qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
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16,532 | <p>I want to export a plot into an eps file. Below I present a sample plot code in order to demonstrate the two issues that appear. </p>
<p>(1). <code>LabelStyle</code>: When I use "Times" or "Helvetica" style all seem to work fine. However, when I choose "Traditional" or "Tahoma" the style of the labels in the exported eps file is horribly deformed. This happens regardless the procedure I use; using the <code>Export</code> command or choosing the plot and then form <code>File -> Save Selection As ...</code> </p>
<p>(2). Capital Greek letters: I use the capital Greek letter <code>Theta</code> for the label of the y-axis. However, I discovered that many capital Greek letters are also deformed in the exported eps or pdf file.</p>
<p>Many thanks in advance and I look forward for your replies and comments</p>
<pre><code>S1 = Plot[Exp[-x]*Sin[7*x], {x, 0, 5}, Axes -> False, Frame -> True,
FrameLabel -> {"x", "Θ(x)"}, RotateLabel -> False, PlotStyle -> Red,
LabelStyle -> Directive[FontSize -> 17, FontFamily -> "Helvetica"],
PlotRange -> All, ImageSize -> 500]
Export["Plot.eps", S1]
</code></pre>
<hr>
<p>Thank you very much for all your replies and I am really very sorry for not responding sooner. I am using the latest version 9.0 of Mathematica in Windows XP SP3. However, the same issues appear in older versions (8,7, ... , 4) which are also installed in my PC. Therefore, I suppose that the version is irrelevant. Below I present three exports using different format (eps, pdf and jpg). I used "Tahoma" style.</p>
<p>First the .jpg output. All seems to be OK.</p>
<p><img src="https://i.stack.imgur.com/k4gh5.jpg" alt="enter image description here"> </p>
<p>Then the .pdf output. Well, the Greek capital letter Theta is deformed.</p>
<p><img src="https://i.stack.imgur.com/zLfGE.png" alt="enter image description here"></p>
<p>And finally, the .eps output loaded in GSView 5.0. Here, not only the Greek letter but also the axis labels are deformed. However, if I convert the exported .eps file to .pdf with Adobe the axis labels are becoming normal as Fig. 2 but still the Greek letter is deformed. </p>
<p><img src="https://i.stack.imgur.com/3RHyC.png" alt="enter image description here"></p>
<p>If I am correct, I encounter an issue regarding fonts. So, I would be very grateful if you could suggest me some solutions to my problem. </p>
| cormullion | 61 | <p>Your plot, exported to EPS, looks OK on my system (<em>Mathematica</em> 9, Mac OS X):</p>
<p>The version using Helvetica: </p>
<p><img src="https://i.stack.imgur.com/b07jt.png" alt="Helvetica"></p>
<p>A version using Georgia:</p>
<p><img src="https://i.stack.imgur.com/HX06l.png" alt="Georgia"></p>
<p>(I used Georgia because it's easier to spot with those lower-case numerals.)</p>
<p>The Θ (theta) in both cases is identical - probably the <code>Mathematica2</code> font.</p>
<p>In your EPS example, the main font has been replaced with what looks like Courier - which is the time-honoured way for font handling to fail in a PostScript-based workflow. (A google search for this issue should show millions of results going back to the 20th century.)</p>
<p>However, the EPS file which Mathematica generated for me contains the fonts embedded within it, and so there's no obvious reason why a PostScript renderer should fail to use them. If your EPS files also contain the needed font resources, which a quick glance with a text editor could confirm, then your problems may be more to do with your system, and the installation of gsview and GhostScript (if it's still called that) than with Mathematica.</p>
|
16,532 | <p>I want to export a plot into an eps file. Below I present a sample plot code in order to demonstrate the two issues that appear. </p>
<p>(1). <code>LabelStyle</code>: When I use "Times" or "Helvetica" style all seem to work fine. However, when I choose "Traditional" or "Tahoma" the style of the labels in the exported eps file is horribly deformed. This happens regardless the procedure I use; using the <code>Export</code> command or choosing the plot and then form <code>File -> Save Selection As ...</code> </p>
<p>(2). Capital Greek letters: I use the capital Greek letter <code>Theta</code> for the label of the y-axis. However, I discovered that many capital Greek letters are also deformed in the exported eps or pdf file.</p>
<p>Many thanks in advance and I look forward for your replies and comments</p>
<pre><code>S1 = Plot[Exp[-x]*Sin[7*x], {x, 0, 5}, Axes -> False, Frame -> True,
FrameLabel -> {"x", "Θ(x)"}, RotateLabel -> False, PlotStyle -> Red,
LabelStyle -> Directive[FontSize -> 17, FontFamily -> "Helvetica"],
PlotRange -> All, ImageSize -> 500]
Export["Plot.eps", S1]
</code></pre>
<hr>
<p>Thank you very much for all your replies and I am really very sorry for not responding sooner. I am using the latest version 9.0 of Mathematica in Windows XP SP3. However, the same issues appear in older versions (8,7, ... , 4) which are also installed in my PC. Therefore, I suppose that the version is irrelevant. Below I present three exports using different format (eps, pdf and jpg). I used "Tahoma" style.</p>
<p>First the .jpg output. All seems to be OK.</p>
<p><img src="https://i.stack.imgur.com/k4gh5.jpg" alt="enter image description here"> </p>
<p>Then the .pdf output. Well, the Greek capital letter Theta is deformed.</p>
<p><img src="https://i.stack.imgur.com/zLfGE.png" alt="enter image description here"></p>
<p>And finally, the .eps output loaded in GSView 5.0. Here, not only the Greek letter but also the axis labels are deformed. However, if I convert the exported .eps file to .pdf with Adobe the axis labels are becoming normal as Fig. 2 but still the Greek letter is deformed. </p>
<p><img src="https://i.stack.imgur.com/3RHyC.png" alt="enter image description here"></p>
<p>If I am correct, I encounter an issue regarding fonts. So, I would be very grateful if you could suggest me some solutions to my problem. </p>
| Jens | 245 | <p>There was another issue in the comment to @cormullion's answer, related to a missing sliver of the exported graphic on the side. If you see that kind of problem, it can help to set the <code>PlotRegion</code> to be slightly less than 100% of the available width or height determined by <code>ImageSize</code>. </p>
<p>To show this, I'll use the answer <a href="https://mathematica.stackexchange.com/a/8413/245">linked here</a> and modify your plot by appending two options at the end:</p>
<pre><code>S1 = Plot[Exp[-x]*Sin[7*x], {x, 0, 5}, Axes -> False, Frame -> True,
FrameLabel -> {"x", "\[CapitalTheta](x)"}, RotateLabel -> False,
PlotStyle -> Red,
LabelStyle -> Directive[FontSize -> 17, FontFamily -> "Helvetica"],
PlotRange -> All, ImageSize -> 500,
PlotRegion -> {{0.04, 0.96}, {0.04, 0.96}},
FrameStyle -> Thickness[.002]]
outlinedExport[name_, gr_, opts : OptionsPattern[]] :=
Export[name,
First@ImportString[ExportString[gr, "PDF"], "PDF",
"TextOutlines" -> True], FilterRules[{opts}, Options[Export]]]
outlinedExport["plot.eps", S1]
(* ==> "plot.eps" *)
</code></pre>
<p>Now the result of <code>Import["plot.eps"]</code> will look like this:</p>
<p><img src="https://i.stack.imgur.com/JSLhV.png" alt="with margins"></p>
<p>where I highlighted the image border to show the extra margins around the frame.</p>
|
424,209 | <p>I am a Computer Science student. While going through some random maths topics I came across Chaos Theory. I wanted to know if there are any applications of it in CS.
I tried searching on the internet about this but ended up only with <a href="https://security.stackexchange.com/questions/31000/does-chaos-theory-have-any-practical-application-in-computer-security">this</a> and <a href="http://www.slideshare.net/konakid/computer-science-student-uses-chaos-theory-to-create-new-routes" rel="nofollow noreferrer">this</a>.
But are there any other domains of computer science where this theory could be applied ?</p>
<p>Please help!</p>
| Fixed Point | 30,261 | <p>Cryptography is another application of chaos to computer science. There are a few chaotic encryption schemes around.</p>
|
1,937,762 | <p>Let's say I have a ratio of polynomials as follows</p>
<p>$P(x)=\frac{a_0x^n+a_1x^{n-2}+a_2x^{n-4}+...}{b_0x^n+b_1x^{n-2}+b_2x^{n-4}+...}$.</p>
<p>The polynomials are finite. Is there a procedure to convert it into a polynomial</p>
<p>$P(x) = A_0 + A_1 f(x) + A_2g(x) + ...$</p>
<p>where $f(x)$ and $g(x)$ are some power functions of x. Maybe there is a way to expand it in series if it is not divisible...</p>
| Dosetsu | 139,331 | <p>This is not possible. Think of $P(x)=1/x$. This does not have a Taylor expansion, so even your second questions, the series expansion is not possible.</p>
|
13,166 | <p>I taught IT in an engineering school during three years in <a href="https://en.wikipedia.org/wiki/Problem-based_learning" rel="nofollow noreferrer">problem based learning</a> (PBL) only. Now I teach maths to pupils between 10 and 15 years old who have a lot of educational difficulties.</p>
<p>I'm thinking to use PBL to increase their motivation and, by the way, increase their learning.</p>
<p>But, I have a question: is PBL adapted to pupils who have already know a lot of failures (in school and elsewhere)?</p>
| Photon | 3,542 | <p>I cannot provide any research results supporting my claim, but intuitively, I'd assume that it can work out if the level of the exercises is low enough to provide success experience. </p>
<p>For example, I had a pupil aged about 15 who had really big problems in maths (and also other subjects because she attended a Montessori school), failed entry exams for a regular school and had to go to a private school in the end. I gave her exercises which were intended for primary school kids, that is, required no prior knowledge but at the same time required some thinking and trying around. I had the impression that they fitted her level quite well and increased her motivation. Also, they motivated the introduction of some mathematical tools (say, systems of linear equations) quite nicely. </p>
<p>But of course, if you are dealing with a group of pupils rather than a single one, it is nearly impossible to fit everybody's level so you might bore some and overwhelm others.</p>
|
3,858,517 | <p>Is it possible to count exactly the number of binary strings of length <span class="math-container">$n$</span> that contain no two adjacent blocks of 1s of the same length? More precisely, if we represent the string as <span class="math-container">$0^{x_1}1^{y_1}0^{x_2}1^{y_2}\cdots 0^{x_{k-1}}1^{y_{k-1}}0^{x_k}$</span> where all <span class="math-container">$x_i,y_i \geq 1$</span> (except perhaps <span class="math-container">$x_1$</span> and <span class="math-container">$x_k$</span> which might be zero if the string starts or ends with a block of 1's), we should count a string as valid if <span class="math-container">$y_i\neq y_{i+1}$</span> for every <span class="math-container">$1\leq i \leq k-2$</span>.</p>
<p>Positive examples : 1101011 (block sizes are 2-1-2), 00011001011 (block sizes are 2-1-2), 1001100011101 (block sizes are 1-2-3-1)</p>
<p>Negative examples : 1100011 (block sizes are <strong>2-2</strong>), 0001010011 (block sizes are <strong>1-1</strong>-2), 1101011011 (block sizes are 2-1-<strong>2-2</strong>)</p>
<p>The sequence for the first <span class="math-container">$16$</span> integers <span class="math-container">$n$</span> is: 2, 4, 7, 13, 24, 45, 83, 154, 285, 528, 979, 1815, 3364, 6235, 11555, 21414. For <span class="math-container">$n=3$</span>, only the string 101 is invalid, whereas for <span class="math-container">$n=4$</span>, the invalid strings are 1010, 0101 and 1001.</p>
| leonbloy | 312 | <p>An aproximation for large <span class="math-container">$n$</span></p>
<p>The runs of <span class="math-container">$0$</span>s and <span class="math-container">$1$</span>s can be approximated by iid geometric random variables (with <span class="math-container">$p=1/2$</span>, mean <span class="math-container">$2$</span>). Hence we have in average <span class="math-container">$n/2$</span> runs, of which <span class="math-container">$n/4$</span> are runs of <span class="math-container">$1$</span>s.</p>
<p>Then, the problem is asymtpotically equivalent to : given <span class="math-container">$m=n/4$</span> iid Geometric variables <span class="math-container">$X_1, X_2 \cdots X_m$</span> find <span class="math-container">$P_m=$</span> probability that <span class="math-container">$X_{i+1} \ne X_i$</span> for all <span class="math-container">$i$</span>.</p>
<p>This does not seem a trivial problem, though (and I haven't found any reference).</p>
<p>A crude aproximation would be to assume that the events <span class="math-container">$X_{i+1} \ne X_i$</span> are independent. Under this assumption we get</p>
<p><span class="math-container">$$P_m \approx P_2^{m-1}= (2/3)^{m-1} \tag 1$$</span></p>
<p>This approximation is not justified, and it does not seem to improve with <span class="math-container">$n$</span> increasing.</p>
<p>The exact value can be obtained by a recursion on the probabilities for each final value, which together with a GF gives me this recursion :</p>
<p><span class="math-container">$$P_m = r(1,m) $$</span></p>
<p><span class="math-container">$$r(z,m)= \frac{1}{2z-1} r(1,m-1) - r(2z,m-1) \tag 2$$</span></p>
<p>with the initial value <span class="math-container">$r(z,1)=\frac{1}{2z-1}$</span></p>
<p>Finally, the total number of valid sequences is <span class="math-container">$C_m = P_m \, 2^n$</span> (<span class="math-container">$n=4m$</span>)</p>
<p>I've not yet found an explicit or asympotic for <span class="math-container">$(2)$</span>.</p>
<p>Some values oc <span class="math-container">$C_m$</span></p>
<pre><code>n m r(2) iid(1) exact
4 1 16 16 13
8 2 170.6 170.6 154
12 3 1950.5 1820.4 1815
16 4 21637.3 19418.1 21414
20 5 243540.2 207126.1 252680
24 6 2720810.9 2209345.3 2981452
28 7 30515606.3 23566350.0 35179282
</code></pre>
|
3,783,878 | <p>Hey everyone can anyone help me in simplifying the following boolean expression with explanation?</p>
<p><span class="math-container">\begin{equation}[((p\land q)\implies r)\implies((q\land r')\implies r')]\land[(p \land q)\implies(q\iff p)]\end{equation}</span></p>
| Hussain-Alqatari | 609,371 | <p>Based on the following table, the given statement is always true for any <span class="math-container">$p,q,r$</span>:</p>
<p><a href="https://i.stack.imgur.com/LGTS7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LGTS7.png" alt="enter image description here" /></a></p>
|
3,197,046 | <p>I'm interested in <span class="math-container">${\bf integer}$</span> solutions of </p>
<p><span class="math-container">$$abcd+1=(ecd-c-d)(fab-a-b)$$</span></p>
<p>subject to <span class="math-container">${\bf a,b,c,d \geq 2}$</span>, and <span class="math-container">${\bf e,f \geq 1}$</span>. </p>
<p><span class="math-container">${\bf Questions:}$</span> Are there finitely many solutions? If no, is there a nice infinite family of solutions?</p>
| Jeff | 562,680 | <p>There are finitely many such solutions.</p>
<p><span class="math-container">${\bf Proof.}$</span></p>
<p>It follows directly from the following (not too hard to verify) statements:</p>
<p>(1) <span class="math-container">$e$</span> or <span class="math-container">$f$</span> is <span class="math-container">$1$</span>. Once <span class="math-container">$f$</span> is fixed to be <span class="math-container">$1$</span>, <span class="math-container">$e \leq 7$</span>.</p>
<p>(2) At least one number from the set <span class="math-container">$S=\{a,b,c,d \}$</span> is <span class="math-container">$\leq 5$</span>. </p>
<p>(3) For a fixed number from the set <span class="math-container">$S$</span>, say <span class="math-container">$a$</span>, there are at most finitely many triples <span class="math-container">$(b,c,d)$</span> satisfying this equation (for a proof see <a href="https://math.stackexchange.com/questions/3198207/integer-points-on-a-surface">this question</a>).</p>
|
2,637,812 | <p>Here is dice game question about probability.</p>
<p>Play a game with $2$ die. What is the probability of getting a sum greater than $7$?</p>
<p>I know how the probability for this one is easy, $\cfrac{1+2+3+4+5}{36}=\cfrac 5{12}$.</p>
<p>I don't know how to solve the follow-up question:</p>
<p>Play a game with $200$ die. What is the probability of getting a sum greater than $700$?</p>
| user284331 | 284,331 | <p>\begin{align*}
\lim_{x\rightarrow 0}\dfrac{x^{2}-\log(1+x^{2})}{x^{2}\sin^{2}x}&=\lim_{x\rightarrow 0}\dfrac{x^{2}-x^{2}+\dfrac{1}{2}x^{4}-\dfrac{1}{3}x^{6}\cdots}{x^{2}\sin^{2}x}\\
&=\lim_{x\rightarrow 0}\dfrac{\dfrac{1}{2}x^{2}-\dfrac{1}{3}x^{4}\cdots}{\sin^{2}x}\\
&=\lim_{x\rightarrow 0}\dfrac{\dfrac{1}{2}x^{2}\left(1-\dfrac{1}{3}x^{2}+\cdots\right)}{\sin^{2}x}\\
&=\lim_{x\rightarrow 0}\left(\dfrac{1}{2}\cdot\dfrac{x^{2}}{\sin^{2}x}\right)\cdot\lim_{x\rightarrow 0}\left(1-\dfrac{1}{3}x^{2}+\cdots\right)\\
&=\dfrac{1}{2}.
\end{align*}</p>
|
2,637,812 | <p>Here is dice game question about probability.</p>
<p>Play a game with $2$ die. What is the probability of getting a sum greater than $7$?</p>
<p>I know how the probability for this one is easy, $\cfrac{1+2+3+4+5}{36}=\cfrac 5{12}$.</p>
<p>I don't know how to solve the follow-up question:</p>
<p>Play a game with $200$ die. What is the probability of getting a sum greater than $700$?</p>
| Barry Cipra | 86,747 | <p>Note that</p>
<p>$${x^2-\log(1+x^2)\over x^2\sin^2x}={x^2\over\sin^2x}\cdot{x^2-\log(1+x^2)\over x^4}={x^2\over\sin^2x}\cdot{u-\log(1+u)\over u^2}$$</p>
<p>where $u=x^2\to0^+$ as $x\to0$. If we take ${x\over\sin x}\to1$ for granted, then L'Hopital takes care of the rest:</p>
<p>$$\lim_{u\to0^+}{u-\log(1+u)\over u^2}=\lim_{u\to0}{1-{1\over1+u}\over2u}=\lim_{u\to0}{1\over2(1+u)}={1\over2}$$</p>
|
396,085 | <p>The length of three medians of a triangle are $9$,$12$ and $15$cm.The area (in sq. cm) of the triangle is</p>
<p>a) $48$</p>
<p>b) $144$</p>
<p>c) $24$</p>
<p>d) $72$</p>
<p>I don't want whole solution just give me the hint how can I solve it.Thanks.</p>
| robjohn | 13,854 | <p><strong>Area of a Triangle from the Medians</strong></p>
<p>A triangle is divided in to $6$ equal areas by its medians:</p>
<p>$\hspace{2cm}$<img src="https://i.stack.imgur.com/c63v0.png" alt="enter image description here"></p>
<p>In the case where the two blue triangles share a common side of the triangle, it is pretty simple to see they share a common altitude (dotted) and equal bases; therefore, equal areas.</p>
<p>In the case where the two red triangles share a common $\frac23$ of a median, the altitudes (dotted) are equal since they are corresponding sides to two right triangles with equal hypotenuses and equal <a href="http://en.wikipedia.org/wiki/Angle#Intersecting_angle_pairs" rel="noreferrer">vertically opposite angles</a>, and they share a common base; therefore, equal areas.</p>
<p>Now duplicate the original triangle (dark outline) by rotating it one-half a revolution on the middle of one of its sides:</p>
<p>$\hspace{3cm}$<img src="https://i.stack.imgur.com/N0wrm.png" alt="enter image description here"></p>
<p>The triangle in green has sides $\frac23a$, $\frac23b$, and $\frac23c$, and by <a href="http://en.wikipedia.org/wiki/Heron%27s_formula" rel="noreferrer">Heron's formula</a> has area
$$
\frac49\sqrt{s(s-a)(s-b)(s-c)}\tag{1}
$$
where $s=(a+b+c)/2$. Thus, each of the $6$ small, equal-area triangles in the original triangle has an area of half of that. Therefore, the area of the original triangle is $3$ times that given in $(1)$:
$$
\frac43\sqrt{s(s-a)(s-b)(s-c)}\tag{2}
$$</p>
|
396,085 | <p>The length of three medians of a triangle are $9$,$12$ and $15$cm.The area (in sq. cm) of the triangle is</p>
<p>a) $48$</p>
<p>b) $144$</p>
<p>c) $24$</p>
<p>d) $72$</p>
<p>I don't want whole solution just give me the hint how can I solve it.Thanks.</p>
| vaasie | 225,294 | <p>The area of a triangle made by the medians taken as sides is 75% of the triangle of which the medians are given. Now you can find the area by heron formula and the area thus you get will be 75% of the area of the triangle of which the medians are given. </p>
|
2,275,785 | <p>I asked a similar question last night asking for an explanation of the statement, however I was unable to find how to prove such a statement, so I have a proof, however I think it is wrong, so I'm just asking for it to be checked and if it is, for it to be corrected, thanks! </p>
<p><strong>Question</strong></p>
<p>Describe the following set, and prove your answer correct. (Here brackets
denote intervals on $\mathbb{R}$.)</p>
<p>$\bigcup_{i=0}^{\infty}[i, i+2]$</p>
<p><strong>Working</strong></p>
<p>$[i,i+2]=${$x\in\mathbb{R}|i\leq x\leq i+2$}</p>
<p>Thus, with the union we have {$x\in\mathbb{R}|i\leq x\leq i+2$}$=[0, \infty)$</p>
<p>Now, note that $\bigcup_{i=0}^{\infty}[i, i+2]=[0, 2]\cup [1, 3]\cup [2,4]\cup...\cup[k, k+2]\cup...$ for some $k\in\mathbb{R}$</p>
<p>Consider $x \notin [0, \infty)$ then $|x|>[0, \infty)$, so $\exists$ some $i\in\mathbb{N}$ such that $i<|x|$. Thus, $x\notin (0, \infty]$</p>
<p>$\therefore x$ is not the union</p>
<p>Thus, we have shown the union of $[i, i+2]$ is on the interval $[0, \infty)$ so;</p>
<p>$\bigcup_{i=0}^{\infty}[i, i+2]=[0, \infty)$</p>
| DeepSea | 101,504 | <p><strong>hint</strong>: You have $ \lfloor x \rfloor \le x \le \lfloor x \rfloor + 1 < \lfloor x \rfloor + 2$</p>
|
2,966,010 | <p>how to show that <span class="math-container">$$\sum_{n=1}^{\infty}(-1)^{n}\dfrac{3n-1}{n^2 + n} = \log\left({32}\right) - 4$$</span>? Can I use the Alternating Series test and how? </p>
| Darío A. Gutiérrez | 353,218 | <p><span class="math-container">$$S_1= \sum_{n=1}^{\infty}(-1)^{n}\dfrac{3n-1}{n^2 + n}$$</span>
<span class="math-container">$$\dfrac{3n-1}{n^2 + n} = \dfrac{3-\frac{1}{n}}{n + 1} = \left(\frac{4}{n+1} - \frac{1}{n}\right)$$</span>
So
<span class="math-container">\begin{align}
S_1 &= \sum_{n=1}^{\infty}(-1)^{n}\left(\frac{4}{n+1} - \frac{1}{n}\right)\\
&= 4\underbrace{\sum_{n=1}^{\infty}(-1)^{n}\frac{1}{n+1}}_{s_2} - \underbrace{\sum_{n=1}^{\infty}(-1)^{n}\frac{1}{n}}_{-\ln(2)}
\end{align}</span>
We know, then
<span class="math-container">$$\ln(1 + x) = \sum_{k=1}^{\infty}(-1)^{k+1}\frac{x^k}{k} = x - \frac{x^2}{2} + \frac{x^2}{3} - \frac{x^2}{4} + \cdots$$</span>
and for <span class="math-container">$x = 1$</span> we have
<span class="math-container">$$\ln(1 + 1) = \ln(2) = \sum_{k=1}^{\infty}(-1)^{k+1}\frac{1^k}{k} = 1 \underbrace{- \frac{1^2}{2} + \frac{1^2}{3} - \frac{1^2}{4} + \cdots}_{s_2} = 1- s_2$$</span></p>
<p>Which implies
<span class="math-container">$$s_2 = \ln(2) -1$$</span>
Therefore
<span class="math-container">\begin{align}
S_1 &= 4(s_2) - (-\ln(2))\\
&= 4(\ln(2) -1) - (-\ln(2))\\
&= 4\ln(2) -4 + \ln(2)\\
&= 5\ln(2) -4\\
&= \ln(32) -4\\
\end{align}</span>
<br>
See:</p>
<p><a href="https://math.stackexchange.com/questions/167155/the-sum-sum-n-1-infty-1n1-frac1n-ln2-using-riemann-integral">the sum: <span class="math-container">$\sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n}=\ln(2)$</span> using Riemann Integral and other methods</a></p>
|
207,243 | <p>I am using FindFit function in order to fit my data and get two parameters: c and m.</p>
<p>The function that I am using has the following form:</p>
<pre><code>function = (m/x*(x/c)^m)*Exp[-1*(x/c)^m];
</code></pre>
<p>The answer should be c = 64.68 and m = 2.47, but I am constantly getting the error message Overflow.</p>
<p>My data and core are here:</p>
<pre><code>data = {{172.345, 0.000710716}, {136.899, 0.00238143}, {108.742,
0.00822192}, {86.3772, 0.0117385}, {68.6119, 0.0123856}, {54.5003,
0.0147863}, {43.2912, 0.0148628}, {34.3874, 0.0131214}, {27.3149,
0.00879578}, {21.697, 0.00634049}, {17.2345, 0.00310614}, {13.6899,
0.00231347}, {10.8742, 0.000858576}, {8.63772, 0.000286827}};
function = (m/x*(x/c)^m)*Exp[-1*(x/c)^m];
params = {c, m};
FindFit[data, function, params, x, MaxIterations -> 10000]
fitedfunc = function /. %
plot1 = ListPlot[data, PlotStyle -> Black, PlotMarkers -> "*"];
plot2 = Plot[fitedfunc, {x, 0, 200}, PlotStyle -> {Red, Thick}];
Show[plot1, plot2, Frame -> True]
</code></pre>
| mikado | 36,788 | <p>You might find the following works. I'll create some points</p>
<pre><code>pts = Table[(5 + t) {Cos[t], Sin[t]}, {t, 0, 5}] // N
(* {{5., 0.}, {3.24181, 5.04883}, {-2.91303, 6.36508}, {-7.91994,
1.12896}, {-5.88279, -6.81122}, {2.83662, -9.58924}} *)
</code></pre>
<p>and extract the cells of the triangulation</p>
<pre><code>MeshCells[DelaunayMesh[pts], 2]
(* {Polygon[{3, 4, 1}], Polygon[{1, 4, 5}], Polygon[{1, 2, 3}],
Polygon[{1, 5, 6}]} *)
</code></pre>
<p>For comparison in <strong>Octave</strong> (similar to MATLAB)</p>
<pre><code> >> pts = [[5., 0.]; [3.24181, 5.04883]; [-2.91303, 6.36508]; [-7.91994, 1.12896];[-5.88279, -6.81122]; [2.83662, -9.58924]];
>> delaunay(pts)
ans =
3 1 4
5 1 4
5 6 1
2 3 1
</code></pre>
<p>So the same triangles, but in a different order.</p>
|
397,040 | <p>What is the domain for $$\dfrac{1}{x}\leq\dfrac{1}{2}$$</p>
<p>according to the rules of taking the reciprocals, $A\leq B \Leftrightarrow \dfrac{1}{A}\geq \dfrac{1}{B}$, then the domain should be simply $$x\geq2$$</p>
<p>however negative numbers less than $-2$ also satisfy the original inequality. When am I missing in my understanding?</p>
| Zev Chonoles | 264 | <p>The equivalence
$$A\leq B\iff \frac{1}{A}\geq\frac{1}{B}$$
only holds for numbers $A$ and $B$ that <strong>have the same sign</strong> (i.e., are both positive or both negative). Remember, when $c>0$, we have
$$A\leq B \iff c A\leq c B$$
and when $c<0$, we have
$$A\leq B\iff cA\geq cB.$$
To go from $A\leq B$ to
$$\frac{1}{A} \mathbin{\fbox{$\leq$ or $\geq$}} \frac{1}{B}$$
you'll multiply both sides by $c=\frac{1}{AB}$. This $c$ is greater than $0$ when $AB>0$, which happens when $A$ and $B$ have the same sign, and it is less than $0$ when $AB<0$, which happens when $A$ and $B$ have opposite signs.</p>
<p>Thus,
$$\frac{1}{x}\leq\frac{1}{2}\iff\begin{cases} x\geq 2 &\text{ if }x\text{ has the same sign as 2},\\
x\leq 2 &\text{ if }x\text{ has the opposite sign as 2}.
\end{cases}$$</p>
|
4,056,073 | <p>I need help with this task, if anyone had a similar problem it would help me !</p>
<p>The task is: Determine the type of interruption at the point x = 0 for the function</p>
<p><span class="math-container">$$f(x)=2^{-\frac{1}{x^{2}}}$$</span></p>
<p>I did:</p>
<p><span class="math-container">$$L=\lim_{x\to 0^{-}} 2^{-\frac{1}{x^{2}}} = 0 $$</span>
<span class="math-container">$$R=\lim_{x\to 0^{+}} 2^{-\frac{1}{x^{2}}} = 0 $$</span>
<span class="math-container">$$L=R=x=0$$</span></p>
<p>And as I concluded the function is continuous at x = 0, but in the solution it says that the break is of the first kind.
So I don’t understand why a breakup is the first kind?</p>
<p>Thanks in advance !</p>
| Adam Latosiński | 653,715 | <p>The ordering of the roots doesn't give you any new information about the polynomial. You have a map
<span class="math-container">$$ {\mathbb C}^n \ni (r_1,r_2,\dots r_n) \mapsto (c_0,c_1\dots c_{n-1}) \in \mathbb{C}^n$$</span>
This map is surjective, but not injective. That kind of things may happen because <span class="math-container">$\mathbb{C}^n$</span> is an infinte set. It is also true for other algebraically closed fields - <a href="https://math.stackexchange.com/questions/56397/do-finite-algebraically-closed-fields-exist">no algebraically closed field is finite</a>.</p>
|
1,627,050 | <p>Let $f:[a,b]\rightarrow R$ be differentiable at $c\in [a,b]$. Show that for every $\epsilon >0$, there is a $\delta(\epsilon) >0$ s.t if $0<|x-y|<\delta(\epsilon)$ and $a\leq x \leq c\leq y \leq b$, then\
$$ |{\frac{f(x)-f(y)}{x-y}-f'(c)}|<\epsilon$$
I can only think of using triangle inequality, but it seems does not work. Can anyone help me?</p>
| David C. Ullrich | 248,223 | <p>Hint: If $x=c<y$ or $x<c=y$ this is immediately immediate from the definition. So assume $x<c<y$. Then $$\frac{f(y)-f(x)}{y-x}-f'(c)
=\frac{y-c}{y-x}\left(\frac{f(y)-f(c)}{y-c}-f'(c)\right)+\frac{c-x}{y-x}\left(\frac{f(c)-f(x)}{c-x}-f'(c)\right).$$</p>
<p>Now you can use the triangle inequality there; <em>when</em> you use the triangle inequality it's going to be very important that the two numbers $(y-c)/(y-x)$ and $(c-x)/(y-x)$ are both positive and add up to $1$.</p>
<p><strong>Added:</strong> Of course user254665's answer is cleaner, applying directly to the case $x\le c\le y$. It seems possible that my version will <em>seem</em> simpler to some readers (and that his will seem simpler to others). A compromise, a version of my version that doesn't require handling the cases $x=c$ and $y=c$ separately:</p>
<p>Define $$S(x)=\begin{cases}\frac{f(x)-f(c)}{x-c}-f'(c),&(x\ne c),
\\0,&(x=c).\end{cases}$$ Then for $x\le c\le y$ we have $$\frac{f(y)-f(x)}{y-x}-f'(c)
=\frac{y-c}{y-x}S(y)+\frac{c-x}{y-x}S(x).$$</p>
<p>[...]</p>
|
1,677,035 | <p>I'm new to this website so I apologize in advance if what I'm going to ask isn't meant to be posted here.</p>
<p>A bit of background though: I haven't been to school in 6 years and the last level I've graduated was Grade 7 due to financial problems, as well as my mom frequently being in and out of the hospital. I am now 18 and I wish to go to college as soon as I can, but I need to be caught up on all the math I've missed (I have been studying these past few years with what's available on the internet, but I don't think it's enough).</p>
<p>So my question is, are there any good, easy to understand, high school math books suited for my situation? I learn better with a teacher who can explain the lesson, but since I don't have one I'd prefer books that aren't too difficult, but at the same time provide everything necessary for high school level math and more. I used to be a bright student so I'm sure I can do this on my own with the right material.</p>
<p>Again, if this question isn't meant to be on this site I'd be more than willing to delete it asap! That's all. Thank you for reading. :)</p>
| H Huang | 604,218 | <p><a href="http://tutorial.math.lamar.edu" rel="nofollow noreferrer">http://tutorial.math.lamar.edu</a> is really good for calculus. I’ve never used the non-calc parts, but from a cursory look, they look of the same quality. Don’t use it exclusively though, since it is designed for a single college course, it has to skip some portions that there is more time to teach in high school.</p>
|
1,643,201 | <p>The spectrum-functor
$$
\operatorname{Spec}: \mathbf{cRng}^{op}\to \mathbf{Set}
$$
sends a (commutative unital) ring $R$ to the set $\operatorname{Spec}(R)=\{\mathfrak{p}\mid \mathfrak{p} \mbox{ is a prime ideal of R}\}$ and a morpshim $f:S\to R$ to the map $\operatorname{Spec}(R)\to \operatorname{Spec}(S)$ with $\mathfrak{p}\mapsto f^{-1}(\mathfrak{p})$. Does this functor send pullback squares
\begin{eqnarray}
S\times_R T&\to& T\\
\downarrow && \downarrow\\
S&\to& R
\end{eqnarray}
of (commutative unital) rings to pushout squares
\begin{eqnarray}
\operatorname{Spec}(R)&\to& \operatorname{Spec}(T)\\
\downarrow && \downarrow\\
\operatorname{Spec}(S)&\to& \operatorname{Spec}(S\times_R T)
\end{eqnarray}
of sets? Put in other words, does the functor $\operatorname{Spec}$ from above preserve pushouts?</p>
| Win Vineeth | 311,216 | <p>You are absolutely right, and that's the shortest way of doing it as far as I know.</p>
|
3,916,092 | <blockquote>
<p>A ball rotates at a rate <strong><span class="math-container">$r$</span></strong> rotations per second and simultaneously revolves around a stationary point <strong><span class="math-container">$O$</span></strong> at a rate <strong><span class="math-container">$R$</span></strong> revolutions per second <strong><span class="math-container">$(R<r)$</span></strong>. The rotation and revolution are in the same sense. A certain point on the ball is in the line of the centre of the ball and point <strong><span class="math-container">$O$</span></strong> at a certain time. This configuration repeats after a time</p>
</blockquote>
<blockquote>
<p><span class="math-container">$(1)\ $</span> <span class="math-container">$\frac{1}{r-R}$</span></p>
</blockquote>
<blockquote>
<p><span class="math-container">$(2)\ $</span> <span class="math-container">$\frac{1}{R}-\frac{1}{r}$</span></p>
</blockquote>
<blockquote>
<p><span class="math-container">$(3)\ $</span> <span class="math-container">$\frac{1}{r+R}$</span></p>
</blockquote>
<blockquote>
<p><span class="math-container">$(4)\ $</span> <span class="math-container">$\frac{1}{R}+\frac{1}{r}$</span></p>
</blockquote>
<p>How do I solve this aptitude question? Thanks for your time.</p>
<p><strong>Source</strong> <span class="math-container">$:$</span> CSIR NET JUNE <span class="math-container">$2019.$</span></p>
| Math Lover | 801,574 | <p><a href="https://i.stack.imgur.com/OlccJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OlccJ.png" alt="enter image description here" /></a></p>
<p>As rotation and revolution are both in the same direction (say, counterclockwise), the point will come in the line connecting centers after one complete rotation. Say that happens after part revolution of angle <span class="math-container">$\theta$</span> against one full revolution (<span class="math-container">$2\pi$</span>). In that time the rotation has completed by <span class="math-container">$(2\pi + \theta)$</span> (more than one rotation of (<span class="math-container">$2\pi$</span>).</p>
<p>One rotation takes <span class="math-container">$\frac{1}{r}$</span> second and one revolution takes <span class="math-container">$\frac{1}{R}$</span> second. Equating the time taken -</p>
<p><span class="math-container">$\displaystyle \frac{2\pi + \theta}{2\pi r} = \frac{\theta}{2\pi R}$</span></p>
<p>Now please find <span class="math-container">$\theta$</span> and then plug in either Right Hand side or in Left Hand side to find the time.</p>
|
3,357,841 | <p>In the diagram (which is not drawn to scale) the small triangles each have the area shown. Find the area of the shaded quadrilateral.</p>
<p><a href="https://i.stack.imgur.com/DK8sn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DK8sn.png" alt="enter image description here"></a></p>
| Jean Marie | 305,862 | <p>There is a simple solution obtained by splitting the area to be found into two areas <span class="math-container">$x$</span> and <span class="math-container">$y$</span> by diagonal <span class="math-container">$CM$</span> (see figure).</p>
<p><a href="https://i.stack.imgur.com/BtR4J.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BtR4J.jpg" alt="enter image description here"></a></p>
<p>This diagonal allows to use the fact that <span class="math-container">$\dfrac{[ACM]}{[MCA']}=\dfrac{[BAM]}{[MAB']}=\dfrac{AM}{MA'}$</span> (ratio of areas equal to the ratio of bases because of a common height) :</p>
<p><span class="math-container">$$\dfrac{7+x}{y}=\dfrac{14}{4}\left(=\dfrac{AM}{MA'}\right)\tag{1}$$</span></p>
<p>For a similar reason, <span class="math-container">$\dfrac{[B'CM]}{[MCB]}=\dfrac{[B'AM]}{[MAB]}=\dfrac{BM}{MB'}$</span> giving :</p>
<p><span class="math-container">$$\dfrac{x}{y+4}=\dfrac{7}{14}\left(=\dfrac{BM}{MB'}\right)\tag{2}$$</span></p>
<p>(1) and (2) give rise to a linear system of 2 equations in the 2 unknowns <span class="math-container">$x$</span> and <span class="math-container">$y$</span> ; its solutions are :</p>
<blockquote>
<p><span class="math-container">$$x=\dfrac72 \ \ \ \text{and} \ \ \ y=3, $$</span>
giving finally total area </p>
<p><span class="math-container">$$x+y=\dfrac{13}{2}.$$</span></p>
</blockquote>
<p>Remark : one can find in book 6 of Euclides : "Triangles sharing the same altitude are between themselves [understand : their resp. areas] as their bases."</p>
|
192,125 | <p>Solve: $$\sqrt{x-4} + 10 = \sqrt{x+4}$$
Little help here? >.<</p>
| Robert Israel | 8,508 | <p>There are no real solutions, nor any complex solutions if you use the principal branch of the square root. Squaring both sides and simplifying gives you $20 \sqrt{x-4} = -92$.</p>
<p>EDIT: More generally, for any $a, b \ge 0$, $\sqrt{a + b} \le \sqrt{a} + \sqrt{b}$. Since
$(x+4) - (x-4) = 8$, the most $\sqrt{x+4} - \sqrt{x-4}$ can be is $\sqrt{8}$. </p>
|
1,323,943 | <p>Identify the parent function </p>
<p>$g(x)=\sqrt {-x+2 }$</p>
<p>1-Identify the parent function</p>
<p>2-Identify the transformations being applied, in appropriate order</p>
<p>3-sketch a graph of the transformed function</p>
<p>my work for first one </p>
<p>from parent function <img src="https://i.stack.imgur.com/cvGi7.png" alt="enter image description here"></p>
<p>the parent function is </p>
<p>$f(x)=\sqrt {x }$</p>
<p>then </p>
<p>$g(x)= f(-x+2)$</p>
<p>need help about second part and graph .please </p>
| wythagoras | 236,048 | <ul>
<li><p>Translation $(-2,0)$ gives $\sqrt{x+2}$</p></li>
<li><p>Multiplication through the y-axis by -1 gives $\sqrt{-x+2}$.</p></li>
</ul>
<p>Graph: The start point is $(2,0)$. Then draw a square root graph to the left. </p>
|
1,323,943 | <p>Identify the parent function </p>
<p>$g(x)=\sqrt {-x+2 }$</p>
<p>1-Identify the parent function</p>
<p>2-Identify the transformations being applied, in appropriate order</p>
<p>3-sketch a graph of the transformed function</p>
<p>my work for first one </p>
<p>from parent function <img src="https://i.stack.imgur.com/cvGi7.png" alt="enter image description here"></p>
<p>the parent function is </p>
<p>$f(x)=\sqrt {x }$</p>
<p>then </p>
<p>$g(x)= f(-x+2)$</p>
<p>need help about second part and graph .please </p>
| John Joy | 140,156 | <p>Sometimes its just easier thinking about the equation inside out looking at $g^{-1}$. Consider
$$\begin{array}{lll}
g(x)=f(a(x-b))\\
f^{-1}(g(x))=f^{-1}(f(a(x-b)))\\
f^{-1}(g(x))=a(x-b)\\
\frac{1}{a}[f^{-1}(g(x))] + b=x\\
\end{array}$$
Which looks somewhat cryptic, but lets look at your particular example.
$$\begin{array}{lll}
y=\sqrt{-x+2}\\
y^2=-x+2 = -(x-2),y\ge0\\
-y^2= x-2\\
-y^2 + 2= x\\
\end{array}$$
If we graph this function with $x$ on the vertical axis and $y$ on the horizontal axis we see that we have a parabola that has been flipped about the horizontal axis and shifted upwards by 2 units. Now when we pull the switcheroo on the x/y-axes (flip the graph diagonally so the $x$ is on the horizontal axis and $y$ is on the veritcal axis) we see that the horizontal flip becomes a vertical flip and the upward shift of 2 units becomes a rightward shift of 2 units.
<img src="https://i.stack.imgur.com/PmTi0.png" alt="enter image description here"></p>
<p>Try copying the image into Word, and print it out. Try holding the printout up to the light to see the "<em>switcheroo</em>" in action. Hold the printout up so that the black $y$ is at the top, then flip it over so that the green $y$ is at the right. Pay special attention to what happens to the position of the vertex and the direction that the parabola opens toward. Note that the dotted red curve is the negative shadow of the square root parent function; it just makes it easier to see the parabola.</p>
|
4,434,832 | <p>I have taken this question from molodovian national MO 2008
The question is as follows</p>
<p>The sequence <span class="math-container">$(a_p)_p\ge 0$</span> is defined as <span class="math-container">$$a_p=\sum_{i=0}^p (-1)^i\frac{\binom{p}{i}}{(i+2)(i+4)}$$</span></p>
<p>Now let's find the limit</p>
<p><span class="math-container">\begin{align*} \lim_{n \to +\infty} a_0+a_1+a_2+.......a_n \end{align*}</span></p>
<p><a href="https://i.stack.imgur.com/CE73S.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CE73S.png" alt="enter image description here" /></a></p>
<p>Can anyone help me to go further.</p>
| Stefan Lafon | 582,769 | <p>Let <span class="math-container">$$\begin{split}
S_{k,p} &= \sum_{i=0}^p\binom{p}{i}\frac{(-1)^i}{i+k}\\
&= \sum_{i=0}^p\binom{p}{i}(-1)^i\int_0^1 x^{i+k-1}dx\\
&= \int_0^1 \sum_{i=0}^p\binom{p}{i}(-1)^ix^{i+k-1}dx\\
&= \int_0^1 x^{k-1}(1-x)^pdx
\end{split}$$</span>
Then since <span class="math-container">$$\frac 1{(i+2)(i+4)}=\frac 1 2 \left(\frac 1 {i+2}-\frac 1 {i+4}\right)$$</span>
we have
<span class="math-container">$$a_p=\frac 1 2 (S_{2,p}-S_{4,p})=\frac 1 2 \int_0^1 (x-x^3)(1-x)^pdx$$</span>
Thus,</p>
<p><span class="math-container">$$\begin{split}
\sum_{p=0}^n &a_p= \frac 1 2 \int_0^1(x-x^3)\sum_{p=0}^n(1-x)^pdx\\
&=\frac 1 2 \int_0^1(x-x^3)\frac{1-(1-x)^{n+1}}{x}dx\\
&=\frac 1 2 \int_0^1 (1-x^2)(1-(1-x)^{n+1})dx\\
\end{split}$$</span></p>
<p>Using <a href="https://en.wikipedia.org/wiki/Dominated_convergence_theorem" rel="nofollow noreferrer">Lebesgue's Dominated Convergence Theorem</a>:
<span class="math-container">$$\lim_{n\rightarrow+\infty}\sum_{p=0}^n a_p = \frac 1 2 \int_0^1 (1-x^2)dx=\frac 1 2 \left (1- \frac 1 3\right) = \frac 1 {3}$$</span></p>
|
351,846 | <p>The following problem was on a math competition that I participated in at my school about a month ago: </p>
<blockquote>
<p>Prove that the equation $\cos(\sin x)=\sin(\cos x)$ has no real solutions.</p>
</blockquote>
<p>I will outline my proof below. I think it has some holes. My approach to the problem was to say that the following equations must have real solution(s) if the above equation has solution(s):</p>
<p>$$
\cos^2(\sin x)=\sin^2(\cos x)\\
1-\cos^2(\sin x)=1-\sin^2(\cos x)\\
\sin^2(\sin x)=\cos^2(\cos x)\\
\sin(\sin x)=\pm\cos(\cos x)\\
$$</p>
<p>I then proceeded to split into cases and use the identity $\cos t = \sin(\frac{\pi}{2} \pm t\pm y2\pi)$ to get </p>
<p>$$
\sin x=\frac{\pi}{2} \pm \cos x\pm y2\pi\\
$$</p>
<p>and the identity $-\cos t = \sin(-\frac{\pi}{2}\pm t\pm y2\pi) $ to get </p>
<p>$$
\sin x=- \frac{\pi}{2}\pm \cos x \pm y2\pi.\\
$$</p>
<p>where $y$ is any integer. I argued that $y=0$ was the only value of $y$ that made any sense (since the values of sine and cosine remain between $-1$ and $1$). Therefore, the above equations become </p>
<p>$$
\sin x=\frac{\pi}{2} \pm \cos x\implies \sin x \pm \cos x=\frac{\pi}{2}\\
$$</p>
<p>and </p>
<p>$$
\sin x=- \frac{\pi}{2}\pm \cos x\implies \cos x\pm\sin x= \frac{\pi}{2}.\\
$$</p>
<p>Then, by a short optimization argument, I showed that these last two equations have no real solutions.</p>
<p>First, does this proof make sense? Second, if my proof makes sense, then I feel that it was not very elegant nor simple. Is my approach the best, or is there a better (i.e., more elegant, shorter, simpler) proof?</p>
| Micah | 30,836 | <p>A possibly-shorter way of getting there would be to write
$$\cos (\sin x) - \sin(\cos x)=\cos (\sin x) - \cos(\pi/2-\cos x)$$
and then use a sum-to-product identity to turn this last expression into:
\begin{eqnarray}
&&−2 \sin \left(\frac{\sin x + \pi/2 - \cos x}{2}\right)\sin\left(\frac{\sin x - \pi/2 + \cos x}{2}\right)\\
&=& −2 \sin \left(\frac{\pi/2 + \sqrt{2}\sin (x-\pi/4)}{2}\right)\sin\left(\frac{- \pi/2 + \sqrt{2}\sin (x + \pi/4)}{2}\right) \, .
\end{eqnarray}
Since $\pi/2$ is not within $\sqrt{2}$ of any multiple of $2\pi$, this last expression never vanishes; thus your equation has no real solutions.</p>
<p>(Really, though, this is equivalent to your solution, except that we're outsourcing a lot of the case analysis to trig identities...)</p>
|
210,658 | <p>$$\sum_{i=1}^n t^{i-1}$$
I am stuck with the proof of this equality. </p>
| Mikasa | 8,581 | <p>Use induction on $n\in \mathbb N, n\ge1$ with $P(n):=\sum_{i=1}^n t^{i-1}=\frac{1-t^n}{1-t}$:
For $n=1$ you have: $$P(1): 1=\frac{1-t}{1-t}$$
which is correct. Suppose that $P(n)$ is true, now check it for $n+1$. You have $$P(n+1): \sum_{i=1}^{n+1} t^{i-1}=\sum_{i=1}^n t^{i-1}+t^{n+1}=\frac{1-t^n}{1-t}+t^{n+1}=\frac{1-t^{n+1}}{1-t}$$ So the statement is true for any $n\in\mathbb N$.</p>
|
210,658 | <p>$$\sum_{i=1}^n t^{i-1}$$
I am stuck with the proof of this equality. </p>
| Per Erik Manne | 33,572 | <p>Hint: Write out the summation
$$\sum_{i=1}^n t^{i-1}=1+t+t^2+\cdots+t^{n-1}$$
Now multiply by $1-t$, and observe which terms will cancel and which will survive.</p>
|
1,387,184 | <p>Can someone show how to compute the residue of this function:
$$\frac{z}{e^z - 1}$$</p>
<p>I think can represent the Taylor series of $e^z$ as
$$e^z = 1 + z + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots$$
Then, we have
$$\frac{z}{e^z - 1} = \frac{z}{(1 +z + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots) -1}$$
$$ = 1 + 1 +\frac{2}{z} + \frac{3!}{z^2} + \cdots - 1 = 1 +\frac{2}{z} + \frac{3!}{z^2} + \cdots$$
and, this function has a pole at $z_0 = 2n \pi i$
Hence, $$\operatorname{Res}_{z_0}f(z) = 0$$</p>
<p>EDIT: </p>
<p>The way I used to solve this problem is $$\textrm{Res}_{z_0}(\frac{f}{g}) = \frac{f(z_0)}{g'(z_0)} = \frac{2n \pi i}{e^{2n \pi i}}$$</p>
<p>Am I correct? If not, can someone please show me how to do it please?</p>
| Pedro | 23,350 | <p>Matrices are Morita equivalent to their underlying ring, and Morita equivalent rings have isomorphic Hochschild cohomology. A field as an algebra over itself has trivial cohomology for nonzero degrees.</p>
|
1,386,683 | <p>I posted early but got a very tough response.</p>
<p>Point $A = 2 + 0i$ and point $B = 2 + i2\sqrt{3}$ find the point $C$ $60$ degrees ($\pm$) such that Triangle $ABC$ is equilateral. </p>
<p>Okay, so I'll begin by converting into polar form:</p>
<p>$A = 2e^{2\pi i}$ and $B = 4e^{\frac{\pi}{3}i}$</p>
<p>$\overline{AB} = \sqrt{13}$</p>
<p>How should I find a point with length $\overline{BC} = \overline{AC} = \sqrt{13}$ and the sufficient angle?</p>
| Empty | 174,970 | <p><strong>From your calculation :</strong></p>
<p>$$=\lim_{\overset{x\to0}{y\to0}}\frac{(x^2-2xyi-y^2)(x-iy)}{x^2+y^2}$$</p>
<p>$$=\lim_{(x,y)\to (0,0)}\frac{x^3-3xy^2}{x^2+y^2}-i\lim_{(x,y)\to (0,0)}\frac{3x^2y-y^3}{x^2+y^2}$$</p>
<p>From here, show that both the limits are <strong>zero</strong> by changing polar form , $x=r\cos \theta$ , $y=r\sin \theta$.</p>
<p><strong>For the first limit,</strong></p>
<p>$$\left|\frac{r^3\cos^3\theta-3r^3\cos \theta\sin^2\theta}{r^2}\right|\le 4r<\epsilon$$whenever, $r^2<\epsilon^2/16$ i,e, whenever $|x|<\epsilon/\sqrt 8=\delta(say)$ , and $|y|<\epsilon/\sqrt 8=\delta(say)$.</p>
<p>Similarly the second limit will be zero and hence the given limit will be $0$.</p>
|
4,495,044 | <p>Edit: There is an answer at the bottom by me explaining what is going on in this post.</p>
<p>Define a function <span class="math-container">$f : R \to R$</span> by <span class="math-container">$f(x) = 1$</span> if <span class="math-container">$x = 0$</span> and <span class="math-container">$f(x) = 0$</span> if <span class="math-container">$x \ne 0$</span>. I was attempting to prove that <span class="math-container">$\lim_{x \to 0; x\in R}f(x)$</span> is undefined. The following is my proof.</p>
<p>Proof: Suppose that <span class="math-container">$\lim_{x\to 0; x \in R}f(x)=L$</span>. Then for every <span class="math-container">$\varepsilon > 0$</span> there exists a <span class="math-container">$\delta > 0$</span> such that for all those <span class="math-container">$x \in R$</span> for which <span class="math-container">$|x-0|<\delta$</span> we have that <span class="math-container">$|f(x)-L|<\varepsilon$</span>. But <span class="math-container">$|0| < \delta$</span> and by the Archimedean property we know that for <span class="math-container">$\delta > 0$</span> there exists an integer <span class="math-container">$n>0$</span> such that <span class="math-container">$0<|\frac 1 n| < \delta$</span>. This is a contradiction, as <span class="math-container">$f(0) = 1$</span> and <span class="math-container">$f(\frac 1 n)= 0$</span> and both are less than <span class="math-container">$\delta$</span>.</p>
<p>Is the proof correct?</p>
<p>Edit: Here a limit is defined using adherent points and not limit points. If we were to use limit points then, <span class="math-container">$\lim_{x \to 0; x\in R\setminus \{0\}}f(x)=0$</span>. I have updated the question with correct notation.</p>
<p>Edit 2: Most textbooks define limits using limit points. In which case you we would have that <span class="math-container">$\lim_{x \to 0}f(x)=\lim_{x \to 0; x\in R\setminus \{0\}}f(x)$</span>. We are considering the definition of the limit where limits are defined using adherent points. Where it really matters whether we are considering <span class="math-container">$lim_{x\to 0; x\in R\setminus \{0\}}f(x)$</span> or <span class="math-container">$lim_{x\to 0; x\in R}f(x)$</span>.</p>
<p>Edit 3: This is the definition of convergence of a function at a point in the book Analysis 1 by Terence Tao. Let <span class="math-container">$X$</span> be a
subset of <span class="math-container">$R$</span>, let <span class="math-container">$f : X → R$</span> be a function, let <span class="math-container">$E$</span> be a subset of <span class="math-container">$X$</span>, <span class="math-container">$x_0$</span>
be an adherent point of <span class="math-container">$E$</span>, and let L be a real number. We say that f
converges to <span class="math-container">$L$</span> at <span class="math-container">$x0$</span> in <span class="math-container">$E$</span>, and write <span class="math-container">$\lim_{x \to x_0;x\in E} f(x) = L$</span>, iff <span class="math-container">$f$</span>, after
restricting to <span class="math-container">$E$</span>, is ε-close to <span class="math-container">$L$</span> near <span class="math-container">$x_0$</span> for every <span class="math-container">$\varepsilon > 0$</span>. If <span class="math-container">$f$</span> does not
converge to any number <span class="math-container">$L$</span> at <span class="math-container">$x_0$</span>, we say that <span class="math-container">$f$</span> diverges at <span class="math-container">$x0$</span>, and leave
<span class="math-container">$\lim_{x\to x_0;x\in E} f(x)$</span> undefined.
In other words, we have <span class="math-container">$lim_{x\to x_0;x\in E} f(x) = L$</span> iff for every <span class="math-container">$\varepsilon > 0$</span>,
there exists a <span class="math-container">$\delta > 0$</span> such that <span class="math-container">$|f(x) − L| ≤ ε$</span> for all <span class="math-container">$x \in E$</span> such that
<span class="math-container">$|x − x0| < \delta$</span>.</p>
<p>There are a two other things he defines that are used in this definition. That of <span class="math-container">$\varepsilon$</span> closeness and local <span class="math-container">$\varepsilon$</span> closeness. For those wanting to read those, <a href="https://lms.umb.sk/pluginfile.php/111477/mod_page/content/5/TerenceTao_Analysis.I.Third.Edition.pdf" rel="nofollow noreferrer">Here is the link</a>. It is on page 221.</p>
<p>Edit 4: It might be useless defining limits without limit points. But that is the definition for which I am trying to prove this.</p>
| AlvinL | 229,673 | <blockquote>
<p>Proof: Suppose that <span class="math-container">$\lim_{x\to 0; x \in R}f(x)=L$</span>. Then for every <span class="math-container">$\varepsilon > 0$</span> there exists a <span class="math-container">$\delta > 0$</span> such that for all those <span class="math-container">$x \in R$</span> for which <span class="math-container">$|x-0|<\delta$</span> we have that <span class="math-container">$|f(x)-L|<\varepsilon$</span>. But <span class="math-container">$|0| < \delta$</span> and by the Archimedean property we know that for <span class="math-container">$\delta > 0$</span> there exists an integer <span class="math-container">$n>0$</span> such that <span class="math-container">$0<|\frac 1 n| < \delta$</span>. This is a contradiction, as <span class="math-container">$f(0) = 1$</span> and <span class="math-container">$f(\frac 1 n)= 0$</span> and both are less than <span class="math-container">$\delta$</span>.</p>
</blockquote>
<p>It's not clear what you are contradicting with. We assume the limit is <span class="math-container">$L$</span>. Taking <span class="math-container">$\varepsilon = \frac{1}{5}$</span>, for instance, it should hold that for some <span class="math-container">$\delta>0$</span>
<span class="math-container">$$ L-\frac{1}{5} < f(x) < L+\frac{1}{5} $$</span>
for any <span class="math-container">$x\in (-\delta,\delta)$</span>. In particular, for sufficiently large <span class="math-container">$N$</span>,
<span class="math-container">$$ -\frac{1}{5} < 1-L < \frac{1}{5} \quad\mbox{and}\quad -\frac{1}{5} < f(1/N)-L = -L < \frac{1}{5} $$</span>
and that does lead to contradiction. You should write it out.</p>
<p>Furthermore, in your argument, you don't mention what <span class="math-container">$\varepsilon$</span> is. It could be <span class="math-container">$\varepsilon = 100$</span>. I don't see how you arrive at a contradiction.</p>
<hr />
<p>Alternatively,</p>
<p>Use Proposition 9.3.9. Take the constant sequence <span class="math-container">$x_n = 0$</span> for every <span class="math-container">$n$</span>. Then <span class="math-container">$f(x_n) \to 1$</span>. For another sequence, <span class="math-container">$x_n = 1/n$</span>, for instance, <span class="math-container">$f(x_n) \to 0$</span>. So <span class="math-container">$\lim _{x\to 0, x\in\mathbb R} f(x)$</span> does not exist.</p>
<hr />
<p>This sort of definition for limit is nonstandard. Mainly because it makes the existence of the limit depend on a specific point. It's very dangerous to define it like this. It's not even clear if it is a well defined concept.</p>
|
2,879,883 | <p>Suppose that $f$ and $g$ are differentiable functions on $(a,b)$ and suppose that $g'(x)=f'(x)$ for all $x \in (a,b)$. Prove that there is some $c \in \mathbb{R}$ such that $g(x) = f(x)+c$.</p>
<p>So far, I started with this:</p>
<p>Let $h'(x)=f'(x)-g'(x)=0$, then MVT implies $\exists$ c $\in \mathbb{R}$ such that $h'(c) = \frac{h(b)-h(a)}{b-a} =0$. Then $h'(c)=0 \implies h(c)=c$</p>
<p>After this i'm not sure where to go, or if this is correct at all, any hints?
This is also my first post in Latex so sorry if there's any mistakes!</p>
| Robert Lewis | 67,071 | <p>If $F(x)$ is a differentiable function on $(a, b)$ with</p>
<p>$F'(x) = 0, \; \forall x \in (a, b), \tag 1$</p>
<p>then </p>
<p>$\exists c \in \Bbb R, \; F(x) = c; \tag 2$</p>
<p>for, picking any $x_1, x_2 \in (a, b)$ with $x_1 < x_2$, we have, by the fundamental theorem of calculus,</p>
<p>$F(x_2) - F(x_1) = \displaystyle \int_{x_1}^{x_2} F'(s) \; ds = \int_{x_1}^{x_2} 0 \; ds = 0; \tag 3$</p>
<p>it follows that</p>
<p>$F(x_1) = F(x_2), \; \forall x_1, x_2 \in (a, b); \tag 4$</p>
<p>thus we may take</p>
<p>$c = F(y) \tag 5$</p>
<p>for any $y \in (a, b)$. Now setting</p>
<p>$F(x) = f(x) - g(x), \tag 6$</p>
<p>we find</p>
<p>$F'(x) = f'(x) - g'(x) = 0, \; \forall x \in (a, b); \tag 7$</p>
<p>we see from the above that</p>
<p>$f(x) - g(x) = F(x) = c, \tag 8$</p>
<p>whence</p>
<p>$f(x) = g(x) + c, \; \forall x \in (a, b). \tag 9$</p>
<p>If one wishes, for whatever reason, to avoid integration and the fundamental theorem, then of course a demonstration based upon the mean value theorem is also possible: again, letting $F(x)$ be differentiable on $(a, b)$, suppose that there were $x_1, x_2 \in (a, b)$, $x_1 < x_2$, with</p>
<p>$F(x_1) \ne F(x_2); \tag{10}$</p>
<p>then by the MVT, there exists $\xi \in (x_1, x_2)$ such that</p>
<p>$F(x_2) - F(x_1) = F'(\xi)(x_2 - x_1), \tag{11}$</p>
<p>whence</p>
<p>$F'(\xi) = \dfrac{F(x_2) - F(x_1)}{x_2 - x_1} \ne 0 \tag{12}$</p>
<p>by virtue of (10), contradicting the hypothesis that $F'(x) = 0$ for $x \in (a, b)$. Thus</p>
<p>$F(x_2) = F(x_1) = c, \; \text{a constant}, \; \forall x_1, x_2 \in (a, b); \tag{13}$ </p>
<p>the desired result now follows by taking $F(x)$ as in (6).</p>
<p>As for our OP Brad Scott's effort to invoke the MVT, though it seems to be headed in the right general direction, it does erroneously conclude that $h(c) = c$ for some $c \in \Bbb R$; it is almost as if the MVT has been confused with the intermediate value theorem here. But as pointed out by my colleagues mfl and Lev Ban in their answers, the correct approach is to use the MVT in the present context is to affirm it implies that $h(x_1) = h(x_2)$ for $x_1, x_2 \in (a, b)$, or as I have phrased it, $F(x) = f(x) - g(x)$ is constant.</p>
|
2,601,088 | <p>I'm new to the group theory and want to get familar with the theorems in it, so I choose a number $52$
to try making some obseveration on all group that has this rank. Below are my thoughts. I don't know if there is any better way to think of these (i.e., an experienced group theorist would think), and I still have some questions not being solved. So I post this.</p>
<p>Let $G$ be a group and $|G|=52$. Since $52$ in not prime, then there is no theorem guarantee that $G$ must be cyclic. On the other hand, since $\Bbb Z_{52}$ is cyclic, so what we can know is only that "there exists a group $G$ with rank $52$ that is cyclic." Besides, $52$ is not of the form $pq$, where $p,~q$ are prime numbers, so again we can not use such theorem to say that $G$ must be cyclic. (Also, $52$ is not of the form $p^2$, so we're unable to conclude that $G$ is abelian anyway.)</p>
<p>So there remains a question: can $G$ be non-cyclic? I don't know how to answer it.</p>
<p>Next, move on to the obseverations on its subgroup. If $G$ is cyclic (i.e., isomorphic to $\Bbb Z_{52}$), then by a theorem that I don't know its name, $G$ has <em>exactly</em> one nontrival proper subgroup of rank $2$ (also $4,~13,~26$, respectively). By Lagrange Theorem we know that these $4$ subgroup, plus the $\{e\}$ and $G$, are the all $6$ subgroups of $G$. However, since I don't know whether $G$ can be non-cyclic, so I discuss such case below. The prime divisors of $52$ are $2,~13$. By Cauchy theorem, there exists at least one element that has order $2$ and $13$ respectively. So we know that there are two subgroups of $G$ with rank $2$ and $13$ respectively (but maybe $G$ has more subgroup with these ranks). However, how about subgroup with rank $26$ and $4$, etc.? Is their a way to tell whether such subgroup exists?</p>
| BallBoy | 512,865 | <p>Are you familiar with the Sylow theorems? They are the most commonly used tool to answer questions about groups of a given order. (For example, they will immediately tell you that a subgroup of order $4$ must exist.) Look up the Sylow theorems, and if you search "groups of order $x$" for many different values of $x$ you can find several discussions of this type (many on this site)></p>
|
2,371,108 | <p>Cubic equations of the form $ax^3+bx^2+cx+d$ can be solved in various ways. Some are easy to easy to factor in a pair, for some the roots can be found out by trial-and-error, some are one-of-a-kind, some can be reduced to a quadratic equation. A compilation of all possible ways to solve cubic equations would be very helpful for students and learners. </p>
| Mathlover | 22,430 | <p>$ax^3+bx^2+cx+d=0$</p>
<ol>
<li>Divide to a to get $x^3$ term 1</li>
</ol>
<p>$$x^3+\frac{b}{a}x^2+\frac{c}{a} x+\frac{d}{a}=0$$</p>
<p>$$x^3+b_1 x^2+c_1 x+d_1=0$$</p>
<ol start="2">
<li>Eliminate $x^2$ term via using $x=z-\frac{b_1}{3}$ transform
Then you will get </li>
</ol>
<p>$$(z-\frac{b_1}{3})^3+b_1(z-\frac{b_1}{3})^2+c_1(z-\frac{b_1}{3})+d_1=0$$</p>
<p>$$z^3+c_2 z+d_2=0$$</p>
<ol start="3">
<li>Use binom expansion
$$(p+y)^3=p^3+3p^2y+3py^2+y^3$$
Reorder the equation as:
$$(p+y)^3-3py(p+y)-(p^3+y^3)=0$$</li>
</ol>
<p>Define:
$p+y=z$ Thus </p>
<p>$$3.p.y=-c_2$$<br>
$$p^3+y^3=-d_2$$ </p>
<p>$$p=-\frac{c_2}{3y}$$</p>
<p>$$-(\frac{c_2}{3y})^3+y^3=-d_2$$</p>
<p>Then solve the quadratic equation with after $y^3=m$</p>
<p>$$-\frac{c^3_2}{27m}+m=-d_2$$</p>
<p>$$m^2+d_2.m-\frac{c^3_2}{27}=0$$</p>
<ol start="4">
<li>The qadratic equation can be solved via $m=s-\frac{d_2}{2}$</li>
</ol>
<p>$$(s-\frac{d_2}{2})^2+d_2.(s-\frac{d_2}{2})-\frac{c^3_2}{27}=0$$</p>
<p>$$s^2=\frac{c^3_2}{27}+\frac{d^2_2}{2}=\Delta$$</p>
<p>Write $$s=\sqrt{\Delta}$$</p>
<p>$$m=\sqrt{\Delta}-\frac{d_2}{2}$$
$$y^3=m$$
$$y=\sqrt[3]{\sqrt{\Delta}-\frac{d_2}{2}}$$</p>
<p>$$p=-\frac{c_2}{3y}$$
$$p=-\frac{c_2}{3\sqrt[3]{\sqrt{\Delta}-\frac{d_2}{2}}}$$
$$p+y=z$$
$$x=\sqrt[3]{\sqrt{\Delta}-\frac{d_2}{2}}-\frac{c_2}{3\sqrt[3]{\sqrt{\Delta}-\frac{d_2}{2}}}-\frac{b_1}{3}$$ </p>
|
1,040,932 | <p>I have a system of congruence equations</p>
<p>$$
\begin{cases}
x \equiv 17 \pmod{15} \\
x \equiv 14 \pmod{33}
\end{cases}
$$</p>
<p>I need to investigate the system and see if they've got any solutions.</p>
<p>I know that I should use the Chinese remainder theorem "in a reverse order" so I think I should split each congruence equation in two new systems of two congruence equations.</p>
<p>From the CRT two congruence equations can be joined in a single congruence equation by</p>
<p>$$
x \equiv b_1 + c n_1 (b_2 - b_1) \pmod{n_1 n_2}
$$</p>
<p>From the first congruence equation I can get these two
$$
b_1 + c n_1 (b_2 - b_1) = 17 \\
n_1 n_2 = 15
$$</p>
<p>and from the second I can get
$$
b_1 + cn_1 (b_2 - b_1) = 14 \\
n_1 n_2 = 33
$$</p>
<p>but the unknown variables are not combined so I cannot just solve the system of four equations.</p>
<p>I need a hint :-)</p>
| user194150 | 194,150 | <p>$
\left\{ \begin{array}{l}
x \equiv 17\left[ {15} \right] \\
x \equiv 14\left[ {33} \right] \\
\end{array} \right. \Rightarrow x \equiv 47\left[ {495} \right]
$</p>
|
1,462,379 | <p>I have been given the task to compute $\int_{-1}^1 \sqrt{1-x^2} dx$ by means of calculus. We got the hint to substitute $x=\sin u$, but that only seems to make things more complicated:</p>
<p>$$\int_{-1}^1 \sqrt{1-x^2}dx = \int_{\arcsin-1}^{\arcsin1} \sqrt{1-\sin^2u} \frac{d\arcsin u}{du} du = \int_{\arcsin-1}^{\arcsin1} \sqrt{\frac{(1-\sin u)(1+\sin u)}{(1-u)(1+u)}} du.$$</p>
<p>I know that the answer should be $\frac\pi2$, because this is the area under one half of the unit circle, but how to arrive there by means of calculus is completely unclear to me. Could someone point me in the right direction?</p>
| Jack D'Aurizio | 44,121 | <p>$$\begin{align*} \int_{-1}^{1}\sqrt{1-x^2}\,dx &= 2\int_{0}^{1}\sqrt{1-x^2}\,dx \\&= 2\int_{0}^{\pi/2}\cos^2\theta\,d\theta\\&=\int_{0}^{\pi/2}(\cos(2\theta)+1)\,d\theta\\&=\int_{0}^{\pi/2}1\,d\theta=\color{red}{\frac{\pi}{2}}\end{align*} $$
as expected. We exploited the parity of the function $\sqrt{1-x^2}$, the substitution $x=\sin\theta$, the duplication formula for the cosine function and the periodicity of the cosine function.</p>
|
3,438,653 | <p>I have this thing written on my notes: let <span class="math-container">${x}, {y}\in\mathbb R^n$</span> be two distinct points, then the set <span class="math-container">$$\{ \lambda x + (1-\lambda){y}\;\lvert\; \lambda \in [0,1] \}$$</span> contains all the points on the line segment that connects <span class="math-container">$x$</span> to <span class="math-container">$y$</span>.
I can't seem to understand why it is so, and why we need to require that <span class="math-container">$\lambda \in [0,1]$</span>.
Thanks for any clarification.</p>
| user | 505,767 | <p>Let consider</p>
<p><span class="math-container">$$P(\lambda)=\lambda x + (1-\lambda){y}$$</span></p>
<p>then </p>
<p><span class="math-container">$$P(0)=y, \quad P(1)=x$$</span></p>
<p>and </p>
<p><span class="math-container">$$P(\lambda)-P(0)=\lambda (x-y)$$</span></p>
<p>therefore for <span class="math-container">$\lambda$</span> varing form <span class="math-container">$0$</span> to <span class="math-container">$1$</span>, <span class="math-container">$P(\lambda)$</span> describes linearly all the ponts between <span class="math-container">$y$</span> and <span class="math-container">$x$</span>.</p>
|
336,834 | <p>It is a well known theorem, that every signed measure can be split into its positive and negative parts (Hahn-Jordan-Decomposition). My question is, if something similar is possible for functionals on Sobolev spaces.</p>
<p>To be precise, let $\Omega \subset \mathbb{R}^n$ be some open domain and $\mu \in H^{-1}(\Omega) = H_0^1(\Omega)^*$. Are there $\mu^+, \mu^- \in H^{-1}(\Omega)$, which are positive in the sense that
$$\langle \mu^+, v \rangle \ge 0 \quad\text{for all } v \in H_0^1(\Omega), v \ge 0,$$
(and the same for $\mu^-$) and $\mu = \mu^+ - \mu^-$?</p>
<p>(By duality arguments, one obtains, that the set of all differences $\mu^+ - \mu^-$ of positive functionals $\mu^+, \mu^-$ is dense in $H^{-1}(\Omega)$.)</p>
<p><strong>Edit:</strong> Found a very related question on MO: <a href="https://mathoverflow.net/questions/149151/is-any-order-bounded-continuous-linear-functionals-a-difference-of-positive-cont">https://mathoverflow.net/questions/149151/is-any-order-bounded-continuous-linear-functionals-a-difference-of-positive-cont</a></p>
| user67888 | 67,888 | <p>No, such a decomposition is not possible. <a href="http://en.wikipedia.org/wiki/Distribution_%28mathematics%29#Functions_as_distributions">The reason is that</a> </p>
<blockquote>
<p>every distribution which is non-negative on non-negative functions is of the form $\varphi\mapsto \int \varphi \, d\mu$ for some (positive) Radon measure.</p>
</blockquote>
<p>Hence, $\mu=\mu^+-\mu^-$ would be given by the difference of two Radon measures. But the elements of $H^{-1}$ can be more singular than measures. For example, take $n=1$, $\Omega=(-1,1)$, and $f(x)=|x|^{-1/3}$. Then $f\in L^2(\Omega)$, which implies that the distributional derivative $f'$, that is, the functional $v\mapsto -\int_\Omega f(x)v'(x)\,dx $, belongs to $H^{-1}(\Omega)$. If this functional were given by a signed measure $\mu$, that is,
$$-\int_\Omega f(x)v'(x)\,dx = \int_\Omega v(x) d\mu \ \text{ for all }\ v\in H^1_0(\Omega)$$ then after integration by parts on the right we would obtain $f(x)=C+\mu((-1,x))$ a.e., which is impossible because $f$ is not a function of locally bounded variation.</p>
<p>It is instructive to try to decompose $f'$ formally: $$f'(x)=-\frac{1}{3}|x|^{-4/3}\operatorname{sign}x = \mu^+-\mu^-$$ where $$\mu_+=\frac{1}{3}|x|^{-4/3}\chi_{(-1,0)}\quad \text{and}\quad \frac{1}{3}|x|^{-4/3}\chi_{(0,1)}$$
The problem is, neither $\mu_+$ nor $\mu_-$ define bounded linear functionals on $H^1$. These are positive measures that are not locally finite, and therefore cannot be integrated even against nice bump functions. The fact that $f'$ can be (in a sense) integrated against nice functions is due to cancellation between $\mu^+$ and $\mu^-$. The positive and negative parts really need each other. </p>
|
355,740 | <p>Today in class we learned that for exponential functions $f(x) = b^x$ and their derivatives $f'(x)$, the ratio is always constant for any $x$. For example for $f(x) = 2^x$ and its derivative $f'(x) = 2^x \cdot \ln 2$</p>
<p>$$\begin{array}{c | c | c | c}
x & f(x) & f'(x) & \frac{f'(x)}{f(x)}\\ \hline
-1 & \frac{1}{2} & 0.346 & 0.693 \\
0 & 1 & 0.693 & 0.693 \\
1 & 2 & 1.38 & 0.693\\
2 & 4 & 2.77 &0.693&
\end{array}$$ </p>
<p>So as you can see, the ratio is the same and this is true for all functions of the form $b^x$ and its derivative. So my question is, why is the ratio always constant? Is there some proof or logic behind it or is <em>just like that</em>? Furthermore, what's the use of knowing this?</p>
<p>EDIT:</p>
<p>It seems that I have missed the fairly simple </p>
<p>$$\require{cancel}\frac{f^\prime(x)}{f(x)}=\frac{\cancel{{b^x}}\ln\,b}{\cancel{{b^x}}}=\ln\,b$$</p>
<p>But what's the use knowing and learning this? Will this reduce a step in the future or help solve a much harder problem more easily? </p>
| Zev Chonoles | 264 | <p>Given a ring $R$, if
$$\underbrace{1_R+\cdots+1_R}_{n\text{ times}}=0_R,$$
then for any $a\in R$, we have (by the distributive property) that
$$\underbrace{a+\cdots+a}_{n\text{ times}}=a\cdot\underbrace{\left(1_R+\cdots+1_R\right)}_{n\text{ times}}=a\cdot 0_R=0_R,$$
so whatever the number is for the multiplicative identity, that is an upper bound on the number for any other element.</p>
<p>In the ring $R=(\mathbb{Z}/n_1\mathbb{Z})\times\cdots\times (\mathbb{Z}/n_k\mathbb{Z})$, the element
$$e_i=(\overline{0},\ldots,\underset{\substack{i\text{th}\\\text{place}}}{\overline{1}},\ldots,\overline{0})$$
has the property that
$$\underbrace{e_i+\cdots+e_i}_{n_i\text{ times}}=(\overline{0},\ldots,\overline{0})=0_R,$$
so you can pretty much make a ring with any collection of such numbers you want.</p>
<hr>
<p>Also, I second spin's suggestion that this be referred to as the "additive order" of an element.</p>
|
20,314 | <p>Hi all.
I'm looking for english books with a good coverage of distribution theory.
I'm a fan of Folland's Real analysis, but it only gives elementary notions on distributions.
Thanks in advance.</p>
| Rasmus | 1,291 | <p>Gel'fand, I. M. and Shilov, G. E.: <em>Generalized Functions</em></p>
|
20,314 | <p>Hi all.
I'm looking for english books with a good coverage of distribution theory.
I'm a fan of Folland's Real analysis, but it only gives elementary notions on distributions.
Thanks in advance.</p>
| Piero D'Ancona | 7,294 | <p>What do you need distributions for? Your request is strange, PDEs are the fundamental application, the origin, and the main source of examples for distribution theory, so no surprise all the books on distributions after a while steer to PDEs.</p>
<p>Thus maybe my advice is misguided since I do not understand your needs. Anyway, in my opinion the best introduction to distributions is a nice little collection of exercises written by Claude Zuily some years ago (Problems in distributions, North Holland). If you finish it you will be familiar with all the basic theory and you'll be ready to delve into the intricacies, which can be challenging (see the first volume of Hormander, which is essentially a treatise on distributions, or the fear-inducing first volume of John Horvath with its fourteen different topologies on spaces in duality :)</p>
|
20,314 | <p>Hi all.
I'm looking for english books with a good coverage of distribution theory.
I'm a fan of Folland's Real analysis, but it only gives elementary notions on distributions.
Thanks in advance.</p>
| Bon Clarke | 11,157 | <p>Two very readable, wide ranging and well motivated accounts are "Generalised Functions and Partial Differential Equations" by Georgi E. Shilov, published by Gordon and Breach 1968, and "Advanced Mathematical Analysis" by Richard Beals, published by Springer 1973 (International student edition). Both are unfortunately out of print and I keep hoping Dover will pick them up so I can recommend them. A recent advanced textbook is "Distributions and Operators" by Gerd Grubb, published by Springer 2009 Vol 252 GTM.</p>
|
4,291,864 | <p>I have the following equation:</p>
<p><span class="math-container">$y=\frac{3x}{x^{2}+1}$</span></p>
<p>and I want to obtain x in terms of y, so far what I have done is the following:</p>
<p><span class="math-container">$3x=y(x^{2}+1)$</span></p>
<p><span class="math-container">$3x=x^{2}y+y$</span></p>
<p><span class="math-container">$3x-x^{2}y=y$</span></p>
<p>and at this point I got stucked, because that cuadratic term would not allow me to just get one x in one side of the equation. Any advice? Thanks</p>
| Henno Brandsma | 4,280 | <p>For every open set <span class="math-container">$O$</span> in a topology <span class="math-container">$(X, \mathcal T)$</span>, we form its complement. These sets together are called the closed sets of<span class="math-container">$X$</span>. So <span class="math-container">$\emptyset = X^\complement$</span> and <span class="math-container">$X = \emptyset^\complement$</span> are closed sets ( in any topology). If <span class="math-container">$O$</span> is any other open set in the cofinite topology, by definition <span class="math-container">$O^\complement$</span> is finite and conversely any finite subset is the complement of an open set and so itself closed.</p>
<p>So all closed sets for the cofinite topology are: all finite sets (including <span class="math-container">$\emptyset$</span>) plus <span class="math-container">$X$</span>. So if <span class="math-container">$A$</span> is any infinite set it only has <strong>one</strong> closed superset namely <span class="math-container">$X$</span>. So the closure of <span class="math-container">$A$</span> is <span class="math-container">$X$</span>, there are no other candidates.</p>
|
662,403 | <p>I'm working on a homework assignment concerning convex optimization and I came across a problem involving the convexity of the function and the convexity of the domain of the function.</p>
<p>Consider the function $f : [0,1]^3 \in R$ with the following form
$$
f(x,y,z) = xlnx + ylnz + zlnz + \alpha ( x + y + z - 1)
$$</p>
<p>The gradient</p>
<p>$$
\nabla f(x,y,z) = (lnx + \alpha + 1)\vec{i} + (lny + \alpha + 1)\vec{j} + (lnz + \alpha + 1)\vec{k}
$$</p>
<p>The hessian matrix:</p>
<p>$$
\left[
\begin{array}{ccc}
\frac{1}{x} & 0 & 0 \\
0 & \frac{1}{y} & 0 \\
0 & 0 & \frac{1}{z} \\
\end{array}
\right]
$$</p>
<p>Since for the interior of the function of which is $[0,1]$ does it follow that the epigraphs that are formed would be positive semidefinite since the hessian matrix is positive definite?</p>
<p>And therefore would the domain of the function also be convex since the cartesian product of the epigraphs would result in a convex set?</p>
| Nick Alger | 3,060 | <ol>
<li><p>This is a really interesting function as it is the Lagrangian for <a href="http://en.wikipedia.org/wiki/Principle_of_maximum_entropy" rel="nofollow">constrained maximum information entropy</a>, of, for example, a probability distribution on a set with 3 points.</p></li>
<li><p>One can see that $f$ is convex since it is the sum of convex functions. </p></li>
<li><p>I'm not clear what you are trying to get at in terms of the Hessian. Positive semi-definiteness of the Hessian at all points in the domain does imply convexity of a function, so that would be an alternative proof of convexity. See Theorem 3.37 <a href="http://books.google.com/books?id=nDYz-NIpIuEC&pg=PT110&lpg=PT110&dq=positive%20definite%20epigraph&source=bl&ots=qLoN0kwhvo&sig=6fWIMeF5UQux4D-6Xmb7jpbC2_E&hl=en&sa=X&ei=y_jvUo-BL-jcyQHJi4CQAQ&ved=0CDQQ6AEwAw#v=onepage&q=positive%20definite%20epigraph&f=false" rel="nofollow">here</a> for example.</p></li>
<li><p>One can show that the domain $[0,1]^3$ directly through the definition of convexity - pick two points, find a third directly inbetween them, and verify that the third point is also in the domain. It's the unit cube.</p></li>
</ol>
<p><a href="http://en.wikipedia.org/wiki/Convex_set" rel="nofollow">Convex Set</a></p>
<p><a href="http://en.wikipedia.org/wiki/Convex_function" rel="nofollow">Convex Function</a></p>
|
1,125,842 | <p>In $\sf ZFC$ we have the axiom of infinity and thus can define the natural numbers $$\mathbb N \equiv \bigcap\{X:\emptyset\in X\land \forall n(n\in X\implies n\cup\{n\}\in X)\}.$$ From this it's not particularly hard (exercises 1.6 and 1.7 in Jech - <em>Set Theory</em>) to prove that, firstly, every $n\in\mathbb N$ obeys foundation and secondly, $\mathbb N$ itself obeys it, all without explicitly using the axiom of foundation. My question is: does this extend to arbitrary finite sets or even other countable sets? If not, are there any other specific important examples for which we can verify foundation?</p>
| hmakholm left over Monica | 14,366 | <p>All the sets we define and use in the development of ordinary mathematics are automatically well-founded. Essentially, any set you can prove <em>exists</em> can also be proven to be well-founded without using the axiom of foundation.</p>
<p>The role of the axiom of foundation is not to tell us anything <em>new</em> about the sets we're already using, but merely to claim that <em>there are no other sets</em> than the well-founded ones. It puts a limit on which kinds of wild, spurious things may exist in a model of set theory without our asking for them, which is sometimes technically convenient -- but it doesn't say anything we don't already know about the sets we <em>want</em> to exist.</p>
|
753,881 | <p>I want to know some typical forms of system of equations generating from practical problems in engineering/economics/physics,etc.</p>
<p>Some examples or research articles would be good.</p>
<p>Specifically, I am looking for some examples of nonlinear system of equations generated from practical problems.</p>
<p>Thanks.</p>
| ml0105 | 135,298 | <p>So you have four odd numbers. They must occupy slots $2, 4, 6, 8$. So you permute $\{1, 1, 1, 3\}$. There are $\dfrac{4!}{3!}$ ways to do this by the multinomial distribution.</p>
<p>Now we permute $\{2, 2, 4, 4\}$ in slots $1, 3, 5, 7$. There are $\dfrac{4!}{2! * 2!}$ ways to do this by the multinomial distribution.</p>
<p>By rule of product, we multiply: $\dfrac{4! * 4!}{3! * 2! * 2!}$.</p>
|
1,762,268 | <p>Let $X$ be a Hausdorff space and let $f:X\to \mathbb{R}$. If grapph of $f$ is compact we have to show that $f$ is continuous. </p>
<p>Since every closed subset of a Hausdorff space is closed, therefore grapph of $f$ is closed. WE know that if $f:X\to Y$ and $Y$ is compact, then graph of $f$ is clsed implies $f$ is continuous. But here $\mathbb R$ is not compact. Please help!</p>
| Forever Mozart | 21,137 | <p>Proof without nets:</p>
<p>Let $G$ be the graph of $f$. </p>
<p>Suppose $G$ is compact. </p>
<p>Let $A$ be closed in $\mathbb R$. We show $f^{-1} [A]$ is closed.</p>
<p>Note that $f^{-1}[A]=\pi_X [G\cap (X\times A)]$. </p>
<p>Since $G\cap (X\times A)$ is compact and the projection $\pi _X$ is continuous, $f^{-1}[A]$ is compact. </p>
<p>$X$ is Hausdorff, so every compact subset of $X$ is closed. DONE! </p>
|
422,799 | <p>Maschke's theorem says that every <em>finite-dimensional</em> representation of a finite group is completely reducible. Is there a simple example of an infinite-dimensional representation of a finite group which is not completely reducible?</p>
<p>EDIT: As mentioned in the answers, there is actually no finite-dimensional caveat in Maschke's theorem. It seems that I just got unlucky, in the that the first couple of references I found included a finite-dimensional assumption.</p>
| Keenan Kidwell | 628 | <p>For a ring $R$, not necessarily commutative, but with a multiplicative identity, a non-zero $R$-module is a direct sum of simple modules if and only if it is a sum of simple modules, if and only if every submodule has a complement. A representation of a finite group $G$ over $\mathbf{C}$ is the same thing as a $\mathbf{C}[G]$-module. Every $\mathbf{C}[G]$-module $M$, whether finite dimensional or not, has the property that every submodule has a complement (the proof of Maschke's theorem does not use that the module is finite dimensional, only that $G$ is finite). So by the quoted result, every $\mathbf{C}[G]$-module is a direct sum of simple submodules, that is, every complex representation of $G$ is completely reducible.</p>
|
422,799 | <p>Maschke's theorem says that every <em>finite-dimensional</em> representation of a finite group is completely reducible. Is there a simple example of an infinite-dimensional representation of a finite group which is not completely reducible?</p>
<p>EDIT: As mentioned in the answers, there is actually no finite-dimensional caveat in Maschke's theorem. It seems that I just got unlucky, in the that the first couple of references I found included a finite-dimensional assumption.</p>
| rschwieb | 29,335 | <p>I was not able to find any reference for Maschke's theorem talking <em>only</em> about the finite dimensional representations.</p>
<p>The "modern" statement of Maschke's theorem (or at lest, the one ring theorists like) is this:</p>
<blockquote>
<p>For any commutative ring $R$ and finite group $G$, the group ring $R[G]$ is <a href="http://en.wikipedia.org/wiki/Semisimple_ring#Semisimple_rings">semisimple</a> iff $R$ is semisimple and the order of $G$ is a unit in $R$.</p>
</blockquote>
<p>Semisimplicity, linked above, is the condition where all right $R$ modules split into sums of simple modules. This is the corresponding fact to the result in representation theory about all representations being completely reducible.</p>
<p>Now if $R$ is a field of characteristic zero, the order of $G$ is going to be a unit no matter what order it has, and furthermore, fields are semisimple. So in particular, you have a corollary:</p>
<blockquote>
<p>For any field $F$ of characteristic 0 and finite group $G$, the group ring $F[G]$ is semisimple, and hence all of its modules (=representations of $G$) are completely reducible.</p>
</blockquote>
<p>When the order of $G$ divides the characteristic of a field $F$, $F[G]$ <em>does</em> have representations that are not completely reducible. The easiest example in that case would have to be $F[G]$ itself, which necessarily has a nonzero Jacobson radical.</p>
<p>As a toy example, you could take the cyclic group of order two $C_2=\{1,c\}$ and the field $F_2$ of order two, and consider its group algebra $F_2(C_2)$. You get a ring of four elements $\{0,1,c,1+c\}$. You can see that $(c+1)^2=0$, so that $\{0,c+1\}$ is the nilradical (which is the Jacobson radical in this case).</p>
|
873,992 | <p>I am having problem with the onto part of this problem.</p>
<p>$\mathbb{N}\rightarrow \mathbb{E}$</p>
<p>My function or pattern is </p>
<p>$x \rightarrow f(x)=2x$ </p>
<p>Which take my natural to even.</p>
<p><strong>One to One</strong></p>
<p>$f(x)=f(y)$</p>
<p>$2x=2y$</p>
<p>$x=y$</p>
<p><strong>Onto</strong></p>
<p>Let</p>
<p>$n\in E$</p>
<p>Not I have to make $x$ equal something when plug into $f(x)$ that will give n ,for all n in even.</p>
<p>And I cannot say let $x=\frac{1}{2}n$</p>
<p>Because the domain is a natural one.</p>
| amWhy | 9,003 | <p>You can approach this literally, more or less. We are being told that there does not exist an $x$ such that $x$ is both naive ($N(x)$) and bad ($B(x)$).</p>
<p>It makes sense to let "all people" to be the domain (or universe) in which $x$ resides. </p>
<p>$$\begin{align} \lnot \exists x(N(x) \land B(x)) & \equiv \forall x \lnot(N(x) \land B(x)) \\ \\&\equiv \forall x( \lnot N(x) \lor \lnot B(x))\\ \\ &\equiv \forall x (N(x) \rightarrow \lnot B(x))\end{align}$$</p>
|
3,077,084 | <blockquote>
<p>If <span class="math-container">$a,b,c>0.$</span> Then minimum value of</p>
<p><span class="math-container">$(8a^2+b^2+c^2)\cdot (a^{-1}+b^{-1}+c^{-1})^2$</span></p>
</blockquote>
<p>Try: Arithmetic geometric inequality</p>
<p><span class="math-container">$8a^2+b^2+c^2\geq 3\cdot 2\sqrt{2}(abc)^{1/3}$</span></p>
<p>and <span class="math-container">$(a^{-1}+b^{-1}+c^{-1})\geq 3(abc)^{-1/3}$</span></p>
<p>so <span class="math-container">$(8a^2+b^2+c^2)\cdot (a^{-1}+b^{-1}+c^{-1})^2\geq 18\sqrt{2}(abc)^{-1/3}$</span></p>
<p>could some help me to solve it. answer is <span class="math-container">$64$</span></p>
| Macavity | 58,320 | <p><strong>Hint:</strong> Another way is to consider <a href="https://en.wikipedia.org/wiki/H%C3%B6lder%27s_inequality#Counting_measure" rel="nofollow noreferrer">Hölder's Inequality</a></p>
<p><span class="math-container">$$(8a^2+b^2+c^2)(a^{-1}+b^{-1}+c^{-1})^2 \geqslant (2+1+1)^3$$</span></p>
<p>Equality is possible when <span class="math-container">$8a^3=b^3=c^3=1$</span> (why?), so this is the minimum.</p>
|
44,562 | <p>The question is motivated from the definition of $C^r(\Omega)$ I learned from S.S.Chern's <em>Lectures on Differential Geometry</em>:</p>
<p>Suppose $f$ is a real-valued function defined on an open set $\Omega\subset{\bf R}^m$. If all the $k$-th order partial derivatives of $f$ exist and are continuous for $k\leq r$, then we say $f\in C^r(\Omega)$. Here $r$ is some positive integer.</p>
<p>While in Folland's <em>Introduction to Partial Differential Equations</em>, $C^r(\Omega)$ denotes the space of functions possessing continuous derivatives up to order $r$ on $\Omega$, where $\Omega$ is an open subset of ${\bf R}^m$ and $k$ is a positive integer.</p>
<p>It's trivial to show that these two definitions are equivalent when $m=1$. So here is my question:</p>
<blockquote>
<p>Are these two definitions equivalent in the higher dimensions? How to prove it?</p>
</blockquote>
<hr>
<p><strong>Edit</strong>: The question was partially answered by Didier <a href="https://math.stackexchange.com/questions/44355/can-being-differentiable-imply-having-continuous-partial-derivatives/44364#44364">here</a>. I do not think it is trivial for me. It boils down to the following one:</p>
<blockquote>
<p>In the higher dimension case, say,
$f:{\bf R}^n\to {\bf R}$($n\geq 2$), what's
the relationship between the higher
order partial derivative of $f$ and
the high order derivative $f^{(k)}$?</p>
</blockquote>
<p>When $1\leq k\leq 2$, $f^{(k)}$ is the <a href="http://en.wikipedia.org/wiki/Gradient" rel="nofollow noreferrer">gradient</a> and the <a href="http://en.wikipedia.org/wiki/Hessian_matrix" rel="nofollow noreferrer">Hessian matrix</a> respectively. The answer to the above questions in these two cases is clear. I have no idea for $k>2$. I don't know much about tensor, I'm not sure if the question is related to the topic of multilinear algebra.</p>
| Asaf Karagila | 622 | <p>Suppose $A\subseteq\mathbb R$ can be well-ordered by the usual $<$, and fix an enumeration of the rationals, i.e. $\mathbb Q = \langle q_n\mid n\in\omega\rangle$.</p>
<p>For $a\in A$ denote $S(a)$ the successor of $a$ in $A$, if $b\in A$ is a maximal element of $A$ then $S(b) = b+1$.</p>
<p>For every $a\in A$ set $q_a$ the least rational $q_n$ in the enumeration, such that $a< q_n< S(a)$. Since $a\neq S(a)$ we have that $\mathbb Q\cap (a,S(a))$ is non-empty, therefore such minimal element exists.</p>
<p>We prove that this is indeed one to one, suppose not then for some $a, b\in A$ such that $a\le b$ we have $q_a=q_b$. By the choice of $q_a$ we have that $a<q_a=q_b<S(a)$, therefore $a\le b<q_b=q_a<S(a)$. Since the $S(a)$ is the least element of $A$ such that $a<S(a)$, we have that $b\le a$ therefore $a=b$.</p>
<p>We found an injective function of $A$ into a countable set, therefore it is countable.</p>
|
44,562 | <p>The question is motivated from the definition of $C^r(\Omega)$ I learned from S.S.Chern's <em>Lectures on Differential Geometry</em>:</p>
<p>Suppose $f$ is a real-valued function defined on an open set $\Omega\subset{\bf R}^m$. If all the $k$-th order partial derivatives of $f$ exist and are continuous for $k\leq r$, then we say $f\in C^r(\Omega)$. Here $r$ is some positive integer.</p>
<p>While in Folland's <em>Introduction to Partial Differential Equations</em>, $C^r(\Omega)$ denotes the space of functions possessing continuous derivatives up to order $r$ on $\Omega$, where $\Omega$ is an open subset of ${\bf R}^m$ and $k$ is a positive integer.</p>
<p>It's trivial to show that these two definitions are equivalent when $m=1$. So here is my question:</p>
<blockquote>
<p>Are these two definitions equivalent in the higher dimensions? How to prove it?</p>
</blockquote>
<hr>
<p><strong>Edit</strong>: The question was partially answered by Didier <a href="https://math.stackexchange.com/questions/44355/can-being-differentiable-imply-having-continuous-partial-derivatives/44364#44364">here</a>. I do not think it is trivial for me. It boils down to the following one:</p>
<blockquote>
<p>In the higher dimension case, say,
$f:{\bf R}^n\to {\bf R}$($n\geq 2$), what's
the relationship between the higher
order partial derivative of $f$ and
the high order derivative $f^{(k)}$?</p>
</blockquote>
<p>When $1\leq k\leq 2$, $f^{(k)}$ is the <a href="http://en.wikipedia.org/wiki/Gradient" rel="nofollow noreferrer">gradient</a> and the <a href="http://en.wikipedia.org/wiki/Hessian_matrix" rel="nofollow noreferrer">Hessian matrix</a> respectively. The answer to the above questions in these two cases is clear. I have no idea for $k>2$. I don't know much about tensor, I'm not sure if the question is related to the topic of multilinear algebra.</p>
| André Nicolas | 6,312 | <p>The following idea uses some set-theoretic machinery, has the advantage of coming from a simple geometric visualization. </p>
<p>Suppose to the contrary that there is an uncountable set $A$ of reals which is well-ordered under the natural order.</p>
<p>Then $A$ is order isomorphic to an uncountable ordinal. It follows that some subset $B$ of $A$ is order isomorphic to the <em>least uncountable ordinal</em> $\omega_1$.</p>
<p>The set $B$ has a least element. By shifting if necessary, we can make that least element $0$. Either $B$ is bounded above or it isn't. If it is bounded above, let $m$ be the least upper bound. Note that since $B$ is order-isomorphic to $\omega_1$, $m$ cannot be in $B$. Then the map that takes $x$ to $x/(m-x)$ is order preserving, and "stretches" $B$ so that for every integer $n$, there is $b\in B$ such that $b>n$. (We have kept the name $B$ for the stretched set.)</p>
<p>So we can assume that $B$ has smallest element $0$, and that for every integer $n$, there is an element of $B$ which is $>n$.</p>
<p>Let $B_n$ be the intersection of $B$ with the interval $[0, n]$. Then $B_n$ is order isomorphic to an initial segment of $\omega_1$. Since $\omega_1$ is the least uncountable ordinal, it follows that $B_n$ is countable.</p>
<p>Since
$$B =\bigcup_{n \in N} B_n$$
we have expressed $B$ as a countable union of countable sets, or equivalently $\omega_1$ as a countable union of countable ordinals. This is impossible.</p>
|
3,652,518 | <p>I am no mathematician but have studied mathematics some 20 years ago. So I know basics of number theory but have lost the skills to solve problems. </p>
<p>I was wondering if the equation <span class="math-container">$3ax^2 + (3a^2+6ac)x-c^3=0$</span> in which <span class="math-container">$a$</span> and <span class="math-container">$c$</span> are arbitrary positive integers and <span class="math-container">$c$</span> is divisible by 6 has an integer solution. </p>
<p>So we need to show there is either no <span class="math-container">$a$</span>, <span class="math-container">$c$</span> and <span class="math-container">$x$</span> all positive that satisfy the equation, or one counter example exist. </p>
<p>Rational root theorem didn't help me. </p>
<p>Thanks. </p>
| Jingeon An-Lacroix | 471,868 | <p>Let <span class="math-container">$$\alpha:=x+\cdots+x^n.$$</span> Then <span class="math-container">$$\alpha x=x^2+\cdots+x^{n+1}.$$</span> Therefore, <span class="math-container">$$\alpha(x-1)=x^{n+1}-x=x(x^n-1).$$</span> If <span class="math-container">$x\neq 1$</span>, divide both sides with <span class="math-container">$x-1$</span>, so <span class="math-container">$$\alpha=x\frac{x^n-1}{x-1},$$</span> the result you want. If <span class="math-container">$x=1$</span>, <span class="math-container">$\alpha=n$</span>.</p>
|
2,706,872 | <p>I'm working on my latest linear algebra assignment and one question is as follows: </p>
<p>In $\mathbb R^3$ let <em>R</em> be the reflection over the null space of the matrix </p>
<p><em>A</em> = [4 4 5]</p>
<p>Find the matrix which represents <em>R</em> using standard coordinates. </p>
<p>I am familiar with the fact that the matrix for the orthogonal projection onto a subspace is given by [<em>P</em>] = <em>A</em>(<em>$A^TA$)$A^T$</em>, where <em>A</em> is a matrix with columns that form a basis for the space in question. However, I am not familiar with the concept of a "reflection over the null space". How does this compare with the matrix for the orthogonal projection onto the null space? Is it a similar process?
Thanks for any help.</p>
| RCT | 424,406 | <p>I could imagine that there's some ugly formula for this, but I don't know it, and it's probably more instructive to solve the problem from basic principles anyway, so let's try that! My first thought when reading this problem is to start by changing to a basis in which it's obvious what the reflection does.</p>
<p>Note that multiplication by $A$ is just dotting with the vector $(4,4,5),$ so the kernel of $A$ is precisely $(4,4,5)^\perp.$ So $(4,4,5)$ is orthogonal to the plane of reflection, and thus the reflection simply negates it. </p>
<p>Now extend $\{(4,4,5)\}$ to a basis for $\mathbb{R}^3$ by picking a basis for the null space itself. For example, $(0,-5,4)$ and $(-5,0,4)$ are two independent vectors killed by $A.$ Since these live <em>in</em> the plane of reflection, they are fixed by the reflection.</p>
<p>So in our new basis, the matrix for the reflection is simply $$R' = \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}.$$</p>
<p>Now to write the matrix in the standard basis, all you have to do is compute the inverse of the change of basis matrix $$P = \begin{bmatrix} 4 & 0 & -5 \\ 4 & -5 & 0 \\ 5 & 4 & 4 \end{bmatrix}$$ that took the standard basis to our preferred basis. It ends up being $$P^{-1} = \frac{1}{285}\begin{bmatrix} 20 & 20 & 25 \\ 16 & -41 & 20 \\ -41 & 16 & 20 \end{bmatrix}.$$</p>
<p>So $R$ is obtained by changing to the new basis, applying $R',$ then changing back to the standard basis, i.e. $R = P^{-1}R'P.$</p>
|
804,483 | <p>The following integrals look like they might have a closed form, but Mathematica could not find one. Can they be calculated, perhaps by differentiating under the integral sign?</p>
<p>$$I_1 = \int_{-\infty }^{\infty } \frac{\sin (x)}{x \cosh (x)} \, dx$$
$$I_2 = \int_{-\infty }^{\infty } \frac{\sin ^2(x)}{x \sinh (x)} \, dx$$</p>
| Urgje | 95,681 | <p>I do not know of a closed form but it seems that the integral can be
converted into an infinite series. Let
$$
I_{1}(k)=\int_{-\infty }^{+\infty }dx\frac{\sin kx}{x\cosh x}
$$
so $I_{1}=I_{1}(1).$ Now (note that $I_{1}(0)$ vanishes)
\begin{eqnarray*}
\partial _{k}I_{1}(k) &=&\int_{-\infty }^{+\infty }dx\frac{\cos kx}{\cosh x}=%
{Re}\int_{-\infty }^{+\infty }dx\frac{\exp [ikx]}{\cosh x}\\&=&2{Re}%
\int_{-\infty }^{+\infty }dx\exp [ikx]\frac{1}{\exp [x]+\exp [-x]} \\
I_{1} &=&\int_{0}^{1}dk\partial _{k}I_{1}(k)
\end{eqnarray*}
If this Fourier transform is known we can try to integrate over $k$ and
obtain $I_{1}(k)$ and hence $I_{1}$. If not we note that $(\exp [z]+\exp
[-z])^{-1}$ has poles $z_{n}=i\pi (1+2n)$ so closing the contour in the
upper half plane results in
$$
\int_{-\infty }^{+\infty }dx\exp [ikx]\frac{1}{\exp [x]+\exp [-x]}=2\pi
i\sum_{n}r_{n}\exp [ikz_{n}]
$$
where the $r_{n}$ are the residues and the sum over $n$ is that with $z_{n}$
in the upper half plane. Now the $k$-integral can be done and we end up with
a sum over $n$.</p>
|
804,483 | <p>The following integrals look like they might have a closed form, but Mathematica could not find one. Can they be calculated, perhaps by differentiating under the integral sign?</p>
<p>$$I_1 = \int_{-\infty }^{\infty } \frac{\sin (x)}{x \cosh (x)} \, dx$$
$$I_2 = \int_{-\infty }^{\infty } \frac{\sin ^2(x)}{x \sinh (x)} \, dx$$</p>
| Felix Marin | 85,343 | <p>$\newcommand{\+}{^{\dagger}}
\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
\newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
\newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
\newcommand{\dd}{{\rm d}}
\newcommand{\down}{\downarrow}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
\newcommand{\fermi}{\,{\rm f}}
\newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
\newcommand{\half}{{1 \over 2}}
\newcommand{\ic}{{\rm i}}
\newcommand{\iff}{\Longleftrightarrow}
\newcommand{\imp}{\Longrightarrow}
\newcommand{\isdiv}{\,\left.\right\vert\,}
\newcommand{\ket}[1]{\left\vert #1\right\rangle}
\newcommand{\ol}[1]{\overline{#1}}
\newcommand{\pars}[1]{\left(\, #1 \,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\pp}{{\cal P}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
\newcommand{\sech}{\,{\rm sech}}
\newcommand{\sgn}{\,{\rm sgn}}
\newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
\newcommand{\ul}[1]{\underline{#1}}
\newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
\newcommand{\wt}[1]{\widetilde{#1}}$</p>
<ol><li>$\ds{\large\color{#000}{\tt I_{1}\equiv
\int_{-\infty }^{\infty }{\sin\pars{x} \over x\cosh\pars{x}} \,\dd x}}$
<blockquote>
\begin{align}I_{1}&\equiv\color{#c00000}{
\int_{-\infty }^{\infty }{\sin\pars{x} \over x\cosh\pars{x}} \,\dd x}
=\int_{-\infty }^{\infty }
{2\expo{x} \over \expo{2x} + 1}\int_{0}^{1}\cos\pars{kx}\,\dd k\,\dd x
\\[3mm]&=2\Re\int_{0}^{1}\
\overbrace{\int_{-\infty}^{\infty}
{\pars{\expo{x}}^{\verts{k}\ic} \over \expo{2x} + 1}\,\expo{x}\,\dd x}
^{\ds{\mbox{Set}\ \expo{x} \equiv t}}\ \,\dd k
=2\int_{0}^{1}\color{#00f}{%
\Re\int_{0}^{\infty}{t^{\verts{k}\ic} \over t^{2} + 1}\,\dd x}\,\dd k\tag{1}
\end{align}
</blockquote>
\begin{align}&\color{#00f}{%
\Re\int_{0}^{\infty}{t^{\verts{k}\ic} \over t^{2} + 1}\,\dd x}
=\Re\bracks{2\pi\ic\,{\pars{\expo{\pi\ic/2}}^{\verts{k}\ic} \over 2\ic}
+ 2\pi\ic\,{\pars{\expo{3\pi\ic/2}}^{\verts{k}\ic} \over -2\ic}
-\int_{\infty}^{0}
{t^{\verts{k}\ic}\pars{\expo{2\pi\ic}}^{\verts{k}\ic} \over t^{2} + 1}\,\dd t}
\\[3mm]&=\pi\expo{-\pi\verts{k}/2}
-\pi\expo{-3\pi\verts{k}/2}
+\expo{-2\pi\verts{k}}\color{#00f}{%
\Re\int_{0}^{\infty}{t^{\verts{k}\ic} \over t^{2} + 1}\,\dd x}
\\[5mm]&\imp\ \color{#00f}{%
\Re\int_{0}^{\infty}{t^{\verts{k}\ic} \over t^{2} + 1}\,\dd x}
={\pi\expo{-\pi\verts{k}/2}
-\pi\expo{-3\pi\verts{k}/2} \over 1 - \expo{-2\pi\verts{k}}}
=\pi\,{\expo{\pi\verts{k}/2}
-\expo{-\pi\verts{k}/2} \over \expo{\pi\verts{k}} - \expo{-\pi\verts{k}}}
\\[3mm]&=\pi\,{\sinh\pars{\pi\verts{k}/2} \over \sinh\pars{\pi\verts{k}}}
={\pi \over 2}\,\sech\pars{{\pi \over 2}\,k}
\end{align}
<blockquote>
Replacing in $\pars{1}$:
\begin{align}
&\color{#66f}{\large\int_{-\infty }^{\infty }{\sin\pars{x} \over x\cosh\pars{x}} \,\dd x}
=2\int_{0}^{1}{\pi \over 2}\,\sech\pars{{\pi \over 2}\,k}\,\dd k
=\left. 4\arctan\pars{\tanh\pars{{\pi \over 4}\,k}}\right\vert_{0}^{1}
\\[3mm]&=\color{#66f}{\large 4\arctan\pars{\tanh\pars{\pi \over 4}}}
\approx {\tt 2.3218}
\end{align}
This result can be expressed in terms of the <a href="http://en.wikipedia.org/wiki/Gudermannian_function" rel="nofollow">Gudermannian Function</a> $\ds{{\rm gd}\pars{z} \equiv 2\arctan\pars{\expo{z}} - {\pi \over 2}}\quad$
as
$\quad\ds{4\arctan\pars{\tanh\pars{\pi \over 4}} = 2\,{\rm gd}\pars{\pi \over 2}}$.
</blockquote>
</li>
<li>$\ds{\large\color{#000}{\tt I_{2}\equiv
\int_{-\infty }^{\infty }{\sin^{2}\pars{x} \over x\sinh\pars{x}} \,\dd x}}$
<blockquote>
$\tt\mbox{It can be evaluated by following a similar technique}$.
</blockquote>
</li>
</ol>
|
685,681 | <p>I want to prove that $\dim V/(X \cap Y)$ in finite, if $V$ be a vector space and $X$, $Y$ two sub spaces of $V$ such that $\dim V/Y$ and $\dim V/X$ are finite.</p>
| ajd | 90,897 | <p>A much easier argument - consider the map $V\to (V/X)\times (V/Y)$ given by the product of the two projection maps. The kernel of this map is $X\cap Y$. Therefore, we have an injection $V/(X\cap Y)\to (V/X)\times (V/Y)$. Therefore, $\dim V/(X\cap Y)\le \dim [(V/X)\times (V/Y)]=\dim(V/X)\dim(V/Y)<\infty$.</p>
|
878,939 | <p>I have found the eigen vaues, I also know that you can find the eigenvectors through a Gausian Jordan.
-- x1, gauss jordan gives me rows(1 -1/3 ,, 0 0 ), so [a, b] = [1,3]
For vector x2, GJ gives (1 -2/5 ,, 0 0 ), I would assume [a,b] = [2,5], but why did they choose to go with [-2,-5]. I don't get it?</p>
<p>A bigger picture is on this webpage if needed;</p>
<p><a href="http://oi59.tinypic.com/2v7unw1.jpg" rel="nofollow noreferrer">http://oi59.tinypic.com/2v7unw1.jpg</a></p>
<p><img src="https://i.stack.imgur.com/5h7Ix.jpg" alt="enter image description here"></p>
| mfl | 148,513 | <p>Every even integer $n$ can be written as $n=2m,$ for some integer $m.$ Now, there are two possibilities: $m$ is even ($m=2q$) or $m$ is odd ($m=2q+1$) for some integer $q.$ In the first case we have</p>
<p>$$n=2m=4q$$</p>
<p>and in the second case</p>
<p>$$n=2m=2(2q+1)=4q+2.$$</p>
<p>Another way to see why we can express any even number as before is consider the remainder when we divide by $4.$ There are only two possibilities: remainder zero ($n=4q$) and remainder two ($n=4q+2.$)</p>
<p>If you try to do the same with a expression of the form $n=5q+r$ then you don't get any advantage, in general, since the remainder of the division by $5$ can take any value from $0$ to $4=5-1.$ Note that
$$1=5\cdot 0 +\color{blue}{1},2=5\cdot 0 +\color{blue}{2},3=5\cdot 0 +\color{blue}{3},4=5\cdot 0 +\color{blue}{4},5=5\cdot 1 +\color{blue}{0}.$$</p>
|
508,791 | <p>I have an integer list that is <code>n</code> long and each value can be ranging from <code>1 .. n</code>.</p>
<p>I need a formula that tells me how many of all possible lists for a given n, that have one or more consecutive sequences of a length of exactly 2 of the same number and no other consecutive sequences that are longer than 2.</p>
<p>For example for n=5:</p>
<p>These two should count: </p>
<pre><code>{ 1, 1, 5, 3, 3 }
{ 2, 3, 2, 5, 5 }
</code></pre>
<p>Where as these should not:</p>
<pre><code>{ 1, 1, 1, 2, 2 }
{ 1, 3, 2, 5, 4 }
</code></pre>
<p>I've been attempting to do this by looking at the possible sequences using the following formula where x = n-1:</p>
<pre><code>(n) x n n = x * n^3
x (n) x n = x^2 * n^2
n x (n) x = x^2 * n^2
n n x (n) = x * n^3
</code></pre>
<p>And sum these four up.</p>
<p>However, these also needs to take overlaps between the four into account. This is where I could use a bit of help..? What would the formulas be for excluding the overlaps?</p>
<p>An alternative approach would also be welcome.</p>
<p>Going trough all sequences counting manually is not an option - I need this to work for for n larger than what makes that approach computationally feasible.</p>
<p>If it helps anyone, I've written a little program that runs trough all the sequences and counts have the following results:</p>
<pre><code>n = 2
L[2]: 2
L[1]: 2
n = 3
L[3]: 3
L[2]: 12
L[1]: 12
n = 4
L[4]: 4
L[3]: 24
L[2]: 120
L[1]: 108
n = 5
L[5]: 5
L[4]: 40
L[3]: 280
L[2]: 1520
L[1]: 1280
</code></pre>
<p>Where <code>n = 5</code>, <code>L[2]: 1520</code> is the result I've asked for a formula to in the above question.</p>
| miracle173 | 11,206 | <p>How many sequences of length $q$ of numbers from $\{1,...,p\}$ are there such that consecutive elements are always different?
For the first element of such a sequence we can selcet one of the $p$ different values of $\{1,...,p\}$. For the following $q-1$ positions we can select
always all values from $\{1,...,p\}$ except the value of its predecessor in the sequence. These are $p-1$ possible values. So we have</p>
<p>$$ p(p-1)^{q-1}$$</p>
<p>different sequences.</p>
<p>$\Omega_n$ is the number of sequences of length $n$ with elements from $\{1,\ldots,n\}$
such that no three consecutive elements have the same value but at least one pair of
consecutive elements have the same value.</p>
<p>$\Omega_{n,k}$ is the number of sequences of length $n$ with elements from $\{1,\ldots,n\}$
such that no three consecutive elements have the same value but exactly $k$ pairs of
consecutive elements have the same value.</p>
<p>For arbitrary $n$ there are $k \le \frac{n}{2}$ different possible values for the number of consecutive element pairs that have equal values. </p>
<p>We have </p>
<p>$$|\Omega_{n}|=\sum_{k=1}^{\lfloor \frac{n}{2}\rfloor }|\Omega_{n,k}|$$</p>
<p>Now we select a $k$. Choose a sequence of $n-k$ values from $\{1,\ldots,n\}$ such that two consecutive values always differ.
There are $n (n-1)^{n-k-1}$ such sequences. For such a sequence we select $k$ of its elements (there are $\binom{n-k}{k}$ such possibilities)
and insert an element with the same value after each selected value. So
$$|\Omega_{n,k}|=\binom{n-k}{k}n(n-1)^{n-k-1}$$
and
$$|\Omega_{n}|=\sum_{k=1}^{\lfloor \frac{n}{2}\rfloor }\binom{n-k}{k}n(n-1)^{n-k-1}$$</p>
<ul>
<li><p>For $n=3$ there are $12$ sequences.</p></li>
<li><p>For $n=4$ there are t $120$ sequences.</p></li>
<li><p>For $n=5$ there are $1520$ sequences.</p></li>
</ul>
<p>The answer to the original question is <a href="https://math.stackexchange.com/questions/508791/number-of-lists-in-which-an-element-is-repeated-consecutively-exactly-twice/509032#508979">deleted</a></p>
|
2,073,410 | <p>If $$ a-(a \bmod x)<b$$ how do I prove that $$c-(c\bmod x)<b \;\forall c<a?$$ </p>
| user64066 | 64,066 | <p>Write $a=k_a+p_a x$ and $c=k_c+p_cx$ where $k_a, k_c\geq 0$. Then we have $p_a x <b$, and since $c<a$ implies $p_c x<p_a x$. Thus, the result follows. </p>
|
3,191,233 | <p>Why is <span class="math-container">$ce^λ=1$</span> equal to <span class="math-container">$c=e^{-λ}$</span>?</p>
| Kwin van der Veen | 76,466 | <p>The answer by Michael Stachowsky is incomplete. It is indeed correct that the time evolution of a continuous time LTI state space model</p>
<p><span class="math-container">$$
\left\{
\begin{align}
\dot{x}(t) &= A\,x(t) + B\,u(t), \\
y(t) &= C\,x(t),
\end{align}
\right. \tag{1}
$$</span></p>
<p>can be expressed using a <a href="https://math.stackexchange.com/a/69494/76466">convolution integral</a></p>
<p><span class="math-container">$$
x(t) = e^{A\,(t-t_0)}\,x(t_0) + \int_{t_0}^t e^{A\,(t-\tau)}\,B\,u(\tau)\,d\tau. \tag{2}
$$</span></p>
<p>However the input <span class="math-container">$u(t)$</span> is also subjected to a delay of <span class="math-container">$\delta=0.13$</span> time units and it is also given that <span class="math-container">$u(t)$</span> is equal to the constant <span class="math-container">$u_k$</span> for <span class="math-container">$k\,T<t\leq(k+1)\,T\ \forall\,k\in\mathbb{Z}$</span>, with <span class="math-container">$T=0.1$</span> time units. Using this <span class="math-container">$(2)$</span> can be written as</p>
<p><span class="math-container">\begin{align}
x_{k+1} &= e^{A\,T}\,x_k + \int_{k\,T}^{(k+1)\,T} e^{A\left((k+1)\,T-\tau\right)}\,B\,u(\tau-\delta)\,d\tau \\
&= e^{A\,T}\,x_k + \int_{k\,T}^{k\,T+\delta-T} e^{A\left((k+1)\,T-\tau\right)}\,B\,u(\tau-\delta)\,d\tau + \int_{k\,T+\delta-T}^{(k+1)\,T} e^{A\left((k+1)\,T-\tau\right)}\,B\,u(\tau-\delta)\,d\tau \\
&= e^{A\,T}\,x_k +
\int_{k\,T}^{k\,T+\delta-T} e^{A\left((k+1)\,T-\tau\right)}\,d\tau\,B\,u_{k-2} +
\int_{k\,T+\delta-T}^{(k+1)\,T} e^{A\left((k+1)\,T-\tau\right)}\,d\tau\,B\,u_{k-1} \\
&= \underbrace{e^{A\,T}}_{A_d}\,x_k +
\underbrace{\int_{0}^{\delta-T} e^{A\left(T-\tau\right)}\,d\tau\,B}_{B_1}\,u_{k-2} +
\underbrace{\int_{\delta-T}^{T} e^{A\left(T-\tau\right)}\,d\tau\,B}_{B_2}\,u_{k-1}
\end{align}</span></p>
<p>with <span class="math-container">$x_k = x(k\,T)$</span>. Here the resulting matrices <span class="math-container">$A_d$</span>, <span class="math-container">$B_1$</span> and <span class="math-container">$B_2$</span> are constant matrices that do not change with time. It can be noted that <span class="math-container">$T<\delta<2\,T$</span>, that is why I split the integral after <span class="math-container">$\delta-T$</span>, since that lies in between the integration interval. In order to be able to reuse older inputs of <span class="math-container">$u_k$</span> they need to be added to the state space, which can be written as</p>
<p><span class="math-container">\begin{align}
\begin{bmatrix}
x_{k+1} \\ u_{k} \\ u_{k-1}
\end{bmatrix} &=
\begin{bmatrix}
A_d & B_2 & B_1 \\
0 & 0 & 0 \\
0 & I & 0
\end{bmatrix}
\begin{bmatrix}
x_{k} \\ u_{k-1} \\ u_{k-2}
\end{bmatrix} +
\begin{bmatrix}
0 \\ I \\ 0
\end{bmatrix}
u_k, \\
y_k &=
\begin{bmatrix}
C & 0 & 0
\end{bmatrix}
\begin{bmatrix}
x_{k} \\ u_{k-1} \\ u_{k-2}
\end{bmatrix}.
\end{align}</span></p>
<p>The values for <span class="math-container">$B_1$</span> and <span class="math-container">$B_2$</span> can be found by using a <a href="https://en.wikipedia.org/wiki/Discretization#Discretization_of_linear_state_space_models" rel="nofollow noreferrer">zero order hold discretization trick</a>, which states that</p>
<p><span class="math-container">$$
B_d = \int_0^T e^{A\,\tau}\,d\tau\,B
$$</span></p>
<p>can also be found with</p>
<p><span class="math-container">$$
e^{
\begin{bmatrix}
A & B \\ 0 & 0
\end{bmatrix}T} =
\begin{bmatrix}
A_d & B_d \\ 0 & I
\end{bmatrix}.
$$</span></p>
<p>The expressions for <span class="math-container">$B_1$</span> and <span class="math-container">$B_2$</span> can be rewritten to</p>
<p><span class="math-container">\begin{align}
B_1 &= \int_{0}^{\delta-T} e^{A\left(T-\tau\right)}\,d\tau\,B \\
&= \int_{2\,T-\delta}^{T} e^{A\,\tau}\,d\tau\,B \\
&= \int_{0}^{T} e^{A\,\tau}\,d\tau\,B - \int_{0}^{2\,T-\delta} e^{A\,\tau}\,d\tau\,B \\
B_2 &= \int_{\delta-T}^{T} e^{A\left(T-\tau\right)}\,d\tau\,B \\
&= \int_{0}^{2\,T-\delta} e^{A\,\tau}\,d\tau\,B
\end{align}</span></p>
<p>Now by using the zero order hold discretization trick it can be shown that</p>
<p><span class="math-container">\begin{align}
\begin{bmatrix}
\mathcal{A}_1 & \mathcal{B}_1 \\ 0 & I
\end{bmatrix} &=
e^{
\begin{bmatrix}
A & B \\ 0 & 0
\end{bmatrix}T} \\
\begin{bmatrix}
\mathcal{A}_2 & \mathcal{B}_2 \\ 0 & I
\end{bmatrix} &=
e^{
\begin{bmatrix}
A & B \\ 0 & 0
\end{bmatrix}(2\,T-\delta)} \\
B_1 &= \mathcal{B}_1 - \mathcal{B}_2 \\
B_2 &= \mathcal{B}_2
\end{align}</span></p>
<p>It can be noted that the matlab command <span class="math-container">$\texttt{absorbDelay(c2d(sys,T))}$</span> should give the same LTI system (with maybe differing by a similarity transformation), but this answer hopefully gives you some more insights of how to actually calculate it.</p>
|
4,408,772 | <p>I'm read about a Lienard System in Perko books, but I don't understand how this applies
<a href="https://i.stack.imgur.com/PFVeR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PFVeR.png" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/vgttf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vgttf.png" alt="enter image description here" /></a></p>
<p>I have managed to understand the proof, the shape of the trajectory is intuitively clear, doing an analysis of the vector field <span class="math-container">$F(x,y)=(y-F(x),-g(x))$</span>, b ut how do I formally justify it? Any suggestion? If anyone knows a book or article where this system and its generalizations are treated in detail, I would be very grateful if you could cite it.</p>
<p>Postscript: The result mentioned in Perko is as follows (Ordinary differential equations by Philip Hartman):
<a href="https://i.stack.imgur.com/HyaYl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HyaYl.png" alt="enter image description here" /></a></p>
| Lutz Lehmann | 115,115 | <p>I too can not see the value of the cited theorem without further context.</p>
<p>Let's fixate what is obviously true:</p>
<ul>
<li><p>As long as <span class="math-container">$x>0$</span> we have <span class="math-container">$\dot y=-g(x)<0$</span>, so <span class="math-container">$y$</span> is at least non-increasing, and with <span class="math-container">$xg(x)\ge cx^2$</span> or a similar coercivity condition, <span class="math-container">$y$</span> will also be strictly decreasing until <span class="math-container">$x$</span> crosses again to negative values.</p>
</li>
<li><p>Due to <span class="math-container">$\dot x=y-F(x)$</span>, <span class="math-container">$x$</span> will be increasing as long as <span class="math-container">$y>F(x)$</span>. As <span class="math-container">$y$</span> is falling and <span class="math-container">$F(x)$</span> eventually increasing towards <span class="math-container">$+\infty$</span>, there will be a cross-over point.</p>
</li>
<li><p>If <span class="math-container">$y\le F_\min-\epsilon$</span>, where <span class="math-container">$F_\min$</span> is the minimum value taken by <span class="math-container">$F$</span> on <span class="math-container">$[0,\infty)$</span>, then from this moment on <span class="math-container">$\dot x\le-\epsilon$</span>, so <span class="math-container">$x$</span> has an upper bound that falls to zero in finite time.</p>
</li>
</ul>
<p>It remains to cover the segment from the crossover <span class="math-container">$y(t_2)=F(x(t_2))$</span> to some time <span class="math-container">$t_3$</span> where <span class="math-container">$y(t_3)<F_\min$</span>. Consider <span class="math-container">$v=\dot x=y-F(x)$</span> with
<span class="math-container">$$
\dot v =-g(x)-f(x)v,
\\
\frac{d}{dt}(e^{F(x)}v)=-g(x)e^{F(x)}.
$$</span>
At the cross-over point <span class="math-container">$t_2$</span> we have <span class="math-container">$v=0$</span> and thus <span class="math-container">$\dot v<0$</span>. With the integrating factor in the second equation we see that <span class="math-container">$e^{F(x)}v$</span> is strictly decreasing as long as <span class="math-container">$x>0$</span>. So as <span class="math-container">$e^{F(x)}v<-\epsilon$</span> for <span class="math-container">$t>t_2+\delta$</span> for some <span class="math-container">$\epsilon,\delta>0$</span>, we also get <span class="math-container">$\dot x=v<-\tilde\epsilon$</span> as <span class="math-container">$F$</span> is bounded on <span class="math-container">$[0,x(t_2)]$</span>. This again gives an upper bound for <span class="math-container">$x$</span> that falls to zero in finite time.</p>
|
3,800,521 | <p>Let <span class="math-container">$x=\tan y$</span>, then
<span class="math-container">$$
\begin{align*}\sin^{-1} (\sin 2y )+\tan^{-1} \tan 2y
&=4y\\
&=4\tan^{-1} (-10)\\\end{align*}$$</span></p>
<p>Given answer is <span class="math-container">$0$</span></p>
<p>What’s wrong here?</p>
| Alexey Burdin | 233,398 | <p><span class="math-container">$$\arcsin(\sin(x))=\left\{
\begin{array}{ll}
x-2\pi n&\hbox{for }-\frac{\pi}{2}+2\pi n\le x\le \frac{\pi}{2}+2\pi n\\
\pi-x-2\pi n&\hbox{for }\frac{\pi}{2}+2\pi n\le x\le \frac{3\pi}{2}+2\pi n\\
\end{array}\right.,$$</span>
<span class="math-container">$$\arctan(\tan(x))=x-\left\lfloor\frac{x+\pi/2}{\pi}\right\rfloor\pi$$</span></p>
|
3,800,521 | <p>Let <span class="math-container">$x=\tan y$</span>, then
<span class="math-container">$$
\begin{align*}\sin^{-1} (\sin 2y )+\tan^{-1} \tan 2y
&=4y\\
&=4\tan^{-1} (-10)\\\end{align*}$$</span></p>
<p>Given answer is <span class="math-container">$0$</span></p>
<p>What’s wrong here?</p>
| lab bhattacharjee | 33,337 | <p>Let <span class="math-container">$\tan^{-1}\dfrac{2x}{1-x^2}=u\implies-\dfrac\pi2<u<\dfrac\pi2$</span></p>
<p><span class="math-container">$\tan u=\dfrac{2x}{1-x^2}$</span></p>
<p><span class="math-container">$\implies\sec u+\sqrt{1+\left(\dfrac{2x}{1-x^2}\right)^2}=\dfrac{1+x^2}{|1-x^2|}$</span></p>
<p><span class="math-container">$\sin u=\dfrac{\tan u}{\sec u}=\text{sign of}(1-x^2)\cdot\dfrac{2x}{1+x^2}$</span></p>
<p><span class="math-container">$\implies u=\sin^{-1}\left(\text{sign of}(1-x^2)\cdot\dfrac{2x}{1+x^2}\right)$</span></p>
<p>So if <span class="math-container">$1-x^2<0\iff x^2>1, u=\sin^{-1}\left(-\dfrac{2x}{1+x^2}\right)=-\sin^{-1}\left(\dfrac{2x}{1+x^2}\right)$</span></p>
|
17,134 | <p>On a very regular basis we see new users that are not accustomed with the use of MathJaX on MSE. Sometimes even some users that aren't that new to the site. Most of us, when this happens, kindly bring to this users attention that there is a <a href="http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference">MathJax basic tutorial and quick reference</a> and give a proper link. </p>
<p>Now this is not a bad habit at all, because you can't expect new users to magically know this. My idea however was that we could promote this a bit more actively. </p>
<p>For instance: a new user is taken by the hand when he/she is taking the <a href="https://math.stackexchange.com/tour">tour</a>. Would it not be to this new user's benefit to point out in the <a href="https://math.stackexchange.com/tour">tour</a>, that we write our maths in MathJaX. For example place a link to <a href="http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference">-MathJax basic tutorial and quick reference-</a> or even a link to a -MathJaX tour page- (which would probably have the same structure as the normal <a href="https://math.stackexchange.com/tour">tour</a> page). And perhaps even -but this is really going ahead of things- award a badge for reading this page too.</p>
<p>This was a thought of mine that I wanted to share. I would like to know how the community feels about this. Or perhaps someone could make clear why this would not be helpfull.</p>
| daOnlyBG | 173,397 | <p>The motivation behind your idea is good; after all, the lack of MathJax editing can stir up confusion for both the asker and the helper. We all know MathJax's benefits.
However, I'm not quite sure that adding a "tutorial" of some sort would really help out- it's easy to skip through the tutorial steps, especially when <strike> someone posts their homework question at 2:30 a.m., the night before their assignment is due</strike> one is eager to present their question in a quick manner. </p>
<p>Furthermore, having the ability to edit others' posts provides an opportunity for newcomers to rack up some reputation points, so that they could take a stronger and more active role in Math.SE. Of course, you could counter that argument with the points awarded rate (2 per edit) isn't very high (though a few edits could have you commenting very quickly), and there will always be an abundance of non-MathJax editing issues to solve (i.e., poor grammar and spelling). </p>
<p>I think a neat idea would be to have a reference page. I personally learned most of MathJax by looking at others' posts with similar symbols that I would need in my post, copying and pasting that code when I click "edit," and then using that code in my answer to a different problem. For example: I needed to answer a factorial sum problem a couple weeks ago. I didn't know how to use the sigma notation on MathJax, so I simply browsed around until I found a thread with a sigma notation in a response. I clicked "edit" of the post that had it, and saw how to code it. </p>
<p>This isn't always sufficient; sometimes there are obscure math symbols that take a little browsing to find. Of course, upon searching for that symbol, one does end up learning a lot of math by casually browsing... </p>
<p>...but yeah, I think a reference page would be a neat idea. </p>
|
17,134 | <p>On a very regular basis we see new users that are not accustomed with the use of MathJaX on MSE. Sometimes even some users that aren't that new to the site. Most of us, when this happens, kindly bring to this users attention that there is a <a href="http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference">MathJax basic tutorial and quick reference</a> and give a proper link. </p>
<p>Now this is not a bad habit at all, because you can't expect new users to magically know this. My idea however was that we could promote this a bit more actively. </p>
<p>For instance: a new user is taken by the hand when he/she is taking the <a href="https://math.stackexchange.com/tour">tour</a>. Would it not be to this new user's benefit to point out in the <a href="https://math.stackexchange.com/tour">tour</a>, that we write our maths in MathJaX. For example place a link to <a href="http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference">-MathJax basic tutorial and quick reference-</a> or even a link to a -MathJaX tour page- (which would probably have the same structure as the normal <a href="https://math.stackexchange.com/tour">tour</a> page). And perhaps even -but this is really going ahead of things- award a badge for reading this page too.</p>
<p>This was a thought of mine that I wanted to share. I would like to know how the community feels about this. Or perhaps someone could make clear why this would not be helpfull.</p>
| ELC | 189,635 | <p>I just joined today, and it took me a few minutes to locate a tutorial for syntax, but it wasn't too difficult. </p>
<p>I think it would be useful to have this tutorial available in a more obvious place or perhaps sent along with the registration email. </p>
|
896,940 | <p>i tried 9 D + (-10 D)</p>
<p>9= 0000 1001</p>
<p>10= 0000 1010</p>
<p>Reverse 10 = 1111 0101 and add 1 become 1111 0110</p>
<p>after that add up 9 D + (-10 D) == 0000 1001 + 1111 0110 but the answer is equal to 1111 1111 whch is 255 in decimal but the answer should be -1 right? anything goes wrong?</p>
<p>Thank you very much. </p>
| Adriano | 76,987 | <p>In two's complement, we deal with signed numbers by checking the leftmost bit. If this leftmost bit is a $0$, then it's positive, so we proceed like we do with unsigned numbers. Otherwise, if the leftmost bit is a $1$, then it's negative, so we have to do the "reverse the bits then add one" trick that you did when you converted $-10_{10}$ to binary. In general:
$$
-x = \overline x + 1 \iff x = -(\overline x + 1)
$$
Hence, since $1111~1111$ starts with a $1$ and we are dealing with two's complement, we have:
\begin{align*}
(1111~1111)_2
&= -((\overline{1111~1111})_2 + 1) \\
&= -((0000~0000)_2 + 1) \\
&= -((0000~0001)_2) \\
&= -1_{10}
\end{align*}</p>
|
6,887 | <p>Let x_1, x_2, ... be iid draws from a laplace distribution with scale parameter b. Is there a relatively nice closed form for x_1+x_2+...x_n? I've seen a derivation floating around for when b=1, but I couldn't figure out a generalisation. </p>
| Andrew M Ross | 2,128 | <p>One way to generate a Laplace random variable is to generate two IID (independent and identically distributed) exponential random variables and then subtract them:
x_i = y_i - z_i
with y_i and z_i ~ exponential(parameter=b), and of course everything independent.
Then the sum of the x_i is simply (sum y_i) - (sum z_i); each of those two sums have Gamma distributions. To be more specific, since we are summing an integer number of terms, they have Erlang distributions. The difference of two Gammas is called "bilateral gamma", and there are a few papers out there on it. A quick search just found:</p>
<p>Bilateral gamma distributions and processes in financial mathematics
Uwe Küchlera, Stefan Tappe</p>
<p>On the shapes of bilateral Gamma densities
Uwe Küchlera, Stefan Tappe</p>
<p>It would be nice if someone would write a Wikipedia article about bilateral Gammas, I guess.</p>
|
329,600 | <p>Let $U,V,W$ be vector spaces over $F$ and $S: U \to V$, and $T: V \to W$ linear maps.</p>
<p>(a) Show that if $S$ and $T$ are isomorphisms, then $T\circ S$ is an isomorphism, too.</p>
<p>(b) Show that if $U$ is isomorphic to $V$ and $V$ is isomorphic to $W$, then $U$ is isomorphic to $W$.</p>
| azimut | 61,691 | <p><strong>Hints:</strong></p>
<ul>
<li><p>For (a): Check that the conditions on an isomorphism hold for $T\circ S$.</p></li>
<li><p>For (b): Apply (a).</p></li>
</ul>
|
2,646,363 | <p>Let $A_1, A_2, \ldots , A_{63}$ be the 63 nonempty subsets of $\{ 1,2,3,4,5,6 \}$. For each of these sets $A_i$, let $\pi(A_i)$ denote the product of all the elements in $A_i$. Then what is the value of $\pi(A_1)+\pi(A_2)+\cdots+\pi(A_{63})$?</p>
<p>Here is the solution </p>
<p>For size 1: sum of the elements, which is 21
For size 2: $ 1 \cdot (2 + 3 + 4 + 5 + 6) = 20 $, $ 2 \cdot (3 + 4 + 5 + 6) = 36 $, $ 3 \cdot (4 + 5 + 6) = 45 $, $ 4 \cdot (5 + 6) = 44 $, $ 5 \cdot 6 = 30 $. Sum is 175.
For size 3: Those with least element 1: $ 6, 8, 10, 12, 12, 15, 18, 20, 24, 30 = 155 $. Those with least element 2: $ 24, 30, 36, 40, 48, 60 = 238 $. Those with least element 3: $ 60 + 72 + 90 = 222 $. Those with least element 4: only one possible subset, which is $ \{4, 5, 6\} $, the $ \pi $ of which is 120. The total sum here is 735.
For size 4: Least element 1: $ 24 + 30 + 36 + 40 + 48 + 60 + 60 + 72 + 90 + 120 = 580 $; least element 2: $ 120 + 144 + 180 + 240 + 360 = 1044 $; least element 3: only one, which is $ 3 \cdot 4 \cdot 5 \cdot 6 = 360 $. The total sum here is 1984.
For size 5: Exclude each one individually to get $ 720 + 360 + 240 + 180 + 144 + 120 = 1764 $
For size 6: $ 6! = 720 $</p>
<p>The final answer is $ 21 + 175 + 735 + 1984 + 1764 + 720 = \boxed{5399} $</p>
<p>Is there any shorter way for doing this ?</p>
<p>Thank a lot </p>
| Jose L. Arregui | 892,303 | <p>The answer is 5039 just because, if you add 1 (let 1 be the product of the elements of the empty set) you must get <span class="math-container">$7!$</span>.</p>
<p>This is a general property: if <span class="math-container">$\pi(A)$</span> is the product of the elements of <span class="math-container">$A$</span> (with the convention <span class="math-container">$\pi(\emptyset)=1$</span>), and <span class="math-container">$[n] = \{1,2,\dots, n\}$</span>, then for every natural <span class="math-container">$n$</span>
<span class="math-container">$$
\sum_{A\subseteq [n]} \pi(A) = (n+1)!\,.
$$</span>
It is an equivalent fact that
<span class="math-container">$$
\sum_{A\subseteq [n]} \frac 1 {\pi(A)} = n+1\,,
$$</span>
as we see by dividing each term by <span class="math-container">$n!$</span> (every term <span class="math-container">$\pi(A)$</span> becomes <span class="math-container">$1/\pi(A^c)$</span>).</p>
<p>Both of them are very easily proved by induction, but an alternative proof and a nice probabilistic interpretation of the second one is as follows:</p>
<p>We are at the <span class="math-container">$(n+1)$</span>-th floor, and we take a lift to get down to the ground floor. But every time we press the button the lift will stop at a random level on the way, with uniform probability.</p>
<p>Let <span class="math-container">$A$</span> be the set of levels (floors) where the lift stops before we are downstairs. If we get there in the first step then <span class="math-container">$A=\emptyset$</span>. When we are at the <span class="math-container">$k$</span>-th floor, we have <span class="math-container">$k$</span> different possible stops in the next stage. Hence the probability of any <span class="math-container">$A$</span> as the set of intermediate points is <span class="math-container">$1/\big((n+1)\,\pi(A)\big)$</span>, and summing up for all <span class="math-container">$A$</span> we get
<span class="math-container">$$
1 = \frac 1 {n+1} \,\sum_{A\subseteq [n]} \frac 1 {\pi(A)}\,.
$$</span></p>
|
1,282,489 | <p>I have a simple problem that I need to solve. Given a height (in blue), and an angle (eg: 60-degrees), I need to determine the length of the line in red, based on where the green line ends. The green line comes from the top of the blue line and is always 90-degrees.</p>
<p>The height of the blue line is variable.
The angle of the blue line is variable.</p>
<p>Also, I do not know the length of the green dashed-line. Is there a way to figure out the length of the red line without knowing the length of the green?</p>
<p>-Adam</p>
<p>Any help would be much appreciated!</p>
<p>-Adam </p>
<p><img src="https://i.stack.imgur.com/3ccLl.png" alt="enter image description here"></p>
| Hagen von Eitzen | 39,174 | <p>Reflect at the green line to obtain a triangle, which must be equilateral.</p>
|
3,101,098 | <p>From 11, 12 in the book Logic in Computer Science by M. Ryan and M. Huth:</p>
<p>**</p>
<blockquote>
<p>"What we are saying is: let’s make the assumption of ¬q. To do this,
we open a box and put ¬q at the top. Then we continue applying other
rules as normal, for example to obtain ¬p. But this still depends on
the assumption of ¬q, so it goes inside the box. Finally, we are ready
to apply →i. It allows us to conclude ¬q → ¬p, but that conclusion no
longer depends on the assumption ¬q. Compare this with saying that ‘If
you are French, then you are European.’ The truth of this sentence
does not depend on whether anybody is French or not. Therefore, we
write the conclusion ¬q → ¬p outside the box."</p>
</blockquote>
<p>**</p>
<p>My question is about the scope of assumptions in propositional logic and proving techniques. I am not sure I fully understand what this text is trying to say. </p>
<p>How can an assumption only have scope inside the box, but once you finish what you want to prove it is no more part of the assumption box and is accessible universally in the proof? WHY is this possible? Why does it not break things in the proof? This looks too convenient and random.</p>
<p>Secondly, I do not understand the French and European example connection to what is written in this text. If somebody could please connect this example to what the author is actually trying to explain through this. </p>
| Graham Kemp | 135,106 | <blockquote>
<p>How can an assumption only have scope inside the box, but once you finish what you want to prove it is no more part of the assumption box and is accessible universally in the proof? </p>
</blockquote>
<p>Assumptions are raised and discharged.</p>
<p>When you raise an assumption, you open a box (rather, a subproof). Until you discharge the assumption (close the box) every statement derived is made in the context (aka scope) of that assumption -- they are statements that are asserted to be true <em>if</em> the assumption is true. If you are using the Fitch system for natural deduction, this us usually highlighted by underlining the assumption to distinguish it from derivations made in its context.</p>
<p>The '<span class="math-container">$\to$</span> introduction' rule is one of the rules used to discharge an assumption. The conditional statement that results is not inside the box -- after all, it is itself basically stating "if this assumption is true, then this conclusion is true too." </p>
<p><span class="math-container">$\def\fitch#1#2{\quad\begin{array}{|l}#1\\\hline#2\end{array}}$</span> <span class="math-container">$$\fitch{}{~\vdots\\\fitch{\lnot q\qquad \text{Assumption raised for the subproof.}}{~\vdots\qquad\quad\text{Derivations made in the context of the assumption}\\\lnot p\qquad\text{Conclusion reached within the context of the assumption}}\\\lnot q\to\lnot p\quad\text{Conditional statement deduced by the subproof}}$$</span></p>
<p><span class="math-container">$$\fitch{}{~\vdots\\\fitch{\text{Assuming you are French}}{\text{Then your native country is in Europe}\\\text{Thus you are European}}\\\text{If you are French, then you are European}}$$</span></p>
|
2,286,540 | <p>$(x,y)=(x,\sqrt{x})$
$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$
$=\sqrt{x^4-5x^2+9}$</p>
<p>$g(x)=x^4-5x^2+9$ , $g'(x)=4x^3-10x=0$ $: x=0, x=+-\sqrt{10}/2$</p>
<p>The question is: Should I put those x-values in g(x) or the orginal graph,$
y=\sqrt{x}$. To me, it's logical to put it into g(x), but in an example it was reverse</p>
| Jack D'Aurizio | 44,121 | <p>I feel an issue shared by many students: to apply algorithms without really understanding what they are doing. What is the question? The question is <em>locate a point such that$\ldots$</em>. So the answer is a point, i.e. a couple of coordinates. $f(x)$ is defined only on $[0,+\infty]$, so the answer is given by $(x_0,\sqrt{x_0})$ where $x_0$ is the positive abscissa you have found by solving the minimization problem.</p>
<p>Did you find it correctly? Let us see. The squared distance of $(x_0,\sqrt{x_0})$ from $(3,0)$ is $(x_0-3)^2+x_0 = \left(x_0-\frac{5}{2}\right)^2+\frac{11}{4}$, so the positive abscissa solving the minimization problem is $x_0=\frac{5}{2}$ and the wanted point is $\color{red}{\left(\frac{5}{2},\sqrt{\frac{5}{2}}\right)}$. The minimum distance is $\sqrt{\frac{11}{4}}=\frac{1}{2}\sqrt{11}$ and we don't even need derivatives, it is enough to complete a square.</p>
|
2,286,540 | <p>$(x,y)=(x,\sqrt{x})$
$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$
$=\sqrt{x^4-5x^2+9}$</p>
<p>$g(x)=x^4-5x^2+9$ , $g'(x)=4x^3-10x=0$ $: x=0, x=+-\sqrt{10}/2$</p>
<p>The question is: Should I put those x-values in g(x) or the orginal graph,$
y=\sqrt{x}$. To me, it's logical to put it into g(x), but in an example it was reverse</p>
| Mercy King | 23,304 | <p>You want to minimize the function
$$
f(x)=\|(x,\sqrt{x})-(3,0)\|^2=\|(x-3,\sqrt{x})\|^2=(x-3)^2+(\sqrt{x})^2=x^2-5x+9
$$
The function $f$ reaches its minimum value at
$$
h=-\dfrac{-5}{2(1)}=\dfrac{5}{2}
$$
and the minimum value is
$$
k=f(h)=\dfrac{11}{4}
$$
Hence the point of $y=\sqrt{x}$ closest to the point $(3,0)$ is the point
$\left(\dfrac{5}{2},\sqrt{\dfrac{5}{2}}\right)$, and its distance to the point $(3,0)$ is $\sqrt{k}=\dfrac{\sqrt{11}}{2}$.</p>
|
573,964 | <blockquote>
<p>Let set $S$ be the set of all functions $f:\mathbb{Z_+} \rightarrow \mathbb{Z_+}$. Define a realtion $R$ on $S$ by $(f,g)\in R$ iff there is a constant $M$ such that $\forall n (\frac{1}{M} < \frac{f(n)}{g(n)}<M). $ Prove that $R$ is an equivalence relation and that there are infinitely mane equivalence classes.</p>
</blockquote>
<p><strong>Attempt</strong>: Would it work if I define $f$ as $f_k(n)=kn$ and g as $g_k(n)=(M-k)n$ where $M>k$. Then, $$\forall n ((\frac{1}{M} < \frac{f(n)}{g(n)}<M)=(\frac{1}{M} < \frac{k}{M-k} <M))$$ is true as long as $M>k$. </p>
<p>So, $R$ is <strong>reflexive</strong>: $(f,f) \in R$
$$\forall n ((\frac{1}{M} < \frac{f(n)}{f(n)}<M)=(\frac{1}{M} < 1 <M)), \space M>1$$</p>
<p>$R$ is <strong>symmetric</strong>: $(f,g)\in R \Rightarrow (g,f)\in R$
$$\forall n ((\frac{1}{M} < \frac{f(n)}{g(n)}<M)=(\frac{1}{M} < \frac{k}{M-k} <M))$$
$$\forall n ((\frac{1}{M} < \frac{g(n)}{f(n)}<M)=(\frac{1}{M} < \frac{M-k}{k} <M))$$</p>
<p>for $M>k$.</p>
<p>$R$ is <strong>transitive</strong>: $(f,g)\in R \wedge (g,h) \in R \Rightarrow (f,h)\in R$
$$\frac{f(n)}{g(n)} \in R, \space \frac{g(n)}{h(n)} \in R \Rightarrow \frac{f(n)g(n)}{h(n)g(n)}=\frac{f(g)}{h(n)}\Rightarrow \forall n ((\frac{1}{M} < \frac{f(n)}{h(n)}<M) $$</p>
<p>Then, $R$ is an equivalence relation.</p>
| sheriff | 49,030 | <p>But $k$ must live in a field, and $\mathbb{Z}$ isn't.</p>
|
2,404,176 | <p>From the days I started to learn Maths, I've have been taught that </p>
<blockquote>
<p>Adding Odd times Odd numbers the Answer always would be Odd; e.g.,
<span class="math-container">$$3 + 5 + 1 = 9$$</span></p>
</blockquote>
<p>OK, but look at this question </p>
<p><a href="https://i.stack.imgur.com/TmYsJ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TmYsJ.jpg" alt="UPSC Question"></a></p>
<p>This question was solved and the answer was 30, how it was possible? Need a valid explanation please.</p>
| Robert Soupe | 149,436 | <p>Do you <em>have</em> to fill in <em>all</em> three boxes? I would just put 15 in <em>two</em> of the boxes, e.g., $$\fbox{ } + \fbox{15} + \fbox{15} = 30,$$ and hope people understand the first box as being an implicit 0.</p>
|
2,854,671 | <p>Consider a $n\times n$ Hankel Matrix</p>
<p>$$
H = \begin{bmatrix}
x_{1} & x_{2} & \dots & x_{n} \\
x_{2} & x_{3} & \dots & x_{n+1} \\
\vdots \\
x_{n} & x_{n+1} & \dots & x_{2n}
\end{bmatrix}
$$
, where all $x_i \in \mathbb{Z}_p = \{ 0,\dots,p-1 \}$, where $p$ is prime.</p>
<p>What is the most efficient way to test whether the matrix is invertible or not. More concretely:
Is there a more efficient than computing the determinant?
If not, is there a more efficient way of computing the determinant of such a matrix?</p>
| littleO | 40,119 | <p>This is not a full answer, but it's too long for a comment. The Hankel matrix
$$
H = \begin{bmatrix} x_1 & x_2 & x_3 & x_4 \\
x_2 & x_3 & x_4 & x_5 \\
x_3 & x_4 & x_5 & x_6 \\
x_4 & x_5 & x_6 & x_7 \\
\end{bmatrix}
$$
can be enlarged to an anti-circulant matrix
$$
\tilde H =
\left[
\begin{array}{c|c}
\begin{array}{c c c c}
x_1 & x_2 & x_3 & x_4 \\
x_2 & x_3 & x_4 & x_5 \\
x_3 & x_4 & x_5 & x_6 \\
x_4 & x_5 & x_6 & x_7 \\
\end{array}
&
\begin{array}{c c c}
x_5 & x_6 & x_7 \\
x_6 & x_7 & x_1 \\
x_7 & x_1 & x_2 \\
x_1 & x_2 & x_3\\
\end{array}
\\
\hline
\begin{array}{c c c c}
x_5 & x_6 & x_7 & x_1 \\
x_6 & x_7 & x_1 & x_2 \\
x_7 & x_1 & x_2 & x_3 \\
\end{array}
&
\begin{array}{c c c}
x_2 & x_3 & x_4 \\
x_3 & x_4 & x_5 \\
x_4 & x_5 & x_6\\
\end{array}
\end{array}
\right].
$$
I think some discrete Fourier transform techniques are available to perform computations with anti-circulant matrices efficiently. See this post for example: <a href="https://math.stackexchange.com/a/1377399/40119">https://math.stackexchange.com/a/1377399/40119</a>
Perhaps this would help with computations involving $H$, though I'm not certain.</p>
|
2,631,342 | <p>$$\lim_{x\rightarrow 14}\frac{\sqrt{x-5}-3}{x-14}$$</p>
<p>How do I evaluate the limit when I put x = 14 and I got 0/0?</p>
| Michael Rozenberg | 190,319 | <p>$$\frac{\sqrt{x-5}-3}{x-14}=\frac{x-14}{(x-14)(\sqrt{x-5}+3)}\rightarrow\frac{1}{6}.$$</p>
|
106,131 | <blockquote>
<p>Define $f:\mathbb{R}\rightarrow\mathbb{R}$ by<br>
$\
f(x) =
\begin{cases}
1/q
& \text{if } x =p/q \space(\mathrm{lowest}\space \mathrm{terms},\space\mathrm{nonzero})\\
0 & \text{if } x = 0\space\mathrm{or}\space x\not\in\mathbb{Q}
\end{cases}
$<br>
Show that f is continuous at 0 and every $x\in\mathbb{R}\setminus\mathbb{Q}$. Show that $f$ is not continuous at any nonzero rational pt. </p>
</blockquote>
<p>Attempt: (1) First I need to show that $f$ is continuous at zero. Then I need to show that $\forall\epsilon>0,\exists\delta>0$ s.t. $y\in B_{\delta}(0)$ implies $f(y)\in B_{\epsilon}(f(0))$. So I need to show $y\in B_{\delta}(0)$ implies $f(y)\in B_{\epsilon}(0)$ for $y\in\mathbb{R}$. Note $f(0)=0$. Pick $\delta =...$</p>
<p>(2) Then, I need to show that $f$ is continuous at every irrational number. Here I need to show that $\forall\epsilon>0,\exists\delta>0$ s.t. $y\in B_{\delta}(0)$ implies $f(y)\in B_{\epsilon}(f(0))$. Note that once again $f(0)=0$. Pick $\delta = ...$</p>
<p>(3) Then, I need to show that $f$ is not continuous at every nonzero rational number. Let $q\in\mathbb{Q}$. Intuitively, because $\mathbb{I}$ is dense in $\mathbb{R}$, we can construct a sequence $x_n\in\mathbb{I}$ such that $x_n\rightarrow q$. Since $x_n\in\mathbb{I}$, $\lim{x_n}=0\neq f(x)$ so clearly $f(x_n)\not\rightarrow f(x)$. </p>
| David Mitra | 18,986 | <p>Hints:</p>
<p>Keep in mind that to show $f$ is continuous at $p=0$ or at $p$ irrational, that $f(p)=0$. So to show continuity at $p$, you need to show that $|f(z)|$ can be made as small as desired by taking $z$ sufficiently close to $p$.</p>
<p>Towards this end, use (and prove) the fact that one can select an open ball centered at $p$ whose radius is so small that the ball excludes all rationals in lowest terms with denominator less than a given integer. So, for example, at the irrational $p$, there is an open ball $U$ centered at $p$ that contains no rational in lowest terms of the form $M/N$ with $N\le 1000$ (or, equivalently, if $q$ is rational, in $U$, and in lowest terms, then its denominator is greater than 1000).</p>
<p><br></p>
<p>A sequential argument, as you have in (3) (but your argument seems off...), is suitable
to show that $f$ is not continuous at a rational $q\ne0$. Here, note that $f(q)>0$ and that every nhood of $q$ contains irrational numbers. (So, your argument in (3) should start with <i>irrationals</i> converging to $x$.)</p>
|
154,722 | <p>Let $A = \pmatrix{1 & 0 \\ \alpha & 1} $ and $ B = \pmatrix{1 & 1 \\ 0 & 1}$, where $\alpha \in \mathbb{C}$ is a complex parameter.</p>
<p>Now consider the family of representations $r_{\alpha}$ of the free group on two generators $F_2 = \langle a,b\rangle$ in $\mathrm{SL}(2, \mathbb{C})$ setting $r_{\alpha}(a) = A$ and $r_{\alpha}(b) = B$. One can see that when $\alpha$ is transcendental over $\mathbb{Q}$, the representation $r_{\alpha}$ is faithful (see T. Church & A. Pixton "Separating twists and the Magnus representation of the Torelli group" Lemma 5.1). </p>
<p>The question I am interested in is the following : when is (or is not) $r_{\alpha}(F_2)$ a discrete subgroup of $\mathrm{SL}(2, \mathbb{C})$ ? </p>
<p>I suppose this is a difficult question of dynamics, I am curious if anyone has ever studied similar questions. </p>
| Alexandre Eremenko | 25,510 | <p>Yes, the answer is complicated, it is related to holomorphic dynamics,
and the question was much studied, see for example:
MR0869581
Lyubich, M. Yu.; Suvorov, V. V.
Free subgroups of SL2(C) with two parabolic generators.</p>
|
154,722 | <p>Let $A = \pmatrix{1 & 0 \\ \alpha & 1} $ and $ B = \pmatrix{1 & 1 \\ 0 & 1}$, where $\alpha \in \mathbb{C}$ is a complex parameter.</p>
<p>Now consider the family of representations $r_{\alpha}$ of the free group on two generators $F_2 = \langle a,b\rangle$ in $\mathrm{SL}(2, \mathbb{C})$ setting $r_{\alpha}(a) = A$ and $r_{\alpha}(b) = B$. One can see that when $\alpha$ is transcendental over $\mathbb{Q}$, the representation $r_{\alpha}$ is faithful (see T. Church & A. Pixton "Separating twists and the Magnus representation of the Torelli group" Lemma 5.1). </p>
<p>The question I am interested in is the following : when is (or is not) $r_{\alpha}(F_2)$ a discrete subgroup of $\mathrm{SL}(2, \mathbb{C})$ ? </p>
<p>I suppose this is a difficult question of dynamics, I am curious if anyone has ever studied similar questions. </p>
| Ian Agol | 1,345 | <p>The free discrete subgroups consist of the closure of the Riley slice of Schottky space (notice that one may assume $Re(\alpha)\geq 0, Im(\alpha)\geq 0$, since $\alpha \mapsto -\alpha, \overline{\alpha}$ preserves discreteness).
Here's a picture of the first quadrant of the Riley slice:
<img src="https://i.stack.imgur.com/IqO9w.jpg" alt="enter image description here"></p>
<p>The exterior of the black fractal represents free discrete groups that are generalized Schottky groups. The colored regions correspond to the combinatorics of the Ford domain. These have been investigated in detail by <a href="http://www.ams.org/mathscinet-getitem?mr=1698107" rel="nofollow noreferrer">Akiyoshi, Sakuma, Wada, and Yamashita</a> (see also <a href="http://f3.tiera.ru/2/M_Mathematics/Mln_Lecture%20notes/Akiyoshi%20H.,%20Sakuma%20M.,%20Wada%20M.,%20et%20al.%20Punctured%20Torus%20Groups%20and%202-Bridge%20Knot%20Groups%201%20%28LNM1909,%20Springer,%202007%29%28ISBN%203540718060%29%28297s%29_Mln_.pdf" rel="nofollow noreferrer">their monograph</a> where the picture was taken from). The black curve consists of geometrically finite groups with a cusp or degenerate groups. By the <a href="http://link.springer.com/article/10.1007/s11511-012-0088-0" rel="nofollow noreferrer">density conjecture</a>, it is known that all free two-parabolic generator groups lie in the boundary of the Riley slice. Moreover, <a href="http://www.ams.org/books/conm/510/10029" rel="nofollow noreferrer">it is known that there is a group realizing each ending lamination</a> parameterized by $\mathbb{R/Z}-\mathbb{Q/Z}$. </p>
<p>In the exterior of the Riley slice, there are many more non-free discrete groups, such as those corresponding to 2-bridge links. In fact, <a href="http://www.ams.org/mathscinet-getitem?mr=1393562" rel="nofollow noreferrer">all of the torsion-free discrete groups correspond to 2-bridge links</a>. Note that for 2-bridge links, although they are known to be generated by 2 parabolics, the precise value of $\alpha$ which gives the parabolics has not been determined. One may see an example of this for the twist knots, for which $\alpha$ has been worked out by <a href="http://www.worldscientific.com/doi/abs/10.1142/S0218216501001049" rel="nofollow noreferrer">Hoste and Shanahan</a>. </p>
<p>For the discrete groups with torsion, there are certain other orbifolds closely related to the 2-bridge links; see some <a href="https://www.dropbox.com/sh/cxdm5tu5hsnho0i/oByjC-AcYN/parabolic" rel="nofollow noreferrer">incomplete notes of mine</a>. Conjecturally, all of the discrete groups are obtained by extending the "pleating locus" for each rational number $\in \mathbb{Q/Z}$ (representing a simple closed curve on the 4-punctured torus) through the cusp, and into the complement of Schottky space. The points where these elements are primitive elliptics should correspond to discrete groups (with some caveats); these pleating rays are displayed in the picture. In fact, the discrete groups in the complement of the Riley slice closure form a discrete set, with limit points at the boundary of the Riley slice. </p>
|
2,721,718 | <blockquote>
<p>$AD, BE$ and $CF$ are three concurrent lines in $\triangle ABC$, meeting the opposite sides in $D, E$ and $F$ respectively. Show that the joins of the midpoints of $BC, CA$ and $AB$ to the midpoints of $AD, BE$ and $CF$ are concurrent.</p>
</blockquote>
<p>Let $D', E'$ and $F'$ be the midpoints of $BC, CA$ and $AB$. Then consider $\triangle ABD$. $E'F' \parallel BC$ and therefore in $\triangle ABD$, $E'F'$ cuts $AD$ at its midpoint (By the midpoint theorem). Similiarly $E'D'$ cuts $CF$ at its midpoint and $D'F'$ cuts $BC$ at its midpoint. </p>
<p>How do I proceed with this?</p>
| Tsemo Aristide | 280,301 | <p>Let $f:X\times Y-A\times B\rightarrow\{0,1\}$ be a continuous function. There exists $x\in X-A, y\in Y-B$, suppose $f(x,y)=0$. We have $f(x\times Y)=f(X\times y)=0$ since $x\times Y$ and $X\times y\subset :X\times Y-A\times B$. Let $(u,v)\in :X\times Y-A\times B$. You have $u$ is not in $X$ or $v$ is not in $Y$. If $u$ is not in $X$, then $u\times Y\subset :X\times Y-A\times B$ and $f(u,v)=f(u\times Y)=f(u,y)=f(X\times y)=f(x,y)=0$, if $v$ is not in $Y$ a similar argument shows that $f(u,v)-0$, so $f$ is constant and $X\times Y-A\times B$ is connected.</p>
|
2,721,718 | <blockquote>
<p>$AD, BE$ and $CF$ are three concurrent lines in $\triangle ABC$, meeting the opposite sides in $D, E$ and $F$ respectively. Show that the joins of the midpoints of $BC, CA$ and $AB$ to the midpoints of $AD, BE$ and $CF$ are concurrent.</p>
</blockquote>
<p>Let $D', E'$ and $F'$ be the midpoints of $BC, CA$ and $AB$. Then consider $\triangle ABD$. $E'F' \parallel BC$ and therefore in $\triangle ABD$, $E'F'$ cuts $AD$ at its midpoint (By the midpoint theorem). Similiarly $E'D'$ cuts $CF$ at its midpoint and $D'F'$ cuts $BC$ at its midpoint. </p>
<p>How do I proceed with this?</p>
| Henno Brandsma | 4,280 | <p>Pick $p \in X \setminus A$, $q \in Y \setminus B$, by properness of inclusions.</p>
<p>For $x \in X$ define $V_x = \{x\} \times Y$ the vertical stalk at $X$, homeomorphic to $Y$, so connected. </p>
<p>Likewise for $y \in Y$, $H_y = X \times \{y\}$, the horizontal "stalk" at $y$, homeomorphic to $X$ and hence connected.</p>
<p>Let $\mathcal{V} = \{ V_x : x \in X \setminus Y\}$, a set of vertical stalks, all of which are subsets of $(X \times Y) \setminus (A \times B)$ as the $x$ coordinate of each stalk lies outside $A$, and also, $V$ already contains $H_p$, so is non-trivial.</p>
<p>Then the set $H_q$ intersects all $V_x$ in $\mathcal{V}$ (namely in $(x,q)$). So by a standard therem on unions of connected subsets we have that $V' = \bigcup \mathcal{V} \cup H_q$ is connected. (the stripe of "glue" $H_q$ glues the vertical stalks together).</p>
<p>Likewise $\mathcal{H} = \{H_y: y \in Y \setminus B\}$ is a set of horizontal stalks (all inside $(X \times Y) \setminus (A \times B)$ again), all of which intersect $V_p$ (and we already have $H_q \in \mathcal{H}$) so again</p>
<p>$H' = \bigcup \mathcal{H} \cup V_p$ is connected.</p>
<p>$H'$ and $V'$ intersect in $(p,q)$ and both are connected and one easily checks that $$(X \times Y) \setminus (A \times B) = H' \cup V'$$ and so our space is connected.</p>
|
4,545,364 | <blockquote>
<p>Solve the quartic polynomial :
<span class="math-container">$$x^4+x^3-2x+1=0$$</span>
where <span class="math-container">$x\in\Bbb C$</span>.</p>
<p>Algebraic, trigonometric and all possible methods are allowed.</p>
</blockquote>
<hr />
<p>I am aware that, there exist a general quartic formula. (Ferrari's formula). But, the author says, this equation doesn't require general formula. We need some substitutions here.</p>
<p>I realized there is no any rational root, by the rational root theorem.</p>
<p>The harder part is, WolframAlpha says the factorisation over <span class="math-container">$\Bbb Q$</span> is impossible.</p>
<p>Another solution method can be considered as the quasi-symmetric equations approach. (divide by <span class="math-container">$x^2$</span>).</p>
<p><span class="math-container">$$x^2+\frac 1{x^2}+x-\frac 2x=0$$</span></p>
<p>But the substitution <span class="math-container">$z=x+\frac 1x$</span> doesn't make any sense.</p>
<p>I want to ask the question here to find possible smarter ways to solve the quartic.</p>
| Parcly Taxel | 357,390 | <p>Substitute <span class="math-container">$x=y+1$</span>, then <span class="math-container">$z=y+1/y$</span> to get
<span class="math-container">$$y^4+5y^3+9y^2+5y+1=0$$</span>
<span class="math-container">$$y^2((y+1/y)^2+5(y+1/y)+7)=0$$</span>
<span class="math-container">$$z^2+5z+7=0$$</span>
Then <span class="math-container">$z=\frac{-5\pm\sqrt{-3}}2$</span> and <span class="math-container">$x=\frac{z\pm\sqrt{z^2-4}}2+1$</span>, where the four solutions are obtained from the different sign choices for <span class="math-container">$z$</span> and <span class="math-container">$x$</span>.</p>
<p>More concretely, let <span class="math-container">$t,u=\frac{-1\pm\sqrt{-3}}2$</span> be the two roots of <span class="math-container">$x^2+x+1$</span>, then
<span class="math-container">$$x^4+x^3-2x+1=(x^2-tx+t)(x^2-ux+u)$$</span></p>
|
2,179,253 | <p>$$n{n-1 \choose 2}={n \choose 2}{(n-2)}$$
Give a conceptual
explanation of why this formula is true.</p>
| Joffan | 206,402 | <p>$$\begin{align}
n{n-1 \choose 2}&=\binom n1 \binom{n-1}{2} \tag {choose 1 then 2}\\[1ex]
&= \binom n2 \binom{n-2}{1} \tag {choose 2 then 1}\\[1ex]
&={n \choose 2}{(n-2)}
\end{align}$$</p>
|
2,068,986 | <p>Consider the function</p>
<p>$$K(u) = \frac 1 {\sqrt {2\pi}} \left( \Bbb e ^{-\frac 1 2 \left( \frac {u-5.3} h \right)^2 } + \Bbb e ^{-\frac 1 2 \left( \frac {u-1.6} h \right)^2 } + \Bbb e ^{-\frac 1 2 \left( \frac {u-2.1} h \right)^2 } + \Bbb e ^{-\frac 1 2 \left( \frac {u-1.7} h \right)^2 } + \Bbb e ^{-\frac 1 2 \left( \frac {u-1.9} h \right)^2 } \right) .$$</p>
<p>I know how to compute the derivative of each term in order to find its extrema, but how should I proceed in order to find the extrema of the whole sum?</p>
| John Hughes | 114,036 | <p>Write $\ln 2$ as a binary number with bits $b_1, b_2, \ldots$. </p>
<p>As others have suggested in briefly-present answers, taking logs is the secret: you want
$$
\lim_{n\to+\infty}\frac{a_n}{2^n}=\ln(2)
$$</p>
<p>and hence
$$a_n=2^n\ln(2).$$
but this choice of $a_n$ is not an integer. On the other hand, the
floor of it is. So we pick</p>
<p>$$a_n=\lfloor 2^n\ln(2) \rfloor$$</p>
<p>which is just the binary "whole number" whose bits (before the decimal point, so to speak) are $b_1, b_2, \ldots, b_n$. </p>
<p>Then the question becomes why the limit in this case is the same as in the case without the floors. </p>
<p>Letting
$$q_n=2^n\ln(2),$$
I'm claiming that
$$
\lim \left(1-\frac{1}{2^{n}}\right)^{a_{n}}=\lim \left(1-\frac{1}{2^{n}}\right)^{q_{n}}.
$$</p>
<p>Taking log of both sides, this is the same as
$$
\lim a_n \log \left(1-\frac{1}{2^{n}}\right)= \lim q_n \log\left(1-\frac{1}{2^{n}}\right)^{q_{n}}
$$
or
$$
\lim (a_n-q_n) \log \left(1-\frac{1}{2^{n}}\right)= 0.
$$
Now $a_n - q_n$ is between 0 and 1, and $\log \left(1-\frac{1}{2^{n}}\right) \approx -\frac{1}{2^{n}}$, so it's clear that the limit is bounded below by $$-2\frac{1}{2^{n}},$$ which goes to 0, so the squeeze lemma applies and the limit is indeed 0. </p>
|
3,164,280 | <p>By accident, I find this summation when I pursue the particular value of <span class="math-container">$-\operatorname{Li_2}(\tfrac1{2})$</span>, which equals to integral <span class="math-container">$\int_{0}^{1} {\frac{\ln(1-x)}{1+x} \mathrm{d}x}$</span>.</p>
<p>Notice this observation</p>
<p><span class="math-container">$$\int_{0}^{1} {\frac{\ln(1-x)}{1+x} \mathrm{d}x} = \int_{0}^{1} {\frac{\ln(1-x^{2})}{1+x} \mathrm{d}x} - \frac{(\ln2)^{2}}{2}$$</span></p>
<p>And using the Taylor series of <span class="math-container">$\ln(1-x^{2})$</span>, I found this summation <span class="math-container">$\sum_{n=1}^{\infty} {\frac1{n} (H_{2n}-H_{n}-\ln2)}$</span>, where <span class="math-container">$H_{n}$</span> is the harmonic-numbers.</p>
<p>If the value of <span class="math-container">$\operatorname{Li_2}(\tfrac1{2})=\tfrac1{2}(\zeta(2)-(\ln2)^{2})$</span> is given, the result can be easily deduced, which is</p>
<blockquote>
<p><span class="math-container">$$\sum_{n=1}^{\infty} {\frac1{n} (H_{2n}-H_{n}-\ln2)} = -\frac{\zeta(2)}{2}+(\ln2)^{2}$$</span></p>
</blockquote>
<p>For the original goal is to calculate <span class="math-container">$\operatorname{Li_2}(\tfrac1{2})$</span>, I expect some other approaches to the summation without using the value of <span class="math-container">$\operatorname{Li_2}(\tfrac1{2})$</span>. I already knew some famous problem like <a href="https://math.stackexchange.com/questions/275643/proving-an-alternating-euler-sum-sum-k-1-infty-frac-1k1-h-kk?noredirect=1&lq=1">Euler's Sum</a>, which holds very similar form to this summation, but still in trouble finding the appropriate path.</p>
| Jack D'Aurizio | 44,121 | <p>Well, ignoring the dilogarithm reflection formula, we still have
<span class="math-container">$$ \sum_{n=1}^{N}\frac{\log(2)}{n}=\log(2)H_N,\qquad \sum_{n=1}^{N}\frac{H_n}{n}\stackrel{\text{sym}}{=}\frac{H_n^2+H_n^{(2)}}{2} $$</span>
and
<span class="math-container">$$ \sum_{n=1}^{N}\frac{H_{2n}}{n}\stackrel{\text{SBP}}{=}H_N H_{2N}-\sum_{n=1}^{N-1}H_n\left(\frac{1}{2n+2}+\frac{1}{2n+1}\right) $$</span>
can be reduced (up to known terms) to
<span class="math-container">$$ \sum_{n=1}^{N}\left[\frac{1}{n}\sum_{k=1}^{n}\frac{1}{2k-1}+\frac{1}{2n-1}\sum_{k=1}^{n}\frac{1}{k}\right]=\sum_{n=1}^{N}\frac{1}{n}\sum_{n=1}^{N}\frac{1}{2n-1}+\sum_{n=1}^{N}\frac{1}{n(2n-1)}. $$</span>
Exploiting <span class="math-container">$H_n^{(2)}=\zeta(2)+o(1)$</span> and <span class="math-container">$H_n = \log(n)+\gamma+o(1)$</span> for <span class="math-container">$n\to +\infty$</span> we end up with the explicit value of <span class="math-container">$\text{Li}_2\left(\frac{1}{2}\right)$</span>. Nice exercise!</p>
|
3,164,280 | <p>By accident, I find this summation when I pursue the particular value of <span class="math-container">$-\operatorname{Li_2}(\tfrac1{2})$</span>, which equals to integral <span class="math-container">$\int_{0}^{1} {\frac{\ln(1-x)}{1+x} \mathrm{d}x}$</span>.</p>
<p>Notice this observation</p>
<p><span class="math-container">$$\int_{0}^{1} {\frac{\ln(1-x)}{1+x} \mathrm{d}x} = \int_{0}^{1} {\frac{\ln(1-x^{2})}{1+x} \mathrm{d}x} - \frac{(\ln2)^{2}}{2}$$</span></p>
<p>And using the Taylor series of <span class="math-container">$\ln(1-x^{2})$</span>, I found this summation <span class="math-container">$\sum_{n=1}^{\infty} {\frac1{n} (H_{2n}-H_{n}-\ln2)}$</span>, where <span class="math-container">$H_{n}$</span> is the harmonic-numbers.</p>
<p>If the value of <span class="math-container">$\operatorname{Li_2}(\tfrac1{2})=\tfrac1{2}(\zeta(2)-(\ln2)^{2})$</span> is given, the result can be easily deduced, which is</p>
<blockquote>
<p><span class="math-container">$$\sum_{n=1}^{\infty} {\frac1{n} (H_{2n}-H_{n}-\ln2)} = -\frac{\zeta(2)}{2}+(\ln2)^{2}$$</span></p>
</blockquote>
<p>For the original goal is to calculate <span class="math-container">$\operatorname{Li_2}(\tfrac1{2})$</span>, I expect some other approaches to the summation without using the value of <span class="math-container">$\operatorname{Li_2}(\tfrac1{2})$</span>. I already knew some famous problem like <a href="https://math.stackexchange.com/questions/275643/proving-an-alternating-euler-sum-sum-k-1-infty-frac-1k1-h-kk?noredirect=1&lq=1">Euler's Sum</a>, which holds very similar form to this summation, but still in trouble finding the appropriate path.</p>
| omegadot | 128,913 | <p>Here is an approach that avoids knowing the value of <span class="math-container">$\operatorname{Li}_2 (\frac{1}{2})$</span>. </p>
<p>Let
<span class="math-container">$$S = \sum_{n = 1}^\infty \frac{1}{n} \left (H_{2n} - H_n - \ln 2 \right ).$$</span>
Observing that
<span class="math-container">$$\int_0^1 \frac{x^{2n}}{1 + x} \, dx = H_n - H_{2n} + \ln 2,$$</span>
your sum can be rewritten as
<span class="math-container">\begin{align}
S &= -\int_0^1 \frac{1}{1 + x} \sum_{n = 1}^\infty \frac{x^{2n}}{n} \, dx\\
&= \int_0^1 \frac{\ln (1 - x^2)}{1 + x} \, dx\\
&= \int_0^1 \frac{\ln (1 + x)}{1 + x} \, dx + \int_0^1 \frac{\ln (1 - x)}{1 + x} \, dx\\
&= I + J.
\end{align}</span>
For the first of the integrals <span class="math-container">$I$</span>, one has
<span class="math-container">$$I = \frac{1}{2} \ln^2 2.$$</span>
Now consider <span class="math-container">$J - I$</span>. Then
<span class="math-container">$$J - I = \int_0^1 \ln \left (\frac{1 - x}{1 + x} \right ) \frac{dx}{1 + x}.$$</span>
Employing a <em>self-similar substitution</em> of <span class="math-container">$t = (1-x)/(1+x)$</span> leads to
<span class="math-container">\begin{align}
J - I &= \int_0^1 \frac{\ln t}{1 + t} \, dt\\
&= \sum_{n = 0}^\infty (-1)^n \int_0^1 t^n \ln t \, dt\\
&= \sum_{n = 0}^\infty (-1)^n \frac{d}{ds} \left [\int_0^1 t^{n + s} \, dt \right ]_{s = 0}\\
&= \sum_{n = 0}^\infty (-1)^n \frac{d}{ds} \left [\frac{1}{n + s + 1} \right ]_{s = 0}\\
&= -\underbrace{\sum_{n = 0}^\infty \frac{(-1)^n}{(n + 1)^2}}_{n \, \mapsto \, n - 1}\\
&= -\sum_{n = 1}^\infty \frac{(-1)^{n+1}}{n^2}\\
&= -\sum_{n = 1}^\infty \frac{1}{n^2} + \frac{1}{2} \sum_{n = 1}^\infty \frac{1}{n^2}\\
&= -\frac{1}{2} \zeta (2).
\end{align}</span>
Thus
<span class="math-container">$$J = I - \frac{1}{2} \zeta (2) = \frac{1}{2} \ln^2 2 - \frac{1}{2} \zeta (2).$$</span>
Since <span class="math-container">$S = I + J$</span>, we immediately see that
<span class="math-container">$$\sum_{n = 1}^\infty \frac{1}{n} \left (H_{2n} - H_n - \ln 2 \right ) = \ln^2 2 - \frac{1}{2} \zeta (2),$$</span>
as desired. </p>
|
1,112,081 | <p>Does $\int_0^\infty e^{-x}\sqrt{x}dx$ converge? Thanks in advance.</p>
| dalastboss | 194,935 | <p>Hint: Note that $\sqrt xe^{-x}$is monotonically decreasing for large enough $x$ and as such the integral converges only if the infinite sum of the same converges. Then apply ratio test. </p>
|
1,890,040 | <p>This is related to <a href="https://math.stackexchange.com/questions/1888881/expanding-a-potential-function-via-the-generating-function-for-legendre-polynomi">this previous question of mine</a> where (with lots of help) I show that $$\sum_{l=0}^\infty \frac{R^l}{a^{l+1}}P_l(\cos\theta)=\frac{1}{\sqrt{a^2-2aR\cos\theta+R^2}}\tag{1}$$ by using the Legendre generating function.</p>
<p>The generating function for Legendre Polynomials is:</p>
<p>$$\Phi(x,h)=(1-2xh+h^2)^{-1/2} \quad\text{for}\quad |h|\lt 1\tag{2}$$
or
$$\Phi(x,h)=\sum_{l=0}^\infty h^l P_l(x)\quad\text{for}\quad |h|\lt 1\tag{3}$$</p>
<p>I need to show that $$\sum_{l=0}^\infty\frac{R^{2l+1}r^{-l-1}P_l(\cos\theta)}{a^{l+1}}=\frac{R}{a\sqrt{(R^2/a)^2-2r(R^2/a)\cos\theta+r^2}}\tag{4}$$ I note the striking similarity between the RHS's of $(1)$ and $(4)$ within the square root of the denominator.</p>
<hr>
<p><code>It is not necessary to read the following if you understand how to prove that the sum can be written as above; But here is some context anyway:</code>
<a href="https://i.stack.imgur.com/DIR6f.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DIR6f.png" alt="Diagram"></a>
<a href="https://i.stack.imgur.com/PkIFq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PkIFq.png" alt="Potential function 1"></a>
<a href="https://i.stack.imgur.com/CtYlI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CtYlI.png" alt="Potential function 2"></a> </p>
<hr>
<p><code>So here is my attempt:</code></p>
<p>From the RHS of equation $(4)$:</p>
<p>$$\displaystyle\begin{align}\frac{R}{a\sqrt{r^2-2r(R^2/a)\cos\theta+(R^2/a)^2}}&=\frac{R}{ar\sqrt{1-2(R^2/ar)\cos\theta+(R^2/ar)^2}}\end{align}$$</p>
<p>Changing variables to match equation $(2)$: Let $h=R^2/ar$ and $x=\cos\theta$</p>
<p>Then
$$\displaystyle\begin{align}\frac{R}{ar\sqrt{1-2(R^2/ar)\cos\theta+(R^2/ar)^2}}&=\frac{R}{ar\sqrt{1-2hx+h^2}}\\&=\frac{R}{ar}(1-2xh+h^2)^{-1/2}\\&=\frac{R}{ar}\Phi(x,h)\\&=\frac{R}{ar}\sum_{l=0}^\infty h^l P_l(x)\\&=\frac{R}{ar}\sum_{l=0}^\infty \frac{R^{2l} P_l(\cos\theta)}{(ar)^l}\\&=\sum_{l=0}^\infty \frac{R^{2l+1} P_l(\cos\theta)}{a^{l+1}r^{l+1}}\\&=\sum_{l=0}^\infty \frac{R^{2l+1}r^{-l-1} P_l(\cos\theta)}{a^{l+1}}\quad\fbox{}\end{align}$$</p>
<hr>
<p>You may be wondering what I'm asking at this point; so here is the question:</p>
<p>In the last line of the extract the textbook says "by summing the series"; and this seems to suggest that one must arrive at the RHS <em>from</em> the LHS (instead of going from RHS to LHS as I did). So is my interpretation correct (LHS $\to$ RHS) and if so <em>can</em> it be done in this order?</p>
<p>Many thanks.</p>
| Toutatis | 360,924 | <p>I found 432 in the book
"Les nombres remarquables" by Francois Le Lionnais (Herman, 1983)
(which is a list of interesting numbers), but there is no reference or proof. Among these 432 parts 192 are bounded.</p>
|
4,324,493 | <p>According to Rick Miranda's Algebraic curves and Riemann surfaces, a hyperelliptic curve is defined as the Riemann surface obtained by gluing two algebraic curves, <span class="math-container">$y^2=h(x)$</span> and <span class="math-container">$w^2 = k(z)$</span> (where <span class="math-container">$h$</span> has distinct roots and <span class="math-container">$k(z) := z^{2g+2} h(1/z)$</span>) through the map <span class="math-container">$(x,y) \mapsto (z,w) := (1/x, y/x^{g+1})$</span>.
He also mentions that this is topologically a genus <span class="math-container">$g$</span> surface; and has a notion of involution <span class="math-container">$\sigma : (x,y)\mapsto(x,-y)$</span></p>
<p>I am trying to visualise how all these actually looks like, and I have a couple of questions:</p>
<ol>
<li><p>Is there any geometric way to visualise the construction? Formally, is there any embedding/immersion of this construction into <span class="math-container">$\mathbb{R}^3$</span>, that helps see, atleast topologically, atleast for specific examples, what parts of genus <span class="math-container">$g$</span> surface are being glued together?</p>
</li>
<li><p>What does the <span class="math-container">$\sigma$</span> 'looks like' as a map on genus <span class="math-container">$g$</span> surface, <span class="math-container">$\Sigma_g$</span>, imagined as embedded in <span class="math-container">$\mathbb{R}^3$</span> in the standard way? Does it look like a reflection? Or maybe a <span class="math-container">$180^\circ$</span> rotation?</p>
</li>
<li><p>I think this brings about some sort of duality between the polynomials <span class="math-container">$h$</span> and <span class="math-container">$k$</span>. Is this somehow important? Specifically, is there anything special about the Riemann surfaces that are obtained by gluing the same algebraic curve? As an example for such a curve, if <span class="math-container">$x\neq 0$</span>, for <span class="math-container">$g=3$</span>, <span class="math-container">$$y^2 = 3x^8 +10x^4 +3 \Leftrightarrow \bigg(\frac{y}{x^{3+1}}\bigg)^2 = 3\bigg(\frac{1}{x}\bigg)^8 +10\bigg(\frac{1}{x}\bigg)^4 +3 $$</span></p>
</li>
</ol>
<p>Sorry if the questions are a bit vague. I am just trying to get more intuition about the subject.</p>
| KReiser | 21,412 | <p>This question has been <a href="https://mathoverflow.net/questions/14024/degree-2-branched-map-from-the-torus-to-the-sphere">addressed on MathOverflow</a>. Here's the accepted answer there, from user <a href="https://mathoverflow.net/users/3384/stankewicz">stankewicz</a>:</p>
<blockquote>
<p>One example: lay your <span class="math-container">$g$</span>-holed torus <span class="math-container">$T$</span> out flat and draw a line the long way through each hole. It hits the torus in <span class="math-container">$2g + 2$</span> points. Consider the 180 degree rotation <span class="math-container">$w$</span> through that line. Now consider the space <span class="math-container">$T/w$</span> formed by identifying two points <span class="math-container">$P$</span> and <span class="math-container">$Q$</span> if <span class="math-container">$P = wQ$</span> (since <span class="math-container">$w^2 = 1$</span> we also have <span class="math-container">$Q = wP$</span>). I claim that it's not too hard to see that <span class="math-container">$T/w$</span> is isomorphic to the projective line, and the <span class="math-container">$2g+2$</span> points which hit the line are the ramification points.</p>
<p>edit: This is of course more general than you were asking, but the picture is completely general when you're talking about topology.</p>
<p>edit 2: A picture of this (albeit approached from the perspective of starting on the projective line and cutting slits) can be found in section 20e of Fulton's Algebraic Topology book.</p>
</blockquote>
|
1,818,764 | <p>I think everything I have done is kosher, but unless I am missing an identity it is a different answer than the online quiz and wolfram alpha give.</p>
<p>I tried to use the trig substitution
$$ x=2\sin(\theta)\Rightarrow dx=2\cos(\theta)$$</p>
<p>Which yields
$$\int\frac{x^2}{\sqrt{4-x^2}}dx=\int\frac{4\sin^2(\theta)}{2\sqrt{1-\sin^2{\theta}}}2\cos(\theta)d\theta=4\int \sin^2(\theta)d\theta\\
=2\int (1-\cos(2\theta))d\theta=2\theta-\sin(2\theta)$$ </p>
<p>By the half angle formula. Then since $x=2\sin(\theta)\Rightarrow \theta=\arcsin(x/2)$ this gives a final answer of </p>
<p>$$\int\frac{x^2}{\sqrt{4-x^2}}dx=2\arcsin(x/2)-\sin(2\arcsin(x/2))+c $$</p>
<p>Is this right? If not, where did I go wrong?</p>
| SchrodingersCat | 278,967 | <p>There are $3$ mistakes that you have made:</p>
<ol>
<li>Missed the $d\theta$ in the first line. </li>
</ol>
<blockquote>
<p>$$ x=2\sin(\theta)\Rightarrow dx=2\cos(\theta)$$</p>
</blockquote>
<ol start="2">
<li>Then you did the integration in the following step as:</li>
</ol>
<blockquote>
<p>$$=2\int (1-\cos(2\theta))d\theta=2\theta-2\sin(2\theta)$$</p>
</blockquote>
<p>But actually, this is wrong. What you should have done is:</p>
<blockquote>
<p>$$=2\int (1-\cos(2\theta))d\theta=2\left(\theta-\frac{\sin 2\theta}{2}\right)=2\theta-\sin 2\theta $$</p>
</blockquote>
<ol start="3">
<li>And finally you missed the constant of integration "<strong><em>c</em></strong>"... <strong>:P</strong></li>
</ol>
<p>Hope this helps.</p>
|
32,137 | <p>I have an equation that I evaluate at some point (let's say $x=1$) that have terms of the form</p>
<pre><code>f[1,y] D[g[1,y],{y,2}]
</code></pre>
<p>Is there an easy way to replace [1,y] by [x,y] with a simple replacement rule? The thing is that </p>
<pre><code>g[1,y] /. g[1,y] -> g[x,y]
</code></pre>
<p>will indeed replace g, but it won't work for any of its derivatives, and there are far too many functions and derivatives for me to make a replacement rule for all of them. </p>
<p>Would there be a clever way to do this? So far, my only solution has been copy-pasting my equations in Word and using the text replacement tool in there. It would be neat not having to go outside of Mathematica to do this though.</p>
| István Zachar | 89 | <p>A more general approach that replaces argument lists <code>[1, y]</code> <em>anywhere</em> (note, that this might result in unwanted replacements compared to Szabolcs's solution):</p>
<pre><code>(f[1, y] D[g[1, y], {y, 2}]) /. h_[1, y] :> h[x, y]
(* ==> f[x, y]*Derivative[0, 2][g][x, y] *)
</code></pre>
|
1,323,845 | <p>For a nonnegative integer $n$, a composition of $n$ means a partition in which the order of the parts matters.</p>
<p>Consider the generating function
$$C(x) = \sum_{n=0}^{\infty} c_nx^n,$$
where $c_n$ is the number of distinct compositions of $n$ (note that $c_0=1$ by convention).</p>
<p>What is the value of $C\left(\tfrac 15\right)$?</p>
<hr>
<p>How can I start this?</p>
| Tim Raczkowski | 192,581 | <p>I don't think this is true. Take two points in an open ball. Since there is a path between the points and the image of the path is uncountable, the open ball is uncountable.</p>
|
1,323,845 | <p>For a nonnegative integer $n$, a composition of $n$ means a partition in which the order of the parts matters.</p>
<p>Consider the generating function
$$C(x) = \sum_{n=0}^{\infty} c_nx^n,$$
where $c_n$ is the number of distinct compositions of $n$ (note that $c_0=1$ by convention).</p>
<p>What is the value of $C\left(\tfrac 15\right)$?</p>
<hr>
<p>How can I start this?</p>
| Matematleta | 138,929 | <p>At least, if the open ball $B$ is connected, this is false because then if $B$ is countable, it has only has one point. To see this, suppose there are two distinct points $a,b\in B$. Then $d$ restricted to $B$ is a metric on $B$. </p>
<p>Consider S= $\left \{ d(a,x) :x\in B\right \}$. If $S$ contains every real number in $(0,d(a,b))$ then $B$ is uncountable, which is a contradiction.</p>
<p>If there is a real number $z\in (0,d(a,b))$ such that $d(a,x)\neq z$ for any $x\in B$, then the sets $\left \{ x\in B:d(a,x) >z\right \}$ and $\left \{ x\in B:d(a,x) <z\right \}$ provide a separation of $B$ and so $B$ cannot be connected. </p>
|
4,128,046 | <p>I am working through a pure maths book as a hobby. This question puzzles me.</p>
<p>The line y=mx intersects the curve <span class="math-container">$y=x^2-1$</span> at the points A and B. Find the equation of the locus of the mid point of AB as m varies.</p>
<p>I have said at intersection:</p>
<p><span class="math-container">$mx = x^2-1 \implies x^2 - mx - 1=0$</span></p>
<p>Completing the square:</p>
<p><span class="math-container">$(x-\frac{m}{2})^2-\frac{m^2}{4}= 1$</span></p>
<p><span class="math-container">$(x-\frac{m}{2})^2 = 1 + \frac{m^2}{4}$</span></p>
<p><span class="math-container">$x-\frac{m}{2} = \sqrt\frac{4+m^2}{4} = \frac{\pm\sqrt(4+m^2)}{2}$</span></p>
<p>x-coordinates for points of intersection are</p>
<p><span class="math-container">$x= \frac{-\sqrt(4+m^2) +m}{2}$</span> and <span class="math-container">$x= \frac{\sqrt(4+m^2)+m}{2}$</span></p>
<p><span class="math-container">$\implies$</span> x-co-ordinate of P is mid-way between the two above points, namely <span class="math-container">$\frac{2m}{2} = m$</span></p>
<p>So <span class="math-container">$x=m, y = mx = m^2\implies y = x^2$</span></p>
<p>But my book says <span class="math-container">$y=2x^2$</span></p>
| Vishu | 751,311 | <p>The <span class="math-container">$x$</span>-coordinate of the midpoint is the the sum of the other two <em>divided by <span class="math-container">$2$</span></em>, i.e. <span class="math-container">$$ x=\frac m2$$</span> And <span class="math-container">$$y=mx =\frac{m^2}{2} $$</span> and <span class="math-container">$$y=2x^2$$</span></p>
|
4,128,046 | <p>I am working through a pure maths book as a hobby. This question puzzles me.</p>
<p>The line y=mx intersects the curve <span class="math-container">$y=x^2-1$</span> at the points A and B. Find the equation of the locus of the mid point of AB as m varies.</p>
<p>I have said at intersection:</p>
<p><span class="math-container">$mx = x^2-1 \implies x^2 - mx - 1=0$</span></p>
<p>Completing the square:</p>
<p><span class="math-container">$(x-\frac{m}{2})^2-\frac{m^2}{4}= 1$</span></p>
<p><span class="math-container">$(x-\frac{m}{2})^2 = 1 + \frac{m^2}{4}$</span></p>
<p><span class="math-container">$x-\frac{m}{2} = \sqrt\frac{4+m^2}{4} = \frac{\pm\sqrt(4+m^2)}{2}$</span></p>
<p>x-coordinates for points of intersection are</p>
<p><span class="math-container">$x= \frac{-\sqrt(4+m^2) +m}{2}$</span> and <span class="math-container">$x= \frac{\sqrt(4+m^2)+m}{2}$</span></p>
<p><span class="math-container">$\implies$</span> x-co-ordinate of P is mid-way between the two above points, namely <span class="math-container">$\frac{2m}{2} = m$</span></p>
<p>So <span class="math-container">$x=m, y = mx = m^2\implies y = x^2$</span></p>
<p>But my book says <span class="math-container">$y=2x^2$</span></p>
| Parcly Taxel | 357,390 | <p>You made a small mistake near the end: the midpoint's <span class="math-container">$x$</span>-coordinate is not the <em>sum</em> of those of the intersection points, but half that sum, which is still <span class="math-container">$m/2$</span>. Then the corresponding <span class="math-container">$y$</span>-coordinate is <span class="math-container">$m^2/2$</span> and <span class="math-container">$y=2x^2$</span>.</p>
|
422,941 | <p>How can we expand the following by the binomial expansion, upto the term including $x^3$? That'll be 4 terms.</p>
<p>This the expression to be expanded: $\sqrt{2+x\over1-x}$</p>
<p>I understand how to do the numerator and denominator individually. Now this is what I'm doing - having expanded the denominator (using the standard expansion formula for $(1+x)^{-1}$), do I now simply need to multiply this expansion once with the numerator $(2+x)$? I'm not getting the correct answer, but is this the correct method?</p>
| André Nicolas | 6,312 | <p>You have tow alternatives, neither completely pleasant.</p>
<p>1) Let our function be $f(x)$, Compute $f(0)$, $f'(0)$, $f''(0)$, and $f'''(0)$. Then the answer is
$$f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3.$$</p>
<p>2) Or else (easier) compute the expansion of $(2+x)^{1/2}$ and $(1-x)^{-1/2}$ up to the $x^3$ terms, and multiply the two series term by term, only worrying about the constant, the term in $x$, the one in $x^2$, and the one in $x^3$.</p>
<p>For computing the expansions of $(2+x)^{1/2}$ and $(1-x)^{-1/2}$, use the "derivatives" technique, which will be much more pleasant for these functions than it is for the original function. </p>
<p><strong>Remark:</strong> You may be able to save time in (2) by using "canned" formulas (the general binomial expansion of $(1+t)^\alpha$.) If you are using such a canned formula, first express $(2+x)^{1/2}$ as $\sqrt{2}\left(1+\frac{x}{2}\right)^{1/2}$. </p>
|
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