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245,049 | <p>I am trying to do a n-round of convolution of a function. The code is posted as below. But it is not working. Is there a solution?</p>
<pre><code>p[x_] := 1/(x + 1)*UnitStep[x]
p1[x_] := Convolve[p[y], p[y], y, x]
p2[x_] := Convolve[p[y], p1[y], y, x]
</code></pre>
<p>p1 succeeded. But the output of p2 only repeats the question as follows:</p>
<pre><code>Convolve[UnitStep[y]/(1 + y),
Convolve[UnitStep[y]/(1 + y), UnitStep[y]/(1 + y), y, y], y, x]
</code></pre>
| bill s | 1,783 | <p>There are two issues: do you have the form right, and is the integral (the convolution) solvable. Let's do the first one by picking a really simple example, like p[x]:=UnitStep[x]. You can see that your code fails. Because of the way things are defined, you need to change the variables, or they get all messed up. Note that this works:</p>
<pre><code>p[x_] := UnitStep[x]
p1[x_] := Convolve[p[y], p[y], y, x]
p2[x_] := Convolve[p[z], p1[z], z, x]
p3[x_] := Convolve[p[w], p2[w], w, x]
{p[x],p1[x],p2[x],p3[x]}
</code></pre>
<p>Now change p[x] to your desired function... you'll see it works fine for p1[x] and p2[x]. I got bored waiting to see if p3[x] would finish. But at some point, this is going to fail because the functions are getting too hard to integrate in closed form.</p>
|
4,277,616 | <blockquote>
<p>In how many different ways can we arrange <span class="math-container">$120$</span> students into <span class="math-container">$6$</span> groups for <span class="math-container">$6$</span> different classes so that the largest group has at most <span class="math-container">$2$</span> members more than the smallest group?</p>
</blockquote>
<p>My initial plan was to use a generating function, but I stumbled across a problem. Let's mark the groups with numbers <span class="math-container">$1$</span> to <span class="math-container">$6$</span> and let <span class="math-container">$n_i,i\in\{1,\ldots,6\}$</span> denote the number of members of the <span class="math-container">$i-$</span> th group in some arrangment. To see where this would lead me, for a moment, I assumed <span class="math-container">$n_1\le n_2\le\cdots\le n_6\le n_1+2$</span> in hope to find some range <span class="math-container">$\{m,\ldots M\}$</span> for <span class="math-container">$n_i$</span>'s and use a generating function <span class="math-container">$f(x)=(x^m+\cdots+x^M)^6$</span> and find <span class="math-container">$\langle x^{120}\rangle-$</span> the coefficient in front of <span class="math-container">$x^{120}$</span>, however students are distinct entities and <span class="math-container">$m$</span> and <span class="math-container">$M$</span> still remained misterious. I then tried figuring out if I was on the somewhat right track by, again taking <span class="math-container">$m=\min\{n_1,\ldots, n_6\}$</span> and write <span class="math-container">$n_i=m+j_i, j_i\in\{0,1,2\}.$</span> I believe, an arrangement with <span class="math-container">$2$</span> groups of <span class="math-container">$19, 2$</span> groups of <span class="math-container">$20$</span> and <span class="math-container">$2$</span> groups of <span class="math-container">$21$</span> people suggests there should be at least <span class="math-container">$19$</span> people in each group.</p>
<p>I also had a look at this problem:</p>
<p><a href="https://math.stackexchange.com/q/2386814/942722">The number of the partition of the set $A$ into $k$ bounded blocks.</a></p>
<p>but, again, I don't have any bounds on the blocks.</p>
<p>How should I proceed?</p>
| Salmon Fish | 955,791 | <p>For an another approach : When we check over the question , we see that it is distributing distingusiable objects into distinguishable boxes question. By that reason , using exponential generating function is more suitable than using ordinary generating functions.</p>
<p>It is given that the largest class can have at most <span class="math-container">$2$</span> more students than the smallest. Then , by using the reason given by @Yorch , we see that a group can have at least <span class="math-container">$19$</span> and at most <span class="math-container">$21$</span> students. Then if the groups will be distingusihable then our exponential generating function of each group will be <span class="math-container">$$\bigg(\frac{x^{19}}{19!} + \frac{x^{20}}{20!} +\frac{x^{21}}{21!} \bigg)$$</span></p>
<p>Then ,we should find the coefficient of <span class="math-container">$\frac{x^{120}}{120!}$</span> or find the coefficient of <span class="math-container">$x^{120}$</span> and multiply it by <span class="math-container">$120!$</span> in the expansion of <span class="math-container">$$\bigg(\frac{x^{19}}{19!} + \frac{x^{20}}{20!} +\frac{x^{21}}{21!} \bigg)^6$$</span></p>
<p>However , if the groups will be <span class="math-container">$\color{red}{\text{indistinguishable}}$</span> , then multiply the result of <span class="math-container">$$[x^{120}]\bigg(\frac{x^{19}}{19!} + \frac{x^{20}}{20!} +\frac{x^{21}}{21!} \bigg)^6$$</span>
by <span class="math-container">$\frac{1}{6!}$</span></p>
<p>Then , the result is <span class="math-container">$$\frac{1}{6!} \times [x^{120}]\bigg(\frac{x^{19}}{19!} + \frac{x^{20}}{20!} +\frac{x^{21}}{21!} \bigg)^6$$</span></p>
<p><em><a href="https://www.wolframalpha.com/input/?i=coefficient+of+%28x%5E120++%29+in+%28x%5E19+%2F19%21+%2B+x%5E20+%2F20%21+%2B+x%5E21+%2F21%21%29%5E6" rel="nofollow noreferrer">Wolfram -calculation</a></em></p>
<p>When we calculate the result using given coefficient in the link , the answer is <span class="math-container">$5.756410166785662e+87$</span></p>
|
2,838,312 | <p>What is the Fourier transform of $\mathrm{e}^{ik|x|}$? Here, $k > 0$ is real.</p>
<p>I use the definition $$ F(\omega) = \int_{-\infty}^\infty \mathrm{e}^{-i\omega x} f(x) \mathrm{d}x.$$</p>
<p>Thanks!</p>
| José Carlos Santos | 446,262 | <p>Note that if $|z|=r$, then\begin{align}|a_{n-1}z^{n-1}+\cdots+a_1z+a_0|&\leqslant|a_{n-1}|r^{n-1}+\cdots+|a_1|r+|a_0|\\&\leqslant(r-1)(r^{n-1}+\cdots+r+1)\\&=r^n-1\\&<r^n\\&=|z|^n.\end{align}Therefore, by Rouché's theorem, $z^n$ and $p(z)$ have the same number of zeros in that region. But $z^n$ has $n$ zeros there (counting with multiplicities).</p>
|
1,281,627 | <p>Today I completed the chapter of '**Limits **' in my school, and I found this chapter very fascinating. But the only problem I have with limits and Derivatives is that I don't know How can I use it in my daily life. (Any Book Recommendation?)</p>
| user157227 | 157,227 | <p>You shouldn't view limits as a tool to solve problems. Instead, you should view limits as a way to describe situations (or ask more interesting problems).</p>
<p>The derivative is a perfect example of this. If you want to express the idea of "instantaneous rate of change," you are going to use limits to do this.</p>
<p>For another example, suppose you have a biased coin and you want to know how often you will get heads when you flip this coin. You know that if you flip the coin many times you will be able to approximate the probability of heads with some degree of confidence. Furthermore, you know that the larger the number of times you flip, the more confident you can be and you know that if you flip enough times you can be as confident as you want.</p>
<p>This is a scenario that would make a lot of sense to describe using limits.</p>
|
997,999 | <p>I read that integration is the opposite of differentiation AND at the same time is a summation process to find the area under a curve. But I can't understand how these things combine together and actually an integral can be the same time those two things. If the integration is the opposite of differentiation, then the result of the integration should be the initial function from which we derived the derivative function. How can this be at the same time the area under the curve of the (which?) function?</p>
| Ivo Terek | 118,056 | <p>Integration is not exactly the opposite as differentiation. And the interpretation of the integral as the area under a curve is good for positive functions only, otherwise you must keep your attention for signs. These things combine together by the fundamental theorem of calculus, which states that if $f$ is a continuous function, then: $$\int_a^b f(x) \ \mathrm{d}x = F(b) - F(a),$$ where $F$ is any primitive of $f$.</p>
|
2,214,236 | <p>The question:</p>
<blockquote>
<p>An object is dropped from a cliff. How far does the object fall in the 3rd second?"</p>
</blockquote>
<p>I calculated that a ball dropped from rest from a cliff will fall $45\text{ m}$ in $3 \text{ s}$, assuming $g$ is $10\text{ m/s}^2$.</p>
<p>$$s = (0 \times 3) + \frac{1}{2}\cdot 10\cdot (3\times 3) = 45\text{ m}$$</p>
<p>But my teacher is telling me $25\text{ m}$! </p>
<p>EDITS: His reasoning was that from $t=0$ to $t=1$, $s=10\text{ m}$, and from $t=1$ to $t =2$, $s=20$...</p>
<p>The mark scheme also says $25\text{ m}$</p>
| Chen Mo | 431,584 | <p>You are correct. If the question is exactly how you phrased it, the displacement should be 45m down. </p>
<p>We know that:</p>
<p>a (acceleration) = $10m/s^2$</p>
<p>t (time) = 3 seconds</p>
<p>u (initial velocity) = $0m/s$</p>
<p>Hence, using the formula $s=ut-(1/2)at^2$ the answer should be 45 m.</p>
|
30,918 | <p>Judging by some of the posts on meta<sup>1</sup> and comments posted there it seems that there are users who try to improve the posts by correcting spelling mistakes. Of course, there are other ways to improve the posts via editing, some of them probably more important than grammar and spelling.<sup>2</sup> Still this is certainly an improvement, a post with correct spelling is easier to read and using English correctly also increases the professional appearance of this site. Moreover, it helps with searching. (Posts which contain only the misspelled form of the words are less likely to be found when searching.)</p>
<p>Maybe it could be useful to <em>collect some commonly misspelled words</em>. This could make it easier for users, who have some spare time and want to do a few such edits, to find posts which have this problem.</p>
<p>For this reason it might be useful also include link to a search (or a <a href="https://math.meta.stackexchange.com/tags/data-explorer/info">SEDE query</a>) which returns those posts. Some minor comments on searching:</p>
<ul>
<li>For some words, the search also returns posts which contain minor variants of the words. Since we are searching for misspellings, in such cases adding quotation marks might help to find the misspelled ones. (For example, searching for <a href="https://math.stackexchange.com/search?q=occurence">occurence</a> returns many results, but we are more interested in the ones which are found when we search for <a href="https://math.stackexchange.com/search?q="occurence"">"occurence"</a> and <a href="https://math.stackexchange.com/search?q="occurences"">"occurences"</a>.)</li>
<li>A simple SEDE query searching for some string can return words which contain this string as a some part. So sometimes we might want to look only at results containing exactly the given word. (Compare the results <a href="https://data.stackexchange.com/math/query/1157659/posts-containing-a-given-text-case-insensitive?num=5000&text=banch?num=5000&text=teh" rel="nofollow noreferrer">searching for teh</a> with the results where <a href="https://data.stackexchange.com/math/query/1157664/search-for-exact-string-case-insensitive?text=teh" rel="nofollow noreferrer">teh is preceded and followed by a non-letter</a>).</li>
<li>It is better to use case insensitive version of SEDE queries. (So that we catch also misspelled word at the beginning of a sentence. This is especially important for names.)</li>
<li>With SEDE you can find several words which contain the same substring. For example, words containing "anali" might often be typos (such as analisis, analise, analitic, etc.} and you can search for <a href="https://data.stackexchange.com/math/query/1157659/posts-containing-a-given-text-case-insensitive?num=5000&text=anali" rel="nofollow noreferrer">anali</a>.</li>
</ul>
<p>A few comments about editing the misspellings. (Some of them apply to editing in general.)</p>
<ul>
<li>It is better to avoid <a href="https://math.meta.stackexchange.com/questions/5068/how-much-bumping-is-too-much">bumping too many old posts at the same time</a>. (Frontpage is considered a precious commodity.)</li>
<li>For this reason it is better to concentrate on questions which are new or have been recently bumped for some other reason.</li>
<li>If you edit some post, try to check whether there are also other improvements to do. (Since the post was bumped by correcting the misspelled word, it is better to improve also other things which are worth editing than to bump the same post by additional edits later.)</li>
<li>Difference between British/American spelling should not be considered a misspelling, as discussed before: <a href="https://math.meta.stackexchange.com/q/30335">Is English (US) vs English (UK) grounds for an edit?</a></li>
</ul>
<p>Some names of mathematicians which are commonly misspelled can also be found in this answer: <a href="https://math.meta.stackexchange.com/q/8698#8709">Searching for accented characters is too strict</a>. (Although that answer is primarily about alternative spellings.) This might also help when creating list of common misspellings.</p>
<p><sup>1</sup>Posts such as: <a href="https://math.meta.stackexchange.com/q/13730">Edit session for wrong spelling of mathematicians and mathematical concepts</a> or
<a href="https://math.meta.stackexchange.com/q/30912">Wrong spelling of “occurrence”</a>. Some users have mentioned in chat that they do this kind of edits, for example, <a href="https://chat.stackexchange.com/transcript/message/2443651#2443651">Srivatsan</a>.</p>
<p><sup>2</sup>Just to name a few common problems with posts on this site which could be helped by editing: <a href="https://math.meta.stackexchange.com/questions/9959/how-to-ask-a-good-question#10144">non-descriptive titles</a>, incorrectly tagged question, various problems with MathJax in the post or <a href="https://math.meta.stackexchange.com/questions/9687/guidelines-for-good-use-of-rm-latex-in-question-titles">in the title</a>, etc. (Still, I'd guess that posts with misspellings are slightly more likely to have also some other problems - so perhaps checking them from time to time might be useful also for other purposes.)</p>
<p><sup>3</sup>These SEDE queries are more for people who are curious (since they do not actually help with editing), but they still might be interesting anyway. You can find <a href="https://data.stackexchange.com/math/query/1157683/posts-which-contained-the-text-in-some-revision-and-it-was-later-removed-case-in?word=haussdorff" rel="nofollow noreferrer">posts where a specific word has already been corrected</a>. You can also search in comments <a href="https://data.stackexchange.com/math/query/556789/comments-containing-given-keyword-with-text-and-author?Word=occurence" rel="nofollow noreferrer">for some string</a> or for <a href="https://data.stackexchange.com/math/query/1157639/comments-containing-exact-string-with-text-and-author?text=teh" rel="nofollow noreferrer">an exact word</a>.</p>
| Martin Sleziak | 8,297 | <p>Strings which most likely appear in the words which are
misspelled:</p>
<ul>
<li>alegb*; SEDE <a href="//data.stackexchange.com/math/query/1157659/posts-containing-a-given-text-case-insensitive?num=5000&text=alegb" rel="nofollow noreferrer">anywhere</a>, <a href="//data.stackexchange.com/math/query/1157688/search-for-string-at-the-beginning-of-a-word-case-insensitive?text=alegb" rel="nofollow noreferrer">beginning</a></li>
<li>anali*; SEDE <a href="//data.stackexchange.com/math/query/1157659/posts-containing-a-given-text-case-insensitive?num=5000&text=anali" rel="nofollow noreferrer">anywhere</a>, <a href="//data.stackexchange.com/math/query/1157688/search-for-string-at-the-beginning-of-a-word-case-insensitive?text=anali" rel="nofollow noreferrer">beginning</a></li>
<li>anay*; SEDE <a href="//data.stackexchange.com/math/query/1157659/posts-containing-a-given-text-case-insensitive?num=5000&text=anay" rel="nofollow noreferrer">anywhere</a>, <a href="//data.stackexchange.com/math/query/1157688/search-for-string-at-the-beginning-of-a-word-case-insensitive?text=anay" rel="nofollow noreferrer">beginning</a></li>
<li>ananl*; SEDE <a href="//data.stackexchange.com/math/query/1157659/posts-containing-a-given-text-case-insensitive?num=5000&text=ananl" rel="nofollow noreferrer">anywhere</a>, <a href="//data.stackexchange.com/math/query/1157688/search-for-string-at-the-beginning-of-a-word-case-insensitive?text=ananl" rel="nofollow noreferrer">beginning</a></li>
<li>assymp*; SEDE: <a href="//data.stackexchange.com/math/query/1157659/posts-containing-a-given-text-case-insensitive?num=5000&text=assymp" rel="nofollow noreferrer">anywhere</a>, <a href="//data.stackexchange.com/math/query/1157688/search-for-string-at-the-beginning-of-a-word-case-insensitive?text=assymp" rel="nofollow noreferrer">beginning</a></li>
<li>asypto*; SEDE: <a href="//data.stackexchange.com/math/query/1157659/posts-containing-a-given-text-case-insensitive?num=5000&text=asypto" rel="nofollow noreferrer">anywhere</a>, <a href="//data.stackexchange.com/math/query/1157688/search-for-string-at-the-beginning-of-a-word-case-insensitive?text=asypto" rel="nofollow noreferrer">beginning</a></li>
<li>caracter*; SEDE: <a href="//data.stackexchange.com/math/query/1157659/posts-containing-a-given-text-case-insensitive?num=5000&text=caracter" rel="nofollow noreferrer">anywhere</a>, <a href="//data.stackexchange.com/math/query/1157688/search-for-string-at-the-beginning-of-a-word-case-insensitive?text=caracter" rel="nofollow noreferrer">beginning</a></li>
<li>contino*; SEDE: <a href="//data.stackexchange.com/math/query/1157659/posts-containing-a-given-text-case-insensitive?num=5000&text=contino" rel="nofollow noreferrer">anywhere</a>, <a href="//data.stackexchange.com/math/query/1157688/search-for-string-at-the-beginning-of-a-word-case-insensitive?text=contino" rel="nofollow noreferrer">beginning</a></li>
<li>defint*; SEDE: <a href="//data.stackexchange.com/math/query/1157659/posts-containing-a-given-text-case-insensitive?num=5000&text=defint" rel="nofollow noreferrer">anywhere</a>, <a href="//data.stackexchange.com/math/query/1157688/search-for-string-at-the-beginning-of-a-word-case-insensitive?text=defint" rel="nofollow noreferrer">beginning</a></li>
<li>definito*; SEDE: <a href="//data.stackexchange.com/math/query/1157659/posts-containing-a-given-text-case-insensitive?num=5000&text=definito" rel="nofollow noreferrer">anywhere</a>, <a href="//data.stackexchange.com/math/query/1157688/search-for-string-at-the-beginning-of-a-word-case-insensitive?text=definito" rel="nofollow noreferrer">beginning</a></li>
<li>defni*; SEDE: <a href="//data.stackexchange.com/math/query/1157659/posts-containing-a-given-text-case-insensitive?num=5000&text=defni" rel="nofollow noreferrer">anywhere</a>, <a href="//data.stackexchange.com/math/query/1157688/search-for-string-at-the-beginning-of-a-word-case-insensitive?text=defni" rel="nofollow noreferrer">beginning</a></li>
<li>deriav*; SEDE: <a href="//data.stackexchange.com/math/query/1157659/posts-containing-a-given-text-case-insensitive?num=5000&text=deriav" rel="nofollow noreferrer">anywhere</a>, <a href="//data.stackexchange.com/math/query/1157688/search-for-string-at-the-beginning-of-a-word-case-insensitive?text=deriav" rel="nofollow noreferrer">beginning</a></li>
<li>difer*; SEDE: <a href="//data.stackexchange.com/math/query/1157659/posts-containing-a-given-text-case-insensitive?num=5000&text=difer" rel="nofollow noreferrer">anywhere</a>, <a href="//data.stackexchange.com/math/query/1157688/search-for-string-at-the-beginning-of-a-word-case-insensitive?text=difer" rel="nofollow noreferrer">beginning</a></li>
<li>evalau*; SEDE: <a href="//data.stackexchange.com/math/query/1157659/posts-containing-a-given-text-case-insensitive?num=5000&text=evalau" rel="nofollow noreferrer">anywhere</a>, <a href="//data.stackexchange.com/math/query/1157688/search-for-string-at-the-beginning-of-a-word-case-insensitive?text=evalau" rel="nofollow noreferrer">beginning</a></li>
<li>fucn*; SEDE: <a href="//data.stackexchange.com/math/query/1157659/posts-containing-a-given-text-case-insensitive?num=5000&text=fucn" rel="nofollow noreferrer">anywhere</a>, <a href="//data.stackexchange.com/math/query/1157688/search-for-string-at-the-beginning-of-a-word-case-insensitive?text=fucn" rel="nofollow noreferrer">beginning</a></li>
<li>funcion*; SEDE: <a href="//data.stackexchange.com/math/query/1157659/posts-containing-a-given-text-case-insensitive?num=5000&text=funcion" rel="nofollow noreferrer">anywhere</a>, <a href="//data.stackexchange.com/math/query/1157688/search-for-string-at-the-beginning-of-a-word-case-insensitive?text=funcion" rel="nofollow noreferrer">beginning</a></li>
<li>funtion*; SEDE: <a href="//data.stackexchange.com/math/query/1157659/posts-containing-a-given-text-case-insensitive?num=5000&text=funtion" rel="nofollow noreferrer">anywhere</a>, <a href="//data.stackexchange.com/math/query/1157688/search-for-string-at-the-beginning-of-a-word-case-insensitive?text=funtion" rel="nofollow noreferrer">beginning</a></li>
<li>intega*; SEDE: <a href="//data.stackexchange.com/math/query/1157659/posts-containing-a-given-text-case-insensitive?num=5000&text=intega" rel="nofollow noreferrer">anywhere</a>, <a href="//data.stackexchange.com/math/query/1157688/search-for-string-at-the-beginning-of-a-word-case-insensitive?text=intega" rel="nofollow noreferrer">beginning</a></li>
<li>integt*; SEDE: <a href="//data.stackexchange.com/math/query/1157659/posts-containing-a-given-text-case-insensitive?num=5000&text=integt" rel="nofollow noreferrer">anywhere</a>, <a href="//data.stackexchange.com/math/query/1157688/search-for-string-at-the-beginning-of-a-word-case-insensitive?text=integt" rel="nofollow noreferrer">beginning</a></li>
<li>interg*; SEDE: <a href="//data.stackexchange.com/math/query/1157659/posts-containing-a-given-text-case-insensitive?num=5000&text=interg" rel="nofollow noreferrer">anywhere</a>, <a href="//data.stackexchange.com/math/query/1157688/search-for-string-at-the-beginning-of-a-word-case-insensitive?text=interg" rel="nofollow noreferrer">beginning</a></li>
<li>mimin*; SEDE: <a href="//data.stackexchange.com/math/query/1157659/posts-containing-a-given-text-case-insensitive?num=5000&text=mimin" rel="nofollow noreferrer">anywhere</a>, <a href="//data.stackexchange.com/math/query/1157688/search-for-string-at-the-beginning-of-a-word-case-insensitive?text=mimin" rel="nofollow noreferrer">beginning</a></li>
<li>minum*; SEDE: <a href="//data.stackexchange.com/math/query/1157659/posts-containing-a-given-text-case-insensitive?num=5000&text=minum" rel="nofollow noreferrer">anywhere</a>,
<a href="//data.stackexchange.com/math/query/1157688/search-for-string-at-the-beginning-of-a-word-case-insensitive?text=minum" rel="nofollow noreferrer">beginning</a></li>
<li>polino*; SEDE: <a href="//data.stackexchange.com/math/query/1157659/posts-containing-a-given-text-case-insensitive?num=5000&text=polino" rel="nofollow noreferrer">anywhere</a>, <a href="//data.stackexchange.com/math/query/1157688/search-for-string-at-the-beginning-of-a-word-case-insensitive?text=polino" rel="nofollow noreferrer">beginning</a></li>
<li>subsit*; SEDE: <a href="//data.stackexchange.com/math/query/1157659/posts-containing-a-given-text-case-insensitive?num=5000&text=subsit" rel="nofollow noreferrer">anywhere</a>, <a href="//data.stackexchange.com/math/query/1157688/search-for-string-at-the-beginning-of-a-word-case-insensitive?text=subsit" rel="nofollow noreferrer">beginning</a></li>
<li>toplog*; SEDE: <a href="//data.stackexchange.com/math/query/1157659/posts-containing-a-given-text-case-insensitive?num=5000&text=toplog" rel="nofollow noreferrer">anywhere</a>, <a href="//data.stackexchange.com/math/query/1157688/search-for-string-at-the-beginning-of-a-word-case-insensitive?text=toplog" rel="nofollow noreferrer">beginning</a></li>
<li>varit*; SEDE: <a href="//data.stackexchange.com/math/query/1157659/posts-containing-a-given-text-case-insensitive?num=5000&text=varit" rel="nofollow noreferrer">anywhere</a>, <a href="//data.stackexchange.com/math/query/1157688/search-for-string-at-the-beginning-of-a-word-case-insensitive?text=varit" rel="nofollow noreferrer">beginning</a></li>
<li>*morophism; SEDE: <a href="//data.stackexchange.com/math/query/1157659/posts-containing-a-given-text-case-insensitive?num=5000&text=morophism" rel="nofollow noreferrer">anywhere</a>, <a href="//data.stackexchange.com/math/query/1158167/search-for-string-at-the-end-of-a-word-case-insensitive?text=morophism" rel="nofollow noreferrer">end</a></li>
<li>*morophi*; SEDE: <a href="//data.stackexchange.com/math/query/1157659/posts-containing-a-given-text-case-insensitive?num=5000&text=morophi" rel="nofollow noreferrer">anywhere</a></li>
<li>*finte*; SEDE: <a href="//data.stackexchange.com/math/query/1157659/posts-containing-a-given-text-case-insensitive?num=5000&text=finte" rel="nofollow noreferrer">anywhere</a></li>
</ul>
<p>Phrases (consisting of more than one word) which might often indicate a mistake or misspelling:</p>
<ul>
<li>a dice; <a href="//math.stackexchange.com/search?tab=active&q=%22a+dice%22">search</a>, <a href="//data.stackexchange.com/math/query/1157664/search-for-exact-string-case-insensitive?num=5000&text=a%20dice" rel="nofollow noreferrer">SEDE</a></li>
<li>Baye's theorem; <a href="//math.stackexchange.com/search?tab=active&q=%22baye%27s+theorem%22">search</a>, <a href="//data.stackexchange.com/math/query/1157664/search-for-exact-string-case-insensitive?num=5000&text=baye%27%27s%20theorem" rel="nofollow noreferrer">SEDE</a></li>
<li>I thing; <a href="//math.stackexchange.com/search?tab=active&q=%22I+thing%22">search</a>, <a href="//data.stackexchange.com/math/query/1157664/search-for-exact-string-case-insensitive?num=5000&text=i%20thing" rel="nofollow noreferrer">SEDE</a></li>
<li>linear application; <a href="//math.stackexchange.com/search?tab=active&q=%22linear+application%22">search</a>, <a href="//data.stackexchange.com/math/query/1157664/search-for-exact-string-case-insensitive?num=5000&text=linear%20application" rel="nofollow noreferrer">SEDE</a></li>
<li>Stoke's theorem; <a href="//math.stackexchange.com/search?tab=active&q=%22stoke%27s+theorem%22">search</a>, <a href="//data.stackexchange.com/math/query/1157664/search-for-exact-string-case-insensitive?num=5000&text=stoke%27%27s%20theorem" rel="nofollow noreferrer">SEDE</a></li>
<li>to proof; <a href="//math.stackexchange.com/search?tab=active&q=%22to+proof%22">search</a>, <a href="//data.stackexchange.com/math/query/1157664/search-for-exact-string-case-insensitive?num=5000&text=to%20proof" rel="nofollow noreferrer">SEDE</a></li>
</ul>
|
948,862 | <p>Consider the <a href="http://en.wikipedia.org/wiki/Fenchel's_duality_theorem" rel="noreferrer">Fenchel dual</a> and the <a href="http://en.wikipedia.org/wiki/Duality_(optimization)" rel="noreferrer">Lagrangian dual</a>. </p>
<p>Are these duals equivalent? In other words, is using one of the these duals (say for solving an optimization), would give the same answer as using the other one? </p>
<p>I think the answer is no, but I am not sure. One reason for saying that, is that, in the Lagrange dual, we have a relatively straightforward way to add the constraints into the objective function. But what about the Fenchel? I have not seen any. </p>
<p>But I have seen some problems in which both of these give same answers. So, I would assume that, on a subset of problems, these two dualities, are exactly the same. </p>
<p>And also, if they are different, how would you choose which one to use on your problem? </p>
| Pushpendre | 52,858 | <p>Let the fenchel dual of a function $f : \mathbb{R}^n \to \mathbb{R}$ be $D_f$. Consider the convex program $P^{+} = \min f(x)$ s.t. $x \ge \boldsymbol{0}$ and let $D_l$ be this program's lagrangian dual then for $\boldsymbol{\lambda} \ge \boldsymbol{0}$ $D_f(\lambda) = -D_l(\lambda)$. So on the positive orthant the fenchel dual agrees with the lagrangian dual of $P^{+}$.</p>
<p>Similarly on the negative orthant $D_f$ agrees with the dual of $P^{-} = \min f(x)$ s.t. $x \le \boldsymbol{0}$. </p>
<p>The general case is let $P^{A,b} = \min f(x)$ s.t. $Ax \le b$ and let $D_l^{A,b}$ be its dual, then for $\boldsymbol{\lambda} \ge \boldsymbol{0}$ $D_f(-A^T\boldsymbol{\lambda}) + D_l^{A,b}(\boldsymbol{\lambda}) + \boldsymbol{\lambda}^Tb=0$. </p>
<p>So the Fenchel dual is stitched together from lots of lagrange duals.
Or conversely the lagrange dual of convex problems on polyhedrons can be constructed from the fenchel dual.</p>
|
948,862 | <p>Consider the <a href="http://en.wikipedia.org/wiki/Fenchel's_duality_theorem" rel="noreferrer">Fenchel dual</a> and the <a href="http://en.wikipedia.org/wiki/Duality_(optimization)" rel="noreferrer">Lagrangian dual</a>. </p>
<p>Are these duals equivalent? In other words, is using one of the these duals (say for solving an optimization), would give the same answer as using the other one? </p>
<p>I think the answer is no, but I am not sure. One reason for saying that, is that, in the Lagrange dual, we have a relatively straightforward way to add the constraints into the objective function. But what about the Fenchel? I have not seen any. </p>
<p>But I have seen some problems in which both of these give same answers. So, I would assume that, on a subset of problems, these two dualities, are exactly the same. </p>
<p>And also, if they are different, how would you choose which one to use on your problem? </p>
| theV0ID | 82,092 | <p>From your comments, I understand that your confusion is about the relation of Fenchel and Lagrange duality. So I will focus on that relation. Particularly, I would like to contradict the comment which stated, that Lagrange and Fenchel duality are two categorically different concepts.</p>
<p>Consider the <em>primal problem</em>
$$\nu(0) = \inf_{x \in \mathbb R^n} \varphi(x,0) = \inf_{x \in \mathbb R^n} f(x)\text,$$
where $\varphi(x,u)$ is the <a href="https://math.stackexchange.com/a/624633/82092"><em>perturbation function</em></a> of $f$ with $\varphi(x,0) = f(x)$.</p>
<p>Using Fenchel conjugates, one can show that
$$\nu^{\ast\ast}(0) = \sup_{p\in\mathbb R^m} -\varphi^\ast(0,p)\text,$$
which is known as the <em>dual problem</em>.</p>
<p>The relation $\nu^{\ast\ast} \leq \nu$ is known as <em>weak duality</em>, and it is immediately seen from $\nu^{\ast\ast}$ being the Fenchel biconjugate of $\nu$. The difference $\nu(0) - \nu^{\ast\ast}(0)$ is the famous <em>duality gap</em>. As you probably know, strong duality refers to the special case $\nu^{\ast\ast}=\nu$, which occurs if $f$ is convex and some other mathematical requirements are satisfied.</p>
<p>Why am I recalling all this? Because you should note, that all these concepts exist despite that we haven't even mentioned Lagrangians yet. These concepts are part of the <em>Fenchel duality</em>.</p>
<p>Now, consider the <strong>special case</strong> that $$f(x) = f_0(x) + g(F(x))$$
with the perturbation function $$\varphi(x,u) = f_0(x) + g(F(x) + u)\text,$$ such that $\varphi(x,0) = f(x)$ is still satisfied.</p>
<p>With $f_0 : \mathbb R^n \to \overline{\mathbb R}$, $g : \mathbb R^m \to \overline{\mathbb R}$, and $F : \mathbb R^n \to \mathbb R^m$, where $$\overline{\mathbb R} = \mathbb R \cup \{-\infty, +\infty\}$$ is the extension of the real line, this primal problem implicitly induces the feasibility set $\mathcal G = \{ x \in \operatorname{dom} f_0 : F(x) \in \operatorname{dom} g \}$, where $\operatorname{dom}$ denotes the <em>effective domain</em>, that is the set of arguments to a function, where the function is not infinity-valued.</p>
<p>By calculating the Fenchel conjugate $\varphi^\ast$, we obtain the dual problem $$\nu^{\ast\ast}(0) = \sup_{p \in \mathbb R^m} \inf_{x \in \mathbb R^n} L(x,p) - g^\ast(p) \text,$$
where $$L(x,p) = f_0(x) + \langle F(x), p \rangle$$ is called the <em>Lagrangian</em>.</p>
<p>Now, we specialize further. Consider the prominent case, that our primal problem was to minimize a convex function $f_0(x)$ with $\operatorname{dom} f_0 = \mathbb R^n$ subject to $f_i(x) \leq 0$ for all $i = 1, \dots, k$ and $f_i(x) = 0$ for the remaining i = $k+1, \dots, m$. Using the notation of an <em>indicator function</em> $$\delta_K(p) = \begin{cases} 0 & \text{if } p \in K \\ +\infty & \text{else,} \end{cases}$$ we can write the primal problem like $$\inf_{x \in \mathbb R^n} f_0(x) + \delta_K(F(x))\text,$$ where we set $g = \delta_K$, choose $F^{\mathsf T} = \begin{bmatrix}f_1, \dots, f_m\end{bmatrix}$, and $K = \mathbb R^k_- \times \{0\}^{m-k}$ correspondingly. Since the Fenchel conjugate of the indicator function of a closed convex cone $K$ equals to the indicator function of the <em>polar cone</em> $K^\ast$, that is, $\delta_K^\ast = \delta_{K^\ast}$, the dual problem reads $$\nu^{\ast\ast}(0) = \sup_{p \in \mathbb R^m} \inf_{x \in \mathbb R^n} L(x,p) - \delta_{K^\ast}(p) = \sup_{p \in K^\ast} \inf_{x \in \mathbb R^n} L(x,p)\text.$$</p>
<p>Note that by its formal definition, the the polar cone of $K = \mathbb R^k_- \times \{0\}^{m-k}$ $$K^\ast = \left\{ p \in \mathbb R^m \middle| \langle p,q \rangle \leq 0 \forall q \in K \right\}$$
translates to $$K^\ast = \left\{ p \in \mathbb R^m \middle| p_i \geq 0 \forall i = 1, \dots, k \right\} = \mathbb R^k_+ \times \mathbb R^{m-k} \text,$$
which are precisely the constraints that we have in mind, when we choose the Lagrangian multipliers, namely non-negative factors for the inequalities $f_1, \dots, f_k$ and arbitrary factors for the equality constraints $f_{k+1}, \dots, f_m$. Hence, our dual problem is to maximize $$\inf_{x \in \mathbb R^n} L(x,\left[\begin{smallmatrix}\lambda\\\mu\end{smallmatrix}\right]) = \inf_{x \in \mathbb R^n} f_0(x) + \sum_{i=1}^k \lambda_i f_i(x) + \sum_{i=1}^{m-k} \mu_i f_{k+i}(x)$$
w.r.t. the dual variables $\lambda \in \mathbb R^k$ and $\mu \in \mathbb R^{m-k}$, and subject to $\lambda \geq 0$.</p>
<p>So ultimately, we obtain the famous Lagrangian dual problem as a special case of Fenchel duality. To put it more precisely in view of your original question: Lagrangian duality is a <em>result</em> of Fenchel duality, the latter being a more general concept. Thus, of course, they are not equivalent.</p>
|
2,831,199 | <p>What is the probability of getting $6$ $K$ times in a row when rolling a dice N times?</p>
<p>I thought it's $(1/6)^k*(5/6)^{n-k}$ and that times $N-K+1$ since there are $N-K+1$ ways to place an array of consecutive elements to $N$ places.</p>
| drhab | 75,923 | <p>In the following for convenience I write $p$ for $\frac{1}{6}$ and $q$ for $\frac{5}{6}$</p>
<p>Let $A_{K,N}$ denote the event that among the first $N$ throws there
is a consecutive row of $K$ sixes.</p>
<p>Let it be that the $F$-th throw is the first throw that does not
give a six, so that $P\left(F=r\right)=p^{r-1}q$.</p>
<p>Then $P\left(A_{K,N}\mid F=r\right)=1$ if $r>K$ and $P\left(A_{K,N}\mid F=r\right)=P\left(A_{K,N-r}\right)$
otherwise. </p>
<p>Further $P\left(A_{K,N-r}\right)=0$ if $K>N-r$.</p>
<p>So for $K\leq N$ we find:</p>
<p>$$P\left(A_{K,N}\right)=\sum_{r=1}^{\infty}P\left(A_{K,N}\mid F=r\right)P\left(F=r\right)=\sum_{r=1}^{K}P\left(A_{K,N-r}\right)p^{r-1}q+p^{K}$$</p>
<p>Based on this equality with induction it can be proved that:</p>
<p>$$P\left(A_{K,N}\right)=p^{K}\left(1+\left(N-K\right)q\right)\text{ for }
N\in\left\{ K,K+1,\dots,2K\right\} $$(which is not valid for $N>2K$).</p>
<hr>
<p>See also this <a href="https://math.stackexchange.com/q/504693/75923">question</a> that was posed by me once at a time.</p>
|
320,378 | <p>Given a number of normal distributions <span class="math-container">$N(\mu_1, \sigma^2), N(\mu_2, \sigma^2), ..., N(\mu_n, \sigma^2)$</span> with fixed variance <span class="math-container">$\sigma^2$</span>, but not necessary equal means. My question is how to approximate the variance given a number of samples of the normal distributions. Hence given samples
<span class="math-container">$$
X^1_1, X^1_2, ..., X^1_{m_1} \sim N(\mu_1, \sigma^2), \\
X^2_1, X^2_2, ..., X^2_{m_2} \sim N(\mu_2, \sigma^2), \\
\vdots \\
X^n_1, X^n_2, ..., X^n_{m_n} \sim N(\mu_n, \sigma^2).
$$</span>
Where <span class="math-container">$m_1, m_2, ..., m_n > 0$</span>, but again not necessary equal. What is a good way to approximate the variance <span class="math-container">$\sigma^2$</span>?</p>
<p>With <em>good way to approximate</em> I mean the following. I could take <span class="math-container">$m_i$</span> such that <span class="math-container">$m_i \geq m_1, m_2, ..., m_n$</span> and approximate <span class="math-container">$\sigma^2$</span> with
<span class="math-container">$$
\frac{1}{m_i} \sum_{j=1}^{m_i} (X^i_j - E(X^i))^2
$$</span>
where <span class="math-container">$E(X^i) = \frac{1}{m_i} \sum_k X^i_k$</span> is the average over the samples <span class="math-container">$X^i_1, ..., X^i_{m_i}$</span>. Can we do better? A good way to approximate <span class="math-container">$\sigma^2$</span> would be an estimation method that approximate <span class="math-container">$\sigma^2$</span> with better precision (on average) than the method described above. I also want to know how one would prove that one estimation method for the variance is better than another method.</p>
<p>For example, my gut feeling is telling me that a weighted average over all variances would approximate better, i.e.
<span class="math-container">$$
\frac{1}{m_1 + ... + m_n} \sum_{i=0}^{n} \sum_{j=1}^{m_i} (X^i_j - E(X^i))^2,
$$</span>
but I don't know how to prove this. Also I'm worried that one of the variance could be skewed if one of the normal distributions has way less samples than all the others.</p>
| David G. Stork | 89,654 | <p>You can solve this precisely computing the basic integrals. </p>
<p>For two normal distributions <span class="math-container">$N(\mu_1, \sigma_1)$</span> and <span class="math-container">$N(\mu_2, \sigma_2)$</span> the variance is:</p>
<p><span class="math-container">$$\frac{1}{4} \left(\text{$\mu $1}^2-2 \text{$\mu $1} \text{$\mu $2}+\text{$\mu
$2}^2+2 \text{$\sigma $1}^2+2 \text{$\sigma $2}^2\right)$$</span></p>
<p>For three distributions:</p>
<p><span class="math-container">$$\frac{1}{9} \left(2 \text{$\mu $1}^2-2 \text{$\mu $1} \text{$\mu $2}-2 \text{$\mu
$1} \text{$\mu $3}+2 \text{$\mu $2}^2-2 \text{$\mu $2} \text{$\mu $3}+2
\text{$\mu $3}^2+3 \text{$\sigma $1}^2+3 \text{$\sigma $2}^2+3 \text{$\sigma
$3}^2\right)$$</span></p>
<p>And so forth.</p>
<p>If they have the same standard deviation (or variance):</p>
<p><span class="math-container">$$\frac{1}{9} \left(2 \text{$\mu $1}^2-2 \text{$\mu $1} \text{$\mu $2}-2 \text{$\mu
$1} \text{$\mu $3}+2 \text{$\mu $2}^2-2 \text{$\mu $2} \text{$\mu $3}+2
\text{$\mu $3}^2+9 \sigma ^2\right)$$</span></p>
|
2,951,242 | <p>1) <span class="math-container">$cl(\mathbb R) $</span></p>
<p>2) <span class="math-container">$int ([1, \infty) \cup $</span> {3})</p>
<p>3) <span class="math-container">$ \partial (-1,\infty ) \cap $</span> {-3}
it’s a boundary</p>
<p>My solution:
1) it’s same <span class="math-container">$\mathbb R $</span>
2) <span class="math-container">$(1,\infty ) \cap $</span> {3}
3) {-1,-3}
Correct or no ?</p>
| dmtri | 482,116 | <p>In case 2) why do you write that the interior is the intersection of interiors? This may lead you to false , if for example instead of <span class="math-container">$3$</span> were <span class="math-container">$-2$</span>.</p>
|
2,640,477 | <p>According to <a href="https://rads.stackoverflow.com/amzn/click/0073383090" rel="nofollow noreferrer">Rosen</a>, an infinite set A is countable if $|A|= |\mathbb{Z}^+|$ which in turn can be established by finding a bijection from A to $\mathbb{Z}^+$.</p>
<p>Also, a sequence is defined as a function from $\mathbb{Z}^+$ (or $\{0\} \cup \mathbb{Z}^+$) to some set.</p>
<p>With the above, a sequence is certainly enumerable. However, it need not be a bijection, e.g. Fibonacci(1) = Fibonacci(2) = 1.</p>
<p>This implies that not every sequence is countable which seems counterintuitive. Are there any results in this regard? Is there a mistake in the reasoning above?</p>
| Community | -1 | <p>As a sequence is a set indexed by the natural numbers, there exists a surjection from the naturals to the set. Let <span class="math-container">$A$</span> be the set and g: <span class="math-container">$\mathbb{N}\rightarrow A$</span> the surjection. Then we can define a map <span class="math-container">$f:A \rightarrow \mathbb{N}$</span> as <span class="math-container">$f(a)=$</span> the minimum natural number n such that <span class="math-container">$g(n)=a$</span>. Then f is injective and thus A must be countably infinite or finite. Further, you can similarly show that every set indexed by a countably infinite set is either finite or countably infinite.</p>
|
678,073 | <p>I am working with the multiplicative ring of integers modulo $2^{127}$.</p>
<p>Consider the set $E=\{(k,l) \mid 5^k \cdot 3^l \equiv 1\mod 2^{127}, k > 0, l> 0\}$.
I wonder if anybody knows or has an idea where to look for a result related to a lower bound for $M=\min\{k+l \mid (k,l)\in E \}$.</p>
<p>We have that $0<M\leq \mathrm{ord}_{\mathbb{Z}_{2^{127}}}(5)+\mathrm{ord}_{\mathbb{Z}_{2^{127}}}(3)$ where $\mathrm{ord}_{\mathbb{Z}_{2^{127}}}(5)=2^{125}$ and $\mathrm{ord}_{\mathbb{Z}_{2^{127}}}(3)=2^{125}$ (orders of these primes in the multiplicative ring $\mathbb{Z}_{2^{127}}$).</p>
<p>I also would like to generalize the above for primes other than 5 and 3.</p>
<p>Is there a result about a tighter lower bound for $M$?</p>
| coffeemath | 30,316 | <p>Let $k=l=n$ so that $5^k3^l=15^n=(16-1)^n=(2^4-1)^n$. Now take $n=2^{123}$ and apply the binomial theorem, which gives all but the last two terms clearly divisible by $2^{127}$ and the last two terms are
$$-2^{123}\cdot 2^4 +1,$$
which is $1$ mod $2^{127}.$ So here $k+l=2^{124},$ which is a fourth of the value ord(3)+ord(5).</p>
<p>Maybe a lower value can be obtained on using other imposed relations between the exponents $k,l.$</p>
|
3,464,282 | <p>I have a heat type equation
<span class="math-container">$$\frac{d}{dt}V + \frac{1}{2} \sigma^{2} S^{2} \frac{d^{2}}{dS^{2}}V + (r-D) S \frac{d}{ds} V - rV = 0$$</span></p>
<p>Am asked to prove the solution is separable
<span class="math-container">$$V=A(t) B(s)$$</span>
and that A(t) is 1st order diff eq and B(S) 2nd order diff eq.</p>
<p>I did
<span class="math-container">$$\frac{d}{dt}V=B(s) \frac{d}{dt}A(t)$$</span>
and
<span class="math-container">$$\frac{d}{dS}V=A(t) \frac{d}{dS}B(S)$$</span>
and
<span class="math-container">$$\frac{d^2}{dS^2}V=A(t) \frac{d^2}{dS^2}B(S)$$</span></p>
<p>Plugged it in, got</p>
<p><span class="math-container">$$\frac{\frac{d}{dt} A(t)}{A(t)} + \frac{1}{2} \sigma^{2} S^{2} \frac{\frac{d^{2}}{dS^{2}} B(S)}{B(S)} + (r-D) S \frac{\frac{d}{dS} B(S)}{B(S)} - r = 0$$</span></p>
<p>but don't know where to go from here?</p>
| Donald Splutterwit | 404,247 | <p>We have
<span class="math-container">\begin{eqnarray*}
-\frac{\frac{d}{dt} A(t)}{A(t)} = \frac{1}{2} \sigma^{2} S^{2} \frac{\frac{d^{2}}{dS^{2}} B(S)}{B(S)} + (r-D) S \frac{\frac{d}{dS} B(S)}{B(S)} - r.
\end{eqnarray*}</span>
The RHS is a function of <span class="math-container">$t$</span> only and the LHS is a function of <span class="math-container">$S$</span> only, so they are both constant. So we have
<span class="math-container">\begin{eqnarray*}
-\frac{\frac{d}{dt} A(t)}{A(t)} = k_i \\
\frac{1}{2} \sigma^{2} S^{2} \frac{\frac{d^{2}}{dS^{2}} B(S)}{B(S)} + (r-D) S \frac{\frac{d}{dS} B(S)}{B(S)} - r= k_i \\
\end{eqnarray*}</span>
where <span class="math-container">$k_i$</span> is a constant and a general solution will be of the form
<span class="math-container">\begin{eqnarray*}
V=\sum_{k_i} A_{k_i} (t) B_{k_i}(S) .
\end{eqnarray*}</span></p>
|
1,439,850 | <p>So the problem states that the centre of the circle is in the first quadrant and that circle passes through $x$ axis, $y$ axis and the following line: $3x-4y=12$. I have only one question. The answer denotes $r$ as the radius of the circle and then assumes that centre is at $(r,r)$ because of the fact that the circle passes through $x$ and $y$ axis. I was thinking that this single fact does not permit one to assume that centre must be at $(r,r)$, simply because the centre may be positioned in such a manner that the distance to $y$ and $x$ axis is not the same and not necessarily $r$. Is my thinking correct? If not, why? </p>
| haqnatural | 247,767 | <p>$${ x }^{ 2 }\frac { dy }{ dx } +3xy=1\\ \\ { x }^{ 2 }\frac { dy }{ dx } +3xy=0\\ \frac { dy }{ dx } =-3\frac { y }{ x } \\ \int { \frac { dy }{ y } } =-3\int { \frac { 1 }{ x } dx } \\ \ln { \left| y \right| =-3\ln { \left| cx \right| } } \\ y=\frac { C }{ { x }^{ 3 } } \\ y=\frac { C\left( x \right) }{ { x }^{ 3 } } \\ \frac { dy }{ dx } =\frac { { C }^{ \prime }\left( x \right) { x }^{ 3 }-3{ x }^{ 2 }C\left( x \right) }{ { x }^{ 6 } } \\ { x }^{ 2 }\frac { { C }^{ \prime }\left( x \right) { x }^{ 3 }-3{ x }^{ 2 }C\left( x \right) }{ { x }^{ 6 } } +3x\frac { C }{ { x }^{ 3 } } =1\\ { C }^{ \prime }\left( x \right) =x\\ { C }\left( x \right) =\frac { { x }^{ 2 } }{ 2 } +C_{ 1 }\\ y=\frac { \frac { { x }^{ 2 } }{ 2 } +C_{ 1 } }{ { x }^{ 3 } } =\frac { 1 }{ 2{ x } } +\frac { C_{ 1 } }{ { x }^{ 3 } } \\ \\ $$</p>
|
2,986,515 | <p>Can anyone help me with this problem? </p>
<p>Prove that for any real number <span class="math-container">$x > 0$</span> and for any <span class="math-container">$M > 0$</span> there is <span class="math-container">$N ∈ \mathbb N$</span> so that if <span class="math-container">$n > N$</span> then <span class="math-container">$(1 + x)^n > M.$</span></p>
<p>A sequence <span class="math-container">${a_n}$</span> diverges to <span class="math-container">$+\infty$</span> if for any <span class="math-container">$M > 0$</span> there is <span class="math-container">$N ∈\mathbb N$</span> so that if <span class="math-container">$n > N$</span> then <span class="math-container">$a_n > M$</span>. Suppose <span class="math-container">${a_n}$</span> diverges to <span class="math-container">$+\infty$</span> and <span class="math-container">$a_n \ne 0$</span> for all <span class="math-container">$n$</span>. Prove that in this case <span class="math-container">$\frac{1}{a_n}$</span> converges to <span class="math-container">$0$</span>.</p>
<p>How do I approach these proofs?</p>
| Cesareo | 397,348 | <p>Hint.</p>
<p><span class="math-container">$$
\left(a_{n}\right)_{n \in \mathbb{N}} = \dfrac{3n^2 - 9n + 6}{n^3 + 5n^2 + 8n + 4} = \frac{3(n-2)(n-1)}{(n+1)(n+2)^2}
$$</span></p>
|
1,821,186 | <p>Why is the solution of $|1+3x|<6x$ only $x>1/3$? After applying the properties of modulus, I get $-6x<1+3x<6x$. And after solving each inequality, I get $x>-1/9$ and $x>1/3$, but why is $x>-1/9$ rejected? </p>
| Robert Israel | 8,508 | <p>Try balls of radius $1/m$ centred at the points $x_n$ of your dense countable subset.</p>
<p>Edit: to see how this works, consider a ball of radius $r$ around any point $p$. What conditions on $m$ and $x_n$ would guarantee that the ball of radius $1/m$ around $x_n$ contains $p$ and is contained in this ball?</p>
|
428,415 | <p>I tried using integration by parts twice, the same way we do for $\int \sin {(\sqrt{x})}$
but in the second integral, I'm not getting an expression that is equal to $\int x\sin {(\sqrt{x})}$.</p>
<p>I let $\sqrt x = t$ thus,
$$\int t^2 \cdot \sin({t})\cdot 2t dt = 2\int t^3\sin(t)dt = 2[(-\cos(t)\cdot t^3 + \int 3t^2\cos(t))] = 2[-\cos(t)\cdot t^3+(\sin(t)\cdot 3t^3 - \int 6t \cdot \sin(t))]]$$</p>
<p>which I can't find useful. </p>
| Sri Krishna | 74,110 | <p>you have done till this - $ 2[-\cos(t)\cdot t^3+(\sin(t)\cdot 3t^3 - \int 6t \cdot \sin(t))]]$</p>
<p>again use parts to get </p>
<p>$ 2[-\cos(t)\cdot t^3+(\sin(t)\cdot 3t^3 - 6(t(-\cos t) +\sin(t))$ = </p>
<p>$ 2[-\cos(t)\cdot t^3+(\sin(t)\cdot 3t^3 + 6(t(\cos t) -6\sin(t))$ .
giving you your answer .</p>
<p>just put $ \sqrt{x} = t $ and you are through.</p>
|
428,415 | <p>I tried using integration by parts twice, the same way we do for $\int \sin {(\sqrt{x})}$
but in the second integral, I'm not getting an expression that is equal to $\int x\sin {(\sqrt{x})}$.</p>
<p>I let $\sqrt x = t$ thus,
$$\int t^2 \cdot \sin({t})\cdot 2t dt = 2\int t^3\sin(t)dt = 2[(-\cos(t)\cdot t^3 + \int 3t^2\cos(t))] = 2[-\cos(t)\cdot t^3+(\sin(t)\cdot 3t^3 - \int 6t \cdot \sin(t))]]$$</p>
<p>which I can't find useful. </p>
| amWhy | 9,003 | <p>Yes, indeed, continue as you did in the comments, treating $\int 6t\sin t \,dt\;$ as a separate integral, use integration by parts, and add (or subtract, if appropriate) that result to your earlier work, and you will end with an expression with no integrals remaining!:</p>
<p>$$\int t^2 \cdot \sin({t})\cdot 2t dt = $$</p>
<p>$$= 2[-\cos(\sqrt x) \cdot x(\sqrt x) + \sin(\sqrt x)\cdot 3x -(\cos(\sqrt x)\cdot6\sqrt x+\sin(\sqrt x)\cdot \sqrt x + \cos (\sqrt x))] + C$$</p>
<p>after substituting $\sqrt x$ for $t$, though I'd suggest finding a way to simplify (combining like terms, etc.)</p>
|
2,988,089 | <p>Let A, B, C, and D be sets. Prove or disprove the following:</p>
<pre><code> (A ∩ B) ∪ (C ∩ D)= (A ∩ D) ∪ (C ∩ B)
</code></pre>
<p>I am just wondering can i simply prove it using a membership table ( seems to easy ) or do i have to use setbuilder notation?</p>
<p>Thank you!</p>
| Melody | 598,521 | <p>Well it's sometimes true. Example, let all of the sets be <span class="math-container">$\{1\}$</span>.</p>
<p>To show it doesn't always hold we could consider <span class="math-container">$A,B=\{1\}$</span> and <span class="math-container">$C,D=\{0\}$</span>. Then <span class="math-container">$A\cap B=\{1\}$</span> and <span class="math-container">$C\cap D=\{0\}$</span>, so <span class="math-container">$(A\cap B)\cup(C\cap D)=\{0,1\}$</span>. Notice however that <span class="math-container">$A\cap D=C\cap B=\emptyset$</span>, therefore <span class="math-container">$(A\cap D)\cup(C\cap B)=\emptyset$</span>. Since <span class="math-container">$\emptyset\not=\{0,1\}$</span> we have a counterexample.</p>
|
262,745 | <p>I need to find the normal vector of the form Ax+By+C=0 of the plane that includes the point (6.82,1,5.56) and the line (7.82,6.82,6.56) +t(6,12,-6), with A=1.</p>
<p>Of course, this is easy to do by hand, using the cross product of two lines and the point. There's supposed to be an automated way of doing it, though, and I can't find it.
Any ideas on an efficient way of doing it?</p>
| cvgmt | 72,111 | <p><code>RegionWithin</code> can be use to express "the plane that includes the point and the line".</p>
<pre><code>Clear["Global`*"];
pt = {6.82, 1, 5.56};
lineeq = {7.82, 6.82, 6.56} + t {6, 12, -6};
line = ParametricRegion[lineeq // Rationalize, t];
plane = Hyperplane[{a, b, c}, d];
eqs = RegionWithin[plane, line] &&
RegionWithin[plane, Point[Rationalize@pt]];
sol = Reduce[eqs, Reals];
instance = FindInstance[sol && a == 1, {a, b, c, d}, Reals, 1][[1]];
normal = {a, b, c} /. instance
Show[Region[line, BaseStyle -> {Red}],
Graphics3D[{Yellow, plane /. instance, Blue, PointSize[Large],
Point[pt], Green, Arrow[{pt, pt + 20 normal}]}],
ViewPoint -> {2.5, -.5, -1.5}]
</code></pre>
<blockquote>
<p><code>{1, -(100/391), 191/391}</code></p>
</blockquote>
<p><a href="https://i.stack.imgur.com/eAiTe.png" rel="noreferrer"><img src="https://i.stack.imgur.com/eAiTe.png" alt="enter image description here" /></a></p>
<pre><code>SubtractSides[RegionConvert[plane /. instance, "Implicit"][[1]]] /.
Thread[{\[FormalX], \[FormalY], \[FormalZ]} -> {x, y, z}]
</code></pre>
<p>The plane equation is</p>
<blockquote>
<p><code>181429 - 19550 x + 5000 y - 9550 z == 0</code></p>
</blockquote>
<ul>
<li>Replace <code>plane</code> and <code>eqs</code> as below also work.</li>
</ul>
<pre><code>plane=ImplicitRegion[a*x+b*y+c*z==d,{x,y,z}]
eqs=Rationalize@pt∈plane&&ForAll[{x,y,z},{x,y,z}∈line,{x,y,z}∈plane];
</code></pre>
|
3,789,676 | <p>I am try to calculate the derivative of cross-entropy, when the softmax layer has the temperature T. That is:
<span class="math-container">\begin{equation}
p_j = \frac{e^{o_j/T}}{\sum_k e^{o_k/T}}
\end{equation}</span></p>
<p>This question here was answered at T=1: <a href="https://math.stackexchange.com/questions/945871/derivative-of-softmax-loss-function">Derivative of Softmax loss function</a></p>
<p>Now what would be the final derivative in terms of <span class="math-container">$p_i$</span>, <span class="math-container">$q_i$</span>, and T? Please see the linked question for the notations.</p>
<p>Edit: Thanks to Alex for pointing out a typo</p>
| samirzach | 262,548 | <p>The cross-entropy loss for softmax outputs assumes that the set of target values are one-hot encoded rather than a fully defined probability distribution at <span class="math-container">$T=1$</span>, which is why the usual derivation does not include the second <span class="math-container">$1/T$</span> term.</p>
<p>The following is from <a href="https://peterroelants.github.io/posts/cross-entropy-softmax/#Derivative-of-the-cross-entropy-loss-function-for-the-softmax-function" rel="nofollow noreferrer">this</a> elegantly written article:</p>
<p><span class="math-container">\begin{split}
\frac{\partial \xi}{\partial z_i} & = - \sum_{j=1}^C \frac{\partial t_j \log(y_j)}{\partial z_i}{}
= - \sum_{j=1}^C t_j \frac{\partial \log(y_j)}{\partial z_i}
= - \sum_{j=1}^C t_j \frac{1}{y_j} \frac{\partial y_j}{\partial z_i} \\
& = - \frac{t_i}{y_i} \frac{\partial y_i}{\partial z_i} - \sum_{j \neq i}^C \frac{t_j}{y_j} \frac{\partial y_j}{\partial z_i}
= - \frac{t_i}{y_i} y_i (1-y_i) - \sum_{j \neq i}^C \frac{t_j}{y_j} (-y_j y_i) \\
& = - t_i + t_i y_i + \sum_{j \neq i}^C t_j y_i = - t_i + \sum_{j = 1}^C t_j y_i
= -t_i + y_i \sum_{j = 1}^C t_j \\
& = y_i - t_i
\end{split}</span></p>
<p>where <span class="math-container">$C$</span> is the number of output classes. The above derivation neither assumes the <span class="math-container">$T \ne 1$</span> condition nor that the target distribution is also a softmax output. So in order to find out what the gradient looks like when we plug in these two missing assumptions into the derivation, let's first see what we get when we plug in the <span class="math-container">$T \ne 1$</span> assumption:</p>
<p><span class="math-container">\begin{split}
\frac{\partial \xi}{\partial z_i} & = - \sum_{j=1}^C \frac{\partial t_j \log(y_j)}{\partial z_i}{}
= - \sum_{j=1}^C t_j \frac{\partial \log(y_j)}{\partial z_i}
= - \sum_{j=1}^C t_j \frac{1}{y_j} \frac{\partial y_j}{\partial z_i} \\
& = - \frac{t_i}{y_i} \frac{\partial y_i}{\partial z_i} - \sum_{j \neq i}^C \frac{t_j}{y_j} \frac{\partial y_j}{\partial z_i}
= - \frac{t_i}{y_i} \frac{1}{T} y_i (1-y_i) - \sum_{j \neq i}^C \frac{t_j}{y_j} \frac{1}{T} (-y_j y_i) \\
& = -\frac{1}{T} t_i + \frac{1}{T} t_i y_i + \frac{1}{T} \sum_{j \neq i}^C t_j y_i = - \frac{1}{T} t_i + \frac{1}{T} \sum_{j = 1}^C t_j y_i
= -\frac{1}{T} t_i + \frac{1}{T} y_i \sum_{j = 1}^C t_j \\
& = \frac{1}{T} (y_i - t_i)
\end{split}</span></p>
<p>The last part, where the assumption that the targets are soft as well is also injected into the derivation, is beautifully summarized in section 2.1 of Prof. Hinton's 2015 <a href="https://arxiv.org/abs/1503.02531" rel="nofollow noreferrer">paper</a> titled 'Distilling the Knowledge in a Neural Network'. Rewriting the same in the context of the derivation given above, we get:</p>
<p><span class="math-container">\begin{split}
\frac{\partial \xi}{\partial z_i} & = \frac{1}{T} (y_i - t_i)
= \frac{1}{T} (\frac{e^{z_i/T}}{\sum_{d=1}^C e^{z_d/T}} - \frac{e^{v_i/T}}{\sum_{d=1}^C e^{v_d/T}})
\end{split}</span></p>
<p>If the temperature is high compared with the magnitude of the logits, we can approximate:
<span class="math-container">\begin{split}
\frac{\partial \xi}{\partial z_i} & \approx \frac{1}{T} (\frac{1 + z_i/T}{C + \sum_{d=1}^C z_d/T} - \frac{1 + v_i/T}{C + \sum_{d=1}^C v_d/T})
\end{split}</span></p>
<p>since, we can indeed approximate <span class="math-container">$e^{very small value}$</span> with <span class="math-container">$1 + {very small value}$</span> (The denominator terms are nothing but a straightforward generalization of these values when summed up). If we now assume that the logits have been zero-meaned separately for each transfer case so that <span class="math-container">$\sum_{d} z_d = \sum_{d} v_d = 0$</span>, then the above equation simplifies to:
<span class="math-container">\begin{split}
\frac{\partial \xi}{\partial z_i} & \approx \frac{1}{CT^2} (z_i - v_i)
\end{split}</span></p>
<p>This is when we arrive at the <span class="math-container">$1 / T^2$</span> term. Here 'transfer set' refers to the dataset that is used to train the to-be-distilled student model, labelled using soft targets produced via the softmax outputs of the cumbersome teacher model(s).</p>
|
4,243,030 | <p>I tried to evaluate the integral <span class="math-container">$$ \oint_c\dfrac{dz}{\sin^2 z}$$</span> where <span class="math-container">$c$</span> is a circle <span class="math-container">$|z|=1/2$</span>. The only pole within <span class="math-container">$c$</span> is <span class="math-container">$z=0$</span> and the residue at <span class="math-container">$z=0$</span> is found as <span class="math-container">$$\lim_{z\to 0}\dfrac{d}{dz}\left (\dfrac{z^2}{\sin^2 z}\right )=0$$</span> so that integral is zero. Where did I go wrong as it violates Cauchy Goursat theorem in complex integration.</p>
| Greg Martin | 16,078 | <p>You seem to be confusing Cauchy's theorem with its converse. Your example shows that the converse of Cauchy's theorem is false, but that doesn't refute Cauchy's theorem itself.</p>
|
2,502,161 | <p>I'm wondering if it's valid to write the follwing: <span class="math-container">$$\lim_{x \rightarrow \infty}\frac{2}{x^r}=2\lim_{x \rightarrow \infty}\frac{1}{x^r}=2.\frac{1}{\infty}=2.0=0$$</span></p>
<p>I know it's valid to say that <span class="math-container">$\frac{1}{\infty}=0$</span> in limits but I'm not suring if it would be valid to say <span class="math-container">$2.\frac{1}{\infty}=2.0=0$</span></p>
| Hans | 64,809 | <p>$$y''=\frac{dy'}{dy}\frac{dy}{dx},$$
$$y''y'y=y'^2\frac{dy'}{dy}y=\frac13\frac{dy'^3}{d\ln y}.$$
You should be able to finish the rest.</p>
|
58,870 | <p>I am teaching a introductory course on differentiable manifolds next term. The course is aimed at fourth year US undergraduate students and first year US graduate students who have done basic coursework in
point-set topology and multivariable calculus, but may not know the definition of differentiable manifold. I am following the textbook <a href="http://rads.stackoverflow.com/amzn/click/0132126052">Differential Topology</a> by
Guillemin and Pollack, supplemented by Milnor's <a href="http://rads.stackoverflow.com/amzn/click/0691048339">book</a>.</p>
<p>My question is: <strong>What are good topics to cover that are not in assigned textbooks?</strong> </p>
| agt | 12,617 | <p>In a first course aiming to introduce differentiable manifolds as the spaces on which do calculus, you could give to the students the notion of connection at least on vector bundles.</p>
<p>In order to reflect on the reason for this choice, I report the words of S.S.Chern closing the introduction of Global Differential Geometry, MAA Studies in Math.27, 1989:</p>
<blockquote>
<p>The Editor is convinced that the notion of a connection in a
vector bundle will soon find its way into a class on advanced
calculus, as it is a fundamental notion and its applications are
wide-spread. His chapter, "Vector Bundles with a Connection,"
hopefully will show that it is basically an elementary concept. </p>
</blockquote>
|
58,870 | <p>I am teaching a introductory course on differentiable manifolds next term. The course is aimed at fourth year US undergraduate students and first year US graduate students who have done basic coursework in
point-set topology and multivariable calculus, but may not know the definition of differentiable manifold. I am following the textbook <a href="http://rads.stackoverflow.com/amzn/click/0132126052">Differential Topology</a> by
Guillemin and Pollack, supplemented by Milnor's <a href="http://rads.stackoverflow.com/amzn/click/0691048339">book</a>.</p>
<p>My question is: <strong>What are good topics to cover that are not in assigned textbooks?</strong> </p>
| David MJC | 10,106 | <p>Differential forms.</p>
<p>Books by Darling (Differential forms and connections) and Madsen-Tornehave (From calculus to cohomology: de Rham cohomology and characteristic classes) may help.</p>
|
857,801 | <p>"If in the obvious equalities $(k+1)^3−k^3=3k^2+3k+1$, for the different values $k=1,2,…,n−1$, we add the left and the right sides separately, we obtain the equation $n^3−1=3σ_n+\frac{3(n−1)n}{2}+n−1$, where $σ_n=1^2+2^2+…+(n−1)^2$."</p>
<p>I'm stuck trying to understand what the author has done in the paragraph below. Is it possible to explain it using high school-level math? For instance, what is meant with "for the different values $k=1,2,…,n−1$, we add the left and the right sides separately"? I don't understand what he does after that either.</p>
| Surb | 154,545 | <p>This equality </p>
<p>$$(k+1)^3−k^3=3k^2+3k+1,$$</p>
<p>is true for every $k$, in particular it is true when $k=1$,$k=2$,etc... Writing all these equations, we get</p>
<p>$$\begin{array}{rcl}
(1+1)^3−1^3&=&3\cdot 1^2+3\cdot 1+1\\
(2+1)^3−2^3&=&3\cdot 2^2+3\cdot 2+1\\
&\vdots &\\
((n-2)+1)^3−(n-2)^3&=&3\cdot (n-2)^2+3\cdot (n-2)+1\\
((n-1)+1)^3−(n-1)^3&=&3\cdot (n-1)^2+3\cdot (n-1)+1\\
\end{array}$$
Then you just sum all these equations (right hand sides with right hand sides, left hand sides with left hand sides) to find </p>
<p>$$\color{red}{2^3}-1^3+\color{blue}{3^3}-\color{red}{2^3}+4^3-\color{blue}{3^3}+\ldots +\color{green}{(n-1)^3}−(n-2)^3+n^3−\color{green}{(n-1)^3} = n^3-1$$</p>
<p>for the left hand side (note that this is called a telescopic sum). Now for the other side of the equation, note that
$$3\cdot 1^2 + 3 \cdot 2^2+ \ldots +3\cdot (n-2)^2+3\cdot (n-1)^2 = 3 \sigma_n$$
where $σ_n=1^2+2^2+…+(n−1)^2.$ Furthermore
$$3\cdot 1 + 3 \cdot 2+ \ldots +3\cdot (n-2)+3\cdot (n-1)= 3(1+2+\ldots +(n-1)) = 3\frac{n(n-1)}{2}. $$
(see the note for a proof of this fact) and
$$\underbrace{1+ 1+\ldots +1 + 1}_{(n-1) \text{ times}} = n-1.$$ So you finally get
$$n^3−1=3σ_n+\frac{3(n−1)n}{2}+n−1$$
<strong>NOTE:</strong> For any $k \in \mathbb{N},$ we have
$$\begin{array}{rcl}2(1+2+\ldots+(k-1)+k) &=& (1+2+\ldots+k)+(k+(k-1)+\ldots+2+1)\\ &=& (1+k)+(2+(k-1))+\ldots +((k-1)+2)+(k+1)\\ &=& k(k+1)\end{array}$$</p>
|
1,877,558 | <p>For instance, let $(\mathbb{R}, \mathfrak{T})$ be $\mathbb{R}$ with the usual topology. </p>
<p>Why is that $\mathfrak{T} \times \mathfrak{T}$ is a basis on $\mathbb{R} \times \mathbb{R}$ instead of topology?</p>
<p>It seems that people just take $\mathfrak{T} \times \mathfrak{T}$ as a basis by definition. There must be some open sets in the product that cannot be represented by Cartesian product of $\mathfrak{T} \times \mathfrak{T}$, but I don't have any examples handy. </p>
<p>Can someone please instruct?</p>
<p><em>I am still not convinced because the examples so far are not "constructive" :(</em></p>
| drhab | 75,923 | <p>If $X=(|X|,\tau)$ denotes a topological space with underlying set $|X|$ and topology $\tau$ then the collection $\tau\times\tau:=\{U\times V\mid U,V\in\tau\}$ <strong>is by definition</strong> a <a href="https://en.wikipedia.org/wiki/Base_(topology)#Definition_and_basic_properties" rel="nofollow">base</a> for a topology on set $|X|\times|X|$ so that arguments cannot play any role in this.</p>
<p>It is evident that the collection covers the set and that it is closed under finite intersections. </p>
<p>That's why "people just take $\tau \times \tau$ as a basis by definition".</p>
<p>Is it also a topology? </p>
<p>For that it must also be closed under arbitrary unions. To come to the answer "no" in an informal sense it is enough allready to find two squares (sets $U\times U\in\tau\times\tau$) such that the union of these squares is not a square.</p>
<p>As an exercise: find out when $\tau\times\tau$ is indeed a topology (special case).</p>
|
2,782,109 | <blockquote>
<p>If a positive integer $m$ was increased by $20$%, decreased by $25$%, and then increased by $60$%, the resulting number would be what percent of $m$?</p>
</blockquote>
<p>A common step-by-step calculation will take time.</p>
<p>After $20$% increase, $6m/5$.<br>
After $25$% decrease, $9m/10$.<br>
After $60$% increase, $144m/100$.<br>
Finally, $m \cdot \frac{x}{100} = \frac{144m}{100} = 144$%</p>
<p>what is the faster (or, fastest) method to solve this?</p>
| Love Invariants | 551,019 | <p>There is no shorter method. Only thing you can do to make it short is do all calculations in the end. Log would've worked but it is for only even increases/decreases and for a long period of time<br>
OR<br>
For an alternative take the number as 1, it doesn't change anything except making expression simpler.</p>
|
2,782,109 | <blockquote>
<p>If a positive integer $m$ was increased by $20$%, decreased by $25$%, and then increased by $60$%, the resulting number would be what percent of $m$?</p>
</blockquote>
<p>A common step-by-step calculation will take time.</p>
<p>After $20$% increase, $6m/5$.<br>
After $25$% decrease, $9m/10$.<br>
After $60$% increase, $144m/100$.<br>
Finally, $m \cdot \frac{x}{100} = \frac{144m}{100} = 144$%</p>
<p>what is the faster (or, fastest) method to solve this?</p>
| G Tony Jacobs | 92,129 | <p>The quickest way to do this is to dispense with algebra, and instead of calling your original amount $m$, call it $100$.</p>
<p>A $20\%$ increase turns $100$ into $120$.</p>
<p>A $25\%$ decrease takes off $1/4$ of that, leaving $90$.</p>
<p>A $60\%$ increase adds $6/10$ of $90$, which is $6\times 9$, which is $54$. Thus, $90+54=144$.</p>
<p>Since you started with $100$ and ended with $144$, that's a $44\%$ increase.</p>
|
2,221,033 | <p>My question is due to <a href="https://en.wikipedia.org/w/index.php?title=Imaginary_number&diff=prev&oldid=175488747" rel="noreferrer">an edit</a> to the Wikipedia article: <a href="https://en.m.wikipedia.org/wiki/Imaginary_number" rel="noreferrer">Imaginary number</a>.</p>
<p>The funny thing is, I couldn't find (in three of my old textbooks) a clear definition of an "imaginary number". (Though they were pretty good at defining "imaginary component", etc.)</p>
<p>I understand that the number zero lies on both the real and imaginary axes.<br>
<em>But is $\it 0$ both a real number and an imaginary number?</em></p>
<p>We know certainly, that there are complex numbers that are neither purely real, nor purely imaginary. But I've always previously considered, that a purely imaginary number had to have a square that is a real and negative number (not just non-positive). </p>
<p>Clearly we can (re)define a real number as a complex number with an imaginary component that is zero (meaning that $0$ is a real number), but if one were to define an imaginary number as a complex number with real component zero, then that would also include $0$ among the pure imaginaries. </p>
<p><em>What is the complete and formal definition of an "imaginary number" (outside of the Wikipedia reference or anything derived from it)?</em></p>
| Ethan Bolker | 72,858 | <p>I don't think there is a </p>
<blockquote>
<p>complete and formal definition of "imaginary number"</p>
</blockquote>
<p>It's a useful term sometimes. It's an author's responsibility to make clear what he or she means in any particular context where precision matters. If $0$ should count, or not, then the text must say so.</p>
<p>Your question shows clearly that you understand the structure of the complex numbers, so you should be able to make sense of any passage you encounter.</p>
|
947,358 | <p>Okay $g(x)= \sqrt{x^2-9}$</p>
<p>thus, $x^2 -9 \ge 0$</p>
<p>equals $x \ge +3$ and $x \ge -3$</p>
<p>thus the domains should be $[3,+\infty) \cup [-3,\infty)$ how come the answer key in my book is stating $(−\infty, −3] \cup[3,\infty)$. </p>
| layman | 131,740 | <p>When you get to the point of $x^{2} - 9 \geq 0$, you are definitely allowed to add $9$ to both sides to get $x^{2} \geq 9$, but you are <strong>not</strong> allowed to then take the square root of both sides.</p>
<p>When you have inequalities (like $< , \le, >, \ge$), you can't take the square root of both sides like you can when you have an equality ($=$).</p>
<p>If we have $x^{2} - 9 = 0$, then you know how to solve this. First add $9$ to both sides and then take the square root of both sides, and get $x = 3$ and $x = -3$ as solutions.</p>
<p>But if you can't take the square root of both sides in $x^{2} \geq 9$, then how the heck do you solve it?? Well, there is a special technique we can use for inequalities. First, don't add $9$ to both sides. Instead, keep it as $x^{2} - 9 \geq 0$.</p>
<p>Now, you should notice that $x^{2} - 9$ is the difference of squares. I can rewrite it as $x^{2} - 3^{2}$ since $3^{2} = 9$. And you should know how to factor the difference of squares. $x^{2} - a^{2}$ becomes $(x - a)(x + a)$. So, that means we can rewrite $x^{2} - 9$ as $(x - 3)(x + 3)$.</p>
<p>Great, so since we wanted to solve $x^{2} - 9 \geq 0$, this becomes $(x - 3)(x + 3) \geq 0$. Now what?</p>
<p>Well, now you need to find which values of $x$ make $(x - 3)(x + 3)$ <em>greater than or equal to</em> $0$. Here is a trick that will always help you solve this: Draw a number line and mark it with the numbers that make the factors on the left hand size $0$. So $3$ makes $x-3$ equal to $0$, and $-3$ makes $x + 3$ equal to $0$, so I will draw a number line and mark it with $3$ and $-3$.</p>
<p>------------|---------------|-----------</p>
<p>...............-3....................3...............</p>
<p>Now, you need to pick test points. Pick a number less than $-3$, like $-4$, and plug it into $(x - 3)(x + 3)$. If after you simplify you get something greater than or equal to $0$, then part of our solution will be $(-\infty, -3)$. If you get something negative, then we don't want $(-\infty, -3)$ in our solution.</p>
<p>Now you must test the marking $-3$. Plug it into $(x - 3)(x + 3)$ and you will get $0$. But $0$ is greater than or equal to $0$, so we want to include $-3$ in our solution.</p>
<p>Now pick an easy number between $-3$ and $3$ (like $0$), and plug it into $(x - 3)(x + 3)$. Again, if we get something greater than or equal to $0$, we will want to include the interval $(-3,3)$ in our solution. If we get something negative, we don't want this interval to be in our solution.</p>
<p>Now test the number $3$. Plug it into $(x - 3)(x + 3)$ and you will get $0$. But $0$ is greater than or equal to $0$, so we want to include $3$ in our solution.</p>
<p>Finally, pick a number greater than $3$ (like $4$), and plug it into $(x - 3)(x + 3)$. Again, if we get something greater than or equal to $0$, we will want to include the interval $(3, \infty)$ in our solution. If we get something negative, we don't want this interval to be in our solution.</p>
<p>So, after doing all of that, you should find that the final answer is $(-\infty, -3] \cup [3, \infty)$.</p>
|
1,320,469 | <p>I am working on the following problem</p>
<blockquote>
<p>[R. Vakil] Exercise 19.8.B: Suppose $C$ is a curve of genus $g>1$ over
a field $k$ that is not algebraically closed. Show that $C$ has a
closed point of degree at most $2g-2$ over the base field.</p>
</blockquote>
<p>I have no idea how to do this question. This is what I know: Since $g>1$, so the sections of the dualising sheaf $\omega$ defines a morphism $\varphi: C\rightarrow\mathbb{P}^{g-1}$. Hence it defines a morphism $\varphi:C\rightarrow C'$ from $C$ onto its image curve $C'$. The degree of this morphism is $2g-2$, which corresponds to the degree of the extension of the function field $[k(C):k(C')]$. </p>
<p>How should I go on from here? i.e. Where to find that closed point? I tried to break down the cases where either $\varphi$ is a closed embedding or it is hyperelliptic but they don't seem to help.</p>
| Bananeen | 165,356 | <p>Pick a point $p\in C$. Consider invertible sheaf $\omega_C (-p)$. In the process of exercise 19.8.A one shows using Riemann-Roch that $h^0 (C, \omega(-p))=g-1$ and so under assumption $g>1$ it has a global section $s$. By the description of exercise 14.2.J, $s$ corresponds to a rational section $t$ of $\omega_C $, s.t. $div (t) - p \geq 0$. By Hartog's lemma, a section without poles lifts to a global section $t$ of $\omega_C $ with a zero at $p$. Now the degree of $\omega_C $ is $2g-2$ and the formula $2g -2 = \sum_q val_q(t) \cdot \deg q$ with $val_q(t) \geq 0$ shows $deg \ p \leq 2g-2$, since for $p$, $val_p(t) > 0$.</p>
|
417,896 | <p>Given a connected smooth manifold <span class="math-container">$M$</span> of dimension <span class="math-container">$m>1$</span>, points <span class="math-container">$p_1,\dots,p_n\in M$</span> and positive values <span class="math-container">$\{d_{i,j};1\leq i<j\leq n\}$</span> satisfying the strict triangle inequalities <span class="math-container">$d_{i,j}<d_{i,k}+d_{k,j}$</span>,</p>
<p>Can we give <span class="math-container">$M$</span> a complete riemannian metric <span class="math-container">$g$</span> so that <span class="math-container">$d_g(p_i,p_j)=d_{i,j}$</span>, where <span class="math-container">$d$</span> is the geodesic distance?</p>
<p>This can fail in dimension <span class="math-container">$2$</span>, as shown in the <a href="https://mathoverflow.net/a/417951">answer</a> by André Henriques. I'm pretty sure it has to be true for <span class="math-container">$m\geq3$</span>, but I have not been able to prove it.</p>
<p>Some comments:</p>
<ul>
<li><p>This occurred to me while answering <a href="https://mathoverflow.net/questions/417712/equidistant-points-on-a-compact-riemannian-manifold">Equidistant points on a compact Riemannian manifold</a>, <a href="https://mathoverflow.net/questions/417712/equidistant-points-on-a-compact-riemannian-manifold/417722#417722">my answer</a> to that question contains the ideas I tried for <span class="math-container">$m\geq3$</span>.</p>
</li>
<li><p>By homogeneity of manifolds you can suppose the points <span class="math-container">$P_1,\dotsc,P_n$</span> are any set of <span class="math-container">$n$</span> points of <span class="math-container">$M$</span>, and using that it is not hard to reduce the problem to the case of <span class="math-container">$M$</span> being diffeomorphic to <span class="math-container">$\mathbb{R}^m$</span>. In particular if you prove it for <span class="math-container">$\mathbb{R}^3$</span> you will have proved it for any manifold of dimension <span class="math-container">$\geq 3$</span>.</p>
</li>
<li><p>One of the first ideas which come to mind is trying to somehow imbed <span class="math-container">$M$</span> in <span class="math-container">$\mathbb{R}^N$</span> for some big <span class="math-container">$N$</span>, but <a href="https://en.wikipedia.org/wiki/Distance_geometry#Characterization_via_Cayley%E2%80%93Menger_determinants" rel="noreferrer">triangle inequalities are not sufficient</a> for a finite set to be isometrically imbedded in some <span class="math-container">$\mathbb{R}^N$</span>.</p>
</li>
<li><p>What if we change the strict triangle inequalities for the usual ones?</p>
</li>
</ul>
| André Henriques | 5,690 | <p>It is not possible to find <span class="math-container">$5$</span> points <span class="math-container">$x_1,\ldots,x_5$</span> on a genus zero Riemannian 2-manifold (a sphere) such that <span class="math-container">$d(x_i,x_j)=1$</span> for all <span class="math-container">$i,j$</span>.</p>
<p>The reason is that the complete graph <span class="math-container">$K_5$</span> is not planar.</p>
<p>Assume by contradiction that we have <span class="math-container">$5$</span> points <span class="math-container">$x_1,\ldots,x_5$</span> with <span class="math-container">$d(x_i,x_j)=1$</span>.
Up to permuting the points, we may assume that the minimal geodesic connecting <span class="math-container">$x_1$</span> and <span class="math-container">$x_3$</span> crosses the minimal geodesic connecting <span class="math-container">$x_2$</span> and <span class="math-container">$x_4$</span>.</p>
<p>Let <span class="math-container">$y$</span> be the point at which these two geodesics intersect.
Then
<span class="math-container">\begin{align*}
2&=d(x_1,x_3)+d(x_2,x_4)\\
&=d(x_1,y)+d(y,x_3)+d(x_2,y)+d(y,x_4)\\
&=\tfrac12\big((d(x_1,y)+d(y,x_2))\\
&\qquad(d(x_2,y)+d(y,x_3))+\\
&\qquad(d(x_3,y)+d(y,x_4))+\\
&\qquad(d(x_4,y)+d(y,x_1))
\big)\\
&\ge\tfrac12\big(d(x_1,x_2)+d(x_2,x_3)+d(x_3,x_4)+d(x_4,x_1)\big)=2
\end{align*}</span>
with equality iff <span class="math-container">$y$</span> lies on all six geodesics (between <span class="math-container">$x_i$</span> and <span class="math-container">$x_j$</span> <span class="math-container">$\forall i,j\in\{1,2,3,4\})$</span>. But if <span class="math-container">$y$</span> lies on all six geodesics, then these six geodesics are all part of a single geodesic line (i.e. the points are "aligned"), which is clearly impossible.</p>
<p>The crucial thing that I'm using here is the fact that geodesics admit unique extensions. If the ambient space was a graph, then my argument for deriving a contradiction wouldn't work as I wouldn't be able to conclude that the points are "aligned".</p>
<hr>
<p>In higher dimensions, the answer is yes.</p>
<p>Take the complete graph <span class="math-container">$K_n$</span> on your set of points. Embed it in <span class="math-container">$M$</span>. Then put a metric on <span class="math-container">$M$</span> that agrees with your desired metric in a neighbourhood of the graph, and which is extremely huge away from the graph. Then minimal geodesics will essentially follow the graph.</p>
<p>This solves the problem "up to <span class="math-container">$\varepsilon$</span>", as the geodesics don't exactly follow the graph, but do so only approximately.</p>
<p>To finish the argument, do the same thing in families, and invoke some version of the intermediate value theorem.
Here's how the argument goes. Let <span class="math-container">$D$</span> be the space of metrics on your fixed finite set.
Instead of doing the above construction for a single choice <span class="math-container">$d\in D$</span> of distances between the points <span class="math-container">$x_i$</span>, imagine that we adapt it to instead construct a <em>family</em> of Riemannian metrics on <span class="math-container">$M$</span> parametrised by the space <span class="math-container">$D$</span>. Starting from <span class="math-container">$d\in D$</span>, the geodesic distance between the <span class="math-container">$x_i$</span> produces another element <span class="math-container">$d'\in D$</span>. So we get a self-map <span class="math-container">$D\to D$</span> which is <span class="math-container">$\varepsilon$</span>-away from the identity map on <span class="math-container">$D$</span>. Now, <span class="math-container">$D$</span> is itself a manifold, and any self-map that's <span class="math-container">$\varepsilon$</span>-away from the identity is surjective.</p>
<p><b>[added later: the answer is no]</b><br>
Error in the above argument: <span class="math-container">$D$</span> is in fact a manifold with boundary. My argument works for metrics <span class="math-container">$d\in D\setminus \partial D$</span>. I.e., metrics where the triangle inequality holds strictly.</p>
<p>A counterexample is provided by the metric on <span class="math-container">$\{x_1,x_2,x_3,x_4\}$</span> given by
<span class="math-container">$d(x_1,x_i)=1$</span>, <span class="math-container">$d(x_i,x_j)=2$</span> (where <span class="math-container">$i,j\in\{2,3,4\}$</span>)</p>
<p><b>[added even later: all is good]</b><br> Ha ha! I hadn't noticed that you had assumed the strict triangle inequality to hold. So all is good, and this is a valid argument.</p>
|
216,021 | <p>Suppose $A$ and $B$ are $n \times n$ matrices. Assume $AB=I$. Prove that $A$ and $B$ are invertible and that $B=A^{-1}$.</p>
<p>Please let me know whether my proof is correct and if there are any improvements to be made.</p>
<p>Assume $AB=I$. Then $(AB)A=IA=A$. So, $A(BA)=AI=A$. Then $BA=I$. Therefore $AB=BA=I$. Thus $A$ and $B$ are invertible. And by definition $B=A^{-1}$, so $AB=AA^{-1}=I$.</p>
<p>It doesn't seem quite correct. Thanks in advance for any advice.</p>
| Brian M. Scott | 12,042 | <p>A good first step would be to look at some of the answers to <a href="https://math.stackexchange.com/questions/3852/if-ab-i-then-ba-i">this question</a>. The accepted one, by Davidac897, is pretty elementary and is probably the place to start. You’re almost certainly not yet ready for Martin Brandenberg’s answer, and I’d also skip Bill Dubuque’s answers for now: they’re also aimed at someone with more background. The proof given by falagar, on the other hand, is well worth a look, and you should certainly look at Blue’s answer, which is deliberately very elementary.</p>
|
216,021 | <p>Suppose $A$ and $B$ are $n \times n$ matrices. Assume $AB=I$. Prove that $A$ and $B$ are invertible and that $B=A^{-1}$.</p>
<p>Please let me know whether my proof is correct and if there are any improvements to be made.</p>
<p>Assume $AB=I$. Then $(AB)A=IA=A$. So, $A(BA)=AI=A$. Then $BA=I$. Therefore $AB=BA=I$. Thus $A$ and $B$ are invertible. And by definition $B=A^{-1}$, so $AB=AA^{-1}=I$.</p>
<p>It doesn't seem quite correct. Thanks in advance for any advice.</p>
| Martin Argerami | 22,857 | <p><strong>Proof #1:</strong> (along the lines mentioned in the comments)</p>
<p>As $AB=I$, you know that $A$ is onto as a linear transformation, because $x=Ix=ABx=A(Bx)$ for any $x\in\mathbb{R}^n$. This implies that $A$ is bijective, being a surjective linear transformation in a finite-dimensional space. So there exists $A^{-1}$. Now
$$
A^{-1}=A^{-1}I=A^{-1}AB=B.
$$</p>
<p><strong>Proof #2</strong> (using determinants)</p>
<p>Since $AB=I$, we have
$$
1=\det I=\det AB=\det A\,\det B.
$$
So $\det A\ne0$ and $A$ is invertible, and again we can do
$$
B=IB=A^{-1}AB=A^{-1}I=A^{-1}.
$$</p>
|
1,596 | <p><a href="https://en.wikipedia.org/wiki/Lenna" rel="nofollow noreferrer">Lenna</a> is commonly used as an example placeholder image. I also recently used it in an <a href="https://mathematica.stackexchange.com/questions/87693/how-to-put-an-imported-image-in-a-disk/87699#87699">answer on the site</a>. However, as the user <a href="https://mathematica.stackexchange.com/users/21825/lightness-races-in-orbit">Lightness Races in Orbit</a> pointed out, since Lenna is actually a Playboy centerfold image some people might find the usage of the image offensive, or unwelcoming to women (I don't see why women should be more offended by this than men, but let's keep that aside for now).</p>
<p>I tend to agree, and indeed, I changed the picture to a somewhat less offensive picture of Jeb Bush. However, it made me think, that we might want, as a community, to discourage the usage of Lenna, and even maybe add a functionality to the site (or to all SE sites, for that matter) that notifies users who use <code>ExampleData[{"TestImage", "Lena"}]</code> that this is discouraged, and suggest an alternative text image, for example, this one:</p>
<p><img src="https://i.stack.imgur.com/LfVGhm.jpg" alt="enter image description here"></p>
<p>Any thoughts or suggestions will be welcome.</p>
| Niki Estner | 242 | <p>First of all, I see @Lightness Races in Orbit's point. (If you don't, imagine you're the only male person in e.g. a painting class, so you feel somewhat isolated and awkward anyway, when you find out what you'll be painting is a nude shot of Burt Reynolds. Wearing a feather boa ;-) You probably wouldn't complain, but you might not feel more welcome, either.)</p>
<p>Second, this isn't about liberty of speech. Suggesting one (or a small set of) sample images makes answers more comparable. That's the point why there <em>are</em> standard test images like "Lena" in the first place.</p>
<p>On the other hand, from a purely technical standpoint, "Lena" is a <em>great</em> test image (or it would have been replaced long ago). It contains a face (humans can detect small errors in images of faces much more easily than e.g. artifacts, animals or aerial images), both sharp and blurry edges, straight edges, round edges and corners, non-uniform background... If you want to test e.g. a sharpening, denoising, inpainting, resizing or compression algorithm, those are all things you want in a test image.</p>
<p>So in a nutshell: I don't mind replacing Lena, I just think it's difficult to find a good replacement.</p>
<p>To make this rant constructive, my suggested alternative would be the "Barbara" test image:</p>
<p><img src="https://i.stack.imgur.com/Ips0H.jpg" alt="enter image description here"></p>
|
1,596 | <p><a href="https://en.wikipedia.org/wiki/Lenna" rel="nofollow noreferrer">Lenna</a> is commonly used as an example placeholder image. I also recently used it in an <a href="https://mathematica.stackexchange.com/questions/87693/how-to-put-an-imported-image-in-a-disk/87699#87699">answer on the site</a>. However, as the user <a href="https://mathematica.stackexchange.com/users/21825/lightness-races-in-orbit">Lightness Races in Orbit</a> pointed out, since Lenna is actually a Playboy centerfold image some people might find the usage of the image offensive, or unwelcoming to women (I don't see why women should be more offended by this than men, but let's keep that aside for now).</p>
<p>I tend to agree, and indeed, I changed the picture to a somewhat less offensive picture of Jeb Bush. However, it made me think, that we might want, as a community, to discourage the usage of Lenna, and even maybe add a functionality to the site (or to all SE sites, for that matter) that notifies users who use <code>ExampleData[{"TestImage", "Lena"}]</code> that this is discouraged, and suggest an alternative text image, for example, this one:</p>
<p><img src="https://i.stack.imgur.com/LfVGhm.jpg" alt="enter image description here"></p>
<p>Any thoughts or suggestions will be welcome.</p>
| Mr.Wizard | 121 | <p>Seeking the original for context I conclude:</p>
<ol>
<li><p>It seems to me a fairly tasteful photograph. There is similar nudity in many pieces of classical artwork that ostensibly have broad appeal. Not only is the cropped headshot only that but it does not allude to anything that a "reasonable observer" would find despicable, in my opinion.</p></li>
<li><p><code>ExampleData[{"TestImage", "Lena"}]</code> exhibits low image quality compared to available sources.</p></li>
</ol>
<p>For those who agree with (1) and wish to remedy (2) <a href="https://i.stack.imgur.com/o1z7p.jpg" rel="nofollow noreferrer">here is a copy of better quality</a>.</p>
|
2,170,278 | <p>The answer is $\binom {13}1 \binom42 \binom{12}3 \binom 41^3$</p>
<p>I want to break the last term and see what happens. [Struggling with the concept so trying to work with it as much as possible].</p>
<p>$\binom 41^3$ means that $\heartsuit \diamondsuit \spadesuit$ is different from $\diamondsuit \heartsuit \spadesuit$ and $\heartsuit \spadesuit \diamondsuit $ </p>
<p>$\binom 42 \binom 41$ means that $\heartsuit \diamondsuit \spadesuit$ is is the same as $\diamondsuit \heartsuit \spadesuit$</p>
<p>$\binom 43$ means that $\heartsuit \diamondsuit \spadesuit$, $\diamondsuit \heartsuit \spadesuit$, $\heartsuit \spadesuit \diamondsuit $ are all the same.</p>
<p>Basically, in the first case all three suits are ordered, in the second case first two suitss are unordered but the third one is and in the third case none of the suits are ordered.</p>
<p>Does that make sense?</p>
<p>Clarification:</p>
<p>A one pair consists of five cards where two are of the same kind and the other three are of different kinds. How many such hands are possible? The correct answer is $\binom {13}1 \binom42 \binom{12}3 \binom 41^3$. What I am doing is choosing the suits in different ways to see how that affects the correct answer. Want to see if my reasoning holds up.</p>
| Fabio Somenzi | 123,852 | <p>I think the problem is that in writing $P(k+1) \rightarrow P(k)$ you are implicitly assuming that $k+1 > 0$. That is, a full statement could be written as</p>
<p>$$ \forall k \in \mathbb{N} \,.\, k > 0 \rightarrow (P(k) \rightarrow P(k-1)) \enspace. $$</p>
<p>Once you get to $0$, you are "off the hook." For infinite descent, however, this is not the case. If $P(n)$ is true, there's got to be $m < n$ such that $P(n)$ is true. Once you get to $0$, you run out of $m$'s. You cannot satisfy the existential requirement, which forces you to conclude that $P(n)$ must be false for all $n$.</p>
|
178,028 | <p>I am given $G = \{x + y \sqrt7 \mid x^2 - 7y^2 = 1; x,y \in \mathbb Q\}$ and the task is to determine the nature of $(G, \cdot)$, where $\cdot$ is multiplication. I'm having trouble finding the inverse element (I have found the neutral and proven the associative rule.</p>
| Jacob | 37,079 | <p>Here's a hint: try using the difference of squares formula on the left hand side of the equation $x^2 -7y^2 =1$.</p>
|
178,028 | <p>I am given $G = \{x + y \sqrt7 \mid x^2 - 7y^2 = 1; x,y \in \mathbb Q\}$ and the task is to determine the nature of $(G, \cdot)$, where $\cdot$ is multiplication. I'm having trouble finding the inverse element (I have found the neutral and proven the associative rule.</p>
| Clive Newstead | 19,542 | <p>For $a+b\sqrt{7}$ we seek $x+y\sqrt{7}$ such that $(a+b\sqrt{7})(x+y\sqrt{7})=1$. Expanding these brackets and comparing coefficients gives</p>
<p>$$\begin{align} ax+7by &= 1 \\ bx+ay &= 0 \end{align}$$</p>
<p>It is then just a task of solving for $x$ and $y$.</p>
<p>This can be done using matrices:
$$\begin{align} \begin{pmatrix} a & 7b \\ b & a \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} &= \begin{pmatrix} 1 \\ 0\end{pmatrix} \\ \Rightarrow \begin{pmatrix} x \\ y \end{pmatrix} &= \begin{pmatrix} a & 7b \\ b & a \end{pmatrix}^{-1} \begin{pmatrix} 1 \\ 0 \end{pmatrix} \end{align}$$</p>
<p>Note that $a^2-7b^2 \ne 0$ for any $a,b \in \mathbb{Q}$ and so this matrix <em>is</em> invertible. It comes out from here in a couple of lines; then $x+y\sqrt{7} = (a+b\sqrt{7})^{-1}$.</p>
|
1,261,825 | <p>How can I find the inverse function of $f(x) = x^x$? I cannot seem to find the inverse of this function, or any function in which there is both an $x$ in the exponent as well as the base. I have tried using logs, differentiating, etc, etc, but to no avail. </p>
| wythagoras | 236,048 | <p>You have to use the Lambert W function, the inverse of $x e^x$. Using this function, one can find that the inverse of $y=x^x$ is $x=e^{W(\ln(y))}$</p>
|
1,445,702 | <p>i'm a little confused. </p>
<p>1)Which axis is which in 3 Dimensional system?</p>
<p>2)Does it matter if I switch the x-axis to y-axis?</p>
| Senex Ægypti Parvi | 89,020 | <p>The way I understand it is:<br>
$(0\mid 0\mid 0)$ is said to be "the origin."</p>
<p>[b is any non-zero number]
$(b\mid 0\mid 0),(0\mid b\mid 0),(0\mid 0\mid b)$ are said to be on the x-axis, the y-axis and the z-axis, respectively. </p>
<p>[b and c are any non-zero numbers]<br>
$(0\mid b\mid c),(b\mid 0\mid c),(b\mid c\mid 0)$ are said to be on the yz-plane, the xz-plane and the xy-plane, respectively.</p>
<p>[b, c and d are any non-zero numbers]
$(b\mid c\mid d)$ is said not to be in any of the above categories. </p>
<p>The x-, y- and z-coördinates are referred to by the terms "abscissa," "ordinate" and "applicate," respectively.</p>
<p>There are two orientations of the three axes, called "right-hand" and "left-hand," where one would associate a positive axis with the thumb, index finger and insult finger as a mnemonic.</p>
<p>Good hunting!</p>
|
366,401 | <p>Let <span class="math-container">$\nu$</span> be the uniform measure on the unit circle <span class="math-container">$\mathbb{S}^1 \subset \mathbb{R}^2$</span>, normalised so that <span class="math-container">$\nu(\mathbb{S}^1) = 1$</span>. Suppose <span class="math-container">$\mu$</span> is a Borel probability measure on <span class="math-container">$\mathbb{S}^1$</span> which is absolutely continuous w.r.t. <span class="math-container">$\nu$</span>, that is <span class="math-container">$\mu \ll \nu$</span>. Let <span class="math-container">$\{f_n\}_{n\geq 1}$</span> be an orthonormal basis for <span class="math-container">$L^2(\mathbb{S}^1,\mu)$</span>. Is it true that for <span class="math-container">$g \in C^k(\mathbb{S}^1)$</span>
<span class="math-container">$$
\int_{[0,2\pi]} f_n(\theta) g(\theta)d\mu(\theta) = o(1/n^k).
$$</span>
Or is it possible to <em>choose</em> an ONB such that the above holds? My question is motivated by the case when <span class="math-container">$\mu = \nu$</span> and the ONB is <span class="math-container">$\{1,z,\overline{z},z^2,\overline{z^2},\ldots\}$</span>, where it is known to be true (See <a href="https://math.stackexchange.com/questions/52260/asymptotic-estimate-for-riemann-lebesgue-lemma">this</a>.)</p>
<p>Thanks!</p>
| Pierre PC | 129,074 | <p>Here is a second example where a given basis fails to do the job. If <span class="math-container">$\mu$</span> has, say, a continuous positive density, then there a homeomorphism <span class="math-container">$h$</span> that sends <span class="math-container">$(\mathbb S^1,\mu)$</span> to <span class="math-container">$(\mathbb S^1,\nu)$</span> (uniform), in the sense <span class="math-container">$h^*\nu=\mu$</span>. Now because of the result you cite, the usual basis <span class="math-container">$(\phi_n)$</span> in <span class="math-container">$(\mathbb S^1,\nu)$</span> detects <span class="math-container">$\mathcal C^k$</span> functions according to your criterion, hence its preimage <span class="math-container">$(\phi_n\circ h)$</span> in <span class="math-container">$(\mathbb S^1,\mu)$</span> detects functions <span class="math-container">$f$</span> such that <span class="math-container">$f\circ h$</span> is <span class="math-container">$\mathcal C^k$</span>. If <span class="math-container">$h$</span> is not <span class="math-container">$\mathcal C^k$</span> (i.e. the density of <span class="math-container">$\mu$</span> is not <span class="math-container">$\mathcal C^{k-1}$</span>, I suspect), then you will have functions satisfying your criterion but are not smooth.</p>
<p>If the density <span class="math-container">$\rho$</span> such that <span class="math-container">$\mathrm d\mu=\rho\mathrm d\nu$</span> is bounded above and below, then I believe <span class="math-container">$(I_n\cdot\phi_n/\rho)$</span> will do the trick, with <span class="math-container">$1/I_n=\|\phi_n/\rho\|^2$</span>.</p>
|
88,145 | <p>A couple of recent questions on MO have involved the characters or the orders of specific finite groups of the form $G(\mathbb{Z}/n\mathbb{Z})$ for a familiar algebraic group $G$ defined over $\mathbb{Z}$ (implicitly as a group scheme) especially when $n$ is a prime power: <a href="https://mathoverflow.net/questions/87254/">here</a> and
<a href="https://mathoverflow.net/questions/87904/">here</a>.</p>
<p>It's reasonable to ask in what generality questions of the second kind can be treated uniformly. Here for example the question doesn't seem to require any knowledge of the structure of symplectic matrices, once one knows the orders of the finite symplectic groups over fields. (These have been written down in numerous papers and books.) In particular:</p>
<blockquote>
<p>For group schemes $G$ over $\mathbb{Z}$, in what natural generality is there a uniform procedure for computation of the orders of finite groups $G(\mathbb{Z}/p^k\mathbb{Z})$ as $p$ ranges over the prime numbrs?</p>
</blockquote>
<p>An old computation having a similar flavor occurs for the multiplicative group scheme (which is reductive but not simple) in the determination of the group of units in the ring $\mathbb{Z}_p$ of $p$-adic integers, written as an inverse limit of finite groups with successive quotients of order $p-1$ or $p$. Here the additive group scheme (viewed as the Lie algebra of the multiplicative group scheme) enters the picture, providing an iterated extension of the group of units of the residue field $\mathbb{F}_p$. See for example the book by Serre <em>A Course in Arithmetic</em>, II, section 3.</p>
<p>As George McNinch points out in his comment on Scott Carnahan's direct computation of the order for finite symplectic groups, a uniform approach is suggested by a paper of Serre, "Exemples de plongements des groupes PSL$_2(\mathbb{F}_p)$ dans des groupes de Lie simples", <em>Invent. Math.</em> 124 (1996), 3.1. Here <em>G</em> is a connected simple algebraic group over an algebraically closed field, essentially treated as a group scheme. Serre refers in turn to Demazure-Gabriel, <em>Groupes Algebriques</em> (1970), II, section 4, no. 3, a treatise influenced by the earlier Demazure-Grothendieck seminar SGA3. </p>
<p>In a Bourbaki talk, Chevalley showed how to view the simple adjoint groups, which he had constructed in a uniform way in 1955, as group schemes over $\mathbb{Z}$, a theme refined further by Kostant, and more recently by Lusztig in <em>J. Amer. Math. Soc.</em> 22 (2009). That seems to be a good setting for the question I've raised, though perhaps one can go further in the direction of reductive groups?</p>
<p>ADDED: The answers and comments are very interesting, but while I think further about them and the literature I should clarify that I'm taking for granted the standard (though nontrivial) formulas for the group orders over finite fields. And while it's natural (at least for Chevalley groups) to start over the ring of integers, there is certainly a passage to local rings implied here. The specific prime stays in the background, since unlike many questions in Lie theory this kind of computation doesn't distinguish "good" and "bad" primes: whatever is done should apply uniformly to all primes. Meanwhile a result in Appendix A.5 of the recent book by Conrad-Gabber-Prasad on pseudo-reductive groups has been pointed out to me. This suggests that groups aren't so essential to my question, but only a well-behaved class of schemes (again assuming that one can already count their points over finite fields). </p>
| A Stasinski | 2,381 | <p>Let $G$ be a smooth group scheme of finite type over $\mathbb{Z}/p^{r}$
(one can also do this over any Artinian local ring with perfect residue
field, but then some details below change). For example, $G$ could
be a reductive group scheme over $\mathbb{Z}/p^{r}$. This also includes
Chevalley groups over $\mathbb{Z}$ after extension of scalars to
$\mathbb{Z}/p^{r}$. In this setting the Greenberg functor $\mathcal{F}$
transforms $G$ into a smooth group scheme $\mathcal{F}G$ of finite
type over $\mathbb{F}_{p}$, such that $G(\mathbb{Z}/p^{r})\cong(\mathcal{F}G)(\mathbb{F}_{p})$.
The smoothness of $G$ implies that the natural map
<code>\[
G(\mathbb{Z}/p^{r})\longrightarrow G(\mathbb{F}_{p})
\]</code>
is surjective. The key is now Greenberg's structure theorem (see the
end of Greenberg's paper Schemata over local rings II), which implies
that for any $i\geq1$, the kernel of the natural map (which is a surjection, thanks to the smoothness of $G$)
<code>\[
G(\mathbb{Z}/p^{i+1})\longrightarrow G(\mathbb{Z}/p^{i})
\]</code>
is isomorphic to the $\mathbb{F}_{p}$-points of an algebraic group
over $\mathbb{F}_{p}$ whose underlying variety is affine $d$-space
where $d=\dim G_{\mathbb{F}_{p}}$. Greenberg conjectured that the kernel is always isomorphic to the vector group $\mathbb{G}_a^d$. If I've understood things correctly, this would imply that
$$
|G(\mathbb{Z}/p^{r})|=p^{d(r-1)}|G(\mathbb{F}_{p})|.
$$
I do not know about the status of Greenberg's conjecture or the stronger conjecture right at the end of his paper referred to above. It would be very interesting to know if any progress has been made on this since Greenberg's paper appeared.</p>
<p><strong>Edit:</strong> It may not be necessary to go via Greenberg's conjecture. It should be enough to show that the $\mathbb{F}_{p}$-rational structure on the kernel (viewed as a connected unipotent group) is sufficiently simple so that the order of the finite kernel is $p^{d}$.</p>
|
1,738,153 | <p>I know the definition is given as follows:</p>
<p>A map $p: G \rightarrow GL(V)$ such that $p(g_1g_2)=p(g_1)p(g_2)$ but I still do not really understand what this means</p>
<p>Can someone help me gain some intuition for this - perhaps a basic example?</p>
<p>Thanks</p>
| Nate | 91,364 | <p>I have a favorite example, or rather pair of examples that I like to use to motivate looking at representations assuming a bit of familiarity with groups.</p>
<p>The first thing I want to consider is the full symmetry group of a regular tetrahedron, where by full I mean "allowing reflections". In this case it's easy to see that we can order the four vertices in any way and the full symmetry group is the symmetric group $S_4$.</p>
<p>Next let's consider the group of rotational symmetries of a cube, where by rotational I mean, you guessed it, "not allowing reflections". It's easy to see that there are 24 such symmetries (by looking at say the "top face"), with a little more work you can also see that this group is again $S_4$ (hint: look at the space diagonals).</p>
<p>So somehow we have the same group arising as the symmetries of objects in two ways that feel very different in some sense, for one thing one has reflections and the other does not. If instead of talking about symmetries of a tetrahedron/cube we talk of symmetries of $\mathbb{R}^3$ that preserve the tetrahedron/cube, then we have two copies of $S_4$ in $GL_3(\mathbb{R})$ (or even $O(3)$) which we can think of as maps (homomorphisms) $S_4 \to GL_3(\mathbb{R})$. Such maps into $GL_n$ are called representations. The point of this example is that while the underlying groups here are the same, the representations are not.</p>
|
151,937 | <p>In <code>FindGraphCommunities</code>, how can one find the vertices associated with the edges that are found to connect one or more communities?</p>
| Szabolcs | 12 | <p>We'll use this example graph:</p>
<pre><code>g = ExampleData[{"NetworkGraph", "LesMiserables"}]
</code></pre>
<p>Find the edges which are <em>within</em> communities:</p>
<pre><code>subgraphs = EdgeList@Subgraph[g, #] & /@ FindGraphCommunities[g];
</code></pre>
<p>Visualize them:</p>
<pre><code>HighlightGraph[Graph[g, GraphStyle -> "BasicBlack"], subgraphs]
</code></pre>
<p><a href="https://i.stack.imgur.com/6OWBq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6OWBq.png" alt="enter image description here"></a></p>
<p>You want the edges which are not within a community but connecting different communities, i.e. the black ones in the visualization above.</p>
<pre><code>connectingEdges = Complement[EdgeList[g], Union @@ subgraphs];
</code></pre>
<p>Now get the vertices incident to these edges.</p>
<pre><code>Union@Level[connectingEdges, {2}]
(* {"Babet", "Bahorel", "Bamatabois", "Bossuet", "Brevet", \
"Brujon", "Champmathieu", "Chenildieu", "Claquesous", "Cochepaille", \
"Combeferre", "Cosette", "Courfeyrac", "Enjolras", "Eponine", \
"Fantine", "Feuilly", "Gavroche", "Gueulemer", "Javert", "Joly", \
"Judge", "Mabeuf", "Marguerite", "Marius", "Mlle Baptistine", "Mme. \
Magloire", "Mme. Thenardier", "Montparnasse", "Myriel", "Simplice", \
"Thenardier", "Tholomyes", "Valjean"} *)
</code></pre>
|
922,731 | <p>I am stuck on this question, not sure what I am doing wrong:
\begin{align*}
f(x) &=(5-x^2)(\sqrt{x}) \\
\\
f'(x) &=(5-x^2)(x)^{1/2} \\
&=(-2x)(x)^{1/2}+(-x^2+5)(\dfrac{1}{2})(x)^{-1/2}\\
&=-2x^{3/2}+\dfrac{1}{2}(-x^{3/2}+5x^{-1/2})\\
&=\dfrac{1}{2}x^{-1/2}\bigg(x^2-2x^2+10\bigg)\\
&=\dfrac{1}{2}x^{-1/2}\bigg(-x^2+10\bigg)
\end{align*}</p>
| Satish Ramanathan | 99,745 | <p>Answer</p>
<p>In your fourth step:</p>
<p>$f'(x) = \frac{-5}{2}x^{\frac{3}{2}} + \frac{5}{2}x^{\frac{-1}{2}}$</p>
<p>$f'(x) =-\frac{5}{2}\left(\frac{x^2 -1}{\sqrt{x}}\right)$</p>
|
922,731 | <p>I am stuck on this question, not sure what I am doing wrong:
\begin{align*}
f(x) &=(5-x^2)(\sqrt{x}) \\
\\
f'(x) &=(5-x^2)(x)^{1/2} \\
&=(-2x)(x)^{1/2}+(-x^2+5)(\dfrac{1}{2})(x)^{-1/2}\\
&=-2x^{3/2}+\dfrac{1}{2}(-x^{3/2}+5x^{-1/2})\\
&=\dfrac{1}{2}x^{-1/2}\bigg(x^2-2x^2+10\bigg)\\
&=\dfrac{1}{2}x^{-1/2}\bigg(-x^2+10\bigg)
\end{align*}</p>
| mathlove | 78,967 | <p>Note that
$$-2x^{3/2}+\dfrac{1}{2}(-x^{3/2}+5x^{-1/2})\not =\dfrac{1}{2}x^{-1/2}\bigg(x^2-2x^2+10\bigg).$$</p>
<p>We have
$$f'(x)=\cdots=-2x^{3/2}+\dfrac{1}{2}(-x^{3/2}+5x^{-1/2})=\frac 12x^{-\frac 12}(-4x^2-x^2+5)=\frac{5(1-x^2)}{2\sqrt x}.$$</p>
|
3,037,296 | <p>I'm confused of what <span class="math-container">$\sqrt {3 + 4i}$</span> would be after I used quadratic formula to simplify <span class="math-container">$z^2 + iz - (1 + i)$</span></p>
| KCd | 619 | <p>If you want to understand <em>where</em> a formula for the real and imaginary parts of square roots comes from, set <span class="math-container">$(x+yi)^2 = a+bi$</span>, where we know <span class="math-container">$a$</span> and <span class="math-container">$b$</span> we want to find <span class="math-container">$x$</span> and <span class="math-container">$y$</span>. Assume <span class="math-container">$b \not= 0$</span> (otherwise <span class="math-container">$a + bi = a$</span> is a real number and you know what to do whether <span class="math-container">$a > 0$</span> or <span class="math-container">$a < 0$</span>, or <span class="math-container">$a = 0$</span>). Then <span class="math-container">$x^2 - y^2 = a$</span> and <span class="math-container">$2xy = b$</span>. Since <span class="math-container">$b \not= 0$</span>, <span class="math-container">$x$</span> and <span class="math-container">$y$</span> are nonzero and we can write <span class="math-container">$y = b/(2x)$</span>. Then <span class="math-container">$x^2 - b^2/(4x^2) = a$</span>, so <span class="math-container">$4x^4 - 4ax^2 - b^2 = 0$</span>. </p>
<p>Actually, we could have derived that same equation even if <span class="math-container">$b = 0$</span>: multiply <span class="math-container">$x^2 - y^2 = a$</span> on both sides by <span class="math-container">$x^2$</span> to get <span class="math-container">$x^4 - (xy)^2 = ax^2$</span>, so <span class="math-container">$x^4 - b^2/4 = ax^2$</span>. Multiplying through by <span class="math-container">$4$</span> and bringing all the terms to the left, we get <span class="math-container">$4x^4 - 4ax^2 - b^2 = 0$</span> again. Thus this quartic in <span class="math-container">$x$</span> is valid no matter what <span class="math-container">$b$</span> is.</p>
<p>The quartic in <span class="math-container">$x$</span> is quadratic in <span class="math-container">$x^2$</span>, so
<span class="math-container">$$
x^2 = \frac{4a \pm \sqrt{16a^2 + 16b^2}}{2\cdot 4} = \frac{a \pm \sqrt{a^2 + b^2}}{2}.
$$</span>
For <span class="math-container">$x$</span> to be real, <span class="math-container">$x^2$</span> needs to be positive so we <em>must</em> use the <span class="math-container">$+$</span> sign on the right. Thus
<span class="math-container">$$
x^2 = \frac{a + \sqrt{a^2 + b^2}}{2} \geq 0,
$$</span>
so <span class="math-container">$x$</span> is the positive or negative square root of this real number. The roles of <span class="math-container">$x$</span> and <span class="math-container">$y$</span> are nearly symmetric: we can write <span class="math-container">$y^2 - x^2 = -a$</span> and <span class="math-container">$2yx = b$</span>, so running through the same argument with <span class="math-container">$a$</span> changed to <span class="math-container">$-a$</span>, we have
<span class="math-container">$$
y^2 = \frac{-a + \sqrt{a^2 + b^2}}{2} \geq 0,
$$</span>
so <span class="math-container">$y$</span> is the positive or negative square root of this real number. </p>
<p>Watch out: we gave two values for <span class="math-container">$x$</span> and two for <span class="math-container">$y$</span>, but those choices are not independent (otherwise we'd have <em>four</em> square roots of <span class="math-container">$a+bi$</span> in <span class="math-container">$\mathbf C$</span>). They are linked by the relation <span class="math-container">$2xy = b$</span>, and this linkage means the signs of <span class="math-container">$x$</span> and <span class="math-container">$y$</span> (as square roots) match if <span class="math-container">$b > 0$</span> and they are opposite if <span class="math-container">$b < 0$</span> (figure out what happens if <span class="math-container">$b = 0$</span>).</p>
|
1,538,496 | <p>I came across this riddle during a job interview and thought it was worth sharing with the community as I thought it was clever:</p>
<blockquote>
<p>Suppose you are sitting at a perfectly round table with an adversary about to play a game. Next to each of you is an infinitely large bag of pennies. The goal of the game is to be the player who is able to put the last penny on the table. Pennies cannot be moved once placed and cannot be stacked on top of each other; also, players place 1 penny per turn. There is a strategy to win this game every time. Do you move first or second, and what is your strategy?</p>
</blockquote>
<p>JMoravitz has provided the answer (hidden in spoilers) below in case you are frustrated!</p>
| mvw | 86,776 | <p>Update: OP changed rules at <a href="https://math.stackexchange.com/posts/1538496/revisions">revision 3</a>.</p>
<p>My attempt as spoiler below.</p>
<blockquote class="spoiler">
<p>I move first and cover the whole table with pennies.</p>
</blockquote>
|
1,335,842 | <p>The smallest solution to the above equation for various primes are:</p>
<p>$(p=2)$ $3^2 = 2*2^2 +1$</p>
<p>$(p=3)$ $2^2 = 2*1^2 +1$</p>
<p>$(p=5)$ $9^2 = 5*4^2 +1$</p>
<p>$(p=7)$ $8^2 = 7*3^2 +1$</p>
<p>Is there at least one solution for each prime?
If there is one solution, there are infinite.</p>
| orangeskid | 168,051 | <p>Here is the recipe for finding a (all the ) solution(s) of the equation
$$a^2 - d b^2 = 1$$
where $d>0$ is a square free integer. </p>
<p>Consider the continued fraction expansion of $\sqrt{d}$. It will be of the form
$$\sqrt{d} = [q,\overline{q_1, q_2 , \ldots q_r}]$$
with $q = [\sqrt{d}]$, with the sequence $q_1 q_2 \ldots q_r$ repeating indefinitely ( a (mixed) periodic continued fraction). Assume $r$ is minimal with this property. Then the minimal solution of the equation will be $(a_1,b_1)$, where
$$\frac{a_1}{b_1} = [q,q_1, q_2 , \ldots q_{\epsilon r-1}] $$
where $\epsilon =1$ if $r$ even and $\epsilon=2$ otherwise.
All the solutions will be $(a_k, b_k)$ where
$$\frac{a_k}{b_k} = [q,q_1, q_2 , \ldots q_{k\epsilon r-1}]$$
Alternatively, $$(a_k + \sqrt{d} b_k ) = (a_1 + \sqrt{d} b_1)^k$$</p>
<p>Let's consider the example $d = 43= p_{14}$. We have the continued fraction expansion of $\sqrt{43}$:
$$\sqrt{43} = [6,\overline{1,1,3,1,5,1,3,1,1,12}]$$
Calculate
$$[6,1,1,3,1,5,1,3,1,1] = \frac{3482}{531}$$
Fair enough, we see that
$$3482^2 - 43 \cdot 531^2 = 12\,124\,324 - 43 \cdot 281\,961 =12\,124\,324 - 12\,124\,323=1$$</p>
<p>But then also
$$[6, 1, 1, 3, 1, 5, 1, 3, 1, 1, 12, 1, 1, 3, 1, 5, 1, 3, 1, 1, 12, 1, \
1, 3, 1, 5, 1, 3, 1, 1] = \frac{168\,867\,574\,226}{25\,752\,063\,645}$$</p>
<p>and one can also check that
$$168\,867\,574\,226^2 - 43\cdot 25\,752\,063\,645^2 =1$$</p>
<p>Of course, we have
$$(3482 + \sqrt{43}\cdot 531)^3 = 168\,867\,574\,226 + \sqrt{43}\cdot25\,752\,063\,645$$</p>
<p>so the only surprise might be that all the solutions are coming from the continued fraction, which is true. </p>
<p>$\bf{Added:}$ Now that I looked at the example of @Travis, let's see whether the theory works:</p>
<p>$$\sqrt{61} = [7, \overline{1, 4, 3, 1, 2, 2, 1, 3, 4, 1, 14}]$$
The period $r =11$, odd, so we need to take chunks of lenght even multiple of $r$ $2kr$. The first one is
$$\frac{a_1}{b_1} = [7, 1, 4, 3, 1, 2, 2, 1, 3, 4, 1, 14, 1, 4, 3, 1, 2, 2, 1, 3, 4, 1]=\frac{1766319049}{226153980}$$
confirmed.</p>
<p>Note that if we take the chunk of lenght $11$ instead
$$\frac{a}{b} = [7, 1, 4, 3, 1, 2, 2, 1, 3, 4, 1] = \frac{29718}{3805}$$
we get
$$29\,718- 61 \cdot 3\,805^2 = -1 $$</p>
|
1,961,727 | <p>As far as I understood <a href="https://en.wikipedia.org/wiki/Gram%E2%80%93Schmidt_process">Gram–Schmidt orthogonalization</a> starts with a set of linearly independent vectors and produces a set of mutually orthonormal vectors that spans the same space that starting vectors did.</p>
<p>I have no problem understanding the algorithm, but here is a thing I fail to get. Why do I need to do all these calculations? For example, instead of doing the calculations provided in that wiki page in example section, why can't I just grab two basis vectors $w_1 = (1, 0)'$ and $w_2 = (0, 1)'$? They are clearly orthonormal and span the same subspace as the original vectors $v_1 = (3, 1)'$, $v_2 = (2, 2)'$.</p>
<p>It is clear that I'm missing something important, but I can't see what exactly.</p>
| erfink | 376,021 | <p>If the entire vector space we wanted a basis for all of $\mathbb{R}^n$ or a basis for, say, the $xy$-plane in $\mathbb{R}^3$, then we could certainly do this. The problem arises when we're looking for an orthonormal basis for subspaces that are more complicated.</p>
<p>For example, let take our subspace to be the plane $x+y+z=0$ inside of $\mathbb{R}^3$. This subspace can be written as</p>
<p>$$\mathrm{span} \left\{ \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} , \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} \right\}$$</p>
<p>Running Gram-Schmidt on this gives the basis</p>
<p>$$\left\{
\begin{bmatrix}
\frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} \\ 0 \end{bmatrix} ~,~ \begin{bmatrix}
\frac{1}{\sqrt{6}} \\\frac{1}{\sqrt{6}}\\ -\sqrt{\frac{2}{3}} \end{bmatrix}
\right\}$$
which <em>generates the same subspace of $\mathbb{R}^3$</em>. However, if we just tried to take the basis $\left\{ \vec{e}_1 = [1~ 0~ 0]^T, ~ \vec{e}_2 = [ 0 ~ 1 ~ 0]^T \right\}$, we would certainly have an ortho-normal basis, but of a different subspace. </p>
<p>TL;DR: Gram-Schmidt is designed to turn a basis into an ortho-normal basis <em>without altering the subspace that it spans</em>.</p>
<p><strong>Edit</strong> Gram-Schmidt is also important in that it preserves the orientation of given basis (roughly speaking, the order in which the basis elements are introduced). Somewhat higher level and beyond the scope of an introductory course to linear algebra, but worth tucking away for later.</p>
|
1,961,727 | <p>As far as I understood <a href="https://en.wikipedia.org/wiki/Gram%E2%80%93Schmidt_process">Gram–Schmidt orthogonalization</a> starts with a set of linearly independent vectors and produces a set of mutually orthonormal vectors that spans the same space that starting vectors did.</p>
<p>I have no problem understanding the algorithm, but here is a thing I fail to get. Why do I need to do all these calculations? For example, instead of doing the calculations provided in that wiki page in example section, why can't I just grab two basis vectors $w_1 = (1, 0)'$ and $w_2 = (0, 1)'$? They are clearly orthonormal and span the same subspace as the original vectors $v_1 = (3, 1)'$, $v_2 = (2, 2)'$.</p>
<p>It is clear that I'm missing something important, but I can't see what exactly.</p>
| Wrzlprmft | 65,502 | <p>To give a somewhat practical example, let’s consider the following iterative sequence:</p>
<p>$$ x_{i+1} = \frac{A x_i}{\left |A x_i \right|},$$</p>
<p>Where the $x_i$ are vectors and $A$ is a quadratic matrix with a matching dimension. The first vector $x_1$ shall be random¹.</p>
<p>When $i$ is increased, $x_i$ will more and more be aligned to an eigenvector corresponding to the largest eigenvalue of $A$ and quickly become such an eigenvector for all practical purposes. So we can use this sequence to get a numerical approximation for the largest eigenvalue of $A$ and a corresponding eigenvector (of length 1). This approach is actually used for this purpose.</p>
<p>Now, if we want to obtain the second largest eigenvalue of $A$ (and the corresponding eigenvector) in a similar way, we have to remove the influence of the first eigenvalue, as it would inevitably dominate our sequence otherwise. Hence we have to work in a subspace that is orthogonal to the eigenvector for the largest eigenvalue. Gram–Schmidt orthonormalisation provides a way to do this: We start with two random vectors $x_0$ and $y_0$ and then we do a Gram–Schmidt after every step:</p>
<p>$$\begin{align}
x_{i+1} &= \frac{A x_i}{\left |A x_i \right|},\\
y_{i+1} &= \frac{A y_i - \left \langle A y_i, x_{i+1} \right \rangle}{\left | A y_i - \left \langle A y_i, x_{i+1} \right \rangle \right |}
\end{align}
$$</p>
<p>Due to this, $x_i$ will align with the first eigenvector and $y_i$ will align with the second one. We can extend this scheme to further eigenvectors if we like.</p>
<p>The crucial feature of the Gram–Schmidt process that we exploit here is that the first $k$ vectors of its result span the same subspace as the first $k$ vectors of its input <strong>for any $k$</strong>. A consequence of this is that the $k$th output vector is orthogonal to all previous output vectors. Obviously, this would not work with any basis.</p>
<p>Now in many cases, you can determine the first eigenvectors more easily, but there are analogous problems, where you need to go the way described above. For example, it is used for numerical calculation of <a href="http://en.wikipedia.org/wiki/Lyapunov_exponent#Numerical_calculation" rel="nofollow">Lyapunov exponents</a>, where $A$, $x$, and $y$ are subject to a complex temporal evolution.</p>
<hr>
<p><sup>¹ and thus we can assume that it is not orthogonal to any eigenvector of $A$</sup></p>
|
3,238,670 | <p>Could someone explain <strong>how to get from: <span class="math-container">$x-\frac{1}{x}=A$</span> to <span class="math-container">$x+\frac{1}{x}=\sqrt{A^2+4}$</span></strong> ? It is one of the Algebra II tricks.</p>
<p>Thanks.</p>
| BladeofDagger | 676,262 | <p>Ok, I got it!</p>
<p><span class="math-container">$x-\frac{1}{x}=A$</span></p>
<p><span class="math-container">$\Bigl(x-\frac{1}{x}\Bigl)^2=A^2$</span></p>
<p><span class="math-container">$x^2+\frac{1}{x^2}-2=A^2$</span></p>
<p><span class="math-container">$x^2+\frac{1}{x^2}=A^2+2$</span></p>
<p><span class="math-container">$\Bigl(x+\frac{1}{x}\Bigl)^2-2=A^2+2$</span></p>
<p><span class="math-container">$\Bigl(x+\frac{1}{x}\Bigl)^2=A^2+4$</span></p>
<p><span class="math-container">$\sqrt{\Bigl(x+\frac{1}{x}\Bigl)^2}=\sqrt{A^2+4}$</span></p>
<p><span class="math-container">$x+\frac{1}{x}=\sqrt{A^2+4}$</span></p>
|
3,238,670 | <p>Could someone explain <strong>how to get from: <span class="math-container">$x-\frac{1}{x}=A$</span> to <span class="math-container">$x+\frac{1}{x}=\sqrt{A^2+4}$</span></strong> ? It is one of the Algebra II tricks.</p>
<p>Thanks.</p>
| zwim | 399,263 | <p>Remark that <span class="math-container">$x-\frac 1x=A\iff x^2-Ax-1=0$</span></p>
<p>Which is a quadratics with roots <span class="math-container">$x$</span> and <span class="math-container">$-\frac 1x$</span>.</p>
<p>Calculating <span class="math-container">$\Delta=A^2+4$</span> and roots <span class="math-container">$\dfrac{A\pm\sqrt{\Delta}}{2}$</span></p>
<p>It is now obvious why their sum is <span class="math-container">$A$</span> and their difference is <span class="math-container">$\sqrt{\Delta}$</span>.</p>
|
2,939,585 | <p>I want to prove that if <span class="math-container">$ \gamma$</span> is a closed path and <span class="math-container">$\gamma\subseteq B_R(0) $</span> then <span class="math-container">$\mathbb{C}\setminus B_R(0)\subseteq \operatorname{Ext}_\gamma$</span> where <span class="math-container">$ \operatorname{Ext}_\gamma=\{a\not \in \gamma : \operatorname{Ind}_\gamma a=0\} $</span> and <span class="math-container">$ \operatorname{Ind}_\gamma a=\frac{1}{2\pi i}\oint_\gamma \frac{dz}{z-a}$</span>.</p>
<p>I think that one have to bound the index like this:
<span class="math-container">$$ \left|\operatorname{Ind}_\gamma a\right|=\left|\frac{1}{2\pi i}\oint_\gamma \frac{dz}{z-a}\right| \leq \operatorname{lenght}(\gamma)\frac{1}{2\pi } \frac{1}{\min\{|z-a|:z\in \gamma\}}$$</span>
and then use the fact that I can make <span class="math-container">$ \min\{|z-a|:z\in \gamma\}$</span> as larger as I want, but I don't know how to proceed.</p>
| Paul Frost | 349,785 | <p>It is well-known that the winding number is <em>homotopy invariant</em>, i.e. if <span class="math-container">$\gamma_0, \gamma_1 : [0,1] \to \mathbb{C} \setminus \{ a\}$</span> are homotopic closed paths (which means that there exists a continuous map <span class="math-container">$\Gamma : [0,1] \times [0,1] \to \mathbb{C} \setminus \{ a\}$</span> such that <span class="math-container">$\Gamma(t,i) = \gamma_i(t)$</span> for all <span class="math-container">$t \in [0,1]$</span> and <span class="math-container">$i = 0, 1$</span> and <span class="math-container">$\Gamma(0,s) = \Gamma(1,s)$</span> for all <span class="math-container">$s \in [0,1]$</span>), then <span class="math-container">$\text{Ind}_{\gamma_0}a = \text{Ind}_{\gamma_1}a$</span>.</p>
<p>Let <span class="math-container">$a \in \mathbb{C} \setminus B_R(0)$</span>. Then <span class="math-container">$\gamma$</span> and the constant path <span class="math-container">$c_0(t) \equiv 0$</span> are homotopic closed paths in <span class="math-container">$\mathbb{C} \setminus \{ a\}$</span>, as can be seen via <span class="math-container">$\Gamma : [0,1] \times [0,1] \to B_R(0) \subset \mathbb{C} \setminus \{ a\}, \Gamma(t,s) = (1-s)\gamma(t)$</span>. Hence <span class="math-container">$\text{Ind}_{\gamma}a = \text{Ind}_{c_0}a = 0$</span> because <span class="math-container">$c_0$</span> has length <span class="math-container">$0$</span>.</p>
|
999,147 | <p>I'm looking to gain a better understanding of how the cofinite topology applies to R.
I know the definition for this topology but I'm specifically looking to find some properties such as the closure, interior, set of limit points, or the boundary set and how these change based on whether a subset A in R is closed, open, or clopen. </p>
<p>Any help would be appreciated. </p>
<p>Note: I have only the most basic of definitions for closure, interior, etc. </p>
<p>Thank you! </p>
| Clive Newstead | 19,542 | <p>In the cofinite topology, '$A$ is closed' means '$A$ is finite or is $\mathbb{R}$' and '$A$ is open' means '$A$ has finite complement or is empty'. In particular...</p>
<ul>
<li>If $A \subseteq \mathbb{R}$ is infinite then the only closed set containing $A$ is $\mathbb{R}$, and hence $\operatorname{cl} A = \mathbb{R}$. In particular, if $A$ is infinite, every point of $\mathbb{R} - A$ is a limit point of $A$. Hence if $A$ is infinite then $\partial A = \mathbb{R} - A$.</li>
<li>If $A \subseteq \mathbb{R}$ has infinite compliment then its only open subset is $\varnothing$, so $\operatorname{int} A = \varnothing$.</li>
<li>The only clopen sets are $\mathbb{R}$ and $\varnothing$.</li>
</ul>
|
9,934 | <p>I have been given some code with the following line</p>
<pre><code>PeriodicExtension[g_, x_] := If[Abs[x] < Pi, g[x], PeriodicExtension[g, x - 2 Sign[x] Pi]]
</code></pre>
<p>I do not understand the syntax. I would appreciate if someone can explain what this
code does for different values of <code>x</code>.</p>
| J. M.'s persistent exhaustion | 50 | <p>I'm not terribly fond of the use of a recursive solution when a non-recursive approach is easier to look at, so here's a more compact and more general implementation based on <a href="https://math.stackexchange.com/a/63463">this math.SE answer</a> I wrote:</p>
<pre><code>PeriodicExtension[g_, {a_?NumericQ, b_?NumericQ}, x_?NumericQ] := g[Mod[x, b - a, a]]
</code></pre>
<p>Here, we use <em>Mathematica</em>'s shifted version of <code>Mod[]</code>, <code>Mod[x, m, h] == h + Mod[x - h, m]</code>.</p>
<p>To use Verde's example:</p>
<pre><code>Plot[PeriodicExtension[Identity, {-π, π}, x], {x, 0, 4 π}]
</code></pre>
<p><img src="https://i.stack.imgur.com/uC2nr.png" alt="periodic extension"></p>
|
3,534,364 | <blockquote>
<p><span class="math-container">$x^2y'^2 + 3xyy' +2y^2 = 0 $</span></p>
</blockquote>
<p>Usually, to solve an ODE with respect to <span class="math-container">$y'=p$</span>, we first isolate the <span class="math-container">$y$</span>, to get <span class="math-container">$y = f(x,p)$</span> and then differentiate with respect to <span class="math-container">$x$</span> to get an expression that only depends on <span class="math-container">$x$</span> and <span class="math-container">$p$</span>. Then, we can write <span class="math-container">$x$</span> and <span class="math-container">$y$</span> in terms of <span class="math-container">$p$</span> and get to a solution.</p>
<p>But what do we do if, like in this particular example, we can't isolate the y? Is there another method for these kind of ODEs?</p>
<p>My manual lists the solutions for this equation as <span class="math-container">$xy=c$</span> or <span class="math-container">$yx^2 =c$</span>, but I have no clue as to how they come to that conclusion. </p>
| user577215664 | 475,762 | <p><span class="math-container">$$x^2y'^2 + 3xyy' +2y^2 = 0$$</span>
I suppose <span class="math-container">$y \ne 0$</span>
<span class="math-container">$$x^2 \frac {y'^2}{y^2} + 3x\frac {y'}{y} +2 = 0$$</span>
<span class="math-container">$$x^2 \left ( \frac {y'}{y} \right )^2 + 3x \left ( \frac {y'}{y} \right ) +2 = 0$$</span>
<span class="math-container">$$x^2 \left ( (\ln y)' \right )^2 + 3x(\ln y )'+2 = 0$$</span>
It's a quadtratic equation.
<span class="math-container">$$ \left ( x(\ln y)' \right )^2 + 3(x(\ln y )')+2 = 0$$</span></p>
|
345,094 | <p>If $f(x-1)+f(x-2) = 5x^2 - 2x + 9$</p>
<p>and</p>
<p>$f(x)= ax^2 + bx + c$</p>
<p>what would be the value of $a+b+c$?</p>
<p>I was doing</p>
<p>$f(x-1)+f(x-2)= f(x-3)$
then
$f(x)$</p>
<pre><code>a = 5
b = -2
c = 9
</code></pre>
<p>$(5-3)+(-2-3)+(9-3)$</p>
<p>But do not think is is correct</p>
<p>What would be correct approach?</p>
| Clayton | 43,239 | <p>We know $$f(x-1)=a(x-1)^2+b(x-1)+c\quad\text{and}\quad f(x-2)=a(x-2)^2+b(x-2)+c.$$ Expand these terms, add them, and combine like terms via powers of $x$. Now you can get three equations in three variables by equating the coefficients of the left with the right since you know $$f(x-1)+f(x-2)=5x^2-2x+9.$$</p>
|
345,094 | <p>If $f(x-1)+f(x-2) = 5x^2 - 2x + 9$</p>
<p>and</p>
<p>$f(x)= ax^2 + bx + c$</p>
<p>what would be the value of $a+b+c$?</p>
<p>I was doing</p>
<p>$f(x-1)+f(x-2)= f(x-3)$
then
$f(x)$</p>
<pre><code>a = 5
b = -2
c = 9
</code></pre>
<p>$(5-3)+(-2-3)+(9-3)$</p>
<p>But do not think is is correct</p>
<p>What would be correct approach?</p>
| cloned | 53,024 | <p>$$f''(x)=2a, 4a=f''(x-1)+f''(x-2)=10, a=2.5$$</p>
<p>$$f'(x)=2ax+b=5x+b, 5(x-1)+5(x-2)+2b$$ </p>
<p>$$f'(x-1)+f'(x-2)=10x-2, -15+2b=-2, b=6.5$$</p>
<p>$$f(x)=ax^2+bx+c, 2.5(x-1)^2+6.5(x-1)+2.5(x-2)^2+6.5(x-2)+2c=f(x-1)+f(x-2)=5x^2-2x+9$$</p>
<p>let $x=0$ on both side, $2.5-6.5+10-13+2c=9, c=8$</p>
<p>so $a+b+c=2.5+6.5+8=17$</p>
|
3,568,693 | <p>I am trying to solve <span class="math-container">$n! = 10^6$</span> for <span class="math-container">$n$</span>. I thought to do this using the <a href="https://en.wikipedia.org/wiki/Gamma_function" rel="nofollow noreferrer">gamma function</a>:</p>
<p><span class="math-container">$$(n - 1)! = \Gamma(n) = \int_0^\infty x^{n - 1}e^{-x} \ dx$$</span></p>
<p>So I have that </p>
<p><span class="math-container">$$\Gamma(n + 1) = \int_0^\infty x^n e^{-x} \ dx = 10^6$$</span></p>
<p>I thought to solve this using integration by parts:</p>
<p><span class="math-container">$$\begin{align} \Gamma(n + 1) &= \int_0^\infty x^n e^{-x} \ dx \\ &= [-x^n e^{-x}]^{\infty}_0 + \int_0^{\infty} nx^{n - 1}e^{-x} \ dx \end{align}$$</span></p>
<p>But, as you can see, unless I have made a mistake, we get the term <span class="math-container">$\int_0^{\infty} nx^{n - 1}e^{-x} \ dx$</span>, which, as far as I can tell, means that we get stuck in an infinite loop of integration by parts.</p>
<p>So how do I solve this? </p>
<p>Thank you.</p>
| Claude Leibovici | 82,404 | <p>If you look at <a href="https://math.stackexchange.com/questions/430167/is-there-an-inverse-to-stirlings-approximation/461207#461207">this question</a>, you will find a very good approximation of the inverse of the factorial function.</p>
<p>Applied to your case
<span class="math-container">$$n! =y$$</span>it write</p>
<p><span class="math-container">$$n \sim \frac{{\log \left( {\frac{y}{{\sqrt {2\pi } }}} \right)}}{{W\left( {\frac{1}{e}\log \left( {\frac{y}{{\sqrt {2\pi } }}} \right)} \right)}} -\frac{1}{2}$$</span> and, for <span class="math-container">$y=10^6$</span>, it will give <span class="math-container">$n=9.44379$</span> while the exact solution is <span class="math-container">$9.44561$</span>.</p>
|
1,644,845 | <blockquote>
<p>Show that $\lim_{z \to 0} \frac{\Re(z)}{z}$ doesn't exist.</p>
</blockquote>
<p>Let $z=r(\cos(\theta)+i \sin(\theta))$. So $\frac{\Re(z)}{z} =\cos ^2(\theta) - i \cos(\theta)\sin(\theta) $, and $$\lim_{z \to 0} \frac{\Re(z)}{z} = \lim_{r \to 0} (\cos ^2(\theta) - i \cos(\theta)\sin(\theta)) = (\cos ^2(\theta) - i \cos(\theta)\sin(\theta))$$</p>
<p>So the limit cannot exist, because it depends of $\theta$.</p>
<p>Am I right? If not, is there existed another approach?</p>
| blizzard22 | 549,349 | <p>I believe you have the denominators switched. Also, say $c_1 = \chi^2_{1, 1-\alpha/2} $ and $c_2 =\chi^2_{1, \alpha/2 } $
then for (c), the expected width of the interval is
$$ \mathbb{E} \left[ n (\bar{X} - \mu )^2 ( \frac{1}{c_2} - \frac{1}{c_1}) \right] =
n ( \frac{1}{c_2} - \frac{1}{c_1}) \times \mathbb{E}(\bar{X} - \mu )^2 =
n ( \frac{1}{c_2} - \frac{1}{c_1}) \; \; \times \frac{\sigma^2}{n} = ( \frac{1}{c_2} - \frac{1}{c_1}) \sigma^2 $$</p>
<p>since $ \quad \quad \frac{n}{\sigma^2} ( \bar{X} - \mu ) \sim \chi^2_1 $ </p>
<p>and thus $ \quad \mathbb{E}(\bar{X} - \mu )^2 = \frac{\sigma^2}{n} $</p>
<p>strange the expected width does not depend on $n$...</p>
|
3,356,544 | <p>A lot of calculators actually agree with me saying that it is defined and the result equals 1, which makes sense to me because:</p>
<p><span class="math-container">$$ (-1)^{2.16} = (-1)^2 \cdot (-1)^{0.16} = (-1)^2\cdot\sqrt[100]{(-1)^{16}}\\
= (-1)^2 \cdot \sqrt[100]{1} = (-1)^2 \cdot 1 = 1$$</span></p>
<p>However, there are certain calculators (WolframAlpha among them) which contest this answer, and instead claim it is equal to:</p>
<p><a href="https://i.stack.imgur.com/XB8nG.png" rel="noreferrer"><img src="https://i.stack.imgur.com/XB8nG.png" alt="enter image description here"></a></p>
<p>Graphing this as an exponential function was not possible.</p>
<p>What's going on?</p>
| zwim | 399,263 | <p>If we stay strictly in the real domain, then exponentiation is defined by the formula <span class="math-container">$x^y=\exp(y\ln(x))$</span>.</p>
<p>Since the logarithm is defined only on positive numbers, exponentiation is also well defined only on positive bases <span class="math-container">$x$</span>.</p>
<p>With this definition <span class="math-container">$(-1)^{2.16}$</span> is undefined.</p>
<hr>
<p>However, since it also comes from rational exponentiation by continuity, which is merely an extension of integer exponentiation where things like <span class="math-container">$(-1)^n$</span> are perfectly defined, it is possible to extend exponentiation to negative real numbers as well, but with limitation to rational exponents.</p>
<p>If we call <span class="math-container">$\mathbb Q_{odd}=\{\frac pq\in\mathbb Q\mid p\in\mathbb Z, q\in\mathbb N^*, \gcd(p,q)=1\text{ AND } q\text{ odd}\}$</span></p>
<p>Then for <span class="math-container">$x<0$</span> we have <span class="math-container">$x^r$</span> defined for any <span class="math-container">$r\in\mathbb Q_{odd}$</span> by
<span class="math-container">$$x^r=x^{\frac pq}=\sqrt[q]{x^p}=\left({\sqrt[q]{x}}\right)^p$$</span></p>
<p>Since the <span class="math-container">$q-$</span>root is an odd function, it is perfectly defined for negative numbers.</p>
<p>Note that we explicitly require that <span class="math-container">$\gcd(p,q)=1$</span> to avoid contradictions like <span class="math-container">$-1=(-1)^{\frac 13}=(-1)^{\frac 26}=\sqrt[6]{(-1)^2}=1$</span></p>
<p>Considering this <span class="math-container">$(-1)^{2.16}=(-1)^{\frac {54}{25}}=1$</span> would be a reasonable answer.</p>
<hr>
<p>Now it is also possible to take some high, and examine the situation on the complex domain.</p>
<p><span class="math-container">$z_1^{z_2}=\exp(z_2\ln(z_1))$</span> is defined by the same formula that in the real case, but this time the logarithm has an extended definition.</p>
<p>The price to pay is that since <span class="math-container">$\exp$</span> is a <span class="math-container">$2i\pi-$</span>periodic function, its inverse, the logarithm becomes multi-valued, meaning there will be more than one value per arithmetic expression.</p>
<p>We define <span class="math-container">$$\ln(z)=\ln|z|+i(\arg(z)+2k\pi)\quad k\in\mathbb Z$$</span></p>
<p><br>
Let's examine the consequences for <span class="math-container">$(-1)^{2.16}$</span></p>
<p><span class="math-container">$\ln(-1)=\underbrace{\ln(1)}_0+i(\underbrace{\arg(-1)}_{\pi\text{ from }e^{i\pi}=-1}+2k\pi)=(2k+1)i\pi$</span></p>
<p>Also since <span class="math-container">$2.16$</span> is in fact rational <span class="math-container">$\frac {54}{25}$</span> as we already seen, we won't get an infinite number of values for the expression, yet still <span class="math-container">$25$</span> different ones.</p>
<p>Let's call them for <span class="math-container">$k=0\cdots24$</span>
<span class="math-container">$$z_{[k]}=\exp(\frac{54(2k+1)}{25}i\pi)=\exp(\frac{108k}{25}i\pi+\frac{54}{25}i\pi)=z_{[0]}\,\phi^k$$</span></p>
<p><br>
With <span class="math-container">$\begin{cases}z_{[0]}=e^{(2.16i\pi)} & \text{is called the principal value at }k=0\\\phi=\exp(\frac{108}{25}i\pi) & \text{is a cyclic multiplicative factor (in this case a rotation)}\end{cases}$</span></p>
<p><br>
As noted by Conrad, when <span class="math-container">$k=12$</span> then <span class="math-container">$z_{[12]}=1$</span> the answer found in the rational exponentiation case is effectively among the results, although it is not the principal value.</p>
|
1,998,938 | <p>How can I solve \begin{cases}
u_t-u_{xx}=0,&\text{if $0<x<1, t>0$}\\
u(0,t)=u(1,t)=0, & \text{if $t>0$}\\u(x,0)=u_0(x), &\text{if $x\in(0,1)$} \end{cases}</p>
<p>where $$u_0=min(x,1-x)$$</p>
| reuns | 276,986 | <p>Let $$F_k(s) = s\int_0^\infty \{x\}^k x^{-s-1}dx = s \int_0^\infty \left\{\frac1t\right\}^k t^{s-1}dt$$</p>
<p>Show that $$\int_0^x \{t\}^kdt = \frac{\lfloor x \rfloor +\{x\}^{k+1}}{k+1}$$</p>
<p><a href="https://en.wikipedia.org/wiki/Abel%27s_summation_formula" rel="nofollow noreferrer">And that (where it converges)</a> $$\zeta(s) = s\int_0^\infty \lfloor x \rfloor x^{-s-1}dx = \frac{s}{s-1}-s\int_1^\infty \{x\} x^{-s-1}dx, \qquad\quad F_1(s) = -\zeta(s)$$</p>
<p>Integrating by parts
$$F_k(s) = s(s+1)\int_0^\infty \frac{\lfloor x \rfloor +\{x\}^{k+1}}{k+1} x^{-s-2}dx
= s\frac{\zeta(s+1)+F_{k+1}(s+1)}{k+1}$$
i.e.
$$\boxed{F_k(s) = \frac{k}{s-1} F_{k-1}(s-1)-\zeta(s)= -\sum_{m=0}^{k-1} \zeta(s-m)\prod_{l=0}^{m-1} \frac{k-l}{s-l-1}}$$</p>
|
1,998,938 | <p>How can I solve \begin{cases}
u_t-u_{xx}=0,&\text{if $0<x<1, t>0$}\\
u(0,t)=u(1,t)=0, & \text{if $t>0$}\\u(x,0)=u_0(x), &\text{if $x\in(0,1)$} \end{cases}</p>
<p>where $$u_0=min(x,1-x)$$</p>
| Felix Marin | 85,343 | <p>$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$</p>
<blockquote>
<p>$\ds{\int_{0}^{\infty}\braces{1 \over t}^{kn}t^{s - 1}\,\dd t:\ ?.\qquad
k,n \in \mathbb{N}_{\ \geq\ 0}\,,\quad\Re\pars{s} \in \pars{0,1}}$.</p>
</blockquote>
<p>\begin{align}
\int_{0}^{\infty}\braces{1 \over t}^{kn}t^{s - 1}\,\dd t &
\,\,\,\stackrel{t\ \mapsto\ 1/t}{=}\,\,\,
\int_{\infty}^{0}\braces{t}^{kn}\,\pars{1 \over t}^{s - 1}
\pars{-\,{\dd t \over t^{2}}} =
\int_{0}^{\infty}\braces{t}^{kn}\, t^{-s - 1}\,\dd t
\\[5mm] & =
\int_{0}^{\infty}\pars{t - \left\lfloor t\right\rfloor}^{kn}\, t^{-s - 1}\,\dd t
\end{align}
<hr>
With $\ds{N \in \mathbb{N}_{\ >\ 0}}$:
\begin{align}
&\int_{0}^{N}\pars{t - \left\lfloor t\right\rfloor}^{kn}\, t^{-s - 1}\,\dd t
\\[5mm] = &\
\int_{0}^{1}\pars{t - 0}^{kn}t^{-s - 1}\,\dd t +
\int_{1}^{2}\pars{t - 1}^{kn}\, t^{-s - 1}\,\dd t + \cdots +
\int_{N - 1}^{N}\pars{t - N + 1}^{kn}\, t^{-s - 1}\,\dd t
\\[5mm] = &\
\int_{0}^{1}t^{kn}t^{-s - 1}\,\dd t +
\int_{0}^{1}t^{kn}\, \pars{t + 1}^{-s - 1}\,\dd t + \cdots +
\int_{0}^{1}t^{kn}\, \pars{t + N - 1}^{-s - 1}\,\dd t
\\[5mm] = &\
\int_{0}^{1}t^{kn}\sum_{k = 0}^{N - 1}{1 \over \pars{k + t}^{s + 1}}\,\dd t
\end{align}
<hr>
When $\ds{N \to \infty}$, it becomes:
\begin{align}
\int_{0}^{\infty}\braces{1 \over t}^{kn}t^{s - 1}\,\dd t & =
\int_{0}^{1}t^{kn}\,\zeta\pars{s + 1,t}\,\dd t\label{1}\tag{1}
\end{align}
where $\ds{\zeta\pars{a,b}}$ is the
<a href="https://en.wikipedia.org/wiki/Hurwitz_zeta_function" rel="nofollow noreferrer">Hurwitz Zeta function</a>.</p>
<blockquote>
<p>So far, there are many representations of the <em>Hurwitz Zeta Function</em> which can be 'inserted' in the above result \eqref{1} to yield an expression as a series but a 'closed form' doesn't seem feasible at this time.
<br><br>
In particular, CAS doesn't help at all, either.</p>
</blockquote>
|
3,366,064 | <p>I have a baking recipe that calls for 1/2 tsp of vanilla extract, but I only have a 1 tsp measuring spoon available, since the dishwasher is running. The measuring spoon is very nearly a perfect hemisphere. </p>
<p>My question is, to what depth (as a percentage of hemisphere radius) must I fill my teaspoon with vanilla such that it contains precisely 1/2 tsp of vanilla? Due to the shape, I obviously have to fill it more than halfway, but how much more?</p>
<p>(I nearly posted this in the Cooking forum, but I have a feeling the answer will involve more math knowledge than baking knowledge.)</p>
| J. W. Tanner | 615,567 | <p>Assuming the spoon is a hemisphere with radius <span class="math-container">$R$</span>, </p>
<p>let <span class="math-container">$x$</span> be the height from the bottom of the spoon, and let <span class="math-container">$h$</span> range from <span class="math-container">$0$</span> to <span class="math-container">$x$</span>. </p>
<p>The radius <span class="math-container">$r$</span> of the circle at height <span class="math-container">$h$</span> satisfies <span class="math-container">$r^2=R^2-(R-h)^2=2hR-h^2$</span>.</p>
<p>The volume of liquid in the spoon when it is filled to height <span class="math-container">$x$</span> is <span class="math-container">$$\int_0^x\pi r^2 dh=\int_0^x\pi(2hR-h^2)dh=\pi Rh^2-\frac13\pi h^3\mid_0^x=\pi Rx^2-\frac13\pi x^3.$$</span></p>
<p>(As a check, when the spoon is full, <span class="math-container">$x=R$</span> and the volume is <span class="math-container">$\frac23\pi R^3,$</span> that of a hemisphere.)</p>
<p>The spoon is half full when <span class="math-container">$\pi Rx^2-\frac13\pi x^3=\frac13\pi R^3;$</span> i.e., <span class="math-container">$3Rx^2-x^3=R^3;$</span> </p>
<p>i.e., <span class="math-container">$a^3-3a^2+1=0$</span>, where <span class="math-container">$a=x/R$</span>. </p>
<p>The only physically meaningful solution of this cubic equation is <span class="math-container">$a\approx 65\%.$</span></p>
|
3,366,064 | <p>I have a baking recipe that calls for 1/2 tsp of vanilla extract, but I only have a 1 tsp measuring spoon available, since the dishwasher is running. The measuring spoon is very nearly a perfect hemisphere. </p>
<p>My question is, to what depth (as a percentage of hemisphere radius) must I fill my teaspoon with vanilla such that it contains precisely 1/2 tsp of vanilla? Due to the shape, I obviously have to fill it more than halfway, but how much more?</p>
<p>(I nearly posted this in the Cooking forum, but I have a feeling the answer will involve more math knowledge than baking knowledge.)</p>
| Mohammad Riazi-Kermani | 514,496 | <p>Without loss of generality we assume the radius of the sphere to be <span class="math-container">$1$</span></p>
<p>The volume of the liquid is found by an integral <span class="math-container">$$V= \int _{-1}^{-1+h} \pi (1-y^2 )dy$$</span></p>
<p>and you want the volume of the liquid to be half of the hemisphere which is <span class="math-container">$\pi/3$</span></p>
<p>After evaluating the integral and solving the equation I have found <span class="math-container">$$h=0.65270365$$</span>
That is a little bit more than half as expected. </p>
|
3,987,470 | <p>I am reading about how a wrong formulation of the tower of Hanoi and the inductive hypothesis can lead to a dead-end.<br />
The example I am reading states the following:</p>
<blockquote>
<p>The task is to move N discs from a <em>specific</em> pole to another
<em>specific</em> pole. Assume there are poles <span class="math-container">$A$</span>, <span class="math-container">$B$</span> and <span class="math-container">$C$</span>. The <strong>base case</strong> is that when the number of discs is 0 then no steps are needed
to complete the tasks. For the <strong>inductive step</strong> assume that we
can move <span class="math-container">$n$</span> discs from pole <span class="math-container">$A$</span> to pole <span class="math-container">$B$</span> and we are required to
show how to move <span class="math-container">$n + 1$</span> discs from <span class="math-container">$A$</span> to <span class="math-container">$B$</span></p>
</blockquote>
<p>Then it highlights that this definition is a dead-end since the only 2 ways to use the induction hypothesis as set don't lead anywhere.<br />
Specifically the options are:</p>
<blockquote>
<ol>
<li>Move the top <span class="math-container">$n$</span> discs from pole <span class="math-container">$A$</span> to <span class="math-container">$B$</span>. After this point all possibilities of using the induction hypothesis have been exhausted
since <span class="math-container">$n$</span> discs are on pole <span class="math-container">$B$</span> and we do not have hypothesis about
moving discs from that pole.</li>
<li>Move the smallest disc from pole <span class="math-container">$A$</span> to <span class="math-container">$C$</span>. Then move the remaining <span class="math-container">$n$</span> discs from <span class="math-container">$A$</span> to <span class="math-container">$B$</span>. Once again we have exhausted all
possibilities of using the induction hypothesis, because <span class="math-container">$n$</span> discs are
now on pole <span class="math-container">$B$</span>, and we have no hypothesis about moving discs from
this pole.</li>
</ol>
</blockquote>
<p>The reasoning for <span class="math-container">$1$</span> is clear to me. If we move the top <span class="math-container">$n$</span> discs i.e. the <span class="math-container">$n$</span> smaller discs to pole <span class="math-container">$B$</span> then at some point we would have to move them again since we would need to place the <span class="math-container">$nth + 1$</span> remaining largest disc bellow them. And there is no inductive hypothesis for that so I think I get it.</p>
<p>I need help with the second part. How I understand it is we move the smallest disc to pole <span class="math-container">$C$</span>. Then we use the inductive hypothesis to move the remaining <span class="math-container">$n$</span> larger discs from pole <span class="math-container">$A$</span> to <span class="math-container">$B$</span>. (At this part I am not sure how this could happen if pole <span class="math-container">$C$</span> is occupied by the smallest disc but I guess it is part of the logic of using induction in proof.)<br />
Then at that state it seems to me that the only thing pending would be to move the smallest disc from <span class="math-container">$C$</span> to <span class="math-container">$B$</span> and finish the task.<br />
Why does it state that we would need to move the <span class="math-container">$n$</span> discs from pole <span class="math-container">$B$</span> and that is not possible since we don't have an induction hypothesis?<br />
Am I misunderstanding something about statements on induction here?</p>
| Brian M. Scott | 12,042 | <p>You are correct: the author’s explanation of the problem with the second part is rubbish. He should instead have pointed to the problems that you saw:</p>
<ul>
<li>The induction hypothesis is incomplete, since it pretty clearly must include a requirement that pole <span class="math-container">$C$</span> be available for the disks being transferred from <span class="math-container">$A$</span> to <span class="math-container">$B$</span>.</li>
<li>If the transfer of the bottom <span class="math-container">$n$</span> disks from <span class="math-container">$A$</span> to <span class="math-container">$B$</span> actually <strong>were</strong> possible, we clearly could complete the job of transferring the whole stack by moving the smallest disk from <span class="math-container">$C$</span> to <span class="math-container">$B$</span>.</li>
</ul>
<p>The problem here isn’t really that the induction hypothesis leads to a dead end: it’s that it is too strong and allows a fallacious argument if this problem isn’t noticed in time.</p>
|
4,227,536 | <blockquote>
<p>Let <span class="math-container">$X$</span> be the product space <span class="math-container">$\Bbb R^{\Bbb R}$</span>. Let <span class="math-container">$A \subset X $</span> be the set of all characteristic functions of finite sets. Show that the constant map <span class="math-container">$g, g(x) = 1$</span> belongs to the closure of <span class="math-container">$A$</span>.</p>
</blockquote>
<p>In order to show that some <span class="math-container">$x \in \overline{A}$</span> I need to show that for every open nbdh <span class="math-container">$U_x$</span> the property <span class="math-container">$A \cap U_x \ne \emptyset$</span> holds.</p>
<p>Now I don’t know exactly how to approach the problem is it so that I would need to show that for every nbdh of the constant map <span class="math-container">$g$</span> I need to satisfy the property I stated preivously?</p>
| cohomonoid | 586,016 | <p>Let's assume the question is in the context of the product topology on the Cartesian product <span class="math-container">$\mathbb{R}^\mathbb{R} = \prod_{x\in\mathbb{R}} \mathbb{R}$</span> based on the usual metric topology on <span class="math-container">$\mathbb{R}$</span>. Then convergence in <span class="math-container">$\mathbb{R}^\mathbb{R}$</span> is just pointwise convergence (i.e. convergence of each projection), so done.</p>
|
2,912,570 | <p>Let $X$ and $Y$ be two standard normal distributions with correlation $-0.72$. Compute $E(3X+Y\mid X-Y=1)$.</p>
<p>My solution: Conditioning on $X-Y=1$, we have $E(3X+Y\mid X-Y=1) = E(4Y+3\mid X-Y=1) = 3+4E(Y\mid X-Y=1) = 3$.</p>
<p>(1) Is my solution correct? My intuition is that the conditional density of $Y$ remains symmetric about 0 conditioning on $X-Y=1$.</p>
<p>(2) How to solve $E(Y\mid X-Y=1)$ more rigorously?</p>
<p>Thank you, guys!</p>
| grand_chat | 215,011 | <p>A hint for your question (2): Define variables $U:=X+Y$ and $V:=X-Y$. Check that $\operatorname{Cov}(U,V)=0$. Since $U$ and $V$ are jointly Gaussian, a covariance of zero implies that $U$ and $V$ are independent. Now express $E(Y\mid X-Y)$ in terms of $U$ and $V$: $$E(Y\mid X-Y)=E\left(\textstyle\frac12(U-V)\mid V\right)$$ Simplifying the RHS will yield an expression in terms of $U$ and $V$, which you can re-express in terms of $X$ and $Y$.</p>
|
627,258 | <p>Helly everybody,<br>
I'm trying to find another approach to topology in order to justify the axiomatization of topology. My idea was as follows:</p>
<p>Given an <strong>arbitrary</strong> collection of subsets of some space: $\mathcal{C}\in\mathcal{P}^2(\Omega)$<br>
Define a closure operator by: $\overline{A}:=\bigcap_{A\subseteq C\in\mathcal{C}}C$<br>
This gives rise to a topology apart from the space itself being open.<br>
However, considering the space as being equipped with a notion of being close, all topological question can be studied - as in topological spaces.<br>
<em>(I left out the details as being part of my research)</em></p>
<p>So my question is:<br>
<em>What could go BADLY! wrong if a collection would satisfy all axioms for open sets but the entire space not necessarily being open?</em></p>
<p>Thanks for your help! Cheers Alex</p>
| user119107 | 119,107 | <p>If the whole space $X$ is not open, then constant functions might not be continuous, depending on the definition chosen for "continuous."</p>
|
496,255 | <p>Let $u$ be an integer of the form $4n+3$, where $n$ is a positive integer. Can we find integers $a$ and $b$ such that $u = a^2 + b^2$? If not, how to establish this for a fact? </p>
| lhf | 589 | <p>No, integers of the form $4n+3$ cannot be written as a sum of two squares.</p>
<p>To prove this, consider $z=x^2+y^2$ modulo $4$ and you'll see that you cannot get $3$.</p>
|
1,985,402 | <p>I wrote down $$12 \times 0 = 0$$ Then, I divided both sides by $0$ like so:
$$12 = \frac {0}{0}$$ I know that $$ \frac {0}{x} = 0, x \in R$$
Therefore, $$12=0$$ which is a false statement. Where did I go wrong?</p>
| Lolas-Poaras | 382,736 | <p>You cannot divide anything by 0. That is just not possible unless you want an undefined answer.</p>
|
1,218,140 | <p>I am reading Hartshorne's proof of $\mathbb{P}^1$ being simply connected as a scheme. It seems one ingredient of the proof is that if $X\rightarrow\mathbb{P}^1$ is an étale covering, then X has only finitely many connected components. But I do not see why.</p>
<p>Thanks in advance.</p>
| Keenan Kidwell | 628 | <p>Since $f$ is finite, in particular of finite type, $X$ is a Noetherian scheme. Noetherian schemes are locally connected (see <a href="http://stacks.math.columbia.edu/tag/04MF" rel="nofollow">http://stacks.math.columbia.edu/tag/04MF</a>), so the connected components of $X$ are open. They therefore form a covering of $X$ by pairwise disjoint open sets. Because $X$ is quasi-compact, this covering must be finite, i.e., $X$ must have finitely many connected components. </p>
|
3,079,493 | <p>Let <span class="math-container">$$D_6=\langle a,b| a^6=b^2=1, ab=ba^{-1}\rangle$$</span> <span class="math-container">$$D_6=\{1,a,a^2,a^3,a^4,a^5,b,ab,a^2b,a^3b,a^4b,a^5b\}$$</span></p>
<p>I would like to compute its character table and its irreducible representations.</p>
<p>I will explain what I have done so far and I will add some doubts I had while doing this.</p>
<p><strong>MY ATTEMPT</strong></p>
<ol>
<li><p>Compute conjugacy classes.
<span class="math-container">$$C_1=\{e\}, C_2=\{a,a^5\},C_3=\{a^2,a^4\}$$</span><span class="math-container">$$C_4=\{a^3\},C_5=\{b,a^2b,a^4b\},C_6=\{ab,a^3b,a^5b\}$$</span></p></li>
<li><p>Find <span class="math-container">$1$</span>-dimensional representations. Since <span class="math-container">$D_6/\{a,a^5\}\cong \mathbb{Z}_2$</span>, we have one more representation apart from <span class="math-container">$\alpha_1=id$</span>. That is <span class="math-container">$$\alpha_2: G \longrightarrow \mathbb{C}: a \mapsto 1, b\mapsto -1$$</span>
Again using irreducible representations from quotient group by normal subgroup, I considered <span class="math-container">$G/\{\overline{1},\overline{a},\overline{b},\overline{ab}\}\cong \mathbb{Z}_2\times\mathbb{Z}_2$</span> (since it is abelian). Then from here I obtained
<span class="math-container">$$\alpha_3:G\longrightarrow \mathbb{C}: a\mapsto -1, b\mapsto 1$$</span>
<span class="math-container">$$\alpha_4:G\longrightarrow \mathbb{C}: a\mapsto -1, b\mapsto -1$$</span></p></li>
<li><p>Find <span class="math-container">$2$</span>-dimensional representations. I have seen in my notes that for <span class="math-container">$D_n$</span> we can define <span class="math-container">$2$</span>-dimensional representations:
<span class="math-container">$$\alpha_5: G\longrightarrow GL_2(\mathbb{C}): a\mapsto \begin{bmatrix}cos(\frac{2\pi}{n}) & -sin(\frac{2\pi}{n})\\ sin(\frac{2\pi}{n}) &cos(\frac{2\pi}{n})\end{bmatrix}, b\mapsto \begin{bmatrix}1 & 0\\ 0 &-1\end{bmatrix}$$</span>
Hence my <span class="math-container">$$\alpha_5: G\longrightarrow GL_2(\mathbb{C}): a\mapsto \begin{bmatrix}\frac{1}{2} & -\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2} &\frac{1}{2}\end{bmatrix}, b\mapsto \begin{bmatrix}1 & 0\\ 0 &-1\end{bmatrix}$$</span></p></li>
<li><p>Build my character table.
<span class="math-container">\begin{array}{|c|c|c|c|}
\hline
& C_1 & C_2 &C_3 &C_4 &C_5 &C_6 \\ \hline
\chi_1& 1 & 1 &1 &1 &1&1 \\ \hline
\chi_2& 1 & 1 &1 &1 &-1 &-1 \\ \hline
\chi_3& 1 & -1 &1 &-1 &1 &-1 \\ \hline
\chi_4& 1 & -1 &1 &-1 &-1 &1 \\ \hline
\chi_5& 2 & 1 &-1 &-2 &0 &0 \\ \hline
\chi_6& 2 & -1 &-1 &2 &0 &0 \\ \hline
\end{array}</span></p></li>
</ol>
<p>where I have computed <span class="math-container">$\chi_6$</span> by the orthogonality formula <span class="math-container">$(\chi_6|\chi_j)=\delta_{6,j}$</span>.</p>
<p><strong>QUESTIONS</strong></p>
<ol>
<li>Is <span class="math-container">$D_6/\{a^2,a^4\}$</span> really abelian? I can not see it clearly.</li>
<li><p>My first question comes when I have to find <span class="math-container">$2$</span>-dimensional irreducible representations. I have find them because I have seen it in my notes. But how could I get <span class="math-container">$\alpha_5$</span> and <span class="math-container">$\alpha_6$</span> without knowing the special case of <span class="math-container">$D_n$</span>. I know that I also could get it from <span class="math-container">$S_3$</span> (one of them). But I have again the same problem, if you are looking for <span class="math-container">$2$</span>-dimensional irreducible representations of <span class="math-container">$S_3$</span>, how do you find them? (Both).</p></li>
<li><p>Now consider <span class="math-container">$X$</span> to be the set of the vertices of a regular <span class="math-container">$6$</span>-gon and consider the action of <span class="math-container">$D_6$</span> on the set <span class="math-container">$X$</span> by restricting the usual action of <span class="math-container">$D_6$</span> on the <span class="math-container">$6$</span>-gon to the set of vertices <span class="math-container">$X$</span>. Let <span class="math-container">$\phi$</span> be the induced permutation representation (over <span class="math-container">$\mathbb{C}$</span>) of <span class="math-container">$D_6$</span>. I would like to write it as a sum of irreducible representations by computing the in-product of the irreducible characters with <span class="math-container">$\chi_{\phi}$</span>. What should I do? I do not understand this induced permutation representation. Any help?</p></li>
</ol>
| user3482749 | 226,174 | <ol>
<li><p>It has four elements, so yes. </p></li>
<li><p>Get the characters by messing around with orthogonality, then come up with a representation that does that. </p></li>
<li><p>Label the vertices of your hexagon <span class="math-container">$a_1$</span> through <span class="math-container">$a_6$</span>. Then take the vector space <span class="math-container">$V$</span> over <span class="math-container">$\mathbb{C}$</span> to be the set of formal <span class="math-container">$\mathbb{C}$</span>-weighted sums of <span class="math-container">$a_1$</span> through <span class="math-container">$a_6$</span>. Then we construct a representation of <span class="math-container">$D_6$</span> by an action on this vector space given by defining, for <span class="math-container">$\sigma \in D_6$</span>, the action of <span class="math-container">$\sigma$</span> to be that which sends each <span class="math-container">$a_i$</span> to the <span class="math-container">$a_j$</span> that <span class="math-container">$\sigma$</span> sends <span class="math-container">$a_i$</span> to in the action on the hexagon. Calculate the character of this representation, then proceed as the question tells you to.</p></li>
</ol>
|
3,265,835 | <p>I came across some equation in physics which had a different kind of integration. Like it should have <span class="math-container">$dx2$</span> but had <span class="math-container">$d2x$</span> . And I did some substitution for solving it like putting <span class="math-container">$x= u^2$</span> and then double differentiating it to get <span class="math-container">$d2x = 2.du2$</span> . So I thought it could be right but my teachers said we can not do this and no other proof is coming into my mind.!(<a href="https://i.stack.imgur.com/C5Bg8.jpg" rel="nofollow noreferrer">https://i.stack.imgur.com/C5Bg8.jpg</a>)</p>
| user66081 | 66,081 | <p>I guess we are dealing here with <span class="math-container">$\int \int \mathbb{x} \, d^2 \mathbb{x}$</span>, where <span class="math-container">$\mathbb{x}$</span> is a vector. Another notation for this would be
<span class="math-container">$$
\int \int \left( x \atop y \right) dx dy
.
$$</span></p>
<p>This quantity is related to the <a href="https://en.wikipedia.org/wiki/Center_of_mass" rel="nofollow noreferrer">center of mass</a>.</p>
<p>The wiki page on <a href="https://en.wikipedia.org/wiki/Multiple_integral" rel="nofollow noreferrer">multiple integral</a> mentions that notation in the section "Mathematical definition".</p>
|
3,750,747 | <blockquote>
<p><span class="math-container">$$\frac{\cos^2\left(\dfrac\pi2 \cos\theta\right)}{\sin^2\theta} = 0.5$$</span></p>
</blockquote>
<p>I want to solve the above equation for <span class="math-container">$\theta$</span> in order to find its value, but I am stuck.</p>
<p>Could anyone enlighten me by a method to solve it?</p>
| Claude Leibovici | 82,404 | <p>This is a transcendental equation; then no analytical solutions and numerical methods are required.</p>
<p>Making the problem more general, you want to solve for <span class="math-container">$x$</span> the equation
<span class="math-container">$$y=\cos ^2\left(\frac{\pi}{2} \cos (x)\right) \csc ^2(x)\qquad \text{where} \qquad 0 \leq y \leq 1$$</span></p>
<p>As usual with trigonometric equations, there is an infinite number of solutions.</p>
<p>Suppose that you are concerned by the first root (we shall only consider positive solutions since the function is even). For an approximation, compose Taylor series around <span class="math-container">$x=0$</span> (I let you the intermediate steps to do). You will get
<span class="math-container">$$y=\frac{\pi ^2 }{16}x^2+\frac{\pi ^2 }{96}x^4+\left(\frac{17 \pi ^2}{11520}-\frac{\pi
^4}{768}\right) x^6+O\left(x^8\right)$$</span> Use series reversion to obtain
<span class="math-container">$$x=\frac{4 }{\pi }y^{1/2}-\frac{16 }{3 \pi ^3}y^{3/2}+\frac{32 \left(6+5 \pi
^2\right) }{15 \pi ^5}y^{5/2}+O\left(y^{7/2}\right)$$</span> Using <span class="math-container">$y=\frac 12$</span>, this would give as an <em>approximation</em>
<span class="math-container">$$x \sim \frac{2 \sqrt{2} \left(4+5 \pi ^4\right)}{5 \pi ^5}\approx 0.90771$$</span> while the "exact" solution is <span class="math-container">$0.88944$</span>; this is not too bad.</p>
<p>If you want to polish the solution, use Newton method and the iterates will be
<span class="math-container">$$\left(
\begin{array}{cc}
n & x_n \\
0 & 0.9077104212 \\
1 & 0.8894960222 \\
2 & 0.8894396939 \\
3 & 0.8894396932
\end{array}
\right)$$</span></p>
<p><strong>Edit</strong></p>
<p>After @Quanto's answer, the specific solution of
<span class="math-container">$$\cos(\pi x) +x^2=0$$</span> could have been <em>approximated</em> <span class="math-container">$\color{red}{1,400}$</span> years ago using
<span class="math-container">$$\cos(t) \simeq\frac{\pi ^2-4t^2}{\pi ^2+t^2}\qquad (-\frac \pi 2 \leq t\leq\frac \pi 2)$$</span> This would give
<span class="math-container">$$x^2+\frac{1-4 x^2}{x^2+1}=0$$</span> which is quadratic in <span class="math-container">$x^2$</span> leading to the beautiful
<span class="math-container">$$x \sim \frac{1}{\phi }=0.618$$</span> To keep this beauty, one iteration of Newton method gives
<span class="math-container">$$x=\frac{1}{\phi }+\frac{\phi ^2 \cos \left(\frac{\pi }{\phi }\right)+1}{\phi \left(\pi \phi \sin
\left(\frac{\pi }{\phi }\right)-2\right)}=0.629613$$</span></p>
<p>Another (very accurate) approximation is
<span class="math-container">$$x \sim \frac{638 \pi ^2-269 \pi-487}{434 \pi ^2 +1123 \pi +71}$$</span> which gives <span class="math-container">$18$</span> significant figures.</p>
|
3,750,747 | <blockquote>
<p><span class="math-container">$$\frac{\cos^2\left(\dfrac\pi2 \cos\theta\right)}{\sin^2\theta} = 0.5$$</span></p>
</blockquote>
<p>I want to solve the above equation for <span class="math-container">$\theta$</span> in order to find its value, but I am stuck.</p>
<p>Could anyone enlighten me by a method to solve it?</p>
| Quanto | 686,284 | <p>Let <span class="math-container">$x =\cos\theta$</span> to simplify the equation to</p>
<p><span class="math-container">$$\cos(\pi x) +x^2=0$$</span></p>
<p>which has the trivial roots <span class="math-container">$\pm 1$</span> (excluded due to <span class="math-container">$\sin\theta \ne 0$</span>), as well as the root that can be approximated with <span class="math-container">$\frac\pi2-\pi x+x^2 =0$</span>, i.e.</p>
<p><span class="math-container">$$x=\left(1+\sqrt{1-\frac2\pi} \right)^{-1}= 0.6239$$</span></p>
<p>(vs. the exact <span class="math-container">$ 0.6298$</span>). Thus, the solutions are
<span class="math-container">$$\theta = 2\pi k\pm \text{arcsec} \left(1+\sqrt{1-\frac2\pi} \right) $$</span></p>
|
4,321,604 | <p>I have come across an expression like this,</p>
<p><span class="math-container">$$ \frac{f(x) + f(a)}{2\sqrt{f(x)f(a)}}\,\delta(x-a), $$</span></p>
<p>where I expected to find just <span class="math-container">$\delta(x-a)$</span>. When I thought about it, though, I realised maybe... they are identical? Because both yield <span class="math-container">$1$</span> when integrating over a domain that contains <span class="math-container">$a$</span> and <span class="math-container">$0$</span> otherwise, so both distributions behave identically.</p>
<p>So, can I say
<span class="math-container">$$ \frac{f(x) + f(a)}{2\sqrt{f(x)f(a)}}\,\delta(x-a) = \delta(x-a) $$</span>
or can I not?</p>
| PrincessEev | 597,568 | <p>Not quite. Bear in mind that</p>
<p><span class="math-container">$$\frac{f(x) + f(a)}{2\sqrt{f(x)f(a)}} \Bigg|_{x=a} = \frac{2 f(a)}{2 \sqrt{ f(a)^2 }} = \frac{f(a)}{|f(a)|} = \begin{cases}
1 & f(a) > 0 \\
-1 & f(a) < 0\end{cases}$$</span></p>
<p>since <span class="math-container">$\sqrt{z^2} = |z|$</span>. So the most you can claim is that</p>
<p><span class="math-container">$$ \frac{f(x) + f(a)}{2\sqrt{f(x)f(a)}}\,\delta(x-a) = \begin{cases}
\delta(x-a) & f(a) > 0 \\
-\delta(x-a) & f(a) < 0\end{cases}$$</span></p>
|
2,877,833 | <p><a href="https://i.stack.imgur.com/9wqM5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9wqM5.png" alt="enter image description here"></a></p>
<p>Look at this part:</p>
<blockquote>
<p>Define the vector $p = -\nabla f(x^*)$ and note that $p^T\nabla f(x^*)
= -||\nabla f(x^*)||^2 <0$. Because $f$ is continuous near $x^*$, there is a scalar $T>0$ such that </p>
<p>$p^T\nabla f(x^*+tp) <0, \forall t\in [0,T]$</p>
</blockquote>
<p>Why the continuity of the gradient imply that? I understand that because the gradient is continuous, we can move around smoothly and retain the signal. But I'd suppose it works for $\nabla f$ only. Why it works for $p^T\nabla f(x^*+tp)$?</p>
<p>Also, what if I chose $p = \nabla f(x^*)$ instead of the negative?</p>
| Theoretical Economist | 388,944 | <p>Let $p \in \mathbb R^n$, and $g:\mathbb R^n \to \mathbb R$, so that we can write $g(x) = (g_1(x),g_2(x),\ldots,g_n(x))$. Write $$G(x) = p^Tg(x) = p_1g_1(x) + p_2g_2(x) + \cdots + p_ng_n(x).$$</p>
<p>If $g$ is continuous, then so each each $g_i$. Hence, $G$ inherits continuity from $G$. Now take $p$ as defined in your textbook, and choose $g = \nabla f$.</p>
<p>You can choose $p = \nabla f(x^*)$ if you wish: you would just need to rejig the proof a little bit to make it work. It reads nicely as is, however, so there's no reason to do this.</p>
|
313,030 | <p>I often find myself writing a definition which requires a proof. You are defining a term and, contextually, need to prove that the definition makes sense. </p>
<p>How can you express that? What about a definition with a proof?</p>
<p>Sometime one can write the definition and then the theorem. But often happens that many definition which should stay together need to be split
because a theorem is required in between.</p>
<p>A tentative example:</p>
<p><strong>Definition</strong> (rational numbers)
Let <span class="math-container">$\sim$</span> be the equivalence relation on <span class="math-container">$\mathbb Z^*\times \mathbb Z$</span> given
by
<span class="math-container">$$
(q,p) \sim (q',p') \iff pq' = p'q.
$$</span>
We define <span class="math-container">$\mathbb Q= (\mathbb Z^*\times \mathbb Z)/\sim$</span>.
On <span class="math-container">$\mathbb Q$</span> we define addition and multiplication as follows
<span class="math-container">$$
[(q,p)] + [(q',p')] = [(qq',pq'+p'q)] \\
[(q,p)] \cdot[(q',p')] = [(qq',pp')]
$$</span>
With these operations and choosing
<span class="math-container">$0_\mathbb Q=[(1,0)]$</span>, and <span class="math-container">$1_\mathbb Q=[(1,1)]$</span>
turns out that <span class="math-container">$\mathbb Q$</span> is a field.</p>
<p><strong>Proof.</strong>
We are going to prove that <span class="math-container">$\sim$</span> is indeed an equivalence relation,
that addition and multiplication are well defined and that the resulting
set is a field. [...]</p>
| Nik Weaver | 23,141 | <p>I disagree with the implication that it's always necessary, when trying to write something to be understood, that every sentence has to be a logical consequence of previous sentences or assumed background knowledge.</p>
<p>If you're careful about it, and you say this is what you're doing, I think it can be pedagogical to state a definition first and then show that it makes sense. Especially if the verification is routine and you don't want the definition to be buried among a mass of trivialities. I can give examples of this from my own writing.</p>
|
356,353 | <p>I learned in my Intro Algebraic Number Theory class that there exist infinitely many integer pairs $(x,y)$ that satisfy the hyperbola $x^2-ny^2=1$; just consider that there are infinitely many units in $\mathcal{O}_{\mathbb{Q}(\sqrt{n})}$, and their norms satisfy the desired equation. Although this is a nice connection, I was wondering if it is possible to reach the solution without using high-powered Algebraic Number Theory. And more generally, does the same result hold true for $x^2-ny^2=k$ where $k$ is any integer? And how would one solve that?</p>
| Will Jagy | 10,400 | <p>EDIT: there have been many comments on my answers asking about the use of the word <strong>automorph</strong>. This is a real thing! I did not just make up a word. For a fixed quadratic form, you get a <strong>group of integral automorphs</strong>. In just two variables, there is a good recipe for finding all. In three or more variables, it is a mess. Sometimes this is called the <strong>orthogonal group</strong> of the form, or the <strong>isometry group</strong>. If you think about finding all real solutions of the basic matrix equation, $A^T F A = F,$ where $F$ is a symmetric matrix associated with a quadratic form, the group part may be clearer, especially when $F=I.$ If $F$ has all rational entries, it is reasonable to solve this with rational or $p$-adic entries in $A.$ Finally, when $F,$ or at least $2 F,$ has integer entries, it is reasonable to ask about $A$ with integer entries. </p>
<p>I would like people to know more about this. In dimension 2 this has considerable overlap with algebraic number theory for quadratic fields. In dimension 3 or more, the theory of quadratic forms, in this case indefinite forms, separates from algebraic number theory to a considerable extent. </p>
<p>ORIGINAL:Given nontrivial $\tau^2 - n \sigma^2 = 1,$ we get</p>
<p>$$ \left( \begin{array}{cc}
\tau & \sigma \\
n \sigma & \tau
\end{array}
\right)
\left( \begin{array}{cc}
1 & 0 \\
0 & -n
\end{array}
\right)
\left( \begin{array}{cc}
\tau & n \sigma \\
\sigma & \tau
\end{array}
\right) =
\left( \begin{array}{cc}
1 & 0 \\
0 & -n
\end{array}
\right).
$$ </p>
<p>As a result, if $x^2 - n y^2 = k,$ then we get the same $k$ for
$$ \left( \begin{array}{cc}
\tau & n \sigma \\
\sigma & \tau
\end{array}
\right)
\left( \begin{array}{c}
x \\
y
\end{array}
\right)
=
\left( \begin{array}{c}
\tau x + n \sigma y \\
\sigma x + \tau y
\end{array}
\right).
$$ </p>
<p>This 2 by 2 matrix is called an automorph of the quadratic form. </p>
<p>Every indefinite form $f(x,y) = a x^2 + b x y + c y^2$ where $\Delta = b^2 - 4 a c$ is positive but not a square, has such an automorph, leading to infinitely many solutions. Indeed, given $\tau^2 - \Delta \sigma^2 = 4,$ we get</p>
<p>$$ \left( \begin{array}{cc}
\frac{\tau - b \sigma}{2} & a \sigma \\
-c \sigma & \frac{\tau + b \sigma}{2}
\end{array}
\right)
\left( \begin{array}{cc}
a & \frac{b}{2} \\
\frac{b}{2} & c
\end{array}
\right)
\left( \begin{array}{cc}
\frac{\tau - b \sigma}{2} & -c \sigma \\
a \sigma & \frac{\tau + b \sigma}{2}
\end{array}
\right) =
\left( \begin{array}{cc}
a & \frac{b}{2} \\
\frac{b}{2} & c
\end{array}
\right).
$$
Therefore, if we have $ a x^2 + b x y + c y^2 = k,$ we have another with
$$
\left( \begin{array}{cc}
\frac{\tau - b \sigma}{2} & -c \sigma \\
a \sigma & \frac{\tau + b \sigma}{2}
\end{array}
\right)
\left( \begin{array}{c}
x \\
y
\end{array}
\right)
=
\left( \begin{array}{c}
\frac{\tau - b \sigma}{2} x - c \sigma y \\
a \sigma x + \frac{\tau + b \sigma}{2} y
\end{array}
\right).
$$ </p>
<p>For previous answers in which I show how to use an automorph, see
<a href="https://math.stackexchange.com/questions/181758/solve-the-diophantine-equation-3x2-2y2-1">Solve the Diophantine equation $ 3x^2 - 2y^2 =1 $</a> </p>
<p><a href="https://math.stackexchange.com/questions/228356/how-to-find-solutions-of-x2-3y2-2">How to find solutions of $x^2-3y^2=-2$?</a></p>
<p>Books: H.E.Rose, <em>A Course in Number Theory</em>, chapter 9, section 3, especially pages 162-164 in the first edition. </p>
<p>Thomas W. Cusick and Mary E. Flahive, <em>The Markoff and Lagrange Spectra</em> appendix 3 on pages 91-92.</p>
<p>Duncan A. Buell, <em>Binary Quadratic Forms</em> chapter 3, section 2, pages 31-34. </p>
<p>William J. LeVeque, <em>Topics in Number Theory</em>, volume 2, pages 24-29. The two volumes are available as a one volume paperback, <a href="http://store.doverpublications.com/0486425398.html" rel="nofollow noreferrer">LeVeque Book</a> </p>
<p>Leonard Eugene Dickson, <em>Introduction to the Theory of Numbers</em>, especially pages 111-112.</p>
|
1,315,744 | <p>Already I know that harmonic series, $$\sum_{k=1}^n\frac1k $$ is divergent series.</p>
<p>And, it is also divergent by Abel Sum or Cesaro Sum.</p>
<p>However, I do not know how to prove it is divergent by concept of Abel or Cesaro.</p>
<p>Abel Sum or Cesaro Sum do not exist in this problem.</p>
<p>But, how can I prove it?</p>
<p>I have tried to prove for long time...</p>
<p>Can you please show your work?</p>
| JignEsh Shingod | 887,397 | <p>Since the term of the Harmonic series are of O(1/n). Suppose, if the Harmonic series is Cesaro summable, then by Hardy's theorem, the series is convergent, but Harmonic series is divergent. So, Harmonic series is not Cesaro summable.</p>
|
2,781,153 | <p>I've a right triangle that is inscribed in a circle with radius $r$ the hypotunese of the triangle is equal to the diameter of the circle and the two other sides of the triangle are equal to eachother.</p>
<blockquote>
<p>Prove that when you divide the area of the circle by the area of the triangle that you will get $\pi$.</p>
</blockquote>
<p>This is what I did:</p>
<p>The area of a triangle is $\frac{height\times width}{2}$ and the area of a circle is $\pi r^2$. Now I do not know how to continue.</p>
| Love Invariants | 551,019 | <p>Hint: Find the side of the triangle using pythagoras theorem.<br>
$hypotenuse=2r$</p>
|
2,781,153 | <p>I've a right triangle that is inscribed in a circle with radius $r$ the hypotunese of the triangle is equal to the diameter of the circle and the two other sides of the triangle are equal to eachother.</p>
<blockquote>
<p>Prove that when you divide the area of the circle by the area of the triangle that you will get $\pi$.</p>
</blockquote>
<p>This is what I did:</p>
<p>The area of a triangle is $\frac{height\times width}{2}$ and the area of a circle is $\pi r^2$. Now I do not know how to continue.</p>
| Henry | 6,460 | <p>Hints:</p>
<ul>
<li><p>You need to calculate the height and width in terms of $r$: there are two ways of doing this which will give the same area</p></li>
<li><p>You need to divide the area of the circle by the area of the triangle </p></li>
</ul>
<p><a href="https://i.stack.imgur.com/ZGA1C.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZGA1C.png" alt="![enter image description here"></a></p>
|
2,781,153 | <p>I've a right triangle that is inscribed in a circle with radius $r$ the hypotunese of the triangle is equal to the diameter of the circle and the two other sides of the triangle are equal to eachother.</p>
<blockquote>
<p>Prove that when you divide the area of the circle by the area of the triangle that you will get $\pi$.</p>
</blockquote>
<p>This is what I did:</p>
<p>The area of a triangle is $\frac{height\times width}{2}$ and the area of a circle is $\pi r^2$. Now I do not know how to continue.</p>
| Jan Eerland | 226,665 | <p>Well, first of all let's summarise the things we can say about this problem and after that we can set up a system of equations.</p>
<blockquote>
<ol>
<li>By the <a href="https://en.wikipedia.org/wiki/Pythagorean_theorem" rel="nofollow noreferrer">Pythagorean theorem</a>, we can write:
<span class="math-container">$$\left|\text{AB}\right|^2=\left|\text{AC}\right|^2+\left|\text{BC}\right|^2\tag1$$</span>
2. The <a href="https://en.wikipedia.org/wiki/Circle#Area_enclosed" rel="nofollow noreferrer">area enclosed</a> by a <a href="https://en.wikipedia.org/wiki/Circle" rel="nofollow noreferrer">circle</a> is given by:
<span class="math-container">$$\mathcal{A}_{\space\circ}=\pi\cdot\text{r}_{\space\circ}^2\tag2$$</span></li>
</ol>
<p>Where <span class="math-container">$\text{r}_{\space\circ}$</span> is the radius of the circle.</p>
<ol start="3">
<li>The connection between the <a href="https://en.wikipedia.org/wiki/Radius" rel="nofollow noreferrer">radius</a> and the diameter of a circle is given by:
<span class="math-container">$$\text{d}_{\space\circ}=2\cdot\text{r}_{\space\circ}\tag3$$</span>
4. The area of a <a href="https://en.wikipedia.org/wiki/Triangle#Computing_the_area_of_a_triangle%5C" rel="nofollow noreferrer">triangle</a> is given by:
<span class="math-container">$$\mathcal{A}_{\space\triangle}=\frac{1}{2}\cdot\text{b}_{\space\triangle}\cdot\text{h}_{\space\triangle}\tag4$$</span>
Where <span class="math-container">$\text{b}_{\space\triangle}$</span> is the length of the base of the triangle, and <span class="math-container">$\text{h}_{\space\triangle}$</span> is the height or altitude of the triangle.</li>
</ol>
</blockquote>
<hr>
<p>Now, from your question we know that:</p>
<p> - The <a href="https://en.wikipedia.org/wiki/Hypotenuse" rel="nofollow noreferrer">Hypotenuse</a> of the triangle is equal to the diameter of the circle:
<span class="math-container">$$\left|\text{AB}\right|=\text{d}_{\space\circ}\tag5$$</span>
- The two other sides of the <a href="https://en.wikipedia.org/wiki/Triangle#Right_triangles" rel="nofollow noreferrer">right triangle</a>, <span class="math-container">$\triangle\text{ABC}$</span>, are equal to each other:
<span class="math-container">$$\left|\text{AC}\right|=\left|\text{BC}\right|\tag6$$</span>
- The <span class="math-container">$\text{b}_{\space\triangle}$</span> and <span class="math-container">$\text{h}_{\space\triangle}$</span> of <span class="math-container">$\triangle\text{ABC}$</span> are equal to each other and equal to the two other sides of the right triangle:
<span class="math-container">$$\text{b}_{\space\triangle}=\text{h}_{\space\triangle}=\left|\text{AC}\right|=\left|\text{BC}\right|\tag7$$</span></p>
<p>So, we can write:</p>
<p><span class="math-container">$$
\begin{cases}
\left|\text{AB}\right|=\text{d}_{\space\circ}\\
\\
\text{d}_{\space\circ}=2\cdot\text{r}_{\space\circ}\\
\\
\mathcal{A}_{\space\circ}=\pi\cdot\text{r}_{\space\circ}^2\\
\\
\left|\text{AB}\right|^2=\left|\text{AC}\right|^2+\left|\text{AC}\right|^2=2\cdot\left|\text{AC}\right|^2\\
\\
\mathcal{A}_{\space\triangle}=\frac{1}{2}\cdot\text{b}_{\space\triangle}\cdot\text{h}_{\space\triangle}\\
\\
\text{b}_{\space\triangle}=\text{h}_{\space\triangle}=\left|\text{AC}\right|=\left|\text{BC}\right|
\end{cases}\tag8
$$</span></p>
<p>Which give for <span class="math-container">$\mathcal{A}_{\space\triangle}$</span>:</p>
<p><span class="math-container">$$\mathcal{A}_{\space\triangle}=\frac{\mathcal{A}_{\space\circ}}{\pi}\space\Longleftrightarrow\space\frac{\mathcal{A}_{\space\circ}}{\mathcal{A}_{\space\triangle}}=\pi\tag9$$</span></p>
|
1,137,079 | <p>I'm new to the concept of complex plane. I found this exercise:</p>
<blockquote>
<p>Let $z,z_1,z_2\in\mathbb C$ such that $z=z_1/z_2$. Show that the length of $z$ is the quotient of the length of $z_1$ and $z_2$.</p>
</blockquote>
<p>If $z_1=x_1+iy_1$ and $z_2=x_2+iy_2$ then $|z_1|=\sqrt{x_1^2+y_1^2}$ and $|z_2|=\sqrt{x_2^2+y_2^2}$, which yields $|z_1|/|z_2|=\sqrt{\dfrac{x_1^2+y_1^2}{x_2^2+y_2^2}}$.</p>
<p>Now, $z=\dfrac{z_1}{z_2}=\dfrac{x_1+iy_1}{x_2+iy_2} $. The first issue is to try and separate the imaginary part from the real one. I did this by:
$$\dfrac{x_1+iy_1}{x_2+iy_2}=\dfrac{x_1+iy_1}{x_2+iy_2}\times\frac{x_2-iy_2}{x_2-iy_2}=\frac{x_1x_2-ix_1y_2+ix_2y_1+y_1y_2}{x_2^2+y_2^2}\\ =\frac{x_1x_2+y_1y_2}{x_2^2+y_2^2}+i\frac{x_2y_1-x_1y_2}{x_2^2+y_2^2}.$$
Hence $|z|=\sqrt{\left(\dfrac{x_1x_2+y_1y_2}{x_2^2+y_2^2}\right)^2+\left(\dfrac{x_2y_1-x_1y_2}{x_2^2+y_2^2}\right)^2}=\sqrt{\dfrac{(x_1x_2+y_1y_2)^2+(x_2y_1-x_1y_2)^2}{(x_2^2+y_2^2)^2}}$. Continuing I get
$$\sqrt{\frac{(x_1x_2)^2+(y_1y_2)^2+(x_2y_1)^2+(x_1y_2)^2}{(x_2^2+y_2^2)^2}},$$
but I can't see any future here. Is there any mistake? How can I simplify the expression for $|z|$? I appreciate your help.</p>
| JMP | 210,189 | <p>Let $z_1=r.e^{i\theta}$ and $z_2=R.e^{i\alpha}$. The result follows easily.</p>
|
757,049 | <p>as the title suggests, I need help proving that the cardinality of $(0,1)$ and $[0,1]$ are the same. </p>
<p>Here is my work: </p>
<p>$f:[0,1] \rightarrow (0,1)$</p>
<p>Let $n\in N$</p>
<p>Let $A=\{\frac{1}{2}, \frac{1}{3}, \frac{1}{4}....\}\cup \{0\}$</p>
<p>On $[0,1]\in A: f(x)=x$</p>
<p>On $A: f(0)=\frac{1}{3}$</p>
<p>$f(1)=\frac{1}{2}$</p>
<p>$f(\frac{1}{n})=\frac{1}{n+2}, n>2$</p>
<p>Now I will prove that $f$ is a bijective function.
Let $f(n)=f(m)$ for $n,m \in N$ and $n,m>2$. Then $\frac{1}{n+2}=\frac{1}{m+2}$. We multiply both sides of the equation by $(n+2)(m+2)$ and obtain $m+2=n+2 \rightarrow m=n$. Thus $f$ is injective.</p>
<p>From here on out, I am kind of shakey. I know the gist of this proof, but I don't know how to set up the sequencing correctly. For instance, I know that I need to map $x_1$ to $0$ and $x_2$ to $1$ and $x_{n}$ to $x_{n+2}$, but I am confused on how to do so. </p>
| Michael Weiss | 79,741 | <p>I think you're getting a little too bogged down in the details. The basic idea is that the sets $$A_0=\{a_0,a_1,a_2\ldots,a_n,\ldots\}\text{ and }A_1=\{a_1,a_2,a_3\ldots,a_n,\ldots\}$$ have the same cardinality, where $\{a_0,a_1,a_2\ldots\}$ is any sequence of distinct numbers. So $$B\cup A_0\text{ and }B\cup A_1$$ have the same cardinality, where $B$ is any set disjoint from $A_0$ (and hence also disjoint from $A_1$). This shows that adding one element to the set $B\cup A_1$ doesn't change the cardinality.</p>
<p>So if you <em>start</em> with some infinite set $S$ and some element $a_0$ not in the set, the trick is to express $S$ as $B\cup A_1$. Well, just pick an infinite sequence in $S$ to serve as $A_1$, and let $B=S-A_1$.</p>
<p>The sets $(0,1)$ is an example of such an infinite set $S$, so it follows that $[0,1)$ has the same cardinality as $(0,1)$. Repeat to get $[0,1]$ with the same cardinality. Or modify the previous argument to show that adding two elements doesn't change the cardinality of an infinite set: $$A_0=\{a_0,a_1,a_2\ldots,a_n,\ldots\}\text{ and }A_2=\{a_2,a_3,a_4\ldots,a_n,\ldots\}$$ have the same cardinality, etc.</p>
<p>Now your proof follows this idea; you've just made a particular choice for the sequence $\{a_2,a_3,\ldots\}$, namely $\{\tfrac{1}{2}, \tfrac{1}{3},\ldots\}$. Also you are trying to construction the one-one correspondence explicitly.</p>
<p>This isn't hard. We want $f:B\cup A_0\rightarrow B\cup A_2$ one-one and onto. To begin with, as you indicated, we want $f(a_0)=a_2$, $f(a_1)=a_3$, etc.; i.e., $f(0)=\tfrac{1}{2}$, $f(1)=\tfrac{1}{3}$, and $f(\tfrac{1}{n})=\tfrac{1}{n+2}$ for $n\geq 2$. Then for $x\not\in \{a_0,a_1,\ldots\}$, let $f(x)=x$.</p>
<p>(I'm assuming when you wrote "On $[0,1]\in A$", that was a typo for "On $[0,1]-A$". Also, I'm using $-$ for set-difference, what some people denote by $\smallsetminus$.)</p>
|
1,141,632 | <p>I need to answer a question on fractals from the book <em>Fractals Everywhere</em> by M. Barsley and I have been struggling with it for a while:</p>
<p>Use collage theorem to help you find an IFS consisting of two affine maps in $\mathbb{R}^2$ whose attractor is close to this set:
<img src="https://i.stack.imgur.com/8H7l9.jpg" alt="enter image description here"></p>
<p>It would be great to have some hints/help - I feel like it should be really simple but I'm having a hard tome to wrap my head around it. Thanks!</p>
| Mark McClure | 21,361 | <p>If you have Mathematica, you can generate a good approximation to the set like so:</p>
<pre><code>Import["https://sites.google.com/a/unca.edu/mark-mcclure/IteratedFunctionSystems.m"]
IFS = {
{{{0.980019, -0.153438}, {0.25526, 0.88033}}, {54.4695, -50.3965}},
{{{-0.131103, -0.0796567}, {0.123349, -0.136077}}, {97.7113, 313.055}}
};
ShowIFSStochastic[IFS, 100000, Colors -> True, ImageSize -> 500]
</code></pre>
<p><img src="https://i.stack.imgur.com/pT3Ig.png" alt="enter image description here"></p>
<p>Even if you don't have Mathematica, you should be able to read the IFS you seek right out of the code. The package represents an affine transformation $T:\mathbb R^2 \rightarrow \mathbb R^2$ given by
$$T(x,y) = \left(
\begin{array}{cc}
a & b\\
c & d
\end{array}
\right)\left(
\begin{array}{c}
x \\ y
\end{array}\right)
+\left(
\begin{array}{c}
e \\ f
\end{array}\right)
$$
in the form</p>
<pre><code>{ {{a,b},{c,d}}, {e,f} }
</code></pre>
<p>and you can see two such transformations in the code.</p>
<p>The idea behind the collage theorem is to find transformations that map the whole image onto the two, colored sub-parts. You can automate the procedure by mapping key points in the whole to the corresponding key points in the sub-part. To map onto the red sub-portion, for example, you might consider the following picture:</p>
<p><img src="https://i.stack.imgur.com/PUKaK.png" alt="enter image description here"></p>
<p>The arrows indicate how certain key points in the whole map to their corresponding key point in the red sub-part. The central dot in the middle of the red spiral is fixed. Since we know the function has a particular form dependent on the parameters <code>{a,b,c,d,e,f}</code>, we can use a least squares procedure to find a very good approximation to the function. That is, for each pair consisting of a key point $(x_1,y_1)$ and its known image $(x_2,y_2)$, we form the squared norm
$$\|T(x_1,y_1)-(x_2,y_2)\|^2 =
\left\|\left(
\begin{array}{cc}
a & b\\
c & d
\end{array}
\right)\left(
\begin{array}{c}
x_1 \\ y_1
\end{array}\right)
+\left(
\begin{array}{c}
e \\ f
\end{array}\right) -
\left(
\begin{array}{c}
x_2 \\ y_2
\end{array}\right)
\right\|^2.
$$
We then total these and minimize over the parameters <code>{a,b,c,d,e,f}</code> to obtain a function.</p>
<p>The other transformation can be estimated in similar fashion. There, we attempt to map the green key points in the following image to the red key points in the smaller figure as shown here:</p>
<p><img src="https://i.stack.imgur.com/LgRmB.png" alt="enter image description here"></p>
<p>The process should be pretty good. Here's the generated image overlayed on top of the original scanned image.</p>
<p><img src="https://i.stack.imgur.com/GVrcG.png" alt="enter image description here"></p>
<p>Could be better, actually but still, it's pretty good. I picked the points using a point and click tool that certainly had a little error.</p>
|
3,464,291 | <blockquote>
<p>If <span class="math-container">$x,y,z>0.$</span> Then minimum value of</p>
<p><span class="math-container">$x^{\ln(y)-\ln(z)}+y^{\ln(z)-\ln(x)}+z^{\ln(x)-\ln(y)}$</span></p>
</blockquote>
<p>what i try</p>
<p>Let <span class="math-container">$\ln(x)=a,\ln(y)=b.\ln(z)=c$</span></p>
<p>So <span class="math-container">$x=e^{a},y=e^{b},z=e^{c}$</span></p>
<p>How do i solve it help me</p>
| gt6989b | 16,192 | <p>After your substitution you get what seems like a more manageable
<span class="math-container">$$
f(a,b,c) = e^{a(b-c)} + e^{b(c-a)} + e^{c(a-b)}
$$</span>
and you can now minimize easily using the standard techniques.</p>
|
2,794,715 | <p>Is it right that</p>
<p><strong>$$\sqrt[a]{2^{2^n}+1}$$</strong></p>
<p>for every $$a>1,n \in \mathbb N $$ </p>
<p>is always irrational?</p>
| Rhys Hughes | 487,658 | <p>$$a=1\to 2^{2^n}\in \Bbb Q$$
$$a=2\to \sqrt{2^{2^n}} \in \Bbb Q$$</p>
|
3,991,105 | <p>I understand the concept of one coordinate moving while the rest don't change, however I can't make up the exact mapping that would prove this. Can anyone give me the concrete mapping?</p>
| Anonymath | 875,272 | <p>Let <span class="math-container">$(a,b,c)$</span>, <span class="math-container">$(d,e,f)$</span> be two points in <span class="math-container">$\mathbb{R}^3\setminus\mathbb{Q}^3$</span>. Then at least one of <span class="math-container">$a,b,c$</span> is not rational. By symmetry, we may assume WLOG that it is <span class="math-container">$a$</span>.</p>
<p>If <span class="math-container">$d$</span> is not rational, we can follow the following path which is entirely contained in <span class="math-container">$\mathbb{R}^3\setminus\mathbb{Q}^3$</span>:</p>
<p><span class="math-container">$$(a,b,c) \to (a,\sqrt2,c)\to(d,\sqrt2,c)\to(d,e,f)$$</span>
where <span class="math-container">$P\to Q$</span> signifies a straight line path from <span class="math-container">$P$</span> to <span class="math-container">$Q$</span>.</p>
<p>If <span class="math-container">$e$</span> is not rational, we follow:
<span class="math-container">$$(a,b,c) \to (a,e,c) \to (d,e,f)$$</span></p>
<p>If <span class="math-container">$f$</span> is not rational, we follow:
<span class="math-container">$$(a,b,c) \to (a,b,f) \to (d,e,f)$$</span></p>
|
3,991,105 | <p>I understand the concept of one coordinate moving while the rest don't change, however I can't make up the exact mapping that would prove this. Can anyone give me the concrete mapping?</p>
| José Carlos Santos | 446,262 | <p>Take <span class="math-container">$(a_1,a_2,a_3),(b_1,b_2,b_3)\in\Bbb R^3\setminus\Bbb Q^3$</span>. Assume, for instance, that <span class="math-container">$a_1,b_3\notin\Bbb Q$</span>. Then consider<span class="math-container">$$\begin{array}{rccc}\gamma\colon&[0,2]&\longrightarrow&\Bbb R^3\\&t&\mapsto&\begin{cases}\bigl(a_1,tb_2+(1-t)a_2,tb_3+(1-t)a_3\bigr)&\text{ if }t\in[0,1]\\\bigl((t-1)b_1+(2-t)a_1,b_2,b_3\bigr)&\text{ if }t\in[1,2].\end{cases}\end{array}$$</span>Since <span class="math-container">$a_1,b_3\notin\Bbb Q$</span>, <span class="math-container">$(\forall t\in[0,2]):\gamma(t)\in\Bbb R^3\setminus\Bbb Q^3$</span>.</p>
|
2,261,410 | <blockquote>
<p>The generating function for a Bessel equation is:</p>
<p>$$g(x,t) = e^{(x/2)(t-1/t))}$$</p>
<p>Using the product $g(x,t)\cdot g(x,-t)$ show that:</p>
<p>a) $$[J_0(x)]^2 + 2[J_1(x)]^2 + 2[J_2(x)]^2 + \cdots = 1$$</p>
<p>and consequently:</p>
<p>b)</p>
<p>$$|J_0(x)|\le 1, \forall x$$</p>
<p>c) $$|J_n(x)| \le \frac{1}{\sqrt{2}}; n=1,2,3,\cdots$$</p>
</blockquote>
<p>For a) I tried the product:</p>
<p>$$e^{(x/2)(t-1/t))}\cdot e^{(x/2)(-t+1/t))} = 1$$</p>
<p>I at least arrived at the right side of the equation. Since this generates the bessel functions, I should arrive at something related to $J_n$ in the left side. I know that</p>
<p>$$e^{(x/2)(t-1/t))} = \sum_{n=0\infty}^{\infty} J_n(x)t^n$$</p>
<p>$$e^{(x/2)(-t+1/t))} = \sum_{n=0\infty}^{\infty} J_n(x)(-t)^n$$</p>
<p>but it's not just a matter of multiplying coefficients from the two infinite series, right?</p>
<p>For $b$, I tried to use <a href="https://math.stackexchange.com/questions/2261372/proving-bessel-equation-j-0uv-j-0u-cdot-j-0v2-sum-s-1-inftyj-s">Proving Bessel equation $J_{0}(u+v) = J_0(u)\cdot J_0(v)+2\sum_{s=1}^{\infty}J_s(u)\cdot J_{-s}(v)$</a> but it's not as obvious</p>
<p>I have no idea how to deal with $c$, could somebody help me?</p>
| Paramanand Singh | 72,031 | <p>I gave some comments on the accepted answer to help clarify some doubts raised by OP. I think it is proper to avoid too much discussion in comments and hence I put all that explanation into an answer.</p>
<p>The generating function $g(x, t) = e^{(x/2)(t-(1/t))}$ has the property that $g(x, - t) =g(x, 1/t)$ and this means that $$\sum_{n=-\infty} ^{\infty} J_{n} (x)(-t) ^{n} =\sum_{n=-\infty} ^{\infty} J_{n} (x)(1/t)^{n}$$ or $$\sum_{n=-\infty} ^{\infty} (-1)^{n}J_{n}(x)t^{n}=\sum_{n=-\infty}^{\infty}J_{n}(x)t^{-n}=\sum_{n=-\infty}^{\infty}J_{-n}(x)t^{n}$$ where the last equality in above equation is a result of changing index of summation $n$ to $-n$ (this does not affect the sum because as $n$ takes all integer values, $-n$ also takes all integer values). Now equating coefficients of $t^{n} $ in first and last expressions of above equation we get $J_{-n}(x) =(-1)^{n}J_{n}(x)$. This relation will be needed later on. </p>
<p>Next we note that $g(x, t) g(x, 1/t)=1$ and hence $$\left(\sum_{n=-\infty}^{\infty}J_{n}(x)t^{n}\right)\left(\sum_{n=-\infty}^{\infty}J_{n}(x)t^{-n}\right)=1$$ and we can express the sums by splitting positive and negative indexes to get $$\left(J_{0}(x)+\sum_{n=1}^{\infty}J_{n}(x)t^{n}+\sum_{n=1}^{\infty}J_{-n}(x)t^{-n}\right)\left(J_{0}(x)+\sum_{n=1}^{\infty}J_{-n}(x)t^{n}+\sum_{n=1}^{\infty}J_{n}(x)t^{-n}\right)=1$$ It is the next step where you are facing difficulty. Note that RHS of the above equation does not involve variable $t$, but the LHS does involve $t$. So we need to find out terms after multiplication in LHS which don't involve $t$. The way to multiply two series is multiplying each term of one series with every term of other series. The Cauchy product also does the same, but it also groups the resulting terms in a particular way to make a new series. Here we don't need that grouping. Just individual terms of the product which don't involve $t$ are needed.</p>
<p>Clearly one such term is obtained by multiplying first term in both series namely $J_{0}(x)J_{0}(x)=J_{0}^{2}(x)$. Next we mark the four sums involving $\sum$ notation as $A, B, C, D$ respectively. If we multiply the term with $t^{n} $ in $A$ and the term with $t^{-n} $ in $D$ then we get $J_{n} ^{2}(x)$ (the $t$'s cancel out). Similarly we multiply term with $t^{-n} $ in $B$ and term with $t^{n} $ in $C$ to get $J_{-n}^{2}(x)=(-1)^{2n}J_{n}^{2}(x)=J_{n}^{2}(x)$. Note that any other terms in the product must involve $t$. Thus the terms without $t$ in the product on LHS are $$J_{0}^{2}(x)+2J_{1}^{2}(x)+2J_{2}^{2}(x)+\cdots$$ and this is equal to $1$ by comparison with RHS. </p>
|
4,309,247 | <p>My question comes from an exercise in Shilov's <em>Linear Algebra</em>. His hint is to use induction, but I'm struggling to get anywhere. I looked through the book and couldn't find any theorem that seemed useful, so I'm guessing there is some sort of manipulation I must be missing? A good first step to take would be much appreciated.</p>
<p>Thank you for any help you can provide!</p>
| DreamAR | 983,565 | <p><span class="math-container">$AB=I+BA,$</span> so <span class="math-container">$A^2B=A+(AB)A=2A+BA^2.$</span> This gives you the proof when <span class="math-container">$m=2.$</span></p>
<p>Assume you has the proposition when <span class="math-container">$m=k-1,$</span> try to prove the case when <span class="math-container">$m=k$</span> by similar idea.</p>
<p>If you want to use induction to prove something, try to solve it in the simple case first. The idea in proving normal case will be implied in the simple one.</p>
|
3,060,742 | <p><span class="math-container">$\sum_{n=1}^\infty \frac{1}{n^2} = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots = 1.644934$</span> or <span class="math-container">$\frac{\pi^2}{6}$</span></p>
<p>What if we take every 3rd term and add them up? </p>
<p>A = <span class="math-container">$ \frac{1}{3^2} + \frac{1}{6^2} + \frac{1}{9^2} + \cdots = ??$</span></p>
<p>How to take every 3rd-1 term and add them up?</p>
<p>B = <span class="math-container">$ \frac{1}{2^2} + \frac{1}{5^2} + \frac{1}{8^2} + \cdots = ??$</span></p>
<p>How to take every 3rd-2 term and add them up?</p>
<p>C = <span class="math-container">$ \frac{1}{1^2} + \frac{1}{4^2} + \frac{1}{7^2} + \cdots = ??$</span></p>
<p>I am not sure how to adapt Eulers methods as he used the power series of sin for his arguments: <a href="https://en.wikipedia.org/wiki/Basel_problem" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Basel_problem</a></p>
| User3910 | 9,235 | <p>KM101 deleted his hint... not sure why.</p>
<p><span class="math-container">$\frac{1}{3^2}+\frac{1}{6^2}+\frac{1}{9^2}+\dots=\frac{1}{3^2 1^2}+\frac{1}{3^2 2^2}+\frac{1}{3^2 3^3}+\dots=\frac{1}{9}(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\dots)= \frac{\pi^2}{54}$</span></p>
<p><span class="math-container">$\frac{1}{2^2}+\frac{1}{5^2}+\frac{1}{8^2}+\dots=\frac{1}{2^2 1^2}+\frac{1}{3^2?? 1^2}+\frac{1}{3^2 ??1^2}$</span></p>
|
642,443 | <p>Let $\{a_n\}_{n\ge1}^{\infty}=\bigg\{\cfrac{1}{1\cdot3}+\cfrac{1}{2\cdot4}+\dots+\cfrac{1}{n\cdot(n+2)}\bigg\}$. Find $\lim_{n\to \infty}{a_n}$.</p>
<p>I write:
$$\lim_{n\to \infty}{a_n}=\sum_{n=1}^{\infty}{\frac{1}{n\cdot(n+2)}}=\sum_{n=1}^{\infty}{\frac{1}{n^2+2n}}\approx\sum_{n=1}^{\infty}{\cfrac{1}{n^2}}$$</p>
<p>I put some values of $n$ for finding a pattern:
$$
\begin{array}{c|lcr}
n & \text{1}&\text{2}&\text{3}&\text{4}\\
\hline
\sum &\cfrac{1}{3}&\cfrac{11}{24}&\cfrac{21}{40}&\cfrac{17}{30}
\end{array}
$$</p>
<p>... but no hope.
I know that the limit/series converges, because $\forall n\in\mathbb N^*$:</p>
<ol>
<li><p>$\{a_n\}$ is increasing by the test of monothony : $a_{n+1}-a_{n}=\cfrac{1}{(n+1)(n+3)}>0$</p></li>
<li><p>$\{a_n\}$ is bounded : $0\le a_1=\frac{1}{3}\le a_2=\frac{11}{24}\le \dots \le a_n \le1$</p></li>
</ol>
<p>Wolfram says that the summation can be written as follows:
$$\cfrac{3}{4}-\cfrac{2n+3}{2(n+1)(n+2)}$$
How did it end up at this formula? Thank you.</p>
| Hagen von Eitzen | 39,174 | <p>Telescope! Note that $$ \frac1{n(n+2)}=\frac12\cdot\left(\frac1n-\frac1{n+2}\right)$$</p>
|
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