qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
136,264 | <p>I have a question concerning the stability analysis for a kind of differential equation taking the form
$$\dot x=Ax+Bw,$$
where $A\in \mathbb{R}^{n \times n}$, $B\in \mathbb{R}^{n \times m}$ are constant matrices and
$w \in \mathbb{R}^m$ is a normal random variable, i.e., $w\sim \mathcal{N}(0,W)$ with $W$ be a symmetric and positive definite matrix.</p>
<p>Since the zero is not an equilibrium of the system, the Lyapunov analysis does not make sense. When the input-to-state stability analysis is considered, the robust control theory does not apply due to the unboundness of $w$. By resorting to stochastic stability in the sense of mean square or almost surely, the Ito formula seems to be invalid.</p>
<p>HOW to carry out the stability analysis of this kind of systems? Any pointer will be helpful. Thanks!</p>
| juan | 7,402 | <p>I myself am not very fun of rankings. But when the Library of my University
decided to cut down some of the journal subscriptions
(due to the budget crisis in the Eurozone), I gave my personal viewpoint
by means of creating a unbiased ranking of Mathematical Journals.</p>
<p>Mainly what I wanted to measure was the impact of the Maths published
by each Journal throughout its life on today's Math. I took the raw data from MathSciNet.
The result can be consulted in the web page of the society journal of
the Spanish Math Society (the "Gaceta de la Real Sociedad Matematica Española"):</p>
<p><a href="http://gaceta.rsme.es/adicional.php?id=1215" rel="noreferrer">http://gaceta.rsme.es/adicional.php?id=1215</a></p>
<p>and also in my personal web page:</p>
<p><a href="http://personal.us.es/arias/V17N3_439.pdf" rel="noreferrer">http://personal.us.es/arias/V17N3_439.pdf</a></p>
<p>The paper, where I explain the procedure devised for creating the ranking my procedure, is in Spanish,
but at the end you can find the ranking, which is easily understood.</p>
<p>My ranking treats on the same footing all journals in applied, pure and statistical math.</p>
|
3,555,084 | <blockquote>
<p>Let
<span class="math-container">$$f(z) = e^z (1+\cos\sqrt{z} ) $$</span>
<span class="math-container">$\Omega=\{z\in\Bbb C: |z|\gt r\}$</span>, <span class="math-container">$r\gt 0$</span>. What is <span class="math-container">$f(\Omega)$</span>?</p>
<p>where <span class="math-container">$\sqrt{z}=\exp{(\text{Log }z/2)}, \text{Log }z=\log|z|+i\arg z,\arg z\in(-\pi,\pi]$</span></p>
</blockquote>
<p><span class="math-container">$\infty$</span> is an essential singularity. Picard's Great theorem , <span class="math-container">$\Bbb C\setminus f(\Omega) $</span> contains at most one point. <span class="math-container">$f(\Omega)$</span> is <span class="math-container">$\Bbb C$</span> ? </p>
<p>When <span class="math-container">$ z\in\Bbb R $</span>, <span class="math-container">$f(z)\geq0$</span>. According to <a href="https://en.wikipedia.org/wiki/Schwarz_reflection_principle" rel="nofollow noreferrer">Schwarz reflection principle</a> and Picard's Great theorem, all <span class="math-container">$ x+iy(y\ne0), \in\Bbb f(\Omega) $</span>. How to show that all negative real numbers belong to <span class="math-container">$f(\Omega)$</span>? thanks a lot!</p>
<p><strong>Is there a simple, more Elementary Proof</strong> than Conard's ? </p>
| LHF | 744,207 | <p>To write the complete solution, as per the hint in the comments I gave:</p>
<p><span class="math-container">$$
\begin{aligned}
\lim_{x \to \infty}x^2\left(x^2 - x \cdot \sqrt{x^2 + 6} + 3\right) &= \lim_{x \to \infty}x^2\left(\sqrt{(x^2+3)^2} - \sqrt{x^4 + 6x^2}\right)\\
&=\lim_{x \to \infty}x^2\left(\frac{(x^2+3)^2-x^4-6x^2}{\sqrt{(x^2+3)^2} + \sqrt{x^4 + 6x^2}}\right)\\
&=\lim_{x \to \infty}\frac{9x^2}{x^2+3 + \sqrt{x^4 + 6x^2}}\\
&=\lim_{x \to \infty}\frac{9}{1+\frac{3}{x^2} + \sqrt{1 + \frac{6}{x^2}}}\\
&=\frac{9}{2}\\
\end{aligned}
$$</span></p>
|
2,614,969 | <p>I wonder whether there is a general method for accurately estimating the limit of the sequence:</p>
<p>\begin{equation}
x_{n+1} = x_n - x_{n}^{n+1}, \forall x_1 \in (0,1)
\end{equation}</p>
<p>After showing that the limit exists, since $ x_n $ is decreasing and bounded, I managed to derive a lower-bound. In particular, I used the fact that:</p>
<p>\begin{equation}
\frac{x_{n+1}}{x_n} = 1-x_{n}^n \tag{1}
\end{equation}</p>
<p>Using $(1)$ we obtain:</p>
<p>\begin{equation}
\frac{x_N}{x_{N-1}}...\frac{x_2}{x_1}=\prod_{n=1}^{N} (1-x_{n}^n)=\frac{x_N}{x_1} \tag{2}
\end{equation}</p>
<p>From this we deduce:</p>
<p>\begin{equation}
\begin{split}
\lim_{N \to \infty} x_N & = \lim_{N \to \infty}x_1 \prod_{n=1}^{N} (1-x_{n}^n) \\ & = x_1 (\lim_{N \to \infty} \prod_{n=1}^{N} e^{\ln (1-x_{n}^n)}) \\ & = x_1 (\lim_{N \to \infty} e^{\sum_{n=1}^N\ln (1-x_{n}^n)})
\end{split}
\tag{3}\end{equation}</p>
<p>Using the following facts:</p>
<p>\begin{cases}
\sum_{n=1}^{N} \ln(1-x_{n}^n) \geq \sum_{n=1}^{N} \ln(1-x_{1}^n),\\
x \approx 0 \implies \ln(1+x) \approx x \\
\tag{4}\end{cases}</p>
<p>We may deduce that for $M$ sufficiently large:</p>
<p>\begin{equation}
\sum_{n=1}^{\infty}\ln (1-x_{n}^n) \geq \sum_{n=1}^{M} \ln(1-x_{1}^n)-\sum_{n=M}^\infty x_{1}^n \tag{5}
\end{equation}</p>
<p>And using $(5)$ we have a useful lower-bound. However, I wonder whether there's a more direct integration technique which can give me a good approximation to $(3)$. </p>
| Community | -1 | <p>The observation of Fede Poncio that the limit of $x_n$ is well-approximated by a quadratic polynomial in terms of $x_1$ is very useful. With a few lines of Python3 code we observe:</p>
<pre><code>import numpy as np
import matplotlib.pyplot as plt
def f(x,N):
X = np.zeros(N,dtype='float64')
X[0] = x
for i in range(1,N):
X[i] = X[i-1] - X[i-1]**(i+1)
return X, X[-1]
z= np.linspace(0,1,100)
q = [f(z[i],1000)[1] for i in range(100)]
plt.xlabel('unit interval')
plt.ylabel('approximate limit')
plt.plot(z,q,color='steelblue')
</code></pre>
<p><a href="https://i.stack.imgur.com/jI1tU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jI1tU.png" alt="enter image description here"></a></p>
<p>Indeed, this observation may be used to derive a very good approximation. </p>
<p>By expanding $x_{N+1}$ we obtain:</p>
<p>\begin{equation}
\begin{split}
x_{N+1} & = x_1-x_1^2-x_2^3-x_3^4-...-x_N^{N+1} \\
& = (x_1-x_1^2)-\sum_{n=2}^N x_n^{n+1} < x_1-x_1^2=x_2 \\
\end{split}
\tag{1} \end{equation}</p>
<p>Now, given that $x_n$ is decreasing:</p>
<p>\begin{equation}
\sum_{n=2}^{N} x_n^{n+1} < x_2 \sum_{n=2}^{\infty} x_2^n=x_2\big(\frac{x_2}{1-x_2}-x_2 \big)=\frac{x_2^3}{1-x_2}<2x_2^3
\tag{2}
\end{equation}</p>
<p>In fact, we may show that:</p>
<p>\begin{equation}
\lim_{n\to\infty} x_n \sim x_1-x_1^2
\tag{3}
\end{equation}</p>
<p>The quality of this approximation can be checked using $(1)$ and $(2)$ as follows:</p>
<p>\begin{equation}
\big\lVert 1-\frac{\lim_{n\to\infty} x_n}{x_1-x_1^2} \big\rVert \leq \frac{2x_2^3}{x_2}=2x_2^2 \leq \frac{2}{4^2}=12.5 \%
\tag{4}
\end{equation}</p>
<p>Moreover, if we calculate the expected value of $(4)$ we find:</p>
<p>\begin{equation}
\mathbb{E}\big[\big\lVert 1-\frac{\lim_{n\to\infty} x_n}{x_1-x_1^2} \big\rVert\big] \leq \frac{1}{15} \sim 6 \%
\tag{5}
\end{equation}</p>
|
2,573,487 | <p>I have given this set</p>
<blockquote>
<p>$$ M = \{ x \in [1,2]\times [3,4] ~|~ x\in\mathbb{Q}^2 \} \subset \mathbb{R}^2 $$</p>
</blockquote>
<p>First I have to identify the boundary $\partial M$ and then tell if it is open or closed.</p>
<p>I think that $$ \partial M = \{ (x,y) ~|~ x\not\in\mathbb{Q}^2, 1\leq x \leq 2, 3\leq y\leq 4 \} $$</p>
<p>is the boundary, but I am not sure about it. Is this correct? If not, what is the boundary?</p>
<p>Also: I am pretty sure that the set is open, because you can for sure find a series (e.g. for Pi) that converges to an irrational number, but with fraction series values in $\mathbb{Q}$.</p>
| ajotatxe | 132,456 | <p>The set is not open or closed, because any open ball contains rational and irrational points.</p>
<p>To solve the problem, show that $\overline M=[1,2]\times[3,4]$ and $\overline{M^c}=\Bbb R^2$.</p>
<p>Then $\partial M=[1,2]\times[3,4]\cap \Bbb R^2$</p>
|
93,063 | <p>There is a story I read about tiling the plane with convex pentagons.</p>
<p>You can read about it in this <a href="http://www.ivanrival.com/docs/picturepuzzling_2.pdf">article</a> on pages 1 and 2.</p>
<p>Summary of the story:
A guy showed in his doctorate work all classes of tiling the plane with convex pentagons, and proved that they are indeed all possible cases.
Later, riddle was published in popular science magazine to find all these classes. One of the readers found a tiling that did not belong to any of the classes, and so the claim and the proof turned out to be wrong.</p>
<hr>
<p>Reading this made me think about some questions.</p>
<p>Is it rare when a theorem was proved and the proof was published, and later it turned out that the theorem is wrong?</p>
<p>Can we somehow guess how many theorems out there that we think are right, but actually are wrong? I bet that if in our case, the theorem was about tiling in $R^3$ nobody would ever notice.</p>
<p>What can be the effect of such theorems on mathematics in general? Can it be a serious issue for mathematics even if the wrong theorems were not very important, but still, some stuff was based on them?</p>
| Carl Mummert | 630 | <p>It is not rare, but is uncommon, for false theorems to be published. It is somewhat more common, although still not frequent, for flawed proofs to be published, even though the theorem as stated is correct. One way to find these is to search for "correction", "corrigendum", or "retraction" on MathSciNet. Peer review can catch some of these errors, but mathematics is a human field in the end. </p>
<p>Every specialty has its own anecdotes about flawed proofs that are used to scare some caution into graduate students. For example, I would guess many logicians and analysts have heard how Lebesgue falsely claimed in print that every analytic set is Borel (in 1905).</p>
<p>In principle, the discovery of a flawed proof could mean that people have to re-check many other results. But in practice the effect is usually localized. Many researchers are cautious about using new results "blindly", once they realize that errors are not rare. In particular, being cautious means making sure you understand how to prove the results you use in your own proofs, whenever possible, so that nobody can later say your proof is flawed. By the time you are working on research papers, it becomes extremely unsatisfying to use a result of someone else as a "black box" without understanding it. On the other hand, it would be perfectly possible for a flawed proof to linger for a while if nobody else needs to use the result. </p>
<p>One of the roles of monographs and textbooks is to give another vetting to the theorems. A result that is proved in secondary books is somehow more reliable than a result that can only be cited to one research paper. This is another reason that errors tend to cause only localized problems, because by the time a result becomes standard in its field, a large number of mathematicians will have looked at it. </p>
|
3,247,563 | <p>Given a segment AB on the plane let <span class="math-container">$r$</span> and <span class="math-container">$s$</span> parallel lines with <span class="math-container">$A \in r$</span> and <span class="math-container">$B \in s $</span>. What is the locus of the circles tangent to <span class="math-container">$AB$</span>, <span class="math-container">$r$</span> and <span class="math-container">$s$</span>?
How could I solve this analytically?</p>
| Angina Seng | 436,618 | <p>How about
<span class="math-container">$$|\ln(1-x)|=\left|-\sum_{n=1}^\infty\frac{x^n}n\right|
\le\sum_{n=1}^\infty\frac{|x|^n}n\le\sum_{n=1}^\infty|x|^n=\frac{|x|}{1-|x|}$$</span>
etc.?</p>
|
2,217,338 | <p>I am trying to define a simple function that is first concave and then convex as shown in the picture below. Since the resulting equation have to be explained/used by a non-technical audience, the function should be ideally as simple as possible, but I have been unable to find any simple form that matches the requirements below.</p>
<p><a href="https://i.stack.imgur.com/7c1II.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7c1II.jpg" alt="enter image description here"></a></p>
<p>Any help of ideas would be greatly appreciated.</p>
<p><strong>EDIT</strong>: to further clarify</p>
<ol>
<li>The dashed red line is the constant line <span class="math-container">$Y=X$</span> </li>
<li><span class="math-container">$X_1$</span> and <span class="math-container">$X_2$</span> should be ideally points that I can control within the function.</li>
</ol>
| Marco Cantarini | 171,547 | <p>We have that <span class="math-container">$$\begin{align} I\left(a\right) = & \int_{0}^{x}x\sin^{2}\left(ax\right)e^{-x^{2}}dx \\ = &-\frac{1}{4}\int_{0}^{\infty}xe^{-x^{2}-2iax}dx-\frac{1}{4}\int_{0}^{\infty}xe^{-x^{2}+2iax}dx+\frac{1}{2}\int_{0}^{\infty}xe^{-x^{2}}dx \end{align}$$</span> so let us analyze <span class="math-container">$$ I_{1}\left(a\right)= -\frac{1}{4}\int_{0}^{\infty}xe^{-x^{2}-2iax}dx.$$</span>It is not so hard to see that <span class="math-container">$$ \begin{align} I_{1}\left(a\right)= & -\frac{e^{-a^{2}}}{4}\int_{0}^{\infty}xe^{-\left(x+ia\right)^{2}}dx \\ = & -\frac{e^{-a^{2}}}{4}\int_{0}^{\infty}\left(x+ia\right)e^{-\left(x+ia\right)^{2}}dx+\frac{e^{-a^{2}}ia}{4}\int_{0}^{\infty}e^{-\left(x+ia\right)^{2}}dx \\ = & -\frac{1}{8}+\frac{e^{-a^{2}}ai\sqrt{\pi}\left(1-i\textrm{erfi}\left(a\right)\right)}{8} \end{align}$$</span> where <span class="math-container">$\textrm{erfi}\left(z\right)$</span> is the <a href="http://mathworld.wolfram.com/Erfi.html" rel="noreferrer">imaginary error function</a>. In a similar manner <span class="math-container">$$\begin{align} I_{2}\left(a\right)= & -\frac{1}{4}\int_{0}^{\infty}xe^{-x^{2}+2iax}dx \\ = & -\frac{e^{-a^{2}}}{4}\int_{0}^{\infty}xe^{-\left(x-ia\right)^{2}}dx \\= & -\frac{e^{-a^{2}}}{4}\int_{0}^{\infty}\left(x-ia\right)e^{-\left(x-ia\right)^{2}}dx-\frac{e^{-a^{2}}ia}{4}\int_{0}^{\infty}e^{-\left(x-ia\right)^{2}}dx \\ = & -\frac{1}{8}-\frac{e^{-a^{2}}ai\sqrt{\pi}\left(1+i\textrm{erfi}\left(a\right)\right)}{8} \end{align}$$</span> and obiously <span class="math-container">$$I_{3}=\frac{1}{2}\int_{0}^{\infty}xe^{-x^{2}}dx=\frac{1}{4}.$$</span> So finally <span class="math-container">$$I\left(a\right)=I_{1}\left(a\right)+I_{2}\left(a\right)+I_{3}=\frac{e^{-a^{2}}a\sqrt{\pi}\textrm{erfi}\left(a\right)}{4}=\color{red}{\frac{aF\left(a\right)}{2}}$$</span> as wanted.</p>
|
3,137,160 | <blockquote>
<p>Determine which of the following are subspaces of <span class="math-container">$3 \times 3$</span> matrix <span class="math-container">$M$</span>
all <span class="math-container">$3 \times 3$</span> matrices <span class="math-container">$A$</span> such that the trace of <span class="math-container">$A$</span> is <span class="math-container">$\mbox{tr}(A) = 0$</span>.</p>
</blockquote>
<p>What does trace mean?</p>
| Shri | 442,962 | <p>Take <span class="math-container">$xH \in G/H$</span> of order p by Cauchy theorem then <span class="math-container">$x^p \in H$</span>.</p>
|
2,331,191 | <p>Use either direct proof, proof by contrapositive, or proof by contradiction.</p>
<p>Using proof by contradiction method</p>
<blockquote>
<p>Assume n is a perfect square and n+3 is a perfect square (proof by
contradiction)</p>
<p>There exists integers x and y such that <span class="math-container">$n = x^2$</span> and <span class="math-container">$n+3 = y^2$</span></p>
<p>Then <span class="math-container">$x^2 + 3 = y^2$</span></p>
<p>Then <span class="math-container">$3 = y^2 - x^2$</span></p>
<p>Then <span class="math-container">$3 = (y-x)(y+x)$</span></p>
<p>Then <span class="math-container">$y+x = 3$</span> and <span class="math-container">$y-x=1$</span></p>
<p>Then <span class="math-container">$x = 1, y = 2$</span></p>
<p>Since <span class="math-container">$x = 1$</span>, that implies <span class="math-container">$n = 1$</span></p>
<p><em><strong>this is how far I got</strong></em></p>
</blockquote>
<p>Anyone know what I should do now?</p>
| Paolo Leonetti | 45,736 | <p>$$
3=y^2-x^2 \ge (x+1)^2-x^2= 2x+1
$$
which is false if $x>1$.</p>
|
461 | <p>There is a function on $\mathbb{Z}/2\mathbb{Z}$-cohomology called <em>Steenrod squaring</em>: $Sq^i:H^k(X,\mathbb{Z}/2\mathbb{Z}) \to H^{k+i}(X,\mathbb{Z}/2\mathbb{Z})$. (Coefficient group suppressed from here on out.) Its notable axiom (besides things like naturality), and the reason for its name, is that if $a\in H^k(X)$, then $Sq^k(a)=a \cup a \in H^{2k}(X)$ (this is the cup product). A particularly interesting application which I've come across is that, for a vector bundle $E$, the $i^{th}$ Stiefel-Whitney class is given by $w_i(E)=\phi^{-1} \circ Sq^i \circ \phi(1)$, where $\phi$ is the Thom isomorphism.</p>
<p>I haven't found much more than an axiomatic characterization for these squaring maps, and I'm having trouble getting a real grip on what they're doing. I've been told that $Sq^1$ corresponds to the "Bockstein homomorphism" of the exact sequence $0 \to \mathbb{Z}/2\mathbb{Z} \to \mathbb{Z}/4\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0$. Explicitly, if we denote by $C$ the chain group of the space $X$, we apply the exact covariant functor $Hom(C,-)$ to this short exact sequence, take cohomology, then the connecting homomorphisms $H^i(X)\to H^i(X)$ are exactly $Sq^1$. This is nice, but still somewhat mysterious to me. Does anyone have any good ideas or references for how to think about these maps?</p>
| Eric Wofsey | 75 | <p>Here's one way to understand them. The external cup square $a \otimes a \in H^{2n}(X \times X)$ of $a \in H^n(X)$ induces a map $f:X \times X \to K(Z_2, 2n)$. It can be show that this map factors through a map $g:(X \times X) \times_{Z_2} EZ_2 \to K(2n)$, where $Z_2$ acts on the product by permuting the factors and $EZ_2$ can be taken to just be $S^\infty$. If you unravel what this means, it says that our original map $f$ was homotopic to the map obtained by first switching the coordinates and then applying $f$. It also says that this homotopy, when applied twice to get a homotopy from $f$ to itself, is homotopic to the identity homotopy, and we similarly have a whole series higher "coherence" homotopies. Now $X \times BZ_2$ maps to $(X \times X) \times_{Z_2} EZ_2$ as the diagonal, so we get a map $X \times BZ2 \to K(2n)$. But $BZ_2$'s cohomology is just $Z_2[t]$, so this gives a cohomology class $Sq(a) \in H^*(X)[t]$ of degree $2n$. If we write $Sq(a)=\sum s(i) t^i$, it can be shown that $s(i)=Sq^{n-i}a$.</p>
<p>What does this mean? Well, if our map $f$ actually <em>was</em> invariant under switching the factors (which you might think it ought to be, given that it appears to be defined symmetrically in the two factors), we could take $g$ to be just the projection onto $X \times X$ followed by $f$. This would mean that $Sq(a)$ comes from just projecting away the $BZ_2$ and then using $a^2$, i.e. $Sq^n(a)=a^2$ and $Sq^i(a)=0$ for all other $i$. Thus the nonvanishing of the lower Steenrod squares somehow measures how the cup product, while <em>homotopy</em>-commutative (in terms of the induced maps to Eilenberg-MacLane spaces), cannot be straightened to be actually commutative. Indeed, in the universal example $X=K(Z_2,n)$, the map $f$ is exactly the universal map representing the cup product of two cohomology classes of degree $n$.</p>
<p><strike>Some somewhat terse notes on this can be found <a href="http://my.harvard.edu/icb/icb.do?keyword=k54584&pageid=icb.page237558" rel="noreferrer">here</a>; see particularly part III.</strike> (Sorry, the link is now dead.)</p>
|
461 | <p>There is a function on $\mathbb{Z}/2\mathbb{Z}$-cohomology called <em>Steenrod squaring</em>: $Sq^i:H^k(X,\mathbb{Z}/2\mathbb{Z}) \to H^{k+i}(X,\mathbb{Z}/2\mathbb{Z})$. (Coefficient group suppressed from here on out.) Its notable axiom (besides things like naturality), and the reason for its name, is that if $a\in H^k(X)$, then $Sq^k(a)=a \cup a \in H^{2k}(X)$ (this is the cup product). A particularly interesting application which I've come across is that, for a vector bundle $E$, the $i^{th}$ Stiefel-Whitney class is given by $w_i(E)=\phi^{-1} \circ Sq^i \circ \phi(1)$, where $\phi$ is the Thom isomorphism.</p>
<p>I haven't found much more than an axiomatic characterization for these squaring maps, and I'm having trouble getting a real grip on what they're doing. I've been told that $Sq^1$ corresponds to the "Bockstein homomorphism" of the exact sequence $0 \to \mathbb{Z}/2\mathbb{Z} \to \mathbb{Z}/4\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0$. Explicitly, if we denote by $C$ the chain group of the space $X$, we apply the exact covariant functor $Hom(C,-)$ to this short exact sequence, take cohomology, then the connecting homomorphisms $H^i(X)\to H^i(X)$ are exactly $Sq^1$. This is nice, but still somewhat mysterious to me. Does anyone have any good ideas or references for how to think about these maps?</p>
| Andrew Stacey | 45 | <p>The Steenrod square is an example of a cohomology operation. Cohomology operations are natural transformations from the cohomology functor to itself. There are a few different types, but the most general is an <strong>unstable</strong> cohomology operation. This is simply a natural transformation from $E^k(-)$ to $E^l(-)$ for some fixed $k$ and $l$. Here, one regards the graded cohomology functors as a family of set-valued functors so the functions induced by these unstable operations do not necessarily respect any of the structure of $E^k(X)$.</p>
<p>Some do, however. In particular, there are <strong>additive</strong> cohomology operations. These are unstable operations which are homomorphisms of abelian groups.</p>
<p>In particular, for any multiplicative cohomology theory (in particular, ordinary cohomology or ordinary cohomology with $\mathbb{Z}/2\mathbb{Z}$ coefficients) there are the <em>power</em> operations: $x \to x^k$. These are additive if the coefficient ring has the right characteristic. In particular, squaring is additive in $\mathbb{Z}/2\mathbb{Z}$ cohomology.</p>
<p>Given an unstable cohomology operation $r: E^k(-) \to E^l(-)$ there is a way to manufacture a new operation $\Omega r: \tilde{E}^{k-1} \to \tilde{E}^{l-1}(-)$ using the suspension isomorphism (where the tilde denotes that these are reduced groups):</p>
<p>$E^{k-1}(X) \cong E^k(\Sigma X) \to E^l(\Sigma X) \cong E^{l-1}(X)$</p>
<p>This is quite straightforward and is a cheap way of producing more operations. When applied to the power operations it produces almost nothing since the ring structure on the cohomology of a suspension is trivial: apart from the inclusion of the coefficient ring all products are zero.</p>
<p>What is an interesting question is whether or not this looping can be reversed. Namely, if $r$ is an unstable operation, when is there another operation $s$ such that $\Omega s = r$? And how many such are there? Most interesting is the question of when there is an infinite chain of operations, $(r_k)$ such that $\Omega r_k = r_{k-1}$. When this happens, we say that $r$ comes from a <strong>stable</strong> operation (there is a slight ambiguity here as to when the sequence $(r_k)$ <em>is</em> a stable operation or merely comes from a stable operation).</p>
<p>One necessary condition is that $r$ be additive. This is not, in general, sufficient. For example, the Adams operations in $K$-theory are additive but all but two are not stable.</p>
<p>However, for ordinary cohomology with coefficients in a field, additive is sufficient for an operation to come from a stable operation. Moreover, there is a unique sequence for each additive operation. This means that the squaring operation in $\mathbb{Z}/2\mathbb{Z}$ cohomology has a sequence of "higher" operations which loop down to squaring. These are the Steenrod squares.</p>
<p>The sequence stops with the actual squaring (rather, becomes zero after that point) because, as remarked above, the power operations loop to zero.</p>
<p>One important feature of these operations is that they give necessary conditions for a spectrum to be a suspension spectrum of a space. If a spectrum is such a suspension spectrum then its $\mathbb{Z}/2\mathbb{Z}$-cohomology must be a ring. That's not enough, however, it must also have the property that, in the right dimensions, the Steenrod operations act by squaring. (Of course, this is necessary but not sufficient.)</p>
|
1,643,579 | <p>I've an homework problem that i'm unable to find the right answer.</p>
<p>The problem is:</p>
<p>The line $tx + sy = 2$ goes through point $(2,1)$ and is parallel to line $y = 8 -3x$, find the value of $t^2 + s^2$. </p>
<p>$ A. {32\over49}$ $B.{18\over49}$ $C.{36\over49}$ $D.{25\over49} $ $E.{40\over49} $</p>
<p>I was able to find a parallel line at $ y = -3x + 7 $ but i'm unable to find any of the possible answers to be right.</p>
| ncmathsadist | 4,154 | <p>Use L'Hospital and the fact that
$$- \left(\prod \cos(kx)\right)' =
\sum_{k=1}^n k\sin(kx)\prod_{j\not=k} \cos(jx).$$</p>
|
602,852 | <p>Following Halmos's Naive Set Theory, Russell's Paradox emerges from using the axiom of specification (that for every set $A$ and property $\phi$ there exists a set $Y$ whose elements are those elements $x$ of $A$ for which this property holds) with the property $x \notin x$.</p>
<p>We find then that for any set $A$, there is a set $B$ which is not in $A$. How does this - that everything contains something outside it - entail that 'nothing contains everything', as Halmos says? And further, that there is therefore no 'universe of discourse in set theory'? Does this come as a result of the unrestricted application of the axiom of specification, or in spite of it?</p>
| MJD | 25,554 | <blockquote>
<p>We find then that for any set A, there is a set B which is not in A. How does this - that everything contains something outside it - entail that 'nothing contains everything',</p>
</blockquote>
<p>Suppose there were a set $U$ that we thought <em>did</em> contain absolutely everything. But we showed, as you said, that for any set $A$, there is a set $B_A$ which is not in $A$. We can apply this construction to $U$ and obtain a set $B_U$ that is not in $U$. But then $U$ does <em>not</em> contain everything, since it does not contain $B_U$. </p>
<p>Halmos's claim that there is no "universe of discourse in set theory" is a little too strong; it depends on what kind of set theory one chooses. The claim is true, however, for ZF set theory, which is the set theory most commonly used in foundational mathematics.</p>
<blockquote>
<p>Does this come as a result of the unrestricted application of the axiom of specification, or in spite of it?</p>
</blockquote>
<p>I'm not sure here what you're asking. Truly unrestricted comprehension is logically incoherent, as the Russell paradox shows, because unrestricted comprehension says that for any property $\phi$ there is some set $S_\phi$ that consists of precisely the objects for which $\phi$ is true. But the Russell paradox shows that this is simply false; there are properties for which there is no such set, the simplest such being the property of not belonging to oneself. If $\rho(x)$ is the property $x\notin x$, then there simply is no set of $S_\rho$ of all $x$ for which $\rho$ holds.</p>
<p>Since in mathematics we usually try to avoid choosing axioms that are false, the unrestricted comprehension axiom won't do, and we make do instead with one of several restricted form of set comprehension.</p>
|
602,852 | <p>Following Halmos's Naive Set Theory, Russell's Paradox emerges from using the axiom of specification (that for every set $A$ and property $\phi$ there exists a set $Y$ whose elements are those elements $x$ of $A$ for which this property holds) with the property $x \notin x$.</p>
<p>We find then that for any set $A$, there is a set $B$ which is not in $A$. How does this - that everything contains something outside it - entail that 'nothing contains everything', as Halmos says? And further, that there is therefore no 'universe of discourse in set theory'? Does this come as a result of the unrestricted application of the axiom of specification, or in spite of it?</p>
| Dan Christensen | 3,515 | <p>You may be confusing Russell's paradox with the paradox of the universal set. </p>
<p>From the Russell's paradox, we obtain:</p>
<p>$\neg\exists r: \forall x: [x\in r \iff x\notin x]$ </p>
<p>We used this result to resolve the paradox of the universal set, obtaining:</p>
<p>$\neg\exists U: \forall x: x\in U$</p>
<p>Switching quantifiers, we have: $\forall U:\exists x: x\notin U$</p>
|
3,815,640 | <p>what is the most efficient way to calculate the argument of
<span class="math-container">$$
\frac{e^{i5\pi/6}-e^{-i\pi/3}}{e^{i\pi/2}-e^{-i\pi/3}}
$$</span> without calculator ?</p>
<p>i tried to use <span class="math-container">$\arg z_1-\arg z_2$</span> but the argument of <span class="math-container">$e^{i5\pi/6}-e^{-i\pi/3}$</span> take some time .
is there a formula to calculate the argument of that kind of complex numbers ?</p>
| Stinking Bishop | 700,480 | <p><strong>Hint</strong>: Sketch the points <span class="math-container">$A=e^{i5\pi/6}$</span>, <span class="math-container">$B=e^{-i\pi/3}$</span> and <span class="math-container">$C=e^{i\pi/2}$</span> on the unit circle, and use geometrical arguments to calculate the angle between <span class="math-container">$\vec{BA}$</span> and <span class="math-container">$\vec{BC}$</span>.</p>
<p>Should boil down to a very simple use of the Central Angle theorem.</p>
|
4,458,863 | <p>Let <span class="math-container">$z_1,\;z_2,\;z_3\;$</span> be complex number such that <span class="math-container">$|z_1|=|z_2|=|z_3|=|z_1+z_2+z_3|=2\;\;$</span>. If <span class="math-container">$|z_1-z_3|=|z_1-z_2|\; \;$</span> and <span class="math-container">$z_2 \neq z_3.\; \; $</span> Then Find value of <span class="math-container">$|z_1+z_2||z_1+z_3|$</span>.</p>
<p><strong>My Thinking:</strong></p>
<p>All I could think is that <span class="math-container">$z_1, z_2, z_3 \;$</span> lies on a circle of radius <span class="math-container">$2$</span> with origin as center. Can anyone help me in how to process further.</p>
| Dan | 1,374 | <p><a href="https://www.wolframalpha.com/input?i=integral+of+sqrt%28cosh%28x%29%29" rel="nofollow noreferrer">WolframAlpha</a> provides links to <a href="https://reference.wolfram.com/language/ref/EllipticE.html" rel="nofollow noreferrer">EllipticE</a> and <a href="https://reference.wolfram.com/language/ref/Hypergeometric2F1.html" rel="nofollow noreferrer">Hypergeometric2F1</a>. Not sure how helpful they are for your purpose.</p>
<p>I don't think there's any “nice” way to express this integral. I'd recommend evaluating it numerically.</p>
<p>Here's a Taylor series for <span class="math-container">$x$</span> near 0:</p>
<p><span class="math-container">$$x + \frac{x^3}{12} - \frac{x^5}{480} + \frac{19 x^7}{40320} - \frac{559 x^9}{5806080} + \frac{2651 x^{11}}{116121600} - \frac{2368081x^{13}}{398529331200} + O(x^{14})$$</span></p>
<p>For larger <span class="math-container">$x$</span>, the <span class="math-container">$e^x$</span> term will dominate the <span class="math-container">$e^{-x}$</span> term, and you can just approximate the integral as <span class="math-container">$\sqrt{2}e^{x/2}$</span>.</p>
|
2,762,428 | <p>Excuse me, please, for the initial post, I did not know that someone would apprehend this rudely, and also excuse my English ...<br>
Task:
Find the equivalent to a function, when $ t \to +\infty $<br>
$$ f(t) = \int \limits_{t}^{2t} \frac{x^2}{e^{x^2}} dx $$</p>
<p>My ideas:<br>
1) I tried to find such a function to integrate by parts, I would get some function in a closed form and an asymptotically small function.<br>
The problem is that it is not so easy to find it by choosing it, but in general terms it is not clear how to do it.<br>
2) One can try to use the mean value theorem if we find an equivalent function for the integrand<br>
The problem is that there seems to be no equivalent ...<br>
3) You can try to guess the answer if you use the rule of l'Hospital for the original function and function, which we do not yet know.
The problem is that in an adequate form I could not do it.</p>
<p>P.S. I came up with this task myself based on some examples given to me at the lecture. I really do not understand why people do not like my questions ...</p>
| Somos | 438,089 | <p>You want the <a href="https://oeis.org/A007153" rel="nofollow noreferrer">Dedekind numbers (OEIS sequence A007153)</a>. The first few terms of the sequence: $(1, 4, 18, 166, 7579,\dots).$ The key is in the title "number of monotone Boolean functions". This is equivalent to being constructible from AND and OR. Read the sequence entry for more details and examples.</p>
<p>The part of the question about MAX and MIN makes sense in the context of a <a href="https://en.wikipedia.org/wiki/Lattice_(order)" rel="nofollow noreferrer">Lattice</a> in the sense of an ordered structure. The Wikipedia article has some details. In particular, the two operations are usually denoted by $\;\vee\;$ and $\;\wedge\;$ and sometimes called "join" and "meet". The Boolean lattice is equivalent to the lattice of subsets of a fixed set using union and intersection as lattice operators.</p>
|
2,762,428 | <p>Excuse me, please, for the initial post, I did not know that someone would apprehend this rudely, and also excuse my English ...<br>
Task:
Find the equivalent to a function, when $ t \to +\infty $<br>
$$ f(t) = \int \limits_{t}^{2t} \frac{x^2}{e^{x^2}} dx $$</p>
<p>My ideas:<br>
1) I tried to find such a function to integrate by parts, I would get some function in a closed form and an asymptotically small function.<br>
The problem is that it is not so easy to find it by choosing it, but in general terms it is not clear how to do it.<br>
2) One can try to use the mean value theorem if we find an equivalent function for the integrand<br>
The problem is that there seems to be no equivalent ...<br>
3) You can try to guess the answer if you use the rule of l'Hospital for the original function and function, which we do not yet know.
The problem is that in an adequate form I could not do it.</p>
<p>P.S. I came up with this task myself based on some examples given to me at the lecture. I really do not understand why people do not like my questions ...</p>
| Bram28 | 256,001 | <p>Any such formula can be put into DNF by repeated distribution of AND over OR. And, since by Absorption we have that $p + pq=p$, this means that we just need to find the number of ways we can form a collection of terms where no term is part of another term. </p>
<p>For example, for $n=3$ (Say we have $p$, $q$, and $r$) we can create the following collections of terms:</p>
<p>$p$</p>
<p>$q$</p>
<p>$r$</p>
<p>$p+q$</p>
<p>$p+r$</p>
<p>$q+r$</p>
<p>$p+q+r$</p>
<p>$p+qr$</p>
<p>$q+pr$</p>
<p>$r+pq$</p>
<p>$pq$</p>
<p>$pr$</p>
<p>$qr$</p>
<p>$pq+pr$</p>
<p>$pq+qr$</p>
<p>$pr+qr$</p>
<p>$pq+pr+qr$</p>
<p>$pqr$</p>
<p>As such, the problem is equivalent to finding the number of sets of subsets of a set of $n$ elements such that no two elements of any such set of subsets (i.e. no two subsets of the original set) are a subset each other.</p>
|
802,014 | <p>For all sets $A$, $B$, $C$, if $A$ is subset of $B$, $B$ is subset of $C$, and $C$ is subset of $A$, then $A = B = C$.</p>
<p>This is a true statement and I need to provide a proof? Thus, when a statement is false I need to provide it with counterexample whereas if it is true then it has to be provided by a proof?</p>
| Ittay Weiss | 30,953 | <p>Yes, the claim you state is correct, and you thus need to prove it. </p>
<p>In general whatever you claim you need to supply argument for. Whether you call it a proof or a counterexample depends on the situation. To provide a counterexample is to prove that a claim is wrong by exhibiting that it is wrong for a particular case. To provide a proof is to give a sequence of arguments that establish the statement is correct. If you think logically about what you need to do, then you won't get confused with such irrelevant issues. </p>
<p>For instance, if you want to prove that it is not true that for all real numbers $x$: $x^2> 0$, then you can do it by providing the counterexample $x=0$, since then $x^2=0$, not bigger than $0$. It's a counter example since you showed the claim is incorrect by exhibiting a particular case. </p>
<p>On the other hand, if you want to prove that for all real numbers $x$: $x^2\ge 0$, then you can not do it by observing that for $x=1$ you have $x^2=1>0$, since that still leaves infinitely many other values for you to verify the claim for. Thus, you need to supply a general proof in this case.</p>
<p>I hope this helps. </p>
|
1,449,776 | <p>I have always known that $a^n=a*a*a*.....$(n times)</p>
<p>Then what exactly is the meaning if $a^0$ and why will it be equal to $1$?</p>
<p>I have checked it in the internet but everywhere the solution is based on the principle that $a^m*a^n=a^{m+n}$ and when $n=0$ it will be $a^m$ and clearly $a^0$ is equal to $1$. </p>
<p>But what exactly does $a^0$ mean does it mean $a*a*a*...$(zero times)?</p>
<p>Any help is highly appreciated.</p>
| Brian Tung | 224,454 | <p>It's defined that way (except, usually, for $a = 0$) because it's most consistent to do so. The <em>empty product</em> is defined to be $1$ because $1$ is the multiplicative identity, much as the empty sum is defined to be $0$ because $0$ is the additive identity. Both of these definitions allow for the pattern that arises from successive multiplication to be "extended backward" to a product (or sum) with zero terms.</p>
<p>The extension of exponents to zero, to negative numbers, to the rationals, to the reals, and to complex numbers, in each case continues a pattern identified in the previous, "smaller" domain. Those patterns are useful; that is why they are defined that way. Nothing stops you from defining them differently—nothing, that is, except that they generally are less useful that way.</p>
|
4,072,386 | <p>I assume this is a simple proof but i'm stuck here.</p>
<p>I need to prove that if <span class="math-container">$A^3B-B$</span> is invertible then <span class="math-container">$BA-B$</span> is invertible.</p>
<p>So <span class="math-container">$A^3B-B=(A^3-I)B$</span> and then both <span class="math-container">$(A^3-I)$</span> and <span class="math-container">$B$</span> are invertible.</p>
<p>So I need to show that <span class="math-container">$BA-B=B(A-I)$</span> are both invertible.</p>
<p>I thought maybe I could somehow show that if <span class="math-container">$(A^3-I)$</span> is invertible then <span class="math-container">$(A-I)$</span> is invertible but I guess there's a simpler way to get there. What am I missing?</p>
<p>Thanks!</p>
| Igor Rivin | 109,865 | <p>Hint <span class="math-container">$$A^3-I = (A-I)(A^2 + A + I).$$</span></p>
|
1,407,700 | <p>I am stuck on solving the following systems of equations with 3 variables. The textbook asks to use the addition method so can we please stick to that.</p>
<p>${5x -y = 3}$</p>
<p>${3x + z = 11}$</p>
<p>${y - 2z = -3}$</p>
<p>I am used to systems of equations where each equation has at least one instance of the variable e.g. ${x + y + z = 1}$ but in each of the above, one of the variables is omitted in each equation.</p>
<p>Could somebody explain what to do in this situation? Should I multiply both sides by one of the variables to balance it up? </p>
| Riemann | 27,899 | <p>Take <span class="math-container">$$s_n=n+\frac{1}{n},\quad t_n=n,$$</span>
then <span class="math-container">$\lim_{n\to\infty}(s_n-t_n)=0$</span>, but
<span class="math-container">$$\lim_{n\to\infty}(f(s_n)-f(t_n))
=\lim_{n\to\infty}\left[\left(n+\frac{1}{n}\right)^3\sin\frac{1}{n+\frac{1}{n}}
-n^3\sin\frac1n\right]=2.$$</span>
This implies <span class="math-container">$f$</span> is not uniform continuous on <span class="math-container">$(0,\infty)$</span>.</p>
|
1,217,557 | <p>I was tasked with drawing the contour lines of $ z = \sqrt{xy} $, which I find a bit problematic since I can see no way in which one can plot (by hand, and not with wolfram and others....) the $ z = \sqrt{xy} $ graph in $R^2( x-, y- $ projection} to begin with for this surface...</p>
<p>How can one draw this contour graph manually? </p>
| Jam | 131,857 | <p>You are right .
$$L(x)$$
is not linear try not to confuse it because its graph is a line. IF one of the linearity properties do not apply then it is not.IN general
$$F(x)=ax+b$$
is linear iff b=0</p>
|
1,402,953 | <ol>
<li>How can one intuitively understand the definition of a bilinear map? Is there some way of looking at it geometrically? I found the following definition:</li>
</ol>
<p>Let $\mathit{A}$,$\mathit{B}$,$\mathit{C}$ be vector spaces. A map $f:\mathit{A}\times \mathit{B}\to C$ is said to be bilinear if for each fixed element $b\in \mathit{B}$, $f(.,b):\mathit{A}\to\mathit{C}$ is a linear map. Similarly, for each fixed element of $\mathit{A}$.</p>
<p>Matrix multiplication is an example of a bilinear map.Following my definition, I can prove that it is a bilinear map, but I don't understand the intuitive idea behind it. In my opinion, it is simply a linear map with one element fixed.</p>
<ol start="2">
<li>Is there some formal definition of a bilinear algorithm? I could find an explanation for it only in the context of matrix multiplication: <a href="http://www.issac-symposium.org/2014/tutorials/ISSAC2014_tutorial_handout_LeGall.pdf" rel="nofollow">http://www.issac-symposium.org/2014/tutorials/ISSAC2014_tutorial_handout_LeGall.pdf</a></li>
</ol>
<p>Kindly help me with these questions.
Thanks!</p>
| Sempliner | 122,727 | <p>I think there are many explanations of what it is, but as far as intuition is concerned I think the following is by far the most satisfying. We see bilinear maps as generalizations of the properties of a product, for instance if $K$ is a field or even a ring then the map $\times: K \times K \to K$ is a bilinear map, and this is the starting point, in a sense, of a very robust class of bilinear maps that occur in mathematics. The rules that define the tensor product (the unique vector space where its linear maps are the same as bilinear maps on the corresponding product) are all just simple properties of this pairing.</p>
<p>Edit: I was trying to point out that a motivating example for bilinear maps is to look at the multiplication map $\times$, given any ring, the distributive property means that the map $(a, b) \to ab$ gives a bilinear map from $R \times R \to R$. This has always struck me as the symbological inspiration for writing a tensor product of two modules as $V \otimes W$ to show the relation to the product. Further all of the axiomatic properties that describe $V \otimes W$ as a quotient of $V \times W$ can be interpreted as the relation that a product in a non-commutative ring would satisfy, complete with some properties of multiplication by scalars to make the tensor product an $R$-module.</p>
<p>This makes it clear that the tensor product of vector spaces or modules is intended to be some kind of "vector-space generalization" of the multiplication of numbers, which is made clear by its dimension formula. By the same token the direct sum is supposed to be a generalization of addition in some sense.</p>
|
2,852,550 | <p>I was wondering, Is there Cauchy sequence that are not bounded ? Of course, in complete spaces is not possible. I have a theorem that says that if $(A,d)$ is not complete, then $(\bar A,d)$ is complete. But are there spaces s.t. indeed for all $\varepsilon>0$, there is $N$ s.t. $d(x_n,x_{m})<\varepsilon$ for all $n\geq N$, and all $r\geq 0$, but the ball also move to $+\infty $ ? </p>
| Hagen von Eitzen | 39,174 | <p>Note that $d(x_n,x_N)<\epsilon$ for all $n\ge N$ and that only finitely many $n<N$ are to be considered beyond that.</p>
|
374,881 | <p>I'd like to know how I can recursively (iteratively) compute variance, so that I may calculate the standard deviation of a very large dataset in javascript. The input is a sorted array of positive integers.</p>
| DolphinDream | 80,405 | <p>Here are the iterative formulas (with derivations) for the <strong>population</strong> (<span class="math-container">$N$</span> normalized) and <strong>sample</strong> (<span class="math-container">$N-1$</span> normalized) standard deviations, which express the <span class="math-container">$\sigma_{n+1}$</span> (<span class="math-container">$s_{n+1}$</span> for sample) for the <span class="math-container">$n+1$</span> value set in terms of <span class="math-container">$\sigma_{n}$</span> (<span class="math-container">$s_{n}$</span> for sample), <span class="math-container">$\bar x_{n}$</span> of the <span class="math-container">$n$</span> value set plus the new value <span class="math-container">$x_{n+1}$</span> added to the set.</p>
<p>Essentially we need to find:</p>
<p><span class="math-container">$$\bar x_{n+1} = f(n, \bar x_n, x_{n+1})$$</span>
and
<span class="math-container">$$\sigma_{n+1} = g(n, \sigma_n, \bar x_n, x_{n+1})$$</span></p>
<p><strong>Derivation for the Average</strong></p>
<p>For both cases, the average for <span class="math-container">$n\geqslant1$</span> is,</p>
<p>for <span class="math-container">$n$</span> values:
<span class="math-container">$$
\bar x_n=\frac1n\sum_{k=1}^nx_k
$$</span></p>
<p>for <span class="math-container">$n+1$</span> values:
<span class="math-container">$$
\bar x_{n+1}=\frac1{n+1}\sum_{k=1}^{n+1}x_k = \frac1{n+1}(n\bar x_n + x_{n+1}) \leftarrow f(n, \bar x_n, x_{n+1})
$$</span></p>
<p><strong>Derivation for the Standard Deviation</strong></p>
<p>The standard deviation formulas for <strong>population</strong> and <strong>sample</strong> are:</p>
<p><span class="math-container">\begin{aligned}
\sigma_{n} &= \sqrt {\frac1{n} \sum_{k=1}^{n}(x_k - \bar x_{n})^2 } && \textit{for} \textbf{ population } \textit{Standard Deviation}\\
\\
s_{n} &= \sqrt {\frac1{n-1} \sum_{k=1}^{n}(x_k - \bar x_{n})^2 } && \textit{for} \textbf{ sample } \textit{Standard Deviation } \\
\end{aligned}</span></p>
<p>To consolidate the derivations for both <strong>population</strong> and <strong>sample</strong> formulas we'll write the standard deviation using a generic factor <span class="math-container">$\alpha_{n}$</span> and replace it at the end to get the population and sample formulas.</p>
<p>with:</p>
<p><span class="math-container">\begin{equation}
\alpha_{n} =
\begin{cases}
n & \textit{for} \textbf{ population } \textit{Standard Deviation} \\
n-1 & \textit{for} \textbf{ sample } \textit{Standard Deviation } \\
\end{cases}
\end{equation}</span></p>
<p>the equation for the standard deviation for the <span class="math-container">$n$</span> values can be written as:</p>
<p><span class="math-container">\begin{equation}
\tag{1}
\begin{aligned}
\alpha_{n}\sigma^2_{n}
& = \sum_{k=1}^{n}(x_k - \bar x_{n})^2 \\
& = \sum_{k=1}^{n}\big(x_k^2 - 2 x_k \bar x_n + (\bar x_{n})^2\big) \\
& = \sum_{k=1}^{n}x_k^2 - 2 \bar x_n \sum_{k=1}^{n}x_{k} + (\bar x_{n})^2\sum_{k=1}^{n}1 \\
& = \sum_{k=1}^{n}x_k^2 - 2 \bar x_n n \bar x_n + n (\bar x_{n})^2 \\
& = \sum_{k=1}^{n}x_k^2 - n (\bar x_{n})^2
\end{aligned}
\end{equation}</span></p>
<p>Thus, the same equation for <span class="math-container">$n+1$</span> values is:</p>
<p><span class="math-container">\begin{equation}
\begin{aligned}
\alpha_{n+1}\sigma^2_{n+1}
& = \sum_{k=1}^{n+1}(x_k-\bar x_{n+1})^2 \\
& = \sum_{k=1}^{n+1}x_k^2 - (n+1)(\bar x_{n+1})^2 \\
& = \sum_{k=1}^{n}x_k^2 + (x_{n+1})^2 - (n+1)(\bar x_{n+1})^2 \\
& = \sum_{k=1}^{n}x_k^2 + (x_{n+1})^2 - (n+1) \big(\frac1{n+1}(n\bar x_{n} + x_{n+1}) \big)^2 \\
& = \sum_{k=1}^{n}x_k^2 + (x_{n+1})^2 - \frac1{n+1} \big(n^2(\bar x_{n})^2 + 2 n \bar x_{n} x_{n+1} + (x_{n+1})^2 \big) \\
\end{aligned}
\end{equation}</span></p>
<p>from the equation <span class="math-container">$(1)$</span> we substitute <span class="math-container">$\sum_{k=1}^{n}x_k^2$</span> with <span class="math-container">$\alpha_{n}\sigma^2_{n} + n (\bar x_{n})^2$</span> and get:</p>
<p><span class="math-container">\begin{equation}
\begin{aligned}
\alpha_{n+1}\sigma^2_{n+1}
& = \alpha_{n}\sigma^2_{n} + n (\bar x_{n})^2 + (x_{n+1})^2 - \frac1{n+1} \big(n^2(\bar x_{n})^2 + 2 n \bar x_{n} x_{n+1} + (x_{n+1})^2 \big) \\
\end{aligned}
\end{equation}</span></p>
<p>arranging the terms and simplifying we get:</p>
<p><span class="math-container">$$
\sigma_{n+1}
= \sqrt { \Big( \sigma^2_{n} + \frac{n}{n+1} \frac1{\alpha_n} (\bar x_n - x_{n+1})^2 \Big) \frac{\alpha_{n}}{\alpha_{n+1}} }
\leftarrow g(n, \sigma_n, \bar x_n, x_{n+1})
$$</span></p>
<p>Replacing the <span class="math-container">$\alpha$</span> values, the specific iterative formulas for <strong>population</strong> and <strong>sample</strong> standard deviations are:</p>
<p><span class="math-container">\begin{equation}
\begin{aligned}
\sigma_{n+1} &= \sqrt{ \Big( \sigma^2_{n} + \frac{1}{n+1}(\bar x_n - x_{n+1})^2 \Big) \frac{n}{n+1} } &&\textit{population STD} \\
\\
s_{n+1} &= \sqrt{ \Big( s^2_{n} + \frac{n}{n^2-1}(\bar x_n - x_{n+1})^2 \Big) \frac{n-1}{n} } &&\textit{sample STD} \\
\end{aligned}
\end{equation}</span></p>
|
19,996 | <p>In 1556, Tartaglia claimed that the sums<br>
1 + 2 + 4<br>
1 + 2 + 4 + 8<br>
1 + 2 + 4 + 8 + 16<br>
are alternative prime and composite. Show that his conjecture is false. </p>
<p>With a simple counter example, $1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 = 255$, apparently it's false. However, I want to prove it in general case instead of using a specific counter example, but I got stuck :( !
I tried:<br>
The sum $\sum_{i=0}^n 2^i$ is equal to $2^{n+1} - 1$. I assumed that $2^{n+1} - 1$ is prime, then we must show that $2^{n+1} - 1 + 2^{n+1} = 2^{n+2} - 1$ is not composite. Or we assume $2^{n+1}$ is composite and we must show that $2^{n+2} - 1$ is not prime.
But I have no clue how $2^{n+2} - 1$ relates to its previous prime/composite. Any hint?</p>
| Brandon Carter | 1,016 | <p>There is simply no such correlation in general. If that were the case, Mersenne primes would not be as interesting to study as they are. You can not prove that $2^n-1$ composite $\Rightarrow 2^{n+1} - 1$ prime, because it isn't true. As Ross mentioned, $2^n-1$ can only be prime when $n$ is prime, so for any prime $p > 2$, if $2^p-1$ is prime, then $2^{p+1}-1$ must be composite, as $2 \mid (p+1)$.</p>
|
2,521,710 | <p>I am trying to do a proof for convergence. But I am stuck in my proof not getting any further... What is missing to finish that proof?</p>
<p>$$a_n = \frac{1}{(n+1)^2}$$
Show that: $$\lim_{n \to \infty}a_n=0$$</p>
<p>Let $e > 0$ and $\forall n \ge n_0 = \lceil \frac{1}{\sqrt{\epsilon}}\rceil+1 \in \mathbb Z^+:$</p>
<p>$\begin{align}
|a_n-0| &\equiv |a_n| \\
&\equiv | \frac{1}{(n+1)^2}|\\
&\equiv|\frac{1}{(n+1)} \cdot \frac{1}{(n+1)}| \\
&\equiv |\frac{1}{n+1}| \cdot |\frac{1}{n+1}| \\
&< |\frac{1}{n}| \cdot |\frac{1}{n}| \\
&(\text{because } n \in \mathbb N) \\
&= \frac{1}{n^2} < \epsilon \text{ (by definition of convergence) }
\end{align}$</p>
<p>$\begin{align}
n \ge n_0 &= \lceil \frac{1}{\sqrt{\epsilon}}\rceil +1 \\
&> \lceil \frac{1}{\sqrt{\epsilon}} \rceil \\
&\ge \frac{1}{\sqrt{\epsilon}}
\end{align}$</p>
<p>thus</p>
<p>$\begin{align}
n &> \frac{1}{\sqrt{\epsilon}} \\
&\equiv \frac{1}{n} > \sqrt{\epsilon} \\
&\equiv \frac{1}{n^2} > \epsilon
\end{align}$</p>
<p>But the line that should follow: </p>
<p>$\begin{align}
\frac{1}{n^2} &< \epsilon
\equiv \frac{1}{n^2} < \frac{1}{n^2}
\end{align}$</p>
<p>which is wrong.. </p>
| Community | -1 | <p>You are doing way too much here.</p>
<p>Let $\epsilon > 0$ and choose $n_0 \in \mathbb{N}$ such that $\frac{1}{n_0} < \epsilon$ (note that we used the archimedian property of the real numbers here!)</p>
<p>Then, for $n \geq n_0$, we have:</p>
<p>$$\left|\frac{1}{(n+1)²}\right| = \frac{1}{(n+1)²} \leq \frac{1}{n+1} \leq \frac{1}{n} \leq \frac{1}{n_0} < \epsilon$$</p>
<p>Hence, we have shown that $$\frac{1}{(n+1)²}\rightarrow 0$$</p>
<p>as desired.</p>
|
367,116 | <p>I have a question about vacuous true and it always make me confused. If I want to prove that the empty set is the subset of all the set A, the proof is as following:
if x is in empty set, then x is in A. since x is in empty set is always false,, so the conditional statement is always true~
my question is why x is in empty set is always false, what if x=unicorn, since unicorn is in empty set so x is in empty set is true, right?
the question goes to why I can not plug a non-exist thing( like unicorn)into the variable x?</p>
| Austin Mohr | 11,245 | <p>You want to prove something like the following:</p>
<p>If $x \in \emptyset$, then $x \in A$.</p>
<p>If you don't like the vacuous nature of the hypothesis, consider the contrapositive (which is logically equivalent to the original statement):</p>
<p>If $x \notin A$, then $x \notin \emptyset$.</p>
<p>Select any $x$ that does not belong to $A$. Surely it also does not belong to the empty set, since the empty set contains no elements at all. (Notice the vacuous hypothesis in the original statement corresponds to a tautological conclusion in the contrapositive.) Having proved the contrapositive, we can also conclude the original statement.</p>
|
1,179,843 | <p>Proving $\sum_{n=1}^\infty \frac{\xi ^n}{n}$ is not uniformly convergent for $\xi \in (0,1)$.</p>
<p>I am trying to do the above. I have attempted to show it is not a cauchy sequence by considering $||\frac{\xi ^n}{n} ||_{\sup}$ but no avail. Any help please</p>
| user2566092 | 87,313 | <p>Let $f_n$ be the $n$th partial sum function, and consider $|f_{2n} - f_n|_1$. This is equal to</p>
<p>$$H_{2n} - H_n$$</p>
<p>where $H_n = \sum_{k=1}^n 1/k $ is the $n$th harmonic series partial sum. As $n \to \infty,$ we have $H_n \to \ln n + \lambda$ for some constant $\lambda$ independent of $n$ (look up Euler constant if you haven't seen this before). Then</p>
<p>$$H_{2n} - H_n \simeq \ln 2n - \ln n = \ln 2$$.</p>
|
275,310 | <p>I am a bit confused. What is the difference between a linear and affine function? Any suggestions will be appreciated</p>
| William Balthes | 231,063 | <p>$(1)$<strong>Linear continuous Functional equations of the Form</strong>:
$$F(\alpha x +\delta y)= \alpha F(x) + \delta F(y)$$;</p>
<p>or rather 1- point homogeneity </p>
<p>$(1a)$ (often with Cauchy's equation as well), as in the above post.
However, sometimes Cauchy's equation is not needed in addition to $(1a)$ if one can get to it $(1a)$ (for all reals) directly, which is not generally the case.</p>
<p>$(2)$<strong>Affine of the form</strong>; </p>
<p><strong>Continuous Function Form</strong>: $$\forall (x,y)\in \mathbb{R}F((1-t) x + t y)= tF(y) +(1-t)F(y); t\in [0,1]$$
ie:'concave and convex' . </p>
<p>*Or is it $\forall (x,y)\in $ $\text{dom(F)}$ or $\forall (x,y)\in$[0,1]$?</p>
<p>For(2) <strong>Function Form</strong>: $$F(x)=Ax+B$$ </p>
<p>Where $A$ is an arbitrary constant</p>
<p>(1a)$$\forall x\in \mathbb{R},\forall \delta \in \mathbb{R}:F(\delta x)=\delta F(x)$$
(1/1a) Linear Function: $$F(x)=Ax$$(in this case $A=F(1)$)</p>
<p>One always (or nearly always) needs to derive Cauchy's equation beforehand, to derive $(1a)$ for all rational numbers. And then a weak regularity requirement, generally, given Cauchy's equation, to extend homogeneity to all reals ie to get to equation $(1a)$)</p>
<p>Sometimes on can extend it to the algebraic irrationals given the field auto-morph-ism equations, although they 'apparently' already specify the function and grant continuity I have some issues with that (but that another story; vis a vis- the trans-transcendental numbers).</p>
<p>In both cases, I may have accidentally may be restricted domain to $[0,1]$ but are in their continuous forms, so that in some sense they are now function rather than functional equations . Or rather in a functional form from which the function should be directly derivable; as opposed to 'the general (often continuous) solution to Cauchy equation or Jensen's equation etc). </p>
<p>Although that may not be quite correct.</p>
<p>I think that sometimes convex functions (presumably not convex and concave ones but I might be wrong) can have trouble with continuity at the end points . However, I presume they are ruled as degenerate or not possible in the current system, ie not Lebesgue measurable.</p>
<p>And often that is when one is speaking of midpoint convexity or Jensen's equation rather than their,already, or (possibly) allegedly already, continuous versions $(1)$ and $(2)$.</p>
<p>I am just being tentative about $(2)$ here. I am not disputing anything, I just want to be careful. </p>
<p>Generally unless it is not restricted to an interval (or real line version). </p>
<p>interval (there is presumably a real line generalization). Notice that delta is restricted here, one cannot solve for the origin directly. </p>
<p>I presume that $(2)$ is just the real line version $(2)$ of convexity and concavity perhaps. once the origin has been fixed.</p>
<p>In (2) One cannot solve for $F(0)=0$,or directly solve for $F(1)=a$; for $F(x)=ax$, but only that $F(0)=b$ for $F(x)=ax+b$.</p>
<p>Although I think that both $(1)$ and $(2)$ are the restricted versions to the unit range, but with $F(0)$ being directly incorporated into $(1)$. </p>
<p>As one can set $a=1$ and $b=1$,$x=0$, $y=0$ </p>
<p>To get $[F(0)=2F(0)]\,\Rightarrow\,F(0)=0$ in $(1)$</p>
<p>But one cannot do so in $(2)$. </p>
<p>As there is only one free parameter, so that one always gets $F(0)=F(0)$ </p>
<p>I suppose one gets that (I think/perhaps)</p>
<p>(2a)$$F(tA)= t(F(A)-F(0)] +F(0)$$</p>
<p>Where $$F(A)$$ is the F(max element of domain; when $A$ positive or increasing) or
$$F(A)=F(1)-F(0)$$ here.</p>
<p>I am not sure if $(2)$, in this form, $F$ is defined only the domain $[0,1]$ .</p>
<p>However, its unclear whether $F(x)=ax+b$ falls out/(is derivable) of $(2)$ As directly as $F(x)=Ax$ (is derivable)/falls out of $(1)$.</p>
<p>I am probably using the wrong version confined to that domain $[0,1]$ domain). </p>
|
1,530,874 | <p>Is there a case where a function $f$ that is not differentiable at $0$ and a function $g$ that is differentiable at $0$ where $f+g$ is differentiable at $0$?</p>
| Emilio Novati | 187,568 | <p>Hint:</p>
<p>let $f+g=h$, if $h$ is differentiable and $g$ is differentiable what can we say about $f=h-g$ ?</p>
|
77,089 | <p>Fix a field $k$. For a singular variety $X$, I understand that the Grothendieck group $K^0(X)$ of vector bundles on $X$ is not necessarily isomorphic to the Grothendieck group $K_0(X)$ of coherent sheaves on $X$. </p>
<p>I am curious to learn what is known about these two groups in one family of examples: $\mathbb P^n_{D}$, where $D$ is the dual numbers $D=k[\epsilon]/(\epsilon^2)$. </p>
<p>References would be especially appreciated, as I know very little about K-theory.</p>
| Steven Landsburg | 10,503 | <p>The identity map from $k$ to itself factors through $D$. Thus, if $K$ represents either $K^0$ or $K_0$, $K(\mathbb P^n_{k})$ is a direct summand of $K(\mathbb P^n_{D})$.</p>
<p>For $M$ a coherent $\mathbb P^n_{D}$-module, we have an exact sequence</p>
<p>$$0\rightarrow \epsilon M \rightarrow M \rightarrow M/\epsilon M\rightarrow 0$$</p>
<p>The classes of the modules on the left and right, and therefore the class of the module in the middle, all come from the $K_0(\mathbb P^n_{k})$ direct summand. This shows that
$K_0(\mathbb P^n_{D})=K_0(\mathbb P^n_{k})$.</p>
<p>For $K^0$, you can apply Nakayama's Lemma.</p>
|
935,707 | <p>I'm having trouble determining this problem.</p>
<p>I need to find the integers in the set {1, ... , 100} that are divisible by 2 or 3 but not both.</p>
<p>The way I tried to approach it was:</p>
<p>If a number is divisible by both 2 and 3 then we can say it is divisible by 6. So we need to exclude integers divisible by 6. From here am I supposed to just go through each integer? Or is there a better way to approach this?</p>
<p>Thanks</p>
| Adriano | 76,987 | <p><strong>Hint:</strong> In the set $\{1,\cdots, 100\}$, count the number of multiples of $2$. Then count the number of multiples of $3$, and add the two numbers together. Then subtract twice the number of multiples of $6$. Note that the number of multiples of $6$ is $\lfloor 100/6 \rfloor = 16$, since they are:
\begin{align*}
6(1) &= 6, \\
6(2) &= 12, \\
&~~\vdots \\
6(16) &= 96
\end{align*}</p>
|
935,707 | <p>I'm having trouble determining this problem.</p>
<p>I need to find the integers in the set {1, ... , 100} that are divisible by 2 or 3 but not both.</p>
<p>The way I tried to approach it was:</p>
<p>If a number is divisible by both 2 and 3 then we can say it is divisible by 6. So we need to exclude integers divisible by 6. From here am I supposed to just go through each integer? Or is there a better way to approach this?</p>
<p>Thanks</p>
| Community | -1 | <p>Every integer can be written in the form $n=6q+r$ by a division by $6$ ($r$ is the remainder of the division). The term $6q$ is a multiple of both $2$ and $3$, so it suffices to reason on the divisibility of the remainder.</p>
<p>The remainders are periodic,</p>
<p>$$\color{blue}{1,2,3,4,5,0},1,2,3,4,5,0,\color{blue}{1,2,3,4,5,0},1,2,3,4,5,0,\color{blue}{1,2,3,4,5,0,}\cdots$$</p>
<p>The then truth values of the condition "divisible by 2 or 3 but not both" are</p>
<p>$$\color{blue}{f,t,t,t,f,f,}f,t,t,t,f,f,\color{blue}{f,t,t,t,f,f,}f,t,t,t,f,f,\color{blue}{f,t,t,t,f,f,}\dots$$</p>
<p>You can summarize by saying, "skip $1$, then repeatedly take $3$ and skip $3$".</p>
|
1,475,235 | <p>Why doesn't $e^x$ have an inverse in the complex plane? Can someone please clarify it?</p>
| AnatolyVorobey | 155,893 | <p>Among reals, only $0$ has the property that $e^0 = 1$, but among complex numbers, there are many $z$ such that $e^z=1$, for example, $2\pi i$, $4\pi i$, $6\pi i$ etc. But since $e^{z+w} = e^z*e^w$, you could add any of those numbers to any exponent $w$ and the value of $e^w$ doesn't change. Therefore $e^w$ is not one-to-one and so cannot have an inverse.</p>
|
281,717 | <p>Suppose that $\beta \mathbb{R}$ is Stone–Čech compactification of $\mathbb{R}$. What is the closure of $\mathbb{Q}$? </p>
| Martin Argerami | 22,857 | <p>I would guess it is your first option. A usual terminology in calculus is about absolute and relative (or local) maxima and minima. </p>
<p>The absolute maximum would be then $\max\{f(x):\ x\in[-2,4]\}$. </p>
<p>The phrase "absolute maximum <em>value</em>" probably has to do with the fact that when looking at extrema of functions, one usually focus on where they are (i.e. $x=\ldots$) rather than what they are (i.e. $f(x)=\ldots$). The latter is the value, so saying "absolute maximum value" one wants the answer "$28$" as opposed "$x=4$". </p>
|
1,406,535 | <p>Let $ f$ be a function such that $|f(u)-f(v)|\leq|u-v|$ for all real $u$ and $v$ in an interval $[a,b]$.Then:<br>
$(i)$Prove that $f$ is continuous at each point of $[a,b]$.<br></p>
<p>$(ii)$Assume that $f$ is integrable on $[a,b]$.Prove that,$|\int_{a}^{b}f(x)dx-(b-a)f(c)|\leq\frac{(b-a)^2}{2}$,where $a\leq c \leq b$<br></p>
<p>I tried to solve second part,First part i could not get idea.<br>
$|\int_{a}^{b}f(x)dx-(b-a)f(c)|=|\int_{a}^{b}f(x)-f(c)dx|=\int_{a}^{b}|f(x)-f(c)|dx\leq\int_{a}^{b}|x-c|dx\leq\int_{a}^{c}(c-x)dx+\int_{c}^{b}(x-c)dx$<br></p>
<p>But i am not getting desired result,what have i done wrong in this?Or is there another method to prove it.Please help.</p>
| PITTALUGA | 94,471 | <p>I believe you got confused. Also in Fermat's little theorem you must put the condition $\gcd(a,p)=1$, otherwise if $p\mid a$ then
$$
a^{p-1}\equiv 0 \mod{p}\;.
$$
Of course also in it's generalization, Euler's theorem, only if $\gcd(a,n)=1$ the you can say
$$
a^{\varphi(n)}\equiv 1 \mod n\;.
$$
I suggest you to look also at this link, <a href="https://en.wikipedia.org/wiki/Carmichael_function" rel="nofollow">https://en.wikipedia.org/wiki/Carmichael_function</a>: you have to understand we are interested in "the exponent of the multiplicative group of integers modulo n".</p>
<p>Last but not least: if you want to check if $n$ is prime or not and you find out that $\gcd(a,n)>1$ for a certain $a<n$, you already know that $n$ is not prime, no more tests needed...</p>
|
2,146,508 | <blockquote>
<p>Let $K$ be the algebraic closure of a finite field $k$. Prove that $Gal(K/k) \cong \hat{\mathbb{Z}}$.</p>
</blockquote>
<p>From the definition in the book, here is how $\hat{\mathbb{Z}}$ is defined:
Let $D = Cr(\mathbb{Z}_{p} | \; p \; prime)$, let $\delta: \mathbb{Z} \rightarrow D$ be the map taking $x \in \mathbb{Z}$ to the vector with all coordinates equal to $x$. Then the group $D$ together with the map $\delta$ is the profinite completion of $\mathbb{Z}$, denoted $\hat{\mathbb{Z}}$.</p>
<p>There seem to be many sources online that cite this result as true, but I'm having trouble finding anywhere that shows a proof. This question is from Profinite Groups (Wilson), so I doubt that the solution is all that straight-forward. Could anyone offer me a solution or perhaps some insight on how to tackle this problem?</p>
| Dietrich Burde | 83,966 | <p>A detailed proof is, for example, given in James S. Milne's lecture notes on <a href="http://www.jmilne.org/math/CourseNotes/FT.pdf" rel="nofollow noreferrer">Fields and Galois Theory</a>, EXAMPLE 7.16., page $97$. Ingredients are the canonical Frobenius element $\sigma:a\mapsto a^p$, the profinite completion of $\mathbb{Z}$, and the isomorphism $\widehat{\mathbb{Z}}\rightarrow Gal(\overline{\mathbb{F}_p}/\mathbb{F}_p)$.</p>
|
3,106,550 | <p>Determine the values of a for which the following system of linear equations has no solutions, a unique solution, or infinitely many solutions. You can select 'always', 'never', '<span class="math-container">$a = $</span>', or '<span class="math-container">$a \neq$</span>', then specify a value or comma-separated list of values.
<span class="math-container">$$
\begin{align}
2x−6y−4z &= 16\\
ax−y−4z &= 6\\
2x−3y−4z &= 10
\end{align}
$$</span>
This is a problem for my Linear Algebra class, and I can't seem to figure out how to get it into RREF and evaluate.</p>
| Dr. Sonnhard Graubner | 175,066 | <p>With <span class="math-container">$$z=\frac{1}{2}(x-3y-8)$$</span> (from the first equation) we get
<span class="math-container">$$x(a-2)+5y=-10$$</span> and <span class="math-container">$$y=-2$$</span> in the third equation, so we obtain
<span class="math-container">$$x(a-2)=0$$</span> Can you finish?</p>
|
3,106,550 | <p>Determine the values of a for which the following system of linear equations has no solutions, a unique solution, or infinitely many solutions. You can select 'always', 'never', '<span class="math-container">$a = $</span>', or '<span class="math-container">$a \neq$</span>', then specify a value or comma-separated list of values.
<span class="math-container">$$
\begin{align}
2x−6y−4z &= 16\\
ax−y−4z &= 6\\
2x−3y−4z &= 10
\end{align}
$$</span>
This is a problem for my Linear Algebra class, and I can't seem to figure out how to get it into RREF and evaluate.</p>
| Rhys Hughes | 487,658 | <p>Hint:</p>
<p>Let the equations be <span class="math-container">$(1), (2), (3)$</span> in the order you show.</p>
<p>Perform the following three calculations:</p>
<p><span class="math-container">$(1)-(2)$</span></p>
<p><span class="math-container">$(1)-(3)$</span></p>
<p><span class="math-container">$(3)-(2)$</span></p>
<p>The rest should fall out quite nicely.</p>
|
1,580,270 | <p>Consider the groups $G = \{0,1,2\} = \mathbb Z_3$ and $H = \{a,b,c\}$
given by the following multiplication tables:</p>
<p><a href="https://i.stack.imgur.com/hXgBb.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hXgBb.jpg" alt="enter image description here"></a></p>
<p>The first one isn't really multiplication but in my notes it said it doesn't really matter.</p>
<p>So how do I show an isomorphism? The groups have the same size so they can be bijective right? But it just seems so abstract to show if there's an isomorphism... Exactly what do we have to check.</p>
| XPenguen | 293,908 | <p><span class="math-container">$O$</span> is the identity in G and <span class="math-container">$b$</span> is the identity in H. Another thing to notice is that the tables are symmetric. Hence <span class="math-container">$G,H$</span> are commutative groups.<br><br></p>
<p>Take a look at <span class="math-container">$\psi:H\rightarrow G$</span><br>
<span class="math-container">$\psi(b)=0$</span><br>
<span class="math-container">$\psi(a)=1$</span><br>
<span class="math-container">$\psi(c)=2$</span><br>
<br>
<span class="math-container">$\psi(a*c)=\psi(b)=\psi(a)+\psi(c)...$</span><br></p>
|
659,254 | <p>Say $X_1, X_2, \ldots, X_n$ are independent and identically distributed uniform random variables on the interval $(0,1)$.</p>
<p>What is the product distribution of two of such random variables, e.g.,
$Z_2 = X_1 \cdot X_2$?</p>
<p>What if there are 3; $Z_3 = X_1 \cdot X_2 \cdot X_3$?</p>
<p>What if there are $n$ of such uniform variables?
$Z_n = X_1 \cdot X_2 \cdot \ldots \cdot X_n$?</p>
| robjohn | 13,854 | <p>An adaptation of <a href="https://math.stackexchange.com/a/2812234">this answer</a> is given here.</p>
<hr>
<p><strong>PDF of a Function of a Random Variable</strong></p>
<p>If $P(X\le x)=F(x)$ is the CDF of $X$ and $P(Y\le y)=G(y)$ is the CDF of $Y$ where $Y=f(X)$, then
$$
F(x)=P(X\le x)=P(Y\le f(x))=G(f(x))\tag1
$$
Taking the derivative of $(1)$, we get
$$
F'(x)=G'(f(x))\,f'(x)\tag2
$$
where $F'$ is the PDF of $X$ and $G'$ is the PDF of $Y$.</p>
<hr>
<p><strong>PDF of the Product of Independent Uniform Random Variables</strong></p>
<p>If $[0\le x\le1]$ is the PDF for $X$ and $Y=\log(X)$, then by $(2)$ the PDF of $Y$ is $e^y[y\le0]$. The PDF for the sum of $n$ samples of $Y$ is the $n$-fold convolution of $e^y[y\le0]$ with itself. The Fourier Transform of this $n$-fold convolution is the $n^\text{th}$ power of the Fourier Transform of $e^y[y\le0]$, which is
$$
\int_{-\infty}^0 e^{-2\pi iyt}e^y\,\mathrm{d}y=\frac1{1-2\pi it}\tag3
$$
Thus, the PDF for the sum of $n$ samples of $Y$ is
$$
\begin{align}
\sigma_n(y)
&=\int_{-\infty}^\infty\frac{e^{2\pi iyt}}{(1-2\pi it)^n}\,\mathrm{d}t\tag{4a}\\
&=\frac{e^y}{2\pi i}\int_{1-i\infty}^{1+i\infty}\frac{e^{-yz}}{z^n}\,\mathrm{d}z\tag{4b}\\
&=e^y\frac{(-y)^{n-1}}{(n-1)!}\,[y\le0]\tag{4c}
\end{align}
$$
Explanation:<br>
$\text{(4a)}$: take the inverse Fourier Transform<br>
$\text{(4b)}$: substitute $t=\frac{1-z}{2\pi i}$<br>
$\text{(4c)}$: if $y\gt0$, close the contour on the right half-plane, missing the singularity at $z=0$<br>
$\phantom{\text{(4c):}}$ if $y\le0$, close the contour on the left half-plane, enclosing the singularity at $z=0$ </p>
<p>We can get the PDF for the product of $n$ samples of $X$ by applying $(2)$ to $(4)$
$$
\bbox[5px,border:2px solid #C0A000]{\pi_n(x)=\frac{(-\log(x))^{n-1}}{(n-1)!}\,[0\le x\le1]}\tag5
$$
<a href="https://i.stack.imgur.com/4dXby.png" rel="noreferrer"><img src="https://i.stack.imgur.com/4dXby.png" alt="enter image description here"></a></p>
|
2,955,780 | <p>The midpoint of a chord of length <span class="math-container">$2a$</span> is at a distance <span class="math-container">$d$</span> from the midpoint of the minor arc it cuts out from the circle. Show that the diameter of the circle is <span class="math-container">$\frac{a^2+d^2}{d}$</span> .</p>
<p>I know I have to find similar triangles, I cannot see them...</p>
| Catalin Zara | 317,861 | <p>If the chord is <span class="math-container">$AB$</span> with midpoint <span class="math-container">$M$</span>, the midpoint of the arc is <span class="math-container">$P$</span> and the point diametral opposite to <span class="math-container">$P$</span> is <span class="math-container">$Q$</span>, then the triangles <span class="math-container">$AMP$</span> and <span class="math-container">$QAP$</span> are similar.</p>
|
1,495,256 | <p>What are the algebraic rules for solving inequalities with negative signs and fractions like:</p>
<p>$$\frac{1}{x}<-\frac{1}{5}$$</p>
| Yes | 155,328 | <p>If $x \in \mathbb{R}$, then
$\frac{1}{x} < -\frac{1}{5}$ iff $x > -5$ and iff $-x < 5$.</p>
<p>If you are seeking after the rule of thumb, here it is: whenever we take inverse of both sides, the inequality sign reverses; whenever we multiply both sides by $-1$, the inequality sign reverses too.</p>
|
1,495,256 | <p>What are the algebraic rules for solving inequalities with negative signs and fractions like:</p>
<p>$$\frac{1}{x}<-\frac{1}{5}$$</p>
| GambitSquared | 245,761 | <ol>
<li><p>Whenever we take the inverse, and both sides are either positive or both sides are negative then the inequality sign reverses; (if one is side is positive and one side is negative, then the sign stays the same.)</p></li>
<li><p>Whenever we multiply both sides by $-1$, the inequality sign reverses too.</p></li>
</ol>
<p>Since you don´t know whether x is positive or negative, you need to consider both scenarios.</p>
|
1,134,510 | <p>Regarding My Background I have covered stuff like </p>
<p>1.Single Variable Calculus</p>
<p>2.Multivariable Calculus (Multiple Integration,Vector Calculus etc) (Thomas Finney)</p>
<p>3.Basic Linear Algebra Course (Containing Vector spaces,Linear Transformation)</p>
<p>4.Ordinary Differential Equation</p>
<p>5.Real Analysis (Sequences And series)</p>
<p>I am interested In Number theory and i am big fan of Ramanujan .I have not been through rigorous proofs before . But i want to dig deep into number theory especially area where Ramanujan was working .Anyone researcher,Professor can advice me about which preparations are needed for going into number theory and which books should i need .I will be highly obliged .I am asking this for self study or you can say pursuing research at home .</p>
<p>Note - All stuff i have covered is with help of youtube videos and self study .Thanks </p>
| barak manos | 131,263 | <p>The number of ways to choose $2$ out of $26$ letters is:</p>
<p>$$\binom{26}{2}=325$$</p>
<p>The number of ways to choose $3$ places for the $1$st letter and $4$ places for the $2$nd letter is:</p>
<p>$$\binom{7}{3}\cdot\binom{4}{4}=35$$</p>
<p>The number of ways to choose $4$ places for the $1$st letter and $3$ places for the $2$nd letter is:</p>
<p>$$\binom{7}{4}\cdot\binom{3}{3}=35$$</p>
<p>So the total number of passwords is:</p>
<p>$$325\cdot(35+35)=22750$$</p>
|
304 | <p>Per <a href="http://blog.stackoverflow.com/2010/07/moderator-pro-tempore/">this post on the SO/SE blog</a> (which, curiously, does not include math.SE in its graphic list), it looks like the admins will choose moderators pro tempore at about 7 days into the public beta. In the roughly 24 hours that we've been in public beta, I've wondered several times: should we push to have the admins choose moderators pro tempore sooner (e.g. <em>now</em>)?</p>
<p><em>edit</em>: to try to get a bit more clarity in response to this question, I've created CW answers "YES" and "NO" below--please up/down vote those as you see fit.</p>
<p><em>edit2</em>: see also: <a href="https://math.meta.stackexchange.com/questions/150/elect-our-provisional-moderators">elect our (Provisional) Moderators</a></p>
| Robert Cartaino | 69 | <p>The Moderator Pro Tem program is just about complete. <strong>I don't have an objection</strong> to bumping up Math.SE in the schedule a few days (would have been on Tuesday, 8/3 anyway). The meta activity has been pretty solid. That's assuming there isn't any wide-spread objection from here or higher-on up.</p>
<p><sup><em>Sorry to break in on your YES/NO thread.</em></sup></p>
|
1,830,989 | <p>so while playing around with circles and triangles I found 2-3 limits to calculate the value of $ \pi $ using the <em>sin, cos and tan</em> functions, I am not posting the formula for obvious reasons.<br>
My question is that is there another infinite series or another way to define the trig functions when the value of the angle is in <strong>degrees</strong> without converting it to radians, I know of the <em>taylor series</em> but it takes the value of x in radians and to convert the angle to radians you obviously need $ \pi $, So is there another way to convert the or maybe find the angles in radians without using $ \pi $ or maybe a series for the trig functions which uses degrees? Also I know as a rule of thumb you always use radians in calculus can anyone explain to me why??<br>
Sorry if i asked a really dumb question.<br>
Regards,<br>
Kinshuk </p>
| John Hughes | 114,036 | <p>Well...you could avoid the $\pi$ in the previous answers by substituting for it one of the many infinite series formulas for $\pi$. That'd get rid of the $\pi$, but wouldn't be much practical use, which may be why you're not seeing it all over the internet. :) </p>
<p>I actually <em>do</em> occasionally want to know the sine of an angle in degrees. Fortunately, 60 degrees is close enough to 1 radian that the formula
$$
\sin^{\circ} (x) = \frac{x}{60}
$$
for $x$ an angle in degrees, no more than about $\pm 10$ --- based on $\sin x \approx x$ for small $x$
--- works pretty well. Rule of thumb navigators (and lifeboat navigators) have used this idea for many years. </p>
|
431,690 | <p>As far as I know, for any $A$:
$$\mathbf{x}^{T}A\mathbf{y}=0;\forall\mathbf{x},\mathbf{y}\in R^n\Rightarrow A=0$$</p>
<p>Does it mean that
$$\mathbf{x}^{T}A\mathbf{x}=0;\forall\mathbf{x}\in R^n\Rightarrow A=0$$</p>
<p>The condition of the first claim $\forall\mathbf{x},\mathbf{y}\in R^n$ implies that we could take $y=x$, and, therefore the second claim should hold. Correct?</p>
| Alex Becker | 8,173 | <p>As pointed out in the comments, your conclusion is false. The problem with your reasoning is that the first claim reads (in plain English):</p>
<blockquote>
<p>If, for all $x$ and $y$, $x^TAy=0$, then $A=0$.</p>
</blockquote>
<p>When you set $y=x$ then you are no longer considering all $x$ and $y$, only the pairs $x,y$ such that $y=x$.</p>
|
67,630 | <p>I know of a theorem from Axler's <em>Linear Algebra Done Right</em> which says that if $T$ is a linear operator on a complex finite dimensional vector space $V$, then there exists a basis $B$ for $V$ such that the matrix of $T$ with respect to the basis $B$ is upper triangular.</p>
<p>The proof of this theorem is by induction on the dimension of $V$. For dim $V = 1$ the result clearly holds, so suppose that the result holds for vector spaces of dimension less than $V$. Let $\lambda$ be an eigenvalue of $T$, which we know exists for $V$ is a complex vector space.</p>
<p>Consider $U = $ Im $(T - \lambda I)$. It is not hard to show that Im $(T - \lambda I)$ is an invariant subspace under $T$ of dimension less than $V$.</p>
<p>So by the induction hypothesis, $T|_U$ is an upper triangular matrix. So let $u_1 \ldots u_n$ be a basis for $U$. Extending this to a basis $u_1 \ldots u_n, v_1 \ldots v_m$ of $V$, the proof is complete by noting that for each $k$ such that $1 \leq k \leq m$, $T(v_k) \in $ span $\{u_1 \ldots u_n, v_1 \ldots v_k\}$.</p>
<p>The proof of this theorem seems to be only using the hypothesis that $T$ has at least one eigenvalue (in a complex vector space). So I turned to the following example of a linear operator $T$ on a <em>real</em> vector space $(\mathbb{R}^3)$ instead that has one real eigenvalue:</p>
<p>$T(x,y,z) = (x, -z ,y)$. </p>
<p>In the standard basis of $\mathbb{R}^3$ the matrix of $T$ is </p>
<p>$\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{array}\right]$</p>
<p>The only real eigenvalue of this matrix is $1$ with corresponding eigenvector </p>
<p>$\left[\begin{array}{c}1 \\ 0 \\ 0 \end{array}\right]$.</p>
<p>This is where I run into trouble: If I just extend this to any basis of $\mathbb{R}^3$, then using the standard basis again will not put the matrix of $T$ in upper triangular form. Why can't the method used in the proof above be used to put the matrix of $T$ in upper triangular form?</p>
<p>I know this is not possible for if $T$ were to be put in upper triangular form, then this would mean all its eigenvalues are real which contradicts it having eigenvalues $\pm i$ as well.</p>
| Harry Altman | 2,884 | <p>The problem is that it doesn't just use the fact that T has an eigenvalue -- it uses that, plus the inductive hypothesis. And in order to prove it for this smaller invariant subspace, you need the fact that T has an eigenvalue on that as well. So there's the problem -- it's not enough to have an eigenvalue, you need an eigenvalue on every invariant subspace. (Well, perhaps you don't need one on <em>every</em> invariant subspace.) In the complex case you have this, since any linear transform on <em>any</em> finite-dimensional vector space has an eigenvalue. Here you don't -- the image of T-1 is the plane x=0, and though T preserves that plane, it acts as a quarter-turn rotation on it and thus has no eigenvalue when restricted.</p>
|
34,795 | <p>Consider torsion free modules over the germ of a fixed isolated algebraic hypersurface singularity {$f=0$}$\subset\mathbb{C}^n$.
There are natural functors (using categories of finitely generated modules): </p>
<p>modules over $\mathbb{C}[x_1,..,x_n]_{(x_1,\dots,x_n)}/(f)$--> modules over $\mathbb{C}${$x_1,..,x_n$}$/(f)$--> modules over $\mathbb{C}[[x_1,..,x_n]]/(f)$.</p>
<p>Are they faithful, surjective? I know they are not surjective for an arbitrary local ring, but isolated hypersurface singularity is quite special.</p>
<p>upd: </p>
<ol>
<li><p>The example of $x^2=y^2+y^3$ certainly counts, but can you suggest smth similar in the case of a locally irreducible analytic hypersurface?</p></li>
<li><p>Sorry I'm outsider in algebra. By surjectivity I meant smth like: every formal module over an analytic hypersurface arises from a locally analytic module.
(Or maybe weaker: if a formal module has a submodule of the same rank that arises from locally analytic category, then the initial formal module arises from locally analytic category.)</p></li>
</ol>
<p>up2: Thanks to everybody, sorry for delay</p>
| Hailong Dao | 2,083 | <p>The most relevant result I am aware of is the following: suppose <span class="math-container">$R$</span> is a local Gorenstein ring, essentially of finite type over a field, with an isolated singularity (so certainly include your case) then the stable categories of maximal Cohen-Macaulay modules over <span class="math-container">$R$</span> and the completion <span class="math-container">$\hat R$</span> are equivalent up to direct summand via <span class="math-container">$\hat R \otimes_R?$</span>. See Proposition 1.6 of <a href="https://arxiv.org/abs/0803.0720" rel="nofollow noreferrer">this paper</a> by Keller-Murfet-Van den Bergh.</p>
<p>In everyday English, this means that every maximal Cohen-Macaulay (MCM) module over <span class="math-container">$\hat R$</span> (which are torsion-free, and include high enough syzygies of any given module) is a direct summand of a completion of some MCM module over <span class="math-container">$R$</span>. So restricting to MCM modules, the functor: (MCM modules over) <span class="math-container">$\mathbb C[x_1,\cdots,x_n]_{(x_1,\cdots,x_n)}/(f) \to \mathbb C[[x_1,\cdots,x_n]]/(f)$</span> is "surjective" up to direct summand.</p>
<p>In general, as BCnrd remarked, one can only hope to descend to the henselization of <span class="math-container">$R$</span>. There is a big literature on this topic, see for example the references <a href="http://arxiv.org/abs/math/0612311" rel="nofollow noreferrer">here</a>.</p>
<p>ADDED: The nodal curve example by Manish is discussed in example A.5 of the Keller-Murfet-Van den Bergh paper. The completion is isomorphic to <span class="math-container">$S=\mathbb C[[u,v]]/(uv)$</span> and each <span class="math-container">$S/(u), S/(v)$</span> is not extended, but their direct sum is.</p>
<p>For your updated question 1 and Manish's comment: I think one can build higher dimension examples from Manish's original example using a result known as Knörrer's periodicity theorem (see Invent. Math., 88 (1987), 153–-164), namely that there is an equivalence of the stable categories of MCM modules over <span class="math-container">$\mathbb C[[x_1,\cdots,x_n]]/(f)$</span> and <span class="math-container">$\mathbb C[[x_1,\cdots,x_n,u,v]]/(f+uv)$</span>. In particular, there should be an example for <span class="math-container">$f=y^2-x^2-x^3+uv$</span>, which will be analytically irreducible.</p>
|
34,795 | <p>Consider torsion free modules over the germ of a fixed isolated algebraic hypersurface singularity {$f=0$}$\subset\mathbb{C}^n$.
There are natural functors (using categories of finitely generated modules): </p>
<p>modules over $\mathbb{C}[x_1,..,x_n]_{(x_1,\dots,x_n)}/(f)$--> modules over $\mathbb{C}${$x_1,..,x_n$}$/(f)$--> modules over $\mathbb{C}[[x_1,..,x_n]]/(f)$.</p>
<p>Are they faithful, surjective? I know they are not surjective for an arbitrary local ring, but isolated hypersurface singularity is quite special.</p>
<p>upd: </p>
<ol>
<li><p>The example of $x^2=y^2+y^3$ certainly counts, but can you suggest smth similar in the case of a locally irreducible analytic hypersurface?</p></li>
<li><p>Sorry I'm outsider in algebra. By surjectivity I meant smth like: every formal module over an analytic hypersurface arises from a locally analytic module.
(Or maybe weaker: if a formal module has a submodule of the same rank that arises from locally analytic category, then the initial formal module arises from locally analytic category.)</p></li>
</ol>
<p>up2: Thanks to everybody, sorry for delay</p>
| Mohan | 9,502 | <p>There are well studied hypersurfaces of dimension 2 which are UFDs whose completions are not. So, any ideal representing a non-trivial element in the class group of the completion will not come from the algebraic ring.</p>
|
34,795 | <p>Consider torsion free modules over the germ of a fixed isolated algebraic hypersurface singularity {$f=0$}$\subset\mathbb{C}^n$.
There are natural functors (using categories of finitely generated modules): </p>
<p>modules over $\mathbb{C}[x_1,..,x_n]_{(x_1,\dots,x_n)}/(f)$--> modules over $\mathbb{C}${$x_1,..,x_n$}$/(f)$--> modules over $\mathbb{C}[[x_1,..,x_n]]/(f)$.</p>
<p>Are they faithful, surjective? I know they are not surjective for an arbitrary local ring, but isolated hypersurface singularity is quite special.</p>
<p>upd: </p>
<ol>
<li><p>The example of $x^2=y^2+y^3$ certainly counts, but can you suggest smth similar in the case of a locally irreducible analytic hypersurface?</p></li>
<li><p>Sorry I'm outsider in algebra. By surjectivity I meant smth like: every formal module over an analytic hypersurface arises from a locally analytic module.
(Or maybe weaker: if a formal module has a submodule of the same rank that arises from locally analytic category, then the initial formal module arises from locally analytic category.)</p></li>
</ol>
<p>up2: Thanks to everybody, sorry for delay</p>
| Mohan | 9,502 | <p>Dear Hailong (and all others reading his), I apologize for using `answer' for a comment, since my openid does not work (I am traveling) or I have forgotten it and MO does not give me a way to comment. </p>
<p>Yes, there are examples over any field. If my memory serves me right, the first example I saw was in Samuel's TIFR lecture notes on UFD's. There are also examples (as an aside) in my paper on rational double points which appeared in the 80s in the Inventiones.</p>
<p>For the question of Brian, in general class group is used for (over normal domains) rank one torsion free finitely generated modules upto isomorphism modulo free modules. If one wants to look at invertible modules, nowadays it seems more common (especially people with an alg. geom. bend) to use Picard group. The examples I mention above are `algebraic' and not Henselizations.</p>
|
1,098,253 | <p>I have got some trouble with proving that for $x\neq 0$:
$$
\frac{\arctan x}{x }< 1
$$
I tried doing something like $x = \tan t$ and playing with this with no success.</p>
| miracle173 | 11,206 | <p>A basic proof without calculus:</p>
<p>We assume that $x>0$. Then we have to show that</p>
<p>$$\arctan(x)<x \tag{1}$$</p>
<p>But we can concluded this from </p>
<p>$$x \lt \tan(x), \;\; \forall x \in (0,\pi/2) \tag{2}$$
If we substitute all occurrences of $x$ by $\arctan(x)$ in $(2)$ we get</p>
<p>$$\arctan(x) \lt \tan(\arctan(x)) \tag{3a}$$
The rhs of $(3a)$ can be simplified to $x$ because $\arctan$ is the inverse of $\tan$ and so we get $(1)$ from $(3)$.</p>
<hr>
<p>Another possibility is to apply $\arctan$ to the equation $(2)$. Because $\arctan$ is a strictly increasing function we get
$$\arctan(x) \lt \arctan(\tan(x)) \tag{3b}$$
and from this we get $(1)$ again.</p>
<hr>
<p>The validness of $(2)$ can be seen in the following picture:</p>
<p><img src="https://i.stack.imgur.com/9xbLp.png" alt="enter image description here"></p>
<p>The blue arc is the angle $x$ , the area of the yellow segment of the circle is therefore
$$\pi \frac{x}{2\pi}=\frac{x}{2}$$
The green line is $\tan(x)$ and the (yellow + red) area of the triangle is
$$\frac{1\cdot \tan x}{2}=\frac{\tan x}{2}$$
So we have
$$ \frac{x}{2}<\frac{\tan x}{2} \tag{4}$$
and therefore $(2)$.</p>
<p>The $\arctan$ of the length of the green line is the length of the blue arc.
We see that the $\arctan$ of a value greater $0$ is a value between $0$ and $\frac{\pi}{2}$</p>
<p>From the picture we see that
$$\arctan(-x)=-x \tag{5}$$
and so we have
$$\frac{\arctan(x)}{x}=\frac{\arctan(-x)}{-x} \lt 1 \tag{6}$$
if $x<0$. </p>
|
1,098,253 | <p>I have got some trouble with proving that for $x\neq 0$:
$$
\frac{\arctan x}{x }< 1
$$
I tried doing something like $x = \tan t$ and playing with this with no success.</p>
| Felix Marin | 85,343 | <p>$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
\newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
\newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
\newcommand{\dd}{{\rm d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}}
\newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
\newcommand{\fermi}{\,{\rm f}}
\newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
\newcommand{\half}{{1 \over 2}}
\newcommand{\ic}{{\rm i}}
\newcommand{\iff}{\Longleftrightarrow}
\newcommand{\imp}{\Longrightarrow}
\newcommand{\Li}[1]{\,{\rm Li}_{#1}}
\newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert}
\newcommand{\pars}[1]{\left(\, #1 \,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\pp}{{\cal P}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
\newcommand{\sech}{\,{\rm sech}}
\newcommand{\sgn}{\,{\rm sgn}}
\newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
\newcommand{\ul}[1]{\underline{#1}}
\newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
By <i>Mean Value Theorem</i>: $\ds{\forall\ x\ \not=\ 0\,;\quad\exists\ \xi\ \mid\ 0\ <\ \xi\ <\ \verts{x}}$
and
\begin{align}
{\arctan\pars{x} \over x}={1 \over \xi^{2} + 1} < 1
\end{align}</p>
|
1,599,890 | <blockquote>
<p>Let $a_n$ be the number of those permutation $\sigma $ on $\{1,2,...,n\}$ such that $\sigma $ is a product of exactly two disjoint cycles. Then find $a_4$ and $a_5$.</p>
</blockquote>
<p>Calculating $a_4$: Possible cases which can happen are $(12)(34),(13)(24),(14)(23)$, any cycle of the form $(123)$ or $(12)$ i.e. two-cycles and three cycles thus we have in total $3+\frac{1}{3}4P_3+\frac{1}{2}4P_2=3+8+6=17$ but the correct answer is given to be either $11$ or $14$.</p>
<p>Where am I wrong? Please help.</p>
| Christian Blatter | 1,303 | <p>Let $r\in[1\>..\>n-1]$ be the length of the cycle containing the number $n$. Begin the listing of this cycle with $n$. There are $(n-1)(n-2)\cdots(n-r+1)$ ways to choose the remaining entries of this cycle. Now there will be $n-r\geq1$ numbers left over, one of them the largest. Begin the listing of the second cycle with this largest left-over number, and you will then have $(n-r-1)!$ ways to choose the remaining entries of the second cycle. In all we have ${(n-1)!\over n-r}$ permutations of this kind. It follows that there are
$$a_n=(n-1)!\sum_{k=1}^{n-1}{1\over k}\tag{1}$$
elements of $S_n$ having exactly two cycles. In particular this gives $a_4=11$. – Maybe the formula $(1)$ can be brought into a closed form, using some combinatorial identity.</p>
<p>Computing the $a_n$ we obtain the sequence $(1, 3, 11, 50, 274, 1764, 13068, 109584, 1026576,\ldots)$ which OEIS identifies as <a href="http://oeis.org/A000254" rel="noreferrer">A000254</a>.</p>
|
3,159,884 | <p>Prove that if <span class="math-container">$|z+w|=|z-w|$</span> then <span class="math-container">$z\overline{w}$</span> is purely imaginary.</p>
<p>To start off, I said let <span class="math-container">$z=a+bi$</span> and let <span class="math-container">$w=p+qi$</span>. Not sure where to go from here after subbing in those for <span class="math-container">$z$</span> and <span class="math-container">$w$</span>.</p>
| DINEDINE | 506,164 | <p><span class="math-container">$$|z+\omega|^2=|z-\omega|^2 \iff$$</span>
<span class="math-container">$$(z+\omega)(\bar{z}+\bar{\omega})= (z-\omega)(\bar{z}-\bar{\omega})\iff$$</span>
<span class="math-container">$$z\bar{z}+{\omega}\bar{z}+z\bar{\omega}+\bar{\omega}\omega= z\bar{z}-\omega\bar{z}-z\bar{\omega}+\bar{\omega}\omega\iff$$</span>
<span class="math-container">$$2\omega\bar{z}=-2\bar{\omega}z\iff$$</span>
<span class="math-container">$$\omega\bar{z}=-\overline{\omega\bar{z}}$$</span></p>
|
298,912 | <p>I was reading some basic information from Wiki about category theory and honestly speaking I have a very weak knowledge about it. As it sounds interesting, I will go into the theory to learn more if it is actually useful in practice.</p>
<p>My question is to know if category theory has some applications in practice, namely in engineering problems.</p>
<p>I have already read this <a href="https://math.stackexchange.com/questions/280166/applications-of-category-theory-and-topoi-topos-theory-in-reality">Applications of category theory and topoi/topos theory in reality</a> </p>
<p>and the answers are only about programming which are not very interesting from my point of view.</p>
<p>Any comments are welcomed, thanks in advance.</p>
| yoknapatawpha | 108,381 | <p>I know this answer is (very) late, but I think this may be of interest. <a href="http://ames.tamu.edu/A%20categorical%20theory.pdf" rel="noreferrer">This</a> is the Ph.D. thesis of Professor Aaron Ames of Texas A&M which he wrote while a student at UC Berkeley. It applies category theory to hybrid systems and specifically uses category theory for the purposes of model reduction and analyzing stability in hybrid systems. It also presents some results on networked systems that are rooted in category theory. </p>
|
41,836 | <p>Nakayama's lemma is as follows:</p>
<blockquote>
<p>Let <span class="math-container">$A$</span> be a ring, and <span class="math-container">$\frak{a}$</span> an ideal such that <span class="math-container">$\frak{a}$</span> is contained in every maximal ideal. Let <span class="math-container">$M$</span> be a finitely generated <span class="math-container">$A$</span>-module. Then if <span class="math-container">$\frak{a}$$M=M$</span>, we have that <span class="math-container">$M = 0$</span>.</p>
</blockquote>
<p>Most proofs of this result that I've seen in books use some non-trivial linear algebra results (like Cramer's rule), and I had come to believe that these were certainly necessary. However, in Lang's Algebraic Number Theory book, I came across a quick proof using only the definitions and induction. I felt initially like something must be wrong--I thought perhaps the proof is simpler because Lang is assuming throughout that all rings are integral domains, but he doesn't use this in the proof he gives, as far as I can see.</p>
<p>Here is the proof, verbatim: We do induction on the number of generators of <span class="math-container">$M$</span>. Say M is generated by <span class="math-container">$w_1, \cdots, w_m$</span>. There exists an expression <span class="math-container">$$w_1 = a_1w_1 + \cdots + a_mw_m$$</span> with <span class="math-container">$a_i \in \frak{a}$</span>. Hence <span class="math-container">$$(1-a_1)w_1 = a_2w_2 + \cdots +a_mw_m$$</span> If <span class="math-container">$(1-a_1)$</span> is not a unit in A, then it is contained in a maximal ideal <span class="math-container">$\frak{p}$</span>. Since <span class="math-container">$a_1 \in \frak{p}$</span> by hypothesis, we have a contradiction. Hence <span class="math-container">$1-a_1$</span> is a unit, and dividing by it shows that <span class="math-container">$M$</span> can be generated by <span class="math-container">$m-1$</span> elements, thereby concluding the proof.</p>
<p>Is the fact that <span class="math-container">$A$</span> is assumed to be a domain being smuggled in here in some way that I missed? Or is this really an elementary proof of Nakayama's lemma, in full generality?</p>
| Martin Brandenburg | 2,841 | <p>There are various forms of the Nakayama lemma. Here is a rather general one; note that it does <em>not</em> involve maximal ideals and is a constructive theorem (Atiyah-MacDonald, Commutative Algebra, Prop. 2.4 ff).</p>
<blockquote>
<p>Let $M$ be a finitely generated $A$-module, $\mathfrak{a} \subseteq A$ be an ideal and $\phi \in End_A(M)$ such that $\phi(M) \subseteq \mathfrak{a} M$. Then there is an equation of the form
$\phi^n + r_1 \phi^{n-1} + ... + r_n = 0$,
where the $r_i$ are in $\mathfrak{a}$.</p>
</blockquote>
<p>The proof uses the equality $adj(X) \cdot X = \text{det}(X)$ for quadratic matrices over a ring. I call this an elementary linear algebra fact. Of course, there you only prove it for fields but using function fields implies the result for general rings. If we take $\phi=\text{id}_M$, we get the following form:</p>
<blockquote>
<p>Let $M$ be a finitely generated $A$-module and let $\mathfrak{a} \subseteq A$ be an ideal such that $\mathfrak{a} M = M$. Then there exists some $r \in A$ such that $rM = 0$ and $r \equiv 1$ mod $\mathfrak{a}$.</p>
</blockquote>
<p>In particular, we get:</p>
<blockquote>
<p>Let $M$ be a finitely generated $A$-module and let $\mathfrak{a} \subseteq A$ be an ideal such that $\mathfrak{a} M = M$ and $\mathfrak{a}$ lies in every maximal ideal of $A$. Then $M=0$.</p>
</blockquote>
<p>Observe that this argument uses Zorn's lemma (namely that every non-unit is contained in a maximal ideal) and is thus nonconstructive. Which is of course not surprising since without Zorn's lemma it is consistent that there are nontrivial rings without any maximal ideals at all. This should convince you that the first form of the Nakayama lemma is the most easy and elementary one. The last form has another short proof, which is standard and given in the question above.</p>
<p>Here is another short well-known proof for the last form, which also works if $A$ is noncommutative (then we have to replace "maximal ideal" by "maximal left ideal"): Assume $M \neq 0$. Since $M$ is finitely generated, an application of Zorn's lemma shows that $M$ has a maximal proper submodule $N$. Then $M/N$ is simple, thus isomorphic to $A/\mathfrak{m}$ for some maximal left ideal $\mathfrak{m}$. Then $N = \mathfrak{m} M = M$, contradiction.</p>
<p>By the way, I don't know if the first form is true if $A$ is noncommutative. The theory of determinants is not really prosperous over noncommutative rings. Hints?</p>
<p>In many texts about algebraic geometry only the last form of the Nakayama lemma is needed. But the first one is stronger and is used in many results in commutative algebra.</p>
|
275,974 | <blockquote>
<p><strong>Possible Duplicate:</strong><br>
<a href="https://math.stackexchange.com/questions/264889/how-is-this-called-rationals-and-irrationals">How is this called? Rationals and irrationals</a> </p>
</blockquote>
<p>Please help me prove, that
$$\underset{n\rightarrow\infty}{\lim}\left(\underset{k\rightarrow\infty}{\lim}(\cos(|n!\pi x|)^{2k})\right)=\begin{cases}
1 & \iff x\in\mathbb{Q}\\
0 & \iff x\notin\mathbb{Q}
\end{cases}$$</p>
<p>Seems very complicated, but it's on calc I. I've tried use series expansions of cos, but it don't lead to answer. Thanks in advance!</p>
<p><strong>Edit</strong></p>
<p>Please don't use too much advanced techniques.</p>
| zrbecker | 19,536 | <p>Baby Rudin has this as an example.</p>
<p><img src="https://i.stack.imgur.com/DDSHn.png" alt="See Baby Rudin page 145"></p>
|
3,689,096 | <p>This was the Question:- Find all positive integers <span class="math-container">$n$</span> such that <span class="math-container">$\varphi(n)$</span> divides <span class="math-container">$n^2 + 3$</span></p>
<p>What I tried:-</p>
<p>I knew the solution and explanation of all positive integers <span class="math-container">$n$</span> such that <span class="math-container">$\varphi(n)\mid n$</span> .
<a href="https://math.stackexchange.com/q/135756/11619">That answer</a> was when <span class="math-container">$n = 1$</span>, or <span class="math-container">$n$</span> is of the form of <span class="math-container">$2^a$</span> or <span class="math-container">$2^a3^b$</span> . </p>
<p>I tried to relate this fact with this problem in many ways, but couldn't get to a possible solution.</p>
<p>Any hints or suggestions will be greatly appreciated</p>
| AT1089 | 758,289 | <p>Note that <span class="math-container">$\phi(n)=1$</span> if and only if <span class="math-container">$n=1,2$</span>, and even if and only if <span class="math-container">$n>2$</span>. Therefore, <span class="math-container">$\boxed{n=1,2}$</span> are both solutions to</p>
<p><span class="math-container">$$ \phi(n) \mid (n^2+3). \quad \ldots \quad (1) $$</span></p>
<p>Henceforth, assume <span class="math-container">$n>2$</span>. Then <span class="math-container">$\phi(n)$</span> is even, and so <span class="math-container">$n$</span> must be odd. But then <span class="math-container">$n^2+3 \equiv 4\pmod{8}$</span>, so <span class="math-container">$n^2+3=4m$</span>, with <span class="math-container">$m$</span> odd.</p>
<p>Since <span class="math-container">$\phi$</span> is multiplicative:</p>
<p><span class="math-container">$$ \phi(mn) = \phi(m) \cdot \phi(n) $$</span></p>
<p>whenever <span class="math-container">$\gcd(m,n)=1$</span>, and</p>
<p><span class="math-container">$$ \phi(p^{\alpha}) = p^{\alpha-1}(p-1), $$</span></p>
<p>we have</p>
<p><span class="math-container">$$ \phi(n) = \prod_{p^{\alpha}\,\mid\mid\,n} p^{\alpha-1}(p-1). $$</span></p>
<p>Each prime factor <span class="math-container">$p$</span> of <span class="math-container">$n$</span> contributes <span class="math-container">$p-1$</span> to <span class="math-container">$\phi(n)$</span>. Since <span class="math-container">$p-1$</span> is even, <span class="math-container">$\phi(n)$</span> is divisible by <span class="math-container">$2^k$</span>, where <span class="math-container">$k$</span> equals the number of distinct prime divisors of <span class="math-container">$n$</span>. Since <span class="math-container">$n^2+3=4m$</span>, with <span class="math-container">$m$</span> odd, <span class="math-container">$k=1$</span> or <span class="math-container">$2$</span>.</p>
<p><span class="math-container">$\bullet$</span> Suppose <span class="math-container">$k=1$</span>, and write <span class="math-container">$n=p^{\alpha}$</span>, <span class="math-container">$p$</span> prime. Then eqn.<span class="math-container">$(1)$</span> gives</p>
<p><span class="math-container">$$ p^{\alpha-1}(p-1) \mid (p^{2\alpha}+3). \quad \ldots \quad (2) $$</span></p>
<p>If <span class="math-container">$\alpha=1$</span>, then <span class="math-container">$(p-1) \mid (p^2+3)$</span>, and so <span class="math-container">$(p-1) \mid \big((p^2+3)-(p^2-1)\big)$</span>. Thus, <span class="math-container">$p-1 \in \{1,2,4\}$</span>, and <span class="math-container">$p=3$</span> or <span class="math-container">$5$</span>; so <span class="math-container">$n=3,5$</span> are solutions.</p>
<p>If <span class="math-container">$\alpha>1$</span>, then <span class="math-container">$p \mid (p^{2\alpha}+3)$</span>, and so <span class="math-container">$p \mid 3$</span>. This implies <span class="math-container">$p=3$</span>, and eqn.<span class="math-container">$(2)$</span> gives</p>
<p><span class="math-container">$$ 2 \cdot 3^{\alpha-1} \mid 3\big(3^{2\alpha-1}+1\big). $$</span></p>
<p>Since the highest power of <span class="math-container">$3$</span> dividing the RHS is <span class="math-container">$1$</span>, we can only have <span class="math-container">$\alpha=2$</span>. We note that <span class="math-container">$\phi(3^2) \mid (3^4+3)$</span>; so <span class="math-container">$n=3^2$</span> is a solution.</p>
<p>So the three solutions in this case are <span class="math-container">$\boxed{n=3,5,3^2}$</span>.</p>
<p><span class="math-container">$\bullet$</span> Suppose <span class="math-container">$k=2$</span>, and write <span class="math-container">$n=p^{\alpha}q^{\beta}$</span>, <span class="math-container">$p,q$</span> primes, <span class="math-container">$\alpha \ge \beta \ge 1$</span>. Note that since <span class="math-container">$4$</span> is the highest power of <span class="math-container">$2$</span> dividing the LHS of eqn.<span class="math-container">$(1)$</span>, we must have <span class="math-container">$p \equiv q \equiv 3\pmod{4}$</span>.</p>
<p>Now eqn.<span class="math-container">$(1)$</span> gives</p>
<p><span class="math-container">$$ p^{\alpha-1} q^{\beta-1} (p-1)(q-1) \mid (p^{2\alpha}q^{2\beta}+3). \quad \ldots \quad (3) $$</span></p>
<p>If <span class="math-container">$\alpha=1$</span>, then <span class="math-container">$\beta=1$</span>, and eqn.<span class="math-container">$(3)$</span> gives</p>
<p><span class="math-container">$$ (p-1)(q-1) \mid (p^2q^2+3). \quad \ldots \quad (4) $$</span></p>
<p>Thus, <span class="math-container">$(p-1) \mid \big((p^2q^2+3)-(p^2-1)q^2\big)=(q^2+3)$</span>; similarly, <span class="math-container">$(q-1) \mid (p^2+3)$</span>.</p>
<p>If <span class="math-container">$p=3$</span>, this gives <span class="math-container">$(q-1) \mid 12$</span>. Since <span class="math-container">$q-1 \equiv 2\pmod{4}$</span>, <span class="math-container">$q>p$</span>, we have <span class="math-container">$q=7$</span>. Note that <span class="math-container">$\phi(3 \cdot 7) \mid (21^2+3)$</span>; so <span class="math-container">$n=3 \cdot 7$</span> is a solution.</p>
<p>Now suppose <span class="math-container">$3<p<q$</span>. Then <span class="math-container">$p^2+3 \equiv q^2+3 \equiv 1\pmod{3}$</span>, so that <span class="math-container">$p \equiv q \equiv 2\pmod{3}$</span> since <span class="math-container">$(p-1) \mid (q^2+3)$</span> and <span class="math-container">$(q-1) \mid (p^2+3)$</span>. So both <span class="math-container">$p+1$</span> and <span class="math-container">$q+1$</span> are multiples of <span class="math-container">$3$</span> and <span class="math-container">$4$</span>, and hence <span class="math-container">$p,q \equiv -1\pmod{12}$</span>.</p>
<p>Now suppose <span class="math-container">$\ell$</span> is a prime divisor of <span class="math-container">$(pq)^2+3$</span>, <span class="math-container">$\ell>3$</span>. Then <span class="math-container">$-3$</span> is a quadratic residue modulo <span class="math-container">$\ell$</span>, and so we have</p>
<p><span class="math-container">$$ \left(\frac{-1}{\ell}\right) = \left(\frac{3}{\ell}\right) = \pm 1. $$</span></p>
<p>If each is <span class="math-container">$+1$</span>, then <span class="math-container">$\ell \equiv 1\pmod{4}$</span> and so <span class="math-container">$1=\left(\frac{\ell}{3}\right)$</span>. This implies <span class="math-container">$\ell \equiv 1\pmod{3}$</span>. Together we get <span class="math-container">$\ell \equiv 1\pmod{12}$</span>.</p>
<p>If each is <span class="math-container">$-1$</span>, then <span class="math-container">$\ell \equiv -1\pmod{4}$</span> and so <span class="math-container">$1=\left(\frac{\ell}{3}\right)$</span>. Again <span class="math-container">$\ell \equiv 1\pmod{3}$</span>, and we get <span class="math-container">$\ell \equiv 7\pmod{12}$</span>.</p>
<p>Hence, <span class="math-container">$\ell \equiv 1\pmod{6}$</span>. Since <span class="math-container">$p,q \equiv -1\pmod{12}$</span>, <span class="math-container">$3 \nmid (p-1)$</span> and <span class="math-container">$3 \nmid (q-1)$</span>. So both <span class="math-container">$p-1,q-1$</span> are of the form <span class="math-container">$2t$</span>, where <span class="math-container">$t$</span> is a (possibly empty) product of primes of the form <span class="math-container">$6\lambda+1$</span>.</p>
<p>The empty product gives <span class="math-container">$p-1=2$</span>, and has been dealt with. Otherwise, both <span class="math-container">$p-1$</span> and <span class="math-container">$q-1$</span> are of the form <span class="math-container">$12\lambda+2$</span>, which contradicts the fact that <span class="math-container">$p,q \equiv -1\pmod{12}$</span>.</p>
<p>We conclude that there is no solution with <span class="math-container">$3<p<q$</span>.</p>
<p>If <span class="math-container">$\alpha>1$</span>, then <span class="math-container">$p$</span> must divide <span class="math-container">$p^{2\alpha}q^{2\beta}+3$</span>. Hence, <span class="math-container">$p \mid 3$</span>, and so <span class="math-container">$p=3$</span>. Now eqn.<span class="math-container">$(3)$</span> gives</p>
<p><span class="math-container">$$ 2(q-1)3^{\alpha-1}q^{\beta} \mid 3\big(3^{2\alpha-1}q^{2\beta}+1\big). $$</span></p>
<p>Since the highest power of <span class="math-container">$3$</span> dividing the RHS is <span class="math-container">$1$</span>, we can only have <span class="math-container">$\alpha=2$</span>. Thus,</p>
<p><span class="math-container">$$ 2(q-1)q^{\beta} \mid (27q^{2\beta}+1). $$</span></p>
<p>However, <span class="math-container">$q$</span> divides the LHS but not the RHS, so we have no further solutions.</p>
<p>The only solution in this case is <span class="math-container">$\boxed{n=3 \cdot 7}$</span>.</p>
<p>Therefore,</p>
<p><span class="math-container">$$ \phi(n) \mid (n^2+3) \Longleftrightarrow n \in \{1,2,3,5,9,21\}. $$</span></p>
|
2,164,823 | <p>Could someone please explain mathematical explanation behind this?</p>
<p>You can win three kinds of basketball points, 1 point, 2 points, and 3 points. Given a total score $n$, print out all the combination to compose $n$.</p>
<p>Examples:<br>
For n = 1, the program should print following:<br>
1 </p>
<p>For n = 2, the program should print following:<br>
1 1<br>
2 </p>
<p>For n = 3, the program should print following:<br>
1 1 1<br>
1 2<br>
2 1<br>
3 </p>
<p>For n = 4, the program should print following:<br>
1 1 1 1<br>
1 1 2<br>
1 2 1<br>
1 3<br>
2 1 1<br>
2 2<br>
3 1 </p>
<p>Algorithm:</p>
<ul>
<li>At first position we can have three numbers 1 or 2 or 3. First put 1 at first position and recursively call for n-1. </li>
<li>Then put 2 at first position and recursively call for n-2. </li>
<li>Then put 3 at first position and recursively call for n-3. </li>
<li>If n becomes 0 then we have formed a combination that compose n, so print the current combination. </li>
</ul>
<p>I've solved in using JS as per below. But I don't quite understand the mathematical reasoning behind it.</p>
<p><a href="https://jsfiddle.net/d2Lft7d1/" rel="nofollow noreferrer">https://jsfiddle.net/d2Lft7d1/</a></p>
| Hazem Orabi | 367,051 | <p>It is a <a href="https://en.wikipedia.org/wiki/Fibonacci_number" rel="nofollow noreferrer">Fibonacci</a> like sequence. <br/>
Let $\,A_{\small n}\,$ be the total number of $\,(1,\,2,\,3)\,$ combinations that compose $\,n\,$, Then:
$$ A_{\small 1}=1,\,A_{\small 2}=2,\,A_{\small 3}=4,\quad\color{red}{A_{\small n}=A_{\small n-1}+A_{\small n-2}+A_{\small n-3}} \\[4mm] \Rightarrow\quad \left\{A_{\small n}\right\}=\left\{1,\,2,\,4,\,7,\,13,\,24,\,44,\,\cdots\right\} $$
And the idea behind that for $\,n\gt3\,$, you will have the ability to add a Most Significant Digit (MSD) equals $\,1,\,2,\text{ or } \,3\,$. This should left you with $\,n-1,\,n-2,\text{ and } \,n-3\,$ respectively. For Example:
$$ \begin{align}
n &=5 \\[2mm]
\text{MSD} &=\color{red}{1} \quad\Rightarrow\text{ The comination of }\,(n-1=4)= \begin{cases} \color{red}{1}\,1\,1\,1\,1 \\ \color{red}{1}\,1\,1\,2 \\ \color{red}{1}\,1\,2\,1 \\ \color{red}{1}\,1\,3 \\ \color{red}{1}\,2\,1\,1 \\ \color{red}{1}\,2\,2 \\ \color{red}{1}\,3\,1 \end{cases} \\[2mm]
\text{MSD} &=\color{blue}{2} \quad\Rightarrow\text{ The comination of }\,(n-2=3)= \begin{cases} \color{blue}{2}\,1\,1\,1 \\ \color{blue}{2}\,1\,2 \\ \color{blue}{2}\,2\,1 \\ \color{blue}{2}\,3 \end{cases} \\[2mm]
\text{MSD} &=\color{Green}{3} \quad\Rightarrow\text{ The comination of }\,(n-3=2)= \begin{cases} \color{Green}{3}\,1\,1 \\ \color{Green}{3}\,2 \end{cases} \\[2mm]
A_{\small5} &= \color{red}{A_{\small4}}+\color{blue}{A_{\small3}}+\color{green}{A_{\small2}} = \color{red}{7}+\color{blue}{4}+\color{green}{2} = 13
\end{align} $$ </p>
<hr>
<p>For other similar combination $\,\left({\small\text{e.g }}\,(1,2)\,,(1,2,4)\,,\cdots\right)\,$, we start by computing the first required terms, then we apply the concept of Fibonacci sequence and Most Significant Digit (MSD). <br/><br>
$\underline{\bf(1,2)}$:
$$ \begin{align}
(n=1) &\rightarrow \begin{cases} \color{red}{1} \end{cases} \qquad\Rightarrow\, A_{\small 1}=1 \\[2mm]
(n=2) &\rightarrow \begin{cases} \color{blue}{1}\,\color{red}{1} \\ \color{blue}{2} \end{cases} \quad\Rightarrow\, A_{\small 2}=2 \\[2mm]
A_{\small n} &= A_{\small n-1}+A_{\small n-2} = \left\{1,\,2,\,3,\,5,\,8,\,13,\,21,\,\cdots\right\}
\end{align} $$
$\underline{\bf(1,2,4)}$:
$$ \begin{align}
(n=1) &\rightarrow \begin{cases} \color{red}{1} \end{cases} \qquad\qquad\Rightarrow\, A_{\small 1}=1 \\[2mm]
(n=2) &\rightarrow \begin{cases} \color{blue}{1}\,\color{red}{1} \\ \color{blue}{2} \end{cases} \quad\qquad\Rightarrow\, A_{\small 2}=2 \\[2mm]
(n=3) &\rightarrow \begin{cases} \color{green}{1}\,\color{blue}{1}\,\color{red}{1} \\ \color{green}{1}\,\color{blue}{2} \\ \color{green}{2}\,\color{red}{1} \end{cases} \qquad\Rightarrow\, A_{\small 3}=3 \\[2mm]
(n=4) &\rightarrow \begin{cases} 1\,\color{green}{1}\,\color{blue}{1}\,\color{red}{1} \\ 1\,\color{green}{1}\,\color{blue}{2} \\ 1\,\color{green}{2}\,\color{red}{1} \\ 2\,\color{blue}{1}\,\color{red}{1} \\ 2\,\color{blue}{2} \\ 4 \end{cases} \quad\Rightarrow\, A_{\small 4}=6 \\[2mm]
A_{\small n} &= A_{\small n-1}+A_{\small n-2}+A_{\small n-4} = \left\{1,\,2,\,3,\,6,\,10,\,18,\,31,\,\cdots\right\}
\end{align} $$</p>
|
28,892 | <p>I was searching on MathSciNet recently for a certain paper by two mathematicians. As I often do, I just typed in the names of the two authors, figuring that would give me a short enough list. My strategy was rather dramatically unsuccessful in this case: the two mathematicians I listed have written 80 papers together!</p>
<p>So this motivates my (rather frivolous, to be sure) question: which pair of mathematicians has the most joint papers? </p>
<p>A good meta-question would be: can MathSciNet search for this sort of thing automatically? The best technique I could come up with was to think of one mathematician that was both prolific and collaborative, go to their "profile" page on MathSciNet (a relatively new feature), where their most frequent collaborators are listed, alphabetically, but with the wordle-esque feature that the font size is proportional to the number of joint papers. </p>
<p>Trying this, to my surpise I couldn't beat the 80 joint papers I've already found. Erdos' most frequent collaborator was Sarkozy: 62 papers (and conversely Sarkozy's most frequent collaborator was Erdos). Ronald Graham's most frequent collaborator is Fan Chung: 76 papers (and conversely).</p>
<p>I would also be interested to hear about triples, quadruples and so forth, down to the point where there is no small set of winners.</p>
<hr>
<p><b>Addendum</b>: All right, multiple people seem to want to know. The 80 collaboration pair I stumbled upon is Blair Spearman and Kenneth Williams. </p>
| Community | -1 | <p>Sergio Albeverio and Raphael Hoegh-Krohn have <a href="http://www.ams.org/mathscinet/search/publications.html?pg1=INDI&s1=24435%2520and%252086835" rel="noreferrer">98 papers together</a>
according to MathSciNet. </p>
|
28,892 | <p>I was searching on MathSciNet recently for a certain paper by two mathematicians. As I often do, I just typed in the names of the two authors, figuring that would give me a short enough list. My strategy was rather dramatically unsuccessful in this case: the two mathematicians I listed have written 80 papers together!</p>
<p>So this motivates my (rather frivolous, to be sure) question: which pair of mathematicians has the most joint papers? </p>
<p>A good meta-question would be: can MathSciNet search for this sort of thing automatically? The best technique I could come up with was to think of one mathematician that was both prolific and collaborative, go to their "profile" page on MathSciNet (a relatively new feature), where their most frequent collaborators are listed, alphabetically, but with the wordle-esque feature that the font size is proportional to the number of joint papers. </p>
<p>Trying this, to my surpise I couldn't beat the 80 joint papers I've already found. Erdos' most frequent collaborator was Sarkozy: 62 papers (and conversely Sarkozy's most frequent collaborator was Erdos). Ronald Graham's most frequent collaborator is Fan Chung: 76 papers (and conversely).</p>
<p>I would also be interested to hear about triples, quadruples and so forth, down to the point where there is no small set of winners.</p>
<hr>
<p><b>Addendum</b>: All right, multiple people seem to want to know. The 80 collaboration pair I stumbled upon is Blair Spearman and Kenneth Williams. </p>
| S. Okada | 36,665 | <p>We get 135 matches for "Author=(Jimbo, Michio and Miwa, Tetsuji)" in mathscinet.</p>
|
606,843 | <p><strong>Definitions:</strong></p>
<ol>
<li><p>$[x]$ is the integer value of $x$. For example: $[4.3]=4$.</p></li>
<li><p>$\{x\}=x-[x]$.</p></li>
</ol>
<p>Can someone help me calculate the derivative of the following functions, and determine where the derivative are not defined?</p>
<ol>
<li><p>$[x^2]\sin^2(\pi x)$</p></li>
<li><p>$\{x^2\}\sin^2(\pi x)$</p></li>
</ol>
<p>Thank you!!</p>
| jimbo | 115,363 | <p>Derivate Fourier expansion </p>
<p>$$\{x\}= \frac{1}{2} - \frac{1}{\pi} \sum_{k=1}^\infty \frac{\sin(2 \pi k x)} {k}$$</p>
|
374,619 | <p>In <a href="https://math.stackexchange.com/a/373935/752">this recent answer</a> to <a href="https://math.stackexchange.com/q/373918/752">this question</a> by Eesu, Vladimir
Reshetnikov proved that
$$
\begin{equation}
\left( 26+15\sqrt{3}\right) ^{1/3}+\left( 26-15\sqrt{3}\right) ^{1/3}=4.\tag{1}
\end{equation}
$$</p>
<p>I would like to know if this result can be <em>generalized</em> to other triples of
natural numbers. </p>
<blockquote>
<p><strong>Question</strong>. What are the solutions of the following equation? $$ \begin{equation} \left( p+q\sqrt{3}\right) ^{1/3}+\left(
p-q\sqrt{3}\right) ^{1/3}=n,\qquad \text{where }\left( p,q,n\right)
\in \mathbb{N} ^{3}.\tag{2} \end{equation} $$</p>
</blockquote>
<p>For $(1)$ we could write $26+15\sqrt{3}$ in the form $(a+b\sqrt{3})^{3}$ </p>
<p>$$
26+15\sqrt{3}=(a+b\sqrt{3})^{3}=a^{3}+9ab^{2}+3( a^{2}b+b^{3})
\sqrt{3}
$$</p>
<p>and solve the system
$$
\left\{
\begin{array}{c}
a^{3}+9ab^{2}=26 \\
a^{2}b+b^{3}=5.
\end{array}
\right.
$$</p>
<p>A solution is $(a,b)=(2,1)$. Hence $26+15\sqrt{3}=(2+\sqrt{3})^3 $. Using the same method to $26-15\sqrt{3}$, we find $26-15\sqrt{3}=(2-\sqrt{3})^3 $, thus proving $(1)$.</p>
<p>For $(2)$ the very same idea yields</p>
<p>$$
p+q\sqrt{3}=(a+b\sqrt{3})^{3}=a^{3}+9ab^{2}+3( a^{2}b+b^{3}) \tag{3}
\sqrt{3}
$$</p>
<p>and</p>
<p>$$
\left\{
\begin{array}{c}
a^{3}+9ab^{2}=p \\
3( a^{2}b+b^{3}) =q.
\end{array}
\right. \tag{4}
$$</p>
<p>I tried to solve this system for $a,b$ but since the solution is of the form</p>
<p>$$
(a,b)=\Big(x^{3},\frac{3qx^{3}}{8x^{9}+p}\Big),\tag{5}
$$</p>
<p>where $x$ satisfies the <em>cubic</em> equation
$$
64x^{3}-48x^{2}p+( -15p^{2}+81q^{2}) x-p^{3}=0,\tag{6}
$$
would be very difficult to succeed, using this naive approach. </p>
<blockquote>
<p>Is this problem solvable, at least partially?</p>
</blockquote>
<p>Is $\sqrt[3]{p+q\sqrt{3}}+\sqrt[3]{p-q\sqrt{3}}=n$, $(p,q,n)\in\mathbb{N} ^3$ solvable?</p>
| Joseph G. | 73,740 | <p>Using the fact that: $$x^3+y^3=(x+y)(x^2-xy+y^2)$$ and letting $x=m^{\frac{1}{3}}$ and $y=n^{\frac{1}{3}}$, we get the statement: $$m+n=(m^{\frac{1}{3}}+n^{\frac{1}{3}})(m^\frac{2}{3}-(mn)^{\frac{1}{3}}+n^{\frac{2}{3}})$$Maybe this will help.</p>
|
374,619 | <p>In <a href="https://math.stackexchange.com/a/373935/752">this recent answer</a> to <a href="https://math.stackexchange.com/q/373918/752">this question</a> by Eesu, Vladimir
Reshetnikov proved that
$$
\begin{equation}
\left( 26+15\sqrt{3}\right) ^{1/3}+\left( 26-15\sqrt{3}\right) ^{1/3}=4.\tag{1}
\end{equation}
$$</p>
<p>I would like to know if this result can be <em>generalized</em> to other triples of
natural numbers. </p>
<blockquote>
<p><strong>Question</strong>. What are the solutions of the following equation? $$ \begin{equation} \left( p+q\sqrt{3}\right) ^{1/3}+\left(
p-q\sqrt{3}\right) ^{1/3}=n,\qquad \text{where }\left( p,q,n\right)
\in \mathbb{N} ^{3}.\tag{2} \end{equation} $$</p>
</blockquote>
<p>For $(1)$ we could write $26+15\sqrt{3}$ in the form $(a+b\sqrt{3})^{3}$ </p>
<p>$$
26+15\sqrt{3}=(a+b\sqrt{3})^{3}=a^{3}+9ab^{2}+3( a^{2}b+b^{3})
\sqrt{3}
$$</p>
<p>and solve the system
$$
\left\{
\begin{array}{c}
a^{3}+9ab^{2}=26 \\
a^{2}b+b^{3}=5.
\end{array}
\right.
$$</p>
<p>A solution is $(a,b)=(2,1)$. Hence $26+15\sqrt{3}=(2+\sqrt{3})^3 $. Using the same method to $26-15\sqrt{3}$, we find $26-15\sqrt{3}=(2-\sqrt{3})^3 $, thus proving $(1)$.</p>
<p>For $(2)$ the very same idea yields</p>
<p>$$
p+q\sqrt{3}=(a+b\sqrt{3})^{3}=a^{3}+9ab^{2}+3( a^{2}b+b^{3}) \tag{3}
\sqrt{3}
$$</p>
<p>and</p>
<p>$$
\left\{
\begin{array}{c}
a^{3}+9ab^{2}=p \\
3( a^{2}b+b^{3}) =q.
\end{array}
\right. \tag{4}
$$</p>
<p>I tried to solve this system for $a,b$ but since the solution is of the form</p>
<p>$$
(a,b)=\Big(x^{3},\frac{3qx^{3}}{8x^{9}+p}\Big),\tag{5}
$$</p>
<p>where $x$ satisfies the <em>cubic</em> equation
$$
64x^{3}-48x^{2}p+( -15p^{2}+81q^{2}) x-p^{3}=0,\tag{6}
$$
would be very difficult to succeed, using this naive approach. </p>
<blockquote>
<p>Is this problem solvable, at least partially?</p>
</blockquote>
<p>Is $\sqrt[3]{p+q\sqrt{3}}+\sqrt[3]{p-q\sqrt{3}}=n$, $(p,q,n)\in\mathbb{N} ^3$ solvable?</p>
| N. S. | 9,176 | <p>Yikes. I am hopind I didn't do any mistake, but that would be a miracle :)</p>
<p>Let $x= \left( p+q\sqrt{3}\right) ^{1/3}\,;\, y= \left(
p-q\sqrt{3}\right) ^{1/3}$.</p>
<p>Then </p>
<p>$$xy= (p^2-3q^2)^\frac{1}{3} $$</p>
<p>Hence</p>
<p>$$2p=x^3+y^3=(x+y)^3-3xy(x+y)=n^3-3n\sqrt[3]{p^2-3q^2}$$</p>
<p>This shows that $\sqrt[3]{p^2-3q^2}$ must be rational, hence integer.</p>
<p>Let $p^2-3q^2=k^3 (*)$. Then $y=\frac{k}{x}$ and thus</p>
<p>$$x+\frac{k}{x} =n \Rightarrow x^2-nx+k=0 \Rightarrow x= \frac{n \pm \sqrt{n^2-4k}}{2} \,.$$</p>
<p>The two roots of this equation must be $x$ and $y$, and thus we get:</p>
<p>$$x=\frac{n + \sqrt{n^2-4k}}{2} \,;\, y= \frac{n - \sqrt{n^2-4k}}{2} \,.$$</p>
<p>Then</p>
<p>$$p+q\sqrt{3}=\left(\frac{n + \sqrt{n^2-4k}}{2} \right)^3 \,.$$</p>
<p>From here, we get that $\sqrt{n^2-4k} \notin \mathbb Q$ and hence</p>
<p>$$p=\frac{n^3+3n^3-12nk}{8}=\frac{n^3-3nk}{2}$$
$$q\sqrt{3}=\frac{3n^2+n^2-4k}{8}\sqrt{n^2-4k}=\frac{n^2-k}{2}\sqrt{n^2-4k}$$</p>
<p>Thus, $n^2-4k=3u^2 $, for some $u$, hence</p>
<p>$$k=\frac{n^2-3u^2}{4} (***)$$</p>
<p>Thus, we get:</p>
<p>$$p=\frac{n^3+9nu^2}{8}$$
$$q=\frac{3n^2+3u^2}{8}u=\frac{3n^2u+3u^3}{8}$$</p>
<p>With this choice we have</p>
<p>$$p+q\sqrt{3}=(\frac{n+u\sqrt{3}}{2})^3$$
$$p-q\sqrt{3}=(\frac{n-u\sqrt{3}}{2})^3$$</p>
<p>Note that $p,q$ integers if and only if $8|n(n^2+u^2)$ and $8|u(n^2+u^2)$. It is easy to check that the first one cannot happen if $n$ is odd. Thus, $n$ must be even and $ 8|u(n^2+u^2) \Rightarrow 2|u^3 \Rightarrow u$ even.</p>
<p>If $n=2s, u=2t$ we get the general solution</p>
<p>$$p=s^3+9st^2$$
$$q=3s^2t+3t^3$$
$$n=2s$$</p>
<p>Note that in this case, </p>
<p>$$\sqrt[3]{p+q\sqrt{3} }=s+t\sqrt{3}$$
$$\sqrt[3]{p-q\sqrt{3} }=s-t\sqrt{3}$$</p>
|
2,666,409 | <blockquote>
<p>If $SL(2)=\{A\in \mathcal{M}_2: \det(A)=1\}$, find a parametrization around the identity matrix $I_2$ and find the first fundamental form. </p>
</blockquote>
<p>I've proved that $SL(2)$ is an hypersurface in $\mathbb{R}^4$, so it has dimension $3$. However, I don't understand very well what does "around the identity" means. I think that if I have the parametrization I'll be able to compute the first fundamental form. </p>
<p>Thank you. </p>
| K B Dave | 534,616 | <p>Since $\mathrm{SL}_2$ is a quadric hypersurface, one way to get a parametrization near the point $I_2$ is to <a href="https://en.wikipedia.org/wiki/Stereographic_projection#Generalizations" rel="nofollow noreferrer">stereographically project</a> $\mathrm{SL}_2$ onto the tangent hyperplane at $I_2$ from some other point (for instance, $-I_2$).</p>
|
685,918 | <p>I'm doing exercises in Real Analysis of Folland and got stuck on this problem. I don't know how to calculate limit with the variable on the upper bound of the integral. Hope some one can help me solve this. I really appreciate.</p>
<blockquote>
<blockquote>
<p>Show that $\lim\limits_{k\rightarrow\infty}\int_0^kx^n(1-k^{-1}x)^k~dx=n!$</p>
</blockquote>
</blockquote>
<p>Thanks so much for your consideration.</p>
| Community | -1 | <p>We have</p>
<p>$$\int_0^k x^n(1-k^{-1}x)^kdx=\int_0^\infty x^n(1-k^{-1}x)^k\chi_{(0,k)}dx$$
then since
$$x^n(1-k^{-1}x)^k\chi_{(0,k)}\le x^n e^{-x},\;\forall k$$
and the function
$$x\mapsto x^n e^{-x}$$ is integrable on the interval $(0,\infty)$ then by the dominated convergence theorem we have
$$\lim_{k\to\infty}\int_0^k x^n(1-k^{-1}x)^kdx=\int_0^\infty x^ne^{-x}dx=\Gamma(n+1)=n!$$</p>
|
226,449 | <p>Many counting formulas involving factorials can make sense for the case $n= 0$ if we define $0!=1 $; e.g., Catalan number and the number of trees with a given number of vetrices. Now here is my question:</p>
<blockquote>
<p>If $A$ is an associative and commutative ring, then we can define an
unary operation on the set of all the finite subsets of our ring,
denoted by $+ \left(A\right) $ and $\times \left(A\right)$. While it
is intuitive to define $+ \left( \emptyset \right) =0$, why should
the product of zero number of elements be $1$? Does the fact that $0! =1$ have anything to do with 1 being the multiplication unity of integers?</p>
</blockquote>
| ncmathsadist | 4,154 | <p>Here are two resons.</p>
<ol>
<li><p>We can define $n!$ to be the number of rearrangements of $n$ distinct objects in a list. The empty list has one rearrangement: itself. </p></li>
<li><p>We can define $n!$ as the product of all positive integers $k$ with $1\le k \le n$. If $n$ is zero, we have an empty product. An empty product must be neutral for multiplication, so it must be 1. </p></li>
</ol>
<p>Take your pick.</p>
|
1,873,194 | <p>Entropy of random variable is defined as:</p>
<p>$$H(X)= \sum_{i=1}^n p_i \log_2(p_i)$$</p>
<p>Which as far as I understand can be interpreted as how many yes/no questions one would have to ask on average, to find out the value of the random variable $X$.</p>
<p>But what if the log base is changed to for example e? What would the interpretation be then? Is there an intuitive explanation?</p>
| Olivier Oloa | 118,798 | <p>One has
$$\log_a(x)=\frac{\log b}{\log a}\:\log_b(x)$$
thus
$$H(X)= \sum_{i=1}^n p_i \log_a(p_i)=\frac{\log b}{\log a}\sum_{i=1}^n p_i \log_b(p_i)$$ or
$$
H_a(X)=\frac{\log b}{\log a}H_b(X).
$$</p>
|
3,104,051 | <p>I have the work of the proof done, but at the end after showing
<span class="math-container">$3^{2(n+1)} - 1=9(3^{2n} - 1)+8$</span>
I make the statement that since <span class="math-container">$9(3^{2n} - 1)$</span> is a multiple of 8 and 8 is a multiple of 8 then <span class="math-container">$3^{2n} - 1$</span> is a multiple of 8. I need to quote the facts about divisibility I use in this statement and I'm not sure exactly what to put for that, I have already included the definition of divisibility in the proof</p>
| B. Goddard | 362,009 | <p>If you want to add a detail, you're assuming <span class="math-container">$3^{2n}-1$</span> is a divisible by <span class="math-container">$8$</span>, so it equals <span class="math-container">$8w$</span> for some integer <span class="math-container">$w$</span>. Then your last expression is</p>
<p><span class="math-container">$$9(3^{2n}-1)+8 = 9(8w)+8 = 8(9w+1),$$</span></p>
<p>which is a multiple of <span class="math-container">$8$</span>.</p>
|
537,968 | <p>Let $|\alpha|<1$ and $\psi_{\alpha}(z)=(\alpha-z)/(1-\bar\alpha z)$. I want to prove that $$\frac 1 {\pi} \int\int_{\mathbb{D}}|{\psi_{\alpha}}^{'}|dxdy = \frac{1-|\alpha|^2}{|\alpha|^2}\log\frac{1}{1-|\alpha|^2}$$</p>
<p>I calculated ${\psi_\alpha}^{'}(z)=(|\alpha|^2-1)/(1-\bar\alpha z)^2$. I substituted it and use $z=re^{i \theta}$ and need to integrate $1/|1-\bar\alpha re^{i \theta}|^2$ along circle of radius r (fixed). But how can I do this?</p>
| Daniel Fischer | 83,702 | <p>Without loss of generality, you can assume $0 < \alpha < 1$. Let us call $\rho := \alpha r$. Then</p>
<p>$$\frac{1}{\lvert 1 - \rho e^{i\theta}\rvert^2} = \frac{1}{1+\rho^2 - 2\rho\cos\theta},$$</p>
<p>and you can integrate that using one of the classical methods, like the residue theorem.</p>
|
4,552,955 | <p>I'm solving a probability problem, and I've ended up with this sum:</p>
<p><span class="math-container">$$\sum\limits_{k=0}^{n-a-b}\binom{n-a-b}{k}(a+k-1)!(n-a-k)!$$</span></p>
<p>WolframAlpha says I should get the answer <span class="math-container">$\frac{n!}{a\binom{a+b}{a}}$</span>, but I don't see how to get there. I tried to get to something containing <span class="math-container">$\binom{a+k-1}{k}$</span> so that I could use the Hockey Stick Theorem, but I wasn't successful.</p>
<p>So any hints would be very welcome, thanks for any help</p>
| Marko Riedel | 44,883 | <p>We seek to find a closed form of</p>
<p><span class="math-container">$$(n-1)! \sum_{k=0}^{n-a-b}
{n-a-b\choose k} {n-1\choose n-a-k}^{-1}$$</span></p>
<p>where <span class="math-container">$n\gt a+b$</span> and <span class="math-container">$a,b\ge 1.$</span></p>
<p>Recall from <a href="https://math.stackexchange.com/questions/4316307/">MSE
4316307</a> the
following identity which was proved there: with <span class="math-container">$1\le k\le n$</span></p>
<p><span class="math-container">$$\frac{1}{k} {n\choose k}^{-1}
= [z^n] \log\frac{1}{1-z} (z-1)^{n-k}.$$</span></p>
<p>We get for our sum</p>
<p><span class="math-container">$$(n-1)! \sum_{k=0}^{n-a-b} {n-a-b\choose k}
(n-a-k) [z^{n-1}] \log\frac{1}{1-z} (z-1)^{a-1+k}
\\ = (n-1)! \sum_{k=0}^{n-a-b} {n-a-b\choose k}
(k+b) [z^{n-1}] \log\frac{1}{1-z} (z-1)^{n-1-b-k}.$$</span></p>
<p>We get two pieces, the first is</p>
<p><span class="math-container">$$b [z^{n-1}] \log\frac{1}{1-z} (z-1)^{n-1-b}
\sum_{k=0}^{n-a-b} {n-a-b\choose k} (z-1)^{-k}
\\ = b [z^{n-1}] \log\frac{1}{1-z} (z-1)^{n-1-b}
\left[1+\frac{1}{z-1}\right]^{n-a-b}
\\ = b [z^{a+b-1}] \log\frac{1}{1-z} (z-1)^{a-1}
\\ = {a+b-1\choose b}^{-1}.$$</span></p>
<p>The second is</p>
<p><span class="math-container">$$ (n-a-b) [z^{n-1}] \log\frac{1}{1-z} (z-1)^{n-1-b}
\sum_{k=1}^{n-a-b} {n-a-b-1\choose k-1} (z-1)^{-k}
\\ = (n-a-b) [z^{n-1}] \log\frac{1}{1-z} (z-1)^{n-2-b}
\sum_{k=0}^{n-a-b-1} {n-a-b-1\choose k} (z-1)^{-k}
\\ = (n-a-b) [z^{n-1}] \log\frac{1}{1-z} (z-1)^{n-2-b}
\left[1+\frac{1}{z-1}\right]^{n-a-b-1}
\\ = (n-a-b) [z^{a+b}] \log\frac{1}{1-z} (z-1)^{a-1}
\\ = \frac{n-a-b}{b+1} {a+b\choose b+1}^{-1}.$$</span></p>
<p>Collecting everything we find</p>
<p><span class="math-container">$$(n-1)! {a+b\choose b}^{-1}
\left[ \frac{a+b}{a} + \frac{n-a-b}{b+1} \frac{b+1}{a} \right]
\\ = \frac{n!}{a} {a+b\choose b}^{-1}.$$</span></p>
<p>This is the claim.</p>
|
2,329,600 | <p>I haven't studied any maths since I was at university 20 years ago. Yesterday, however, I came across a pair of equations in an online article about gaming and I couldn't understand how they'd been derived. </p>
<p>Here's the scenario. If we make a single trial of generating a number between 1 and 20, there's an even 5% chance of getting any given number. Therefore to get "at least", say, a 11, you can just add up the percentages and subtract them from 100: 100 - (5% x 10) (because 1 is the minimum, not zero) = 50%. </p>
<p>What happens, though, if you make two trials and take either the highest or the lowest number? How then do you calculate the chance of getting "at least" a certain number? For this, I was given the following equations:</p>
<ul>
<li>If taking the highest, it's <em>1-(1-P)^2</em></li>
<li>If taking the lowest it's <em>P^2</em> </li>
</ul>
<p>It's clear these are correct. Rolling "at least" an 11 when taking the lowest values is therefore 0.5 * 0.5 ... 0,25. But what I want to understand is how someone arrived at these equations without using trial and error.</p>
<p>From the limited maths I can recall, P^2 looks not unlike the binomial distribution formula of μ = np - but of course it's raising the probability to the power of two rather than multiplying by two for two trials.</p>
<p>Can someone please explain to me where these come from?</p>
| Matti P. | 432,405 | <p>If I interpret correctly, the question you're asking is: Choosing randomly two numbers between 1 and 20, what is the probability that at least one of them is at least 11?</p>
<p>To answer this, we can look at the three cases where this occurs:</p>
<ul>
<li>The first number is less than $11$ and the second is at least 11. The probability of this happening is $\frac{10}{20}\times \frac{10}{20} = \frac{1}{4}$</li>
<li>The first number is larger than 11 and the second is less than 11. This has the same probability as the previous point, $\frac{1}{4}$</li>
<li>Both of the numbers are at least 11. The probability is, by a coincidence, $\frac{10}{20} \times \frac{10}{20} = \frac{1}{4}$</li>
</ul>
<p>All of these events are separate, so the total probability is the sum of these, or $\frac{3}{4}$.
Do you feel you can generalize the case for arbitrary values?</p>
|
2,150,832 | <p>I don't understand this equation $\int_0^t ds \int_0^{t'} ds' \delta(s-s')= \min(t,t')$.
I tried to work with the property of the dirac delta function that $\int_a^b \delta(x-c)dx = 1$ if $c \in [a,b]$, but I can't see how I can obtain the minimum. Can someone help me? </p>
<p>Thank you in advance!</p>
| David Holden | 79,543 | <p>let $\chi_t(x)$ be the indicator function for the interval $[0,t]$. then, as you point out:
$$
\int_0^{t'}ds'\delta(s-s') = \chi_{t'}(s)
$$
but now:
$$
\int_0^t\chi_{t'}(s) ds = \int_0^{\infty} \chi_{t}(s)\chi_{t'}(s) ds = \int_0^{\infty} \chi_{\min(t,t')}(s) ds = \min(t,t')
$$</p>
|
307,545 | <p>If $\gcd(a,b)=1$, how can I find the values that $\gcd(a+b,a^2+b^2)$ can possibly take? I can't find a way to use any of the elemental divisibility and gcd theorems to find them. </p>
| Pedro | 23,350 | <p>Note that $a^2+b^2=(a+b)^2-2ab$. Thus $a^2+b^2\equiv -2ab\mod a+b$ and you need to find $(a+b,-2ab)=(a+b,2ab)$. Can you move on?</p>
<p><strong>ADD</strong> Note that $a,b$ cannot be both even. If one is odd and the other is even, then $2\not\mid a+b$, so $(a+b,2ab)=(a+b,ab)$. But if $p>2$ is a prime with $p\mid a+b,ab$ then $p\mid a$ or $p\mid b$, since $p\mid ab$. But in any case, since $p\mid a+b$, this would give $p \mid b$ (if $p\mid a$) or $p \mid a$ (if $p\mid b)$, contrary to $(a,b)=1$. Thus $(a+b,ab)=1$. </p>
<p>If $a,b$ are both odd, then $a+b$ is even and $2\mid (a+b,2ab)$. And again, if $p$ is an odd prime factor $p\mid 2ab\implies p\mid ab$ and the same argument above goes through. Thus whenever $(a,b)=1$, $$(a+b,a^2+b^2)=\begin{cases} 1 &\text{if one of } a,b \text{ is odd}\\2&\text{if both} a,b\text{ are odd}\end{cases}$$</p>
|
7,223 | <p>I want to produce a <em>Mathematica</em> Computable Document in which <code>N</code> appears as a variable in my formulae. But <code>N</code> is a reserved word in the <em>Mathematica</em> language. Is there a way round this other than using a different symbol? It seems a severe limitation if you cannot use <em>Mathematica</em> to generate papers in which <code>N</code> is employed as a variable.</p>
| Rojo | 109 | <p>Ok, I'm late here, but the first thing I would have answered hasn't been answered already.</p>
<p>If it's a one-cell-er, you can use <code>Module</code></p>
<pre><code>Module[{N},
N = 8;
N + 3]
</code></pre>
<p>This can never bring you trouble with internal definitions since that N you see is not actually N.</p>
<p>If it comprises multiple cells, you can use contexts, and interpret the RED N as a reminder that you're doing weird things. You can always access the regular N function by <code>System`N</code>, and return to normal by either <code>Remove[N]</code>, or <code>$ContextPath=Rest@$ContextPath</code></p>
<pre><code>prvt`N;
PrependTo[$ContextPath, "prvt`"];
</code></pre>
|
2,454,895 | <p>I don't know how to solve this equation:$$(1)\quad e^ {-x} = -\ln x$$</p>
<p>$x$ should be the abscissa of the point $P$ where the two functions meet on the plan and $$ P \in f(x) :y=x$$</p>
<p>so $(1)$ should be equal to $$ e^{-x}=x=-\ln x$$ </p>
<p>How do I solve this?</p>
| Stu | 460,772 | <p>$\quad e^ {-x} = -\ln x\iff e^ {-x}+\ln x=0 $</p>
<p>Let $g(x)=e^ {-x}+\ln x$</p>
<p>$g(1)=\dfrac{1}{e}>0$ and $\displaystyle \lim_{x\to 0} g(x)=-\infty$</p>
<p>The Intermediate value theorem guaranties there exists $c\in (0,1)$ such that $g(c)=0 \iff \exists c\in (0,1) $ such that $e^ {-c} = -\ln c$</p>
|
2,291,852 | <p>$$\int_\pi^\infty{\frac{x \cos x}{x^2-1}dx}$$</p>
<p>So the only think I came up with was to take an absolute value of ${\frac{x \cos x}{x^2-1}}$ and by comparison test the integral does not converge. </p>
<p>But I see it's not very close to the solution, so what should I do?</p>
| Joe | 107,639 | <p>Let's call your set $A$.</p>
<p>Being your $A\subseteq\Bbb R$, you know that it is compact iff it is closed and bounded.</p>
<p>Now $A$ is closed since it is counterimage of the singleton $\{1\}$ (which is closed) thru the continous function $f(x)=x^2+e^x$.</p>
<p>Finally, $A$ is bounded since</p>
<p>$$
\lim_{x\to\pm\infty}f(x)=+\infty\;\;.
$$</p>
|
2,291,852 | <p>$$\int_\pi^\infty{\frac{x \cos x}{x^2-1}dx}$$</p>
<p>So the only think I came up with was to take an absolute value of ${\frac{x \cos x}{x^2-1}}$ and by comparison test the integral does not converge. </p>
<p>But I see it's not very close to the solution, so what should I do?</p>
| MooS | 211,913 | <p>Since you also asked to prove that the set is finite (which is of course much stronger than compact):</p>
<p>If $f(x)=x^2+e^x-1$ has infinitely many roots, then by Rolle's theorem the same holds for $f'$ and thus also $f''$. But $f''(x)=2+e^x$ has no roots at all.</p>
|
2,163,494 | <p>Let $f: A\to B; \ g,h:B\to A$ and $f\circ g = I_B$ and $h \circ f = I_A$</p>
<p>I want to simply state that for any function $f$ if $f \circ h = I_A$ then it must be that $h = f^{-1}$ but that seems incomplete to me. What can I do for fixing this?</p>
| Nigel Overmars | 96,700 | <p>To put the other answer a bit differently: We have that $A \subseteq \overline{A}$, and since $A$ is bounded, $\overline{A}$ is compact. It follows that $f(\overline{A})$ is compact and hence bounded. Since we have that $f(A) \subseteq f(\overline{A})$, the result follows.</p>
|
3,518,719 | <p>Evaluate <span class="math-container">$$\lim_{n \to \infty} \sqrt[n^2]{2^n+4^{n^2}}$$</span></p>
<p>We know that as <span class="math-container">$n\to \infty$</span> we have <span class="math-container">$2^n<<2^{2n^2}$</span> and therefore the limit is <span class="math-container">$4$</span></p>
<p>In a more formal way I started with:</p>
<p><span class="math-container">$$\log(L)=\lim_{n \to \infty} \log(2^n+4^{n^2})^{\frac{1}{n^2}}=\lim_{n \to \infty}\frac{1}{n^2}\log(2^n+2^{2n^2})$$</span></p>
<p>Continuing to <span class="math-container">$$\log(L)=\lim_{n \to \infty}\frac{1}{n^2}\log\left[2^n(1+2^{2n})\right]$$</span></p>
<p>Did not help much</p>
<p>As I arrived to <span class="math-container">$$\log(L)=\lim_{n \to \infty}\frac{1}{n^2}\left[\log(2^n)+\log(1+2^{2n})\right]=\lim_{n \to \infty}\frac{1}{n^2}\log(2^n)+\lim_{n \to \infty}\frac{1}{n^2}\log(1+2^{2n})=0+\lim_{n \to \infty}\frac{1}{n^2}\log(1+2^{2n})$$</span></p>
| Ted Shifrin | 71,348 | <p>Of course it's true for a plane curve that <span class="math-container">$\dot b = -\tau n$</span>. Since <span class="math-container">$b$</span> is constant, we deduce that <span class="math-container">$\tau = 0$</span>. There's no contradiction here.</p>
|
2,452,777 | <p>Let X be a topological space, $\mathcal{U} = \{U_\alpha\}_\alpha$ an open cover of $X$ and $\mathcal{F}$ a presheaf of abelian groups on $X$. Then one can define the Čech cohomology groups of $\mathcal{U}$ with values in $\mathcal{F}$:
\begin{equation}
\check{H}^k(\mathcal{U}, \mathcal{F})
\end{equation}
The Čech cohomology groups of $X$ with values in $\mathcal{F}$ are usually defined by the inductive limit over the set of open covers ordered by refinement:
\begin{equation}
\check{H}^k(X, \mathcal{F}) = \varinjlim_{\mathcal{U}} \check{H}^k(\mathcal{U}, \mathcal{F})
\end{equation}</p>
<p>But I am confused with the notion of open cover. Indeed one can consider the open cover as a set $\{U_\alpha\}_\alpha\subset \mathcal{P}(X)$ but also as a map $A\to \mathcal{P}(X)$ for some index set $A$. The open cover as a set is the range of the open cover as a map. But the distinction is real, since in the second case you can consider repeated open sets in the cover.</p>
<p>It seems to me that it is the second notion that is used to define Čech cohomology but then one cannot talk about "the set of open covers ordered by refinement" and then take the inductive limit since in the second case you don't even have a set of open overs, but rather a class.</p>
<p>So which alternative is used for open covers?</p>
| Daniel Robert-Nicoud | 60,713 | <p>You can give the index set $A$ a poset structure by $a\le b$ if $U_a\subseteq U_b$. Then taking the limit over the associated category $A$ of the functor
$$\mathcal{U}:A\longrightarrow P(X)$$
given by $\mathcal{U}(a):=U_a$ is the same as taking the limit over the cover ordered by refinement.</p>
|
300,531 | <p>Prove that : $$ \gamma=-\int_0^{1}\ln \ln \left ( \frac{1}{x} \right) \ \mathrm{d}x.$$</p>
<p>where $\gamma$ is Euler's constant ($\gamma \approx 0.57721$).</p>
<hr>
<p>This integral was mentioned in <a href="http://en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant">Wikipedia</a> as in <a href="http://mathworld.wolfram.com/Euler-MascheroniConstant.html">Mathworld</a> , but the solutions I've got uses corollaries from <a href="http://en.wikipedia.org/wiki/Harmonic_number">this theorem</a>. Can you give me a simple solution (not using much advanced theorems) or at least some hints.</p>
| Chappers | 221,811 | <p>Another way, from another definition: by the Dominated Convergence Theorem,
$$ \int_0^{\infty} e^{-u} \log{u} \, du = \lim_{n \to \infty} \int_0^n \left( 1 - \frac{u}{n} \right)^{n-1} \log{u} \, du. $$
Then, changing variables to $v=1-u/n$,
$$ \begin{align}
\int_0^n \left( 1 - \frac{u}{n} \right)^{n-1} \log{u} \, du &= n \int_0^1 v^{n-1} \log{( n(1-v))} \, dv \\
&= n\log{n} \int_0^1 v^{n-1} \, dv + (n+1) \int_0^1 v^{n-1} \log{(1-v)} \, dv \\
&= \log{n} - n \int_0^1 \sum_{k=1}^{\infty} \frac{v^{k+n-1}}{k} \, dv \\
&= \log{n} - n \sum_{k=1}^{\infty} \int_0^1 \frac{v^{k+n-1}}{k} \, dv \\
&= \log{n} - n \sum_{k=1}^{\infty} \frac{1}{k(k+n)} \\
&= \log{n} - \sum_{k=1}^{\infty} \left( \frac{1}{k} - \frac{1}{n+k} \right) \\
&= \log{n} - \sum_{k=1}^{n} \frac{1}{k},
\end{align} $$
using uniform convergence and partial fractions. But this is precisely the definition
$$ \gamma = \lim_{n \to \infty} \sum_{k=1}^{n} \frac{1}{k} - \log{n}. $$</p>
|
1,722,443 | <p>I am unsure how to solve the following problem. I was able to find similar questions, but had trouble understanding them since they did not show full solutions.</p>
<p>The question:</p>
<p>Find ALL solutions (between $1$ & $40$) to the equation $25x \equiv 10 \pmod{40}$.</p>
| N. F. Taussig | 173,070 | <p>Let's use the definition of congruence. $a \equiv b \pmod{n} \iff a = b + kn$ for some integer $k$. Hence, $25x \equiv 10 \pmod{40}$ means $$25x = 10 + 40k$$ for some integer $k$. Dividing each side of the equation $25x = 10 + 40k$ by $5$ yields $$5x = 2 + 8k$$
for some integer $k$. Thus,
$$5x \equiv 2 \pmod{8}$$
Since $\gcd(5, 8) = 1$, $5$ has a multiplicative inverse modulo $8$. To isolate $x$, we must multiply both sides of the congruence $5x \equiv 2 \pmod{8}$ by the multiplicative inverse of $5$ modulo $8$. To find the multiplicative inverse, we use the <a href="https://en.wikipedia.org/wiki/Extended_Euclidean_algorithm" rel="nofollow">extended Eucldean algorithm</a>.<br>
\begin{align*}
8 & = 5 + 3\\
5 & = 3 + 2\\
3 & = 2 + 1\\
2 & = 2 \cdot 1
\end{align*}
Working backwards through this partial sequence of <a href="https://en.wikipedia.org/wiki/Fibonacci_number" rel="nofollow">Fibonacci numbers</a> to solve for $1$ as a linear combination of $5$ and $8$ yields
\begin{align*}
1 & = 3 - 2\\
& = 3 - (5 - 3)\\
& = 2 \cdot 3 - 5\\
& = 2(8 - 5) - 5\\
& = 2 \cdot 8 - 3 \cdot 5
\end{align*}
Therefore, $1 \equiv -3 \cdot 5 \pmod{8}$. Hence, $-3 \equiv 5^{-1} \pmod{8}$. Since $-3 \equiv 5 \pmod{8}$, we have $5 \equiv 5^{-1} \pmod{8}$. Thus, $5 \cdot 5x \equiv x \pmod{8}$. Hence,
\begin{align*}
5x & \equiv 2 \pmod{8}\\
5 \cdot 5x & \equiv 5 \cdot 2 \pmod{8}\\
x & \equiv 10 \pmod{8}\\
x & \equiv 2 \pmod{8}
\end{align*}
What remains is for you to find the solutions of the congruence $x \equiv 2 \pmod{8}$ such that $0 \leq x < 40$. </p>
|
1,908,820 | <p>The question is: </p>
<p>If $$\int_3^9f(x) dx = 7$$
evaluate
$$\int_3^9 2f(x)+1 dx$$</p>
<p>I know that you can factor the 2 outside of the integral. But, then I am still left with a '+ 1' inside the integral that when I take the integral of becomes $x$. So then would I proceed to stating this:
$$=2\int_3^9 f(x) +1 dx$$
$$=2\int_3^9 f(x)dx +\int_3^91 dx$$
$$= 2\left((7)+\left[x\right]_3^9\right)$$
$$= 2(7+(9-3))$$
$$=26$$</p>
<p>Is this right because the answer in my book says 20, which leads me to believe that they did not multiply the 2 to everything but only to the 7.</p>
<p>$$= 2(7)+\left[x\right]_3^9$$
$$=2(7) + 6$$
$$= 14 +6$$
$$=20$$</p>
<p>Is my interpretation of the textbook's answer correct? If so, if I factor the 2 out of the integral, why does it only apply to the 7? If not, then what am I doing wrong in my first solution?</p>
| ncmathsadist | 4,154 | <p>Look at your algebra; you have extended the parenthesis too far. The factor of $2$ does not extend to the second integral.</p>
|
1,908,820 | <p>The question is: </p>
<p>If $$\int_3^9f(x) dx = 7$$
evaluate
$$\int_3^9 2f(x)+1 dx$$</p>
<p>I know that you can factor the 2 outside of the integral. But, then I am still left with a '+ 1' inside the integral that when I take the integral of becomes $x$. So then would I proceed to stating this:
$$=2\int_3^9 f(x) +1 dx$$
$$=2\int_3^9 f(x)dx +\int_3^91 dx$$
$$= 2\left((7)+\left[x\right]_3^9\right)$$
$$= 2(7+(9-3))$$
$$=26$$</p>
<p>Is this right because the answer in my book says 20, which leads me to believe that they did not multiply the 2 to everything but only to the 7.</p>
<p>$$= 2(7)+\left[x\right]_3^9$$
$$=2(7) + 6$$
$$= 14 +6$$
$$=20$$</p>
<p>Is my interpretation of the textbook's answer correct? If so, if I factor the 2 out of the integral, why does it only apply to the 7? If not, then what am I doing wrong in my first solution?</p>
| Enrico M. | 266,764 | <p>$$\int_3^9 2\ f(x) + 1\ \text{d}x = 2\int_3^9 f(x)\ \text{d}x + \int_3^9 1\ \text{d}x$$</p>
<p>Since you already know what the $f(x)$ integral is, you just have to evaluate the other trivial one.</p>
<p>$$ = 2\cdot 7 + x\bigg|_3^9 = 14 + (9 - 3) = 20$$</p>
<p>The $2$ factor does multiply $f(x)$ only.</p>
<p>You have</p>
<p>$$2\ f(x) + 1$$</p>
<p>Not</p>
<p>$$2\left(f(x) + 1\right)$$</p>
|
3,499,352 | <p>Define a function <span class="math-container">$f:X\to Y$</span> to be a strict contraction, if there exists a <span class="math-container">$c\in [0,1[$</span> such that
<span class="math-container">$$\forall x,y \in X: d_Y\left( f(x),f(y)\right)\le cd_X(x,y)$$</span>
Now, consider a metric space <span class="math-container">$(X,d)$</span> and a sequence <span class="math-container">$(f_n)_n$</span> of strict contractions <span class="math-container">$f_n:X \to X$</span>, which is converging to a function <span class="math-container">$f$</span>. My question is under what conditions will the limit function be a strict contraction again?</p>
<p>I made the following estimate
<span class="math-container">$$d(f(x),f(y))\le d(f(x),f_n(x))+d(f_n(x),f_n(y))+d(f_n(x),f(y))$$</span>
Now each <span class="math-container">$f_n$</span> has its own contraction factor <span class="math-container">$c_n$</span> such that this can be rewritten as
<span class="math-container">$$d(f(x),f(y))\le d(f(x),f_n(x))+c_nd(x,y)+d(f_n(x),f(y))$$</span>
for all <span class="math-container">$n\in \mathbb{N}$</span>. Now by taking the limit <span class="math-container">$n\to\infty$</span> we obtain
<span class="math-container">$$d(f(x),f(y))\le c d(x,y)$$</span>
where <span class="math-container">$c = \sup\{c_n|n\in\mathbb{N}\}$</span>. This <span class="math-container">$c$</span> is not necessarily strictly smaller than <span class="math-container">$1$</span>.</p>
<p>So I was looking for additional conditions such that we are guaranteed the limit function to be a strict contraction (i.e. <span class="math-container">$c<1$</span>). The easiest way, would be to demand that all <span class="math-container">$f_n$</span> have the same contraction factor. But that's quite a strong constrain, so my suggestions would be to demand the convergence <span class="math-container">$f_n\to f$</span> to be uniformly and <span class="math-container">$(X,d)$</span> to be a complete metric space. </p>
<p><strong>My sketch of the proof</strong>:
Each <span class="math-container">$f_n$</span> has fixed point <span class="math-container">$x_n^*$</span>. In addition we have uniform convergence, so there exists some <span class="math-container">$n_0$</span> such that <span class="math-container">$$d(f_n(x_n^*),f_m(x_m^*))=d(x^*_n,x^*_m)< \epsilon$$</span>
for all <span class="math-container">$n,m\ge n_0$</span>. So <span class="math-container">$(x^*_n)_n$</span> is Cauchy sequence in complete X, so it converges to some <span class="math-container">$x^*$</span>. Using the fact that uniform convergence, we have
<span class="math-container">$$x^*=\lim_{n\to\infty}x^*_n = \lim_{n\to \infty}f_n(x_n^*) = f(x^*)$$</span>
This shows that <span class="math-container">$f$</span> has a unique fixed point in a complete metric space.</p>
<p><strong>Edit:</strong> But the existence of a fixed point does not imply that <span class="math-container">$f$</span> will be a strict contraction. So how can I make sure <span class="math-container">$f$</span> to be a strict contraction? </p>
| Fimpellizzeri | 173,410 | <p><span class="math-container">$f_c(x) = cx$</span> is a strict contraction whenever <span class="math-container">$|c|<1$</span>.
Let <span class="math-container">$c_n = 1-1/n$</span> so that <span class="math-container">$c_n\to 1$</span>.</p>
<p>With <span class="math-container">$f(x) = x$</span> being the limit of <span class="math-container">$f_n = f_{c_n}$</span>, we have <span class="math-container">$|f_n(x) - f(x)| = |x|/n$</span> so the convergence is uniform on bounded sets but the limit function is not a strict contraction.</p>
|
64,881 | <p>I am having trouble with this problem from my latest homework.</p>
<p>Prove the arithmetic-geometric mean inequality. That is, for two positive real
numbers $x,y$, we have
$$ \sqrt{xy}≤ \frac{x+y}{2} .$$
Furthermore, equality occurs if and only if $x = y$.</p>
<p>Any and all help would be appreciated.</p>
| lhf | 589 | <p>$\phantom{Proof without words.........}$
<a href="https://i.stack.imgur.com/9OXeY.png" rel="noreferrer"><img src="https://i.stack.imgur.com/9OXeY.png" alt="enter image description here"></a></p>
|
2,864,585 | <p>I tried to calculate the Hessian matrix of linear least squares problem (L-2 norm), in particular:</p>
<p>$$f(x) = \|AX - B \|_2$$
where $f:{\rm I\!R}^{11\times 2}\rightarrow {\rm I\!R}$</p>
<p>Can someone help me?<br>
Thanks a lot.</p>
| mathreadler | 213,607 | <p>Yep squared norm is better. </p>
<p>$$\|AX-B\|_F^2 = (AX-B)^T(AX-B) = \Big/\text{ simplify }\Big/ = X^TA^TAX + \text{linear & const terms}$$</p>
<p>Now you should see what the Hessian is. If you still don't you can check out <a href="https://en.wikipedia.org/wiki/Hessian_matrix#Use_in_optimization" rel="nofollow noreferrer">Hessian matrix - use in optimization.</a></p>
<hr>
<p>If linear problem then the Hessian is directly in the second order term, if non-linear problem solved by <em>trust-region</em> approach it is matrix on second term of Taylor expansion around <em>trust region</em>.</p>
|
2,864,585 | <p>I tried to calculate the Hessian matrix of linear least squares problem (L-2 norm), in particular:</p>
<p>$$f(x) = \|AX - B \|_2$$
where $f:{\rm I\!R}^{11\times 2}\rightarrow {\rm I\!R}$</p>
<p>Can someone help me?<br>
Thanks a lot.</p>
| Rodrigo de Azevedo | 339,790 | <p>Let $f : \mathbb R^{m \times n} \to \mathbb R$ be defined by</p>
<p>$$f (\mathrm X) := \frac 12 \| \mathrm A \mathrm X - \mathrm B \|_{\text{F}}^2 = \frac 12 \| (\mathrm I_n \otimes \mathrm A) \, \mbox{vec} (\mathrm X) - \mbox{vec} (\mathrm B) \|_2^2$$</p>
<p>where $\mbox{vec}$ is the <a href="https://en.wikipedia.org/wiki/Vectorization_(mathematics)" rel="nofollow noreferrer">vectorization</a> operator and $\otimes$ is the <a href="https://en.wikipedia.org/wiki/Kronecker_product" rel="nofollow noreferrer">Kronecker</a> product. Thus, the Hessian of $f$ is </p>
<p>$$(\mathrm I_n \otimes \mathrm A)^\top (\mathrm I_n \otimes \mathrm A) = \mathrm I_n \otimes \mathrm A^\top \mathrm A$$</p>
|
858,716 | <p>I'm self-studying real analysis using Abbott's text "Understanding Analysis." I'm trying to think out/prove as much on my own as I can, so I am working on proving the Nested Interval Property (Theorem 1.4.1 in the book) using "just" the Axiom of Completeness. The author does prove it in the book, but as I say, I like to try to prove things myself before reading the author's proof.</p>
<p>So I have what I think is a convincing proof, but I would just like anyone who's willing to take a look and tell me if this is a convincing proof or not. Please don't give me a proof that works. It is best if you just tell me if you are not convinced and why.</p>
<p>Here is the theorem:</p>
<p>For each natural number n, assume we've been given a closed interval $I_n = [a_n, b_n] = \{x \in \mathbb R : a_n \le x \le b_n\}$. Assume also that each $I_n$ contains $I_{n+1}$. Then the resulting nested sequence of closed intervals has a nonempty intersection.</p>
<p>Here is my proof:</p>
<p>For any natural number n, consider the set $\{a_1, a_2, a_3, ...\}$. (I drew a diagram to work this out. Hopefully it is not unclear.) This set is bounded above, i.e. by any member of the set $\{b_1, b_2, b_3, ..., b_n\}$. By the axiom of completeness, therefore, the set has a supremum. Now suppose the intersection of the sequence of nested intervals is empty. Then for some natural number $n$, $[a_n, b_n]$ is empty. This means there can be no such number as $a_n$, for since $[a_n, b_n]$ is defined to include $a_n$, if $a_n$ existed, the set would not be empty. However, if there is no such number $a_n$, then the nonempty, bounded set $\{a_1, a_2, a_3, ...\}$ can have no supremum. This contradicts the axiom of completeness. Therefore the intersection cannot be empty. QED. </p>
<p>Right now I'm not so concerned with finer points of the proof, except insofar as they contribute to whether or not it is convincing. </p>
<p>Thank you.</p>
<p>EDIT: I believe I now have a convincing proof. To prove that the intersection of closed, nested intervals is non-empty, I need to show that for every natural number n, there is some real number x such that $a_n \le x \le b_n$ I broke this into two parts to make it easier for my simple brain. I need to show that there is some number x that is greater than or equal to a_n for every n, and I need to show that there is some number y that is less than or equal to b_n for every n. These need not be the same number. </p>
<p>So now I form the sets A = {$a_1, a_2, ..., a_n$} and B = {$b_1, b_2, ..., b_n$}. Since the intervals are nested, every member of B is an upper bound for A. Therefore, by the axiom of completeness, A has a supremum. Call it s. By definition of supremum $s \ge a$ for every a in A. Likewise, by definition of supremum and because every member of B is an upper bound of A, we have $s \le b$ for every b in B. This shows that for every a and for every b, there is a real number s such that $a \le s \le b$ This is enough to show that the intersection of the nested intervals always contains at least one real number, i.e that it is non-empty. QED.</p>
<p>My only concern with this is that the sets A and B have infinite number of members. But I don't think this matters, since they are both bounded. But it makes me less than comfortable with the "for every" quantifier, knowing that there are infinitely many. I considered induction but I don't see a remotely elegant way of doing that. </p>
| Nannes | 153,221 | <p>$f(x,y)=x/(x-y)$</p>
<p>Just use the definition</p>
<p>$\bigtriangledown f(x,y)=(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y})$</p>
<p>$\frac{\partial f}{\partial x}=-y/(x-y)^2$</p>
<p>$\frac{\partial f}{\partial y}=x/(x-y)^2$</p>
<p>so</p>
<p>$\bigtriangledown f(x,y)=(-y/(x-y)^2,x/(x-y)^2)$</p>
|
252,767 | <p>I'm looking for a tangible example of a free abelian group whose quotient with a subgroup is not free abelian. There's a theorem that says that every abelian group is a quotient of some free group, but I'm looking for a more exact example.</p>
| Cameron Buie | 28,900 | <p>The most obvious example of a free group is $\Bbb Z$--which is the free group of one generator. (Can you see why?)</p>
<p>Now consider any finite cyclic group--necessarily abelian but not a free abelian group, and (isomorphic with) a quotient of $\Bbb Z$.</p>
|
4,302,855 | <p>I have tried setting up multiple systems of equations using many known volumes but I always seem to come up short. My last attempt was a hollow cylinder but that leaves you with three unknowns in only two sim. equations (for V and S.A). Can anyone help?</p>
| Second Person Shooter | 3,901 | <p>There should be many closed curves satisfying the given constraints.</p>
<p><a href="https://i.stack.imgur.com/PRaRn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PRaRn.png" alt="enter image description here" /></a>
If <span class="math-container">$A$</span>, <span class="math-container">$C$</span>, <span class="math-container">$R_a$</span>, and <span class="math-container">$R_c$</span> are the area, circumference, distance from axis to area centroid, and distance from axis to circumference centroid, respectively, we have
<span class="math-container">\begin{align}
R_a A=36\\
R_c C=18\\
\text{no part of red region cuts the axis}
\end{align}</span></p>
<p>I have tried choosing the red region as a circle as well as a rectangle, they are not possible. Other configurations are left for others. Calculus of variations might help.</p>
|
186,240 | <p>I need some notion about topology(I'm very interested in boundary points, open sets) and few examples of solved exercises about limits of functions($f:\mathbb{R}^{n}\rightarrow \mathbb{R}^m$) using $\epsilon, \delta$ and also some theory for continous functions.
Please give me some links or name of the books which can help me. </p>
<p>Thanks :) </p>
| rschwieb | 29,335 | <p><a href="http://www.math.psu.edu/katok_a/TOPOLOGY/Chapter1.pdf" rel="nofollow">Here is a fast free internet resource</a> to get you started, while your text is in the mail.</p>
|
2,308,770 | <p>I have an equation of the form $A*i*j + B*i +C*j = N$
where I have the values of $A,B,C$ and $N$ and I want to solve for integer values of $i$ and $j$.</p>
<p>How would I approach this? I could try trial and error but the numbers I'm working with are relatively large (eg $>10^{40}$). But I'm also happy to work on understanding the process with much smaller numbers. I'm not sure if it matters, but it's also possible there will not be a solution.</p>
| CY Aries | 268,334 | <p>Multiply the whole equation by $A$.</p>
<p>\begin{align}
A^2ij+ABi+ACj&=AN\\
(Ai+C)(Aj+B)&=AN+BC
\end{align}</p>
<p>So $Ai+C$ and $Aj+B$ are factors of $AN+BC$.</p>
<p>For example, consider $2ij+3i+7j=336$. the equation can be written as</p>
<p>\begin{align}
2ij+3i+7j&=336\\
4ij+6i+14j&=672\\
(2i+7)(2j+3)&=693
\end{align}</p>
<p>$693=3^2\cdot7\cdot11$.</p>
<p>Factors of $693$ has the form $\pm3^a\cdot 7^b\cdot 11^c$, where $a\in\{0,1,2\}$ and $b,c\in\{0,1\}$. The numner of possible factors is $2\times3\times2\times2=24$.</p>
<p>S0, all possible $2i+7$ are $\pm1$, $\pm 3$, $\pm9$, $\pm7$, $\pm 21$, $\pm63$, $\pm 11$, $\pm 33$, $\pm 99$, $\pm77$, $\pm231$ and $\pm693$.</p>
<p>All possible $(i,j)$ are $(-3,345)$, $(-2,114)$, $(1,37)$, $(0,48)$, $(7,15)$, $(28,4)$, $(2,30)$, $(13,9)$, $(46,2)$, $(35,3)$, $(112,0)$, $(343,-1)$, $(-4,-348)$, $(-5,-117)$, $(-8,-40)$, $(-7,-51)$, $(-14,-18)$, $(-35,-7)$, $(-9,-33)$, $(-20,-12)$, $(-53,-5)$, $(-42,-6)$, $(-119,-3)$, $(-350,-2)$
.</p>
|
253,359 | <p>I'm trying to prove by induction the following statement without success:<br>
$$\forall n \ge 2, \;\forall d \ge 2 : d \mid n(n+1)(n+2)...(n+d-1) $$</p>
<p>For the base case: $n = 2$, $d = 2$<br>
$2\mid 2(2+1)$ which is true.<br></p>
<p>Now, the confusion begins! I assume I would need to use the second induction principle to proof this because $P(n)$ and $P(n+1)$ are not related at all. It is also the first time I am dealing with more than one variable so it makes it harder for me.</p>
<p>I tried the following:<br>
- Trying to prove by simple induction. I did not go very far.<br>
- Trying to split my induction step in 2 parts: If $d\mid n$, I'm done. <br></p>
<p>If $d$ does not divide $n$, then I would need to do a second proof and this is where I'm blocked.</p>
<p>Anyone could tell me what's wrong in the approach I take to solve this problem?<br></p>
<p>Any help would be appreciated!</p>
| wisefool | 51,807 | <p>Fixed $d$, it is easy to show that the statement holds for every $n$, by induction. The inductive step is as follows:
suppose $d|n(n+1)\cdots(n+d-1)$, then put $m=gcd(d,n)$ and observe that $gcd(d,n+d)=m$ as well, therefore, by assumption
$$\frac{d}{m}\vert (n+1)\cdots(n+d-1)$$
so
$$d=\frac{d}{m}m\vert (n+1)\cdots(n+d-1)(n+d)\;.$$
Now you just have to show that the statement is true for $n=2$ and every $d$, but this is easy: $2\leq d\leq d+2-1=d+1$, therefore $d$ divides $2\cdot3\cdots(d+1)=(d+1)!$.</p>
|
1,458,975 | <p>I'm having a issue with solving this problem. I know that the answer is $ a=3, b=1 $. But i'm not sure how to get to that conclusion.</p>
<p>Given that $(a+i)(2-bi)=7-i$, find the value of $a$ and of $b$, where $a,b \in \mathbb{Z}$.</p>
| John Hughes | 114,036 | <p>You're halfway on the right track. But you're not going to be able to prove it's rational. </p>
<p>Suppose that you DO have $\sqrt{3} = a + b \sqrt{2}$. Try squaring both sides and simplifying. Then do something like that again, until you've got no square roots. And then argue ( by looking at factors of 2 and 3 in $a^2, b^2, ab$, and other such numbers) that the resulting equation can't have rational solutions. </p>
<p>Reread the proof that $\sqrt{2}$ is irrational for inspiration. </p>
<p>When you've done all that, report back and maybe we can help you out some more. </p>
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.