x + y)$</span> is equivalent to <span class="math-container">$\forall t {\in}\mathbb{R} \;P(t)$</span>. Therefore, the original statement can be expressed as</p>
<p><span class="math-container">$$
\forall x {\in} \mathbb{R} \left(P(x) \implies \left(\forall t {\in}\mathbb{R} \;P(t) \right)\right)
$$</span></p>
<p>which implies</p>
<p><span class="math-container">$$
\left(\forall x {\in}\mathbb{R}\; P(x)\right) \text{ or } \left(\forall x {\in}\mathbb{R}\;\lnot P(x)\right)
$$</span></p>
<p>Either there is no real number <span class="math-container">$x$</span> such that <span class="math-container">$P(x)$</span> is true, or there is at least one, but in that case, <span class="math-container">$P(x)$</span> is true for every real number <span class="math-container">$x$</span>.</p>
<hr />
<p>There are however problems with your reasoning:</p>
<blockquote>
<p>First we show that <span class="math-container">$\exists x \in \mathbb{R}:\, P(x)$</span></p>
</blockquote>
<p>You can't do this, since there could be no <span class="math-container">$x$</span> such that <span class="math-container">$P(x)$</span> is true.</p>
<blockquote>
<p>We wish to show that <span class="math-container">$\forall x \in \mathbb{R},\, P(x)$</span></p>
</blockquote>
<p>This is not what you want to show (because it could be false). For an example of statement <span class="math-container">$P$</span> where <span class="math-container">$P(x)$</span> is always false but <span class="math-container">$\forall x {\in} \mathbb{R} \left(P(x) \implies \left(\forall y {\in}\mathbb{R} \;P(x + y) \right)\right)$</span> is true, consider <span class="math-container">$P(x)$</span>: "<span class="math-container">$x>x+1$</span>".</p>
<p>Indeed, let <span class="math-container">$x$</span> be a real number and assume that <span class="math-container">$P(x)$</span> is true (that is, <span class="math-container">$x>x+1$</span>). Then consider <span class="math-container">$y\in \mathbb R$</span>. We have <span class="math-container">$x>x+1$</span> so <span class="math-container">$x+y>x+y+1$</span>. This means that <span class="math-container">$P(x+y)$</span> is also true.</p>
|
467,301 | <p>I'm reading Intro to Topology by Mendelson.</p>
<p>The problem statement is in the title.</p>
<p>My attempt at the proof is:</p>
<p>Since $X$ is a compact metric space, for each $n\in\mathbb{N}$, there exists $\{x_1^n,\dots,x_p^n\}$ such that $X\subset\bigcup\limits_{i=1}^p B(x_i^n;\frac{1}{n})$. Let $K=\frac{2p}{n}$. Then for each $x,y\in X$, $x\in B(x_i^n;\frac{1}{n})$ and $y\in B(x_j^n;\frac{1}{n})$ for some $i,j=1,\dots,p$. Thus, $d(x,y)\leq\frac{2p}{n}$.</p>
<p>The approach I was taking is taking $K$ to be the addition of the diameters of each open ball in the covering for $X$, that way, for any two elements in $X$, the distance between them must be less than the overall length of the covering. Did I say this mathematically or are there holes I need to fill in?</p>
<p>Thanks for any help or feedback!</p>
| Dan Rust | 29,059 | <p>Suppose no such $K$ exists. Let $x$ be in $X$. Then for all $n\geq 1$, there exists an $N$ such that $B_N(x)\setminus B_n(x)\neq\emptyset$. let $\Lambda=\{B_n(x)\mid n\in\mathbb{N}\}$. $\Lambda$ is an open cover of $X$ but no finite subset of $\Lambda$ will cover $X$ because a finite subset has as union its largest ball, which we already know does not contain some element in some larger ball.</p>
|
982,938 | <p>I was doing the integration
$$
\int\frac{1}{(u^2+a^2)^2}du
$$
and I had a look at the lecturer's steps where I got stuck in the following step:
$$
\int\frac{1}{(u^2+a^2)^2}du=\frac{1}{2a^2}\left(\frac{u}{u^2+a^2}+\int\frac{1}{u^2+a^2}du\right).
$$
I guess it is integrating this by parts, but I could't see the trick. Could somebody help me with this please?</p>
<p>I had some work after I asked this question here. Here is my approach. </p>
<p>First note
$$
d\left(\frac{a^2}{u^2+a^2}\right)=-\frac{2a^2u}{(u^2+a^2)^2}du,
$$
then we have
$$
\int\frac{1}{(u^2+a^2)^2}du=\int\left(-\frac{1}{2a^2u}\right)\left(-\frac{2a^2u}{(u^2+a^2)^2}\right)du=\frac{1}{2a^2}\int\left(-\frac{1}{u}\right)d\left(\frac{a^2}{u^2+a^2}\right).
$$
Then integrate by parts, we have
$$
\int\left(-\frac{1}{u}\right)d\left(\frac{a^2}{u^2+a^2}\right)=-\frac{a^2}{u(u^2+a^2)}-\int\frac{a^2}{u^2(u^2+a^2)}du.
$$
But notice
$$
\frac{a^2}{u^2(u^2+a^2)}=\frac{1}{u^2}-\frac{1}{u^2+a^2},
$$
then we get
$$
\int\frac{a^2}{u^2(u^2+a^2)}du=-\frac{1}{u}-\int\frac{1}{u^2+a^2}du,
$$
then
$$
\int\left(-\frac{1}{u}\right)d\left(\frac{a^2}{u^2+a^2}\right)=-\frac{a^2}{u(u^2+a^2)}+\frac{1}{u}+\int\frac{1}{u^2+a^2}du.
$$
But
$$
\frac{1}{u}-\frac{a^2}{u(u^2+a^2)}=\frac{u}{u^2+a^2},
$$
then
$$
\int\frac{1}{(u^2+a^2)^2}du=\frac{1}{2a^2}\int\left(-\frac{1}{u}\right)d\left(\frac{a^2}{u^2+a^2}\right)=\frac{1}{2a^2}\left(\frac{u}{u^2+a^2}+\int\frac{1}{u^2+a^2}du\right).
$$</p>
<p>Is this approach okay? I am positive with this but not sure it is comprehensive. </p>
| Community | -1 | <p>The trick is using the substitution $u = a\tan x $, Then we have</p>
<p>$$ \int \frac{du}{(u^2 +a^2)^2 } = \int \frac{du}{(a^2 \tan^2 x + a^2)^2}= \int \frac{du}{(a \sec x)^2}$$</p>
<p>and $ du = a \sec^2 x dx $</p>
|
846,980 | <p>Assume we need to choose $k$ numbers out of $[1...n]$ so no two numbers are consecutive. I know the number of such combinations is given by$\binom {n-k+1}{k}$. Assume the numbers are given by $r_1<r_2...<r_k$ when the difference between $r_{i+1}$ and $r_i$ is always higher than $2$. Now, I want to restrict the combinations so that for $p=1$ $(r_{i+1},r_i)$ pairs, the difference between them will be at least $t=3$ (and not at least 2).</p>
<p><strong>Example:</strong> for $p=1$ and $t=3$ The number of combinations is given by $\binom {n-k+1}{k}$-$(n-2k+2)$ because we subtract all the combinations in which all the differences are exactly 2. </p>
<p>How can it be generalized to any $p$ and/or any $t$?</p>
<p>For $t=3$ I thought about counting the combinations by $\binom {n-k+1-p}{k}$ so we can insert the $p$ "dummies" in each combination but it seems it doesn't work and doesn't coincide with the example for $p=1$. </p>
<p>Thank you</p>
| Asinomás | 33,907 | <p>We will show the particular case when $p=k-1$. </p>
<p>Any selection of $k$ numbers from $\{1,2,\dots ,n-(t-1)(k-1)\}$ gives you a suitable combination when you add $t-1$ to the second smallest element, $2(t-1)$ to the third smallest element$\dots$ $(t-1)(k-1)$ to the largest alement. Using the inverse process (substracting instead of adding) on a suitable sequence will always give you a set in $\{1,2,3,\dots ,(n-(k-1)(t-1)\}$</p>
|
846,980 | <p>Assume we need to choose $k$ numbers out of $[1...n]$ so no two numbers are consecutive. I know the number of such combinations is given by$\binom {n-k+1}{k}$. Assume the numbers are given by $r_1<r_2...<r_k$ when the difference between $r_{i+1}$ and $r_i$ is always higher than $2$. Now, I want to restrict the combinations so that for $p=1$ $(r_{i+1},r_i)$ pairs, the difference between them will be at least $t=3$ (and not at least 2).</p>
<p><strong>Example:</strong> for $p=1$ and $t=3$ The number of combinations is given by $\binom {n-k+1}{k}$-$(n-2k+2)$ because we subtract all the combinations in which all the differences are exactly 2. </p>
<p>How can it be generalized to any $p$ and/or any $t$?</p>
<p>For $t=3$ I thought about counting the combinations by $\binom {n-k+1-p}{k}$ so we can insert the $p$ "dummies" in each combination but it seems it doesn't work and doesn't coincide with the example for $p=1$. </p>
<p>Thank you</p>
| Geoffrey Critzer | 139,257 | <p>Table[Table[Binomial[n - 2 (k - 1), k], {k, 0, Floor[(n + 2)/3]}], {n,
0, 10}] // Grid</p>
<p>Table[Table[
Length[Select[Tuples[{0, 1}, n],
FreeQ[#, {<strong>_, 1, 0, 1, <em></strong>}] && FreeQ[#, {<strong></em>, 1, 1, _</strong>}] &&
Count[#, 1] == k &]], {k, 0, Floor[n + 2]/3}], {n, 0,
10}] // Grid</p>
<p>nn = 10; CoefficientList[
Series[1/(1 - x)/(1 - y x^3/(1 - x)), {x, 0, nn}], {x, y}] // Grid</p>
<p>These three Mathematica codes return the same triangular array of numbers, T(n,k): the number of size k subsets of {1,2,...,n} such that the difference between any two elements is at least 3.</p>
|
396,713 | <p>I'm attempting to derive a formula for the sum of all elements of an arithmetic series, given the first term, the limiting term (the number that no number in the sequence is higher than), and the difference between each term; however, I am unable to find one that works. Here is what I have so far:</p>
<p>Let $a_0$ be the first term, $a_n$ be the last term, and $x$ be the difference between each term.</p>
<p>If we have the sequence $a_0, a_0 + x, a_0 + 2x ... a_n$, then we can add the first and last term, the second and second last term, etc., to quickly find the sum based on the number of terms. Thus, the sum of these terms is $n\frac{(a_n + a_0)}{2}$, where $n$ is the total number of terms.</p>
<p>The number of terms must be the number of times the first term was increased by $x$ plus one (to account for the first term), and so $n = \frac{(a_n - a_0)}{x} + 1$.</p>
<p>Thus, the sum is equal to $(\frac{(a_n - a_0)}{x} + 1)\frac{(a_n + a_0)}{2}$.</p>
<p>However, I am unable to integrate the limiting term in the place of $a_n$; any ideas for how to make this work?</p>
<p>In case my definition of a limiting term is ambiguous; an example would be if there was a set $3, 6, 9$; I'd like to be able to replace $a_n$ (which is $9$ in this case) with any number above $9$, and below $12$, and still get the same answer.</p>
| Sandkar | 232,349 | <p>This may be a bit late but I stumbled across this question myself and I thought about this not too long ago, so I might as well post what I managed to derive on here.</p>
<p>We know that a term, can be expressed thusly:</p>
<p>$$a_n = a_{n-1} + d$$
$$a_{n-1} = a_{n-2} + d$$
$$...$$
$$a_{n-k} = a_{n-k-1} + d$$</p>
<p>Where $a_n$ is the $n_{th}$ term, $d$ is the difference between two consecutive terms and $k$ is some integer constant. We can now substitute "lower" degrees (where $k$ is larger) of the recursive definition into the original one to get:</p>
<p>$$a_n = a_{n-2} +d + d$$
$$a_n = a_{n-3} +d + d + d$$
$$...$$</p>
<p>This goes on until $a_{n-k} = a_1$, we can now see that:</p>
<p>$$a_n = a_1 + d(n-1)$$</p>
<p>Because we went from $n-1$ to $1$ this means that we added together $d, $ $n-1$ number of times. Something quite elementary as well is that the sum of the series is all of the terms added together:</p>
<p>$$S = \sum_{i=1}^{n}a_i = \sum_{i=1}^{n}(a_1 + d(n-i))$$</p>
<p>Break out the terms:</p>
<p>$$S = \sum_{i=1}^{n}a_1 + \sum_{i=1}^{n}d(n-i)$$</p>
<p>We can see that when $i = n$ then $n - i = 0$, this means that we can just change the second summations upper bound from $n$ to $n-1$:
$$S = \sum_{i=1}^{n}a_1 + \sum_{i=1}^{n-1}d(n-i)$$
$$S = na_1 + \sum_{i=1}^{n-1}dn - \sum_{i=1}^{n-1}di$$
The sum for all natural numbers up to and including $n-1$ is $\frac{n(n-1)}{2}$, use this on the third summation:</p>
<p>$$S = na_1 + dn(n-1) - \frac{dn(n-1)}{2}$$
$$S = na_1 + \frac{2dn(n-1) - dn(n-1)}{2}$$
$$S = na_1 + \frac{dn(n-1)}{2}$$</p>
<p><em>Q.E.D</em> </p>
|
42,617 | <p>Let $f(x,y)$ define a surface $S$
in $\mathbb{R}^3$ with a unique local minimum at $b \in S$.
Suppose <a href="http://en.wikipedia.org/wiki/Gradient_descent" rel="noreferrer">gradient descent</a> from any start point $a \in S$
follows a geodesic on $S$ from $a$ to $b$.
(<b>Q1</b>.)
What is the class of functions/surfaces
whose gradient-descent paths are geodesics?</p>
<p>Certainly if $S$ is a surface of revolution
about a $z$-vertical line through $b$,
its "meridians"
are geodesics, and these would be the paths followed
by gradient descent down to $b$.
So the class of surfaces includes surfaces of
revolution. But surely it is wider than that?</p>
<p>(<b>Q2</b>.)
One could ask the same question about paths followed by
<a href="http://en.wikipedia.org/wiki/Newton%27s_method_in_optimization" rel="noreferrer">Newton's method</a>, which in general are different from gradient-descent
paths, as indicated in this Wikipedia image:
<br />
<img src="https://upload.wikimedia.org/wikipedia/commons/thumb/d/da/Newton_optimization_vs_grad_descent.svg/220px-Newton_optimization_vs_grad_descent.svg.png" alt="Newton's vs. Gradient">
<em>Gradient descent: green.
Newton's method: red.</em>
<br /></p>
<p>(<b>Q3</b>.) These questions make sense in arbitrary dimensions,
although my primary interest is for surfaces in $\mathbb{R}^3$.</p>
<p>Any ideas on how to formulate my question as constraints on $f(\;)$,
or pointers to relevant literature,
would be appreciated. Thanks!</p>
| Joseph O'Rourke | 6,094 | <p>Here is a function $f(x,y)$ which is 0 inside the square $C=[\pm1,\pm1]$,
and outside that square
has value equal to the Euclidean distance $d( p, C )$ from $p=(x,y)$ to the boundary of $C$.
[I am trying to follow Pietro Majer's suggestion, as far as I understand it.]
It is not a surface of revolution
(but it is centrally symmetric).
Are its gradient descent paths geodesics?
I think so...
<br />
<img src="https://i.stack.imgur.com/mwvrR.jpg" alt="Function, Contours">
<br />
<em>Left above</em>: $f(x,y)$. <em>Right above</em>: Level sets of $f$.
<em>Below</em>: $\nabla f$.
<br />
<img src="https://i.stack.imgur.com/6Sp7s.jpg" alt="Gradient">
<br />
And here (<em>below</em>) is a closeup of the function defined using squared distance
$[d( p, C )]^2$, as per Willie Wong's suggestion:
<br /><img src="https://i.stack.imgur.com/A5gpI.jpg" alt="WW"></p>
|
280,156 | <p>I have a code as below:</p>
<pre><code>countpar = 10;
randomA = RandomReal[{1, 10}, {countpar, countpar}];
randomconst = RandomInteger[{0, 1}, {countpar, 1}];
For[i = 1, i < countpar + 1, i++,
If[randomconst[[i, 1]] != 0,
randomA[[All, i]] = 0.; randomA[[i, All]] = 0.;
randomA[[i, i]] = 1;
];
];
</code></pre>
<p>The problem is when I have changed countpar, lets say, i.e, countpar=1000, then the computational time of the for loop increases dramatically. Is there a way to decrease this time from an expert eye ?</p>
<p>Best Regards,</p>
<p>Ahmet</p>
| E. Chan-López | 53,427 | <p>Following Syed's idea, another way to do this is to use <code>FoldList</code>:</p>
<pre><code>With[{x = 2, y = 1, n = 2}, FoldList[(#^(x - 1) + n)/(#^(x - 1) + #^(x - 2)) &, y, Range[6]]]
(*{1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169}*)
</code></pre>
<p>Also, you can use <code>FixedPointList</code>:</p>
<pre><code>With[{x = 2, y = 1, n = 2}, FixedPointList[(#^(x - 1) + n)/(#^(x - 1) + #^(x - 2)) &, y, 6]]
(*{1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169}*)
</code></pre>
|
2,563,966 | <p>I'm reading a proof in a linear algebra book. It mentions
$$p(x) -p(c)= (x - c) h(x),$$
where $c$ is a constant, and $p(x)$ and $h(x)$ are polynomials.</p>
<p>Can we always factor $p(x) - p(c)$ in this way? </p>
<p>Please give a proof.</p>
| Chirantan Chowdhury | 337,567 | <p>Let $ p(x) = \sum_{i=0}^n p_ix^i $ Now $$p(x) - p(a) = \sum_{i=o}^n p_i(x^i-a^i)$$ Use the formula $$x^i-a^i= (x-a)( \sum_{j=0}^{i-i}x^ja^{i-1-j} ) $$(This comes from a geometric series) Thus now you can see from each term a factor of $x-a$ comes out . Thus it is proved.</p>
|
2,555,463 | <p>Given a line $l$ and two points $p_1$ and $p_2$, identify the point $v$ which is equidistant from $l$, $p_1$, and $p_2$, assuming it exists.</p>
<p>My idea is to: (1) identify the parabolas containing all points equidistant from each point and the line, then (2) intersect these parabolas. As $v$ is equidistant from all three and each parabola contains all points equidistant from $l$ and each point, the intersection of these parabolas must be $v$. However, I have had no luck in finding a way to compute, much less represent, these parabolas.</p>
| Nominal Animal | 318,422 | <p>I personally prefer to attack such problems first using basic vector algebra, mostly because they tend to be easier to implement for numerical computations.</p>
<p>In other words, this answer assumes the underlying problem to be solved, is to find and implement a stable numerical algorithm for solving specific problems of this type; and not something like a geometric exercise or educational investigation.</p>
<hr>
<p>First, we want to simplify the situation. The obvious operation is to transform the system so that the line $l$ is on the $x$ axis.</p>
<p>Let's define the line $l$ in vector form:
$$\vec{l}(t) = \vec{l}_0 + t \hat{l}_n \tag{1}\label{NA1}$$
where $\vec{l}_0 = (x_0 ,\, y_0)$ is any point the line passes through, and $\hat{l}_n = (x_n ,\, y_n)$ is an unit vector ($x_n^2 + y_n^2 = 1$).</p>
<p>If we define rotation matrix $\mathbf{R}$,
$$\mathbf{R} = \left [ \begin{array}{cc} x_n & y_n \\ y_n & -x_n \end{array} \right ] \tag{2}\label{NA2}$$
then the transformation we want is
$$\vec{p}' = \mathbf{R}\left(\vec{p} - \vec{l}_0\right)\tag{3a}\label{NA3a}$$
i.e. point $\vec{p} = (x, y)$ becomes
$$\left\lbrace\begin{array}{l}
x' = x_n ( x - x_0 ) + y_n ( y - y_0 ) \\
y' = y_n ( x - x_0 ) - x_n ( y - y_0 )
\end{array}\right.\tag{3b}\label{NA3b}$$</p>
<p>The inverse operation is $$\vec{p} = \vec{l}_0 + \mathbf{R}^T \vec{p}'$$
and because $\mathbf{R}^T = \mathbf{R}$ in this particular case, we can write it as
$$\left\lbrace\begin{array}{l}
x = x_0 + x_n x' + y_n y' \\
y = y_0 + y_n x' - x_n y' \end{array}\right.\tag{3c}\label{NA3c}$$</p>
<hr>
<p>Next, we can consider this much simpler situation.</p>
<p>Let's say the two points (as transformed by $\eqref{NA3b}$) are $\vec{p}_1 = ( x_1 ,\, y_1 )$ and $\vec{p}_2 = ( x_2 ,\, y_2 )$.</p>
<p>The requirement for the point $\vec{p} = ( x ,\, y )$ we seek is that it is as far from $\vec{p}_1$ and $\vec{p}_2$ as it is from the line $l$, i.e. from $y = 0$.
$$\begin{cases}
\lvert y \rvert = \sqrt{ (x - x_1)^2 + (y - y_1)^2 } \\
\lvert y \rvert = \sqrt{ (x - x_2)^2 + (y - y_2)^2 } \end{cases}$$
Because both sides are nonnegative, we can square them, to get the pair of equations in a more tractable form,
$$\begin{cases}
y^2 = (x - x_1)^2 + (y - y_1)^2 \\
y^2 = (x - x_2)^2 + (y - y_2)^2 \end{cases} \iff \begin{cases}
(x - x_1)^2 + y_1^2 - 2 y y_1 = 0 \\
(x - x_2)^2 + y_2^2 - 2 y y_2 = 0 \end{cases} \tag{4}\label{NA4}$$
After finding the solution or solutions $(x, y)$, use $\eqref{NA3c}$ to transform them back to the original coordinate system.</p>
<hr>
<p>Let's consider the numerical solutions to $\eqref{NA4}$, and the cases where there is no solution at all. (I shall quietly assume you are aware of the pitfalls when using floating-point numbers; especially allowing for rounding error, <em>"machine epsilon"</em>, when comparing values for equality.)</p>
<ol>
<li><p>If $y_1 y_2 \lt 0$, there can be no result, because the given points are on different sides of the line: the distance from the tentative result point to at least one of the given points is always greater than the distance from the tentative result point to the line.</p></li>
<li><p>If $y_1 \ne y_2$, but $y_1 = 0$ or $y_2 = 0$ (or $y_1 y_2 = 0$), there can be no result, because only one of the given points is on the line $l$, and the common distance cannot be zero and nonzero at the same time.</p></li>
<li><p>If $y_1 = y_2 = 0$ but $x_1 \ne x_2$, there is no solution, because the given points are separate but both on the line $l$, and the common distance cannot be both zero and nonzero at the same time.</p></li>
<li><p>If $y_1 = y_2 = 0$ and $x_1 = x_2$, the given points are at the same point on the line $l$, so the common distance is zero. The result is the position of the given points:
$$\begin{cases}x = x_1 = x_2 \\ y = y_1 = y_2 = 0 \end{cases}$$</p></li>
<li><p>If $y_1 = y_2 \ne 0$ and $x_1 = x_2$, the given points are the same, and the result is at midway between the given points and the line:
$$\begin{cases}x = x_1 = x_2 \\ y = \frac{y_1}{2} = \frac{y_2}{2} \end{cases}$$</p></li>
<li><p>If $y_1 = y_2 \ne 0$ and $x_1 \ne x_2$, there is a single solution:
$$\begin{cases}x = \frac{x_1 + x_2}{2} \\ y = \frac{y_1}{2} + \frac{(x_2 - x_1)^2}{8 y_1} \end{cases}$$</p></li>
<li><p>If $y_1 \ne y_2$, $y_1 y_2 \gt 0$, and $x_1 = x_2$, there given points are on the line perpendicular to line $l$, on the same side of line $l$, and there is a pair of solutions:
$$\begin{cases}
x = x_1 \pm \sqrt{y_1 y_2} \\
y = \frac{y_1 + y_2}{2} \end{cases}$$</p></li>
<li><p>Otherwise, $y_1 \ne y_2$, $y_1 y_2 \gt 0$, and $x_1 \ne x_2$, and there are two solutions:
$$\begin{cases}
x = \frac{x_1 y_2 - x_2 y_1 + \sqrt{y_1 y_2 \left ( (x_2 - x_1)^2 + (y_2 - y_1)^2 \right )}}{y_2 - y_1} \\
y = \frac{(y_1 + y_2) \left ( (x_2 - x_1)^2 + (y_2 - y_1)^2 \right ) - (x_2 - x_1)\sqrt{y_1 y_2 \left ( (x_2 - x_1)^2 + (y_2 - y_1)^2 \right )}}{(y_2 - y_1)^2 } \end{cases}$$
and
$$\begin{cases}
x = \frac{x_1 y_2 - x_2 y_1 - \sqrt{y_1 y_2 \left ( (x_2 - x_1)^2 + (y_2 - y_1)^2 \right )}}{y_2 - y_1} \\
y = \frac{(y_1 + y_2) \left ( (x_2 - x_1)^2 + (y_2 - y_1)^2 \right ) + (x_2 - x_1)\sqrt{y_1 y_2 \left ( (x_2 - x_1)^2 + (y_2 - y_1)^2 \right )}}{(y_2 - y_1)^2 } \end{cases}$$
If you need to choose only one of the solutions, I recommend choosing the one with minimum $y$ in magnitude, $\lvert y \rvert$, as that solution corresponds to the smaller common distance.</p></li>
</ol>
<p>If there is a solution, you can use $\eqref{NA3c}$ to convert back to the original coordinate system.</p>
<p>Note that the above cases are not in recommended implementation order; if you draw the cases into a diagram, you'll see you have a simple decision tree. As conditionals tend to be slow on current hardware, implementing a decision tree (to find the solution in fewest average and maximum conditional tests) is recommended.</p>
<p>Also note that I obtained the above solutions using free and open source Sagemath, but there may be transcription errors. If you find an error, or a suspicious point, please let me know in a comment so I can verify and fix.</p>
|
2,178,395 | <p>In how many ways can the letters of the word CHROMATIC be arranged,</p>
<p>find the probability that the string of letters begins with the letter M</p>
<p>I don't understand how to single out M so the possibilities would only begin with M?</p>
| Graham Kemp | 135,106 | <p>There are two methods.</p>
<p>As joeb states, you can compare the size of the favoured event to the sample space by <a href="https://math.stackexchange.com/questions/2176973/in-how-many-ways-can-the-letters-of-the-word-chromatic-be-arranged">counting arrangements of all the letters</a>.</p>
<p>Alternatively, only the placement of the letter $M$ is relevant. The arrangement of the other eight letters among themselves is independent of this.</p>
<p>If you are not <em>required</em> to count arrangements before finding the probability, the second method is much more direct.</p>
<hr>
<p>Method 1(a): Count ways to place $M$ in first place, then arrange the remaining $8$ symbols of which exactly $2$ are identical. Compare to the count of ways to arrange all $9$ symbols of which exactly $2$ are identical.</p>
<p>Method 1(b): Count ways to place $M$ in first place, then arrange the remaining $8$ individual symbols (treating them as distinct). Compare to the count of ways to arrange all $9$ individual symbols (treating them likewise).</p>
<p>Method 2: Count the ways to place $M$ in first place. Compare to ways to place $M$ in any of the $9$ positions. The remaining letters are irrelevant.</p>
|
18,493 | <p>Is there a way to split an external file using Mathematica? (Other than importing it, dividing it internally and exporting the parts)</p>
<p>For instance like some File Split utility do.</p>
| Sjoerd C. de Vries | 57 | <p>From the <a href="http://reference.wolfram.com/mathematica/howto/CreateAComputableDocumentFormatFile.html" rel="nofollow">docs</a>: "Because CDF Player cannot load custom data at runtime, you must ensure that all necessary information is embedded within the interactive elements of your .cdf file. This can be done with either Initialization or SaveDefinitions; both are options to Manipulate."</p>
|
18,493 | <p>Is there a way to split an external file using Mathematica? (Other than importing it, dividing it internally and exporting the parts)</p>
<p>For instance like some File Split utility do.</p>
| Murta | 2,266 | <p>If you use Enterprise Mathematica edition, or some kind of signed version, you can do what you want.
In this example I use the CDF with package to hide a database connection, so the package can be distributed without someone discover your connection user name and password.</p>
<p>There is the package, that should be saved together with the CDF file (same directory):</p>
<pre><code>BeginPackage["connectionPack`",{"DatabaseLink`"}]
marcheConn::usage="myConn[] connect to database";
Begin["`Private`"]
myConn[] :=
OpenSQLConnection[
JDBC["Microsoft SQL Server(jTDS)", "myIPAddress"],
"Username" -> "userName",
"Password" -> "password"
];
End[]
EndPackage[]
</code></pre>
<p>Here is the code for the test CDF</p>
<pre><code>testCDF=DynamicModule[{},
Column[{
Button["Random Sample from Database",output=test[]],
Button["Current directory",output=Directory[]],
Dynamic[output//TableForm]
}]
,Initialization:>(
SetDirectory[NotebookDirectory[]];
Needs["connectionPack`",NotebookDirectory[]<>"connectionPack.m"];
test[]:=Module[{conn,r},
conn=connectionPack`myConn[];
r=SQLExecute[conn,"select * from someTable"];
CloseSQLConnection[conn];
RandomSample[r,5]
];
)
]
</code></pre>
<p>Now you can export this using:</p>
<pre><code>CDFDeploy["test.cdf", testCDF, "Method" -> "Standalone", "Target" -> "PlayerPro"]
</code></pre>
<p><strong>Important points:</strong></p>
<ul>
<li>When using <code>Needs</code> inside <code>Initialization</code> use the full package directory in the second argument.</li>
<li>When using the functions from your package inside <code>Initialization</code> you should use it full name as connectionPack`myConn[]</li>
<li>Just for completeness of this example. If you wand to protect you connection string you could Encode and use <code>SetAttributes[myConn, {ReadProtected,Locked}];</code>. This way user can not retrive it using <code>??myConn</code>.</li>
</ul>
<h2>update</h2>
<p>I discovered that the encryption do not solve the conn security.
When you create the connection, you can retrieve all connections parameters just executing conn. The only advantage is that the user needs Mathematica to do this, and user and pass will not be in plain text.</p>
|
1,840,485 | <p>I am an undergraduate really passionate about the mathematics and microbiology. I have few big problems in learning which I would like to seek your advice. </p>
<p>Whenever I study mathematical books (Rudin, Hoffman/Kunze, etc.), I always try to prove every theorem, lemma, corollary, and their relationships in the book. Unfortunately, that determination has been demanding huge time consumption; sometimes, it takes me days to fully understand and able to prove materials in the few pages of book. I am willing to devote my time to understand the topics, but I also wanted to devote time to my undergraduate research projects and other courses. Recently, I started to depend a lot more to the proofs presented in books and websites (like MSE), which has been causing a huge guilt and fear that I am not making the knowledge into my own.</p>
<p>Despite my effort to prove/solve every problem per chapter, I found myself to skip some of the problems and move on to the next chapter, which resulted huge fear as that means I did not fully understand the materials..</p>
<ul>
<li>How do you read the mathematics books and make knowledge on your own?</li>
<li>Is it absolutely recommended to prove everything and solve every problems in the book? </li>
<li>Also is it recommended to devote more time to the problems than exposition preceding the problems? I found myself devoting a lot time to the actual expositions in the book as I like to play around with definitions and theorems, try to come up with my own ideas, and formulate my own problems (I actually found that making my own problems is much more fun than problems presented in the book).</li>
</ul>
| Will R | 254,628 | <p><strong>Overall answer: no.</strong></p>
<p>I struggled with the same problem as you for quite a long time and, in hindsight, I think I could have spent my time more wisely. Here are my current general guidelines at the time of this post. They may or may not work for you. The overall philosophy I employ is that exercises are usually there to get you comfortable with the material: you are probably usually expected to remember the theorems and (mostly) forget the exercises once you have done them. I expound on this below.</p>
<p><strong>How do you read the mathematics books and make knowledge on your own?</strong></p>
<p>Nowadays I spend more time thinking about the material in the text than doing exercises. I treat the theorems as problems and try to see how far I can get in proving them before reading the proof. Importantly, I don't spend all day doing this! A good strategy might be to set yourself a goal, such as "I want to get to such-and-such page by the end of the week", and pace yourself accordingly; if you spend thirty to forty-five minutes thinking hard about a theorem and having no ideas, perhaps take a peek at the proof and continue from there (alternatively, skip that proof and come back to it later).</p>
<p>If the book has examples in the text, read and understand as many as you can. Otherwise, be sure to allocate generous amounts of time towards coming up with your own examples. Note that this is not necessarily easy, and is arguably the most important stage.</p>
<p>Finally, if the book has exercises, be sure to do at least some of them and try to understand the general idea of the rest. Don't be afraid to ask and try to answer your own questions which might be inspired by some of the exercises.</p>
<p>One downside to this method is that it doesn't work for all books and all subject matters. Some authors place key theorems in the exercises and proceed to use them later, expecting you to have proven them yourself. The hope is that for each topic there exists a book for which this method is not useless.</p>
<p>Another downside is that, in rare cases, too many of the exercises are interesting (this appears to be the case with Rudin, among other books)! In this case it's definitely up to you how much time you spend on the exercises. Allocate time according to how interesting you find the exercises and how comfortable you are with the material in the text.</p>
<p><strong>Is it absolutely recommended to prove everything and solve every problems in the book?</strong></p>
<p>If you <em>can</em> do this and still lead a comfortable life, then by all means do so (if it doesn't hurt, then it can only help, right?). Unfortunately, attempting to do this will probably make life less than comfortable for you, so I would advice against being this extreme.</p>
<p>However, the point of this "advice" is that you should get as comfortable with the theorems and the examples as much as you possibly can, and this <em>is</em> good advice. I would advise just thinking about stuff as much as possible. Thinking about maths in the shower, on the way to the shops, while cooking dinner, etc. will get you used to thinking about the topics that interest you.</p>
<p><strong>Also is it recommended to devote more time to the problems than exposition preceding the problems?</strong></p>
<p>This heavily depends on both the book and the reader. However, if you prefer to come up with your own examples and fiddle around with the theorems, and if the book you are working from is not expecting you to prove key theorems in the exercises, then I would say this approach is a good substitute for doing exercises.</p>
|
1,107,250 | <p><img src="https://i.stack.imgur.com/ILg7L.png" alt="enter image description here"></p>
<p>In above lemma, why $|a'| \leq 1$ still holds? I didn't see how it relates to "algebraic conjugate of a root of unity is also a root of unity", since $a$ is the sum of unity.</p>
<p>(definition of algebraic conjugate: <a href="http://planetmath.org/algebraicconjugates" rel="nofollow noreferrer">http://planetmath.org/algebraicconjugates</a>)</p>
| Empy2 | 81,790 | <p>The functions which are $f(-1)=-1,f(1)=1$ and all other derivatives are zero are</p>
<p>$$C_n\int_0^x (1-t^2)^n dt$$ </p>
<p>It follows that, pointwise, they reach constants as $n\to\infty$ because the bulk of $(1-t^2)^n$ narrows towards $x=0$.</p>
<p>The OP's functions are </p>
<p>$$D_n\int_0^x (t-t^2)^n \, dt$$ </p>
<p>Let $I_n=\int_0^1(t-t^2)^n \, dt=1/D_n$. Apply integration by parts, but let the integral of $1$ be $t-1/2$ instead of $t$, to use the symmetry between $0$ and $1$.</p>
<p>$$
\begin{align}
& I_n=\left.(t-1/2)(t-t^2)^n\vphantom{\frac11}\right|_0^1-\int_0^1(t-1/2)n(t-t^2)^{n-1}(1-2t) \, dt\\
= {} & 0-n\int_0^1(-2t^2+2t-1/2)(t-t^2)^{n-1}dt\\
= {} & -2nI_n+n/2I_{n-1}\\[6pt]
& (1+2n)I_n=(n/2)I_{n-1}\\[6pt]
& (2n+1)D_{n-1}=(n/2)D_n\\[6pt]
& (2n+1)(2n)D_{n-1}=n^2D_n
\end{align}
$$</p>
<p>Now David's formula for $D_n$ follows by induction.</p>
|
2,356,553 | <p>Why can't we write a simple equation where if we give the value of $x$ as input, we get the value of $\sin(x)$ as output?</p>
<p>By simple, I mean an equation involving just addition, division, subtraction and multiplication and exponentiation and keeping it in the realm of real numbers.</p>
<p>And I'm not necessarily asking for an equation, I'm asking if one does not exist, why is it so?</p>
<p>Even an intuitive explanation would work.</p>
<p>(It is my first question here so sorry for not being rigorous enough)</p>
| Daniel | 221,735 | <p>It can $sin(x)=\sum^{\infty}_{k=0}\frac{-(1)^kx^{2k+1}}{2k+1}$, for further information look at <a href="https://en.wikipedia.org/wiki/Taylor_series" rel="nofollow noreferrer">Taylor Series</a></p>
|
2,356,553 | <p>Why can't we write a simple equation where if we give the value of $x$ as input, we get the value of $\sin(x)$ as output?</p>
<p>By simple, I mean an equation involving just addition, division, subtraction and multiplication and exponentiation and keeping it in the realm of real numbers.</p>
<p>And I'm not necessarily asking for an equation, I'm asking if one does not exist, why is it so?</p>
<p>Even an intuitive explanation would work.</p>
<p>(It is my first question here so sorry for not being rigorous enough)</p>
| Pedro | 70,305 | <p>Because $\sin(x)$ is not an <a href="https://en.wikipedia.org/wiki/Algebraic_function" rel="nofollow noreferrer"><em>algebraic function</em></a>. It is instead a <a href="https://en.wikipedia.org/wiki/Transcendental_function" rel="nofollow noreferrer"><em>transcendental function</em></a>.</p>
<p><a href="https://math.stackexchange.com/a/1670527/70305">Here</a> there is a proof.</p>
|
192,821 | <p>I am using <a href="https://reference.wolfram.com/language/ref/TransformedField.html" rel="noreferrer"><code>TransformedField</code></a> to convert a system of ODEs from Cartesian to polar coordinates:</p>
<pre><code>TransformedField[
"Cartesian" -> "Polar",
{μ x1 - x2 - σ x1 (x1^2 + x2^2), x1 + μ x2 - σ x2 (x1^2 + x2^2)},
{x1, x2} -> {r, θ}
] // Simplify
</code></pre>
<p>and I get the result</p>
<pre><code>{r μ - r^3 σ, r}
</code></pre>
<p>but I am pretty sure that the right answer should be</p>
<pre><code>{r μ - r^3 σ, 1}
</code></pre>
<p>Where is the error?</p>
| Moo | 36,141 | <p>We can define our own functions.</p>
<blockquote>
<p>From <span class="math-container">$x',y'$</span> to <span class="math-container">$r',\theta'$</span>, we derive:
<span class="math-container">$$
r' = \left(\sqrt{x^2 +y^2} \right)'
= \frac{(x^2 +y^2)'}{2
\sqrt{x^2 +y^2}}=\frac{xx' +yy'}{r}
$$</span>
and
<span class="math-container">$$
\theta' = \left(\arctan \frac{y}{x} \right)'
= \frac{(y/x)'}{1+(y/x)^2} = \frac{y' x -x' y}{r^2}.
$$</span></p>
</blockquote>
<p>First, we define</p>
<pre><code>rdot[x1_, x2_] := (x1 (μ x1 - x2 - σ x1 (x1^2 + x2^2)) + x2 (x1 + μ x2 - σ x2 (x1^2 + x2^2)))/r
</code></pre>
<p>We now make the substitution and simplify</p>
<pre><code>rdot[r Cos[t], r Sin[t]] // FullSimplify
</code></pre>
<p>This yields (matches Mathematica)</p>
<p><span class="math-container">$$r' = \mu r-r^3 \sigma$$</span></p>
<p>We now do the same for the other</p>
<pre><code>thetadot[x1_,x2_]:=(x1 (x1+μ x2-σ x2 (x1^2+x2^2)) - x2(μ x1-x2-σ x1 (x1^2+x2^2)))/r^2
</code></pre>
<p>We now make the substitution and simplify</p>
<pre><code>thetadot[r Cos[t], r Sin[t]] // FullSimplify
</code></pre>
<p>This yields (does not match Mathematica, but see accepted answer)</p>
<p><span class="math-container">$$\theta'= 1$$</span></p>
<p>I have asked this question before on this site in two different ways and have never gotten an answer that resolves the matter, but that could just be my denseness as the accepted answer now shows! </p>
<p><strong>Update</strong></p>
<p>I have received a response from Wolfram Support and wanted to post it as others may find it as helpful as I did.</p>
<p>Thank you for contacting Wolfram Technical Support.</p>
<p>I want to highlight a couple more pieces of information that you might find useful.</p>
<p>There is a more comprehensive tutorial on how Mathematica handles coordinate transformations, and particularly how it handles basis transformations for vectors, available at </p>
<p><a href="https://reference.wolfram.com/language/tutorial/ChangingCoordinateSystems.html" rel="nofollow noreferrer">https://reference.wolfram.com/language/tutorial/ChangingCoordinateSystems.html</a></p>
<p>Under the section "Relating Orthonormal Bases", the tutorial highlights that the transformation of vectors is given by an orthonormal rotation matrix. In particular, this guarantees that a vector will have the same norm in any coordinate system. So, the vector {0,1,0} in the {r, th, phi} coordinate system must have a norm 1 in the {x,y,z} coordinate system. </p>
<p>In the question you posted on StackExchange, the norm of the original vector is </p>
<pre><code>r Sqrt[1 + (\[Mu] - r^2 \[Sigma])^2]
</code></pre>
<p>(after the change of variables from {x1,x2} to {r,theta} has been made).</p>
<p>This highlights that the proposed solution</p>
<pre><code>{r ? - r^3 ?, 1}
</code></pre>
<p>cannot be correct, as it has a different norm. On the other hand,</p>
<pre><code>{r ? - r^3 ?, r}
</code></pre>
<p>has the same norm. </p>
<p>Please let me know if you have any further questions.</p>
<h2>Sincerely,</h2>
<p>Wolfram Technology Group
<a href="http://www.wolfram.com/support/" rel="nofollow noreferrer">http://www.wolfram.com/support/</a></p>
|
139,417 | <p>I have a polygon defined by a list of nodes (x,y). I want to cut the polygon by a horizontal line at position y = a and get the new polygon above the position y = a. I am using the RegionIntersect function, but it seems very slow if I want to combine the function with Manipulate function as well. Is there any way to improve my code to get better speed?</p>
<pre><code>R2 = Polygon[{{0, 0}, {300, 0}, {300, 500}, {0, 750}}] ;
Manipulate[
R1 = ImplicitRegion[{0 <= x <= 300, a <= y <= 700}, {x, y} ];
R3 = RegionIntersection[R1, R2];
RegionPlot[R3], {a, 1, 499}]
</code></pre>
| Jason B. | 9,490 | <p>It's a lot faster to find the intersection with a <code>Polygon</code> than with an <code>ImplicitRegion</code>. In your case you can write your region succinctly as a simple <code>Polygon</code></p>
<pre><code>DynamicModule[{R1, R2, R3},
R2 = Polygon[{{0, 0}, {300, 0}, {300, 500}, {0, 750}}];
Manipulate[
R1 = Polygon[{{0, a}, {300, a}, {300, 700}, {0, 700}}];
R3 = RegionIntersection[R1, R2];
RegionPlot[R3,
PlotRange -> MinMax /@ Thread[First@R2]],
{a, 1, 499}]
]
</code></pre>
<p><a href="https://i.stack.imgur.com/HOu4N.gif" rel="noreferrer"><img src="https://i.stack.imgur.com/HOu4N.gif" alt="enter image description here"></a></p>
<p>If you want it faster, you can skip using <code>RegionPlot</code>, since your <code>Region</code> is just another <code>Polygon</code>. Something like this gives a good approximation of <code>RegionPlot</code></p>
<pre><code>DynamicModule[{R1, R2, R3},
R2 = Polygon[{{0, 0}, {300, 0}, {300, 500}, {0, 750}}];
Manipulate[
R1 = Polygon[{{0, a}, {300, a}, {300, 700}, {0, 700}}];
R3 = RegionIntersection[R1, R2];
Graphics[{
RGBColor[0.36, 0.5, 0.7],
EdgeForm[Directive[Thick, RGBColor[0.36, .5, .7]]],
Opacity[0.4],
R3},
PlotRange -> MinMax /@ Thread@First@R2,
Frame -> True, AspectRatio -> 5/6],
{a, 1, 499}]
]
</code></pre>
|
4,069,185 | <p>I got stuck here:</p>
<p>The probability that it will rain today is that it did not rain in the previous two days is <span class="math-container">$0.3$</span>, but if it rained in one of the last two days then the probability of rain today is <span class="math-container">$0.6$</span>.</p>
<p><span class="math-container">$W(n)$</span> is a random variable that receives the value <span class="math-container">$1$</span> if in the <span class="math-container">$n\ge1$</span> day it rained, and the value <span class="math-container">$0$</span>, if it's not.</p>
<p>I need to explain why the series <span class="math-container">$(W(n))_{n\ge 0}$</span> is not a Markov chain, but when they define <span class="math-container">$X_n=(W_n,W_{n-1})$</span>, so <span class="math-container">$(W(n))_{n\ge 0}$</span> it is a Markov chain!</p>
| cos_dm_math21 | 679,837 | <p>First, notice that <span class="math-container">$\text{Ker}(T^i) \subseteq \text{Ker}(T^{i+1})$</span>, for any <span class="math-container">$i\ge 0$</span>.</p>
<p>Remark that if <span class="math-container">$\text{Ker}(T^k) = \text{Ker}(T^{k+1})$</span> for some <span class="math-container">$k\ge 0$</span> then one has <span class="math-container">$\text{Ker}(T^i) = \text{Ker}(T^{i+1})$</span> for every <span class="math-container">$i\ge k$</span>. This fact can be proved easily by an induction.</p>
<p>By the hypothesis, <span class="math-container">$\text{Ker}(T^{n-2}) \subsetneq \text{Ker}(T^{n-1})$</span>, hence there is no <span class="math-container">$0\le k \le n-2$</span> such that <span class="math-container">$\text{Ker}(T^k) = \text{Ker}(T^{k+1})$</span>. Therefore, we get the following strictly ascending chain:
<span class="math-container">$$ 0 = \text{Ker}(T^0) \subsetneq \text{Ker}(T^1) \subsetneq \ldots \subsetneq \text{Ker}(T^{n-1})$$</span>
Hence, <span class="math-container">$\text{dim} \ \text{Ker}(T^{n-1})$</span> can be either <span class="math-container">$n-1$</span> or <span class="math-container">$n$</span>.</p>
<p>If <span class="math-container">$\text{dim} \ \text{Ker}(T^{n-1}) = n$</span>, then actually <span class="math-container">$T^{n-1} = 0$</span>, so all the eigenvalues of <span class="math-container">$T$</span> are <span class="math-container">$0$</span>.</p>
<p>We remain with the case <span class="math-container">$\text{dim} \ \text{Ker}(T^{n-1}) = n-1$</span>. By the Rank-Nullity theorem, <span class="math-container">$\text{dim} \ \text{Im}(T^{n-1}) = 1$</span>. Since <span class="math-container">$0 = \text{Ker}(T^0) \subsetneq \text{Ker}(T^1)$</span>, we get that <span class="math-container">$0$</span> is an eigenvalue of <span class="math-container">$T$</span>. Assume that <span class="math-container">$T$</span> has two nonzero eigenvalues <span class="math-container">$\lambda_1 \neq \lambda_2$</span>. Consider <span class="math-container">$v_1$</span> and <span class="math-container">$v_2$</span> the corresponding eigenvectors. Since <span class="math-container">$\lambda_1 \neq \lambda_2$</span>, the vectors <span class="math-container">$v_1$</span> and <span class="math-container">$v_2$</span> are linearly independent. (It is a pretty well-known fact. If you don't know it, try to prove it as an exercise.) Now, just note that <span class="math-container">$T(v_i) = \lambda_i v_i$</span>, implies that <span class="math-container">$T^{n-1}(v_i) = \lambda_i^{n-1} v_i$</span>, with <span class="math-container">$i=1,2$</span>. Thus, <span class="math-container">$\text{Im}(T^{n-1})$</span> contains two linearly independent vectors, which is a contradiction to the fact that <span class="math-container">$\text{dim} \ \text{Im}(T^{n-1}) = 1$</span>.</p>
|
2,978,605 | <p>Is there a sophisticate way to proof that:</p>
<p><span class="math-container">$$\frac{n}{N}\cdot\frac{n-1}{N-1}\cdot\frac{n-2}{N-2}\cdot\ldots\cdot \frac{1}{N-n+1} = \frac{1}{{N\choose n}}$$</span></p>
<p>where <span class="math-container">${N\choose n}$</span> denotes combinations.</p>
<p>When replacing <span class="math-container">$N$</span> and <span class="math-container">$n$</span> with values, both produce the same result.</p>
| phaedo | 549,654 | <p>This is a definition: <span class="math-container">$ {N \choose n} = \frac{N!}{n!(N-n)!} $</span></p>
<p>Expand all factorials and simplify the tail (N-n)!</p>
|
843,634 | <p>I am wondering whether for any two lines $\mathfrak{L}, \mathfrak{L'}$ and any point $\mathfrak{P}$ in $\mathbf{P}^3$ there is a line having nonempty intersection with all of $\mathfrak{L}, \mathfrak{L'}$, $\mathfrak{P}$. I don't really know how to approach this, because I was never taught thinking about such a problem, not even related ones. I think the answer should be no, but have no means to justify it. Perhaps use the Klein representation of lines in 3 space? Could someone also recommend a book where similar problems are solved or at least posed as exercises? Also feel free to give an algebraic geometry perspective, but I don't really know how to approach this with algebraic geometry.</p>
| lhf | 589 | <p>Over the reals (or any infinite field), a polynomial induces the zero function iff it is the zero polynomial because a polynomial cannot have an infinite number of roots.</p>
<p>Things are different in finite fields. If the field has $q$ elements, then $X^{q}-X$ induces the zero function but is not the zero polynomial.</p>
|
2,131,679 | <p>Let $f: \mathbb{R} \to \mathbb{R}$ be continuous and $D \subset \mathbb{R}$ be a dense subset of $\mathbb{R}$. Furthermore, $\forall y_1,y_2 \in D \ f(y_1)=f(y_2)$. Should $f$ be a constant function?</p>
<p>My attempt:
Since $f$ is continuous
$$\forall x_0 \ \forall \varepsilon >0 \ \exists \delta>0 \ \forall x \in \mathbb{R} \ \left(|x-x_0|<\delta \Longrightarrow |f(x)-f(x_0)|<\varepsilon \right)$$
Let $f$ be non-constant function.
Since $D$ is dense $\exists x_1 \in (x_0-\delta, x_0+\delta) \ : \ x_1 \in D$.
Let's take $x_2 \in (x_0-\delta, x_0+\delta)$ such that $f(x_2) \ne f(x_1)$.
Let $\varepsilon = \frac{|f(x_2)-f(x_1)|}{2}>0$.
Therefore, we have
$$|f(x_1)-f(x_0)|<\frac{|f(x_2)-f(x_1)|}{2} \ \ \ |f(x_2)-f(x_0)|<\frac{|f(x_2)-f(x_1)|}{2}$$
Adding the expressions above, we obtain
$$|f(x_2)-f(x_1)|\le |f(x_1)-f(x_0)|+|f(x_2)-f(x_0)|<|f(x_2)-f(x_1)|$$
what is the contradiction.
Are my mussings correct?</p>
| Stefan4024 | 67,746 | <p>Extend $AO$ to cut $BC$ in $F$ and also let $G$ be the altitude from $C$ to $BC$. Now from Ceva's Theorem we have that $BF:FC = 1:4$. From Menelaus' Theorem on $\triangle BCE$ and $A-O-F$ we have:</p>
<p>$$1 = \frac{BF}{FC} \times \frac{CO}{OE} \times \frac{EA}{BA} \implies \frac{CO}{OE} = 6$$</p>
<p>Now we'll prove that $\angle EDC = 90$. Draw the altitude from $B$ to $AC$ and let it's foot be $X$. It's easy to see that $\frac{AD}{DX} = 2$. So as $\frac{AD}{DX} =2 = \frac{AE}{EB}$ we get that $ED \parallel BX$, so $\angle EDC = 90$.</p>
<p>Finally we have:</p>
<p>$$\frac{AG}{GE} \times \frac{EO}{OC} \times \frac{CD}{DA} = 3 \cdot \frac 16 \cdot 2 = 1$$</p>
<p>Hence $CG, ED, AO$ intersect and they do this at the orthocenter of $\triangle CEA$, as $ED, CG$ are altitudes in the triangle. Hence $AO$ is altitude in $\triangle CEA$ and $\angle AOC = 90$</p>
|
726,574 | <p>Ant stands at the end of a rubber string which has 1km of length. Ant starts going to the other end at speed 1cm/s. Every second the string becomes 1km longer. </p>
<p>For readers from countries where people use imperial system: 1km = 1000m = 100 000cm</p>
<p><strong>Will the ant ever reach the end of the string? But how to explain it.</strong> </p>
<p>I know that yes.</p>
<p>Let :
<code>a</code> - distance covered by ant
<code>d</code> - length of string
<code>c</code> - constant by which the string is extended</p>
<p>The distance covered by ant in second <code>i</code> is <code>a[i] = (a[i-1] + 1)* (d + c)/d</code></p>
<p>I even did computer simulation in microscale where the string is 10cm long and extends by 10cm every second and the ant reaches the end:</p>
<pre><code>public class Mrowka {
public final static double DISTANCE_IN_CM = 10;
public static void main(String[] args) {
double ant = 0;//ants distance
double d = DISTANCE_IN_CM;//length of string
double dLeft = d - ant;//distance left
int i = 0;
while(dLeft > 0){
ant++;
ant = ant * (d + DISTANCE_IN_CM)/d;
d = d + DISTANCE_IN_CM;
dLeft = d - ant;
System.out.println(i + ". Ant distance " + ant +"\t Length of string " + d + " distance left " + dLeft);
i++;
}
System.out.println("end");
}
}
</code></pre>
<p><strong>Output:</strong></p>
<pre><code>0. Ant distance 2.0 Length of string 20.0 distance left 18.0
1. Ant distance 4.5 Length of string 30.0 distance left 25.5
2. Ant distance 7.333333333333333 Length of string 40.0 distance left 32.666666666666664
.....
12364. Ant distance 123658.53192119849 Length of string 123660.0 distance left 1.4680788015102735
12365. Ant distance 123669.5318833464 Length of string 123670.0 distance left 0.46811665360291954
12366. Ant distance 123680.53192635468 Length of string 123680.0 distance left -0.5319263546844013
end
</code></pre>
<p>EDIT:</p>
<p>I think that I need to calculate this <code>a[n] = (a[n-1] + 1)*(1 + 1/(1+n))</code> when <code>n->+oo</code></p>
| Fly by Night | 38,495 | <p>As you rightly say, the gradient of the line $2y-x=7$ is $\frac{1}{2}$ because we may rearrange $2y-x=7$ to give $y=\frac{1}{2}x+\frac{7}{2}$. The line you want to find is parallel to this and so also has gradient $\frac{1}{2}$.</p>
<p>Again, you rightly say that the bisector of $(3,1)$ and $(1,-5)$ passes through the midpoint, i.e.
$$\left( \frac{3+1}{2},\frac{1-5}{2}\right)=(2,-2)$$
We want the line that passes through $(2,-2)$ and has gradient $\frac{1}{2}$. </p>
<p>We have the line $y=\frac{1}{2}x+c$ which passes through $(2,-2)$. We need to find $c$. Putting $x=2$ and $y=-2$ gives $-2=\frac{1}{2}(2)+c$, i.e. $-2=1+c$, i.e. $c=-3$. The equation is then
$$y = \frac{1}{2}x - 3$$</p>
|
4,241 | <p>I was preparing for an area exam in analysis and came across a problem in the book <em>Real Analysis</em> by Haaser & Sullivan. From p.34 Q 2.4.3, If the field <em>F</em> is isomorphic to the subset <em>S'</em> of <em>F'</em>, show that <em>S'</em> is a subfield of <em>F'</em>. I would appreciate any hints on how to solve this problem as I'm stuck, but that's not my actual question.</p>
<p>I understand that for finite fields this implies that two sets of the same cardinality must have the same field structure, if any exists. The classification of finite fields answers the above question in a constructive manner.</p>
<p>What got me curious is the infinite case. Even in the finite case it's surprising to me that the field axioms are so "restrictive", in a sense, that alternate field structures are simply not possible on sets of equal cardinality. I then started looking for examples of fields with characteristic zero while thinking about this problem. I didn't find many. So far, I listed the rationals, algebraic numbers, real numbers, complex numbers and the p-adic fields. What are other examples? Is there an analogous classification for fields of characteristic zero?</p>
| Jonas Kibelbek | 1,461 | <p>Yuval just posted the function field construction, which is an important example for your question.</p>
<p>You mentioned the algebraic numbers, but just to make it explicit, there are many algebraic number fields (often called just "number fields"). We have $ \bar Q $, the field of <em>all</em> algebraic numbers, but we also have $Q(\sqrt{2})$, $Q(\sqrt[5]{7})$, $Q(i,\sqrt{2},\sqrt[5]{7})$, and any other field you wish to generate over $Q$ by some tower of finite extensions. </p>
<p>(If we take $Q(\pi)$, we get a field isomorphic to the function field $Q(t)$, since $\pi$ is not algebraic.)</p>
<p>You can also take infinitely many different algebraic extensions of the p-adic fields, like $Q_3(i)$ or $Q_3(\sqrt[4]{2})$. </p>
<p>If you like big constructions, you can take the algebraic closure $\bar {Q_p}$of a p-adic field (which turns out not to be p-adically complete), then take the completion $\hat { \bar {Q_p}}$ of that to get a p-adcially complete and algebraically complete field which is isomorphic to the field of complex numbers (but not in any canonical way).</p>
|
36,756 | <p>In a class, 18 students like to play chess, 23 like to play soccer, 21 like biking, and 17 like jogging. The number of those who like to play both chess and soccer is 9. We also know
that 7 students like chess and biking, 6 students like chess and jogging, 12 like soccer and
biking, 9 like soccer and jogging, and finally 12 students like biking and jogging. There
are 4 students who like chess, soccer, and biking, 3 who like chess, soccer, and jogging, 5
who like chess, biking, and jogging, and 7 who like soccer, biking, and jogging. Finally,
there are 3 students who like all four activities. In addition, we know that every student
likes at least one of these activities. How many students are there in the class? </p>
<p>I know this is a case of using the inclusion-exclusion principle, but I'm a little overwhelmed, given that there are 4 sets. Can someone please explain this to me? Thanks!</p>
| yoyo | 6,925 | <p>given sets $X_1, X_2, X_3, X_4$ we have
$$|\cup X_i|=\sum_i|X_i|-\sum_{i\neq j}|X_i\cap X_j|+\sum_{i\neq j\neq k}|X_i\cap X_j\cap X_k|-|\cap X_i|$$
so in your example this is
$$
(18+23+21+17)-(9+7+6+12+9+12)+(4+3+5+7)-3=40
$$</p>
|
157,374 | <p>I'm new to Wolfram Mathematica.
I have an initial image:</p>
<p><a href="https://i.stack.imgur.com/XekQw.png" rel="noreferrer"><img src="https://i.stack.imgur.com/XekQw.png" alt="enter image description here"></a></p>
<p>My goal is to find a radius of the water drop, like so:</p>
<p><a href="https://i.stack.imgur.com/MLHH2.png" rel="noreferrer"><img src="https://i.stack.imgur.com/MLHH2.png" alt="enter image description here"></a></p>
<p>With that in mind, I've changed the initial photo:</p>
<pre><code>resultImage = EdgeDetect[ImageResize[Binarize[Sharpen[//..here goes the image]], 2000]]
</code></pre>
<p>which produced:
<a href="https://i.stack.imgur.com/S9sup.png" rel="noreferrer"><img src="https://i.stack.imgur.com/S9sup.png" alt="enter image description here"></a></p>
<p>To solve the initial problem, I suspect that I can use <code>ImageData</code> to convert the latter result into the corresponding matrix and manually process it row by row. But is there a more slick and convenient solution? What would you suggest? Thanks.</p>
| b3m2a1 | 38,205 | <p>Here's a way that might work for you:</p>
<pre><code>img = Import["https://i.stack.imgur.com/XekQw.png"]
imClip = ImageClip[img, {.9, 1}, {0, 1}];
coords = Position[MorphologicalComponents[imClip], 0];
possibilities = MinMax[#[[All, 2]]] & /@ GatherBy[coords, #[[1]] &];
Subtract @@
Reverse@MaximalBy[possibilities, Abs@*Apply[Subtract]][[1]]
303
</code></pre>
<p>I had to actually check what number <a href="http://reference.wolfram.com/language/ref/MorphologicalComponents.html" rel="noreferrer"><code>MorphologicalComponents</code></a> assigned to the black portions of the clipped image. That's where the 0 comes from.</p>
<p>Past that it's a pretty simple filtering, dependent on that bulge part being wider than any of the other parts.</p>
<p>Hopefully that's easier than what you had in mind.</p>
|
157,374 | <p>I'm new to Wolfram Mathematica.
I have an initial image:</p>
<p><a href="https://i.stack.imgur.com/XekQw.png" rel="noreferrer"><img src="https://i.stack.imgur.com/XekQw.png" alt="enter image description here"></a></p>
<p>My goal is to find a radius of the water drop, like so:</p>
<p><a href="https://i.stack.imgur.com/MLHH2.png" rel="noreferrer"><img src="https://i.stack.imgur.com/MLHH2.png" alt="enter image description here"></a></p>
<p>With that in mind, I've changed the initial photo:</p>
<pre><code>resultImage = EdgeDetect[ImageResize[Binarize[Sharpen[//..here goes the image]], 2000]]
</code></pre>
<p>which produced:
<a href="https://i.stack.imgur.com/S9sup.png" rel="noreferrer"><img src="https://i.stack.imgur.com/S9sup.png" alt="enter image description here"></a></p>
<p>To solve the initial problem, I suspect that I can use <code>ImageData</code> to convert the latter result into the corresponding matrix and manually process it row by row. But is there a more slick and convenient solution? What would you suggest? Thanks.</p>
| bill s | 1,783 | <p>One way to proceed (that doesn't require searching through the rows) is to binarize the image, remove the top portion, and then crop. The width of the cropped image is the maximum width of the water drop.</p>
<pre><code>img = Import["https://i.stack.imgur.com/XekQw.png"];
ImageCrop[ImageTake[Binarize[img], -550]] // ImageDimensions
{303, 387}
</code></pre>
<p>So it is 303 pixels wide.</p>
<p>(Thanks to Alexey for hint regarding an extraneous use of FillingTransform.)</p>
|
705,744 | <p>Hello everyone. I have a couple questions this time, but I think if I understand how to do this one, I'll understand the others.</p>
<p>A particular online banking system uses the following rules for its passwords:<br/>
a. Passwords must be 6-8 characters in length<br/>
b. Passwords must use only alphabetical and numeric characters, and must have at least one alpha and one numeric character.<br/>
c. Letters are case sensitive.</p><p>Under these rules, how many different passwords are possible?</p>
| André Nicolas | 6,312 | <p>Count the number of legal passwords of length $6$, $7$, $8$ separately, and then add up.</p>
<p>We do the length $7$ case.</p>
<p>If we are using the standard alphabet, there are $26$ lower case characters, $26$ upper case characters, and $10$ digits, for a total of $62$.</p>
<p>There are $62^7$ words of length $7$ made up by choosing symbols from our $62$-element symbol set.</p>
<p>This is because the first symbol of the word can be chosen in $62$ ways, and for <strong>each</strong> of these ways the second symbol can be chosen in $62$ ways, and so on. </p>
<p>However, some of these words are <strong>forbidden</strong>. We count the forbidden words.</p>
<p>There are $52^7$ "all-letter" words. There are $10^7$ "all-digit" words. So there are $52^7+10^7$ forbidden passwords of length $7$. This leaves $62^7 -52^7-10^7$ allowed passwords of length $7$. </p>
|
3,162,338 | <p>Consider <span class="math-container">$ x_1, x_2, ..., x_n \in \mathbb{R}$</span>.</p>
<p>We have to prove that each <span class="math-container">$\sqrt x $</span> is rational if the sum of <span class="math-container">$\sqrt x_1 + \ldots + \sqrt x_n $</span> is rational. </p>
<p>I think that I could prove it using fact, that only the sum of the opposite irrational number is rational for example <span class="math-container">$ \sqrt 2 + (2-\sqrt 3) = 2 $</span>, because if <span class="math-container">$ x_1, x_2, ..., x_n \in \mathbb{R} $</span> then <span class="math-container">$ \sqrt (-1) \neq - \sqrt 1$</span>. </p>
| jmerry | 619,637 | <p>As stated, with real numbers <span class="math-container">$x_i$</span>, it's false. After all, we could take something like <span class="math-container">$x_1=(\sqrt{2})^2$</span>, <span class="math-container">$x_2=(2-\sqrt{2})^2$</span>. For this to make sense, those <span class="math-container">$x_i$</span> must all be rational.</p>
<p>For <span class="math-container">$n=2$</span>, this is easy to do in an elementary way. We can solve for one of the square roots and then square:
<span class="math-container">\begin{align*}x_1+x_2 &= S\\ \sqrt{x_1} &= S - \sqrt{x_2}\\ x_1 &= S^2 + x_2 - 2S\sqrt{x_2}\\
\sqrt{x_2} &= \frac{S^2+x_2-x_1}{2S}\end{align*}</span>
With one of the square roots rational, the other has to be as well.</p>
<p>This does not generalize well, at least at this level of understanding. With larger numbers of square roots involved, squaring the other side leads to the square roots of the products getting involved, increasing the total number of square roots involved. We can work through it, but it requires the language of linear algebra and field extensions to make sense of it in general.</p>
<p>See also <a href="https://math.stackexchange.com/questions/30687/the-square-roots-of-different-primes-are-linearly-independent-over-the-field-of">here</a> for a proof of the general case.</p>
|
72,201 | <p>Two people play a game. They play a series of points, each producing a winner and a loser, until one player has won at least 4 points and has won at least 2 more points than the other. Anne wins each point with probability p.
What is the probability that she wins the game on the kth point played for k=4,5,6,...</p>
<p>Ok so I started with k=4
P(win on 4th point) can happen 3 ways 4-0,4-1,4-2
=p^4(1 + (4 choose 1)q +(5 choose 2)q^2), which is a geometric series for q. I'm not sure what to do with the changing binomial coefficients though.
Similarly, P(win on 5th point can happen 4 ways) 5-0, 5-1, 5-2, 5-3
=p^5(1 + (5 choose 1)q + (6 choose 2) q^2 +(7 choose 3)q^3)</p>
<p>So it seems like we will have two geometric series one of powers of p, each containing one of powers of q. I'm not sure the exact pattern and how to formalize this. Thank you!</p>
| Dilip Sarwate | 15,941 | <p>I am wondering if the interpretation used by caligirl11 and @joriki is the one that was intended by whomsoever devised the problem. </p>
<blockquote>
<p>What is the probability that she wins the game on the kth point played for k=4,5,6,...</p>
</blockquote>
<p>I suggest that the $k$-th point <em>played</em> is <em>not</em> the same
as the $k$-th point <em>won</em> by Anne.</p>
<blockquote>
<p>Ok so I started with k=4 P(win on 4th point) can happen 3 ways 4-0,4-1,4-2 =p^4(1 + (4 choose 1)q +(5 choose 2)q^2)</p>
</blockquote>
<p>I would say here that Anne wins the game on the $4$th point
<em>played</em> with probability $p^4$; the probability that she
wins on the $5$th point <em>played</em> is $\binom{4}{1}p^4q$; and
the probability that she wins on the $6$th point <em>played</em>
is $\binom{5}{2}p^4q^2$. The corresponding win probabilities
for Betty are obtained by interchanging $p$ and $q$ in the
previous sentence.</p>
<p>If $6$ points have been played and the game is not over,
the score must be $3$-$3$ (deuce) and this event has probability
$\binom{6}{3}p^3q^3 = 20p^3q^3$. The game can now end only
when an even number of <em>additional</em> points have been played,
and the probability of Anne winning on the $k$-th point played
(where $k = 6+2n, n>0$) is $\binom{6}{3}p^3q^3\cdot p^2(2pq)^{n-1}$.
To get the probability that Anne wins the game, sum the
probabilities that she wins in $4$, $5$, $6$, $8$, $10, \ldots$
points; a geometric series is involved.</p>
<p>If $f(p)$ denotes the probability that Anne wins the game,
then $f(0.5) = 0.5$ but $\left . \frac{d}{dp}f(p)\right\vert_{p = 0.5} = 2.5$
so that $f(0.5 + \epsilon) = 0.5 + 2.5\epsilon + \cdots$,
that is, a small difference $p-q = 2\epsilon$ in the
<em>point win</em> probabilities
for Anne and Betty is amplified by the rules of tennis
into a greater difference in the <em>game win</em> probabilities. See
"The Drunken Tennis Player" in Ian Stewart's <em>Game, Set, and Math</em>,
Penguin Books, 1981.</p>
|
1,171,911 | <p><img src="https://i.stack.imgur.com/3q5iO.png" alt="Taken from khan academy ">
Hi, so this question is taken straight from khan academy help exercises, i know how to do it dynamically meaning using the determinant and the adjugate how i was trying to do it using guass bla bla way with help of RREF but i somehow never managed to find the inverse.
my second question would be is there anyway that i can find the whether or not the matrix is invertable without trying to find the determinant i mean also using Gauss bla bla way
i use the word bla bla because i dont know what it is actually called :p</p>
| Ben Grossmann | 81,360 | <p>Using Gauss-Jordan elimination:
$$
\left[\begin{array}{ccc|ccc}
0&1&2 &1&0&0\\
1&0&1 &0&1&0\\
0&1&0 &0&0&1
\end{array}\right] \to \\
\left[\begin{array}{ccc|ccc}
1&0&1 &0&1&0\\
0&1&0 &0&0&1\\
0&1&2 &1&0&0\\
\end{array}\right] \to\\
\left[\begin{array}{ccc|ccc}
1&0&1 &0&1&0\\
0&1&0 &0&0&1\\
0&0&2 &1&0&-1\\
\end{array}\right] \to \\
\left[\begin{array}{ccc|ccc}
1&0&1 &0&1&0\\
0&1&0 &0&0&1\\
0&0&1 &1/2&0&-1/2\\
\end{array}\right] \to \\
\left[\begin{array}{ccc|ccc}
1&0&0 &-1/2&1&1/2\\
0&1&0 &0&0&1\\
0&0&1 &1/2&0&-1/2\\
\end{array}\right]
$$
So, we conclude
$$
D^{-1} =
\pmatrix{
-1/2&1&1/2\\
0&0&1\\
1/2&0&-1/2}
$$</p>
|
97,261 | <p>This semester, I will be taking a senior undergrad course in advanced calculus "real analysis of several variables", and we will be covering topics like: </p>
<p>-Differentiability.
-Open mapping theorem.
-Implicit function theorem.
-Lagrange multipliers. Submanifolds.
-Integrals.
-Integration on surfaces.
-Stokes theorem, Gauss theorem.</p>
<p>I need to know if anyone of you guys know good textbooks that contain practice problems with full solutions or hints that can be used to understand the material. Most of the textbooks I found are covering only the material with few examples.</p>
| Community | -1 | <p>Here are some free options:</p>
<p><a href="http://www.whitman.edu/mathematics/multivariable/" rel="nofollow">http://www.whitman.edu/mathematics/multivariable/</a></p>
<p><a href="http://synechism.org/drupal/cfsv/" rel="nofollow">http://synechism.org/drupal/cfsv/</a></p>
<p><a href="http://www.mecmath.net/" rel="nofollow">http://www.mecmath.net/</a></p>
<p>I maintain a catalog of free books at <a href="http://theassayer.org" rel="nofollow">http://theassayer.org</a> , and you could poke around there for further possibilities.</p>
|
1,350,837 | <p>Find all integer numbers $n$, such that, $$\sqrt{\frac{11n-5}{n+4}}\in \mathbb{N}$$</p>
<p>I really tried but I couldn't guys, help please.</p>
| DeepSea | 101,504 | <p><strong>hint</strong>: $\dfrac{11n-5}{n+4} = k^2\to 11n-5=nk^2+4k^2\to n(11-k^2)=5+4k^2= 49+4(k^2-11)\to n = \dfrac{-49}{k^2-11}-4\to k^2-11 = \pm1,\pm7,\pm49.$ From this you can find all possible values of $k$, and then find $n$.</p>
|
526,627 | <p>What is the answer for factoring:</p>
<p>$$10r^2 - 31r + 15$$</p>
<p>I have tried to solve it. This was my prior attempt:</p>
<p>$$10r^2 - 31r + 15\\
= (10r^2 - 25r) (-6r + 15)\\
= -5r(-2r+5) -3 (2r-5) $$ </p>
| Ross Millikan | 1,827 | <p>You lost a $+$ sign in the first line starting with $=$. Then note that in the next line, the expressions inside the parentheses are the same except for multiplying by $-1$. So change the sign inside one set of parentheses and the leading coefficient.
$$-5r(-2r+5)-3(2r-5)=5r(2r-5)-3(24-5)=(5r-3)(2r-5)$$</p>
|
1,053,065 | <p>I have a function called $P(t)$ that is the number of the population at time $t$. $t$ being in days.</p>
<p>We know the growth rate is $P'(t) = 2t + 6$</p>
<p>We also know that $P(0) = 100$. How many days till the population doubles?</p>
<p>edit: $P(t) = t^2 + 6t$
edit: $P(t) = t^2 + 6t = 200$
edit: $t^2 + 6t - 200 = 0$</p>
| Dietrich Burde | 83,966 | <p>A bound for the number of divisors of $n$ is given here: <a href="https://math.stackexchange.com/questions/63687/bound-for-divisor-function">Bound for divisor function</a>. We also have an effective upper bound as follows:
$$
d(n)\le n^{\frac{1.5379 \log(2)}{\log(\log(n))}},
$$
for all $n\ge 3$. This is much better than what you want (for $n$ not too small, at least, i.e., $n\ge 41$; the case $n\le 40$ is settled by inspection).</p>
|
2,529,533 | <p>Let $f:\mathbf{R}^n \to \mathbf{R}$ be differentiable, $\sum_{i=1}^n y_i \frac{\partial f}{\partial x_i}(y)\geq 0$ for all $y=(y_1,...,y_n)\in \mathbf{R}^n$. How do I show that $f$ is bounded from below by $f(0)$?</p>
| Netivolu | 440,984 | <p>Isn't
$$
f:\mathbb{R}^2 \rightarrow \mathbb{R}: (x_1,x_2) \mapsto x_1 + x_2
$$
a counterexample?</p>
|
3,985,302 | <p>Let <span class="math-container">$L/K$</span> be an extension of local fields. We can find <span class="math-container">$\alpha$</span> such that <span class="math-container">$\mathcal{O}_L=\mathcal{O}_K[\alpha]$</span>. What do we know about this generating element? I think that this <span class="math-container">$\alpha$</span> can be selected in such a way that, in addition to the above property, it is also a uniformizing parameter at the same time. But I can not prove it.</p>
| reuns | 276,986 | <p><span class="math-container">$O_L=O_K[\pi_L]$</span> iff <span class="math-container">$f(L/K)=1$</span> and <span class="math-container">$O_L=O_K[\zeta_{q_L-1}]$</span> iff <span class="math-container">$e(L/K)=1$</span>. In general it is <span class="math-container">$$O_L=O_K[\zeta_{q_L-1}+\pi_L]$$</span></p>
<p>Hensel lemma is needed to construct the root unity, then (in the non-trivial case <span class="math-container">$e\ne 1,f\ne 1$</span>) we need a closed-ness and density argument, investigating <span class="math-container">$\zeta_{q_L-1}+\pi_L-(\zeta_{q_L-1}+\pi_L)^q$</span></p>
|
3,985,302 | <p>Let <span class="math-container">$L/K$</span> be an extension of local fields. We can find <span class="math-container">$\alpha$</span> such that <span class="math-container">$\mathcal{O}_L=\mathcal{O}_K[\alpha]$</span>. What do we know about this generating element? I think that this <span class="math-container">$\alpha$</span> can be selected in such a way that, in addition to the above property, it is also a uniformizing parameter at the same time. But I can not prove it.</p>
| Torsten Schoeneberg | 96,384 | <p>Hint: If <span class="math-container">$\mathcal{O}_L = \mathcal{O}_K[\alpha]$</span> with <span class="math-container">$\alpha$</span> a uniformiser of <span class="math-container">$\mathcal{O}_L$</span>, then the canonical inclusion of their residue fields <span class="math-container">$\mathcal{O}_K/(\pi_K) \hookrightarrow \mathcal{O}_L/(\alpha)$</span> is surjective.</p>
<p>Conversely, if the extension is totally ramified (which is what the above says), then ...</p>
<p>reuns' answer says the same and more.</p>
|
607,264 | <p>Let the directional derivative of a function $f(x,y)$ at a point $P$ in the direction of $(1/\sqrt{5})\mathbf{i}+(2/\sqrt{5})\mathbf{j}$ be $16/\sqrt{5}$ and the partial derivative $\partial f / \partial x$ evaluated at $P$ be $6$. Then what is the directional derivative in the direction of $\mathbf{i}-\mathbf{j}$?</p>
<p>I got $1$ as the answer but this is incorrect. It should come out to $1/\sqrt{2}$.</p>
<p>My work so far:</p>
<p>$D(P)=(f_x,f_y)\cdot(1/\sqrt{5},2/\sqrt{5}) $</p>
<p>if $\partial f/ \partial x=6$ then $6+2f_y=16 \implies f_y=5$</p>
<p>so $(1\cdot6)+(5\cdot(-1))=1$.</p>
| Berci | 41,488 | <p>The differential of $f$ at the given point $P$ is a linear map, respresented by a $1\times 2$ matrix, namely $f'(P)=\left(\displaystyle\frac{df}{dx}(P),\ \frac{df}{dy}(P)\right)$. Now, by assumption,
$df/dx(P)=6$, and let $b:=df/dy(P)$ the missing coordinate.</p>
<p>The other piece of information tells us that
$$\pmatrix{6&b}\cdot\pmatrix{1/\sqrt5 \\ 2/\sqrt5} = \frac{16}{\sqrt5}\,.$$
From this you can express $b$ and calculate
$$\pmatrix{6&b}\cdot\pmatrix{1 \\ -1}\,.$$
(Or <em>normalize</em> the vector $(1,\,-1)^T$, dividing it by its length, $\sqrt2$.)</p>
|
355,552 | <p>How would you compute the first $k$ digits of the first $n$th Fibonacci numbers (say, calculate the first 10 digits of the first 10000 Fibonacci numbers) without computing (storing) the whole numbers ?</p>
<p>A trivial approach would be to store the exact value of all the numbers (with approximately $0.2n$ digits for the $n$th number) but this requires performing additions over numbers with <em>many</em> digits (and also a lot of storage), even if $k$ is small. Perhaps there's a way to accomplish this using only smart approximations that lead to precise results for the first $k$ digits.</p>
<p>Thanks in advance.</p>
| Ross Millikan | 1,827 | <p>How about <a href="http://en.wikipedia.org/wiki/Jacques_Philippe_Marie_Binet" rel="nofollow">Binet's formula</a>? If you want the first $k$ digits, you need to calculate $\sqrt 5$ to not many more than $k$ digits (use the continued fraction, for example) and you are there. The subtraction doesn't cancel, and if $n$ is large you can ignore the second term in the numerator-it is tiny.</p>
|
60,322 | <p>I'm interested in Lie theory and its connections to dynamical systems theory. I am starting my studies and would like references to articles on the subject.</p>
| mathphysicist | 2,149 | <p>One of the most important connections of the two fields can be found in the theory of Hamiltonian dynamical systems with Lie groups being the symmetry groups. The interplay of these leads to many interesting concepts (including, inter alia, the classical R-matrix) and results. For starters you can try the books <a href="https://books.google.com/books?id=sI2bAxgLMXYC" rel="nofollow noreferrer">Applications of Lie Groups to Differential Equations</a> by Olver and <a href="https://rads.stackoverflow.com/amzn/click/com/0817623361" rel="nofollow noreferrer" rel="nofollow noreferrer">Integrable Systems of Classical Mechanics and Lie Algebras</a> by Perelomov, and the survey paper <a href="https://arxiv.org/abs/nlin/0209057" rel="nofollow noreferrer">Integrable Systems and Factorization Problems</a> by Semenov-Tian-Shansky.</p>
|
3,459,106 | <p>I have a function
<span class="math-container">$$
\frac{\ln x}{x}
$$</span> and I wonder, is <span class="math-container">$y=0$</span> an asymptote? I mean it is kinda strange that graph is in some place is going through that asymptote. I know it meets the criterium of asymptote, but its kinda strange if you understand me. :D</p>
| Kavi Rama Murthy | 142,385 | <p>Without continuity of <span class="math-container">$f_n$</span>'s this is false. Take any bounded function <span class="math-container">$f$</span> which is not continuous and let <span class="math-container">$f_n =f $</span>. Then <span class="math-container">$f_n \to f$</span> uniformly. </p>
|
3,873,071 | <p>This is my first post and I apologize in advance if I'm not using the right formatting/approach.</p>
<p><strong>Problem</strong></p>
<p>A coin, having probability <span class="math-container">$p$</span> of landing heads, is continually flipped until at least one head and one tail have been flipped.</p>
<p>Find the expected number of flips needed.</p>
<p>typical examples: “HT”, X = 2; “TTTTH”, X = 5.</p>
<p><strong>Solution Begin</strong></p>
<p>Denote X: # of flips needed. Y: outcome of 1st flip.</p>
<p><span class="math-container">$$\operatorname E[X] = \operatorname E[X\mid Y = H]P(Y = H) + \operatorname E[X\mid Y = T]P(Y = T)$$</span></p>
<p><span class="math-container">$$E[X\mid Y = H] = 1 + \operatorname E[\text{additional flips needed}] = 1 + \frac1{1-p}$$</span></p>
<p><strong>Question</strong></p>
<p>This is regarding <span class="math-container">$$1 + \frac1{1-p}$$</span></p>
<p>I understand that <span class="math-container">$1$</span> is for the failed trial but why is the <span class="math-container">$1/(1-p)$</span> there? Given the conditional probability/expectation, I thought the denominator would be the <span class="math-container">$P(Y=H)$</span> which is <span class="math-container">$p.$</span> I just don't understand the overall reason for <span class="math-container">$1/(1-p).$</span> Could someone help me understand or point me in the right direction?</p>
| lab bhattacharjee | 33,337 | <p>Hint:</p>
<p><span class="math-container">$$\dfrac{(x^4-1)\sqrt{x^4+1}}{x^8+1}=x\cdot\dfrac{\left(1-\dfrac1{x^4}\right)\sqrt{x^2+\dfrac1{x^2}}}{x^4+\dfrac1{x^4}}$$</span></p>
<p>Set <span class="math-container">$x^2=y$</span> to find</p>
<p><span class="math-container">$$\int\dfrac{(x^4-1)\sqrt{x^4+1}}{x^8+1}dx=\dfrac12\int\dfrac{\left(1-\dfrac1{y^2}\right)\sqrt{y+\dfrac1y}}{\left(y+\dfrac1y\right)^2-2}dy$$</span></p>
<p>Now choose <span class="math-container">$\sqrt{y+\dfrac1y}=u\implies y+\dfrac1y=u^2$</span></p>
|
470,617 | <ol>
<li><p>Two competitors won $n$ votes each.
How many ways are there to count the $2n$ votes, in a way that one competitor is always ahead of the other?</p></li>
<li><p>One competitor won $a$ votes, and the other won $b$ votes. $a>b$.
How many ways are there to count the votes, in a way that the first competitor is always ahead of the other?
(They can have the same amount of votes along the way)</p></li>
</ol>
<p>I know that the first question is the same as the number of different legal strings built of brackets, which is equal to the Catalan number; $\frac{1}{n+1}{2n\choose n}$, by the proof with the grid.</p>
<p>I am unsure about how to go about solving the second problem.</p>
| lab bhattacharjee | 33,337 | <p>If $7x=\pi,4x=\pi-3x$</p>
<p>$\implies \sin4x=\sin(\pi-3x)=\sin3x$</p>
<p>$\implies 2\sin2x\cos2x=3\sin x-4\sin^3x$</p>
<p>$\implies 4\sin x\cos x\cos2x=3\sin x-4\sin^3x$</p>
<p>If $\sin x\ne0,$ we have $4\cos x\cos2x=3-4\sin^2x\implies 4\cos x(1-2\sin^2x)=3-4\sin^2x$</p>
<p>On squaring & rearrangement, $64(\sin^2x)^3-112(\sin^2x)^2+56\sin^2x-7=0$ </p>
<p>which is a cubic equation in $\sin^2x$ with roots being $\sin^2\frac{r\pi}7$<br>
where $r=(1$ or $6),(2$ or $5)$ and $(3$ or $4)$ as $\sin\frac{(7-r)\pi}7=\sin (\pi-\frac{r\pi}7)=\sin\frac{r\pi}7$</p>
<p>Using Vieta's Formula we have, </p>
<p>$\displaystyle\sin^2\frac{\pi}7\sin^2\frac{2\pi}7\sin^2\frac{4\pi}7=\frac7{64}$ and $\displaystyle\sin^2\frac{\pi}7\sin^2\frac{2\pi}7+\sin^2\frac{2\pi}7\sin^2\frac{4\pi}7+\sin^2\frac{4\pi}7\sin^2\frac{\pi}7=\frac{56}{64}$</p>
<p>We need to find </p>
<p>$\displaystyle\frac1{\sin^2\frac{\pi}7}+\frac1{\sin^2\frac{2\pi}7}+\frac1{\sin^2\frac{4\pi}7}=\displaystyle\frac{\sin^2\frac{\pi}7\sin^2\frac{2\pi}7+\sin^2\frac{2\pi}7\sin^2\frac{4\pi}7+\sin^2\frac{4\pi}7\sin^2\frac{\pi}7}{\sin^2\frac{\pi}7\sin^2\frac{2\pi}7\sin^2\frac{4\pi}7}=\frac{\frac{56}{64}}{\frac7{64}}=\frac{56}7=8$</p>
|
112,096 | <p>Does the inequality $2 \langle x , y \rangle \leqslant \langle x , x \rangle + \langle y , y \rangle $, where $$ \langle \cdot, \cdot \rangle $$ denotes scalar product, have a name? </p>
<p>I've tried looking at several inequalities on wikipedia but I didn't find this one. And of course googling doesn't work for this purpose.</p>
| Beni Bogosel | 7,327 | <p>It is the development of
$$ \langle x-y,x-y \rangle \geq 0$$
and it follows from the positive definitness of the scalar product.</p>
<hr>
<p>Apart of the above proof of the inequality, and as a response to the comments to the question, here are a few reasons as to why this inequality should be true, at a first glance:</p>
<ul>
<li><p>a scalar product has the properties of the multiplication on the real line, so the inequality $2xy\leq x^2+y^2$ should pop up while looking at the given inequality;</p></li>
<li><p>Cauchy Schwarz immediatley implies the inequality:
$$2\langle x,y \rangle \leq 2 \|x\|\|y\| \leq\|x\|^2+\|y\|^2 $$</p></li>
</ul>
|
685,567 | <p>For the function, $f(x,y,z)=\sqrt{x^2+y^2+z^2}$, do directional derivatives exist at the origin? If I use the definition $$lim_{h\to 0}\frac{f(x+hv)-f(x)}{h},$$ then I get $$\frac{|h|}{h}$$ which is without limit. But in some places, I keep reading that the directional derivative is 1. </p>
<p>Also, if I were to write the function in spherical coordinates, it would be simply $f(\rho\theta)=\rho$, which is differentiable in $\rho$, irrespective of $\theta$. </p>
<p>What is the source of this ambiguity and is there a convention for this? </p>
| Vadim | 26,767 | <p><strong>Hint</strong> Consider the PGF for this distribution. It is</p>
<p>$$pgf(z)=\mathbb{E}z^X=\sum_{k=0}^\infty\left(e^{-\lambda}\frac{\lambda^k}{k!}\right)z^k=e^{\lambda(z-1)}$$</p>
<p>Then, assuming they are independent, the sum has PGF</p>
<p>$$pgf_n(z)=(pgf(z))^n=e^{n\lambda(z-1)}=\sum_{k=0}^\infty\left(e^{-n\lambda}\frac{n^k\lambda^k}{k!}\right)z^k$$</p>
<p>which gives you the probability distribution: Poisson with the parameter $n\lambda$, and you can also check the mean:</p>
<p>$$\mathbb{E}(X_1+\ldots+X_n)=f'(1)=n\lambda.$$</p>
|
3,559,167 | <p>I know that it is a bounded below set and the infimum is 4, but I'm unsure of going about how to prove that it is indeed bounded. Any help would be greatly appreciated! </p>
| RishiNandha Vanchi | 633,100 | <p>Consider Positive Values for x alone for now,</p>
<p>AM-GM</p>
<p><span class="math-container">$x + \frac 4x \geq 2 \sqrt {x \frac 4x}$</span></p>
<p><span class="math-container">$f(x) \geq 4$</span></p>
<p>It is Easy to see that it is a monotonically increasing function beyond x = 2 by differentiating,</p>
<p>Thus it suffices to show that Limit at Infinity Is not Infinity, Easy to see that It is,</p>
<p>Noe generalize this to -ve side as it is an odd function. This gives you the Range <span class="math-container">$(-\infty,-2] \ U \ [2,\infty]$</span> and thus it's unbounded.</p>
|
3,564,476 | <p>I'm stuck on the following problem: </p>
<p><span class="math-container">$$\int_{\frac\pi{12}}^{\frac\pi2}(1-\cos4x)\cos2x\>dx$$</span></p>
<p>I think I can use the double angle formulas here but I'm not sure how to apply it, or even if it's the right approach. I'm also not sure if 1-cos4x can be translated into anything. </p>
<p><a href="https://i.stack.imgur.com/8oKQW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8oKQW.png" alt="enter image description here"></a></p>
<p>A step-through would be greatly appreciated. </p>
| Bernard | 202,857 | <p><strong>Hint</strong>:</p>
<p>For this type of integrals, use the linearisation formulæ:
<span class="math-container">$$2\sin a\sin b=\cos(a-b)-\cos(a+b),\qquad2\cos a\cos b=\cos(a-b)+\cos(a+b),$$</span>
<span class="math-container">$$2\sin a\cos b=\sin(a-b)+\sin(a+b.$$</span></p>
<p><em>Some details</em>:</p>
<p><span class="math-container">\begin{align}
\int_{\frac\pi{12}}^{\frac\pi2}(1-\cos4x)\cos2x\,\mathrm dx&=\int_{\frac\pi{12}}^{\frac\pi2}(\cos 2x-\cos4x\cos2x)\,\mathrm dx\\
&=\int_{\frac\pi{12}}^{\frac\pi2}\bigl(\cos 2x-\frac12(\cos2x-\cos6x)\bigr)\,\mathrm dx \\
&= \frac12\int_{\frac\pi{12}}^{\frac\pi2}\bigl(\cos 2x+\cos6x)\,\mathrm dx = \frac14\sin 2x+\frac1{12}\sin 6x\,\Biggr|_{\frac\pi{12}}^{\frac\pi2}
\end{align}</span></p>
|
24,927 | <p>Does this mean that the first homotopy group in some sense contains more information than the higher homotopy groups? Is there another generalization of the fundamental group that can give rise to non-commutative groups in such a way that these groups contain more information than the higher homotopy groups? </p>
| Qiaochu Yuan | 232 | <p>Thinking about the higher homotopy groups as just groups is in some sense missing the point. The higher homotopy groups are not just abelian groups: they are $\pi_1$-modules, for one thing.</p>
<p>More loftily, from the <a href="http://ncatlab.org/nlab/show/nPOV">n-categorical point of view</a>, the homotopy groups are really just a convenient stand-in for a more fundamental structure, the <a href="http://ncatlab.org/nlab/show/fundamental+infinity-groupoid">fundamental $\infty$-groupoid</a> of a space. Roughly speaking, the fundamental $\infty$-groupoid is a gadget that incorporates information about the paths between points, homotopies between paths, homotopies between homotopies between paths, and so forth.</p>
<p>It is possible to truncate the fundamental $\infty$-groupoid into a collection of easier-to-understand objects, the fundamental $n$-groupoids $\Pi_n$:</p>
<ul>
<li>The fundamental $0$-groupoid $\Pi_0$ is just the set of connected components. </li>
<li>The fundamental $1$-groupoid is the <a href="http://ncatlab.org/nlab/show/groupoid">groupoid</a> of homotopy classes of paths between points; it is a generalization of the fundamental group that is independent of basepoint. If the space is connected, the fundamental $1$-groupoid is equivalent to the category with a single object whose morphisms are the elements of the fundamental group $\pi_1$.</li>
<li>The fundamental $2$-groupoid is the <a href="http://nlab.mathforge.org/nlab/show/2-groupoid">$2$-groupoid</a> of paths and homotopy classes of homotopies between them; it is a generalization of the action of $\pi_1$ on $\pi_2$ that is independent of basepoint. </li>
</ul>
<p>And so forth: more generally the fundamental $n$-groupoid is a generalization of the relationship between the first $n$ homotopy groups. Unfortunately I can't think of a nice reference to these ideas off the top of my head; I've gleaned them from several sources. The references in Baez and Shulman's <a href="http://arxiv.org/abs/math/0608420">Lectures on n-categories and cohomology</a> might be a good start. </p>
|
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