qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
2,856,121 | <p>Before I start explaining what I will say regarding the question, read the question before going any further.</p>
<p>Question: <em>Mariam and Leena both live on the bank of the same river, a distance apart. When Mariam visits Leena, she swims downstream (with the current), but then must swim upstream (against the current) to return back home. When the nearby dam is closed, the water between there homes is still and Mariam can visit Leena by swimming in both directions without the help (or hindrance) of any current. Is Mariam’s round trip between her house and Leena’s house quicker when the dam is closed or when it is open?</em></p>
<p>As you can from the question, it is asking is the round trip faster when the dam is closed or when it is open. I stated that the round trip's time should be the same regardless of the dam is open or closed. If the dam is closed, Mariam's round trip is $2t$ seconds. If the dam is open, the current decrease the amount of time required for Mariam to reach Leena; Mariam saves $x$ seconds; her time will be $t-x$ seconds. But when Mariam goes back home, she uses the $x$ she saved. Her time should be $t+x$ seconds on the way back to home. In theory, the round trip's time should be the same regardless of the dam is open or closed.</p>
<p>Problem: The problem is that I was told this is a good theory, but it is wrong. The person wasn't able to explain properly why it is wrong and as a result, I still remain confused on why it is wrong.</p>
| Ross Millikan | 1,827 | <p>The point is that current changes the velocity of swimming, not the time spent. Let the distance be $d$, the still water swimming speed be $v$ and the current speed be $c$. Swimming with the current is at speed $v+c$ and against the current is at speed $v-c$. If the dam is closed the round trip swimming time is $2\frac dv$. If the dam is open the round trip swimming time is $\frac d{v+c}+\frac d{v-c}$ We have
$$\frac d{v+c}+\frac d{v-c}=\frac {d(v-c)+d(v+c)}{(v+c)(v-c)}\\
=\frac {2dv}{v^2-c^2}\\=2\frac dv\frac 1{1-\frac {c^2}{v^2}}$$
As the final factor is greater than $1$ the time swimming is longer when the dam is open. If $c \gt v$ she can't do the round trip at all with the dam open.</p>
|
1,844,846 | <p>I had an exam, and there was a question - <code>is cosh(z) injective?</code><br>
I presented it as cos(iz) (or cos(-iz), doesn't matter because it's an even function), but it didn't help because I don't know if <em>this</em> function is injective either.</p>
| shai horowitz | 339,356 | <p>If $\cos(iz)=\cos(-iz)$ then by definition its not injective from $C$ to $C$.</p>
|
1,609,845 | <p><strong>Problem</strong>: Let $\alpha, \beta$ be non-negative numbers.
Suppose the number of strictly increasing sequences $a_0, a_1, a_2 \cdots a_{2014}$ satisfying $0 \leq 3m$ is $2^{\alpha}(2\beta+1)$.
Find $\alpha$.</p>
<p>My attempt:</p>
<p>I tried to find an explicit form for the $nth$ of the sequence using recursions. </p>
<p>Let $P_{n,k}$ be the number of sequences of length $n$ that end in $k$, i.e, $a_n=k$.</p>
<p>Then it's easy to say that the following recursions hold:</p>
<p>$P_{n+1, n+1}=P_{n, n}$</p>
<p>$P_{n+1, n+2}=P_{n, n}+P_{n, n+1}$</p>
<p>$P_{n+1, n+3}=P_{n, n}+P_{n, n+1}+P_{n, n+2}$</p>
<p>And so on.. and the last equation is,</p>
<p>$P_{n+1, 3n+3}=P_{n, n}+P_{n, n+1}+P_{n, n+2}+\cdots + P_{n, 3n}$ </p>
<p>And of course, we want $P_{n+1, n+1}+ P_{n+1, n+2}+P_{n+1, n+3}+ \cdots + P_{n+1, 3n+3}$</p>
<p>But i can't make any more progress.
So, any help will be appreciated.</p>
| Random Variable | 16,033 | <p>To supplement Dave's awesome answer, I'm adding a proof of the integral <span class="math-container">$$\zeta(s)-\frac{1}{s-1} = \frac{1}{\Gamma(s)} \int_0^{\infty} \left(\frac{1}{e^x-1}+1-\frac{1}{x} \right) x^{s-1} e^{-x} \, \mathrm dx \ , \quad \Re (s)>0. $$</span></p>
<p>All we need are the well-known integrals <span class="math-container">$$\Gamma(s) = \int_{0}^{\infty} x^{s-1}e^{-x} \, \mathrm dx \ , \quad \Re(s)>0 \tag{1} $$</span> and <span class="math-container">$$\Gamma(s) \zeta(s) = \int_{0}^{\infty} \frac{x^{s-1}}{e^x-1} \, \mathrm dx \ , \quad \Re(s)>1. \tag{2}$$</span></p>
<p>Subtracting <span class="math-container">$(1)$</span> from <span class="math-container">$(2)$</span>, we get <span class="math-container">$$ \Gamma(s) \zeta(s) - \Gamma(s)=\int_{0}^{\infty} x^{s-1} \left(\frac{1}{e^{x}-1} - e^{-x} \right) \, \mathrm dx = \int_{0}^{\infty} \frac{x^{s-1}e^{-x}}{e^{x}-1} \, \mathrm dx \ , \quad \Re(s)>1.$$</span></p>
<p>The integral on the right side of the equation can be rewritten as
<span class="math-container">$$ \begin{align} \ \int_{0}^{\infty} \frac{x^{s-1}e^{-x}}{e^{x}-1} \, \mathrm dx &= \int_{0}^{\infty} \left(\frac{1}{e^x-1}-\frac{1}{x} \right) x^{s-1} e^{-x} \, \mathrm dx+ \int_{0}^{\infty} x^{s-2} e^{-x} \, \mathrm dx \\ &= \int_{0}^{\infty} \left(\frac{1}{e^x-1}-\frac{1}{x} \right) x^{s-1} e^{-x} \, \mathrm dx + \Gamma(s-1) , \end{align} $$</span> where <span class="math-container">$\frac{1}{x}$</span> is the first term of the Laurent series expansion of <span class="math-container">$\frac{1}{e^x-1}$</span> at the origin.</p>
<p>Dividing both sides of the equation by <span class="math-container">$\Gamma(s)$</span>, we get <span class="math-container">$$\zeta(s) - 1 = \frac{1}{\Gamma(s)} \int_0^{\infty} \left(\frac{1}{e^x-1}- \frac{1}{x}\right) x^{s-1} e^{-x} \, \mathrm dx + \frac{1}{s-1} , \quad \Re(s) >0. \tag{3}$$</span></p>
<p>To get the integral in the form given in Dave's answer, add <span class="math-container">$1$</span> to both sides of the equation and use the fact that <span class="math-container">$$1 = \frac{1}{\Gamma(s)} \int_{0}^{\infty} x^{s-1}e^{-x} \, \mathrm dx.$$</span></p>
|
3,873,630 | <p>Hi Guys can some one help with this question?</p>
<p>Let <span class="math-container">$\mathcal{H}$</span> be a Hilbert space and F a linear bounded functional in <span class="math-container">$\mathcal{H}^∗$</span>
such that <span class="math-container">$F \neq 0$</span>. Prove that dim(kerF)<span class="math-container">$^{\perp} = 1$</span></p>
| Clement Yung | 620,517 | <p><span class="math-container">$
\renewcommand{\c}[1]{\langle #1 \rangle}
\renewcommand{\bb}[1]{\left( #1 \right)}
\renewcommand{\In}{\mathrm{In}}
$</span>
This proof is motivated by @M477's answer, and serves to fill in the details. Let <span class="math-container">$F,G_1,\dots,G_k$</span> be functions. We say that <span class="math-container">$F$</span> is a <strong>recursive function of <span class="math-container">$G_1,\dots,G_k$</span></strong> if there exists a recursive function <span class="math-container">$H$</span> such that:<span class="math-container">\begin{align*}
F(a,\mathfrak{a}) = H(G_1(a,\mathfrak{a}),\dots,G_k(a,\mathfrak{a}),a,\mathfrak{a})\end{align*}</span></p>
<p>Note that:</p>
<ol>
<li><p>If <span class="math-container">$G_1,\dots,G_k$</span> are all recursive and <span class="math-container">$F$</span> is a recursive function of <span class="math-container">$G_1,\dots,G_k$</span>, then <span class="math-container">$F$</span> is recursive by <strong>R2</strong>.</p>
</li>
<li><p>If <span class="math-container">$F$</span> is a recursive function of <span class="math-container">$G_1,\dots,G_k$</span> and each <span class="math-container">$G_i$</span> is a recursive function of <span class="math-container">$H_1,\dots,H_\ell$</span>, then <span class="math-container">$F$</span> is a recursive function of <span class="math-container">$H_1,\dots,H_\ell$</span> by <strong>R2</strong>.</p>
</li>
<li><p>If <span class="math-container">$F$</span> is a recursive function of <span class="math-container">$\bar{F}$</span>, then <span class="math-container">$F$</span> is recursive by <strong>R14</strong>.</p>
</li>
</ol>
<hr />
<p>Following the hint, given <span class="math-container">$a,\mathfrak{a}$</span>, define <span class="math-container">$L(a,\mathfrak{a}) := \c{F(a,\mathfrak{a}),G(a,\mathfrak{a})}$</span>. To show that <span class="math-container">$L$</span> is recursive, we proceed in a few steps. The proof will proceed in several steps.</p>
<p><strong>Step 1 - Show that <span class="math-container">$\bar{F}$</span>,<span class="math-container">$\bar{G}$</span> are both recursive functions of <span class="math-container">$\bar{L}$</span>.</strong> For <span class="math-container">$\bar{F}$</span>, we have:
<span class="math-container">\begin{align*}
\bar{F}(a,\mathfrak{a}) &= \mu x\bb{lh(x) = a \wedge \forall i_{i < a}\bb{(x)_i = F(i,a)}} \\
&= \mu x\bb{lh(x) = a \wedge \forall i_{i < a}\bb{(x)_i = ((\bar{L}(a,\mathfrak{a}))_i)_0}}
\end{align*}</span>
and similarly for <span class="math-container">$\bar{G}$</span> as well.</p>
<p><strong>Step 2 - Show that <span class="math-container">$F$</span> is a recursive function of <span class="math-container">$\bar{L}$</span>.</strong> By definition, we see that <span class="math-container">$F$</span> is a recursive function of <span class="math-container">$\bar{F},\bar{G}$</span>, and is hence a recursive function of <span class="math-container">$\bar{L}$</span> by (2) above.</p>
<p><strong>Step 3 - Show that <span class="math-container">$(a,\mathfrak{a}) \mapsto \bar{F}(a+1,\mathfrak{a})$</span> is a recursive function of <span class="math-container">$\bar{L}$</span>.</strong> Observe that:
<span class="math-container">\begin{align*}
\bar{F}(a+1,\mathfrak{a}) &= \c{F(0,a),\dots,F(a,\mathfrak{a})} \\
&= \mu x\bb{lh(x) = a + 1 \wedge \In(x,a) = \bar{F}(a,\bar{a}) \wedge (x)_a = F(a,\mathfrak{a})}
\end{align*}</span>
Thus, <span class="math-container">$(a,\mathfrak{a}) \mapsto F(a+1,\mathfrak{a})$</span> is a recursive function of <span class="math-container">$F,\bar{F}$</span>, and is thus a recursive function of <span class="math-container">$\bar{L}$</span> by (2).</p>
<p><strong>Step 4 - Show that <span class="math-container">$G$</span> is a recursive function of <span class="math-container">$\bar{L}$</span>.</strong> Similarly, by definition we see that <span class="math-container">$G$</span> is a recursive function of <span class="math-container">$(a,\mathfrak{a}) \mapsto \bar{F}(a+1,\mathfrak{a}),\bar{G}$</span>, and is hence a recursive function of <span class="math-container">$\bar{L}$</span> by the previous steps and (2) above.</p>
<p><strong>Step 5 - Show that <span class="math-container">$L$</span> is recursive.</strong> Finally, since <span class="math-container">$L(a,\mathfrak{a}) = \c{F(a,\mathfrak{a}),G(a,\mathfrak{a})}$</span> and both <span class="math-container">$F$</span> and <span class="math-container">$G$</span> are recursive functions of <span class="math-container">$\bar{L}$</span>, we have that <span class="math-container">$L$</span> is a recursive function of <span class="math-container">$\bar{L}$</span>, and is hence recursive.</p>
|
1,787,072 | <p>Boolean algebras aren't algebras (to the best of my understanding).</p>
<p>So why are they called algebras?</p>
<p>Wouldn't it make more sense to call them a "Boolean system" or a "Boology" or something else like that?</p>
| Noah Schweber | 28,111 | <p>They are algebras, in the sense of <em>universal algebra</em> (where "algebra" is basically synonymous with "first-order structure", except that it requires the language to have no relation symbols).</p>
<p>In fact, I think this notion of algebra = algebraic structure long preceded the definition of algebra as (roughly) a module with multiplication. And it was in this context that Boolean algebras were so named - that is, just because of their "algebraic" nature. </p>
|
167,049 | <p>$p(z),t(z)$ are two mutually defined functional equations, while $\widehat{G}(z)$ is the exponential generation function of <a href="https://oeis.org/A182173" rel="nofollow noreferrer">A182173</a> (maybe, I am not sure...lol):</p>
<p>$$\begin{cases}
p(z)=e^{t(z)}-t(z)+2 z-1\\
t(z)=2\left(e^{p(z)}-e^{p(z)/2}+z\right)-p(z)\\
\widehat{G}(z)=p(z)+t(z)-2 z
\end{cases}$$</p>
<p><strong>Is there any way to get series coefficients effectively?</strong></p>
<pre><code>p[z]->2z+E^t[z]-1-t[z]
t[z]->2*( z+ E^p[z] -E^(p[z]/2) ) - p[z]
Ge=p[z]+t[z]-2z
A182173[n_]:=n!SeriesCoefficient[Ge,{z,0,n}]
</code></pre>
| Akku14 | 34,287 | <p>Find <code>Ge[z]</code>:</p>
<pre><code>r1 = p[z] -> 2 z + E^t[z] - 1 - t[z]
r2 = t[z] -> 2*(z + E^p[z] - E^(p[z]/2)) - p[z]
Ge[z_] = p[z] + t[z] - 2 z /. r1 /. r2 // FullSimplify
(* -1 + E^(2 (-E^((p[z]/2)) + E^p[z] + z) - p[z]) *)
</code></pre>
<p><code>SeriesCoefficient</code> produces derivatives of <code>p</code> at <code>z == 0</code>. Therefore we have to solve for the <code>p[0]</code>, <code>p'[0]</code>, <code>p''[0]</code> ...</p>
<pre><code>der1[i_] := Derivative[i][p][0] == Derivative[i][2 # + E^t[#] - 1 - t[#] &][0]
der2[i_] :=
Derivative[i][t][0] == Derivative[i][2*(# + E^p[#] - E^(p[#]/2)) - p[#] &][0]
sol[0] = First @
Solve[{der1[0], der2[0]}, {Derivative[0][p][0],
Derivative[0][t][0]}, Reals]
(* {p[0] -> 0, t[0] -> 0} *)
</code></pre>
<p>Lower derivatives are needed to solve for higher ones:</p>
<pre><code>sol[j_ /; j > 0] := sol[j]=
First@Solve[{der1[j], der2[j]} //.
Flatten[Table[sol[k], {k, 0, j - 1}]], {Derivative[j][p][0],
Derivative[j][t][0]}, Reals]
</code></pre>
<p>Inserting these found derivatives of <code>p</code> into the series coefficients gives the desired <code>A</code> numbers.</p>
<pre><code>A[n_] := n! SeriesCoefficient[Ge[z], {z, 0, n}] //.
Flatten[Table[sol[i], {i, 0, n}]]
Table[A[i], {i, 1, 10}]
(* {2, 10, 94, 1466, 31814, 887650, 30259198, 1218864842, 56644903958,
2983300619410} *)
</code></pre>
|
176,386 | <p>I stumbled across this "dictionary for noncommutative topology" <a href="http://planetmath.org/noncommutativetopology" rel="nofollow">http://planetmath.org/noncommutativetopology</a>
and I would be very interested in learning more on the subject, particularly I'd like to see why these results are true. </p>
<p>What is the reference to the results in section 3 of the above linked page?</p>
| Michael | 33,290 | <p>K-Theory and C*-Algebras: A Friendly Approach (Oxford Science Publications)
by N.E. Wegge-Olsen. </p>
|
3,489,898 | <p>In order to show that a metric space <span class="math-container">$(X, d)$</span> is not complete one may apply the definition and look for a Cauchy sequence <span class="math-container">$\{x_n\}\subset X$</span> which does not converge with respect to the metric <span class="math-container">$d$</span>. Now I have often seen (on books, e.g.) another approach: one may show that a sequence <span class="math-container">$\{x_n\}\subset X$</span> converges with respect to the metric <span class="math-container">$d$</span> to a limit <span class="math-container">$x$</span> which is not contained in <span class="math-container">$X$</span>. </p>
<p>A common example may be the following: since <span class="math-container">$x_n:= (1+1/n)^n\in \mathbb{Q}$</span> for every <span class="math-container">$n \in \mathbb{N}$</span> and <span class="math-container">$x_n \to e$</span>, but <span class="math-container">$e \notin \mathbb{Q}$</span>, one can conclude that <span class="math-container">$\mathbb{Q}$</span> is not complete. </p>
<p>I've always considered this to be obvious but I now realize I can't explain why this works. The quantity <span class="math-container">$d(x_n, x)$</span> itself need not be well-defined, in general, if <span class="math-container">$x \notin X$</span>. So my question is: why (and under which conditions) this criterion for not-completeness of a metric space ("limit is not in the same space as the sequence") can be used? </p>
| Marios Gretsas | 359,315 | <p>For the second method you mentioned note that <span class="math-container">$(1+1/n)^n$</span> is a Cauchy sequence on the rationals with respect to the usual metric restricted on <span class="math-container">$\Bbb{Q}$</span></p>
<p>Even if the sequence <span class="math-container">$x_n$</span> converged to a point <span class="math-container">$x \notin Y \subseteq X$</span>, it is still Cauchy with the metric restricted on <span class="math-container">$Y$</span>(name it <span class="math-container">$d_Y$</span>)</p>
<p>If <span class="math-container">$(Y,d_Y)$</span> was complete then would exist <span class="math-container">$z \in Y$</span> such that <span class="math-container">$x_n \to z$</span>.</p>
<p>But every metric space is Hausdorff and thus the limit o a sequence is unique,so <span class="math-container">$z=x \notin Y$</span> which is a contradiction. </p>
|
648,946 | <p>I want to find
$$ \lim_{x\to0} \frac{(e^{-x^2}-1)\sin x }{x\ln(1+x^2)}$$
using a Maclaurin series and not using the l'Hôpital's rule.</p>
<p>However I can't seem to get it right.</p>
<p>Thanks for any possible answers.</p>
| Mr.Fry | 68,477 | <p>$(1)$ $z=(1+i\sqrt{3})^{-1+i} \iff z=e^{\ (-1+i) \log(1+i\sqrt{3})}$</p>
<p>$(2)$ Use $Log(z):= \ln |z| + i \arg(z)$</p>
<p>$(3)$ Here $|z|=2$ and $\arg(z)=\frac{\pi}{3}+2\pi n$ for any integer $n$</p>
<p>$(4)$ So, $z= e^{ (1+i)(2+i(\frac{\pi}{3}+2\pi n))}$</p>
<p>We can simplify this further, but I think this answers your question.</p>
|
244,718 | <p>How can I plot the function</p>
<p><code>F[x_,a_]:=Integrate[Abs[x - z]^(-1-2a), {z, -Infinity, -1}] + Integrate[Abs[x - z]^(-1-2a), {z, 1, Infinity}]</code></p>
<p>for x in (-1,1) and a in (0,1)?
I have been using Wolfram Cloud to do
<code>Plot[F[x,1], {x,-1,1}]</code>,
but I get that the computational time is exceeded
So my questions are:</p>
<ul>
<li>What does the plot look like? Is this an error or just a long computation?</li>
<li>If I wrote <code>Plot[F[x,a], {x,-1,1}, {a,0,1}]</code>
would it work to plot in the same picture the function F[x] for the various values of a? What if I just want to plot some values of a, for example a = 1/4, 1/3, 1/2, 2/3, 3/4, 1?</li>
</ul>
| Andreas | 69,887 | <p>You can do numerical integration and a 3D plot like:</p>
<pre><code>Plot3D[NIntegrate[Abs[x - z]^(-1 - 2 a), {z, -Infinity, -1}] +
NIntegrate[Abs[x - z]^(-1 - 2 a), {z, 1, Infinity}], {x, -1, 1}, {a,0,1}]
</code></pre>
<p><a href="https://i.stack.imgur.com/HfFCf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HfFCf.png" alt="enter image description here" /></a></p>
<p>a 1-d plot could look like</p>
<pre><code>Plot[Evaluate[
Table[((1 - x)^(-2*a) + (1 + x)^(-2*a))/(2*a), {a, 0.1, 1, 0.2}]],
{x, -1, 1}]
</code></pre>
<p><a href="https://i.stack.imgur.com/RPDCo.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RPDCo.png" alt="enter image description here" /></a></p>
|
244,718 | <p>How can I plot the function</p>
<p><code>F[x_,a_]:=Integrate[Abs[x - z]^(-1-2a), {z, -Infinity, -1}] + Integrate[Abs[x - z]^(-1-2a), {z, 1, Infinity}]</code></p>
<p>for x in (-1,1) and a in (0,1)?
I have been using Wolfram Cloud to do
<code>Plot[F[x,1], {x,-1,1}]</code>,
but I get that the computational time is exceeded
So my questions are:</p>
<ul>
<li>What does the plot look like? Is this an error or just a long computation?</li>
<li>If I wrote <code>Plot[F[x,a], {x,-1,1}, {a,0,1}]</code>
would it work to plot in the same picture the function F[x] for the various values of a? What if I just want to plot some values of a, for example a = 1/4, 1/3, 1/2, 2/3, 3/4, 1?</li>
</ul>
| Roman | 26,598 | <p>Using some <a href="https://reference.wolfram.com/language/ref/Assuming.html" rel="nofollow noreferrer">assumptions</a> and an <a href="https://reference.wolfram.com/language/ref/Set.html" rel="nofollow noreferrer">immediate assignment</a> gives an analytic integral, which is easy to plot:</p>
<pre><code>F[x_, a_] = Assuming[-1 <= x <= 1 && a > 0,
Integrate[Abs[x - z]^(-1 - 2 a), {z, -Infinity, -1}] +
Integrate[Abs[x - z]^(-1 - 2 a), {z, 1, Infinity}]]
(* (1 - x)^(-2 a)/(2 a) + (1 + x)^(-2 a)/(2 a) *)
</code></pre>
<p>The plot looks exactly like the one of <a href="https://mathematica.stackexchange.com/a/244720/26598">@Andreas' answer</a>.</p>
|
1,752,468 | <p>I've tried some values of $m$, and found that the equality $C^{m}_{2m}=\Sigma_{i=1}^{m}C^{i}_{m+1}C^{i-1}_{m-1}$ holds. But I can't give it a proof. Can anybody give some suggestions?</p>
| Patrick Sole | 266,100 | <p>There is an obvious graphical picture of VanderMonde equality. Imagine a 2m set split between an <span class="math-container">$m+1$</span>--set A and an <span class="math-container">$m-1$</span>--set B. An <span class="math-container">$m$</span>--set E is then split itself into <span class="math-container">$E \cap A$</span> of size <span class="math-container">$i$</span> and <span class="math-container">$E \cap B$</span> of size <span class="math-container">$m-i$</span>. Summing over <span class="math-container">$i$</span> the result follows. Note that <span class="math-container">${m-1 \choose m-i}={m-1 \choose i-1}.$</span> </p>
|
4,468,883 | <p>In the book "<em>Primes of the form <span class="math-container">$x^2+ny^2$</span></em>", David Cox had shown that:
<span class="math-container">$$p=x^2+14y^2 \Longleftrightarrow (-14/p)=1 \;\text{and}\; (x^2+1)^2=8 \mod p \; \text{has an integer solution.} $$</span>
Is this imply that there are infinitely many primes of the form <span class="math-container">$x^2+14y^2$</span>? It is easy to see that there are infinitely many primes that the equation <span class="math-container">$(x^2+1)^2=8 \mod p$</span> has an integer solution but I don't have any clue to check if some of them can take <span class="math-container">$-14$</span> as their quadratic residue.</p>
| franz lemmermeyer | 23,365 | <p>According to a theorem due to Dirichlet (proved later by various mathematicians such as A. Meyer and H. Weber), every primitive binary quadratic form represents infinitely many prime numbers.</p>
<p>In the case of <span class="math-container">$x^2 + 14y^2$</span>, this follows from the observation above in addition with the fact that there are infinitely many primes that split completely in a given number field (a very special case of the standard density theorems in algebraic number theory). In your case, the number field is the Hilbert class field of <span class="math-container">$k = {\mathbb Q}(\sqrt{-14})$</span> defined by the roots of <span class="math-container">$(x^2+1)^2 = 8$</span> over <span class="math-container">$k$</span>.</p>
|
1,048,260 | <p>I'm trying to read algebraic geometry on my own by doing homeworks on course hold in 2003. One of the problem is the following:</p>
<p>Let $k$ be a field, $S=k[T_0,\ldots,T_r]$, $\mathbb{P}=\mathbb{P}_k^r=\operatorname{Proj}(S)$ and $\mathcal O$ the structure sheaf of $\mathbb{P}$. Let $r=3$ and let $f\in S_d,g\in S_e$ be two relatively prime homogeneous forms. Let $X=V_+(f)\cap V_+(g)$. Compute the characteristic $\chi (O_X)$.</p>
<p>But what does the symbols $S_d,S_e$ means and how to do the problem?</p>
| jflipp | 187,123 | <p>$S_d \subseteq S$ is the space of homogeneous polynomials of degree $d,$ i.e. the $k$-span of all monomials $T_0^{a_0}\cdots T_r^{a_r}$ with $a_0,\ldots,a_r \in \mathbb N_0$ and $a_0 +\cdots+a_r =d.$</p>
|
1,048,260 | <p>I'm trying to read algebraic geometry on my own by doing homeworks on course hold in 2003. One of the problem is the following:</p>
<p>Let $k$ be a field, $S=k[T_0,\ldots,T_r]$, $\mathbb{P}=\mathbb{P}_k^r=\operatorname{Proj}(S)$ and $\mathcal O$ the structure sheaf of $\mathbb{P}$. Let $r=3$ and let $f\in S_d,g\in S_e$ be two relatively prime homogeneous forms. Let $X=V_+(f)\cap V_+(g)$. Compute the characteristic $\chi (O_X)$.</p>
<p>But what does the symbols $S_d,S_e$ means and how to do the problem?</p>
| Georges Elencwajg | 3,217 | <p>Now that the the meaning of the notations is clarified, here is a solution to the actual problem.<br>
We have a short exact sequence of coherent sheaves on $P=\mathbb P^3_k$: </p>
<p>$$0\to\mathcal O_P (-d-e)\to \mathcal O_P (-d)\oplus\mathcal O_P (-e) \to \mathcal O_P\to\mathcal O_Z\to 0$$ The first non trivial map is $h\mapsto (hg,-hf)$ and the second is $(u,v)\mapsto uf+vg$<br>
Exactness at $\mathcal O_P (-d)\oplus\mathcal O_P (-e)$ results from $f,g$ being relatively prime. </p>
<p>Taking Euler characteristic, which is additive in short exact sequences, we get: $$\chi(Z, \mathcal O_Z) =\chi(P,\mathcal O_P (-d-e))- \chi(P,\mathcal O_P (-d))-\chi(P,\mathcal O_P (-e))+\chi(P,\mathcal O_P )$$
Finally, remembering that $\chi(P,\mathcal O_P (k))=\binom {k+3}{3}$, we obtain the desired formula $$\chi(Z, \mathcal O_Z) = \binom {-d-e+3}{3}-\binom {-d+3}{3}-\binom {-e+3}{3}+1$$ </p>
|
3,645,742 | <p>The number <span class="math-container">$2012 \cdot 2013 \cdot 2014 + 2013$</span> is the cube of </p>
<p>a) <span class="math-container">$2012$</span><br>
b) <span class="math-container">$2013$</span><br>
c) <span class="math-container">$2014$</span><br>
d) <span class="math-container">$2112$</span><br>
e) <span class="math-container">$2113$</span></p>
| Stackman | 595,519 | <p>As suggested by Andrew Chin in the comments to the question, let <span class="math-container">$n=2013,$</span> so that <span class="math-container">$2012=n-1$</span> and <span class="math-container">$2014=n+1$</span>. Then writing the given quantity in terms of <span class="math-container">$n$</span> we see that
<span class="math-container">$$(n-1)n(n+1)+n=n[(n-1)(n+1)+1]=n[n^2-1+1]=n^3.$$</span>
Therefore the answer is <span class="math-container">$2013.$</span><br>
Note: above we used the formula <span class="math-container">$a^2-b^2=(a-b)(a+b).$</span></p>
|
3,827,422 | <p>Given a finite group <span class="math-container">$G$</span>, how do we know that there exists a map <span class="math-container">$\rho: G \rightarrow GL(V)$</span> such that <span class="math-container">$\rho(g_1\circ g_2) = \rho(g_1).\rho(g_2)$</span> for any <span class="math-container">$g_1, g_2\in G$</span>?</p>
<p>Intuitively, why does matrix multiplication always capture the properties of a group?</p>
| Alekos Robotis | 252,284 | <p>Let <span class="math-container">$G$</span> be any finite group. We can form a free vector space on the group <span class="math-container">$G$</span> by <span class="math-container">$k[G]$</span>. That is, we define a basis <span class="math-container">$\{e_g\}_{g\in G}$</span>. Now, we can define a linear action of <span class="math-container">$G$</span> on <span class="math-container">$k[G]$</span> by left translation. That is, let <span class="math-container">$g\in G$</span> act on the left by <span class="math-container">$g\cdot e_h=e_{gh}$</span> for all <span class="math-container">$g,h\in G$</span>. We extend this by linearity to a linear map <span class="math-container">$T_g:k[G]\to k[G]$</span>. It is not hard to see that <span class="math-container">$g\mapsto T_g$</span> defines a group homomorphism <span class="math-container">$G\to \text{GL}(k[G])$</span>.</p>
<p>Indeed, <span class="math-container">$T_g\circ T_{g'}(e_h)=e_{gg'h}=T_{gg'}(e_h).$</span> This formula also shows that <span class="math-container">$T_{g^{-1}}=T_g^{-1}$</span> so that the <span class="math-container">$T_g$</span> are linear isomorphisms.</p>
|
3,306,209 | <p>I was given this problem and told to find it's derivative: <span class="math-container">$$f(x)=\int^{x^2}_1 t^2+t+1\ dt$$</span>I thought the derivative was simply the inside function in terms of x - <span class="math-container">$x^2+x+1$</span>. But, this is not correct. Where am I going wrong? What piece am I missing? </p>
| gt6989b | 16,192 | <p>If you think of
<span class="math-container">$$
F(t) = \int \left(t^2+t+1\right)dt
$$</span>
then
<span class="math-container">$$
f(x) = F\left(x^2\right) - F(1),
$$</span>
therefore
<span class="math-container">$$
f'(x)
= \frac{d}{dx} F\left(x^2\right)
= f\left(x^2\right) \frac{d\left[x^2\right]}{dx}
$$</span></p>
<p>Can you finish this now?</p>
|
57,508 | <p>While learning commutative algebra and basic algebraic geometry and trying to understand the structure of results (i.e. what should be proven first and what next) I came to the following question: </p>
<p>Is it possible to prove that $\mathbb A^2-point$ is not an affine variety, if you don't know that the polynomial ring is a unique factorisation domain?</p>
<p>It seems to me, that this question has some meaning, since when we define affine variety, we don't need to use the fact that the polynomial ring is an UFD. Don't we?</p>
| Guillermo Mantilla | 2,089 | <p>I'd say yes. I'll work over $\mathbb{C}$ to make my life easier. First it is enough to show that the ring of regular functions on $Y:=\mathbb{A}^2(\mathbb{C})−0$ is isomorphic $\mathbb{C}[x,y]$. If Y were affine, the identity map from $\mathbb{C}[x,y]$ to itself would induce an isomorphism between $Y$ and $X:=\mathbb{C}^{2}$ which is impossible.(Polynomial maps are continuous in the usual topology, but $X$ is simply connected and $Y$ is not.)</p>
<p>Now let $Y_{1}:=\mathbb{A}^2(\mathbb{C}) \setminus \{x=0\}$ and $Y_{2}:=\mathbb{A}^2(\mathbb{C}) \setminus \{y=0\}$. Then $\mathcal{O}(Y)=\mathcal{O}(Y_1)\cap \mathcal{O}(Y_2)=\mathbb{C}[x,y]_{(x)} \cap \mathbb{C}[x,y]_{(y)}$.</p>
<p>On the other hand $\mathbb{C}[x,y]_{(x)} \cap \mathbb{C}[x,y]_{(y)}=\mathbb{C}[x,y]$. At first I thought that one needs unique factorization for the last equality, but the only thing one needs is that $\displaystyle (x^{i}y^{j})_{i,j \in \mathbb{N}}$ is a linearly independent subset of $\mathbb{C}[x,y]$. So my point is that $\left( x^{m}y^{n}=x^{p}y^{q} \rightarrow (m,n)=(p,q)\right)$ follows if we know the ring is a UFD, but it also follows trivially from linearly independence. </p>
|
1,106,320 | <p>I know the chance of hitting either red or black on the first roll is about 48%
1/37 chance on European 0
But shouldn't the chance of getting it right on either the first or second roll be high</p>
<p>like 66% or 75%</p>
<p>so to bet 25 1st roll and 50 second roll should have about 75% chance of a win?</p>
| 123 | 172,170 | <p>Things to consider: </p>
<ul>
<li><p>Each spin of the roulette wheel is an independent even. It is a common fallacy for a gambler to reason in the following manner: Because the ball has landed black three times consecutively, I should now bet red since the odds of the ball landing on black a fourth time are very small. This is false.</p></li>
<li><p>You could consider a geometric distribution if you want to calculate the chance of the $nth$ spin of the wheel producing your first success. Rather than reproduce every bit of information on this distribution, I will provide <a href="http://en.wikipedia.org/wiki/Geometric_distribution" rel="nofollow">this link</a></p></li>
<li><p>You can use a binomial distribution if you are interested in examining the probability of winning $n$ times at the roulette table in $m$ spins. <a href="http://en.wikipedia.org/wiki/Binomial_distribution" rel="nofollow">Here is information</a> on the binomial distribution. </p></li>
</ul>
|
2,439,069 | <p>$$\int_0^{\infty}\frac{dx}{2+\cosh (x)}$$</p>
<p>I'm not sure how to approach this problem. I tried substituting $\cosh (x) = \frac{e^x + e^{-x}}{2}$, and following the substitution $y = e^x$, I ended up with the integral </p>
<p>$$2 \cdot \int_1^{\infty} \frac{1}{(y+2)^2-3} dy$$</p>
<p>Which I wasn't able to evaluate, nor am I sure it is the best form to proceed. </p>
<p>Thanks!</p>
| Angina Seng | 436,618 | <p>Using your substitution I get
$$2\int_1^\infty\frac{dy}{y^2+4y+1}=
2\int_3^\infty\frac{du}{u^2-3}.$$
This can be done by partial fractions.</p>
|
2,795,230 | <blockquote>
<p>Let $V$ be a vector space of continuous functions on $[-1,1]$ over $\Bbb R$. Let $u_1,u_2,u_3,u_4\in V$ defined as $u_1(x)=x,u_2(x)=|x|,u_3(x)=x^2,u_4(x)=x|x|$ then </p>
<ol>
<li>$\{u_1,u_2\}$ is linearly dependent.</li>
<li>$\{u_1,u_3,u_4\}$ is linearly dependent.</li>
<li>$\{u_1,u_2,u_4\}$ is linearly dependent.</li>
<li>None of the above.</li>
</ol>
</blockquote>
<p>I think none of the above is correct: Plugging $x=1$ in $ax=|x|$ gives us $a=1$, but then for $x=-1$ we don't get $ax=|x|$. So, (1) fails. </p>
<p>Plugging $x=1,-1$ in $x=ax^2+bx|x|$ gives $a=0,b=1$, but for $x=\frac12$, $x=ax^2+bx|x|$ fails.</p>
<p>Plugging $x=1,\frac12$ in $x=a|x|+bx|x|$ gives $a=1,b=0$, but for $x=-\frac12$, $x=a|x|+bx|x|$ fails.</p>
<p>so, none of the above given set is linearly dependent, right?</p>
| Adriano | 76,987 | <p>Correct. More generally, the four functions form a linearly independent set, because if:
$$
ax + b|x| + cx^2 + dx|x| = 0
$$</p>
<p>then it can be shown that $a = b = c = d = 0$. Just evaluate the equation at $x = \pm 1, \pm \frac{1}{2}$ and solve the resulting system of four equations in four unknowns.</p>
|
2,795,230 | <blockquote>
<p>Let $V$ be a vector space of continuous functions on $[-1,1]$ over $\Bbb R$. Let $u_1,u_2,u_3,u_4\in V$ defined as $u_1(x)=x,u_2(x)=|x|,u_3(x)=x^2,u_4(x)=x|x|$ then </p>
<ol>
<li>$\{u_1,u_2\}$ is linearly dependent.</li>
<li>$\{u_1,u_3,u_4\}$ is linearly dependent.</li>
<li>$\{u_1,u_2,u_4\}$ is linearly dependent.</li>
<li>None of the above.</li>
</ol>
</blockquote>
<p>I think none of the above is correct: Plugging $x=1$ in $ax=|x|$ gives us $a=1$, but then for $x=-1$ we don't get $ax=|x|$. So, (1) fails. </p>
<p>Plugging $x=1,-1$ in $x=ax^2+bx|x|$ gives $a=0,b=1$, but for $x=\frac12$, $x=ax^2+bx|x|$ fails.</p>
<p>Plugging $x=1,\frac12$ in $x=a|x|+bx|x|$ gives $a=1,b=0$, but for $x=-\frac12$, $x=a|x|+bx|x|$ fails.</p>
<p>so, none of the above given set is linearly dependent, right?</p>
| copper.hat | 27,978 | <p>Suppose $f(x) = \sum_k \alpha_k u_k(x) = 0$.</p>
<p>Note that $f$ is smooth for $x \neq 0$.</p>
<p>Note that $f''(x) = 2 \alpha_3 u_3''(x) + 2 \alpha_4 u_4''(x)$. Evaluating at $x=\pm 1$
shows that $\alpha_3 = \alpha_4 = 0$.</p>
<p>Then $f'(x) = \alpha_1 u_1'(x) + \alpha_2 u_2'(x)$, evaluating at $x= \pm 1$ shows that $\alpha_1 = \alpha_2 = 0$.</p>
<p>Hence $u_1,...,u_4$ are linearly independent.</p>
|
2,061,695 | <p>I'm examining function slope and determine relative extrema of function (local minimum and local maximum)</p>
<p>The example is as following:
$
y = x^8 e^{-5x}
$</p>
<p>To do this I have to determine derivative of this function, which is:
$$
y' = 8x^7e^{-5x}+x^8(-5e^{-5x}) = -e^{-5x}x^7(5x-8)
$$</p>
<p>Then according to my notes I need to solve equation from derivative:
$
-e^{-5x}x^7(5x-8)=0
$</p>
<p>How to solve this equation?</p>
| Ross Millikan | 1,827 | <p>When you have a product that equals zero, at least one of the factors has to equal zero. $e^{-5x}$ is never zero, so you can divide by it. You are left with $x^7(5x-8)=0$ Now take each factor and set it to zero, which you should be able to solve. The solution is the union of these.</p>
|
2,061,695 | <p>I'm examining function slope and determine relative extrema of function (local minimum and local maximum)</p>
<p>The example is as following:
$
y = x^8 e^{-5x}
$</p>
<p>To do this I have to determine derivative of this function, which is:
$$
y' = 8x^7e^{-5x}+x^8(-5e^{-5x}) = -e^{-5x}x^7(5x-8)
$$</p>
<p>Then according to my notes I need to solve equation from derivative:
$
-e^{-5x}x^7(5x-8)=0
$</p>
<p>How to solve this equation?</p>
| Simply Beautiful Art | 272,831 | <p>Well, we have three cases</p>
<p>$$\begin{cases}0=-e^{-5x}\\0=x^7\\0=5x-8\end{cases}$$</p>
<p>Clearly, the first case is not possible. For the second case, the solution is $x=0$. For the last case, the solution is $x=\frac85$.</p>
|
2,737,830 | <p>For some fixed b, r > 0, place b blue balls and r red balls in a box. Select balls from the box, one after the other, without replacement. Prove that the probability that the kth ball taken from the box is blue is the same for 1 ≤ k ≤ b.</p>
<p>Does this just come from the fact that if you do not know what you are pulling out, the probability remains the same? Like if I had a deck of cards face down in a line, the probability that any kth card is an Ace is 4/52. How do you create a proof on this?</p>
| Hisoka Moroh | 459,561 | <p>We have $b$ blue balls and $r$ red balls, making for $b+r$ total balls in the box. Let $B_k$ be the event the $k^{th}$ ball drawn is blue.</p>
<p>Clearly, $P(B_1) = \frac{b}{b+r}$, because of the total number of balls in the box, we have $b$ blue balls to chose from. </p>
<p>What is $P(B_2)$? This requires a little more thought. Since we are drawing as second ball, this means we have already drawn a ball from the box, meaning we are now drawing from a box of $b+r - 1$ balls. Also, we must consider the cases where the already drawn ball is either blue or red.</p>
<p>In the first case, there are $b-1$ balls to choose from out of the $b+r-1$ balls. This case happens with $\frac{b}{b+r}$ probability. In the other case, we drew red and there are still $b$ balls to choose from. This case happens with $\frac{r}{b+r}$ probability. So, we have:</p>
<p>$P(B_2) = \frac{b}{b+r}\cdot\frac{b-1}{b+r-1} + \frac{r}{b+r}\frac{b}{b+r-1}$
$=\frac{b(b+r-1)}{(b+r)(b+r-1)} = \frac{b}{b+r}$</p>
<p>Using this information, can you figure out a way we can frame the expression for the general case of $P(B_k)$? (Hint: conditional probabilities, like we used for the case of $B_2$).</p>
|
176,600 | <p>The question is about factoring extremely large integers but you can have a look at <a href="https://stackoverflow.com/questions/11699464/mathematically-navigating-a-large-2d-numeric-grid-in-c-sharp">this question</a> to see the context if it helps. Please note that I am not very familiar with mathematical notation so would appreciate a verbose description of equations.</p>
<p><strong>The Problem:</strong></p>
<p>The integer in question will ALWAYS be a power of N and will be known to us to calculate against. So let's assume N = 2 for example. That will give us a sequence of numbers like:</p>
<pre><code>2, 4, 8, 16... up to hundreds of thousands of digits.
</code></pre>
<p>I need to find all possible factors (odd, even, prime, etc.) as efficiently as possible.</p>
<p><strong>The Question:</strong></p>
<p>What is the solution and how could I understand this from mathematical and computational perspectives?</p>
<p><strong>EDIT:</strong></p>
<p>Does the fact that each number to be factored is a power of 2 help in eliminating any complexity or computational time?</p>
| MJD | 25,554 | <p>If the number in question is known to be a power of 2, you are just trying to find $n$ such that $N=2^n$. If the input number is in decimal notation, just count the digits, subtract 1, multiply by 3.3219, and round up. Then add 1 if the initial digit is 2 or 3, 2 if the initial digit is 4, 5, 6, or 7, and 3 if the initial digit is 8 or 9. </p>
<p>For example, suppose $N=1267650600228229401496703205376$. This has 31 digits; $30\cdot3.3219 = 99.657$, so $N=2^{100}$. Or take $N=$</p>
<pre><code>43699499387321412970609716695670835099367888141129535719972915195176
79444176163354392285807163181819981286546206512408458617685052043667
09906692902245553277900892247131030458103436298545516643924637451297
481464347472084863384057367177715867713536
</code></pre>
<p>which has 246 digits. $245\cdot3.3219 = 813.872383$; we round up to 814, add 2 because the first digit is 4, so this number is $2^{816}$.</p>
<p>The magic constant 3.3219 is actually $\log 10 / \log 2$. For input numbers in the hundreds of thousands of digits you will need a more accurate version, say 3.3219281.</p>
|
3,184,887 | <p>I want to divide a prize, say 1000, among an unknown number of contestants, as determined by their rank in a marathon race. I want the smallest portion to be above zero (so it is worthwhile to compete). So the higher the rank: the higher the prize.</p>
<p>PS, feel free to explain in detail as I am a math noob. Thanks.</p>
<p>PPS, I just made the marathon up to make the problem more familiar, but I guess real marathons actually are not even like that. My application is for a drag and drop stack of user interface elements, for which I want to have a gradient of values according to their sequence, above being higher value, below being lower value.</p>
<p>Update: this was simpler than I thought, and based on the comments, I came to a formula: </p>
<p>n = number of contestants<br>
P = prize : 1000<br>
r = current contestant rank<br>
p = unknown portion </p>
<p>p = P(r/(n(n+1)/2)) </p>
<p>But I would like to simplify this formula please if possible, cause I'll use it in programming</p>
<p>edit 2:
I got this far: P(r/(n(n+1)/2)) = 2Pr/(n(n+1))</p>
| MediaMaker | 663,415 | <p>Following my understanding of Chris Moorhead's comments, I'll try to answer this myself:</p>
<p>for 6 contestants</p>
<p>1+2+3+4+5+6</p>
<p>= 21</p>
<p>1/21 = .0476<br>
2/21 = .0952<br>
3/21 = .1428<br>
4/21 = .1905<br>
5/21 = .2381<br>
6/21 = .2857<br>
6/21+5/21+4/21+3/21+2/21+1/21 = 1
= sweet!</p>
<p>Then just multiply those decimal numbers against 1000 to get the portion of the prize for that contestant:</p>
<p>1000 * .0476 = 47.6<br>
1000 * .0952 = 95.2<br>
1000 * .1428 = 142.8<br>
1000 * .1905 = 190.5<br>
1000 * .2381 = 238.1<br>
1000 * .2857 = 285.7 </p>
<p>then check it adds up again:<br>
47.6 + 95.2 + 142.8 + 190.5 + 238.1 + 285.7 = 999.9<br>
close enough I think :)</p>
|
1,321,704 | <p>For example: $3^\sqrt5$ versus $5^\sqrt3$</p>
<p>I tried to write numbers as this:</p>
<p>$3^{5^{\frac{1}{2}}}$ and then as
$3^{\frac{1}{2}^5}$</p>
<p>But this method gives the wrong answer because $a^{(b^c)} \ne a^{bc}$</p>
| Fermat | 83,272 | <p>Hint: Consider the function $$f(x)={\ln x\over \sqrt x}$$ and see where it is increasing or decreasing.</p>
|
1,321,704 | <p>For example: $3^\sqrt5$ versus $5^\sqrt3$</p>
<p>I tried to write numbers as this:</p>
<p>$3^{5^{\frac{1}{2}}}$ and then as
$3^{\frac{1}{2}^5}$</p>
<p>But this method gives the wrong answer because $a^{(b^c)} \ne a^{bc}$</p>
| YawarRaza7349 | 156,463 | <p>user134824's approach, but with the $\sqrt{3}$ instead, is a bit nicer:</p>
<p>$$
\begin{align}
3^\sqrt{5} &\text{ vs. } 5^\sqrt{3}\\
(3^\sqrt{5})^\sqrt{3} &\text{ vs. } (5^\sqrt{3})^\sqrt{3}\\
3^{\sqrt{5}\sqrt{3}} &\text{ vs. } 5^{\sqrt{3}\sqrt{3}}\\
3^\sqrt{15} &\text{ vs. } 5^3\\
\end{align}
$$</p>
<p>Now notice that since $\sqrt{15} < 4$, $3^\sqrt{15} < 3^4$ (via monotonicity of $3^x$), and since $3^4 < 5^3$ ($81 < 125$), we can say $3^\sqrt{15}<5^3$ (via transitivity), and subsequently $3^\sqrt{5} < 5^\sqrt{3}$ (via monotonicity of $x^\dfrac{1}{\sqrt{3}}, x \ge 0$).</p>
|
540 | <p>A user has been using multiple unregistered accounts (with the same name and the same avatar). For reference, the accounts I've been able to find are 8280, 8281, 8284, 8289, 8294, 8308, 8309, 8310, 8311, 8312, 8313, 8314, 8315, 8316, 8317, 8318, and 8319, but I expect more to be added soon.</p>
<p>If these were registered accounts, then a moderator could send an e-mail to the user and explain the correct use of the site (which is to use only a single account). But since these are unregistered accounts, I do not know how the user can be contacted privately.</p>
<p>I've been leaving some comments on a few of the user's answers saying that multiple accounts should not be used, but I don't think my comments had any effect.</p>
<p>My questions are:</p>
<blockquote>
<p>Is this creation of at least 17 accounts (in the span of 8 days), presumably by the same user, an abuse of how the site is used? If so, then what can be done to stop it?</p>
</blockquote>
| Chris Cunningham | 11 | <p>I helped [guest] combine two non-answers into an answer on the question here: <a href="https://matheducators.stackexchange.com/questions/12924/how-can-we-focus-students-on-the-various-data-types-in-multivariable-calculus/12941#12941">How can we focus students on the various data types in multivariable calculus?</a> </p>
<p>Unfortunately [guest] now believes that it is reasonable to post two separate posts as [guest] and have someone friendly come by and combine the answers together, as seen here: <a href="https://matheducators.stackexchange.com/questions/12945/a-question-about-vector-analysis-problems">A question about Vector Analysis problems</a>.</p>
<p>To avoid further "if you give a mouse a cookie" style developments, I would recommend not being too much more helpful to this user, instead insisting on the creation of an account which would allow editing of answers and full participation in comment threads.</p>
|
746,751 | <p>I'm studying the theory of computation, and I know there are pumping lemmas for regular and context-free languages, but why not for recursively enumerable languages? Is there something about a Turing machine that would make a pumping lemma impossible? My guess is that no matter how you pump a string, a Turing machine can always recognize it, but I am uncertain.</p>
| ml0105 | 135,298 | <p>Suppose I have $L = \{ <M>,w :$ M is a TM that halts on w $\}$. This is clearly recursively enumerable. It's not regular though, nor is it context free. The pumping lemmas for regular and context free languages tell you that if you have a string in the language, building the string but maintaining the pattern will keep the new string within the language.</p>
<p>So if I have a FSA, I can simply keep looping or going back and forth between a couple states to build up a string of desired length. Regular languages are context free grammars, so the intuition is the same. A Turing Machine is more sophisticated in what it can accept. Keep in mind that programming a Turing Machine is like writing machine code (not assembly, but machine code). </p>
|
2,088,564 | <blockquote>
<p>Sum of binomial product $\displaystyle \sum^{n}_{r=0}\binom{p}{r}\binom{q}{r}\binom{n+r}{p+q}$</p>
</blockquote>
<p>Simplifying $\displaystyle \frac{p!}{r!\cdot (p-r)!} \cdot \frac{q!}{r!\cdot (q-r)!}\cdot \frac{(n+r)!}{(p+q)! \cdot (n+r-p-q)!}$.</p>
<p>Could some help me with this, thanks</p>
| G Cab | 317,234 | <p>The proposed identity can be derived (putting $r=s$) from this more general one
$$
\sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left( \matrix{
m - \left( {r - s} \right) \cr
k \cr} \right)\left( \matrix{
n + \left( {r - s} \right) \cr
n - k \cr} \right)\left( \matrix{
r + k \cr
m + n \cr} \right)} = \left( \matrix{
r \cr
m \cr} \right)\left( \matrix{
s \cr
n \cr} \right)\quad \quad \left| \matrix{
\,{\rm 0} \le {\rm integer \, }m,n \hfill \cr
\;{\rm real}\;r,s \hfill \cr} \right.
$$
reported in <em>"Concrete Mathematics" - pag. 171</em>, and which is demonstrated through the following passages:
$$
\begin{gathered}
\sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,n} \right)} {\left( \begin{gathered}
m - \left( {r - s} \right) \\
k \\
\end{gathered} \right)\left( \begin{gathered}
n + \left( {r - s} \right) \\
n - k \\
\end{gathered} \right)\left( \begin{gathered}
r + k \\
m + n \\
\end{gathered} \right)} = \hfill \\
= \sum\limits_{\begin{subarray}{l}
\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,n} \right) \\
\left( {0\, \leqslant } \right)\,j\,\left( { \leqslant \,n} \right)
\end{subarray}} {\left( \begin{gathered}
m - \left( {r - s} \right) \\
k \\
\end{gathered} \right)\left( \begin{gathered}
n + \left( {r - s} \right) \\
n - k \\
\end{gathered} \right)\left( \begin{gathered}
r \\
m + n - j \\
\end{gathered} \right)\left( \begin{gathered}
k \\
j \\
\end{gathered} \right)} = \hfill \\
= \sum\limits_{\begin{subarray}{l}
\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,n} \right) \\
\left( {0\, \leqslant } \right)\,j\,\left( { \leqslant \,m + n} \right)
\end{subarray}} {\left( \begin{gathered}
m - \left( {r - s} \right) \\
k \\
\end{gathered} \right)\left( \begin{gathered}
k \\
j \\
\end{gathered} \right)\left( \begin{gathered}
n + \left( {r - s} \right) \\
n - k \\
\end{gathered} \right)\left( \begin{gathered}
r \\
m + n - j \\
\end{gathered} \right)} = \hfill \\
= \sum\limits_{\begin{subarray}{l}
\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,n} \right) \\
\left( {0\, \leqslant } \right)\,j\,\left( { \leqslant \,m + n} \right)
\end{subarray}} {\left( \begin{gathered}
m - \left( {r - s} \right) \\
j \\
\end{gathered} \right)\left( \begin{gathered}
m - \left( {r - s} \right) - j \\
k - j \\
\end{gathered} \right)\left( \begin{gathered}
n + \left( {r - s} \right) \\
n - k \\
\end{gathered} \right)\left( \begin{gathered}
r \\
m + n - j \\
\end{gathered} \right)} = \hfill \\
= \sum\limits_{\left( {0\, \leqslant } \right)\,j\,\left( { \leqslant \,m + n} \right)} {\left( \begin{gathered}
m - \left( {r - s} \right) \\
j \\
\end{gathered} \right)\left( \begin{gathered}
m + n - j \\
n - j \\
\end{gathered} \right)\left( \begin{gathered}
r \\
m + n - j \\
\end{gathered} \right)} = \hfill \\
= \sum\limits_{\left( {0\, \leqslant } \right)\,j\,\left( { \leqslant \,m + n} \right)} {\left( \begin{gathered}
m - \left( {r - s} \right) \\
j \\
\end{gathered} \right)\left( \begin{gathered}
m + n - j \\
m \\
\end{gathered} \right)\left( \begin{gathered}
r \\
m + n - j \\
\end{gathered} \right)} = \hfill \\
= \sum\limits_{\left( {0\, \leqslant } \right)\,j\,\left( { \leqslant \,m + n} \right)} {\left( \begin{gathered}
m - \left( {r - s} \right) \\
j \\
\end{gathered} \right)\left( \begin{gathered}
r \\
m \\
\end{gathered} \right)\left( \begin{gathered}
r - m \\
n - j \\
\end{gathered} \right)} = \hfill \\
= \left( \begin{gathered}
r \\
m \\
\end{gathered} \right)\left( \begin{gathered}
s \\
n \\
\end{gathered} \right) \hfill \\
\end{gathered}
$$
consisting in: </p>
<ul>
<li>Vandermonde de-convolution</li>
<li>shift of the 4th binomial</li>
<li>"trinomial revision"</li>
<li>sum on $k$ by Vandermonde convolution</li>
<li>symmetry</li>
<li>"trinomial revision"</li>
<li>sum on $j$ by Vandermonde convolution</li>
</ul>
|
1,550,736 | <p>Consider Z a Normal (Gaussian) random variable with mean 0 and variance 1.<br>
It has density
$$f_Z(z)=\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}} \text{for all x real numbers}$$
We consider $X=2Z+1$. Write the CDF of X in terms of the one of Z and take the derivative to get that the density of X is
$$f_X(x)=\frac{1}{2\sqrt{2\pi}}e^{\frac{-(x-1)^2}{8}} \text{for all x real numbers}$$<br>
I know that I have to take the integral of $f_Z(z)$ in terms of X to get the cdf, I just do not know how to get it in terms of X. The bounds of the integral should be $-\infty$ to $\infty$ right?</p>
| Ian | 83,396 | <p>Hint: let $x \in \mathbb{R}^q$ and write it as a linear combination of right singular vectors of $A$. Notice that $Ax$ is a linear combination of the corresponding <em>left</em> singular vectors of $A$. But these are orthogonal, so you have the Pythagorean theorem, which makes it straightforward to solve the minimization problem. In particular you should find that if $x=\sum c_i v_i$ then $Ax=\sum c_i \sigma_i u_i$ so $\| Ax \|_2^2 = \sum c_i^2 \sigma_i^2$.</p>
|
159,495 | <p>I have a text file where each line is a data point in the form:</p>
<pre><code>[ -495.01172, -158.35966, 2705.0 ]
[ -489.15576, -127.229675, 2673.0 ]
[ -487.6918, -97.679855, 2665.0 ]
[ -487.32578, -68.4594, 2663.0 ]
[ -485.86182, -39.19415, 2655.0 ]
[ -485.3128, -10.12311, 2652.0 ]
[ -484.03183, 18.853745, 2645.0 ]
[ -482.75082, 47.677364, 2638.0 ]
[ -481.6528, 76.37677, 2632.0 ]
[ -481.6528, 105.184616, 2632.0 ]
...
</code></pre>
<p>Each line represents <code>[x,y,z]</code>. I need to 3D plot these, but I am getting errors. Below is what I've tried along with the resulting error.</p>
<pre><code>data = Import[
"C:\\Users\\user\\Desktop\\out.txt", "text"]
data2 = List[
StringReplace[
data, {"[" -> "{", "]" -> "}", "\n" -> ",", " " -> ""}]]
ListPointPlot3D[data2]
</code></pre>
<p>The first two lines run successfully. The last line returns:</p>
<pre><code>...{1161.5232,-887.44867,1677.0}} must be a valid array or a list of valid arrays >>
</code></pre>
| Jack LaVigne | 10,917 | <p>I copied the 10 lines in your question and pasted them into a text file called <strong><em>out.txt</em></strong>.</p>
<pre><code>data = Import["D:\\at_work\\mathematica\\stack_exchange\out.txt",
"text"]
</code></pre>
<p>The basic problem with the way you were working is that what appears to be numbers are strings and you have to convert them.</p>
<p>I did that as follows. First I split them into separate lines.</p>
<pre><code>dataN = StringSplit[data, EndOfLine]
</code></pre>
<p>This produces a list of strings of the form you indicated for each line, one per line.</p>
<p>Now I used <code>StringCases</code> on each of these lines converting the strings to numbers.</p>
<pre><code>data2 = Map[
StringCases[#,
"[" ~~ Whitespace ~~ x__ ~~ "," ~~ Whitespace ~~ y__ ~~ "," ~~
Whitespace ~~ z__ ~~ Whitespace ~~ "]" :>
Sequence[ToExpression[x], ToExpression[y], ToExpression[z]]] &,
dataN
]
(* {{-495.012, -158.36, 2705.}, {-489.156, -127.23,
2673.}, {-487.692, -97.6799, 2665.}, {-487.326, -68.4594,
2663.}, {-485.862, -39.1942, 2655.}, {-485.313, -10.1231,
2652.}, {-484.032, 18.8537, 2645.}, {-482.751, 47.6774,
2638.}, {-481.653, 76.3768, 2632.}, {-481.653, 105.185, 2632.}} *)
</code></pre>
<p>The output is a list of lists with three real numbers in each sub-list.</p>
<pre><code>ListPlot3D[data2]
</code></pre>
<p><img src="https://i.stack.imgur.com/bq7lm.png" alt="Mathematica graphics"></p>
|
50,406 | <p>This is a question about support of modules under extension of scalars.</p>
<p>Let $f \colon A \to B$ be a homomorphism of commutative rings (with unity), and let $M$ be a finitely generated $A$-module.
Recall that the <em>support</em> of $M$ is the set of prime ideals $\mathfrak{p}$ of $A$ such that the localization $M_{\mathfrak{p}}$
is nonzero.
Then
$\mathrm{Supp} _B(B \otimes_A M) = f^{*-1}(\mathrm{Supp}_A(M))$,
the set of prime ideals of $B$ whose contractions are in the support of $M$. </p>
<p>The $\subseteq$ containment is true for any $M$.
What's an obvious example of a non-finitely generated module where the other containment doesn't hold?</p>
| S.Hamid Hassanzadeh | 47,945 | <p>The question is amount to the statement that if <span class="math-container">$f:(A,\mathfrak{m})\to (B,\mathfrak{n})$</span> is a local homomorphism and <span class="math-container">$M$</span> is a non-zero <span class="math-container">$A$</span>-module then <span class="math-container">$M\otimes _AB$</span> is also non-zero. Notice that this statement is true if <span class="math-container">$M$</span> as an <span class="math-container">$A$</span>-module possesses a minimal generating set. A proof would be as follows:
In a presentation of <span class="math-container">$M$</span> over <span class="math-container">$A$</span> the entries of the representing matrix belong to <span class="math-container">$\mathfrak{m}$</span>. After tensoring into <span class="math-container">$B$</span> over <span class="math-container">$A$</span>, the entries go to <span class="math-container">$\mathfrak{n}$</span>; so that the cokernel of the new matrix (which is isomorphic to <span class="math-container">$M\otimes _AB$</span>) won't be zero. </p>
<p>In the example, by Angelo, <span class="math-container">$\mathbb{Q}$</span> does not have a minimal generating set as a <span class="math-container">$\mathbb{Z}$</span>-module.</p>
|
85,651 | <p>Let $\Gamma$ be one of the classical congruence subgroups $\Gamma_0(N)$, $\Gamma_1(N)$ and $\Gamma(N)$ of $SL(2, \mathbb{Z})$.</p>
<p>How does the lower bound for the length of primitive geodesics on $\Gamma \backslash \mathbb{H}$ depending on $N \rightarrow \infty$?</p>
<p>Any suggestions?</p>
| S. Carnahan | 121 | <p>I don't have an answer, but here is a suggestion. From the environs of exercise 20 in section 3.7 of Terras's <em>Harmonic analysis on symmetric spaces I</em>, we find that a primitive geodesic whose beginning and end are joined by the hyperbolic transformation $\gamma$ has length $\log N(\gamma)$, where $N(\gamma) = a^2$ and $a$ is a real number satisfying $|a| > 1$, such that the Jordan form of $\gamma$ is $\left(\begin{smallmatrix} a & 0 \\ 0 & 1/a \end{smallmatrix} \right)$. Now you just need to find how norms of hyperbolic elements grow with level.</p>
|
85,651 | <p>Let $\Gamma$ be one of the classical congruence subgroups $\Gamma_0(N)$, $\Gamma_1(N)$ and $\Gamma(N)$ of $SL(2, \mathbb{Z})$.</p>
<p>How does the lower bound for the length of primitive geodesics on $\Gamma \backslash \mathbb{H}$ depending on $N \rightarrow \infty$?</p>
<p>Any suggestions?</p>
| Igor Rivin | 11,142 | <p>For more general results and extensive references, see <a href="http://u.math.biu.ac.il/~vishne/publications/JDG-76-3-399-422.pdf" rel="nofollow">Katz, Schaps, Vishne, Logarithmic growth of systole of arithmetic riemann surfaces along congruence subgroups (JDG, 2007)</a></p>
|
2,475,054 | <p>I'm wondering if someone could give me some hints as to how to approach this question, or some theory to understand what it's actually asking:</p>
<p><a href="https://i.stack.imgur.com/FXvsz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FXvsz.png" alt="enter image description here"></a></p>
<p>Thank you!</p>
| Robert Z | 299,698 | <p>The tangent plane at $(x_0,y_0,z_0)$ to a surface given in an implicit equation $F(x,y,z)=0$ is given by
$$F_x(x_0,y_0,z_0)(x-x_0)+F_y(x_0,y_0,z_0)(y-y_0)+F_z(x_0,y_0,z_0)(z-z_0)=0.$$
Can you take it from here? How do you recognize an horizontal plane?</p>
|
575,069 | <p>Let $\mathbb R_n[x]$ the vector space of polynomials with degree less or equal $n$ and we consider the linear transformation $f$ defined by
$$\forall P\in \mathbb R_n[x]\quad f(P)=(x^2-1)P''+2xP'$$
I proved that $f$ has the spectrum
$$\mathrm{sp}(f)=\{k(k+1),\ k=0,\ldots,n\}$$
I'm stuck in this question: Prove that there's a unique basis $(P_0,\ldots,P_n)$ of $\mathbb R_n[x]$ such that:
$$\forall k=0,\ldots,n\quad P_k \ \text{is a monic polynomial with degree }\ k\ \text{which's an eigenvector of }\ f $$
Any help would be appreciated.</p>
| Zhoe | 99,231 | <p>We know $$n^2 > (n - 1)^2$$ </p>
<p>Inverting the fractions the sign changes and we get </p>
<p>$$n^2 > (n - 1)^2 \implies \frac{1}{n^2} < \frac{1}{(n - 1)^2}$$</p>
<p>Take $b_n = \frac{1}{(n - 1)^2}$</p>
<p>Hence $a_n < b_n$ is satisfied.</p>
<p>P-series are of the form $$f(x) = \frac{1}{x^p}$$</p>
<p>And when $p > 1$ the series converges; $p \le 1$ the series diverges.</p>
<p>Since $b_n$ is a p-series with $p > 1 (= 2)$, $b_n$ converges.</p>
<p>Since $b_n$ converges and $b_n > a_n$; $a_n$ converges as well.</p>
|
58,710 | <p>I've got a bunch of readings at various dates and times. I'd like to be able to interpolate and then integrate over, say days. I can do this by converting my date/time to <code>AbosluteTime</code> and get the interpolating function that way. Then by integrating and plugging in seconds as my limits of integration I get what I want.</p>
<p>Also, since seconds in really finer than I need, I convert all of the <code>AbsoluteTimes</code> to the number of hours since the first reading.</p>
<p>I was wondering if there might be a way to do this directly using dates. For what it's worth, my readings last for about 2 months and the values are between about 90 and 300.</p>
<p><strong>SAMPLE DATA</strong></p>
<pre><code> x[[1 ;; 5]]
</code></pre>
<blockquote>
<p>{{{2014, 8, 4, 10, 36, 0.}, 257.},{{2014, 8, 4, 16, 28, 0.}, 385.}, </p>
<p>{{2014, 8, 4, 22, 53, 0.}, 176.},
{{2014, 8, 5, 6, 52, 0.}, 148.}, {{2014, 8, 5, 11, 19, 0.}, 192.}}</p>
</blockquote>
| kale | 1,560 | <p>There's a couple ways to do this:</p>
<pre><code>data = {{{2014, 8, 4, 10, 36, 0.}, 257.}, {{2014, 8, 4, 16, 28, 0.}, 385.},
{{2014, 8, 4, 22, 53, 0.}, 176.}, {{2014, 8, 5, 6, 52, 0.}, 148.},
{{2014, 8, 5, 11, 19, 0.}, 192.}};
</code></pre>
<p>1) Convert dates to absolute times:</p>
<pre><code>data[[All, 1]] = AbsoluteTime /@ data[[All, 1]];
f1 = Interpolation@data;
f1[AbsoluteTime@{2014,8,4,10,40}]
</code></pre>
<blockquote>
<p><code>261.669</code></p>
</blockquote>
<p>2) With <code>TimeSeries</code> (v10)</p>
<pre><code>ts = TimeSeries[data];
ts[{2014, 8, 4, 10, 40}]
</code></pre>
<blockquote>
<p><code>258.455</code></p>
</blockquote>
<p>The <code>TimeSeries</code> interpolation uses first order by default but can be changed by changing the <code>ResamplingMethod -> {"Interpolation", opts}</code> option in the <code>TimeSeries</code> function.</p>
<pre><code>ts = TimeSeries[data, ResamplingMethod -> {"Interpolation", InterpolationOrder -> 3}];
ts[{2014, 8, 4, 10, 40}]
</code></pre>
<blockquote>
<p><code>261.669</code></p>
</blockquote>
<p>To integrate over:</p>
<pre><code>NIntegrate[ts[t], {t, AbsoluteTime[{2014, 8, 4, 10, 40}], AbsoluteTime[{2014, 8, 4,11}]}]
</code></pre>
|
1,872,369 | <p>Prove that the equation $z^{3}e^{z}=1$ has infinitely many complex solutions.How many of them are real?</p>
<p>Use the argument principle,I choose a disk centered at $0$ with radius $R$ and get $\int_{\partial D}\dfrac {3z^{2}e^{z}+z^{3}e^{z}}{z^{3}e^{z}-1}dz$,and I don't how to do with this integral.Additionally,I use the monotonicity when z $z$ takes value in real number to find that there could be one real solution.</p>
| Zach Blumenstein | 199,290 | <p>Set $f(z)=z^3e^z-1$, and let $g(z)$ be the meromorphic function defined by $g(z)=f(1/z)$. It is easy to check, by looking at a power series expansion, that $g$ has an essential singularity at $0$. So by Picard’s Theorem, for every neighborhood $U$ of $0$, there exists $w \in \mathbb{C}$ so that $g(U \setminus \{0\}) = \mathbb{C} \setminus \{w\}$. Now, for each $n > 0$, set $U_n = \{0 < |z| < 1/n\}$. We know $-1 \notin g(U_n)$, since $g(z)=-1$ implies $f(1/z)+1=0$ and thus $1/z=0$, an impossibility. So we must have $g(U_n) = \mathbb{C} \setminus \{-1\}$. In particular, there exists $z_n$ with $g(z_n) = 0$. We then have $\{1/z_n\}_{n \in \mathbb{N}}$ providing a set of solutions, and it is necessarily infinite, since $|1/z_n| > n$ for each $n$. (I apologize if Picard’s Theorem is too heavy-duty for what you’re looking for. I’m not sure how to do it any other way.)</p>
<p>Now consider when $z=x \in \mathbb{R}$. If $x \leq 0$, then $x^3e^x \leq 0$, so $f(x)$ is negative. On the other hand, $f'(x) = (x^3+3x^2)e^x$, so $f'(x) > 0$ for all $x > 0$. This means $f$ crosses the $x$-axis at most once. And, indeed, it crosses at least once, since $f(0)=-1$ and $f(1)>0$. So $f$ has a zero in $(0,\infty)$, hence one real zero overall.</p>
|
505,067 | <p>I am writing a computer program which generate 5 digit alpha-numeric codes. Each character can contain a-z,A-Z and 0-9 (62 different possibility per character).</p>
<p>So if there are 5 character how many possibility are there?</p>
| Shobhit | 79,894 | <p>$(\frac{p}q)^n$ cannot be an integer for $p,q$ and $n$ integers and $n\geq 1$ and $p$ is not divisible by $q$, as in your case. For if $p$ is not divisible by $q$, then so is the case for $p^n$ and $q^n$, hence it cannot be an integer for if you want it to be an integer the denominator must be $\pm 1$, or $q$ must divide $p$. Also it trivially holds for $n=0$. Also for $n<0$ and $p$ does not divide $q$, then $(\frac{p}q)^n$ will be an integer for $p=\pm 1$ and any integer $q \neq 0 $. </p>
|
11,172 | <p>As a TA who led calculus* 1 and 2 discussion section and holds office hour** in the previous year, I heard the following (wrong) arguments several times.</p>
<blockquote>
<ol>
<li><p><span class="math-container">$\displaystyle \lim_{x\to \infty} \sqrt{x+1}-\sqrt{x}=0$</span> because <span class="math-container">$\infty-\infty=0$</span>.</p>
</li>
<li><p><span class="math-container">$\displaystyle \lim_{x\to \infty} x^{1/x}=1$</span> because <span class="math-container">$\infty^0=1$</span>.</p>
</li>
<li><p><span class="math-container">$\int_1^{\infty}f(x)dx$</span> and <span class="math-container">$\int_1^{\infty}g(x)dx$</span> both diverge so <span class="math-container">$\int_1^{\infty}f(x)+g(x)dx$</span> diverge.</p>
</li>
</ol>
</blockquote>
<p>I usually explain the arguments are not true in general by providing a (very trivial) counter example, for example,</p>
<blockquote>
<ol>
<li><p><span class="math-container">$\displaystyle \lim_{x\to \infty} f(x)=\infty$</span> and <span class="math-container">$\displaystyle \lim_{x\to \infty} g(x)=\infty$</span> does not guarantee <span class="math-container">$\displaystyle \lim_{x\to \infty} f(x)-g(x)=0$</span>, for example, <span class="math-container">$f(x)=x+1$</span> and <span class="math-container">$g(x)=x$</span>.</p>
</li>
<li><p><span class="math-container">$\displaystyle \lim_{x\to \infty} f(x)=\infty$</span> and <span class="math-container">$\displaystyle \lim_{x\to \infty} g(x)=0$</span> does not guarantee <span class="math-container">$\displaystyle \lim_{x\to \infty} f(x)^{g(x)}=1$</span>, for example, <span class="math-container">$f(x)=2^x$</span> and <span class="math-container">$g(x)=1/x$</span>.</p>
</li>
<li><p>False in general, for example <span class="math-container">$f(x)=-g(x)=1$</span></p>
</li>
</ol>
</blockquote>
<p>After giving explanations like that I sometime heard "But in your examples you can cancel the expression/formula..." and I was not sure how to continue. I tried the following methods, non of them seem to work very well.</p>
<p>a. Provide a much more complicated counter example which requires a few minutes of calculation to get the answer. This often leads to further confusion.</p>
<p>b. Just say that is the wrong way to do it. It sounds like "I'm the teacher so believe me." and doesn't do too much.</p>
<p>c. Show them the correct way to do their problems. This is almost like b (Why is your way the right way and mine is the wrong way?).</p>
<p>I'm looking for a better way to deal with questions like these.</p>
<p>*<span class="math-container">$\epsilon-\delta$</span> definition is not introduced.
** Office hour is in tutoring center where I'm also responsible for students take the class from the professors I'm not TA'ing for.</p>
| Math Misery | 6,898 | <p>Wanted to write this as a comment, but I'm not strong enough [reputation]. This may be a bit heavy handed or you're already doing it, but what about challenging the very notion of arithmetic when dealing with infinities before going towards counterexamples? That's usually what the fuss tends to be about.</p>
<p>As an aside, when teaching probability theory, the Cauchy distribution tends to flummox students because we run into problems when trying to compute the mean even though the distribution is symmetric. Why does there have to be a central tendency? It challenges a notion that they've held to be implicitly true. When they can't answer this general question of "why", it often helps to break the cognitive dissonance and they are more receptive to the idea that, yes, perhaps the mean doesn't have to exist.</p>
<p>Similarly, with limits involving infinities, I would ask students what rule says that $\infty - \infty = 0$ every time? Then, trivial examples as you give can help to seal the deal and disavow them of such a notion. Perhaps? Basically and I'm not sure if you already do this, but I would first challenge, philosophically, what students have considered to be axiomatically / dogmatically true before showing counterexamples.</p>
|
11,172 | <p>As a TA who led calculus* 1 and 2 discussion section and holds office hour** in the previous year, I heard the following (wrong) arguments several times.</p>
<blockquote>
<ol>
<li><p><span class="math-container">$\displaystyle \lim_{x\to \infty} \sqrt{x+1}-\sqrt{x}=0$</span> because <span class="math-container">$\infty-\infty=0$</span>.</p>
</li>
<li><p><span class="math-container">$\displaystyle \lim_{x\to \infty} x^{1/x}=1$</span> because <span class="math-container">$\infty^0=1$</span>.</p>
</li>
<li><p><span class="math-container">$\int_1^{\infty}f(x)dx$</span> and <span class="math-container">$\int_1^{\infty}g(x)dx$</span> both diverge so <span class="math-container">$\int_1^{\infty}f(x)+g(x)dx$</span> diverge.</p>
</li>
</ol>
</blockquote>
<p>I usually explain the arguments are not true in general by providing a (very trivial) counter example, for example,</p>
<blockquote>
<ol>
<li><p><span class="math-container">$\displaystyle \lim_{x\to \infty} f(x)=\infty$</span> and <span class="math-container">$\displaystyle \lim_{x\to \infty} g(x)=\infty$</span> does not guarantee <span class="math-container">$\displaystyle \lim_{x\to \infty} f(x)-g(x)=0$</span>, for example, <span class="math-container">$f(x)=x+1$</span> and <span class="math-container">$g(x)=x$</span>.</p>
</li>
<li><p><span class="math-container">$\displaystyle \lim_{x\to \infty} f(x)=\infty$</span> and <span class="math-container">$\displaystyle \lim_{x\to \infty} g(x)=0$</span> does not guarantee <span class="math-container">$\displaystyle \lim_{x\to \infty} f(x)^{g(x)}=1$</span>, for example, <span class="math-container">$f(x)=2^x$</span> and <span class="math-container">$g(x)=1/x$</span>.</p>
</li>
<li><p>False in general, for example <span class="math-container">$f(x)=-g(x)=1$</span></p>
</li>
</ol>
</blockquote>
<p>After giving explanations like that I sometime heard "But in your examples you can cancel the expression/formula..." and I was not sure how to continue. I tried the following methods, non of them seem to work very well.</p>
<p>a. Provide a much more complicated counter example which requires a few minutes of calculation to get the answer. This often leads to further confusion.</p>
<p>b. Just say that is the wrong way to do it. It sounds like "I'm the teacher so believe me." and doesn't do too much.</p>
<p>c. Show them the correct way to do their problems. This is almost like b (Why is your way the right way and mine is the wrong way?).</p>
<p>I'm looking for a better way to deal with questions like these.</p>
<p>*<span class="math-container">$\epsilon-\delta$</span> definition is not introduced.
** Office hour is in tutoring center where I'm also responsible for students take the class from the professors I'm not TA'ing for.</p>
| Henry Towsner | 62 | <p>This is a difficult problem, because the students you're dealing with aren't really prepared to think about logical reasoning and the role of counterexamples. (For example, I find students at that level often don't distinguish between examples and arguments.)</p>
<p>I'm of the view that you can't actually expect students to understand material at a level much deeper than it's being taught. If an algorithm works on all problems they'll see then, in a real sense, the algorithm is correct for purposes of the class.</p>
<p>As it happens, the examples you're describing are probably <em>not</em> correct, even "locally", because they're going to be expected to deal (if not immediately then later in the course) with problems for which that algorithm doesn't work, so you do need to correct these misconceptions.</p>
<p>My suggestion would be:</p>
<ol>
<li><p>Focus on giving counterexamples. It's important to avoid trivial problems. Students treat algorithms as heuristics, and have a hierarchy from easier to harder heuristics; you have to choose problems which trigger the faulty heuristic, and on which it fails. There's a natural supply of these: the problems which students will later have to solve using more sophisticated methods (L'Hospital's rule problems for a and b, other integrals for c). You don't need to work out the solution; you can say "Okay then what's the solution to this problem?" and then when they get the wrong answer, say "Actually, we're going to learn later that it's really ...". Instead of doing the calculation, you can draw (or even better, use a computer to generate) a graph of the function which supports your answer (don't present it as a proof, of course, just emphasize that it's suggestive, and the rigorous method will come later).</p></li>
<li><p>Another meaningful kind of counterexample, if you can use it, is problems where the shortcut doesn't apply. In other words, don't give a counterexample to its accuracy, give a counterexample to its usefulness. You don't want to say "this is right because I'm the teacher", but it's reasonable to say "we need this method to solve hard problems, so we're going to practice it on easy problems, even if there's a shortcut".</p></li>
</ol>
|
2,895,923 | <p>I am trying to solve this simple geometry problem but I am always tangled in so many equations it makes my head spin. I tried solving it via similar triangles but i cant seem to eliminate all the unwanted variables. Please help.</p>
<p>I have to prove $ r_1\times r_3=(r_2)^2$</p>
<p><a href="https://i.stack.imgur.com/5KQru.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5KQru.png" alt="Question Image"></a></p>
<p>Thank you</p>
| Community | -1 | <p>Due to the way they are constructed, the triangles $ABO,ACP,ADQ$ are similar, the figures $ABCPO$, $ACDQP$ are homothetic and $\dfrac{r_2}{r_1}=\dfrac{r_3}{r_2}.$</p>
|
3,497,554 | <blockquote>
<p>Show that <span class="math-container">$\frac{(2n-1)!}{(n)!(n-1)!}$</span> is odd or even according as <span class="math-container">$n$</span> is or is not a power of <span class="math-container">$2$</span>.</p>
</blockquote>
<p>I know that the index of the highest power of <span class="math-container">$2$</span> contained in <span class="math-container">$n!$</span> is <span class="math-container">$n-1$</span> when <span class="math-container">$n$</span> is a power of <span class="math-container">$2$</span> and <span class="math-container">$n-r$</span> when <span class="math-container">$n$</span> is equal to <span class="math-container">$2^r-1$</span>. </p>
<p>I've expanded the above to the result that the index of the highest power of <span class="math-container">$2$</span> contained in <span class="math-container">$n!$</span> is <span class="math-container">$2^r-1$</span> when <span class="math-container">$n$</span> is equal to <span class="math-container">$2^r+1$</span>. <em>(If you're wondering how I derived it, refer to the formula given in <a href="http://math.stackexchange.com/q/3497392/710663">this</a> question and correct me if I'm wrong)</em>. Using that, I've got that the highest power of <span class="math-container">$2$</span> in the term when <span class="math-container">$n$</span> is a power of <span class="math-container">$2$</span> is <span class="math-container">$\frac{2^{r+1}-(r+2)}{(2^r-1)(2^r-(r+1))}$</span>.</p>
<p>Now here, if <span class="math-container">$r$</span> is odd, then the numerator is odd, so the whole term is odd and everything is fine. But if <span class="math-container">$r$</span> is odd, then the numerator is even and the denominator is odd, so the term is even which contradicts the above statement.</p>
<p>Putting <span class="math-container">$n=2^r+1$</span> the term equals <span class="math-container">$\frac{(2^{r+1}+1)}{(2^r+1)(2^r)}$</span>. Finding the highest powers of <span class="math-container">$2$</span> I've got <span class="math-container">$\frac{(2^{r+1}-1)}{(2^r-1)(2^r-1)}$</span> which is odd.</p>
<p>I'm getting nearly opposite results.</p>
<p><strong>Am I doing something very wrong?</strong></p>
<p>Any help would be highly appreciated.</p>
| lulu | 252,071 | <p>This follows from <a href="https://en.wikipedia.org/wiki/Lucas%27s_theorem" rel="nofollow noreferrer">Lucas' Theorem</a>.</p>
<p>that theorem tells us that <span class="math-container">$\binom ab$</span> is divisible by a prime <span class="math-container">$p$</span> if and only if one of the base <span class="math-container">$p$</span> digits of <span class="math-container">$b$</span> exceeds the corresponding digit of <span class="math-container">$a$</span>.</p>
<p>If <span class="math-container">$n$</span> is not a power of <span class="math-container">$2$</span> then we can write <span class="math-container">$$n=2^{k_1}+\cdots + 2^{k_{r-1}}+2^{k_r}$$</span> where the <span class="math-container">$k_i$</span> are strictly decreasing and <span class="math-container">$r$</span> is at least <span class="math-container">$2$</span>. But then <span class="math-container">$$2n-1=2^{k_1+1}+\cdots + 2^{k_{r-1}+1}+2^{k_r+1}-1=2^{k_1+1}+\cdots + 2^{k_{r-1}+1}+2^{k_r}+\cdots +1$$</span></p>
<p>We observe that <span class="math-container">$2^{k_{r-1}}$</span> is present in the expansion of <span class="math-container">$n$</span> but not in the expansion of <span class="math-container">$2n-1$</span> so Lucas tells us that <span class="math-container">$2\,\big |\,\binom {2n-1}n$</span> when <span class="math-container">$n$</span> is not a power of <span class="math-container">$2$</span>.</p>
<p>The case where <span class="math-container">$n$</span> is a power of <span class="math-container">$2$</span> is similar and straight forward.</p>
|
3,497,554 | <blockquote>
<p>Show that <span class="math-container">$\frac{(2n-1)!}{(n)!(n-1)!}$</span> is odd or even according as <span class="math-container">$n$</span> is or is not a power of <span class="math-container">$2$</span>.</p>
</blockquote>
<p>I know that the index of the highest power of <span class="math-container">$2$</span> contained in <span class="math-container">$n!$</span> is <span class="math-container">$n-1$</span> when <span class="math-container">$n$</span> is a power of <span class="math-container">$2$</span> and <span class="math-container">$n-r$</span> when <span class="math-container">$n$</span> is equal to <span class="math-container">$2^r-1$</span>. </p>
<p>I've expanded the above to the result that the index of the highest power of <span class="math-container">$2$</span> contained in <span class="math-container">$n!$</span> is <span class="math-container">$2^r-1$</span> when <span class="math-container">$n$</span> is equal to <span class="math-container">$2^r+1$</span>. <em>(If you're wondering how I derived it, refer to the formula given in <a href="http://math.stackexchange.com/q/3497392/710663">this</a> question and correct me if I'm wrong)</em>. Using that, I've got that the highest power of <span class="math-container">$2$</span> in the term when <span class="math-container">$n$</span> is a power of <span class="math-container">$2$</span> is <span class="math-container">$\frac{2^{r+1}-(r+2)}{(2^r-1)(2^r-(r+1))}$</span>.</p>
<p>Now here, if <span class="math-container">$r$</span> is odd, then the numerator is odd, so the whole term is odd and everything is fine. But if <span class="math-container">$r$</span> is odd, then the numerator is even and the denominator is odd, so the term is even which contradicts the above statement.</p>
<p>Putting <span class="math-container">$n=2^r+1$</span> the term equals <span class="math-container">$\frac{(2^{r+1}+1)}{(2^r+1)(2^r)}$</span>. Finding the highest powers of <span class="math-container">$2$</span> I've got <span class="math-container">$\frac{(2^{r+1}-1)}{(2^r-1)(2^r-1)}$</span> which is odd.</p>
<p>I'm getting nearly opposite results.</p>
<p><strong>Am I doing something very wrong?</strong></p>
<p>Any help would be highly appreciated.</p>
| Nεo Pλατo | 719,444 | <p>So my breakthrough discovery is for when n is a power of 2.</p>
<p>Let's do some rewriting.</p>
<p><span class="math-container">$\displaystyle\dfrac{(2n-1)!}{n!(n-1)!}=\dfrac{\dfrac{(2n)!}{2n}}{n! \cdot \dfrac{n!}{n}}=\dfrac{(2n)!}{2 \cdot (n!)^2}!$</span></p>
<p>And I'll replace <span class="math-container">$n=2^k$</span> to yield</p>
<p><span class="math-container">$\dfrac{(2^{k+1})!}{2 \cdot ((2^k)!)^2}$</span></p>
<p>Now the best option is to deduce how many groups of 2 are in the numerator and denominator.</p>
<p>So for a number <span class="math-container">$\alpha!$</span>, half of the numbers are divisible by 2, half of which are divisible by 4 and so on.</p>
<p>We can make a similar pattern for ourselves.</p>
<p><span class="math-container">$2^{k+1}$</span> has <span class="math-container">$2^k$</span> numbers divisible by 2, <span class="math-container">$2^{k-1}$</span> divisible by 4 until only one number is divisible by <span class="math-container">$2^k$</span>. I hope this pattern is easy to follow since it's not easy to type MathJax on a phone.</p>
<p>At the top the number of groups of 2 are</p>
<p><span class="math-container">$2^{k+1}+2^k \ldots +1=(2^{k+1}-1)\text{groups}$</span></p>
<p>The pattern can be repeated for <span class="math-container">$2^k$</span> to get <span class="math-container">$(2^k-1)\text{groups}$</span>. However that denominator is guided by special conditions. There's a power of 2 which makes it double and a factor of 2 which adds 1 more power. This leads to </p>
<p><span class="math-container">$2(2^k-1)+1=(2^{k+1}-1)\text{groups}$</span></p>
<p>See that! The denominator and numerator have the same number of 2s. They therefore cancel out leaving no pairs for our poor fraction.</p>
<p>This thought process took about 24 hours. Peace.</p>
<p>P.S I can assume for the other case that there will always be a 2 somewhere to spare. I hope I can prove it sometime this week.</p>
|
3,274,461 | <p>Suppose I have a recurrence relation like : </p>
<p><span class="math-container">\begin{equation}
f(n)=\begin{cases}
2*f(n-1) , & \text{n%2 = 1}.\\
f(n-1) + 1, & \text{n%2 = 0}.
\end{cases}
\end{equation}</span></p>
<p><span class="math-container">\begin{equation}
f(0) = f(1) = 1
\end{equation}</span></p>
<p>How would I go about solving this? What do you call such a recurrence relation?</p>
<p>Is there a way I can get <span class="math-container">$f(n)$</span> that will solve it for both cases?</p>
<p>I tried using the WolframAlpha <a href="https://reference.wolfram.com/language/ref/RSolve.html?q=RSolve" rel="nofollow noreferrer">RSolve</a> function and tried to use <a href="https://reference.wolfram.com/language/ref/Mod.html" rel="nofollow noreferrer">Mod</a> to write a single equation hoping it could solve it. But it kept throwing a syntax error.</p>
<p>I am fine even if you provide a solution using WolframAlpha.</p>
| resound | 575,412 | <p>Refer this link : <a href="https://codeforces.com/blog/entry/53202" rel="nofollow noreferrer">Codeforces</a> and check Yeputons' second answer(the part where he breaks the solution). Has a nice way of breaking the function into two different functions. Maybe you can proceed with it.</p>
|
3,810,944 | <p>Consider a positive integer <span class="math-container">$n$</span> and the function
<span class="math-container">$f:\mathbb{N}\to \mathbb{N}$</span> (<span class="math-container">$\mathbb N$</span> includes <span class="math-container">$0$</span>) by</p>
<p><span class="math-container">$$f(x) = \begin{cases} \frac{x}{2}
& \text{if } x \text{ is even} \\ \frac{x-1}{2} + 2^{n-1} & \text{if } x \text{ is odd} \end{cases} $$</span>
Determine the set</p>
<p><span class="math-container">$$ A = \{ x\in \mathbb{N} \mid \underbrace{\left( f\circ f\circ ....\circ f \right)}_{n\ f\text{'s}}\left( x \right)=x \}. $$</span></p>
<p>(Romania NMO 2013)</p>
<p>The solution starts by stating that <span class="math-container">$f(x)<x, \quad\forall x\ge 2^n-1$</span>. This was easy enough to understand. However, they continue by saying this implies that <span class="math-container">$A\subset\{0,1,\dots,2^n-1\}$</span>. Why is that?</p>
<p>Please help me understand! Thanks in advance!</p>
| Tortar | 704,856 | <p>Stating <span class="math-container">$f(x) < x,\forall x > 2^n-1$</span> can be seen as the base step for an induction argument.</p>
<p>We can finish by induction proving :</p>
<p>if <span class="math-container">$\underbrace{\left( f\circ f\circ ....\circ f \right)}_{k\ f\text{'s}}\left( x \right) < x,\forall x > 2^n-1$</span> then <span class="math-container">$\underbrace{\left( f\circ f\circ ....\circ f \right)}_{k+1\ f\text{'s}}\left( x \right) <x,\forall x > 2^n-1$</span>.</p>
<p>First of all put <span class="math-container">$\underbrace{\left( f\circ f\circ ....\circ f \right)}_{k\ f\text{'s}}\left( x \right) = t_1$</span> and <span class="math-container">$\underbrace{\left( f\circ f\circ ....\circ f \right)}_{k+1\ f\text{'s}}\left( x \right) = t_2$</span> , now:</p>
<ul>
<li><p>if <span class="math-container">$t_1$</span> is even then <span class="math-container">$t_2 = \frac{t_1}{2} < t_1< x$</span>.</p>
</li>
<li><p>if <span class="math-container">$t_1$</span> is odd, then <span class="math-container">$t_2 = \frac{t_1-1}{2}+2^{n-1}$</span>, then we have to show:<span class="math-container">$$\frac{t_1-1}{2}+2^{n-1} < x \iff t_1 <x+x-(2^n-1)$$</span>but we know that <span class="math-container">$t_1<x$</span> and that <span class="math-container">$x-(2^n-1)>0$</span>, so <span class="math-container">$t_1 <x+x-(2^n-1)$</span> is true.</p>
</li>
</ul>
<p>So clearly <span class="math-container">$A$</span> is a subset of <span class="math-container">$\{1,2,...,2^n-1\}$</span>.</p>
|
3,810,944 | <p>Consider a positive integer <span class="math-container">$n$</span> and the function
<span class="math-container">$f:\mathbb{N}\to \mathbb{N}$</span> (<span class="math-container">$\mathbb N$</span> includes <span class="math-container">$0$</span>) by</p>
<p><span class="math-container">$$f(x) = \begin{cases} \frac{x}{2}
& \text{if } x \text{ is even} \\ \frac{x-1}{2} + 2^{n-1} & \text{if } x \text{ is odd} \end{cases} $$</span>
Determine the set</p>
<p><span class="math-container">$$ A = \{ x\in \mathbb{N} \mid \underbrace{\left( f\circ f\circ ....\circ f \right)}_{n\ f\text{'s}}\left( x \right)=x \}. $$</span></p>
<p>(Romania NMO 2013)</p>
<p>The solution starts by stating that <span class="math-container">$f(x)<x, \quad\forall x\ge 2^n-1$</span>. This was easy enough to understand. However, they continue by saying this implies that <span class="math-container">$A\subset\{0,1,\dots,2^n-1\}$</span>. Why is that?</p>
<p>Please help me understand! Thanks in advance!</p>
| Sebastian Spindler | 793,374 | <p>Note first that the inequality <span class="math-container">$f(x) < x$</span> only holds for <span class="math-container">$x \geq 2^n$</span> since <span class="math-container">$f(2^n - 1) = 2^n -1$</span>.</p>
<p>If there is an element <span class="math-container">$x \in A$</span> with <span class="math-container">$x \geq 2^n$</span>, then we can choose such an <span class="math-container">$x$</span> minimally, i.e. such that every <span class="math-container">$y \in A$</span> with <span class="math-container">$y < x$</span> satisfies <span class="math-container">$y \leq 2^n - 1$</span>. Now we have <span class="math-container">$f^n(x) = x$</span> and hence <span class="math-container">$$f^n(f(x)) = f^{n+1}(x) = f(f^n(x)) = f(x),$$</span> so <span class="math-container">$f(x) \in A$</span> and the minimality of <span class="math-container">$x$</span> together with <span class="math-container">$f(x) < x$</span> implies <span class="math-container">$f(x) \leq 2^n - 1$</span>.</p>
<p>Now check that this implies <span class="math-container">$f^k(f(x)) \leq 2^n-1$</span> for all <span class="math-container">$k \in \mathbb{N}$</span> (as hinted at in Daniel's comment) and in particular <span class="math-container">$x = f^n(x) = f^{n-1}(f(x)) \leq 2^n-1$</span>, contradicting the initial choice of <span class="math-container">$x$</span>.</p>
|
2,710,900 | <blockquote>
<p>Let $S=\{1,2t,-2+4t^2,-12t+8t^3\}$ be a set of polynomials in $P_3$.
Show that these polynomials make up a basis for $P_3$ and determine
the coordinates for $p=7-12t-8t^2+12t^3$ in this basis.</p>
</blockquote>
<p>The first part of the problem was easy, showing that they make up a basis by showing that the polynomials are linearly independant and since $\dim(P_3)=\dim(S)=4,$ they can span $P_3$. </p>
<p>But how do I determine the coordinates for $p=7-12t-8t^2+12t^3$ in this basis?</p>
| Arthur | 15,500 | <p>If you take a calculator and press <code>2^3^2</code>, whether you get $2^9$ or $2^6$ (or an error) will be entirely up to the manufacturer. There is no convention for this, and to human readers it is entirely ambiguous.</p>
<p>When writing, you should specify either $2^{3^2}$ or $(2^3)^2$, and when using a calculator, you ought to use parentheses and type either <code>2^(3^2)</code> or <code>(2^3)^2</code>. The same goes for <code>2^2^3</code>, of course.</p>
|
2,710,900 | <blockquote>
<p>Let $S=\{1,2t,-2+4t^2,-12t+8t^3\}$ be a set of polynomials in $P_3$.
Show that these polynomials make up a basis for $P_3$ and determine
the coordinates for $p=7-12t-8t^2+12t^3$ in this basis.</p>
</blockquote>
<p>The first part of the problem was easy, showing that they make up a basis by showing that the polynomials are linearly independant and since $\dim(P_3)=\dim(S)=4,$ they can span $P_3$. </p>
<p>But how do I determine the coordinates for $p=7-12t-8t^2+12t^3$ in this basis?</p>
| JMoravitz | 179,297 | <p>Be careful. Writing <code>2^2^3</code> can be thought of as ambiguous. It is standard to read power-towers from top down. That is to say, without parentheses, <code>a^b^c</code> should be interpreted as $a^{(b^c)}$, not as $(a^b)^c$</p>
<p>The property you think you are remembering is that $(a^b)^c=a^{b\times c}$, in which case yes, $(2^2)^3=(2^3)^2=2^6=64$</p>
<p><code>2^2^3</code> is actually to be interpreted as $2^{(2^3)}=2^8=256$</p>
<p>On the other hand, <code>2^3^2</code> is to be interpreted as $2^{(3^2)}=2^9=512$</p>
|
2,710,900 | <blockquote>
<p>Let $S=\{1,2t,-2+4t^2,-12t+8t^3\}$ be a set of polynomials in $P_3$.
Show that these polynomials make up a basis for $P_3$ and determine
the coordinates for $p=7-12t-8t^2+12t^3$ in this basis.</p>
</blockquote>
<p>The first part of the problem was easy, showing that they make up a basis by showing that the polynomials are linearly independant and since $\dim(P_3)=\dim(S)=4,$ they can span $P_3$. </p>
<p>But how do I determine the coordinates for $p=7-12t-8t^2+12t^3$ in this basis?</p>
| Rob Arthan | 23,171 | <p>Your calculator is correctly implementing the (hotly disputed) correct answers to <a href="https://math.stackexchange.com/questions/1633790/what-is-the-order-when-doing-xyz-and-why/1633800#1633800">What is the order when doing $x^{y^z}$ and why?</a></p>
|
2,764,447 | <p>In the following question:</p>
<p><a href="https://i.stack.imgur.com/Rgzjr.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Rgzjr.jpg" alt="enter image description here" /></a>
I'm getting the answer as LHL=-infinity while RHL= infinity</p>
<p>The answer given to this question is D.</p>
<p>However as far as I know when a limit tends to infinity it does exist. I don't know if I'm wrong, could someone please help.</p>
| Hagen von Eitzen | 39,174 | <p>Speaking of sequences instead of functions (for which the limit is ultimately defined via sequences), we say that $a$ is the limit of $\{a_n\}$, if for every $\epsilon>0$, at most finitely many $a_k$ are outside the $\epsilon$-neighbourhood of $a$.</p>
<p>$$\tag1 a_n\to a:\iff \forall \epsilon>0\;\exists N\in\Bbb N\;\forall n>N\;|a_n-a|<\epsilon.$$
If an $a$ that is the limit of $a:n$ exists, we say that $\{a_n\}$ is convergent; otherwise it is divergent.
Can $\infty$ be the limit? No, because there is no way to have $|a_n-\infty|<\epsilon$.
However, the concept of a sequence tending to $+\infty$ or $-\infty$ id still of interest. It just doesnt't fit into the pattern $(1)$ and requires a separate definition
$$\tag2 a_n\to+\infty:\iff\forall M\in\Bbb R\; \exists N\in\Bbb N\;\forall n>N\;a_n>M$$
$$\tag3 a_n\to+\infty:\iff\forall M\in\Bbb R\; \exists N\in\Bbb N\;\forall n>N\;a_n<M$$
Thus the notion of "less than arbitrarily small $\epsilon$ away" is replaced with "above arbitrarily large $M$".
By what we previously said, we cannot (or at least should not) say that $a_n$ <em>converges</em> to $\infty$. Instead, in English one usually says that it <em>tends</em> to $\infty$.</p>
|
882,540 | <p>Does anyone know of a non-trivial (i.e. cardinality $\geq 2)$ algebraic structure $(X,+,-)$ satisfying the following identities?</p>
<ol>
<li><p>$(x+a)-a=x$</p></li>
<li><p>$(x-a)+a=x$</p></li>
<li><p>$(x+y)+a = (x+a)+(y+a)$</p></li>
<li><p>$(x-y)+a = (x+a)-(y+a)$</p></li>
</ol>
<p><em>Remark.</em> The Abelian group of order $2$ doesn't satisfy the last two conditions.</p>
<p><strong>Motivation.</strong> I think its cool that if $X$ is such an algebraic structure, then for every $a \in X$, the functions $$x \mapsto x+a, \qquad x \mapsto x-a$$</p>
<p>are automorphism of $X$. This mean that if $a \in X$ and $f \in \mathrm{Aut}(X)$, then $f+a \in \mathrm{Aut}(X)$ and $f-a \in \mathrm{Aut}(X).$</p>
| Clinton Curry | 166,978 | <p>Below I will demonstrate that, if $+$ is associative, then $x+a = x$ and $x-a = x$ for all $x$ and $a$. This is enough, I think, to qualify as a "trivial" algebraic structure, even though the underlying set can be as large as you like.</p>
<p>Beginning from (3):</p>
<p>\begin{align*}
(x+y)+a &= (x+a) + (y+a)\\
(x+y)+a &= ((x+a)+y) + a & \text{by associativity}\\
x+y &= (x+a)+y & \text{by (1)}\\
x &= x+a & \text{by (1)}
\end{align*}</p>
<p>Further, using (2) we find that $x = (x+a) - a = x - a$.</p>
|
882,540 | <p>Does anyone know of a non-trivial (i.e. cardinality $\geq 2)$ algebraic structure $(X,+,-)$ satisfying the following identities?</p>
<ol>
<li><p>$(x+a)-a=x$</p></li>
<li><p>$(x-a)+a=x$</p></li>
<li><p>$(x+y)+a = (x+a)+(y+a)$</p></li>
<li><p>$(x-y)+a = (x+a)-(y+a)$</p></li>
</ol>
<p><em>Remark.</em> The Abelian group of order $2$ doesn't satisfy the last two conditions.</p>
<p><strong>Motivation.</strong> I think its cool that if $X$ is such an algebraic structure, then for every $a \in X$, the functions $$x \mapsto x+a, \qquad x \mapsto x-a$$</p>
<p>are automorphism of $X$. This mean that if $a \in X$ and $f \in \mathrm{Aut}(X)$, then $f+a \in \mathrm{Aut}(X)$ and $f-a \in \mathrm{Aut}(X).$</p>
| James | 751 | <p>A structure with these properties is called an involuntary <a href="http://en.wikipedia.org/wiki/Racks_and_quandles" rel="nofollow">quandle</a>, or kei. They arise as important algebraic invariants of knots. See <a href="http://www.esotericka.org/cmc/quandles.html" rel="nofollow">here</a>.</p>
|
3,238,306 | <p>Hello I have the differential equation <span class="math-container">$$x'=\frac{8t+10x}{17t+x}$$</span>
Brought it in a eqation where I can substitute <span class="math-container">$$ u=\frac{x}{t}$$</span>
and after some transformations I got the equation
<span class="math-container">$$ \frac{17+u}{8-7u-u^2}du=\frac{1}{t}dt $$</span>
with the following equation
<span class="math-container">$$ -2ln(u-1)+ln(u+8)=ln(t)+c$$</span>
Then I tried to multiply it with <span class="math-container">$$e$$</span> to make the <span class="math-container">$$ln()$$</span> disappear
then I got the equation <span class="math-container">$$ \frac{u+8}{(u-1)^2}=te^c$$</span> but now I dont know how to resubstitute or to break the fracture.Thanks,Ciwan.</p>
| Mohammad Riazi-Kermani | 514,496 | <p><span class="math-container">$$ ab\ge 2a$$</span></p>
<p><span class="math-container">$$ab\ge 2b$$</span></p>
<p>Add both sides</p>
<p><span class="math-container">$$2ab \ge 2a+2b$$</span>
<span class="math-container">$$ ab\ge a+b $$</span></p>
|
290,173 | <p>Suppose $d$ is a bounded metric on $X$, i.e. $d(x,y)< K<\infty$ for all $x,y\in X$. Is there a standard way to convert $d$ into another metric $\widetilde{d}$ on $X$ with the property that $\widetilde{d}(x,y)\to\infty$ if and only if $d(x,y)\to K$? One way would be to find some function $f$ such that $\widetilde{d}(x,y)=f(d(x,y))$ satisfies the given conditions, but it is not obvious that this is always possible.</p>
<p>The following properties are additionally useful, but not necessary: </p>
<ul>
<li>$\widetilde{d}$ preserves the topology of $(X,d)$</li>
<li>$d(x,y)>d(x',y')\implies \widetilde{d}(x,y)>\widetilde{d}(x',y')$</li>
</ul>
<p>Also, if this is possible when $(X,d)$ satisfies certain extra assumptions but not in general, answers in this direction are welcome.</p>
| pteromys | 116,623 | <p>Not unless you're willing to lose the triangle inequality.</p>
<p>Nik Weaver's counterexample in the comments doesn't require $\tilde{d}$ to be of the form $f \circ d$: If $(X,d) = [0,1]$ with the usual metric, then you'd be requiring $\tilde{d}(0,1) = \infty$ while $\tilde{d}(0,0.5)$ and $\tilde{d}(0.5,1)$ are finite.</p>
<p>For the specific case of $f \circ d$ on $[0,1]$, the triangle inequality <a href="https://math.stackexchange.com/questions/987602/" title="Composition of a function with a metric">appears to be related to concavity</a> of $f$.</p>
<p>The closest thing I can think of that respects the triangle inequality is dividing a Riemannian metric by the distance* to some set that you're about to delete.</p>
<p>*If you want the result to be Riemannian too, you still have to worry about smoothness, which probably requires constructing a different function to divide by; e.g. assume the deleted set is a submanifold and modify the metric only on a tubular neighborhood.</p>
|
290,173 | <p>Suppose $d$ is a bounded metric on $X$, i.e. $d(x,y)< K<\infty$ for all $x,y\in X$. Is there a standard way to convert $d$ into another metric $\widetilde{d}$ on $X$ with the property that $\widetilde{d}(x,y)\to\infty$ if and only if $d(x,y)\to K$? One way would be to find some function $f$ such that $\widetilde{d}(x,y)=f(d(x,y))$ satisfies the given conditions, but it is not obvious that this is always possible.</p>
<p>The following properties are additionally useful, but not necessary: </p>
<ul>
<li>$\widetilde{d}$ preserves the topology of $(X,d)$</li>
<li>$d(x,y)>d(x',y')\implies \widetilde{d}(x,y)>\widetilde{d}(x',y')$</li>
</ul>
<p>Also, if this is possible when $(X,d)$ satisfies certain extra assumptions but not in general, answers in this direction are welcome.</p>
| Nik Weaver | 23,141 | <p>I don't think there's any standard procedure, but here is one way to do this if there is a "base point" $e\in X$ satisfying $d(e,p) < K/2$ for all $p\in X$. I am not sure whether such a point can always be added. Wlog take $K=2$ and fix $e$. For any $p,q \in X$ find $p', q'$ in the open unit disc equipped with the euclidean metric such that the distances between $p'$, $q'$, and $0$ equal the distances between $p$, $q$, and $e$. There should be only one way to do this up to rotations and reflections. Then define $\tilde{d}(p,q)$ to be the distance between $p'$ and $q'$ for the hyperbolic metric on the disc.</p>
|
2,481,463 | <p>I am self-learning elliptic functions and modular form but struggling with the very basics.
I have programmed the Eisenstein series
$$E_6 = \sum_{m,n \in \mathbb{Z}} \frac{1}{(m +n \tau)^6} $$
and produced the following image (color denotes the argument)</p>
<p><a href="https://i.stack.imgur.com/NZeZb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NZeZb.png" alt="enter image description here"></a>
which is similar to that on <a href="https://en.wikipedia.org/wiki/Eisenstein_series" rel="nofollow noreferrer">wiki</a>.</p>
<p><em>Eisenstein series and automorphic representations</em> by P. Fleig introduced the fundamental domain, as the figure below.
<a href="https://i.stack.imgur.com/YWFa8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YWFa8.png" alt="enter image description here"></a></p>
<p>A similar figure finds in L. Vepstak’s <em>The Minkowski question Mark, GL(2,Z) and the Modular Group</em>.
<a href="https://i.stack.imgur.com/OyRSW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OyRSW.png" alt="enter image description here"></a></p>
<p>My question is how to transform from the fundamental domain
$$
|Re(z)|<\frac{1}{2} \quad and \quad |z|>1| \quad and \quad Im(z)\geqslant 0
$$
to other domains (appears as distorted triangles), or how to map a point in the fundamental domain to a corresponding point in another domain?</p>
<p>I guess that SL(2, $\mathbb{Z}$)-invariance should associate with the symmetries between the triangular domain
$$
G_{2k}\left( \frac{az+b}{cz+d} \right) = (cz+d)^{2k}G_{2k}(z)
$$
but no clues how this equation divides into triangles that correspond to the fundamental domain.</p>
| Kenny Wong | 301,805 | <p>To answer your question, and to understand the picture of the "fundamental domains in the upper half plane" (your third picture), it's probably best if we forget about $E_6$ and $G_{2k}$ for the time being. They are not directly relevant to your question, although we will return to them.</p>
<p>Instead, I would recommend that we concentrate on what the picture is actually showing. </p>
<p>The picture is showing the "fundamental domains" of the modular group. The modular group is a group of transformations on the (extended) upper half plane:
$$ \left\{ z \mapsto\frac{az + b}{cz+d} \ : \ a, b, c, d \in \mathbb Z, \ \ ad-bc=1\right\}.$$
In the picture, the upper half plane is divided into many regions (also known as "fundamental domains"). These regions obey a nice property: Given any two regions $R_1$, $R_2$, there exists a unique transformation $g$ in the group of modular transformations that maps $R_1$ onto $R_2$.</p>
<hr>
<p>Shall we work through an explicit example? Let's consider the transformation
$$ g : z \mapsto - \frac 1 {z}$$
(i.e. $a = 0, b = 1, c = - 1, d = 0$). And let's consider the region
$$ R_1 = \left\{ |{\rm Re}(z) | \leq \tfrac 1 2, \ \ |z| \geq 1 \right\}.$$
Can we work out the image of $R_1$ under the action of $g$?</p>
<p>Well, $g$ is a Mobius transformation, and we know that Mobius transformations map (generalised) circles to (generalised) circles. The boundary of $R_1$ is made up of segments of generalised circles. So the boundary of $g(R_1)$ should be made up of segments of generalised circles too. To be more specific:</p>
<ul>
<li><p>The part of the boundary of $R_1$ that is a line segment containing the points $\frac 1 2 + \frac{i \sqrt 3}{2}$, $\frac 1 2 + i$ and $\infty$ is mapped to a circle segment containing the points $-\frac 1 2 + \frac{i\sqrt 3}{2}$, $- \frac 2 5 + \frac{4i}{5}$ and $0$.</p></li>
<li><p>The part of the boundary of $R_1$ that is a line segment containing the points $-\frac 1 2 + \frac{i \sqrt 3}{2}$, $-\frac 1 2 + i$ and $\infty$ is mapped to a circle segment containing the points $\frac 1 2 + \frac{i\sqrt 3}{2}$, $ \frac 2 5 + \frac{4i}{5}$ and $0$. </p></li>
<li><p>The part of the boundary of $R_1$ that is a circle segment containing the points $- \frac 1 2 + \frac{i\sqrt 3}{2}$ and $i$, $-\frac 1 2 + \frac{i\sqrt 3}{2}$ is mapped to itself.</p></li>
</ul>
<p>Thus $g$ maps our region $R_1$ to the region in your diagram with vertices at $0$, $- \frac 1 2 + \frac{i \sqrt 3}{2}$ and $\frac 1 2 + \frac{i \sqrt 3}{2}$. And for the remainder of this answer, I'll call this new region $R_2$.</p>
<hr>
<p>So what does all this have to do with your Eisenstein series? Well, the Eisenstein series obey the relation
$$ G_{2k} \left( \frac{az+b}{cz+d}\right) = (cz+d)^{2k}G_{2k}(z).$$
You may like to think of this equation as a simple relationship between the values of the function $G_{2k}$ in one region of your picture and the values in another region.</p>
<p>For example, we've seen that $g : z \mapsto - 1 / z$ maps the region $R_1$ to the region $R_2$. The equation
$$ G_{2k} \left( - \frac 1 z \right) = z^{2k} G_{2k}(z)$$
is then a nice relationship between the values that $G_{2k}(z)$ takes in the region $R_1$ and the values it takes in the region $R_2$: if you calculate the values of $G_{2k}$ in one region, you get the values of $G_{2k}$ in the other region for free.</p>
|
103,084 | <p><strong>Bug introduced in 8.0 or earlier and fixed in 11.0.0</strong></p>
<p>Case number: 305932.</p>
<hr>
<p>In the front-end notebook (Mathematica 9.0 and 10.2):</p>
<pre><code>g[x: Optional[_,default]]:=x;
</code></pre>
<p>has no problems, and works correctly (<code>g[]</code> outputs <code>default</code>, and <code>g[1]</code> outputs <code>1</code>).</p>
<p>However, when the following package is constructed in an <code>.m</code> file called <code>Dummy.m</code>:</p>
<pre><code>BeginPackage["Dummy`"];
f
g; Thing;
Begin["`Private`"];
g[x: Optional[_,default]]:=x; (*Minimal case causing problem*)
f[x: Optional[Thing->{_},Thing->{1}]]:=x; (*I need this in my package*)
End[];
EndPackage[];
</code></pre>
<p>The same line of code throws a <code>General::patop</code> error upon package initialization. (The definition for <code>f</code> is the structure of the optional argument I need to achieve in my package.)</p>
<pre><code><<Dummy`
</code></pre>
<p><a href="https://i.stack.imgur.com/VidwS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VidwS.png" alt="enter image description here"></a></p>
<p>Is this a bug? and I can I define <code>f</code> in my package?</p>
| jkuczm | 14,303 | <p>String <code>"x : Optional[_, default]"</code> in <code>StandardForm</code> is interpreted as:</p>
<pre><code>ToExpression["x : Optional[_,default]", StandardForm, HoldComplete] // FullForm
(* HoldComplete[Optional[Pattern[x, Blank[]], default]] *)
</code></pre>
<p>and in <code>InputForm</code> as:</p>
<pre><code>ToExpression["x : Optional[_,default]", InputForm, HoldComplete] // FullForm
(* HoldComplete[Pattern[x, Optional[Blank[], default]]] *)
</code></pre>
<p>The latter is done when reading a package file, and similar thing to the former is done in the front-end.</p>
<p>As per <a href="https://mathematica.stackexchange.com/questions/103084/optional-throws-error-when-called-from-a-package-workaround#comment280049_103099">comment of Itai Seggev</a> in the front-end <code>x : Optional[_,default]</code> is represented by following boxes, which are interpreted as following expression:</p>
<pre><code>RowBox[{"x", ":", RowBox[{"Optional", "[", RowBox[{"_", ",", "default"}], "]"}]}] // MakeExpression // FullForm
(* HoldComplete[Optional[Pattern[x, Blank[]], default]] *)
</code></pre>
<p>instead of <code>HoldComplete[Pattern[x, Optional[Blank[], default]]]</code>, which is the surprising (possibly buggy) fact.</p>
<p>I think it should be reported to WRI.</p>
<hr>
<p>As can be seen in <a href="http://reference.wolfram.com/language/ref/Optional.html#12623" rel="nofollow noreferrer">Details section of <code>Optional</code> documentation</a>, what you called a workaround, <a href="https://mathematica.stackexchange.com/questions/103084/optional-throws-error-when-called-from-a-package-workaround#comment279915_103084">in a comment</a>, is actually the correct, documented way, which in full form is:</p>
<pre><code>Optional[Pattern[name, something], defaultValue]
</code></pre>
<p>In a package you can use full form directly:</p>
<pre><code>Optional[Pattern[x, Thing->{_}], Thing->{1}]
</code></pre>
<p>use infix form of <code>Pattern</code>:</p>
<pre><code>Optional[x : (Thing->{_}), Thing->{1}]
</code></pre>
<p>or infix form of both <code>Pattern</code> and <code>Optional</code>:</p>
<pre><code>x : (Thing->{_}) : (Thing->{1})
</code></pre>
|
1,403,228 | <blockquote>
<p><strong>Question:</strong></p>
<p>Let <span class="math-container">$m, n, q, r \in \mathbb Z$</span>. If <span class="math-container">$m = qn + r$</span>, show that <span class="math-container">$\gcd(m, n) = \gcd(n, r)$</span>. Hence justify the Euclidean Algorithm.</p>
</blockquote>
<p>I found this question in a past test paper, but cannot seem to find a reference in my textbook that indicates how I can go about "proving" the above statement. Can anyone please point me in the right direction?</p>
| Manish Kumar Singh | 234,705 | <p><strong>To Prove $\gcd(m, n) = \gcd (n, r)$ if $m = qn + r$</strong></p>
<p>Let $\gcd(m, n) = d$.</p>
<p>So $d \mid m$ and $d \mid n$ implies $d \mid r$ (read $d$ divides...)</p>
<p>Similarly if $n = q_1r + r_1$ and $d \mid n$ and $d \mid r$ implies $d \mid r_1$.</p>
<p>Note $r_i$ are reducing by each successive terms, hence this algorithm is guaranteed to terminate.</p>
<p>Now suppose the last expression (say algorithm takes $n$ steps) reads like this:</p>
<p>$$r_{n - 3} = q_{n - 3}r_{n - 2} + r_{n - 1}$$
$$r_{n - 2} = q_{n - 2}r_{n - 1} + r_n$$</p>
<p>For uniqueness by the above logic $d \mid r_n$, let $d_1 \mid r_n$ and $d_1 \mid r_{n - 2}$) [$d_1$ is some divisor of $n, r$], therefore $d_1 \mid (q_{n - 2}r_{n - 1})$.</p>
<p>Now $d_1 \mid r_{n - 3}$ because $d_1 \mid r_{n - 2}$ and $d_1 \mid r_{n - 1}$ and so on $d_1 \mid m$ and $d_1 \mid n$.</p>
<p>Therefore $m = nd$ and $m = nd_1$ but by definition $d$ is the greatest divisor, hence $d > d_1$ and we know that since the choice of $d_1$ was arbitrary, we are sure that $d \mid n$ and $d \mid r$ is the greatest possible $d$, hence the proof.</p>
|
1,050,141 | <p>Find the roots of the equation $3^{x+2}$+$3^{-x}$=10 . By inspection the roots are $x=0$ and $x=-2$. But how can I solve this equation otherwise? </p>
| MPW | 113,214 | <p><strong>Hint:</strong> Multiply both sides by the positive quantity $3^x$ to clear the fraction implied by the negative exponent, and rearrange a bit:
$$3^{x+2} + 3^{-x}=10$$
$$3^x\cdot3^{x+2} + 3^x\cdot 3^{-x} = 3^x\cdot 10$$
$$3^{2x+2} + 1 = 3^x\cdot 10$$
$$3^{2x}\cdot 3^2 + 1 = 3^x\cdot 10$$
$$(3^x)^2\cdot 9 + 1 = 3^x\cdot 10$$
$$9\cdot(3^x)^2 - 10\cdot(3^x) + 1 = 0$$
Now think of this as a quadratic equation in $3^x$:
$$9u^2 - 10u + 1 = 0$$
where $u = 3^x$.
Solve the quadratic to get the value(s) of $u$, say $u=A$ and $u=B$. Then remember that $u$ stands for $3^x$, so now you have to solve the equations $3^x=A$ and $3^x=B$, where $A$ and $B$ are the solutions to your quadratic.</p>
<p>Note that you will only have real solutions if $A>0$ or $B>0$, since $3^x$ can never be negative or zero if $x$ is real.</p>
|
189,618 | <p>I'm trying to plot 2D vectors in Mathematica. Built in functions don't really work for me because I want to plot vectors of matrices from the origin to the their coordinates with an arrow on their tips. I made a function </p>
<pre><code>plotMatrixVectors[mat_List] :=
Graphics[Table[Arrow[{origin, i}], {i, mat}], Axes -> True],
</code></pre>
<p>and it worked just fine, until it suddenly stopped working and instead of a graph started reporting this to me:</p>
<pre><code>{{plotMatrixVectors[1], plotMatrixVectors[2]}, {plotMatrixVectors[3],
plotMatrixVectors[1]}}
</code></pre>
<p>Note: these numbers in square bracket have nothing to the with coordinates in the lower example, this message came up when I tried to plot different vectors, but you get the point.</p>
<p>I rewrote the function because I thought I unintentionally messed it up somehow, but it still printed the same thing. However, when I copypasted the body of the function and just plugged in a 2x2 matrix, say this one </p>
<pre><code>plot2D1 = {{1, 5}, {-6, 4}};
Graphics[Table[Arrow[{origin, i}], {i, plot2D1}], Axes -> True]
</code></pre>
<p>it gives me this, which is good.
<a href="https://i.stack.imgur.com/VLyUf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VLyUf.png" alt="enter image description here"></a></p>
<p>I have 2 questions here.
1) Why did my function stopped working but the same thing works when it's typed outside of the function? I restarted Mathematica and it still didn't work.
2) Is there any way to strecth those axes so they go beyond the limits of matrix vectors coordinates?</p>
<p>Thanks a lot!</p>
| Rainb | 59,208 | <p>I'd like to show this code, which is essentially the same as from m_goldberg, but it uses Standard colors. And adds label, the reason I did this, is because I needed this on my own plots and wanted to share.</p>
<pre class="lang-mathematica prettyprint-override"><code>plotVector[vectors : {{_?NumberQ, _?NumberQ} ..},
OptionsPattern[{"Label" -> False}]] :=
Block[{arrows, labels},
arrows =
MapIndexed[
Style[Arrow[{{0, 0}, #1}], ColorData[97][#2[[1]]], Thick] &,
vectors];
labels =
MapIndexed[
Style[Text[#1, #1 + {0.2, 0.2}, {1, 1}],
ColorData[97][#2[[1]]]] &, vectors];
Graphics[
Join[{Arrowheads[0.05]}, arrows,
If[OptionValue["Label"], labels, {}]], Axes -> True]];
</code></pre>
<p>This function takes a list of 2D vectors as input and plots them as arrows on a 2D plot, with the origin <code>{0, 0}</code> as the starting point and the vector as the end point. The arrows are styled using the Style function, with the color generated using the <code>ColorData</code> function and the thickness specified using the Thick option. The x-axis and y-axis are displayed on the plot using the Axes option, and the size of the arrowheads is specified using the Arrowheads option.</p>
<p>You use it like this:
<code>plotVector[{{1, 1}, {2, 3}, {0, 1}, {1, 0}}, Label -> True]</code>, in order to get:</p>
<p><a href="https://i.stack.imgur.com/FNlDz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FNlDz.png" alt="Mathematica screenshot plotting the result of plotVector[{{1, 1}, {2, 3}, {0, 1}, {1, 0}}, Label -> True]" /></a></p>
|
478,620 | <p>Let $S$ be a set that satisfies the following property:</p>
<blockquote>
<p>For every countable family $\{U_{i}\}_{i\in \mathbb{N}}$ of subsets of $S$ for which $$S=\bigcup_{i\in \mathbb{N}}U_{i}$$there is a finite set $I\subset \mathbb{N}$ for which $$S=\bigcup_{i \in I}U_{i}$$</p>
</blockquote>
<p>Does it follow that $S$ is finite?</p>
| A. H. | 160,657 | <p>To answer your question specifically, your idea is not correct because after you eliminate the integers that contain 5, you no longer have a 1:1 ratio of even:odd integers, so you can't simply divide by 2 to get your "number of odd integers that do not contain the digit 5."</p>
|
759,683 | <p>Given an undirected connected simple graph $G=(V,E)$ there are $2^{|E|}$ orientations. How many of these orientations are acyclic? </p>
| ml0105 | 135,298 | <p>I came across a research paper from 1972 which addresses this question. Let $\chi(G, \lambda)$ be the chromatic polynomial. Then $(-1)^{V} \chi(G, -1)$ is the number of acyclic orientations of $G$, your graph. I let $V$ be the vertex count of $G$. This is Theorem 1.3 of the paper.</p>
<p><a href="http://www-math.mit.edu/~rstan/pubs/pubfiles/18.pdf" rel="nofollow">http://www-math.mit.edu/~rstan/pubs/pubfiles/18.pdf</a></p>
|
380,696 | <p>Show that for any odd $n$ it follows that $n^2 \equiv 1 \mod 8$ and for uneven primes $p\neq 3$ we have $p^2 \equiv 1\mod 24$.</p>
<p>My workings so far: I proceeded by induction. Obviously $1^2 \equiv 1 \mod 8$. Then assume that for a certain uneven number $k$ we have $k^2 \equiv 1 \mod 8$. Then the the next uneven number is $k+2$ and $(k+2)^2 = k^2 + 4k + 4 \equiv 4k + 5 \mod 8$. by the induction hypothesis. Now since $k$ is uneven we can write it as $k=2j+1$ and thus $4k+5 =8j+9 \equiv 1 \mod 8$ and we have shown what was asked for by induction.</p>
<p>However, in the case of a prime number $n$ I am not so certain how to proceed because I can't use induction. It comes down to proving that $24|(p^2-1)$. This is certainly the case for the first uneven prime $p \neq 3$, namely $p=5$ such that $p^2-1 =24$. How would I proceed from here, or how should I approach the problem differently? Many thanks in advance!</p>
| Martin Kochanski | 340,970 | <p>Never use induction at all. It is far too heavy a weapon for this kind of question.</p>
<ol>
<li>An odd number is a number of the form $4n±1$. Square it, and you get $16n^2±8n+1$, which clearly equals 1 modulo 8.</li>
<li>Primes are irrelevant: this is true for all odd numbers which are not multiples of 3. Such a number is of the form $3n±1$. Square it, and you get $p^2=9n^2±6n+1$, which clearly equals 1 modulo 3. If this doesn't make it obvious to you that $p^2$ is 1 modulo 24, consider $p^2-1$. It is a multiple of 8, and it is also a multiple of 3, so it must be a multiple of 24.</li>
</ol>
|
2,478,026 | <p><a href="https://i.stack.imgur.com/TyhHX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TyhHX.png" alt="enter image description here"></a></p>
<p>It doesn't make sense to me that |E[x] - m| = |E[X-m]|</p>
| heropup | 118,193 | <p>I think the underlying issue is an apparent confusion regarding the meaning of "median" in this question. In the present context, the median refers to a property of the distribution of the random variable $X$, rather than a statistic of a sample drawn from such a distribution. Both are called "median" (for brevity) but the latter is more precisely called the <em>sample</em> median.</p>
<p>To recap, $m$ is a <strong>median</strong> of a real-valued random variable $X$ when $$ \quad \Pr[X \ge m] \ge \frac{1}{2} \quad \textrm{AND} \quad \Pr[X \le m] \ge \frac{1}{2}.$$ Presumably, the question implies that $m$ is unique, in which case equality is observed. Such a value is not stochastic: it is a deterministic quantity that depends on the parametric distribution of $X$, just as $\mu$ and $\sigma$ are (when these exist). Just because these may not be assigned a known numeric value does not mean they are random.</p>
<p>This is in contrast to the definition of a sample median, which is a statistic and therefore itself a random variable.</p>
|
2,916,168 | <p>I am a beginner in proofs and, unfortunately, I cannot wrap my mind around how to prove the simplest things, so I need a bit of help getting started. This is the proof that I am dealing with:</p>
<p>$\text{If }x< y< 0\text{, then }x^{2}> y^{2}\text{.}$</p>
<p>Thank you in advance.</p>
| abc... | 470,090 | <p>$x^2>y^2\Rightarrow|x|>|y|\Rightarrow y>x$ for $x,y<0.$</p>
|
3,129,192 | <p>I've been trying to integrate
<span class="math-container">$$\int\frac{dx}{(x-a)(x-b)}$$</span></p>
<p>By using the substitution
<span class="math-container">$$x=a \cos^2 \theta + b \sin^2 \theta$$</span></p>
<p>The only problem here is I arrived at the result
<span class="math-container">$$\frac{2}{a-b} \ln |\csc 2\theta - \cot 2\theta|+c$$</span></p>
<p>and I having trouble on how to substitute from <span class="math-container">$\theta$</span> to <span class="math-container">$x$</span>.</p>
| user555203 | 447,605 | <p>Note: throughout, if <span class="math-container">$p\in I^2$</span> or <span class="math-container">$p\in A$</span>, <span class="math-container">$[p]$</span> will denote the equivalence class of <span class="math-container">$p$</span> in the respective quotient space.</p>
<p>You can first map <span class="math-container">$I^2$</span> to <span class="math-container">$A$</span> by
<span class="math-container">$$f(x,y) = ((y+1)\cos(2\pi x), (y+1)\sin(2\pi x)).$$</span>
(To visualize this, think of <span class="math-container">$A \subset \mathbb{C}$</span> in the obvious way, and think of <span class="math-container">$f$</span> as <span class="math-container">$f(x,y) = (y+1)e^{2\pi i x}$</span>.)
As you see, <span class="math-container">$\{x=0\}$</span> and <span class="math-container">$\{x=1\}$</span> end up in the same place; in fact, <span class="math-container">$(0,y)$</span> and <span class="math-container">$(1,y)$</span> end up in the same place for all <span class="math-container">$y$</span>. (You can see this way that if you only identify <span class="math-container">$\{x=0\}$</span> and <span class="math-container">$\{x=1\}$</span> in <span class="math-container">$I^2$</span>, you are already homeomorphic to <span class="math-container">$A$</span>.) And moreover, <span class="math-container">$\{y=0\}$</span> and <span class="math-container">$\{y=1\}$</span> end up as the inner and outer circles of <span class="math-container">$A$</span>, with <span class="math-container">$(x,0)$</span> and <span class="math-container">$(x,1)$</span> at the same angle for all <span class="math-container">$x$</span>.</p>
<p>So now if <span class="math-container">$\pi : A \to T_A$</span> and <span class="math-container">$\rho:I^2 \to T_I$</span> are the quotient maps, <span class="math-container">$\pi \circ f$</span> factors through <span class="math-container">$\rho$</span> because of how the equivalences match up, so that there is <span class="math-container">$g:T_I \to T_A$</span> with <span class="math-container">$\pi \circ f = g \circ \rho$</span>, so really we have a formula for <span class="math-container">$g$</span>, namely <span class="math-container">$g([(x,y)]) = \pi(f(x,y))$</span>.</p>
<p>You can check that <span class="math-container">$g$</span> is a homeomorphism, given by
<span class="math-container">$$g([(x,y)]) = [((y+1)\cos(2\pi x), (y+1)\sin(2\pi x))].$$</span></p>
|
3,129,192 | <p>I've been trying to integrate
<span class="math-container">$$\int\frac{dx}{(x-a)(x-b)}$$</span></p>
<p>By using the substitution
<span class="math-container">$$x=a \cos^2 \theta + b \sin^2 \theta$$</span></p>
<p>The only problem here is I arrived at the result
<span class="math-container">$$\frac{2}{a-b} \ln |\csc 2\theta - \cot 2\theta|+c$$</span></p>
<p>and I having trouble on how to substitute from <span class="math-container">$\theta$</span> to <span class="math-container">$x$</span>.</p>
| mechanodroid | 144,766 | <p>Let's show that the map <span class="math-container">$g : I^2/_\sim \to A/_{\approx}$</span> from @csprun's answer is indeed a homeomorphism:
<span class="math-container">$$g([(x,y)]) = [f(x,y)] = [((y+1)\cos(2\pi x), (y+1)\sin(2\pi x))].$$</span></p>
<p>First note that <span class="math-container">$g$</span> is well-defined:</p>
<ul>
<li>if <span class="math-container">$x,y \in (0,1)$</span> then <span class="math-container">$(x,y)$</span> is the only member of <span class="math-container">$[(x,y)]$</span></li>
<li>if <span class="math-container">$x = 0$</span>, then <span class="math-container">$(0,y) \sim (1,y)$</span> but <span class="math-container">$$f(0,y) = (y+1)(\cos(0),\sin(0)) = (y+1)(1,0) = (y+1)(\cos(2\pi), \sin(2\pi)) = f(1,y)$$</span>
so in particular <span class="math-container">$[f(0,y)] = [f(1,y)]$</span>.</li>
<li>if <span class="math-container">$y = 0$</span> then <span class="math-container">$(x,0) \sim (x,1)$</span> but
<span class="math-container">$$[f(x,0)] = [(\cos(2\pi x), \sin(2\pi x))] = [2(\cos(2\pi x), \sin(2\pi x))] = [f(x,1)]$$</span></li>
</ul>
<p>Now we show that <span class="math-container">$g$</span> is continuous. As in @csprun's answer, denote the respective quotient maps by <span class="math-container">$\pi : A \to A/_\approx$</span> and <span class="math-container">$\rho : I^2 \to I^2/_\sim$</span>. Recall that quotient topologies are given by <span class="math-container">$$\{U \subseteq A/_\approx : \pi^{-1}(U) \text{ open in } A\}, \quad \{V \subseteq I^2/_\sim : \rho^{-1}(V) \text{ open in } I^2\}$$</span></p>
<p>Notice that <span class="math-container">$g$</span> satisfies the identity <span class="math-container">$g \circ \rho = \pi \circ f$</span>. The function <span class="math-container">$\pi \circ f : I^2 \to A/_\approx$</span> is continuous as a composition of two continuous functions. We claim that <span class="math-container">$g$</span> is continuous. Let <span class="math-container">$W \subseteq A/_\approx$</span> be open, we claim that <span class="math-container">$g^{-1}(W)$</span> is open in <span class="math-container">$I^2/\sim$</span>. For that it is sufficient to show that <span class="math-container">$\rho^{-1}(g^{-1}(W))$</span> is open in <span class="math-container">$I^2$</span>, but this is true since <span class="math-container">$$\rho^{-1}(g^{-1}(W)) = (g \circ \rho)^{-1}(W) = (\pi \circ f)^{-1}(W), \text{ and this is open in } I^2$$</span></p>
<p>Now onto the inverse of <span class="math-container">$g$</span>. We will provide an explicit inverse <span class="math-container">$g^{-1} : A/_\approx \to I^2/_\sim$</span> and show that it is continuous. First notice that the inverse of <span class="math-container">$f$</span> is given by <span class="math-container">$f^{-1} : A \to I^2$</span>
<span class="math-container">$$f^{-1}(x,y) = \left(\frac1{2\pi}\omega(x,y), \|(x,y)\| - 1\right)$$</span>
where <span class="math-container">$\omega : A \to [0,2\pi]$</span> is the unique angle which the point <span class="math-container">$(x,y)$</span> closes with the positive <span class="math-container">$x$</span>-axis:
<span class="math-container">$$\omega(x,y) = \begin{cases}
0, &\text{ if } y = 0, x > 0\\
\operatorname{arccot}\frac{x}y, &\text{ if } y > 0\\
\pi, &\text{ if } y = 0, x < 0\\
\pi + \operatorname{arccot}\frac{x}y, &\text{ if } y < 0\\
\end{cases}$$</span>
Note (because of <span class="math-container">$\omega$</span>) that <span class="math-container">$f^{-1}$</span> is not continuous on the line segment <span class="math-container">$\{x > 0, y = 0\} \cap A$</span>, but it is continuous elsewhere on <span class="math-container">$A$</span>.</p>
<p>Now we claim that <span class="math-container">$g^{-1}$</span> is given by <span class="math-container">$$g^{-1}([(x,y)]) = [f^{-1}(x,y)]$$</span>
Indeed, this is well defined as for <span class="math-container">$(x,y)$</span> on the unit circle we have
<span class="math-container">$$[f^{-1}(x,y)] = \left[\left(\frac1{2\pi}\omega(x,y), 0\right)\right] = \left[\left(\frac1{2\pi}\omega(2x,2y), 1\right)\right] = [f^{-1}(2x,2y)]$$</span>
Notice that <span class="math-container">$g^{-1}$</span> satisfies <span class="math-container">$g^{-1} \circ \pi = \rho \circ f^{-1}$</span> so we have
<span class="math-container">$$(g^{-1} \circ g) \circ \rho = g^{-1} \circ (g \circ \rho) = g^{-1} \circ (\pi \circ f) = (g^{-1} \circ \pi) \circ f = (\rho \circ f^{-1}) \circ f = \rho \circ (f^{-1} \circ f) = \rho$$</span>
so <span class="math-container">$g^{-1} \circ g = \operatorname{id}_{I^2/_\sim}$</span>. Similarly we show <span class="math-container">$(g \circ g^{-1}) \circ \pi = \pi$</span> so <span class="math-container">$g \circ g^{-1} = \operatorname{id}_{A/_\approx}$</span>. Hence <span class="math-container">$g^{-1}$</span> is really the inverse of <span class="math-container">$g$</span>. It remains to show that <span class="math-container">$g^{-1}$</span> is continuous.</p>
<p>First notice that <span class="math-container">$\rho \circ f^{-1} : A \to I^2/_\sim$</span> is continuous. We know that <span class="math-container">$f^{-1}$</span> is continuous on <span class="math-container">$A\setminus \{x > 0, y = 0\}$</span>, so <span class="math-container">$\rho \circ f^{-1}$</span> is continuous there as well. For <span class="math-container">$(x,0) \in A$</span>, <span class="math-container">$x > 0$</span> let <span class="math-container">$((x_\lambda, y_\lambda))_{\lambda \in \Lambda}$</span> be a net in <span class="math-container">$A$</span> which converges to <span class="math-container">$(x,0)$</span>. We can assume <span class="math-container">$x_\lambda > 0, \forall \lambda \in \Lambda$</span>.
We have
<span class="math-container">$$(\rho \circ f^{-1})(x_\lambda, y_\lambda) = \begin{cases} \left[\left(\frac1{2\pi}\operatorname{arccot} \frac{x_\lambda}{y_\lambda}, \|(x_\lambda, y_\lambda)\| - 1\right)\right], &\text{ if } y_\lambda > 0 \\
\left[\left(\frac12 + \frac1{2\pi}\operatorname{arccot} \frac{x_\lambda}{y_\lambda}, \|(x_\lambda, y_\lambda)\| - 1\right)\right], &\text{ if } y_\lambda < 0 \\
\left[\left(0, \|(x_\lambda, y_\lambda)\| - 1\right)\right], &\text{ if } y_\lambda = 0 \\
\end{cases} \to \begin{cases} \left[\left(0, \|(x,y)\| - 1\right)\right], &\text{ if } y_\lambda > 0 \\
\left[\left(1, \|(x,y)\| - 1\right)\right], &\text{ if } y_\lambda < 0 \\
\left[\left(0, \|(x,y)\| - 1\right)\right], &\text{ if } y_\lambda = 0 \\
\end{cases} = \left[\left(0, \|(x,y)\| - 1\right)\right] = (\rho \circ f^{-1})(x,y)$$</span>
We conclude that <span class="math-container">$\rho \circ f^{-1}$</span> is continuous at <span class="math-container">$(x,0)$</span>.
Finally, let <span class="math-container">$W \subseteq I^2/_\sim$</span> be an open set and we claim that <span class="math-container">$(g^{-1})^{-1}(W)$</span> is open in <span class="math-container">$A/_\approx$</span>. It suffices to verify that <span class="math-container">$\pi^{-1}((g^{-1})^{-1}(W))$</span> is open in <span class="math-container">$A$</span>, which is true:
<span class="math-container">$$\pi^{-1}((g^{-1})^{-1}(W)) = (g^{-1} \circ \pi)^{-1}(W) = (\rho \circ f^{-1})^{-1}(W), \text{ and this is open in } A$$</span>
Therefore <span class="math-container">$g^{-1}$</span> is continuous so we conclude that <span class="math-container">$g$</span> is a homeomorphism.</p>
|
1,326,525 | <p>Suppose I have some functions $f,g$ such that
$$f:\Bbb{R} \mapsto \Bbb{R}^2$$
$$g:\Bbb{R}^2 \mapsto \Bbb{R}^n$$</p>
<p><strong>My Question:</strong></p>
<p>For some $c \in \Bbb{R}$, is $g(f(c))$ a function of one variable? If so, why is the fact $g$ requires two arguments irrelevant? </p>
<p>EDIT: </p>
<p>Specifically, I am considering a function
$$\mathbf{F}(\mathbf{y}) = \nabla f(\mathbf{x}) -\nabla (\lambda g(\mathbf{x}))$$
where $$\mathbf{y} = [\lambda,\mathbf{x}]$$
I want to write
$$\mathbf{F}(c,\mathbf{y}) = 0$$
and use the implicit function theorem to show that $\lambda$ and $\mathbf{x}$ can be parametrized by $c$, but I am confused why/if I can write $\mathbf{F}$ in terms of $c$ and $\mathbf{y}$ given that the first derivative of my Lagrangian multiplier equation isn't a function of $c$. </p>
| RowanS | 246,774 | <p>P( not cat | dog) = P( dog and not cat) /P (dog)</p>
|
1,212,177 | <p>Does $i\arg(e^{2z})=2iy?$ If it does I have solved my problem, and hence it seems like it must be the case, but I don't see it.</p>
<p>$$i\arg(e^{2z})=i\arg(e^{2x+2iy})=i\arg(e^{2x}e^{2iy})\implies \theta=2y(?)$$ Why does the $2x$ get 'ignored'?</p>
| Trajan | 119,537 | <p>$$\begin{align}i\arg(e^{2z})&=i\arg(e^{2x+2iy}) \\
&=i\arg(e^{2x}e^{2iy}) \\
&=i\arg(e^{2x}(\cos2y+i\sin2y)) \\
&=i\arg((\cos2y+i\sin2y)) \ \ \text{ as $e^{2x} \in \mathbb{R}$} \\
&=i(2y) \ \ \ \ \ \ \ \text{ by definition} \\
\end{align}$$</p>
|
61,174 | <p>I was just thinking about this recently, and I thought of a possible bijection between the natural numbers and the real numbers. First, take the numbers between zero and one, exclusive. The following sequence of real numbers is suggested so that we have bijection. </p>
<p>0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 0.01, 0.02, ... , 0.09, 0.10, 0.11, ... , 0.99, 0.001, 0.002, ... , 0.999, 0.0001, etc.</p>
<p>Obviously, this includes repeats, but this set is countable. Therefore, the set of all numbers between zero and one is a subset of the above countable set, and is thus countable. Then we simply extend this to all real numbers and all the whole numbers themselves, and since the real numbers, as demonstrated above, between any two whole numbers is countable, the real numbers are the union of countably many countable sets, and thus the real numbers are countable. </p>
<p>Please help me with this. I understand the diagonalization argument by Cantor, but I am curious specifically about this proof which I thought of and its strengths and flaws.</p>
<p>Thanks.</p>
| Vhailor | 887 | <p>If you know Cantor's diagonalization argument, then you should be able to find a counterexample to your problem using it. Say the method we use to make the $k$-th digit of our new real number different from the $k$-th digit of the $k$-th number in your list is adding 1 to it (and make it 0 if it's 9). Then Cantor's argument gives the number $0.21111111111111...$ Which is not in your list because (as said below) you only have real numbers with finite decimal expansions in your list.</p>
<p>(was previously a comment, but made an answer following suggestion)</p>
|
660,629 | <p>According to Wikipedia, factorial only is defined for non-negative integers.
How come Spotlight, the Windows calculator and the Google search engine come up with $\sqrt\pi$ if you try to solve $(-0.5)!$ ?</p>
| David H | 55,051 | <p>The area of the quadrilateral $BEDC$ is what's leftover if you subtract the triangle $ADE$ from triangle $ABC$. Since both triangles are right triangles, </p>
<p>$$area(BEDC)=area(ABC)-area(ADE)\\
=\frac12\|AB\|\cdot\|BC\|-\frac12\|AD\|\cdot\|DE\|\\
=\frac{24\cdot10}{2}-\frac{13}{2}\|DE\|.$$</p>
<p>To find $\|DE\|$, note that $\triangle ADE$ is similar to the $5,12,13$ right triangle, so</p>
<p>$$\frac{\|DE\|}{\|DA\|}=\frac{\|DE\|}{13}=\frac{5}{12}\implies \|DE\|=\frac{65}{12}.$$</p>
<p>Now just complete the arithmetic.</p>
|
660,629 | <p>According to Wikipedia, factorial only is defined for non-negative integers.
How come Spotlight, the Windows calculator and the Google search engine come up with $\sqrt\pi$ if you try to solve $(-0.5)!$ ?</p>
| Sawarnik | 93,616 | <p><img src="https://i.stack.imgur.com/1nQyf.png" alt="enter image description here"></p>
<p>Let us look at $\triangle AED$ and $\triangle CED$:</p>
<ul>
<li>$\angle EDA = \angle EDC = 90\,^{\circ}$</li>
<li>$DC= DA = 13$ cm</li>
<li>$ED$ is common</li>
</ul>
<p>So, by $SAS$ congruence, those two triangles are congruent, and hence $EC = AE$. Now let $BE$ be $x$. So, by Pythagoras Theorem, $(EC)^2=x^2+100$. But, $(EC)^2= (AE)^2 = (24-x)^2$. Therefore, we get an equation, $(24-x)^2= 576 + x^2 - 48x = x^2+100$. Which upon solving, we get, $x = \frac{119}{12}$ .</p>
<p>Back to $\triangle AED$, we see that $(ED)^2 = (24-x)^2 - 169$. So, we know $x$ and we can get $ED$ as well. The final touch, $$Area(BEDC) = Area(ABC) - Area(EDA) = \frac{24 \cdot 10}{2} - \frac{ED \cdot 13}{2}$$</p>
<p><strong>Edit</strong>: Details of calculation:
$(ED)^2 = (24 - x)^2 - 169 = (24 - \frac{119}{12})^2 - 169 = (\frac{169}{12})^2 - 169 = 169(\frac{169}{144}- 1)= \frac{169 \cdot 25 }{144}$
Therefore, $ED = \sqrt{\frac{169 \cdot 25 }{144}} = \frac{13 \cdot 5 }{12}$.</p>
<p>So, $$Area(BEDC) = \frac{24 \cdot 10}{2} - \frac{ED \cdot 13}{2} = 120 - \frac{13 \cdot 13 \cdot 5}{24} = 120- \frac{845}{24} = \frac{2035}{24} = 84.791666...$$</p>
|
2,939,605 | <p>The given task goes as follows:</p>
<blockquote>
<p>Show that <span class="math-container">$ f: \mathbb{R} \longrightarrow \mathbb{R}$</span> defined by <span class="math-container">$f(x) = \sqrt{1 + x^2} $</span> is not a polynomial function.</p>
</blockquote>
<p>I tried this approach - if <span class="math-container">$f(x)$</span> is a <span class="math-container">$n$</span>-degree polynomial function, then the <span class="math-container">$(n+1)$</span>-st derivative equals to 0 and I was trying to determine the <span class="math-container">$k$</span>-th derivative of <span class="math-container">$f(x)$</span> (and show it differs from 0 for any <span class="math-container">$k$</span>) but without success. Since <span class="math-container">$f(x)$</span> is continuous and defined over whole R domain, I have no idea how to carry on. Any ideas? </p>
| Angel Moreno | 327,493 | <p>Another demonstration:</p>
<p>A polynomial of degree one or higher when squared, should result in a polynomial that has all its roots double or more than doubles.</p>
<p><span class="math-container">$f ^ 2 = x ^ 2 + 1 = (x + i) (x-i)$</span> </p>
<p>That has two roots but they are not equal. And by the fundamental theorem of algebra, that factorization is unique.</p>
<p>Therefore f can not be a polynomial.</p>
|
2,846,700 | <p>Consider the functions $f(x)$, $g(x)$, $h(x)$, where $f(x)$ is neither odd nor even, $g(x)$ is even and $h(x)$ is odd. Is it possible for $f(x) + g(x)$ to be </p>
<ol>
<li>even;</li>
<li>odd?</li>
</ol>
<p>For the second case I can imagine for example $f(x) = x - 1$ and $g(x) = 1$. Then $f$ is neither even nor odd and $g$ is even but their sum is odd, hence it's possible to get odd function from the sum of neither odd nor even and even function.</p>
<p>It feels like $f(x) + g(x)$ can never be even, but I couldn't manage to prove that.</p>
<p>I've tried to do it the following way:
Let $f(x) = - g(x) - h(x)$, which doesn't contradict the initial statement. Then we can express $g(x)$ and $h(x)$ and see whether the facts that they are either even or odd holds, but this always leads to valid equations:</p>
<p>$$
h(x) = \frac{f(-x) - f(x)}{2} \;\;\; \text{is an odd function} \\
g(x) = \frac{-f(x) - f(-x)}{2} \;\;\; \text{is an even function}
$$</p>
<p>I'm stuck at that point.</p>
<blockquote>
<p>How can I prove/disprove that $f(x) + g(x)$ may be even?</p>
</blockquote>
| Hagen von Eitzen | 39,174 | <p>If $f+g$ is even, then $(f+g)-g$ is the difference of even functions, hence even</p>
|
873,224 | <p>I'm not mathematically inclined, so please be patient with my question.</p>
<p>Given </p>
<ul>
<li><p>$(x_0, y_0)$ and $(x_1, y_1)$ as the endpoints of a cubic Bezier curve.</p></li>
<li><p>$(c_x, c_y)$ and r as the centerpoint and the radius of a circle.</p></li>
<li><p>$(x_0, y_0)$ and $(x_1, y_1)$ are on the circle.</p></li>
<li><p>if it makes the calculation simpler, it's safe to assume the arc is less than or equal to $\frac{\pi}{2}$.</p></li>
</ul>
<p>How do I calculate the two control points of the Bezier curve that best fits the arc of the circle from $(x_0, y_0)$ to $(x_1, y_1)$?</p>
| me22 | 875,771 | <p>I got inspired to attempt yet another way to interpret "best fit": most-constrained <em>curvature</em>.</p>
<p>Let's look at the first quarter-circle, using the four control points <span class="math-container">$<1,0>$</span>, <span class="math-container">$<1,α>$</span>, <span class="math-container">$<α,1>$</span>, <span class="math-container">$<1,0>$</span>.</p>
<p><span class="math-container">$$
\begin{align}
x(t)
&= (1-t)^3 + 3 (1-t)^2 t + 3 α (1-t) t^2 \\
y(t)
&= 3 α (1-t)^2 t + 3 (1-t) t^2 + t^3 \\
&= x(1-t) \\
k(t)
&= \frac{ x' y'' - y' x'' }{ (x'^2 + y'^2)^{3/2} } \\
&= \frac{ -18(2α(t - 1)t + αt^2 - 2(t - 1)t)(2α(t - 1) + αt - 2t + 1) + 18(α(t - 1)^2 + 2α(t - 1)t - 2(t - 1)t)(α(t - 1) + 2αt - 2t + 1) }
{ 27((α(t - 1)^2 + 2α(t - 1)t - 2(t - 1)t)^2 + (2α(t - 1)t + αt^2 - 2(t - 1)t)^2)^{3/2} }
\end{align}
$$</span></p>
<p>Where <span class="math-container">$k(t)$</span> is the <em>signed</em> curvature, but since we're going clockwise it'll be positive.</p>
<p>As is usually the case, the curvature expression is horrible, so we need a more tractable way forward.</p>
<p>Thinking about the extreme cases, a lower value of <span class="math-container">$α$</span> (like <span class="math-container">$0.1$</span>) will make the ends turn sharply to get a flat middle while a high value of <span class="math-container">$α$</span> (like <span class="math-container">$1$</span>) will make flat ends thus causing a pointy middle.</p>
<p>Graphing the curvature for a few values of <span class="math-container">$t$</span> helps confirm that intuition (<a href="https://sagecell.sagemath.org/?z=eJx1UktugzAQ3UfKHUbKwjZ1EgxNNxWrHiNKKic4EYIAAieC7nuIHqP7dtMD9BA9SWdwyKcficEz8958_OSDrjhLmYS42Okkj1hldMbE_XAwHKzMU2KqkFsJpY-m0AK0UEAEXI2tWIZe6cMNhJ71XCLwSuUS6HYprwwwYYkaDgcNdUupwUV31X2pBJ8GtydKwxWMwUWjywK_Y1OZ23RUZoXl8665mgZCQnt210XNrVcmXVAneR8sJBx7KQQyszV5_JjplcmiOWuYZC0aFuMfq9hC0KQ62eYmfthXB233lSEhIE42G95IwLt2bitpZytw91N8hJoeEjC9qlySSGc2hmLJQ9yShrrb_RjNSYTuRtdpNb37D7j9G_AnMwJIj9TpMZn9VsRGPj4TPL6eX5zz8XY83xnW6sbUjlwj-_OVsHU_gwj1WmcmYrXZJVmxRXGh3eGLmyhydBMpX3wDliq5XA==&lang=sage&interacts=eJyLjgUAARUAuQ==" rel="nofollow noreferrer">source</a>)
<a href="https://i.stack.imgur.com/8oSlb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8oSlb.png" alt="curvature for various values of t" /></a></p>
<p>(<span class="math-container">$t > 1/2$</span> not included on the chart due to symmetry.)</p>
<p>Since we're approximating a unit circle we're hoping to get <span class="math-container">$k(t) ≈ 1$</span>, and those lines seem to intersect around there, at roughly <span class="math-container">$t ≈ 0.55$</span>.</p>
<p>To balance out the extremes, we can pick <span class="math-container">$α$</span> such that ends and the middle have the same curvature. To avoid weird curves we'll require <span class="math-container">$α \in (0, 1)$</span>, which simplifies a few things</p>
<p><span class="math-container">$$
\begin{align}
k(0) = k(1) &= k(1/2) \\
\frac{-2(α-1)}{3α^2} &= \frac{ 8\sqrt{2} α }{ 3(α-2)^2 } \\
{-2 (α-1) }{ 3(α-2)^2 } &= { 8\sqrt{2} α }{3α^2} \\
α &≈ 0.550581172753306
\end{align}
$$</span></p>
<p>That value gives curvature within about <span class="math-container">$±1\%$</span> for <span class="math-container">$t \in [0, 1]$</span> (<a href="https://sagecell.sagemath.org/?z=eJx9UktugzAQ3SNxB0tZ2KZOYuO6iVSx6rJHiJKKBCdCoQSBE0FXPU33vUgP0ZN0bOdDqrQSg2bmzZs3M3BIa4K3mKFs95rmZYJrnRaYPoZBGCz1W65rSQxDFQcTYDGYpChBRAwNXcio4ugOychEPhFHlfAJcF0qqmJIGFsqw6C13ba2Qa-7cM-WIW6Fu3NJSwQaIh8N-gTuqi3NTzqoip0hM9dcjGPKUHdxV7uGmKjKXdDk5SmYM3TsJQAo9EaX2UuRLnWRzHCLGe7AgAxvYOE5tUpNvil19rSvD6nZ19oeAmX5ek1ahmBX53bMzmwozH6Oj1B7gigaXzEX9kiXagjpgkiY0n-IxjyDFB8pxdVUiEk8UVLyhzDwi_-ayq01UvfKLXwLUvw25JT-Zqn_j_b1mVhV-Ju8q3jPVdPv949LrDC0SlvdeHYDdGPR1UkTDv4DnGG6FA==&lang=sage&interacts=eJyLjgUAARUAuQ==" rel="nofollow noreferrer">source</a>)
<a href="https://i.stack.imgur.com/f2ruw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/f2ruw.png" alt="enter image description here" /></a></p>
<p><em>Exercise for the reader</em>: Convince a CAS to compute</p>
<p><span class="math-container">$$
\int_0^1 (k(t) - 1)^2 \mathrm{d}t
$$</span></p>
<p>or</p>
<p><span class="math-container">$$
\int_0^1 log(k(t))^2 \mathrm{d}t
$$</span></p>
<p>and find the value of <span class="math-container">$α$</span> that actually minimizes the error in curvature -- it probably gives a slightly different value from the one I computed here.</p>
|
104,328 | <p>This question is related to <a href="https://mathematica.stackexchange.com/questions/104310/export-bug-in-combination-with-paralleltable">Export bug in combination with ParallelTable?</a>.</p>
<p>The following code shows an even stranger bug than in the upper question.</p>
<p><strong>The y-axis should always be logarithmic, but in the resulting plot files it jumps randomly between logarithmic and linear</strong>. The linear axis numbers are wrong.</p>
<pre><code>ChoiceDialog[{FileNameSetter[Dynamic[outputDir], "Directory"], Dynamic[outputDir]}];
SetDirectory[outputDir];
image = ColorConvert[ExampleData[{"TestImage", "Lena"}], "Grayscale"];
levels = ImageLevels[image, "Byte"];
hist = Histogram[WeightedData @@ Transpose[levels], 256,
ScalingFunctions -> "Log", ImageSize -> 600];
ParallelTable[
fileName = StringJoin[outputDir, "\\histogram_parallel_table_", ToString[i], ".png"];
Export[fileName, hist, "PNG"],
{i, 1, 10}
];
</code></pre>
<p><strong>This error does not occur when <code>Table</code> is used or when logarithmic scaling is removed.</strong></p>
<p>For one run I got e.g. the following two different plots (there are more):</p>
<p><code>i=1</code>:</p>
<p><a href="https://i.stack.imgur.com/CNwd7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CNwd7.png" alt="enter image description here"></a></p>
<p><code>i=6</code>:</p>
<p><a href="https://i.stack.imgur.com/EHAcy.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EHAcy.png" alt="enter image description here"></a></p>
<p>I am programming with Mathematica 10.3.1.0 on Windows 10 Professional 64 Bit and have an i7-4940-MX 3,1 GHz processor (4 cores).</p>
| Ymareth | 880 | <p>Confirmation rather than answer. I'm not sure the original example is entirely functional, at least it was not for me. Mine is...</p>
<pre><code>SetDirectory["F:\\Temp"]; (* Adjust to suit your environment *)
hist = Histogram[RandomVariate[NormalDistribution[0, 1], 10000], ScalingFunctions -> "Log", ImageSize -> 600]
</code></pre>
<p>Note log scale on y-axis in resulting histogram.</p>
<pre><code>DistributeDefinitions[hist]; (*missing in original - seems necessary or remote kernels don't know what hist is.*)
ParallelEvaluate[Export[StringJoin["histogram_parallel_table_",ToString[$KernelID],".png"],hist,"PNG"]];
</code></pre>
<p>This gives me a log scale in all 8 versions produced BUT only the one from kernel 1 has the scaled labelled with log intervals (1,10,100,1000). 2-8 are labelled (1,2,3,4,5,6,7,8); Ln(count).</p>
<p>From Kernel 1...</p>
<p><a href="https://i.stack.imgur.com/yGJiw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yGJiw.png" alt="From kernel 1"></a></p>
<p>From Kernels 2-8...</p>
<p><a href="https://i.stack.imgur.com/vImXH.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vImXH.png" alt="enter image description here"></a></p>
|
2,655,178 | <p>I am asked to find the equation of a cubic function that passes through the origin. It also passes through the points $(1, 3), (2, 6),$ and $(-1, 10)$. </p>
<p>I have walked through many answers for similar questions that suggest to use a substitution method by subbing in all the points and writing in terms of variables. I have tried that but I don't really know where to take it from there or what variables to write it as. </p>
<p>If anyone could provide their working out for this problem it would be extremely enlightening. </p>
| Siong Thye Goh | 306,553 | <p>Guide:</p>
<p>Let the equation be $y=ax^3+bx^2+cx+d$, since it passes through $(1,3)$, we have</p>
<p>$$a(1)^3+b(1)^2+c(1)+d=3$$</p>
<p>Do the same thing for the other $3$ points. </p>
<p>Hence you will obtain $4$ linear equation with $4$ variables.</p>
<p>You can then solve it using elementary row operations to recover $a,b,c,d$.</p>
|
2,329,003 | <p>Parametrise the circle centered at $ \ (1,1,-1) \ $ with radius equal to $ 3 $ in the plane $ x+y+z=1 $ with positive orientation . $$ $$ I have thought the parametriation: </p>
<p>\begin{align} x(t)=1+ 3 \cos (t) \hat j +3 \sin (t) \hat k \\ y(t)=1+3 \cos (t) \hat i+3 \sin (t) \hat k \\ z(t)=-1+3 \cos (t) \hat i +3 \sin (t) \hat j , \ \ 0 \leq t \leq 2 \pi \end{align} But I am not sure . <strong>Any help is there ?</strong></p>
| Doug M | 317,162 | <p>$\mathbf u = (\frac {\sqrt 2}{2} \mathbf i - \frac {\sqrt 2}{2} \mathbf j)\\
\mathbf v = (\frac {\sqrt 6}{6} \mathbf i + \frac {\sqrt 6}{6} \mathbf j -\frac {\sqrt 6}{3} \mathbf k)$</p>
<p>$(x,y,z) = (1,1,-1) + 3\mathbf u \cos t + 3\mathbf v \sin t\\
x = 1 + 3\frac {\sqrt 2}{2} \cos t + \frac{\sqrt {6}}{2} \sin t\\
y = 1 - 3\frac {\sqrt 2}{2} \cos t + \frac {\sqrt {6}}{2} \sin t\\
z = -1 - \sqrt {6} \sin t$</p>
|
212,466 | <p>Consider the list <code>list={1,2,3,4,5,6,7,8}</code> and imagine I wanted to change it to <code>{{1,2,3},{4,5,6},{7,8}}</code>. How can I do this?</p>
<p>Using <code>ArrayReshape</code> in the usual manner</p>
<pre><code>list = {1, 2, 3, 4, 5, 6, 7, 8};
ArrayReshape[list, {3, 3}]
</code></pre>
<p>yields <code>{{1, 2, 3}, {4, 5, 6}, {7, 8, 0}}</code>. Is it possible to use <code>ArrayReshape</code> and avoid that extra 0 at the end? Essentially, I want to reshape my list according to its size.</p>
| kglr | 125 | <p>It seems that using <code>ArrayReshape</code> and modifying, if needed, the last entry to remove trailing zeros is faster than using <code>Partition</code> or using <code>Inactive @ Nothing</code> padding:</p>
<pre><code>ClearAll[f1, f2, f3, f4]
f1 = Partition[#, UpTo[#2]] &; (* from Okkes's answer *)
f2 = Partition[#, #2, #2, 1, {}] &;
f3 = Activate @ ArrayReshape[#,
{Ceiling[Length[#]/#2], #2}, Inactive @ Nothing] &; (* from MarcoB's answer *)
f4 = Module[{p = ArrayReshape[#, {Floor[Length[#]/#2], #2}], m = Mod[Length@#, #2]},
If[m == 0, p, Join @@ {p, {#[[-m ;;]]}}]] &;
</code></pre>
<p><strong><em>Timings:</em></strong></p>
<pre><code>n = 999999;
k = 199;
list = Range[n];
First[RepeatedTiming[r1 = f1[list, k]]]
</code></pre>
<blockquote>
<p>0.078</p>
</blockquote>
<pre><code>First[RepeatedTiming[r2 = f2[list, k]]]
</code></pre>
<blockquote>
<p>0.075</p>
</blockquote>
<pre><code>First[RepeatedTiming[r3 = f3[list, k]]]
</code></pre>
<blockquote>
<p>0.094</p>
</blockquote>
<pre><code>First[RepeatedTiming[r4 = f4[list, k]]]
</code></pre>
<blockquote>
<p>0.012</p>
</blockquote>
<pre><code>r1 == r2 == r3 == r4
</code></pre>
<blockquote>
<p>True</p>
</blockquote>
<p>When <code>n</code> is divisible by <code>k</code> (so that <code>ArrayReshape</code> produces a rectangular array with no paddings):</p>
<pre><code>k = 99;
First[RepeatedTiming[r1 = f1[list, k]]]
</code></pre>
<blockquote>
<p>0.083</p>
</blockquote>
<pre><code>First[RepeatedTiming[r2 = f2[list, k]]]
</code></pre>
<blockquote>
<p>0.075</p>
</blockquote>
<pre><code>First[RepeatedTiming[r3 = f3[list, k]]]
</code></pre>
<blockquote>
<p>0.0072</p>
</blockquote>
<pre><code>First[RepeatedTiming[r4 = f4[list, k]]]
</code></pre>
<blockquote>
<p>0.0071</p>
</blockquote>
<pre><code>r1 == r2 == r3 == r4
</code></pre>
<blockquote>
<p>True</p>
</blockquote>
|
212,466 | <p>Consider the list <code>list={1,2,3,4,5,6,7,8}</code> and imagine I wanted to change it to <code>{{1,2,3},{4,5,6},{7,8}}</code>. How can I do this?</p>
<p>Using <code>ArrayReshape</code> in the usual manner</p>
<pre><code>list = {1, 2, 3, 4, 5, 6, 7, 8};
ArrayReshape[list, {3, 3}]
</code></pre>
<p>yields <code>{{1, 2, 3}, {4, 5, 6}, {7, 8, 0}}</code>. Is it possible to use <code>ArrayReshape</code> and avoid that extra 0 at the end? Essentially, I want to reshape my list according to its size.</p>
| Mr.Wizard | 121 | <p><a href="https://reference.wolfram.com/language/ref/ArrayReshape.html" rel="nofollow noreferrer"><code>ArrayReshape</code></a> is implemented using <code>Internal`Deflatten</code> (at least in version 10.1). This function requires <span class="math-container">$n \times m$</span> elements, or at least behaves strangely when this is not provided. I am guessing this relates to memory pointers in the low-level implementation.</p>
<pre><code>Internal`Deflatten[Range@8, {3, 3}]
Internal`Deflatten[Range@8, {3, 3}]
Internal`Deflatten[Range@8, {3, 3}]
Internal`Deflatten[Range@8, {3, 3}]
</code></pre>
<blockquote>
<pre><code>{{1, 2, 3}, {4, 5, 6}, {7, 8, 281500746514433}}
{{1, 2, 3}, {4, 5, 6}, {7, 8, 300647710721}}
{{1, 2, 3}, {4, 5, 6}, {7, 8, 411062144}}
{{1, 2, 3}, {4, 5, 6}, {7, 8, 7164775599194924370}}
</code></pre>
</blockquote>
<p><em>Note:</em> If we unpack the list it doesn't work at all:</p>
<pre><code>list = List @@ Range@8;
Developer`PackedArrayQ[list]
Internal`Deflatten[list, {3, 3}]
</code></pre>
<blockquote>
<pre><code>False
Internal`Deflatten[{1, 2, 3, 4, 5, 6, 7, 8}, {3, 3}]
</code></pre>
</blockquote>
<p>This is ultimately why something like this also fails:</p>
<pre><code> ArrayReshape[Range@8, {3, 3}, Unevaluated @ Unevaluated[## &[]] ] (* does not work *)
</code></pre>
<p>To avoid this any input to <code>ArrayReshape</code> is first padded with <a href="https://reference.wolfram.com/language/ref/ArrayPad.html" rel="nofollow noreferrer"><code>ArrayPad</code></a> or trimmed to <span class="math-container">$length = n \times m$</span>.</p>
<p><a href="https://reference.wolfram.com/language/ref/Partition.html" rel="nofollow noreferrer"><code>Partition</code></a> drops in speed when using <code>{}</code> padding, at least in some cases, probably due to working around a similar low-level implementation. I used a method similar to kglr's but for <code>Partition</code> here:</p>
<ul>
<li><a href="https://mathematica.stackexchange.com/q/28416/121#28428">Reversibly merging sets of $k$ adjacent elements in an array</a></li>
<li><a href="https://mathematica.stackexchange.com/q/128133/121">Partition array without unpacking</a></li>
</ul>
<p>Closely related:</p>
<ul>
<li><a href="https://mathematica.stackexchange.com/q/58215/121">UnFlatten an array into a ragged matrix</a></li>
</ul>
<hr>
<h2>Silly games</h2>
<p>Just having fun using the above knowledge to break <code>ArrayReshape</code>:</p>
<pre><code>ClearAll[foo]
foo /: {x__Integer, foo[] ..} := Developer`ToPackedArray[{x}]
ArrayReshape[{1, 2, 3, 4, 5}, {3, 3}, foo[]]
</code></pre>
<blockquote>
<pre><code>{{1, 2, 3}, {4, 5, 470483008}, {36571776, 8, 5100273664}}
</code></pre>
</blockquote>
|
2,054,338 | <p>I had this question on my exam and I thought I could solve it using the row echelon method. Well, I couldn't, and I still don't know how to solve it. We were asked to "determine $a$ such the equation system has a unique solution" and to determine $a$ itself. The system had three equations: </p>
<p>$2x - y + az = 3$ </p>
<p>$3x - 4y + 2az = 1$</p>
<p>$x + y - z = 2$.</p>
<p>Could you provide me with a step-by-step of how to solve this?</p>
<p>Thank you</p>
| Mitchell Spector | 350,214 | <p>No calculus needed:</p>
<p>\begin{align}
\sum_{k=1}^\infty k(1-a)^{k-1}a &= \sum_{k=1}^\infty \sum_{j=1}^k (1-a)^{k-1}a
\\&= \sum_{j=1}^\infty \sum_{k=j}^\infty (1-a)^{k-1}a
\\&= \sum_{j=1}^\infty \frac{(1-a)^{j-1}}{1-(1-a)}a
\\&= \sum_{j=1}^\infty (1-a)^{j-1}
\\&=\frac1{1-(1-a))}
\\&=\frac1{a}.
\end{align}</p>
<p>The rearrangements are justified by the absolute convergence of the geometric series in question since $0\lt a \lt 1.$</p>
|
2,044,362 | <p>(<em>This summarizes scattered results from <a href="https://math.stackexchange.com/questions/879089/prove-2f-1-left-frac13-frac13-frac56-27-right-stackrel-color808080">here</a>, <a href="https://math.stackexchange.com/questions/879854/prove-large-int-11-fracdx-sqrt394-sqrt5-x-left1-x2-right">here</a>, <a href="https://math.stackexchange.com/questions/1326557/integral-large-int-0-infty-fracdx-sqrt47-cosh-x">here</a> and elsewhere. See also this <a href="https://math.stackexchange.com/questions/2043030/closed-forms-for-int-0-infty-fracdx-sqrt355-cosh-x-and-int-0-inf">older post</a></em>.)</p>
<blockquote>
<p><strong>I. Cubic</strong></p>
</blockquote>
<p>Define $\beta= \tfrac{\Gamma\big(\tfrac56\big)}{\Gamma\big(\tfrac13\big)\sqrt{\pi}}= \frac{1}{48^{1/4}\,K(k_3)}$. Then we have the nice evaluations,</p>
<p>$$\begin{aligned}\frac{3}{5^{5/6}} &=\,_2F_1\big(\tfrac{1}{3},\tfrac{1}{3};\tfrac{5}{6};-4\big)\\
&=\beta\,\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt[3]{x^2+4x^3}}\\[1.7mm]
&=\beta\,\int_{-1}^1\frac{dx}{\left(1-x^2\right)^{\small2/3} \sqrt[3]{\color{blue}{9+4\sqrt{5}}\,x}}\\[1.7mm]
&=2^{1/3}\,\beta\,\int_0^\infty\frac{dx}{\sqrt[3]{9+\cosh x}}
\end{aligned}\tag1$$
and,
$$\begin{aligned}\frac{4}{7} &=\,_2F_1\big(\tfrac{1}{3},\tfrac{1}{3};\tfrac{5}{6};-27\big)\\
&=\beta\,\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt[3]{x^2+27x^3}}\\[1.7mm]
&=\beta\,\int_{-1}^1\frac{dx}{\left(1-x^2\right)^{\small2/3} \sqrt[3]{\color{blue}{55+12\sqrt{21}}\,x}}\\[1.7mm]
&=2^{1/3}\,\beta\,\int_0^\infty\frac{dx}{\sqrt[3]{55+\cosh x}}
\end{aligned}\tag2$$
Note the powers of <em><a href="http://mathworld.wolfram.com/FundamentalUnit.html" rel="nofollow noreferrer">fundamental units</a></em>,
$$U_{5}^6 = \big(\tfrac{1+\sqrt{5}}{2}\big)^6=\color{blue}{9+4\sqrt{5}}$$
$$U_{21}^3 = \big(\tfrac{5+\sqrt{21}}{2}\big)^3=\color{blue}{55+12\sqrt{21}}$$
<em>Those two instances can't be coincidence.</em></p>
<blockquote>
<p><strong>II. Quartic</strong></p>
</blockquote>
<p>Define $\gamma= \tfrac{\sqrt{2\pi}}{\Gamma^2\big(\tfrac14\big)}= \frac{1}{2\sqrt2\,K(k_1)}=\frac1{2L}$ with <em><a href="http://mathworld.wolfram.com/LemniscateConstant.html" rel="nofollow noreferrer">lemniscate constant</a></em> $L$. Then we have the nice,</p>
<p>$$\begin{aligned}\frac{2}{3^{3/4}} &=\,_2F_1\big(\tfrac{1}{4},\tfrac{1}{4};\tfrac{3}{4};-3\big)\\
&=\gamma\,\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt[4]{x^3+3x^4}}\\[1.7mm]
&\overset{\color{red}?}=\gamma\,\int_{-1}^1\frac{dx}{\left(1-x^2\right)^{\small3/4} \sqrt[4]{\color{blue}{7+4\sqrt{3}}\,x}}\\[1.7mm]
&=2^{1/4}\,\gamma\,\int_0^\infty\frac{dx}{\sqrt[4]{7+\cosh x}}
\end{aligned}\tag3$$
and,
$$\begin{aligned}\frac{3}{5}&=\,_2F_1\big(\tfrac{1}{4},\tfrac{1}{4};\tfrac{3}{4};-80\big)\\
&=\gamma\,\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt[4]{x^3+80x^4}}\\[1.7mm]
&\overset{\color{red}?}=\gamma\,\int_{-1}^1\frac{dx}{\left(1-x^2\right)^{\small3/4} \sqrt[4]{\color{blue}{161+72\sqrt{5}}\,x}}\\[1.7mm]
&=2^{1/4}\,\gamma\,\int_0^\infty\frac{dx}{\sqrt[4]{161+\cosh x}}
\end{aligned}\tag4$$</p>
<p>with $a=161$ given by Noam Elkies in this <a href="https://math.stackexchange.com/questions/1326557/integral-large-int-0-infty-fracdx-sqrt47-cosh-x?noredirect=1&lq=1#comment2706992_1326557">comment</a>. (For $4$th roots, I just assumed the equality using the blue radicals based on the ones for cube roots.) Note again the powers of fundamental units,
$$U_{3}^2 = \big(2+\sqrt3\big)^2=\color{blue}{7+4\sqrt{3}}$$
$$U_{5}^{12} = \big(\tfrac{1+\sqrt{5}}{2}\big)^{12}=\color{blue}{161+72\sqrt{5}}$$
<em>Just like for the cube roots version, these can't be coincidence.</em></p>
<blockquote>
<p><strong>Questions:</strong></p>
</blockquote>
<p>Is it true these observations can be explained by, let $b=2a+1$, then,</p>
<p>$$\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt[3]{x^2+ax^3}}=\int_{-1}^1\frac{dx}{\left(1-x^2\right)^{\small2/3} \sqrt[3]{b+\sqrt{b^2-1}\,x}}=2^{1/3}\int_0^\infty\frac{dx}{\sqrt[3]{b+\cosh x}}$$</p>
<p>$$\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt[4]{x^3+ax^4}}=\int_{-1}^1\frac{dx}{\left(1-x^2\right)^{\small3/4} \sqrt[4]{b+\sqrt{b^2-1}\,x}}=2^{1/4}\int_0^\infty\frac{dx}{\sqrt[4]{b+\cosh x}}$$</p>
| Lucian | 93,448 | <p><strong>Too long for a comment :</strong> In general, for strictly positive values of <em>n</em> we have </p>
<blockquote>
<p>$$\begin{align}
\sqrt[n]2\int_0^\infty\frac{dx}{\sqrt[n]{\cosh2t~+~\cosh x}}
~&=~
\int_0^1\frac{dx}{\sqrt{1-x}\cdot\sqrt[n]{x^{n-1}~+~x^n\cdot\sinh^2t}}
\\\\
~&=~
\int_{-1}^1\frac{dx}{\sqrt[n]{(1-x^2)^{n-1}}\cdot\sqrt[n]{\cosh2t~+~x\cdot\sinh2t}}
\end{align}$$</p>
</blockquote>
|
2,411,811 | <p>Determine the limit:</p>
<p>$$\lim_{h\to 0}\frac{1-\cos(2h)}{h}$$</p>
<p>I know that the answer is $0$, I am just unsure of how to solve this. Thanks.</p>
| farruhota | 425,072 | <p>The third method is to multiply and divide by conjugate of numerator:
$$\lim_{h\to 0}\frac{1-\cos(2h)}{h}=\lim_{h\to 0}\frac{1-\cos(2h)}{h} \cdot \frac{1+\cos(2h)}{1+\cos(2h)}=\lim_{h\to 0}\frac{\sin^2(2h)}{h(1+\cos(2h))}=$$
$$\lim_{h\to 0}\frac{\sin^2(2h)}{2h}=\lim_{h\to 0}\frac{\sin(2h)}{2h}\cdot \lim_{h\to 0} \sin(2h)=1\cdot0=0.$$
The fourth method is to use L'Hospital's rule:
$$\lim_{h\to 0}\frac{1-\cos(2h)}{h}=\lim_{h\to 0}\frac{2\sin(2h)}{1}=0.$$</p>
|
17,819 | <p>$\mathfrak{sl}_2(\mathbb{C})$ is usually given a basis $H, X, Y$ satisfying $[H, X] = 2X, [H, Y] = -2Y, [X, Y] = H$. What is the origin of the use of the letter $H$? (It certainly doesn't stand for "Cartan.") My guess, based on similarities between these commutator relations and ones I have seen mentioned when people talk about physics, is that $H$ stands for "Hamiltonian." Is this true? Even if it's not, is there a connection? </p>
| LSpice | 2,383 | <p>I always thought that $H$ <em>did</em> stand for Cartan—at least, for ‘Henri’. However, I seem to recall that I had this discussion with <a href="https://mathoverflow.net/users/3927/brian-conrad">Brian Conrad</a>, and that he said it was actually Élie, not Henri, after whom the subgroups were named.</p>
<p>For what it's worth, the $(X, Y, H)$ convention (in preference to $(E, F, H)$) is the one to which I'm accustomed; Carter uses it, for example, in his discussion of Jacobson–Morosov triples. Embarrassingly, I don't know where the Jacobson–Morosov theorem was proven; but that's where I'd look for the history of the name.</p>
|
1,612,859 | <p>Take $i(G)$ to be the independence number of $G$, i.e. the maximum
number of
pairwise nonadjacent vertices in $G.$ I want to show that if $G$ has
$n$ vertices and $\frac{nk}{2}$ edges where $k \geq 1,$
$i(G) \geq \frac{n}{k+1}.$ I'm having difficulties making the
combinatorial argument on this graph. I get that if the graph has
$\frac{nk}{2}$ edges, $2|n \vee 2|k.$ However, beyond
this, I am having few insights. I am wondering, how might I best
represent pairwise nonadjacent vertices using this knowledge of
the number of edges in this graph?</p>
| Brian M. Scott | 12,042 | <p>I'm assuming that the word <em>then</em> is missing after $k\ge 1$. Suppose that $G$ is a counterexample: $G$ has $n$ vertices and $\frac{nk}2$ edges, and $i(G)<\frac{n}{k+1}$. Let $V_1$ be a maximal independent set of vertices in $G$, $V_2$ a maximal independent set of vertices in $G-V_1$, and so on through $V_m$ for some $m\in\Bbb Z^+$. Note that $m>k+1$; why?</p>
<p>Clearly $|V_j|<\frac{n}{k+1}$ for $j=1,\ldots,m$. Moreover, the maximality of each $V_j$ ensures that if $1\le j<\ell\le m$, and $v\in V_\ell$, then there is at least one edge between $v$ and some vertex in $V_j$.</p>
<p>There are $n-|V_1|>n-\frac{n}{k+1}=\frac{kn}{k+1}$ vertices in $\bigcup_{j=2}^mV_j$, each of which is connected by an edge to a vertex in $V_1$; this accounts for more than $\frac{ kn}{k+1}$ edges of $G$. Similarly, there are more than $\frac{(k-1)n}{k+1}$ vertices in $\bigcup_{j=3}^mV_j$, each of which is connected by an edge to a vertex in $V_2$; this accounts for more than another $\frac{(k-1)n}{k+1}$ edges of $G$. </p>
<p>Continuing in this fashion, we find that $G$ has more than</p>
<p>$$\sum_{j=1}^k\frac{jn}{k+1}=\frac{n}{k+1}\sum_{j=1}^kj$$</p>
<p>edges; can you see why this is a contradiction?</p>
|
1,836,785 | <p>Following are the two theorems that Hardy and Wright state in their book</p>
<blockquote>
<p><strong>Theorem A:</strong> The number of primes not exceeding <span class="math-container">$x$</span> is given by <span class="math-container">$\pi(x) \sim \frac{x}{\log{x}}$</span>.</p>
<p><strong>Theorem B</strong>: The order of magnitude of <span class="math-container">$\pi(x)$</span> is <span class="math-container">$\pi(x) \asymp \frac{x}{\log{x}}$</span>.</p>
</blockquote>
<p>where,</p>
<ol>
<li><span class="math-container">$f \sim \phi$</span> iff <span class="math-container">$\; \frac{f}{\phi} \to 1\;$</span>, ie. two the two functions are asymptotically similar.</li>
<li><span class="math-container">$f \asymp \phi$</span> iff <span class="math-container">$\;A^{'}\phi < f < A \phi\;$</span> ie. <span class="math-container">$\;f\;$</span> is of same order of magnitude as <span class="math-container">$\phi$</span> ( <span class="math-container">$A$</span> and <span class="math-container">$A^{'}$</span> are constants ).</li>
</ol>
<p>How are the two theorems different ? Am I missing something trivial ?</p>
| Community | -1 | <p>They are slightly different. The first gives a limit asymptotic constant equal to $1$, but no bracketing. The second doesn't give a single asymptotic constant, but a bracketing (for $x$ large enough). Obviously, $A'\le1\le A$.</p>
|
1,041,623 | <p>I have been suggested to read the <a href="http://press.princeton.edu/chapters/gowers/gowers_VIII_6.pdf"><em>Advice to a Young Mathematician</em> section </a> of the <em>Princeton Companion to Mathematics</em>, the short paper <a href="http://alumni.media.mit.edu/~cahn/life/gian-carlo-rota-10-lessons.html"><em>Ten Lessons I wish I had been Taught</em></a> by Gian-Carlo Rota, and the <a href="http://terrytao.wordpress.com/career-advice/"><em>Career Advice</em></a> section of Terence Tao's blog, and I am amazed by the intelligence of the pieces of advice given in these pages. </p>
<p>Now, I ask to the many accomplished mathematicians who are active on this website if they would mind adding some of their own contributions to these already rich set of advice to novice mathematicians. </p>
<p>I realize that this question may be seen as extremely opinion-based. However, I hope that it will be well-received (and well-answered) because, as Timothy Gowers put it,</p>
<blockquote>
<p>"The most important thing that a young mathematician needs to learn is
of course mathematics. However, it can also be very valuable to learn
from the experiences of other mathematicians. The five contributors to
this article were asked to draw on their experiences of mathematical
life and research, and to offer advice that they might have liked to
receive when they were just setting out on their careers."</p>
</blockquote>
| domotorp | 88,597 | <p>While there are many excellent suggestions, I would like to add that it is crucial to go abroad/other cities to get acquainted with many people, from whom you can learn a lot in many ways. This is something that's not good to delay, as later you might have no chance for family reasons!</p>
|
1,041,623 | <p>I have been suggested to read the <a href="http://press.princeton.edu/chapters/gowers/gowers_VIII_6.pdf"><em>Advice to a Young Mathematician</em> section </a> of the <em>Princeton Companion to Mathematics</em>, the short paper <a href="http://alumni.media.mit.edu/~cahn/life/gian-carlo-rota-10-lessons.html"><em>Ten Lessons I wish I had been Taught</em></a> by Gian-Carlo Rota, and the <a href="http://terrytao.wordpress.com/career-advice/"><em>Career Advice</em></a> section of Terence Tao's blog, and I am amazed by the intelligence of the pieces of advice given in these pages. </p>
<p>Now, I ask to the many accomplished mathematicians who are active on this website if they would mind adding some of their own contributions to these already rich set of advice to novice mathematicians. </p>
<p>I realize that this question may be seen as extremely opinion-based. However, I hope that it will be well-received (and well-answered) because, as Timothy Gowers put it,</p>
<blockquote>
<p>"The most important thing that a young mathematician needs to learn is
of course mathematics. However, it can also be very valuable to learn
from the experiences of other mathematicians. The five contributors to
this article were asked to draw on their experiences of mathematical
life and research, and to offer advice that they might have liked to
receive when they were just setting out on their careers."</p>
</blockquote>
| fosco | 685 | <p>I don't know how many of these advices are already present in the pdfs, but I found these really valuable pieces of advice.</p>
<ol start="0">
<li>Choose a subject, an area of mathematics, which is "your favourite one". Live there as it was your home.</li>
<li>Relentlessly go back to the very basic fundamentals of that subject. Re-study everything from scratch once a year, re-do things you know using all you've learned in the last months. Do what professional basketball players do: fundamentals, all the time.</li>
<li>Don't wait for others to learn what you want to learn. The question "Hi, I took only a course in algebra, but I want to have an idea of what the hell is Galois theory." is perfectly legitimate, and it's your teacher's fault if they can't give you a simple, well posed and enlightening elementary example.</li>
<li>Recall yourself that old mathematics done in a deeper and more elegant way is new mathematics. This might be a very opinion-based piece of advice, and yet.</li>
<li>Don't fear to travel outside your preferred field. Your home will look the same, but totally different after each trip.</li>
<li>Don't indulge in the thought that you don't want to check if an idea is a good idea because it might be wrong and spoil your last month's work. We are already ignorant about almost everything in mathematics, there is no need to be also coward.</li>
</ol>
|
2,377,139 | <p>My question:</p>
<p>Characterize those power series $\sum_{k=0}^\infty a_{k}(x-c)^{k}$ that converge uniformly on ($-\infty, \,\infty$).
What does it mean to characterize a power series?</p>
<p>Let {$a_{k}$} be a sequence of coefficients for a power series. By definition, the radius of convergence $R = \infty$ if $\limsup\sqrt[k]{|a_{k}|} = 0$. </p>
| lhf | 589 | <p>A <em>characterization</em> in this case is a condition on the series (that is, on its coefficients) that is equivalent to uniform convergence in the whole real line.</p>
|
979,947 | <p>If $(A, B, C)$ are distinct integers $> 1$, and $$f(A, B, C) = \frac{\frac{A^2-1}{A} + \frac{B^2-1}{B}}{\frac{C^2-1}{C}},$$ then for what (if any) triplets $(A, B, C)$ is $f(A, B, C)$ an integer?</p>
<p>UPDATE: I've done some more work and have come up with the following: Based on the equations I combined to arrive at $f(A,B,C)$, it follows that: $$f(A,B,C)>2\Rightarrow A>B>C>1$$ and $$f(A,B,C)=2\Rightarrow A>C>B>1$$</p>
<p>My further work: </p>
<p>Let $f(A,B,C)=k$ where $k>1$ is an integer. Rewrite $f(A,B,C)$ as $$k=\frac{A-\frac{1}{A}+B-\frac{1}{B}}{C-\frac{1}{C}}$$</p>
<p>Thus $$A-\frac{1}{A}+B-\frac{1}{B}-k(C-\frac{1}{C})=A-\frac{1}{A}+B-\frac{1}{B}-kC+\frac{k}{C}=0$$</p>
<p>Note that $\frac{1}{A}+\frac{1}{B}<1$ for all integer $A,B$ such that $A>B>1$.</p>
<p>Now we consider three cases: $1)$ $k<C$, $2)$ $k=C$, $3)$ $k>C$.</p>
<p>Case $1$: $k<C$</p>
<p>Since $\frac{k}{C}<1$ it follows that these equations hold:$$A+B=kC$$ $$\frac{1}{A}+\frac{1}{B}=\frac{A+B}{AB}=\frac{k}{C}$$ </p>
<p>By substitution $$\frac{kC}{AB}=\frac{k}{C}\Rightarrow AB=C^2$$</p>
<p>This is only true if $A=np^2$, $B=nq^2$, and $C=npq$ for integers $n,p,q\geq 1$ and $p>q$. From this we have $$p^2>pq; pq>q^2\Rightarrow p^2>pq>q^2\Rightarrow np^2>npq>nq^2\Rightarrow A>C>B>1\Rightarrow k\leq2$$ This last condition, $k\leq2$, follows from the conditions stated above and leaves us with two possibilities: $k=2$ or $k=1$.</p>
<p>If $k=1$ we have $A+B=C$ which violates the condition $A>C$.</p>
<p>If $k=2$ we have $A+B=2C\Rightarrow \frac{A+B}{2}=C$. By substitution we have $$\frac{A+B}{AB}=\frac{2}{\frac{A+B}{2}}$$ $$4AB=(A+B)^2\Rightarrow A^2-2AB+B^2=0\Rightarrow (A-B)^2=0$$</p>
<p>However, this leads to the result that $A=B$ which violates that stated conditions.</p>
<p>Case $1$, $k<C$, fails.</p>
<p>Case $2$: $k=C$</p>
<p>Since $\frac{k}{C}=1$ it follows that $$A+B+1=C^2$$ $$\frac{1}{A}+\frac{1}{B}=0$$</p>
<p>This later equation is clearly false and thus Case $2$, $k=C$ fails.</p>
<p>Case $3$: $k>C$</p>
<p>Let $p,q$ be integers with $C>q\geq0$ and $p>1$. From this we can write $k$ as $k=pC+q$.</p>
<p>Note that $q=0 \Rightarrow A+B+p=C^2$ and $\frac{1}{A}+\frac{1}{B}=0$ (as above) which is clearly a contradiction, so $q\geq1$.</p>
<p>Substituting in $k$ above we have $$A-\frac{1}{A}+B-\frac{1}{B}-pC^2-qC+p+\frac{q}{C}=0$$ </p>
<p>Applying the same logic as above we have the following equations $$\frac{A+B}{AB}=\frac{q}{C}$$ $$A+B=pC^2+qC-p$$</p>
<p>By substitution we have the following system of equations (one cubic and one quadratic) $$pC^3+qC^2-pC=qAB\Rightarrow pC^3+qC^2-pC-qAB=0$$ $$pC^2+qC-p-A-B=0$$</p>
<p>Since at least one $C$ that works must solve both equations we denote the roots of the cubic as $x_1,x_2,x_3$ and the roots of the quadratic as $x_1,y_2$.</p>
<p>By Vieta's formulas we have $$x_1+x_2+x_3=-\frac{q}{p}$$ $$x_1 x_2+x_1 x_3+x_2 x_3=-1$$ $$x_1 x_2 x_3=\frac{qAB}{p}$$</p>
<p>and $$x_1+y_2=-\frac{q}{p}$$ $$x_1 y_2= -\frac{p+A+B}{p}$$</p>
<p>From the first equations (setting $x_1+x_2+x_3=x_1+y_2$) we have $y_2=x_2+x_3$.</p>
<p>Thus $$x_1 y_2=x_1 x_2+x_1 x_3=-\frac{p+A+B}{p}\Rightarrow x_2 x_3-\frac{p+A+B}{p}=-1\Rightarrow x_2 x_3=\frac{A+B}{p}$$ $$x_2 x_3=\frac{A+B}{p}\Rightarrow x_1\frac{A+B}{p}=\frac{qAB}{p}\Rightarrow x_1=\frac{qAB}{A+B}$$</p>
<p>Since $x_1=C$ and $B>C$ we have $$B>\frac{qAB}{A+B}\Rightarrow B^2>(q-1)AB$$</p>
<p>Since $A>B$ it follows $AB>B^2$ and so $q-1<1\Rightarrow q<2$. Since $q\neq0$ (see above) we have $q=1$ and so $C=\frac{AB}{A+B}$.</p>
<p>Substituting for $C$ in the quadratic (although the cubic has the same result) we have $$0=p(\frac{AB}{A+B})^2+\frac{AB}{A+B}-p-A-B$$ Which, after a fair bit of rearranging, becomes $$p=\frac{(A+B)^3-AB(A+B)}{(AB)^2-(A+B)^2}$$ Thus $f(A,B,C)$ is an integer (with $A,B,C$ constrained as above) iff $$p=\frac{(A+B)^3-AB(A+B)}{(AB)^2-(A+B)^2}$$ has integer solutions $p>0$ and $A>B>2$.</p>
<p>Any help with solving this last bit would be greatly appreciated. So far guesswork and Wolfram Alpha have failed to produce results.</p>
| Mathink | 166,778 | <p>You shouldn't guess for two reasons:
1-If you're wrong, you're giving proof that you haven't learned your lessons, rules, theorems...
2-You want to know your real level through SAT. If by guessing, you get some correct answers, then the score you'll have won't reflect your real value and skills.</p>
|
4,380,862 | <p>How can I solve the tangent point and <span class="math-container">$a$</span> when <span class="math-container">$f(x)=\left(\log_{a}{x}\right)^2$</span> is tangent to <span class="math-container">$g(x)=-ax+2$</span>?</p>
<p>Although this can be solved by substituting <span class="math-container">$a=e^2$</span> and <span class="math-container">$x=e^{-2}$</span>, then</p>
<p><span class="math-container">$f\left(e^{-2}\right)=g\left(e^{-2}\right)$</span> and <span class="math-container">$f^\prime\left(e^{-2}\right)=g^\prime\left(e^{-2}\right)$</span> can be proved,</p>
<p>is there any general solution for this, rather than just substituting random numbers? Maybe using Lambert-W could help?</p>
| IV_ | 292,527 | <p>for real <span class="math-container">$a,x$</span></p>
<p><span class="math-container">$$\frac{d}{dx}\left(\log_a(x)^2\right)=\frac{d}{dx}\left(-ax+2\right)$$</span>
<span class="math-container">$$\frac{2\ln(x)}{\ln(a)^2x}=-a$$</span>
<span class="math-container">$a\to e^t$</span>:
<span class="math-container">$$t^2e^t=-\frac{2\ln(x)}{x}$$</span>
<span class="math-container">$$\sqrt{t^2e^t}=\sqrt{-2\frac{\ln(x)}{x}}$$</span>
<span class="math-container">$$te^{\frac{1}{2}t}=\pm\sqrt{-2\frac{\ln(x)}{x}}$$</span>
<span class="math-container">$$\frac{1}{2}te^{\frac{1}{2}t}=\pm\frac{1}{2}\sqrt{-2\frac{\ln(x)}{x}}$$</span>
<span class="math-container">$k\in\{-1,0\}$</span>:
<span class="math-container">$$\frac{1}{2}t=W_k\left(\pm\frac{1}{2}\sqrt{-2\frac{\ln(x)}{x}}\right)$$</span>
<span class="math-container">$$t=2W_k\left(\pm\frac{1}{2}\sqrt{-2\frac{\ln(x)}{x}}\right)$$</span>
<span class="math-container">$$a=e^{2W_k\left(\pm\frac{1}{2}\sqrt{-2\frac{\ln(x)}{x}}\right)}$$</span>
<span class="math-container">$$a=-\frac{1}{2}\frac{\ln(x)}{x\ W_k\left(\pm\frac{1}{2}\sqrt{-2\frac{\ln(x)}{x}}\right)^2}$$</span>
<span class="math-container">$$a=-\frac{1}{2}\frac{\ln(x)}{x\ W_k\left(\pm\frac{1}{\sqrt{-\frac{2x}{\ln(x)}}}\right)^2}$$</span></p>
|
278,637 | <p>After spending more than 1 hr on this, and looking at many questions, I give up as I am not able to figure a solution.</p>
<p>I have my current package in the standard location given by <code>FileNameJoin[{$UserBaseDirectory, "Applications"}]</code> which on windows is</p>
<pre><code>C:\Users\Owner\AppData\Roaming\Mathematica\Applications
</code></pre>
<p>My package is <code>nma</code>, so I have in the above the standard set up of <code>nma.m</code> and <code>kernel\init.m</code> where init.m loads all the files using <code>Get</code>.</p>
<pre><code>C:\Users\Owner\AppData\Roaming\Mathematica\Applications\nma\nma.m
C:\Users\Owner\AppData\Roaming\Mathematica\Applications\nma\A.m
C:\Users\Owner\AppData\Roaming\Mathematica\Applications\nma\B.m
C:\Users\Owner\AppData\Roaming\Mathematica\Applications\nma\kernel\init.m
</code></pre>
<p>The above is all working fine. I can do</p>
<pre><code> <<nma`
</code></pre>
<p>From any notebook, and all the files in the package are loaded just fine.</p>
<p>But I do not like to keep my software on the C drive. My backup software only backups another drive. So I wanted to move <code>C:\Users\Owner\AppData\Roaming\Mathematica\Applications\nma\</code> to say <code>G:\</code> drive and the whole tree to say <code>G:\mathematica_version\code\nma\</code>. But when I did this, and made sure to open the notebook in same folder and made sure to do <code>SetDirectory[NotebookDirectory[]]</code> so that current directory is the above where all the files including <code>kernel</code> folder are, now when I do</p>
<pre><code> <<nma`
</code></pre>
<p>It no longer loads all the files in the package, it only loads <code>nma.m</code>, because I assume it does not use <code>init.m</code> in the <code>kernel</code> folder anymore, where <code>init.m</code> was loading all the other .m files.</p>
<p>It seems <code>application_name/kernel/init.m</code> is only used when the package lives in the standard location <code>C:\Users\Owner\AppData\Roaming\Mathematica\Applications</code> ? Is this true?</p>
<p><strong>My question is</strong>, how to copy <code>Applications\nma\...</code> from C drive to any other location and have</p>
<pre><code><<nma`
</code></pre>
<p>work the same as before? i.e. as if the package was in the standard location?</p>
<p>I tried my other things, like <code>AppendTo[$Path,"G:\\mathematica_version\\code\\nma"]</code> but this had no effect.</p>
<p>So currently after moving the package to the different location, I have to now manually do a <code>Get</code> on each file. This is something that <code>init.m</code> was doing before. i.e. I am doing this now</p>
<pre><code> SetDirectory[NotebookDirectory[]]
Get["nma.m"]
Get["A.m"]
Get["B.m"]
etc...
</code></pre>
<p>for each .m file. I would prefer to have init.m do this as before if possible.</p>
<p><strong>Update</strong></p>
<p>I also tried the following. Changed <code>$UserBaseDirectory</code> to point to the location in the other disk. Removed all the <code>Application\</code> folder and moved it to the new location. But this did not work. When I did</p>
<pre><code><<nma`
</code></pre>
<p>It did not load the package. To change <code>$UserBaseDirectory</code> I had to do the following</p>
<pre><code> Unprotect[<span class="math-container">$UserBaseDirectory]
$</span>UserBaseDirectory="new path here"
Protect[$UserBaseDirectory]
</code></pre>
<p>So I copied the Application folder back to where it was. I thought may be by moving the whole Application tree and changing <code>$UserBaseDirectory</code> will make it work. May be I did not do it right, I looked at option inspector and did not see where <code>$UserBaseDirectory</code> is defined. Is it ok to change <code>$UserBaseDirectory</code> to new location? If so, what is the correct way to do it?</p>
| Vitaliy Kaurov | 13 | <p>Perhaps something like that:</p>
<pre><code>Scalarize[x_List]:=If[Length[Flatten[x]]==1,Identity@@Flatten[x],x]
</code></pre>
|
631,348 | <p>Let $\sum a_n$ be a series of non-negative terms and let $$L = \lim_{n\to\infty}n\left(1-\frac{a_{n+1}}{a_n}\right)$$
Prove that the series converges (resp. diverges) if $L > 1$ (resp. $L<1$). I've tried, for example, that when $L<1$, $$n\left(1-\frac{a_{n+1}}{a_n}-L\right)= \frac{n(a_n-a_{n+1}-La_n)}{a_n}\ge\frac{n(a_n-a_{n+1}-a_n)}{a_n}=\frac{-na_{n+1}}{a_n}$$ and then using epsilons and the sort, but I can't get anywhere. Any tips?</p>
<p>P.S. using Kummer's test doesn't count</p>
| QA Ngô | 122,043 | <p>The following is a new argument for the convergence part. Assume that <span class="math-container">$L>1$</span>. Choose <span class="math-container">$\varepsilon > 0$</span> small enough such that
<span class="math-container">$$L-\varepsilon > 1.$$</span>
There exists some <span class="math-container">$1 \ll N=N(\varepsilon)$</span> such that
<span class="math-container">$$n\Big( 1- \frac {a_{n+1}}{a_n} \Big) > L-\varepsilon$$</span>
for any <span class="math-container">$n \geq N$</span>, namely
<span class="math-container">$$\frac {a_{n+1}}{a_n} < 1 - \frac{L-\varepsilon}n = \frac{n-(L-\varepsilon)}n$$</span>
for any <span class="math-container">$n \geq N$</span>. Since
<span class="math-container">$L-\varepsilon>1$</span>, one can always choose <span class="math-container">$\alpha >1$</span> in such a way that
<span class="math-container">$$\frac{n-(L-\varepsilon)}n < \Big( \frac{n-1}n \Big)^\alpha$$</span>
for any <span class="math-container">$n \gg 1$</span>, say <span class="math-container">$n \geq M > N$</span>. Hence for large <span class="math-container">$n$</span>, we obtain
<span class="math-container">$$ a_{n+1} \leq \Big( \frac{n-1}n \Big)^\alpha a_n \leq \Big( \frac{n-2}n \Big)^\alpha a_{n-1} \leq \cdots \leq \Big( \frac{M-1}n \Big)^\alpha a_M.$$</span>
Hence <span class="math-container">$\sum a_n$</span> converges by the comparison test.</p>
|
2,226,386 | <p>Solve $$\frac{dx}{y+z}=\frac{dy}{z+x}=\frac{dz}{x+y}.$$ Solve the similtaneous equation by using method of multipliers. How can we choose these multipliers? Is there any specific method to know these multipliers?</p>
| Cye Waldman | 424,641 | <p>I think the difficulty you are having is the misconception that a complex number, say <em>z</em>, is a point in the complex plane. Rather, think of it as a vector (hence all those arrows). And they add and subtract just like vectors and they have scalar and vector products as well. For example, given two complex numbers, say $z_1$ and $z_2$, then the complex product $z_1z_2^*$, where * denotes the conjugate gives both the scalar and vector products. Specifically,</p>
<p>$$\Re\{z_1z_2^*\}=|z_1| \cdot |z_2| \cos(\zeta)=\frac{1}{2} (z_1z_2^*+z_1^*z_2) \\
\Im\{z_1z_2^*\}=|z_1| \cdot |z_2| \sin(\zeta)=\frac{1}{2} (z_1z_2^*-z_1^*z_2)$$</p>
<p>where $\zeta$ is angle between the two vectors.</p>
<p>Of course, complex numbers have many more properties. I'm just indicating a new way for to you to think about them.</p>
|
3,370,061 | <p>Could anyone clarify why this Boolean expression AB'+AB'AC' = AB'? I did not understand what happened to the C'</p>
| Andrei | 331,661 | <p>If <span class="math-container">$AB'$</span> is true, the <span class="math-container">$+$</span> will not change that. </p>
<p>If <span class="math-container">$AB'$</span> is false, then <span class="math-container">$(AB')(AC')$</span> is also false, no matter what <span class="math-container">$C'$</span> is. Then the entire expression is false, which is equal to <span class="math-container">$AB'$</span></p>
|
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