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1,021,631 |
<p>Does anybody know how to solve the equation</p>
<p>$\mathbf{a} + \mathbf{b} \times \hat{\mathbf{v}} = c \hat{\mathbf{v}},$</p>
<p>where $\mathbf{a}$ and $\mathbf{b}$ are given real vectors, for the unit vector $\hat{\mathbf{v}}$ and the real number $c$?</p>
|
JP McCarthy
| 19,352 |
<p>What you need is the <a href="http://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality" rel="nofollow">Cauchy Schwarz Inequality</a>.</p>
|
1,045,941 |
<p>Usually this is just given as a straight up definition in a calculus course. I am wondering how you prove it?
I tried using the limit definition, $$\lim\limits_{h\rightarrow 0} \dfrac{\log(x+h)-\log(x)}{h}$$
but this led to no developments.</p>
|
Hernandez
| 165,046 |
<p>Set $y = \log_e(x).$ Rearranging, we get $e^{y} = x\,(*).$ Differentiating both sides implicitly with respect to $x$:
$$e^{y}\cdot\frac{dy}{dx} = 1.$$ It follows that $$\frac{dy}{dx} = \frac{1}{e^y}$$ and hence that the derivative is $\frac{1}{x}$ from $(*).$</p>
|
243,377 |
<p>Consider a random walk on the real time, starting from $0$. But this time assume that we can decide, for each step $i$, a step size $t_i>0$ to the left or the right with equal probabilities. </p>
<p>To formalize this, we have $(X_n)_{n\geq 0}$ such that $X_0=0$, $Pr[X_n=t_n]=0.5$ and $Pr[X_n=-t_n]=0.5$ (hence still $E(X_i)=0$). Let $I=[-1,1]$ and $$p=Pr[\sum_{i=1}^n X_i \in I. \forall n\geq 0]$$ </p>
<p>The question is: does there exist a sequence $(t_i)$ such that $t_i>0$ and $\sum_{i=0}^\infty=\infty$ (i.e. $(t_i)$ diverges), to ensure $p>0$. </p>
<p>Clearly such a sequence does not exist to let $p=1$, and intuitively it seems to me that neither exists $(t_i)$ to force $p>0$, but I am looking for a rigorous proof or disproof. Moreover, I am not exactly sure whether this kind of questions has been studied, so references would be appreciated. </p>
|
Ori Gurel-Gurevich
| 1,061 |
<p>The crucial requirement is that $\sum_{i=0}^\infty t_i^2 < \infty$. See <a href="https://en.wikipedia.org/wiki/Kolmogorov%27s_two-series_theorem">Kolmogorov's two-series theorem</a> and also the more general <a href="https://en.wikipedia.org/wiki/Kolmogorov%27s_three-series_theorem">Kolmogorov's three-series theorem</a>.</p>
|
1,926,383 |
<p>Recently I baked a spherical cake ($3$ cm radius) and invited over a few friends, $6$ of them, for dinner. When done with main course, I thought of serving this spherical cake and to avoid uninvited disagreements over the size of the shares, I took my egg slicer with parallel wedges(and designed to cut $6$ slices at a go; my slicer has $5$ wedges) and placed my spherical cake right in the exact middle of it before pressing uniformly upon the slicer.</p>
<p>What should the relative placement of my wedges of the slicer be like so that I am able to equally distribute the cake among my $6$ friends?The set-up of the egg slicer looks something like this :
<a href="https://i.stack.imgur.com/Mx0xL.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Mx0xL.jpg" alt="enter image description here"></a></p>
|
G Cab
| 317,234 |
<p>Let me recast <em>Lovsovs</em>' answer in another perspective.<br>
Consider at first to split an <strong>hemi-sphere</strong> into $n$ parts of equal volume.
Let designate as $h(k,n)$ the relative position of the $k$-th cut along the radius of the emisphere starting from the base circle
(corresponding to $k=0$).
$$
0 \leqslant h(k,n) \leqslant 1\quad \left| {\;0 \leqslant k \leqslant n} \right.
$$
Thus we shall have
$$
\frac{2}
{3}\pi R^{\,3} \frac{k}
{n} = \pi \int_{y\; = \;0}^{h\,R} {\left( {R^{\,2} - y^{\,2} } \right)dy} = \pi \left( {R^{\,2} R\,h - \frac{1}
{3}R^{\,3} h^{\,3} } \right)
$$
that means
$$
h^{\,3} - 3h + 2\frac{k}{n} = 0
$$
We will proceed and solve such <em>depressed cubic</em> equation
according to the method indicated in <a href="http://ebook-kings.com/pdf/risoluzione-delle-equazioni-di-terzo-e-quarto-5800324.html" rel="nofollow noreferrer">this work by Alessandra Cauli</a>, refer also to the <a href="https://math.stackexchange.com/questions/1928332/how-to-show-that-the-roots-of-x33x-left2-frac4n-right-0-are-real-a/1928624#1928624">answer to this post</a>.<br>
Putting apart the case $k=0$, for $\;1 \leqslant k \leqslant n$ we define
$$
u = \sqrt[{3\,}]{{ - \frac{q}
{2} + \sqrt {\frac{{q^{\,2} }}
{4} + \frac{{p^{\,3} }}
{{27}}} }}
$$
where $p$ and $q$ are respectively the coefficients of $h^1$ and $h^0$.
The 2nd radical is
$$
\frac{{q^{\,2} }}
{4} + \frac{{p^{\,3} }}
{{27}} = \left( {\frac{k}
{n}} \right)^{\,2} - 1 = - \frac{{n^{\,2} - k^{\,2} }}
{{n^{\,2} }} \leqslant 0
$$
which being non-positive tells us that there are three real solutions.<br>
Completing the calculation for $u$
$$
\begin{gathered}
u = \sqrt[{3\,}]{{ - \frac{k}
{n} + i\frac{1}
{n}\sqrt {n^{\,2} - k^{\,2} } }} = \frac{1}
{{\sqrt[{3\,}]{n}}}\sqrt[{3\,}]{{n\,e^{\,i\,\alpha } }} = \hfill \\
= e^{\,\,i\,\alpha \,/\,3} \quad \left| \begin{gathered}
\;1 \leqslant k \leqslant n \hfill \\
\;\alpha = \arctan _{\text{4Q}} \left( { - k,\sqrt {n^{\,2} - k^{\,2} } } \right) = \pi - \arctan \left( {\sqrt {\left( {\frac{n}
{k}} \right)^{\,2} - 1} } \right) = \hfill \\
\;\;\;\; = \pi - \beta \hfill \\
\end{gathered} \right. \hfill \\
\end{gathered}
$$
We take then
$$
v = - \frac{p}
{{3\,u}} = \frac{1}
{u}\quad \quad \omega = e^{\,i\,\frac{{2\pi }}
{3}} = 1/2\left( { - 1 + i\sqrt 3 } \right)
$$
and arrive to obtain the three solutions as
$$
\begin{gathered}
1 \leqslant k \leqslant n\quad 0 \leqslant \beta = \arctan \left( {\sqrt {\left( {\frac{n}
{k}} \right)^{\,2} - 1} } \right) < \frac{\pi }
{2} \hfill \\
\left\{ \begin{gathered}
h_{\,1} = e^{\,i\,\alpha /3} + e^{\, - \,i\,\alpha /3} = 2\cos \left( {\frac{\pi }
{3} - \frac{\beta }
{3}} \right) \hfill \\
h_{\,2} = e^{\,i\,\alpha /3 + 2\pi /3} + e^{\, - \,i\,\alpha /3 - 2\pi /3} = - 2\cos \left( {\frac{\beta }
{3}} \right) \hfill \\
h_{\,3} = e^{\,i\,\alpha /3 - 2\pi /3} + e^{\, - \,i\,\alpha /3 + 2\pi /3} = 2\cos \left( {\frac{\pi }
{3} + \frac{\beta }
{3}} \right) \hfill \\
\end{gathered} \right. \hfill \\
\end{gathered}
$$
of which we can easily determine that only $h_3$
respect the physical conditions of our problem.</p>
<p>Passing now to split the <strong>entire sphere</strong>, imagine it consisting of the two halves, placed base-to-base, one on the positive and one on the negative $h$ axis.<br>
So, when $n$ is even we will place the cuts at $k = 0,\; \pm 2, \cdots ,\; \pm \left( {n - 2} \right)$,<br>
while for $n$ odd at $k = \pm 1,\; \pm 3, \cdots ,\; \pm \left( {n - 2} \right)$.<br>
In conclusion, always as a ratio to the radius, the cutting blades shall be positioned at:
$$
\left\{ \begin{gathered}
2 \leqslant n\quad 0 \leqslant k \leqslant \left\lfloor {\frac{n}
{2}} \right\rfloor - 1 \hfill \\
\beta \left( {k,n} \right) = \arctan \left( {\sqrt {\left( {\frac{n}
{k}} \right)^{\,2} - 1} } \right) \hfill \\
h_{\,T} (k,n) = \pm 2\cos \left( {\frac{\pi }
{3} + \frac{{\beta (2k + \bmod (n,2),\,n)}}
{3}} \right) \hfill \\
\end{gathered} \right.
$$
(if you can accept that$\pm 0=0$).</p>
|
3,914,365 |
<p>I'm pretty stuck with this exercise. Hope somebody can help me:</p>
<p>show that
<span class="math-container">$$
\sum_{n=2^{k}+1}^{2^{k+1}} \frac{1}{n} \geq\left(2^{k+1}-2^{k}\right) \frac{1}{2^{k+1}}=\frac{1}{2}
$$</span>
and use this to show that the harmonic series is divergent.</p>
<p>First I can't figure out how to show the inequality.
Second I'm not quite sure how to use the inequality to show divergence. Because it doesn't really help letting k go to infinity.</p>
<hr />
<p>Writing out the series gave me this, but where do I go from here:
<span class="math-container">$$
\sum_{n=2^{k}+1}^{2^{k+1}} \frac{1}{n}=\frac{1}{2^{k}+1}+\frac{1}{2^{k}+2}+\cdots \frac{1}{2^{k}+2^{k}}=\frac{1}{2^{k}+1}+\frac{1}{2^{k}+2}+\cdots+\frac{1}{2^{k+1}}
$$</span></p>
<hr />
<p><strong>Edit:</strong>
So I understand how to show the inequality now. Thanks!
But I'm still lost with how to use the inequality for the divergence proof.</p>
|
Surb
| 154,545 |
<p>(I am summarizing in this answer the steps discussed with OP in the comments)</p>
<p>Let <span class="math-container">$k\geq 0$</span> be an integer.</p>
<p>First note that the sum <span class="math-container">$\sum_{n=2^{k}+1}^{2^{k+1}} \frac{1}{n}$</span> has <span class="math-container">$2^{k+1}-(2^{k}+1)+1= 2^{k+1}-2^k = 2^k$</span> terms, that is
<span class="math-container">$$\sum_{n=2^{k}+1}^{2^{k+1}} 1 = 2^k\tag{$*$}$$</span></p>
<p>Second, note that
<span class="math-container">$$\frac{1}{n}\geq \frac{1}{2^{k+1}},\qquad \forall n\, \text{ s.t. }\,2^{k}+1\leq n\leq 2^{k+1}\tag{$**$}$$</span></p>
<p>Combining the above observations, we have
<span class="math-container">$$\sum_{n=2^{k}+1}^{2^{k+1}} \frac{1}{n}\,\overset{(**)}\geq\,\sum_{n=2^{k}+1}^{2^{k+1}}\frac{1}{2^{k+1}}\,=\,\frac{1}{2^{k+1}}\sum_{n=2^{k}+1}^{2^{k+1}}1\,\overset{(*)}{=}\, \frac{2^k}{2^{k+1}}\,=\,\frac{1}{2}\tag{$*\!*\!*$}$$</span></p>
<p>Now, note that</p>
<p><span class="math-container">$$\sum_{n=1}^{2^{k+1}}\frac{1}{n}\,\geq\, \sum_{n=2}^{2^{k+1}}\frac{1}{n}\, =\, \sum_{m=0}^k\, \sum_{n=2^{m}+1}^{2^{m+1}}\frac{1}{n} \, \overset{(***)}{\geq}\, \sum_{m=0}^k \frac{1}{2}\, = \, \frac{k+1}{2}$$</span></p>
<p>As the above holds for every <span class="math-container">$k\geq 0$</span>, by letting <span class="math-container">$k\to \infty$</span>, we conclude that the harmonic series diverges.</p>
|
1,946,535 |
<p>A Liouville number is an irrational number $x$ with the property that, for every positive integer $n$, there exist integers $p$ and $q$ with $q > 1$ and such that $0 < \mid x - \frac{p}{q} \mid < \frac{1}{q^n} $.</p>
<p>I'm looking for either hints or a complete proof for the fact that $e$ is not a Liouville number. I can prove that $e$ is irrational and even that it is transcendental, but I'm a bit stuck here.</p>
<p>Here's my research:</p>
<p><a href="https://en.wikipedia.org/wiki/Liouville_number" rel="noreferrer">The wikipedia article about Liouville numbers</a> states:</p>
<blockquote>
<p>[...] not every transcendental number is a Liouville number. The terms in the continued fraction expansion of every Liouville number are unbounded; using a counting argument, one can then show that there must be uncountably many transcendental numbers which are not Liouville. Using the explicit continued fraction expansion of $e$, one can show that e is an example of a transcendental number that is not Liouville.</p>
</blockquote>
<p>However, theres clearly more to the argument then the boundedness of the continued fraction of $e$, because the terms of $e$'s continued fraction expansion <em>are</em> unbounded and yet it is not a Liouville number. Also, if possible, i would like to avoid using continued fractions at all.</p>
<p><a href="https://books.google.de/books?id=R70lCQAAQBAJ&pg=PA43&lpg=PA43&dq=e%20not%20a%20liouville%20number%20proof&source=bl&ots=DyJ4TDhjaR&sig=8Cjvcnj9fOXhLBda9I9DMKNe4TE&hl=de&sa=X&ved=0ahUKEwj45q7pubTPAhUmQpoKHeSjDQY4ChDoAQgbMAA#v=onepage&q=e%20not%20a%20liouville%20number%20proof&f=false" rel="noreferrer">This book</a> has the following as an exercise:</p>
<blockquote>
<p>Prove that $e$ is not a Liouville number. (Hint: Follow the irrationality proof of $e^n$ given in the supplements to Chapter 1.)</p>
</blockquote>
<p>Unfortunately, the supplements to Chapter 1 are not publically available in the sample and I do not want to buy that book.</p>
<p><a href="https://books.google.de/books?id=TRNxKFMuNPkC&pg=PA25&lpg=PA25&dq=e%20not%20a%20liouville%20number%20proof&source=bl&ots=wyHE_EcTUE&sig=-MsBOxHYKLh-K_xGKBIB2JthBM8&hl=de&sa=X&ved=0ahUKEwjSvP2S0LTPAhWlNJoKHcR3Dwg4FBDoAQgjMAE#v=onepage&q=e%20not%20a%20liouville%20number%20proof&f=false" rel="noreferrer">This book</a> states:</p>
<blockquote>
<p>Given any $\varepsilon > 0$, there exists a constant $c(e,\varepsilon) > 0$ such that for all $p/q$ there holds $\frac{c(e,\varepsilon)}{q^{2+\varepsilon}} < \mid e - \frac{p}{q} \mid$. [...] Using [this] inequality, show that $e$ is not a Liouville number.</p>
</blockquote>
<p>Which, given the inequality, I managed to do. But I do not have any idea of how one would go about proving that inequality.</p>
<p>I greatly appreciate any help!</p>
|
Thomas Andrews
| 7,933 |
<p>In continued fractions for $\alpha=[a_0,a_1,a_2,\dots,a_n,\dots]$ we have $$\left|\frac{P_n}{Q_n} -\alpha\right|>\frac{1}{2}\left|\frac{P_n}{Q_n}-\frac{P_{n+1}}{Q_{n+1}}\right|=\frac{1}{2Q_nQ_{n+1}}\geq\frac{1}{2Q_n^2(a_n+1)}$$.</p>
<p>So the fact that $Q_n\geq F_n$ (with $F_n$ the $n$th Fibonacci number) and $a_n=O(n)$ when $\alpha=e$ means that $2Q_n^2(a_n+1)$ can't get bigger than $Q_n^3$ for large $n$.</p>
<p>This provides at least one necessary condition for $\alpha$ being Liouville - that there are for any $k$ infinitely man $n$ with $a_n\geq Q_n^k/2$.</p>
|
3,843,807 |
<p>Let <span class="math-container">$d_1$</span> be a metric on <span class="math-container">$X$</span>. If there exists a strictly increasing, continuous, and subadditive <span class="math-container">$ f:\mathbb {R} \to \mathbb {R} _{+}$</span> such that <span class="math-container">$d_{2}=f\circ d_{1}$</span>. Then <span class="math-container">$d_1,d_2$</span> are topologically equivalent.</p>
<hr />
<p>Note : I found the proposition here:<a href="https://en.wikipedia.org/wiki/Equivalence_of_metrics" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Equivalence_of_metrics</a></p>
|
Amirhossein
| 456,050 |
<p>I came up with a solution, and I will write it below; it seems the conditions of being subadditive is surplus. I think you needed the subadditivity if you wanted to show that <span class="math-container">$d_2$</span> is a metric. But when it is given, you won't need it.</p>
<p>Let <span class="math-container">$B^{1}_{r}(x) = \{ y \in X | d_1(x,y) < r\}$</span>. We will show that <span class="math-container">$B^{1}_{r}(X)$</span> is open in the topology induced by the metric <span class="math-container">$d_2$</span>. let <span class="math-container">$ y \in B^{1}_{r}(x)$</span>. Then you have that : <span class="math-container">$d_2(x,y) =f \circ d_1(x,y) < f(r)$</span>. So <span class="math-container">$y \in B^{2}_{f(r)}(x)$</span>. now let <span class="math-container">$ z \in B^{2}_{f(r)}(x)$</span> then <span class="math-container">$ d_2(x,z) < f(r)$</span>. which means that <span class="math-container">$f (d_1(x,z)) < f(r)$</span>. As f is strictly increasing, we get <span class="math-container">$d_1(x,z) < r$</span>. So <span class="math-container">$z \in B^{1}_{r}(x)$</span>. Until now we have that the topology induced by <span class="math-container">$d_2$</span> is finer than the one from <span class="math-container">$d_1$</span>.</p>
<p>Now let <span class="math-container">$r^{\prime} \in \Bbb R_+$</span>. and consider the the open ball <span class="math-container">$B^{2}_{r^{\prime}}(x)=\{z \in X| d_2(x,z) < r^{\prime} \}$</span>. For simplicity, suppose that : <span class="math-container">$$r^{\prime} < \sup f(d_1(X \times X)).$$</span> and I do not use the continuity condition in my proof. So it might be that my solution is incorrect.</p>
<p>Now let <span class="math-container">$z \in B^{2}_{r^{\prime}}(x)$</span>. Thus we have :
<span class="math-container">$d_2(x,z) < r^{\prime}$</span>. which means <span class="math-container">$ d_1(x,z) < f^{-1}(r^{\prime})$</span>(we can take inverse from <span class="math-container">$r^{\prime}$</span> because <span class="math-container">$f$</span> is continuous and strictly increasing. Now let <span class="math-container">$ d_1(x , y) < f^{-1}(r)$</span>. Then we have <span class="math-container">$d_2(x,y) < r^{\prime}$</span>.</p>
<p>You should consider that <span class="math-container">$X, \emptyset$</span> are open in both of the topologies.</p>
<p>We have shown that the topology of <span class="math-container">$d_1$</span> is finer than <span class="math-container">$d_2$</span>. Thus these metrics are topologically equivalent.</p>
|
262,773 |
<p>In the curve obtained with following</p>
<pre><code> Plot[{1/(3 x*Sqrt[1 - x^2])}, {x, 0, 1}, PlotRange -> {0, 1},
GridLines -> {{0.35, 0.94}, {}}]
</code></pre>
<p>how can one fill the top and bottom with different colors or patterns such that two regions are perfectly visible in a black-n-white printout?</p>
|
Ulrich Neumann
| 53,677 |
<p>You have to plot it twice:</p>
<pre><code>Show[{
Plot[{1/(3 x*Sqrt[1 - x^2]), 1/(3 x*Sqrt[1 - x^2])}, {x, 0, 1},PlotRange -> {0, 1}, GridLines -> {{0.35, 0.94}, {}}, Filling -> {Top }, FillingStyle -> {Red }],
Plot[{1/(3 x*Sqrt[1 - x^2]), 1/(3 x*Sqrt[1 - x^2])}, {x, 0, 1},PlotRange -> {0, 1}, GridLines -> {{0.35, 0.94}, {}}, Filling -> {Bottom}, FillingStyle -> {Blue },RegionFunction -> Function[x, 0.35 < x < 0.94]]
}]
</code></pre>
<p><a href="https://i.stack.imgur.com/xCm9u.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xCm9u.png" alt="enter image description here" /></a></p>
|
262,773 |
<p>In the curve obtained with following</p>
<pre><code> Plot[{1/(3 x*Sqrt[1 - x^2])}, {x, 0, 1}, PlotRange -> {0, 1},
GridLines -> {{0.35, 0.94}, {}}]
</code></pre>
<p>how can one fill the top and bottom with different colors or patterns such that two regions are perfectly visible in a black-n-white printout?</p>
|
Bob Hanlon
| 9,362 |
<p>Using <a href="https://reference.wolfram.com/language/ref/RegionPlot.html" rel="nofollow noreferrer"><code>RegionPlot</code></a></p>
<pre><code>RegionPlot[
{0.35 <= x <= 0.94 && y > 1/(3 x*Sqrt[1 - x^2]),
0.35 <= x <= 0.94 && y < 1/(3 x*Sqrt[1 - x^2])},
{x, 0, 1}, {y, 0, 1},
PlotStyle -> {
Lighter[Gray, 0.7],
Lighter[Gray, 0.9]},
BoundaryStyle ->
Directive[AbsoluteThickness[1], Gray],
PlotPoints -> 75,
AspectRatio -> 1,
ImageSize -> 288]
</code></pre>
<p><a href="https://i.stack.imgur.com/id6e3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/id6e3.png" alt="enter image description here" /></a></p>
|
1,268 |
<p>There has long been debate about whether a first year undergraduate course in discrete mathematics would be better for students than the traditional calculus sequence. The purpose of this question is not to further that debate, but to inquire whether any text has tried to teach calculus by emphasizing probability as a primary motivating example.</p>
<p>It seems that such a text would introduce integrals before derivatives, which I know many authors have tried.</p>
<p>I am asking this question because it seems <a href="https://mathoverflow.net/questions/162125/what-areas-of-pure-mathematics-research-are-best-for-a-post-phd-transition-to-in">industry has use for people proficient with probability</a>, and pure mathematics makes great use of these ideas as well (think of probabilistic methods in number theory, for example). It seems that a good course of the above type would be cosmopolitan enough to attract students with tastes in either pure or applied mathematics to major in mathematics.</p>
<p>EDIT: In the absence of a perfect text for this, it may be worth a collaborative effort by mathematicians to write such a text. Is there a way to "crowdsource" writing a text like this as a wiki?</p>
|
kcrisman
| 1,608 |
<p>This isn't exactly a full answer for you, but <a href="http://www.ams.org/bookstore-getitem/item=acalc" rel="nofollow">Approximately Calculus</a> has a large bit on the prime number theorem as a motivation, which was motivated by Gauss' thinking of prime distribution probabilistically and where using integrals doesn't make sense unless you are thinking of integrating a pdf.</p>
|
4,204,282 |
<p>How do I calculate <span class="math-container">$$\int_{-\infty}^{\infty} \frac{dw}{1+iw^3}$$</span> using complex path integrals?</p>
<p>I just need a hint on how to start, not the actual computation, because I need to understand how to deal with similar questions.</p>
<p><strong>Edit:</strong> Following @Tavish's comments, I used the residue theorem:</p>
<p>The function has poles at <span class="math-container">$w=\pm 0.866 +0.5i$</span> and <span class="math-container">$w=i$</span>. Now the integral is equal to <span class="math-container">$$2\pi i [Res(f,0.866 +0.5i)+Res(f,-0.866 +0.5i)]=\frac{2\pi}3.$$</span></p>
<p>However, I'm more interested in understanding the steps here than finding the answer. For example, how does one know which contour to take? Moreover, is there an alternate way to find this integral?</p>
|
mjw
| 655,367 |
<p>Consider the integral <span class="math-container">$$\int_{-\infty}^\infty \frac{dw}{p(w)},$$</span> where <span class="math-container">$p(w)$</span> is a polynomial of degree <span class="math-container">$\ge 2$</span>. If <span class="math-container">$p(w)$</span> has no poles on the real axis (this can be dealt with later), then consider the function <span class="math-container">$\displaystyle f(z) = \frac{1}{p(z)}$</span> integrated along the contour consisting of a segment on the real axis from <span class="math-container">$-R$</span> to <span class="math-container">$R$</span> and closed in the upper half plane by a semicircle of radius <span class="math-container">$R$</span>. As <span class="math-container">$R\to \infty$</span>, <span class="math-container">$\int_{\Gamma_R} f(z)\,dz$</span> can easily be shown to go to zero, and thus</p>
<p><span class="math-container">$$\int_{-\infty}^\infty \frac{dw}{p(w)} = 2\pi i\sum_{z_j \in \text{zeros of }{p(z)}\text{ in upper half plane}} \text{res}_{z=z_j} f(z).$$</span></p>
<p>If you like, you could just as easily complete the contour in the lower half plane.</p>
<p>On the semicircle, eventually, for large enough <span class="math-container">$R$</span>,</p>
<p><span class="math-container">$$\left| \int_{\Gamma_R} f(z) \, dz \right| \le \int_0^\pi
\left|\frac{R i e^{i\theta}}{a_n R^n-a_{n-1}R^{n-1}- \cdots- a_0}\right|\,d\theta \le \int_0^\pi
\frac{R }{\frac{a_n}{2} R^n}\,d\theta =\frac{2\pi}{a_n R^{n-1}} \to 0.$$</span></p>
|
347,009 |
<p>I am looking for a generalisation of a modular form that transforms as something like:</p>
<p><span class="math-container">$f(\frac{a \tau+b}{c \tau+d}) = (c \tau+d)^k c^k f(\tau)$</span></p>
<p>I understand this cannot be literally true, as the multiplier c^k is not a root of unity, but does something like this arise in the context of modular forms? (or generalisations of those). Thanks!</p>
|
Matt F.
| 44,143 |
<p>Here are three conjectures, in the hopes of showing that we can say something non-trivial.</p>
<p>[<strong>UPDATE</strong>: This is essentially all in a paper from Erdős, "On Sets of Distances of <span class="math-container">$n$</span> Points", from <a href="https://www.jstor.org/stable/2305092?seq=1" rel="nofollow noreferrer">1946</a>. The conjecture on maxima is in his theorem 3, though he called that part well-known. The conjecture on minima is a later theorem of Harborth, cited <a href="https://arxiv.org/abs/1210.5756" rel="nofollow noreferrer">here</a>, but the bound in less precise forms is also in Erdős's theorem 3. The conjecture on modes is made after Erdős's theorem 2, in the speculation that <span class="math-container">$g(n)<n^{1+\epsilon}$</span>, after a proof that <span class="math-container">$g(n)<n^{3/2}$</span>. Finally Erdős's theorem 1 gives bounds on the number of distinct distances, between <span class="math-container">$(n-3/4)^{1/2}-1/2$</span> and <span class="math-container">$cn/(\log n)^{1/2}$</span>.]</p>
<p><strong>Maxima</strong>: Among <span class="math-container">$n$</span> points, the maximal distance can be achieved by at most <span class="math-container">$2n$</span> ordered pairs. This bound will be achieved when all the points are on a Reuleaux triangle, with the three 60-degree circle arcs centered on three of the points. [Thanks to Yoav Kallus for the idea for this construction.]</p>
<p><a href="https://i.stack.imgur.com/oR4Yu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oR4Yu.png" alt="enter image description here"></a></p>
<p><strong>Minima</strong>: Among <span class="math-container">$n$</span> points, the minimal distance can be achieved by at most <span class="math-container">$6n-2\sqrt{12n-3}$</span> ordered pairs. This bound will be achieved when <span class="math-container">$n=3k^2-3k+1$</span> and the points are in a hexagonal lattice with <span class="math-container">$k$</span> points on each side of the hexagon.</p>
<p><a href="https://i.stack.imgur.com/cR3Bz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cR3Bz.png" alt="enter image description here"></a></p>
<p><strong>Modes</strong>: For arbitrarily large <span class="math-container">$k$</span> and <span class="math-container">$n$</span>, there are configurations of <span class="math-container">$n$</span> points in which the <em>modal</em> distance is achieved by at least <span class="math-container">$kn$</span> ordered pairs. For example, using square lattices:</p>
<ul>
<li>If <span class="math-container">$k=7$</span>, the distance of <span class="math-container">$\sqrt{5}=\sqrt{2^2+1^2}$</span> can be achieved by at least <span class="math-container">$7n$</span>, and almost <span class="math-container">$8n$</span> ordered pairs. </li>
<li>If <span class="math-container">$k=15$</span>, the distance of <span class="math-container">$\sqrt{65}=\sqrt{8^2+1^2}=\sqrt{7^2+4^2}$</span> can be achieved by at least <span class="math-container">$15n$</span>, and almost <span class="math-container">$16n$</span> ordered pairs. </li>
<li>If <span class="math-container">$k=23$</span>, the distance of <span class="math-container">$\sqrt{325}=\sqrt{18^2+1^2}=\sqrt{17^2+6^2}=\sqrt{15^2+10^2}$</span> can be achieved by at least <span class="math-container">$23n$</span>, and almost <span class="math-container">$24n$</span> ordered pairs.</li>
</ul>
<p>But the claim would be false if we replaced <span class="math-container">$kn$</span> by <span class="math-container">$kn^{1+\epsilon}$</span>.</p>
<p><a href="https://i.stack.imgur.com/aq8E9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aq8E9.png" alt="enter image description here"></a></p>
<p>Again, these are only conjectures; I'd be happy to see proofs or alternatively configurations where the minimal or maximal or modal distance is achieved more often.</p>
|
745,876 |
<p>$$\exists! x : A(x) \Rightarrow \exists x : A(x)$$
Assuming that $A(x)$ is an open sentence. I'm new to abstract mathematics and proofs, so I came here to ask for some simplification. Thanks</p>
|
Community
| -1 |
<p>The problem states that:</p>
<blockquote>
<p>If there exists exactly one $x$ such that $A(x)$ is true, then there exists at least one $x$ such that $A(x)$ is true.</p>
</blockquote>
<p>The proof should be trivial.</p>
|
4,230,782 |
<blockquote>
<p>Compute <span class="math-container">$\frac 17$</span> in <span class="math-container">$\Bbb{Z}_3.$</span></p>
</blockquote>
<p>We will have to solve <span class="math-container">$7x\equiv 1\pmod p,~~p=3.$</span></p>
<ul>
<li>We get <span class="math-container">$x\equiv 1\pmod 3.$</span></li>
<li>Then <span class="math-container">$x\equiv 1+3a_1\pmod 9,$</span> so <span class="math-container">$7(1+3a_1)\equiv 1 \pmod 9$</span> basically lifting the exponent of <span class="math-container">$p=3,$</span> we get <span class="math-container">$1+3a_1\equiv 4\pmod 9\implies a_1\equiv 1\pmod 3.$</span></li>
<li>So let <span class="math-container">$$x\equiv 1+3\cdot 1+3^2\cdot a_2 \pmod 2\implies 7(4+3^2\cdot a_2)\equiv 1\pmod {27}\implies 4+3^2\cdot a_2\equiv 4\pmod {27}\implies a_2\equiv 0 \pmod 3.$$</span></li>
<li>So let <span class="math-container">$$ x\equiv 1+3\cdot 1+3^2\cdot 0+ 3^3\cdot a_3 \pmod {81}\implies 7(4+3^2\cdot 0+3^3\cdot a_3)\equiv 1\pmod {81}\implies 4+3^3\cdot a_3\equiv 58\pmod {81}\implies a_2\equiv 2 \pmod 3.$$</span></li>
<li>So let <span class="math-container">$$ x\equiv 1+3\cdot 1+3^2\cdot 0+ 3^3\cdot 2+3^4\cdot a_4 \pmod {243}\implies 7(4+3^2\cdot 0+3^3\cdot 2+3^4\cdot a_4)\equiv 1\pmod {243}\implies 1+3+54+3^4\cdot a_4\equiv 139\pmod {243}\implies a_4\equiv 1 \pmod 3.$$</span></li>
<li>So let <span class="math-container">$$ x\equiv 1+3\cdot 1+3^2\cdot 0+ 3^3\cdot 2+3^4\cdot 1+ 3^5\cdot a_5 \pmod {729}\implies 7(4+3^2\cdot 0+3^3\cdot 2+3^4\cdot 1+3^5\cdot a_5)\equiv 1\pmod {729}\implies 1+3+54+81\equiv 625\pmod {243}\implies a_5\equiv 2 \pmod 3.$$</span></li>
</ul>
<p>I haven't worked out but I think <span class="math-container">$a_6$</span> is <span class="math-container">$0.$</span></p>
<p>So the sequence we are getting is <span class="math-container">$(a_0,a_1,a_2,a_3,a_4,\dots)=(1,1,0,2,1,2,\dots).$</span></p>
<p>But I am not sure if it's correct, since it's not being periodic. Any help?</p>
|
Sangchul Lee
| 9,340 |
<p>You are doing correct.</p>
<p>For a slightly more efficient approach, note that <span class="math-container">$7\mid 3^6-1$</span> by Fermat's little theorem. Then by noting that <span class="math-container">$3^6-1=7\cdot104$</span>, we have</p>
<p><span class="math-container">\begin{align*}
\frac{1}{7}
&= -\frac{104}{1 - 3^6} \\
&= -\sum_{n=0}^{\infty}104 \cdot 3^{6n} \\
&= 1 + \sum_{n=0}^{\infty} (3^{6} - 1)3^{6n} - \sum_{n=0}^{\infty}104 \cdot 3^{6n} \\
&= 1 + \sum_{n=0}^{\infty}(3^6-105)3^{6n}
\end{align*}</span></p>
<p>in <span class="math-container">$\mathbb{Z}_3$</span>, the ring of <span class="math-container">$3$</span>-adic integers. Then by using <span class="math-container">$3^6-105 = 212010_{(3)}$</span>, it follows that</p>
<p><span class="math-container">$$ \frac{1}{7} = \overline{212010}212011_{(3)} = \overline{021201}1_{(3)}. $$</span></p>
|
1,792,402 |
<p>I am trying to solve <a href="https://en.wikipedia.org/wiki/Mean_value_theorem" rel="nofollow">Mean Value theorem</a> but i ran into a road block trying to solve the question. So the question is:
Assuming</p>
<p>$$ \frac{f(b) − f(a)}{b − a} = f'(c) \quad \text{for some } c \in (a, b)$$</p>
<p>Let $f(x) = \sqrt x$ and $[a, b] = [0, 4]$. Find the point(s) $c$ specified by the theorem.</p>
<p>so i attempted to solve this by first getting the derivative of f which gave me $\frac {1}{2\sqrt{x}}$. After plugging in into the equation $\frac{\frac{1}{2\sqrt{4}} - \frac{1}{2\sqrt{0}}}{4- 0}$ gives me $\frac{1}{0}$ which equates to infinity.<br>
Now i am convinced that my answer is incorrect because the theorem states <strong>f(x) is continuous function on the (closed and finite) interval</strong>. </p>
<p>Can someone guide me in the right direction with this equation. </p>
|
Michael Hardy
| 11,667 |
<p>Since $f'(x) = \dfrac 1 {2\sqrt x}$, you need
$$
\frac{\sqrt 4 - \sqrt 0}{4-0} = \frac{f(4) - f(0)}{4-0} = \frac{f(b)-f(a)}{b-a} = f'(c) = \frac 1 {2\sqrt c}.
$$
If $\dfrac{\sqrt 4} 4 = \dfrac 1 {2\sqrt c}$ then what number is $c$?</p>
|
1,792,402 |
<p>I am trying to solve <a href="https://en.wikipedia.org/wiki/Mean_value_theorem" rel="nofollow">Mean Value theorem</a> but i ran into a road block trying to solve the question. So the question is:
Assuming</p>
<p>$$ \frac{f(b) − f(a)}{b − a} = f'(c) \quad \text{for some } c \in (a, b)$$</p>
<p>Let $f(x) = \sqrt x$ and $[a, b] = [0, 4]$. Find the point(s) $c$ specified by the theorem.</p>
<p>so i attempted to solve this by first getting the derivative of f which gave me $\frac {1}{2\sqrt{x}}$. After plugging in into the equation $\frac{\frac{1}{2\sqrt{4}} - \frac{1}{2\sqrt{0}}}{4- 0}$ gives me $\frac{1}{0}$ which equates to infinity.<br>
Now i am convinced that my answer is incorrect because the theorem states <strong>f(x) is continuous function on the (closed and finite) interval</strong>. </p>
<p>Can someone guide me in the right direction with this equation. </p>
|
Martin Argerami
| 22,857 |
<p>$$
\frac1{2\sqrt c}=\frac{\sqrt 4-\sqrt 0}{4-0}=\frac24=\frac12.
$$
Then
$
\sqrt c=1
$, so $c=1$. </p>
|
2,513,509 |
<p>If $f,g,h$ are functions defined on $\mathbb{R},$ the function $(f\circ g \circ h)$ is even:</p>
<p>i) If $f$ is even.</p>
<p>ii) If $g$ is even.</p>
<p>iii) If $h$ is even.</p>
<p>iv) Only if all the functions $f,g,h$ are even. </p>
<p>Shall I take different examples of functions and see?</p>
<p>Could anyone explain this for me please?</p>
|
egreg
| 62,967 |
<p>You may want to compute
$$
\lim_{x\to0}\frac{\arcsin x-\arctan x}{\sin x-\tan x} \tag{*}
$$
which your limit will be equal to, provided (*) exists.</p>
<p><sup>The limit of the sequence might exist also if the limit of the function doesn't. But if the limit of the function exists, also does the limit of the sequence and they're equal.</sup></p>
<p>Evaluating the order of infinitesimal of the denominator helps:
$$
\sin x-\tan x=\sin x\frac{\cos x-1}{\cos x}
$$
With Taylor expansion:
$$
\sin x\frac{\cos x-1}{\cos x}=
\frac{1}{\cos x}(x+o(x))\left(-\frac{x^2}{2}+o(x^2)\right)=
\frac{1}{\cos x}\left(-\frac{x^3}{2}+o(x^3)\right)
$$
The cosine can be left as is, because it doesn't contribute.</p>
<p>Thus we know that we have to expand the numerator up to degree $3$. The Taylor expansion of the arctangent is
$$
\arctan x=x-\frac{x^3}{3}+o(x^3)
$$
For the arcsine, we can consider
$$
\arcsin'x=(1-x^2)^{-1/2}=1+\frac{x^2}{2}+o(x^2)
$$
and so
$$
\arcsin x=x+\frac{x^3}{6}+o(x^3)
$$
Then
$$
\lim_{x\to0}\frac{\arcsin x-\arctan x}{\sin x-\tan x}=
\lim_{x\to0}\cos x\frac{(x+x^3/6)-(x-x^3/3)+o(x^3)}{-x^3/2+o(x^3)}=-1
$$
because
$$
\frac{1}{6}+\frac{1}{3}=\frac{1}{2}
$$</p>
|
267,753 |
<p>A number of seemingly unrelated elementary notions can be defined uniformly with help of (iterated) Quillen lifting property
(a category-theoretic construction I define below) "starting" to a single (counter)example or a simple class of morphisms,
for example a finite group being nilpotent, solvable, p-group, a topological space being compact, discrete, T4 (normal).</p>
<p>I would like to see more examplees, to help me understand if there is a bigger picture behind. </p>
<p>Let me give the definitions.</p>
<p>For a property $C$ of arrows (morphisms) in a category, define
its {\em left and right orthogonals} as</p>
<p>$$ C^\perp := \{ f :\text{ for each }g \in C\ f \,\rightthreetimes\, g \} $$
$$ {}^\perp C := \{ g :\text{ for each }f \in C\ f \,\rightthreetimes\, g \} $$</p>
<p>here $f \,\rightthreetimes\, g$ reads " $f$ has the left lifting property wrt $g$ ",
" $f$ is (left) orthogonal to $g$ ",
i.e. for $f:A\longrightarrow B$, $g:X\longrightarrow Y$,
$f \,\rightthreetimes\, g$ iff for each $i:A\longrightarrow X$, $j:B\longrightarrow Y$ such that $ig=fj$ ("the square commutes"),
there is $j':B\longrightarrow X$ such that $fj'=i$ and $j'g=j$ ("there is a diagonal
making the diagram commute").</p>
<p>Examples:</p>
<p>In the category Sets of sets the right orthogonal
${}^\perp \{\emptyset \longrightarrow \{*\}\}$ of the simplest non-surjection
$\emptyset \longrightarrow \{*\}$ is the class of surjections.
The triple left orthogonal $ ((\{\emptyset \longrightarrow \{*\}\}^\perp)^\perp)^\perp$ is the class of functions which split.</p>
<p>The left and right orthogonals of $ \{x_{1},x_{2}\}\longrightarrow \{*\} $, the simplest non-injection, are both precisely the class of injections.</p>
<p>A finite group $H$ is nilpotent iff $H\longrightarrow H\times H$ is in ${}^\perp(\{ 0\longrightarrow G : G\text{ arbitrary} \}^\perp)$ </p>
<p>A Hausdorff space $K$ is compact iff $K\longrightarrow \{*\}$ is in ${}^\perp({}^\perp(\{a\}\longrightarrow \{a{<}b\})_{<5})^{\perp})$;
here $^\perp(\{a\}\longrightarrow \{a{<}b\})_{<5}$ denotes maps in $^\perp(\{a\}\longrightarrow \{a{<}b\})$
between spaces of size less than 5.</p>
<p>I give more examples in the answers to my own question I posted,
as they require some notation.</p>
|
user108780
| 108,780 |
<p>To improve readability of iterated orthogonals, I write $C^l$ instead of $C^\perp$ and $C^r$ instead of $C^r$.</p>
<p>(i) $(\emptyset\longrightarrow \{*\})^r$, $(0\longrightarrow R)^r$, and $\{0\longrightarrow \Bbb Z\}^r$ are the classes of surjections in
Sets, R-modules, and Groups, resp,
(where $\{*\}$ is the one-element set, and in the category of groups, $0$ denotes the trivial group)</p>
<p>(ii) $(\{a,b\}\longrightarrow \{*\})^l=(\{a,b\}\longrightarrow \{*\})^r$, $(R\longrightarrow 0)^r$, $\{\Bbb Z\longrightarrow 0\}^r$ are the classes of injections in Sets, R-modules, and Groups, resp</p>
<p>(iii) in R-mod, a module $P$ is projective iff $0\longrightarrow P$ is in $(0\longrightarrow R)^{rl}$</p>
<p>a module $I$ is injective iff $I\longrightarrow 0$ is in $(R\longrightarrow 0)^{rr}$</p>
<p>(iv) in the category of groups,
a finite group $H$ is nilpotent iff
$H\longrightarrow H\times H$ is in $\{\, 0\longrightarrow G : G\text{ arbitrary} \}^{lr}$ </p>
<p>a finite group $H$ is solvable iff $0\longrightarrow H$ is in $\{\, 0\longrightarrow A : A\text{ abelian }\}^{lr}= \{\, [G,G]\longrightarrow {G} : G\text{ arbitrary }\}^{lr}$</p>
<p>a finite group ${H}$ is of order prime to $p$ iff $H\longrightarrow 0$ is in $\{\Bbb Z/p\Bbb Z\longrightarrow 0\}^{rr}$</p>
<p>a finite group $H$ is a p-group iff $H\longrightarrow 0$ is in $\{\Bbb Z/p\Bbb Z\longrightarrow 0\}^{rr}$</p>
<p>a group $H$ is torsion-free iff $0\longrightarrow H$ is in $\{ n\Bbb Z\longrightarrow \Bbb Z: n>0 \}^r$</p>
<p>(v) in the category of metric spaces and uniformly continuous maps,
a metric space $X$ is complete iff $\{1/n\}_n\longrightarrow \{1/n\}_n\cup \{0\} \,\rightthreetimes\, X\longrightarrow \{0\}$
where the metric on $\{1/n\}_n$ and $\{1/n\}_n\cup \{0\}$ is induced from the real line</p>
<p>a subset $A \subset X$ is closed iff $\{1/n\}_n\longrightarrow \{1/n\}_n\cup \{0\} \,\rightthreetimes\, A\longrightarrow X$</p>
<p>for a connected topological space X, each function on $X$ is bounded
iff $ \emptyset\longrightarrow X \,\rightthreetimes\, \cup_n (-n,n) \longrightarrow \Bbb R$ (disjoint union)</p>
|
1,304,948 |
<p>If I solve $Tx=0$ where $T$ is some square matrix
then if I multiply both sides by $T$ and solve for $T^2x=0$, will my x be the same?
In other words if I were to multiply to both sides of the equation $Tx=0$ to any order of $T$, will my x be the same? or at least satisfy the first equation?</p>
|
Adren
| 405,819 |
<p>It's worth to note that if <span class="math-container">$V=W\oplus W^\perp$</span> (which is necessary true if <span class="math-container">$W$</span> is finite dimensional and, in particular, if <span class="math-container">$V$</span> is finite dimensional), we have <span class="math-container">$(W^\perp)^\perp=W$</span>.</p>
<p>To prove this, we only have to check that <span class="math-container">$(W^\perp)^\perp\subset W$</span>, since the reverse inclusion is true in general (that is, without any assumption on <span class="math-container">$W$</span> and even if <span class="math-container">$W$</span> is not a subspace of <span class="math-container">$V$</span>).</p>
<p>Let <span class="math-container">$x\in W^\perp$</span>. There exists <span class="math-container">$a\in W$</span> and <span class="math-container">$b\in W^\perp$</span> such that <span class="math-container">$x=a+b$</span>. By difference : <span class="math-container">$a=x-b\in W^\perp$</span>, hence <span class="math-container">$a\in W\cap W^\perp$</span>, so that <span class="math-container">$a=0$</span>. This proves that <span class="math-container">$x\in W$</span>.</p>
|
1,972,252 |
<blockquote>
<p>If $v^TAv = v^TBv$ for all constant vectors $v$ and $A,B$ are matrices of size $n$ by $n$, is it true that $A=B$? </p>
</blockquote>
<p>I have thought about using basis vectors but cannot get system of equations uniquely. Is there a trick here?</p>
|
H. H. Rugh
| 355,946 |
<p>If you assume that the matrices are symmetric then it is true. Otherwise not. In fact, $M=A-B$ may be any anti-symmetric matrix, i.e. $M^T=-M$.</p>
<p>If $M$ is symmetric on the other hand it is diagonaliable with real eigenvalues and in an orthonormal basis. Picking an eigenvector $v$ for eigenvalue $\lambda$ you see that $0=v^T M v = \lambda v^T v$ so $\lambda=0$. Thus, all eigenvalues are zero and for a diagonalizable matrix this implies that the matrix is the zero matrix.</p>
|
2,375,023 |
<p>So I have to find an interval (in the real numbers) such that it contains all roots of the following function:
$$f(x)=x^5+x^4+x^3+x^2+1$$</p>
<p>I've tried to work with the derivatives of the function but it doesn't give any information about the interval, only how many possible roots the function might have.</p>
|
Angina Seng
| 436,618 |
<p>Can you find a positive number $N$ such that
$N^5>N^4+N^3+N^2+1$? If $|z|\ge N$ then one cannot
have $z^5=-z^4-z^3-z^2-1$.</p>
|
2,375,023 |
<p>So I have to find an interval (in the real numbers) such that it contains all roots of the following function:
$$f(x)=x^5+x^4+x^3+x^2+1$$</p>
<p>I've tried to work with the derivatives of the function but it doesn't give any information about the interval, only how many possible roots the function might have.</p>
|
Andreas
| 317,854 |
<p>EDITED for correct function.</p>
<p>$f(x) = 1+x^2+x^3+x^4+x^5$
is positive for x>0.</p>
<p>So consider $x < 0$:</p>
<p>One can write $f(x) = (1+x^2)(1+x^3) + x^4$ which is positive at least for $(1+x^3) >0$, i.e. $x > -1$.</p>
<p>$f'(x) = 2x+3x^2+4x^3+5x^4 = x(2+3x)+x^3(4+5x)$
so this is positive at least for $x < -4/5 $ since then all terms are positive.</p>
<p>This means that $f(x) = 0$ can only happen at $x < -1$ and since $f(x)$ is rising there, we have that there can only be one real root, which is at $x < -1$.</p>
<p>Now this root can be further locked in. Observing e.g. that $f(-1.5) = -85/32 <0$, we can give an interval for the root at (-1.5 , -1). This can obviously be improved, knowing that there is only <em>one</em> real root and that f(x) is monotonously increasing in this interval.</p>
|
2,375,023 |
<p>So I have to find an interval (in the real numbers) such that it contains all roots of the following function:
$$f(x)=x^5+x^4+x^3+x^2+1$$</p>
<p>I've tried to work with the derivatives of the function but it doesn't give any information about the interval, only how many possible roots the function might have.</p>
|
Ovi
| 64,460 |
<p>The common factoring formula $x^n-1=(x-1)(x^{n-1}+x^{n-2}+ \cdots + x^2+1)$ tells us that your $f(x)$ is equivalent to $g(x)=\dfrac {x^6-1}{x-1}$ as long as $x \not = 1$. Setting $g(x)=0$ yields $x^6-1=0$. Clearly the only real solutions to this are $\pm1$. But only $-1$ is a zero of $g$ and of $f$, so $f$ has only one real zero.</p>
|
1,929,977 |
<p>Let $f\colon\mathbb R\to\mathbb R$ satisfy the Lipschitz condition: there exists $K\geq 0$ such that for all $x,y\in\mathbb R$, we have $|f(x)-f(y)|\leq K\cdot |x-y|$. Is it true that $f$ has one-sided derivatives everywhere? I.e., that the limits
$$\lim_{h\nearrow 0}\frac{f(x+h)-f(x)}{h}\quad\text{and}\quad\lim_{h\searrow 0}\frac{f(x+h)-f(x)}{h}$$
exist for every $x$?</p>
<p>I know that both exist a.e. since the Lipschitz condition implies that $f$ is differentiable a.e., but I would like to know if it is not only true a.e., but <em>everywhere</em>. After working with this for some time, I got the feeling that it might be.</p>
|
Hagen von Eitzen
| 39,174 |
<p>Let
$$ f(x)=\begin{cases}x\sin\frac1x&0<|x|<\frac1\pi\\0&\text{otherwise}\end{cases}$$
Then $f$ is Lipschitz, but none of the one-sided derivatives exists at $x=0$.</p>
|
2,538,553 |
<p>For example in statistics we learn that mean = E(x) of a function which is defined as</p>
<p>$$\mu = \int_a^b xf(x) \,dx$$</p>
<p>however in calculus we learn that</p>
<p>$$\mu = \frac {1}{b-a}\int_a^b f(x) \,dx $$ </p>
<p>What is the difference between the means in statistics and calculus and why don't they give the same answer?</p>
<p>thank you</p>
|
Hilton Fernandes
| 499,387 |
<p>Maybe you can consider the second form of mean as a <strong>sample mean</strong>, analogue to $\bar{x} = \frac{1}{n} \sum_{i=1}^n x_i$ in the discrete case.</p>
<p>That is: the function $f(x)$ in this case would be the <strong>realization</strong> of a random variable.</p>
<p>Please consider the notion of a stochastic process, where a collection of random variables $X$ is indexed, or associated with a continuous deterministic variable $t$. For instance, $X(t)$ could model the weather temperature in a given moment of time $t$, during the interval $[ a, b]$. So, $X(a)$ would not be a value, but a whole random variable, with a given mean, variance, etc. And when you write $X(t)$ you have a collection of random variables, one for each $t \in [a, b]$.</p>
<p>If all these variables have the same properties, one say $X(t)$ is <strong>stationary</strong>, and therefore, $\mu$, the mean of $X(t)$, can be estimated using the series of values of a realization of $X(t)$ -- in our case, some actual measurements of the weather temperature in a given range of time. In that case, your function $f(x)$ would be renamed $x(t)$ (with a lowercase $x$) to indicate one realization of the stochastic process $X(t)$. And then</p>
<p>$$\bar{x} = \frac{1}{b - a} \int_a^b x(t)\,dt$$</p>
<p>would be an estimator of $\mu$, the actual mean of the stationary process $X(t)$.</p>
|
2,289,813 |
<p>Let$\ f(x,y,z)=4xz -y^2 +z^2$ be a differentiable function, let$\ P=(0,1,1)$ and$\ Q=(1,3,2)$. Find$\ T$ such that $\ f(P)-f(Q) = \nabla f(T)(P-Q)$</p>
<p>What I did:</p>
<p>$\ f(P)=0$, $\ f(Q)=3$, $\ (P-Q)=(-1,-2,-1)$, $\nabla f(P-Q)=(-4z, 4y, -4x+2z)$</p>
<p>Let $\ T=(a,b,c)$ then$\ f(P)-f(Q) = \nabla f(T)(P-Q)$ if and only if </p>
<p>$\ -3 = (-4c, 4b, -4a-2c)$, I'm stuck at this point, how can I determine this? </p>
<p>Looking for some advice, thanks!</p>
|
DavidTomahawk
| 430,305 |
<p>Following from what I did:</p>
<p>Let $\ g(t)=(1, 3, 2)+t(-1, -2, -1) = (1-t, 3-2t, 2-t)$</p>
<p>$\nabla f(1-t, 3-2t, 2-t) = (-4t+8, 4t-6, -6t+8)$</p>
<p>$\ -3 = (-4t+8, 4t-6, -6t+8)(-1, -2, -1) = (4t-8, -8t+12, 6t-8)$ if and only if $\ t=1/2$, substitute$\ t$ in the function$\ g $ and </p>
<p>$\ T=g(t)=(\frac{1}{2}, 2,\frac{3}{2})$ is the answer</p>
|
1,031,464 |
<p>I am supposed to simplify this:</p>
<p>$$(x^2-1)^2 (x^3+1) (3x^2) + (x^3+1)^2 (x^2-1) (2x)$$</p>
<p>The answer is supposed to be this, but I can not seem to get to it:</p>
<p>$$x(x^2-1)(x^3+1)(5x^3-3x+2)$$</p>
<p>Thanks</p>
|
Aaron Maroja
| 143,413 |
<p>Try this</p>
<p>$$x(x^2-1)(x^3+1)\Bigg[3x(x^2-1)+2(x^3+1)\Bigg]$$</p>
|
734,203 |
<blockquote>
<p>Find all vectors of $V_{3}$ which are perpendicular to the vector $(7,0,-7)$ and belong to the subspace $L((0,-1,4), (6,-3,0)$.</p>
</blockquote>
<p>As a note, this is an extra question of a long exercise, the vectors found above are replaced by results I found at the above questions.</p>
<p>Let $\overline{x}$ = {${ x_{1}, x_{2}, x_{3} } $} then since it's perpendicular to {${7,0, -7}$}</p>
<p>from there I get </p>
<p>$<\overline{x}, 2\overline{u_{2}} + \overline{u_{3}} > = 0$</p>
<p>$7x_{1} + 7x_{3} = 0$</p>
<p>$x_{1} = x_{3}$</p>
<p>So $\overline{x} = \left \{ \left. x_{1}, x_{2}, x_{1} \right \} \right.$ , $x_{i} \epsilon \mathbb{R}$, $i=1,2,3$</p>
<p>How do I proceed from here?</p>
|
Supersingularity
| 139,189 |
<p>I misunderstood the question and will edit this later with (hopefully) a response.</p>
<p>Given any $x \in \mathbb{C}$, there exists $n$ possible $y \in \mathbb{C}$ such that $y^n = P(x)$ (possibly with multiplicity, if say $P(x) = 0$). However, determining whether the system has a solution (and how many) depends completely on the structure of $Q(x,y)$. Consider the examples below.</p>
<p>If $Q(x,y) = y^n - P(x)$, then any pair $(x,y)$ from above satisfies the system, so we get uncountably many solutions.</p>
<p>If $Q(x,y) = y^n - P(x) +1$, then the first relation forces $y^n - P(x) = 0$, so that $Q(x,y) = 1$ for all $(x,y)$ satisfying the first. Hence the system has no solutions at all in this case.</p>
|
3,371,694 |
<p>Let <span class="math-container">$ \forall n \in \mathbf{N}, ~ u_{n+1} = \frac{1}{\tanh^2(u_n)} - \frac{1}{u^2_n} $</span> with <span class="math-container">$ u_0 = a > 0 $</span>.</p>
<p>What is the limit of <span class="math-container">$(u_n)$</span> ?</p>
<p>I tried to find a fixed point of this sequence but the equation is impossible to solve algebraically.</p>
|
Secret
| 709,089 |
<p>You have that <span class="math-container">$X = (-a, \sqrt{(b^{2} - a^{2})})$</span> and <span class="math-container">$Y = (a,-\sqrt{(b^{2} - a^{2})})$</span></p>
<p>Then <span class="math-container">$\|X-Y\| = \|(-2a, 2\sqrt{(b^{2} - a^{2})})\| = $</span> <span class="math-container">$\sqrt{4a^{2} + 4(b^{2} - a^{2})} = \sqrt{4b^{2}} = 2b$</span></p>
|
2,421,344 |
<p>I'm studying the weak Mordell-Weil theorem in Silverman's "Arithmetic of elliptic curves" and I can't understand the proof of proposition 1.6 of chapter 8.</p>
<p>Here $K$ is a number field, $S$ a finite set of places containing the archimedean ones and $m>2$ an integer. I want to prove that the maximal abelian extension of exponent $m$ unramified outside $S$ is finite. </p>
<p>We can enlarge $K$ so we add the $m^{th} $ roots of unit to use kummer. We can also enlarge $S$ to make the ring o $S$-integers $R_S=\lbrace a\in K : v(a) \geq 0 \text{ for all} v\notin S\rbrace$ a PID. </p>
<p>The first question is, how can I do it? It says to add to $S$ the valuation corresponding to the prime dividing the integral representant of the ideal classes. Why does it make $R_S$ a PID? </p>
<p>Then we apply kummer and we deduce that the extension we are looking for is generated by the $m^{th} $ root of elements $a\in K^*/(K^*) ^m$ such that $\text{ord}_v(a) =0$ mod $m$ for every $v\notin S$.</p>
<p>We have to prove that this set called $T_S$ is finite by showing a surjective map from $R_S^*/(R_S^*) ^m$ which is finite thanks to Dirichlet theorem. </p>
<p>So we take a representant $a\in T_S$. $aR_S$ is an ideal (fractional?) which is the $m$ power of an ideal in $R_S$ given that the prime in $R_S$ correspond to the valuation not in $S$. Obviously the set of element $x\in R_S$ such that $v(x) >0$ is a prime but how can I show that every prime is in this form? </p>
<p>The last question could make the previous easier to answer. Given a set as $S$ and a $v_0\notin S$ is there an element $x\in K$ such that $v_0(x)=1$ and $v(x) =0$ for every other $v \notin S$? </p>
|
Robert Z
| 299,698 |
<p>I think that here the eggs are drawn without replacement. Hence it should be an <a href="https://en.wikipedia.org/wiki/Hypergeometric_distribution#Combinatorial_identities" rel="nofollow noreferrer">Hypergeometric distribution</a>. So, if $3$ eggs are drawn then, for $0\leq k\leq 3$,
$$p_k:=P(\mbox{number of bad eggs}=k)=\frac{\binom{3}{k}\binom{7}{3-k}}{\binom{10}{3}}.$$
It follows that
$$E(X)=\sum_{k=0}^3 kp_k=\frac{21}{10}\quad\mbox{and}\quad
\mbox{Var}(X)=\sum_{k=0}^3 (k-E(X))^2p_k=\frac{49}{100}.$$</p>
|
3,460,204 |
<p>If <span class="math-container">$X\subseteq \mathbb R^n$</span> and <span class="math-container">$x\in X$</span> is an isolated point, then <span class="math-container">$x$</span> is a cluster point of <span class="math-container">$X^c$</span>.</p>
<p>I don't know if that is true or not, can someone help me, please?</p>
|
Adam Martens
| 691,077 |
<p><span class="math-container">$x\in X$</span> is isolated means that there exists open <span class="math-container">$U\ni x$</span> such that <span class="math-container">$U\cap X=\{x\}$</span>. But <span class="math-container">$U\subset \mathbb{R}^n$</span> is open implies there exits <span class="math-container">$r>0$</span> such that <span class="math-container">$B_r(x)\subset U$</span>. But <span class="math-container">$B_r(x)\cap (\mathbb{R}^n\setminus X)=B_r(x)\setminus \{x\}$</span>. </p>
|
3,781,968 |
<p>Let <span class="math-container">$d\in\mathbb N$</span>, <span class="math-container">$U\subseteq\mathbb R^d$</span> be open and <span class="math-container">$M\subseteq U$</span> be a <span class="math-container">$k$</span>-dimensional embedded <span class="math-container">$C^1$</span>-submanifold of <span class="math-container">$\mathbb R^d$</span></p>
<blockquote>
<p>Let <span class="math-container">$f\in\mathcal L^1(U)$</span> and <span class="math-container">$\sigma_M$</span> denote the surface measure on <span class="math-container">$\mathcal B(M)$</span>. Are we able to show that <span class="math-container">$\left.f\right|_M\in\mathcal L^1(\sigma_M)$</span>?</p>
</blockquote>
<p>Let <span class="math-container">$\lambda$</span> denote the Lebesgue measure on <span class="math-container">$\mathcal B(\mathbb R)$</span>. Maybe we can show <span class="math-container">$$\sigma_M(B)\le\lambda^{\otimes d}(B)\;\;\;\text{for all }B\in\mathcal B(M)\tag1$$</span> and use this to conclude the desired claim.</p>
<p>In this regard, we may note that, trivially, <span class="math-container">$U$</span> is a <span class="math-container">$d$</span>-dimensional embedded <span class="math-container">$C^1$</span>-submanifold of <span class="math-container">$\mathbb R^d$</span> and <span class="math-container">$$\sigma_U=\left.\lambda^{\otimes d}\right|_U\tag2.$$</span></p>
<p><em>Remark</em>: It might be useful to note that there is the following characterization of the surface measure: <span class="math-container">$\sigma_M$</span> is the unique measure on <span class="math-container">$\mathcal B(M)$</span> with <span class="math-container">$$\left.\sigma_M\right|_\Omega=\sigma_\Omega\tag3$$</span> for every open subset (in the subspace topology) <span class="math-container">$\Omega$</span> of <span class="math-container">$M$</span>.</p>
|
supinf
| 168,859 |
<p><em>sketch of a counterexample</em>:</p>
<p>Let <span class="math-container">$d=2$</span>, <span class="math-container">$U=(-9,9)\times (-9,9)$</span> and let <span class="math-container">$M$</span> be the <span class="math-container">$1$</span>-dimensional submanifold
which is described by the circle with radius <span class="math-container">$1$</span> and center <span class="math-container">$(1,0)$</span>.
For the function <span class="math-container">$f$</span> we choose
<span class="math-container">$$
f(x,y)=
\begin{cases}
x^{-2/3} &: x>0
\\
0 &: x\leq 0
\end{cases},
$$</span>
where <span class="math-container">$(x,y)\in\mathbb R^2$</span>.</p>
<p>Then one can show that <span class="math-container">$f\in L^1(U)$</span>, but not <span class="math-container">$f\in L^1(\sigma_M)$</span>.</p>
<p>(<em>hint</em>: for showing that <span class="math-container">$f\not\in L^1(\sigma_M)$</span>
estimates of the form <span class="math-container">$c_1s^2\leq 1-\cos s \leq c_2 s^2$</span>
can be useful for <span class="math-container">$s\in\mathbb R$</span> where <span class="math-container">$|s|$</span> is small.)</p>
|
2,111,734 |
<p>a) Show that $f^2 - (f')^2 = 0$</p>
<p>I tried to solve by doing $f=\pm f'$ not sure what to do from here.</p>
|
marwalix
| 441 |
<p>Consider $g=f^2-(f')^2$. One has</p>
<p>$$g'(x)=2f(x)f'(x)-2f''(x)f'(x)=2f'(x)\left(f(x)-f''(x)\right)=0$$</p>
<p>So $g$ is constant and by assumption $g(0)=0$ so $f^2-(f')^2=0$</p>
<p>This means $f'(x)=f(x)$ or $f'(x)=-f(x)$.</p>
<p>Solving these two ODE one gets $f(x)=f(0)e^x=0$ or $f(x)=f(0)e^{-x}=0$</p>
|
3,266,367 |
<p>Find the solution set of <span class="math-container">$\frac{3\sqrt{2-x}}{x-1}<2$</span></p>
<p>Start by squaring both sides
<span class="math-container">$$\frac{-4x^2-x+14}{(x-1)^2}<0$$</span>
Factoring and multiplied both sides with -1
<span class="math-container">$$\frac{(4x-7)(x+2)}{(x-1)^2}>0$$</span>
I got
<span class="math-container">$$(-\infty,-2)\cup \left(\frac{7}{4},\infty\right)$$</span>
Since <span class="math-container">$x\leq2$</span> then
<span class="math-container">$$(-\infty,-2)\cup \left(\frac{7}{4},2\right]$$</span></p>
<p>But the answer should be <span class="math-container">$(-\infty,1)\cup \left(\frac{7}{4},2\right]$</span>. Did I missed something?</p>
|
J. W. Tanner
| 615,567 |
<p>For <span class="math-container">$\dfrac{3\sqrt{2-x}}{x-1}$</span> to be defined, <span class="math-container">$x\le2$</span> and <span class="math-container">$x\ne1$</span>.</p>
<p>If <span class="math-container">$x<1,$</span> then the expression is negative (i.e., <span class="math-container">$<0$</span>), so of course it is <span class="math-container">$< 2$</span>.</p>
<p>If <span class="math-container">$x>1,$</span> then, as you showed, the inequality holds when <span class="math-container">$x>\dfrac74$</span>.</p>
<p>Therefore, the solution set is <span class="math-container">$x<1$</span> or <span class="math-container">$\dfrac74<x\le2$</span>.</p>
|
106,560 |
<p>Mochizuki has recently announced a proof of the ABC conjecture. It is far too early to judge its correctness, but it builds on many years of work by him. Can someone briefly explain the philosophy behind his work and comment on why it might be expected to shed light on questions like the ABC conjecture?</p>
|
grp
| 26,145 |
<p>[The answer below is a response to an earlier version of the question that was rather different in certain respects. Minhyong Kim's answer gives excellent insight into ideas that Mochizuki had back in 2000 and that provide essential building blocks for the more recent work. But I still believe that it is too premature for a non-expert to seek insight into the <em>new</em> work, for reasons explained below, given that many top experts are presently trying to absorb the ideas Mochizuki developed back in 2000.]</p>
<p>This question appears to be inspired by an historical fallacy: the only "vision" of a proof of the Weil Conjectures that Grothendieck had when he began developing ideas related to his work on the problem (i.e., etale cohomology) was the one laid out in Weil's original paper. The yoga around the standard conjectures came much later. </p>
<p>That being said, although the new ABC developments are potentially very exciting, and it is understandable to want to "share in the excitement", for reasons specific to this situation it seems to be much too premature to ask for a sketch on MO or in a blog of Mochizuki's vision/proof with <em>an expectation of insight</em> into the new work. Let me try to indicate why this is the case.</p>
<p>As has been explained clearly by JSE elsewhere, there are plenty of top experts in arithmetic geometry who are presently struggling to get <em>even a small handle</em> on what is really going on in Mochizuki's papers (due entirely to the experts' lack of prior study of these ideas; Mochizuki's writing is extremely precise, detailed, thorough, and full of intuitive asides!). So the situation seems to be rather different from that of other tremendous advances in recent decades (by Perelman, Faltings, Wiles, etc.), for which the deep new work took place within a context that was already somewhat familiar to a good-sized community of experts in the field (who could then use their experience and expertise to quickly disseminate a "bird's eye view" to others of some of the key new ideas). </p>
<p>Because of the rather unique circumstances of this case, as just indicated, I believe that quid's initial urging of patience (if one isn't going to be directly engaged with the struggle to read the actual papers and the prior work upon which they depend) is appropriate. </p>
<p>But to end on a semi-positive note, let me explain why quid's mention of Mochizuki's survey papers is very apt. Some of those surveys are relatively short (e.g., less than 20 pages), and if you find them difficult to grok then you will get a real sense of the difficulties that a lot of top experts are current facing in their efforts to try to understand what Mochizuki has achieved. Please be patient! As quid has noted, in due time, as experts eventually come to acquire a genuine understanding of the overall structure of the arguments in these papers, plenty of expositions for wide dissemination of the ideas will emerge. Mochizuki has put a lot of effort into providing indications of his motivation and insights throughout his papers (which are a serious challenge even for top experts to absorb), and to respect his remarkable effort it seems best to engage with it directly (whether through reading the surveys or the main papers). </p>
|
106,560 |
<p>Mochizuki has recently announced a proof of the ABC conjecture. It is far too early to judge its correctness, but it builds on many years of work by him. Can someone briefly explain the philosophy behind his work and comment on why it might be expected to shed light on questions like the ABC conjecture?</p>
|
Marty
| 3,545 |
<p>I'll take a stab at answering this controversial question in a way that might satisfy the OP and benefit the mathematical community. I also want to give some opinions that contrast with or at least complement grp. Like others, I must give the caveats: I do not understand Mochizuki's claimed proof, his other work, and I make no claims about the veracity of his recent work.</p>
<p>First, some background which might satisfy the OP. For years, Mochizuki has been working on things related to Grothendieck's anabelian program. Here is why one might hope this is useful in attacking problems like ABC:</p>
<p>Begin with the Neukirch-Uchida theorem. See "Über die absoluten Galoisgruppen algebraischer Zahlkörper," by J. Neukirch, Journées Arithmétiques de Caen (Univ. Caen, Caen, 1976), pp. 67–79. Asterisque, No. 41-42, Soc. Math. France, Paris, 1977. Also "Isomorphisms of Galois groups," by K. Uchida, J. Math. Soc. Japan 28 (1976), no. 4, 617–620. </p>
<p>The main result of these papers is that a number field is determined by its absolute Galois group in the following sense: fix an algebraic closure $\bar Q / Q$, and two number fields $K$ and $L$ in $\bar Q$. Then if $\sigma: Gal(\bar Q / K) \rightarrow Gal(\bar Q / L)$ is a topological isomorphism of groups, then $\sigma$ extends to an <em>inner</em> automorphism $Int(\tau): g \mapsto \tau g \tau^{-1}$ of $Gal(\bar Q / Q)$. Thus $\tau$ conjugates the number field $K$ to the number field $L$, and they are isomorphic.</p>
<p>So while class field theory guarantees that the absolute Galois group $Gal(\bar Q / K)$ determines (the profinite completion of) the multiplicative group $K^\times$, the Neukirch-Uchida theorem guarantees that the entire field structure is determined by the profinite group structure of the Galois group. Figuring out how to recover aspects of the field structure of $K$ from the profinite group structure of $Gal(\bar Q / K)$ is a difficult corner of number theory.</p>
<p>Next, consider a (smooth) curve $X$ over $Q$; suppose that the fundamental group $\pi_1(X({\mathbb C}))$ is nonabelian. Let $\pi_1^{geo}(X)$ be the profinite completion of this nonabelian group. Basic properties of the etale fundamental group give a short exact sequence:
$$1 \rightarrow \pi_1^{geo}(X) \rightarrow \pi_1^{et}(X) \rightarrow Gal(\bar Q / Q) \rightarrow 1.$$</p>
<p>Now, just as one can ask about recovering a number field from its absolute Galois group ($Gal(\bar Q / K)$ is isomorphic to $\pi_1^{et}(K)$), one can ask how much one can recover about the curve $X$ from its etale fundamental group. Any $Q$-point $x$ of $X$, i.e. map of schemes from $Spec(Q)$ to $Spec(X)$ gives a section $s_x: Gal(\bar Q / Q) \rightarrow \pi_1^{et}(X)$. </p>
<p>One case of the famous "section conjecture" of Grothendieck states that this gives a bijection from $X(Q)$ to the set of homomorphisms $Gal(\bar Q / Q) \rightarrow \pi_1^{et}(X)$ splitting the above exact sequence. One hopes, more generally, to recover the structure of $X$ as a curve over $Q$ from the induced <em>outer</em> action of $Gal(\bar Q / Q)$ on $\pi_1^{geo}(X)$. (take an element $\gamma \in Gal(\bar Q / Q)$, lift it to $\tilde \gamma \in \pi_1^{et}(X)$, and look at conjugation of the normal subgroup $\pi_1^{geo}(X)$ by $\tilde \gamma$, well-defined up to inner automorphism independently of the lift.)</p>
<p>As in the case of the Neukirch-Uchida theorem, there is an active and difficult corner of number theory devoted to recovering properties of rational points of (hyperbolic) curves from etale fundamental groups. Here are two dramatically difficult problems in the same spirit:</p>
<ol>
<li><p>How can you describe the regulator of a number field $K$ from the structure of the profinite group $Gal(\bar Q / K)$?</p></li>
<li><p>Given a section $s: Gal(\bar Q / Q) \rightarrow \pi_1^{et}(X)$, how can one describe the height of the corresponding point in $X(Q)$?</p></li>
</ol>
<p>I would place Mochizuki's work in this anabelian corner of number theory; I have always kept a safe and respectful distance from this corner.</p>
<p>Now, to say something not quite as ancient that I gleaned from flipping through Mochizuki's recent work:</p>
<p>Many people here on MO and elsewhere have been following research on the field with one element. It is a tempting object to seek, because analogies between number fields and function fields break down quickly when you realize there is no "base scheme" beneath $Spec(Z)$. But I see Mochizuki's work as an anabelian approach to this problem, and I'll try to describe my understanding of this below.</p>
<p>Consider a smooth curve $X$ over a function field $F_p(T)$. The anabelian approach suggests looking at the short exact sequence
$$1 \rightarrow \pi_1^{et}(X_{\overline{F_p(T)}}) \rightarrow \pi_1^{et}(X) \rightarrow Gal(\overline{F_p(T)} / F_p(T)) \rightarrow 1.$$
But much more profitable is to look instead at $X$ as a surface over $F_p$ which corresponds in the anabelian perspective to studying
$$1 \rightarrow \pi_1^{et}(X_{\bar F_p}) \rightarrow \pi_1^{et}(X) \rightarrow Gal(\bar F_p / F_p) \rightarrow 1.$$
But this is pretty close to looking at $\pi_1^{et}(X)$ by itself; there's just a little profinite $\hat Z$ quotient floating around, but this can be characterized (I think) group theoretically within the study of $\pi_1^{et}(X)$ itself.</p>
<p>I would understand (after reading Mochizuki) that looking at curves $X$ over function fields $F_p(T)$ as surfaces over $F_p$ is like looking at only the etale fundamental group $\pi_1^{et}(X)$ without worrying about the map to $Gal(\overline{F_p(T)} / F_p(T))$.</p>
<p>So, the natural number field analogue would be the following. Consider a smooth curve $X$ over $Q$. In fact, let's make $X = E - \{ 0 \}$ be a once-punctured elliptic curve over $Q$. Then the absolute anabelian geometry suggests that to study $X$, it should be profitable to study the etale fundamental group $\pi_1^{et}(X)$ <em>all by itself</em> as a profinite group. This is the anabelian analogue of what others might call "studying (a $Z$-model of) $X$ as a surface over the field with one element". </p>
<p>Without understanding any of the proofs in Mochizuki, I think that his work arises from this absolute anabelian perspective of understanding the arithmetic of once-punctured elliptic curves over $Q$ from their etale fundamental groups. The ABC conjecture is equivalent to <a href="http://en.wikipedia.org/wiki/Szpiro's_conjecture" rel="noreferrer">Szpiro's conjecture</a> which is a conjecture about the arithmetic of elliptic curves over $Q$.</p>
<p>Now here is a suggestion for number theorists who, like myself, have unfortunately ignored this anabelian corner. Let's try to read the papers of Neukirch and/or Uchida to get a start, and let's try to understand Minhyong Kim's work on Siegel's Theorem ("The motivic fundamental group of $P^1 \backslash ( 0, 1, \infty )$ and the theorem of Siegel," Invent. Math. 161 (2005), no. 3, 629–656.)</p>
<p>It would be wonderful if, while we're waiting for the experts to weight in on Mochizuki's work, we took some time to revisit some great results in the anabelian program. If anyone wants to start a reading group / discussion blog on these papers, I would enjoy attending and discussing.</p>
|
106,560 |
<p>Mochizuki has recently announced a proof of the ABC conjecture. It is far too early to judge its correctness, but it builds on many years of work by him. Can someone briefly explain the philosophy behind his work and comment on why it might be expected to shed light on questions like the ABC conjecture?</p>
|
Pasten
| 26,218 |
<p>I want to point out a bibliographical information that perhaps is not very well-known and can be taken as "evidence" for the possibility of applying anabelian geometry to the ABC conjecture successfully. However, I am not claiming that this is related in any sort of way to Mochizuki's work. </p>
<p>Here is the fact: There is a $\pi_1$ proof of the function field Szpiro conjecture (over the complex numbers, as far as I know). The proof is indeed easy and conceptually clear, you can find a nice exposition of it in some (expository) paper of Zhang, whose title is lost somewhere in my memories. (EDIT: the paper is "Geometry of algebraic points").</p>
<p>Anyway, I can tell you what is the key point of the argument. Let E be an elliptic fibration over the projective line L over the complex numbers. Assume that E has only multiplicative bad reduction. You can read the order of the discriminant at a point of L from the Kodaira type of the fibre, which in turn can be recovered in terms of monodromy representations of the fundamental group of L minus the points with bad fibres: smooth fibres have trivial monodromy, and the monodromy of singular fibres is determined by Dehn twists (assuming multiplicative reduction). You can look at all these local representations at once, after choosing loops to link bad points to some generic point p of L and then study the image of the global monodromy representation on the homology of the fibre above p. Choosing loops appropriately gives the usual commutator relation which in the image of the global representation gives a relation R=1 among generators of the local reps (and they "know" what the discriminant is). Everything here is inside $SL_2(Z)=Aut(Z^2)=Aut(H_1(E_p,Z))$ which acts on the real plane, and up to scalars it acts on the projective real line whose universal covering you already know (yes, the real line). One can lift the relation R=1 to a relation among automorphisms of the real line to get a relation R'=1' where now 1' knows the number of terms appearing on R, that is to say the number of singular fibres, which is the conductor of E in this setting. Then the Szpiro bound can be recovered from the relation R'=1'. </p>
<p>And there you have, a derivative-free proof of the Szpiro conjecture for function fields (a bit shocking at least for me the first time I saw it). All the diophantine information being supplied by fundamental groups.</p>
|
106,560 |
<p>Mochizuki has recently announced a proof of the ABC conjecture. It is far too early to judge its correctness, but it builds on many years of work by him. Can someone briefly explain the philosophy behind his work and comment on why it might be expected to shed light on questions like the ABC conjecture?</p>
|
Vesselin Dimitrov
| 26,522 |
<p>Let me also try to give, in a modest complement to Minhyong Kim's great post, some additional remarks on Mochizuki's strategy. The idea that has led to the development of "Inter-universal Teichmuller theory for number fields" is certainly very beautiful, and was known to Mochizuki, along with the nature of the final estimate, already in 2000. (But let us recall, as a sane reminder of just how elusive the ABC conjecture has been, Miyaoka's flawed proof: did the idea of a Bogomolov-Miyaoka-Yau type bound involving arithmetic Chern numbers in Arakelov theory not seem equally beautiful, exciting, and promising?) </p>
<p>In brief, the main idea behind the IUTT-series is to construct, outside the rigid confines of algebraic geometry, a subtle object simulating a rank-1, <em>Galois-stable</em> quotient of $E[\ell]$. Here, $E/\mathbb{Q}$ is a (pretty much) <em>arbitrary</em> rational elliptic curve (and this is the main point: such a Galois-stable quotient will almost never exist!); and $\ell \geq 5$ is an auxiliary prime, generic for $E$ in a <em>very</em> mild sense, but otherwise free to optimize until the very final estimate. This is then applied to construct, in the non-linear discretized "Hodge-Arakelov theory," a comparison isomorphism between $(E^{\dagger}, <\ell)$ and $E[\ell]$, <em>which is free of Gaussian poles</em> at the bad places of $E$. For this then leads to a promising Galois-theoretic "Kodaira-Spencer map," as explained in Minhyong Kim's post, hopefully leading in the familiar way to the arithmetic Szpiro inequality for this very same elliptic curve: $\log{|\Delta_{\mathrm{min}}(E)|} \leq (6+\varepsilon) \log{N_E} + O_{\varepsilon}(1)$.</p>
<p>Let me, however, disagree with one point from M. Kim's post. My impression is that what Mochizuki calls an "initial $\Theta$-datum" - and which is, essentially, the pair of the rational elliptic curve $E$ (or equivalently, the $abc$-triple from the ABC-conjecture!) and, until the very final estimate in Ch. 2 of the fourth paper, the prime level $\ell$ - are <em>fixed for good</em> throughout the entire series of IUTT-papers. The <em>deformation</em> flavor of "Teichmuller theory" refers to <em>dismantling</em> the underlying number field, and <em>not</em> to the elliptic curve enhancement (indeed, in Mochizuki's dictionary with his own <em>$p$-adic Teichmuller theory</em>, it is the number field that corresponds to a hyperbolic curve; the elliptic curve enhancement corresponds to an "indigenous bundle" over the hyperbolic curve, and invites the anabelian philosophy via the \'etale fundamental group of the once-punctured elliptic curve). All the "Hodge theaters" associated to the initial $\Theta$-datum are isomorphic to one another, and form a vastly complicated $2$-dimensional non-commutative array - the "$\mathfrak{log}-\Theta$ lattice" - of non-ring theoretic translations between one another. <em>What Mochizuki writes on p. 10 of IUTT-I</em> is that the theory of $\Theta$-Hodge theaters "may be regarded as a sort of <em>solution</em> to the problem of constructing the <em>global</em> quotient $E[\ell] \twoheadrightarrow Q$" [needed for the application to arithmetic Kodaira-Spencer]. He does not seem to suggest that this is done by "moving the initial $E$ to a single elliptic curve via the intermediate case of an <em>elliptic curve in general position</em>," as M. Kim writes. (The term "elliptic curves in general position" indeed figures in Mochizuki's fourth paper, but it has a different, not-so-essential significance that comes through his entirely self-contained paper [GenEll], and whose purely technical purpose is to reduce the general ABC conjecture to the restricted version of Szpiro's inequality for $E$, in Thm. 1.10 of IUTT-IV, coming from the estimate in IUTT-III).</p>
<p>In particular, in sharp contrast to the Thue-Siegel-Roth tradition of Diophantine approximations, Mochizuki's program <em>does not</em> seem to compare different elliptic curves / $abc$-triples, all the way through to the key estimate
$$
(*) \hspace{3cm} \log{|\Delta_{\mathrm{min}}(E)|} \leq \big(6 + \varepsilon + 200/\ell\big)\log{N_E} + 12\log(\ell\varepsilon^{-7})
$$
of IUTT-IV [asserted for <em>all</em> primes $\ell \geq 5$ that are generic for $E$ in a <em>rather mild</em> sense: essentially, $\ell$ has to be prime to the degenerate places and the $q$-parameters of $E$. Also, $\varepsilon \in (0,\epsilon_0)$ is <em>arbitrary</em>, with $\epsilon_0$ a numerical value. ] In this sense, Mochizuki's approach - nevermind the vast technical difficulties precipitated by the non-ring theoretic simulation of a global quotient $E[\ell] \twoheadrightarrow Q$ - is entirely direct and, consequently, effective.</p>
<p><em>So what does Mochizuki actually (claim to) prove?</em> </p>
<p>Start with an $abc$-triple (co-prime rational integers with $a+b+c=0$). Since the discriminant $(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)$ of a cubic polynomial $x^3 + \cdots$ encapsulates exactly this equation, it is a profitable, traditional idea to interpret the $abc$-datum as the giving of the rational elliptic curve $E = E_{a,b,c}$ defined by the equation $y^2 = x(x-a)(x+b)$. The (apparently weaker, but virtually as powerful) ABC conjecture $abc < K_{\varepsilon}\cdot\mathrm{rad(abc)}^{3+\varepsilon}$ then translates into Szpiro's inequality: $\log{|\Delta_{\min}(E)|} \leq (6+\varepsilon)\log{N_E} + O_{\varepsilon}(1)$ between the minimal discriminant $\Delta_{\min}(E)$ and conductor $N_E$ of $E$ (which are, <em>essentially</em>, $(abc)^2$ and $\mathrm{rad}(abc)$). Pick the "auxiliary prime" $\ell \geq 5$ to be generic for $E$ in the sense that, <em>essentially</em>: (1) $\ell \nmid abc$; (2) $\ell$ does not divide the prime exponents in $abc$; (3) for $F := \mathbb{Q}( \sqrt{-1}, E[15] )$, the Galois representation of $G_F$ on $E[\ell]$ has full image $\mathrm{GL}_2 (\mathbb{Z}/\ell)$. [<em>Conjecturally, the last condition should only exclude a finite list of primes, independent of $E$</em>!] Then Mochizuki [IUTT-IV, Thm. 1.10] claims that (*) should hold for any $\varepsilon < \epsilon_0$.</p>
<p>This is the essential Diophantine estimate. Anything further than that [i.e., the deduction of the full ABC conjecture in IUTT-IV, Section 2] consists of standard, and relatively straightforward reductions [such as, e.g., the use of non-critical Belyi maps] elaborated in Mochizuki's self-contained paper [GenEll]: "Arithmetic elliptic curves in general position." Mochizuki indeed writes, in his first paper, that the auxiliary prime level $\ell \geq 5$ from the Hodge-Arakelov discretized non-linear comparison isomorphisms/correspondences $(E^{\dagger}, < \ell) \leftrightarrow E[\ell]$, will be chosen in the Diophantine application to be <em>large, roughly on the order of the height of $E$</em>. But this comes entirely through Theorem 3.8 in [GenEll]: there, the various non-divisibility properties are ensured by simply taking $\ell$ to exceed all the primes of bad reduction / all the $q$-parameters (also, the full Galois action is ensured unconditionally). In (*), $\ell$ could be any prime satisfying the mentioned non-divisibility conditions. (This, by the way, is what I considered highly disturbing).</p>
<p><em>My apology if I have misunderstood - and misrepresented - the points from Mochizuki's papers that I have alluded to.</em></p>
|
106,560 |
<p>Mochizuki has recently announced a proof of the ABC conjecture. It is far too early to judge its correctness, but it builds on many years of work by him. Can someone briefly explain the philosophy behind his work and comment on why it might be expected to shed light on questions like the ABC conjecture?</p>
|
user30304
| 30,304 |
<p>NEW !! (2013-02-21)</p>
<p>A Panoramic Overview of Inter-universal Teichmüller
Theory
By
Shinichi Mochizuki</p>
<p><a href="http://www.kurims.kyoto-u.ac.jp/~motizuki/Panoramic%20Overview%20of%20Inter-universal%20Teichmuller%20Theory.pdf">http://www.kurims.kyoto-u.ac.jp/~motizuki/Panoramic%20Overview%20of%20Inter-universal%20Teichmuller%20Theory.pdf</a></p>
|
320,019 |
<p>Assume $Y$ is non negative random variable. Prove that $X+Y$ is stochastically greater than $X$ for any random variable $X$.</p>
<p>We have to prove there that $\Pr(X+Y > x) \geq \Pr(X>x) $ for all $x$</p>
|
gnometorule
| 21,386 |
<p>Recalling that $A \subset B \Rightarrow P(A) \leq P(B)$, and that $B := \{ X > x -\epsilon \} \supset \{ X > x \} := A$ for any $\epsilon \geq 0$, and that by positivity of $Y$, $x - Y = x - \epsilon$ for some $\epsilon \geq 0$ (for every sample point $\omega$): </p>
<p>$$P(X+Y > x) = P(X > x - Y) \geq P(X>x).$$</p>
|
2,591,976 |
<p>If $f(x-1)=2x^2-10x+3$, find $f(x)$. I tried the problem and received the answer $f(x)=2x^2-14x-5$, is this right? </p>
|
fleablood
| 280,126 |
<blockquote>
<p>$f(x)=2x^2-14x-5$ is this right?</p>
</blockquote>
<p>Let's see. $f(x-1) = 2(x-1)^2 - 14(x-1) - 5 = 2(x^2 -2x +1)-14x + 14 - 5$</p>
<p>$= 2x^2 - 4x +2 - 14x + 9 = 2x^2-18x +11$</p>
<p>No, that doesn't seem to be right.</p>
<p>.....</p>
<p>Instead try.</p>
<p>$x = (x+1) - 1$</p>
<p>Plug $(x+1)-1$ into $f$ and you get:</p>
<p>$f(x) = f([x+1]-1) = 2([x+1])^2 - 10([x+1]) + 3$. </p>
<p>$= 2(x^2+2x + 1) - 10x -10 + 3=$</p>
<p>$2x^2 +4x +2 - 10x -7 =$</p>
<p>$2x^2 - 6x -5$</p>
<p>[And we test it $f(x-1) = 2(x-1)^2 -6(x-1) -5 = 2x^2 -4x + 2 - 6x + 6 -5=2x^2 - 10x +3$.</p>
<p>]</p>
<p>.... But really this easiest is to say</p>
<p>let $x = y-1$ and so $y = x + 1$</p>
<p>then $f(x) = f(y-1) = 2y^2 -10y + 3$. But $y =x+1$ so </p>
<p>$f(x) = f(y-1) = 2y^2 -10y + 3 = 2(x+1)^2 - 10(x+1) + 3 =$</p>
<p>and .... so on.</p>
|
2,591,976 |
<p>If $f(x-1)=2x^2-10x+3$, find $f(x)$. I tried the problem and received the answer $f(x)=2x^2-14x-5$, is this right? </p>
|
Michael Rozenberg
| 190,319 |
<p>$$f(x)=2(x+1)^2-10(x+1)+3=2x^2-6x-5.$$</p>
|
2,486,334 |
<p>What is the plane graph of $|z-1|+|z-5| < 4$ ?</p>
<p>What i know is that there is nothing for $y\geq4$ or $y \leq -4$ or $x \geq 5$ or $x \leq 1$.</p>
<p>Trying to let $z=x+y i$ such that $x,y \in \mathbb{R}$ did not help either.</p>
|
jonsno
| 310,635 |
<p>Consider two points in cartesian plane, $A(1,0)$ and $B(5,0)$. You have to find a locus of point $P$ such that <em>sum of distance from these two points is less than</em> $4$. </p>
<p>$$PA+PB < 4$$</p>
<p>Since the points themselves are at a distance $4$, ie $AB = 4$ , you can never have $PA+PB < 4$, because <strong>triangle inequality</strong> tells :</p>
<p>$$PA+PB \ge AB$$</p>
|
2,441,660 |
<p>Suppose I have $n$ observations, which are all normally distributed with the same mean (which is unknown) but each has a different variance (the different variances could be called $v_1,...,v_n$ for example, which are all assumed to be known).</p>
<p>What is the best estimate of the mean? And further, how does one compute the variance of the mean?</p>
<p>Clearly, the sample average would be the best estimate if iid, but intuitively, an observation which has a relatively low variance would provide more information about the mean than the others, suggesting that the sample mean would not be the best estimate. My problem is that I don't know how to 'quantify' this intuition to generate an estimate of the mean and it's variance. Any ideas on how to do this? Thanks.</p>
|
N. F. Taussig
| 173,070 |
<p>We must find the number of solutions of the equation
<span class="math-container">$$x_1 + x_2 + x_3 = 9 \tag{1}$$</span>
in the positive integers subject to the restrictions that <span class="math-container">$x_1, x_2, x_3 \leq 6$</span>.</p>
<p>A particular solution of equation 1 in the positive integers corresponds to the placement of <span class="math-container">$3 - 1 = 2$</span> addition signs in the <span class="math-container">$9 - 1 = 8$</span> spaces between successive ones in a row of nine ones.
<span class="math-container">$$1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1$$</span>
For instance, placing addition signs in the third and seventh spaces yields
<span class="math-container">$$1 1 1 + 1 1 1 1 + 1 1$$</span>
which corresponds to the solution <span class="math-container">$x_1 = 3$</span>, <span class="math-container">$x_2 = 4$</span>, <span class="math-container">$x_3 = 2$</span>. </p>
<p>The number of solutions of equation 1 in the positive integers is
<span class="math-container">$$\binom{9 - 1}{3 - 1} = \binom{8}{2}$$</span>
since we must choose which two of the eight spaces between successive ones to fill with addition signs. </p>
<p>However, we have included solutions in which some <span class="math-container">$x_i > 6$</span>. Since the summands are positive integers, this can only occur if one of the <span class="math-container">$x_i$</span>'s is <span class="math-container">$7$</span> and the other two <span class="math-container">$x_i$</span>'s are <span class="math-container">$1$</span>. There are three choices for which <span class="math-container">$x_i$</span> is equal to <span class="math-container">$7$</span>. Hence, the number of ways the sum of the numbers on the three dice could add up to <span class="math-container">$9$</span> is
<span class="math-container">$$\binom{8}{2} - \binom{3}{1}$$</span></p>
<p>We can formalize the argument about the restrictions as follows. Suppose <span class="math-container">$x_1 > 6$</span>. Then <span class="math-container">$x_1' = x_1 - 6$</span> is a positive integer. Substituting <span class="math-container">$x_1' + 6$</span> for <span class="math-container">$x_1$</span> in equation 1 yields
<span class="math-container">\begin{align*}
x_1' + 6 + x_2 + x_3 & = 9\\
x_1' + x_2 + x_2 & = 3 \tag{2}
\end{align*}</span>
Equation 2 is an equation in the positive integers with
<span class="math-container">$$\binom{3 - 1}{3 - 1} = \binom{2}{2}$$</span>
solutions. By symmetry, the number of solutions of equation 1 in which <span class="math-container">$x_1 > 6$</span> is equal to the number of solutions in which <span class="math-container">$x_2 > 6$</span> and to the number of solutions in which <span class="math-container">$x_3 > 6$</span>. Hence,
<span class="math-container">$$\binom{3}{1}\binom{2}{2}$$</span>
solutions of equation 1 violate the restrictions. Therefore, the number of admissible solutions is
<span class="math-container">$$\binom{8}{2} - \binom{3}{1}\binom{2}{2}$$</span></p>
|
2,232,531 |
<p>I have a dash camera on my windshield, and I'm wondering if there is any way to determine my vehicle's speed using the video footage from the dash cam. Please help!</p>
|
dantopa
| 206,581 |
<p>One strategy may be to use the dashed lines in the lane dividers.</p>
<p>A source like <a href="http://commons.bcit.ca/civil/edufacts/highway_paint.html" rel="nofollow noreferrer">this</a> can provide size and spacing. Then mean velocity $v$ can be expressed in terms of, $d$, distance (on the video) and time, $t$, (a watch your arm):
$$
\bar{v} = \frac{d}{t}
$$</p>
|
2,232,531 |
<p>I have a dash camera on my windshield, and I'm wondering if there is any way to determine my vehicle's speed using the video footage from the dash cam. Please help!</p>
|
Corey Levinson
| 299,500 |
<p>The federal guideline for the white dashed lines on roads is 10 feet, and the empty spaces in between are 30 feet. If you measure the time it takes to move from the beginning of one white dashed line to the next, that is the time it takes your car to move 40 feet, which is a measure of speed.</p>
|
4,624,695 |
<p>I have an exercise with text: With lines</p>
<p><span class="math-container">$$
p_1 \ldots \frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-4}{1}, \quad p_2 \ldots \frac{x-5}{2}=\frac{y-1}{-1}=\frac{z-2}{1} .
$$</span>
Determine one plane <span class="math-container">$\pi$</span> with respect to which the lines <span class="math-container">$p_1$</span> and <span class="math-container">$p_2$</span> are symmetric.</p>
<p>How to approach this? I should somehow conclude that lines go through dots A1(1,-1,4) A2(5,1,2).</p>
|
GReyes
| 633,848 |
<p>Your function is odd with respect to <span class="math-container">$x=a$</span>. That is, the area from <span class="math-container">$0$</span> to <span class="math-container">$a$</span> and that from <span class="math-container">$a$</span> to <span class="math-container">$2a$</span> are equal in magnitude and have opposite sign, so they cancel, giving a zero integral. Similar to having the integral of an odd function over a symmetric interval.</p>
<p>An example would be <span class="math-container">$f(x)=(x-a)^3$</span>.</p>
|
2,822,126 |
<blockquote>
<p>If $0⩽x⩽y⩽z⩽w⩽u$ and $x+y+z+w+u=1$, prove$$
xw+wz+zy+yu+ux⩽\frac15.
$$</p>
</blockquote>
<p>I have tried using AM-GM, rearrangement, and Cauchy-Schwarz inequalities, but I always end up with squared terms. For example, applying AM-GM to each pair directly gives$$
x^2+y^2+z^2+w^2+u^2 ⩾ xw + wz + zy + yu + ux,
$$
but I cannot seem to continue from here or use $x + y + z + w + u = 1$. Other inequalities like Chebyshev's rely on the multiplied pairs to be in order from least to greatest or vice versa, so I am stuck here.</p>
|
Cesareo
| 397,348 |
<p>It is quite easy to solve this problem with the Lagrange multipliers</p>
<p>Calling </p>
<p>$$
L(x,y,z,w,u,\lambda) = \frac 15 -(xw+wz+zy+yu+ux)+\lambda(x+y+z+w+u-1)
$$</p>
<p>The stationary points are the solutions for</p>
<p>$$
\nabla L = (L_x,L_y,L_z,L_w,L_u,L_{\lambda})=0
$$</p>
<p>or</p>
<p>$$
\lambda -u-w = 0\\
\lambda -u-z = 0\\
\lambda -w-y = 0\\
\lambda -x-z = 0\\
\lambda -x-y = 0\\
u+w+x+y+z-1 = 0
$$</p>
<p>Solving this linear system we obtain</p>
<p>$$
x = y = z = w = u = \frac 15,\;\lambda = \frac25
$$</p>
<p>This point is the only tangent point between the surface $g(x,y,z,w,u)=\frac 15 -(xw+wz+zy+yu+ux)$ and the hyperplane $\Pi =x+y+z+w+u-1 = 0$</p>
<p>and $g(x,y,z,w,u) \ge 0$ is located into one of the semi-spaces delimited by $\Pi$ and also at the tangency point we have $g = 0$ with the values found before.</p>
<p>NOTE</p>
<p>This formulation is a short hand for</p>
<p>$$
L(x,y,z,w,u,\lambda,\epsilon) = \frac 15 -(xw+wz+zy+yu+ux)+\lambda_1(x+y+z+w+u-1)+\lambda_2(x-\epsilon_1^2)+\lambda_2(y-x-\epsilon_2^2)+\lambda_4(z-y-\epsilon_3^2)+\lambda_5(w-z-\epsilon_4^2)+\lambda_6(u-w-\epsilon_5^2)
$$</p>
<p>and the result should be the same as you can verify with a little patience.</p>
|
2,822,126 |
<blockquote>
<p>If $0⩽x⩽y⩽z⩽w⩽u$ and $x+y+z+w+u=1$, prove$$
xw+wz+zy+yu+ux⩽\frac15.
$$</p>
</blockquote>
<p>I have tried using AM-GM, rearrangement, and Cauchy-Schwarz inequalities, but I always end up with squared terms. For example, applying AM-GM to each pair directly gives$$
x^2+y^2+z^2+w^2+u^2 ⩾ xw + wz + zy + yu + ux,
$$
but I cannot seem to continue from here or use $x + y + z + w + u = 1$. Other inequalities like Chebyshev's rely on the multiplied pairs to be in order from least to greatest or vice versa, so I am stuck here.</p>
|
Michael Rozenberg
| 190,319 |
<p>Let $y=x+a$, $z=x+a+b$, $w=x+a+b+c$ and $u=x+a+b+c+d$.</p>
<p>Thus, $a$, $b$, $c$ and $d$ are non-negatives and we need to prove that:
$$(x+y+z+u+w)^2\geq5(xw+wz+zy+yu+ux)$$ or
$$(5x+4a+3b+2c+d)^2\geq5(x(x+a+b+c)+(x+a+b+c)(x+a+b)+(x+a+b)(x+a)+(x+a)(x+a+b+c+d)+(x+a+b+c+d)x)$$ or
$$(5x+4a+3b+2c+d)^2\geq5(5x^2+(8a+6b+4c+2d)x+(a+b+c)(a+b)+(a+b)a+a(a+b+c+d))$$ or
$$(4a+3b+2c+d)^2\geq5((a+b+c)(a+b)+(a+b)a+a(a+b+c+d)),$$
which is obvious.</p>
<p>Done!</p>
|
216,473 |
<p>Let me begin with some background: I used to enjoy mathematics immensely in school, and wanted to pursue higher studies. However, everyone around me at that time told me it was a stupid area (that I should focus on earning as soon as possible), and so instead I opted for an engineering degree. While in college, I used to find myself fascinated by higher math and other stuff, and used to score near-perfect all the time (but only in mathematics!). I must make it clear that I don't have an extraordinary talent for math; it's just that I enjoy exploring it very much, and then describing it to others in interesting ways. Anyway, it so happened that I went on changing job after job and was never satisfied. Now at the age of 27, I'm sitting home and realizing that I should have given mathematics a thought. So I'm thinking of taking up a graduate course and picking up where I left off.</p>
<p>HOWEVER . . .</p>
<p>I dream of becoming a teacher cum researcher some day, even if it's at school-level only. Before my graduate course begins (in 2-3 months), I've started reviewing math from my school books. Now the point is that I expect myself to be much more mature and smarter by now. That means I should be able to solve any problem and prove any theorem given in school math. But I can't, and it's shattering me. I mean, if I can't even prove basic theorems related to Euclid's geometry, how can I ever hope to do some authentic research like the mathematicians I so admire? This makes me wonder if I’ll ever be fit for teaching. If, at the age of 27, I can’t even master basic school mathematics with certainty, how can I ever hope to tackle problems in calculus and polynomials that students bring me tomorrow? It’s as if I’m cheating myself and those who’ll come to me for instruction.</p>
<p>Am I being too hard on myself? Am I expecting too much too soon? Does there come a point in a person’s math studies when he is able to discern properties and theorems all on his own? Or are all students of mathematics struggling and hiding their weaknesses? Or I really have no talent for math and it’s just an idle indulgence of mine. I mean, I’m not sure “how good” I’m supposed to be in order to feel confident that I can pull it off. In general, I find myself wondering how much do the others know math. Do all the teachers have perfect knowledge of, say, geometry, and can tackle every problem? If not, what gives them the right to call themselves teachers?</p>
<p>I have a feeling most of these questions are absurd, but I’ll be very thankful if someone can put me out of my misery.</p>
|
Euler....IS_ALIVE
| 38,265 |
<p>As a grad student, I think that you should stick to engineering. The jobs are just so more plentiful.</p>
|
2,908,186 |
<p>There's an old-school pocket calculator trick to calculate <span class="math-container">$x^n$</span> on a pocket calculator, where both, <span class="math-container">$x$</span> and <span class="math-container">$n$</span> are real numbers. So, things like <span class="math-container">$\,0.751^{3.2131}$</span> can be calculated, which is awesome.</p>
<p>This provides endless possibilities, including calculating nth roots on a simple pocket calculator.</p>
<p><strong>The trick goes like this:</strong></p>
<ol>
<li>Type <span class="math-container">$x$</span> in the calculator</li>
<li>Take the square root twelve times</li>
<li>Subtract one</li>
<li>Multiply by <span class="math-container">$n$</span></li>
<li>Add one</li>
<li>Raise the number to the 2nd power twelve times (press <code>*</code> and <code>=</code> key eleven times)</li>
</ol>
<p><strong>Example:</strong></p>
<p>I want to calculate <span class="math-container">$\sqrt[3]{20}$</span> which is the same as <span class="math-container">$20^{1/3}$</span>. So <span class="math-container">$x=20$</span> and <span class="math-container">$n=0.3333333$</span>. After each of the six steps, the display on the calculator will look like this:</p>
<ol>
<li><span class="math-container">$\;\;\;20$</span></li>
<li><span class="math-container">$\;\;\;1.0007315$</span></li>
<li><span class="math-container">$\;\;\;0.0007315$</span></li>
<li><span class="math-container">$\;\;\;0.0002438$</span></li>
<li><span class="math-container">$\;\;\;1.0002438$</span></li>
<li><span class="math-container">$\;\;\;2.7136203$</span></li>
</ol>
<p>The actual answer is <span class="math-container">$20^{1/3}\approx2.7144176$</span>. So, our trick worked to three significant figures. It's not perfect, because of the errors accumulated from the calculator's 8 digit limit, but it's good enough for most situations.</p>
<blockquote>
<p><strong>Question:</strong></p>
<p>So the question is now, why does this trick work? More specifically, how do we prove that:
<span class="math-container">$$x^n\approx \Big(n(x^{1/4096}-1)+1\Big)^{4096}$$</span></p>
</blockquote>
<p>Note: <span class="math-container">$4096=2^{12}$</span>.</p>
<p>I sat in front of a piece of paper trying to manipulate the expression in different ways, but got nowhere.</p>
<p>I also noticed that if we take the square root in step 1 more than twelve times, but on a better-precision calculator, and respectively square the number more than twelve times in the sixth step, then the result tends to the actual value we are trying to get, i.e.:</p>
<blockquote>
<p><span class="math-container">$$\lim_{a\to\infty}\Big(n(x^{1/2^a}-1)+1\Big)^{(2^a)}=x^n$$</span></p>
</blockquote>
<p>This, of course doesn't mean that doing this more times is encouraged on a pocket calculator, because the error from the limited precision propagates with every operation. <span class="math-container">$a=12$</span> is found to be the optimal value for most calculations of this type i.e. the best possible answer taking all errors into consideration. Even though 12 is the optimal value on a pocket calculator, taking the limit with <span class="math-container">$a\to\infty$</span> can be useful in proving why this trick works, however I still can't of think a formal proof for this.</p>
<p><strong>Thank you for your time :)</strong></p>
|
Xiangxiang Xu
| 86,009 |
<p>For fixed $x > 0$ and $n$, let $t = 1/2^a \to 0$. Then we need to prove that
$$
\lim_{t \to 0} \left( n (x^t - 1) + 1 \right)^{1/t} = x^n.
$$
In fact, we have
$$
\ln \left[\lim_{t \to 0} \left( n (x^t - 1) + 1 \right)^{1/t}\right] =
\lim_{t \to 0} \frac{\ln (1 + n(x^t - 1))}{t} = \lim_{t \to 0} \frac{n(x^t - 1)}{t} = n\ln x,
$$
where the first equality follows from the continuity of $\ln(x)$, and the second equality has used the fact that $\ln(1 + x) \sim x$ when $x \to 0$.</p>
|
256,138 |
<p>I need to generate four positive random values in the range [.1, .6] with (at most) two significant digits to the right of the decimal, and which sum to exactly 1. Here are three attempts that do not work.</p>
<pre><code>x = {.15, .35, .1, .4}; While[Total[x] != 1,
x = Table[Round[RandomReal[{.1, .6}], .010], 4]];
x = {.25, .25, .25, .25}; While[Total[x] == 1,
x = Table[Round[RandomReal[{.1, .6}], .010], 4]];
NestWhileList[Total[x],
x = Table[Round[RandomReal[{.1, .6}], .010], 4],
Plus @@ x == 1][[1]]
</code></pre>
|
kglr
| 125 |
<pre><code>SeedRandom[1]
list = {##, 1 - +##} & @@@ Round[RandomPoint[Simplex[3], 10], .01];
Grid[Prepend[{"4-tuple", "total"}][{#, Total@#} & /@ list ],
Dividers -> {False, {True, True, {False}}}]
</code></pre>
<p><a href="https://i.stack.imgur.com/Axq8k.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Axq8k.png" alt="enter image description here" /></a></p>
|
256,138 |
<p>I need to generate four positive random values in the range [.1, .6] with (at most) two significant digits to the right of the decimal, and which sum to exactly 1. Here are three attempts that do not work.</p>
<pre><code>x = {.15, .35, .1, .4}; While[Total[x] != 1,
x = Table[Round[RandomReal[{.1, .6}], .010], 4]];
x = {.25, .25, .25, .25}; While[Total[x] == 1,
x = Table[Round[RandomReal[{.1, .6}], .010], 4]];
NestWhileList[Total[x],
x = Table[Round[RandomReal[{.1, .6}], .010], 4],
Plus @@ x == 1][[1]]
</code></pre>
|
Joshua Schrier
| 63,709 |
<p>Building on @kglr's response—the way to do this is to incorporate some of the additional constraints into the definition of the region from which sampling occurs (it is a bit more complicated than a Simplex). We can define a region where the 3 variables go over the appropriate range. The implicit fourth variable means that the sum of the 3 cannot be less than 0.4 nor greater than 0.9 (i.e., the fourth variable also goes from 0.1 to 0.6)</p>
<pre><code>h = ImplicitRegion[
0.4 <= x1 + x2 + x3 <= 0.9,
{{x1, 0.1, 0.6}, {x2, 0.1, 0.6}, {x3, 0.1, 0.6}}];
RegionPlot3D[h] (*take a look before we proceed*)
</code></pre>
<p><a href="https://i.stack.imgur.com/oOZlC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oOZlC.png" alt="the region" /></a></p>
<p>Notice how this extra constraint cuts off some corners that would have been present in the simplex...</p>
<p>Now we can continue with @kglr's solution:</p>
<pre><code>list = {##, 1 - +##} & @@@ Round[RandomPoint[h, 10], 0.01];
Grid[
Prepend[{"4-tuple", "total"}][{#, Total@#} & /@ list],
Dividers -> {False, {True, True, {False}}}]
</code></pre>
<p><a href="https://i.stack.imgur.com/YVCBy.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YVCBy.png" alt="table of results" /></a></p>
|
256,138 |
<p>I need to generate four positive random values in the range [.1, .6] with (at most) two significant digits to the right of the decimal, and which sum to exactly 1. Here are three attempts that do not work.</p>
<pre><code>x = {.15, .35, .1, .4}; While[Total[x] != 1,
x = Table[Round[RandomReal[{.1, .6}], .010], 4]];
x = {.25, .25, .25, .25}; While[Total[x] == 1,
x = Table[Round[RandomReal[{.1, .6}], .010], 4]];
NestWhileList[Total[x],
x = Table[Round[RandomReal[{.1, .6}], .010], 4],
Plus @@ x == 1][[1]]
</code></pre>
|
ubpdqn
| 1,997 |
<p>You could use<a href="https://reference.wolfram.com/language/ref/DirichletDistribution.html" rel="nofollow noreferrer"><code>DirichletDistribution</code></a></p>
<p>For example:</p>
<pre><code>rv = Select[{##, 1 - Total[{##}]} & @@@
RandomVariate[DirichletDistribution[{1, 1, 1, 1}], 100000],
Min[#] > 0.1 && Max[#] < 0.6 &];
</code></pre>
<p>Illustrating the truncated region alluded to in other posts for first three components:</p>
<pre><code>Show[ListPointPlot3D[rv[[All, {1, 2, 3}]],
PlotRange -> Table[{0, 1}, 3]],
RegionPlot3D[x + y + z < 1, {x, 0, 1}, {y, 0, 1}, {z, 0, 1},
PlotStyle -> Opacity[0.1], Mesh -> None]]
Grid[{##, Total[{##}]} & @@@ rv[[1 ;; 10]],
Dividers -> {{{False}, True, False}, None}]
</code></pre>
<p><a href="https://i.stack.imgur.com/vCCQI.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vCCQI.jpg" alt="enter image description here" /></a></p>
|
4,221,530 |
<p>Assume terms <span class="math-container">$a_n$</span> and <span class="math-container">$a_{n+1}$</span> share some common factor <span class="math-container">$x$</span> so that <span class="math-container">$a_n = xm$</span> for some integer <span class="math-container">$m$</span> and <span class="math-container">$a_{n+1} = xk$</span> for some integer <span class="math-container">$k$</span>.</p>
<p>Since <span class="math-container">$a_{n-1}$</span> = <span class="math-container">$a_{n+1} - a_n = xk - xm = x(k-m)$</span>, <span class="math-container">$a_{n-1}$</span> is a multiple of <span class="math-container">$x$</span> as well. Then <span class="math-container">$a_{n-2}$</span> must also be a multiple of <span class="math-container">$x$</span> because <span class="math-container">$a_{n-2} = a_{n} - a_{n-1} = xm - (xk - xm) = x(2m-k)$</span>. Since each lower or upper term can be found by subtracting or adding one multiple of <span class="math-container">$x$</span> to another, we can say that every term in the sequence must then be a multiple of <span class="math-container">$x$</span>.</p>
<p>Therefore, if any two consecutive terms share a common factor, then all terms must share a common factor. However, if <span class="math-container">$a_1=1$</span> we can see the counterexample of <span class="math-container">$a_3 = 2, a_4=3$</span> where <span class="math-container">$2$</span> and <span class="math-container">$3$</span> are consecutive terms with no common factors. Since not all terms then share a common factor, no two consecutive terms can share a common factor.</p>
|
Lion Heart
| 809,481 |
<p><span class="math-container">$$\angle BAM =\angle AFT = 120^\circ-90^\circ=30^\circ $$</span></p>
<p><span class="math-container">$$\angle ABM = \angle AMB = \angle FAT = \angle FTA = \frac{180^\circ-30^\circ}{2}=75^\circ$$</span></p>
<p><span class="math-container">$$\angle MAT = 120^\circ- (75^\circ+30^\circ)=15^\circ$$</span></p>
<p><span class="math-container">$$\angle AMB =x+15^\circ=75^\circ$$</span></p>
<p><span class="math-container">$$x=60^\circ$$</span></p>
|
3,441,438 |
<p>I was wondering if anyone could get me started on solving the following:
<span class="math-container">$$(u')^2-u\cdot u''=0$$</span></p>
<p>I have tried letting <span class="math-container">$v=u'$</span>, but I don't seem to make progress with such a substitution.</p>
|
user0102
| 322,814 |
<p>Notice that, for a polynomial with real coefficients, if <span class="math-container">$z\in\mathbb{C}$</span> is a root, then <span class="math-container">$\overline{z}$</span> is also a root. At your case, <span class="math-container">$z = -2+i\sqrt{5}$</span> is known to be a root. Then divide the fourth degree polynomial by <span class="math-container">$(x-z)(x-\overline{z})$</span>, from whence you obtain a second degree equation. Can you take it from here? </p>
|
554,825 |
<p>Here's a question from my homework. First 2 questions I solved (but would appreciate any input you can give on my solution) and the last question I'm just completely stumped. It's quite complicated.</p>
<p>On the shelf there are 5 math books, 3 science fiction books and 2 thrillers (<strong>all of the books are different</strong>)</p>
<p>1) In how many different combinations can you organize the books on the shelf, without any limitation? **My answer: $10!$ since all the books are different, it's just organizing 10 books.</p>
<p>2) In how many different combinations can you organize the books so that books of the same kind are next to each other? **My answer: $3!*5!*3!*2!$ - first imagine all the math books are just 1 block, all the sci fi books are 1 block, and that the 2 thrillers are 1 block. I have $3!$ ways to organize these blocks on the shelf. The $5!*3!*2!$ is due to the order of the books inside the block.</p>
<p>3) How many different combinations are there to organize the books such that the sci fi books are together, and there is at least 1 book inbetween the thrillers?
**My answer: I don't know. It's too complex. I thought maybe simplying it by saying at least 1 book in between = combinations with 1 book in between + combinations with 2 books etc but even that doesn't make it any simpler...</p>
<p>Help? :)</p>
|
Sebastian
| 25,717 |
<p>The statement is correct if both $R$ and $S$ are local rings and if $f$ is a homomorphism of local rings, i.e. $f(\mathfrak{m}_{R}) \subset \mathfrak{m}_{S}$, where $\mathfrak{m}_{R}, \mathfrak{m}_{S}$ denotes the maximal ideal of $R$, $S$, respectively.</p>
<p>Proof: The set of non-units in $R$ is the maximal ideal $\mathfrak{m}_{R}$. Since $f(\mathfrak{m}_{R}) \subset \mathfrak{m}_{S}$, it hence follows that if $a$ is a non-unit then $f(a)$ is a non-unit. Let $b\in S$ be irreducible. Suppose that there exists an $x\in f^{-1}(b)$ which is reducible. Then there exist $y, z \in \mathfrak{m}_{R}$ so that $x = y \cdot z$, which yields a factorization of $b= f(x) = f(y) \cdot f(z)$, and since neither $f(y)$ nore $f(z)$ is a unit, this gives contradiction to the irreducibility of $b$.</p>
|
112,394 |
<p>Could someone clarify why the first of these <code>MatchQ</code> finds a match whereas the second does not? (I'm using version 10.0, in case that matters.)</p>
<pre><code>MatchQ[Hold[x + 2 y], Hold[x + 2 _]]
(*True*)
MatchQ[Hold[x + 2 y + 0], Hold[x + 2 _ + 0]]
(*False*)
</code></pre>
<p>EDIT: The conclusion below is that when encountering an <code>Orderless</code> function (here <code>Plus</code>), the pattern matcher does not test for all possible orderings of arguments but rather sorts the pattern's constant arguments and allows blanks <code>_</code> to appear in arbitrary places within that list. This reordering disregards <code>Hold</code>. On the other hand <code>Hold</code> does keep the order of the arguments of the expression fixed, so <code>Hold[x+2y+0]</code> is compared (with fixed argument order) to <code>Hold[0+x+2_]</code>, <code>Hold[0+2_+x]</code> and <code>Hold[2_+0+x]</code>, none of which match.</p>
|
M.R.
| 403 |
<p>You can put a <code>Verbatim</code> on the Plus:</p>
<pre><code>MatchQ[Hold[x + 2 y + 0], Hold[Verbatim[Plus][x, 2 _, 0]]] (* True *)
</code></pre>
<p>Another way:</p>
<pre><code>expr = Inactivate[x + 2 y + 1];
form = Inactivate[x + 2 _ + 1];
MatchQ[expr, IgnoringInactive@form] (* True *)
</code></pre>
|
4,014,756 |
<p>I was reading the book "Quantum Computing Since Democritus".</p>
<blockquote>
<p>"The set of ordinal numbers has the important property of being well
ordered,which means that every subset has a minimum element. This is
unlike the integers or the positive real numbers, where any element
has another that comes before it."</p>
</blockquote>
<p>Unlike integers? Let's consider a set <span class="math-container">$\{1,2,3\}$</span> This has a minimum element.</p>
<p>Do you get what does the author wants to say here?</p>
|
Evangelopoulos Phoevos
| 739,818 |
<p>Hint: show that <span class="math-container">$\prod C_i= \bigcap π_i^{-1}(C_i)$</span> where <span class="math-container">$π_i: \prod X_i \to X_i $</span> is the i-th projection</p>
|
2,133,870 |
<p>I'm having a little trouble solving this contour integral. I've found the singularities to be at 0, 2, and $1\pm i$. I'm not quite sure how to evaluate the integral when there are two poles on the contour instead of just one. Any help is appreciated.</p>
|
Jack D'Aurizio
| 44,121 |
<p>$$\text{PV}\int_{-\infty}^{+\infty}\frac{x}{(x-1)^4-1}\,dx = \text{PV}\int_{-\infty}^{+\infty}\frac{x+1}{x^4-1}\,dx = \text{PV}\int_{-\infty}^{+\infty}\frac{dx}{(x-1)(x^2+1)} $$
equals, by partial fraction decomposition,
$$ \frac{1}{2}\,\text{PV}\int_{-\infty}^{+\infty}\frac{dx}{x-1}-\frac{1}{2}\,\text{PV}\int_{-\infty}^{+\infty}\frac{x+1}{x^2+1}\,dx = \color{red}{-\frac{\pi}{2}}.$$</p>
|
1,791,849 |
<p>On the <a href="https://en.wikipedia.org/wiki/Pentagon" rel="nofollow">Wikipedia Page about Pentagons</a>, I noticed a statement in their work saying that $\sqrt{25+10\sqrt{5}}=5\tan(54^{\circ})$ and $\sqrt{5-2\sqrt{5}}=\tan(\frac {\pi}{5})$</p>
<p>My question is: How would you justify that? My goal is to simplify $\sqrt{25+10\sqrt{5}}$ and $\sqrt{5-2\sqrt{5}}$ without knowledge on the detested radical!</p>
|
lulu
| 252,071 |
<p>Thanks to @N.F.Taussig for pointing out some errors.</p>
<p>Suppose you have a regular pentagon, with vertices $\{A,B,C,D,E\}$ (labelled cyclically so $A$ is neighbor to $B$ and $E$). Let us take each side length to be $1$. Let $X$ be the length of a diagonal, say $AC$. Let's first compute $X$. </p>
<p><a href="https://i.stack.imgur.com/2t8Eb.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2t8Eb.jpg" alt="regular_pentagon"></a></p>
<p>To do it, let $P$ be the intersection of $AC$ and $BE$. Now we do some easy angle chasing: $\angle{BCA}=36=\angle{BAC}$, $\angle{CBP}=72=\angle{CPB}$. $\angle{ABP}=36$. Of course $\Delta BPC$ is isosceles (though not similar) and this implies that $PC$ has length $1$ We also see that $\Delta ABP$ is isosceles with angles $108-36-36$. Thus it is similar to $\Delta ACB$. Similarity then lets us compute $X$: $$\frac {X-1}1=\frac 1X\implies X^2-X-1=0\implies X=\frac 12(1+\sqrt5)$$</p>
<p>Now drop the perpendicular from $P$ to $AB$, let $Q$ denote its foot. Clearly $\Delta PAQ$ is a right triangle with angles $90-36-54$. We know one leg ($\frac 12$) and the hypotenuse ($\frac 12 (\sqrt 5 -1 )$). That suffices to solve the triangle and gives you the answer you seek.</p>
|
852,404 |
<p>How to prove that there exist an infinite number of prime $n$ for which $n^2=p+8$ for some prime $p$?</p>
<p>Verification of the form $n^2=p+8$ where $n$ and $p$ are some $p$.</p>
<p>$$\begin{array}{|c|c|} \hline
n & n^2 = 8 + p \\
\hline
11 & 121 = 8 + 113 \\
23 & 529 = 8 + 521 \\
31 & 961 = 8 + 953 \\
37 & 1369= 8 + 1361 \\
\hline
\end{array}
$$</p>
|
Will Jagy
| 10,400 |
<p>No proof is available. There are no proofs that any polynomial in a single integer variable, of degree at least two, such as your $n^2 - 8,$ take on an infinite number of prime values. In comparison, $n^2 - 8 m^2$ certainly does, but this is a function of two integer variables. </p>
<p>The best known is $n^2 + 1$ . People suspect there are an infinite number of primes of this form, but no proof. One of the most efficient such polynomials is $n^2 - n + 41,$ the value is prime for $1 \leq n \leq 40,$ then it seems to give relatively frequent primes thereafter as long as $n \neq 0,1 \pmod {41},$ but we cannot be sure about infinite sets of primes with this one either. See <a href="http://en.wikipedia.org/wiki/Formula_for_primes#Prime_formulas_and_polynomial_functions" rel="nofollow">http://en.wikipedia.org/wiki/Formula_for_primes#Prime_formulas_and_polynomial_functions</a> </p>
<p>Alright, needs emphasis; people have suspicions about one-variable polynomials and primes, <a href="http://en.wikipedia.org/wiki/Bunyakovsky_conjecture" rel="nofollow">http://en.wikipedia.org/wiki/Bunyakovsky_conjecture</a> but no proofs are yet available for <strong>any</strong> such problem. No proofs are expected for the foreseeable future. This is not the same as saying something is unprovable, I know of no conjectures in day-to-day number theory that have been shown to be unprovable. There is a tiny collection of statements in set theory, the one I remember being The Continuum Hypothesis, where there can be no proof because the statement is independent of the other axioms. Well, the guy got the Fields Medal for that, it does not happen every day. I suppose it is valid to say that The Parallel Postulate is independent of the other axioms of plane geometry and unprovable, as there is the hyperbolic/non-Euclidean plane. </p>
|
1,905,186 |
<blockquote>
<p>Let <span class="math-container">$R$</span> be a commutative Noetherian ring (with unity), and let <span class="math-container">$I$</span> be an ideal of <span class="math-container">$R$</span> such that <span class="math-container">$R/I \cong R$</span>. Then is it true that <span class="math-container">$I=(0)$</span> ?</p>
</blockquote>
<p>I know that a surjective ring endomorphism of a Noetherian ring is also injective, and since there is a natural surjection from <span class="math-container">$R$</span> onto <span class="math-container">$R/I$</span> we get a surjection from <span class="math-container">$R$</span> onto <span class="math-container">$R$</span>, but the problem is I can not determine the map explicitly and I am not sure about the statement. Please help. Thanks in advance.</p>
|
quid
| 85,306 |
<p>Assume that $R/I$ and $R$ are isomorphic. Let us denote the isomophism by $f:R/I \to R$. </p>
<p>Let $\pi:R \to R/I$ denote the usual map $x \mapsto x + I$.
This is a of course a surjective ring homomorphism. </p>
<p>The composition $f \circ \pi : R \to R$ is thus a surjective ring endomorphism (the composition of surjections is a surjection). </p>
<p>By the result quoted in the question $f \circ \pi$ is an isomorphism, in particular it is injective. It follows that $\pi$ is injective, otherwise the composition could not be injective. </p>
<p>The kernel of $\pi$ is thus $\{0\}$; it is also is $I$. Thus $I = \{0\}$</p>
|
1,905,186 |
<blockquote>
<p>Let <span class="math-container">$R$</span> be a commutative Noetherian ring (with unity), and let <span class="math-container">$I$</span> be an ideal of <span class="math-container">$R$</span> such that <span class="math-container">$R/I \cong R$</span>. Then is it true that <span class="math-container">$I=(0)$</span> ?</p>
</blockquote>
<p>I know that a surjective ring endomorphism of a Noetherian ring is also injective, and since there is a natural surjection from <span class="math-container">$R$</span> onto <span class="math-container">$R/I$</span> we get a surjection from <span class="math-container">$R$</span> onto <span class="math-container">$R$</span>, but the problem is I can not determine the map explicitly and I am not sure about the statement. Please help. Thanks in advance.</p>
|
Aragogh
| 665,469 |
<p>Assume <span class="math-container">$a$</span> is a proper ideal. Suppose they were isomorphic. Then <span class="math-container">$\varphi: A \to A/a$</span> is some arbitrary isomorphism, and correspondingly <span class="math-container">$\varphi(a) := I_{1} \subset A/a$</span> is an ideal (proper inclusion as <span class="math-container">$a \subset A$</span> is a proper inclusion). By the correspondence principle, <span class="math-container">$I_{1} \subset A/a$</span> pulls back to an ideal <span class="math-container">$a \subset I'_{1} \subset A$</span> which are all proper inclusions (as <span class="math-container">$\varphi(a)$</span> is nonzero in <span class="math-container">$A/a$</span>), where <span class="math-container">$I'_{1} = \pi^{-1}(I_{1})$</span> for <span class="math-container">$\pi: A \to A/a$</span> the canonical projection map. </p>
<p>The key step is that <span class="math-container">$\varphi$</span> induces an isomorphism <span class="math-container">$\overline{\varphi}: A/a \to A/I_{1}$</span>. But then <span class="math-container">$\overline{\varphi}(I_{1}):= I_{2}$</span> pulls back to an ideal <span class="math-container">$I'_{2} \subset A$</span> (proper inclusion using the same rationale as above) via <span class="math-container">$\pi_{1}: A \to A/I_{1}$</span> the canonical projection, such that <span class="math-container">$a \subset I'_{1} \subset I'_{2}$</span> in <span class="math-container">$A$</span> are all proper inclusions. Now iterate this process to yield an ascending chain which does not stabilize. Contradiction!</p>
|
496,178 |
<p>Let $t$ be a positive real number. Differentiate the function</p>
<blockquote>
<p>$$g(x)=t^x x^t.$$</p>
</blockquote>
<p>Your answer should be an expression in $x$ and $t$.</p>
<p>came up with the answer </p>
<blockquote>
<p>$$(x/t)+(t/x)\ln(t^x)(x^t)=\ln(t^x)+\ln(x^t)=x\ln t+t\ln x .$$</p>
</blockquote>
<p>and the derivative to that is $(x/t)+(t/x)$. Not sure if I've done it right.</p>
|
Brian M. Scott
| 12,042 |
<p>I’m not sure just what you were doing; it looks as if you might have been trying to use logarithmic differentiation and getting a bit confused in the process. In any case there’s no need for anything fancy.</p>
<p>Since $g(x)$ is a product, your first step should be to use the product rule:</p>
<p>$$g'(x)=t^x(x^t)'+\left(t^x\right)'x^t\;,$$</p>
<p>where the primes indicate derivatives with respect to $x$. Differentiating $x^t$ is just a matter of using the power rule: $(x^t)'=tx^{t-1}$. Differentiating the exponential is like differentiating $e^x$, provided that you remember to compensate for the fact that $t$ isn’t (ncessarily) $e$: $(t^x)'=t^x\ln t$. Thus,</p>
<p>$$\begin{align*}
g'(x)&=t^x(tx^{t-1})+(t^x\ln t)x^t\\
&=t^{x+1}x^{t-1}+t^xx^t\ln t\;;
\end{align*}$$</p>
<p>if you like, you can pull out some common factors and write</p>
<p>$$g'(x)=t^xx^{t-1}\left(t+x\ln t\right)\;.$$</p>
|
1,465,989 |
<p>I'm trying to find the column space of $\begin{bmatrix}a&0\\b&0\\c&0\end{bmatrix}$, which I think is $span\left(~\begin{bmatrix}a\\b\\c\end{bmatrix}~\begin{bmatrix}0\\0\\0\end{bmatrix}~\right)$. Since by definition a span needs to include the null vector, this is redundant. Would it also be correct to say that this is redundant because $\begin{bmatrix}0\\0\\0\end{bmatrix}$ is linearly dependent on $\begin{bmatrix}a\\b\\c\end{bmatrix}$ (can multiply it by a scalar of 0 to reach the null vector)? Or is that a misapplication of the concept of linear dependence?</p>
|
Martin Argerami
| 22,857 |
<p>It is exactly as you say: in any vector space, the null vector belongs to the span of any vector. </p>
|
2,533,186 |
<p>How would you find all pairs $(z,w)$ that satisfy $zw=-1$ </p>
<p>Im stuck on this problem and the best way I can think of would be to guess and check?</p>
|
fleablood
| 280,126 |
<p>Well, $z\ne 0$ and $w = \frac {-1}z$ will do it. The question is what format do you want? </p>
<p>If you want it is $z = a+bi$ and $w= c+di$ where $c$ and $d$ are expressed in terms of $a$ and $b$.</p>
<p>Then $\frac {-1}{a+bi} = \frac {-1(a-bi)}{(a+bi)(a-bi)} = \frac {-a+bi}{a^2 + b^2} = -\frac {a}{a^2 + b^2} + \frac {b}{a^2 + b^2} i$. </p>
<p>But $z = a+bi$ is not the only way to express complex numbers.</p>
<p>$w = \frac {-1}z$ is a <em>perfectly</em> acceptable answer.</p>
|
592,805 |
<p>How to prove $$\limsup\limits_{n \to \infty} \frac{1}{\sqrt n}B(n) = +\infty$$ using Blumenthal zero-one law, where $(B(t))_{t \geq 0}$ is a Brownian motion?</p>
|
Alex R.
| 22,064 |
<p>Here are some hints:</p>
<p>1) Use the fact that $aB_t$ has the same distribution as $B_{a^2t}$. This means that at any given time $n$, your quantity inside the supremum has distribution $B(1)$.</p>
<p>2) For any $a>0$, </p>
<p>$$\mathbb{P}(B(n)>a\sqrt{n} \mbox{ occurs infinitely often})\geq \limsup_{n\rightarrow\infty}\mathbb{P}(B(n)>a\sqrt{n})=P(B(1)>a)>0$$</p>
<p>3) You now want to show that $B(n)>a\sqrt{n}$ occuring infinitely often is a 0-1 event. Do you see how to invoke the Blumenthal 0-1 law to show this? As a hint, let $T$ be the hitting time of $a\sqrt{n}$ and notice that $T$ satisfies the strong Markov property...</p>
|
4,000,459 |
<p>How can we mathematically precisely argue that <span class="math-container">$$\lim_{n \to \infty} \left(1-\frac xn+\frac{x}{n^2}\right)^n = e^{-x}$$</span> holds?</p>
<p>So how can we bring</p>
<p><span class="math-container">$$1-\frac xn+\frac{x}{n^2} = 1- \frac{(n+1)x}{n^2} \approx 1 - \frac xn $$</span>
and
<span class="math-container">$$\lim_{n \to \infty} \left(1-\frac xn\right)^n=e^{-x}$$</span>
together?</p>
<p>Or can we use this form somehow:
<span class="math-container">$$\left(1-\frac xn+\frac{x}{n^2}\right)^n
= \left(\left( 1 - \frac{x}{\frac{n^2}{n+1}} \right)^{\frac{n^2}{n+1} }\right)^{\frac{n+1}{n}}$$</span></p>
|
Z Ahmed
| 671,540 |
<p>When <span class="math-container">$L=\lim_{x\to a} f(x)^{g(x)}\to 1^{\infty}$</span>, then
<span class="math-container">$L=\exp[\lim_{x\to a} g(x) [f(x)-1]$</span>
<span class="math-container">$$K=\lim_{n \to \infty} \left(1-\frac xn+\frac{x}{n^2}\right)^n $$</span>
<span class="math-container">$$K=\exp[ \lim_{x \to \infty} n [1-x/n+x^2/n^2-1]]=\lim_{x \to \infty} -x+x/n=e^{-x}$$</span></p>
|
4,000,459 |
<p>How can we mathematically precisely argue that <span class="math-container">$$\lim_{n \to \infty} \left(1-\frac xn+\frac{x}{n^2}\right)^n = e^{-x}$$</span> holds?</p>
<p>So how can we bring</p>
<p><span class="math-container">$$1-\frac xn+\frac{x}{n^2} = 1- \frac{(n+1)x}{n^2} \approx 1 - \frac xn $$</span>
and
<span class="math-container">$$\lim_{n \to \infty} \left(1-\frac xn\right)^n=e^{-x}$$</span>
together?</p>
<p>Or can we use this form somehow:
<span class="math-container">$$\left(1-\frac xn+\frac{x}{n^2}\right)^n
= \left(\left( 1 - \frac{x}{\frac{n^2}{n+1}} \right)^{\frac{n^2}{n+1} }\right)^{\frac{n+1}{n}}$$</span></p>
|
CHAMSI
| 758,100 |
<p>Notice that <span class="math-container">$ \left(\forall y\in\left(0,1\right)\right) $</span> : <span class="math-container">$$ y\leq-\ln{\left(1-y\right)}\leq\frac{y}{1-y} $$</span></p>
<p>That can be proven either using the main value theorem,, or studiing some functions.</p>
<p>Thus, setting <span class="math-container">$ y_{n}=\frac{x}{n}-\frac{x}{n^{2}} $</span>, where <span class="math-container">$ x\in\mathbb{R} $</span>, and <span class="math-container">$ n\in\mathbb{N}^{*} $</span> being bigger enough so that <span class="math-container">$ y_{n}\in\left(0,1\right) $</span>. We get : <span class="math-container">$$ \ \ \ \ \ \ \ \ \ \ \ \ \frac{x}{n}\left(1-\frac{1}{n}\right)\leq -\ln{\left(1-\frac{x}{n}+\frac{x}{n^{2}}\right)}\leq\frac{\frac{x}{n}\left(1-\frac{1}{n}\right)}{1-\frac{x}{n}+\frac{x}{n^{2}}}\\ \iff \frac{-x\left(1-\frac{1}{n}\right)}{1-\frac{x}{n}+\frac{x}{n^{2}}}\leq\ln{\left(1-\frac{x}{n}+\frac{x}{n^{2}}\right)^{n}}\leq -x\left(1-\frac{1}{n}\right) $$</span></p>
<p>Taking <span class="math-container">$ n $</span> to <span class="math-container">$ +\infty $</span>, we get that : <span class="math-container">$$ \ln{\left(1-\frac{x}{n}+\frac{x}{n^{2}}\right)^{n}}\underset{n\to +\infty}{\longrightarrow}-x $$</span></p>
<p>Which means : <span class="math-container">$$ \left(1-\frac{x}{n}+\frac{x}{n^{2}}\right)^{n}\underset{n\to +\infty}{\longrightarrow}\mathrm{e}^{-x} $$</span></p>
|
4,310,529 |
<p>I need a formula that will give me all points of random ellipse and circle intersection (ok, not fully random, the center of circle is laying on ellipse curve)</p>
<p>I need step by step solution (algorithm how to find it) if this is possible.</p>
|
David Quinn
| 187,299 |
<p>Referring to the standard equation of the ellipse, <span class="math-container">$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$</span>, choose the centre of the circle to correspond to the parameter value <span class="math-container">$\theta$</span>, so the equation of the circle with this centre and radius <span class="math-container">$r$</span> is <span class="math-container">$$(x-a\cos\theta)^2+(y-b\sin\theta)^2=r^2.$$</span></p>
<p>Let points of intersection of the circle with the ellipse have parameter values <span class="math-container">$\phi$</span>, so for a chosen <span class="math-container">$\theta, a, b$</span> and <span class="math-container">$r$</span> you can find values of <span class="math-container">$\phi$</span> by solving
<span class="math-container">$$(a\cos\phi-a\cos\theta)^2+(b\sin\phi-b\sin\theta)^2=r^2.$$</span></p>
|
46,076 |
<p>I'm developing a larger package which includes several subpackages. My problem is, that I can't introduce the symbols in the subpackages to the autocompletion by loading the main package, but by calling a subpackage.</p>
<p>Let's explain this with an example: I have a main package, called <code>main</code> which loads the subpackages <code>sub1</code>, <code>sub2</code> and <code>sub3</code>. The directory should look like</p>
<pre><code>main
├── Kernel
│ └── init.m
├── main.m
├── sub1.m
├── sub2.m
└── sub3.m
</code></pre>
<p>The package <code>main</code> loads all three subpackages in its <code>BeginPackage</code> statement, while the package <code>sub1</code> loads <code>sub2</code> and <code>sub3</code>. The respective <code>init.m</code> looks like:</p>
<pre><code>Get["main`sub3`"]
Get["main`sub2`"]
Get["main`sub1`"]
Get["main`main`"]
</code></pre>
<p>The big problem I have is, that by calling</p>
<pre><code>Needs["main`"]
</code></pre>
<p>I can see all symbols in <code>main</code>, but not those in the subpackages (though, they are usable).</p>
<p>By calling</p>
<pre><code>Needs["main`sub1`"]
</code></pre>
<p>I can see all symbols in the subpackages without any problems but can't load <code>main</code> anymore, because then some definitions are redone, which leads to error messages about protected symbols. The most interesting thing is, that the <code>$ContextPath</code> includes the subpackages in both cases.</p>
<p>Is there a nice possibility to get the symbols of the subpackages into the autocompletion?</p>
<hr>
<p><em>Update</em>: By prepending the path of a subpackage to the <code>$ContextPath</code>, I can load the corresponding symbols of that package into the autocompletion.</p>
<pre><code>$ContextPath = Prepend[$ContextPath, "main`sub1`"]
</code></pre>
<p>Unfortunately, this is not very nice and yields duplicates in the <code>$ContextPath</code>.</p>
<p>Is there a better, automatic way to achieve this?</p>
|
Mr.Wizard
| 121 |
<p>As <a href="https://mathematica.stackexchange.com/users/1356/oska">Öskå</a> notes you can <a href="http://reference.wolfram.com/mathematica/ref/Partition.html" rel="noreferrer"><code>Partition</code></a> your data and then <a href="http://reference.wolfram.com/mathematica/ref/Map.html" rel="noreferrer"><code>Map</code></a> <a href="http://reference.wolfram.com/mathematica/ref/Mean.html" rel="noreferrer"><code>Mean</code></a>:</p>
<pre><code>a = {q, r, s, t, u, v, w, x, y};
Mean /@ Partition[a, 3]
</code></pre>
<blockquote>
<pre><code>{1/3 (q + r + s), 1/3 (t + u + v), 1/3 (w + x + y)}
</code></pre>
</blockquote>
<p>However if performance is a concern I propose using <a href="http://reference.wolfram.com/mathematica/ref/Total.html" rel="noreferrer"><code>Total</code></a> or <a href="http://reference.wolfram.com/mathematica/ref/Dot.html" rel="noreferrer"><code>Dot</code></a>:</p>
<pre><code>blockAverage1[a_List, n_Integer] := a ~Partition~ n ~Total~ {2} / n
blockAverage2[a_List, n_Integer] := Partition[a, n].ConstantArray[1/n, n]
</code></pre>
<p>Timings:</p>
<pre><code>a = RandomReal[9, 5*^7]; (* big list *)
Mean /@ Partition[a, 20] // Timing // First
blockAverage1[a, 20] // Timing // First
blockAverage2[a, 20] // Timing // First
</code></pre>
<blockquote>
<pre><code>1.311
0.0654
0.0306
</code></pre>
</blockquote>
<p>If you want averages of overlapping blocks see also: </p>
<ul>
<li><p><a href="https://mathematica.stackexchange.com/q/28240/121">Taking averages at certain intervals</a></p></li>
<li><p><a href="https://mathematica.stackexchange.com/q/29644/121">Finding the midpoints of an ordered list of numbers</a></p></li>
</ul>
<p>Related: </p>
<ul>
<li><p><a href="https://mathematica.stackexchange.com/q/25054/121">Apply a Function Pairwise</a></p></li>
<li><p><a href="https://mathematica.stackexchange.com/q/31615/121">How to find first list element that differs from average of N previous elements by more than a given amount?</a></p></li>
<li><p><a href="https://mathematica.stackexchange.com/q/45926/121">How to partition three years of data into daily samples</a></p></li>
</ul>
|
3,851,650 |
<p>We want to prove that the number of sequences <span class="math-container">$(a_1,...,a_{2n})$</span> such that
<span class="math-container">$$
• \text{ every } a_i \text{ is equal to} ±1;\\• a_1 + ··· + a_{2n} = 0;\\• \text{ every partial sum satisfies } a_1 + ··· + a_i > −2
$$</span>
is a Catalan number.</p>
<p>I've been trying to form a bijection between this and ballot sequences of size 2(n+1) by showing that you can remove any one +1 and any one -1 to yield our new sequences, but I am not sure if this is the best method. Any help would be super helpful!</p>
|
Dr. Mathva
| 588,272 |
<p><em>Hint.</em> Catalan numbers <span class="math-container">$C_n$</span> can be used to count the number of "monotonic lattice paths along the edges of a grid with <span class="math-container">$n \times n$</span> square cells, which do not pass above the diagonal. A monotonic path is one which starts in the lower left corner, finishes in the upper right corner, and consists entirely of edges pointing rightwards or upwards" (quoted from <a href="https://en.wikipedia.org/wiki/Catalan_number#Applications_in_combinatorics" rel="nofollow noreferrer">Wikipedia</a>). Here you have the case <span class="math-container">$n=4$</span>.</p>
<p><a href="https://i.stack.imgur.com/3mc5C.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3mc5C.png" alt="enter image description here" /></a></p>
<p>In order to describe the movements, let <span class="math-container">$a_i$</span> be the <span class="math-container">$i$</span>-th movement and choose <span class="math-container">$a_i=1$</span> if you move to the right and <span class="math-container">$a_i=-1$</span> if you move upwards. Is it clear why <span class="math-container">$\sum a_i=0$</span>? And why every <span class="math-container">$a_i$</span> is equal to <span class="math-container">$\pm 1$</span>? Why do they satisfy <span class="math-container">$a_1+a_2+\ldots+ a_k<-2$</span> (look at the diagonal)?</p>
|
3,753,143 |
<p>We spend a lot of time learning different theories (for instance, theory of differential forms, sobolev spaces, homology groups, distributions). Although (at least most parts of) these theories are very natural and understandable when we read them from books, they are very difficult to create at the first place: it could take tens of years of effort of a large number of excellent mathematicians.</p>
<p>After learning those theories, we do exercises or solve problems, but most of the time, we are just using the tools stated in the book. Even the chance that we come up with a "new" definition ourselves is rare. (By "new", I mean "have not learnt", even if someone else have created it before.) <strong>So here is my question:</strong></p>
<p><strong>What are some problems which prompts the creation of a new theory?</strong></p>
<p><strong>EDIT:</strong> Just to clarify, I am looking for some problems which give everybody a chance to experience the process of creating new mathematics; so the problem need not be as difficult as Riemann conjecture.</p>
<p>By "new theory", I just mean something that help us formulate the problem in a different way. For example, this <a href="https://www.youtube.com/watch?v=wTJI_WuZSwE" rel="nofollow noreferrer">video on a chess board puzzle</a> has the idea of creating new theories, because <strong>unlike</strong> other less interesting puzzles about chess board which can be solved by just carefully counting the squares, this video mentions a new way of looking at the problem, namely the vertices of a hypercube.</p>
<p>I have also seen other similar puzzles like this. Apparently, almost all of them are on discrete mathematics, so <strong>it would be really interesting if anyone could provide such a "theory creating" problem in other areas of mathematics (e.g. analysis).</strong></p>
<p>Of course, not all theories are created to tackle specific problems, so other ways of experiencing inventing new maths could also be suggested.</p>
|
Favst
| 742,787 |
<p><a href="https://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem" rel="nofollow noreferrer">Fermat's last theorem</a> is certainly an example of a problem that can be stated simply but led to magnificent efforts over hundreds of years and the development of much machinery before it was finally resolved. Wikipedia says</p>
<blockquote>
<p>The unsolved problem stimulated the development of algebraic number theory in the 19th century and the proof of the modularity theorem in the 20th century.</p>
</blockquote>
<p>More generally, I believe you are referring to "acorns" from the following quote of Erdős:</p>
<blockquote>
<p>A well chosen problem can isolate an essential difficulty in a particular area, serving as a benchmark against which progress in this area can be measured. An innocent looking problem often gives no hint as to its true nature. It might be like a 'marshmallow,' serving as a tasty tidbit supplying a few moments of fleeting enjoyment. Or it might be like an 'acorn,' requiring deep and subtle new insights from which a mighty oak can develop...</p>
</blockquote>
<p>It says at <a href="http://www.math.ucsd.edu/%7Eerdosproblems/" rel="nofollow noreferrer">this link</a></p>
<blockquote>
<p>Throughout his career, work on his proposed problems in a variety of areas of mathematics consistently led to advances and discoveries. Much of Erdős' legacy stems from his ability to capture the essence of a deep mathematical issue in a seemingly simple problem.</p>
</blockquote>
<p>So I'm sure that if you look up the problems posed by Erdős throughout his life, you'll find plenty of acorns.</p>
<p>EDIT: I just read more about Erdős in the essay, The Two Cultures of Mathematics, by Timothy Gowers:</p>
<blockquote>
<p>many people who have solved an Erdős problem... will testify that, as they thought harder and harder about it, they have been led in unexpectedly fruitful directions and come to realize that the problem was more than the amusing curiosity that it might at first have seemed.</p>
</blockquote>
|
2,316,514 |
<p>I want to calculate $\lim_{x \to 1}\frac{\sqrt{|x^2 - x|}}{x^2 - 1}$ . I tried to compute limit when $x \to 1^{+}$ and $x \to 1^{-}$ but didn't get any result . </p>
<p>Please help .</p>
<p>Note : I think it doesn't have limit but I can't prove it .</p>
|
Nominal Animal
| 318,422 |
<p>The trick is that instead of using $t$ as a parameter along one axis, you use it as a free parameter, with $t = 0$ at the beginning of the curve, and $t = 1$ at the end of the curve, with $0 \le t \le 1$ specifying the points on the curve.</p>
<p>All cubic curves of a single parameter are of form
$$\begin{cases}
x(t) = X_0 + X_1 t + X_2 t^2 + X_3 t^3 \\
y(t) = Y_0 + Y_1 t + Y_2 t^2 + Y_3 t^3 \\
z(t) = Z_0 + Z_1 t + Z_2 t^2 + Z_3 t^3 \end{cases} \tag{1}\label{1}$$
The tangent of the curve is
$$\begin{cases}
\frac{d x(t)}{d t} = dx(t) = X_1 + 2 X_2 t + 3 X_3 t^2 \\
\frac{d y(t)}{d t} = dy(t) = Y_1 + 2 Y_2 t + 3 Y_3 t^2 \\
\frac{d z(t)}{d t} = dz(t) = Z_1 + 2 Z_2 t + 3 Z_3 t^2 \end{cases} \tag{2}\label{2}$$</p>
<p>Let's say you need a cubic interpolating curve $\vec{p}(t)$, with
$$\begin{cases}
\vec{p}(0) = \vec{p}_0 = ( x_0 , y_0 , z_0 ) \\
\vec{p}(1) = \vec{p}_1 = ( x_1 , y_1 , z_1 ) \\
\left.\frac{d \vec{p}(t)}{d t}\right\rvert_{t=0} = \vec{d}_0 = ( dx_0 , dy_0 , dz_0 ) \\
\left.\frac{d \vec{p}(t)}{d t}\right\rvert_{t=1} = \vec{d}_1 = ( dx_1 , dy_1 , dz_1 ) \end{cases} \tag{3}\label{3}$$
i.e. the curve starts from $\vec{p}_0$ tangent to $\vec{d}_0$, and ends at $\vec{p}_1$ tangent to $\vec{d}_1$.</p>
<p>If we combine $\eqref{1}$, $\eqref{2}$, and $\eqref{3}$, and solve for the curve coefficients $X_0$, $X_1$, ..., $Z_2$, $Z_3$, we get
$$\begin{cases}
X_0 = x_0 \\
X_1 = dx_0 \\
X_2 = 3 ( x_1 - x_0 ) - ( dx_1 + 2 dx_0 ) \\
X_3 = 2 ( x_0 - x_1 ) + dx_1 + dx_0 \\
Y_0 = y_0 \\
Y_1 = dy_0 \\
Y_2 = 3 ( y_1 - y_0 ) - ( dy_1 + 2 dy_0 ) \\
Y_3 = 2 ( y_0 - y_1 ) + dy_1 + dy_0 \\
Z_0 = z_0 \\
Z_1 = dz_0 \\
Z_2 = 3 ( z_1 - z_0 ) - ( dz_1 + 2 dz_0 ) \\
Z_3 = 2 ( z_0 - z_1 ) + dz_1 + dz_0 \end{cases} \tag{4}\label{4}$$</p>
<hr>
<p>Computer graphics, and file formats like SVG and PostScript and PDF, use <a href="https://en.wikipedia.org/wiki/B%C3%A9zier_curve" rel="noreferrer">cubic Bézier curves</a> rather than the form specified in equation $\eqref{1}$ above. (Bézier curves trivially extend to any number of dimensions, although the two-dimensional ones are most common.)</p>
<p>Cubic Bézier curves are defined as
$$\vec{p}(t) = (1 - t)^3 \vec{c}_0 + 3 t (1 - t)^2 \vec{c}_1 + 3 t^2 (1 - t) \vec{c}_2 + t^3 \vec{c}_3 \tag{5}\label{5}$$
By defining $\vec{p}(t) = \left ( x(t) , y(t) , z(t) \right )$ and
$$\begin{cases} \vec{c}_0 = ( x_0 , y_0 , z_0 ) \\
\vec{c}_1 = \left ( x_0 + \frac{dx_0}{3} , y_0 + \frac{dy_0}{3} , z_0 + \frac{dz_0}{3} \right ) \\
\vec{c}_2 = \left ( x_1 - \frac{dx_1}{3} , y_1 - \frac{dy_1}{3} , z_1 - \frac{dz_1}{3} \right ) \\
\vec{c}_3 = ( x_1 , y_1 , z_1 ) \end{cases}$$
we have a Bézier curve that fulfills the requirements in $\eqref{3}$, and is exactly the same as the curve specified by $\eqref{1}$ and $\eqref{4}$.</p>
|
4,563,725 |
<p>Is it possible for two non real complex numbers a and b that are squares of each other? (<span class="math-container">$a^2=b$</span> and <span class="math-container">$b^2=a$</span>)?</p>
<p>My answer is not possible because for <span class="math-container">$a^2$</span> to be equal to <span class="math-container">$b$</span> means that the argument of <span class="math-container">$b$</span> is twice of arg(a) and for <span class="math-container">$b^2$</span> to be equal to <span class="math-container">$a$</span> means that arg(a) = 2.arg(b) but the answer is it is possible.</p>
<p>How is it possible when arg(b) = 2.arg(a) and arg(a) = 2.arg(b) contradict each other?</p>
|
Torsten Schoeneberg
| 96,384 |
<p>As user Henry says in a comment, <span class="math-container">$a^2=b$</span> and <span class="math-container">$b^2=a$</span> imply <span class="math-container">$a^4=a$</span>, i.e. <span class="math-container">$a$</span> is a zero of the polynomial:</p>
<p><span class="math-container">$$p(x)=x^4-x$$</span></p>
<p>And conversely, every zero of that polynomial will give a solution for <span class="math-container">$a$</span>, setting <span class="math-container">$b:=a^2$</span>.</p>
<p>Now <span class="math-container">$p(x)$</span> factors in <span class="math-container">$\mathbb Z[x]$</span> as</p>
<p><span class="math-container">$$p(x) = x \cdot (x-1) \cdot (x^2+x+1)$$</span></p>
<p>Which gives the two real solutions <span class="math-container">$(a,b)=(0,0)$</span> and <span class="math-container">$(a,b)=(1,1)$</span>. The quadratic term does not factor in <span class="math-container">$\mathbb R$</span>, but it does e.g. in <span class="math-container">$\mathbb C$</span>, as</p>
<p><span class="math-container">$$x^2+x+1 = (x+\frac12 -\frac{i}{2}\sqrt{3})(x+\frac12 +\frac{i}{2}\sqrt{3}) = (x-e^{i\frac{\pi}{3}})(x-e^{i\frac{2\pi}{3}})$$</span></p>
<p>The two zeros there are the two non-trivial cube roots of unity, often called <span class="math-container">$\zeta_3$</span> and <span class="math-container">$\bar \zeta_3$</span> (which is <span class="math-container">$ =\zeta_3^2$</span>). They give the non-real solutions</p>
<p><span class="math-container">$$(a,b) = (\zeta_3, \bar \zeta_3) \text{ and } (a,b) = (\bar \zeta_3, \zeta_3).$$</span></p>
|
4,563,725 |
<p>Is it possible for two non real complex numbers a and b that are squares of each other? (<span class="math-container">$a^2=b$</span> and <span class="math-container">$b^2=a$</span>)?</p>
<p>My answer is not possible because for <span class="math-container">$a^2$</span> to be equal to <span class="math-container">$b$</span> means that the argument of <span class="math-container">$b$</span> is twice of arg(a) and for <span class="math-container">$b^2$</span> to be equal to <span class="math-container">$a$</span> means that arg(a) = 2.arg(b) but the answer is it is possible.</p>
<p>How is it possible when arg(b) = 2.arg(a) and arg(a) = 2.arg(b) contradict each other?</p>
|
Piquito
| 219,998 |
<p>There are the number of real solutions of the system
<span class="math-container">$$a^2-b^2=(a^2-b^2)^2-4a^2b^2\\2ab=4ab(a^2-b^2)\tag1$$</span> which come from
the equalities <span class="math-container">$$(a+bi)^2=(a^2-b^2)+2abi\\((a^2-b^2)+2abi)^2=(a^2-b^2)^2-4a^2b^2+4ab(a^2-b^2)i$$</span>
The solution of system <span class="math-container">$(1)$</span> is easy and leaves to the only solutions, the two non real roots of unity.</p>
|
3,267,216 |
<p>An urn contains equal number of green and red balls. Suppose you are playing the following game. You draw one ball at random from the urn and note its colour. The ball is then placed back in the urn, and the selection process is repeated. Each time a green ball is picked you get 1 Rupee. The first time you pick a red ball, you pay 1 Rupee and the game ends. Your expected income from this game is..</p>
<p>The answer given is 0 but shouldn't it be positive as the as the income rises if you consecutively draw the green ball? Please help me clarify this doubt</p>
|
Rohit Singh
| 683,232 |
<p>Apply componendo &dividendo in ratio of AM/GM and then u will get root A +root B ka whole square on top and root a -root B ka whole square in bottom,then u can easily get ratio.i am new so I am not able to upload image to show u</p>
|
369,435 |
<p>My task is as in the topic, I've given function $$f(x)=\frac{1}{1+x+x^2+x^3}$$
My solution is following (when $|x|<1$):$$\frac{1}{1+x+x^2+x^3}=\frac{1}{(x+1)+(x^2+1)}=\frac{1}{1-(-x)}\cdot\frac{1}{1-(-x^2)}=$$$$=\sum_{k=0}^{\infty}(-x)^k\cdot \sum_{k=0}^{\infty}(-x^2)^k$$ Now I try to calculate it the following way:</p>
<p>\begin{align}
& {}\qquad \sum_{k=0}^{\infty}(-x)^k\cdot \sum_{k=0}^{\infty}(-x^2)^k \\[8pt]
& =(-x+x^2-x^3+x^4-x^5+x^6-x^7+x^8-x^9+\cdots)\cdot(-x^2+x^4-x^6+x^8-x^{10}+\cdots) \\[8pt]
& =x^3-x^4+0 \cdot x^5+0 \cdot x^6 +x^7-x^8+0 \cdot x^9 +0 \cdot x^{10} +x^{11}+\cdots
\end{align}</p>
<p>And now I conclude that it is equal to $\sum_{k=0}^{\infty}(x^{3+4 \cdot k}-x^{4+4 \cdot k})$ ($|x|<1$)
Is it correct? Are there any faster ways to solve that types of tasks? Any hints will be appreciated, thanks in advance.</p>
|
Clayton
| 43,239 |
<p><strong>Hint:</strong> Another method to approach the problem (more by brute force than Andre's clever technique) is $$f(x)=\frac{1}{1+x+x^2+x^3}=\frac{1}{(x+1)(x^2+1)}.$$ Now use partial fraction decomposition and geometric series.</p>
|
37,380 |
<p>I do have noisy data and want to smooth them by a Savitzky-Golay filter because I want to keep the magnitude of the signal. </p>
<p>a) Is there a ready-to-use Filter available for that? </p>
<p>b) what are appropriate values for m (the half width) and for the coefficients for 3000-4000 data points?</p>
|
J. M.'s persistent exhaustion
| 50 |
<p>I'm just posting this to record for posterity <a href="http://chat.stackexchange.com/transcript/message/4829800#4829800">something I posted in the chatroom</a> a not-so-long time ago. As I noted there, the following routine will only do smoothing; <del>I had a more general routine for generating the differentiation coefficients, but I still have not been able to find it</del>. (For the more general, but less compact version, see below.) As with Virgil's method (the one in Alexey's answer), this is based on <a href="http://pubs.acs.org/doi/abs/10.1021/ac00205a007" rel="noreferrer">Gorry's procedure</a> (though I have traced the spirit of the algorithm going as far back as <a href="http://books.google.com/books?id=hqucruPBheQC&pg=PA357" rel="noreferrer">Hildebrand's book</a>):</p>
<pre><code>GramP[k_Integer, m_Integer, t_Integer] :=
(-1)^k HypergeometricPFQ[{-k, k + 1, -t - m}, {1, -2 m}, 1]
SavitzkyGolay[n_Integer, m_Integer, t_Integer] :=
Table[Sum[(Binomial[2 m, k]/Binomial[2 m + k + 1, k + 1])
GramP[k, m, i] GramP[k, m, t] (1 + k/(k + 1)), {k, 0, n},
Method -> "Procedural"], {i, -m, m}]
SavitzkyGolay[n_Integer, m_Integer] := Table[SavitzkyGolay[n, m, t], {t, -m, m}]
</code></pre>
<p>Usage is pretty straightforward: <code>n</code> is the order of the polynomial smoothing; <code>2 m + 1</code> is the window size, and <code>t</code> tells how much to shift the window.</p>
<hr>
<p><strong>Added 12/17/2015</strong></p>
<p>Here is a faster routine for evaluating the Gram polynomial, using some undocumented functionality:</p>
<pre><code>GramP[k_Integer, m_Integer, t_Integer] :=
(-1)^k Internal`DCHypergeometricPFQ[k, {-k, k + 1, -m - t}, {1, -2 m}, 1]
</code></pre>
<hr>
<p>I managed to finally recover the general SG routine I once wrote through the kind assistance of a friend. To share my joy, I now release this to you:</p>
<pre><code>Options[SavitzkyGolay] = {Derivative -> 0, WorkingPrecision -> Infinity};
SavitzkyGolay[n_Integer?Positive, m_Integer?Positive, t_Integer,
OptionsPattern[]] /; 1 < n < 2 m + 1 && -m <= t <= m :=
Module[{o = OptionValue[Derivative], c, s, h, p, q, u, v, w},
u = UnitVector[o + 1, 1]; v = ConstantArray[0, o + 1];
c = 1/(2 m + 1); s = Join[{Boole[o == 0] c},
Table[h = 0;
{p, q} = {2 (2 k - 1), (k - 1) (2 m + k)}/(k (2 m - k + 1));
Do[w = u[[j]]; (* evaluate Gram polynomial and derivatives *)
u[[j]] = p (t w + (j - 1) h) - q v[[j]];
v[[j]] = h = w,
{j, Min[k, o] + 1}];
c *= (2 m - k + 1) (1 + 1/k)/(2 m + k + 1);
c (1 + k/(k + 1)) u[[o + 1]],
{k, n}]];
Table[h = s[[n]] + 2 (2 n - 1) (p = s[[n + 1]]) j/(n (2 m - n + 1));
Do[q = p; p = h; (* Clenshaw's recurrence *)
h = s[[k]] + 2 (2 k - 1) p j/(k (2 m - k + 1)) -
k (2 m + k + 1) q/((k + 1) (2 m - k)),
{k, n - 1, 1, -1}];
N[h, OptionValue[WorkingPrecision]], {j, -m, m}] //
Developer`ToPackedArray];
SavitzkyGolay[n_Integer?Positive, m_Integer?Positive, opts___?OptionQ] /;
1 < n < 2 m + 1 :=
Developer`ToPackedArray[Table[SavitzkyGolay[n, m, t, opts], {t, -m, m}]]
</code></pre>
<p>As advertised, it uses no matrices, and instead uses the recurrence relation of the Gram polynomial. If need be, the guts of the routine can be embedded within a <code>Compile[]</code>.</p>
<hr>
<p><strong>Added 12/17/2015</strong></p>
<p>Altho <code>SavitzkyGolayMatrix[]</code> is now built-in in version 10, it is only limited to producing the "central" coefficients, as opposed to the routine <code>SavitzkyGolay[]</code> in this answer that can also generate coefficients for the left and right ends.</p>
<pre><code>SavitzkyGolayMatrix[{2}, 3, 1, WorkingPrecision -> ∞]
{1/12, -2/3, 0, 2/3, -1/12}
SavitzkyGolay[3, 2, 0 (* central *), Derivative -> 1]
{1/12, -2/3, 0, 2/3, -1/12}
</code></pre>
<p>In general, the result of <code>SavitzkyGolayMatrix[]</code> is built from appropriate outer products of coefficient lists.</p>
<pre><code>SavitzkyGolayMatrix[{3, 4}, {2, 3}, WorkingPrecision -> ∞] ===
Outer[Times, SavitzkyGolay[2, 3, 0], SavitzkyGolay[3, 4, 0]]
True
</code></pre>
|
2,523,570 |
<p>$15x^2-4x-4$,
I factored it out to this:
$$5x(3x-2)+2(3x+2).$$
But I don’t know what to do next since the twos in the brackets have opposite signs, or is it still possible to factor them out?</p>
|
Stella Biderman
| 123,230 |
<p>I think you made an arithmetic mistake. I got $$15x^2-4x-4=(5x+2)(3x-2)=5x(x-2)+2(3x-2).$$</p>
|
227,296 |
<p>I happened to ponder about the differentiation of the following function:
$$f(x)=x^{2x^{3x^{4x^{5x^{6x^{7x^{.{^{.^{.}}}}}}}}}}$$
Now, while I do know how to manipulate power towers to a certain extent, and know the general formula to differentiate $g(x)$ wrt $x$, where $$g(x)=f(x)^{f(x)^{f(x)^{f(x)^{f(x)^{f(x)^{f(x)^{.{^{.^{.}}}}}}}}}}$$
I'm still unable to figure out as to how I can adequately manipulate the function to differentiate it within its domain of convergence, if it exists.</p>
<p><strong>Sub-Q: What is the domain over which the function converges? Does it have only a finite domain (meaning there's no necessity to discuss its derivative)?</strong></p>
<hr>
<p>General formula (making domanial assumptions for f(x) of course): $$g'(x)=\frac{g^2(x)f'(x)}{f(x)\left[1-g(x)\ln(f(x))\right]}$$</p>
<hr>
<p><strong>Note</strong>: This has been posted in stackexchange; a look at the (incomplete) answers and their respective comments' sections will give you an insight into what has been looked into so far - [<a href="https://math.stackexchange.com/questions/1592377/what-is-the-derivative-of-fx-x2x3x4x5x6x7x">https://math.stackexchange.com/questions/1592377/what-is-the-derivative-of-fx-x2x3x4x5x6x7x</a> ]</p>
|
juan
| 7,402 |
<p>This is not exactly an answer. I try to give a meaning to $f(x)$.
Not as a limit. I will give a formal
solution as a power series (possibly not converging). </p>
<p>Consider a function of two parameters
$$u(t,x)=(t+1)x^{(t+2)x^{(t+3)x^{(t+4)x^{(t+5)x^{(t+6)x^{(t+7)x^{.{^{.^{.}}}}}}}}}}$$
so that $f(x)=u(0,x)$. Now whatever the meaning we assign to $f(x)$ it appear that
$u(t,x)$ satisfies the functional equation
$$u(t,x)=(t+1) x^{u(t+1,x)}.$$
So that
$$\log u(t,x)=\log(t+1)+u(t+1,x)\log x.$$
It is natural to put $u(t,1)=t+1$. So we try an analytic solution
$$u(t,x)=(t+1)\Bigl(1+\sum_{n=1}^\infty a_n(t) (x-1)^n\Bigr)$$
Then
$$\log\Bigl(1+\sum_{n=1}^\infty a_n(t) (x-1)^n\Bigr)=
(t+2)\Bigl(1+\sum_{n=1}^\infty a_n(t+1) (x-1)^n\Bigr)\log x$$
Expanding
$$a_1(t)(x-1)+\Bigl(a_2(t)-\frac{1}{2}a_1(t)^2\Bigr)(x-1)^2+
\Bigl(\frac13a_1(t)^3-a_1(t)a_2(t)+a_3(t))(x-1)^3+\cdots =$$
$$=
(t+2)\Bigl(1+\sum_{n=1}^\infty a_n(t+1) (x-1)^n\Bigr)\Bigl((x-1)-\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}+\cdots\Bigr)$$
Giving equations
$$a_1(t)=t+2$$
$$a_2(t)-\frac{1}{2}a_1(t)^2=-\frac{t+2}{2}+(t+2)a_1(t+1)$$
So
$$a_2(t)=\frac12(t+2)^2-\frac{t+2}{2}+(t+2)(t+3)=(t+2)\Bigl(\frac{t+2}{2}-\frac12+t+3
\Bigr)=(t+2)\frac{3t+7}{2}. $$
It is easy to show that the equation resulting from equating the coefficients of
$(x-1)^n$ determine $a_n(t)$ in a unique way, as a polynomial of degree $n$. </p>
<p>With Mathematica we obtain
$$\eqalign{
a_1(t)&=\frac{1}{1!}(2+t)\cr
a_2(t)&=\frac{1}{2!}(14+13t+3t^2)\cr
a_3(t)&=\frac{1}{3!}(234+287t+117t^2+16t^3)\cr
a_4(t)&=\frac{1}{4!}(6792+10014 t+5505 t^2+1348 t^3+125 t^4)\cr
a_5(t)&=\frac{1}{5!}(301980+507614 t+338960 t^2+113200 t^3+19030 t^4+1296 t^5)\cr
a_6(t)&=\frac{1}{6!}(18996384+35272036 t+27073062 t^2+11070440 t^3+2559125 t^4+318834 t^5+16807 t^6)\cr
}$$
Therefore the formal solution is the power series
$$f(x)=1+\frac{2}{1!}(x-1)+\frac{14}{2!}(x-1)^2+\frac{234}{3!}(x-1)^3
+\frac{6792}{4!}(x-1)^4+\frac{301980}{5!}(x-1)^5+\frac{18996384}{6!}(x-1)^6+\cdots
$$
The sequence of coefficients do not appear in the OEIS.
The coefficient of $t^n$ in $a_n(t)$ appear to be $(n+1)^{n-1}$. </p>
<p>I have computed the coefficients of the power series for $f(x)$ for $n\le30$.
It appear that $\log a_n(t)\le \frac34 n^{3/2}$, (with approximate equality) if this trend continues, the series
will be divergent except for $x=1$. But Euler summed the
divergent series $\sum n! x^n$.</p>
|
3,960,189 |
<p>An auto insurance company is implementing a new bonus system. In each month, if a policyholder does not have an accident, he or she will receive a $5 cashback bonus from the insurer. Among the 1000 policyholders, 400 are classified as low-risk drivers and 600 are classified as high-risk drivers. In each month, the probability of zero accidents for high-risk drivers is 0.8 and that for low-risk drivers is 0.9.</p>
<p>Calculate the expected bonus payment from the insurer to the 1000 policyholders in one year.</p>
<p><strong>The question is what's wrong with my (following) solution?</strong></p>
<p>Probability of zero accidents <span class="math-container">$= (0.8\times 0.6) + (0.90 \times 0.4) = 0.84$</span></p>
<p>Then, the expected payment is given by <span class="math-container">$$\sum_{x=1}^{12} (0.84^x) \times 1000 \times 5x$$</span></p>
|
Alex
| 38,873 |
<p>OK here's the full solution:</p>
<p>Each risky customer is Binomial(<span class="math-container">$0.8$</span>), so the mean number of accident-free months is <span class="math-container">$\mathbf{E}X=np = 9.6$</span>. Bonus is a function of this rv, <span class="math-container">$^Y=cX$</span>, so <span class="math-container">$\mathbf{E}Y=9.6 \times 5 = 48$</span>. Expectation is linear, so even if <span class="math-container">$Y$</span> are not independent, expected bonus for all risky customers, <span class="math-container">$S_1=\sum_{k=1}^{600}Y_k$</span> is:
<span class="math-container">$$
\mathbf{E}S_1 = \sum_{k=1}^{400}\mathbf{E}Y_k = 600 \times 48=28800
$$</span>
Now, repeat this for non-risky customers,
<span class="math-container">$$
\mathbf{E}S_2 = 400 \times 0.9 \times 5 \times 12 = 21600
$$</span>
In total,
<span class="math-container">$$
\mathbf{E}S = 28800+21600 = 50400
$$</span></p>
|
2,078,142 |
<p>Let $X$ be locally compact metric space which is $\sigma$-compact also. I want to show that $X=\bigcup\limits_{n=1}^{\infty}K_n$, where $K_n$'s are compact subsets of $X$ satisfying $K_n\subset K_{n+1}^{0}$ for all $n\in \mathbb N$. </p>
<p>I know that since $X$ is $\sigma-$compact, therefore there exists a sequence of compact subsets $(C_n)$ of $X$ such that $X=\bigcup\limits_{n=1}^{\infty}C_n$. I am not getting any idea how to construct $K_n$'s. Please help!</p>
|
PatrickR
| 52,912 |
<p>The desired property is called the existence of an <a href="https://en.wikipedia.org/wiki/Exhaustion_by_compact_sets" rel="nofollow noreferrer">exhaustion by compact sets</a> for the space <span class="math-container">$X$</span>. (see Wikipedia)</p>
<p>The answer of @bof holds in the context of locally compact Hausdorff spaces (since the OP mentioned metric spaces). But the result is more generally true for <a href="https://topology.pi-base.org/properties/P000023" rel="nofollow noreferrer">weakly locally compact spaces</a>, that is, <a href="https://en.wikipedia.org/wiki/Locally_compact_space" rel="nofollow noreferrer">locally compact spaces</a> in the weakest sense in that each point has a compact neighborhood.</p>
<blockquote>
<p><strong>Proposition:</strong> The following are equivalent for a topological space <span class="math-container">$X$</span>:</p>
<ol>
<li><span class="math-container">$X$</span> is exhaustible by compact sets.</li>
<li><span class="math-container">$X$</span> is <span class="math-container">$\sigma$</span>-compact and weakly locally compact.</li>
<li><span class="math-container">$X$</span> is <a href="https://en.wikipedia.org/wiki/Lindel%C3%B6f_space" rel="nofollow noreferrer">Lindelöf</a> and weakly locally compact.</li>
</ol>
</blockquote>
<p>We need a preliminary</p>
<blockquote>
<p><strong>Lemma</strong>: In a weakly locally compact space <span class="math-container">$X$</span>, for every compact subset <span class="math-container">$K\subseteq X$</span> there is a compact subset of <span class="math-container">$X$</span> containing <span class="math-container">$K$</span> in its interior.</p>
</blockquote>
<p>To see this, every point <span class="math-container">$x\in K$</span> has a compact nbhd. Since <span class="math-container">$K$</span> is covered by the interiors of all these nbhds, it is covered by a finite number of these interiors, and the union of those finitely many compact nbhds is compact and contains <span class="math-container">$K$</span> in its interior.</p>
<p>Now the proof of the proposition.</p>
<p>(1) implies (2): Suppose <span class="math-container">$X=\bigcup_n K_n$</span> with each <span class="math-container">$K_n$</span> compact and <span class="math-container">$K_n\subseteq\operatorname{int}(K_{n+1})$</span>. Clearly <span class="math-container">$X$</span> is <span class="math-container">$\sigma$</span>-compact. Also the interiors of the <span class="math-container">$K_n$</span> cover <span class="math-container">$X$</span>, so <span class="math-container">$X$</span> is weakly locally compact.</p>
<p>(2) implies (3): because <span class="math-container">$\sigma$</span>-compact implies Lindelöf.</p>
<p>(3) implies (2): Since <span class="math-container">$X$</span> is weakly locally compact, every point is in the interior of some compact set. By the Lindelöf property <span class="math-container">$X$</span> is covered by the interiors of a countable number of these compact sets, and hence by a countable number of these compact sets themselves.</p>
<p>(2) implies (1): This is just a modification of the argument from @bof. Suppose <span class="math-container">$X=\bigcup_n C_n$</span> with each <span class="math-container">$C_n$</span> compact. Take <span class="math-container">$K_1=C_1$</span>. The set <span class="math-container">$K_1\cup C_2$</span> is compact. So by the lemma it is contained in the interior of some compact set <span class="math-container">$K_2$</span>. Continuing like this, at each step choose <span class="math-container">$K_{n+1}$</span> compact such that <span class="math-container">$K_n\cup C_{n+1}\subseteq\operatorname{int}(K_{n+1})$</span>. The result is an increasing sequence of compact sets, each contained in the interior of the next one, and adding up to the whole space, that is, an exhaustion of <span class="math-container">$X$</span> by compact sets.</p>
|
2,969,363 |
<p>I have 3 points in space A, B, and C all with (x,y,z) coordinates, therefore I know the distances between all these points. I wish to find point D(x,y,z) and I know the distances BD and CD, I do NOT know AD.</p>
<p>The method I have attempted to solve this using is first saying that there are two spheres known on points B and C with radius r (distance to point D). The third sphere is found by setting the law of cosines equal to the formula for distance between two vectors ((V1*V2)/(|V1||V2|)) = ((a^2+b^2-c^2)/2ab). </p>
<p>Now point D should be the intersection of these three spheres, but I have not been able to calculate this or find a way to. I either need help finding point D and I can give numeric points for an example, or I need to know if I need more information to solve (another point with a distance to D known).</p>
<p>Ok, now assuming that distance AD is known, how do I calculate point D?
<a href="https://i.stack.imgur.com/vzmbz.png" rel="nofollow noreferrer">This is what it looks like when I graph the two spheres and the one I calculated, as you can see it intersects on point F. (D in this case)</a></p>
|
Ross Millikan
| 1,827 |
<p>Knowing <span class="math-container">$|BD|$</span> and <span class="math-container">$|CD|$</span> is only enough to know that <span class="math-container">$D$</span> is on the intersection of two spheres, one around <span class="math-container">$B$</span> and one around <span class="math-container">$C$</span>. Assuming they intersect at all, the intersection is likely a circle and <span class="math-container">$D$</span> could be anywhere on that circle. <span class="math-container">$A$</span> is not giving you anything. Even if you get <span class="math-container">$|AD|$</span> you will still have an ambiguity between two points unless two of the spheres are tangent.</p>
|
2,969,363 |
<p>I have 3 points in space A, B, and C all with (x,y,z) coordinates, therefore I know the distances between all these points. I wish to find point D(x,y,z) and I know the distances BD and CD, I do NOT know AD.</p>
<p>The method I have attempted to solve this using is first saying that there are two spheres known on points B and C with radius r (distance to point D). The third sphere is found by setting the law of cosines equal to the formula for distance between two vectors ((V1*V2)/(|V1||V2|)) = ((a^2+b^2-c^2)/2ab). </p>
<p>Now point D should be the intersection of these three spheres, but I have not been able to calculate this or find a way to. I either need help finding point D and I can give numeric points for an example, or I need to know if I need more information to solve (another point with a distance to D known).</p>
<p>Ok, now assuming that distance AD is known, how do I calculate point D?
<a href="https://i.stack.imgur.com/vzmbz.png" rel="nofollow noreferrer">This is what it looks like when I graph the two spheres and the one I calculated, as you can see it intersects on point F. (D in this case)</a></p>
|
Nominal Animal
| 318,422 |
<p>(This is not an answer to the stated question per se, but explains how to efficiently do trilateration.)</p>
<p>In 3D, you need four fixed points (that are not all on the same plane), and their distances to the unknown point, to exactly determine the location of the unknown point.</p>
<p>Three fixed points (that are not all on the same line) and their distances to the unknown point will typically give you two possibilities, symmetrically mirrored by the plane formed by the three fixed points.</p>
<p>Let's say the unknown point is at <span class="math-container">$\vec{p} = (x, y, z)$</span>, the three fixed points are at <span class="math-container">$\vec{v}_1 = (x_1 , y_1 , z_1)$</span>, <span class="math-container">$\vec{v}_2 = (x_2 , y_2 , z_2)$</span>, and <span class="math-container">$\vec{v}_3 = (x_3 , y_3 , z_3)$</span>, at distances <span class="math-container">$d_1$</span>, <span class="math-container">$d_2$</span>, and <span class="math-container">$d_3$</span> from the unknown point, respectively. Solving the system of equations
<span class="math-container">$$\left\lbrace\begin{aligned}
\left\lVert \vec{p} - \vec{p}_1 \right\rVert &= \lvert d_1 \rvert \\
\left\lVert \vec{p} - \vec{p}_2 \right\rVert &= \lvert d_2 \rvert \\
\left\lVert \vec{p} - \vec{p}_3 \right\rVert &= \lvert d_3 \rvert \\
\end{aligned}\right . \iff
\left\lbrace\begin{aligned}
(x - x_1)^2 + (y - y_1)^2 + (z - z_1)^2 &= d_1^2 \\
(x - x_2)^2 + (y - y_2)^2 + (z - z_2)^2 &= d_2^2 \\
(x - x_3)^2 + (y - y_3)^2 + (z - z_3)^2 &= d_3^2 \\
\end{aligned}\right.$$</span>
is nontrivial, especially in algebraic form.</p>
<p>Instead, change to a coordinate system where <span class="math-container">$(x_1 , y_1 , z_1)$</span> is at origin, <span class="math-container">$(x_2 , y_2 , z_2)$</span> is at <span class="math-container">$(h , 0 , 0)$</span>, and <span class="math-container">$(x_3 , y_3 , z_3)$</span> is at <span class="math-container">$(i, j, 0)$</span>.
The unit vectors <span class="math-container">$\hat{e}_1 = ( X_1 , Y_1 , Z_1 )$</span>, <span class="math-container">$\hat{e}_2 = ( X_2 , Y_2 , Z_2 )$</span>, and <span class="math-container">$\hat{e}_3 = (X_3 , Y_3 , Z_3 )$</span> are
<span class="math-container">$$\left\lbrace\begin{aligned}
\vec{e}_1 &= \vec{v}_2 - \vec{v}_1 \\
\hat{e}_1 &= \frac{\vec{e}_1}{\left\lVert\vec{e}_1\right\rVert} \\
\vec{e}_2 &= \vec{v}_3 - \vec{v}_1 - \hat{e}_1 \left ( \hat{e}_1 \cdot \left ( \vec{v}_3 - \vec{v}_1 \right ) \right ) \\
\hat{e}_2 &= \frac{\vec{e}_2}{\left\lVert\vec{e}_2\right\rVert} \\
\hat{e}_3 &= \hat{e}_1 \times \hat{e}_2 \\
\end{aligned}\right.$$</span>
and the fixed point coordinates are
<span class="math-container">$$\left\lbrace\begin{aligned}
h &= \left\lVert \vec{v}_2 - \vec{v}_1 \right\rVert = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2 } \\
i &= \hat{e}_1 \cdot \left(\vec{v}_3 - \vec{v}_1\right) = X_1 ( x_3 - x_1 ) + Y_1 ( y_3 - y_1 ) + Z_1 ( Z_3 - Z_1 ) \\
j &= \hat{e}_2 \cdot \left(\vec{v}_3 - \vec{v}_1\right) = X_2 ( x_3 - x_1 ) + Y_2 ( y_2 - y_1 ) + Z_2 ( Z_3 - Z_2 ) \\
\end{aligned}\right.$$</span>
The distances stay the same, but the system of equations is much simpler. For clarity, I'll use <span class="math-container">$(u, v, w)$</span> in these new coordinates instead of <span class="math-container">$(x, y, z)$</span>:
<span class="math-container">$$\left\lbrace\begin{aligned}
u^2 + v^2 + w^2 &= d_1^2 \\
(u - h)^2 + v^2 + w^2 &= d_2^2 \\
(u - i)^2 + (v - j)^2 + w^2 &= d_3^2
\end{aligned}\right.$$</span>
which is easily solved:
<span class="math-container">$$\left\lbrace\begin{aligned}
u &= \frac{d_1^2 - d_2^2 + h^2}{2 h} \\
v &= \frac{d_1^2 - d_3^2 + i^2 + j^2 - 2 i u}{2 j} \\
w &= \pm \sqrt{d_1^2 - u^2 - v^2} \\
\end{aligned}\right.$$</span>
In the original coordinate system,
<span class="math-container">$$\vec{p} = \vec{v}_1 + u \hat{e}_1 + v \hat{e}_2 + w \hat{e}_3 \quad \iff \quad
\left\lbrace\begin{aligned}
x &= x_1 + u X_1 + v X_2 + w X_3 \\
y &= y_1 + u Y_1 + v Y_2 + w Y_3 \\
z &= z_1 + u Z_1 + v Z_2 + w Z_3 \\
\end{aligned}\right.$$</span>
noting that if <span class="math-container">$w$</span> is not a real, then there is no solution; if <span class="math-container">$w \approx 0$</span>, there is one solution; and otherwise there are two solutions, one with positive <span class="math-container">$w$</span>, and the other with negative <span class="math-container">$w$</span>.</p>
<p>If you know the distances to four fixed points, you only really need the fourth point (not coplanar with the three other fixed points) to distinguish which case it is. If the distances contain noise, it might make sense to calculate the result using each unique triplet (<span class="math-container">$123$</span>, <span class="math-container">$124$</span>, <span class="math-container">$134$</span>, and <span class="math-container">$234$</span>), and return their mean.</p>
<p>In pseudocode, for trilateration, you should precalculate the values that only depend on the fixed points:</p>
<pre><code>Let ex1 = x2 - x1
Let ey2 = y2 - y1
Let ez2 = z2 - z1
Let h = sqrt( ex1*ex1 + ey1*ey1 + ez1*ez1 )
If h <= epsilon:
Error: First and second point are too close.
End If
Let ex1 = ex1 / h
Let ey1 = ey1 / h
Let ez1 = ez1 / h
Let i = ex1*(x2 - x1) + ey1*(y2 - y1) + ez1*(z2 - z1)
Let ex2 = x3 - x1 - i*ex1
Let ey2 = y3 - y1 - i*ey1
Let ez2 = z3 - z1 - i*ez1
Let t = sqrt(ex2*ex2 + ey2*ey2 + ez2*ez2)
If t <= epsilon:
Error: the three fixed points are too close to being on the same line.
End If
Let ex2 = ex2 / t
Let ey2 = ey2 / t
Let ez2 = ez2 / t
Let j = ex2*(x3 - x1) + ey2*(y3 - y1) + ez2*(z3 - z1)
If j <= epsilon and j >= -epsilon:
Error: the three fixed points are too close to being on the same line.
End If
Let ex3 = ey1*ez2 - ez1*ey2
Let ey3 = ez1*ex2 - ex1*ez2
Let ez3 = ex1*ey2 - ey2*ex1
</code></pre>
<p>where <code>epsilon</code> is the largest positive number that should be treated as zero, and represents the expected precision in coordinates and distances, for example <code>0.001</code>.</p>
<p>The function that finds the coordinates for the unknown point is then</p>
<pre><code># Fixed points are at (x1,y1,z1), (x2,y2,z2), (x3,y3,z3)
# Function takes the distances to fixed points d1, d2, d3
# Function expects unit vectors (ex1,ey1,ez1), (ex2,ey2,ez2), (ex3,ey3,ez3)
# to be precalculated, with fixed points at (0,0,0), (h,0,0), (i,j,0)
# in that coordinate system, without changing the distances
Function Trilaterate(d1, d2, d3):
Let u = (d1*d1 - d2*d2 + h*h) / (2*h)
Let v = (d1*d1 - d3*d3 + i*(i - 2*u) + j*j) / (2*j)
Let ww = d1*d1 - u*u - v*v
If ww < -epsilon:
Return no solutions
Else
If ww < epsilon:
Return a single solution:
x = x1 + u*ex1 + v*ex2
y = y1 + u*ey1 + v*ey2
z = z1 + u*ez1 + v*ez2
Else:
w = sqrt(ww)
Return two solutions:
x = x1 + u*ex1 + v*ex2 + w*ex3
y = y1 + u*ey1 + v*ey2 + w*ey3
z = z1 + u*ez1 + v*ez2 + w*ez3
And
x = x1 + u*ex1 + v*ex2 - w*ex3
y = y1 + u*ey1 + v*ey2 - w*ey3
z = z1 + u*ez1 + v*ez2 - w*ez3
End If
End Function
</code></pre>
<p>This is simple enough to be done on 8-bit microcontrollers, if necessary.</p>
|
1,213,663 |
<p>If you measure a task & it takes 3 seconds, then the next time you do the same task, it takes you 1 second, is the difference 200% or 67%? </p>
<p>Or would you say the difference is 200% because 3-1=2 or 200% better -- but the percentage of difference is 2/3 or 67%? I'm pretty sure I'm confusing something if not someone. Be that as it may, I need to explain this clearly so that the analysis is clear & credible. The example I would site would be: Let's say you are mesuring system transaction response times & on two separate tests find the response time improvement noted. (Thanks PH)</p>
|
Robert Israel
| 8,508 |
<p>Permuting the rows of $M$ gives you $P M$ where $P$ is a permutation matrix.
The singular values of $PM$ are the square roots of the (nonzero) eigenvalues of $(PM)^* PM = M^* P^* P M = M^* M$, so they are the same as the singular values of $M$.</p>
|
1,691,825 |
<p>My textbook goes from</p>
<p>$$\frac{\left( \frac{6\ln^22x}{2x} \right)}{\left(\frac{3}{2\sqrt{x}}\right)}$$</p>
<p>to:</p>
<p>$$\frac{6\ln^22x}{3\sqrt{x}}$$</p>
<p>I don't see how this is right. Could anyone explain?</p>
|
Jan Eerland
| 226,665 |
<p>$$\frac{\frac{6\ln^2(2x)}{2x}}{\frac{3}{2\sqrt{x}}}=\frac{6\ln^2(2x)}{2x}\cdot\frac{2\sqrt{x}}{3}=\frac{6\ln^2(2x)}{x}\cdot\frac{\sqrt{x}}{3}=$$
$$\frac{2\ln^2(2x)}{x}\cdot\frac{\sqrt{x}}{1}=\frac{2\ln^2(2x)\sqrt{x}}{x}=\frac{2\ln^2(2x)}{\sqrt{x}}$$</p>
|
1,691,825 |
<p>My textbook goes from</p>
<p>$$\frac{\left( \frac{6\ln^22x}{2x} \right)}{\left(\frac{3}{2\sqrt{x}}\right)}$$</p>
<p>to:</p>
<p>$$\frac{6\ln^22x}{3\sqrt{x}}$$</p>
<p>I don't see how this is right. Could anyone explain?</p>
|
Michael Hoppe
| 93,935 |
<p>Just multiply denominator and numerator by $2x=2\sqrt x\cdot\sqrt x$ to obtain the given result.</p>
|
934,353 |
<p>I am a high school student in Calculus, and we are finishing learning basic limits. I am reviewing for a big test tomorrow, and I could do all of the problems correctly except this one.</p>
<p>I have no idea how to solve the problem this problem correctly. I looked up the answer online, but I can't figure out how they got their answer. All of the online tools show the steps using L'Hospital's rule or derivation, but I haven't learned either yet.</p>
<p>This is the problem:</p>
<p>$$\large\lim_{x\rightarrow 0}{\left(\frac{\frac{1}{\sqrt{1+x}}-1}{x}\right)}$$</p>
<p>This is the problem that I did incorrectly. I converted the $-1$ to $\frac{\sqrt{1+x}}{\sqrt{1+x}}$, then subtracted the fraction, and multiplied the result by $\frac{1}{x}$ to remove the double division.</p>
<p>$$\large\frac{1-\sqrt{1+x}}{x\sqrt{1+x}}$$</p>
<p>When I substitute $x$, I get $0$, but the answer is $-\frac{1}{2}$. I am doing something simple incorrectly, but I really cannot figure it out.</p>
|
Mohamed
| 33,307 |
<p>The formula :$$f(x)= \frac{1-\sqrt{1+x}}{x \sqrt{1+x}}$$is nont able to give the limit because it gives the indeterminate form $\frac00$.</p>
<p>Using the conjugate expression of $1-\sqrt{1+x}$, you get: $$f(x)=\frac{-x}{x \sqrt{1+x}(1+\sqrt{1+x})},$$
then: $$f(x)=\frac{-1}{ \sqrt{1+x}(1+\sqrt{1+x})}$$
and the limit is done.</p>
|
934,353 |
<p>I am a high school student in Calculus, and we are finishing learning basic limits. I am reviewing for a big test tomorrow, and I could do all of the problems correctly except this one.</p>
<p>I have no idea how to solve the problem this problem correctly. I looked up the answer online, but I can't figure out how they got their answer. All of the online tools show the steps using L'Hospital's rule or derivation, but I haven't learned either yet.</p>
<p>This is the problem:</p>
<p>$$\large\lim_{x\rightarrow 0}{\left(\frac{\frac{1}{\sqrt{1+x}}-1}{x}\right)}$$</p>
<p>This is the problem that I did incorrectly. I converted the $-1$ to $\frac{\sqrt{1+x}}{\sqrt{1+x}}$, then subtracted the fraction, and multiplied the result by $\frac{1}{x}$ to remove the double division.</p>
<p>$$\large\frac{1-\sqrt{1+x}}{x\sqrt{1+x}}$$</p>
<p>When I substitute $x$, I get $0$, but the answer is $-\frac{1}{2}$. I am doing something simple incorrectly, but I really cannot figure it out.</p>
|
user194150
| 194,150 |
<p>$$
\displaylines{
f\left( x \right) = \frac{1}{{\sqrt {1 + x} }} \Rightarrow f'\left( x \right) = \frac{{ - 1}}{{2\left( {1 + x} \right)\sqrt {1 + x} }} \cr
f'\left( 0 \right) = \mathop {\lim }\limits_{x \to 0} \frac{{f\left( x \right) - f\left( 0 \right)}}{{x - 0}} \cr
= \mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{{\sqrt {1 + x} }} - 1}}{x} = \frac{{ - 1}}{{2\left( {1 + 0} \right)\sqrt {1 + 0} }} \cr
\Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{{\sqrt {1 + x} }} - 1}}{x} = - \frac{1}{2} \cr}
$$</p>
|
33,303 |
<p>Is it possible to attach certain pieces of code to certain controls in a Manipulate? For example, consider the following Manipulate</p>
<pre><code>Manipulate[
data = Table[function[x], {x, -Pi*10, Pi*10, Pi/1000}];
ListPlot[{x, data}, PlotRange -> {{start, stop}, Automatic}]
, {function, {Sin, Cos, Tan}}
, {start, 1, Length[data]}
, {{stop, 300}, 1, Length[data]}
]
</code></pre>
<p>Generation of the data is expensive but it only needs to be done if I change function. So, I'd like the line</p>
<pre><code>data = Table[function[x], {x, -Pi*10, Pi*10, Pi/1000}];
</code></pre>
<p>to only run when I change the function control...i.e. I want to attach that line of code to the 'function' control. As it stands, the data is generated when I move the plot range too which is not what I want. </p>
<p><img src="https://i.stack.imgur.com/OHJeN.png" alt="enter image description here"></p>
|
Michael E2
| 4,999 |
<p>Here's a fairly simple way to fix your <code>Manipulate</code> by applying <code>Dynamic</code> to <code>ListPlot</code>.</p>
<pre><code>Manipulate[
(* Beep[]; *)
data = function @ Range[-Pi*10., Pi*10, Pi/1000];
Dynamic @ ListPlot[data, PlotRange -> {{start, stop}, Automatic}],
{function, {Sin, Cos, Tan}},
{start, 1, Length[data]},
{{stop, 300}, 1, Length[data]},
{data, ControlType -> None}]
</code></pre>
<p><img src="https://i.stack.imgur.com/f9Lii.png" alt="Mathematica graphics"></p>
<p>Uncomment <code>Beep[]</code> to hear when <code>data</code> is reevaluated.</p>
<p>There are several questions on this site whose answers discuss using <code>Dynamic</code> for such a purpose. This one is more general than most: <a href="https://mathematica.stackexchange.com/questions/21625/using-refresh-with-trackedsymbols">Using Refresh[..] with TrackedSymbols</a></p>
|
47,561 |
<p>The Hilbert matrix is the square matrix given by</p>
<p>$$H_{ij}=\frac{1}{i+j-1}$$</p>
<p>Wikipedia states that its inverse is given by</p>
<p>$$(H^{-1})_{ij} = (-1)^{i+j}(i+j-1) {{n+i-1}\choose{n-j}}{{n+j-1}\choose{n-i}}{{i+j-2}\choose{i-1}}^2$$</p>
<p>It follows that the entries in the inverse matrix are all integers.</p>
<p>I was wondering if there is a way to prove that its inverse is an integer matrix without using the formula above.</p>
<p>Also, how would one go about proving the explicit formula for the inverse? Wikipedia refers me to a paper by Choi, but it only includes a brief sketch of the proof.</p>
|
Fedor Petrov
| 4,312 |
<p>It suffices to prove Schechter's formula for Cauchy matrix cited in Wikipedia (see link in Faisal's comment). We need to check $\sum_j b_{ij}a_{jk}=\delta_{ik}$, i.e.
$$
\frac{A(y_i)}{B'(y_i)}\sum_j \frac{f(x_j)}{A'(x_j)}=-\delta_{i,k},
$$
where $f(t)=B(t)/((t-y_i)(t-y_k))$. If $i\ne k$, then $f$ is just a polynomial of degree $n-2$, and the inner sum is its coefficient in $t^{n-1}$ (this follows from Lagrange interpolation on points $x_1,\dots,x_n$). If $i\ne k$, then denote by $F$ the corresponding Lagrange polynomial $F(x)=\sum f(x_j) \frac{A(x)}{(x-x_j)A'(x_j)}$, we are searching for a coefficient of $F$ in $x^{n-1}$. We have $F(x_j)=f(x_j)$, so $F(x)(x-y_i)-\prod_{j\ne i}(x-y_i)$ vanishes for $x=x_1,x_2,\dots,x_n$. So, $F(x)(x-y_i)-\prod_{j\ne i}(x-y_i)=cA(x)$, and we find $c$ substituting $x=y_i$. </p>
|
3,252,076 |
<p>I'm doing some Galois cohomology stuff (specifically, trying to calculate <span class="math-container">$H^1(\mathbb{Q}_3,E[\varphi])$</span>, where <span class="math-container">$\varphi:E\to E'$</span> is an isogeny of elliptic curves), and it involves calculating <span class="math-container">$\mathbb{Q}_3(\sqrt{-6})^{\times}/(\mathbb{Q}_3(\sqrt{-6})^{\times})^3$</span>. Here's what I've done so far.</p>
<p>Let <span class="math-container">$K=\mathbb{Q}(\sqrt{-6})$</span>. As <span class="math-container">$-6\not\equiv 1$</span> (mod 4), we have that <span class="math-container">$\mathcal{O}_K=\mathbb{Z}[\sqrt{-6}]$</span>. Let <span class="math-container">$v$</span> be the finite place of <span class="math-container">$K$</span> corresponding to the (non-principal) prime ideal <span class="math-container">$(3,\sqrt{-6})$</span>. It's fairly easy to see that <span class="math-container">$K_v=\mathbb{Q}_3(\sqrt{-6})$</span>, and that the residue field is <span class="math-container">$k_v\cong\mathcal{O}_{K}/(3,\sqrt{-6})\cong\mathbb{F}_3$</span>. Now, by Hensel's lemma, <span class="math-container">$\sqrt{-2}\in\mathbb{Q}_3$</span>, so it follows that <span class="math-container">$\sqrt{3}\in\mathbb{Q}_2(\sqrt{-6})$</span>. In <span class="math-container">$\mathcal{O}_K$</span>, <span class="math-container">$(3)$</span> decomposes as <span class="math-container">$(3,\sqrt{-6})^2$</span>, so <span class="math-container">$v(3)=2$</span>, and hence <span class="math-container">$v(\sqrt{3})=1$</span>. So we can legitimately choose <span class="math-container">$\sqrt{3}$</span> as a uniformizer of <span class="math-container">$\mathcal{O}_{K_v}$</span>. This means that every element of <span class="math-container">$\mathcal{O}_{K_v}$</span> has a unique representation
<span class="math-container">$$\sum_{n=0}^{\infty}a_n\sqrt{3}^n, \text{ where } a_n\in\{-1,0,1\}.$$</span>
The elements of <span class="math-container">$\mathcal{O}_{K_v}^{\times}$</span> are the ones where <span class="math-container">$a_0=\pm1$</span>. Now, using Hensel's lemma I went ahead and showed that
<span class="math-container">$$(\mathcal{O}_{K_v}^{\times})^3=\{\pm1+\sum_{n=3}^{\infty}a_n\sqrt{3}^n~|~a_n\in\{-1,0,1\}\}.$$</span>
But how do I find distinct representatives for <span class="math-container">$\mathcal{O}_{K_v}^{\times}/(\mathcal{O}_{K_v}^{\times})^3$</span>? Does modding out by the group above mean that I can just look up to sign and ignore everything past <span class="math-container">$\sqrt{3}^3$</span>, so that a set of representatives would be <span class="math-container">$\{1,1\pm\sqrt{3},1\pm3,1\pm\sqrt{3}\pm3\}$</span>, which has size 9 (the 2 <span class="math-container">$\pm$</span>'s are independent in the last expression)? Perhaps my working is not useful, because I've written things additively but the groups are multiplicative. Also, I could just as well have chosen <span class="math-container">$\sqrt{-6}$</span> as my uniformizer. Can I replace <span class="math-container">$\sqrt{3}$</span> with <span class="math-container">$\sqrt{-6}$</span> everywhere and still get a set of representatives for <span class="math-container">$\mathcal{O}_{K_v}^{\times}/(\mathcal{O}_{K_v}^{\times})^3$</span>? I'm very confused!</p>
<p>Of course, once <span class="math-container">$\mathcal{O}_{K_v}^{\times}/(\mathcal{O}_{K_v}^{\times})^3$</span> is determined, finding <span class="math-container">$K_v^{\times}/(K_v^{\times})^3$</span> is easy.</p>
|
nguyen quang do
| 300,700 |
<p>Your problem being local, why do you complicate it by bringing it back to a global one ? I'll keep your notation <span class="math-container">$K_v = \mathbf Q_3 (\sqrt {-6})$</span> and work locally.</p>
<p>For any <span class="math-container">$p$</span>-adic local field <span class="math-container">$K$</span> of degree <span class="math-container">$n$</span> over <span class="math-container">$\mathbf Q_p$</span>, the quotient <span class="math-container">$K^*/{K^*}^p$</span> can be viewed (if written additively) as an <span class="math-container">$\mathbf F_p$</span> - vector space, of dimension <span class="math-container">$n+2$</span> (resp. <span class="math-container">$n+1$</span>) according as <span class="math-container">$K$</span> contains or not a primitive <span class="math-container">$p$</span>-th root of <span class="math-container">$1$</span> (this is a matter of Herbrand quotients, see e.g. Serre's "Local Fields", chap.14, prop.10 and ex.3). Here your <span class="math-container">$K_v$</span> is a quadratic totally ramified extension of <span class="math-container">$\mathbf Q_3$</span>, not containing <span class="math-container">$\mu_3$</span> (because <span class="math-container">$(-3)(-6)=2.3^2$</span> is not a square in <span class="math-container">$\mathbf Q_3$</span>), hence the above dimension is <span class="math-container">$3$</span>, and we only need to find an <span class="math-container">$\mathbf F_3$</span>-basis. A first natural vector, coming from an uniformizer, is <span class="math-container">$\sqrt {-6}$</span> (or <span class="math-container">$\sqrt 3$</span> if you want). It remains only to exhibit two linearly independent vectors in <span class="math-container">$U_1/{U_1}^3$</span>, where <span class="math-container">$U_1$</span> is the group of prinipal units. I found the pair <span class="math-container">$1\pm \sqrt {-6}$</span> (but you must check, I am prone to calculation errors). </p>
|
1,091,653 |
<p>Is this correct ?</p>
<p>$$
\frac{d}{dt} \left( \int_0^t \phi(t)dt \right) = \phi(t)
$$</p>
<p>If not, how can I recover $$ \phi(t) $$ knowing only $$ \int_0^t \phi(t)dt $$ ?</p>
|
MathMajor
| 113,330 |
<p>(i) No, it is not correct. You probably mean
$$
\frac{\text{d}}{\text{d}t} \left( \int_0^t \phi(x) \, \text{d}x \right) = \phi(t)
.$$</p>
<p>This equality is also known as the Fundamental Theorem of Calculus, Part 1.</p>
<p>(ii) You can recover $\phi (t)$ by differentiating, as demonstrated above.</p>
|
1,964,139 |
<p>$D_2n$ is not abelian. However, the group of rotations, denoted $R$, is. I've already shown that $R$ is a normal subgroup of $D_2n$; however I'm stuck at showing the quotient group is abelian.</p>
<p>I know if it is abelian, $xRyR=yRxR$ but I get stuck at $xRyR=(xy)R$.
But $x$ and $y$ are not necessarily commutative. How do I continue? </p>
|
Dietrich Burde
| 83,966 |
<p>Since $R=C_n$ is a subgroup of $D_{2n}$ of index $2$, it is a normal subgroup. The quotient $D_{2n}/C_n$ has exactly $2$ elements, because of
$$
|D_{2n}/C_n|=\frac{|D_{2n}|}{|C_n|}=\frac{2n}{n}=2.
$$
But every group of order $2$ is abelian (there is only one).</p>
|
1,553,440 |
<p>Let , $g(x)=f(x)+f(1-x)$ and $f'(x)<0$ for all $x\in (0,1)$. Then , $g$ is monotone increasing in </p>
<p>(A) $(1/2,1)$.</p>
<p>(B) $(0,1/2)$</p>
<p>(C) $(0,1/2)\cup (1/2,1)$.</p>
<p>(D) none.</p>
<p>We have , $g'(x)=f'(x)-f'(1-x)$. Now $g'(x)>0$ if $f'(x)>f'(1-x)$. As $f'(x)<0$ so , $f$ is monotone decreasing in $(0,1)$. From here how I can conclude ?</p>
|
Thomas Andrews
| 7,933 |
<p>Let $f_1(x)=-x^2$. Then $f_1'(x)=-2x<0$ and $g_1(x)=-x^2-(1-x))^2 = -2x^2+2x-1$ and $g_1'(x)=2-4x$ so $g(x)$ is decreasing when $x>\frac{1}{2}$.</p>
<p>On the other hand, if $f_2(x)=(1-x)^2$. Then $f_2'(x)=-2(1-x)<0$ and $g_2(x)=(1-x)^2+x^2=-g_1(x)$. So $g_2(x)$ is increasing precisely where $g_1(x)$ was not, and visa versa.</p>
<p>Therefore, it is not possible to determine where $g$ is increasing/decreasing. I don't feel like "none" is the right answer, but it would depend on the precise phrasing of the question.</p>
|
1,523,329 |
<p>Can cusps be considered points of inflection?</p>
<p>I'm getting conflicting information but my thought process is that cusps cannot be points of inflection?</p>
<p>Can points of inflection exist when there is a vertical tangent to the graph? Assume there is change in concavity and the function is continuous. </p>
|
abel
| 9,252 |
<p>no. look at the graph of $y = x^{2/3}.$ this has a cusp at $(0,0)$ but concave down on $(-\infty, \infty)$ and $(0,0)$ is certainly not a point of inflection. </p>
|
1,523,329 |
<p>Can cusps be considered points of inflection?</p>
<p>I'm getting conflicting information but my thought process is that cusps cannot be points of inflection?</p>
<p>Can points of inflection exist when there is a vertical tangent to the graph? Assume there is change in concavity and the function is continuous. </p>
|
Carly Vollet
| 989,151 |
<p>I would not consider <span class="math-container">$f(x)=x^{2/3}$</span> to be concave down on <span class="math-container">$(-\infty,\infty)$</span>, but rather on <span class="math-container">$(-\infty, 0)\cup(0,\infty)$</span>. If we use the definition of concave down to be "A function <span class="math-container">$f$</span> is concave down when <span class="math-container">$f'$</span> is decreasing", then we can see this is not the case with <span class="math-container">$f(x)=x^{2/3}$</span>. Its derivative, <span class="math-container">$f'(x)=\frac{2}{3\sqrt[3]{x}}$</span>, is decreasing on its domain, which is <span class="math-container">$(-\infty, 0)\cup(0,\infty)$</span>. So by that definition, <span class="math-container">$f(x)$</span> is not concave down on <span class="math-container">$(-\infty,\infty)$</span>.</p>
<p>While I agree that <span class="math-container">$x^{2/3}$</span> does not have an inflection point at its cusp, that doesn't mean we <em>can't</em> have inflection points at cusps. It really depends on your definition of <strong>inflection point</strong>. You can easily make these types of cusps appear by taking absolute values of functions. For example: <span class="math-container">$g(x)=\left| x^2-1\right|$</span> has cusps at <span class="math-container">$x=\pm 1$</span> and also changes concavity there.</p>
<p><a href="https://i.stack.imgur.com/bUdKg.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bUdKg.png" alt="A graph with a change in concavity at a cusp" /></a></p>
|
142,507 |
<p>Let $\psi(x) := \sum_{n\leq x} \Lambda(n)$ where $\Lambda(n)$ is the Von-Mangoldt function.
I want to show that if $$ \lim_{x \rightarrow \infty} \frac{\psi(x)}{x} =1 $$ then also $$\lim_{x\rightarrow \infty} \frac{\pi(x) \log x }{x}=1.$$</p>
<p>I tried to play a little bit with $\psi$, what I want to show is that:</p>
<p>$$\left| \frac{\pi(x) \log x}{x} -1 \right| \leq \left| \frac{\psi(x)}{x} -1 \right| \rightarrow 0$$</p>
<p>So I tried to develop $\psi$ a little bit, but I got astray.</p>
<p>So I have
$$ \frac{\psi(x)}{x} -1 = \sum_{p^k \leq x , k \geq 1} \frac{\log p}{x} -1 = \frac{1}{x}\left(\sum_{p\leq x} \log p + \sum_{p^2\leq x} \log p + ...+ \sum_{p^k \leq x, p^{k+1} >x} \log p \right) -1 $$
and I want to estimate its aboslute value from below, but I don't have any idea?</p>
<p>Any hints?</p>
<p>Thanks.</p>
|
Arturo Magidin
| 742 |
<p>Do you know how to prove that $\mathbb{R}$ is numerically equivalent to $\mathcal{P}(\mathbb{N})$? Show that $\mathbb{R}$ is numerically equivalent to $(0,1)$, then show (using binary representation; careful with the numbers with dual representation) that there is an embedding $(0,1)\hookrightarrow \mathcal{P}(\mathbb{N})$. Then show that there is an embedding $\mathcal{P}(\mathbb{N})\hookrightarrow (0,1)$, say by looking at decimal representations of numbers that only use two digits, neither of them $0$ or $1$.</p>
<p>Now, if each $A_i$ has exactly two elements, do you see a connection between $\prod A_i$ and $\mathcal{P}(\mathbb{N})$?</p>
|
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