qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
2,646,890 | <blockquote>
<p>If <span class="math-container">$p(x)$</span> is a polynomial of degree <span class="math-container">$n$</span> such that <span class="math-container">$$p(-2)=-15,\ p(-1)=1,\ p(0)=7,\ p(1)=9,\ p(2)=13,\ p(3)=25.$$</span>
Then smalest possible value of <span class="math-container">$n$</span> is</p>
<p>Options <span class="math-container">$(a)\; 2\;\;(b)\; 3\;\; (c)\;\; 4\;\; (d)\; 5$</span></p>
</blockquote>
<p>Try: Tracing curve on coordinate axis, it gave one point of intersection Further <span class="math-container">$p(x)$</span> must be an odd degree polynomial. And slope of function is not same in each interval. So it is not linear. So it must have least degree <span class="math-container">$3$</span>.</p>
<p>Can someone explain me if I am doing right? Thanks.</p>
<p>Otherwise please provide solution.</p>
| trancelocation | 467,003 | <p>You do not need to calculate any interpolation polynomial. Here, it is about the minimum degree. So the given points should have a property that makes it possible to find the degree quickly.</p>
<p>When you look at the points you see that the given $x$-values are consecutive integers.</p>
<p>So, you can make a quick check by calculating the first, second and third differences of the sequence of the $y$-values:</p>
<ul>
<li>sequence: -15,1,7,9,13,25</li>
<li>first differences: 16,6,2,4,12</li>
<li>second differences: -10,-4,2,8</li>
<li>third differences: 6,6,6</li>
</ul>
<p>As the third differences are constant the degree must be 3.</p>
|
95,126 | <p>Consider the finite sum</p>
<pre><code>rs[x_, n_] := x/n Sum[n^2/(i + (n - i) x)^2, {i, 1, n}]
</code></pre>
<p>Is there a way to bring <em>Mathematica</em> to calculate the limit for <code>n -> ∞</code>?</p>
<p>I have tried <code>Limit[]</code> as well as <code>NLimit[]</code> without success.</p>
| Anton Antonov | 34,008 | <p>For a fixed numerical value of x the sum limit can be found using <a href="https://en.wikipedia.org/wiki/Shanks_transformation">Shanks transformation</a> or <a href="https://en.wikipedia.org/wiki/Richardson_extrapolation">Richardson extrapolation</a>. The Richardson extrapolation transformation of the sequence gives faster convergence. The results seem to be in agreement with the plots in the answer by Willinski -- see the attached image.</p>
<p>Here is the code for Shanks:</p>
<pre><code>Clear[Shanks]
Shanks[A_, n_] := (A[n + 2] A[n] - A[n + 1]^2)/(
A[n + 2] + A[n] - 2 A[n + 1]);
Shanks[A_, 1, n_] := Shanks[A, n];
Shanks[A_, k_, n_] := (
Shanks[A, -1 + k, n] Shanks[A, -1 + k, 2 + n] -
Shanks[A, -1 + k, 1 + n]^2)/(
Shanks[A, -1 + k, n] + Shanks[A, -1 + k, 2 + n] -
2 Shanks[A, -1 + k, 1 + n]);
</code></pre>
<p>Here is the code for Richardson:</p>
<pre><code>Clear[Richardson]
Richardson[A_, n_, N_] :=
Total@Table[(A[n + k]*(n + k)^N *If[OddQ[k + N], -1, 1])/(
k! (N - k)!), {k, 0, N}]
</code></pre>
<p>This code makes the table in the image:</p>
<pre><code>rs[x_, n_] := x/n Sum[n^2/(i + (n - i) x)^2, {i, 1, n}]
sf[n_] := rs[1/2, n]
ns = {1, 2, 3, 4, 5, 6, 7, 15, 25, 50, 90};
res = Outer[#1[#2] &, {sf, Shanks[sf, #1] &, Shanks[sf, 3, #1] &,
Richardson[sf, #1, 1] &, Richardson[sf, #1, 6] &}, ns];
TableForm[Transpose[Map[N[#1, 20] &, res, {-1}]],
TableHeadings -> {ns, {"function value", "Shanks-1", "Shanks-3",
"Richardson-1", "Richardson-6"}}]
</code></pre>
<p><a href="https://i.stack.imgur.com/gDy0j.png"><img src="https://i.stack.imgur.com/gDy0j.png" alt="enter image description here"></a></p>
<p>The book by Bender and Orszag <a href="http://rads.stackoverflow.com/amzn/click/0387989315">"Advanced Mathematical Methods for Scientists and Engineers: Asymptotic Methods and Perturbation Theory"</a> has good explanations of the Shanks and Richardson transformations. (See chapter 8.)</p>
|
95,126 | <p>Consider the finite sum</p>
<pre><code>rs[x_, n_] := x/n Sum[n^2/(i + (n - i) x)^2, {i, 1, n}]
</code></pre>
<p>Is there a way to bring <em>Mathematica</em> to calculate the limit for <code>n -> ∞</code>?</p>
<p>I have tried <code>Limit[]</code> as well as <code>NLimit[]</code> without success.</p>
| Dr. Wolfgang Hintze | 16,361 | <p>See EDIT #3 for a valid answer.</p>
<p><strong>First answer</strong></p>
<p>This is not an answer to my question but just a mathematical derivation of the limit the existence of which was even doubted in some comments and answers.
I hope this does not spoil the creativity. The task for Mathematica is still open.</p>
<p>I came to the question considering something very elementary in calculus: the definition of the Riemann integral.</p>
<p>With this hint you surely can do the derivation by yourself. Here is mine:</p>
<p>The limit is equal to 1, as Willinski first calculated numerically.</p>
<p>Proof: consider the intergal</p>
<pre><code>fi = Integrate[1/t^2, {t, 1, x}, Assumptions -> x > 1]
(* Out[395]= (-1 + x)/x *)
</code></pre>
<p>And now consider the definition of the Riemann integral of a function g[x] between x=a and x=b as a limit of this sum </p>
<pre><code>gs[a_, b_, n_] := (b - a)/n Sum[g[a + (b - a) k/n], {k, 0, n - 1}]
</code></pre>
<p>Letting </p>
<pre><code>{a -> 1, b -> x, g[t] -> 1/t^2}
</code></pre>
<p>we have</p>
<pre><code>gs1[x_, n_] := (x - 1)/n Sum[1/(1 + (x - 1) k/n)^2, {k, 0, n - 1}]
</code></pre>
<p>This sum is equal to our sum except for the factor (x-1)/x (and the summand for k=n which, however, is negligible in the limit). Hence our limit is equal to x/(x-1) times fi, that is equal to 1. QED.</p>
<p><strong>EDIT #1</strong></p>
<p>One attempt to help Mathematica calculate the limit could be replacing the sum over i by an integral over i. Not strict but at least a heuristic approach.</p>
<p>Here we go</p>
<pre><code>x/n Integrate[n^2/(i + (n - i) x)^2, {i, 1, n}, Assumptions -> {n > 1, x > 1}]
(*
Out[481]= ((-1 + n) x)/(1 + (-1 + n) x)
*)
Limit[%, n -> \[Infinity]]
(* Out[480]= 1 *)
</code></pre>
<p>We must then justifiy the replacement.</p>
<p><strong>EDIT #2</strong></p>
<p>Let us try to callculate symbolically the sum involved</p>
<pre><code>s[n_, x_] := Sum[1/(i + (n - i) x)^2, {i, 0, n}]
Assuming[x > 1, Timing[s[n, x]]]
(* Out[27]= $Aborted *)
</code></pre>
<p>Mathematica won't do this for real x. Hence we adopt the strategy "guess from integers" which I use frequently in such cases : look for results for some interger values of x, and guess what the general expression might be.</p>
<pre><code>Table[{x, Sum[1/(i + (n - i) x)^2, {i, 0, n}]}, {x, 1, 7}] // Column
</code></pre>
<p>$$\left(
\begin{array}{cc}
2 & \psi ^{(1)}(-2 n)-\psi ^{(1)}(1-n) \\
3 & \frac{1}{4} \left(\psi ^{(1)}\left(-\frac{1}{2} (3 n)\right)-\psi ^{(1)}\left(1-\frac{n}{2}\right)\right) \\
4 & \frac{1}{9} \left(\psi ^{(1)}\left(-\frac{1}{3} (4 n)\right)-\psi ^{(1)}\left(1-\frac{n}{3}\right)\right) \\
5 & \frac{1}{16} \left(\psi ^{(1)}\left(-\frac{1}{4} (5 n)\right)-\psi ^{(1)}\left(1-\frac{n}{4}\right)\right) \\
6 & \frac{1}{25} \left(\psi ^{(1)}\left(-\frac{1}{5} (6 n)\right)-\psi ^{(1)}\left(1-\frac{n}{5}\right)\right) \\
7 & \frac{1}{36} \left(\psi ^{(1)}\left(-\frac{1}{6} (7 n)\right)-\psi ^{(1)}\left(1-\frac{n}{6}\right)\right) \\
\end{array}
\right)$$</p>
<p>Where $\psi ^{(1)}(z)$ = PolyGamma[1, z]</p>
<p>Our guess is</p>
<pre><code>sg[n_, x_] := 1/(x - 1)^2 (-PolyGamma[1, 1 - n/(x - 1)] + PolyGamma[1, -((x n)/(x - 1))])
</code></pre>
<p>But still: no results for the limit</p>
<p>For real x > 1</p>
<pre><code>Timing[Limit[n x sg[n, x], {n -> \[Infinity]}]]
(*
Out[26]= {211.21, {Limit[(
n x (-PolyGamma[1, 1 - n/(-1 + x)] +
PolyGamma[1, -((n x)/(-1 + x))]))/(-1 + x)^2, n -> \[Infinity]]}}
*)
</code></pre>
<p>And even not for x = 2</p>
<pre><code>Timing[Limit[n x sg[n, x] /. x -> 2, {n -> \[Infinity]}]]
(*
Out[28]= {18.1741, {Limit[2 n (-PolyGamma[1, 1 - n] + PolyGamma[1, -2 n]),
n -> \[Infinity]]}}
*)
</code></pre>
<p>Tough stuff for Mathematica!</p>
<p><strong>EDIT #3</strong></p>
<p>Now I've found a direct way to an answer my original question, viz. how we can bring Mathematica to calculate the limit.</p>
<p>This method generalizes to similar expressions of which we have to take the limit.</p>
<p>Letting</p>
<pre><code>z -> x - 1;
</code></pre>
<p>our sum becomes</p>
<pre><code>s := x/n Sum[1/(1 + k/n z)^2, {k, 0, n}]
</code></pre>
<p>We have to calculate the limit of s for <code>n->\[Infinity]</code>.</p>
<p>Expanding the summand into a power series</p>
<pre><code>1/(1 + k/n z)^2 -> Sum[Binomial[-2, m] k^m z^m/n^m, {m, 0, \[Infinity]}]
(*
Out[50]= 1/(1 + (k z)/n)^2 -> n^2/(n + k z)^2
*)
</code></pre>
<p>and exchanging the order of summation gives us</p>
<pre><code>s1 := x/n Sum[Binomial[-2, m] z^m c[n, m], {m, 0, \[Infinity]}]
</code></pre>
<p>where</p>
<pre><code>c[m_, n_] = Sum[k^m/n^m, {k, 0, n}]
(*
Out[53]= n^-m (0^m + HarmonicNumber[n, -m])
*)
</code></pre>
<p>Now we take the limit <code>n->\[Infinity]</code>, taking into account the factor <code>1/n</code></p>
<pre><code>clim = Limit[n^-1 c[m, n], n -> \[Infinity], Assumptions -> {m \[Element] Integers, m >= 0}]
(* Out[58]= 1/(1 + m) *)
</code></pre>
<p>and the m-sum becomes</p>
<pre><code>s2 = Sum[x z^m Binomial[-2, m] clim, {m, 0, \[Infinity]}]
(* Out[59]= x/(1 + z) *)
</code></pre>
<p>Finally, replacing z gives</p>
<pre><code>slim = s2 /. z -> x - 1
(* Out[61]= 1 *)
</code></pre>
<p>QED.</p>
|
321,916 | <p>In order to define Lebesgue integral, we have to develop some measure theory. This takes some effort in the classroom, after which we need additional effort of defining Lebesgue integral (which also adds a layer of complexity). Why do we do it this way? </p>
<p>The first question is to what extent are the notions different. I believe that a bounded measurable function can have a non-measurable "area under graph" (it should be doable by transfinite induction), but I am not completely sure, so treat it as a part of my question. (EDIT: I was very wrong. The two notions coincide and the argument is very straightforward, see Nik Weaver's answer and one of the comments).</p>
<p>What are the advantages of the Lebesgue integration over area-under-graph integration? I believe that behaviour under limits may be indeed worse. Is it indeed the main reason? Or maybe we could develop integration with this alternative approach?</p>
<p>Note that if a non-negative function has a measurable area under graph, then the area under the graph is the same as the Lebesgue integral by <a href="https://en.wikipedia.org/wiki/Fubini%27s_theorem" rel="noreferrer">Fubini's theorem</a>, so the two integrals shouldn't behave very differently.</p>
<p>EDIT: I see that my question might be poorly worded. By "area under the graph", I mean the measure of the set of points <span class="math-container">$(x,y) \in E \times \mathbb{R}$</span> where <span class="math-container">$E$</span> is a space with measure and <span class="math-container">$y \leq f(x)$</span>. I assume that <span class="math-container">$f$</span> is non-negative, but this is also assumed in the standard definition of the Lebesuge integral. We extend this to arbitrary function by looking at the positive and the negative part separately.</p>
<p>The motivation for my question concerns mostly teaching. It seems that the struggle to define measurable functions, understand their behaviour, etc. might be really alleviated if directly after defining measure, we define integral without introducing any additional notions.</p>
| Dave L Renfro | 15,780 | <p>The "area under a graph" approach is used in Wheeden/Zygmund's 1977 text <a href="https://rads.stackoverflow.com/amzn/click/com/0824764994" rel="noreferrer" rel="nofollow noreferrer"><strong>Measure and Integral. An Introduction to Real Analysis</strong></a>, a book that was used (among other possible places) in the early 1980s for a 2-semester graduate real analysis course at Indiana University (Bloomington).</p>
<blockquote>
<p><strong>(second sentence of Chapter 5, on p. 64)</strong> <em>The approach we have chosen</em> [for the integral of a nonnegative function <span class="math-container">$f:E \rightarrow [0, +\infty],$</span> where <span class="math-container">$E \subseteq {\mathbb R}^n$</span> is measurable] <em>is based on the notion that the integral of a nonnegative</em> <span class="math-container">$f$</span> <em>should represent the volume of the region under the graph of</em> <span class="math-container">$f.$</span></p>
</blockquote>
<p>I looked in the preface and elsewhere for any historical or literature citations about this approach and did not see anything relevant. Also, later in this book abstract measure and integration theory is developed in one of the standard ways.</p>
|
321,916 | <p>In order to define Lebesgue integral, we have to develop some measure theory. This takes some effort in the classroom, after which we need additional effort of defining Lebesgue integral (which also adds a layer of complexity). Why do we do it this way? </p>
<p>The first question is to what extent are the notions different. I believe that a bounded measurable function can have a non-measurable "area under graph" (it should be doable by transfinite induction), but I am not completely sure, so treat it as a part of my question. (EDIT: I was very wrong. The two notions coincide and the argument is very straightforward, see Nik Weaver's answer and one of the comments).</p>
<p>What are the advantages of the Lebesgue integration over area-under-graph integration? I believe that behaviour under limits may be indeed worse. Is it indeed the main reason? Or maybe we could develop integration with this alternative approach?</p>
<p>Note that if a non-negative function has a measurable area under graph, then the area under the graph is the same as the Lebesgue integral by <a href="https://en.wikipedia.org/wiki/Fubini%27s_theorem" rel="noreferrer">Fubini's theorem</a>, so the two integrals shouldn't behave very differently.</p>
<p>EDIT: I see that my question might be poorly worded. By "area under the graph", I mean the measure of the set of points <span class="math-container">$(x,y) \in E \times \mathbb{R}$</span> where <span class="math-container">$E$</span> is a space with measure and <span class="math-container">$y \leq f(x)$</span>. I assume that <span class="math-container">$f$</span> is non-negative, but this is also assumed in the standard definition of the Lebesuge integral. We extend this to arbitrary function by looking at the positive and the negative part separately.</p>
<p>The motivation for my question concerns mostly teaching. It seems that the struggle to define measurable functions, understand their behaviour, etc. might be really alleviated if directly after defining measure, we define integral without introducing any additional notions.</p>
| Nate Eldredge | 4,832 | <p>Suppose your main interest is in constructing the Lebesgue integral over a general abstract measure space <span class="math-container">$(X,\mu)$</span>. From the usual definitions via simple functions, this is fairly straightforward, and one can prove the standard theorems (dominated convergence, etc) without too much trouble. But if you want to use a definition as "area under the graph", you have to take a detour and construct Lebesgue measure on <span class="math-container">$\mathbb{R}$</span>, which is a fair amount of work.</p>
<p>Indeed, one can sort of see that having Lebesgue measure is overkill for this task, since the whole point of Lebesgue measure is to be able to define the length of very complicated subsets of <span class="math-container">$\mathbb{R}$</span>. But for finding the product measure of the region under the graph of a function <span class="math-container">$f : X \to \mathbb{R}$</span> via a Fubini-like approach, the only subsets of <span class="math-container">$\mathbb{R}$</span> you really need to measure are intervals, of the form <span class="math-container">$[0,f(x)]$</span>. We don't need any fancy measure theory to tell us the length of such sets.</p>
<p>Granted, in most measure theory courses you will eventually want to construct Lebesgue measure on <span class="math-container">$\mathbb{R}$</span> anyway, since an abstract theory should have good motivating examples. But from a certain point of view, having to do it first could obscure the fundamentally simple definition of the integral.</p>
|
321,916 | <p>In order to define Lebesgue integral, we have to develop some measure theory. This takes some effort in the classroom, after which we need additional effort of defining Lebesgue integral (which also adds a layer of complexity). Why do we do it this way? </p>
<p>The first question is to what extent are the notions different. I believe that a bounded measurable function can have a non-measurable "area under graph" (it should be doable by transfinite induction), but I am not completely sure, so treat it as a part of my question. (EDIT: I was very wrong. The two notions coincide and the argument is very straightforward, see Nik Weaver's answer and one of the comments).</p>
<p>What are the advantages of the Lebesgue integration over area-under-graph integration? I believe that behaviour under limits may be indeed worse. Is it indeed the main reason? Or maybe we could develop integration with this alternative approach?</p>
<p>Note that if a non-negative function has a measurable area under graph, then the area under the graph is the same as the Lebesgue integral by <a href="https://en.wikipedia.org/wiki/Fubini%27s_theorem" rel="noreferrer">Fubini's theorem</a>, so the two integrals shouldn't behave very differently.</p>
<p>EDIT: I see that my question might be poorly worded. By "area under the graph", I mean the measure of the set of points <span class="math-container">$(x,y) \in E \times \mathbb{R}$</span> where <span class="math-container">$E$</span> is a space with measure and <span class="math-container">$y \leq f(x)$</span>. I assume that <span class="math-container">$f$</span> is non-negative, but this is also assumed in the standard definition of the Lebesuge integral. We extend this to arbitrary function by looking at the positive and the negative part separately.</p>
<p>The motivation for my question concerns mostly teaching. It seems that the struggle to define measurable functions, understand their behaviour, etc. might be really alleviated if directly after defining measure, we define integral without introducing any additional notions.</p>
| Max M | 375 | <p>There are several references to various books defining the Lebesgue integral this way in the answers, but the first person to define it this way is ... Lebesgue. In his thesis "Intégrale, Longueur, Aire" (1902) he first discusses the Darboux treatment of Riemann's integral, and then defines his own integrals geometrically, and only then goes on to define them analytically. Curiously, it seems to me that he proves that his analytic definition agrees with the geometric definition for Riemann integrable functions, but he does not really say that it's also the same for all (Lebesgue) measurable functions. Here are some quotes (I think "mesurable (J)" means "measurable in the sense of Jordan"):</p>
<ol start="16">
<li>Pour interpréter géométriquement ces nombres, attachons à toute
fonction <span class="math-container">$f$</span> positive définie dans <span class="math-container">$(a, b)$</span> l'ensemble <span class="math-container">$E$</span> des points dont les coordonnées
vérifient à la fois les deux inégalités</li>
</ol>
<p><span class="math-container">$$ a\leq x\leq b \hspace{5mm} 0\leq y \leq f(x) $$</span></p>
<p>[...]</p>
<p>Si la fonction <span class="math-container">$f$</span> est de signe quelconque, nous lui faisons correspondre
l'ensemble <span class="math-container">$E$</span> des points dont les coordonnées vérifient les trois inégalités
<span class="math-container">$$ a\leq x\leq b \hspace{5mm} yf(x)\geq 0 \hspace{5mm} 0\leq y^2 \leq f(x)^2. $$</span></p>
<p>L'ensemble <span class="math-container">$E$</span> est somme de deux ensembles <span class="math-container">$E_1$</span> et <span class="math-container">$E_2$</span> formés des points
à ordonnées positives pour <span class="math-container">$E_1$</span>, et négatives pour <span class="math-container">$E_2$</span>(*). L'intégrale par défaut est l'étendue intérieure de <span class="math-container">$E_1$</span> moins l'étendue extérieure de <span class="math-container">$E_2$</span> l'intégrale par excès est l'étendue extérieure de <span class="math-container">$E_2$</span> moins l'étendue intérieure de <span class="math-container">$E_1$</span>. Si <span class="math-container">$E$</span> est mesurable (J) (auquel cas <span class="math-container">$E_1$</span> et <span class="math-container">$E_2$</span> le sont) la fonction est intégrable, l'intélgrale étant <span class="math-container">$m(E_1) - m (E_2)$</span></p>
<ol start="17">
<li>Ces résultats suggèrent immédiatement la généralisation suivante:
si l'ensemble <span class="math-container">$E$</span> est mesurable, (auquel cas <span class="math-container">$E_1$</span>, et <span class="math-container">$E_2$</span>, le sont) nous appellerons
intégrale définie de <span class="math-container">$f$</span>, prise entre <span class="math-container">$a$</span> et <span class="math-container">$b$</span>, la quantité
<span class="math-container">$$m(E_1)-m(E_2).$$</span>
Les fonctions <span class="math-container">$f$</span> correspondantes seront dites sommables.</li>
</ol>
<p>Relativement aux fonctions non sommables, s'il en existe, nous définirons
les intégrales inférieure et supérieure comme égales à
<span class="math-container">$$m_i (E_1) - m_e (E_2) \hspace{5mm} m_e (E_1) - m_i (E_2).$$</span></p>
<p>Ces deux nombres sont compris entre les intégrales par défaut et par excès.</p>
<ol start="18">
<li>Nous allons définir analytiquement les fonctions sommables.</li>
</ol>
|
230,416 | <p>I'm looking for numerical graph invariants that are bounded by a constant either for a graph $G$ or its complement $\bar{G}$. (The complement graph $\bar{G}$ has the same set of vertices as $G$ but the edges are complemented.) More specifically I’m looking for what numerical “Invariant $X$” is out there for which the following statement is true:</p>
<p>“Invariant $X$ is bounded by a constant $c$ either for $G$ or $\bar{G}$”.</p>
<p>One example is diameter $d(G)$ of a graph $G$ where it is known that:</p>
<p>"Either $d(G) \leq 3$ or $d(\bar{G}) \leq 3$" </p>
<p>(Reference: F. Harary, R. W. Robinson: The diameter of a graph and its complement , The American Mathematical Monthly, Vol. 92, No. 3. (Mar., 1985), pp. 211-212”)</p>
<p>My question is what other such numerical graph invariants are out there? I looked at the list of <a href="https://en.wikipedia.org/wiki/Category:Graph_invariants">invariants</a> on wikipedia but there was no mention of such bound on their respective pages.
Again, all I’m interested is that the condition holds for either $G$ or its complement $\bar{G}$ and not necessarily for both.</p>
| David Eppstein | 440 | <p>Another one: girth. By Ramsey's theorem, for every graph $G$ on six or more vertices, either it or its complement has girth at most three.</p>
|
1,728,595 | <p>Here is the claim I'm trying to understand: Given that $N$ is an integer-valued random variable, why is it true that</p>
<p>$$Var(N) = \sum_{i=1}^\infty Var(1_{N\ge i})$$</p>
<p>For context, this is a step in the answer to exercise 4.5.10 in Rosenthal, <em>A First Look at Rigorous Probability Theory</em>, 2nd ed., p. 53, which I'm trying to work through in self-study.</p>
<p>If one were to substitute expectation for variance, then proposition 4.2.9 would apply, which shows that $\sum_{k=1}^\infty P(X \ge k) = E\lfloor X \rfloor$. And we no longer need the floor function if $X$ is integer valued. Variance is of course defined as an expectation, but even if the original random variable ($N$ above) is integer-valued, its mean doesn't have to be, and that means that the random variable $(N - E(N))^2$ isn't necessarily integer valued, so that theorem doesn't apply directly.</p>
<p>Is there another way to see why the above claim is true? </p>
| MohsenSoltanifar | 330,952 | <p>@ MarshalFarrier, @Jimmy, @ Joriki:</p>
<p>The corrected solution (including correlation) and an additional solution has been posted here:</p>
<p><a href="http://probability.ca/jeff/ftpdir/Ex4.5.10sol.pdf" rel="nofollow">http://probability.ca/jeff/ftpdir/Ex4.5.10sol.pdf</a></p>
|
1,251,537 | <p>$f:[a,b] \to R$ is continuous and $\int_a^b{f(x)g(x)dx}=0$ for every continuous function $g:[a,b]\to R$ with $g(a)=g(b)=0$. Must $f$ vanish identically?</p>
<hr>
<p>Using integration by parts I got the form:
$\int_a^bg(x)f(x)-g'(x)F(x)=0$. Where $F'(x)=f(x)$.</p>
| agha | 118,032 | <p>Suppose that there exists $x_0$ that $f(x_0)=\varepsilon \neq 0$, we can assume that $\varepsilon>0$ without loosing of generality. $f$ is continuous, so there exists $\delta$ such that for $x_1 \in (x_0-\delta,x_0+\delta) \subset (a,b)$ we have $f(x_1) > \frac{\varepsilon}{2}$. Now you can find such a function $g$, that $g(x_2)>1$ for $x_2 \in (x_0-\frac{\delta}{2},x_0+\frac{\delta}{2})$ and $g(x) \geq 0$ (for example piecewise linear), then you have:</p>
<p>$$\int_{a}^{b}f(x)g(x)dx \geq \int_{x_0-\frac{\delta}{2}}^{x_0+\frac{\delta}{2}}f(x)g(x)dx \geq 1 \cdot 2\delta \cdot \frac{\varepsilon}{2}=\delta \varepsilon>0$$</p>
|
4,316,876 | <p>I want to prove that given <span class="math-container">$a,b,c\in\mathbb{R}$</span> we have <span class="math-container">$|a+b|\leq|a|+|b|$</span> using an absurd and reaching a contradiction.</p>
<p>So, I state, by absurd, that <span class="math-container">$|a+b|>|a|+|b|$</span>, but I can't reach the contradiction. It look simple, but I'm afraid that it isn't. In my research I didn't find this proof. Thanks for any contribution!</p>
| soupless | 888,233 | <p>Our claim is that <span class="math-container">$|a| + |b| \geq |a + b|$</span> for real <span class="math-container">$a$</span> and <span class="math-container">$b$</span> and we want to prove it by contradiction. Hence, we assume that <span class="math-container">$|a| + |b| < |a + b|$</span>.</p>
<hr />
<p>Here is a proof.</p>
<p>Case 1: <span class="math-container">$0 < a < b$</span>.</p>
<p>Because both <span class="math-container">$a$</span> and <span class="math-container">$b$</span> are greater than zero, this means that <span class="math-container">$a + b$</span> is also greater than zero. Then <span class="math-container">$|a| = a$</span>, <span class="math-container">$|b| = b$</span>, and <span class="math-container">$|a + b| = a + b$</span>. We then have <span class="math-container">$a + b > a + b$</span> which is a contradiction.</p>
<p>Case 2: <span class="math-container">$a < 0 < b$</span> and <span class="math-container">$|a| < |b|$</span>.</p>
<p>This means that <span class="math-container">$|a| + |b| > 0$</span>. Because of the conditions, we have <span class="math-container">$b > a + b \implies |b| > |a + b|$</span>. Since we have <span class="math-container">$|a| + |b| > 0$</span>, then <span class="math-container">$|a| + |b| > |b| > |a + b| \implies |a| + |b| > |a + b|$</span>. Contradiction.</p>
<p>Case 3: <span class="math-container">$a < 0 < b$</span> and <span class="math-container">$|a| > |b|$</span>.<br />
This also means that <span class="math-container">$|a| + |b| > 0$</span>. Because of the conditions, <span class="math-container">$a < a + b \implies |a| > |a + b|$</span>. Since we have <span class="math-container">$|a| + |b| > 0$</span>, then <span class="math-container">$|a| + |b| > |a| > |a + b| \implies |a| + |b| > |a + b|$</span>. Contradiction.</p>
<p>Case 4: <span class="math-container">$a < b < 0$</span>.</p>
<p>This case can be treated in a similar manner as the first case. Since both <span class="math-container">$a$</span> and <span class="math-container">$b$</span> are less than zero, this means that <span class="math-container">$a + b$</span> is also less than zero. Then <span class="math-container">$|a| = -a$</span>, <span class="math-container">$|b| = -b$</span>, and <span class="math-container">$|a + b| = -a - b$</span> and we have <span class="math-container">$-a - b > -a - b$</span>. Contradiction.</p>
<p>Case 5: <span class="math-container">$a = b$</span>.
This means that <span class="math-container">$|2a| < |2a|$</span> which is a contradiction.</p>
<p>Case 6: <span class="math-container">$a = 0$</span>.
This means that <span class="math-container">$|b| < |b|$</span> which is a contradiction.</p>
<p>Because all cases of <span class="math-container">$a$</span> and <span class="math-container">$b$</span> lead to a contradiction, our assumption must not be true. Therefore, <span class="math-container">$|a + b| \geq |a + b|$</span>.</p>
|
3,696,371 | <p>I was rolling stats for a set of characters (main + backup) with my DM, and he told me I could choose between 3 sets of two rolls. One rolled by him, one rolled by another player, and one rolled by me. Him and the other player use physical dice, rolling three dice, then rerolling the lowest value twice. Myself, I roll electronic dice, and I used a <code>5d6d2</code> command.</p>
<p>When the three of us came up with sets of rolls, the results were very different, to the point that you could easily guess which one was mine (the only that was done using electronic dice):</p>
<p>DM's rolls: {14, 17, 14, 15, 15, 18} & {17, 16, 16, 16, 16, 14}</p>
<p>Player's rolls: {15, 17, 17, 16, 18, 17} & {17, 15, 16, 16, 18, 16}</p>
<p>My rolls: {7, 12, 15, 13, 16, 11} & {16, 16, 10, 14, 14, 17}</p>
<p>This isn't that surprising: in my experience, physical dice have a tendency to roll higher (just by comparing what I roll against them, this is far from the first time when this happens). However, we started a friendly argument regarding the statistics we were using for the rolls, and whether we were actually rolling the same/equivalent thing.</p>
<p>My belief is that the two different roll scenarios are equivalent, while my DM believes it's a situation similar to the <a href="https://en.wikipedia.org/wiki/Monty_Hall_problem" rel="nofollow noreferrer">three doors problem</a>, as he calls it.</p>
<hr>
<h2>My scenario/rolls</h2>
<p>Rolling 5 six-sided dice and taking the three largest values. The <code>5d6D2</code> roll, which using <a href="http://rumkin.com/reference/dnd/diestats.php" rel="nofollow noreferrer">this page</a> we can get a neat graph of the possibilities for each roll.</p>
<p><strong>Example roll:</strong> {3, 2, 2, 1, 1} Result: 7</p>
<h2>My DM's/fellow player's scenario/rolls</h2>
<p>Rolling 3 six-sided dice, then rerolling the lowest value twice.</p>
<p><strong>Example roll:</strong> Roll: {6, 4, 1} Result: 11</p>
<p>Reroll the 1. Roll: {4} Result: 4</p>
<p>Reroll either 4. Roll: {1} Result: 1</p>
<p>Final roll: {6, 4, 4} Result: 14</p>
<hr>
<p>Is there a statistical distribution difference between these two scenarios?</p>
| heropup | 118,193 | <p>There is a bit of ambiguity in Scenario 2. If we interpret the procedure literally, it is possible to roll the maximum outcome of <span class="math-container">$18$</span> before the two re-rolls occur. If the re-rolls are required no matter the earlier result, then we have what I call Scenario <span class="math-container">$2$</span>a. If we stop rolling as soon as a total of <span class="math-container">$18$</span> is attained, I call this Scenario <span class="math-container">$2$</span>b.</p>
<p>In any case, the two scenarios indeed do not have the same distribution.</p>
<p>First, it is easy to obtain an empirical distribution with simulation. With <span class="math-container">$10^6$</span> simulations per scenario, my results were as follows:</p>
<p><span class="math-container">$$
\begin{array}{c|ccc}
\text{Total} & \color{darkblue}{\text{Scenario 1}} & \color{orange}{\text{Scenario 2a}} & \color{green}{\text{Scenario 2b}} \\
\hline
3 & 141 & 139 & 134 \\
4 & 638 & 649 & 690 \\
5 & 1950 & 2633 & 2567 \\
6 & 5393 & 6719 & 6671 \\
7 & 11743 & 15014 & 15304 \\
8 & 21859 & 28780 & 28813 \\
9 & 37989 & 50901 & 50688 \\
10 & 60521 & 79104 & 79556 \\
11 & 85190 & 110311 & 110929 \\
12 & 113285 & 137672 & 137669 \\
13 & 135566 & 151791 & 148344 \\
14 & 148771 & 137877 & 135410 \\
15 & 142639 & 115516 & 113169 \\
16 & 120283 & 87112 & 84080 \\
17 & 78226 & 53661 & 50417 \\
18 & 35806 & 22121 & 35559 \\
\end{array}$$</span>
Although they seem reasonably similar, they are clearly not once plotted.</p>
<p><a href="https://i.stack.imgur.com/JfbBq.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JfbBq.gif" alt="enter image description here"></a></p>
<p>To prove the distributions are not equivalent, we can compute the probability of obtaining a sum of <span class="math-container">$18$</span>. In scenario <span class="math-container">$1$</span>, there are <span class="math-container">$6^5$</span> total elementary outcomes on five dice. Of these, we require at least three of the dice to show <span class="math-container">$6$</span>. There is one outcome in which there are exactly <span class="math-container">$5$</span> sixes; <span class="math-container">$5(5) = 25$</span> outcomes with exactly <span class="math-container">$4$</span> sixes; and <span class="math-container">$\binom{5}{2}(5^2) = 250$</span> outcomes with exactly <span class="math-container">$3$</span> sixes; thus the probability of getting a sum of <span class="math-container">$18$</span> is <span class="math-container">$$\frac{250+25+1}{6^5} = \frac{23}{648} \approx 0.0354938,$$</span> consistent with the simulation.</p>
<p>To count the same probability for Scenario <span class="math-container">$2$</span>a, it is important to note that in the initial roll, at least one die must be a six, since there are only two re-rolls allowed. If all three dice are six, we re-roll any die and re-roll it a second time. The first re-roll is irrelevant, since no matter the value, it will be rolled again. So in the unique initial outcome of <span class="math-container">$3$</span> sixes, the probability of maintaining it is <span class="math-container">$1/6$</span>.</p>
<p>If only two of the three dice are six after the initial roll, the same logic applies as in the previous case, and the probability of getting the last die to be a six is <span class="math-container">$1/6$</span>.</p>
<p>If only one die is a six after the initial roll, we must roll both of the dice that are not six, and each of those re-rolls must result in a six, so the conditional probability of this result is <span class="math-container">$1/36$</span>.</p>
<p>Now we compute the total probability as <span class="math-container">$$\frac{1}{6^3}\cdot\frac{1}{6} + \frac{3(5)}{6^3} \cdot \frac{1}{6} + \frac{3(5^2)}{6^3} \cdot \frac{1}{36} = \frac{19}{864} \approx 0.0219907.$$</span> Again, this is consistent with our simulation. </p>
<p>I leave Scenario <span class="math-container">$2$</span>b to the reader to consider.</p>
|
2,669,524 | <p>I am reading <strong>Algebraic Geometry</strong>, Vol 1, <em>Kenji Ueno</em>. My problem is that
$$k\left[ x,y,t\right]/\left(xy-t\right)\otimes_{k\left[t\right]}k\left[t\right]/\left(t-a\right) \simeq k\left[x,y\right]/\left(xy-a\right) $$
where $k$ is a field and $a$ is an element in $k$. I don't understand how it works. I tried to use the result $$R\otimes_A A/I =R/IR,$$ where $A$ and $R$ are commutative rings and $I$ is an ideal of $A$. Then I obtain
$$LHS=\left[k\left[ x,y,t\right]/\left(xy-t\right)\right]/\left[\left(t-a\right)k\left[ x,y,t\right]/\left(xy-t\right)\right].$$ But now it seems be hard to manage. Please help me. Thank you very much for helps.</p>
| stochastic | 491,395 | <p>An integral is defined by the limit of Riemann sum as <span class="math-container">$$\int_{V_1}^{V_2}f(V) dV = \lim_{N\to\infty}\sum_{\substack{V=V_1\\\text{with increments }\frac{\Delta V}{N}}}^{V_2} f(V)\; \frac{\Delta V}{N},$$</span>
where <span class="math-container">$\Delta V=V_2-V_1$</span>. The founders of calculus picked the notation of integrals based on this definition:</p>
<ul>
<li>The symbol <span class="math-container">$\int$</span> is a short for <span class="math-container">$$\lim_{N\to\infty}\sum_{\substack{V=V_1\\\text{with increments }\frac{\Delta V}{N}}}^{V_2}$$</span></li>
<li>and the symbol <span class="math-container">$dV$</span> is short for <span class="math-container">$$\frac{\Delta V}{N}$$</span>
with the understanding that the limit of <span class="math-container">$N\to\infty$</span> will be taken <strong>after</strong> it is summed over in the above expression.</li>
</ul>
<p>In your case <span class="math-container">$f(V)=1$</span>, and you have
<span class="math-container">$$\int_{V_1}^{V_2}dV = \lim_{N\to\infty}\sum_{\substack{V=V_1\\\text{with increments }\frac{\Delta V}{N}}}^{V_2} \frac{\Delta V}{N} = \lim_{N\to\infty}N \frac{\Delta V}{N} =\Delta V.$$</span></p>
<p>So back to your questions:</p>
<blockquote>
<p>if my understanding is correct, I technically can't assign a "size" to <span class="math-container">$dV$</span> because it's an infinitesimal</p>
</blockquote>
<p>That is true. From the above notation, you can see that <span class="math-container">$dV$</span> is, in fact, <span class="math-container">$\Delta V/N$</span> with the understanding that the limit of <span class="math-container">$N\to \infty$</span> will be taken after it is summed over. You cannot assign a value to it because the limit cannot be evaluated until the sum is computed. That is <span class="math-container">$dV$</span> does not have a meaning when taken out of the integral.</p>
<blockquote>
<p>So it would be incorrect to say <span class="math-container">$dV = \Delta V$</span>.</p>
</blockquote>
<p>Yes, that would be incorrect. </p>
<p>I think there is an unfortunate confusion of notations here: if we had <span class="math-container">$\int_a^b dx$</span>, with the above notation, <span class="math-container">$dx$</span> would be a symbol for <span class="math-container">$(b-a)/N$</span> with the understanding that <span class="math-container">$N\to \infty$</span> limit will be taken later. Sometimes the term <span class="math-container">$(b-a)/N$</span> is written as <span class="math-container">$\Delta x$</span> and the limit of <span class="math-container">$N\to \infty$</span> is expressed as limit of <span class="math-container">$\Delta x\to0.$</span> In this case, <span class="math-container">$dx$</span> is in fact a notation that represents <span class="math-container">$\Delta x$</span> with the understanding that the limit of <span class="math-container">$\Delta x\to 0$</span> will be taken after the sum. Now, in your problem, we already have something called <span class="math-container">$\Delta V$</span> and it is not an infinitesimal change, it is what we called <span class="math-container">$b-a$</span> in this example. I hope I didn't make it more confusing. If I did, just ignore this example.</p>
<blockquote>
<p>if I were using <span class="math-container">$dV$</span> in problem-solving and needed to assign a size to this infinitesimal interval, all that I'd have to do is integrate it using a definite integral, correct?</p>
</blockquote>
<p>Often in physics and engineering, when we need to calculate the total change in something, we would break it down to infinitesimals and then integrate over it. If that is what you mean, then you are correct. Basically, you are writing the total change as a sum of smaller changes of size <span class="math-container">$\Delta V/N$</span>, adding them up, and then taking the limit of <span class="math-container">$N\to\infty$</span>.</p>
|
4,004,978 | <blockquote>
<p>For all <span class="math-container">$a, b, c, d > 0$</span>, prove that
<span class="math-container">$$2\sqrt{a+b+c+d} ≥ \sqrt{a} + \sqrt{b} + \sqrt{c} + \sqrt{d}$$</span></p>
</blockquote>
<p>The idea would be to use AM-GM, but <span class="math-container">$\sqrt{a} + \sqrt{b} + \sqrt{c} + \sqrt{d}$</span> is hard to expand. I also tried squaring both sides, but that hasn't worked either. Using two terms at a time doesn't really work as well. How can I solve this question? Any help is appreciated.</p>
| radekzak | 876,772 | <p>You can use QM-AM for <span class="math-container">$\sqrt{a},\ \sqrt{b},\ \sqrt{c},\ \sqrt{d}$</span>; I'm leaving the details for you.</p>
|
2,429,164 | <p>I'd like to prove
$$\lim_{n\rightarrow+\infty} \int_0^\pi n\sqrt{n^2+x^2-2nx\cos\theta}-n^2 \mathrm{d}\theta = \frac{\pi}{4}x^2$$
for $x\in[0,\infty)$. I checked it numerically and derived it with Matlab Symbolic Toolbox, but cannot prove it by calculus.</p>
<p>I cannot use the Dominated Convergence Theorem since I could not find any $L^1[0,\pi]$ function of $\theta$ dominating $n\sqrt{n^2+x^2-2nx\cos\theta}-n^2$. I also tried integration by parts, but it didn't work.</p>
<p>Has anyone any idea? (change of variable to get rid of $\cos\theta$, perhaps?)
Thanks </p>
<hr>
<p><strong>Update, following Claude's solution</strong></p>
<p>I'm trying to follow Claude's solution, but I've encountered a problem.
I had tried before to transform my integral into the complete elliptic integral of the second kind,</p>
<p>Following your notation,
\begin{equation*}
\begin{array}{rcl}
\displaystyle{\int_0^\pi \sqrt{1-k\cos\theta}~d\theta} &=&
\displaystyle{\int_0^\pi \sqrt{1-k\left(1-2\sin^2\frac{\theta}{2}\right)}}~d\theta \\
&=& \displaystyle{\int_0^\frac{\pi}{2} \sqrt{1-k+2k\sin^2\theta}}~d\theta \\
&=& 2 \sqrt{1-k}\displaystyle{\int_0^\frac{\pi}{2}}\sqrt{1+\frac{2k}{1-k}\sin^2\theta}~d\theta
\end{array}
\end{equation*}
But then I got stuck, because the complete elliptic integral of the second kind is of the form
$$\int_0^\frac{\pi}{2} \sqrt{1-a^2\sin^2\theta}~d\theta$$
with $a\in[0,1]$, right? Where did you get your 2nd equation from?</p>
| Claude Leibovici | 82,404 | <p>It seems that you are entering the world of elliptic integrals.</p>
<p>$$I=\int\sqrt{n^2+x^2-2nx\cos\theta} \,d\theta = \sqrt{n^2+x^2} \int \sqrt{1- \frac {2nx}{n^2+x^2}\cos\theta}\,d\theta $$ and $$\int \sqrt{1-k\cos\theta}\,d\theta=-2 \sqrt{k-1} E\left(\frac{\theta}{2}|\frac{2 k}{k-1}\right)$$ making the definite integral
$$ J=\int_0^\pi n\sqrt{n^2+x^2-2nx\cos\theta}-n^2 \,d\theta = 2 n (n+x)\,E\left(\frac{4 n x}{(n+x)^2}\right)-\pi n^2$$ Now, using the expansion around $t=0$$$E(t)=\frac{\pi }{2}-\frac{\pi t}{8}-\frac{3 \pi t^2}{128}+O\left(t^3\right)$$ will show the limit and how it is approached.
$$J=\frac{\pi x^2}{4}+\frac{\pi x^4}{64 n^2}+O\left(\frac{1}{n^3}\right)$$</p>
|
621,409 | <p>I need some help with the following question:</p>
<p>We have $H$ acting by automorphisms on $N$, and let $\rho:H\to Aut(N)$ the associated representation by automorphisms.</p>
<p>Suppose that $G=H[N]_{\rho}$ is a semidirect product, and $K=\ker(\rho)$.</p>
<p>Prove that $K\unlhd G$ and that $G/K$ is also a semidirect product. </p>
<p>Thanks a lot in advance for any help!</p>
<hr>
<p>Edit: I deleted the part that was unclear (in fact bad formulated). The answers of DonAntonio and user finally solved the question.</p>
| amWhy | 9,003 | <p>Using polar coordinates, and substituting $r^2 = x^2 + y^2$, we have $$\lim_{(x, y) \to (0, 0)}\frac{1-\cos(x^2+y^2)}{(x^2+y^2)^2}=\lim_{r \to 0} \dfrac {1 - \cos(r^2)}{(r^2)^2} = \lim_{r \to 0} \frac{1 - \cos(r^2)}{r^4}$$</p>
<p>You can use l'hopital, now, repeatedly (three times)..</p>
|
748,325 | <p>In order to prove non-uniqueness of singular vectors when a repeated singular value is present, the book (Trefethen), argues as follows: Let $\sigma$ be the first singular value of A, and $v_{1}$ the corresponding singular vector. Let $w$ be another linearly independent vector such that $||Aw||=\sigma$, and construct a third vector $v_{2}$ belonging to span of $v_{1}$ and $w$, and orthogonal to $v_{1}$. All three vectors are unitary, so $w=av_{1}+bv_{2}$ with $|a|^2+|b|^2=1$, and $v_{2}$ is constructed (Gram-Schmidt style) as follows:</p>
<p>$$ {v}_{2}= \dfrac{{w}-({v}_{1}^{T} w ){v}_{1}}{|| {w}_{1}-({v}_{1}^{T} {w} ){v}_{1} ||_{2}}$$</p>
<p>Now, Trefethen says, $||A||=\sigma$, so $||Av_{2}||\le \sigma$ but this must be an equality (and so $v_{2}$ is another singular vector relative to $\sigma$), since otherwise we would have $||Aw||<\sigma$, in contrast with the hypothesis.</p>
<p>How that? I cannot see any elementary application of triangle inequality or Schwarz inequality to prove this claim.</p>
<p>I am pretty well convinced of partial non-uniqueness of SVD in certain situations. Other proofs are feasible, but I wish to undestand this specific algebraic step of this specific proof.</p>
<p>Thanks.</p>
| Ewan Delanoy | 15,381 | <p>This is actually very simple. The main point (seemingly missed by the OP) is that $Av_1$ and $Av_2$ must be orthogonal (this is something obvious to people familiar with proofs of SVD, because the induction decomposes into orthogonal spaces).</p>
<p>For every $z\in{\mathbb C}$, you have </p>
<p>$$
||A(v_2+zv_1)||^2 \leq \sigma^2 ||v_2+zv_1||^2 \tag{1}
$$</p>
<p>Expanding, one obtains</p>
<p>$$
||Av_2||^2+2{\sf Re}\bigg(\bar{z}\big<Av_2,Av_1\big>\bigg)+|z|^2||Av_1||^2 \leq \sigma^2 (||v_2||^2+|z|^2||v_1||^2) \tag{2}
$$</p>
<p>which simplifies to</p>
<p>$$
2{\sf Re}\bigg(\bar{z}\big<Av_2,Av_1\big>\bigg) \leq \sigma^2||v_2||^2-||Av_2||^2 \tag{3}
$$</p>
<p>Letting $z=\big<Av_2,Av_1\big>t$ with $t\in{\mathbb R}$, we deduce
that $2t\Big|\big<Av_2,Av_1\big>\Big|^2 \leq \sigma^2||v_2||^2-||Av_2||^2$, and this is possible
only if $\big<Av_2,Av_1\big>=0$. We then have, for any $a,b$ with
$|a|^2+|b|^2=1$ and $b\neq 0$,</p>
<p>$$
\sigma^2||av_1+bv_2||^2-||A(av_1+bv_2)||^2=
|b|^2 (\sigma^2||v_2||^2-||Av_2||^2) \tag{4}
$$</p>
<p>So $||Aw||=\sigma||w||$ forces $||Av_2||=\sigma||v_2||$.</p>
|
3,197,262 | <p>When I was solving <span class="math-container">$ \operatorname{Cov}(X,E(X\mid Y)) = \operatorname{var}(E(X\mid Y))$</span>, I notice that <span class="math-container">$E(X\mid Y)$</span> was treated as a function of <span class="math-container">$Y$</span>.
My thinking is <span class="math-container">$E(X\mid Y)$</span> is taking values of <span class="math-container">$ \operatorname{Range}(Y) $</span> and for each value of <span class="math-container">$Y$</span>, it maps to the expectation of <span class="math-container">$X$</span>. Is this correct?</p>
| dnqxt | 651,088 | <p>The answer is already given. Here is an example.</p>
<p>Let a discrete r.v. <span class="math-container">$X$</span> have values <span class="math-container">$x\in \{ -6,-3,7,14\}$</span>, and <span class="math-container">$P(X=x)=(0.1,0.2,0.3,0.4)$</span> respectively. Define a r.v. <span class="math-container">$Y$</span> s.t. <span class="math-container">$Y=0$</span> if <span class="math-container">$X \le0$</span>, and <span class="math-container">$Y=1$</span> if <span class="math-container">$X>0$</span>. </p>
<p>Then <span class="math-container">$E(X|Y)$</span> assumes different values for different values of <span class="math-container">$Y$</span> as follows.</p>
<p><span class="math-container">$Y=0:$</span>
<span class="math-container">$$E(X|Y=0)=\sum_{x\in\{-6,-3\}} xP(X=x|Y=0)=-6\times1/3-3\times2/3=-4$$</span></p>
<p><span class="math-container">$Y=1$</span>:
<span class="math-container">$$E(X|Y=1)=\sum_{x\in\{7,14\}} xP(X=x|Y=1)=7\times3/7+14\times4/7=11$$</span></p>
<p>Obviously, <span class="math-container">$E(X|Y)$</span> is a function of <span class="math-container">$Y$</span>. Also, note how the conditional probabilities, <span class="math-container">$P(X=x|X=y)$</span>, change from the original marginal probabilities, <span class="math-container">$P(X)$</span>, for different <span class="math-container">$y$</span>.</p>
|
1,043,094 | <p>I have to find the limit of following</p>
<p><span class="math-container">$$\lim_{x \to 0}\left(\frac{1}{x} - \frac{1}{x^2}\right)$$</span></p>
<p>I have no idea how to start this one off.
How would I do it?</p>
<p>Do I just substitute the <span class="math-container">$0$</span>? It doesn't look that easy and simple. The answer says it's negative infinity.</p>
<p>Please show me a solution without graphing(unless for better explanation).</p>
| Timbuc | 118,527 | <p>An idea. We take $\;x\;$ very close to zero, say $\;|x|<10^{-4}\;$ :</p>
<p>$$x>0:\;\;\frac1x-\frac1{x^2}=\frac{x-1}{x^2}<\frac{-\frac12}{x^2}=-\frac1{2x^2}$$</p>
<p>and now you only have to show the rightmost expresion is unbounded below, which I think is pretty easy.</p>
|
1,043,094 | <p>I have to find the limit of following</p>
<p><span class="math-container">$$\lim_{x \to 0}\left(\frac{1}{x} - \frac{1}{x^2}\right)$$</span></p>
<p>I have no idea how to start this one off.
How would I do it?</p>
<p>Do I just substitute the <span class="math-container">$0$</span>? It doesn't look that easy and simple. The answer says it's negative infinity.</p>
<p>Please show me a solution without graphing(unless for better explanation).</p>
| k170 | 161,538 | <p>Here's another way to prove the following statement
<span class="math-container">$$\lim\limits_{x \to 0}\left(\frac{1}{x} - \frac{1}{x^2}\right)=-\infty$$</span>
Which is equivalent to the following
<span class="math-container">$$\forall N\gt0,\exists\delta\gt 0:0\lt\left|x\right|\lt\delta\Rightarrow\frac{1}{x}-\frac{1}{x^2}\lt -N$$</span>
So whenever <span class="math-container">$0\lt\left|x\right|\lt\delta$</span>, we have
<span class="math-container">$$\frac{1}{x}-\frac{1}{x^2}=\frac{x-1}{x^2}\lt\frac{\delta-1}{x^2}$$</span>
Here we can make <span class="math-container">$\left|x\right|$</span> arbitrarily small by presupposing a bound of <span class="math-container">$\frac12$</span>. We can then use this bound to potentially get a smaller <span class="math-container">$\delta$</span> later. Assuming that <span class="math-container">$|x|\lt\frac12$</span>, we have
<span class="math-container">$$x^2\lt\frac14\Rightarrow-\frac{1}{x^2}\lt-4$$</span>
Which implies that
<span class="math-container">$$\frac{\delta-1}{x^2}=-\frac{1-\delta}{x^2}\lt -4(1-\delta)=-N$$</span>
Now we can set <span class="math-container">$\delta$</span> to the smallest of these two bounds
<span class="math-container">$$\delta =\min\left(\frac12, 1-\frac{N}{4}\right)$$</span>
Putting it all together
<span class="math-container">$$0\lt\left|x\right|\lt\delta\Rightarrow\frac{1}{x}-\frac{1}{x^2}\lt -4(1-\delta)=-N$$</span>
Therefore
<span class="math-container">$$\lim\limits_{x \to 0}\left(\frac{1}{x} - \frac{1}{x^2}\right)=-\infty$$</span></p>
|
864,237 | <p>Let's take a short exact sequence of groups
$$1\rightarrow A\rightarrow B\rightarrow C\rightarrow 1$$
I understand what it says: the image of each homomorphism is the kernel of the next one, so the one between $A$ and $B$ is injective and the one between $B$ and $C$ is surjective. I get it. But other than being a sort of curiosity, what is it really telling me?</p>
| Community | -1 | <p>"The kernel of $B \to C$" is often not a satisfactory description of a group. Having a group $A$ that it is isomorphic to (along with a given isomorphism) is very useful.</p>
<p>"The quotient $B/A$" is often not a satisfactory description of a gruop. Having a group $C$ that it is isomorphic to (along with a given isomorphism) is very useful.</p>
|
949,512 | <p>How do mathematicians define inner product on a vector space. </p>
<p>For example: $a = (x_1,x_2)$ & $ b =(y_1,y_2) $ in $ \mathbb{R}^2.$ </p>
<p>Define $\langle a,b\rangle= x_1y_1-x_2y_1-x_1y_2+4x_2y_2$. It's an inner product.</p>
<p>But how does one motivate this inner product? I think there is some sort of matrix multiplication between some vectors.</p>
| Benjamin | 118,815 | <p>The abstract definition of an inner product of real-valued vectors is a function $\langle \, , \rangle: \mathbb R^n \times \mathbb R^n \to \mathbb R$ satisfying the following axioms, where $\alpha$ and $\beta$ are scalars and the $x$'s and $y$'s are vectors.</p>
<ol>
<li>$\langle \alpha x_1 + \beta x_2, y \rangle = \alpha\langle x_1, y \rangle + \beta\langle x_2, y \rangle$</li>
<li>$\langle x, \alpha y_1 + \beta y_2 \rangle = \alpha\langle x, y_1 \rangle + \beta\langle x, y_2 \rangle$</li>
<li>$\langle x, y \rangle = \langle y, x \rangle$</li>
<li>$\langle x, x \rangle \geq 0$ and $\langle x, x \rangle = 0$ if and only if $x$ is the zero vector.</li>
</ol>
<p>Conditions (1) and (2) tell us that an inner product should be linear in both its arguments.
In addition, condition (2) is not strictly necessary, as conditions (1) and (3) imply it.
This definition also extends to arbitrary vector spaces, not just over $\mathbb R$.</p>
|
949,512 | <p>How do mathematicians define inner product on a vector space. </p>
<p>For example: $a = (x_1,x_2)$ & $ b =(y_1,y_2) $ in $ \mathbb{R}^2.$ </p>
<p>Define $\langle a,b\rangle= x_1y_1-x_2y_1-x_1y_2+4x_2y_2$. It's an inner product.</p>
<p>But how does one motivate this inner product? I think there is some sort of matrix multiplication between some vectors.</p>
| Dan Rust | 29,059 | <p>Every positive-definite matrix $M$ is associated to an inner product when viewed as a quadratic form. In this case the matrix is $M=\begin{pmatrix}1&-1\\-1&4\end{pmatrix}$ and the inner product is given by $\langle x,y\rangle=x^TMy$.</p>
|
633,223 | <p><img src="https://i.stack.imgur.com/xVS2C.png" alt="enter image description here"></p>
<p>This one has a great degree of self-evidence. Paradoxically, I find it difficult to deduce it from primitive propositions. The book only hinted ❋4.21 and ❋4.22.</p>
| Mauro ALLEGRANZA | 108,274 | <p>I think that we may simplify a little bit Albert's proof.</p>
<p>From :</p>
<p>$*4.22. \vdash (p \equiv q) \land (q \equiv r) \supset (p \equiv r)$</p>
<p>applying :</p>
<p>$*3.3.\vdash ((p \land q) \supset r) \supset (p \supset (q \supset r))$</p>
<p>we get directly :</p>
<blockquote>
<p>$\vdash (p \equiv q) \supset ((q \equiv r) \supset (p \equiv r))$.</p>
</blockquote>
<p>Now, we restart from :</p>
<p>$*4.22. \vdash (q \equiv p) \land (p \equiv r) \supset (q \equiv r)$</p>
<p>to get in the same way :</p>
<p>$\vdash (q \equiv p) \supset ((p \equiv r) \supset (q \equiv r))$.</p>
<p>Now, using $*4.21. \vdash (p \equiv q) \equiv (q \equiv p)$ we have :</p>
<blockquote>
<p>$\vdash (p \equiv q) \supset ((p \equiv r) \supset (q \equiv r))$.</p>
</blockquote>
<p>At this point we need </p>
<p>$*3.43. \vdash [(p \supset q) \land (p \supset r)] \supset (p \supset (q \land r))$</p>
<p>to get :</p>
<p>$\vdash (p \equiv q) \supset [((q \equiv r) \supset (p \equiv r)) \land ((p \equiv r) \supset (q \equiv r))]$</p>
<p>and finally apply $*4.01.$ (the def of $\equiv$) to obtain :</p>
<blockquote>
<blockquote>
<p>$*4.86. \vdash (p \equiv q) \supset ((p \equiv r) \equiv (q \equiv r)).$</p>
</blockquote>
</blockquote>
|
1,370,576 | <p>I am working on a trigonometry question at the moment and am very stuck. I have looked through all the tips to solving it and I cant seem to come up with the right answer. The problem is </p>
<blockquote>
<p>What is exact value of<br>
$$\cot \left(\frac{7\pi}{6}\right)? $$</p>
</blockquote>
| Narasimham | 95,860 | <p>$$\cot \left(\frac{7\pi}{6}\right) = \cot \left(\frac{\pi}{6}\right) = \sqrt 3 $$</p>
<p>because $$ \tan ( \theta + \pi) = \tan ( \theta ) $$</p>
|
274,961 | <p>I want to calculate the determinant along the last slice of a 3-dimensional array. So for I do this by slow the <code>Table</code> command. I know that for time reasons I should use <code>Map</code> or <code>Apply</code>, however couldn't successful solve the problem.</p>
<pre><code>m = 2; n = 3; o = 10;
SeedRandom[0];
x = RandomReal[{-1, 1}, {m, n, o}];
Table[Sqrt[Det[x[[;;,;;,i]].Transpose[x[[;;,;;,i]]]]],{i,1,o}]
Map[Sqrt[Det[# . Transpose[#]]] &, {x}, {2}]
(*{1.04663,0.437045,0.479911,0.260814,0.205563,0.171896,1.20112,1.00502,0.855893,0.125758}*)
(*{{5.38795,4.19589}}*)
</code></pre>
| Daniel Huber | 46,318 | <p>You must rearrange the matrix so that the blocks you want the Det from, appear at level 1:</p>
<p>Consider the first block or matrix:</p>
<pre><code>x[[All, All, 1]] // MatrixForm
</code></pre>
<p><a href="https://i.stack.imgur.com/F1AXt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/F1AXt.png" alt="enter image description here" /></a></p>
<p>We can rearrange x so that these matrixes appear at level 1:</p>
<pre><code>Transpose[x, {2, 3, 1}][[1]] // MatrixForm
</code></pre>
<p><a href="https://i.stack.imgur.com/we4uz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/we4uz.png" alt="enter image description here" /></a></p>
<p>With this we can calculate the Det's using Map:</p>
<pre><code>Map[Sqrt[Det[# . Transpose[#]]] &, Transpose[x, {2, 3, 1}], {1}]
(* {1.04663, 0.437045, 0.479911, 0.260814, 0.205563, 0.171896, 1.20112, \
1.00502, 0.855893, 0.125758} *)
</code></pre>
|
650,450 | <p>Suppose that $a_n$ and $b_n$ are Cauchy sequences, and that $a_n < b_n$ for all n. Prove that $\lim_{x \to \infty}a_n \le \lim_{x \to \infty}b_n$ for all n.</p>
<p>Is it sufficient to say that we know both Cauchy sequences must converge to the limit, and since $a_n$ is always less than $b_n$, the limits will follow the desired inequality?</p>
<p>Edit: since this is not true, what would be the appropriate strategy to prove?</p>
| Community | -1 | <p>Let $a_n$ and $b_n$ be your two sequences such that $a_n<b_n$ for all $n$. Consider the
sequence $\{b_n-a_n\}$ which is greater than $0$ for all $n$. This sequence can also be seen to be Cauchy. Of course, all Cauchy sequences in $\mathbb{R}$ converge. Call its limit $x$. Suppose its limit was strictly
less than $0$. Since $x$ is less than $0$, there is a $\varepsilon>0$ such that
$(x-\varepsilon,x+\varepsilon)\subseteq (-\infty,0)$. Now since $\lim_{n\rightarrow\infty} \{b_n-a_n\}=x$, there exists an $N$ such that $b_N-a_N\in(x-\varepsilon,x+\varepsilon)$; that is, $b_N-a_N<0$ which is a contradiction. We must have the limit is at least $0$. In the case
that both limits are finite, it's an easy lemma that $\lim_{n\rightarrow\infty} (b_n-a_n)=\lim_{n\rightarrow\infty} b_n - \lim_{n\rightarrow\infty} a_n$ (which is our case as both $a_n$ and $b_n$ are Cauchy sequences). Thus $0\leq\lim_{n\rightarrow\infty} (b_n-a_n)=\lim_{n\rightarrow\infty} b_n - \lim_{n\rightarrow\infty} a_n$ which implies $\lim_{n\rightarrow\infty} a_n \leq \lim_{n\rightarrow\infty} b_n$.</p>
|
452,803 | <p>Test the convergence of improper integrals :</p>
<p>$$\int_1^2{\sqrt x\over \log x}dx$$</p>
<p>I basically have no idea how to approach a problem in which log appears. Need some hint on solving this type of problems.</p>
| Mhenni Benghorbal | 35,472 | <p>First make the change of variables $\ln(x)=u$</p>
<p>$$\int_1^2{\sqrt x\over \log x}dx =\int _{0}^{\ln \left( 2 \right) }\!{\frac {{{\rm e}^{3/2\,u}}}{u}}{du}.$$</p>
<p>Now, you can see that the integrand behaves as</p>
<p>$$ {\frac {{{\rm e}^{3/2\,u}}}{u}}\sim _{u\to 0} \frac{1}{u} $$</p>
<p>which is not integrable on the given interval. </p>
|
378,953 | <p><strong>Problem:</strong> Give an example of a permutation of the first $n$ natural numbers from which it is impossible to get to the standard permutation $1,2,\ldots,n$ after less than $n-1$ transposition operations (i.e switching the place of two elements).</p>
<p><strong>My attempt</strong></p>
<p>Suppose we have a permutation $T$ and we perform one transposition on $T$ to get $T'$. That would mean $T(i) = j, T(i') = j'$ and $T'(i) = j' , T'(i') = j$ for some $i,j,i',j'$. It is easy to see that the permutation T contains 2 cycles (may be the same): $(i,j,...)$ and $(i',j',\ldots)$. The transposition operation would affect only these two cycles but keep all other cycles intact. Therefore, the number of cycles, if decreased, will not decrease more than $1$.</p>
<p>Now the permutation $[1,2,\ldots,n]$ has n cycles and the permutation $[2,3,\ldots,n,1]$ has 1 cycle only. So it is impossible to use less than $n-1$ operations to get $[2,3,\ldots,n,1]$ from $[1,2,\ldots,n]$. Keeping in mind that getting from permutation A to B is the same as getting from B to A, problem solved.</p>
<p><strong>My question</strong></p>
<p>Is my approach correct and are there any better solutions? Thank you.</p>
| zyx | 14,120 | <p>There is a quantifiable sense in which permutations with more motion require more transpositions to achieve.</p>
<p>For a sequence of transpositions of pairs in a set $S$, build an undirected graph, with vertices $S$ and an edge between every pair that is exchanged at some time in the sequence. </p>
<p>The vertex sets of the connected components of this transposition graph are permuted within themselves. To reach all permutations the entire graph must be a single component. This requires at least $n-1$ edges for $n$ vertices (at minimum, a tree).</p>
<p>This assumes that the transpositions can be chosen to fit the permutation one wants to achieve. If there are $n-1$ transpositions given, then although any permutation can be reached using a chain of those transpositions, the length of the chain can be as high as $n(n-1)/2$, the number of exchanges of adjacent elements needed to reverse the order of an array of $n$ distinct objects.</p>
<p>Same argument shows that $\frac{n-1}{k-1}$ cycles of length $k$ are needed to sort arrays of length $n$. </p>
|
4,330,991 | <p>I do understand that if:</p>
<p><span class="math-container">$a=b \Rightarrow a^2 = b^2 $</span></p>
<p>But clearly, the graph representing these two equations won't be the same. So, (correct me if I'm wrong) this would suggest that if you square both sides of the equation, you essentially get a different set of answers (or graph). What confuses me is this question from my textbook:</p>
<blockquote>
<p>Find and graph all <span class="math-container">$z$</span> such that <span class="math-container">$|z-3| = |z+2i|$</span>.</p>
</blockquote>
<p>The solution goes as such:</p>
<p><span class="math-container">$z=a+bi$</span></p>
<p><span class="math-container">$\sqrt{(a-3)^2+b^2}$</span> = <span class="math-container">$\sqrt{a^2+(b+2)^2}$</span></p>
<p>Squaring both sides then simplifying we end up with the equation:</p>
<p><span class="math-container">$6a + 4b = 5$</span></p>
<p>It proceeds to graph the equation on the complex plane.</p>
<p>How can we claim that the graph of the equation <span class="math-container">$6a + 4b = 5$</span> represents the graph of
<span class="math-container">$\sqrt{(a-3)^2+b^2}$</span> = <span class="math-container">$\sqrt{a^2+(b+2)^2}$</span>
when squaring both sides of the equation was an intermediary step? Doesn't that create extraneous solutions which ends up being graphed? Doesn't this mean that our new graph represents more solutions then what the initial equation was intended for?</p>
<p>I hope that made sense...</p>
<p>It's quite frustrating looking back at some of these concepts you thought you understood and realizing that you didn't.</p>
<p>Anyways, thanks in advance!</p>
| José Carlos Santos | 446,262 | <p>In general, <span class="math-container">$a=b$</span> is not equivalent to <span class="math-container">$a^2=b^2$</span>; for instance <span class="math-container">$1^2=(-1)^2$</span>, but <span class="math-container">$1\ne-1$</span>.</p>
<p>However, if <span class="math-container">$a,b\geqslant0$</span>, then, yes <span class="math-container">$a=b\iff a^2=b^2$</span>. Clearly, as you wrote, <span class="math-container">$a=b\implies a^2=b^2$</span>. And<span class="math-container">$$a^2=b^2\implies\sqrt{a^2}=\sqrt{b^2}\iff a=b,$$</span>since <span class="math-container">$\sqrt{a^2}=a$</span> and <span class="math-container">$\sqrt{b^2}=b$</span>.</p>
<p>Now, use the fact that <span class="math-container">$\sqrt{(a-3)^2+b^2},\sqrt{a^2+(b+2)^2}\geqslant0$</span>.</p>
|
4,140,956 | <p>I'm trying to determine the order of the pole in the complex expression</p>
<p><span class="math-container">$$f(z)=\frac{1}{(1-\cos(z))^2}$$</span></p>
<p>I have determined the pole to be <span class="math-container">$z=2\pi n, n\in \mathbb{Z}$</span>.</p>
<p>However, when I use the equation <span class="math-container">$\lim\limits_{z\rightarrow 2\pi n}[(z-2\pi n)^k f(z)]$</span> with <span class="math-container">$k=1$</span>, it equals <span class="math-container">$\frac{0}{1}=0$</span> or that the function is analytical in the neighborhood. I have used L'Hôpital's rule repeatedly to obtain this result. I checked my answer with Wolfram Alpha, and it's supposed to have a pole of order <span class="math-container">$4$</span> and <span class="math-container">$z=2\pi n$</span>. Where am I going wrong?</p>
| TheSimpliFire | 471,884 | <p>If a function <span class="math-container">$g$</span> has a pole of order <span class="math-container">$k$</span> then <span class="math-container">$g^2$</span> will have a pole of order <span class="math-container">$2k$</span>. Using your method, we have by L'Hopital, <span class="math-container">$$\lim_{z\to2\pi n}\frac{z-2\pi n}{1-\cos z}=\lim_{z\to2\pi n}\frac1{\sin z}=\infty$$</span> but <span class="math-container">$$\lim_{z\to2\pi n}\frac{(z-2\pi n)^2}{1-\cos z}=\lim_{z\to2\pi n}\frac{2(z-2\pi n)}{\sin z}=\lim_{z\to2\pi n}\frac2{\cos z}=2$$</span> so <span class="math-container">$1/(1-\cos z)$</span> has a pole of order <span class="math-container">$2$</span>. Therefore, <span class="math-container">$f(z)=1/(1-\cos z)^2$</span> has a pole of order <span class="math-container">$4$</span>.</p>
<p>In your attempt, the limit with <span class="math-container">$k=1$</span> is in fact of the form <span class="math-container">$1/0$</span> not <span class="math-container">$0/1$</span>, so <span class="math-container">$f$</span> is not analytic as you claim.</p>
|
2,250,469 | <p>Let n $\geq$ 4 be an integer. Determine the number of permutations of
$\{1, 2, . . . , n\}$, in which $1$ and $2$ are next to each other, with $1$ to the left of $2$.<br>
I can't make sense of this problem statement. The way I see it, if $n$ is an integer, then the pair $1,2$ could be formed by any pair with the form $\overline{...x_{i-2}x_{i-1}x_i1}, \overline{2y_{1}y_2y_3...}$ or a number with the form $\overline{...x_{i-2}x_{i-1}x_i12x_{i+1}x_{i+2}x_{i+3}..}$ with $x$'s and $y$'s are some mysterious digits. Can anyone explain this problem?</p>
| user247327 | 247,327 | <p>Think of "12" as a single object, say "a". Then the problem becomes to determine the number of permutations of the n- 1 objects, "a, 3, 4, ..., n". There are (n-1)! such permutations.</p>
|
335,651 | <p>I'm having trouble proving $$\left(\frac{\sin(\frac{n\theta}{2})}{\sin(\frac{\theta}{2})}\right)^2=\left|\sum_{k=1}^{|n|}e^{ik\theta}\right|^2$$ where $n\in\mathbb{Z}$ and $\theta\in\mathbb{R}$. Can anyone suggest a hint?</p>
| prob_noob | 67,596 | <p>Using the sum of sines and cosines with arguments in arithmetic progression as given above: if $\theta\ne0$ and let $\varphi =0$, then we have,
\begin{align} &S =\sin{(\theta)} + \sin{(2\theta)} + \cdots + \sin{(n\theta)} = \frac{\sin{\left(\frac{(n+1) \theta}{2}\right)} \cdot \sin{(\frac{n \theta}{2})}}{\sin{\frac{\theta}{2}}} \quad\hbox{and}\\[10pt] &C =\cos{(\theta)} + \cos{(2\theta)} + \cdots+ \cos{(n\theta)} = \frac{\cos{\left(\frac{(n+1) \theta}{2}\right)} \cdot \sin{(\frac{n \theta}{2})}}{\sin{\frac{\theta}{2}}}. \end{align} </p>
<p>From Euler's Formula and the definition of absolute value of a complex number, we can write $$\left|\sum_{k=1}^{|n|}e^{ik\theta}\right|^2 = C^2 + S^2$$
$$= \frac{\sin{\left(\frac{n \theta}{2}\right)}^2}{\sin{\frac{\theta}{2}}^2}\cdot\left(\cos{(\frac{(n+1) \theta}{2})}^2 + \sin{(\frac{(n+1) \theta}{2})}^2\right)$$</p>
<p>$$ = \left(\frac{\sin{\frac{n \theta}{2}}}{\sin{\frac{\theta}{2}}}\right)^2$$</p>
|
335,651 | <p>I'm having trouble proving $$\left(\frac{\sin(\frac{n\theta}{2})}{\sin(\frac{\theta}{2})}\right)^2=\left|\sum_{k=1}^{|n|}e^{ik\theta}\right|^2$$ where $n\in\mathbb{Z}$ and $\theta\in\mathbb{R}$. Can anyone suggest a hint?</p>
| lab bhattacharjee | 33,337 | <p>Using <a href="http://mathworld.wolfram.com/EulerFormula.html" rel="nofollow">Euler's Formula</a> , $e^{ix}= \cos x+i\sin x$</p>
<p>So, $e^{-ix}= \cos(-x)+i\sin (-x)=\cos x-i\sin x\implies 2i\sin x=e^{ix}-e^{-ix}$ </p>
<p>$$\text{ If }n>0,\sum_{k=1}^{|n|}e^{ik\theta}= \sum_{k=1}^n e^{ik\theta}=e^{i\theta}\left( \frac{e^{in\theta}-1}{e^{i\theta}-1}\right)$$
$$=\frac{e^{in\frac\theta2}}{e^{i\frac\theta2}}\frac{(e^{in\frac\theta2}-e^{-in\frac\theta2})}{(e^{i\frac\theta2}-e^{-i\frac\theta2})}$$
$$=e^{i\frac{(n-1)\theta}2}\frac{2i\sin \frac{n\theta}2}{2i\sin \frac{\theta}2}$$
$$=\left(\cos \frac{(n-1)\theta}2+i\sin \frac{(n-1)\theta}2\right) \frac{\sin \frac{n\theta}2}{\sin \frac{\theta}2}$$</p>
<p>Taking modulus $$\left|\sum_{k=1}^{|n|}e^{ik\theta}\right|$$
$$=\left|\left(\cos \frac{(n-1)\theta}2+i\sin \frac{(n-1)\theta}2\right) \frac{\sin \frac{n\theta}2}{\sin \frac{\theta}2}\right|$$</p>
<p>$$=\left| \cos \frac{(n-1)\theta}2+i\sin \frac{(n-1)\theta}2 \right|\left|\frac{\sin \frac{n\theta}2}{\sin \frac{\theta}2}\right|$$</p>
<p>$$=\left|\frac{\sin \frac{n\theta}2}{\sin \frac{\theta}2}\right|$$</p>
<p>Similarly, for $n<0$</p>
|
2,262,371 | <p>If $a,b,c$ are positive real numbers, prove that
$$\frac{2}{a+b}+\frac{2}{b+c}+ \frac{2}{c+a}≥ \frac{9}{a+b+c}$$</p>
| Darth Geek | 163,930 | <p>Let $a+b+c = s$. Then we have to prove</p>
<p>$$\dfrac{1}{s-a} + \dfrac{1}{s-b} + \dfrac{1}{s-c} \geq \dfrac{9}{2s},$$</p>
<p>or, equivalently, </p>
<p>$$\dfrac{3}{\dfrac{1}{s-a} + \dfrac{1}{s-b} + \dfrac{1}{s-c}} \leq \dfrac{2s}{3}.$$</p>
<p>Note that the LHS is the harmonic mean of $s-a,s-b,s-c$ and the RHS is the arithmetic mean of the same numbers. This inequality is true by the AM-HM inequality.</p>
|
4,004,827 | <p>I need to calculate:
<span class="math-container">$$\displaystyle \lim_{x \to 0^+} \frac{3x + \sqrt{x}}{\sqrt{1- e^{-2x}}}$$</span></p>
<p>I looks like I need to use common limit:
<span class="math-container">$$\displaystyle \lim_{x \to 0} \frac{e^x-1}{x} = 1$$</span></p>
<p>So I take following steps:</p>
<p><span class="math-container">$$\displaystyle \lim_{x \to 0^+} \frac{3x + \sqrt{x}}{\sqrt{1- e^{-2x}}} = \displaystyle \lim_{x \to 0^+} \frac{-3x - \sqrt{x}}{\sqrt{e^{-2x} - 1}}$$</span></p>
<p>And I need to delete root in the denominator and make nominator equal to <span class="math-container">$-2x$</span>. But I don't know how.</p>
| José Carlos Santos | 446,262 | <p>Note that<span class="math-container">\begin{align}\lim_{x\to0^+}\frac{\left(3x+\sqrt x\right)^2}{1-e^{-2x}}&=\lim_{x\to0^+}\frac{9x^2+6x\sqrt x+x}{1-e^{-2x}}\\&=-\lim_{x\to0^+}\frac x{e^{-2x}-1}\left(9x+6\sqrt x+1\right)\\&=-\frac1{\lim_{x\to0^+}\frac{e^{-2x}-1}x}\times\lim_{x\to0^+}\left(9x+6\sqrt x+1\right)\\&=-\frac1{-2}\times1\\&=\frac12.\end{align}</span>Therefore, the limit that you're after is <span class="math-container">$\sqrt{1/2}$</span>.</p>
|
200,093 | <p>I have a BLDC electric motor, I'm currently trying to control via a <code>PIDTune</code>. This is mostly an attempt to reduce (remove) a small run away drift that ends up showing up in the motor signal <code>u[t]</code>.</p>
<p>I've modelled this via:</p>
<pre><code>ssm = StateSpaceModel[\[ScriptCapitalJ] \[Phi]''[t] + \[ScriptCapitalR] \[Phi]'[t] == \[ScriptCapitalT] u[t], {{\[Phi][t], 0}, {\[Phi]'[t], 0}, {u[t], 1}}, u[t], \[Phi]'[t], t]
</code></pre>
<p>And simulated: </p>
<pre><code>params = { \[ScriptCapitalJ] -> 4.63 10^-5, \[ScriptCapitalR] -> 1 10^-5, \[ScriptCapitalT] -> 0.0335};
Plot[Evaluate[OutputResponse[ssm /. params, 1, {t, 0, 12}]], {t, 0, 12}]
</code></pre>
<p><a href="https://i.stack.imgur.com/vSYOC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vSYOC.png" alt="plot"></a></p>
<p>This is a nice model response and mirrors the response of the real motor almost exactly. </p>
<p>So I tried to create a control system to add to the control signal and bring the system relatively quickly back to zero. </p>
<pre><code>control = PIDTune[ssm /. params , {"PID"}]
</code></pre>
<p>But I continue to get the following error:</p>
<pre><code>PIDTune::infgains: Unable to compute finite controller parameters because a denominator in the tuning formula is effectively zero.
</code></pre>
<p>I have tried <em>all</em> tuning methods within the documentation, however I continue to get errors.</p>
<p>Changing to a "PD" control</p>
<pre><code>control = PIDTune[ssm /. params , {"PD"}]
</code></pre>
<p>Gives me control system, however when adding it to the feedback and then seeing the response I get a different error:</p>
<pre><code>simul = SystemsModelFeedbackConnect[ssm, control] /. params
OutputResponse[simul, UnitStep[t - 3], {t, 0, 12}]
OutputResponse::irregss: A solution could not be found for the irregular state-space model with a characteristic equation of zero.
</code></pre>
<p>The error messages don't really make any sense to me...or explain what the issue is with the model...being that it simulates reality quite well....How can I relieve these errors, or create a feedback loop via <code>PIDTune</code> for my system?</p>
<p>Thank you for the help!</p>
<p>There is a similar example with a dcmotor within the documentation for <code>PIDTune</code> for reference which works fine (albeit a different tfm):</p>
<pre><code>dcMotor = TransferFunctionModel[Unevaluated[{{k/(s ((j s + b) (l s + r) + k^2))}}], s, SamplingPeriod ->None, SystemsModelLabels -> {{None}, {None}}] /. pars;
PIDTune[dcMotor, "PID", "PIDData"]
</code></pre>
<p><strong>Update</strong></p>
<p>As per M.K.s suggestion, I have changed the ssm slightly, or rather rewritten it to come directly to the equation of motion for angular velocity omega, instead of the motors angle phi. This change simplifies the ssm and allows <code>PIDTune</code> to come up with a solution.</p>
<p>As a small explanation, the ODE is derived via <a href="http://www.site.uottawa.ca/~rhabash/StateSpaceModelBLDC.pdf" rel="nofollow noreferrer">equation 6 of this paper</a> as a simplified motor for control via amperage of u[t]. Though is is a relatively 'standard' equation used and can be found in many papers. J and R were found via nonlinearfitting of driving the motor at different amperages. As such, the model params, J, T, R are quite accurate. </p>
<pre><code>ssmnew = StateSpaceModel[\[ScriptCapitalJ] \[Omega]'[t] + \[ScriptCapitalR] \[Omega][t] == \[ScriptCapitalT] u[t], {{\[Omega][t], 0}}, {{u[t]}}, {\[Omega][t]}, t]
control = PIDTune[ssmnew /. params, {"PID"}]
loop = SystemsModelFeedbackConnect[ssmnew, control] /. params
test1 OutputResponse[loop, UnitStep[t - 4], {t, 0, 12}]
</code></pre>
<p>or </p>
<pre><code> test2 = OutputResponse[control /. params, UnitStep[t - 3], {t, 0, 10}]
</code></pre>
<p>Unfortunately at this point, I am now getting either new errors, or a response that is completely wrong, using inputs of <code>UnitStep</code> or just <code>1</code></p>
<pre><code>NDSolve::ndsz: At t == 4.000000000000114`, step size is effectively zero; singularity or stiff system suspected.
</code></pre>
<p>or </p>
<pre><code>NDSolve::irfail: Unable to reduce the index of the system to 0 or 1.
</code></pre>
<p><a href="https://i.stack.imgur.com/Chmos.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Chmos.png" alt="plot4"></a></p>
| JJBK | 59,726 | <p>Here's an example of a basic PD controller based on frequency domain tuning. The technique is described in any basic control book.
Basic tuning rule-of-thumbs that have been applied (no further advanced tuning)</p>
<p>Let's write <code>ssm</code> as a transfer function:</p>
<pre><code>tf = TransferFunctionModel[ssm]
</code></pre>
<p>Given some desired crossover <code>w0</code>, we choose</p>
<ul>
<li>kp = 1/3 * 1/tf[w0]</li>
<li>wd = kp/kd = w0/3</li>
</ul>
<p>and we make a PD controller c</p>
<pre><code>c[kp_, wd_] := TransferFunctionModel[kp + kp/wd s, s]
</code></pre>
<p>As an example, we can show responses in a manipulate (as function of the desired crossover frequency).</p>
<pre><code>Manipulate[kp = 1/(3. ((tf /. params)@wc)[[1, 1]]);
wd = wc/3;
ctf = SystemsModelSeriesConnect[c[kp, wd], tf /. params];
Evaluate[{BodePlot[c[kp, wd], {.01, 1000}, PlotLabel -> "Controller",
PlotLegends -> "Expressions"],
BodePlot[tf /. params, {.01, 1000}, PlotLabel -> "Plant"],
BodePlot[ctf, {.01, 1000}, PlotLabel -> "Controller times plant", StabilityMargins -> True],
NyquistPlot[ctf, PlotLabel -> "Nyquist"]}],
{{wc, 100, "Open-loop crossover"}, 10, 1000}]
</code></pre>
<p>Based on your requirements (unknown to me), you can add features to your controller. Hope this helps</p>
|
128,666 | <p>If we start with a number like 1234 and produce the following sum 1234 + 123 + 12 + 1 = 1370.</p>
<p>If we are given the 1370 can I retrieve the 1234? A similar question was migrated over to the Math.SE because the OP did not in any way relate it to MMa. The math given over there is in no way too tough for anyone over here but the solution will still require MMA for anything more than a trivial problem. I wish to do it using MMa. This is what I did:</p>
<p>Suppose we have the non trivial problem of reconstructing the original number from 308460277.</p>
<p>Basically, I just used the structure of the FindInstance command as a template and used MMa commands to fill in the various fields.</p>
<pre><code>n = 9; (*digit length of the number*)
m = 308460277;
var = Table[Subscript[a, k], {k, n, 1, -1}];
f = Sum[1/9 (-1 + 10^k) Subscript[a, k], {k, 1, n}] == m;
s = FindInstance[Append[{f, 10 > Subscript[a, 1] > 0},
Table[10 > Subscript[a, k] >= 0, {k, 2, n}]] // Flatten, var,Integers];
var /. s
(*{{2, 7, 7, 6, 1, 4, 2, 5, 3}}*)
</code></pre>
<p>So the starting number was 277614253</p>
<p>To check:</p>
<p>277614253 + 27761425 + 2776142 + 277614 + 27761 + 2776 + 277 + 27 + 2 = 308460277</p>
<p>This is pretty fast and can do 98766665555567902460 instantaneously.</p>
<p>What is the right way to do this?</p>
<p>Addenda: My solution gags on m = 137174210013717421001371742085. I was worried about FindInstance being able to do larger ones, that is the reason I posted for a better way.</p>
| J. M.'s persistent exhaustion | 50 | <p>A ploddingly procedural implementation:</p>
<pre><code>n = 308460277;
FromDigits[Reap[Do[{q, n} = QuotientRemainder[n, (10^k - 1)/9]; Sow[q],
{k, IntegerLength[n], 1, -1}]][[-1, 1]]]
277614253
</code></pre>
|
2,120,539 | <p>Find the points of local maximum and minimun of the function:
$$f(x)=\sin^{-1}(2x\sqrt{1-x^2})~~~~;~~x\in (-1,1)$$
I know
$$f'(x)=-\frac{2}{\sqrt{1-x^2}}$$</p>
<p>How to find the local maximum and minimum? I have drawn the fig and seen the points of local maximum and minimum. But how to find then analytically?
<a href="https://i.stack.imgur.com/Z2fN6.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Z2fN6.jpg" alt="enter image description here"></a></p>
| Community | -1 | <p><strong>Hint</strong>: We know that $$\arcsin x + \arcsin y = \arcsin (x\sqrt{1-y^2} + y\sqrt {1-x^2}) $$ If $x=y $, then $$2\arcsin x = \arcsin (2x\sqrt {1-x^2}) $$</p>
<p>Hope you can take it from here. </p>
|
2,942,204 | <p>I was reviewing <span class="math-container">$\mathbb{R}-$</span>analisys with a friend and I'm thinking about one of the questions...</p>
<blockquote>
<p>Prove by the <span class="math-container">$\epsilon-\delta$</span> limit definition that <span class="math-container">$\lim_{x\rightarrow 2}\frac{x^2-1}{x-3}=-3$</span>.</p>
</blockquote>
<p>My answer was very long, someone could do a better answer? Does is it OK in this way? Thanks very much.</p>
<p><strong>My attempt</strong></p>
<p>We have <span class="math-container">$|f(x)-(-3)|=|f(x)+3|=|\frac{x^2-1}{x-3}+3|=|\frac{x^2-1+3x-9}{x-3}|=|\frac{x^2+3x-10}{x-3}|=|\frac{(x-2)(x+5)}{x-3}| \qquad (i)$</span></p>
<p>and
<span class="math-container">$|x+5|=|x+7-2|<|x-2|+|7|=|x-2|+7 \qquad (ii)$</span>
<span class="math-container">$ |x-3|=|x-2-1|>|x-2|+|-1|=|x-2|+1\Longrightarrow \dfrac{1}{|x-3|}<\dfrac{1}{|x-2|+1} \qquad(iii)$</span></p>
<p>So, by (ii) and (iii)
<span class="math-container">$|\frac{(x-2)(x+5)}{x-3}|=\dfrac{|x-2||x+5|}{|x-3|}<\dfrac{|x-2|(|x-2|+7)}{|x-2|+1} \qquad (iv)$</span></p>
<p>But also, if <span class="math-container">$|x-2|<\delta \qquad (v)$</span>, we have
<span class="math-container">$|\frac{(x-2)(x+5)}{x-3}|<^{(iv)}\dfrac{|x-2|(|x-2|+7)}{|x-2|+1}<^{(v)}\dfrac{\delta(\delta+7)}{\delta+1} \qquad(vi)$</span></p>
<p>And more than that
<span class="math-container">$\dfrac{\delta(\delta+7)}{\delta+1}<\dfrac{\delta(\delta+1)}{\delta+1}=\delta \qquad (vii)$</span></p>
<p>So, if <span class="math-container">$\delta<\epsilon$</span>, we have
<span class="math-container">$|f(x)-(-3)|=^{(i)}|\frac{(x-2)(x+5)}{x-3}|<^{(vi)}\dfrac{\delta(\delta+7)}{\delta+1}<^{(vii)}\delta<\epsilon$</span>, Q.E.D.</p>
| Hector Blandin | 170,571 | <p>Hint: Recall that <span class="math-container">$0<a<b$</span> implies that <span class="math-container">$\frac{1}{a}>\frac{1}{b}>0$</span>. So,</p>
<p><span class="math-container">$$0<\vert x-2 \vert<\delta$$</span>
<span class="math-container">$$\Longrightarrow \ 1<\vert x-2 \vert+1<\delta+1 $$</span>
<span class="math-container">$$\Longrightarrow \ 1>\frac{1}{\vert x-2\vert+1}>\frac{1}{\delta+1} $$</span></p>
|
2,942,204 | <p>I was reviewing <span class="math-container">$\mathbb{R}-$</span>analisys with a friend and I'm thinking about one of the questions...</p>
<blockquote>
<p>Prove by the <span class="math-container">$\epsilon-\delta$</span> limit definition that <span class="math-container">$\lim_{x\rightarrow 2}\frac{x^2-1}{x-3}=-3$</span>.</p>
</blockquote>
<p>My answer was very long, someone could do a better answer? Does is it OK in this way? Thanks very much.</p>
<p><strong>My attempt</strong></p>
<p>We have <span class="math-container">$|f(x)-(-3)|=|f(x)+3|=|\frac{x^2-1}{x-3}+3|=|\frac{x^2-1+3x-9}{x-3}|=|\frac{x^2+3x-10}{x-3}|=|\frac{(x-2)(x+5)}{x-3}| \qquad (i)$</span></p>
<p>and
<span class="math-container">$|x+5|=|x+7-2|<|x-2|+|7|=|x-2|+7 \qquad (ii)$</span>
<span class="math-container">$ |x-3|=|x-2-1|>|x-2|+|-1|=|x-2|+1\Longrightarrow \dfrac{1}{|x-3|}<\dfrac{1}{|x-2|+1} \qquad(iii)$</span></p>
<p>So, by (ii) and (iii)
<span class="math-container">$|\frac{(x-2)(x+5)}{x-3}|=\dfrac{|x-2||x+5|}{|x-3|}<\dfrac{|x-2|(|x-2|+7)}{|x-2|+1} \qquad (iv)$</span></p>
<p>But also, if <span class="math-container">$|x-2|<\delta \qquad (v)$</span>, we have
<span class="math-container">$|\frac{(x-2)(x+5)}{x-3}|<^{(iv)}\dfrac{|x-2|(|x-2|+7)}{|x-2|+1}<^{(v)}\dfrac{\delta(\delta+7)}{\delta+1} \qquad(vi)$</span></p>
<p>And more than that
<span class="math-container">$\dfrac{\delta(\delta+7)}{\delta+1}<\dfrac{\delta(\delta+1)}{\delta+1}=\delta \qquad (vii)$</span></p>
<p>So, if <span class="math-container">$\delta<\epsilon$</span>, we have
<span class="math-container">$|f(x)-(-3)|=^{(i)}|\frac{(x-2)(x+5)}{x-3}|<^{(vi)}\dfrac{\delta(\delta+7)}{\delta+1}<^{(vii)}\delta<\epsilon$</span>, Q.E.D.</p>
| Nosrati | 108,128 | <p>I think (i) part is correct, and assume we choose <span class="math-container">$x$</span> from <span class="math-container">$|x-2|<\dfrac12$</span> then
<span class="math-container">$$\dfrac32<x<\dfrac52$$</span>
<span class="math-container">$$-\dfrac32<x-3<-\dfrac12$$</span>
<span class="math-container">$$\dfrac{13}{2}<x+5<\dfrac{15}{2}$$</span>
these show
<span class="math-container">$$|\frac{(x-2)(x+5)}{x-3}|<\dfrac{15}{2}|x-2|.2<15\delta$$</span>
so it is sufficient to have <span class="math-container">$\delta\leq\min\{\dfrac{1}{15}\varepsilon,\dfrac12\}$</span>.</p>
|
634,132 | <p>Let $G$ be a cyclic group with $N$ elements. Then it follows that</p>
<p>$$N=\sum_{d|N} \sum_{g\in G,\text{ord}(g)=d} 1.$$</p>
<p>I simply can not understand this equality. I know that for every divisor $d|N$ there is a unique subgroup in $G$ of order $d$ with $\phi(d)$ elements. But how come that when you add all these together you end up with the number of elements in the group $G$. </p>
| Doug Spoonwood | 11,300 | <p>If we assume that $\forall$x(x=x), then yes.</p>
<p>Suppose that b=c.</p>
<p>f(a,b)=f(a,b) since we can derive it from substitution and that $\forall$x(x=x).</p>
<p>Now since b=c we can replace just the second "b" by "c" and obtain</p>
<p>f(a,b)=f(a,c).</p>
|
806,476 | <p>In Milnor's book <em>Topology from the Differentiable Viewpoint</em> there's the following problem:</p>
<p><strong>Problem $6$</strong> (Brouwer). Show that any map $S^n\to S^n$ with degree different from $(-1)^{n+1}$ must have a fixed point.</p>
<p><strong>My solution:</strong> Assume that the map $f:S^n\to S^n$ has no fixed points. Let $a:S^n\to S^n$ denote the antipodal map $a(x)=-x$. Then the map $a\circ f$ is homotopic to the identity as follows:</p>
<p>Since $x$ is never mapped to $-x$ (by assumption), there exist a unique shortest great circle arc from $a(f(x))$ to $x$, simply take the straight line homotopy flowing along such arches to get a homotopy $a\circ f\simeq\operatorname{id}$.</p>
<p>Now we have:
$$1=\deg(\operatorname{id})=\deg(a\circ f)=\deg(a)\deg(f)=(-1)^{n+1}\deg(f)$$
and thus $\deg(f)=(-1)^{n+1}$.</p>
<p>My questions:</p>
<ol>
<li>Is this solution correct?</li>
<li>How to show that the homotopy is continuous? It seems intuitively true to me, but I'm not sure about how to proceed to show it. Maybe saying that it is the flow of some good tangent field?</li>
<li>Are there other (elegant/short/interesting) proofs for this fact?</li>
</ol>
| Behnam Esmayli | 283,487 | <p>Best solution: Notice that $\|f(x)-(a \circ f) (x)\| < 2$ for every $x.$ But then a problem above this one in the book, asks to prove that such maps must be homotopic (smoothly so if maps themselves are smooth.) Then your computation goes thru with nothing on conscience!</p>
|
3,173,242 | <p><strong>Context:</strong></p>
<p>In the context of circuit theory and graph theory, suppose we have a graph <span class="math-container">$G,$</span> then <a href="https://en.wikipedia.org/wiki/Laplacian_matrix" rel="nofollow noreferrer">the Laplacian (Kirchhoff) matrix</a> <span class="math-container">$L$</span> is defined as follows: </p>
<p><span class="math-container">$$
L = D-A \tag{1}
$$</span>
where <span class="math-container">$D$</span> is the degree matrix and <span class="math-container">$A$</span> the adjacency matrix. Alternatively, in terms of the incidence matrix <span class="math-container">$M$</span> it can be also expressed as:</p>
<p><span class="math-container">$$
L = M^T M \tag{2}
$$</span></p>
<p>I am interested in the case where the graph <span class="math-container">$G$</span> is connected. Then if we partition <span class="math-container">$G$</span> into connected subgraphs that are themselves connected, the Laplacian can then be represented as a block matrix. Let's assume we partition the graph into two parts, the boundary nodes <span class="math-container">$B$</span> and the connection nodes <span class="math-container">$C.$</span> Then <span class="math-container">$|V(G)|=|V(G_B)|+|V(G_C)|.$</span> The corresponding Laplacian matrix can be put in the form of the following block matrix:</p>
<p><span class="math-container">$$
L = \begin{bmatrix}
L_{BB} & L_{BC} \\
L_{CB} & L_{CC}
\end{bmatrix}
\tag{3}
$$</span></p>
<p>One often is interested in reducing the size of the network/graph, using a scheme such as the Kron reduction, which relies on taking the <a href="https://en.wikipedia.org/wiki/Schur_complement" rel="nofollow noreferrer">Schur complement</a> (Laplacian matrices are <a href="https://en.wikipedia.org/wiki/M-matrix" rel="nofollow noreferrer">M-Matrices</a>) of <span class="math-container">$L$</span> with respect to one of its blocks. But that is only possible if the submatrix of the chosen block is invertible, and in general this is assumed to follow from the connectedness of the graph.</p>
<p><strong>Questions:</strong></p>
<ol>
<li>Does the fact that <span class="math-container">$G$</span> and its subcomponents (<span class="math-container">$B$</span> and <span class="math-container">$C$</span>) are connected ensure that the submatrices of the block representation of the Laplacian <span class="math-container">$L$</span> are invertible? (does this follow from a known theorem?) And therefore the Schur complements can be safely taken. </li>
<li>In relation to connected graphs and their corresponding incidence matrices <span class="math-container">$M$</span> [*], is there a simple way to see that the kernel of <span class="math-container">$M$</span> is given by <span class="math-container">$\text{ker } M^T=\text{span } \mathbb{1}?$</span></li>
</ol>
<p>[*]: The incidence matrix <span class="math-container">$M$</span> is usually defined with each column corresponding to an edge of <span class="math-container">$G,$</span> where then each column contains exactly one <span class="math-container">$1$</span> at the row corresponding to the tail-node of the edge and a <span class="math-container">$-1$</span> at the row corresponding to the head-node of the edge.</p>
| Romi | 606,615 | <p>For your first question, I think you can easily prove that <span class="math-container">$L_{BB}$</span> and <span class="math-container">$L_{CC}$</span> are
<a href="https://en.wikipedia.org/wiki/Weakly_chained_diagonally_dominant_matrix" rel="nofollow noreferrer">weakly chained diagonally dominant matrices</a> from your connection assumptions:</p>
<p>For each row of <span class="math-container">$L_{BB}$</span> (similar proof holds for <span class="math-container">$L_{CC}$</span>) there is two possible cases:</p>
<ul>
<li>the node associated to the row in <span class="math-container">$B$</span> is connected to some node in <span class="math-container">$C$</span>. Then the row is strictly diagonally dominant (as there are non 0 coefficients in the same row in <span class="math-container">$L_{BC}$</span>).</li>
<li>the node associated to the row in <span class="math-container">$B$</span> is not connected to some node in <span class="math-container">$C$</span> and then diagonally dominant but not strictly. But as you graph is connected, there exists a path from this node to another node <span class="math-container">$n$</span> in <span class="math-container">$B$</span> which is connected to some node in <span class="math-container">$C$</span>. The node <span class="math-container">$n$</span> is then associated to a row of <span class="math-container">$L_{BB}$</span>
This is exactly the definition of a weakly chained diagonally dominant matrix.</li>
</ul>
<p>As these matrices have the property of being non-singular, you can conclude that <span class="math-container">$L_{BB}$</span> and <span class="math-container">$L_{CC}$</span> are invertible.
<span class="math-container">$L_{BC}$</span> (and <span class="math-container">$L_{CB}$</span> which is equal to <span class="math-container">$L_{BC}$</span> for undirected graph which is the case you seem to consider) can be singular, but you don't need nonsingularity of this matrix for applying Schur complement.</p>
|
3,911,548 | <p><strong>If a,b,c,d are real numbers and <span class="math-container">$\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}=17$</span> and <span class="math-container">$\frac{a}{c}+\frac{c}{a}+\frac{b}{d}+\frac{d}{b}=20$</span>, then find the sum of all possible valuse of <span class="math-container">$\frac{a}{b}$</span>+<span class="math-container">$\frac{c}{d}$</span> ?</strong></p>
<p>I tried this problem for a while but made no progress. I don't know how <span class="math-container">$\frac{a}{b}+\frac{c}{d}$</span> can take only certain values. The answer was given to be <span class="math-container">$17$</span>. Can someone help me with this?</p>
| Ng Chung Tak | 299,599 | <p><span class="math-container">\begin{align}
17 &= \frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a} \\
&= \frac{a}{b}+\frac{c}{d}+\frac{b}{c}+\frac{d}{a} \\
&= \color{red}{\frac{ad+bc}{bd}}+\color{blue}{\frac{ab+cd}{ac}} \\
&= \color{red}{x}+\color{blue}{y} \\
20 &= \frac{a}{c}+\frac{c}{a}+\frac{b}{d}+\frac{d}{b} \\
&= \frac{a}{c}+\frac{b}{d}+\frac{d}{b}+\frac{c}{a} \\
&= (ad+bc) \left( \frac{1}{cd}+\frac{1}{ab} \right) \\
&= \frac{(ad+bc)(ab+cd)}{abcd} \\
&= \color{red}{\frac{ad+bc}{bd}} \times \color{blue}{\frac{ab+cd}{ac}} \\
&= \color{red}{x} \color{blue}{y} \\
\end{align}</span></p>
<blockquote class="spoiler">
<p> Now, <span class="math-container">$$20=x(17-x) \implies x^2-17x+20=0$$</span>
Note that <span class="math-container">$x$</span> and <span class="math-container">$y$</span> are conjugate pair and also symmetric in roles.</p>
</blockquote>
|
83,246 | <p>Let H be a separable and infinite-dimensional Hilbert space and let B be the closed ball
of H having unit radius, whose center is at the origin h of H. Suppose one would like to
know how much of B can be "filled up" by any of its compact subsets-since B itself
(although closed and bounded) is not compact. Let E be the set of all positive real
numbers z for which there exists a compact subset C of B such that all points of B lie at
a distance from C (in the metric of H) which is not greater than z. The greatest lower bound of E would be a measure of this "filling up". My question is-what is this greatest
lower bound?
I believe that it is 1 but cannot prove it. Clearly 1 belongs to E since we can take for
C any compact subset of B that contains h. I can prove that no positive real number less
than one-half the square root of 2 belongs to E. But this is as far as I have been able to
get. If 1 is the right answer, it would show that no compact subset of B can "fill up" any
more of B than the set containing only the point h.</p>
| Fabian Lenhardt | 18,256 | <p>Assume $C \subset H$ is compact such that each point of the unit disk has distance at most $1-\epsilon$ from $C$ for some $\epsilon > 0$. There are finitely many points $p_1,...p_n$ such that the balls with radius $\frac{\epsilon}{4}$ around the $p_i$ cover all of $C$. Now pick an orthonormal basis $e_i$ of your Hilbert space and expand the $p_i$ in this basis, say $p_i = \sum\limits_j a_{ij} e_j$. Since there are only finitely many $p_i$, there must be an index $k$ such that $\left\vert a_{ij} \right\vert < \frac{\epsilon}{4}$ for all $i$ and all $j \geq k$. Then $e_k$ has at least distance $1-\frac{\epsilon}{4}$ from all the $p_i$. Since each point of $C$ has distance at most $\frac{\epsilon}{4}$ from one of the $p_i$, no point of $C$ can be closer than $1-\frac{\epsilon}{2}$ to $e_k$, contradicting the original choice of $\epsilon$.</p>
|
192,095 | <p>Suppose I have a convex program which has only two variables, the objective function is strictly convex, and the constraints are linear functions. </p>
<p>I think removing all non-tight constraints doesn't change the optimal solution.</p>
<p>However, when there are more than 2 tight constraints, I am not sure if removing all other tight constraints but only leaving two of them still keep the optimal solution unchanged. </p>
<p>Any advice would be appreciated! </p>
| Dirk | 9,652 | <p>No, you can't. Even in linear optimization with only two variables you may have several linear constraints that form a corner of your feasible domain. Removing some constraints may make the corner less sharp and lead to an unbounded objective. Imagine that removing some constraint adds new feasible directions at the corner (it's simple to draw a picture...). </p>
|
2,713,937 | <p>I need this lemma for another proof I'm doing, but I can't crack it. I want something of the structure:</p>
<p>$$\frac{pq}{(p-1)(q-1)} < \dots = \frac{pq}{\frac{1}{2}pq} = 2,$$
but I can't figure out what to do with the denominator. </p>
| Przemysław Scherwentke | 72,361 | <p>As $p\geq3$, $q\geq5$, we have $p/(p-1)\leq3/2$, $q/(q-1)\leq5/4$.</p>
|
2,713,937 | <p>I need this lemma for another proof I'm doing, but I can't crack it. I want something of the structure:</p>
<p>$$\frac{pq}{(p-1)(q-1)} < \dots = \frac{pq}{\frac{1}{2}pq} = 2,$$
but I can't figure out what to do with the denominator. </p>
| Barry Cipra | 86,747 | <p>If $p\le q$ are odd primes, then $p=3+a$ and $q=5+b$ for some $a,b\ge0$. We have</p>
<p>$$\begin{align}
{pq\over(p-1)(q-1)}={(3+a)(5+b)\over(2+a)(4+b)}\lt2
&\iff15+5a+3b+ab\lt2(8+4a+2b+ab)\\
&\iff0\lt1+3a+b+ab
\end{align}$$</p>
<p>The inequality has very little to do with $p$ and $q$ being either distinct, odd, or prime. </p>
|
1,261,977 | <p>I tried doing this the same way you would find the Fourier transform for $1/(1+x^2)$ but I guess I'm having some trouble dealing with the 2x on top and I could really use some help here.</p>
| Tryss | 216,059 | <p>Another way to solve this. As you know, $$-ix \mathcal{F}[g](x) = \mathcal{F}[g'](x)$$ and $$\mathcal{F}[\mathcal{F}[g]] = cg$$</p>
<p>You have (with some coefficients)</p>
<p>$$ \mathcal{F}[ x \mathcal{F}[g](x) ] (\xi)= \mathcal{F}[i\mathcal{F}[g'] ] (\xi) = i g'(\xi)$$</p>
<p>Now call $f = \mathcal{F}[g]$, and you got</p>
<p>$$ \mathcal{F}[ x f(x) ] (\xi)= i \mathcal{F}[f ]'(\xi)$$</p>
<p>So if you know $\mathcal{F}[f ]$, you can conclude</p>
<p>(I left the exaxct value of the coefficients in the wind, there may be some nasty $2\pi$ and other values here)</p>
|
2,861,867 | <p>Let $A$ be any commutative ring (with $1$) and $x,y \in A$ such that $x+y = 1$. Then it follows that for any $k,l$, there exist $a,b \in A$ such that $ax^k+by^l = 1$. </p>
<p>(Proof: Suppose otherwise. Then, $(x^,k,y^l) \subset \mathfrak p$ for some prime ideal $\mathfrak p$. But then this implies that $x,y \in \mathfrak p$ and therefore $1 = x+y \in \mathfrak p$, contradiction.)</p>
<p>My question is: Can we give a method for constructing the $a,b$ given any $k,l$. Maybe this is too much to ask for in general. What about if I restrict $A$ to be a finitely generated algebra over a field?</p>
| Mohan | 245,104 | <p>You have $(x+y)^{k+l}=1$, so using binomial theorem you get, letting $m=k+l$ (actually $k+l-1$ would do), $\sum \binom{m}{r} x^ry^{m-r}=1$. Thus, you have $\sum_{r\leq k}\binom{m}{r} x^ry^{m-r}+\sum_{r>k} \binom{m}{r} x^ry^{m-r}=1$. Notice that the first sum, every term is a multiple of $y^l$ and the second has all terms multiple of $x^k$. Collecting terms should be easy.</p>
|
3,393,193 | <p>I am asked to answer the following:</p>
<p>Let <span class="math-container">$f:Z->Z$</span> be defined by <span class="math-container">$f(x) = 2x$</span>.</p>
<ul>
<li>Write down infinitely many functions <span class="math-container">$g:Z->Z$</span> such that <span class="math-container">$g(f(x)) = Id_z$</span></li>
</ul>
<p>I thought that the right reasoning was simply to find the inverse of <span class="math-container">$f$</span>, since it is that function such that <span class="math-container">$f(g(x)) = Id_z$</span>.
Hence, <span class="math-container">$g(x) = $$x\over 2$</span>. </p>
<p>However, I'm asked infinitely many functions which made me think of the introduction of a parameter <span class="math-container">$k$</span>. And here is exactly where I am stuck. </p>
<p>Thanks in advance for your help.</p>
| dan_fulea | 550,003 | <p><strong>Discussion, notations and a first solution:</strong></p>
<p>We will use the following (general) stopping strategy, depending on the sets <span class="math-container">$A_1,A_2,A_3\subset\{1,2,3,4,5,6\}=:\Omega_0$</span>. The whole modeling space is <span class="math-container">$\Omega=\Omega_0^{\times 4}$</span>, and we have the usual filtration on it. Let <span class="math-container">$n=(n_1,n_2,n_3,n_4)$</span> be an element in <span class="math-container">$\Omega$</span>. (After the player stops, we can still go on, possibly without telling him the one, two, or three last results.) </p>
<ul>
<li><p>If the player sees <span class="math-container">$n_1\in A_1$</span> he stops. For <span class="math-container">$n_1\not \in A_1$</span> he continues.</p></li>
<li><p>If the player sees <span class="math-container">$n_2\in A_2$</span> he stops. For <span class="math-container">$n_2\not \in A_2$</span> he continues.</p></li>
<li><p>If the player sees <span class="math-container">$n_3\in A_3$</span> he stops. For <span class="math-container">$n_3\not \in A_3$</span> he continues.</p></li>
<li><p>And the player has to accept <span class="math-container">$n_4$</span> as a final result.</p></li>
</ul>
<p>Let <span class="math-container">$p_1,p_2,p_3$</span> be the probabilities to land in <span class="math-container">$A_1$</span>, <span class="math-container">$A_2$</span>, <span class="math-container">$A_3$</span>, so <span class="math-container">$p_k=|A_k|/6$</span>, for <span class="math-container">$k \in \{1,2,3\}$</span>, and let <span class="math-container">$m_k$</span> be the mean of the set <span class="math-container">$A_k$</span>, for <span class="math-container">$k \in \{1,2,3\}$</span>. We have to maximize
<span class="math-container">$$
p_1m_1+(1-p_1)p_2m_2+(1-p_1)(1-p_2)p_3m_3+\frac 72\ .
$$</span>
It is clear now, that <span class="math-container">$A_1$</span> can be taken optimal if it contains the highest values, given its cardinality as fixed. The same applies also for <span class="math-container">$A_2,A_3$</span>. There are thus only <span class="math-container">$6^3$</span> stopping strategies, depending on <span class="math-container">$a_k:=|A_k|$</span>, <span class="math-container">$k\in\{1,2,3\}$</span>.
It makes no sense to take some <span class="math-container">$a_k=0$</span> (instead of using <span class="math-container">$a_k=1$</span>, <span class="math-container">$A_k=\{6\}$</span>),
or <span class="math-container">$a_k=6$</span> (instead using <span class="math-container">$a_k=5$</span>, <span class="math-container">$A_k=\{2,3,4,5,6\}$</span>, and forcing one more turn, where we get at least one as minimal result).
so
<span class="math-container">$a_1,a_2.a_3\in\{1,2,3,4,5\}$</span>.</p>
<p>Then <span class="math-container">$p_k=a_k/6$</span>, the elements of <span class="math-container">$A_k$</span> are <span class="math-container">$6-0,\dots,6-(a_k-1)$</span>,with a mean of
<span class="math-container">$6-\frac 12(a_k-1)$</span>, so we have to maximize:
<span class="math-container">$$
\frac {a_1}6\left( 6-\frac 12(a_1-1)\right)\\
+
\frac {6-a_1}6\cdot\frac {a_2}6\left( 6-\frac 12 (a_2-1)\right)\\
+
\frac {6-a_1}6\cdot\frac {6-a_2}6\cdot\frac {a_3}6\left( 6-\frac 12 (a_3-1)\right)\\
+
\frac 72\ .
$$</span>
The computer gives the best five strategies as follows:</p>
<pre><code>sage: strategies = [] # and we will append
sage: R = [1, 2, 3, 4, 5]
sage: for a1, a2, a3 in cartesian_product( [R, R, R] ):
....: val = f(a1, a2, a3)
....: strategies.append( [val, [a1, a2, a3]] )
....:
sage: strategies.sort()
sage: for val, str in strategies[-5:]:
....: print "Expected final score %7s ~ %.6f for the strategy %s" % (val, val, str)
....:
Expected final score 44/9 ~ 4.888889 for the strategy [2, 3, 4]
Expected final score 265/54 ~ 4.907407 for the strategy [2, 2, 2]
Expected final score 265/54 ~ 4.907407 for the strategy [2, 2, 4]
Expected final score 59/12 ~ 4.916667 for the strategy [2, 3, 3]
Expected final score 89/18 ~ 4.944444 for the strategy [2, 2, 3]
sage:
</code></pre>
<hr>
<p><strong>Second solution:</strong></p>
<p>We can also think recursively. In the present game there are <span class="math-container">$4$</span> steps. (After the <span class="math-container">$4$</span>.th step we must stop.) Let us consider the general game, where there are <span class="math-container">$N$</span> steps. Let <span class="math-container">$W_N\in[1,6]$</span> be the expected final score in a strategy (similar to those discusted above) for the game with <span class="math-container">$N$</span> steps. We can now argue inductively. </p>
<ul>
<li><p>For <span class="math-container">$N=1$</span> we have the expectation <span class="math-container">$W_1=(1+2+3+4+5+6)/6=7/2=3.5$</span>. That's it.</p></li>
<li><p>Now in the <span class="math-container">$N=2$</span> game, we have a similar strategy as described in the first solution, but "it is clear" (by maximizing <span class="math-container">$a\to \frac a6\left(6-\frac 12(a-1)\right) +\frac {6-a}6\cdot W_1$</span>) that we should stop in the first step if we see a number <span class="math-container">$>W_1$</span>, else continue. The numbers bigger <span class="math-container">$W_1$</span> are <span class="math-container">$4,5,6$</span>. The expected final score is now
<span class="math-container">$$
W_2 =\frac 12\cdot 5+\frac 12\cdot W_1 =\frac{17}4=4.25\ .
$$</span></p></li>
<li><p>In the <span class="math-container">$N=3$</span> game,
same thoughts tell us to stop in the first step if we see a number <span class="math-container">$>W_2$</span>, so we stop when getting <span class="math-container">$5$</span> or <span class="math-container">$6$</span>.
The expected final score is now
<span class="math-container">$$
W_3 =\frac 13\cdot \frac{11}2+\frac 23\cdot W_2 =\frac{14}3=4.66666\dots\ .
$$</span></p></li>
<li><p>In the <span class="math-container">$N=4$</span> game,
same thoughts tell us to stop in the first step if we see a number <span class="math-container">$>W_3$</span>, so we stop again when getting <span class="math-container">$5$</span> or <span class="math-container">$6$</span>.
The expected final score is now
<span class="math-container">$$
W_4 =\frac 13\cdot \frac{11}2+\frac 23\cdot W_3 =\frac{89}{18}=4.944444\dots\ .
$$</span></p></li>
<li><p>In the game with <span class="math-container">$N=5$</span> we also stop in the first step, when we see either <span class="math-container">$5$</span> or <span class="math-container">$6$</span>, the corresponding expected final score is
<span class="math-container">$$
W_5 =\frac 13\cdot \frac{11}2+\frac 23\cdot W_4 =\frac{277}{54}=5\ .\ 1\;296\;296\;296\;296\dots\ .
$$</span>
We jump thus over <span class="math-container">$5$</span>, so "next strategies" will stop after the first step only when getting the six. </p></li>
</ul>
|
5,586 | <p>I'm in my last year of highschool. And I'm aiming for a perfect grade in maths. The problem is that this year is the hardest year of maths I have ever faced in my entire life. Especially derivation and limits as its the first time I am studying it. Here are the lessons that are required to study for the first semester:</p>
<ul>
<li>Limit of a function at a point</li>
<li>Limits Theorems</li>
<li>Limits of fractional functions</li>
<li>Limits of Trigonometric functions</li>
<li>Limits at Infinity</li>
<li>Continuity at a point</li>
<li>Continuity on an interval</li>
<li>Rate of Change</li>
<li>First derivative</li>
<li>Continuity and differentiation</li>
<li>Differentiation Rules</li>
<li>Derivatives of Higher Order</li>
<li>The chain rule</li>
<li>Implicit differentiation</li>
<li>Geometric applications of differentiation</li>
<li>Physical applications of differentiation</li>
<li>Related Rates</li>
<li>Increasing and Decreasing functions</li>
<li>Extreme Values.</li>
</ul>
<p>Limits are relatively easy. However, related rates and extreme values are disgustingly difficult, is there any way to make those two lessons easy and routine? Something like a book filled with questions on those two or something.</p>
<p>Thanks.</p>
| JPBurke | 759 | <p>Welcome to the site!</p>
<p>It's great that you're motivated and want to get a perfect grade in your studies! There is no actual formula for getting a perfect grade. One goal among all the goals of your math educators is to deepen your understanding of the mathematics you are studying in class and in whatever books your reading. There are things you can do to address this as well. I will suggest one. Because you seem so motivated, I think this is something that could work for you.</p>
<p>If you haven't already done this, consider forming a study group with other students in your class. It doesn't matter so much whether they are more or less advanced students than you. A mix may actually be good. The aim, primarily, is to gather a few like-minded people who are motivated to think about mathematics together. It's important that you're comfortable talking together about ideas, and especially putting forward suggestions that may not be "correct."</p>
<p>If you can form such a group (even maybe just 2 or 3 people), have meetings around a number of goals and questions such as:</p>
<ul>
<li>Discuss the ideas in a chapter (or chapter section) you all have agreed to read.
<ul>
<li>What was challenging about it?</li>
<li>What seemed obvious about it?</li>
<li>What are you still unclear about?</li>
<li>Where might it be useful?</li>
<li>Did anything seem especially interesting? Exciting? Pointless? Annoying? If so, why?</li>
<li>Do you feel there is a better way to do something presented in the chapter? Why is this other way better? </li>
</ul></li>
<li>Find online sources of information related to things you've already read in the book chapters.
<ul>
<li>Are there differences in how these concepts are presented?</li>
<li>Did anyone find a presentation they found easier to understand? How does that explanation relate to what you read in your textbook?</li>
</ul></li>
<li>Solve problems you've found in other books (or online)
<ul>
<li>What different approaches occur to you? </li>
<li>What are the relative advantages/disadvantages of various approaches?</li>
<li>What is surprising about these problems?</li>
<li>Are there problems you just can't solve? Do people get hung up in the same place? Consider asking a teacher if there is some approach that will help you solve this problem. </li>
</ul></li>
<li>Brainstorm applications for the mathematics you're learning.
<ul>
<li>What contexts do the textbooks and other sources suggest?</li>
<li>Where is this mathematics generally used?</li>
<li>Search online to find what applications the math may relate to. What did different people find? Does it intersect with any of your interests?</li>
<li>Is the math you're learning part of a trajectory leading to some advanced math you know you will be encountering int he future? </li>
</ul></li>
</ul>
<p>You can come up with your own ways of discussing the mathematics you're learning, I'm sure. Obviously, your group could mostly be about discussing and comparing approaches to solutions, or even working problems collaboratively. The point is that there can be a lot to talk about related to your mathematics learning, so a group like this has very few limits. And these discussions could be very productive. </p>
<p>One word of caution: be very mindful of your teacher's rules about homework collaboration. There's nothing wrong with taking on extra problems and working them collaboratively, but your teacher may have good reasons to establish very specific rules on the homework you've been individually assigned. Just be clear on those rules and you'll be fine.</p>
|
5,586 | <p>I'm in my last year of highschool. And I'm aiming for a perfect grade in maths. The problem is that this year is the hardest year of maths I have ever faced in my entire life. Especially derivation and limits as its the first time I am studying it. Here are the lessons that are required to study for the first semester:</p>
<ul>
<li>Limit of a function at a point</li>
<li>Limits Theorems</li>
<li>Limits of fractional functions</li>
<li>Limits of Trigonometric functions</li>
<li>Limits at Infinity</li>
<li>Continuity at a point</li>
<li>Continuity on an interval</li>
<li>Rate of Change</li>
<li>First derivative</li>
<li>Continuity and differentiation</li>
<li>Differentiation Rules</li>
<li>Derivatives of Higher Order</li>
<li>The chain rule</li>
<li>Implicit differentiation</li>
<li>Geometric applications of differentiation</li>
<li>Physical applications of differentiation</li>
<li>Related Rates</li>
<li>Increasing and Decreasing functions</li>
<li>Extreme Values.</li>
</ul>
<p>Limits are relatively easy. However, related rates and extreme values are disgustingly difficult, is there any way to make those two lessons easy and routine? Something like a book filled with questions on those two or something.</p>
<p>Thanks.</p>
| markt | 3,245 | <p>In college I had a friend who was also a Math major, and had been in the Marine Corp. His advice was a little coarse and echoed his military background: "Work problems until you puke, then wipe off the puke and work some more". I followed this advice and was an A student in college. I found that in Calculus in particular, the difficulties I had were due to my weakness in algebra and trigonometry. The equations in Calculus problems will almost always simplify. Hope this helps!</p>
|
649,239 | <p>By <a href="http://en.wikipedia.org/wiki/Post%27s_theorem" rel="nofollow">Post's Theorem</a> we know that a set $A\subseteq\mathbf{N}$ is recursively enumerable iff it is definable by a $\Sigma_1$-formula, i.e. there exists a $\Sigma_1$-formula $\varphi(x)$ with $x$ free such that for every number $n$:
\[
n\in A\longleftrightarrow \mathfrak{N}\vDash\varphi(\overline{n})
\]
where $\mathfrak{N}$ is the standard model of the first-order language of Peano Arithmetic.</p>
<p>I have the following question: given a r.e. set $A$ can we always find a $\Sigma_1$-formula defining it?</p>
| Wouter Stekelenburg | 27,375 | <p>By the axiom of choice there is a function that maps each r.e. set to one of its definitions. We can 'find' the definition in that sense. This function cannot be too nice, however, because that would decide the extensional equivalence of two definitions.</p>
<p>Let $f$ be a function that maps each set to one of its definitions. For each arithmetical statement $p$ there is a set $\{n|p\}$. The equality of definitions $f(\{n|p\}) = f(\mathbb N)$ is way to decide if $p$ is true. G\"odel's incompleteness theorems say we can't do that.</p>
|
4,498,801 | <p>I am trying to deeply understand the similarities between these two theorems; the first being a generalization of the second.</p>
<blockquote>
<p><strong>Theorem 16.13.</strong> If <span class="math-container">$f$</span> is nonnegative, then
<span class="math-container">$$
\int_{\Omega} f(T \omega) \mu(d \omega)=\int_{\Omega^{\prime}} f\left(\omega^{\prime}\right) \mu T^{-1}\left(d \omega^{\prime}\right) .
$$</span>
A function <span class="math-container">$f$</span> (not necessarily nonnegative) is integrable with respect to <span class="math-container">$\mu T^{-1}$</span> if and only if <span class="math-container">$f T$</span> is integrable with respect to <span class="math-container">$\mu$</span>, in which case (16.17) and
<span class="math-container">$$
\int_{T^{-1} A^{\prime}} f(T \omega) \mu(d \omega)=\int_{A^{\prime}} f\left(\omega^{\prime}\right) \mu T^{-1}\left(d \omega^{\prime}\right)
$$</span>
hold. For nonnegative <span class="math-container">$f$</span>, (16.18) always holds.</p>
</blockquote>
<blockquote>
<p><strong>(2.47) Theorem.</strong> Suppose <span class="math-container">$\Omega$</span> is an open set in <span class="math-container">$\mathbf{R}^{n}$</span> and <span class="math-container">$G: \Omega \rightarrow \mathbf{R}^{n}$</span> is a <span class="math-container">$C^{1}$</span> diffeomorphism.
(a) If <span class="math-container">$f$</span> is a Lebesgue measurable function on <span class="math-container">$G(\Omega)$</span>, then <span class="math-container">$f \circ G$</span> is Lebesgue measurable on <span class="math-container">$\Omega$</span>. If <span class="math-container">$f \geq 0$</span> or <span class="math-container">$f \in L^{1}(G(\Omega), m)$</span>, then
<span class="math-container">$$
\int_{G(\Omega)} f(x) d x=\int_{\Omega} f \circ G(x)\left|\operatorname{det} D_{x} G\right| d x
$$</span>
(b) If <span class="math-container">$E \subset \Omega$</span> and <span class="math-container">$E \in \mathscr{L}^{n}$</span>, then <span class="math-container">$G(E) \in \mathscr{L}^{n}$</span> and <span class="math-container">$m(G(E))=$</span> <span class="math-container">$\int_{E}\left|\operatorname{det} D_{x} G\right| d x$</span>.</p>
</blockquote>
<p>Why is the second theorem not written as</p>
<p><span class="math-container">$$
\int_{\Omega} f(G(x)) d x=\int_{G(\Omega)} f(x) \left| \operatorname{det} D_{x} G\right| d x
$$</span></p>
<p>This would make a lot more sense to me as we could think of this as <span class="math-container">$G$</span> is a function that changes the underlying measure space, and we integrate w.r.t. the pushforward measure which turns out to be <span class="math-container">$\left| \operatorname{det} D_{x} G\right| d x$</span>. Otherwise, I cannot see how to make the second version fit within the statement of the first.</p>
| Ruy | 728,080 | <p>I have a lot of sympathy for this question since it bothered me for a long time as well. After meditating about it for
years I eventually settled for the interpretation that these two results cannot be stated in a unified way because they
have a fundamental difference!</p>
<p>We will see that, contrary to what was stated by the OP, none of these Theorems may be seen as generalizing the other!</p>
<p>To put it in perspective let me try to state both results in the most similar way possible. In order not to get lost in
details, I will totally ignore questions about measurability and convergence of integrals and instead focus on their
essential aspects.</p>
<p>In the first result, we are:</p>
<ol>
<li><p>given a measure space <span class="math-container">$(X, \mu )$</span>,</p>
</li>
<li><p>a map <span class="math-container">$T:X\to Y$</span>,</p>
</li>
<li><p>where <span class="math-container">$Y$</span> is another set, and</p>
</li>
<li><p>a function <span class="math-container">$f:Y\to {\mathbb R}$</span>.</p>
</li>
<li><p>We then introduce a new measure on <span class="math-container">$Y$</span>, the RANGE of <span class="math-container">$T$</span>, namely the push-forward measure of <span class="math-container">$\mu $</span> by <span class="math-container">$T$</span>, which I
will denote by <span class="math-container">$\mu _T$</span> (referred to as <span class="math-container">$\mu T^{-1}$</span> by the OP),</p>
</li>
<li><p>and then the conclusion says that
<span class="math-container">$$
\int_Xf\circ T \,d\mu = \int_Y f \,d \mu _T.
$$</span></p>
</li>
</ol>
<p>The second result is similar in the sense that we are now:</p>
<ol>
<li><p>given an open subset <span class="math-container">$X\subseteq {\mathbb R}^n$</span> (called <span class="math-container">$\Omega $</span> by the OP), which incidentally comes with a
canonical measure, namely Lebesgue's measure,</p>
</li>
<li><p>a map <span class="math-container">$T:X\to Y$</span> (called <span class="math-container">$G$</span> by the OP),</p>
</li>
<li><p>where <span class="math-container">$Y=T(X)\subseteq {\mathbb R}^n$</span> is another set, which also comes with the canonical Lebesgue measure, and
finally</p>
</li>
<li><p>a function <span class="math-container">$f:Y\to {\mathbb R}$</span>.</p>
</li>
</ol>
<p>The crucial difference is that here we do not mess with the measure on <span class="math-container">$Y$</span> at all, sticking with Lebesgue measure, which
I will call <span class="math-container">$\mu $</span>, all the time. Instead, we use our ingredients to:</p>
<ol start="5">
<li><p>introduce a new measure on <span class="math-container">$X$</span>, the DOMAIN of <span class="math-container">$T$</span>, namely <span class="math-container">$\mu^T = |\text{det}D_xT|\,dx$</span>.</p>
</li>
<li><p>and then the conclusion says that
<span class="math-container">$$
\int_Xf\circ T \,d\mu ^T = \int_Y f \,d \mu .
$$</span></p>
</li>
</ol>
<p>Summarizing, in the first result, the transformation <span class="math-container">$T$</span> is used to create a new measure <span class="math-container">$\mu _T$</span> on <span class="math-container">$Y$</span>, while in the
second result, <span class="math-container">$T$</span> is used to go the other way, namely introducing a new measure <span class="math-container">$\mu ^T$</span> on <span class="math-container">$X$</span>.</p>
<p>In the second context it may be shown that the push forward measure of <span class="math-container">$\mu ^T$</span> under <span class="math-container">$T$</span> is <span class="math-container">$\mu $</span>, but this is just a
red herring, since it comes only as an afterthought.</p>
<p>It is therefore instructive to consider in which ways can a map <span class="math-container">$T$</span>, as above, be used to give rise to new measures. Of
course we can use <span class="math-container">$T$</span> in a push-forward construction as in the first result above. Now, when all spaces come with
preloaded measures, as is the case of Riemannian manifolds, a differentiable <span class="math-container">$T$</span> leads to its Jacobian, which can be
used as a density function against the default measure to create a new measure on its domain.</p>
|
3,059,695 | <p>Let <span class="math-container">$A$</span> be a subset of a compact topological space such that every point of <span class="math-container">$A$</span> is an isolated point of <span class="math-container">$A$</span>. Is <span class="math-container">$A$</span> necessarily finite?</p>
| TonyK | 1,508 | <p>No. Take for instance the compact space <span class="math-container">$[0,1]$</span>, and let <span class="math-container">$A=\{1/n:n\in\Bbb N_{>0}\}$</span>.</p>
|
530,484 | <p>Let $f:\mathbb{R}\rightarrow\mathbb{R}^2$ be a $C^1$ function. Prove that the image of $f$ contains no open set of $\mathbb{R}^2$.</p>
<p>So say $f(x)=(g(x),h(x))$. Since $f$ is $C^1$, we have that $g'(x),h'(x)$ both exist and are continuous functions in $x$. To show that $f$ contains no open set of $\mathbb{R}^2$, it suffices to show that $f$ does not contain any open ball in $\mathbb{R}^2$. Suppose, for contradiction, that it contains the ball centered at $(a,b)$ with radius $r$. How can I continue?</p>
| copper.hat | 27,978 | <p>Here is a more prosaic approach that relies on $f$ being Lipschitz on compact intervals.</p>
<p>Choose an interval $[t_0,t_1]$. Since $f$ is $C^1$, it is uniformly lipschitz on this interval with some rank $L$.
Let $D = f([t_0,t_1])$.
We can use $L$ to find an upper bound on the measure of $D$.</p>
<p>Let $R(\tau_1,\tau_0) = \{ x | \|x-f(\tau_0) \| \le L |\tau_1-\tau_0| \} = \overline{B}(f(\tau_0), L |\tau_1-\tau_0|)$. We have $f([\tau_0,\tau_1]) \subset R(\tau_1,\tau_0)$, and we can estimate $m R(\tau_1,\tau_0) \le L^2 |\tau_1-\tau_0|^2$.</p>
<p>Hence, for any $n$, we can split the interval $[t_0,t_1]$ into $n$ parts to get $D \subset \cup_{k=0}^{n-1} R(t_0+k\frac{t_1-t_0}{n}, t_0+(k+1)\frac{t_1-t_0}{n})$, and so $mD \le n L^2 \frac{1}{n^2} = \frac{L^2}{n}$. Since $n$ is arbitrary, we have $mD = 0$.</p>
<p>Since $\mathbb{R} = \cup_{m} [-m,m]$, we see that $m f(\mathbb{R}) = 0$.</p>
<p>Since $m B(x,\epsilon) >0$ for any $\epsilon >0$, it follows that $f(\mathbb{R})$ can contain no open ball.</p>
|
2,569,096 | <p>The problem goes as follows:
$$
P=\left(
\begin{matrix}
a & 0.6\\
1-a & 0.4\\
\end{matrix}
\right)
$$</p>
<blockquote>
<p>Determine the value of the parameter $a \in [0,1]$ for which $P$ does <strong>not</strong> have an inverse.</p>
</blockquote>
<p>So then I know the value of $a$ lies between $0$ and $1$, inclusively. And since I don't get information on the current states of the transitional values, this has to be done algebraically. </p>
<p>Is that correct? </p>
<p>In that case, as it is a "political" transition matrix: L = left swing, R = right swing. </p>
<p>$$P \cdot\left(\begin{array}{c} x\\ y\end{array} \right) = \left( \begin{array}{c} L \\ R\end{array} \right)$$</p>
<p>$$ax + .6y = L \\x(1-a) + .4y = R$$</p>
<p>$$ax = L - .6y \\x(1-a)= R - .4y$$ $$a = \frac{L - .6y}{x} \\$$</p>
<p>Or am I lost? </p>
<p>Thanks beforehand for help.</p>
| Community | -1 | <p><strong>Hint:</strong></p>
<p>A matrix does not have an inverse when its <strong>determinant $=0$</strong>. The value of $a$ that does this is: $$a\times 0.4 - (1-a)\times 0.6 =0\implies a=\, ?$$</p>
|
1,701,260 | <p>My textbook states the following:<br>
i) If $ f : [a,b] \rightarrow \mathbb{R} $ is bounded and is continuous at all but finitely many points of $[a,b]$, then it is integrable on $[a,b]$.<br>
ii) Any increasing or decreasing function on $[a,b]$ is integrable on $[a,b]$.</p>
<p>The proof for (i) is clear to me. I followed the entirety of it. My issue is with (ii). Is boundedness and continuity not necessary for (ii), or are they somehow implied by being strictly increasing/decreasing? </p>
| NicNic8 | 24,205 | <p>I suspect that by increasing and decreasing they mean that it's increasing / decreasing on the entire domain, which is <span class="math-container">$\mathbb{R}$</span>. So it's defined everywhere on <span class="math-container">$\mathbb{R}$</span>. Since it's defined everywhere, it's bounded on any compact subset of <span class="math-container">$\mathbb{R}$</span> (since any real function attains its supremum on any compact subset).</p>
<p>Continuity is not required for integrability; however, it is the case that continuity needs to be satisfied on all but a finite set of points in an interval. More specifically, the set of discontinuous points in the interval would have to have measure 0. I think that this must be the case for a function that is either decreasing or increasing.</p>
|
372,064 | <p>can someone explain me why</p>
<p>$\dot{a}\ddot{a}=\frac{1}{2}\frac{d}{dt}\left(\dot{a}^{2}\right)$</p>
<p>Many thanks</p>
| André Nicolas | 6,312 | <p>Use the Product Rule. The derivative of $\dot{a}\dot{a}$ with respect to $t$ is $\dot{a}\ddot{a}+\dot{a}\ddot{a}$, which is<br>
$2\dot{a}\ddot{a}$. </p>
|
54,311 | <p>I found <a href="http://www.rle.mit.edu/dspg/documents/HilbertComplete.pdf" rel="nofollow">this paper</a> on Hilbert Transform, which is a very nice read. I've studied signal processing, but from a more practical than mathematical perspective. Can someone explain to me how we arrive at equation (2) in this paper?</p>
| anon | 11,763 | <p>By symmetry if the first equality of (2) holds then the second equality also holds. Expand the integrand using the series expansions:</p>
<p>$$X(v)H(z/v)v^{-1} = \left(\sum_{n=-\infty}^\infty x(n)v^{-n}\right)\left(\sum_{m=-\infty}^\infty y(m)z^{-m}v^m \right)v^{-1} $$</p>
<p>Note that $\oint_\gamma v^{-k}dv=2\pi i$ if $k=1$ and $0$ otherwise - so all powers of $v$ are irrelevant except for when $m-n=0$ (there is already a $v^{-1}$ factor in the integrand). Hence the contour integral reduces as</p>
<p>$$ \frac{1}{2\pi i}\oint_\gamma \left(\sum_{n=-\infty}^\infty x(n)h(n) z^{-m}\right)v^{-1}dv= \frac{1}{2\pi i} \oint Y(z)v^{-1}dv = Y(z). $$</p>
|
3,829,431 | <blockquote>
<p>If the area of equilateral triangle is <span class="math-container">$3\sqrt3$</span> cm<span class="math-container">$^2$</span> , then what is the height of the equilateral triangle?</p>
</blockquote>
<p>I am stuck with this question
<br>I solved it like this:
<br>Area of equilateral triangle is <span class="math-container">$\frac{a\sqrt3}4$</span>
<br>So, <span class="math-container">$\frac{\sqrt3}4 \cdot a = 3 \sqrt 3$</span>
<br><span class="math-container">$a = \frac{3\sqrt3}{\sqrt3/4} = 3\sqrt3 \cdot \frac4{\sqrt3}$</span>; <span class="math-container">$\sqrt 3$</span>'s cancel
<br><span class="math-container">$a = 3 \cdot 4 = 12$</span>
<br>I found the side is <span class="math-container">$12$</span>.
<br>How do I find the height with the side length?
<br>And also kindly say if I made any mistake in my calculations.</p>
| Georges Elencwajg | 3,217 | <p>According to Zariski's Nullstellensatz (Qing Liu's Corollary 1.12, page 30) the extension <span class="math-container">$\mathbb R\subset A/\mathfrak m$</span> is finite, hence of degree <span class="math-container">$2$</span> (degree <span class="math-container">$1$</span> is excluded because <span class="math-container">$X^2+Y^2+1=0$</span> has no real solution).<br />
Hence there indeed exist <span class="math-container">$a,b,c,d \in \mathbb R$</span> such that <span class="math-container">$x^2+ax+b, y^2+cy+d\in \mathfrak{m}$</span>.
Adding together these expressions we get <span class="math-container">$x^2+y^2+ax+b+cy+d\in \mathfrak m$</span>, hence <span class="math-container">$-1+ax+b+cy+d\in \mathfrak m$</span> and (with obvious notations) we get the required relation <span class="math-container">$f=\alpha x+\beta y+ \gamma \in \mathfrak m$</span>.<br />
Since <span class="math-container">$fA\subset \mathfrak m$</span> to show equality it suffices to prove that <span class="math-container">$fA\subset A$</span> is maximal or that <span class="math-container">$A/fA=\mathbb R[X,Y]/(X^2+Y^2+1, \alpha X+\beta Y+ \gamma)$</span> is a field.<br />
Assuming <span class="math-container">$\beta\neq0$</span> we replace <span class="math-container">$Y$</span> by <span class="math-container">$lX+m$</span> and obtain <span class="math-container">$$A/fA=\mathbb R[X]/(X^2+(lX+m)^2+1)$$</span> which is obviously a field.<br />
And so we have proved that <span class="math-container">$\mathfrak m=fA$</span> is indeed a principal ideal.</p>
<p><strong>Optional Remark</strong><br />
It follows from Qing Liu's exercise that <span class="math-container">$A$</span> is a PID and hence factorial [=UFD].<br />
The complexified algebra <span class="math-container">$A_\mathbb C = \mathbb C[X,Y]/(X^2+Y^2+1)$</span> happens to also be factorial, but beware that this is not obvious <em>a priori</em>.<br />
For example the <span class="math-container">$\mathbb Q$</span>-algebra<br />
<span class="math-container">$B= \mathbb Q[X,Y]/(X^2+2Y^2+1)$</span> is a UFD but the extended <span class="math-container">$\mathbb Q(i)$</span>-algebra <span class="math-container">$B_{\mathbb Q(i)}= \mathbb Q(i)[X,Y]/(X^2+2Y^2+1)$</span> is not a UFD.<br />
On the other hand the the algebra <span class="math-container">$B=\mathbb R[x,y]/(x^2+y^2-1)$</span> is not a UFD although the extended <span class="math-container">$\mathbb C$</span>-algebra
<span class="math-container">$B_\mathbb C=\mathbb C[x,y]/(x^2+y^2-1)$</span> <strong>is</strong> a UFD (<a href="https://mathoverflow.net/a/5596/450">https://mathoverflow.net/a/5596/450</a>)</p>
<p><strong>Moral</strong>: Factoriality is not necessarily preserved under field extension, nor can it be deduced from factoriality after a field extension.</p>
|
2,507,247 | <p>Given a bimatrix game of
<span class="math-container">$$\left(\begin{matrix}(0,-1) & (0,0)\\(-90,-6)&(10, -10)\end{matrix}\right)$$</span>
<a href="https://i.stack.imgur.com/uY44c.jpg" rel="nofollow noreferrer">Source</a></p>
<p>How to find the nash equilibrium strategy for both players?</p>
| quasi | 400,434 | <p>I'll follow the method described in the text
<p>
$\qquad$Thomas$\,-\,$Games, Theory and Applications (1984)
<p>
on pages 59-61.
<p>
Suppose players $1$ and $2$ use mixed strategies $(x,1-x)$ and $(y,1-y)$, respectively, where </p>
<ul>
<li>The probability that player $1$ chooses row $1$ is $x$.$\\[2pt]$
<li>The probability that player $1$ chooses row $2$ is $1-x$.$\\[8pt]$
<li>The probability that player $2$ chooses row $1$ is $y$.$\\[2pt]$
<li>The probability that player $2$ chooses row $2$ is $1-y$.
</ul>
<p>Then the value of the game for player $1$ is
\begin{align*}
v_1(x,y) &= xy(0)+x(1-y)(0)+(1-x)y(-90)+(1-x)(1-y)(10)\\[4pt]
&= (100y-10)x + (10-100 y)\\[4pt]
\end{align*}
and the value of the sgame for player $2$ is
\begin{align*}
v_2(x,y) &= xy(-1)+x(1-y)(0)+(1-x)y(-6)+(1-x)(1-y)(-10)\\[4pt]
&= (4-5x)y + (10x-10)
\end{align*}
<p>
Suppose $(x,y)$ yields a Nash equlibrium.
<p>
From the given payoffs, it's clear that there is no pure strategy equilbrium, hence we must have $0 < x < 1$, and $0 < y < 1$.
<p>
Since $0 < x < 1$, the equation
$$v_1= (100y-10)x + (10-100 y)$$
implies $100y-10 = 0$, else player $1$ can change $x$ slightly, and do better.
<p>
Since $0 < y < 1$, the equation
$$v_2= (4-5x)y + (10x-10)\qquad\;$$
implies $4-5x= 0$, else player $2$ can change $y$ slightly, and do better.
<p>
It follows that the unique Nash equilbrium has $x = {\large{\frac{4}{5}}}$, and $y = {\large{\frac{1}{10}}}$.
<p>
The corresponding values of the game are $v_1 = 0$ for player $1$, and $v_2 = -2$ for player $2$.</p>
|
4,610,394 | <p>Clearly, none of the roots are in <span class="math-container">$\mathbb{Q}$</span> so <span class="math-container">$f(x) = x^4 + 1$</span> does not have any linear factors. Thus, the only thing left to check is to show that <span class="math-container">$f(x)$</span> cannot reduce to two quadratic factors.</p>
<p>My proposed solution was to state that <span class="math-container">$f(x) = x^4 + 1 = (x^2 + i)(x^2 - i)$</span> but <span class="math-container">$\pm i \not\in \mathbb{Q}$</span> so <span class="math-container">$f(x)$</span> is irreducible.</p>
<p>However, I stumbled across this post <a href="https://math.stackexchange.com/questions/1249143/x4-1-reducible-over-mathbbr-is-this-possible">$x^4 + 1$ reducible over $\mathbb{R}$... is this possible?</a> with a comment suggesting that <span class="math-container">$x^4 + 1 = (x^2 + \sqrt{2}x + 1)(x^2 - \sqrt{2}x + 1)$</span> which turns out to be a case that I did not fully consider. It made me realize that <span class="math-container">$\mathbb{Q}[x]$</span> being a UFD only guarantees a unique factorization of irreducible elements in <span class="math-container">$\mathbb{Q}[x]$</span> (which <span class="math-container">$x^2 \pm i$</span> nor <span class="math-container">$x^2 \pm \sqrt{2} x + 1$</span> aren't in <span class="math-container">$\mathbb{Q}[x]$</span>) so checking a single combination of quadratic products is not sufficient.</p>
<p>Therefore, what is the ideal method for checking that <span class="math-container">$x^4 + 1$</span> cannot be reduced to a product of two quadratic polynomials in <span class="math-container">$\mathbb{Q}[x]$</span>? Am I forced to just brute force check solutions of <span class="math-container">$x^4 + 1 = (x^2 + ax + b)(x^2 + cx + d)$</span> don't have rational solutions <span class="math-container">$(a,b,c,d) \in \mathbb{Q}^4$</span>?</p>
| orangeskid | 168,051 | <p>The only non-trivial decomposition over <span class="math-container">$\mathbb{R}$</span> of <span class="math-container">$x^4+1$</span> is <span class="math-container">$x^4 + 1 = (x^2 + \sqrt{2} x + 1)(x^2 - \sqrt{2} x + 1)$</span>. Since <span class="math-container">$\mathbb{Q} \subset \mathbb{R}$</span>, if there is a non-trivial decomposition over <span class="math-container">$\mathbb{Q}$</span>, it's gotta be the above. However, <span class="math-container">$\sqrt{2} \not\in \mathbb{Q}$</span>.</p>
|
4,610,394 | <p>Clearly, none of the roots are in <span class="math-container">$\mathbb{Q}$</span> so <span class="math-container">$f(x) = x^4 + 1$</span> does not have any linear factors. Thus, the only thing left to check is to show that <span class="math-container">$f(x)$</span> cannot reduce to two quadratic factors.</p>
<p>My proposed solution was to state that <span class="math-container">$f(x) = x^4 + 1 = (x^2 + i)(x^2 - i)$</span> but <span class="math-container">$\pm i \not\in \mathbb{Q}$</span> so <span class="math-container">$f(x)$</span> is irreducible.</p>
<p>However, I stumbled across this post <a href="https://math.stackexchange.com/questions/1249143/x4-1-reducible-over-mathbbr-is-this-possible">$x^4 + 1$ reducible over $\mathbb{R}$... is this possible?</a> with a comment suggesting that <span class="math-container">$x^4 + 1 = (x^2 + \sqrt{2}x + 1)(x^2 - \sqrt{2}x + 1)$</span> which turns out to be a case that I did not fully consider. It made me realize that <span class="math-container">$\mathbb{Q}[x]$</span> being a UFD only guarantees a unique factorization of irreducible elements in <span class="math-container">$\mathbb{Q}[x]$</span> (which <span class="math-container">$x^2 \pm i$</span> nor <span class="math-container">$x^2 \pm \sqrt{2} x + 1$</span> aren't in <span class="math-container">$\mathbb{Q}[x]$</span>) so checking a single combination of quadratic products is not sufficient.</p>
<p>Therefore, what is the ideal method for checking that <span class="math-container">$x^4 + 1$</span> cannot be reduced to a product of two quadratic polynomials in <span class="math-container">$\mathbb{Q}[x]$</span>? Am I forced to just brute force check solutions of <span class="math-container">$x^4 + 1 = (x^2 + ax + b)(x^2 + cx + d)$</span> don't have rational solutions <span class="math-container">$(a,b,c,d) \in \mathbb{Q}^4$</span>?</p>
| Bob Dobbs | 221,315 | <p>Clearly, <span class="math-container">$x+a$</span>, <span class="math-container">$a\in\Bbb{Z}$</span> can not be a factor since then <span class="math-container">$a^4=-1$</span>. On the other hand, <span class="math-container">$x^2+ax+b$</span>, <span class="math-container">$a,b\in\Bbb{Z}$</span> can not be a factor since its roots are <span class="math-container">$\frac{a\pm\sqrt{a^2-4b}}{2}$</span> and these roots can not be roots of unity wheras the roots of <span class="math-container">$x^4+1=0$</span> are root of unity. Because taking modulus, we see that <span class="math-container">$b=1$</span>. But, then also considering the argument of the roots <span class="math-container">$\frac{\sqrt{4-a^2}}{a}=\pm1$</span> gives <span class="math-container">$a=\sqrt2$</span>.</p>
|
2,803,398 | <p>We know that in category of $\mathbb{Set}$ the inverse limit is the direct product. But I am looking for specific category in which inverse limit does not exist. Any comments would be highly appreciated.</p>
| David C. Ullrich | 248,223 | <p>Not quite. The correspondence between infinite decimals and elements of $(0,1)$ is not itself a bijection because some numbers have more than one decimal expansion.</p>
<p>You solve that by restricting to the expansion with infinitely many non-zero digits, fine. But now the mapping from $(0,1)^2$ to $(0,1)$ is not surjective. For example there are no $x$ and $y$ that give $r'=0.1101010101...$. Because that would require $x=0.10000...$ and $y=0.111...$. But $x=0.1000...=0.09999...$, so the image of that $(x,y)$ is actually $0.0191919...\ne0.11010101...$.</p>
|
278,368 | <p><strong>Problem:</strong></p>
<p>Assume the number of cars passing a road crossing during an hour satisfies a Poisson distribution with parameter $\mu$, and that the number of passengers in each car satisfies a binomial distribution with parameters $n \in \mathbb{N}$ and $p \in (0,1)$. Let $Y$ denote the total number of passengers passing the road crossing during an hour. Compute $\mathbb{E}[Y]$ and Var$(Y)$.</p>
<p><strong>My thoughts:</strong></p>
<p>If we let $X_i$ be the number of passengers in the $i^{\text{th}}$ car, we have that $$\mathbb{E}[Y]= \sum_{i=1}^N \mathbb{E}[X_i], $$
where $N$ is the total number of cars. Since $X_i$ ~ Binomial$(n,p)$, $\mathbb{E}[X_i]=np$ $\hspace{1mm}$ $\forall$ $i \in \mathbb{N}$. Furthermore, since $N$ ~ Poisson$(\mu)$, we have $\mathbb{E}[N]=\mu$, yielding $\mathbb{E}[Y]=\mu np.$ </p>
<p>This <em>seems</em> right, but I'm dissatisfied with the lack of thoroughness with my argument. Any ideas?</p>
<p>(Edit: This isn't homework.)</p>
| leonbloy | 312 | <p>One way of doing the first is using conditioned probability. $N$ and $X_i$ are both random (independent) variables. Now, what you computed first is actually the conditional expectation:</p>
<p>$$E[Y|N] = \sum_{i=1}^N E(X_i) = N n p$$</p>
<p>but $E[Y] = E[ E[Y | N]]$ (<a href="http://en.wikipedia.org/wiki/Law_of_total_expectation" rel="nofollow">ref</a>; the outer expectation is with respect of N) then we get $E[Y] = E (N n p) = np E(N) = n p \mu$</p>
<p>A similar argument can be used to compute $E(Y^2)$ (and hence the variance). Or you can read
about the Compound Poisson distribution and its <a href="http://en.wikipedia.org/wiki/Compound_Poisson_distribution#Properties" rel="nofollow">properties</a>.</p>
|
3,975,895 | <p>Let <span class="math-container">$a,b,c\in\mathbb{Z}$</span>, <span class="math-container">$1<a<10$</span>, <span class="math-container">$c$</span> is a prime number and <span class="math-container">$f(x)=ax^2+bx+c$</span>. If <span class="math-container">$f(f(1))=f(f(2))=f(f(3))$</span>, find <span class="math-container">$f'(f(1))+f(f'(2))+f'(f(3))$</span></p>
<p>My attempt:
<span class="math-container">\begin{align*}
f'(x)&=2ax+b\\
(f(f(x)))'&=f'(f(x))f'(x)\\
f'(f(x))&=\frac{(f(f(x)))'}{f'(x)}\\
\end{align*}</span></p>
| cosmo5 | 818,799 | <p>(Edited to add details)</p>
<p>Due to the symmetry of vertical parabola, for distinct <span class="math-container">$x_1, x_2,$</span> <span class="math-container">$f(x_1)=f(x_2) \Rightarrow x_1+x_2=-b/a$</span> and <span class="math-container">$f'(x_1)+f'(x_2)=0$</span></p>
<p>For a quadratic, w cannot have <span class="math-container">$f(x_1)=f(x_2)=f(x_3)$</span> for distinct three <span class="math-container">$x_i$</span> since this would imply, <span class="math-container">$f(x)−f(x_1)$</span> has three distinct roots. There arise three conditions of arguments being pairwise equal.</p>
<p><span class="math-container">$f(f(1))=f(f(2))=f(f(3)) \Rightarrow$</span> following cases :</p>
<ul>
<li><span class="math-container">$f(1)=f(2) \Rightarrow 1+2=-b/a$</span>.</li>
</ul>
<p>Now <span class="math-container">$f(1) \neq f(3)$</span>. But <span class="math-container">$f(f(1))=f(f(3))$</span>. <span class="math-container">$\Rightarrow f(1)+f(3)=-b/a=3$</span></p>
<p>Here <span class="math-container">$f(1)+f(2) \neq -b/a$</span> since <span class="math-container">$f(1),f(2)$</span> are identical. Similarly,</p>
<ul>
<li><span class="math-container">$f(2)=f(3) \Rightarrow 2+3=-b/a$$f(1) \neq f(3) \Rightarrow f(1)+f(3)=-b/a=5$</span></li>
<li><span class="math-container">$f(1)=f(3) \Rightarrow 1+3=-b/a$$f(1) \neq f(2) \Rightarrow f(1)+f(2)=-b/a=4$</span></li>
</ul>
<p>But <span class="math-container">$f(1)+f(3)=10a+4b+2c$</span> is even. So only the third case holds: <span class="math-container">$f(1)=f(3)$</span> and <span class="math-container">$f'(1)+f'(3)=0$</span>.</p>
<p>Thus <span class="math-container">$b=-4a$</span>. <span class="math-container">$f(x)=ax^2-4ax+c \Rightarrow f'(x)=2a(x-2)$</span></p>
<p>Also <span class="math-container">$f'(f(x)) = (ay^2-4ay+c)'=2a(y-2)y'=2af'(x)(f(x)-2)$</span></p>
<p>Evaluating, <span class="math-container">$f'(f(1))+f(f'(2)+f'(f(3))=c$</span></p>
<p>where <span class="math-container">$a,c$</span> have been computed as below.</p>
<hr />
<p><strong>Remark :</strong></p>
<p>Inspired from the other answer, even the numerical value of <span class="math-container">$c$</span> can be calculated. Subbing <span class="math-container">$f(x)=ax^2-4ax+c$</span> into <span class="math-container">$f(f(1))=f(f(2))$</span>, one obtains <span class="math-container">$$c=\dfrac{7a+4}{2}$$</span></p>
<p>Quickly checking for <span class="math-container">$a\in \{2,4,6,8\}$</span>, <span class="math-container">$c=23$</span> a prime, only for <span class="math-container">$a=6$</span>.</p>
|
787,894 | <p>Find the values of $x,y$ for which $x^2 + y^2$ takes the minimum value where $(x+5)^2 +(y-12)^2 =14$.</p>
<p>Tried Cauchy-Schwarz and AM - GM , unable to do.</p>
| Community | -1 | <p>HINT: For this case where the curve is a circle, the value you seek is square of the distance of the centre of the circle from origin minus the radius. (Draw a diagram to see why?)
$$x^2+y^2 = (13-\sqrt{14})^2.$$</p>
<p>You can also find the maximum of $x^2+y^2$ using this trick.</p>
|
595,552 | <p>Let $R$ be a ring. Prove that each element of $R$ is either a unit or a nilpotent element iff the ring $R$ has a unique prime ideal.</p>
<p>Help me some hints.</p>
| BIS HD | 73,067 | <p>Hint for $\Leftarrow$:</p>
<p>Every ring with identiy has a prime ideal. Let $P$ be a prime ideal of $R$. Then it contains all the nilpotent elements (why?). It does not contain a unit (why?), so it is in fact the set of nilpotent elements of $R$ and hence because of the condition the unique prime ideal. </p>
|
3,086,218 | <p>The second order differential equation is given by -</p>
<p><span class="math-container">$ \frac{d^{2}y}{dx^{2}} + \sin (x+y) = \sin x$</span> </p>
<p>Is this a homogeneous differential equation <span class="math-container">$?$</span></p>
<p>Well, I guess this is not a homogeneous differential equation since the form of this equation is not <span class="math-container">$a(x)y'' + b(x)y' +c(x)y = 0$</span>.
But the answer is given that it's homogeneous.
How can this equation be homogeneous?</p>
| doraemonpaul | 30,938 | <p>Let <span class="math-container">$u=x+y$</span> ,</p>
<p>Then <span class="math-container">$\dfrac{du}{dx}=1+\dfrac{dy}{dx}$</span></p>
<p><span class="math-container">$\dfrac{d^2u}{dx^2}=\dfrac{d^2y}{dx^2}$</span></p>
<p><span class="math-container">$\therefore\dfrac{d^2u}{dx^2}+\sin u=\sin x$</span></p>
<p>This is a inhomogeneous differential equation.</p>
|
3,224,455 | <p>I derived the volume of a cone using two approaches and compared the results.</p>
<p>First I integrated a circle of radius <span class="math-container">$r$</span> over the height <span class="math-container">$h$</span> to get the expression: <span class="math-container">$$V_1=\frac{1}{3}\pi r^2 h$$</span></p>
<p>Then I considered a polygonal pyramid of infinite sides.</p>
<p>An n-sided polygon with apothem <span class="math-container">$r$</span> has an area of: <span class="math-container">$$A=nr^2\tan{\frac{180°}{n}}$$</span></p>
<p>Integrating this over the height <span class="math-container">$h$</span> gives the expression for the area of the n-sided polygonal pyramid as: <span class="math-container">$$V_2=\frac{1}{3}n\tan{\frac{180°}{n}}r^2 h$$</span></p>
<p>Equating <span class="math-container">$V_1$</span> and <span class="math-container">$V_2$</span> implies that: <span class="math-container">$$ \lim_{n \to \infty} \left(n\tan{\frac{180°}{n}}\right) = \pi $$</span></p>
<p>So is it true to say that: <span class="math-container">$$\infty\tan{\frac{180°}{\infty}} = \pi$$</span> </p>
<p>But: <span class="math-container">$$\tan{\frac{180°}{\infty}}=0$$</span></p>
<p>So: <span class="math-container">$$\infty (0)=\pi$$</span></p>
<p>Can anyone shed some light on this surprising result?</p>
| st.math | 645,735 | <p>You cannot simply replace expressions that tend to infinity by that same symbol; then you would lose information on how fast something tends to infinity, for example.</p>
<p>Take <span class="math-container">$a_n=n$</span>, <span class="math-container">$b_n=2n$</span>. Then clearly, both tend to infinity. But <span class="math-container">$b_n/a_n=2\to2$</span>, and <span class="math-container">$a_n/a_n=1\to1$</span>. Both limits would be "<span class="math-container">$\infty/\infty$</span>", but are still very different. What is key here is the <em>way</em> both expressions tend to infinity. You lose this information when you just replace the sequence with a <span class="math-container">$\infty$</span> symbol.</p>
|
3,224,455 | <p>I derived the volume of a cone using two approaches and compared the results.</p>
<p>First I integrated a circle of radius <span class="math-container">$r$</span> over the height <span class="math-container">$h$</span> to get the expression: <span class="math-container">$$V_1=\frac{1}{3}\pi r^2 h$$</span></p>
<p>Then I considered a polygonal pyramid of infinite sides.</p>
<p>An n-sided polygon with apothem <span class="math-container">$r$</span> has an area of: <span class="math-container">$$A=nr^2\tan{\frac{180°}{n}}$$</span></p>
<p>Integrating this over the height <span class="math-container">$h$</span> gives the expression for the area of the n-sided polygonal pyramid as: <span class="math-container">$$V_2=\frac{1}{3}n\tan{\frac{180°}{n}}r^2 h$$</span></p>
<p>Equating <span class="math-container">$V_1$</span> and <span class="math-container">$V_2$</span> implies that: <span class="math-container">$$ \lim_{n \to \infty} \left(n\tan{\frac{180°}{n}}\right) = \pi $$</span></p>
<p>So is it true to say that: <span class="math-container">$$\infty\tan{\frac{180°}{\infty}} = \pi$$</span> </p>
<p>But: <span class="math-container">$$\tan{\frac{180°}{\infty}}=0$$</span></p>
<p>So: <span class="math-container">$$\infty (0)=\pi$$</span></p>
<p>Can anyone shed some light on this surprising result?</p>
| nmasanta | 623,924 | <p><span class="math-container">$\lim_{n \to \infty} (n\tan{\frac{180°}{n}}) = \pi$</span></p>
<p>Take <span class="math-container">$\frac{1}{n}= x$</span>, then <span class="math-container">${n \to \infty} \implies {x \to 0}$</span></p>
<p>Also we have <span class="math-container">$$\lim_{n \to \infty} {\frac{\tan x}{x}} = 1$$</span></p>
<p>So <span class="math-container">$\lim_{n \to \infty} (n\tan{\frac{180°}{n}}) = \pi \implies \lim_{x \to 0} ({\frac{\tan 180° x}{180° x}}180°) = \pi \implies 180°=\pi$</span></p>
|
1,614,989 | <p>A portion of a $30$m long tree is broken by
tornado and the top struck up the ground
making an angle $30^{\circ}$ with ground
level. The height of the point where the tree
is broken is equal to:</p>
<p>$a.)\ \dfrac{30}{\sqrt{3}}m$ $~~~~~~~~~~$ $\color{green}{b.)\ 10m} \\$
$~~~~~~~~~~$ $c.)\ 30\sqrt{3}m$ $~~~~~~~~~~$ $d.)\ 60m$</p>
<p><a href="https://i.stack.imgur.com/Hohlt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Hohlt.png" alt="enter image description here"></a></p>
<p>My teacher explained that $AB(tree)=30 m$, $BD=x ~~m$</p>
<p>and $DK =(30-x)m$</p>
<p>I didn't understand how come $DK =(30-x)m$ ?</p>
| angryavian | 43,949 | <p>The height of the tree is $30$ meters. You have defined $x$ to be the length of the part of the tree that is still standing, so $30-x$ is the length of the part of the tree that has fallen.</p>
<hr>
<p>You want to solve for $x$. Using the definition of sine, we have $\sin 30^\circ = \frac{x}{30-x}$. You probably learned that $\sin 30^\circ=1/2$ or that the sides of a "$30,60,90$" degree triangle has sides proportional to "$1,\sqrt{3},2$." You can see this by taking half of an equilateral triangle and using the Pythagorean theorem. Solving $\frac{x}{30-x}=\frac{1}{2}$ gives you the answer.</p>
|
3,252,765 | <p>We are trying to codify in terms of modern algorithm the works of the ancient Indian mathematician <em>Udayadivakara</em> (CE 1073). In his work <em>Sundari</em>, he quotes one <em>Acarya Jayadeva</em> who has given methods to solve Pell's equations. In these methods, one can find the the cyclic <em>Chakravala</em> method to deal with <span class="math-container">$X^2-DY^2=1$</span> wrongly attributed to Bhaskara. He also gives the method to solve <span class="math-container">$X^2-DY^2=C$</span> for any integer <span class="math-container">$C$</span>. </p>
<ol>
<li>His algorithm starts off by finding the nearest square integer <span class="math-container">$>D$</span> named <span class="math-container">$P^2$</span>. Then <span class="math-container">$a=P^2-D$</span>.</li>
<li>Now some <span class="math-container">$b$</span> is chosen in such a way that <span class="math-container">$Db^2+Ca$</span> is some perfect square <span class="math-container">$Q^2$</span>. </li>
<li>Then the <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> solutions can be found by using <span class="math-container">$Y=\frac{Q\pm P b}{a}$</span> and <span class="math-container">$X=PY \mp b$</span>.</li>
<li>This procedure can continue indefinitely to find all the solutions. </li>
<li>Coming to the question of the fundamental solution i.e. the solution with which Bhavana has to be performed repeatedly to get other solutions (related to the modern automorphism group of the quadratic form), Prof. K.S. Shukla who first translated the work from Sanskrit to English, in his example says that it should be chosen "appropriately".</li>
<li><strong>Our primary question is then what is the criterion to derive this <em>fundamental solution</em>? Is there a way to derive such a criterion?</strong> </li>
<li>The whole procedure seems to resemble Conway's topograph method which has been posted here several times <a href="https://math.stackexchange.com/questions/1719280/does-the-pell-like-equation-x2-dy2-k-have-a-simple-recursion-like-x2-dy2?noredirect=1&lq=1">Does the Pell-like equation $X^2-dY^2=k$ have a simple recursion like $X^2-dY^2=1$?</a>
It is quite fascinating to think that some wonderful mind came up with this algorithm about 1000 years ago and the optimality of it is equally amazing.</li>
</ol>
<p>P.S: If anyone so wishes, we would be happy to provide a version of the original paper written by Prof. Shukla in 1950 dealing with this!</p>
| Joebloggs | 206,669 | <p>Here is a program in pari-gp which solves all solvable instances of the Pell Equation in reasonably quick time (it's competitive with the best, but a lot simpler), as follows:</p>
<pre><code>Pell_1(N,f)=
</code></pre>
<p>{</p>
<pre><code>/*** Set square root Precision ***/
/*********************************/
/*** Numerator & Denominator ***/
my(numer);
my(denom);
my(d);
/*********************************/
/*** Lower & Upper Limits ***/
my(num_mid);
my(num_lower);
my(num_upper);
my(denom_mid);
my(denom_lower);
my(denom_upper);
/********************************/
/*** Temporary Variables ***/
my(temp_num_low); /* NB: NOT to be confused with temperature! */
my(temp_num_high);
my(temp_denom_low);
my(temp_denom_high);
/********************************/
mid = sqrt(N);
lower = floor(sqrt(N));
upper = ceil(sqrt(N));
my(phi); /* Ratio */
num_upper = upper;
num_lower = lower;
denom_upper = 1;
denom_lower = 1;
numer = num_lower + num_upper;
denom = denom_lower + denom_upper;
phi = numer / denom;
print(numer);
print(denom);
while(numer^2 - N*(denom^2)<>f,
if(phi < mid, temp_num_high = num_upper; num_lower = numer; numer = numer + temp_num_high,
temp_num_low = num_lower; num_upper = numer; numer = numer + temp_num_low);
if(phi < mid, temp_denom_high = denom_upper; denom_lower = denom; denom = denom + temp_denom_high,
temp_denom_low = denom_lower; denom_upper = denom; denom = denom + temp_denom_low); /* Comparison Test */
d = gcd(numer,denom);
numer = numer / d;
denom = denom / d;
phi = numer /denom;
);
print (numer);
print(denom);
</code></pre>
<p>}</p>
|
2,438,111 | <p>I hope my title somehow encapsulates my problem.</p>
<p>Let's say we have a 1-D Grid with the values 2,1,5,8,1,1. Imagine those values are of some physical quantity $\alpha$.
The mean of this would be $(2+1+5+8+1+1)/6 = 3$</p>
<p>Now let's say we have some function $f(x) = x^2$, which computes another quantity $\beta$ out of the initial ones. When we put in our mean it yields $f(3) = 9$. So one could think that the total $\beta$ would be $6 \times 9=54$.</p>
<p>Now let's compute $\beta$ directly for each grid element and sum it up to get the total amount. $\beta_{total} = 4+1+25+64+1+1=96$</p>
<p>$96 \neq 54$.</p>
<p>Intuitively, I'd say 96 is the right result, but I'm kind of at a loss why the mean times the number of values fails.</p>
| Bernard | 202,857 | <p><strong>Hint:</strong></p>
<p>$$\sqrt{3}\cos (x)+\sin(x)=2\biggl(\frac{\sqrt 3}2\cos x+\frac12\sin x\biggr)=\dotsm$$</p>
|
2,346,804 | <p>Please help me finish this problem.</p>
<p>$xy''+(3x-1)y'-(4x+9)y=0$ where $y(0)=0$</p>
<p>$L[xy'']+L[(3x-1)y']-L[(4x+9)y]=L[0]$</p>
<p>$L[xy'']=\frac{d}{dp}(p^2Y)$</p>
<p>$L[(3x-1)y']=-3\frac{d}{dp}(pY)$</p>
<p>$L[(4x+9)y]=-4\frac{dY}{dp}$</p>
<p>$-\frac{d}{dp}(p^2Y)-3\frac{d}{dp}(pY)+4\frac{dY}{dp}=0$</p>
<p>$-\frac{d}{dp}(p^2Y)-(3pY)\frac{d}{dp}+4\frac{dY}{dp}=0$</p>
<p>I don't know what to do from here and would really appreciate some help finishing the problem. </p>
| Community | -1 | <p>You have a population of $10$ where $3$ are true and $7$ are false,</p>
<p>A random draw from this population therefore has $3/10$ chance of being true. </p>
|
2,879,035 | <p>$f(x) = \int_{1}^{\infty} \frac{2}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2} dx$</p>
<p>find $P(X > 1)$</p>
<p>This is $X$ ~ $Norm(0, 1)$.</p>
<p>$P(X > 1) = 1 - P(X \leq 1) = 1 - 2 \phi(1) = 1-2(1-\phi(-1)) = 1 - 2(1-0.1587) = -0.6826$. </p>
<p>Yikes. Negative number. What am I doing wrong? </p>
| Deepesh Meena | 470,829 | <p><a href="https://i.stack.imgur.com/Gd2kt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Gd2kt.png" alt="enter image description here"></a>
thus $$P(X\le1)=\phi(1)$$</p>
<p>$$your answer =1-\phi(1)$$</p>
|
3,244,193 | <p>Here's what I did: <span class="math-container">$$\lim_{n\rightarrow +\infty}(2+3^n)^{\frac{1}{2n}}=\lim_{n\rightarrow +\infty}e^{\frac{1}{2n}\ln(2+3^n)}$$</span>
What should I do next in order to solve it?</p>
| Yiorgos S. Smyrlis | 57,021 | <p>Note that
<span class="math-container">$$
3^n \le 2+3^n\le 2\cdot3^n
$$</span>
Hence
<span class="math-container">$$
3^{1/2} \le (2+3^n)^{1/2n}\le 2^{1/2n}\cdot3^{1/2}\to 3^{1/2}.
$$</span></p>
|
990,930 | <p>Graphing this function is difficult as many overlaps exist and finding a viewing window is hard.</p>
<p>What's a good algebraic method to solve this problem? </p>
| CLAUDE | 118,773 | <p>$2\sin^2(x)-3=3\cos(x)\rightarrow 2(1-\cos^2(x))-3=3\cos(x)\rightarrow 2\cos^2(x)+3\cos(x)+1=0\rightarrow (\cos(x)+1)(2\cos(x)+1)=0\rightarrow$</p>
<p>So we can say $\cos(x)=-1$ or $\cos(x)=\frac{-1}{2}$</p>
<p>$\cos(x)=-1\rightarrow x=2k\pi+\pi\rightarrow x=\pi$</p>
<p>$\cos(x)=\frac{-1}{2}\rightarrow x=\frac{2\pi}{3}$ and $x=\frac{4\pi}{3}$</p>
<p>So totally there are three different values for $x$, satisfying the mentioned equation in the intended interval.</p>
|
1,149,561 | <p>I've tried using mods but nothing is working on this one: solve in positive integers $x,y$ the diophantine equation $7^x=3^y-2$.</p>
| Qiaochu Yuan | 232 | <p>The Euler characteristic is an extremely weak invariant: since it's the alternating sum of the ranks of the homology groups, there's a lot of cancellation possible in that sum, and so there's a lot of ways that two spaces can have the same Euler characteristic. For example, $S^1$ and $S^1 \times S^1$ have the same Euler characteristic ($0 = 1 - 1$ and $0 = 1 - 2 + 1$) but don't even have the same dimension, and also have different $H_1$ as well as different $H_2$. More generally, $S^1$ times any reasonable space (say a compact manifold) has Euler characteristic $0$. </p>
|
3,568,050 | <blockquote>
<p>Let <span class="math-container">$R$</span> be an equivalence relation in the set <span class="math-container">$A$</span> and <span class="math-container">$a,b \in A$</span>. Show that <span class="math-container">$R(a)=R(b)$</span> <strong>iff</strong> <span class="math-container">$aRb$</span>.</p>
</blockquote>
<p>We must prove two implications here. First, that <span class="math-container">$$R(a)=R(b) \Rightarrow aRb$$</span></p>
<p>Note that <span class="math-container">$R(a):= \{b\in A : aRb\}$</span>, and <span class="math-container">$R(b):=\{a\in A : bRa\}$</span>. Also since these sets are equal, they must be subsets of each other <span class="math-container">$$R(a) \subseteq R(b)$$</span></p>
<p>It follows that <span class="math-container">$\forall b \in A$</span> : <span class="math-container">$$b\in R(a) \Rightarrow b \in R(b)$$</span></p>
<p>If <span class="math-container">$b$</span> is an element of both of these equivalence classes, then certainly <span class="math-container">$b=a$</span>. From the fact that <span class="math-container">$$R(b)\subseteq R(a)$$</span></p>
<p>it follows similarly, that <span class="math-container">$a=b$</span>. I'm pretty sure this shows
<span class="math-container">$aRb$</span>, but I'm not sure how I should word it. As for <span class="math-container">$$aRb \Rightarrow R(a)=R(b)$$</span></p>
<p>Since <span class="math-container">$R$</span> is an equivalence relation, we have by symmetry that, <span class="math-container">$\forall a,\forall b \in A$</span> <span class="math-container">$$aRb \Rightarrow bRa$$</span></p>
<p>This is where I am at the moment. I just wanted to check if the proof is ok up to this point.</p>
| Derek Luna | 567,882 | <p>If <span class="math-container">$aRb$</span>, then <span class="math-container">$x \in R(a)$</span> implies that <span class="math-container">$xRa$</span> since <span class="math-container">$R$</span> is symmetric so by transivity <span class="math-container">$xRb$</span> and <span class="math-container">$x \in R(b)$</span>. The other inclusion is similar. We know that <span class="math-container">$bRb$</span> by reflexivity so that if <span class="math-container">$R(a) = R(b)$</span>, then <span class="math-container">$b \in R(a$</span>) and <span class="math-container">$aRb$</span>.</p>
|
3,568,050 | <blockquote>
<p>Let <span class="math-container">$R$</span> be an equivalence relation in the set <span class="math-container">$A$</span> and <span class="math-container">$a,b \in A$</span>. Show that <span class="math-container">$R(a)=R(b)$</span> <strong>iff</strong> <span class="math-container">$aRb$</span>.</p>
</blockquote>
<p>We must prove two implications here. First, that <span class="math-container">$$R(a)=R(b) \Rightarrow aRb$$</span></p>
<p>Note that <span class="math-container">$R(a):= \{b\in A : aRb\}$</span>, and <span class="math-container">$R(b):=\{a\in A : bRa\}$</span>. Also since these sets are equal, they must be subsets of each other <span class="math-container">$$R(a) \subseteq R(b)$$</span></p>
<p>It follows that <span class="math-container">$\forall b \in A$</span> : <span class="math-container">$$b\in R(a) \Rightarrow b \in R(b)$$</span></p>
<p>If <span class="math-container">$b$</span> is an element of both of these equivalence classes, then certainly <span class="math-container">$b=a$</span>. From the fact that <span class="math-container">$$R(b)\subseteq R(a)$$</span></p>
<p>it follows similarly, that <span class="math-container">$a=b$</span>. I'm pretty sure this shows
<span class="math-container">$aRb$</span>, but I'm not sure how I should word it. As for <span class="math-container">$$aRb \Rightarrow R(a)=R(b)$$</span></p>
<p>Since <span class="math-container">$R$</span> is an equivalence relation, we have by symmetry that, <span class="math-container">$\forall a,\forall b \in A$</span> <span class="math-container">$$aRb \Rightarrow bRa$$</span></p>
<p>This is where I am at the moment. I just wanted to check if the proof is ok up to this point.</p>
| fleablood | 280,126 | <blockquote>
<p>If b is an element of both of these equivalency classes, then certainly b=a</p>
</blockquote>
<p>I can not see how you think that. That makes no sense. Unless this class has only one element, then (because they are equal) <em>all</em> the elements are in both and all the elements can't all be equal to each other.</p>
<p>.....</p>
<p>Use the definition of equivalence.</p>
<p>So to prove <span class="math-container">$R(a)=R(b)\to aRb$</span>.....</p>
<p>If <span class="math-container">$R(a) = \{x\in A: aRx\}$</span>....(I advice not using the variable "<span class="math-container">$b$</span>" because you use that variable for another element... oh, maybe that caused your confusion....) </p>
<p>then as <span class="math-container">$R$</span> is an equivalence class then by reflexivity <span class="math-container">$aRa$</span> and so <span class="math-container">$a \in R(a)$</span>.</p>
<p>But if <span class="math-container">$R(a) = R(b) = \{x\in A: bRx\}$</span> it follows that <span class="math-container">$a\in R(b)$</span> so <span class="math-container">$bRa$</span>. </p>
<p>......</p>
<p>Can you finish this?</p>
<blockquote class="spoiler">
<p> but symmetry as <span class="math-container">$bRa$</span> we know <span class="math-container">$aRb$</span>.</p>
</blockquote>
<p>And to prove <span class="math-container">$aRb \to R(a)=R(b)$</span>.</p>
<blockquote class="spoiler">
<p> If <span class="math-container">$c\in R(a)$</span> then <span class="math-container">$cRa$</span> but <span class="math-container">$aRb$</span> so .......</p>
</blockquote>
<p>.........</p>
<blockquote class="spoiler">
<p> .... by transitivity <span class="math-container">$cRb$</span> so ...</p>
</blockquote>
<p>.......</p>
<blockquote class="spoiler">
<p> ..... <span class="math-container">$c \in R(b)$</span> and <span class="math-container">$R(a)\subset R(b)$</span>. If <span class="math-container">$c \in R(b)$</span> then .....</p>
</blockquote>
<p>......</p>
<blockquote class="spoiler">
<p> ...... <span class="math-container">$cRb$</span>. But <span class="math-container">$aRb$</span> so by symmetry <span class="math-container">$bRa$</span> and by transitivity <span class="math-container">$cRa$</span> and be symmettry <span class="math-container">$aRc$</span> so <span class="math-container">$c\in R(a)$</span> so <span class="math-container">$R(b)\subset R(a)$</span> and so <span class="math-container">$R(a) = R(b)$</span>.</p>
</blockquote>
<p>=========</p>
<p>It may help to do an example. <span class="math-container">$\equiv \pmod 5$</span> is an equivalence relation.</p>
<p>So <span class="math-container">$[a] = \{x\in \mathbb Z| x\equiv a \pmod 5\}= \{a + 5k|k\in \mathbb N\}$</span> and <span class="math-container">$[b] = \{x\in \mathbb Z| x\equiv a \pmod 5\}= \{b + 5k|k\in \mathbb N\}$</span>.</p>
<p>the does <span class="math-container">$[a] = [b]$</span> mean that <span class="math-container">$b\equiv a \pmod 5$</span>?</p>
<p>Well, yes....</p>
<p><span class="math-container">$a \equiv a \pmod 5$</span> so <span class="math-container">$a \in [a] = [b]$</span> so <span class="math-container">$a\equiv b\pmod 5$</span> so <span class="math-container">$b \equiv a\pmod 5$</span>.</p>
<p>If if <span class="math-container">$a \equiv b\pmod 5$</span> does that mean <span class="math-container">$[a] = [b]$</span>?</p>
<p>Well, yes.....</p>
<p>If <span class="math-container">$a \equiv \pmod 5$</span> then there is a <span class="math-container">$k$</span> so that <span class="math-container">$a = b + 5k$</span>.</p>
<p>Then <span class="math-container">$c\in [a]\iff c \equiv a \pmod 5\iff $</span> there is a <span class="math-container">$j$</span> so that <span class="math-container">$c = a+5j\iff c=(b+5k)+5j=b+5(k+j)\iff c\equiv b\pmod 5\iff c\in [b]$</span>.</p>
<p>So <span class="math-container">$[a]$</span> and <span class="math-container">$[b]$</span> have the exact same elements.</p>
<p>====</p>
<p>In general. </p>
<p>Suppose <span class="math-container">$R(a) = R(b)$</span>. then <span class="math-container">$bRb\implies b\in R(b)\implies b\in R(a)\implies aRb$</span>.</p>
<p>Suppose <span class="math-container">$aRb$</span>. then <span class="math-container">$c\in R(a)\implies aRc\implies cRa\implies cRa$</span> and <span class="math-container">$aRb\implies cRb \implies c\in R(b)\implies cRb\implies cRb$</span> and <span class="math-container">$ aRb\implies cRb$</span> and <span class="math-container">$bRa\implies cRa\implies c\in R(a)$</span>.</p>
<p>So <span class="math-container">$c\in R(a)\iff c\in R(b)$</span> and so <span class="math-container">$R(a) = R(b)$</span>.</p>
<p>That's all there is to it. </p>
|
3,575,804 | <p>The question goes as follows. My attempts are below it. </p>
<p>A motel has ten rooms, all located on the same side of a single corridor and numbered 1 to 10 in numerical order. The motel always randomly allocates rooms to its guests. There are no other guests besides those mentioned.</p>
<p>a) Friends Molly and Polly have been allocated two separate rooms at the motel. What is the likely number of rooms between their rooms? </p>
<p>b) Molly believes there is a greater than 1/3 chance that at most one room will separate them, but Polly disagrees. Who is right? Explain why.</p>
<p>c) On another occasion, Molly, Polly and a third friend, Ollie, were allocated three separate rooms. Molly believes there is a better than 1/3 chance that they are all within a block of five consecutive rooms. Ollie believes that there is exactly 1/3 chance and Polly believes there is less than 1/3 chance. Who is right? Explain why. </p>
<p>d) Ollie arrived after rooms were allocated to Molly and Polly. There was then a 50% chance he would be in a room adjacent to Molly or Polly or both. In how many ways could a pair of rooms have been allocated to Molly and Polly?</p>
<p>Here I am bit sure of how to work through (c). Do I use permutations or combinations - could you please give me a step by step solution as I am still a bit wobbly on combinatorics.</p>
<p>Also could someone please check my answer for (d) (30)?</p>
<p>Thanks.</p>
| 0Ketrav0 | 766,652 | <p>Now, 10 rooms can be allocated to 3, in 10P3 number of ways, ie. 10*9*8 number of ways; as there are 10 room options for the first person, 9 for the second person and 8 for the third person.</p>
<p>So, all possible cases</p>
<p>=10*9*8</p>
<p>=720.</p>
<p>Now, in 10 rooms, there are 6 combinations of 5 consecutive room blocks. These are {1,2,3,4,5}, {2,3,4,5,6}, {3,4,5,6,7}, {4,5,6,7,8}, {5,6,7,8,9}, and {6,7,8,9,10}.
So, favorable cases</p>
<p>=6*5*4*3</p>
<p>=360</p>
<p>=360/720</p>
<p>=1/2.</p>
<p>So, the probability is 1/2.</p>
<p>Now, 1/2 is greater than 1/3, so among the three friends, Molly is right.</p>
|
1,715,265 | <p>I've tried a method similar to showing that $\mathbb{Q}(\sqrt2, \sqrt3)$ is a primitive field extension, but the cube root of 2 just makes it a nightmare.</p>
<p>Thanks in advance </p>
| Community | -1 | <p>$\mathbb{Q}(\alpha,\beta)$ will equal $\mathbb{Q}(\alpha + s\beta)$ if you choose any rational number $s$ that is not $-\frac{\alpha_i - \alpha}{\beta_j - \beta}$ for any of the conjugates $\alpha_i$ of $\alpha$ and $\beta_j$ of $\beta.$</p>
<p>In your case you can take $s = 1.$</p>
|
1,758,159 | <p>A is symmetric(skew-symmetric) matrix and B is nonsingular matrix .
What can i say about this $$BAB^T$$
???</p>
| Robert Lewis | 67,071 | <p>Note that</p>
<p>$(BAB^T)^T = B^{TT}A^TB^T = BA^TB^T, \tag 1$</p>
<p>since</p>
<p>$B^{TT} = B. \tag 2$</p>
<p>If $A$ is symmetric, $A = A^T$, so</p>
<p>$(BAB^T)^T = BAB^T, \tag 3$</p>
<p>that is, $BAB^T$ is symmetric. Likewise if $A$ is skew-symmetric, $A^T =-A$, so</p>
<p>$(BAB^T)^T = B(-A)B^T= -BAB^T; \tag 4$</p>
<p>$BAB^T$ is skew-symmetric in this case. $BAB^T$ inherits the symmetry or skew-symmetry of $A$. The non-singularity of $B$ plays no role in these conclusions; they hold whether $B$ is singular or not.</p>
|
387,505 | <p>Let <span class="math-container">$f$</span> be a non-invertible bounded outer function on the unit disk. Does <span class="math-container">$f$</span> has radial limit <span class="math-container">$0$</span> somewhere? Note that such a property holds for singular inner functions.</p>
| Mateusz Kwaśnicki | 108,637 | <p>The answer is negative: <span class="math-container">$f$</span> may have a non-zero radial limit everywhere.</p>
<hr />
<p><em>Part 1.</em> Let us start with definitions and notation. A holomorphic function <span class="math-container">$f$</span> in the unit disk <span class="math-container">$D$</span> is an outer function if and only if
<span class="math-container">$$ -\log |f(r e^{i t})| = \int_0^{2\pi} \phi(s) K_r(t - s) ds $$</span>
for some integrable function <span class="math-container">$\phi$</span> on <span class="math-container">$[0, 2\pi)$</span>, where <span class="math-container">$K_r(t)$</span> is the Poisson kernel:
<span class="math-container">$$ K_r(t) = \frac{1}{2 \pi} \frac{1 - r^2}{1 + r^2 - 2 r \cos t} \, . $$</span>
Furthermore, every integrable <span class="math-container">$\phi : [0, 2\pi) \to \mathbb R$</span> corresponds in the above sense to an outer function (a unique one up to multiplication by a constant with modulus <span class="math-container">$1$</span>). For convenience, let us extend <span class="math-container">$\phi$</span> to all of <span class="math-container">$\mathbb R$</span> in such a way that <span class="math-container">$\phi$</span> is <span class="math-container">$2\pi$</span>-periodic.</p>
<p>If <span class="math-container">$\phi$</span> is continuous at <span class="math-container">$t$</span> (or, more generally, if <span class="math-container">$t$</span> is a Lebesgue point of <span class="math-container">$\phi$</span>) and <span class="math-container">$f$</span> corresponds to <span class="math-container">$\phi$</span>, then
<span class="math-container">$$ \lim_{r \to 1^-} (-\log |f(r e^{it})|) = \phi(t) . $$</span>
Due to symmetry of the Poisson kernel, we also have
<span class="math-container">$$ \lim_{r \to 1^-} (-\log |f(r e^{it})|) = \frac{\phi(t^+) + \phi(t^-)}{2} $$</span>
whenever <span class="math-container">$\phi$</span> has one-sided limits <span class="math-container">$\phi(t^+)$</span> and <span class="math-container">$\phi(t^-)$</span> at <span class="math-container">$t$</span>.</p>
<p>(To clarify the notation: <span class="math-container">$e^{-\phi}$</span> is the radial boundary limit of <span class="math-container">$|f|$</span> almost everywhere. Thus, <span class="math-container">$\phi \geqslant 0$</span> a.e. if and only if <span class="math-container">$|f|$</span> is bounded by <span class="math-container">$1$</span>; more generally, <span class="math-container">$\phi \geqslant -\log C$</span> if and only if <span class="math-container">$|f|$</span> is bounded by <span class="math-container">$C$</span>. For <span class="math-container">$p > 1$</span>, <span class="math-container">$f$</span> is in the complex Hardy space <span class="math-container">$H^p$</span> if and only if <span class="math-container">$e^{-\phi}$</span> is in <span class="math-container">$L^p$</span>.)</p>
<hr />
<p><em>Part 2.</em> We are now ready to construct a counter-example. Let
<span class="math-container">$$ \phi(t) = \begin{cases} 2^n & \text{if } t \in [\tfrac{1}{2^n}, \tfrac{1}{2^n} + \tfrac{1}{8^n}] \text{ for some } n = 1, 2, \ldots , \\ 0 & \text{otherwise.} \end{cases} $$</span>
Let <span class="math-container">$f$</span> be an outer function corresponding to <span class="math-container">$f$</span>. We claim that <span class="math-container">$f$</span> is a counter-example: its radial boundary limit is everywhere non-zero, and <span class="math-container">$1/f$</span> is not a bounded outer function.</p>
<p>Indeed, the limit of <span class="math-container">$-\log |f(r e^{it})|$</span> is equal to a finite number <span class="math-container">$\tfrac12(\phi(t^+) + \phi(t^-))$</span> whenever <span class="math-container">$t \in (0, 2 \pi)$</span>, and so the radial boundary limit of <span class="math-container">$|f|$</span> at <span class="math-container">$e^{i t}$</span> is strictly positive (the radial limit of <span class="math-container">$f$</span> may fail to exist, though; see <em>Part 3</em> below for more details). Thus, it remains to see what happens for <span class="math-container">$t = 0$</span>. Clearly,
<span class="math-container">$$ -\log |f(r)| = \int_0^{2\pi} \phi(s) K_r(-s) ds = \int_0^{2\pi} \phi(s) K_r(s) ds . $$</span>
Note that if <span class="math-container">$r \geqslant \tfrac14$</span> and <span class="math-container">$s \in (0, \pi)$</span>, we have
<span class="math-container">$$ \begin{aligned} 0 \leqslant K_r(s) & = \frac{1}{2\pi} \, \frac{(1 + r) (1 - r)}{(1 - r)^2 + 4 r \sin^2 (s/2)} \\ & \leqslant \frac{(1 + r) (1 - r)}{2 \times (1 - r) \times 2 \sqrt{r} \sin(s/2)} \\ & = \frac{1 + r}{4 \sqrt{r} \sin(s/2)} \leqslant \frac{1}{\sin(s/2)} \leqslant \frac{\pi}{s} \, . \end{aligned} $$</span>
Furthermore,
<span class="math-container">$$ \begin{aligned} \int_0^{2\pi} \phi(s) \, \frac{\pi}{s} \, ds & = \sum_{n = 1}^\infty \int_{1/2^n}^{1/2^n+1/8^n} 2^n \times \frac{\pi}{s} \, ds \\ & \leqslant \sum_{n = 1}^\infty \int_{1/2^n}^{1/2^n+1/8^n} 2^n \times 2^n \pi \, ds \\ & = \sum_{n = 1}^\infty \frac{1}{8^n} \times 4^n \pi = \frac{\pi}{3} < \infty . \end{aligned} $$</span>
Thus, by the dominated convergence theorem,
<span class="math-container">$$ \begin{aligned} \lim_{r \to 1^-} (-\log |f(r)|) & = \lim_{r \to 1^-} \int_0^{2\pi} \phi(s) K_r(s) ds \\ & = \int_0^{2\pi} \phi(s) \lim_{r \to 1^-} K_r(s) ds = 0 . \end{aligned} $$</span>
In other words, the radial boundary limit of <span class="math-container">$|f|$</span> at <span class="math-container">$0$</span> is <span class="math-container">$1$</span>.</p>
<p>We have thus proved that at no boundary point the radial boundary limit of <span class="math-container">$f$</span> is equal to zero. On the other hand, the radial boundary limit of <span class="math-container">$-\log |1/f| = \log |f|$</span> at <span class="math-container">$e^{it}$</span> is equal to <span class="math-container">$\phi(t)$</span> for almost all <span class="math-container">$t$</span>, and since <span class="math-container">$\phi$</span> is essentially unbounded, <span class="math-container">$1/f$</span> is not bounded in the unit disk. To be specific: the radial boundary limit of <span class="math-container">$|1/f|$</span> at <span class="math-container">$t = \tfrac{1}{2^n} + \tfrac{1}{2} \tfrac{1}{8^n}$</span> is equal to <span class="math-container">$e^{\phi(t)} = e^{2^n}$</span>. In other words, <span class="math-container">$f$</span> is not invertible in the class of bounded outer functions.</p>
<hr />
<p><em>Part 3.</em> The above <span class="math-container">$f$</span> fails to have a radial boundary limits at <span class="math-container">$t = \tfrac{1}{2^n}$</span> and <span class="math-container">$t = \tfrac{1}{2^n} + \tfrac{1}{8^n}$</span>. If we require <span class="math-container">$f$</span> to have a radial boundary limit everywhere, then we need to modify <span class="math-container">$\phi$</span> appropriately: we choose an auxiliary smooth "bump" function <span class="math-container">$\psi$</span> equal to <span class="math-container">$0$</span> outside of <span class="math-container">$(0, 1)$</span>, equal to <span class="math-container">$1$</span> in <span class="math-container">$[\tfrac13, \tfrac23]$</span>, and taking values in <span class="math-container">$[0, 1]$</span>, and we set
<span class="math-container">$$ \phi(t) = \begin{cases} 2^n \psi(8^n (t - \tfrac{1}{2^n})) & \text{if } t \in [\tfrac{1}{2^n}, \tfrac{1}{2^n} + \tfrac{1}{8^n}] \text{ for some } n = 1, 2, \ldots , \\ 0 & \text{otherwise.} \end{cases} $$</span>
Now <span class="math-container">$\phi$</span> is smooth except at <span class="math-container">$t = 0$</span>, and hence <span class="math-container">$f$</span> is continuous in <span class="math-container">$\overline{D} \setminus \{1\}$</span>. The same calculation as in <em>Part 2</em> shows that <span class="math-container">$-\log |f| = -\Re \log f$</span> again has a radial boundary limit <span class="math-container">$0$</span> at <span class="math-container">$1$</span>. A similar calculation with the conjugate Poisson kernel shows that also <span class="math-container">$-\Im \log f$</span> has a radial boundary limit at <span class="math-container">$1$</span>, and so <span class="math-container">$f$</span> has a (non-zero) radial boundary limit at <span class="math-container">$1$</span>.</p>
|
739,960 | <ol>
<li><p>$ \log_a{b} \times \log_b{a} = $ ?</p></li>
<li><p>$ \log_a{b} + \log_b{a} = \sqrt{29} $</p></li>
</ol>
<p>What is $ \log_a{b} - \log_b{a} = $ ?</p>
<p>3.</p>
<p>What is b in the following:</p>
<p>$$ \log_b{3} + \log_b{11} + \log_b{61} = 1 $$</p>
<p>and</p>
<p>4.</p>
<p>$$ \frac{1}{log_2{x}} + \frac{1}{log_{25}{x}} - \frac{3}{\log_8{x}} = \frac{1}{\log_b{x}} $$
What is b?</p>
<p>Can anyone help me solve these?</p>
| nadia-liza | 113,971 | <p>2 . let $ x=\log_a{b} =\frac{1}{ \log_b{a}} $
$$\log_a{b}+\frac{1}{\log_a{b}}=\sqrt{29}$$
Then $ \log_a{b}$ and $ \log_a{b}$ are solutions of the following equation</p>
<p>$$x+\frac{1}{x}=\sqrt{29}$$</p>
<p>$$x_1=\frac{\sqrt{29}-5}{2}=\log_b{a}$$
$$x_2=\frac{\sqrt{29}+5}{2}=\log_a{b}$$ (We assume that $a<b$ )
Thus $ \log_a{b}- \log_b{a}=\frac{\sqrt{29}+5}{2}-\frac{\sqrt{29}-5}{2}=5 $ </p>
<p>4.</p>
<p>$$ \frac{1}{log_2{x}} + \frac{1}{log_{25}{x}} - \frac{3}{\log_8{x}} = \frac{1}{\log_b{x}} $$
Using formula $ \log_a{b} =\frac{1}{ \log_b{a}} $ we have
$$ \log_x{2} + \log_{x}{25} - 3 log_x{8} = \log_x{b} $$
Use formulas $$\log_x{a} + \log_{x}{c}=\log_{x}{ac}$$ and
$$n \log_x{a} =\log_{x}{a^n}$$</p>
|
1,821,582 | <blockquote>
<p>Find all solutions of $$\{x^3\}+[x^4]=1$$
where $[x]=\lfloor x\rfloor$</p>
</blockquote>
<p>$$$$</p>
<p>I know that $0\le\{x^3\}<1\Rightarrow 0<[x^4]\le 1$. Thus $[x^4]=1$. I couldn't get any further though since I'm having trouble with $x^4$ in the term $[x^4]$. $$$$As an example, in another question, I was given the expression $$[2x]-[x+1]$$ In this, to 'convert' the terms inside the floor function, I split the values of $x$ into 2 cases: $x=[x]+\{x\}, \{x\}\in[0,0.5)$ and $x=[x]+\{x\}, \{x\}\in[0.5,1)$. $$$$In this way, I was able to split the value of $[2x]$ into $2[x]$ and $2[x]+1$ respectively. Hence the expressions within the floor function became easier to deal with as the expression $[2x]-[x+1]$ was converted into $2[x]-[x]-1$ and $2[x]+1-[x]-1$ respectively.$$$$
Is there any way to do something similar with $x^4$ in $[x^4]$ so that the $x^3$ inside $\{x^3\}$ and $x^4$ inside $[x^4]$ are converted into the 'same kind'? If this isn't possible, is there any other way to solve this problem, and other problems in which the expressions within the floor/fractional part function, or the floor/fractional part functions themselves are raised to powers?
$$$$Many thanks in anticipation.</p>
| triple_sec | 87,778 | <p>Suppose that $\{x^3\}+\lfloor x^4\rfloor=1$. Clearly, $\lfloor x^4\rfloor$ is an integer, so that $$\{x^3\}=1-\lfloor x^4\rfloor$$ must be an integer, too. But the fractional part of any number is always contained in $[0,1)$. This implies that $\{x^3\}=0$, so that $x^3$ is an integer.</p>
<p>Going back to the original equation, one has that $\lfloor x^4\rfloor=1$, so that $x^4\in[1,2)$, implying that $x\in[1,2^{1/4})\cup(-2^{1/4},-1]$. Therefore, $x^3\in[1,2^{3/4})\cup(-2^{3/4},-1]$. But, as observed earlier, $x^3$ is an integer, and since $1<2^{3/4}<2$, the only integers in the set $[1,2^{3/4})\cup(-2^{3/4},-1]$ are $1$ and $-1$. It follows that $x^3\in\{-1,1\}$, or $x\in\{-1,1\}$. These are the only two solutions.</p>
|
213,405 | <p>So here's the question:</p>
<blockquote>
<p>Given a collection of points $(x_1,y_1), (x_2,y_2),\ldots,(x_n,y_n)$, let
$x=(x_1,x_2,\ldots,x_n)^T$, $y=(y_1,y_2,\ldots,y_n)^T$,
$\bar{x}=\frac{1}{n} \sum\limits_{i=1}^n x_i$, $\bar{y}=\frac{1}{n} \sum\limits_{i=1}^n y_i$.<br>
Let $y=c_0+c_1x$ be the linear function that gives the best least squares fit to the points. Show that if $\bar{x}=0$, then
$c_0=\bar{y}$ and $c_1=\frac{x^Ty}{x^Tx}$.</p>
</blockquote>
<p>I've managed to do all the problems in this least squares chapter but this one has me completely and utterly stumped. I'm not entirely sure what the question is even telling me in terms of information nor do I get what it's asking. Any ideas on where to start?</p>
| murage kibicho | 873,735 | <p>I saw this in the book, Applied Algebra- Codes, Ciphers and Discrete Logarithms by Darel W.Hardy. It's on page 138 of the second edition, in the section about the Chinese Remainder Theorem and Extended Precision Arithmetic. The accepted answer gives a wonderful summary of the book's content.</p>
<p>In the general case, the author states, "Let <span class="math-container">$m_1,m_2,...,m_r$</span>" be pairwise relatively prime positive integers with the property that <span class="math-container">$m_i^2$</span> can be computed exactly in machine arithmetic. Then any positive integer <span class="math-container">$n$</span> less than the product <span class="math-container">$m_1m_2...m_r$</span> is determined by the numbers <span class="math-container">$$ a_1 = n\: mod\:m_1, a_2 = n\: mod\:m_2,...,a_r = n\: mod\:m_r$$</span> The modular representation of n relative to the bases <span class="math-container">$m_1,m_2,...,m_r$</span> takes the form <span class="math-container">$$ n = [a_1, a_2,...,a_r]$$</span></p>
<p>Knowing this, we can represent 49 using the primes 7,11,13,17.
<span class="math-container">$$49\:mod\:7 = 0$$</span> <span class="math-container">$$49\:mod\:11 = 5$$</span> <span class="math-container">$$49\:mod\:13 = 10$$</span> <span class="math-container">$$49\:mod\:17 = 15$$</span> We get <span class="math-container">$$ 49 = [0, 5,10,15]$$</span></p>
<p>We can get 49 from its modular representation by solving the system of congruences
<span class="math-container">$$x \equiv 0\;mod\: 7$$</span> <span class="math-container">$$x \equiv 5\;mod\: 11$$</span> <span class="math-container">$$x \equiv 10\;mod\: 13$$</span> <span class="math-container">$$x \equiv 15\;mod\: 17$$</span></p>
|
2,191,360 | <blockquote>
<p>Show that
$$ f(x,y)= \begin{cases} \dfrac{xy^2}{x^2+y^4} & (x,y) ≠ (0,0) \\ 0 & (x,y) = (0,0) \end{cases}$$
is bounded.</p>
</blockquote>
<p>I thought about splitting it up into different cases like $x<y$ but it turned out to be too many and I could not cover all of them.
As a hint I got the idea to use the arithmetic geometric inequality.
I hope someone can help me. </p>
| Matthew Leingang | 2,785 | <p>The insight here is that the fraction is of the form $\frac{ab}{a^2+b^2}$, with $a=x$ and $b=y^2$.<br>
\begin{align*}
(x-y^2)^2 &\geq 0 \\
\implies x^2 - 2xy^2 + y^4 &\geq 0 \\
\implies x^2 + y^4 &\geq 2xy^2 \\
\implies\frac{xy^2}{x^2 + y^4} &\leq \frac{1}{2}
\end{align*}</p>
<p>You can also apply the arithmetic mean-geometric mean inequality with $a = x^2$ and $b=y^4$. Then
\begin{align*}
\sqrt{ab} &\leq \frac{1}{2}(a + b) \\
\implies xy^2 &\leq \frac{1}{2}(x^2+y^4) \\
\implies \frac{xy^2}{x^2+y^4} &\leq \frac{1}{2}
\end{align*}</p>
|
3,913,244 | <p>My attempt :</p>
<p><span class="math-container">$A=2^3×5^2×7^3$</span></p>
<p>Let's determine number of numbers primes with A, and
Smaller than A</p>
<p><span class="math-container">$\rho (A) = 2^2 ×4×5×6×7^2 =23520$</span></p>
<p>23520 is a number of numbers primes with A and smaller than A</p>
<p><span class="math-container">$\forall k$</span> <span class="math-container">$ \in {23520} $</span> <span class="math-container">$gcd(A, k)=1$</span></p>
<p>So : number of numbers doesn't prime white A is <span class="math-container">$A-23520=45080$</span></p>
<p><span class="math-container">$\forall k $</span> <span class="math-container">$\in {45080} $</span> <span class="math-container">$gcd(A, k) ≠1$</span></p>
<p>let's search about The number of numbers that divide A</p>
<p><span class="math-container">$A=0[k]$</span>
<span class="math-container">$\Leftrightarrow$</span> <span class="math-container">$68600=0[k]$</span></p>
<p>In the first we can see K can take the following values <span class="math-container">${1,2,4, 5,7,8,10,20,40,50,70,100}$</span></p>
<p>After using this theory <span class="math-container">$(a=b[n] $</span>and <span class="math-container">$a=b[m] $</span> and <span class="math-container">$gcd(m, n) =1$$\Rightarrow$</span> <span class="math-container">$a=b[MN] $</span>)</p>
<p>We can see that</p>
<p><span class="math-container">$K=(1, 2,4,5,7,8,10,20,40,50,70,100,35,56,28,14,140,280,350,700)$</span></p>
<p>So igot 20 value of k</p>
<p>But I know, there are 48 values of k</p>
<p>I want to know how can I continue for find 48 value of k</p>
<p>Note : I don't want any
other solution, I want to develop of my attempt</p>
| Bob Dobbs | 221,315 | <p>Let <span class="math-container">$F(x)=\sum_{k=0}^{m}a_kx^k$</span> where <span class="math-container">$m\leq p-2.$</span>
<span class="math-container">$$\sum_{i=0}^{p-1}F(i)=\sum_{i=0}^{p-1}\sum_{k=0}^{m}a_ki^k=\sum_{k=0}^{m}a_k\sum_{i=0}^{p-1}i^k=\sum_{k=0}^{m}a_k\frac{B_{k+1}(p)-B_{k+1}(0)}{k+1}=\sum_{k=0}^{m}a_k\sum_{j=0}^{k}{k+1 \choose j}B_jp^{k+1-j}$$</span>
by <a href="https://en.wikipedia.org/wiki/Faulhaber%27s_formula" rel="nofollow noreferrer">Faulhaber's formula</a>. Now, we apply another essential theorem: <a href="https://en.wikipedia.org/wiki/Von_Staudt%E2%80%93Clausen_theorem" rel="nofollow noreferrer">Von Staudt-Clausen's theorem</a>. In the last expession above, <span class="math-container">$j\leq k\leq m <p-1$</span>, so <span class="math-container">$p-1$</span> does not divide <span class="math-container">$j$</span>. By Von-Staudt-Clausen's theorem, we conclude that <span class="math-container">$B_jp\equiv 0\pmod p$</span> which implies that <span class="math-container">$B_jp^{k+1-j}\equiv 0\pmod p$</span> for each <span class="math-container">$j,k$</span>. Hence, <span class="math-container">$\sum_{i=0}^{p-1}F(i)\equiv 0\pmod p$</span>.</p>
<p>Dr. İ Dibağ also published an article about Von Staudt-Clausen's theorem: <a href="https://www.sciencedirect.com/science/article/pii/0021869389901804" rel="nofollow noreferrer">Link</a>.</p>
|
572,541 | <blockquote>
<p>Let $L$ be the set of all lines in the plane. Prove that $L$ is uncountable, but only countably many of the lines in $L$ contain more than one rational point.</p>
</blockquote>
<p><strong>Attempt</strong>: Well, I was trying to define $L$ using linear combinations of points since a line is a linear combo of two points. So, I wanted to define $L=\{ax+yb:x+y+z=k,k∈Z_+\}$. But, this does not seem right. Anyway, once I define $L$ I would try to find a function from $L$ to some <strong>uncountable</strong> set D that is onto or a function from D to L that is 1-1. Help appreciated thank you.</p>
| Kile Kasmir Asmussen | 72,934 | <p>All lines in the plane are a superset of all lines perpendicular to a given line.</p>
<p>How is it with perpendicular lines, points on their parent line and the number of points on a line?</p>
|
572,541 | <blockquote>
<p>Let $L$ be the set of all lines in the plane. Prove that $L$ is uncountable, but only countably many of the lines in $L$ contain more than one rational point.</p>
</blockquote>
<p><strong>Attempt</strong>: Well, I was trying to define $L$ using linear combinations of points since a line is a linear combo of two points. So, I wanted to define $L=\{ax+yb:x+y+z=k,k∈Z_+\}$. But, this does not seem right. Anyway, once I define $L$ I would try to find a function from $L$ to some <strong>uncountable</strong> set D that is onto or a function from D to L that is 1-1. Help appreciated thank you.</p>
| GeorgeMoreno | 109,822 | <p>How about this? Let L be a line in the plane and v some vector that does not generates that line. You can prove that for every pair of real numbers (a,b), L+bv and L+av have empty intersection. For the second part, try this idea:
Every line with a irrational slope contains only one rational point (prove it by contradiction).</p>
|
457,977 | <p>I am trying to use residues to compute $$\int_0^\infty\frac{\log x}{(1+x)^3}\,\operatorname d\!x.$$My first attempt involved trying to take a circular contour with the branch cut being the positive real axis, but this ended up cancelling off the term I wanted. I wasn't sure if there was another contour I should use. I also had someone suggest using the substitution $x=e^z$, so the integral becomes $$\int_{-\infty}^\infty\frac{ze^z}{(1+e^z)^3}\,\operatorname d\!z$$so that the poles are the at the odd multiples of $i\pi$. I haven't actually worked this out, but it does not seem like the solution the author was looking for (this question comes from an old preliminary exam).</p>
<p>Any suggestions on how to integrate?</p>
| Tunk-Fey | 123,277 | <h2>This approach is not using residues method but I'd like to post the general solution.</h2>
<hr>
<p>Let
$$\mathcal{I}=\int_0^\infty\frac{x^{m-1}}{(1+x)^{m+n}}\ln x\ dx\tag1$$
Consider <a href="http://en.wikipedia.org/wiki/Beta_function" rel="nofollow">beta function</a>
$$
\text{B}(m,n)=\int_0^\infty\frac{x^{m-1}}{(1+x)^{m+n}}\ dx.\tag2
$$
Differentiating $(2)$ with respect to $m$ yields
\begin{align}
\frac{\partial}{\partial m}\text{B}(m,n)&=\int_0^\infty\frac{\partial}{\partial m}\left(\frac{x^{m-1}}{(1+x)^{m+n}}\right)\ dx\\
(\psi(m)-\psi(m+n))\text{B}(m,n)&=\int_0^\infty\frac{x^{m-1}}{(1+x)^{m+n}}\ln\left(\frac{x}{1+x}\right)\ dx\tag3\\
&=\mathcal{I}-\color{red}{\int_0^\infty\frac{x^{m-1}}{(1+x)^{m+n}}\ln (1+x)\ dx},\tag4
\end{align}
where $\psi(\cdot)$ is the digamma function.</p>
<p>Setting $\color{red}{\displaystyle\ x=\frac1t\;\Rightarrow\;dx=-\frac{dt}{t^2}}$ to the second integral in $(4)$ yields
$$
\int_0^\infty\frac{t^{n-1}}{(1+t)^{n+m}}\ln \left(\frac{1+t}{t}\right)\ dt=(\psi(m+n)-\psi(n))\text{B}(m,n).\tag5
$$
Plugging in $(5)$ to $(4)$ yields
$$
\color{blue}{\int_0^\infty\frac{x^{m-1}}{(1+x)^{m+n}}\ln x\ dx=(\psi(m)-\psi(n))\text{B}(m,n)}.\tag6
$$</p>
<hr>
<p>Thus, using $(6)$ and setting $m=1\; ;\; n=2$, we obtain
$$
\large\int_0^\infty\frac{\ln x}{(1+x)^{3}}\ dx=(\psi(1)-\psi(2))\text{B}(1,2)=\color{blue}{-\frac12},
$$
where $\psi(1)= -\gamma$, $\ \psi(2)= 1-\gamma$, and $\displaystyle\ \text{B}(1,2)=\frac{\Gamma(1)\Gamma(2)}{\Gamma(3)}=\frac12$.</p>
|
1,719,568 | <p>Can we say that $k$ grows faster than $\sqrt{k}$ when term is large? But what is the formal way write it ?</p>
| Gabe Cunningham | 759 | <p>You can factor a $\sqrt{k}$ out of the bottom to get:
$\lim_{k \to \infty} \frac{1}{\sqrt{k}(\sqrt{k}-2)}$.
Now it should be clear that the bottom goes to infinity.</p>
|
1,447,547 | <p>$$x^3>x$$</p>
<p>Steps I took:</p>
<p>$$x^{ 3 }-x>0$$</p>
<p>$$x(x^{ 2 }-1)>0$$</p>
<p>$$x(x-1)(x+1)>0$$</p>
<p>Now I see that all three linear factors must equal a positive value when multiplied. </p>
<p>I took each linear factor and set each to either greater than or less than $0$ since the solution set can either be all positive or two negative and one positive.</p>
<p>$$x>0\quad or\quad x< 0,\quad x>1\quad or\quad x<1,\quad x>-1\quad or\quad x <-1$$ </p>
<p>Now where do I go from here? Or am I doing this all wrong?</p>
| Mark Viola | 218,419 | <p>Either all $3$ factors, $x+1$, $x$, and $x-1$ are positive or one and only one is positive. </p>
<p>That observation facilitates analysis.</p>
<p>Either $x>1$, in which case all $3$ factors are positive or $-1<x<0$, in which case only one, $x+1$, is positive and the other two are negative. In both cases the inequality holds.</p>
|
3,063,651 | <p>i am currently looking out for some possible topics i could study for my research project in high school. Algebra, trigonometry, Pythagoras’ theorem, geometry, circles and their properties, etc. and perhaps combined with a little knowledge from Physics i.e. Kinematics, Gravity, etc. could interest me. </p>
<p>Another particularly interesting one i stumbled upon was the Rubik’s cube since i myself do enjoy solving it. I am not very sure what type of research topics i could come up with that could be geared towards and tied with the math behind solving Rubik’s cube? Any suggestions or advices are greatly appreciated :)</p>
| Wuestenfux | 417,848 | <p>Welcome to MSE! The rubic's cube is an interesting object from the group theory point of view. Below you will find the enumeration of the faces of the cube. The central faces remain stationary. There six basic rotations of 90 degrees as shown: Left, Right, Up, Down, Front, Back.</p>
<p>These rotations can be described by the elements (called permutations) of the symmetric group <span class="math-container">$S_{48}$</span>. The group consists of all bijections <span class="math-container">$[48]\rightarrow[48]$</span>, where <span class="math-container">$[48]=\{1,\ldots,48\}$</span>.</p>
<p><a href="https://i.stack.imgur.com/uCL6U.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uCL6U.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/UqaEX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UqaEX.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/97s29.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/97s29.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/KviHj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KviHj.png" alt="enter image description here" /></a></p>
<p>The basic rotations are</p>
<p>F = (17, 19, 24, 22)(18, 21, 23, 20)(06, 25, 43, 16)
(07, 28, 42, 13)(08, 30, 41, 11)</p>
<p>B = (33, 35, 40, 38)(34, 37, 39, 36)(03, 09, 46, 32)
(02, 12, 47, 29)(01, 14, 48, 27)</p>
<p>L = (09, 11, 16, 14)(10, 13, 15, 12)(01, 17, 41, 40)
(04, 20, 44, 37)(06, 22, 46, 35)</p>
<p>R = (25, 27, 32, 30)(26, 29, 31, 28)(03, 38, 43, 19)
(05, 36, 45, 21)(08, 33, 48, 24)</p>
<p>U = (01, 03, 08, 06)(02, 05, 07, 04)(09, 33, 25, 17)
(10, 34, 26, 18)(11, 35, 27, 19)</p>
<p>D = (41, 43, 48, 46)(42, 45, 47, 44)(14, 22, 30, 38)
(15, 23, 31, 39)(16, 24, 32, 40).</p>
<p>(All these permutations have order 4.)</p>
<p>For instance, in view of F, the cycle (17, 19, 24, 22) says that face 17 mapsto 19, 19 mapsto 24, 24 mapsto 22, and 22 maps to 17.</p>
<p>A basic question would be how to obtain a certain constellation from the unscrambled cube, such as the superflip (all corner stones are at the right position but the edge stones are flipped).</p>
<p><a href="https://i.stack.imgur.com/dHmLf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dHmLf.png" alt="enter image description here" /></a></p>
<p>The answer is <span class="math-container">$UR^2FBRB^2RU^2LB^2RU^{−1}D^{−1}R^2FR^{−1}LB^2U^2F^2$</span></p>
|
718,166 | <p><strong>Question:</strong></p>
<blockquote>
<p>let $a\in(0,1)$, and such $f(x)\geq0$, $x\in R$ is continuous on $R$,</p>
<p>if
$$f(x)-a\int_x^{x+1}f(t)dt,\forall x\in R $$ is constant,</p>
<p>show that</p>
<p>$f(x)$ is constant;</p>
<p>or $$f(x)=Ae^{bx}+B$$ where $A\ge 0,|B|\le A$ and $A,B$ are constant, and the positive number $b$ is such
$\dfrac{b}{e^b-1}=a$</p>
</blockquote>
<p><strong>My try:</strong></p>
<p>let $$f(x)-a\int_x^{x+1}f(t)dt=C$$
then we have
$$f'(x)-af(x+1)+af(x)=0,\forall x\in R$$</p>
<p><strong>other idea:</strong></p>
<p>let $$F(x)=\int_{0}^{x}f(x)dx$$
then
$$f(x)-a\int_x^{x+1}f(t)dt=F'(x)-a[F(x+1)-F(x)]=C$$
then I can't solve this ODE? maybe my idea is not good.</p>
<p>Thank you very much. </p>
| Karl Marx | 144,187 | <p>I need help proving the following result:</p>
<p>If $f(x)-a\int_x^{x+1}f(t)dt$ is constant, then $f(x)$ is constant or $f(x)=Ae^{bx}+B$.</p>
<ol>
<li>Firstly, we transform this ODE to the form:
$$\frac{dy}{dx}+P(x)y=Q(x).$$</li>
</ol>
<p>Such equation has the solution:
$$y=e^{-\int{Pdx}}$$[\int{Qe^{\int{Pdx}}+C}].$$</p>
<ol start="2">
<li>Then,we calculate this solution as the following way:
$$fe^{ax}=\int{Qe^{ax}dx+C}\implies(Ae^{bx}+B)e^{ax}=\int{Qe^{ax}dx+C}\implies Qe^{ax}=Aae^{b}e^{(a+b)x}+Bae^{ax}\implies Q=Aae^{b}e^{bx}+Ba.$$</li>
</ol>
<p>Therefore,
$$f(x+1)=f^{'}(x)\frac{e^{b}}{b}+C\implies Q=f(x+1)$$
where
$Aae^{b}\ge0,$ $Ba\le{Aae^{b}}$.</p>
<p>From my point, this ODE has the solution of $f=Ae^{\lambda{x}}+B$ is obviously, the only thing we should prove is that $\lambda=b=a(e^{b}-1)$.</p>
|
2,904,912 | <p>$$24a(n)=26a(n-1)-9a(n-2)+a(n-3)$$
$$a(0)=46, a(1)=8, a(2)=1$$
$$\sum\limits_{k=3}^{\infty}a(k)=2^{-55}$$
How can I prove it?</p>
| mengdie1982 | 560,634 | <p>Consider the <strong>generating function</strong> $$f(x)=a_0x^0+a_1x^1+a_2x^2+\cdots+a_nx^n+\cdots.\tag1$$</p>
<p>We have $$f(x)\cdot x^3=a_0x^3+a_1x^4+a_2x^5+\cdots+a_nx^{n+3}+\cdots;\tag2$$
$$f(x)\cdot x^2=a_0x^2+a_1x^3+a_2x^4+\cdots+a_nx^{n+2}+\cdots;\tag3$$
$$f(x)\cdot x=a_0x^1+a_1x^2+a_2x^3+\cdots+a_nx^{n+1}+\cdots.\tag4$$
Thus, by $(2)-9\times(3)+26\times(4)-(1)$, we obtain
\begin{align*}
f(x)(x^3-9x^2+26x-1)&=-9a_0x^2+26a_0x^1+26a_1x^2-a_0x^0-a_1x^1-a_2x^2\\
&=-207x^2+1188x-46
\end{align*}</p>
<p>Hence $$\boxed{f(x)=\frac{-207x^2+1188x-46}{x^3-9x^2+26x-1}}.$$</p>
<p>Thus $$\sum_{k=0}^\infty a_k=f(1)=55.$$</p>
<p>As a result, $$\sum_{k=3}^\infty a_k=\sum_{k=0}^\infty a_k-\sum_{k=0}^2 a_k=55-55=0.$$</p>
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