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 |
|---|---|---|---|---|
33,702 | <p>The code that I have written has an unintended consequence that I'm not sure how to get around. I want 3 rotation transforms to be applied simultaneously to 1 graphics object. Instead, I get 3 separate separate copies of the graphics object, one per transformation.</p>
<p>The documentation does state that this will be the outcome of using multiple transformations on a graphics object.</p>
<blockquote>
<p>GeometricTransformation[g, {t1, t2, ...}]<br/>
<strong><em>represents multiple copies</em></strong> of g transformed by a collection of transformations.</p>
</blockquote>
<p>My question is: how is it possible to achieve the the outcome that I described instead of getting multiple copies?</p>
<p>Here is the code I am executing:</p>
<pre><code>Manipulate[
Graphics3D[{
EdgeForm[None],
GeometricTransformation[Cylinder[],
{RotationTransform[a Pi, {1, 0, 0}],
RotationTransform[b Pi, {0, 1, 0}],
RotationTransform[c Pi, {0, 0, 1}]}]}],
{{a, 0}, -1, 1},
{{b, 0}, -1, 1},
{{c, 0}, -1, 1},
SaveDefinitions -> True]
</code></pre>
<p>If anyone could show me a way to accomplish this, I'd appreciate it.</p>
| Alexey Popkov | 280 | <p>The problem arises when <code>function</code> returns a number smaller than <code>$MinMachineNumber</code>:</p>
<pre><code>function[t_] := Exp[-9 t^2];
LogLogPlot[function[t], {t, 8.8718, 8.872}, PlotRange -> All,
GridLines -> {{{Sqrt[Log[1/$MinMachineNumber]]/3,
Directive[Thick, Dashed]}}, None}]
Show[%, Ticks -> Automatic]
</code></pre>
<p><img src="https://i.stack.imgur.com/Q0gmX.png" alt="screenshot"></p>
<p>For some reason <code>LogLogPlot</code> considers numbers smaller than <code>$MinMachineNumber</code> as having logarithm equal to <code>0</code>, which is incorrect of course.</p>
<p>The workaround suggested in the comments is to abandon machine numbers by switching to arbitrary precision arithmetics. Even if one sets <code>WorkingPrecision</code> much smaller than <code>$MachinePrecision</code> this bug completely disappears, <strong>but now <a href="https://mathematica.stackexchange.com/a/15642/280">another bug</a> arises</strong>: the plot produced is not logarithmic in vertical scale and has incorrect vertical tick labels:</p>
<pre><code>LogLogPlot[{function[t], Exp[-9 t^2]}, {t, .01, 100}, WorkingPrecision -> 3]
</code></pre>
<p><img src="https://i.stack.imgur.com/NdcbL.png" alt="screenshot"></p>
<p>It is easy to see that the line computed with non-<code>MachinePrecision</code> <code>WorkingPrecision</code> is identical to the one generated by <code>LogLinearPlot</code>:</p>
<pre><code>LogLogPlot[t^3, {t, 1/100, 100}, WorkingPrecision -> 3][[1]] ===
LogLinearPlot[t^3, {t, 1/100, 100}, WorkingPrecision -> 3][[1]]
</code></pre>
<blockquote>
<p>True</p>
</blockquote>
<hr>
<p>Note that numbers which cannot be expressed as machine numbers are not affected when one tries to convert them to machine numbers:</p>
<pre><code>SetPrecision[function[8.872], MachinePrecision] // Precision
</code></pre>
<blockquote>
<p>12.9256</p>
</blockquote>
<p>So another workaround would be to perform conversion of non-machine-size real numbers manually:</p>
<pre><code>function[t_?NumberQ] :=
If[$MinMachineNumber < Exp[-9 t^2] < $MaxMachineNumber, Exp[-9 t^2],
Indeterminate]
</code></pre>
<p>Now <code>LogLogPlot</code> behaves as expected.</p>
<p>Even better way is to switch off generation of non-machine numbers:</p>
<pre><code>SetSystemOptions["CatchMachineUnderflow" -> False]
</code></pre>
<p>This way completely removes discussed bug and does not introduce new bugs but it is strongly advisable to localize the effect of this setting as explained <a href="https://mathematica.stackexchange.com/a/25668/280">here</a>.</p>
<p>Another workaround is <a href="https://mathematica.stackexchange.com/a/37737/280">suggested by halirutan</a>.</p>
<hr>
<p>And of course it is possible to switch to another function which has no bugs. In this case <code>LogLinearPlot</code> adaptive plotting capabilities can be utilized:</p>
<pre><code>function[t_?NumberQ] := Exp[-9 t^2]
points = Reap[
LogLinearPlot[
Log@Last@Sow[{t, {function[t], Exp[-9 t^2]}}], {t, .01, 100}]][[2,
1]]
ListLogLogPlot[Sort /@ Transpose[Thread /@ points], Joined -> True]
</code></pre>
<p><img src="https://i.stack.imgur.com/tMkvt.png" alt="screenshot"></p>
|
4,213,207 | <p>I am trying to find all homomorphisms from <span class="math-container">$\mathbb{Z}_{20}$</span> to <span class="math-container">$\mathbb{Z}_8$</span>. I understand how to do it - one completely determines any homomorphism, say <span class="math-container">$\phi$</span>, by computing multiples of <span class="math-container">$\phi(1)$</span> or more generally <span class="math-container">$\phi(g)$</span> where <span class="math-container">$g$</span> generates <span class="math-container">$\mathbb{Z}_{20}$</span>. Call <span class="math-container">$\phi_i$</span> the mapping taking <span class="math-container">$1$</span> to <span class="math-container">$i$</span> in <span class="math-container">$\mathbb{Z}_8$</span>. I don’t know how else to determine whether each <span class="math-container">$\phi_i$</span> is a homomorphism besides going through the painstaking process of calculating each <span class="math-container">$\phi_i(x+y)$</span> and comparing it to <span class="math-container">$\phi_i(x)+\phi_i(y)$</span>.</p>
<p>Is there a way to easily rule out certain <span class="math-container">$\phi_i$</span>s?</p>
<p>Any help would be greatly appreciated.</p>
| Dave | 334,366 | <p>You are correct about looking at <span class="math-container">$\phi(1)$</span>: since <span class="math-container">$1$</span> generates the domain (i.e. <span class="math-container">$\mathbb Z_{20}$</span>), you can just look at where <span class="math-container">$1$</span> gets sent under the homomorphism. The only other thing you need to check is that the homomorphism preserves the relations. There is one relation in <span class="math-container">$\mathbb Z_{20}$</span>, namely <span class="math-container">$20=0$</span>. So <span class="math-container">$\phi$</span> better send <span class="math-container">$20$</span> to <span class="math-container">$0$</span> in <span class="math-container">$\mathbb Z_8$</span> (i.e. a multiple of <span class="math-container">$8$</span>). For example, if you have a map which sends <span class="math-container">$1$</span> to <span class="math-container">$3$</span>, then this would send <span class="math-container">$20$</span> to <span class="math-container">$60$</span>, which would mean it's not a homomorphism (since <span class="math-container">$60$</span> is not <span class="math-container">$0$</span> in <span class="math-container">$\mathbb Z_8$</span>).</p>
<p>The real reason behind this is the universal property of quotients. Basically you're looking at group homomorphisms <span class="math-container">$\mathbb Z\to\mathbb Z_8$</span> (which are completely determined by where <span class="math-container">$1$</span> is sent, and you can send <span class="math-container">$1$</span> anywhere) and then asking that the kernel contains the ideal <span class="math-container">$20\mathbb Z\subseteq\mathbb Z$</span> so that it factors as a map <span class="math-container">$\mathbb Z_{20}\to\mathbb Z_8$</span>.</p>
|
4,213,207 | <p>I am trying to find all homomorphisms from <span class="math-container">$\mathbb{Z}_{20}$</span> to <span class="math-container">$\mathbb{Z}_8$</span>. I understand how to do it - one completely determines any homomorphism, say <span class="math-container">$\phi$</span>, by computing multiples of <span class="math-container">$\phi(1)$</span> or more generally <span class="math-container">$\phi(g)$</span> where <span class="math-container">$g$</span> generates <span class="math-container">$\mathbb{Z}_{20}$</span>. Call <span class="math-container">$\phi_i$</span> the mapping taking <span class="math-container">$1$</span> to <span class="math-container">$i$</span> in <span class="math-container">$\mathbb{Z}_8$</span>. I don’t know how else to determine whether each <span class="math-container">$\phi_i$</span> is a homomorphism besides going through the painstaking process of calculating each <span class="math-container">$\phi_i(x+y)$</span> and comparing it to <span class="math-container">$\phi_i(x)+\phi_i(y)$</span>.</p>
<p>Is there a way to easily rule out certain <span class="math-container">$\phi_i$</span>s?</p>
<p>Any help would be greatly appreciated.</p>
| Geoffrey Trang | 684,071 | <p>We need to find how many <span class="math-container">$i \in \mathbb{Z}_{8}$</span> satisfy the relation <span class="math-container">$20i=0$</span>. There are only <span class="math-container">$8$</span> cases to consider:</p>
<ol>
<li><span class="math-container">$i=0$</span> works (<span class="math-container">$0$</span> is divisible by <span class="math-container">$8$</span>)</li>
<li><span class="math-container">$i=1$</span> doesn't work (<span class="math-container">$20$</span> is not divisible by <span class="math-container">$8$</span>)</li>
<li><span class="math-container">$i=2$</span> works (<span class="math-container">$40$</span> is divisible by <span class="math-container">$8$</span>)</li>
<li><span class="math-container">$i=3$</span> doesn't work (<span class="math-container">$60$</span> is not divisible by <span class="math-container">$8$</span>)</li>
<li><span class="math-container">$i=4$</span> works (<span class="math-container">$80$</span> is divisible by <span class="math-container">$8$</span>)</li>
<li><span class="math-container">$i=5$</span> doesn't work (<span class="math-container">$100$</span> is not divisible by <span class="math-container">$8$</span>)</li>
<li><span class="math-container">$i=6$</span> works (<span class="math-container">$120$</span> is divisible by <span class="math-container">$8$</span>)</li>
<li><span class="math-container">$i=7$</span> doesn't work (<span class="math-container">$140$</span> is not divisible by <span class="math-container">$8$</span>)</li>
</ol>
<p>So, there are exactly <span class="math-container">$4$</span> group homomorphisms from <span class="math-container">$\mathbb{Z}_{20}$</span> to <span class="math-container">$\mathbb{Z}_{8}$</span>. Only even values of <span class="math-container">$i$</span> work, and the odd values can be ruled out.</p>
|
2,894,126 | <blockquote>
<p>$$\int \sin^{-1}\sqrt{ \frac{x}{a+x}} dx$$</p>
</blockquote>
<p>We can substitute it as $x=a\tan^2 (\theta)$ . Then:</p>
<p>$$2a\int \theta \tan (\theta)\sec^2 (\theta) d\theta$$</p>
<p>Using integration by parts will be enough here. But I wanted to know if this particular problem can be solved by any other method. Because the above method is quite lengthy. </p>
| lab bhattacharjee | 33,337 | <p>Hint:</p>
<p>Let $\sin^{-1}\sqrt{\dfrac x{x+a}}=y\ge0\implies\tan^2y=\dfrac xa\implies y=\tan^{-1}\sqrt{\dfrac xa}$</p>
<p>$$\int\sin^{-1}\sqrt{\dfrac x{x+a}}dx=\int\tan^{-1}\sqrt{\dfrac xa}dx$$</p>
<p>$$=\tan^{-1}\sqrt{\dfrac xa}\int dx-\int\left(\dfrac{d \tan^{-1}\sqrt{\dfrac xa}}{dx}\int dx\right)dx$$</p>
<p>$$=x\tan^{-1}\sqrt{\dfrac xa}-\int\dfrac{x}{2\sqrt a\sqrt x\left(1+\dfrac xa\right)}dx$$</p>
<p>For the last part, choose $\sqrt x=u$</p>
|
1,729,308 | <p>The sum of the first $n$ $(n>1)$ terms of the A.P. is $153$ and the common difference is $2$. If the first term is an integer , then number of possible values of $n$ is </p>
<p>$a)$ $3$</p>
<p>$b)$ $4$</p>
<p>$c)$ $5$</p>
<p>$d)$ $6$</p>
<p>My approach : I used the formula for the first $n$ terms of an A.P. to arrive at the following quadratic equation
$n^2 + n(a-1) -153 = 0 $</p>
<p>Next up I realised that since we are talking about the number of terms , thus the
possible values which n can take must be whole numbers. That is the discriminant of the above quadratic should yield a whole number in other words </p>
<p>$ (a-1)^2 + 612 = y^2 $ for some y . </p>
<p>However I am stuck at this point , as from here I am unable to figure out the number of such a's ( i.e. the initial terms of an AP ) which will complete the required pythagorean triplet
The answer mentioned is $5$</p>
<p>Please let me know , if I am doing a step wrong somewhere . Or If you have a better solution , that will be welcomed too. </p>
| Steven Alexis Gregory | 75,410 | <p>\begin{align}
n^2 +(a-1)n - 153 &= 0 \\
(n+u)(n-v) &= 0 \\
n^2 + (u-v)n - uv &= 0
\end{align}</p>
<p>So you need two positive integers, say $u>v>0$, such that $uv=153$ and $u-v=a-1$. Which will have solution $n=v$.</p>
<p>There aren't that many possibilities</p>
<p>\begin{array}{|rr|r|rr|}
\hline
u & v & u-v & a & n \\
\hline
153 & 1 & 152 & 153 & 1 \\
51 & 3 & 48 & 49 & 3 \\
17 & 9 & 8 & 9 & 9 \\
\hline
\end{array}</p>
<hr>
<p>Above I was assuming $a > 1$. If you allow $a$ to be a negative number, then the analysis changes a bit.</p>
<p>$$n^2 +(a-1)n - 153 = 0$$</p>
<p>The divisors of $153$ are $1,3,9,17,51,153$ We find the following.</p>
<p>\begin{array}{|l|rr|}
\hline
\text{factors} & a & n\\
\hline
(n-1)(n+153) = n^2+152n - 153 & 153 & 1 \\
(n-3)(n+ 51) = n^2 +48n - 153 & 49 & 3 \\
(n-9)(n+17) = n^2 +8n - 153 & 9 & 9 \\
(n-17)(n+9) = n^2 -8n - 153 & -7 & 17 \\
(n-51)(n+3) = n^2 -48n - 153 & -47 & 51 \\
(n-153)(n+1) = n^2-152n - 153 & -151 & 153 \\
\hline
\end{array}</p>
|
331,654 | <p>After having received Brian M. Scott's permission (see comments in the selected answer) I am integrating his suggestions with my own solutions to form a complete answer to the questions apperaing below. </p>
<blockquote>
<p>Let $\mathscr{T}$ be the collection of subsets of $\Bbb R$ consisting
of $\emptyset, \Bbb R$ and the rays of the form $(r, \infty)$,
where $r \in \Bbb R$.</p>
<p>$(a)$ Exhibit that this, indeed, is a topology on $\Bbb R$.</p>
</blockquote>
<p><em>Proof</em>: Given any finite number of open sets of the form above, $\exists$ a maximal $r_n$. The set $(r_n, \infty)$ is the intersection of these finite sets. Let $\{U_i\}_{i \in I}$ be an arbitrary family of open sets. Let $r_{\min} = \inf\{r_i\}$. Then $(r_{\min}, \infty)$ is the union of the $U_i$ and clearly it is of the desired form to be an element of the topology. If $r_{\min} = -\infty$ then the union of the $U_i$ is the entire $\Bbb R$.</p>
<blockquote>
<p>$(b)$ Show that it fails to be a topology if $r \in \Bbb Q$.</p>
</blockquote>
<p>I don't have the full answer to this. I think that there should be a problem in the union of arbitrarily many open sets. Any help with a counterexample will be very helpful.</p>
<blockquote>
<p>Answer the following questions. Is $(\Bbb R, \mathscr{T})$:</p>
<p>$(c)$ $T_1$?</p>
</blockquote>
<p>No. Let $x_1 \neq x_2 \in \Bbb R$ and assume without loss of generality that $x_1 < x_2$. Then any open set containing $x_1$ will be of the form $x_1 - \epsilon, \infty$ for some $\epsilon > 0$ and will encessarily contain $x_2$.</p>
<blockquote>
<p>$(d)$ Hausdorff</p>
</blockquote>
<p>No. Being Hausdorff (or $T_2$) would imply that it is $T_1$, a contradiction to (a).</p>
<blockquote>
<p>$(e)$ metrizable</p>
</blockquote>
<p>No. If there was a metric, the metric space would have to be Hausdorff, contradicting (b).</p>
<blockquote>
<p>$(f)$ second - countable</p>
</blockquote>
<p>I have no idea on how to go about this one.</p>
<blockquote>
<p>$(g)$ compact</p>
</blockquote>
<p>No. There exists no finite cover of this space. Suppose we are given a finite cover of this space. Then there exists a minimal $r$ such that $(r_{\min}, \infty)$ covers the entire space. Of ,course, this is only possible if $(r_{\min}, \infty) = \Bbb R$.</p>
<blockquote>
<p>$(h)$ locally compact</p>
</blockquote>
<p>Yes. We need to exhibit that any point has a compact neighbourhood. To this end, fix $x \in \Bbb R$. Let $(r, \infty)$ be a neighbourhood of $x$ and let $\{U_i\}_{i \in I}$ be an open cover. Then, there must exist $q \in \Bbb R$ such that $q \ge r$ so that $(q, \infty)$ is a set of the open cover. Clearly, taking $(q, \infty)$ as the subcover completes the proof.</p>
<blockquote>
<p>$(i)$ connected</p>
</blockquote>
<p>In part $(j)$ we prove that $(\Bbb R, \mathcal{T})$ is path-wise connected, hence connected.</p>
<blockquote>
<p>$(j)$ path-wise connected</p>
</blockquote>
<p>$\Bbb R$ is convex. Given $x, y \in \Bbb R$, the path $f: [0,1] \to \Bbb R$ given by $f(t) = (1-t)x + ty$ is continuous and $f(0) = x$ and $f(1) = y$. Since path-wise connected implies connected we also answered $(i)$.</p>
<blockquote>
<p>What is the closure of $\{1\}$ in $(\Bbb R, \mathcal{T})$?</p>
</blockquote>
<p><em>Proof</em>: Since the closure is the smallest closed set containing $\{1\}$, it is clear that it is $(-\infty, 1]$</p>
<p>Any suggestions, corrections, hints and any help, in general, will be tremendously appreciated! Any stylistic improvements in the formatting of the question are also greatly encouraged!</p>
| Brian M. Scott | 12,042 | <p>(b) Consider $\bigcup\{(r,\to):r\in\Bbb Q\text{ and }r>\sqrt2\}$.</p>
<p>(f) Consider the set $\mathscr{B}=\{(r,\to):r\in\Bbb Q\}$: is this a base for $\mathscr{T}$?</p>
<p>(g) A small correction: $\{\Bbb R\}$ is a finite cover of the space, as is any finite set of open sets that includes $\Bbb R$. However, $\{(r,\to):r\in\Bbb R\}$ is certainly an open cover with no finite subcover.</p>
<p>(h) Fix $r\in\Bbb R$, and let $\mathscr{U}=\{(q,\to):q>r\}$; $\mathscr{U}$ is an open cover of $(r,\to)$ with no finite subcover. Thus, <strong>no</strong> member of $\mathscr{T}$ is compact. However, if your definition of <em>neighborhood</em> does not require it to be open, then for any $x\in\Bbb R$ the set $[x-1,\to)$ is a nbhd (though not an open nbhd) of $x$ that <strong>is</strong> compact: if $x-1\in(r,\to)\in\mathscr{T}$, then $(r,\to)\supseteq[x-1,\to)$, so every open cover of $[x-1,\to)$ has a one-element subcover.</p>
<p>(j) You should say <strong>why</strong> $f$ is continuous, though I grant that it’s pretty easy.</p>
|
132,266 | <p>I apologize if this question is trivial, but a couple of days of searching for necessary routines have led me here. </p>
<blockquote>
<p>Does there exist software to compute symmetric powers of Schur polynomials? </p>
</blockquote>
<p>I am seeking such software in the hopes of computing the characters of representations of the simple Lie algebra $A_n$, i.e., $S_\lambda(x_1, \dots, x_n)$ to use the notation of Fulton and Harris, and then applying $\mathrm{Sym}^k$ to the resulting Schur polynomial and writing the result as a sum of Schur polynomials corresponding to differing partitions $\mu$. That is, as an example, I would like to compute something like the following:
$$
\mathrm{Sym}^3(S_\lambda(x_1, \dots, x_4)) = \sum_\mu k_\mu S_\mu(x_1, \dots, x_4)
$$
Where $S_\lambda(x_1, \dots, x_4)$ denotes the character of the irreducible representation of $A_4$ with highest weight $\lambda$. I understand that Mathematica may compute symmetric polynomials, however I have not found any routines for applying $\mathrm{Sym}$ to these polynomials. Regards.</p>
| Per Alexandersson | 1,056 | <p>If I am not mistaken, you are interested in the <a href="https://www2.math.upenn.edu/%7Epeal/polynomials/plethysm.htm" rel="nofollow noreferrer">plethysm</a>
<span class="math-container">$$
h_k[ s_\lambda]
$$</span></p>
<p>If you prefer Mathematica, I have a <a href="https://github.com/PerAlexandersson/Mathematica-packages" rel="nofollow noreferrer">package</a> for this type of computations. For example, <span class="math-container">$Sym^3(S_{21})$</span> is computed as</p>
<pre><code>ToSchurBasis@Plethysm[CompleteHSymbol[3],SchurSymbol[{2,1}]]
</code></pre>
<p>gives
<span class="math-container">$$
s_{63}+s_{333}+s_{432}+s_{441}
+s_{522}+s_{531}+s_{3222}+s
_{3321}+2
s_{4221}+s_{4311}+s_{5211}+
s_{32211}+s_{33111}+s_{4211
1}+s_{411111}
$$</span>
which matches Steven Sam's result (I believe).
Note that the output consists of Schur symmetric functions and not polynomials, so all the Schur functions indexed by partitions with length more than 4 are zero if you only work in <span class="math-container">$x_1,x_2,x_3,x_4$</span>.</p>
|
2,678,077 | <p>I need to solve this differential equation:
$$\frac{du}{dr}=\frac{4+\sqrt{r}}{2+\sqrt{u}}$$</p>
<p>I did it and got
$$u=\frac{2r^{\left(\frac{3}{2}\right)}}{3\sqrt{u}+2}+\frac{4r}{\sqrt{u}+2}+C$$ but my homework system is marking this as wrong. Why is that?</p>
| Dr. Sonnhard Graubner | 175,066 | <p>you must write $$(2+\sqrt{u})du=(4+\sqrt{r})dr$$ and integrate this!
the solution should be $$4\,r+2/3\,{r}^{3/2}-2\,u \left( r \right) -2/3\, \left( u \left( r
\right) \right) ^{3/2}+{\it \_C1}=0
$$</p>
|
4,634,180 | <p><span class="math-container">$$\int \frac{\sin^2(x)dx}{\sin(x)+2\cos(x)}$$</span></p>
<p>I tried to use different substitutions such as <span class="math-container">$t=\cos(x)$</span>, <span class="math-container">$t=\sin(x)$</span>, <span class="math-container">$t=\tan(x)$</span>, and after expressing <span class="math-container">$\sin$</span> and <span class="math-container">$\cos$</span> through <span class="math-container">$\tan(\frac{x}{2})$</span>, I've got <span class="math-container">$ \int -4 \frac{t^2dt}{(1+t^2)^2(t^2-t-1)}.$</span></p>
<p>Rational fractions didn’t work.</p>
| Ryszard Szwarc | 715,896 | <p>We have
<span class="math-container">$$(\sin x+2\cos x)(2\cos x-\sin x)=4-5\sin^2x=5\cos^2x-1$$</span>
Therefore
<span class="math-container">$${\sin^2x\over \sin x+2\cos x} ={\sin^2x(2\cos x-\sin x)\over (\sin x+2\cos x)(2\cos x-\sin x)}\\ =
2{\sin^2x\over 4-5\sin^2x}\cos x-{1-\cos^2x\over 5\cos^2x -1}\sin x$$</span>
Hence substituting <span class="math-container">$s=\sin x$</span> and <span class="math-container">$t=\cos x,$</span> respectively, gives
<span class="math-container">$$\int {\sin^2x\over \sin x+2\cos x}\,dx=2\int{s^2\over 4-5s^2}\,ds +\int{1-t^2\over 5t^2-1}\,dt \\
=-{2\over 5}s+{8\over 5}\int {1\over 4-5s^2}\,ds-{1\over 5}t+{4\over 5}\int{1\over 5t^2-1}\,dt $$</span>
The remaining integrals can be evaluated by easy partial fraction decompositions. At the end we have to get back to <span class="math-container">$x=\arcsin s$</span> and <span class="math-container">$x=\arccos t,$</span> respectively.</p>
|
986,412 | <p>Let $f(x) = \frac1{(1-x)}$.</p>
<p>Define the function $f^r$ to be $f^r(x) = f(f(f(...f(f(x)))))$.</p>
<p>Find $f^{653}(56)$.</p>
<p><strong>What I've done:</strong> </p>
<p>I started with r=1,2,3 and noticed the following pattern:
$$f^r(x)= \left\{
\begin{array}{c}
\frac1{1-x}, when \ r\equiv 1\pmod 3 \\
\frac{x-1}x, when \ r\equiv 2\pmod 3 \\
x, \ when \ r\equiv 0\pmod 3
\end{array}
\right. $$</p>
<p>As $653\equiv 2\pmod 3$, $\\$ $f^{653}(56) = \frac{55}{56}$</p>
<p>BUT how can I prove that I'm right? By induction? I don't know what to do then, when I go from $r$ to $r+1$. </p>
<p>Could you please share with me your reasoning by solving this problem?</p>
<p>PS: This problem is from the book "How to think like a mathematician" by Kevin Houston. </p>
| user84413 | 84,413 | <p>By the Fundamental Theorem of Calculus,</p>
<p>$\displaystyle f^{\prime}(\theta)=\frac{-\sin\theta}{1-\cos^{2}\theta}-\frac{\cos\theta}{\cos^{2}\theta}=-\csc\theta-\sec\theta$,</p>
<p>so $\displaystyle f(\theta)=-\left[\ln(\csc\theta-\cot\theta)+\ln(\sec\theta+\tan\theta)\right]+C=-\left[\ln((\csc\theta-\cot\theta)(\sec\theta+\tan\theta))\right]+C$</p>
<p>$\displaystyle\;\;=-\ln((\csc\theta+1)(\sec\theta-1))+C=-\ln((\csc\theta+1)(\sec\theta-1))=\ln((\sec\theta+1)(\csc\theta-1))$</p>
<p>since $f(\frac{\pi}{4})=0\implies-\ln((\sqrt{2}+1)(\sqrt{2}-1))+C=0\implies C=\ln1=0$</p>
|
2,913,921 | <p>As I know, the general complexity of matrix inversion is $O(n^3)$,
but it is a little bit high.
My matrix is $(I + A)$ , where $I$ is an $n \times n$ identity matrix and $A$ is a hermitian matrix and the norm of all its elements are very small (near $1/100$). Therefore, I think it's a particular matrix that close to an identity matrix. </p>
<p>So, is there any effective algorithm that can solve the inversion of the matrix with low-complexity ? hopefully , it can be reduced to $O(n^2)$. High accuracy is not very necessary. </p>
| Community | -1 | <p>There is something called a <a href="https://www.sciencedirect.com/science/article/pii/S1063520307001364" rel="nofollow noreferrer">fast randomized low-rank SVD</a>. The time complexity is $\mathcal{O}(n^{2}\log(k) +nl^{2}) $ The inversion at the end doesn't take much time. There are other similar matrix decompositions like this. There are likely ones made for Hermitian matrices as well. </p>
|
652,025 | <p>Assume $A$, $B$, and $C$ are three independent predicates. Maybe $A$ stands for "my age is 20," and $B$ "stands for tomorrow is a good day."</p>
<p>So is it true that $(A \lor B) \land C \iff (A \lor C) \land (B \lor C)$?</p>
| amWhy | 9,003 | <p>No, it's NOT true that $$(A \lor B) \land C \equiv (A \lor C) \land (B \lor C)\tag{NOT TRUE}$$</p>
<p>What is true, by the distributive law, is that $$(A\lor B)\land C \equiv (A\land C)\lor (B\land C)\tag{TRUE}$$</p>
<p>Confirm the correct equivalence below:</p>
<p><img src="https://i.stack.imgur.com/bqdET.gif" alt="enter image description here"></p>
<p><img src="https://i.stack.imgur.com/sYO6o.gif" alt="enter image description here"></p>
<p>Compare with the following:</p>
<p><img src="https://i.stack.imgur.com/5ech8.gif" alt="enter image description here"></p>
<p>EDIT: Note that there will be truth value assignments in which it turns out that the truth of $(A \lor B)\land C$ is equal to the truth value of $(A\lor C)\land (B\lor C)$ (and hence the resulting equal probabilities), (for example, see the first rows). But the two expressions are NOT the same under all possible truth value assignments (see for example the second rows, or the second to last rows), and therefore, are not equivalent. </p>
|
640,769 | <p>I'm to prove that every proper ideal is a product of maximal ideals which are uniquely determined up to order.
I have no idea even how to start in the proof to solve this question :(
May anybody help me ? </p>
| Siminore | 29,672 | <p>Write your assumption as $F'(x) \leq c F(x)$. You know that $F(0)=0$. Fix $x_0>0$ and define
$$
M_0 = \sup_{x \in [0,x_0]} F(x), \quad M_1 =\sup_{x \in [0,x_0]} F'(x).
$$
Then, by the Mean Value Theorem, for every $x \in [0,x_0]$, $F(x) \leq M_1 x_0 \leq c x_0 M_0$. If $c x_0 < 1$, then $F(x)=0$ for every $x \in [0,x_0]$. Now you start from $x_0$ and go on.</p>
<p><strong>Bibliography.</strong> W. Rudin. Principles of mathematical analysis. Chapter 5, exercise 26.</p>
|
1,776,260 | <p>After understanding the Cardano's formula for solving the depressed cubic (of the form $x^3+mx=n$, of course), I tried to find the solution of the equation $$x^3+6x=20.$$
After plugging into the formula
$$x=(n/2+\sqrt{ \frac{n^2}{4}+ \frac{m^3}{27} })^{1/3}+(-n/2+\sqrt{ \frac{n^2}{4}+ \frac{m^3}{27} })^{1/3}$$
where $m=6$ and $n=20$, we get
$$x=(10+ \sqrt{108})^{1/3}-(-10+ \sqrt{108})^{1/3}.$$
However, we notice that, without using Cardano's formula, that $x=2$ is the solution for the equation $x^3+6x=20.$
My question is: how does the equation $$x=(10+ \sqrt{108})^{1/3}-(-10+ \sqrt{108})^{1/3}$$ get simplified to $x=2$?</p>
<p>P.S. I understand that it was Niccolo Fontana who first figured out how to solve depressed cubic, to give one the proper credit.</p>
| DeepSea | 101,504 | <p>Let $a^3 = 10+\sqrt{108}=10+6\sqrt{3}, b^3 = -10 + \sqrt{108}=-10+6\sqrt{3}\implies a^3-b^3 = 20 = (a-b)(a^2+ab+b^2)=(a-b)((a-b)^2 + 3ab)$. Observe that $(ab)^3 = 8 \implies ab = 2\implies (a-b)((a-b)^2 + 6) = 20\implies x^3+6x-20 = 0, x = a-b$. Using synthetic division we have: $x^3+6x-20 = 0\implies (x-2)((x+1)^2 + 9) = 0 \implies x = 2$ as claimed.</p>
|
4,602,596 | <p>In an exercise, my teacher asked us to prove that <span class="math-container">$\ell^1$</span> is a Banach space. I was able to do so, but there are two steps in my proof that I'm not quite so sure they are correct. This is what I came up with:</p>
<hr />
<p>let <span class="math-container">$x^n=(x^n_1, x^n_2, ...)$</span> be a cauchy sequence of points in <span class="math-container">$\ell ^1$</span>. Then, for all <span class="math-container">$\varepsilon>0$</span> exists a natural number <span class="math-container">$N$</span> such that:</p>
<p><span class="math-container">$$n,m\geq N \implies\sum_{i=1}^\infty|x^n_i-x^m_i| \leq \varepsilon$$</span>
<span class="math-container">$$\implies |x^n_i-x^m_i| \leq \varepsilon$$</span></p>
<p>So, for every <span class="math-container">$i$</span> the sequence <span class="math-container">$(x^n_i)_{n\in \mathbb N}$</span> is a Cauchy sequence on <span class="math-container">$\mathbb C$</span> and therefore it converges, let's call it's limit <span class="math-container">$y_i \in \mathbb C$</span>.</p>
<p>We want to prove that <span class="math-container">$x^n \to y = (y_1,y_2,...)$</span></p>
<hr />
<p>This is the first step of the proof that I think isn't correct. I tried but failed to prove that <span class="math-container">$y$</span> is indeed a point in <span class="math-container">$\ell^1$</span>, this is:
<span class="math-container">$$\sum_{i=1}^\infty |y_i|\leq\infty$$</span></p>
<hr />
<p>To show that <span class="math-container">$\lim x^n = y$</span>, let <span class="math-container">$\varepsilon>0$</span>. Because <span class="math-container">$\lim_nx^n_i=y_i$</span>, for all <span class="math-container">$i$</span> we have that there is a natural number <span class="math-container">$N_i$</span> such that: <span class="math-container">$$n\geq N_i \implies |x_i^n - y_i|\leq \varepsilon\frac{1}{2^i}$$</span></p>
<p>so <span class="math-container">$$\sum_{i=1}^k|x_i^n - y_i|\leq\varepsilon\sum_{i=1}^k \frac{1}{2^i}\leq \varepsilon$$</span>
for <span class="math-container">$n \geq \max\{N_1,...,N_k\}.$</span></p>
<p>Because this is a bounded increasing sum, we have that it converges and <span class="math-container">$$||x^n - y|| = \lim_{k\to\infty}\sum_{i=1}^k|x_i^n - y_i|\leq \varepsilon$$</span></p>
<p>So <span class="math-container">$\lim x^n = y$</span></p>
<hr />
<p>This last part is the one I'm most unsure of. While I had a finite sum, the condition <span class="math-container">$$\sum_{i=1}^k|x_i^n - y_i|\leq \varepsilon$$</span></p>
<p>Was true for <span class="math-container">$n \geq \max\{N_1,...,N_k\}$</span> but I don't know if this is the case when <span class="math-container">$k\to\infty$</span>.</p>
<p>I thought about changing the definition of <span class="math-container">$n$</span> to <span class="math-container">$n \geq \sup\{N_1,...,N_k\}$</span> to accommodate the case where <span class="math-container">$k\to\infty$</span>, but this solves nothing because either the set is unbounded, or it eventually stays the same at some point for large values of <span class="math-container">$k$</span>.</p>
<p>How can I fix these two issues with my proof?</p>
| Gerd | 960,044 | <p>If <span class="math-container">$f$</span> is bounded (<span class="math-container">$|f(t)| \le b$</span>) then
<span class="math-container">$$
y(t)=e^{-t} \int_{-\infty}^t e^\tau f(\tau)d\tau
$$</span>
is a bounded solution. Note that the integral is convergent since <span class="math-container">$f$</span> is bounded.</p>
<p>We have
<span class="math-container">$$
|y(t)| \le b e^{-t} \int_{-\infty}^t e^\tau d\tau = b \quad (t \in \mathbb{R}).
$$</span>
The proper initial value is therefore <span class="math-container">$\int_{-\infty}^0 e^\tau f(\tau)d\tau$</span>.</p>
|
1,561,563 | <p>Two circles $\Gamma_1,\Gamma_2$ have centers $O_1,O_2$. Let $\Gamma_1\cap\Gamma_2=A,B$, with $A\neq B$. An arbitrary line through $B$ intersects $\Gamma_1$ at $C$ and $\Gamma_2$ at $D$. The tangents to $\Gamma_1$ at $C$ and to $\Gamma_2$ at $D$ intersect at $M$. Let $N=AM\cap CD$. Let $l$ be a line through $N$ parallel to $CM$, and let $l\cap AC=K$. Prove that $BK$ is tangent to $\Gamma_2$.</p>
<hr>
<p>$\qquad\quad$ <img src="https://i.stack.imgur.com/mTLgz.png" alt=""></p>
<hr>
<p>Here is some progress I have:</p>
<p>We are looking to prove $\angle O_2BP=90^{\circ}$, and since $\angle O_2DP=90^{\circ}$, if we could prove $BP=PD$, we would be done by congruent triangles. So we are looking to prove $\dfrac{\sin \angle BDP}{\sin \angle DBP}=1$. Let $AM\cap BK=l$. We have $\angle BDP=\angle QBN$. By the law of sines on $\triangle DNM,\triangle BNQ$, we have $\sin \angle BDP=\sin \angle NDM=\dfrac{NM}{DM}\sin \angle DNM$ and $\sin\angle QBN=\dfrac {NQ}{BQ}\sin\angle BNQ$. Dividing, the sines cancel (since they are supplementary), and we are left with $\dfrac{NM\times BQ}{DM\times NQ}$, so it remains to prove $\dfrac{NM}{DM}=\dfrac{NQ}{BQ}$.</p>
<p>I'm not sure what to do from here. We would be done if we could prove $\triangle NBQ\sim\triangle NDM$, but this would imply $\angle QNB=\angle MND=90^{\circ}$, but from drawing multiple diagrams, it looks like this is not always the case. </p>
<p>As always, any ideas are appreciated!</p>
| Ron Gordon | 53,268 | <p>There's a simple solution somewhere in M.SE which goes like this...</p>
<p>$$I = \int_0^{\pi/2} dt \frac{\cos{t}}{\cos{t}+\sin{t}} $$
$$J = \int_0^{\pi/2} dt \frac{\sin{t}}{\cos{t}+\sin{t}} $$</p>
<p>$$I + J = \frac{\pi}{2}$$
$$I - J = \int_0^{\pi/2} dt \frac{\cos{t}-\sin{t}}{\cos{t}+\sin{t}} = \int_0^{\pi/2} dt \frac{(\cos{t}+\sin{t})'}{\cos{t}+\sin{t}}=[\log{(\cos{t}+\sin{t})}]_0^{\pi/2} = 0$$</p>
<p>Thus $I = J = \frac{\pi}{4}$.</p>
<p><strong>ADDENDUM</strong></p>
<p>The above technique requires the easy computation of the antiderivative, which you should see is</p>
<p>$$\frac12 \left [t + \log{(\cos{t}+\sin{t})} \right ] + C = \frac12 \left [\arcsin{x} + \log{(x+\sqrt{1-x^2})} \right ] + C$$ </p>
|
1,561,563 | <p>Two circles $\Gamma_1,\Gamma_2$ have centers $O_1,O_2$. Let $\Gamma_1\cap\Gamma_2=A,B$, with $A\neq B$. An arbitrary line through $B$ intersects $\Gamma_1$ at $C$ and $\Gamma_2$ at $D$. The tangents to $\Gamma_1$ at $C$ and to $\Gamma_2$ at $D$ intersect at $M$. Let $N=AM\cap CD$. Let $l$ be a line through $N$ parallel to $CM$, and let $l\cap AC=K$. Prove that $BK$ is tangent to $\Gamma_2$.</p>
<hr>
<p>$\qquad\quad$ <img src="https://i.stack.imgur.com/mTLgz.png" alt=""></p>
<hr>
<p>Here is some progress I have:</p>
<p>We are looking to prove $\angle O_2BP=90^{\circ}$, and since $\angle O_2DP=90^{\circ}$, if we could prove $BP=PD$, we would be done by congruent triangles. So we are looking to prove $\dfrac{\sin \angle BDP}{\sin \angle DBP}=1$. Let $AM\cap BK=l$. We have $\angle BDP=\angle QBN$. By the law of sines on $\triangle DNM,\triangle BNQ$, we have $\sin \angle BDP=\sin \angle NDM=\dfrac{NM}{DM}\sin \angle DNM$ and $\sin\angle QBN=\dfrac {NQ}{BQ}\sin\angle BNQ$. Dividing, the sines cancel (since they are supplementary), and we are left with $\dfrac{NM\times BQ}{DM\times NQ}$, so it remains to prove $\dfrac{NM}{DM}=\dfrac{NQ}{BQ}$.</p>
<p>I'm not sure what to do from here. We would be done if we could prove $\triangle NBQ\sim\triangle NDM$, but this would imply $\angle QNB=\angle MND=90^{\circ}$, but from drawing multiple diagrams, it looks like this is not always the case. </p>
<p>As always, any ideas are appreciated!</p>
| Claude Leibovici | 82,404 | <p>Since no one used it, let me consider the case of the antiderivative $$I=\int\frac{\cos x}{\sin x + \cos x} \space dx$$ Now, use the tangent half-angle substitution (Weierstrass substitution) $t=\tan(\frac x 2)$.</p>
<p>We so obtain $$I=\int \frac{2 \left(t^2-1\right)}{t^4-2 t^3-2 t-1}\space dt$$ But $$t^4-2 t^3-2 t-1=(t^2+1)(t-r_1)(t-r_2)\qquad r_{1,2}=1\pm \sqrt 2$$ Using partial fraction decomposition, we have $$\frac{2 \left(t^2-1\right)}{t^4-2 t^3-2 t-1}=\frac{1-t}{t^2+1}+\frac{1}{2 t-2 \sqrt{2}-2}+\frac{1}{2 t+2 \sqrt{2}-2}$$ that is to say $$\frac{2 \left(t^2-1\right)}{t^4-2 t^3-2 t-1}=\frac{1}{1+t^2}-\frac{2t}{1+t^2}+\frac{1}{2 t-2 \sqrt{2}-2}+\frac{1}{2 t+2 \sqrt{2}-2}$$ and each term is easy to integrate. After simplifications, this leads to $$I=\tan ^{-1}(t)+\frac{1}{2} \log \left(\frac{-t^2+2 t+1}{t^2+1}\right)$$ Now, if integration is from $t=0$ to $t=1$, the logarithms disappear and youare just let with $\tan ^{-1}(1)=\frac \pi 4$.</p>
|
2,899,829 | <p>$\newcommand{\d}{\mathrm{d}}$</p>
<blockquote>
<p>Evaluate the integral using the indicated substituion. $$\int \cot x \csc^2x \,\d{x}, \qquad u= \cot x .$$</p>
</blockquote>
<p>Differentiating both sides of $u$, then making the substitution: $$
\begin{align}
u &=
\phantom{-}\cot x, \\
\d u &= -\cot x\csc x \,\d{x}, \\
\d x &= -\frac{\d u}{u \csc x}.
\end{align}$$
$$\int -\frac{u\csc^2 x \,\d{u}}{u\csc x} = \int -\csc x \,\d{u}. $$ </p>
<p>Apparently, this was not an adequate approach, because $x$ is still part of the integrand. What should be done instead?</p>
| Entrepreneur | 571,353 | <p><strong>For alternative way:</strong></p>
<p>$$\int \cot x \csc^2 x dx$$
$$=\int \frac{\cos x dx}{\sin^3 x}$$</p>
<p>Now you can advance taking $\sin x = z$ .</p>
|
1,499,949 | <p>Prove that for all event $A,B$</p>
<p>$P(A\cap B)+P(A\cap \bar B)=P(A)$</p>
<p><strong>My attempt:</strong></p>
<p>Formula: $\color{blue}{P(A\cap B)=P(A)+P(B)-P(A\cup B)}$</p>
<p>$=\overbrace {P(A)+P(B)-P(A\cup B)}^{=P(A\cap B)}+\overbrace {P(A)+P(\bar B)-P(A\cup \bar B}^{=P(A\cap \bar B)})$</p>
<p>$=2P(A)+\underbrace{P(B)+P(\bar B)}_{=1}-P(A\cup B)-P(A\cup \bar B)$</p>
<p>$=1+2P(A)-P(A\cup B)-P(A\cup \bar B)$</p>
| Koro | 266,435 | <p>Yes , there is a slight difference as per one of the conventions. <br>
Consider, $x^{1/2}$ and $\sqrt x$, former is written in a general sense(more appropriate to use) but the latter one is used when $x\ge 0$.So according to this $i^{1/2}$ and not $\sqrt i$.<br>
Hope this helps(not fully but to some extent atleast).<br>
<br>
In nutshell, $x^{1/n}$ is a multi-valued function but $\sqrt[n]{x}$ is a single valued function.<br><br>
As a quick example to this consider $\sqrt[3] 1$,and $1^{1/3}$.The former gives $1$(a single value) but the latter gives "cube roots of unity" (three values).<br>
PS: I posted it as an answer so that others can see it easily while going through this page.<br>
EDIT:THIS DIFFERENCE IS NOT UNIVERSALLY ACCEPTED.Actually I thought it's explicit from my above written sentence -" $x^{1/n}$ is a multi-valued function " but it doesn't seem to be .In multi-valued functions,you can take any branch (for example other than principal branch $(-\pi,\pi]$that you want.Now which one is chosen by you $\implies$ above stated difference is not universal.</p>
|
596,005 | <p>Show that $f:\mathbb{R}^2\to\mathbb{R}$, $f \in C^{2}$ satisfies the equation
$$\frac{\partial^2 f}{\partial x^2} - \frac{\partial^2 f}{\partial y^2} = 0$$
for all points $(x,y) \in \mathbb{R}^2$ if and only if for all $(x,y)\in \mathbb{R}^2$ and $t \in \mathbb{R}$ we have:
$$f(x, y + 2t) + f(x, y) = f(x + t,y + t) + f(x - t, y +t).$$</p>
<p><strong>Note</strong>. In such case, $f$ is said to satisfy the <em>parallelogram's law</em>.</p>
| Yiorgos S. Smyrlis | 57,021 | <p>First assume that (replacing $y$ by $y-t$ in the assumption)
$$
f(x, y + t) + f(x, y) = f(x + t,y) + f(x - t, y -t),
$$
for all $t,x,y\in\mathbb R$. This implies that
$$
\frac{f(x, y + t) -2f(x,y)+ f(x, y-t)}{t^2} = \frac{f(x + t,y)-2f(x,y) + f(x - t, y)}{t^2},
$$
for $x,y\in\mathbb R$ and $t\ne 0$. We shall now explain why, the left hand side tends to $f_{yy}(x,y)$, while the right hand side tends to $f_{xx}(x,y)$, as $t\to 0$. In general, if $g$ is $C^2$, then
\begin{align}
\frac{g(x+h)-2g(x)+g(x-h)}{h^2}&=\frac{\int_0^h g'(x+t)\,dt-\int_0^h g'(x-h+t)\,dt}{h^2}\\
&=\frac{\int_0^h\int_0^h g''(x-h+t+s)\,ds\,dt}{h^2}\to g''(x),
\end{align}
when $h\to 0$, as $g''$ is continuous. </p>
<p>Assume now that $f_{xx}(x,y)=f_{yy}(x,y)$. We shall use the following fact:</p>
<p><strong>Fact.</strong> <em>If $g_x=g_y$, then $g$ is of the form $g(x,y)=h(x+y)$.</em></p>
<p>Let now $F(x,y)=f_x(x,y)+f_y(x,y)$. Clearly,
$$
F_x-F_y=(f_x+f_y)_x-(f_x+f_y)_y=f_{xx}+f_{yx}-f_{xy}-f_{yy}=f_{xx}-f_{yy}=0,
$$
as $f_{xy}=f_{yx}$ since $f$ is $C^2$. Thus, using the above fact, $F(x,y)=H(x+y)$, where $H\in C^1$. Thus
$$
f_x(x,y)+f_y(x,y)=H(x+y).
$$
Now, let see how $f$ behaves along lines of the form $x-y=s$, parametrised as $x=t+s, y=t$. Then
$$
\frac{d}{dt}f\big(x(t),y(t)\big)=\frac{d}{dt}f(t+s,t)=f_x(t+s,t)+f_y(t+s,t)=H(2t+s),
$$
and thus
$$
f(t+s,t)-f(s,0)=\int_0^t H(2\tau+s)\,d\tau,
$$
or, setting $x=s+t$ and $y=t$
$$
f(x,y)=f(x-y,0)+\int_0^{y} H(2\tau+x-y)\,d\tau.
$$
If $g'=H$, then
\begin{align}
f(x,y)&=f(x-y,0)+\int_0^{y} H(2\tau+x-y)\,d\tau =f(x-y,0)+\left.\frac{1}{2}g(2\tau+x-y)\right|_0^y\\ &= f(x-y,0)+\frac{1}{2}g(2y+x-y)-\frac{1}{2}g(x-y)=f_1(x-y)+f_2(x+y),
\end{align}
where
$$
f_1(x)=f(x,0)-\frac{1}{2}g(x) \quad\text{and}\quad f_2(x)=\frac{1}{2}g(x).
$$</p>
<p><em>Proof of the Fact</em>. Fix $s\in\mathbb R$ and Let $\big(x(t),y(t)\big)=(t,s-t)$, $t\in\mathbb R$. Then
$$
\frac{d}{dt}g\big(x(t),y(t)\big)=\frac{d}{dt}g(t,s-t)=g_x-g_y=0.
$$
Thus $g(t,s-t)=g(0,s)$. Now $t=x$ and $s=x+y$, and therefore
$$
g(x,y)=g(t,s-t)=g(0,s)=g(0,x+y)=H(x+y).
$$</p>
|
2,998,189 | <p>I'm looking at a matrix operator in which <span class="math-container">$T \in \mathcal{L}(\mathbb{R}^2)$</span> by <span class="math-container">$T(x,y) = (x, -y)$</span>. So its basis is <span class="math-container">$ \mathcal{M}(T) = \begin{pmatrix}1 & 0\\0 & -1\end{pmatrix}$</span>. </p>
<p>How do I show that <span class="math-container">$T$</span> is self-adjoint? I understand the definition is <span class="math-container">$T=T^*$</span>. I seem to be able to understand it better when the operator is not on matrix... How would I show this? </p>
| lhf | 589 | <p><em>Hint:</em> An operator is self-adjoint iff its matrix is self-adjoint. The adjoint of a matrix is its conjugate transpose.</p>
|
3,371,638 | <p>Measure space <span class="math-container">$(X, \mathcal{A}, ν)$</span> has <span class="math-container">$ν(X) = 1$</span>. Let <span class="math-container">$A_n \in \mathcal{A} $</span> and denote </p>
<p><span class="math-container">$B := \{x : x ∈ A_n$</span> for infinitly many n }.</p>
<p>I want to prove that if <span class="math-container">$ν(A_n) \geq \epsilon > 0$</span> for all n, then <span class="math-container">$ν(B) ≥ \epsilon$</span>.</p>
<p><span class="math-container">$\textbf{My attempt}:$</span></p>
<p><span class="math-container">$$B = \text{limsup} A_n = \bigcap_{n=1}^{\infty}\bigcup_{k=n}^{\infty}A_k$$</span>
taking complement and taking measure from both sides:
<span class="math-container">$$\nu(X)-\nu(B) = \nu\bigg[\bigcup_{n=1}^{\infty} \bigg(\bigcup_{k=n}^{\infty}A_k\bigg)^c\bigg] = \bigcup_{n=1}^{\infty} \nu(B_n)\leq \sum_{n=1}^\infty\nu(B_n)$$</span>
<span class="math-container">$$\nu(B) \geq 1 - \sum_{n=1}^\infty\nu(B_n) $$</span> </p>
<p><span class="math-container">$B_n$</span> is an increasing sequence (i.e.<span class="math-container">$B_n \subset B_{n+1}$</span>) ,right? </p>
<p><span class="math-container">$\sum_{n=1}^\infty\nu(B_n) = \lim_{n \to \infty}\nu(B_n)$</span> so there exist N such that <span class="math-container">$\nu(B_n) \leq \frac{1-\epsilon}{2^n}$</span> thus;</p>
<p><span class="math-container">$$\nu(B) \geq 1 - \sum_{n=1}^\infty\nu(B_n) \leq 1-(1-\epsilon)=\epsilon $$</span> </p>
| Lee Mosher | 26,501 | <p>Let me use the standard terminology <em>disc</em> to refer to a circle union its interior. Also, let me use the standard terminology <em>polygon</em> to refer to a polyhedron in the plane.</p>
<p>Each circle is the boundary of its corresponding disc, and that boundary has the following special relation with the center point of the disc:</p>
<ol>
<li>Given a disc, its center is equidistant from all points on its boundary.</li>
</ol>
<p>On the other hand,</p>
<ol start="2">
<li>Given a polygon, there does not exist any point in the plane which is equidistant from all points on its boundary.</li>
</ol>
<p>Once you are convinced of the truth of 2, then it is clear that a circle is not a polygon, because they do not satisfy the same geometric properties.</p>
<p>To prove 2, one may apply the fact that the boundary of a polygon is a union of finitely many line segments, such that any two of those segments are either disjoint or intersect only at a common endpoint. </p>
<p>Applying that fact, all that remains in order to prove 2 is to prove that for any given line segment, there does not exist any point in the plane which is equidistant from all points on that line segment. I will leave this final proof as an exercise that is easily solved with coordinate geometry (and easily solved with axiomatic Euclidean geometry as well). </p>
|
41,174 | <p>I am trying to find the precise statement of the correspondence between stable Higgs bundles on a Riemann surface $\Sigma$, (irreducible) solutions to Hitchin's self-duality equations on $\Sigma$, and (irreducible) representations of the fundamental group of $\Sigma$. I am finding it a bit difficult to find a reference containing the precise statement. Mainly I'd like to know the statement for the case of stable $GL(n,\mathbb{C})$ Higgs bundles. But if anyone knows the statement for more general Higgs bundles that would be nice too.</p>
<p>Just at the level of say sets and not moduli spaces, I think the statement is that the following 3 things are the same, if I am reading Hitchin's original paper correctly:</p>
<ul>
<li><p>stable $GL(n,\mathbb{C})$ Higgs bundles modulo equivalence,</p></li>
<li><p>irreducible $U(n)$ (or is it $SU(n)$?) solutions of the Hitchin equations modulo equivalence, </p></li>
<li><p>irreducible $SL(n,\mathbb{C})$ (or is it $GL(n,\mathbb{C})$? $PSL$? $PGL$?) representations of $\pi_1$ modulo equivalence. </p></li>
</ul>
<p>Is this correct? Is there a reference?</p>
<p>Hitchin's original paper (titled "Self duality equations on a Riemann surface") does some confusing maneuvers; for example he considers solutions of the self-duality equations for $SO(3)$ rather than for $U(2)$ or $SU(2)$, which would seem more natural to me. Moreover, for instance, he doesn't look at all stable Higgs bundles, but only a certain subset of them - but I think this is just for the purpose of getting a <em>smooth</em> moduli space. And finally, Hitchin looks at $PSL(2,\mathbb{C})$ representations of $\pi_1$ rather than $SL(2,\mathbb{C})$ representations or $GL(2,\mathbb{C})$ representations, which confuses me as well...</p>
<p>Thanks in advance for any help!!</p>
<p><strong>EDIT:</strong> Please note that I am only interested in the case of a Riemann surface. Here it appears that <em>degree zero</em> stable Higgs bundles correspond to $GL(n,\mathbb{C})$ representations. But the question remains: are stable Higgs bundles of <em>arbitrary degree</em> related to representations? If so, which representations, and how are they related? Moreover, I think that general stable Higgs bundles should correspond to solutions of the self-duality equations -- but what's the correct group to take? ("Gauge group"? Is that the correct terminology?) I think it's $U(n)$ but I am not sure.</p>
<p>For example, in Hitchin's paper, he considers the case of rank 2 stable Higgs bundles of <em>odd degree</em> and fixed determinant line bundle, with trace-zero Higgs field (see Theorem 5.7 and Theorem 5.8). As for the self-duality equations, he uses the group $SU(2)/\pm 1$. We get a smooth moduli space. In the discussion following Theorem 9.19, it is shown that this moduli space is a <em>covering</em> of the space of $PSL(2,\mathbb{C})$ representations. It seems that this should generalize...</p>
| Dan Fox | 9,471 | <p>See Ó. Garcia-Prada's appendix to the third edition of R.O. Wells' book Differential Analysis on Complex Manifolds. It addresses most of what you are asking quite explicitly, in the context of Riemann surfaces, and has references to the original papers. Also the book of Lübke and Teleman on the Kobayashi-Hitchin correspondence might be helpful.</p>
|
41,174 | <p>I am trying to find the precise statement of the correspondence between stable Higgs bundles on a Riemann surface $\Sigma$, (irreducible) solutions to Hitchin's self-duality equations on $\Sigma$, and (irreducible) representations of the fundamental group of $\Sigma$. I am finding it a bit difficult to find a reference containing the precise statement. Mainly I'd like to know the statement for the case of stable $GL(n,\mathbb{C})$ Higgs bundles. But if anyone knows the statement for more general Higgs bundles that would be nice too.</p>
<p>Just at the level of say sets and not moduli spaces, I think the statement is that the following 3 things are the same, if I am reading Hitchin's original paper correctly:</p>
<ul>
<li><p>stable $GL(n,\mathbb{C})$ Higgs bundles modulo equivalence,</p></li>
<li><p>irreducible $U(n)$ (or is it $SU(n)$?) solutions of the Hitchin equations modulo equivalence, </p></li>
<li><p>irreducible $SL(n,\mathbb{C})$ (or is it $GL(n,\mathbb{C})$? $PSL$? $PGL$?) representations of $\pi_1$ modulo equivalence. </p></li>
</ul>
<p>Is this correct? Is there a reference?</p>
<p>Hitchin's original paper (titled "Self duality equations on a Riemann surface") does some confusing maneuvers; for example he considers solutions of the self-duality equations for $SO(3)$ rather than for $U(2)$ or $SU(2)$, which would seem more natural to me. Moreover, for instance, he doesn't look at all stable Higgs bundles, but only a certain subset of them - but I think this is just for the purpose of getting a <em>smooth</em> moduli space. And finally, Hitchin looks at $PSL(2,\mathbb{C})$ representations of $\pi_1$ rather than $SL(2,\mathbb{C})$ representations or $GL(2,\mathbb{C})$ representations, which confuses me as well...</p>
<p>Thanks in advance for any help!!</p>
<p><strong>EDIT:</strong> Please note that I am only interested in the case of a Riemann surface. Here it appears that <em>degree zero</em> stable Higgs bundles correspond to $GL(n,\mathbb{C})$ representations. But the question remains: are stable Higgs bundles of <em>arbitrary degree</em> related to representations? If so, which representations, and how are they related? Moreover, I think that general stable Higgs bundles should correspond to solutions of the self-duality equations -- but what's the correct group to take? ("Gauge group"? Is that the correct terminology?) I think it's $U(n)$ but I am not sure.</p>
<p>For example, in Hitchin's paper, he considers the case of rank 2 stable Higgs bundles of <em>odd degree</em> and fixed determinant line bundle, with trace-zero Higgs field (see Theorem 5.7 and Theorem 5.8). As for the self-duality equations, he uses the group $SU(2)/\pm 1$. We get a smooth moduli space. In the discussion following Theorem 9.19, it is shown that this moduli space is a <em>covering</em> of the space of $PSL(2,\mathbb{C})$ representations. It seems that this should generalize...</p>
| Chandan Singh Dalawat | 2,821 | <p>I'm surprised nobody has mentioned Le Potier's Bourbaki exposé :</p>
<p><a href="http://www.numdam.org/item?id=SB_1990-1991__33__221_0">Fibrés de Higgs et systèmes locaux</a>, Séminaire Bourbaki, 33 (1990-1991), Exposé No. 737, 48 p. </p>
|
2,088,229 | <p>Is there a neat way to find the largest integer that divides another integer fully, within a range. As an example, I would like to find the largest integer smaller than 1131 that divides 3500 completely. </p>
<p>So far I have just tried by breaking up 3500 into its prime components and guessing, coming to 875, but is there a more structured way?</p>
<p>EDIT: I guess the problem is somewhat equivalent to get all dividing integers, and then just pick the largest within my range?</p>
| Henry | 6,460 | <p>You could look for the smallest integer greater than $\dfrac{3500}{1131}\approx 3.09$ that divides $3500$ exactly </p>
<p>This is obviously $4$ (there is no smaller integer greater then $3.09$ and $4$ does divide $3500$ as it divides $100$) so the answer to your original question is $\dfrac{3500}{4}=875$</p>
|
148,127 | <p>I am trying to get an analytic expression for this integral:</p>
<pre><code> Integrate[Sign[Cos[q]]/(q + 1), {q, 0, x}, Assumptions -> x > 0]
</code></pre>
<p>Mathematica gives the answer:</p>
<pre><code> Abs[Cos[x]] Log[1 + x] Sec[x]
</code></pre>
<p>However, when I compare it to the numerical integration, I see that the answer is (incorrectly) discontinuous:</p>
<pre><code> Plot[{Abs[Cos[x]] Log[1 + x] Sec[x],
NIntegrate[Sign[Cos[q]]/(q + 1), {q, 0, x}]}, {x, 0, 4 Pi}]
</code></pre>
<p><a href="https://i.stack.imgur.com/ruX8o.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ruX8o.png" alt="enter image description here"></a></p>
<p>Is there a way to solve this issue and get the correct answer? Thank you</p>
| Alan | 19,530 | <p>Use <code>Labeled</code>.</p>
<pre><code>f = Function[{x, y, z}, x z + y z - x y z]
tmp = StringTemplate["C=``"]
plots = Labeled[
ContourPlot3D[f[x, y, z] == #, {x, 0, 2}, {y, 0, 2}, {z, 0, 2}],
tmp[#]] & /@ {0.0, 0.2, 0.4};
Grid[{plots}]
</code></pre>
<p><a href="https://i.stack.imgur.com/3y8ue.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3y8ue.png" alt="three plot"></a></p>
|
3,975,832 | <p>I think the following claim is clearly correct, but I cannot prove it.</p>
<blockquote>
<p>Let <span class="math-container">$A$</span> and <span class="math-container">$B$</span> be sets. If <span class="math-container">$f:A \times B \to \mathbb{R}$</span> satisfies <span class="math-container">$f(a, b) \leq C_a$</span> and <span class="math-container">$f(a, b) \leq C_b$</span> for all <span class="math-container">$(a, b) \in A \times B$</span> , then <span class="math-container">$f$</span> is bounded above. Remark: <span class="math-container">$C_x$</span> is a constant that depends only on <span class="math-container">$x$</span>.</p>
</blockquote>
<p>Background of this question:
I asked the following question, and this can be solved immediately if this original question is correct. (But it was false.)
<a href="https://math.stackexchange.com/questions/3976112/boundedness-of-a-bilinear-operator?noredirect=1#3976154">Boundedness of a bilinear operator</a></p>
<p>This original question is interesting in the following sense. If <span class="math-container">$f:A \times B \to \mathbb{R}$</span> satisfies <span class="math-container">$f(a, b) \leq C_a$</span>, <span class="math-container">$f$</span> has a upper bound that is independent of <span class="math-container">$b \in B$</span>. Similarly, if <span class="math-container">$f(a, b) \leq C_b$</span>, <span class="math-container">$f$</span> has a upper bound that is independent of <span class="math-container">$a \in A$</span>. Therefore, it seems to be that <span class="math-container">$f$</span> has a upper bound that is independent of <span class="math-container">$(a, b) \in A \times B$</span>, namely, <span class="math-container">$f$</span> is bounded above. However, it was false actually.</p>
| Kavi Rama Murthy | 142,385 | <p>Even an upper bound need not exist. Take <span class="math-container">$f:\mathbb R \times \mathbb R \to \mathbb R$</span> defined by <span class="math-container">$f(x,y)=x$</span> if <span class="math-container">$x=y$</span> and <span class="math-container">$0$</span> if <span class="math-container">$x \neq y$</span>.</p>
|
209,842 | <p>I was looking for a general way of formulating solutions for work and time problems.</p>
<p>For example,</p>
<p>30 soldiers can dig 10 trenches of size 8*3*3 ft in half a day working 8 hours per day. How many hours will 20 soldiers take to dig 18 trenches of size 6*2*2 ft working 10 hours per day?</p>
<p>Now i know that Work = Efficiency * Time, but i get confused sometimes in selecting which factor in the given problem will contribute directly to the work i.e. increase it and which factors will result in the work being done faster.</p>
<p>I've seen the method in which one uses a table to write all the parameters given in the problem e.g. making columns titled work, number of soldiers, volume of trench, number of days required, number of hours per day, efficiency, wages etc and uses the direct and inverse proportionality to write the equation for solving a given unknown. However i face the same problem i.e. finding out which factors are directly related and which are in an inverse relation.</p>
<p>Is there a simple way of solving these general problems?</p>
| Brian M. Scott | 12,042 | <p>Identify the basic assumptions. In this problem it’s clear that soldiers are considered interchangeable: they all work at the same rate. (Recall that in some problems we have different workers working at different rates and have to keep track of the work rate of each worker. And it’s the work rates that are important, because they’re additive: if $A$ and $B$ have work rates of $r$ and $s$ amount of work per unit of time, then $A$ and $B$ working together have a combined work rate of $r+s$.) It’s also clear the amount of work is being measured in cubic feet dug, and that cubic feet are to be considered interchangeable: they’re all equally hard to dig. Finally, the basic unit of time here is the hour, since we’re dealing with working days of two different lengths in hours.</p>
<p>Now convert the initial data to basic units. We have $30$ soldiers doing the work. They dig $10$ trenches of $8\cdot3\cdot3$ cubic feet each, for a total of $720$ cubic feet of earth dug, and they do it in half of an $8$-hour day, or $4$ hours.</p>
<p>At this point we can calculate their combined work rate: $720$ cubic feet in $4$ hours is $\frac{720}4=180$ cubic feet per hour. But that’s the combined work rate of $30$ interchangeable soldiers, so each soldier is digging only $\frac1{30}$-th of that, or $\frac{180}{30}=6$ cubic feet per hour. </p>
<p>Now let’s take a look at the question (which we should have had in the backs of our minds all along as a guide to what’s relevant and what’s likely to be useful). We have $20$ soldiers available. We want to dig $18$ trenches, each $6\cdot2\cdot2$ cubic feet in size, so we want to move $18\cdot6\cdot2\cdot2=432$ cubic feet of earth. Finally, we want to know how long this will take. We know that one soldier digs $6$ cubic feet per hour, so $20$ soldiers dig $20\cdot6=120$ cubic feet per hour. It will therefore take them $\frac{432}{120}=3.6$ hours, considerably less than one $10$-hour working day.</p>
|
840,700 | <p>I have to show that the following function $f:[0,1]\rightarrow\mathbb{R}$ is Riemann Integrable:</p>
<p>$$f(x) =
\left\{
\begin{array}{ll}
1 & \mbox{if } x = \frac{1}{n} \\
0 & \mbox{otherwise}
\end{array}
\right.$$</p>
<p>For the upper and lower Riemann sum I am using the following definitions:</p>
<p>$$S_{l}(f,V)=\sum^{n}_{j=1}\inf_{I(j)}(f)(x_j-x_{j-1})$$</p>
<p>With $I(j)$ denoting the interval $[x_{j-1},x_j$] and $V$ is a partition $V=\{0,x_1,...,1\}$. The upper sum is defined with the supremum. I have shown that for any partition on $[0,1]$ the lower sum is $0$. But now I need to prove that for every $\epsilon>0$ there is a partition $V$ such that $S_{u}(f,V)<\epsilon$. Completing the proof is easy. I see that any partition on $[0,1]$ will only contain a limited number of points of the set $\{\frac{1}{n}:n\in\mathbb{N}\}$. But I can't make the proof concrete. Could anybody help me out?</p>
| Badoe | 97,642 | <p>$ou have the right idea. It's all about the technicalities. One possible partition set-up is as follows.</p>
<p>For every $\epsilon>0$, there exists an integer $N_0$ such that $1/N_0<\epsilon$. Let $N=\max\{N_0,5\}$.</p>
<p>Partition $[0,1]$ with $V=\{0,x_1, x_2,\ldots,x_{4N-5}\}$ where $x_1=\frac{1}{2N}$, $x_{4N-5}=1$, $x_{2k+1}-x_{2k}=\frac{1}{N^3}$, and $x_{2k}<\frac{1}{2N-k}<x_{2k+1}$.</p>
<p>Then, $$S_u(f,V)= 1\cdot \frac{1}{2N} + \frac{1}{N^3} \cdot (2N-1)<\frac{1}{2N}+\frac{2N}{N^3}=\frac{1}{2N}+\frac{2}{N^2}$$.</p>
<p>Since $N\ge 5$, $\frac{2}{N^2}<\frac{1}{2N}$. Therefore, $S_u(f,V)<\frac{1}{N}<\epsilon$.</p>
|
4,124,777 | <p>I'm trying to find out whether <span class="math-container">$\sum _{n=0}^{\infty }\left(\cos^n\left(\frac{1}{\sqrt{n}}\right)-\frac{1}{\sqrt{e}}\right)$</span> converges or not. I've tried with taylor series but it doesn't lead me anywhere except with the fact that <span class="math-container">$\lim_{n \to \infty}\cos^n\left(\frac{1}{\sqrt{n}}\right)-\frac{1}{\sqrt{e}}=0$</span> and therefore it has "a chance" to converge.</p>
<p>Any hint?</p>
| Riemann'sPointyNose | 794,524 | <p><strong>EDIT</strong>: I decided to edit my answer to make it more explicit.</p>
<p>To straight up answer the question: is there a difference between <span class="math-container">${\mathbb{R}}$</span> and <span class="math-container">${\mathbb{R}^1}$</span>? Yes, technically there is a difference.</p>
<p>When we speak about <span class="math-container">${\mathbb{R}^n}$</span>, we are talking about a vector space of dimension <span class="math-container">$n$</span> made of <span class="math-container">${n}$</span>-tuples over the complete ordered field <span class="math-container">${\mathbb{R}}$</span>. If you plug in <span class="math-container">${n=1}$</span>, you end up with a <span class="math-container">$1$</span>-dimensional vector space made of <span class="math-container">$1$</span>-tuples over <span class="math-container">${\mathbb{R}}$</span>.</p>
<p>You mentioned in the comments "<em>isn't <span class="math-container">${\mathbb{R}}$</span> a vector space over <span class="math-container">${\mathbb{R}}$</span>?</em>" - you have to be careful with what you mean by this. The notation <span class="math-container">${\mathbb{R}}$</span> is just referencing a complete ordered field, the notation <span class="math-container">${\mathbb{R}^1}$</span> is when we are building a vector space out of <span class="math-container">${\mathbb{R}}$</span> over itself, and we call it <span class="math-container">${\mathbb{R}^1}$</span>. There are structural differences between <span class="math-container">${\mathbb{R}^1}$</span> and <span class="math-container">${\mathbb{R}}$</span>. We have no order defined on <span class="math-container">${\mathbb{R}^1}$</span> by default, and also no multiplication defined. Well, we have the dot product, but this doesn't take us from <span class="math-container">${\mathbb{R}^1\times \mathbb{R}^1\to \mathbb{R}^1}$</span>, it takes us from <span class="math-container">${\mathbb{R}^1\times \mathbb{R}^1\to \mathbb{R}}$</span>:
<span class="math-container">$$
(x)\cdot (y) = xy \in \mathbb{R}
$$</span>
so we need to first fix this:
<span class="math-container">$$
(x)\times (y) := (xy) \in \mathbb{R}^1
$$</span>
and now we need to define an order, since by default <span class="math-container">${\mathbb{R}^n}$</span> has no usual "order" attached to it. Define
<span class="math-container">$$
(x) < (y) \Leftrightarrow x < y
$$</span>
and now if we define the map
<span class="math-container">$$
\phi : \mathbb{R} \to \mathbb{R}^1
$$</span>
by
<span class="math-container">$$
\phi(x) = (x) \in \mathbb{R}^1
$$</span>
it's an easy exercise now to verify that <span class="math-container">${\phi}$</span> is an isomorphism with respect to addition, our fixed multiplication, and the defined order. So now for our "fixed" <span class="math-container">${\mathbb{R}^1}$</span>, <span class="math-container">${\mathbb{R}^1\cong \mathbb{R}}$</span>.</p>
<p>In many texts, however, you will see <span class="math-container">${\mathbb{R}^1}$</span> and <span class="math-container">${\mathbb{R}}$</span> mostly considered as the same thing. Implicitly in the background, the fixing of multiplication and creation of an order is done, but we don't ever explicitly do it since it's rather trivial to do. Ultimately, it's just a matter of semantics when it comes to looking at the difference between <span class="math-container">${\mathbb{R}^1}$</span> and <span class="math-container">${\mathbb{R}}$</span>.</p>
<p>Hope that helps clear things up.</p>
|
1,177,721 | <p>A fair $6$-sided die is rolled $6$ times independently. For any outcome, this is the set of numbers that showed up at least once in the different rolls. For example, the outcome is $(2,3,3,3,5,5)$, the element set is $\{2,3,5\}$. What is the probability the element set has exactly $2$ elements? how about $3$ elements? </p>
<p>I know the sample space is $6^6$. The counting for $2$ elements is $6C2 \cdot 6!$? I would really appreciate the help! :)</p>
| Jeny | 221,588 | <p>Why can't we use the multinomial formula? We want to have a sequence of 2 repeated elements 3 each. This sounds like $\frac{6!}{3!*3!}$, right?
For example we have {1,1,1,2,2,2} or a combination of this. Then we get $\frac{6!}{3!*3!}$ since we have 2 sets of 3 repeated elements and then we divide by the sample space times 6 (there are six possibilities). The result seems to be incorrect using this method. I tried using a 3^3 tree and it seemed ok. Any thoughts?</p>
|
3,091,353 | <p>There are 2 definitions of <strong><em>Connected Space</em></strong> in my lecture notes, I understand the first one but not the second. The first one is:</p>
<blockquote>
<p>A topological space <span class="math-container">$(X,\mathcal{T})$</span> is connected if there does not exist
<span class="math-container">$U,V\in\mathcal{T}$</span> such that <span class="math-container">$U\neq\emptyset$</span>, <span class="math-container">$V\neq\emptyset$</span>, <span class="math-container">$U\cap V=\emptyset$</span> and <span class="math-container">$X=U\cup V$</span></p>
</blockquote>
<p>which makes sense. It is saying that connected spaces can't be cut up into parts that have nothing to do with eachother.</p>
<p>The second definition is: </p>
<blockquote>
<p>A topological space <span class="math-container">$(X,\mathcal{T})$</span> is connected if <span class="math-container">$\emptyset$</span> and <span class="math-container">$X$</span> are the only subsets of <span class="math-container">$X$</span> which are closed and open</p>
</blockquote>
<p>which makes no intuitive sense to me, especially as a definition of connectedness. </p>
<p>Any intuitive explanation behind this second definition?</p>
| Chris Culter | 87,023 | <p>Hint: Consider the following definition, halfway between the first and second:</p>
<p><span class="math-container">$$\mbox{A topological space } (X,\mathcal{T}) \mbox{ is connected if there does not exist } U\in\mathcal{T} \mbox{ such that } U\neq\emptyset, (X\setminus U)\neq\emptyset, (X\setminus U)\in\mathcal{T} $$</span></p>
<p>Can you see how this is equivalent to the first definition?</p>
<p>Can you see how this is equivalent to the second definition?</p>
|
1,992,143 | <p>I'm trying to determine if $\sum \limits_{n=1}^{\infty} \sin(n\pi + \frac{1}{2n})$ absolutely converges or not.</p>
<p>Help me check it. I don't know how to do it. Advance thanks. :)</p>
| hamam_Abdallah | 369,188 | <p>For each $n>0$</p>
<p>$\sin(n\pi+\frac{1}{2n})=(-1)^n\sin(\frac{1}{2n})$</p>
<p>and</p>
<p>$$|\sin(n\pi+\frac{1}{2n})|=\sin(\frac{1}{2n}).$$</p>
<p>but</p>
<p>$$\sin(\frac{1}{2n})=\frac{1}{2n}(1+(2n\sin(\frac{1}{2n})-1))$$</p>
<p>$$=\frac{1}{2n}(1+\epsilon(n))$$ with</p>
<p>$$\lim_{n\to+\infty}\epsilon(n)=0.$$</p>
<p>thus for large enough $n$,</p>
<p>$$\frac{1}{2}\frac{1}{2n}<\sin(\frac{1}{2n})$$</p>
<p>and by comparison with harmonic series, $$\sum \sin(\frac{1}{2n})$$ is divergent.</p>
|
1,992,143 | <p>I'm trying to determine if $\sum \limits_{n=1}^{\infty} \sin(n\pi + \frac{1}{2n})$ absolutely converges or not.</p>
<p>Help me check it. I don't know how to do it. Advance thanks. :)</p>
| Astyx | 377,528 | <p>Notice that $|\sin(n\pi +{1\over2n})| = \sin({1\over2n}) \sim {1\over2n}$ and therefore $\sum \sin(n\pi +{1\over2n})$ is absolutely divergent (because $\sum{1\over n}$ is)</p>
|
3,601,865 | <p>What does the correspondence theorem (or 4th isomorphism theorem for rings) for rings mean and how is it used? That is, why do we care about it?</p>
<hr>
<p>Edit:</p>
<p>My version of the correspondence theorem:</p>
<blockquote>
<p>Let <span class="math-container">$R$</span> be a ring and <span class="math-container">$I$</span> be an ideal in <span class="math-container">$R$</span>. Let <span class="math-container">$K$</span> be the set of ideals in <span class="math-container">$R$</span> containing <span class="math-container">$I$</span>. Let <span class="math-container">$L$</span> be the set of ideals in <span class="math-container">$R/I$</span>. Then there is bijection from <span class="math-container">$K$</span> to <span class="math-container">$L$</span> given by <span class="math-container">$\phi(I’)= \{ x+ I : x \in I’ \}$</span>.</p>
</blockquote>
| Alex Kruckman | 7,062 | <p>One of the most basic applications of this theorem is the result that a quotient ring <span class="math-container">$R/I$</span> is a field if and only if <span class="math-container">$I$</span> is a maximal ideal. (For me, all rings are commutative with unity.) </p>
<p>Every non-zero ring <span class="math-container">$R$</span> has at least two ideals, namely <span class="math-container">$\{0\}$</span> and <span class="math-container">$R$</span>. And <span class="math-container">$R$</span> is a field if and only if these are the only two ideals in <span class="math-container">$R$</span>.</p>
<p>If <span class="math-container">$I\subseteq R$</span> is a proper ideal, then there are at least two ideals in <span class="math-container">$R$</span> containing <span class="math-container">$I$</span>, namely <span class="math-container">$I$</span> and <span class="math-container">$R$</span>. And <span class="math-container">$I$</span> is maximal if and only if these are the only two ideals in <span class="math-container">$R$</span> containing <span class="math-container">$I$</span>. </p>
<p>Now the set of ideals in <span class="math-container">$R$</span> containing <span class="math-container">$I$</span> is in bijection with the set of ideals of <span class="math-container">$R/I$</span>. So <span class="math-container">$I$</span> is maximal if and only if there are exactly two ideals in <span class="math-container">$R$</span> containing <span class="math-container">$I$</span> if and only if there are exactly two ideals in <span class="math-container">$R/I$</span> if and only if <span class="math-container">$R/I$</span> is a field. </p>
|
3,717,506 | <p>I am reading some text about even functions and found this snippet:</p>
<blockquote>
<p>Let <span class="math-container">$f(x)$</span> be an integrable even function. Then,</p>
<p><span class="math-container">$$\int_{-a}^0f(x)dx = \int_0^af(x)dx, \forall a \in \mathbb{R}$$</span></p>
<p>and therefore,</p>
<p><span class="math-container">$$\int_{-a}^af(x)dx = 2\int_0^af(x), \forall a \in \mathbb{R}$$</span></p>
</blockquote>
<p>Why does <span class="math-container">$dx$</span> disappear from <span class="math-container">$2\int_0^af(x)$</span>? Is it just a notation convention?</p>
| Ross Millikan | 1,827 | <p>The limit that you hit any number <span class="math-container">$k$</span>, positive or negative, goes to <span class="math-container">$1$</span> as <span class="math-container">$n \to \infty$</span>. Say we want the chance of hitting <span class="math-container">$k=27$</span>. This is higher than the chance we get a <span class="math-container">$0$</span> and then <span class="math-container">$9\ +3$</span>'s in a row because there are other ways to get to <span class="math-container">$27$</span>, but the chance you get that string in <span class="math-container">$n$</span> throws is <span class="math-container">$1-6^{9-n}$</span>. This goes to <span class="math-container">$1$</span> as <span class="math-container">$n \to \infty$</span>. The reset to <span class="math-container">$0$</span> allows us to overcome the upward bias of <span class="math-container">$\frac 12$</span> per throw.</p>
|
4,043,787 | <p>I have <span class="math-container">$$f_n(x)=\begin{cases}
\frac{1}{n} & |x|\leq n, \\
0 & |x|>n .
\end{cases}$$</span></p>
<p>Why cannot be dominated by an integrable function <span class="math-container">$g$</span> by the Dominated Convergence Theorem? I am also wondering what exactly it means for a function <span class="math-container">$g$</span> to dominate a sequence of functions, since I believe my definition of this is what I am understanding incorrectly.</p>
| RobertTheTutor | 883,326 | <p>In order to dominate a sequence of functions, it has to be larger than ALL of the functions in the entire sequence. There is no such <span class="math-container">$g$</span> for your sequence of functions, because no matter how large you set <span class="math-container">$g$</span> near <span class="math-container">$0$</span>, it will eventually be exceeded by all the terms in the sequence beyond some point.</p>
|
4,498,498 | <p>Lets say you have two sequences of non negative integers each of length <span class="math-container">$n$</span>.</p>
<p>ie <span class="math-container">$(a_1,a_2,...,a_n)$</span> and <span class="math-container">$(b_1,b_2,...,b_n)$</span> such that
<span class="math-container">$\max(a_i) < k$</span>
and
<span class="math-container">$\max(b_i) < k$</span></p>
<p><strong>Game rule:</strong></p>
<p>You can edit both sequence with <span class="math-container">$\mathrm{swap}(a_i, b_i)$</span> for <span class="math-container">$1 ≤ i≤ n$</span>,</p>
<p><strong>Goal:</strong></p>
<p><span class="math-container">$a_i ≤ a_j$</span> for all <span class="math-container">$i ≤ j$</span>
and
<span class="math-container">$b_i ≤b_j$</span> for all <span class="math-container">$i ≤ j$</span></p>
<p>But not all initial sequence <span class="math-container">$a$</span> and <span class="math-container">$b$</span> can be solved. For example <span class="math-container">$(2,0)$</span> and <span class="math-container">$(0,1)$</span> is a pair of sequence that can't be solved.</p>
<p>Now given <span class="math-container">$n$</span> and <span class="math-container">$k$</span>, count number of different pair of initial sequence <span class="math-container">$(a,b)$</span> that can be solved with game described above.</p>
<p><strong>Example:</strong></p>
<p>for <span class="math-container">$n=1$</span>,<span class="math-container">$k=2$</span>:
These are the cases: <span class="math-container">${[(0),(0)],[(0),(1)],[(1),(0)],[(1),(1)]}$</span>.
Hence answer would be <span class="math-container">$4$</span>.</p>
| ploosu2 | 111,594 | <p><em>Disclaimer:</em> This is just a computational answer but it gives a formula for at least for the case <span class="math-container">$k=2$</span> (and other small cases too). And it's too long for a comment.</p>
<p>I get these values for <span class="math-container">$f(k, n) = $</span> the number of these solvable sequence pairs</p>
<p><span class="math-container">$$
\displaystyle \left(\begin{array}{rrrrrrrrr}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
4 & 11 & 26 & 57 & 120 & 247 & 502 & 1013 & 2036 \\
9 & 46 & 180 & 603 & 1827 & 5164 & 13878 & 35905 & 90189 \\
16 & 130 & 750 & 3507 & 14224 & 52068 & 176430 & 562925 & 1711776 \\
25 & 295 & 2345 & 14518 & 75558 & 346050 & 1436820 & 5520295 & 19916039 \\
36 & 581 & 6076 & 48006 & 311136 & 1739166 & 8665866 & 39387491 & 166049884 \\
49 & 1036 & 13776 & 135114 & 1065834 & 7132620 & 41957058 & 222437215 & 1082436355 \\
64 & 1716 & 28260 & 336666 & 3173808 & 25034724 & 171535650 & 1048436675 & 5829137600 \\
81 & 2685 & 53625 & 762399 & 8461167 & 77655435 & 612837225 & 4275850150 & 26924807910
\end{array}\right)
$$</span></p>
<p>Calculation done with a Markov chain (<a href="https://pastebin.com/SyY3UZij" rel="nofollow noreferrer">code here</a>).</p>
<p><strong>The idea:</strong> We start building the two sequences by adding a term to each at a time. We keep as a state <span class="math-container">$(m, M)$</span> where <span class="math-container">$m$</span> is the maximum of the minimums and <span class="math-container">$M$</span> is the maximum of the maximums in the sequences so far. An example: let the sequences be</p>
<p><span class="math-container">$$
(2,1,3,7) \\
(0,2,2,4)
$$</span></p>
<p>Then the path of the <span class="math-container">$(m, M)$</span>'s is <span class="math-container">$(0,0), (0,2), (1,2), (2,3), (4,7)$</span>. It always starts from <span class="math-container">$(0,0)$</span> (when the sequences are empty). Then we append <span class="math-container">$x_1=2, x_2=0$</span>, we get <span class="math-container">$m=\min(x_1, x_2) = 0$</span> and <span class="math-container">$M=\max(x_1, x_2) = 2$</span>. Then append <span class="math-container">$x_1=1, x_2=2$</span> (notice, these need to satisfy <span class="math-container">$\min(x_1, x_2) \geq m$</span> and <span class="math-container">$\max(x_1, x_2) \geq M$</span>) and get <span class="math-container">$(m,M) = (2,3)$</span>. And so on.</p>
<p>This leads to the following: there is a transition from <span class="math-container">$(m_1, M_1)$</span> to <span class="math-container">$(m_2, M_2)$</span> if <span class="math-container">$m_1\leq m_2$</span> and <span class="math-container">$M_1\leq M_2$</span>. And it is of weight <span class="math-container">$1$</span> if <span class="math-container">$m_2=M_2$</span>, because in that case only <span class="math-container">$x_1=x_2$</span> is possible, otherwise we can flip them and get weight <span class="math-container">$2$</span> transition (weight meaning the number of ways that lead to that transition). Also the "transition" -matrix <span class="math-container">$A$</span> has a nice block structure with respect to the blocks determined by value of <span class="math-container">$m$</span> in the state. And I wonder if that can be used for faster calculation of <span class="math-container">$f(k, n) = e_1^T A^n \bf 1$</span>.</p>
<p>For example for <span class="math-container">$k=3$</span> we get the following directed graph (with edge labels indicating how many pairs of numbers lead to that transition)</p>
<p><a href="https://i.stack.imgur.com/7KGEL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7KGEL.png" alt="enter image description here" /></a></p>
<p>To get the value <span class="math-container">$f(k, n)$</span> we count the number of <span class="math-container">$n$</span>-walks on the graph starting from <span class="math-container">$(0,0)$</span>. (Regard as there being multiple edges between the vertices when edge label <span class="math-container">$>1$</span>)</p>
<p>For <span class="math-container">$k=2$</span> the graph is particularly simple and we get that <span class="math-container">$f(2, n)$</span> is the sum of the first row of</p>
<p><span class="math-container">$$
\displaystyle \left(\begin{array}{rrr}
1 & 2 & 1 \\
0 & 2 & 1 \\
0 & 0 & 1
\end{array}\right)^n
$$</span></p>
<p>and that equals <span class="math-container">$2^{n+2}-n-3$</span>.</p>
<p>Finding the Jordan normal form for the involved matrix (<a href="https://pastebin.com/p649U2Sr" rel="nofollow noreferrer">code here</a>), I was able to find the formulas</p>
<p><span class="math-container">$$\begin{align}
f(3, n) &= \frac{1}{2}(n+2)(2^{n+2}(n-1)+n+5) \\
f(4, n) &= \frac{1}{6}(n+2)(n+3)(2^{n}(2n^2-2n+8) -n -7 ) \\
f(5, n) &= \frac{1}{72}(n+2)(n+3)(n+4)(2^n(2n^3+22n-24)+3n+27) \\
f(6,n) &= \frac{1}{720} (n + 2) \dots (n + 5) (2^n(n^{4} + 2n^{3} + 23 n^{2} - 26 n +72) - 6 n - 66)
\end{align}$$</span></p>
<p>These seem to indicate some sort of pattern.</p>
<p>For <span class="math-container">$k=7,8,\dots, 12$</span> we have that <span class="math-container">$\frac{f(k, n)}{n+k-1\choose k-2}$</span> is</p>
<p><span class="math-container">$$ \begin{aligned}
&\frac{1}{180} (2^n(n^{5} + 5 n^{4} + 45 n^{3} - 5 n^{2} + 314 n -360) + 30 n + 390) \\
&\frac{1}{1260} (2^n(n^{6} + 9 n^{5} + 85 n^{4} + 135 n^{3} + 994 n^{2} - 1224 n + 2880) - 180 n - 2700) \\
&\frac{1}{10080} (2^n(n^{7} + 14 n^{6} + 154 n^{5} + 560 n^{4} + 2989 n^{3} - 574 n^{2} + 17016 n - 20160 ) + 1260 n + 21420) \\
&\frac{1}{90720} (2^n(n^{8} + 20 n^{7} + 266 n^{6} + 1568 n^{5} + 8729 n^{4} + 11900 n^{3} + 71644 n^{2} - 94128 n + 201600) - 10080 n - 191520) \\
&\frac{1}{907200} (2^n(n^{9} + 27 n^{8} + 438 n^{7} + 3654 n^{6} + 23961 n^{5} + 71883 n^{4} + 294272 n^{3} - 75564 n^{2} + 1495728 n - 1814400) + 90720 n + 1905120) \\
&\frac{1}{9979200} (2^n(n^{10} + 35 n^{9} + 690 n^{8} + 7590 n^{7} + 60753 n^{6} + 281715 n^{5} + 1193660 n^{4} + 1453060 n^{3} + 7816896 n^{2} - 10814400 n + 21772800) - 907200 n - 20865600)
\end{aligned}
$$</span></p>
<p><strong>EDIT:</strong></p>
<p>Looking at the block structure of the transition matrix, here's for example <span class="math-container">$k=4$</span>:</p>
<p><span class="math-container">$$A_4 = \left(\begin{array}{rrrr|rrr|rr|r}
1 & 2 & 2 & 2 & 1 & 2 & 2 & 1 & 2 & 1 \\
0 & 2 & 2 & 2 & 1 & 2 & 2 & 1 & 2 & 1 \\
0 & 0 & 2 & 2 & 0 & 2 & 2 & 1 & 2 & 1 \\
0 & 0 & 0 & 2 & 0 & 0 & 2 & 0 & 2 & 1 \\
\hline
0 & 0 & 0 & 0 & 1 & 2 & 2 & 1 & 2 & 1 \\
0 & 0 & 0 & 0 & 0 & 2 & 2 & 1 & 2 & 1 \\
0 & 0 & 0 & 0 & 0 & 0 & 2 & 0 & 2 & 1 \\
\hline
0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 2 & 1 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & 1 \\
\hline
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1
\end{array}\right)
$$</span></p>
<p>I was able to come up with <a href="https://pastebin.com/ydLFEnAX" rel="nofollow noreferrer">this <span class="math-container">$O(nk^2)$</span> algorithm</a>.</p>
<p>We need to find <span class="math-container">$A^n \bf 1$</span> (its first component is the solution). Do this by initializing the vector <span class="math-container">$v_0 = \bf 1$</span> and iteratively computing <span class="math-container">$v_{j+1} = Av_j$</span>. Start the computation of <span class="math-container">$v_{j+1}$</span> from the last component upwards and notice that the row <span class="math-container">$i$</span> of the matrix is mostly equal (in the blocks to the left of the diagoal one) to the corresponding row one block-level below. (To see why this is true look at the states <span class="math-container">$(m, M)$</span> and <span class="math-container">$(m, M+1)$</span>). Now to do the computation, keep a running total for the current diagonal block and add the rest from the corresponding (already calculated) value from <span class="math-container">$v$</span>.</p>
|
374,380 | <p>I am having trouble understanding the factor group, $\mathbb{R}$/$\mathbb{Z}$, or maybe i'm not. Here's what I am thinking.</p>
<p>Okay, so i have a group $G=(\mathbb{R},+)$, and I have a subgroup $N=(\mathbb{Z},+)$. Then I form $G/N$. So this thing identifies any real number $x$ with the integers that are exactly 1 unit step away. So if $x=\frac{3}{4}$, then $[x]=({...,\frac{-5}{4},\frac{-1}{4},\frac{3}{4},\frac{7}{4},...})$ and i can do this for any real number. So therefore, my cosets are unit intervals $[0,1)+k$, for integers $k$. Herstein calls this thing a circle and I was not sure why, but here's my intuition. The unit interval is essentially closed and since every real number plus an integer identifies with itself, these "circles" keep piling up on top of each other as if its one closed interval. Since it's closed it is a circle. Does that make sense? </p>
<p>Now how do I extend this intuition to this?<br>
$G'=[(a,b)|a,b\in{\mathbb{R}}], N'=[(a,b)|a,b\in{\mathbb{Z}}].$ What is $G'/N'$? How is this a torus? I can't get an intuitive picture in my head...</p>
<p>EDIT: Actually, are the cosets just simply $[x]=[x\in{\mathbb{R}}|x+k,k\in{\mathbb{Z}}]?$</p>
| Mikasa | 8,581 | <p>You can also use the following nice facts. I hope you are inspired by them.</p>
<p>$$\mathbb R/\mathbb Z\cong T\cong\prod_p\mathbb Z(p^{\infty})\cong\mathbb R\oplus(\mathbb Q/\mathbb Z)\cong\mathbb C^{\times}$$</p>
|
374,380 | <p>I am having trouble understanding the factor group, $\mathbb{R}$/$\mathbb{Z}$, or maybe i'm not. Here's what I am thinking.</p>
<p>Okay, so i have a group $G=(\mathbb{R},+)$, and I have a subgroup $N=(\mathbb{Z},+)$. Then I form $G/N$. So this thing identifies any real number $x$ with the integers that are exactly 1 unit step away. So if $x=\frac{3}{4}$, then $[x]=({...,\frac{-5}{4},\frac{-1}{4},\frac{3}{4},\frac{7}{4},...})$ and i can do this for any real number. So therefore, my cosets are unit intervals $[0,1)+k$, for integers $k$. Herstein calls this thing a circle and I was not sure why, but here's my intuition. The unit interval is essentially closed and since every real number plus an integer identifies with itself, these "circles" keep piling up on top of each other as if its one closed interval. Since it's closed it is a circle. Does that make sense? </p>
<p>Now how do I extend this intuition to this?<br>
$G'=[(a,b)|a,b\in{\mathbb{R}}], N'=[(a,b)|a,b\in{\mathbb{Z}}].$ What is $G'/N'$? How is this a torus? I can't get an intuitive picture in my head...</p>
<p>EDIT: Actually, are the cosets just simply $[x]=[x\in{\mathbb{R}}|x+k,k\in{\mathbb{Z}}]?$</p>
| Xi Wu | 24,367 | <p>Just try to add some intuitive answers:</p>
<ol>
<li><p>The essential intuition that <span class="math-container">$\mathbb{R}/\mathbb{Z}$</span> is isomorphic to unit circle in the complex plane is that we can map a real number <span class="math-container">$x$</span> to the radian so we get <span class="math-container">$e^{2\pi i x}$</span>, and so if two number differ by integers they map to the same radian (we circle back).</p>
</li>
<li><p>The same intuition carries over to the isomorphism that <span class="math-container">$\mathbb{R}^2/\mathbb{Z^2}$</span> is isomorphic to the torus: We just need to tweak how to generate the coordinates: It turns out that any coordinate on a torus can be indexed by two radians. See <a href="https://youtu.be/0H5_h-RB0T8?t=33" rel="nofollow noreferrer">this video at 0:33</a>, there is one horizontal circle, and one vertical circle, so you can think of the first real number as a radian that maps you to some position in the horizontal circle, and the second number maps you to some position in the vertical circle. Again, differing by integers does not matter any more.</p>
</li>
</ol>
|
2,258,697 | <p>I recently encountered this question and have been stuck for a while. Any help would be appreciated!</p>
<p>Q: Given that
$$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{5} \tag{1} \label{eq:1}$$
$$abc = 5 \tag{2} \label{eq:2}$$
Find $a^3 + b^3 + c^3$. It wasn't specified in the question but I think it can be assumed that $a, b, c$ are real numbers.</p>
<p>My approach:
$$ ab + ac + bc = \frac{1}{5} abc = 1 $$
$$ a^3 + b^3 + c^3 = (a+b+c)^3 - 3[(a + b + c)(ab + ac + bc) - abc] $$
$$ a^3 + b^3 + c^3 = (a+b+c)^3 - 3(a+b+c) + 15 $$
From there, I'm not sure how to go about solving for $a + b + c$.
Something else I tried was letting $x = \frac{1}{a}, y = \frac{1}{b}, z = \frac{1}{c}$, so we get $$ xyz = x + y + z = \frac{1}{5} $$Similarly, I'm not sure how to continue from there. </p>
| Martin Argerami | 22,857 | <p>The matrix is a unitary conjugate (via permutations) of
$$\begin{bmatrix}1&1&0&0\\1&1&0&0\\0&0&1&1\\0&0&1&1\end{bmatrix} $$(concretely, use the permutation corresponding to the exchange of the second and third elements of the canonical basis). So the characteristic polynomial is the square of the characteristic polynomial of the $2\times2$ matrix with all ones, which has eigenvalues $0$ and $2$. We get, then, $$\lambda ^2 (2-\lambda)^2. $$</p>
|
2,231,003 | <p>I am trying to prove that the function $f_a(x) = e^{-\frac{1}{(x-a)}}$ is differentiable for all $x>a$. However, I do not know how to show $|\frac{e^{-\frac{1}{(x-a)}}-e^{-\frac{1}{(p-a)}}}{x-p} - (-\frac{e^{-\frac{1}{(p-a)}}}{(p-a)^2})| < \epsilon$ if $0<|x-p|<\delta$. </p>
<p>My next idea was to apply the natural logarithmic function to $e^{-\frac{1}{(x-a)}}$ to get $\hat{f_a}(x)=-\frac{1}{(x-a)}$. From here, I can prove that $\hat{f_a}(x)$ is differentiable for any $x>a$, but I do not know if this implies that $f_a(x)$ is differentiable for any $x>a$. </p>
<p>I know that $f_a(x):(-\infty,\infty) \rightarrow (0,\infty)$ and $\hat{f_a}(x):(-\infty,\infty) \rightarrow (-\infty,\infty)$. </p>
<p>Can you please help me understand how I can prove $f_a(x)$ is differentiable for all $x>a$? Thank you for your time and help. </p>
| marshal craft | 167,793 | <p>There are plenty of proofs for the chain rule, that is if $f(x)$ and $g(x)$ are differentiable, then so is $g(f(x))$. You can take it upon yourself to verify this, as it is typically standard for students to prove these statements they use every day in calculus. As such you can find proofs of such, I believe I have answered one before. Then $\frac{d}{dx} e^x= e^x$ and $\frac{d}{dx} \frac{-1}{a-x}=\frac{-1}{(a-x)^2}$ consider now this is defined for all $x \neq a $ because the denominator is zero.</p>
<p>To discuss how to prove this, you can seek to prove it directly or you can prove greater generality about all functions and composition of functions, then plug this application into that.</p>
|
241,903 | <p>Suppose $f: \mathbb{D}\to \mathbb{C}$ is a univalent function with $$f(z)=z+a_2z^2+a_3z^3+\cdots.$$ The Bieberbach conjecture/de Branges' theorem asserts that $|a_n|\leq n$ with equality for the Koebe function, which has an unbounded image. Suppose we restrict to the class of univalent functions whose image is actually bounded. Is there a better bound than $|a_n|\leq n$ ?</p>
| Alexandre Eremenko | 25,510 | <p>The answer is yes, even the rate of increase of coefficients is much smaller. However even the precise rate is not known (not speaking of the exact estimate).
The strongest results in this direction are here:</p>
<p>MR1162188
Carleson, Lennart; Jones, Peter W.
On coefficient problems for univalent functions and conformal dimension.
Duke Math. J. 66 (1992), no. 2, 169–206. </p>
|
53,188 | <p>Recently I read the chapter "Doctrines in Categorical Logic" by Kock, and Reyes in the Handbook of Mathematical Logic. And I was quite impressed with the entire chapter. However it is very short, and considering that this copy was published in 1977, possibly a bit out of date. </p>
<p>My curiosity has been sparked (especially given the possibilities described in the aforementioned chapter) and would like to know of some more modern texts, and articles written in this same style (that is to say coming from a logic point of view with a strong emphasis on analogies with normal mathematical logic.)</p>
<p>However, that being said, <em>I am stubborn as hell, and game for anything.</em></p>
<p>So, Recommendations?</p>
<p>Side Note: More interested in the preservation of structure, and the production of models than with any sort of proposed foundational paradigm </p>
| Tom Leinster | 586 | <p><em>Introduction to Higher Order Categorical Logic</em> by Lambek and Scott might fit the bill. </p>
|
53,188 | <p>Recently I read the chapter "Doctrines in Categorical Logic" by Kock, and Reyes in the Handbook of Mathematical Logic. And I was quite impressed with the entire chapter. However it is very short, and considering that this copy was published in 1977, possibly a bit out of date. </p>
<p>My curiosity has been sparked (especially given the possibilities described in the aforementioned chapter) and would like to know of some more modern texts, and articles written in this same style (that is to say coming from a logic point of view with a strong emphasis on analogies with normal mathematical logic.)</p>
<p>However, that being said, <em>I am stubborn as hell, and game for anything.</em></p>
<p>So, Recommendations?</p>
<p>Side Note: More interested in the preservation of structure, and the production of models than with any sort of proposed foundational paradigm </p>
| Paul Taylor | 2,733 | <p>Well, for the record, there is also my
<a href="http://paultaylor.eu/prafm/"><i>Practical Foundations of Mathematics</i></a> (CUP, 1999).</p>
|
3,063,742 | <p>Consider the closed interval <span class="math-container">$[0, 1]$</span> in the real line <span class="math-container">$\mathbb{R}$</span> and the product space <span class="math-container">$([0, 1]^{\mathbb{N}}, τ ),$</span></p>
<p>where <span class="math-container">$τ$</span> is a topology on <span class="math-container">$[0, 1]^\mathbb{N} $</span>. Let <span class="math-container">$ D : [0, 1] \rightarrow [0, 1]^\mathbb{N} $</span> be the map defined by
<span class="math-container">$D(x) := (x, x, · · · , x, · · ·)$</span> for <span class="math-container">$x \in [0, 1].$</span>
. The map <span class="math-container">$D$</span> is</p>
<p>choose the correct satements</p>
<p><span class="math-container">$(a)$</span> not continuous if <span class="math-container">$τ$</span> is the box topology and also not continuous if <span class="math-container">$τ$</span> is the product
topology.</p>
<p><span class="math-container">$(b)$</span> continuous if <span class="math-container">$τ$</span> is the product topology and also continuous if <span class="math-container">$τ$</span> is the box
topology.</p>
<p><span class="math-container">$(c)$</span> continuous if <span class="math-container">$τ$</span> is the box topology and not continuous if <span class="math-container">$τ$</span> is the product
topology.</p>
<p><span class="math-container">$(d)$</span> continuous if <span class="math-container">$τ$</span> is the product topology and not continuous if <span class="math-container">$τ$</span> is the box
topology</p>
<p>My attempt : i thinks option <span class="math-container">$c)$</span> is true because box topology is finer then product</p>
<p>Is its true ?</p>
<p>Any hints/solution</p>
| sranthrop | 95,054 | <p>Note that both <span class="math-container">$A$</span> and <span class="math-container">$B$</span> are closed in <span class="math-container">$\mathbb R^2$</span> (why?), and so <span class="math-container">$M:=\mathbb R^2\backslash B$</span> and <span class="math-container">$N:=\mathbb R^2\backslash A$</span> are open in <span class="math-container">$\mathbb R^2$</span>. Consequently, since <span class="math-container">$X=A\cup B$</span> and <span class="math-container">$A\cap B=\emptyset$</span>, we have <span class="math-container">$A=X\cap M$</span> and <span class="math-container">$B=X\cap N$</span>, showing that both <span class="math-container">$A$</span> and <span class="math-container">$B$</span> are also open in <span class="math-container">$X$</span>. </p>
<p>Note that this implies that <span class="math-container">$A$</span> and <span class="math-container">$B$</span> are also closed in <span class="math-container">$X$</span>, since <span class="math-container">$A=X\backslash B$</span> and <span class="math-container">$B=X\backslash A$</span>.</p>
|
1,365,489 | <p>What is the value of the following expression?</p>
<p>$$\sqrt[3]{\ 17\sqrt{5}+38} - \sqrt[3]{17\sqrt{5}-38}$$</p>
| Jan Eerland | 226,665 | <p>$$\sqrt[3]{\ 17\sqrt{5}+38} - \sqrt[3]{17\sqrt{5}-38}=$$
$$\sqrt[3]{8+12\sqrt{5}+30+5\sqrt{5}} - \sqrt[3]{-8+12\sqrt{5}-30+5\sqrt{5}}=$$
$$\sqrt[3]{8+12\sqrt{5}+6\left(\sqrt{5}\right)^2+\left(\sqrt{5}\right)^3} - \sqrt[3]{-8+12\sqrt{5}-6\left(\sqrt{5}\right)^2+\left(\sqrt{5}\right)^3}=$$
$$\sqrt[3]{\left(\sqrt{5}+2\right)^3} - \sqrt[3]{\left(\sqrt{5}-2\right)^3}=$$
$$\left({\left(\sqrt{5}+2\right)^3}\right)^{\frac{1}{3}} - \left({\left(\sqrt{5}-2\right)^3}\right)^{\frac{1}{3}}=$$
$$\left(\sqrt{5}+2\right) -\left(\sqrt{5}-2\right)=$$
$$\left(\sqrt{5}+2\right) +\left(2-\sqrt{5}\right)=$$
$$2+2+\sqrt{5}-\sqrt{5}=2+2=4$$</p>
|
1,866,931 | <p>I would like to see a proof to this fact.</p>
<blockquote>
<p>If $A$ is an invertible matrix and $B \in \mathcal{L}(\mathbb{R}^n,\mathbb{R}^n)$, that is an bounded linear opertor in $\mathbb{R}^n$. Then, if there holds
$$
\|B-A\| \|A^{-1}\| <1,
$$
we have that B is invertible.</p>
</blockquote>
<p>Moreover, if possible, how to use it to prove that the aplication $A \rightarrow A^{-1}$ is continuous?</p>
| be5tan | 339,513 | <p>I think you have to start with with given elements $x, y, z \in A: xRy, yRz$. Now you have to show that $xRz$. However you started with an elements $x, z\in A: (x, z) \in R\circ R$, which is the wrong antecedent for the implication you want to prove.</p>
<p>BS</p>
|
755,571 | <p>$$a_n=3a_{n-1}+1; a_0=1$$</p>
<p>The book has the answer as: $$\frac{3^{n+1}-1}{2}$$</p>
<p>However, I have the answer as: $$\frac{3^{n}-1}{2}$$</p>
<p>Based on:</p>
<p><img src="https://i.stack.imgur.com/4vJrQ.png" alt="enter image description here"></p>
<p>Which one is correct?</p>
<p>Using backwards substitution iteration, the end of this will be
$$3^{n-1}a_0+3^{n-2}+3^{n-3}+...+3+1$$</p>
<p>which is $$=3^{n-1}+3^{n-2}+3^{n-3}+...+3+1=\sum_{i=0}^{n-1}3^i$$</p>
<p>Which according to the theorem should be $$\frac{3^{(n-1)+1}-1}{(3-1)}=\frac{3^{n}-1}{2}$$</p>
| David | 119,775 | <p>First note that "backwards substitution" is not a great way to solve this sort of thing, there are much better methods which I assume you will learn at some stage in your course.</p>
<p>However if you want to do it this way you can, only your working is not accurate. Try this:
$$\eqalign{a_n
&=3a_{n-1}+1\cr
&=3(3a_{n-2}+1)+1\cr
&=3^2a_{n-2}+3+1\cr
&=\cdots\cr
&=3^ka_{n-k}+3^{k-1}+\cdots+3+1\cr
&=\cdots\cr
&=3^na_0+3^{n-1}+3^{n-2}+\cdots+3+1\cr
&=3^n+3^{n-1}+3^{n-2}+\cdots+3+1\cr
&=\frac{3^{n+1}-1}{2}\ .\cr}$$
The third last line is where you went wrong, you stopped one step too soon.</p>
|
1,268,598 | <p>I thought the <code>|</code> symbol meant "divides by", but in set theory it seems that it means "such that." However, I thought we wrote "such that" as <code>:</code>.</p>
<p>Can anybody elaborate?</p>
| Community | -1 | <p>Any of the following might mean "such that":</p>
<p>$: \ ; \ , \ / \ | \ s.t.$ </p>
<p>I don't know if there are others.</p>
|
1,268,598 | <p>I thought the <code>|</code> symbol meant "divides by", but in set theory it seems that it means "such that." However, I thought we wrote "such that" as <code>:</code>.</p>
<p>Can anybody elaborate?</p>
| Asaf Karagila | 622 | <p>This shouldn't be surprising. There are only so many letters that we can easily access.</p>
<ol>
<li><p>$\Bbb{P,Q,R}$ can denote a forcing notion, and have nothing to do with probability, the rationals or the reals. </p></li>
<li><p>$\pi$ denotes a real number, the ratio between a diameter and circumference of a circle, but in many places it also denotes a projection map, or an automorphism, or just a function.</p></li>
<li><p>$\Bbb N$ sometimes denote $\{1,2,3,\ldots\}$ and sometimes $\{0,1,2,\ldots\}$.</p></li>
<li><p>Since set theory is rarely worried with the natural numbers and their arithmetic, there's little point in having a useful symbol like $|$ denoting something like divisibility relation.</p></li>
<li><p>The same can be said about terminology. Regular and normal objects appear everywhere. Saying that something is weakly compact in set theory is not the same as saying it is weakly compact in functional analysis; a critical point in set theory is not the same as a critical point in real analysis; and so on and so forth.</p></li>
</ol>
<p>Different fields have different conventions. The important thing is that (1) you're consistent with your notation throughout a particular text; and (2) the reader knows what you mean when you write mathematical notation.</p>
|
949,664 | <p>Consider the Laplace transform $\int_{0}^{\infty} e^{-px}f(x)\,dx$ <br/>
Assume $f(x)=1$ , then the Laplace transform is $\frac {1}{p}$. <br/>
Assume $f(x)=x$ , then the Laplace transform is $\frac {1}{p^2}$.<br/>
The question is, what will happen to the $f(x)$ after getting transformed?<br/>
Why should the function be transformed and what aspect of initial function will remain in the Laplace transform that makes it so important?<br/>
If someone can give geometric intuition of it, it will be a plus!<br/>
Thanks</p>
| Paul Sundheim | 88,038 | <p>One major advantage lies in the fact that many differential equations become algebraic equations when the Laplace transform is applied. We can then solve the algebraic equations and take the inverse Laplace transform (the transform is one to one and so has an inverse) to arrive at a solution to the differential equation.</p>
<p>To see how the transform relates to infinite series and other connections see <a href="http://en.wikipedia.org/wiki/Laplace_transform#Relation_to_power_series" rel="nofollow">http://en.wikipedia.org/wiki/Laplace_transform#Relation_to_power_series</a></p>
|
1,955,393 | <p>I have been trying to evaluate this limit:</p>
<p>$$\lim_{n\to\infty}{\sqrt[n]{4^n + 5^n}}$$</p>
<p>What methods should I try in order to proceed?</p>
<p>I was advised to use "Limit Chain Rule", but I believe there is a different approach.</p>
| Simply Beautiful Art | 272,831 | <p>By the root test for a power series, this equivalent to the reciprocal of the radius of convergence of the following:</p>
<p>$$\sum_{n=0}^\infty(4^n+5^n)x^n$$</p>
<p>Which is a basic geometric series with radius of convergence $1/5$, so the limit is</p>
<p>$$\lim_{n\to\infty}\sqrt[n]{4^n+5^n}=1/R=5$$</p>
|
11,609 | <p>What's the code for multiple alignment in MathJAX? An analogous question for Latex is at <a href="https://tex.stackexchange.com/questions/43464/multiple-alignment-in-equations">https://tex.stackexchange.com/questions/43464/multiple-alignment-in-equations</a>, but it doesn't appear to function here. Thank you.</p>
<blockquote>
<p>Objective:<br>
$A = B$<br>
$\qquad B = C$<br>
$\qquad \qquad C = D$<br>
$\qquad \qquad \qquad E = F$</p>
</blockquote>
<p>Attempt:
$$\begin{align*} A = B\\\
&B = C\\\
&&C =D\\\
&&&E= F
\end{align*}$$</p>
<p>Code for Attempt:
<code>$$\begin{align*} A = B\\\<br>
&B = C\\\<br>
&&C =D\\\<br>
&&&E= F<br>
\end{align*}$$</code></p>
| Zev Chonoles | 264 | <p>This is pretty hack-ish, but </p>
<pre>
$$\newcommand{\myeq}{\!\!=\!\!}
\begin{array}{ccccccccc}
A & \myeq & B \\
& & B & \myeq & C \\
& & & & C & \myeq & D \\
& & & & & & D & \myeq & E
\end{array}$$
</pre>
<p>produces</p>
<p>$$\newcommand{\myeq}{\!\!=\!\!}
\begin{array}{ccccccccc}
A & \myeq & B \\
& & B & \myeq & C \\
& & & & C & \myeq & D \\
& & & & & & D & \myeq & E
\end{array}$$
Is that like what you want?</p>
|
3,454,146 | <p>I'm trying to wrap my head around this new subject. I have to determine the validity of this argument (using a truth table): </p>
<p>"If Steve went to the movies then Maria's sister would not have stayed home. Either Steve went to the movies or Maria or both. If Maria went, then Maria’s sister would have stayed home. Both Maria's sister and Steve’s sister stayed home. So, Steve did not go to the movies."</p>
<p>So far, I've come up with this, as a first step. I doubt it's correct. Please correct and guide me, as I'm still new to all of this. Thank you </p>
<p>P = Steve went to the movies<br>
Q = Maria's sister stayed home<br>
R = Maria went to the movies<br>
S = Steve's sister stayed home</p>
<p><span class="math-container">$(P \rightarrow \lnot Q), ((P \lor R) \lor (P \land R)), (R \rightarrow Q), (Q \land S) ∴ \lnot P$</span></p>
| RyRy the Fly Guy | 412,727 | <p>Your translation into propositional logic looks great! Note that the inclusive disjunction <span class="math-container">$\vee$</span> is true whenever one or both operands are true, so the statement "either Steve went to the movies or Maria or both" may be written as <span class="math-container">$P \vee R$</span> as opposed to <span class="math-container">$(P \vee R) \vee (P \wedge R)$</span>. However, the latter is not incorrect; it is actually logically equivalent, so if you wish to leave it the way it is, then you may do so.</p>
<p>Now that you have correctly translated your premises and conclusion into propositional logic, you may evaluate the validity of the argument with a truth table using either of the following approaches: </p>
<p><strong>APPROACH I</strong></p>
<p><span class="math-container">$1$</span>. Construct a truth table including all premises as well as the conclusion. Be sure to demarcate the premises and conclusion from one another. For example, you may place a slash <span class="math-container">$/$</span> between premises and a double slash <span class="math-container">$//$</span> to separate the conclusion from the premises.</p>
<p><span class="math-container">$2$</span>. Fill in the truth table with its truth values, and compute the truth value of each statement.</p>
<p><span class="math-container">$3$</span>. Scan each line of the truth table to ensure the following: if the truth values of all the premises are true, then the truth value of the conclusion is also true. If this is the case, then the argument is valid. If this is not the case, then the argument is not valid. In other words, if there exists at least one line in the truth table such that the truth values of all the premises are true and the truth value of the conclusion is false, then the argument is not valid. This criteria follows by definition of validity.</p>
<p><strong>APPROACH II</strong></p>
<p>Alternatively, you may check and see whether the corresponding conditional of the argument is a tautology. In other words, simply create a truth table for the corresponding conditional and determine whether the statement is always true. If so, the argument is valid; otherwise, it is not valid. </p>
|
898,543 | <p>I have the random vector $(X,Y)$ with density function $8x^{2}y$ for $0 < x < 1$, $0 < y < \sqrt{x}$ I am trying to find the marginal distributions of $X$ and $Y$. For $X$ this seems to be simply the integral $\int_{0}^{\sqrt{x}}8x^{2}y = 4x^{3}$, which is also the given solution, and follows the general formula I've gotten, where you find marginal distributions of a variable by integrating the joint PDF of all other variables over their supports. However, this seems to fail in the case of $Y$, where I try the integral $\int_{0}^{1}8x^{2}y = \frac{8y}{3}$, conflicting with the given answer of $\frac{8y}{3}(1-y^{6})$. What am I misunderstanding here? This seems painfully simple, and I have never had issues finding a marginal distribution like this before.</p>
| Umberto P. | 67,536 | <p>Use $P(A | B) = P(A \cap B)/P(B)$. You are given $P(L|B) = \frac 17$ and $P(B|L) = \frac 13$. Thus $P(B) = 7 P(B \cap L)$ and $P(L) = 3 P(B \cap L)$.</p>
<p>You are also given $P(B \cup L) = \frac 15$ since this is the probability of having either attribute. Since $P(B \cup L) + P(B \cap L) = P(B) + P(L)$ you obtain $$P(B \cap L) + \frac 15 = 10 P(B \cap L).$$</p>
|
200,876 | <p>Is there a topological space $(C,\tau_C)$ and two points $c_0\neq c_1\in C$ such that the following holds?</p>
<blockquote>
<blockquote>
<p>A space $(X,\tau)$ is connected if and only if for all $x,y\in X$ there is a continuous map $f:C\to X$ such that $f(c_0) = x$ and $f(c_1) = y$.</p>
</blockquote>
</blockquote>
<p>Is there also a Hausdorff space satisfying the above?</p>
| Helene Sigloch | 62,434 | <p>First, there can't be a path from $c_0$ to $c_1$, else a continuous map would give a path from $x$ to $y$. By the same argument, $C$ is not allowed to have finitely many or countably many path components. If one can build something like the topologist's sine curve on a very long line of arbitrarily large cardinality, the number of path components of $C$ has to be larger or equal than any cardinal number. There is no largest cardinal number. Hence, the set of points of $C$ is no set.</p>
<p>Thus: If there exist long lines in every cardinality, the answer to your question is no.</p>
<hr>
<p>Edit: OK, if $x$ and $y$ are uncountably far apart, there is no path from $x$ to $y$. Call the set of points between $x$ and $y$ the "long path" from $x$ to $y$. It contains an arbitrarily large number of connected non-path-connected, pairwise disjoint sub"paths". Say the number is $\kappa '$. The connected components of their preimages under $f$ also have to be non-path-connected and pairwise disjoint. Hence, the universal space $C$ that we are looking for needs to have at least $\kappa '$ path components, hence at least $\kappa '$ points. If the "very long line" exists for every cardinality, the universal space $(C, c_0,c_1)$ can't exist.</p>
<hr>
<p>Edit: Goldstern's argument is much simpler than this. Just for completeness: Very long lines exist (see Goldstern's comment) and so the answer is definitely No.</p>
<hr>
<p>Edit: The concept of homotopy theory with respect to a big interval is actually worked with. I just stumbled over a <a href="http://arxiv.org/pdf/1410.3088v3.pdf" rel="nofollow">preprint by Penrod</a> in the field and the foundations seem to be laid by Cannon-Gonner [J. Cannon and G. Conner, The big fundamental group, big Hawaiian earrings, and the big free groups, Topology Appl. 106 (2000)].</p>
|
2,130,776 | <p>Let $(L,R)$ be a pair of adjoint functor. </p>
<p>How to show that the commutativity of the left diagram induces the commutativity of the right one?</p>
<p><a href="https://i.stack.imgur.com/lOffl.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lOffl.jpg" alt="diagram"></a></p>
| Community | -1 | <p>Suppose $L : \mathcal{C} \to \mathcal{D}$ is left adjoint to $R : \mathcal{D} \to \mathcal{C}$, with unit $\eta : 1_\mathcal{C} \to RL$ and counit $\epsilon : LR \to 1_\mathcal{D}$.</p>
<p>We can prove the following, more general statement: let $J$ be a category, and let $[J, X]$ be the category of functors from $J$ to $X$. Let $L_*$ denote the composition functor $[J, \mathcal{C}] \to [J, \mathcal{D}]$, and similarly for $R_*$. </p>
<blockquote>
<p><strong>Theorem:</strong> $L_*$ is left adjoint to $R_*$</p>
</blockquote>
<p>The unit $\eta_*$ and counit $\epsilon_*$ are given by composition as well: e.g. given a functor $F : J \to \mathcal{C}$, the component $ \eta_{*F} : F \to RLF$ is the horizontal composite $\eta F$.</p>
<p>That the triangle identities hold for $\eta_*$ and $\epsilon_*$ follows nearly immediately from the fact they hold for $\eta$ and $\epsilon$.</p>
<p>The whole thing can sort of be interpreted as taking the horizontal composite of an adjunction with a functor or natural transformation.</p>
<hr>
<p>The given problem is the case where $J$ is the arrow category $\bullet \to \bullet$, so that a natural transformation between two functors from $J$ is a commutative square. Let $F$ be the diagram $ A \to B$ and $G$ be the diagram $X \to Y$. Then, we have a natural isomorphism</p>
<p>$$\hom_{[J, \mathcal{D}]}(LF, G) \cong \hom_{[J, \mathcal{C}]}(F, RG) $$</p>
|
2,087,084 | <p>Fix an arbitrary abelian category $\mathscr{A}$, and let $$0\to A\xrightarrow{f}B\xrightarrow{g}C\to 0$$ be a short exact sequence in the category of chains $\mathscr{A}_\bullet$, where $A$, $B$, and $C$ have chain maps $\varphi^A_n:A_n\to A_{n-1}$, $\varphi^B_n:B_n\to B_{n-1}$, $\varphi^C_n:C_n\to C_{n-1}$ respectively, and let $A$ and $B$ be exact. I claim that $C$ is exact.</p>
<p>My current approach is to try to show that $$\ker\varphi^C_{n-1} = \mathrm{coker\,}(\ker\varphi^C_n\hookrightarrow C_n) = \mathrm{coim\,}\varphi_n^C = \mathrm{im\,}\varphi_n^C.$$</p>
<p>So, for an arbitrary object $M\in\mathscr{A}$ and morphism $\psi:C_n\to M$ such that $$\left(\ker\varphi^C_n\hookrightarrow C_n\xrightarrow{\psi}M\right) = 0$$ I wish to show that there exists a unique $\ker\varphi^C_{n-1}\to M$ such that $$\left(C_n\twoheadrightarrow\mathrm{im\,}\varphi^C_n\hookrightarrow\ker\varphi^C_{n-1}\to M\right) = \left(C_n\xrightarrow{\psi}M\right).$$ However, despite playing around a lot with commutative diagrams, kernels, and cokernels, I haven't found a good way of doing this. What have I missed?</p>
| Pedro | 23,350 | <p>The <em>long exact homology sequence</em> is <strong>the</strong> result to prove that if two out of three complexes in a SEC are exact, so is the third. Prove this, and you're done. To do so, if suffices you prove the well known <em>snake lemma</em> for an arbitrary abelian category. </p>
|
2,762,953 | <p>I've studied Markov Process with 2x2 matrices.
Using the linear algebra and calculus procedures is clear to me how a Markov chain works.</p>
<p>However, i'm still not able to grasp the intuitive and immediate meaning of a Markov chain.
Why intuitively, for $n\rightarrow +\infty $, the state of the system is independent of both initial state and the states reached during the process ?</p>
| Siong Thye Goh | 306,553 | <p>It does depends on the initial state. Consider the Markov chain which consist of two disjoint state, that is $$\begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}$$</p>
<p>Then if you start in state $1$, you will stuck in state $1$ forever as it is an absorbing state. Similarly if you start in state $2$. </p>
<p>Those example that you may have encounter might have the property that they only contain a unique stationary distribution, hence no matter where you begins, it ends up there.</p>
|
4,357,484 | <p>Suppose the following series:
<span class="math-container">\begin{eqnarray}
\sum_{k'}k'f_{k'}
\end{eqnarray}</span>
where <span class="math-container">$f_{k'}$</span> are some Fourier coefficients that result from a periodic function <span class="math-container">$f(t+T)=f(t)$</span>:
<span class="math-container">\begin{eqnarray}
f_{k'}=\frac{1}{T}\int_{0}^{T}dt e^{ik'2\pi t/T}f(t).
\end{eqnarray}</span>
Is it true then that:
<span class="math-container">\begin{eqnarray}
\sum_{k'}k'f_{k'}=\frac{1}{T}\int_{0}^{T}dt e^{ik'2\pi t/T}f(t)\left(\sum_{k'}k'e^{ik'2\pi t/T}\right)=0,
\end{eqnarray}</span>
that is, given that:
<span class="math-container">\begin{eqnarray}
g(t)=\left(\sum_{k'}k'e^{ik'2\pi t/T}\right)=2i\sum_{k'=1}^{+\infty}k'\sin(2\pi k' t/T)=0
\end{eqnarray}</span>
Is the above identity correct?</p>
<p><strong>EDIT:</strong> Wolfram Alpha gives the wrong answer for the sum <span class="math-container">$g(t)$</span>. So I would like to know how to determine <span class="math-container">$g(t)$</span> in closed form, if possible.</p>
| Captain Lama | 318,467 | <p>To provide an alternative (faster) proof than the method suggested by tomasz, you may note that <span class="math-container">$A$</span> being integral means that for any <span class="math-container">$a\neq 0$</span>, the application <span class="math-container">$A\to A$</span> defined <span class="math-container">$x\mapsto ax$</span> has a trivial kernel. Then you can think about what that implies when <span class="math-container">$A$</span> is finite-dimensional.</p>
|
3,394,378 | <p>I am stuck with this Precalculus problem about polynomial functions. The problem:</p>
<blockquote>
<p>Consider <span class="math-container">$f(x)=x^2+ax+b$</span> with <span class="math-container">$a^2-4b>0$</span>. Let <span class="math-container">$\alpha$</span> and <span class="math-container">$\beta$</span> be the roots of <span class="math-container">$f$</span>. Assume that <span class="math-container">$f(x)$</span> divides <span class="math-container">$f(x+2)f(x-2)$</span>. Then</p>
<ol>
<li><p>Show that <span class="math-container">$\alpha=\beta+2$</span> or <span class="math-container">$\alpha=\beta-2$</span></p>
</li>
<li><p>Using <span class="math-container">$1.$</span>, find the minimum value of <span class="math-container">$f(x)$</span>.</p>
</li>
</ol>
</blockquote>
<p>Part <span class="math-container">$1$</span> is easy: write <span class="math-container">$f(x+2)f(x-2)=f(x)g(x)$</span> and substitute <span class="math-container">$x=\alpha$</span> to obtain <span class="math-container">$f(\alpha+2)f(\alpha-2)=0$</span>, so <span class="math-container">$\alpha+2$</span> or <span class="math-container">$\alpha-2$</span> is a root of <span class="math-container">$f$</span>. This root cannot be <span class="math-container">$\alpha$</span>, so <span class="math-container">$\alpha+2=\beta$</span> or <span class="math-container">$\alpha-2=\beta$</span>.</p>
<p>I am stuck with part <span class="math-container">$2$</span>. <strong>Any idea</strong>?</p>
<p>The minimum is obtained at <span class="math-container">$-\frac{a}{2}=\frac{\alpha+\beta}{2}=\beta+1$</span> or <span class="math-container">$\beta-1$</span> but I don't know how to continue.</p>
| October15 | 715,022 | <p>By parallely translating the graph of the given <span class="math-container">$f(x)$</span>, we can get the graph of any <span class="math-container">$g(x)=x^2+cx+d$</span>. So it is sufficient to find the minimum of any <span class="math-container">$g(x)$</span> whose solutions is 2 apart: for example, <span class="math-container">$x(x-2)$</span>.</p>
|
281,504 | <p>Is there a trick for easily solving a matrix polynomial like
$$
p(A) = \left( 7\cdot A^4 - 4\cdot A^3 + 6\cdot A - 5\cdot E \right)
, A = \left(\begin{matrix}2 & -1 \\ 3 & 5\end{matrix}\right)
$$
or is it really just step-by-step calculation of
$$
7\cdot A\cdot A\cdot A\cdot A\cdot A\cdot A\cdot A-4\cdot A\cdot A\cdot A+6\cdot A-5\cdot E
$$</p>
| Ross Millikan | 1,827 | <p>Yes, if by reflect you mean to draw a perpendicular from the point to the line and continue it the same distance on the other side. If your line is $(p_x,p_y,p_z)+t(q_x,q_y,q_z)$ (is this what you mean by vector equation?) and the point is $(r_x,r_y,r_z)$ the point on the line where the perpendicular hits can be found by the condition that the dot product with the direction vector is zero. We want to find $t$ such that $r_x(p_x+tq_x)+r_y(p_y+tq_y)+r_z(p_z+tq_z)=0$. This is a linear equation that can be solved $t=-\frac {\vec r \cdot \vec p}{\vec r \cdot \vec q}$ The perpendicular point is then point is then $\vec s=\vec p+t \vec q$ and the reflected point is then $2\vec s-\vec r$.</p>
|
2,069,573 | <p>I'm trying to solve the following inequality $\dfrac{(\log_2 (8x) \times \log_{x/8} 2)}{\log_{x/2} 16} \leq 0.25$ </p>
<p>Wolfram alpha gives the answer $(0, 0.5], [1,8)$ but surely $x \not= 2$ since log to base $1$ is undefined. But is the fact that it basically shrinks the fraction down to $0$ sufficient to satisfy this inequality? Could someone clear this up for me?</p>
<p><a href="http://m.wolframalpha.com/input/?i=%28log_2+%288x%29*+log_%28x%2F8%29+2%29%2Flog_%28x%2F2%29+16+%5Cleq+0.25&x=0&y=0" rel="nofollow noreferrer">Wolfram link</a> is here.</p>
| Arnaldo | 391,612 | <p>Put everything on the base $2$:</p>
<p>$$\dfrac{(\log_2 (8x) \times \log_{x/8} 2)}{\log_{x/2} 16} \leq 0.25$$</p>
<p>$$\dfrac{\log_2 (8x) }{\log_{x/2} 16}\times \frac{\log_{2} 2}{\log_{2} (x/8)} \leq 0.25$$</p>
<p>$$\log_2(8x)\times\dfrac{\log_2 (x/2) }{\log_{2} 16}\times \frac{\log_{2} 2}{\log_{2} (x/8)} \leq 0.25$$</p>
<p>$$\log_2(8x)\times\dfrac{\log_2 (x/2) }{4}\times \frac{1}{\log_{2} (x/8)} \leq 0.25$$</p>
<p>$$(3+\log_2 x)\cdot (\log_2 x -1)\cdot \frac{1}{\log_{2} x -3} \leq 1$$</p>
<p>write $\log_{2} x=y$ and then</p>
<p>$$\frac{(y+3)(y-1)}{y-3}\le 1 \Rightarrow \frac{y^2+y}{y-3} \le 0$$</p>
<p>So, $y \le -1$ or $0 \le y <3$.</p>
<p>$$y \le -1 \Rightarrow \log_2 x \le -1 \Rightarrow x \le 1/2 \quad (1)$$</p>
<p>$$0 \le y <3 \Rightarrow 0 \le \log_2 x <3 \Rightarrow 1 \le x < 8 \quad (2)$$</p>
<p>But now we have to check the boundary for the problem. Which are:</p>
<p>$$8x >0 \Rightarrow x>0 \quad (3)$$</p>
<p>$$x/8 >0 \Rightarrow x>0 \quad x/8 \ne 1 \Rightarrow x \ne 8 \quad (4)$$</p>
<p>$$x/2 >0 \Rightarrow x>0 \quad x/2 \ne 1 \Rightarrow x \ne 2 \quad (5)$$</p>
<p>Making the intersection of $(1),(2),(3),(4),(5)$ we get:</p>
<p>$$0<x \le 1/2 \quad \text{or} \quad 1 \le x <8 \quad \text{and} \quad x \ne 2$$</p>
<p>or writing in another way:</p>
<p>$$0<x \le 1/2 \quad \text{or} \quad 1 \le x <2 \quad \text{or} \quad 2<x<8$$</p>
|
1,448,363 | <p>I have gotten to the next stage where you write it as $\frac{1}{\left(\frac 34\right)}$ to the power of $3$, now I am stuck</p>
<p>I've got it now, thanks everyone.</p>
| Mark | 270,016 | <p>$$=\left(\frac{3}{4}\right)^{-3}=\left(\frac{1}{\frac{3}{4}}\right)^3=\left(\frac{4}{3}\right)^3=\frac{4^3}{3^3}=\frac{64}{27}\approx2.37$$</p>
|
2,491,818 | <p>My professor showed us that the Cauchy distribution does not have an expected value, that is, the integral $\int_{-\infty}^{\infty} x f(x) \text{d} x$ is not defined ($f(x)$ is the p.d.f. of the Cauchy distribution). I find that very counterintuitive. What does it actually mean, in the context of probability, to not have a defined expected value?</p>
| Math1000 | 38,584 | <p>For a random variable $X$, write $X=X^+-X^-$, where $X^+=\max\{X,0\}$ and $X^-=\max\{-X,0\}$. Then $|X|=X^++X^-$, so $\mathbb E[|X|]<\infty$ if and only if $\mathbb E[X^+]<\infty$ and $\mathbb E[X^-]<\infty$. When this condition holds, we define
$$
\mathbb E[X] = \mathbb E[X^+-X^-]= \mathbb E[X^+]-\mathbb E[X^-].
$$
If $X$ has standard Cauchy distribution with density $f(x) = \frac1{\pi(1+x^2)}$ then for each $L>0$ we have
\begin{align}
\mathbb E\left[X^+\mathsf 1_{(0,L]}\right](X) = \int_0^L \frac x{\pi(1+x^2)}\ \mathsf dx = \frac{\log\left(1+L^2\right)}{2\pi}\stackrel{L\to\infty}\longrightarrow +\infty,
\end{align}
so that $\mathbb E[X^+]=+\infty$. By symmetry, $\mathbb E[X^-] = -\infty$. It follows that the expression
$\mathbb E[X^+]-\mathbb E[X^-]$ is not well-defined, and neither is $\mathbb E[X]$.</p>
|
2,617,235 | <p>Given a triangle $\Delta$ABC, how to draw any inscribed equilateral triangle whose vertices lie on different sides of $\Delta$ABC?</p>
| Alex | 293,781 | <p><strong>Construction</strong></p>
<ol>
<li><p>Use <a href="https://en.wikipedia.org/wiki/Morley%27s_trisector_theorem" rel="nofollow noreferrer">Morley's theorem</a> to construct an equilateral triangle $\triangle XYZ$ that lies strictly inside $\triangle ABC$.</p></li>
<li><p>Through the vertices of $\triangle XYZ$, draw three lines parallel to the sides of your original $\triangle ABC$. These lines form a new triangle $\triangle A'B'C'$ that is similar to $\triangle ABC$ and lies strictly inside it. (Thus the equilateral triangle $\triangle XYZ$ is now inscribed in the newly constructed $\triangle A'B'C'$.)</p></li>
<li><p>Enlarge the figure so that $\triangle A'B'C'$ becomes the size of $\triangle ABC$. The enlarged image of $\triangle XYZ$ is the sought-for inscribed triangle.</p></li>
</ol>
|
1,242,760 | <p>Now proving by induction is fairly simple. However, this is a multiple choice problem whose answers don't make any sense to me. The actual problem goes as follows:</p>
<p><em>To prove by induction that $n^2 - 7n - 2$ is divisible by $2$ is true for all positive integers $n$, we assume $k^2 - 7k - 2$ is divisible by $2$ is true for some positive integers $k$ and we show that $k^2 - 7k - 2 + A$ is divisible by $2$ where $A$ is:</em></p>
<p>Now I would've assumed that $A$ would be $(k+1)^2 - 7(k + 1) - 2$, but I checked the answer and it is actually $2(k-3)$ which makes no sense to me. I tried to factor, reduce, and anything I could think of but I can't get $(k+1)^2 - 7(k + 1) - 2$. Does anyone know where I am going wrong?</p>
| Bill Dubuque | 242 | <p><strong>Hint</strong> $ $ If $\,f(0)\,$ is even and $\,f(n\!+\!1)-f(n)=2j\,$ is even then by induction $\,f(n)\,$ is even for all $\,n,\,$ the inductive step being: $\ \,f(n+1) = f(n) +2j = $ even + even = even.</p>
<p><strong>Remark</strong> $\ $ mod $\,2\,$ it is simply: $ $ if $f\,$ is constant $\,f(n\!+\!1)\equiv f(n)\,$ then $\,f(n)\equiv f(0)\,$ for all $\,n$.</p>
|
1,003,379 | <p>I've been working problems all day so maybe I'm just confusing myself but in order to do this. I have to the take the integral along each contour $C_1-C_4$. My issue is how to convert to parametric functions in order to this so that I can integrate</p>
<p><img src="https://i.stack.imgur.com/HWRoM.jpg" alt="enter image description here"></p>
| Robert Israel | 8,508 | <p>Hint: $1/z$ has a pole at $z=0$.</p>
|
690,621 | <p>Consider the Quotient ring $\mathbb{Z}[x]/(x^2+3,5)$. </p>
<p>Solution: I first tried to take care of $(5)$ in the above ring. Therefor we can consider $\mathbb{Z_5}[x]/(x^2+3)$. Now and interesting point to note here is $(5) \subset (x^2+3)$. So, we can consider $\mathbb{Z_5}[x]/(5)$. But this is just $\mathbb{Z_5}[x]$. Thus, am I on the right track? Is there a rigorous way to prove the above?</p>
| Elchanan Solomon | 647 | <p>Let's start with an example. You have a table with an assortment of different items. You bring in three people: somebody with good eyesight, somebody with okay eyesight, and somebody who has terrible eyesight. You ask the three people to distinguish the items on the table. The person with good eyesight sees that every object appears different, so she lists as many different items as there are objects on the table. The next person has weaker eyesight: to her, many things appear the same. When she lists the object on the table, she groups together many different objects that appear the same to her. To her, there are fewer types of things on the table. The last person, who has very poor eyesight, cannot distinguish any of the objects on the table. To her, everything is the same: there is only one kind of object on the table.</p>
<p>Slightly more formally, you start with a set $S$ of different elements. You put on a pair of glasses that make you less able to distinguish certain features. For example, maybe your set is the integers, and your glasses only let you see if a number is even or odd. Then you will say: "this set only has two objects". The glasses are the equivalence relation: they identify two things that are similar in a consistent way. If two things look the same, they are in the same equivalence class. If two things don't look the same, they are in distinct equivalence classes. Returning to the example of the integers and the odd-even glasses, $1$ and $2$ look different, but $2$ and $4$ look the same. So $1$ and $2$ are in distinct equivalence classes, but $2$ and $4$ are in the same equivalence class, because when I put on those glasses I can distinguish the former pair, but I can't distinguish the latter pair.</p>
|
1,350,085 | <p>Consider $f(t)$, continuous on $[0,1]$, and $\alpha > 1$, and:</p>
<p>$$\int_0^1 \frac{f(t)}{t^{\alpha + 1}} \ dt$$</p>
<p>How can we tell this integral diverges? Basically since $f$ is continuous it reaches it's minimum at $[0,1]$ so we could make a comparison with $\int_0^1 \frac{f(x_{min})}{t^{\alpha+1}}\ dt$, but $f$ isn't nessecarily non-negative.</p>
<p>Questions:
1. Could look at $g(x) = f(x) + f(x_{min}) \ge 0$?<br>
2. Is there a more convinient way showing it diverges? </p>
<p>Thanks.</p>
| lulu | 252,071 | <p>Well, there are $2^{18}$ subsets all in all. Of these, $\dfrac{18!}{9!\cdot 9!}$ have exactly $9$ elements. The rest either have more than $9$ or fewer than $9$. Now, by symmetry, the number of subsets with more than $9$ elements equals the number of subsets with fewer than $9$ (pair each subset with its complement). Thus the number of subsets with more than $9$ elements is $\dfrac{1}{2} \times \left(2^{18} - \dfrac{18!}{9!\cdot 9!}\right)$. Adding the number of subsets with exactly $9$ elements gives the desired result.</p>
|
1,791,837 | <p>Recently I have been very fascinated by the claim and impact of Godel's incompleteness theorem. To understand the proof given by Godel, I felt the need to read an introductory book in logic to begin with. I have started reading the book titled "A Mathematical Introduction to Logic" by Herbert Enderton. As mentioned by the author, the aim of this book is to formalize the concept of: <em>logical reasoning, truth, proof</em>, etc. In the first chapter of this book the author introduces Sentential Logic. Further, to prove a few results in this chapter, the author provides proofs which are based on basic logical deductions. </p>
<p>My doubt is the following: Our aim of introducing sentential logic was to formalizing certain aspects of logical reasoning, however, in proving a few results about this new system (sentential logic) we use intuitive logic itself. But can we use a circular argument like this for proving?</p>
<p>Please let me know as to where my understanding about sentential logic lacks. I'm just starting with the field of mathematical logic and am currently stuck at this philosophical doubt, hence any help would be very welcome.</p>
| Thomas | 284,057 | <p>Write $S^n$ as the union of two open sets $U= S^n-N$, $V=S^n-S $where $N,S$ are the north and south pole. Note that the intersection $U\cap V$ is connected if $n\geq 2$. $U,V$ are contracile (diffeomorphic to $R^,$, and on each of these sets the form $\alpha$ is exact and andmits primitive $f_U, f_V$. On the intersection $d(f_U-f_V)=0$ and by connexity $f_U-f_V$ must be a constant $c$. So $f_V=f_U+c$ one the intersection. As a primitive of $\alpha$ one can take $f_U$ on $U$ and $f_U +c$ on $V$. The main point is that the intersection is connected.</p>
|
3,301,696 | <p>Prove that if <span class="math-container">$R$</span> is a non-commutative ring with <span class="math-container">$1$</span> and if <span class="math-container">$a,b \in R$</span> and if <span class="math-container">$ab =1 $</span> but <span class="math-container">$ba \neq 1$</span> then <span class="math-container">$R$</span> is infinite</p>
<p>I'm not sure about this one.</p>
<p>I noticed that <span class="math-container">$ba$</span> is an idempotent, and thus <span class="math-container">$(1-ba)$</span> is idenmpotent... So far I have 5 elements wooh! Only need <span class="math-container">$\infty$</span> more! Haha but seriously i'm pretty confused, insight appreciated!</p>
| lisyarus | 135,314 | <p>Consider the sequence of left ideals <span class="math-container">$R=Ra^0 \supset Ra^1 \supset Ra^2 \supset Ra^3 \supset\dots$</span> - we are going to show that all the inclusions are proper, meaning that there is in infinite number of ideals in <span class="math-container">$R$</span>, thus <span class="math-container">$R$</span> itself cannot be finite.</p>
<p>Suppose that at some point the inclusion is <em>not</em> proper, i.e. <span class="math-container">$Ra^n = Ra^{n+1}$</span>. Since <span class="math-container">$1 \in R$</span>, <span class="math-container">$1 \cdot a^n \in R a^{n+1}$</span>, meaning that there is <span class="math-container">$c \in R$</span> such that <span class="math-container">$a^n = ca^{n+1}$</span>. Multiply this by <span class="math-container">$b^n$</span>: <span class="math-container">$a^nb^n=ca^{n+1}b^n \Rightarrow 1 = ca$</span>. So, <span class="math-container">$a$</span> posesses a left inverse <span class="math-container">$c$</span>, and by assumption <span class="math-container">$b$</span> is the right inverse of <span class="math-container">$a$</span>. If <span class="math-container">$a$</span> has both-sided inverses, then these inverses are equal <span class="math-container">$c = b$</span>, so <span class="math-container">$ba=1$</span>, contradicting the assumption. So, all inclusions are proper.</p>
<hr>
<p>This actually proves a stronger result: the situation in question cannot happen in Artinian rings. This proves that, e.g., for matrices <span class="math-container">$ab=1$</span> implies <span class="math-container">$ba=1$</span>.</p>
|
362,716 | <p>Let <span class="math-container">$E$</span> be a separable <span class="math-container">$\mathbb R$</span>-Banach space, <span class="math-container">$\rho_r$</span> be a metric on <span class="math-container">$E$</span> for <span class="math-container">$r\in(0,1]$</span> with <span class="math-container">$\rho_r\le\rho_s$</span> for all <span class="math-container">$0<r\le s\le1$</span>, <span class="math-container">$\rho:=\rho_1$</span>, <span class="math-container">$$d_{r,\:\delta,\:\beta}:=1\wedge\frac{\rho_r}\delta+\beta\rho\;\;\;\text{for }(r,\delta,\beta)\in[0,1]\times(0,\infty)\times[0,\infty)$$</span> and <span class="math-container">$(\kappa_t)_{t\ge0}$</span> be a Markov semigroup on <span class="math-container">$(E,\mathcal B(E))$</span>.</p>
<blockquote>
<p>Assume we arbe able to show that for all <span class="math-container">$n\in\mathbb N$</span> there is a <span class="math-container">$\alpha\in[0,1)$</span> and <span class="math-container">$(r,\delta,\beta)\in[0,1]\times(0,\infty)\times(0,1)$</span> with<span class="math-container">$^1$</span> <span class="math-container">$$\operatorname W_{d_{r,\:\delta,\:\beta}}\left(\delta_x\kappa_n,\delta_y\kappa_n\right)\le\alpha\operatorname W_{d_{r,\:\delta,\:\beta}}\left(\delta_x,\delta_y\right)\tag1$$</span> for all <span class="math-container">$x,y\in E$</span>, where <span class="math-container">$\delta_x$</span> denotes the Dirac measure on <span class="math-container">$(E,\mathcal B(E))$</span> at <span class="math-container">$x\in E$</span>. Why are we able to conclude that there is a <span class="math-container">$(c,\lambda\in[0,\infty)^2$</span> with <span class="math-container">$$\operatorname W_\rho\left(\nu_1\kappa_t,\nu_2\kappa_t\right)\le ce^{-\lambda t}\operatorname W_\rho\left(\nu_1,\nu_2\right)\tag2$$</span> for all <span class="math-container">$\nu_1,\nu_2\in\mathcal M_1(E)$</span> and <span class="math-container">$t\ge0$</span>?</p>
</blockquote>
<p>It's clear to me that if <span class="math-container">$\kappa$</span> is any Markov kernel on <span class="math-container">$(E,\mathcal B(E))$</span> and <span class="math-container">$d$</span> is any metric on <span class="math-container">$E$</span> such that there is a <span class="math-container">$\alpha\ge0$</span> with <span class="math-container">$\operatorname W_d\left(\delta_x\kappa,\delta_y\kappa\right)\le\alpha\operatorname W_d\left(\delta_x,\delta_y\right)$</span> for all <span class="math-container">$x,y\in E$</span>, then this extends to <span class="math-container">$\operatorname W_d(\mu\kappa,\nu\kappa)\le\alpha\operatorname W_d(\mu,\nu)$</span> for all <span class="math-container">$\mu,\nu\in\mathcal M_1(E)$</span>. Moreover, it's clear that <span class="math-container">$\operatorname W_d\left(\delta_x,\delta_y\right)=d(x,y)$</span>.</p>
<p>Note that for any choice of <span class="math-container">$(r,\delta,\beta)\in[0,1]\times(0,\infty)\times[0,\infty)$</span>, it holds <span class="math-container">$$\beta\rho\le d_{r,\:\delta,\:\beta}\le\left(\frac1\delta+\beta\right)\rho.\tag3$$</span></p>
<p><em>Remark</em>: The desired claim seems to be used in the proof of Theorem 3.4 in <a href="https://arxiv.org/pdf/math/0602479.pdf" rel="nofollow noreferrer">https://arxiv.org/pdf/math/0602479.pdf</a>.</p>
<hr>
<p><span class="math-container">$^1$</span> If <span class="math-container">$(E,d)$</span> is a complete separable metric space and <span class="math-container">$\mathcal M_1(E)$</span> is the space of probability measures on <span class="math-container">$\mathcal B(E)$</span>, then the Wasserstein metric <span class="math-container">$\operatorname W_d$</span> on <span class="math-container">$\mathcal M_1(E)$</span> satisfies the identity <span class="math-container">$$\operatorname W_d(\mu,\nu)=\sup_{\substack{f\::\:E\:\to\:\mathbb R\\|f|_{\operatorname{Lip}(d)}\:\le\:1}}(\mu-\nu)f\;\;\;\text{or all }\mu,\nu\in\mathcal M_1(E),$$</span> where <span class="math-container">$$|f|_{\operatorname{Lip}(d)}:=\sup_{\substack{x,\:y\:\in\:E\\x\:\ne\:y}}\frac{|f(x)-f(y)|}{d(x,y)}\;\;\;\text{for }f:E\to\mathbb R$$</span> and <span class="math-container">$\mu f:=\int f\:{\rm d}\mu$</span> for <span class="math-container">$\mu$</span>-integrable <span class="math-container">$f:E\to\mathbb R$</span>.</p>
| 0xbadf00d | 91,890 | <p>Building up on <a href="https://mathoverflow.net/a/362732/91890">Benoît Kloeckner's answer</a>, consider the following simplified scenerio: Let <span class="math-container">$(E,d)$</span> be a complete separable metric space, <span class="math-container">$(\kappa_t)_{\ge0}$</span> be a Markov semigroup on <span class="math-container">$(E,\mathcal B(E))$</span> with <span class="math-container">$$\operatorname W_d(\delta_x\kappa_t,\delta_y\kappa_t)\le c\operatorname W_d(\delta_x,\delta_y)\;\;\;\text{for all }x,y\in E\text{ and }t\in[0,1)\tag{10}$$</span> for some <span class="math-container">$c\ge0$</span> and <span class="math-container">$$\operatorname W_d(\delta_x\kappa_1,\delta_y\kappa_1)\le\alpha\operatorname W_d(\delta_x,\delta_y)\tag{11}$$</span> for some <span class="math-container">$\alpha\in(0,1)$</span>.</p>
<p>From <span class="math-container">$(11)$</span>, we easily deduce <span class="math-container">$$\operatorname W_d\left(\delta_x\kappa_n,\delta_y\kappa_n\right)\le\alpha^n\operatorname W_d\left(\delta_x,\delta_y\right)\tag{12}$$</span> for all <span class="math-container">$x,y\in\mathbb N$</span> and <span class="math-container">$n\in\mathbb N_0$</span>. If <span class="math-container">$t>0$</span>, we may write <span class="math-container">$t=n+r$</span> for some <span class="math-container">$n\in\mathbb N_0$</span> and <span class="math-container">$r\in[0,1)$</span> so that <span class="math-container">$$\operatorname W_d\left(\delta_x\kappa_t,\delta_y\kappa_t\right)\le\alpha^n\operatorname W_d\left(\delta_x\kappa_r,\delta_y\kappa_r\right)\le c\alpha^n\operatorname W_d\left(\delta_x,\delta_y\right)\tag{13}$$</span> for all <span class="math-container">$x,y\in E$</span> by <span class="math-container">$(12)$</span> and <span class="math-container">$(10)$</span>.</p>
<p>Now we only need to note that <span class="math-container">$$c\alpha^n=\frac c\alpha\alpha^{n+1}\le\frac c\alpha\alpha^t\tag{14}$$</span> (the last "<span class="math-container">$\le$</span>" is actually a "<span class="math-container">$<$</span>" as long as <span class="math-container">$c\ne0$</span>) and hence we obtain <span class="math-container">$$\operatorname W_d\left(\mu\kappa_t,\nu\kappa_t\right)\le\tilde ce^{-\lambda t}\operatorname W_d(\mu,\nu)\tag{15}$$</span> for all <span class="math-container">$\mu,\nu\in\mathcal M_1(E)$</span>, where <span class="math-container">$$\tilde c:=\frac c\alpha$$</span> and <span class="math-container">$$\lambda:=-\ln\alpha.$$</span></p>
<h2><strong>Remark</strong></h2>
<p>I'd still be interested in the question whether this result still holds when <span class="math-container">$(10)$</span> and <span class="math-container">$(11)$</span> are replaced by the following assumption: There is a <span class="math-container">$t_0>0$</span> with <span class="math-container">$$\operatorname W_d(\delta_x\kappa_t,\delta_y\kappa_t)\le c\operatorname W_d(\delta_x,\delta_y)\;\;\;\text{for all }x,y\in E\text{ and }t\in[0,t_0)\tag{10'}$$</span> and <span class="math-container">$$\operatorname W_d(\delta_x\kappa_{t_0},\delta_y\kappa_{t_0})\le\alpha\operatorname W_d(\delta_x,\delta_y)\tag{11'}$$</span> for some <span class="math-container">$\alpha\ge0$</span>.</p>
<p>(The original statement in this answer is the particular case <span class="math-container">$t_0=1$</span>.)</p>
|
1,497 | <p>If (C,tensor,1) is a symmetric monoidal category and f:A-->B is a morphism of PROPs (or monoidal cats = colored PROPs), one gets a forgetful functor f^*:B-Alg(C)-->A-Alg(C) (where B-Alg(C)=tensor-preserving functors from B to C) defined by precomposing with f.</p>
<p>Does anyone conditions on A,B,C under which this functor has a left or a right adjoint?
(e.g. if C has the monoidal structure coming from products, it has a left adjoint, is there more to say?)</p>
| Tom Leinster | 586 | <p>I don't know the answer to that question, but I know the answer to some similar questions.</p>
<p>The simplest (which you probably know about) is that if f: A --> B is a map of (small) categories and C has all small colimits (respectively, limits) then f^* has a left (respectively, right) adjoint. These adjoints are called left and right Kan extension along f. </p>
<p>A more complex setting is finite product categories. If f: A --> B is a finite-product preserving functor between categories with finite products, then it's a fact that f^*: FP(B, Set) --> FP(A, Set) has a left adjoint. (Here FP(A, C) means the category of finite-product preserving functors from A to C, and all natural transformations between them.) Looking at the proof, it seems to work if we replace Set by any category with finite products and small colimits such that the former distribute over the latter. </p>
<p>Perhaps it's true that for monoidal categories, f^* has a left adjoint as long as A and B are small, C has small colimits, and tensor in C distributes over small colimits. (And a dual statement should hold for right adjoints.) But that's only a guess.</p>
|
3,482,138 | <p><a href="https://i.stack.imgur.com/OUlV2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OUlV2.png" alt="enter image description here"></a></p>
<ol>
<li>I have found two planes trough the origin that meet the given plane
at right angles.</li>
</ol>
<p>I found three points in the plane, getting the vectors between those points, and using that as a normal vector for the equation of the plane I am looking to find. </p>
<p>I don't understand how I am to describe all of these planes though. Would finding a basis of all vectors lying in the plane and using an arbitrary linear combination of these as our normal to the plane we are looking for be the right way? </p>
<ol start="2">
<li>I took <span class="math-container">$(4, -1, 1)$</span> and <span class="math-container">$(1, -2, -3)$</span> as normal vectors to each plane.
Their cross product is <span class="math-container">$(5, 13, -7)$</span> is in the direction of the line
of intersection of the two planes, thus parallel to both.</li>
</ol>
<p><span class="math-container">$(x, y, z) = (x_0 + 5t, y_0 + 13t, z_0 -7t)$</span> is a parametric equation for all lines parallel to both. Is this correct? And what do they mean by refining this such that each line is listed once?</p>
<p>Any help is greatly appreciated!</p>
| dfnu | 480,425 | <p><strong>A fast approach using Euclidean geometry</strong></p>
<p><a href="https://i.stack.imgur.com/36iTB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/36iTB.png" alt="enter image description here"></a></p>
<ol>
<li><p>Since <span class="math-container">$\mathcal C_1$</span> and <span class="math-container">$\mathcal C_2$</span> have the same radius,
<span class="math-container">$\triangle BEC$</span> is equilateral.</p></li>
<li><p><span class="math-container">$BE$</span> subtends on <span class="math-container">$\mathcal C_2$</span> the angle <span class="math-container">$\angle BDE$</span>, which therefore has measure <span class="math-container">$30^\circ$</span>, and <span class="math-container">\begin{eqnarray}\angle ADE &=& \frac12 \angle ADC + 30^\circ\\ &=& \frac12 \angle ABC + 30^\circ.\end{eqnarray}</span></p></li>
<li>On the other hand, <span class="math-container">$AE$</span> subtends on <span class="math-container">$\mathcal C_1$</span> the angle <span class="math-container">$\angle AFE$</span>, by which we have <span class="math-container">\begin{eqnarray}180^\circ- \angle AFE &=& \frac12 \angle EBA=\\ &=& \frac12 \left(\angle ABC + 60^\circ\right).\end{eqnarray}</span></li>
<li>By 2. and 3. <span class="math-container">$\triangle AFD$</span> is isosceles, with <span class="math-container">$AD \cong AF$</span> and the thesis follows from the fact that <span class="math-container">$AB \cong AD \cong BF$</span>.</li>
</ol>
<p><strong>EDIT</strong> (thanks to <a href="https://i.stack.imgur.com/36iTB.png" rel="nofollow noreferrer">bjorn93</a> for his useful comment.)
The situation when <span class="math-container">$ED<EF$</span> is shown below. However, as you can notice, the relationships found above do not change.</p>
<p><a href="https://i.stack.imgur.com/C99cv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/C99cv.png" alt="enter image description here"></a></p>
|
34,959 | <p>$F(x) = \int_{x-1}^{x+1}f(t)dt$ for x an element of the reals.</p>
<p>Show that $F$ is differentiable on Reals, and compute $F^\prime$.</p>
<p>I am unsure about how to showing $F$ is differentiable. I know that I need to use the fundamental theorem of calculus, but can someone please explain how to do so?</p>
| El Moro | 10,026 | <p>If you haven't done any measure theory , a simple answer would be: </p>
<p>If f is continuous then it has a primitive (the integral is supposed to be rieman one) If 0 belongs to the domain of f, let us then call G(x)=integral from 0 to x f(t)dt such a primitive function wich is differentiable (as f is its derivative and f is continuous).</p>
<p>F(x)=G(x+1)-G(x-1).
F'(x)=(x+1)'f(x+1)-(x-1)'f(x-1).</p>
<p>Thus</p>
<p>F'(x)=f(x+1)-f(x-1).</p>
|
34,959 | <p>$F(x) = \int_{x-1}^{x+1}f(t)dt$ for x an element of the reals.</p>
<p>Show that $F$ is differentiable on Reals, and compute $F^\prime$.</p>
<p>I am unsure about how to showing $F$ is differentiable. I know that I need to use the fundamental theorem of calculus, but can someone please explain how to do so?</p>
| Aryabhata | 1,102 | <p>If $f$ is not continuous, then $F(x) = \int_{x-1}^{x+1} f(t) \text{d}t$ need not be differentiable at the points of discontinuity of $f$.</p>
<p>For instance I believe the following is a counter-example:</p>
<p>$ f(x) = \begin{cases} 0 & x \le 3 \\ 1 & x \gt 3 \end{cases}$</p>
<p>I believe we can show that $F(x) = \int_{x-1}^{x+1} f(t) \text{d}t$ is not differentiable $2$.</p>
<p>However, it is a well known theorem that at any point of continuity $c$ of $f$, the function $G(x) = \int_{a}^{x} f(t) \text{d}t$ is differentiable and $G'(c) = f(c)$.</p>
|
637,061 | <p>I have a problem:</p>
<blockquote>
<p>For a system of linear equations:
$$x_i=\sum_{j=1}^{n}a_{ij}x_j+b_i,\ \ i=1,2, \ldots , n \tag 1$$
Prove that, if<br>
$$\sum_{i=1}^{n}\sum_{j=1}^{n}a_{ij}^2 \le q<1$$
then $(1)$ has a unique solution.</p>
</blockquote>
<p>==================================</p>
<p>My teacher said that We need to use the <a href="http://en.wikipedia.org/wiki/Banach%27s_contraction_principle" rel="nofollow">Banach's Contraction Principle</a>, but I have trouble when I do it...</p>
<p>Any help will be appreciated! Thanks!</p>
| Ulrik | 53,012 | <p>Define $\pmb{A}$ to be the matrix with entries $a_{i,j}$ for $i,j = 1, \ldots, n$, $\pmb{x} = (x_1, \ldots, x_n)$ and $\pmb{b} = (b_1, \ldots, b_n)$. Then your system is equivalent to
$$\pmb{x} = \pmb{A}\pmb{x} + \pmb{b}$$
Define $f: \mathbb{R}^n \to \mathbb{R}^n$ by $f(\pmb{x}) = \pmb{A}\pmb{x} + \pmb{b}$. Equip $\mathbb{R}^n$ with the Euclidean metric.</p>
<p>What does it mean for $f$ to be a contraction in this metric space, and what does Banach's theorem say in that case?</p>
|
2,529,387 | <p><a href="https://i.stack.imgur.com/3M7iS.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3M7iS.jpg" alt="enter image description here"></a></p>
<p>The fourth question is how can I do?</p>
<blockquote>
<p>Q: Given that $a, b \in \mathbb{Z}$ and $a$ is even, show that if $a^2b$ is not divisible by $8$, then $b$ is odd.</p>
</blockquote>
| Mr. Brooks | 162,538 | <p>Is $a$ singly even or doubly even? If the former, that means $a = 4k + 2$, while the latter means $a = 4k$. If $a$ is singly even, then $a^2$ is doubly even, since $a^2 = (4k + 2)(4k + 2) = 16k^2 + 8k + 4$; however, $a^2$ is not divisible by $8$.</p>
<p>If $a$ is doubly even, then so is $a^2$, and what's more, it's also divisible by $8$, in which case the parity of $b$ is irrelevant, $8$ will be a divisor of $a^2 b$.</p>
<p>But if $a$ is singly even and $b$ is odd, then $a^2 b$ will be divisible by $4$ but not by $8$.</p>
<p>A couple of concrete examples: suppose $a = 10$ and $b = 3$. Then $300$ is divisible by $4$ but not by $8$. Suppose $b = 2$ instead. Then $200$ is divisible by $4$ <em>and</em> by $8$.</p>
|
2,764,073 | <p>I recently was working on a question posted in an AP calculus BC multiple choice sheet which asked:</p>
<p>Let f(x) be a positive, continuous deceasing function. If $\int_1^∞ f(x)dx$ = 5, then which of the following statements must be true about the series $\sum_1^∞f(n)$?</p>
<p>(a) $\sum_1^∞f(n)$ = 0</p>
<p>(b) $\sum_1^∞f(n)$ converges, and $\sum_1^∞f(n)$ < 5</p>
<p>(c) $\sum_1^∞f(n)$ = 5</p>
<p>(d) $\sum_1^∞f(n)$ converges, and $\sum_1^∞f(n)$ > 5</p>
<p>(e) $\sum_1^∞f(n)$ diverges.</p>
<p>I assumed, due to the integral test, the answer would be (b). However, the answer sheet claimed (d). I thought that this was merely a mistake, however I tried to find a function that met these conditions ( f(x) is strictly positive, f '(x) is negative, f(x) is continuous $\forall$x $\in$ [1,∞) and $\int_1^∞ f(x)dx$ = 5) and seemed to have found one (where the conditions were confirmed via online sources) and (d) seems to be true, thus I'm asking where this function fails to meet the above conditions or how (d) can be true given the integral test contradicting it (from my understanding).</p>
<p>If F(x) = $\int f(x)dx$, then $\lim_{x\to ∞}$F(x) - F(1) must equal five.
Let F(x) = (x - 1)$\biggr(\frac{5}{x+2}-\frac{1}{(x+4)^2}\biggr)$</p>
<p>Then f(x), which we'll let equal $\dfrac{d}{dx}$F(x), equals $\frac{5}{(x+2)}$ - $\frac{1}{(x+4)^2}$ + (x - 1)$\biggr(\frac{-5}{(x+2)^2} +\frac{2}{(x+4)^3}\biggr)$. By plugging f(x) and its derivative into an online graphing calculator, I find that f(x) is strictly positive and its derivative strictly negative. Thus $\sum_1^∞f(n)$ < $\int_1^∞f(x)dx$, but when I use an online source to find these values I find the opposite result:</p>
<p>f(x) in a graphing calculator (Desmos) (note only from 1 to a large number)
<a href="https://www.desmos.com/calculator/kfeqcphona" rel="nofollow noreferrer">https://www.desmos.com/calculator/kfeqcphona</a></p>
<p>f'(x) in a graphing calculator (Desmos) (I just put f'(x) in a form where f(x) is derived term by term)
<a href="https://www.desmos.com/calculator/fnroc3w9rm" rel="nofollow noreferrer">https://www.desmos.com/calculator/fnroc3w9rm</a></p>
<p>$\sum_1^∞f(n)$ = $\sum_1^∞\frac{5}{n+2}-\frac{1}{(n+4)^2}+(n-1)(\frac{-5}{(n+2)^2}+\frac{2}{(n+4)^3})$ = −10ζ(3)−$\frac{7255}{864}$+8(π)23≈5.90138529723494 </p>
<p>According to <a href="https://www.emathhelp.net/calculators/calculus-2/series-calculator/" rel="nofollow noreferrer">https://www.emathhelp.net/calculators/calculus-2/series-calculator/</a>.</p>
<p>Where am I going wrong?</p>
| Sri-Amirthan Theivendran | 302,692 | <p>Observe that
$$
\int_{1}^nf(x)\,dx=\sum_{i=1}^{n-1}\int_{i}^{i+1}f(x)\,dx<\sum_{i=1}^{n-1}\tag{1}
f(i)$$
for any integer $n\ge 1$ where we have using the fact that $f$ is decreasing for the inequality. Let $n\to\infty$ in (1) to get that
$$
5\leq\sum_{i=1}^\infty f(i)
$$</p>
|
2,764,073 | <p>I recently was working on a question posted in an AP calculus BC multiple choice sheet which asked:</p>
<p>Let f(x) be a positive, continuous deceasing function. If $\int_1^∞ f(x)dx$ = 5, then which of the following statements must be true about the series $\sum_1^∞f(n)$?</p>
<p>(a) $\sum_1^∞f(n)$ = 0</p>
<p>(b) $\sum_1^∞f(n)$ converges, and $\sum_1^∞f(n)$ < 5</p>
<p>(c) $\sum_1^∞f(n)$ = 5</p>
<p>(d) $\sum_1^∞f(n)$ converges, and $\sum_1^∞f(n)$ > 5</p>
<p>(e) $\sum_1^∞f(n)$ diverges.</p>
<p>I assumed, due to the integral test, the answer would be (b). However, the answer sheet claimed (d). I thought that this was merely a mistake, however I tried to find a function that met these conditions ( f(x) is strictly positive, f '(x) is negative, f(x) is continuous $\forall$x $\in$ [1,∞) and $\int_1^∞ f(x)dx$ = 5) and seemed to have found one (where the conditions were confirmed via online sources) and (d) seems to be true, thus I'm asking where this function fails to meet the above conditions or how (d) can be true given the integral test contradicting it (from my understanding).</p>
<p>If F(x) = $\int f(x)dx$, then $\lim_{x\to ∞}$F(x) - F(1) must equal five.
Let F(x) = (x - 1)$\biggr(\frac{5}{x+2}-\frac{1}{(x+4)^2}\biggr)$</p>
<p>Then f(x), which we'll let equal $\dfrac{d}{dx}$F(x), equals $\frac{5}{(x+2)}$ - $\frac{1}{(x+4)^2}$ + (x - 1)$\biggr(\frac{-5}{(x+2)^2} +\frac{2}{(x+4)^3}\biggr)$. By plugging f(x) and its derivative into an online graphing calculator, I find that f(x) is strictly positive and its derivative strictly negative. Thus $\sum_1^∞f(n)$ < $\int_1^∞f(x)dx$, but when I use an online source to find these values I find the opposite result:</p>
<p>f(x) in a graphing calculator (Desmos) (note only from 1 to a large number)
<a href="https://www.desmos.com/calculator/kfeqcphona" rel="nofollow noreferrer">https://www.desmos.com/calculator/kfeqcphona</a></p>
<p>f'(x) in a graphing calculator (Desmos) (I just put f'(x) in a form where f(x) is derived term by term)
<a href="https://www.desmos.com/calculator/fnroc3w9rm" rel="nofollow noreferrer">https://www.desmos.com/calculator/fnroc3w9rm</a></p>
<p>$\sum_1^∞f(n)$ = $\sum_1^∞\frac{5}{n+2}-\frac{1}{(n+4)^2}+(n-1)(\frac{-5}{(n+2)^2}+\frac{2}{(n+4)^3})$ = −10ζ(3)−$\frac{7255}{864}$+8(π)23≈5.90138529723494 </p>
<p>According to <a href="https://www.emathhelp.net/calculators/calculus-2/series-calculator/" rel="nofollow noreferrer">https://www.emathhelp.net/calculators/calculus-2/series-calculator/</a>.</p>
<p>Where am I going wrong?</p>
| Math Helper | 374,223 | <p>As the other answers point out quite well, because your function is positive and decreasing the sum will be larger than the integral. To get at your question though: no, this does not defy the integral test. The integral test does not give an upper bound on the sum, but merely says if one (either the sum or integral) converges then so will the other (either the integral or the sum), which is the case here. We need more information to compare the sizes of the two. </p>
|
2,764,073 | <p>I recently was working on a question posted in an AP calculus BC multiple choice sheet which asked:</p>
<p>Let f(x) be a positive, continuous deceasing function. If $\int_1^∞ f(x)dx$ = 5, then which of the following statements must be true about the series $\sum_1^∞f(n)$?</p>
<p>(a) $\sum_1^∞f(n)$ = 0</p>
<p>(b) $\sum_1^∞f(n)$ converges, and $\sum_1^∞f(n)$ < 5</p>
<p>(c) $\sum_1^∞f(n)$ = 5</p>
<p>(d) $\sum_1^∞f(n)$ converges, and $\sum_1^∞f(n)$ > 5</p>
<p>(e) $\sum_1^∞f(n)$ diverges.</p>
<p>I assumed, due to the integral test, the answer would be (b). However, the answer sheet claimed (d). I thought that this was merely a mistake, however I tried to find a function that met these conditions ( f(x) is strictly positive, f '(x) is negative, f(x) is continuous $\forall$x $\in$ [1,∞) and $\int_1^∞ f(x)dx$ = 5) and seemed to have found one (where the conditions were confirmed via online sources) and (d) seems to be true, thus I'm asking where this function fails to meet the above conditions or how (d) can be true given the integral test contradicting it (from my understanding).</p>
<p>If F(x) = $\int f(x)dx$, then $\lim_{x\to ∞}$F(x) - F(1) must equal five.
Let F(x) = (x - 1)$\biggr(\frac{5}{x+2}-\frac{1}{(x+4)^2}\biggr)$</p>
<p>Then f(x), which we'll let equal $\dfrac{d}{dx}$F(x), equals $\frac{5}{(x+2)}$ - $\frac{1}{(x+4)^2}$ + (x - 1)$\biggr(\frac{-5}{(x+2)^2} +\frac{2}{(x+4)^3}\biggr)$. By plugging f(x) and its derivative into an online graphing calculator, I find that f(x) is strictly positive and its derivative strictly negative. Thus $\sum_1^∞f(n)$ < $\int_1^∞f(x)dx$, but when I use an online source to find these values I find the opposite result:</p>
<p>f(x) in a graphing calculator (Desmos) (note only from 1 to a large number)
<a href="https://www.desmos.com/calculator/kfeqcphona" rel="nofollow noreferrer">https://www.desmos.com/calculator/kfeqcphona</a></p>
<p>f'(x) in a graphing calculator (Desmos) (I just put f'(x) in a form where f(x) is derived term by term)
<a href="https://www.desmos.com/calculator/fnroc3w9rm" rel="nofollow noreferrer">https://www.desmos.com/calculator/fnroc3w9rm</a></p>
<p>$\sum_1^∞f(n)$ = $\sum_1^∞\frac{5}{n+2}-\frac{1}{(n+4)^2}+(n-1)(\frac{-5}{(n+2)^2}+\frac{2}{(n+4)^3})$ = −10ζ(3)−$\frac{7255}{864}$+8(π)23≈5.90138529723494 </p>
<p>According to <a href="https://www.emathhelp.net/calculators/calculus-2/series-calculator/" rel="nofollow noreferrer">https://www.emathhelp.net/calculators/calculus-2/series-calculator/</a>.</p>
<p>Where am I going wrong?</p>
| user2661923 | 464,411 | <p>For visualization purposes, explore $f(x) = (1/x) \Rightarrow \int_{1}^n f(x)dx = ln(n).$<br>
$\sum_{i=1}^{n-1} (1/i) > ln(n) > \sum_{i=2}^n (1/i).$</p>
|
3,091,090 | <p>I came across this question the other day and have been trying to solve it by using some simple algebraic manipulation without really delving into L'Hospital's Rule or the Power Series as I have just started learning limit calculations.
We needed to find :
<span class="math-container">$$\lim_{x \to 0} \frac {x\cos x - \sin x}{x^2\sin x}$$</span>
I approached this problem in two different ways and know what the flaw is, however I have been unable to justify why this is so.</p>
<p>Let <span class="math-container">$$f(x) = \frac {x\cos x - \sin x}{x^2\sin x}$$</span>
Therefore, dividing by <span class="math-container">$x$</span>,
<span class="math-container">$$f(x) = \frac {\cos x - \frac{\sin x}{x}}{x\sin x}$$</span>
Using standard limit properties,
<span class="math-container">$$\lim_{x \to 0}f(x) = \frac{\lim_{x \to 0}\cos x - \lim_{x \to 0}\frac{\sin x}{x}}{\lim_{x \to 0}x\sin x}$$</span>
Since <span class="math-container">$$\lim_{x \to 0} \frac {\sin x}{x}=1$$</span>
<span class="math-container">$$\lim_{x \to 0}f(x)= \frac{\lim_{x \to 0}\cos x-1}{\lim_{x \to 0}x\sin x}$$</span></p>
<p>Rewriting the above as <span class="math-container">$$\lim_{x \to 0}\frac{(\cos x -1)x}{x^2\sin x}$$</span> and using the fact that <span class="math-container">$\lim_{x \to 0} \frac {\sin x}{x}=1$</span> and <span class="math-container">$\lim_{x \to 0}\frac{\cos x -1}{x^2}= -\frac{1}{2}$</span>,
we get <span class="math-container">$$\lim_{x \to 0}f(x)=-\frac{1}{2}$$</span></p>
<p>I know that the answer is wrong although I am not able to understand why.
I believe it is because I cannot combine the numerator and denominator into a single limit function. Using a similar trick, I also obtained the limit to be <span class="math-container">$-\frac{3}{8}$</span>.</p>
<p>Questions:</p>
<p>1) Could someone please explain why combining the numerator and denominator into a single limit is wrong? (The reason I even went ahead with such a manipulation was, we are allowed to separate the numerator and denominator while expanding the limit of a rational function so I felt that the reverse should also work). </p>
<p>2) As you can notice, I have not used L'Hospital's Rules or Power Series expansion of <span class="math-container">$\sin x $</span>and <span class="math-container">$\cos x$</span>. When I used L'Hospital's Rule, I noticed that I needed to go upto the third or fourth derivative to get rid of the <span class="math-container">$\frac{0}{0}$</span> indeterminate form. So would there be a better way of approaching such limits? </p>
<p>Thank You.</p>
| Paramanand Singh | 72,031 | <p>Your approach does not use the laws of limits in proper manner. Each of those laws which deal with limit of sum/difference/product/quotient of two functions has certain restrictions which must be checked before applying them.</p>
<p>In the current question the issue is with application of the quotient rule. You should note that the rule applies only when the limit of denominator is non-zero. But here limit of denominator <span class="math-container">$x\sin x$</span> is <span class="math-container">$0$</span> and thus your first step is wrong.</p>
<p>Further note that the question does require the use of L'Hospital's Rule or Taylor series. Using the limit <span class="math-container">$\lim\limits _{x\to 0}\dfrac{\sin x} {x} =1$</span> you can replace the denominator <span class="math-container">$x^2\sin x$</span> (as given in original problem) by <span class="math-container">$x^3$</span> and the numerator should be split as <span class="math-container">$$x\cos x-x+(x-\sin x) $$</span> so that the fraction under limit can be replaced by <span class="math-container">$$\frac{x-\sin x} {x^3}-\frac{1-\cos x} {x^2}$$</span> The second fraction tends to <span class="math-container">$1/2$</span> (via algebraic manipulation) and the first one tends to <span class="math-container">$1/6$</span> (via L'Hospital's Rule or Taylor series) and thus the desired limit is <span class="math-container">$1/6-1/2=-1/3$</span>. </p>
|
563,161 | <p>For something I'm working on, I have a matrix $A$ with another matrix $U$ which is unitary ($U^*U = I$), and I'm trying to show that, for the Frobenius norm, $\|A\| =\|UA\|$. Now, I can do this pretty easily if an inner product space exists. For example, $\|A\| = \sqrt{\langle A,A\rangle}$ and $\|UA\| = \sqrt{\langle UA,UA\rangle} = \sqrt{\langle A,U^*UA\rangle} = \sqrt{\langle A,A\rangle} = \|A\|$. However, I'm not sure if I evoke the inner product space if I'm just told that the Frobenius norm exists. Is this the appropriate approach?</p>
| Elias Costa | 19,266 | <p>By definition of Frobenius norm of a matrix $A\in\mathbb{C}^{n\times n}$,
$$
\| A\|_F:=\sqrt{\mbox{trace}(A^*A)}=\ldots =\sqrt{\sum_{j=i}^n\sum_{i=1}^n\overline{A_{ji}}\cdot A_{ij}}
$$
This norm is a norm defined by an inner product $\langle \cdot , \cdot \rangle$ (<em>i.e.</em> $\|A\|_F:=\sqrt[2]{\langle A , A\rangle}$ ). In this case, for $A=(A_{\,\alpha\,\beta})_{n\times n}\in\mathbb{C}^{n\times n}$ and $B=(B_{\,u\,v})_{n\times n}\in\mathbb{C}^{n\times n}$ we have
\begin{align}
\langle A, B\rangle_F := & \mbox{trace}\bigg(A^* B\bigg)\\
\end{align}
Remenber that, for all $X=(X_{\,i\,j} )_{n\times n}\in\mathbb{C}^{n\times n}$, $\mbox{trace}(X)=\sum_{\ell=1}^{n}X_{\ell\ell}$ and
\begin{align}
C^*\cdot D=&(C_{ij})_{n\times n}^*\cdot(D_{uv})_{n\times n} \\
=&(\overline{C_{ji}})_{n\times n}\cdot(D_{uv})_{n\times n} \qquad \mbox{ definition of $*$ } \\
=& \bigg(\sum_{k=1}^{n}C_{jk}\cdot D_{kv} \bigg)_{n\times n}\qquad \mbox{ rule of matrix product }
\end{align}
Now, your question
\begin{align}
\| A\|:=&\sqrt{\mbox{trace}(A^*A)}\\
=&\sqrt{\mbox{trace}(A^*IA)}\\
= & \sqrt{\mbox{trace}(A^*[U^*U]A)}\\
= & \sqrt{\mbox{trace}([A^*U^*][UA])}\\
= & \sqrt{\mbox{trace}([UA]^*[UA])}\\
=& \| UA\|
\end{align}</p>
|
2,312,096 | <p>How do I compute the integration for $a^2<1$,
$$\int_0^{2\pi} \dfrac{\cos 2x}{1-2a\cos x+a^2}dx=? $$
I think that:
$$\cos2x =\dfrac{e^{i2x}+e^{-2ix}}{2},
\qquad\cos x =\dfrac{e^{ix}+e^{-ix}}{2}$$
But I cannot. Please help me.</p>
| Robert Z | 299,698 | <p>You are on the right track. By letting $z=e^{ix}$ we have that
$$\int_0^{2\pi} \dfrac{\cos 2x}{1-2a\cos x+a^2}\,dx=
\frac{1}{2}\int_{|z|=1}\dfrac{z^2+1/z^2}{1-a(z+1/z)+a^2}\cdot \frac{dz}{iz}\\=\frac{1}{2i}\int_{|z|=1}\dfrac{z^4+1}{z^2(1-za)(z-a)}\,dz
=\pi\left(\mbox{Res}(f,0)+\mbox{Res}(f,a)\right)$$
where we used the <a href="https://en.wikipedia.org/wiki/Residue_theorem" rel="nofollow noreferrer">Residue Theorem</a> with $f(z)=\dfrac{z^4+1}{z^2(1-za)(z-a)}$ (note that the poles of $f$ inside $|z|<1$ are $0$ and $a$).</p>
<p>Can you take it from here?</p>
<p>P.S. The final result should be $\frac{2\pi a^2}{1-a^2}$.</p>
|
346,198 | <p>Recently I was playing around with some numbers and I stumbled across the following formal power series:</p>
<p><span class="math-container">$$\sum_{k=0}^\infty\frac{x^{ak}}{(ak)!}\biggl(\sum_{l=0}^k\binom{ak}{al}\biggr)$$</span></p>
<p>I was able to "simplify" the above expression for <span class="math-container">$a=1$</span>:</p>
<p><span class="math-container">$$\sum_{k=0}^\infty\frac{x^k}{k!}\cdot2^k=e^{2x}$$</span>
I also managed to simplify the expression for <span class="math-container">$a=2$</span> with the identity <span class="math-container">$\sum_{i=0}^\infty\frac{x^{2k}}{(2k)!}=\cosh(x)$</span>:</p>
<p><span class="math-container">$$\sum_{k=0}^\infty\frac{x^{2k}}{(2k)!}\biggl(\sum_{l=0}^k\binom{2k}{2l}\biggr)=\mathbf[\cdots\mathbf]=\frac{1}{4}\cdot(e^{2x}+e^{-2x})+\frac{1}{2}=\frac{1}{2}\cdot(\cosh(2x)+1)$$</span></p>
<p>However, I couldn't come up with a general method for all <span class="math-container">$a\in\Bbb{N}$</span>. I would be very thankful if someone could either guide me towards simplifying this expression or post his solution here. </p>
| Ivan Meir | 7,113 | <p>Actually the true generalisation for your <span class="math-container">$a=1$</span>, <span class="math-container">$a=2$</span> cases is this expression for your sum: <span class="math-container">$$\frac{1}{a^2}\sum_{s=0}^{a-1}\sum_{r=0}^{a-1}e^{x\omega^r(1+\omega^s)}.$$</span> where <span class="math-container">$\omega$</span> is an <span class="math-container">$a$</span>-th root of unity (<span class="math-container">$\omega^a=1$</span>, <span class="math-container">$a \neq 1$</span>). </p>
<p>When <span class="math-container">$a=1$</span> we have <span class="math-container">$\omega=1$</span> and your sum is <span class="math-container">$e^{2x}$</span>. For <span class="math-container">$a=2$</span> we have <span class="math-container">$\omega=-1$</span> and your sum is <span class="math-container">$\frac{1}{4}(e^{2x}+e^{-2x}+e^0+e^0)=\frac{1}{4}(e^{2x}+e^{-2x}+2)$</span>. Both agreeing with your observations.</p>
<p>Proof:
Using the fact that <span class="math-container">$\sum_{l=0}^{a-1}\omega^{lr}=0$</span> unless <span class="math-container">$a|l$</span> we can show:
<span class="math-container">$$\sum_{l=0}^k\binom{ak}{al}=\frac{1}{a}\sum_{s=0}^{a-1}(1+\omega^s)^{ak}$$</span> and<span class="math-container">$$\sum_{k=0}^\infty\frac{x^{ak}}{(ak)!}=\frac{1}{a}\sum_{r=0}^{a-1}e^{\omega^r x}.$$</span></p>
<p>Substituting the above expressions into your sum we obtain:</p>
<p><span class="math-container">$$\sum_{k=0}^\infty\frac{x^{ak}}{(ak)!}\biggl(\sum_{l=0}^k\binom{ak}{al}\biggr)=\sum_{k=0}^\infty\frac{x^{ak}}{(ak)!}\frac{1}{a}\sum_{s=0}^{a-1}(1+\omega^s)^{ak}=\frac{1}{a}\sum_{s=0}^{a-1}\sum_{k=0}^\infty\frac{(x(1+\omega^s))^{ak}}{(ak)!}$$</span><span class="math-container">$$=\frac{1}{a}\sum_{s=0}^{a-1}(\frac{1}{a}\sum_{r=0}^{a-1}e^{x\omega^r(1+\omega^s)})=\frac{1}{a^2}\sum_{s=0}^{a-1}\sum_{r=0}^{a-1}e^{x\omega^r(1+\omega^s)}.$$</span></p>
|
2,701,128 | <p>I have read multiple definitions so far but something is not clicking.</p>
<p>My most naive understanding is that $|G:H|$ is a "number" (could be infinite) that represents how many times $H$ is in $G$.</p>
<p>But even this doesn't seem fully correct.</p>
<p>I would like a general non-formal explanation and perhaps an example to understand the intuition.</p>
| Community | -1 | <p>Upgrading my comment to an answer:</p>
<p>You have the right intuition, but I would word it slightly differently.</p>
<p>$|G:H|$ is the number of (left or right) cosets of $H$. A coset is not a copy of $H$; indeed, a coset of $H$ isn't even a subgroup unless it's $H$ itself.</p>
<p>It would be more correct to call the coset $aH$ a <em>translate</em> of $H$ by the element $a$. Distinct cosets are disjoint and form a partition of $G$. So one might say that the cosets of $H$ constitute a "tiling" of $G$, and $|G:H|$ is the number of tiles.</p>
<p>To take a concrete example, let $G$ be the additive group of integers modulo $12$, and let $H$ be the subgroup generated by $4$, so $H = \{0, 4, 8\}$. Then there are four distinct cosets of $H$, namely:
$$\begin{aligned}
H &= \{0, 4, 8\} \\
1+H &= \{1, 5, 9\} \\
2+H &= \{2, 6, 10\} \\
3+H &= \{3, 7, 11\} \\
\end{aligned}$$
In other words, the cosets are $H$ and its translates $1+H$, $2+H$, and $3+H$.</p>
<p>Together these four cosets contain all the elements of $G$, so they constitute a partition (tiling) of $G$. The number of tiles is the number of cosets, which is $|G:H| = |G|/|H| = 12/3 = 4$.</p>
|
2,917,742 | <p>$\left\{ 1 + \left( \frac { d y } { d x } \right) ^ { 2 } \right\} ^ { \frac { 3 } { 2 } } = \frac { d ^ { 2 } y } { d x ^ { 2 } }$</p>
<p>what is the degree and order for above equation </p>
<p>well according to my knowledge the order be should $2$</p>
<p>and degree should be $\frac { 3 } { 2 }$</p>
<p>is my answer right or wrong ?</p>
| JJacquelin | 108,514 | <p><span class="math-container">$$\left\{ 1 + \left( \frac { d y } { d x } \right) ^ { 2 } \right\} ^ { \frac { 3 } { 2 } } = \frac { d ^ { 2 } y } { d x ^ { 2 } } \tag 1$$</span>
The highest order derivative is <span class="math-container">$\frac { d ^ { 2 } y } { d x ^ { 2 } }$</span> . So the order of the ODE is two.</p>
<p>In the definition of the degree, the key point is that the differential equation must be a polynomial equation in derivatives. The given differential equation is not a polynomial equation in its derivatives (because of the fractional power, 3/2, to which the term on the left hand side is raised) and so, strictly speaking, its degree is not defined.</p>
<p>However, in a less strict sense, the degree considered is the degree of the highest order derivative which is one. So the degree of Eq.<span class="math-container">$1$</span> is one, without forgetting that the definition of degree is not strictly respected.</p>
<p>If we take the liberty to transform the ODE so that all exponents be integers,
<span class="math-container">$$\left( 1 + \left( \frac { d y } { d x } \right) ^ { 2 } \right) ^ 3 = \left(\frac { d ^ { 2 } y } { d x ^ { 2 } }\right)^2 \tag 2$$</span>
the differential equation becomes a polynomial equation in its derivatives and the degree can be defined. Commonly, the degree considered is the degree of the highest order derivative. So, the degree of Eq.<span class="math-container">$2$</span> is two. (Without forgetting that Eq.<span class="math-container">$2$</span> is not strictly equivalent to Eq.<span class="math-container">$1$</span> ).</p>
|
304,259 | <p>I am stuck on this problem and I'm not sure how to approach it. Can anyone help me out with figuring how to approach the proof?</p>
<p>My task is to:</p>
<blockquote>
<p>Prove that it is impossible to find integers $\,x,\, y\,$ such that $\;2^x = 4y + 3$. </p>
</blockquote>
<p>I assumed a proof by cases would be the way to go?</p>
<p>Any input? Thanks in advance!</p>
| Valtteri | 49,808 | <p>$2^x$ is even and $4 y+3$ is odd...</p>
|
377,471 | <p>If $F[x]$ is a polynomial ring, and $f(x), g(x), h(x)$ and $r(x)$ are four polynomials in it, then is it always true that $f(x)=h(x)g(x)+r(x)$ where $deg(r(x))<deg(g(x))$, or is this true only when $F[x]$ is a Euclidean domain?</p>
<p>Please note that this question has been edited heavily to make it more coherent. </p>
<p>Thanks in advance.</p>
| Emmanuel | 85,851 | <p>If F is an Euclidean domain one can always write f(x)=q(x)g(x)+r(x), with r(x)=0 or deg(r(x)) less than deg(g(x)); for it you can read papers about Gröbner basis over Euclidean rings, or simply read the division algorithm in the paper of Franz Pauer name's "Gröbner bases with coefficients in rings".</p>
|
3,051,643 | <blockquote>
<p>Find the volume of intersection of the cylinder<br>
{<span class="math-container">$ x^2 +
y^2 \leq 1 $</span>} , {<span class="math-container">$ x^2 + z^2 \leq 1$</span>}, {<span class="math-container">$ y^2 + z^2 \leq 1$</span>}.</p>
</blockquote>
<p>i am having tough time finding the volume how do i solve this kind of questions ? .</p>
<p>my trial :
i will move to the cylinder coordinates of the xy cylinder let :</p>
<p><span class="math-container">$x^2 + y^2 = r^2 $</span></p>
<p><span class="math-container">$ z = z $</span></p>
<p><span class="math-container">$0\leq\theta \leq 2\pi$</span></p>
<p>solving the inequalties i get :</p>
<p><span class="math-container">$ 0 \leq r^2 \leq 1$</span> </p>
<p><span class="math-container">$ -\sqrt{1-\frac{r^2}{2}}\leq z \leq \sqrt{1-\frac{r^2}{2}}$</span></p>
<p><span class="math-container">$0\leq\theta \leq 2\pi$</span></p>
<p>the integral is :</p>
<p><span class="math-container">$ \int_{z=-\sqrt{1-\frac{r^2}{2}}}^{z=\sqrt{1-\frac{r^2}{2}}}\int_{r=0}^{r=1}\int_0^{2\pi} dz \ dr \ d{\theta}$</span> = <span class="math-container">$ \frac{4\pi}{\sqrt{2}}\frac{(2-r^2)^{\frac{3}{2}}}{-3} |_{r=0}^{r=1}$</span></p>
| heropup | 118,193 | <p>Perhaps surprisingly, it is not necessary to use polar coordinates, nor even multiple integration or trigonometry, to find the volume.</p>
<p>What we need to do is consider the volume of a solid whose cross-sections perpendicular to the <span class="math-container">$x$</span>-axis are rectangles whose width is <span class="math-container">$x$</span> and whose height is <span class="math-container">$\sqrt{1-x^2} - 1/\sqrt{2}$</span>, over the interval <span class="math-container">$x \in [0,1/\sqrt{2}]$</span>. This is simply <span class="math-container">$$\int_{x=0}^{1/\sqrt{2}} x \left(\sqrt{1 - x^2} - 1/\sqrt{2}\right) \, dx = \left[-\frac{(1-x^2)^{3/2}}{3} - \frac{x^2}{2\sqrt{2}}\right]_{x=0}^{1/\sqrt{2}} = \frac{8 - 5 \sqrt{2}}{24}.$$</span> Then the desired volume is <span class="math-container">$48$</span> times this volume, plus the volume of the inscribed cube, which gives <span class="math-container">$$2\sqrt{2} + 2(8-5\sqrt{2}) = 8(2 - \sqrt{2}).$$</span> I have left the reasoning as an exercise for the reader.</p>
<p>An alternative computation can be set up that includes the inscribed cube in the integral, simply by changing the height of the perpendicular cross section to <span class="math-container">$\sqrt{1-x^2} - x$</span>, giving <span class="math-container">$$\frac{V}{48} = \int_{x=0}^{1/\sqrt{2}} x \left(\sqrt{1-x^2} - x\right) \, dx.$$</span> Again, the details are left for the reader.</p>
|
11,916 | <p>In <a href="https://mathoverflow.net/questions/11845/theory-mainly-concerned-with-lambda-calculus/11861#11861">Theory mainly concerned with lambda-calculus?</a>, F. G. Dorais wrote, of the idea that the lambda-calulus defines a domain of mathematics:</p>
<blockquote>
<p>That would never stick unless there's another good reason. Besides, the schism between cs and math is very recent, I would contend that "functional programming" is actually a math term, historically speaking. More importantly, it would be wrong to use a term different than those who use it most, namely theoretical computer scientists, who are very competent mathematicians by the way. </p>
</blockquote>
<p>The idea, I think, is that the overlap between the kind of constructive mathematics that follows the formulae-as-types correspondence, and pure functional programming is so substantial that the core of the two topics is essentially the same subject.</p>
<p>Is this true?</p>
| Neel Krishnaswami | 1,610 | <p>So, I'm a computer scientist working in this area, and my sense is the following: </p>
<p>You cannot do good work on functional programming if you are ignorant of the logical connection, period. However, while "proofs-as-programs" is a wonderful slogan, which captures a vitally important fact about the large-scale structure of the subject, it doesn't capture <em>all</em> the facts about the programming aspects. </p>
<p>The reason is that when we look at a lambda-term as a program, we additionally care about its <em>intensional</em> properties (such as space usage and runtime), whereas in mathematics we only care about the <em>extensional</em> properties (i.e., the relation it computes).[*] Bubble sort and merge sort are the same function under the lens of $\beta\eta$ equality, but no computer scientist can believe these are the same algorithm.</p>
<p>I hope, of course, that one day these intensional factors will gain logical significance in the same way that the extensional properties already have. But we're not there yet.</p>
<p>[*] FYI: Here I'm using "intensional" and "extensional" in a difference sense than is used in Martin-Lof type theory. </p>
|
1,793,318 | <p>[limit question][1]</p>
<p>Let $x_{k}$ be a sequence of strictly positive real numbers with $\lim \limits_{k \to \infty}\dfrac{x_{k}}{x_{k+ 1}} >1$. Prove that $x_{k}$ is convergent and calculate $\lim \limits_{k \to \infty} x_{k}$.</p>
<p><strong>Attempted answer attached as picture.</strong></p>
<p>I am not sure if I'm actually answering the question properly. Also would I do the same steps to prove that limit is less than 1 and divergent? </p>
<p>Thank you inadvanced for any help. </p>
| user341050 | 341,050 | <p>The problem is that $$\lim_{x \rightarrow 0}\; \sum_{n=1}^{\infty} \frac{x}{1 + n^2 x^2}$$ is not $0$, so the function defined by the series isn't continuous. You can see that by evaluating at $x = 1/k:$ $$\sum_{n=1}^{\infty} \frac{1/k}{1 + n^2 / k^2}$$ is a Riemann sum for the integral $$\int_0^{\infty} \frac{1}{1+y^2} \, \mathrm{d}y$$ and so it tends to $\pi/2$ as $k \rightarrow \infty.$</p>
|
2,574,768 | <p>There are three planes, <strong>A</strong>, <strong>B</strong>, and <strong>C</strong>, all of which intersect at a single point, <strong>P</strong>. The angles between the planes are given: $$\angle\mathbf{AB}=\alpha$$ $$\angle\mathbf{BC}=\beta$$ $$\angle\mathbf{CA}=\gamma$$ $$0\lt\alpha,\beta,\gamma\le\frac{\pi}{2}$$</p>
<p>The intersection of any two of these planes form lines. The intersection of <strong>AB</strong> is $\mathbf{\overline{AB}}$, the intersection of <strong>BC</strong> is $\mathbf{\overline{BC}}$, and the intersection of <strong>CA</strong> is $\mathbf{\overline{CA}}$. It is given that none of these lines are parallel to each other and that they all intersect at the same single point, <strong>P</strong>.</p>
<p>Please express the lesser of the two angles formed by the intersection of these lines in terms of $\alpha$, $\beta$, and $\gamma$:
$$\angle\mathbf{\overline{AB}\;\overline{BC}}=?$$ $$\angle\mathbf{\overline{AB}\;\overline{CA}}=?$$ $$\angle\mathbf{\overline{BC}\;\overline{CA}}=?$$</p>
| Intelligenti pauca | 255,730 | <p>If $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$, are unit vectors perpendicular to $\mathbf{A}$, $\mathbf{B}$, $\mathbf{C}$, then $\mathbf{a}\times\mathbf{b}$ is a vector of magnitude $\sin\alpha$ directed along line $\overline{\mathbf{A}\mathbf{B}}$, and so on. The quantity $(\mathbf{a}\times\mathbf{b})\cdot(\mathbf{b}\times\mathbf{c})$ is then proportional to the cosine of $\angle\overline{\mathbf{A}\mathbf{B}}\ \overline{\mathbf{B}\mathbf{C}}$:
$$
(\mathbf{a}\times\mathbf{b})\cdot(\mathbf{b}\times\mathbf{c})=
\sin\alpha\sin\beta
\cos(\angle\overline{\mathbf{A}\mathbf{B}}\ \overline{\mathbf{B}\mathbf{C}}).
$$
On the other hand, according to <a href="https://en.wikipedia.org/wiki/Vector_calculus_identities#Addition_and_multiplication" rel="nofollow noreferrer">a well-know vector identity</a>:
$$
(\mathbf{a}\times\mathbf{b})\cdot(\mathbf{b}\times\mathbf{c})=
(\mathbf{a}\cdot\mathbf{b})\cdot(\mathbf{b}\cdot\mathbf{c})-
(\mathbf{a}\cdot\mathbf{c})\cdot(\mathbf{b}\cdot\mathbf{b})=
\cos\alpha\cos\beta-\cos\gamma.
$$
It follows that
$$
\cos(\angle\overline{\mathbf{A}\mathbf{B}}\ \overline{\mathbf{B}\mathbf{C}})=
\left|{\cos\alpha\cos\beta-\cos\gamma\over\sin\alpha\sin\beta}\right|,
$$
where I inserted the absolute value in order to pick the lower angle.</p>
|
3,941,815 | <p><span class="math-container">$$\sqrt[4]{16} = ?$$</span></p>
<p><span class="math-container">$$16 = |16|(\cos(0)+i\sin(\pi))$$</span>
the equality above holds.</p>
<p><span class="math-container">$$\sqrt[4]{16} = |\sqrt[4]{16}|(\cos(0 + \frac{2k\pi}{4})+i\sin(\pi+\frac{2k\pi}{4})) $$</span>
using De Moivre"s formula we get:
<span class="math-container">$$ \sqrt[4]{16} = 2(\cos(0 + \frac{2k\pi}{4})+i\sin(\pi+\frac{2k\pi}{4}))$$</span>
<span class="math-container">$$\text{for k = 0, we get one of the correct solutions, but for k = 1 we get $-i\sqrt{2}$ which is not a solution.}$$</span></p>
<p>Why does this kind of approach not work?</p>
| Deepak | 151,732 | <p>The proper way to express a complex number in polar coordinates is <span class="math-container">$z = |z| (\cos(\mathrm{arg \ z}) + i\sin (\mathrm{arg \ z}))$</span>. Crucially, note that both the cosine and sine are of the same angle, the argument of the complex number.</p>
<p>What you did is wrong because you have <span class="math-container">$0$</span> as one angle and <span class="math-container">$\pi$</span> as the other. You cannot apply De Moivre's to that expression because that is not a proper polar form.</p>
<p>Instead, start with <span class="math-container">$16 = 16 (\cos 2k\pi + i \sin 2k\pi)$</span>.</p>
|
255,483 | <p>How to transform this infinite sum</p>
<p>$$1+\sum_{i\geq1}\frac{x^i}{(1-x)(1-x^2)\cdots(1-x^i)}$$</p>
<p>to an infinite product</p>
<p>$$\prod_{i\geq1}\frac{1}{1-x^i}$$</p>
| Paramanand Singh | 72,031 | <p>The usual technique is to study the function $$f(a, q) = \prod_{n = 1}^{\infty}\frac{1}{1 - aq^{n - 1}}\tag{1}$$ and then express it as a series $$f(a, q) = \sum_{n = 0}^{\infty}x_{n}a^{n}\tag{2}$$ Now we can see that $$f(aq, q) = (1 - a)f(a, q)\tag{3}$$ so that $$\sum_{n = 0}^{\infty}x_{n}a^{n}q^{n} = (1 - a)\sum_{n = 0}^{\infty}x_{n}a^{n}\tag{4}$$ and thus we have $$x_{n}q^{n} = x_{n} - x_{n - 1}$$ or $$x_{n} = \frac{x_{n - 1}}{1 - q^{n}} = \frac{x_{0}}{(1 - q)(1 - q^{2})\cdots(1 - q^{n})}$$ Putting $a = 0$ we can see that $x_{0} = 1$ and hence we have $$\prod_{n = 1}^{\infty}\frac{1}{1 - aq^{n - 1}} = \sum_{n = 0}^{\infty}\frac{a^{n}}{(1 - q)(1 - q^{2})\cdots(1 - q^{n})}\tag{5}$$ Putting $a = q$ we get $$\prod_{n = 1}^{\infty}\frac{1}{1 - q^{n}} = \sum_{n = 0}^{\infty}\frac{q^{n}}{(1 - q)(1 - q^{2})\cdots(1 - q^{n})}$$ which is what we need to prove (question uses $x$ in place of $q$).</p>
<p>The identity $(5)$ is a special case of the $q$-binomial theorem which says that $$\sum_{n = 0}^{\infty}\frac{(a;q)_{n}}{(q;q)_{n}}z^{n} = \frac{(az; q)_{\infty}}{(z; q)_{\infty}}\tag{6}$$ where $$(a;q)_{n} = \prod_{i = 1}^{n}(1 - aq^{i - 1}),\, (a;q)_{\infty} = \prod_{n = 1}^{\infty}(1 - aq^{n - 1})$$ Identity $(5)$ is derived when we put $a = 0$.</p>
<p>The above technique is a formal one (i.e. it does not deal with issues of convergence) which was perhaps championed by Euler and it can be used to prove many identities related to $q$-series.</p>
|
450,410 | <p>I'm trying to teach myself how to do $\epsilon$-$\delta$ proofs and would like to know if I solved this proof correctly. The answer given (Spivak, but in the solutions book) was very different.</p>
<hr>
<p><strong>Exercise:</strong> Prove $\lim_{x \to 1} \sqrt{x} = 1$ using $\epsilon$-$\delta$.</p>
<p><strong>My Proof:</strong></p>
<p>We have that $0 < |x-1| < \delta $.</p>
<p>Also, $|x - 1| = \bigl|(\sqrt{x}-1)(\sqrt{x}+1)\bigr| = |\sqrt{x}-1||\sqrt{x}+1| < \delta$.</p>
<p>$\therefore |\sqrt{x}-1|< \frac{\delta}{|\sqrt{x}+1|}$</p>
<p>Now we let $\delta = 1$. Then
\begin{array}{l}
-1<x-1<1 \\
\therefore 0 < x < 2 \\
\therefore 1 < \sqrt{x} + 1<\sqrt{2} + 1 \\
\therefore \frac{1}{\sqrt{x} + 1}<1.
\end{array}</p>
<p>We had that $$|\sqrt{x}-1|< \frac{\delta}{|\sqrt{x}+1|} \therefore |\sqrt{x}-1|<\delta$$</p>
<p>By letting $\delta=\min(1, \epsilon)$, we get that $|\sqrt{x}-1|<\epsilon$ if $0 < |x-1| < \delta $.</p>
<p>Thus, $\lim_{x \to 1} \sqrt{x} = 1$.</p>
<hr>
<p>Is my proof correct? Is there a better way to do it (still using $\epsilon-\delta$)?</p>
| awllower | 6,792 | <p>Your proof is correct.<br>
We can also adopt the following:<br>
Since $|\sqrt x-1|\lt \epsilon$ is equivalent with $1-2\epsilon+\epsilon^2\lt x\lt 1+2\epsilon+\epsilon^2$, we can choose $\delta$ so that $0\lt\delta\lt \min\{|-2\epsilon+\epsilon^2|,|2\epsilon+\epsilon^2|\}$.<br>
Hope this helps. </p>
|
920,050 | <p>The answer is $\frac1{500}$ but I don't understand why that is so. </p>
<p>I am given the fact that the summation of $x^{n}$ from $n=0$ to infinity is $\frac1{1-x}$. So if that's the case then I have that $x=\frac15$ and plugging in the values I have $\frac1{1-(\frac15)}= \frac54$.</p>
| ReverseFlow | 109,908 | <p>The problem you have is that you do not know why the formula works to begin with. If you did the situation would be clear. Here's the thing:</p>
<p>$$\sum_{n=0}^{\infty}r^n=\frac{1}{1-r} \; \;\;\;\;\; |r|<1.$$</p>
<p>Let $r=\frac{1}{5}$, then you really have the following situation:</p>
<p>$$\left(\frac{1}{5}\right)^0+\left(\frac{1}{5}\right)^1+\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^3+\left(\frac{1}{5}\right)^4+\left(\frac{1}{5}\right)^5+....$$
Let's call this infinite sum $S$ and proceed as follows,</p>
<p>$$S=\left(\frac{1}{5}\right)^0+\left(\frac{1}{5}\right)^1+\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^3+\left(\frac{1}{5}\right)^4+\left(\frac{1}{5}\right)^5+....$$
then
$$ \left(\frac{1}{5}\right)S=\left(\frac{1}{5}\right)^1+\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^3+\left(\frac{1}{5}\right)^4+\left(\frac{1}{5}\right)^5+\left(\frac{1}{5}\right)^6....\;\;\;\;$$</p>
<p>Subtract the second from the first,</p>
<p>$$S-\left(\frac{1}{5}\right)S=1$$
$$S(1-\left(\frac{1}{5}\right))=1$$
$$S=\frac{1}{1-\left(\frac{1}{5}\right)}$$
$$S=\frac{5}{4}.$$</p>
<p>Now recall what $S$ was and realize,
$$S=\left(\frac{1}{5}\right)^0+\left(\frac{1}{5}\right)^1+\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^3+\left(\frac{1}{5}\right)^4+\left(\frac{1}{5}\right)^5+....=\frac{5}{4}$$
But, you want the powers to start at $n=4$, so subtract the first 4 powers(0,1,2,3) to get,
$$S-\left(1+\frac{1}{5}+\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^3\right)$$
which means
$$\left(\frac{1}{5}\right)^4+\left(\frac{1}{5}\right)^5+....=\frac{5}{4}-\left(1+\frac{1}{5}+\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^3\right)=\frac{1}{500}.$$</p>
|
29,177 | <p>I had a quick look around here and google before, and didn't find any answer to this particular question, and it's beginning to really irritate me now, so I'm asking here:</p>
<p>How is one supposed to write l (little L), 1 (one) and | (pipe) so that they don't all look the same? One of my teachers draws them all as vertical lines and I have seen things like |l| written as 3 vertical lines.</p>
<p>I tried writing a cursive l, but they always look like e when I write them, which is a whole new problem.</p>
<p>So is there a "best way" to write all of these on paper, or a "least confusing" way?</p>
<p>Thanks in advance guys ^_^</p>
| Isaac | 72 | <p>Personally, I always use a script lower-case L and never a print lower-case L in mathematical expressions (my handwriting is such that my lower-case E is sufficiently short and non-script-like that confusing the two is not an issue). As to one versus vertical bar, I typically make my vertical bars extend slightly above and below the line of text, but I also will draw the serifs on the one (the crossbar at the bottom and the hook at the top) if there's the potential for confusion.</p>
|
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