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 |
|---|---|---|---|---|
3,865,954 | <p>Suppose that I have the following sum:
<span class="math-container">$\sum_{m=0}^{\infty}(e^{it}(1-p))^{m}$</span>, where <span class="math-container">$i^2 = -1$</span>.</p>
<p>This is a geometric series, but involving the complex number <span class="math-container">$i$</span>. Can I just apply the geometric series formula and conclude that the sum if <span class="math-container">$\frac{1}{1-e^{it}(1-p)}$</span>?</p>
<p>Thank you,</p>
| MathieuRund | 837,198 | <p>The geometric series formula you are referring to holds for real numbers as long as the common ratio is less than 1. Analogously, the formula holds for complex numbers when the common ratio has modulus less than 1.</p>
|
200,322 | <p>Is there a compact topological space $(X,\tau)$ such that for no cardinal $\kappa$ there is a surjective continuous map $e:\{0,1\}^\kappa \to X$? </p>
<p>(We assume that $\{0,1\}$ is endowed with the discrete topology, and $\{0,1\}^\kappa$ has the product topology.)</p>
| Joseph Van Name | 22,277 | <p>It is well known that the space $\{0,1\}^{\kappa}$ satisfies the countable chain condition. Recall that a topological space $X$ satisfies the countable chain condition if and only if every collection $\mathcal{A}$ of pairwise disjoint open sets is countable. However, it is easy to show that the surjective continuous image of a space satisfying the countable chain condition must also satisfy the countable chain condition, so the only possible spaces which are images of some $\{0,1\}^{\kappa}$ satisfy the countable chain condition. However, there are plenty of compact Hausdorff spaces that do not satisfy the countable chain condition such as $\beta\mathbb{N}\setminus\mathbb{N}$ or $[0,1]\times[0,1]$ with the order topology inherited from the lexicographic ordering.</p>
<p>To see that $\{0,1\}^{\kappa}$ satisfies the countable chain condition, endow $\{0,1\}^{\kappa}$ with the infinite product measure $m$ where each $\{0,1\}$ is given the measure $\mu$ such that $\mu(\{0\})=\frac{1}{2}$ and $\mu(\{1\})=\frac{1}{2}$. Then $m$ extends to a Borel measure $\overline{m}$ on $\{0,1\}^{\kappa}$ where $\overline{m}(U)>0$ for each non-empty open set $U$. However, there cannot be an uncountable collection $\mathcal{A}$ of disjoint open subsets of $X$ since the union of any uncountable pairwise disjoint collection of open subsets of $X$ would have infinite measure.</p>
|
4,017,964 | <p>Write down in roster notation a set of cardinality 3, of which all elements are sets, and which
satisfies the following property:
∀, ∈ ( ⊆ ⇒ ⊆ ).
No justification is needed. Do not use ellipses (“…”) in your answer.</p>
<p>What I understands from this question is that all the 3 sets have to be the same to fulfill the statement.Example of an answer would be { {1,2,3},{1,2,3},{1,2,3}} am i right?</p>
| Henno Brandsma | 4,280 | <p>An example could be <span class="math-container">$U=\{\{1\}, \{2,3\}, \{4,5,6\}\}$</span>.</p>
<p>It has three elements, all three are sets. And there is no inclusion relation between any of them. So it’s voidly true that <strong>if</strong> one is a subset of the other, the reverse also holds, as your property states. Otherwise put, there is no counterexample among them so the statement holds.</p>
<p>The element sets could have been any size.</p>
|
3,186,627 | <p>Proposition: Let A be a subset of R which is bounded below. Let B be a subset of R which is bounded above. If <span class="math-container">$\inf(A) < \sup (B) $</span> then there is some <span class="math-container">$a \in A$</span> and <span class="math-container">$b \in B$</span> such that <span class="math-container">$a<b$</span>.</p>
<p>Proof: </p>
<p>Let <span class="math-container">$\inf(A) = I$</span> and <span class="math-container">$\sup(B) = S$</span> </p>
<p>So <span class="math-container">$I<S$</span></p>
<p>By definition, <span class="math-container">$\forall a \in A, I\leq a$</span> and <span class="math-container">$\forall b \in B, b\leq S$</span> </p>
<p>Case 1: If I <span class="math-container">$\in A$</span> and S <span class="math-container">$\in B$</span>, we are done.</p>
<p>Case 2: If I <span class="math-container">$\notin A$</span> and S <span class="math-container">$\notin B$</span>, then assume BWOC <span class="math-container">$\nexists a \in A $</span> and <span class="math-container">$b\in B$</span> such that <span class="math-container">$a<b$</span></p>
<p>So <span class="math-container">$\forall a \in A$</span> and <span class="math-container">$\forall b \in B, b \leq a$</span></p>
<p>So <span class="math-container">$\forall a \in A$</span>, <span class="math-container">$a$</span> is an upper bound for B.</p>
<p>But since S is the least upper bound for B, <span class="math-container">$b\leq S<a$</span></p>
<p><span class="math-container">$S<a$</span></p>
<p>So <span class="math-container">$S$</span> is also a lower bound for <span class="math-container">$A.$</span></p>
<p>But since <span class="math-container">$I$</span> is the greatest lower bound for <span class="math-container">$A$</span>, <span class="math-container">$S<I$</span>, which is a contradiction.</p>
<p>Is my proof valid? If not, what am I missing? Would I have to show that this holds for other cases as well? </p>
| Bernard | 202,857 | <p>I think it can be proved without examining different cases:</p>
<p>By definition of <span class="math-container">$\;\inf A$</span>, for any <span class="math-container">$\varepsilon >0$</span>, there exists <span class="math-container">$a\in A$</span> such that
<span class="math-container">$$I\le a <I+\varepsilon.$$</span></p>
<p>Similarly, for any <span class="math-container">$\varepsilon >0$</span>, there exists <span class="math-container">$b\in B$</span> such that
<span class="math-container">$$S-\varepsilon < b \le S.$$</span>
Now, if <span class="math-container">$\;I+\varepsilon <S-\varepsilon$</span>, you're done. Do you see how this can be realised?</p>
|
687,179 | <p>This is a question that asks the reader for a $strategy$ to solve a particular problem. I cannot solve this problem myself so I am looking around for general methods one might use to confront it with. Suppose
$$f(x)=a_0+a_1x+..., g(x)=b_0+b_1x_...$$
and given
$$\lim\limits_{x\to 1^-}\frac{f^{(n)}(x)}{g^{(n)}(x)}=1$$
With the additional
$$\lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}=1$$ and $$\lim\limits_{n\to\infty}\frac{b_{n+1}}{b_n}=1 $$
for all natural $n$. Prove
$$\lim\limits_{n\to \infty}\frac{a_n}{b_n}=1$$</p>
<p>I would like to know your general thoughts about approaching this type of problem. Any insights however small will be much appreciated. Although this is seemingly a problem in real analysis, I have tagged complex analysis because solutions may well involve it.</p>
| sirfoga | 83,083 | <p>Firstly suppose that $f(x)$ and $g(x)$ are two polynomials, and let their degree be $c$ and $d$, respectively. Suppose $ d \gt c$, you get
$\lim\limits_{x\to 1^-}\frac{f^d(x)}{g^d(x)} = \frac{0}{d!\cdot b_d} = 0 \ne 1$, so it's impossible that $ d \gt c$; obviously you can use the same proof to prove that it's impossible that $ c\gt d$.</p>
<p>Now, since $c=d$, $\lim\limits_{x\to 1^-}\frac{f^c(x)}{g^c(x)} = \frac{c!\cdot a_c}{c!\cdot b_c} = \frac{a_c}{b_c} = 1$ , so $a_c$ = $b_c$.</p>
<p>Similarly, $\lim\limits_{x\to 1^-}\frac{f^{c-1}(x)}{g^{c-1}(x)} = \frac{(c-1)!\cdot a_{c-1}+c!\cdot x\cdot a_c}{(c-1)!\cdot b_{c-1}+c!\cdot x\cdot b_c} = \frac{a_{c-1}+a_c}{b_{c-1}+b_c} = 1$, which means that </p>
<p>$a_{c-1}+a_c = b_{c-1}+b_c$</p>
<p>$a_{c-1} = b_{c-1}$.</p>
<p>Clearly you go on with the same "strategy", and you'll obtain that $a_i$ = $b_i$ for all $i$ such that $0\le i \le c$, so $\lim\limits_{n\to \infty}\frac{a_n}{b_n}=1$.</p>
|
79,542 | <p>The <em>polarization identity</em> expresses a symmetric bilinear form on a vector space in terms of its associated quadratic form:
$$
\langle v,w\rangle = \frac{1}{2}(Q(v+w) - Q(v) - Q(w)),
$$
where $Q(v) = \langle v,v\rangle$. More generally (over fields of characteristic $0$), for any homogeneous polynomial
$h(x_1,\dots,x_n)$ of degree $d$ in $n$ variables, there is a unique symmetric $d$-multilinear polynomial $F({\mathbf x}_1,\dots,{\mathbf x}_d)$, where each ${\mathbf x}_i$ consists of $n$ indeterminates, such that $h(x_1,\dots,x_n) = F({\mathbf x},\dots,{\mathbf x})$, where ${\mathbf x} = (x_1,\dots,x_n)$. There is a formula which expresses $F({\mathbf x}_1,\dots,{\mathbf x}_d)$ in terms of $h$, generalizing the above formula for a bilinear form in terms of a quadratic form, and it is also called a polarization identity.</p>
<p>Where did the meaning of "polarization", in this context, come from? Weyl uses it in his book <em>The classical groups</em> (see pp. 5 and 6 on Google books) but I don't know if this is the first place it appeared. Jeff Miller's extensive math etymology website doesn't include this term. See <a href="http://jeff560.tripod.com/p.html">http://jeff560.tripod.com/p.html</a>.</p>
| r.e.s. | 16,397 | <blockquote>
<p>Where did the meaning of "polarization", in this context, come from?
Weyl uses it in his book The classical groups (see pp. 5 and 6 on
Google books) but I don't know if this is the first place it appeared.</p>
</blockquote>
<p>A few things I've managed to find ...</p>
<p>The term <em>polarization</em> in this context did not originate with Weyl (1939). The book <a href="http://books.google.com/books?id=_eQOSvtw7Z8C&dq=isbn%3A0521449030" rel="noreferrer">Hilbert's Invariant Theory Papers</a> is an English translation of four papers by David Hilbert, and the term "polarization" appears in the first two of them (published 1885 and 1887), evidently in the sense you have in mind. In the fourth paper (published 1893), Hilbert uses an expression that translates as "Aronhold process" for what Hawkins' <a href="http://books.google.com.na/books/about/Emergence_of_the_theory_of_Lie_groups.html?id=QbB7_YOrruoC" rel="noreferrer">Emergence of the theory of Lie groups: an essay in the history of mathematics, 1869-1926</a> terms the "Aronhold polarization process". Also, <a href="http://books.google.com/books/about/Vorlesungen_%C3%BCber_Invariantentheorie.html?id=bz5tAAAAMAAJ" rel="noreferrer">Gordan</a> (1885) refers to this same "Aronhold process", which was apparently <a href="http://resolver.sub.uni-goettingen.de/purl?GDZPPN00215028X" rel="noreferrer">published</a> in 1838 (if not also earlier) by Aronhold.</p>
<p>In the above-cited works, the meaning of <em>polarization</em> appears to derive from that of the terms <a href="http://en.wikipedia.org/wiki/Pole_and_polar" rel="noreferrer"><em>pole</em> and <em>polar</em></a> as used in <a href="http://en.wikipedia.org/wiki/Projective_geometry" rel="noreferrer">projective geometry</a>. The entry for "POLE and POLAR" on the <a href="http://jeff560.tripod.com/p.html" rel="noreferrer">webpage by Jeff Miller</a>, mentioned in the question, says the term <em>pôle</em> in this sense was introduced by François Joseph Servois in 1811, and that the term <em>polar</em> (polaire) was introduced by Joseph-Diez Gergonne in the modern geometric sense in 1813.</p>
|
65,223 | <p>The least rational number greater than $\sqrt{2}$ that can be written as a ratio of integers $x/y$ with $y\le10^{100}$ can be found in a moment using a little Python program. Can anyone write a program that finds, in hours rather than centuries, the least rational greater than $\sqrt{2}$ of the form $x/y^2$ with $y^2\le 10^{100}$? </p>
<p>More generally, my question is whether the following computation is known to be feasible or not feasible:</p>
<p>Given $N$, find the least rational greater than $\sqrt{2}$ of the form $x/y^2$,
with $x$ and $y$ integers and $y^2\le N$. For definiteness, let's say that the output should be the required rational written in lowest form.</p>
<p>By a feasible computation I mean one that can be done in $O((\log N)^k)$ bit operations for some constant $k$. </p>
<p>Of course the square root of 2 is not essential here. Any irrational would do, as long as comparisons with rationals are feasible. I don't know of any such irrational for which I can answer the question I've posed.</p>
| Roland Bacher | 4,556 | <p>The following "aglorithm" gives not necessarily the best solution but yields
fairly "good" solutions. </p>
<p>Start with a $n$ "bad" rational approximations $x_1/y_1,\dots,x_n/y_n$ of $2^{1/4}$
(obtained eg. by considering a few convergents of $2^{1/4}$) such that $y_1 \dots y_n < N$ and consider a linear
combination $\sum_{i=1}^n a_i(x_i/y_i)^2=x/(y_1\cdots y_n)^2$ with $a_i\in \mathbb Z,\sum_{i=1}^na_i=1$ which is slightly larger than $\sqrt 2$. One way to get coefficients
$a_1,\dots,a_n\in \mathbb Z$ with $\sum a_i(x_i/y_i)^2$ close to $\sqrt{2}$
is by using the LLL-algorithm: Consider the $(n+1)-$dimensional sublattice $\Lambda$
of $\mathbb R^{n+2}$ spanned by
$f_1=(1,0,0,\dots,0,A(x_1/y_1)^2)$, $f_2=(0,1,0,\dots,0,A(x_2/y)^2),\dots$,
$f_n=(0,0,\dots,1,0,A(x_n/y_n)^2)$, $f_{n+1}=(0,\dots,0,1,A\sqrt 2)$ where $A$ is some huge real number
(one can also work with an integral lattice by rounding off the last coordinate to the nearest integer for a fixed large real number $A$). A short vector of the form $(a_1,\dots,a_n,1)$ in $\Lambda$ yields a good rational approximation $\sum_{i=1}^n a_i(x_i/y_i)^2$ of $\sqrt 2$ . About half of the time, such an approximation should have the correct sign. A few LLL runs for various large constant values of $A$ (which should be larger than $\max (y_i^2)$, perhaps $A\sim \sqrt{N}$ is interesting) and various
finite sets $x_1,\dots,x_n$ (with $n$ also varying) should give interesting approximations.</p>
|
4,041,641 | <p>This is the answer given:</p>
<p><a href="https://i.stack.imgur.com/6dyU4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6dyU4.png" alt="enter image description here" /></a></p>
<p>I think this answer is not correct because the line given by y=-2x/3 makes an angle that is below the x axis, so the order is incorrect right? We need to start with a counterclockwise rotation first and end with the clockwise rotation, but here, we started with the clockwise rotation first.</p>
| trancelocation | 467,003 | <p>You are right. The given answer is not correct.</p>
<p>But what is more:
The way how to find the reflection matrix is far too cumbersome.</p>
<p>The standard way is as follows:</p>
<p>Let <span class="math-container">$d= \begin{pmatrix} 3\\ -2 \end{pmatrix}$</span> be the direction vector of the line. Then, the projection <span class="math-container">$P_d$</span> of any point with positional vector <span class="math-container">$v$</span> onto the line is given by:</p>
<p><span class="math-container">$$P_d(v) = \frac{d\cdot v}{|d|^2}v$$</span></p>
<p>So, you get the reflection <span class="math-container">$T$</span> of <span class="math-container">$v$</span> in your line by flipping the sign of the part of <span class="math-container">$v$</span> which is orthogonal to <span class="math-container">$d$</span>:</p>
<p><span class="math-container">$$v = P_dv \color{blue}{+} (v-P_dv) {\longrightarrow} Tv = P_dv \color{blue}{-} (v-P_dv) = 2P_dv - v = (2P_d-I)v$$</span></p>
<p>So, you only need the matrix representation of <span class="math-container">$P_d$</span> which is quickly done by plugging in the unit vectors <span class="math-container">$e_1 = \begin{pmatrix} 1\\ 0 \end{pmatrix}$</span> and <span class="math-container">$ e_2 =\begin{pmatrix} 0\\ 1 \end{pmatrix}$</span>:</p>
<p><span class="math-container">$$P_de_1 = \frac 1{13}\begin{pmatrix} 9\\ -6 \end{pmatrix},P_de_2 = \frac 1{13}\begin{pmatrix} -6\\ 4\end{pmatrix} \Rightarrow P_d =\frac 1{13}\begin{pmatrix} 9 & -6\\ -6 & 4\end{pmatrix} $$</span></p>
<p>Plug this into</p>
<p><span class="math-container">$$T = 2P_d - I \Rightarrow T = \frac 1{13}\begin{pmatrix} 5 & -12\\ -12 & -5\end{pmatrix}$$</span></p>
<p>I recommend to check - for example - with Geogebra that this matrix nicely reflects points in the given line.</p>
|
285,034 | <p>I currently don't see how to solve the following integral:</p>
<p>$$\int_{-1/2}^{1/2} \cos(x)\ln\left(\frac{1+x}{1-x}\right) \,dx$$</p>
<p>I tried to solve it with integration by parts and with a Taylor series, but nothing did help me so far.</p>
| user 1591719 | 32,016 | <p>Letting $x=-u$ in the second integral you're done
$$\int_{-1/2}^{1/2} \cos(x)\ln(1+x) \ dx-\int_{-1/2}^{1/2} \cos(x)\ln(1-x) \ dx$$</p>
|
285,034 | <p>I currently don't see how to solve the following integral:</p>
<p>$$\int_{-1/2}^{1/2} \cos(x)\ln\left(\frac{1+x}{1-x}\right) \,dx$$</p>
<p>I tried to solve it with integration by parts and with a Taylor series, but nothing did help me so far.</p>
| L. F. | 56,837 | <p>As pointed out, the integrand is odd. To make this clear, let $x\to -x:$</p>
<p>$$I=\int_{-1/2}^{1/2} \cos x\ln\left(\frac{1+x}{1-x}\right)dx=\int_{-1/2}^{1/2} \cos x\ln\left(\frac{1-x}{1+x}\right)dx$$</p>
<p>Adding both:</p>
<p>$$I=\frac{1}{2}\int_{-1/2}^{1/2} \cos x\ln\left(1\right)dx=0$$</p>
<p>For future reference, whenever you see symmetry in the bounds, check whether the integrand is even/odd before embarking on anything else!</p>
|
1,640,653 | <p>I have this matrix below and I'm trying to find it's inverse, I know I augment it with I<sub>2</sub> but I don't know where to go from that.</p>
<p>\begin{bmatrix}
2&1\\
a&a
\end{bmatrix}</p>
| Osai Egharevba | 310,771 | <p>First, you need to find its determinant. By inspection, I can see that the determinant is <em>a</em>.</p>
<p>Then, multiply the reciprocal of the determinant of the matrix by the following</p>
<p>\begin{bmatrix}a&-1\\-a&2\end{bmatrix}</p>
|
1,454,344 | <p>Does span=(2,-1,1,2), (-2,1,-1,-2) represent a line, plane or hyperplane in R4?</p>
<p>We haven't learned matrices yet either </p>
| Mark Watson | 272,109 | <p>Yes, let <strong>x</strong> be one of the two vectors, then the other vector is -<strong>x</strong>. This means they lie on the same line which is a subspace of $\mathbb{R}^{4}.$</p>
|
3,878,723 | <blockquote>
<p>Find the value of <span class="math-container">$k$</span> if the curve <span class="math-container">$y = x^2 - 2x$</span> is tangent to the line <span class="math-container">$y = 4x + k$</span></p>
</blockquote>
<p>I have looked at the solution to this question and the first step is the "equate the two functions":<br />
<span class="math-container">$ x^2 - 2x = 4x + k$</span></p>
<p>Why? How does that help solve the equation? And how can I use what I get from equating the two functions to find the solution?</p>
| Steven Alexis Gregory | 75,410 | <p>The important line is "if the curve <span class="math-container">$y=x^2-2x$</span> <strong>is tangent</strong> to the line <span class="math-container">$y=4x+k$</span>".</p>
<p>If the line is tangent to the curve, then they will intersect at exactly one point.</p>
<p>You DO NOT NEED TO prove that the line IS TANGENT. You just need to show that, if the line is tangent, then <span class="math-container">$k$</span> must be equal to the value of <span class="math-container">$k$</span> that makes the equation <span class="math-container">$x^2-2x = 4x+k$</span> have exactly one solution.</p>
|
3,878,723 | <blockquote>
<p>Find the value of <span class="math-container">$k$</span> if the curve <span class="math-container">$y = x^2 - 2x$</span> is tangent to the line <span class="math-container">$y = 4x + k$</span></p>
</blockquote>
<p>I have looked at the solution to this question and the first step is the "equate the two functions":<br />
<span class="math-container">$ x^2 - 2x = 4x + k$</span></p>
<p>Why? How does that help solve the equation? And how can I use what I get from equating the two functions to find the solution?</p>
| emekadavid | 813,301 | <p>When they say that a line is tangent to a curve at a point, that means at that point, f(x) for the curve and the line are equal. That is why you are told to equate the two equations. If you equate them and solve for x, you would eventually get something like this:
<span class="math-container">$$x = \sqrt{k+9} + 3$$</span>
Then back substitute for x into any of the equations.</p>
|
4,415,907 | <p>I need to evaluate the Fourier inverse integral</p>
<p><span class="math-container">$\displaystyle \int_{-\infty}^{\infty}\frac{\sinh\left(y\sqrt{\alpha^2-\omega^2}\right)}{\sinh\left(H\sqrt{\alpha^2-\omega^2}\right)}e^{i\alpha x}d\alpha \tag*{}$</span></p>
<p>which arose while solving a PDE.</p>
<p>Here, <span class="math-container">$H>0,x\in\mathbb{R},y\in[0,H]$</span>. The domain of <span class="math-container">$\omega$</span> was not given in the original problem, but I am going to assume <span class="math-container">$\omega>0$</span> for simplicity.</p>
<p>The problem asks us to introduce proper branch cuts for the square root function before evaluating the integral.</p>
<p>For reference, <a href="https://imgur.com/a/U341iyL" rel="nofollow noreferrer">this is the original question</a>. My attempt up until I get the integral is also shown in the link.</p>
<p><strong>My Attempt</strong></p>
<p>The branch points of the square root functions are at <span class="math-container">$\alpha =\pm\omega$</span>. So, I considered the following branch cuts and contours.</p>
<p><a href="https://i.stack.imgur.com/YtLxf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YtLxf.png" alt="enter image description here" /></a></p>
<p>(Here <span class="math-container">$\omega$</span> has an absolute value, but you can ignore it and assume <span class="math-container">$|\omega|=\omega$</span>)</p>
<p>We, firstly, need to find the poles of the integrand in the upper half-plane. Those are given by the equation</p>
<p><span class="math-container">$H\sqrt{\alpha^2-\omega^2}=n\pi i\tag*{}$</span></p>
<p>Solving this, we obtain</p>
<p><span class="math-container">$\displaystyle \alpha =\pm\sqrt{\omega^2-\frac{n^2\pi^2}{H^2}}\tag*{} $</span></p>
<p>where <span class="math-container">$n=0,1,2,\cdots$</span>.</p>
<p>The problem is that some of those poles are on the branch cuts depending on the parameters. I have been told this is not permissible, so I am not sure how to proceed.</p>
<p>Edit: The statement that "some of those poles are on the branch cuts" is not correct. It is on the real axis but between <span class="math-container">$[-\omega,\omega]$</span>.</p>
<p>Edit2: <em>Tables of Fourier Transforms and Fourier Transforms of Distributions</em> by Fritz Oberhettinger states(p37,7.48) that if <span class="math-container">$a<b$</span>, we have</p>
<p><span class="math-container">$\displaystyle \int_{0}^{\infty} \frac{\sinh{(a\sqrt{k^2+x^2}})}{\sinh{(b\sqrt{k^2+x^2}})}\cos{xy}dx = -\pi b^{-1} \sum_{n=0}^{\infty}(-1)^nc_n\sin{(ac_n)}v_n^{-1}e^{-yv_n}$</span></p>
<p>where <span class="math-container">$c_n=n\pi/b$</span>, <span class="math-container">$v_n = (k^2+c_n^2)^{1/2}$</span>. I would guess our integral would have a similar form.</p>
| Diger | 427,553 | <p>Even though the branch cut <span class="math-container">$(-\infty,0)$</span> of the square-root transfers to <span class="math-container">$\sinh(\sqrt{z})$</span>, the integral actually does not possess any branch cut. The only branch-cut that could possibly occur, would be <span class="math-container">$(-\omega,\omega)$</span>. However, the phase-jump cancels out in ratio: if for some <span class="math-container">$\Re(\alpha) \in (-\omega,\omega)$</span>, <span class="math-container">$\alpha$</span> goes from the upper half-plane into the lower half-plane, <span class="math-container">$\sqrt{\alpha^2-\omega^2}$</span> changes from being of the form <span class="math-container">$\pm i\cdot r$</span> to <span class="math-container">$\mp i\cdot r$</span>, for some real number <span class="math-container">$r$</span>, but <span class="math-container">$$\sinh(\pm i r)=\pm i \sin(r) \, .$$</span>
Hence, no jump i.e. no branch-cut.</p>
<p>Since you already know how to obtain the result from the residue theorem, here is a proof that shows the vanishing of the semicircle. However, you do not need to use a semi-circle. You can also take a box of side-length <span class="math-container">$R$</span>, which turns out to be a bit easier.</p>
<p>Writing <span class="math-container">$\alpha=r+is$</span>, the contour for <span class="math-container">$x>0$</span> is
<span class="math-container">$$C_R=\{r=R,s\in(0,R)\} \cup\{r\in(R,-R),s=R\} \cup \{r=-R,s=(R,0)\} \, ,$$</span>
from which we'll find
<span class="math-container">$$\left|\int_{C_R}\frac{\sinh\left(y\sqrt{\alpha^2-\omega^2}\right)}{\sinh\left(H\sqrt{\alpha^2-\omega^2}\right)} \, e^{i\alpha x} \, d\alpha \right| \leq \int_{0}^R \left|\frac{\sinh\left(y\sqrt{(R+is)^2-\omega^2}\right)}{\sinh\left(H\sqrt{(R+is)^2-\omega^2}\right)} \right| \, e^{-xs} \, ds \\
+\int_{-R}^{R} \left|\frac{\sinh\left(y\sqrt{(r+iR)^2-\omega^2}\right)}{\sinh\left(H\sqrt{(r+iR)^2-\omega^2}\right)} \right| \, e^{-xR} \, dr + \int_{0}^R \left|\frac{\sinh\left(y\sqrt{(-R+is)^2-\omega^2}\right)}{\sinh\left(H\sqrt{(-R+is)^2-\omega^2}\right)} \right| \, e^{-xs} \, ds \\
= I_1 + I_2 + I_3 \, .$$</span>
Now we notice:</p>
<ol>
<li>The first and third integral are the same, since the arguments of the absolute value function are just complex-conjugate to each other.</li>
<li>As <span class="math-container">$|R+is|$</span> is large, i.e.
<span class="math-container">$$\sqrt{(R+is)^2-\omega^2} = R + is + O\left( \frac{1}{|R+is|} \right)$$</span>
we can remove the square-root. The error made will be small compared to the main term, since <span class="math-container">$$\sinh(R+is+\epsilon) = \sinh(R+is) \left\{ \cosh(\epsilon) + \coth(R+is) \sinh(\epsilon) \right\} \\
= \sinh(R+is) \left\{1+O(\epsilon)\right\}$$</span>
where <span class="math-container">$\epsilon=O\left(|R+is|^{-1}\right)$</span> is small.</li>
<li>The following identity holds for all <span class="math-container">$x,y \in \mathbb{R}$</span>
<span class="math-container">$$\left| \sinh\left(x+iy\right) \right|^2 = \sinh^2(x) + \sin^2(y) \, .$$</span></li>
<li>In estimating the contribution along the box, we need to avoid the poles of <span class="math-container">$1/\sinh$</span> on the imaginary axis at <span class="math-container">$in\pi$</span>. As <span class="math-container">$R\rightarrow \infty$</span>, we can assume <span class="math-container">$HR=(n+1/2)\pi$</span> as <span class="math-container">$n \rightarrow \infty$</span>, for integers <span class="math-container">$n$</span>.</li>
</ol>
<p>Using these observations, we can estimate as follows:
<span class="math-container">$$I_1 \leq \int_0^R \sqrt{\frac{\sinh^2(yR)+\sin^2(ys)}{\sinh^2(HR)+\sin^2(Hs)}} \, e^{-xs} \, {\rm d}s \leq \int_0^R {\frac{\cosh(yR)}{\sinh(HR)}} \, e^{-xs} \, {\rm d}s \leq \frac{\cosh(yR)}{\sinh(HR)} \frac{1}{x} \\
I_2 \leq \int_{-R}^R \sqrt{\frac{\sinh^2(yr)+\sin^2(yR)}{\sinh^2(Hr)+\sin^2(HR)}} \, e^{-xR} \, {\rm d}r \leq \int_{-R}^R {\frac{\cosh(yr)}{\cosh(Hr)}} \, e^{-xR} \, {\rm d}r \leq 2R e^{-xR} \, .$$</span></p>
|
47,926 | <p>Is there any known two-dimensional Conway's game of life variation where each cell can not be just on/off but able to hold more states, maybe 4 or 5?</p>
| Mikhail | 12,631 | <p>There are many: check out this website.
<a href="http://golly.sourceforge.net/Help/Algorithms/RuleTable.html" rel="nofollow">http://golly.sourceforge.net/Help/Algorithms/RuleTable.html</a> </p>
<p>Often scientists will write their own variations to model certain real world scenarios. </p>
|
491,748 | <p>I got stuck on this problem and can't find any hint to solve this. Hope some one can help me. I really appreciate.</p>
<blockquote>
<blockquote>
<p>Give an example of a collection of sets $A$ that is not locally finite, such that the collection $B = \{\bar{X} | X \in A\}$ is locally finite.</p>
</blockquote>
</blockquote>
<p>Note: Every element in $B$ must be unique, so maybe there exist 2 distinct sets $A_{1}, A_{2} \in A$, but have $\bar{A_{1}} = \bar{A_{2}}$. So that's why this problem would be right even though $A \subset \bar{A}$.</p>
<p>Thanks everybody.</p>
| Syd Henderson | 50,986 | <p>Let Q be the rationals in R, and A be the collection of sets of the form {a+√p, a ∈ Q}, p ∈ N.</p>
|
491,748 | <p>I got stuck on this problem and can't find any hint to solve this. Hope some one can help me. I really appreciate.</p>
<blockquote>
<blockquote>
<p>Give an example of a collection of sets $A$ that is not locally finite, such that the collection $B = \{\bar{X} | X \in A\}$ is locally finite.</p>
</blockquote>
</blockquote>
<p>Note: Every element in $B$ must be unique, so maybe there exist 2 distinct sets $A_{1}, A_{2} \in A$, but have $\bar{A_{1}} = \bar{A_{2}}$. So that's why this problem would be right even though $A \subset \bar{A}$.</p>
<p>Thanks everybody.</p>
| Syd Henderson | 50,986 | <p>Sorry, typo, now corrected. All elements of A are dense in R.</p>
|
491,748 | <p>I got stuck on this problem and can't find any hint to solve this. Hope some one can help me. I really appreciate.</p>
<blockquote>
<blockquote>
<p>Give an example of a collection of sets $A$ that is not locally finite, such that the collection $B = \{\bar{X} | X \in A\}$ is locally finite.</p>
</blockquote>
</blockquote>
<p>Note: Every element in $B$ must be unique, so maybe there exist 2 distinct sets $A_{1}, A_{2} \in A$, but have $\bar{A_{1}} = \bar{A_{2}}$. So that's why this problem would be right even though $A \subset \bar{A}$.</p>
<p>Thanks everybody.</p>
| Joshua Doucette | 738,394 | <p>If someone wants to check my easy proof for metric spaces that would be good.</p>
<p>Let the topological space be a metric space. Here I will use X = <span class="math-container">$(0,1)$</span> with the basis of (0,1) being the standard basis of R consisting of open balls, but those balls not contained in (0,1) are removed. Use the absolute value function as the metric for simplicity. (|x-y| = d(x,y) for x,y <span class="math-container">$\in$</span> R).</p>
<p>Then let <span class="math-container">$A_{N}$</span> = <span class="math-container">$\bigcup_{i=0}^{N-1}(\frac{i}{N}, \frac{i+1}{N})$</span> and A = <span class="math-container">$\{A_{N}\}_{N=1}^{N=\infty}$</span></p>
<p>It should be easy to see that <span class="math-container">$A_{N} = (0, \frac{1}{N})\bigcup(\frac{1}{N}, \frac{2}{N})\bigcup...\bigcup(\frac{N-1}{N}, 1)$</span></p>
<p><strong>1.) Choose <span class="math-container">$x_{0} \in (0,1)$</span>. Then a neighborhood U containing <span class="math-container">$x_{0}$</span> intersects (0,1), in which case U will intersect infinitely many <span class="math-container">$A_{N} \in A$</span>.</strong></p>
<p>This is because an open set U containing <span class="math-container">$x_{0}$</span> will be the union of open balls B contained in U. Some ball B in U contains <span class="math-container">$x_{0}$</span>. This ball has a radius <span class="math-container">$r > 0$</span>.</p>
<p>The existence of a ball B in U containing <span class="math-container">$x_{0}$</span> with positive radius r implies that some <span class="math-container">$N_{0} \in \mathbb{Z^{+}}$</span> satisfies <span class="math-container">$\frac{1}{N_{0}} < r$</span>.</p>
<p>Thus B intersects at least 2 balls of radius <span class="math-container">$\frac{1}{N}$</span> in <span class="math-container">$A_{N}$</span> for all <span class="math-container">$N > N_{0}$</span> no matter where x lies in (0,1). Thus B intersects all <span class="math-container">$A_{N}$</span> which have <span class="math-container">$N > N_{0}$</span>. Since B is contained within U, U must intersect the same two balls. As N > <span class="math-container">$N_{0}$</span> goes to infinity, any neighborhood U of <span class="math-container">$x_{0}$</span> intersects 2 balls of infinitely many <span class="math-container">$A_{N}$</span>.</p>
<p>Here we note that A is not locally finite since no open set U of X intersects only finitely many <span class="math-container">$A_{N}$</span>.</p>
<p><strong>2.) Now take the closure of each <span class="math-container">$A_{N}$</span>.</strong></p>
<p><strong>3.) Then <span class="math-container">$\overline{A_{N}}$</span> = (0,1) for each N. (Note 0 and 1 are not elements of X = (0,1)).</strong></p>
<p>Hence the neighborhood U containing <span class="math-container">$x_{0}$</span> intersects the only member of <span class="math-container">$\overline{A} = (0,1)$</span></p>
<hr />
<p><strong>Now, to expand this to any metric space X: first expand this proof to a bounded subset of the metric space X. Then use a limiting process to expand to the entire space. Should work great.</strong></p>
<p>Ill do R^n here for example: Just use a non-empty bounded open ball B of radius R' in R^n, instead of (0,1) in R. Approximate the set from the inside with almost disjoint closed cubes of side length r' = R'/N. (r = R/2, R/3, R/4, etc, etc...) Now take the interiors of these cubes to get a collection of disjoint open cubes. Call the collection <span class="math-container">$C_{N}$</span>. The interior of <span class="math-container">$\overline{\bigcup{C_{N}}}$</span> can now be exhausted with almost disjoint closed cubes <span class="math-container">$C_{N+1}$</span> such that the <span class="math-container">$\overline{\bigcup{C_{N}}} = \overline{\bigcup{C_{N+1}}}$</span> = V. Then as above, any ball B containing x in U will intersect 2 cubes of each <span class="math-container">$C_{N}$</span> as long as N is large enough. Such as when R'/N = r' is smaller than 2 times the radius r of B containing x. Then do some sort of limiting process on R' to exhaust the entire set <span class="math-container">$R^{n}$</span>; <span class="math-container">$\lim_{R' \to \infty} B \supset X$</span>.</p>
|
88,363 | <p>It is easy to truncate Series upto some order, say $n$. My question is how do I remove low orders? Let us say my series is a power series in $x$. I want to remove the terms with negative powers because they diverge at $x = 0$. I can simply write</p>
<p>s1-s2, where</p>
<p>s1=Normal[Series[blah, {x, 0, n}]</p>
<p>s2=Normal[Series[blah, {x, 0, -1}]</p>
<p>but Mathematica does not understand to cancel the removed terms because they are complicated. The solution would be to use Collect[s1-s2, x, Simplify], but this is horribly slow as I increase $n$ above even 2. I suppose I could simply delete the terms by hand, but the outputs are very messy, and there must exist a proper way to do this.</p>
| Bill | 18,890 | <p>Is this as simple as</p>
<pre><code>DeleteCases[s1, _*x^c_ /; c<0]
</code></pre>
<p>That is going to find all the terms in your series with negative exponents and simply delete them.</p>
|
176,260 | <blockquote>
<p>Let $\left\{ f_{n}\right\} $ denote the set of functions on
$[0,\infty) $ given by $f_{1}\left(x\right)=\sqrt{x} $ and
$f_{n+1}\left(x\right)=\sqrt{x+f_{n}\left(x\right)} $ for $n\ge1 $.
Prove that this sequence is convergent and find the limit function.</p>
</blockquote>
<p>We can easily show that this sequence is is nondecreasing. Originally, I was trying to apply the fact that “every bounded monotonic sequence must converge” but then it hit me this is true for $\mathbb{R}^{n} $. Does this fact still apply on $C[0,\infty) $, the set of continuous functions on $[0,\infty) $. If yes, what bound would we use?</p>
| user29999 | 29,999 | <p>You know that the sequence is increasing. Now notice that if $f(x)$ is the positive solution of $y^2=x+y$, we have $f_n(x)<f(x) \ \forall n \in \mathbb{N}$. In fact, notice that this hold for $n=1$ and assuming for $n-1$ we have
$$f^{2}_{n+1}(x) = x +f_n(x) < x+f(x) = f^{2}(x).$$ Then there exist $\lim_n f_n(x)$ and taking the limit in $f^{2}_{n+1}(x) = x + f_n(x)$, we find that $\lim_n f_n(x)= f(x)$.</p>
|
176,260 | <blockquote>
<p>Let $\left\{ f_{n}\right\} $ denote the set of functions on
$[0,\infty) $ given by $f_{1}\left(x\right)=\sqrt{x} $ and
$f_{n+1}\left(x\right)=\sqrt{x+f_{n}\left(x\right)} $ for $n\ge1 $.
Prove that this sequence is convergent and find the limit function.</p>
</blockquote>
<p>We can easily show that this sequence is is nondecreasing. Originally, I was trying to apply the fact that “every bounded monotonic sequence must converge” but then it hit me this is true for $\mathbb{R}^{n} $. Does this fact still apply on $C[0,\infty) $, the set of continuous functions on $[0,\infty) $. If yes, what bound would we use?</p>
| Mercy King | 23,304 | <p>Clearly, for each $x \ge 0$ the sequence $\{f_n(x)\}_n$ is increasing.</p>
<p>Let
$$
g:[0,\infty) \to \mathbb{R},\ g(x)=
\left\{
\begin{array}{cc}
0 & \text{ if } x=0\cr
\frac{1+\sqrt{1+4x}}{2} & \text{ if } x>0
\end{array}
\right..
$$</p>
<p>I claim that $f_n(x) \le g(x)$ for every $x \ge 0$. Indeed $f_n(0)=0=g(0)$ for every $n$, and $f_1(x) \le g(x)$ for every $x > 0$. If I suppose that $f_n(x) \le g(x)$ for $n \ge 1$ and $x > 0$, then
$$
f_{n+1}^2(x)-g^2(x)=x+f_n(x)-\frac{1+2x+\sqrt{1+4x}}{2}=f_n(x)-g(x)\le 0,
$$
i.e. $f_{n+1}(x)\le g(x)$. Hence $f_n(x)\le g(x)$ for every $n \ge 1$ and $x > 0$.</p>
<p>Thus for each $x \ge 0$, the sequence $\{f_n(x)\}_n$ is convergent (being increasing and bounded from above), and its (pointwise) limit $f: [0,\infty) \to \mathbb{R}, x \mapsto f(x)$ satisfies $f(x)=\sqrt{x+f(x)}$, i.e. $f(x)=g(x)$ for every $x \ge 0$.</p>
|
1,579,521 | <p>Find the value of
<span class="math-container">$$ \iint_{\Sigma} \langle x, y^3, -z\rangle. d\vec{S} $$</span>
where <span class="math-container">$ \Sigma $</span> is the sphere <span class="math-container">$ x^2 + y^2 + z^2 = 1 $</span> oriented outward by using the divergence theorem.</p>
<p>So I calculate <span class="math-container">$\operatorname{div}\vec{F} = 3y^2 $</span> and then I convert <span class="math-container">$ x, y, z $</span> into <span class="math-container">$ x = p\sin \phi \cos \theta, y = p\sin \phi \sin \theta, z = p\cos \phi $</span> but then I got stuck from that point.</p>
| Simon S | 21,495 | <p>You therefore want the volume integral</p>
<p>$$\iiint_V 3y^2 \ dV = \int_0^1 \int_0^{2\pi} \int_0^\pi 3(p\sin\phi \sin\theta)^2 . p^2 \sin\phi \ dp \ d\theta \ d\phi = \ \cdots$$</p>
<p>Since the volume of integration is a rectangular box in $(p,\theta, \phi)$ coordinates, you can separate out the integrating variables to make it easy to evaluate the overall integral.</p>
|
874,697 | <p>I'm trying to understand the supremem of a sequence of functions so I came up with a trivial case as follows -</p>
<p>Let $(f_n(x))$ be a sequence of functions with $n$ having a value of either $1$ or $2$. Ie. A sequence with only two elements.</p>
<p>Now if we define $f_1(x)=1$ and $f_2(x)=x$ what is the sup $(f(x))$?</p>
| MPW | 113,214 | <p>"Supremum" generalizes the idea of "maximum". So when you are considering the supremum of a finite collection, it is the same as the maximum. Specifically,</p>
<p>$$\sup \{f_1,f_2\} =\max\{f_1,f_2\}$$</p>
<p>which is meant to be interpreted pointwise. That is, for a specific $x$,</p>
<p>$$(\sup \{f_1,f_2\})(x) = \sup \{f_1(x),f_2(x)\} = \max \{f_1(x),f_2(x)\}
$$</p>
<p>since there are only two values in the collection.</p>
|
75,795 | <p><strong>The problem:</strong></p>
<p>If we have</p>
<blockquote>
<p>$P(H_\eta|E_1,E_2,...,E_e)(1 \leq \eta \leq \mathbb{H})$</p>
</blockquote>
<p>and</p>
<blockquote>
<p>$P(E_1,E_2,...,E_e)$</p>
</blockquote>
<p>for all <strong>True</strong> and <strong>False</strong> values of $E_\epsilon(1 \leq \epsilon \leq e)$ and $H_\eta(1 \leq \eta \leq \mathbb{H})$.</p>
<p>Can we find</p>
<blockquote>
<p>$P(H_h)$, $P(E_\epsilon|H_h) (1 \leq \epsilon \leq e)$ and $P(E_\epsilon) (1 \leq \epsilon \leq e)$</p>
</blockquote>
<p>??</p>
| Craig | 15,279 | <p>Yes. $P(H_{\eta}) = \sum P(H_{\eta} | E_1, E_2, \ldots, E_e) P(E_1, E_2, \ldots, E_e)$ where the sum is over all possible values of $E_1, E_2, \ldots, E_e$.</p>
<blockquote>
<p>$P(E_{\epsilon}) = \sum_{ { E_1, \ldots , E_{\epsilon -1}, E_{\epsilon +1}, \ldots , E_e } } P(E_1, \ldots, E_e)$</p>
</blockquote>
<p>The last one you need to use Bayes' Law: $P(E_{\epsilon} | H_{\eta}) = P(E_{\epsilon}, H_{\eta}) / P(H_{\eta})$. We've determined $P(H_{\eta}$ already, so we just have to get $P(E_{\epsilon}, H_{\eta})$. </p>
<blockquote>
<p>$ P(H_{\eta}, E_{\epsilon}) = \sum_{ \{ E_1, \ldots ,E_{\epsilon - 1}, E_{\epsilon + 1}, \ldots ,E_e \} } P(H_{\eta}, E_1 , \ldots E_e ) = \sum P(H_{\eta} | E_1 ,\ldots , E_e) P(E_1 ,\ldots E_e)$</p>
</blockquote>
|
2,152,872 | <p>I am working on a problem in Artin's Algebra related to the algebraic geometry talked in Chapter 11. The problem number is 9.2., F.Y.I.</p>
<p>Here goes the problem:</p>
<blockquote>
<p>Let $f_1, \dots, f_r$ be complex polynomials in the variables $x_1, \dots, x_n$, let $V$ be the variety of their common zeros, and let $I$ be the ideal of the polynomial ring $R = \mathbb{C}\left [ x_1, \dots, x_n \right ]$ they generate. Define a ring homomorphism from the quotient ring $\bar{R} = R/I$ to the ring $\mathcal{R}$ of continuous, complex-valued functions on $V$.</p>
</blockquote>
<p>I attempted to use the correspondence theorem w.r.t. the variety of a set of polynomials, i.e. the maximal ideals bijectively correspond to the point in $V$ and we may somehow define the continuous functions there. However I cannot come up with any idea further. Also, the term 'continuous' here seems redundant since I expect the homomorphism will carry polynomials to polynomials. </p>
<p>I appreciate your participation and will be thankful to anything from hints to full solution. </p>
| Rebecca J. Stones | 91,818 | <p>We're not completely empty handed.</p>
<p>Sanity checks:</p>
<ul>
<li><p>If a sequence has a number greater than $n-1$, where $n$ is the number of coordinates (vertices), then it cannot be constructed (as there's not enough vertices to form the neighbors).</p></li>
<li><p>If a sequence sums to an odd number, it would violate the <a href="https://en.wikipedia.org/wiki/Handshaking_lemma" rel="nofollow noreferrer">Handshaking Lemma</a>, and can't be constructed.</p></li>
<li><p>If there's a lot of vertices of degree $1$, and two or more vertices of high degree, sometimes this can make it impossible (if we were to attempt to create one, we'd end up with too many vertices of degree $2$ or more).</p></li>
</ul>
<p>Proving existence:</p>
<ul>
<li>We can check if such a graph exists using the <a href="https://en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Gallai_theorem" rel="nofollow noreferrer">Erdős-Gallai Theorem</a>.</li>
</ul>
<p>Construct example:</p>
<ul>
<li>We can construct an example of such a graph using the <a href="https://en.wikipedia.org/wiki/Havel%E2%80%93Hakimi_algorithm" rel="nofollow noreferrer">Havel-Hakimi Algorithm</a>.</li>
</ul>
<p>Construct all non-isomorphic examples:</p>
<ul>
<li>I tend to just use <code>geng</code> (which comes with Nauty), with the command line <code>geng n e:e</code> where $n$ is the number of vertices (in this case 5) and $e$ is the number of edges (in this case $\tfrac{1}{2}(1+1+2+3+3)=5$). Afterwards, I filter out the ones which don't have the degree sequence using a <a href="http://www.gap-system.org/" rel="nofollow noreferrer">GAP</a> script.</li>
</ul>
<p>In this example, there is a unique graph (with vertices marked with their degree):</p>
<p><a href="https://i.stack.imgur.com/GxLKD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GxLKD.png" alt="enter image description here"></a></p>
|
23,566 | <p>I love math, and I used to be very good at it. The correct answers came fast and intuitively. I never studied, and redid the demonstration live for the tests (sometimes inventing new ones). I was the one who answered the tricky questions in class (8 hours of math/week in high school)... You get the idea.</p>
<p>As such I used to have a lot of confidence in my math abilities, and didn't think twice about saying the first idea that came to mind when answering a question.</p>
<p>That was more than 10 years ago, and I (almost) haven't done any math since then. I've graduated in a scientific field that requires little of it (I prefer not to give details) and worked for some time.</p>
<p>Now I'm back at school (master of statistics) and I need to do math, once again. I make mistakes upon blunders with the same confidence I used to have when I was good, which is extremely embarrassing when it happens in class. </p>
<p>I feel like a tone deaf musician and an ataxic painter at the same time.</p>
<p>One factor that probably plays a role is that I've learnt math in my mother tongue, and I'm now using it English, but I wouldn't expect it to make such a difference. </p>
<p>I know that it will require practice and hard work, but I need direction.</p>
<p>Any help is welcome.</p>
<p>Kind regards,</p>
<p>-- Mathemastov</p>
| zellyn | 7,512 | <p>Similar situation, except I didn't get as far originally - just a CS undergrad 13 years ago. Now trying to study machine learning (Stanford online class), and finding everything from Linear Algebra back to even simple derivatives and integration has atrophied immensely. All I can say is keep slogging away at it - if I compare how slow my brain felt three or four years ago to now, it's already night and day. It comes back, but it feels more like a gradual waking up again of the math parts of my brain than a sudden flash. Good luck! :-)</p>
|
23,566 | <p>I love math, and I used to be very good at it. The correct answers came fast and intuitively. I never studied, and redid the demonstration live for the tests (sometimes inventing new ones). I was the one who answered the tricky questions in class (8 hours of math/week in high school)... You get the idea.</p>
<p>As such I used to have a lot of confidence in my math abilities, and didn't think twice about saying the first idea that came to mind when answering a question.</p>
<p>That was more than 10 years ago, and I (almost) haven't done any math since then. I've graduated in a scientific field that requires little of it (I prefer not to give details) and worked for some time.</p>
<p>Now I'm back at school (master of statistics) and I need to do math, once again. I make mistakes upon blunders with the same confidence I used to have when I was good, which is extremely embarrassing when it happens in class. </p>
<p>I feel like a tone deaf musician and an ataxic painter at the same time.</p>
<p>One factor that probably plays a role is that I've learnt math in my mother tongue, and I'm now using it English, but I wouldn't expect it to make such a difference. </p>
<p>I know that it will require practice and hard work, but I need direction.</p>
<p>Any help is welcome.</p>
<p>Kind regards,</p>
<p>-- Mathemastov</p>
| John Hammond | 7,346 | <p>I don't mean to echo @Bill and @Derek, but certainly the hard work of practice, and making many mistakes along the way is absolutely required to get back your mathematical skills.</p>
<p>I want to add to the discussion that you shouldn't underestimate the value of social learning. If you're having trouble with a proof (or even not having problems), grab a piece of chalk and work through the problem on the chalkboard with your fellow grad students. It's an exceptional way to learn. Having to vocalize your arguments really helps to formalize your thoughts.</p>
|
23,566 | <p>I love math, and I used to be very good at it. The correct answers came fast and intuitively. I never studied, and redid the demonstration live for the tests (sometimes inventing new ones). I was the one who answered the tricky questions in class (8 hours of math/week in high school)... You get the idea.</p>
<p>As such I used to have a lot of confidence in my math abilities, and didn't think twice about saying the first idea that came to mind when answering a question.</p>
<p>That was more than 10 years ago, and I (almost) haven't done any math since then. I've graduated in a scientific field that requires little of it (I prefer not to give details) and worked for some time.</p>
<p>Now I'm back at school (master of statistics) and I need to do math, once again. I make mistakes upon blunders with the same confidence I used to have when I was good, which is extremely embarrassing when it happens in class. </p>
<p>I feel like a tone deaf musician and an ataxic painter at the same time.</p>
<p>One factor that probably plays a role is that I've learnt math in my mother tongue, and I'm now using it English, but I wouldn't expect it to make such a difference. </p>
<p>I know that it will require practice and hard work, but I need direction.</p>
<p>Any help is welcome.</p>
<p>Kind regards,</p>
<p>-- Mathemastov</p>
| user31284 | 31,284 | <p>I think that having study groups, peer tutoring, watching youtube videos, iPhone applications, and just going through textbooks will serve you to increase your math skill strength.</p>
|
2,407,820 | <p>$$\ln^q (1+x) \le \frac{q}{p} x^p \quad (x \ge 0, \; 0 < p \le q)$$</p>
<p>For $p=q$ this reduces to the familiar $\ln(1+x) \le x$. Otherwise I haven't had much success in proving it. General suggestions would be appreciated.</p>
| shalop | 224,467 | <p>This is false. If it were true, then for $q \geq 1$ we can set $p=1$ and we would have that $$\log(1+x) \leq q^{1/q} x^{1/q}$$ Then we can let $q \to \infty$ on both sides and we get that $\log(1+x) \leq 1$, which clearly becomes false as soon as $x >e-1$.</p>
<p>For an explicit counterexample, Wolfram alpha tells me that $\log^{50} 3 = 110.2...$ while $\frac{50}{1} \cdot 2^1 =\frac{50}{2} \cdot 2^2= 100$.</p>
|
120,260 | <p>Let $X$ be a simply connected smooth projective variety, whose Picard group is generated by the classes of the irreducible codimension 1 loci $D_1, \ldots, D_k$. Let $E_1, \ldots, E_r$ be other irreducible codimension 1 loci, and suppose that $X^0$ is the complement in $X$ of the divisors $D_i$ and $E_j$.</p>
<p>Suppose now that $X_0$ is the complement of $n$ irreducible loci of codimension $1$ in $Y$, a smooth projective variety.</p>
<p>Question: Can I conclude that the Picard group of $Y$ has rank $n-r$?</p>
<p>I can answer the question affirmatively over $\mathbb{C}$, by using the long exact sequence with compact support associated with the inclusion $Y \setminus X^0 \to Y$, but I would like to know if there is an algebraic proof of this (valid over any algebraically closed field $k$).</p>
<p>EDIT: As pointed out in the answer, I am actually assuming that the Picard group of $X$ is FREELY generated by the $D_1, \ldots, D_k$.</p>
| M.B | 8,419 | <p>Recently several fundamental works have been done in the area of the geometry of numbers. Beside Bhargava's revolutionary ideas, and of course the contribution of his students, Ergodic theory is a new idea that plays an important role in modern geometry of numbers which seems to me that several ideas are coming from Margulis and E. Lindenstrauss's works. </p>
<p>Here is a list of papers in this area which I think are extremely interesting and important. </p>
<ol>
<li><p>Minkowski's theorem for random lattices. Margulis (Russian) Problemy Peredachi Informatsii 47 (2011), no. 4, 104--108; translation in
Probl. Inf. Transm. 47 (2011), no. 4, 398–402 </p></li>
<li><p><a href="http://www.ams.org/mathscinet/search/publdoc.html?arg3=&co4=AND&co5=AND&co6=AND&co7=AND&dr=all&pg4=AUCN&pg5=TI&pg6=PC&pg7=ALLF&pg8=ET&review_format=html&s4=Margulis&s5=&s6=&s7=&s8=All&vfpref=html&yearRangeFirst=&yearRangeSecond=&yrop=eq&r=8&mx-pid=2538473" rel="nofollow">Logarithm laws for unipotent flows</a>. I, Athreya; Margulis</p></li>
<li><p><a href="http://www.ams.org/mathscinet/search/publdoc.html?amp=&loc=refcit&refcit=2538473&vfpref=html&r=1&mx-pid=2861748" rel="nofollow">On the probability of a random lattice avoiding a large convex set</a>, Strömbergsson </p></li>
<li><p><a href="http://math.stanford.edu/~akshay/research/sp.pdf" rel="nofollow">A note on sphere packings in high dimension</a>, Venkatesh</p></li>
<li><p><a href="http://www.ams.org/mathscinet/search/publdoc.html?pg1=INDI&pg6=PC&s1=931224&s6=11&vfpref=html&r=1&mx-pid=2855800" rel="nofollow">On the distribution of angles between the $N$ shortest vectors in a random lattice</a>, Södergren, Anders</p></li>
<li><p><a href="http://www.technion.ac.il/~ushapira/Papers/EuclideanMin.pdf" rel="nofollow">Remarks on Euclidean Minima</a>, Uri Shapira, Zhiren Wang</p></li>
<li><p><a href="http://www.technion.ac.il/~ushapira/Papers/gruber6.pdf" rel="nofollow">On the Mordell-Gruber spectrum</a>, Uri Shapira, Barak Weiss</p></li>
<li><p><a href="http://www.technion.ac.il/~ushapira/Papers/GDP5.pdf" rel="nofollow">A solution to a problem of Cassels and Diophantine properties of cubic numbers</a>, Uri Shapira</p></li>
</ol>
|
2,633,720 | <blockquote>
<p>Prove by induction that $$ (k + 2)^{k + 1} \leq (k+1)^{k +2}$$ for $ k > 3 .$</p>
</blockquote>
<p>I have been trying to solve this, but I am not getting the sufficient insight. </p>
<p>For example, $(k + 2)^{k + 1} = (k +2)^k (k +2) , (k+1)^{k +2}= (k+1)^k(k +1)^2.$</p>
<p>$(k +2) < (k +1)^2 $ but $(k+1)^k < (k +2)^k$ so what I want would clearly not be immediate from using something like If $ 0 < a < b, 0<c<d $ then $0 < ac < bd $. THe formula is valid for n = 4, So if it is valid for $n = k$ I would have to use</p>
<p>$ (k + 2)^{k + 1} \leq (k+1)^{k +2} $ somewhere in order to get that $ (k + 3)^{k + 2} \leq (k+2)^{k +3} $ is also valid. This seems tricky.</p>
<p>I also tried expanding $(k +2)^k $ using the binomial formula and multiplying this by $(k + 2)$, and I expanded $(k+1)^k$ and multiplied it by $(k + 1)^2 $ term by term. I tried to compare these sums, but it also gets tricky. I would appreaciate a hint for this problem, thanks. </p>
| Michael Rozenberg | 190,319 | <p>For $k=4$ it's true.</p>
<p>Let $(k+2)^{k+1}\leq(k+1)^{k+2}.$
Thus, $$((k+2)^2)^{k+1}\leq(k+1)^{k+2}(k+2)^{k+1}$$ or
$$((k+1)(k+3)+1)^{k+1}\leq(k+1)^{k+2}(k+2)^{k+1},$$ which gives
$$((k+1)(k+3))^{k+1}\leq(k+1)^{k+2}(k+2)^{k+1}$$ or
$$(k+3)^{k+1}\leq(k+1)(k+2)^{k+1}.$$
Thus, $$(k+3)^{k+2}\leq(k+3)(k+1)(k+2)^{k+1}=$$
$$=(k^2+4k+3)(k+2)^{k+1}\leq(k+2)^2(k+2)^{k+1}=(k+2)^{k+3}$$ and we are done!</p>
|
322,302 | <p>Conjectures play important role in development of mathematics.
Mathoverflow gives an interaction platform for mathematicians from various fields, while in general it is not always easy to get in touch with what happens in the other fields.</p>
<p><strong>Question</strong> What are the conjectures in your field proved or disproved (counterexample found) in recent years, which are noteworthy, but not so famous outside your field?</p>
<p>Answering the question you are welcome to give some comment for outsiders of your field which would help to appreciate the result.</p>
<p>Asking the question I keep in mind by "recent years" something like a dozen years before now, by a "conjecture" something which was known as an open problem for something like at least dozen years before it was proved and I would say the result for which the Fields medal was awarded like a proof of <a href="https://en.wikipedia.org/wiki/Fundamental_lemma_(Langlands_program)" rel="noreferrer">fundamental lemma</a> would not fit "not so famous", but on the other hand these might not be considered as strict criteria, and let us "assume a good will" of the answerer.</p>
| Jonny Evans | 10,839 | <p>The Audin conjecture in symplectic topology, posed in 1988 by Audin in her famous paper on Lagrangian immersions, asserts that all Lagrangian tori in the standard symplectic vector space have minimal Maslov number 2. This was recently proven by Cieliebak and Mohnke:</p>
<p><a href="https://arxiv.org/abs/1411.1870" rel="nofollow noreferrer">https://arxiv.org/abs/1411.1870</a></p>
<p>That paper nicely summarises the history of the conjecture:</p>
<p>"This question was answered earlier for n = 2 by Viterbo [57] and Polterovich [54], in the monotone case for n ≤ 24 by Oh [52], and in the monotone case for general n by Buhovsky [12] and by Fukaya, Oh, Ohta and Ono [28, Theorem 6.4.35], see also Damian [22]. A different approach has been outlined by Fukaya [27]. The scheme to prove Audin’s conjecture using punctured holomorphic curves was suggested by Y. Eliashberg around 2001. The reason it took over 10 years to complete this paper are transversality problems in the non-monotone case."</p>
<p>Edit: It occurred to me that this paper is probably now published; indeed it appeared in Inventiones in 2017. Here is the DOI:</p>
<p><a href="https://doi.org/10.1007/s00222-017-0767-8" rel="nofollow noreferrer">https://doi.org/10.1007/s00222-017-0767-8</a></p>
|
322,302 | <p>Conjectures play important role in development of mathematics.
Mathoverflow gives an interaction platform for mathematicians from various fields, while in general it is not always easy to get in touch with what happens in the other fields.</p>
<p><strong>Question</strong> What are the conjectures in your field proved or disproved (counterexample found) in recent years, which are noteworthy, but not so famous outside your field?</p>
<p>Answering the question you are welcome to give some comment for outsiders of your field which would help to appreciate the result.</p>
<p>Asking the question I keep in mind by "recent years" something like a dozen years before now, by a "conjecture" something which was known as an open problem for something like at least dozen years before it was proved and I would say the result for which the Fields medal was awarded like a proof of <a href="https://en.wikipedia.org/wiki/Fundamental_lemma_(Langlands_program)" rel="noreferrer">fundamental lemma</a> would not fit "not so famous", but on the other hand these might not be considered as strict criteria, and let us "assume a good will" of the answerer.</p>
| Joseph O'Rourke | 6,094 | <p><a href="https://en.wikipedia.org/wiki/Hinged_dissection" rel="noreferrer">Hinged dissections</a> exist.
(See <a href="https://mathoverflow.net/q/80101/6094">3-piece dissection of square to equilateral triangle?</a> for an animation of Dudeney's famous equilateral-triangle-to-square hinged dissection.)</p>
<blockquote>
<p>Abbott, Timothy G., Zachary Abel, David Charlton, Erik D. Demaine, Martin L. Demaine, and Scott Duke Kominers. "Hinged dissections exist." <em>Discrete & Computational Geometry</em> 47, no. 1 (2012): 150-186.
<a href="https://link.springer.com/article/10.1007/s00454-010-9305-9" rel="noreferrer">Springer link</a>.</p>
</blockquote>
<p>"<em>Abstract</em>. We prove that any finite collection of polygons of equal area has a common hinged dissection. That
is, for any such collection of polygons there exists a chain of polygons hinged at vertices that can be
folded in the plane continuously without self-intersection to form any polygon in the collection. This
result settles the open problem about the existence of hinged dissections between pairs of polygons that
goes back implicitly to 1864 and has been studied extensively in the past ten years."
<hr />
<a href="https://i.stack.imgur.com/8EfyE.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/8EfyE.jpg" alt="HingedFig6"></a>
<hr />
The proof is not simple—as hinted by the above figure—but it is constructive.</p>
|
322,302 | <p>Conjectures play important role in development of mathematics.
Mathoverflow gives an interaction platform for mathematicians from various fields, while in general it is not always easy to get in touch with what happens in the other fields.</p>
<p><strong>Question</strong> What are the conjectures in your field proved or disproved (counterexample found) in recent years, which are noteworthy, but not so famous outside your field?</p>
<p>Answering the question you are welcome to give some comment for outsiders of your field which would help to appreciate the result.</p>
<p>Asking the question I keep in mind by "recent years" something like a dozen years before now, by a "conjecture" something which was known as an open problem for something like at least dozen years before it was proved and I would say the result for which the Fields medal was awarded like a proof of <a href="https://en.wikipedia.org/wiki/Fundamental_lemma_(Langlands_program)" rel="noreferrer">fundamental lemma</a> would not fit "not so famous", but on the other hand these might not be considered as strict criteria, and let us "assume a good will" of the answerer.</p>
| Carl-Fredrik Nyberg Brodda | 120,914 | <p>The Hall-Paige conjecture, first posed in 1955 by Marshall Hall and L. J. Paige, is the following: </p>
<blockquote>
<p>A finite group <span class="math-container">$G$</span> has a complete mapping if and only if its Sylow <span class="math-container">$2$</span>-subgroups are not cyclic.</p>
</blockquote>
<p>Note that a complete mapping is a bijection <span class="math-container">$\phi : G \to G$</span> such that the function given by <span class="math-container">$\psi(g) = g \phi(g)$</span> is also a bijection. The above statement was shown to be necessary by Hall and Paige, but its sufficiency remained open until very recently; in 2009, it was shown to be sufficient to only check the cases when <span class="math-container">$G$</span> is a finite simple group, and the same year all finite simple groups except for <span class="math-container">$J_4$</span> were shown to satisfy the conjecture. John Bray then dealt with this final case in unpublished work, and Peter Cameron was able to convince him (see <a href="https://cameroncounts.wordpress.com/2018/12/18/the-hall-paige-conjecture/" rel="noreferrer">this</a>) to publish these noteworthy calculations many years later; the final proof of the Hall-Paige conjecture, together with some consequences of it regarding synchronicity in groups, was written up in 2018 and can be found as a preprint on <a href="https://arxiv.org/abs/1811.12671" rel="noreferrer">the arXiv</a>.</p>
|
322,302 | <p>Conjectures play important role in development of mathematics.
Mathoverflow gives an interaction platform for mathematicians from various fields, while in general it is not always easy to get in touch with what happens in the other fields.</p>
<p><strong>Question</strong> What are the conjectures in your field proved or disproved (counterexample found) in recent years, which are noteworthy, but not so famous outside your field?</p>
<p>Answering the question you are welcome to give some comment for outsiders of your field which would help to appreciate the result.</p>
<p>Asking the question I keep in mind by "recent years" something like a dozen years before now, by a "conjecture" something which was known as an open problem for something like at least dozen years before it was proved and I would say the result for which the Fields medal was awarded like a proof of <a href="https://en.wikipedia.org/wiki/Fundamental_lemma_(Langlands_program)" rel="noreferrer">fundamental lemma</a> would not fit "not so famous", but on the other hand these might not be considered as strict criteria, and let us "assume a good will" of the answerer.</p>
| Zach Teitler | 88,133 | <p>A tensor <span class="math-container">$T \in V^{\otimes n} = V \otimes V \otimes \dotsb \otimes V$</span> is <em>symmetric</em> if <span class="math-container">$T$</span> is invariant under permutations of the factors <span class="math-container">$V$</span> in the tensor product (here <span class="math-container">$V$</span> is a finite-dimensional vector space). The <em>tensor rank</em> of <span class="math-container">$T$</span> is the least number of terms in a decomposition of <span class="math-container">$T$</span> as a sum of simple (aka decomposable, aka pure, aka rank one...) terms, i.e., ones of the form <span class="math-container">$v_1 \otimes \dotsb \otimes v_n$</span>. The <em>symmetric tensor rank</em> of <span class="math-container">$T$</span> is the least number of terms in a decomposition of <span class="math-container">$T$</span> as a sum of terms of the form <span class="math-container">$v^{\otimes n} = v \otimes v \otimes \dotsb \otimes v$</span>, i.e., <span class="math-container">$v_1 = \dotsb = v_n = v$</span>. Evidently the symmetric tensor rank of a symmetric tensor is greater than or equal to its tensor rank.</p>
<p>Comon raised the question of whether every symmetric tensor has symmetric tensor rank equal to its tensor rank, say over the complex numbers. It is true for <span class="math-container">$n=2$</span>, which is the statement that the (ordinary matrix) rank of a symmetric matrix is equal to the rank of the quadratic form represented by that matrix. "Comon's conjecture" is the assertion that the answer to Comon's question is positive; it's not completely clear to me whether Comon made this assertion (as opposed to just asking the question), but anyway that name has been used. Besides <span class="math-container">$n=2$</span>, some other cases were known.</p>
<p>Recently Shitov has given a counterexample over the complex numbers.</p>
<p><em>Shitov, Yaroslav</em>, <a href="http://dx.doi.org/10.1137/17M1131970" rel="noreferrer"><strong>A counterexample to Comon’s conjecture</strong></a>, SIAM J. Appl. Algebra Geom. 2, No. 3, 428-443 (2018). <a href="https://zbmath.org/?q=an:1401.15004" rel="noreferrer">ZBL1401.15004</a>.</p>
<p>The counterexample is fairly complicated. Shitov takes <span class="math-container">$n=3$</span> and <span class="math-container">$\dim V = 800$</span>, and constructs a symmetric tensor whose rank is <span class="math-container">$903$</span>, but whose symmetric rank is greater than <span class="math-container">$903$</span>. (The symmetric rank of Shitov's example is not known.) Shitov's counterexample works over the complex numbers. No counterexample is known over the real numbers, or any other field. So Comon's conjecture is still open for the reals and other fields, but I wouldn't bet on it holding.</p>
|
322,302 | <p>Conjectures play important role in development of mathematics.
Mathoverflow gives an interaction platform for mathematicians from various fields, while in general it is not always easy to get in touch with what happens in the other fields.</p>
<p><strong>Question</strong> What are the conjectures in your field proved or disproved (counterexample found) in recent years, which are noteworthy, but not so famous outside your field?</p>
<p>Answering the question you are welcome to give some comment for outsiders of your field which would help to appreciate the result.</p>
<p>Asking the question I keep in mind by "recent years" something like a dozen years before now, by a "conjecture" something which was known as an open problem for something like at least dozen years before it was proved and I would say the result for which the Fields medal was awarded like a proof of <a href="https://en.wikipedia.org/wiki/Fundamental_lemma_(Langlands_program)" rel="noreferrer">fundamental lemma</a> would not fit "not so famous", but on the other hand these might not be considered as strict criteria, and let us "assume a good will" of the answerer.</p>
| Thomas Sauvaget | 469 | <p>The following <a href="https://twitter.com/gabrielpeyre/status/1091990679897718784" rel="noreferrer">comes directly</a> from Gabriel Peyré's excellent <a href="https://twitter.com/gabrielpeyre" rel="noreferrer">twitter feed</a>:</p>
<blockquote>
<p>The <a href="https://en.wikipedia.org/wiki/Weierstrass_function" rel="noreferrer">Weierstrass function</a> is continuous if <span class="math-container">$a<1$</span> but nowhere differentiable if <span class="math-container">$ab>1$</span>. The Hausdorff dimension of its graph was conjectured by Mandelbrot in 1977 and <a href="https://arxiv.org/abs/1505.03986" rel="noreferrer">proved</a> by Shen in 2016. </p>
</blockquote>
|
322,302 | <p>Conjectures play important role in development of mathematics.
Mathoverflow gives an interaction platform for mathematicians from various fields, while in general it is not always easy to get in touch with what happens in the other fields.</p>
<p><strong>Question</strong> What are the conjectures in your field proved or disproved (counterexample found) in recent years, which are noteworthy, but not so famous outside your field?</p>
<p>Answering the question you are welcome to give some comment for outsiders of your field which would help to appreciate the result.</p>
<p>Asking the question I keep in mind by "recent years" something like a dozen years before now, by a "conjecture" something which was known as an open problem for something like at least dozen years before it was proved and I would say the result for which the Fields medal was awarded like a proof of <a href="https://en.wikipedia.org/wiki/Fundamental_lemma_(Langlands_program)" rel="noreferrer">fundamental lemma</a> would not fit "not so famous", but on the other hand these might not be considered as strict criteria, and let us "assume a good will" of the answerer.</p>
| EGME | 130,113 | <p><a href="https://arxiv.org/abs/1808.02923" rel="nofollow noreferrer">“Conway’s knot is not slice”</a>, by Lisa Piccirillo, Annals of Mathematics 191-2 (2020) 581-591, settled a problem (we can call it a conjecture) which was at least four decades old. Read in informal account of the result here: <a href="https://www.quantamagazine.org/graduate-student-solves-decades-old-conway-knot-problem-20200519/" rel="nofollow noreferrer">https://www.quantamagazine.org/graduate-student-solves-decades-old-conway-knot-problem-20200519/</a></p>
|
322,302 | <p>Conjectures play important role in development of mathematics.
Mathoverflow gives an interaction platform for mathematicians from various fields, while in general it is not always easy to get in touch with what happens in the other fields.</p>
<p><strong>Question</strong> What are the conjectures in your field proved or disproved (counterexample found) in recent years, which are noteworthy, but not so famous outside your field?</p>
<p>Answering the question you are welcome to give some comment for outsiders of your field which would help to appreciate the result.</p>
<p>Asking the question I keep in mind by "recent years" something like a dozen years before now, by a "conjecture" something which was known as an open problem for something like at least dozen years before it was proved and I would say the result for which the Fields medal was awarded like a proof of <a href="https://en.wikipedia.org/wiki/Fundamental_lemma_(Langlands_program)" rel="noreferrer">fundamental lemma</a> would not fit "not so famous", but on the other hand these might not be considered as strict criteria, and let us "assume a good will" of the answerer.</p>
| Aleksandar Milivojević | 104,342 | <p>In the paper</p>
<ul>
<li>Friedrich Hirzebruch, <em>Some Problems on Differentiable and Complex Manifolds</em>, Annals of Mathematics Second Series, Vol. 60, No. 2 (1954) pp. 213-236, doi:<a href="https://doi.org/10.2307/1969629" rel="nofollow noreferrer">10.2307/1969629</a></li>
</ul>
<p>Hirzebruch collected problems and questions on smooth and complex manifolds presented at a conference the year prior.</p>
<p>For an almost complex manifold <span class="math-container">$M$</span> equipped with a hermitian metric, one can form the Laplacian <span class="math-container">$\Delta_{\bar{\partial}}$</span>. Even though <span class="math-container">$\bar{\partial}$</span> need not square to zero on an almost complex manifold, this Laplacian is an elliptic operator and so the kernel of the Laplacian is finite dimensional if <span class="math-container">$M$</span> is compact. Denote the dimension of this kernel restricted to <span class="math-container">$(p,q)$</span> forms by <span class="math-container">$h^{p,q}$</span>.</p>
<p>Problem 20 in Hirzebruch's list, attributed to Kodaira and Spencer, asks the following about compact almost complex manifolds:</p>
<p>Let <span class="math-container">$M^n$</span> be an almost-complex manifold. Choose an Hermitian structure and consider the numbers <span class="math-container">$h^{p,q}$</span> defined as above.
Is <span class="math-container">$h^{p,q}$</span> independent of the choice of the Hermitian structure? If not, give some other definition of the <span class="math-container">$h^{p,q}$</span> of <span class="math-container">$M^n$</span> which depends only on the almost-complex structure and which generalizes the <span class="math-container">$h^{p,q}$</span> of a complex manifold.</p>
<p>(Note that in the case of an integrable complex structure, the numbers <span class="math-container">$h^{p,q}$</span> are metric independent as they are the dimension of the Dolbeault cohomology group <span class="math-container">$H_{\bar{\partial}}^{p,q}$</span>.)</p>
<p>In 2020, Holt and Zhang posted a preprint (update: the paper is now published in Advances in Mathematics)</p>
<ul>
<li>Tom Holt, Weiyi Zhang, <em>Harmonic Forms on the Kodaira–Thurston Manifold</em>, arXiv:<a href="https://arxiv.org/abs/2001.10962" rel="nofollow noreferrer">2001.10962</a>,</li>
</ul>
<p>showing that the numbers <span class="math-container">$h^{p,q}$</span> are in general metric-dependent. The underlying manifold they work with is the four-dimensional Kodaira–Thurston nilmanifold.</p>
|
322,302 | <p>Conjectures play important role in development of mathematics.
Mathoverflow gives an interaction platform for mathematicians from various fields, while in general it is not always easy to get in touch with what happens in the other fields.</p>
<p><strong>Question</strong> What are the conjectures in your field proved or disproved (counterexample found) in recent years, which are noteworthy, but not so famous outside your field?</p>
<p>Answering the question you are welcome to give some comment for outsiders of your field which would help to appreciate the result.</p>
<p>Asking the question I keep in mind by "recent years" something like a dozen years before now, by a "conjecture" something which was known as an open problem for something like at least dozen years before it was proved and I would say the result for which the Fields medal was awarded like a proof of <a href="https://en.wikipedia.org/wiki/Fundamental_lemma_(Langlands_program)" rel="noreferrer">fundamental lemma</a> would not fit "not so famous", but on the other hand these might not be considered as strict criteria, and let us "assume a good will" of the answerer.</p>
| Andreas Blass | 6,794 | <p>Maryanthe Malliaris and Saharon Shelah proved that the cardinal characteristics <span class="math-container">$\mathfrak p$</span> and <span class="math-container">$\mathfrak t$</span> are equal, answering a question that goes back at least to the 1970's and probably (with different formulation but the same content) to the 1940's:</p>
<ul>
<li>"Cofinality spectrum theorems in model theory, set theory, and general topology." J. Amer. Math. Soc. <strong>29</strong> (2016), 237–297, doi:<a href="https://doi.org/10.1090/jams830" rel="noreferrer">10.1090/jams830</a>, arXiv:<a href="https://arxiv.org/abs/1208.5424" rel="noreferrer">1208.5424</a>, Shelah archive:<a href="https://shelah.logic.at/papers/998/" rel="noreferrer">paper 998</a>).</li>
</ul>
<p>The definition of <span class="math-container">$\mathfrak t$</span> is the shortest possible length <span class="math-container">$\lambda$</span> of a well-ordered sequence <span class="math-container">$(A_\xi:\xi<\lambda)$</span> of infinite subsets of <span class="math-container">$\mathbb N$</span> such that (1) <span class="math-container">$A_\eta-A_\xi$</span> is finite whenever <span class="math-container">$\xi<\eta<\lambda$</span> and (2) there is no infinite <span class="math-container">$B\subseteq\mathbb N$</span> with <span class="math-container">$B-A_\xi$</span> finite for all <span class="math-container">$\xi<\lambda$</span>. That is, <span class="math-container">$\mathfrak t$</span> is the smallest length of any inextendible decreasing-mod-finite sequence of inifinte subsets of <span class="math-container">$\mathbb N$</span>.</p>
<p><span class="math-container">$\mathfrak p$</span> is defined similarly except that, instead of requiring the sequence to decreasing mod finite, one requires only that every finitely many of the <span class="math-container">$A$</span>'s have an infinite intersection.</p>
<p>It is easy to check that <span class="math-container">$\aleph_1\leq\mathfrak p\leq\mathfrak t\leq2^{\aleph_0}$</span>. It was also known previously that if <span class="math-container">$\mathfrak p=\aleph_1$</span> then <span class="math-container">$\mathfrak p=\mathfrak t$</span>. (I believe this result is due to Judith Roitman, but I can't find a reference now, not even in my chapter of the "Handbook of Set Theory" where this result is Theorem 6.25. Mea culpa.) I think it was expected that <span class="math-container">$\mathfrak p<\mathfrak t$</span> would turn out to be consistent with ZFC, until Malliaris and Shelah proved otherwise. Not only the theorem but the method of proof was surprising, as it involved ideas from model theory, even though the result is purely set-theoretic.</p>
|
322,302 | <p>Conjectures play important role in development of mathematics.
Mathoverflow gives an interaction platform for mathematicians from various fields, while in general it is not always easy to get in touch with what happens in the other fields.</p>
<p><strong>Question</strong> What are the conjectures in your field proved or disproved (counterexample found) in recent years, which are noteworthy, but not so famous outside your field?</p>
<p>Answering the question you are welcome to give some comment for outsiders of your field which would help to appreciate the result.</p>
<p>Asking the question I keep in mind by "recent years" something like a dozen years before now, by a "conjecture" something which was known as an open problem for something like at least dozen years before it was proved and I would say the result for which the Fields medal was awarded like a proof of <a href="https://en.wikipedia.org/wiki/Fundamental_lemma_(Langlands_program)" rel="noreferrer">fundamental lemma</a> would not fit "not so famous", but on the other hand these might not be considered as strict criteria, and let us "assume a good will" of the answerer.</p>
| Martin Väth | 165,275 | <p>Not sure whether this counts as recent enough:</p>
<p>Robert Cauty proved 2001 the Schauder conjecture that every continuous map of a nonempty compact convex subset of a topological vector space (not necessarily locally convex!) has a fixed point:</p>
<ul>
<li>Cauty, Robert, <em>Solution du problème de point fixe de Schauder</em>, Fundamenta Mathematica 170, 2001, 231-246.</li>
</ul>
<p>Although some problems have been found in the original paper, it seems that they could all be fixed.</p>
|
1,156,130 | <p>We know that any Abelian group is a $\mathbb{Z}$-module. Is the converse true?</p>
| hmakholm left over Monica | 14,366 | <p>Yes, there is a <em>bijective</em> correspondence between $\mathbb Z$-modules and abelian groups.</p>
<p>From module to group, just forget the scalar multiplication; the module laws directly require that the module's addition constitutes an abelian group.</p>
<p>From group to $\mathbb Z$-module, declare $n\cdot a$ to be $\underbrace{a+a+\cdots+a}_n$ and so forth.</p>
|
1,156,130 | <p>We know that any Abelian group is a $\mathbb{Z}$-module. Is the converse true?</p>
| Olórin | 187,521 | <p>Let <span class="math-container">$A$</span> be a commutative ring with unit. To give yourself an <span class="math-container">$A$</span>-module <span class="math-container">$M$</span> is equivalent to giving yourself an abelian group <span class="math-container">$M$</span> plus an <em>action</em> of <span class="math-container">$A$</span> on <span class="math-container">$M$</span>, that is, a ring morphism <span class="math-container">$\varphi : A \rightarrow \textrm{Hom}_{\textrm{Ab.}}(M,M)$</span> where <span class="math-container">$\textrm{Hom}_{\textrm{Ab.}}(N',N)$</span> is the set of morphism of abelian groups from <span class="math-container">$N'$</span> to <span class="math-container">$N$</span> if <span class="math-container">$N,N'$</span> are abelian groups.</p>
<p>Note that <span class="math-container">$\textrm{Hom}_{\textrm{Ab.}}(M,M)$</span>, also noted <span class="math-container">$\textrm{End}_{\textrm{Ab.}}(M)$</span>, is a ring with unit, but is not necessarily commutative. Now as you know, for any ring <span class="math-container">$R$</span> with unit, you have a unique ring morphism <span class="math-container">$\mathbf{Z}\rightarrow R$</span>. So that in particular if <span class="math-container">$M$</span> is an abelian group, you have a unique ring morphism <span class="math-container">$\varphi : \mathbf{Z} \rightarrow \textrm{Hom}_{\textrm{Ab.}}(M,M)$</span>, that is, a unique structure of <span class="math-container">$\mathbf{Z}$</span>-module on <span class="math-container">$M$</span>. Inversely, having one structure of <span class="math-container">$\mathbf{Z}$</span>-module on <span class="math-container">$M$</span> does not give very much, that is, not more that the already existing unique ring morphism <span class="math-container">$\varphi : \mathbf{Z} \rightarrow \textrm{Hom}_{\textrm{Ab.}}(M,M)$</span>.</p>
<p>In a pedantic way : the forgetful functor from the category of <span class="math-container">$\mathbf{Z}$</span>-modules to the category of abelian groups is an equivalence of category (and indeed worth mentionning, even an isomorphism of categories).</p>
|
3,680,536 | <p>In Eric Lengyel's book, <em>Mathematics for 3D Game Programming and Computer Graphics, 3<sup>rd</sup> Edition</em>, there is a theorem that states</p>
<blockquote>
<p>An <span class="math-container">$n \times n$</span> matrix M is invertible if and only if the rows of M form a linearly independent set of vectors.</p>
</blockquote>
<p>Two proofs were provided, each corresponding to if and only if of the theorem. I understand the first proof, but I don't fully understand the second one. I have some understanding of it, but it's not enough to fully grasp the proof. The proof, slightly modified for brevity, goes as follows:</p>
<blockquote>
<p>Let the rows of M be denoted by <span class="math-container">$R_{1}^{T}, R_{2}^{T}, \ldots, R_{n}^{T}$</span>. Now assume that the rows of M are a linearly independent set of vectors. We first observe that performing elementary row operations on a matrix does not alter the property of linear independence within the rows. Running through Algorithm 3.12, if step C fails, then rows <em>j</em> through <em>n</em> of the matrix at that point form a linearly dependent set since the number of columns for which the rows <span class="math-container">$R_{j}^{T}$</span> through <span class="math-container">$R_{n}^{T}$</span> have at least one nonzero entry is less than the number of rows itself. This is a contradiction, so step C of the algorithm cannot fail, and M must be invertible.</p>
</blockquote>
<p>Note: A screenshot from the book of Algorithm 3.12 is available at the bottom.</p>
<p>What I don't fully understand is this snippet of the proof:</p>
<blockquote>
<p>[...] rows <em>j</em> through <em>n</em> of the matrix at that point form a linearly dependent set since the number of columns for which the rows <span class="math-container">$R_{j}^{T}$</span> through <span class="math-container">$R_{n}^{T}$</span> have at least one nonzero entry is less than the number of rows itself. [...]</p>
</blockquote>
<p>Why do the rows form a linearly dependent set based on the idea that "the number of columns for which the rows <span class="math-container">$R_{j}^{T}$</span> through <span class="math-container">$R_{n}^{T}$</span> have at least one nonzero entry is less than the number of rows itself."?</p>
<hr>
<h3>Appendix</h3>
<p>Algorithm 3.12</p>
<p><a href="https://i.stack.imgur.com/8no1T.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8no1T.png" alt="Description of Algorithm 3.12"></a></p>
| Community | -1 | <p>I will prove the same statement for independent column vectors which is an equivalent statement.</p>
<p>If the columns of <span class="math-container">$A$</span> are linearly independent it is clear that <span class="math-container">$Ax=0$</span> implies <span class="math-container">$x=0$</span> and hence, the nullspace <span class="math-container">$N(A)$</span>={<span class="math-container">$0$</span>}.</p>
<p>That implies (since <span class="math-container">$dimN(A)+dim\Re(A)=n$</span>) that <span class="math-container">$dim\Re(A)=n$</span> and hence every vector in <span class="math-container">$\mathbb{R}^{n}$</span> can be written as <span class="math-container">$Ay$</span>. Therefore <span class="math-container">$e_{1}=Ay_{1},e_{2}=Ay_{2},...e_{n}=Ay_{n}$</span> which gives <span class="math-container">$AY=I$</span>.</p>
<p>Also, <span class="math-container">$N(A)^{\perp}=\mathbb{R}^{n}$</span>, but <span class="math-container">$N(A)^{\perp}$</span>=<span class="math-container">$\Re(A^{T})$</span> which gives that every vector in <span class="math-container">$\mathbb{R}^{n}$</span> can be written as <span class="math-container">$A^{T}z$</span>. Thus <span class="math-container">$e_{1}=A^{T}z_{1},e_{2}=A^{T}z_{2},...e_{n}=A^{T}z_{n}$</span></p>
<p>and hence <span class="math-container">$A^{T}Z=I$</span> i.e. <span class="math-container">$Z^{T}A=I$</span>. Therefore <span class="math-container">$Z^{T}AY=Z^{T} \Rightarrow Y=Z^{T}$</span>.</p>
<p>Thus there is an <span class="math-container">$Y$</span> such that <span class="math-container">$AY=YA=I$</span> which is the definition of the inverse of <span class="math-container">$A$</span>. i.e. <span class="math-container">$A^{-1}=Y$</span>. That completes the proof!!</p>
|
2,836,724 | <p>Since $\mathbb{Z} \subset \mathbb{R},$ all integers $n$ are also a real number. However, $\mathbb{Z}$ and $\mathbb{R}$ is defined by Peano's axiom and Dedekind cut, two very different methods. Does this imply that $n$ as an integer is different as $n$ as a real number? If so, how? Is $n$ as a rational number also different?</p>
| mathcounterexamples.net | 187,663 | <p>What comes at play here is the notion of <a href="https://en.m.wikipedia.org/wiki/Embedding" rel="nofollow noreferrer">embedding</a>, which is an injective map $f : X \to Y$ preserving some structure.</p>
<p>For example, the integers $\mathbb Z$ are defined using equivalence classes on $\mathbb N \times N$ by the equivalence relation $(a,b) \sim (c,d)$ if and only if $a+b=c=d$.</p>
<p>You can <strong>embed</strong> $\mathbb N$ into $\mathbb Z$ by associating $n \in \mathbb N$ to the class $f(n) = \widetilde{(n,0)}$. This is injective, and you have interesting properties preserved like </p>
<p>$$f(n+m)=f(n)+f(m)$$</p>
<p>At the end, $\mathbb N$ is not a subset of $\mathbb Z$ but the image of $\mathbb N$ by the embedding is.</p>
<p>You can proceed with similar embedding to embed $\mathbb Z$ into $\mathbb Q$ and $\mathbb Q$ into $\mathbb R$.</p>
|
2,836,724 | <p>Since $\mathbb{Z} \subset \mathbb{R},$ all integers $n$ are also a real number. However, $\mathbb{Z}$ and $\mathbb{R}$ is defined by Peano's axiom and Dedekind cut, two very different methods. Does this imply that $n$ as an integer is different as $n$ as a real number? If so, how? Is $n$ as a rational number also different?</p>
| Joonas Ilmavirta | 166,535 | <p>I read your question as one about intuition on numbers, and I will answer from that point of view.</p>
<p>There are many equivalent ways to construct differ kinds of numbers.
For example, you can build the reals with Dedekind cuts or with Cauchy sequences.
If you construct the natural numbers first and then expand to integers, rationals, and reals, you have a large number of options.</p>
<p>As others have pointed out, at each step the previous kind of numbers can be seen as a subset of the new ones via an embedding.
The exact embedding depends on your choice of constructions.</p>
<p>I see all of this as formalization or rigorous definition of the various kinds of numbers, not the essence of numbers.
The way I see numbers is different.
I might think of numbers as quantities, points on the geometrical real line, or something else.
I think of them flexibly in different ways.
To me Dedekind cuts and the like come from the intuitive meaning of a number and are secondary.
While rigorous definitions are crucial for building a sound theory, on an intuitive level I see definitions as descriptions.</p>
<p>The number two can be thought of as a Dedekind cut, a supremum of a set of rationals, the formal limit of a Cauchy sequence, the number of elements in some finite sets, the set consisting of the empty set and the set containing the empty set, the successor of one, the ratio or difference of two naturals or integers, or something yet different.
It is useful to see the same thing in many ways.</p>
|
1,327,944 | <p>I'm a hobbyist working on a mechanical sorting machine to sort magic the gathering cards. I'm by no means a mathematician though, and I was wondering if you all wouldn't mind helping me out with a math puzzle to determine the best route to go with my machine.</p>
<p>The average Magic: the Gathering set of cards contains 250 unique cards. My machine will sort those cards alphabetically placing the cards into multiple piles repeatedly to filter the cards down into the correct position. My question is, how many piles would be needed to sort 250 cards with only three passes. Here is some further info more clarification of the question.</p>
<p>Example: a card passes through the scanner, and the computer determines that the cards position in the alphabetized set is within the first 50 positions of the alphabetized set. 50 positions is %20 percent of a 250 card set, which I've chosen that number. It just seems like an easy place to start.</p>
<p>Then the next card passes through, and the scanner determines that the cards position is in the 3rd 50 positions of the alphabetized set (positions 101-150) so it places this in pile three. It keeps doing this until it has 250 cards broken down into 5 piles.</p>
<p>The cards are then scanned again for further refinement until they are in the correct order. So the question is, how many piles would make it possible to on average sort a 250 card set in 3 passes. I only chose 5 piles to illustrate the question, so I hope that's not confusing.</p>
<p>What is the most optimal number of piles needed?</p>
| Brian Tung | 224,454 | <p>In the abstract, it seems you have $250$ cards, each one identified by its rank in some kind of canonical sorting. Presumably, you have mechanical limitations, because otherwise, your best bet would be to simply sort in one pass, using a trivial hash sort that just puts them in the right order. This essentially creates $250$ piles. Since you're not doing that, I'm assuming that $250$ piles is not a practical solution.</p>
<p>If you want to sort in three passes, one way to do that is to represent each card by its rank as a three-digit, base $7$ number. We choose $7$ because $7^3 = 343 > 250$. Thus, for instance, the $192$nd card is $363$ in base $7$, because $3\cdot 7^2+6\cdot 7+3 = 147+42+3 = 192$. Some cards with lower-numbered ranks will have leading zeros in this representation—for instance, card number $4$ is $004$, and card number $19$ is $025$.</p>
<p>Then, in the first pass, you sort each card by its <em>last</em> digit, placing each card whose last digit is $0$ in the first pile, whose last digit is $1$ in the second pile, and so on. Collect all the piles, at the end of the first pass, keeping each pile internally intact, and you have a deck which is sorted in order of last digit.</p>
<p>In the second pass, you sort each card by its <em>middle</em> digit, placing each card whose middle digit is $0$ in the first pile, whose middle digit is $1$ in the second pile, and so on. Collect all the piles at the end of this second pass, and you have a deck which is sorted first in order of middle digit, and then in order of last digit.</p>
<p>In the last pass, you repeat the procedure with the <em>first</em> digit. Collect all the piles at the end of this last pass, and you have a deck which is sorted in the proper lexical order.</p>
<p>If you're doing this mechanically, you will also want to make sure that the order is preserved and not reversed, as often happens when cards are placed on top of one another. For example, in the second pass, as you're collecting all the cards whose middle digit is $0$ in the first pile, make sure that the cards are placed so that those that end $00$ are still in front of those that end $01$, which in turn should be in front of those that end $02$, and so forth. Depending on the mechanics of the device, this could be accomplished by turning each card upside-down before placing it in the proper pile.</p>
<p>This is essentially a merge sort. The reason we do this in reverse order of digits is that the last pass provides the high-order digit. Since the high-order digit is lexically first, we run the passes in reverse order.</p>
|
1,849,758 | <p>Does there exist a name for the class of commutative rings with identity that satisfy the following:</p>
<p>For any 2 ideals $I_1,I_2$ of R,we have : $I_1 I_2= (I_1\cap I_2)(I_1+I_2) $</p>
<p>I would also like to see an example of a ring not satisfying the above property.</p>
<p>Thank you</p>
| MooS | 211,913 | <p>In a (noetherian) factorial ring, which is not a PID, we can always find $f,g$, such that $\operatorname{gcd}(f,g) =1$ but the ideal $(f,g)$ is not the unit-ideal. Then we have</p>
<p>$$((f)+(g))((f) \cap (g))=(f,g)(fg) \subsetneq (fg).$$</p>
<p>For instance, take $k[x,y]$ and the ideals $(x)$ and $(y)$.</p>
|
4,062,242 | <p>Is <span class="math-container">$$\int_1^\infty \frac{\log x}{x^2}dx$$</span>
finite? How to solve this?</p>
| Alann Rosas | 743,337 | <p>We integrate by parts</p>
<p><span class="math-container">\begin{align*}
\int_{1}^{\infty}\frac{\ln(x)}{x^2}dx &= \lim\limits_{t\to\infty}\int_{1}^{t}\frac{\ln(x)}{x^2}dx\\
&= \lim\limits_{t\to\infty}\left(\left[-\frac{\ln(x)}{x}\right]_{1}^{t}-\int_{1}^{t}-\frac{1}{x}\cdot\frac{1}{x}\text{ }dx\right)\\
&= \lim\limits_{t\to\infty}\left(\frac{\ln(1)}{1}-\frac{\ln(t)}{t}-\int_{1}^{t}-\frac{1}{x^2}dx\right)\\
&= \lim\limits_{t\to\infty}\left(-\frac{\ln(t)}{t}-\left[\frac{1}{x}\right]_{1}^{t}\right)\\
&= \lim\limits_{t\to\infty}\left(-\frac{\ln(t)}{t}-\left(\frac{1}{t}-\frac{1}{1}\right)\right)\\
&=\lim\limits_{t\to\infty}\left(1-\frac{\ln(t)}{t}-\frac{1}{t}\right)
\end{align*}</span></p>
<p>As <span class="math-container">$t\to\infty$</span>, <span class="math-container">$\frac{\ln(t)}{t}$</span> and <span class="math-container">$\frac{1}{t}$</span> both tend to <span class="math-container">$0$</span>, so</p>
<p><span class="math-container">$$\int_{1}^{\infty}\frac{\ln(x)}{x^2}dx=\lim\limits_{t\to\infty}\left(1-\frac{\ln(t)}{t}-\frac{1}{t}\right)=1$$</span></p>
|
4,062,242 | <p>Is <span class="math-container">$$\int_1^\infty \frac{\log x}{x^2}dx$$</span>
finite? How to solve this?</p>
| Tharindu Gimras | 892,567 | <p>Let <span class="math-container">$t= \ln x.$</span> When we differentiate this we get <span class="math-container">$dt=\frac{1}{x}dx$</span>. Also <span class="math-container">$e^{t} = x.\Rightarrow \color{red}A$</span></p>
<p>By <span class="math-container">$\color{red}A$</span> we get <span class="math-container">$e^{t}dt=dx\Rightarrow \color{red}B$</span>.
So by substituting <span class="math-container">$\color{red}B$</span> we get,</p>
<p><span class="math-container">$$\int_{0}^{\infty} te^{-t}dt=\int_{0 }^{\infty} t\frac{d\left [ -e^{-t} \right ]}{dt}dt$$</span></p>
<p>By integration by parts we get,
<span class="math-container">$$\int_{0 }^{\infty} te^{-t}dt=\left [ -te^{-t} \right ]_{0}^{\infty}-\int_{0}^{\infty}-e^{-t}dt=\left [ -te^{-t} \right ]_{0 }^{\infty}+\int_{0}^{\infty}e^{-t}dt$$</span></p>
<p><span class="math-container">$$\int_{0 }^{\infty} te^{-t}dt=\left [ -te^{-t} \right ]_{1}^{\infty}+\left [ -e^{-t} \right ]_{0}^{\infty}=\left [ -te^{-t}-e^{-t} \right ]_{0}^{\infty}={\infty}=\left [ -e^{-t}(t+1) \right ]_{0}^{\infty}$$</span></p>
<p>So we have to find <span class="math-container">$\lim_{t\to\infty}e^{-t}(t+1)$</span> value,</p>
<p><span class="math-container">$$\lim_{t\to\infty}e^{-t}(t+1)=\lim_{t\to\infty}\frac{t+1}{e^{t}}$$</span></p>
<p>So by L'hopital rule,</p>
<p><span class="math-container">$$\lim_{t\to\infty}e^{-t}(t+1)=\lim_{t\to\infty}\frac{t+1}{e^{t}}=\lim_{t\to\infty}\frac{1}{e^{t}}=0$$</span></p>
<p><span class="math-container">$$\lim_{t\to \infty} \int_{0 }^{t} te^{-t}dt=\lim_{t \to \infty} \left [ -e^{-t}(t+1) \right ]_{0}^{t}$$</span>
So we get <span class="math-container">$$\lim_{t\to \infty} \int_{0 }^{t} te^{-t}dt=0+1=1$$</span></p>
<p>So as our final answer we get,</p>
<p><span class="math-container">$$\int_1^\infty \frac{\log x}{x^2}dx=1$$</span></p>
<p>I got this as my answer. If there is any miscalculations. Please correct me.</p>
|
2,855,411 | <p>Find all real number(s) $x$ satisfying the equation $\{(x +1)^3\}$ = $x^3$ , where $\{y\}$ denotes the fractional part of $y$ , for example $\{3.1416\ldots\}=0.1416\ldots$.</p>
<p>I am trying all positive real numbers from $1,2,\dots$ but I didn't get any decimals.</p>
<p>Is there a smarter way to solve this problem? ... Please advise.</p>
| Stan Tendijck | 526,717 | <p>Well, first of all you're checking only 'all' integers and not all reals which include also $1/2$ or $\pi$ for example. I will try to give you an outline of how I would approach it.</p>
<p>Second of all, note that the left-hand side is always between $0$ and $1$. So, what is implied about the range $x$ might be in?<br>
Once, you determined that you should be splitting up the range of possible $x$ into smaller segments such that for each segment you can find a natural number $n$ such that
$$\{(x+1)^3\} = (x+1)^3 - n.$$
Hint: there are $7$ segments that you should consider. Then for each segment it boils down in solving
$$ (x+1)^3 - n = x^3.$$
This looks difficult at first sight but I guarantee it will simplify to a quadratic polynomial (which you can solve). Please don't forget that the solutions of this equation should still be in the range on which we consider solutions to be part of.</p>
<p>I hope this helps and if you have any questions about any of these steps, feel free to ask them.</p>
|
2,747,896 | <p>I am a bit confused with cardinality at the moment. I know that the cardinality of $\mathbb{R}$ is equal to $\lvert(0,1)\rvert$, but does that mean they are equal to infinity, if not what are they equal to ?</p>
| Jack Lee | 1,421 | <p>"Infinity" is not the name of a cardinality. It makes sense to say that a set is <strong><em>infinite</em></strong>, meaning that it's not finite. But infinite sets can have many different cardinalities. There's not really a special name for the cardinality of $\mathbb R$, other than "the cardinality of $\mathbb R$," or more traditionally "<a href="https://en.wikipedia.org/wiki/Cardinality_of_the_continuum" rel="nofollow noreferrer">the cardinality of the continuum</a>." It's sometimes denoted by $\mathfrak c$.</p>
|
3,299,661 | <p>I am familiar with all 3 of the entities I have listed in my question. I know the definitions of "reflexive", "symmetric", and "transitive". However, I am afraid I do not mechanistically understand the "flow" of how we ultimately generate equivalence classes from a particular relation that exhibits the 3 properties of equivalence.</p>
<p>To help illustrate my confusion, consider the following example:</p>
<p><span class="math-container">$S=\{1,2,3,4,5,6\}$</span></p>
<p>Let <span class="math-container">$R_1$</span> be a relation on <span class="math-container">$S$</span> such that <span class="math-container">$x-y$</span> is divisible by <span class="math-container">$3$</span></p>
<p>So, firstly, from what I understand about relations, I am going to find all of the order pairs that satisfy this (these ordered pairs are a subset of the cartesian product <span class="math-container">$S$</span> x <span class="math-container">$S$</span>). </p>
<p><span class="math-container">$R_1 = \{(1,4) (2,5) (3,6) (4,1) (5, 2) (6, 3)(1,1)(2,2)(3,3)(4,4)(5,5)(6,6)\}$</span></p>
<p>Ok cool. These are all of the ordered pairs that "satisfy" or "make the <span class="math-container">$R_1$</span> relation true".</p>
<p>For this given relation, I can observe the following:</p>
<p><strong>1 -</strong> The reflexive property is satisfied because of the presence of <span class="math-container">$(1,1), (2,2),\ etc$</span></p>
<p><strong>1st question</strong>: If, for example (6,6) was not in this set, <span class="math-container">$R_1$</span> could not be deemed reflexive because <span class="math-container">$(3,6)$</span> and <span class="math-container">$(6,3)$</span> are present, correct? (i.e. because the element "6" shows up as as an ordered pair, <span class="math-container">$(6,6)$</span> MUST show up as well in order to declare this relation reflexive)</p>
<p><strong>2 -</strong> The symmetric property is satisfied because of the presence of <span class="math-container">$(1,4) \& (4,1)$</span>, <span class="math-container">$(2,5)\&(5,2),\ etc$</span></p>
<p><strong>3 -</strong> The transitive property is satisfied because...</p>
<p><strong>2nd question</strong>: I actually do not immediately see why the transitive property is satisfied (I believe that the transitive property <em>should be</em> satisfied because the "congruence modulo n" relation is an equivalence relation...and I'm fairly certain that the relation <span class="math-container">$R_1$</span> that I described is of that form). Is it just because my set is too small to see the transitive property in its stereotypical form?</p>
<p>So, assuming that this relation IS an equivalence relation (I believe that it is...for the reason mentioned above), I really do not understand how we go from this single set of ordered pairs to equivalence classes. From example videos I have seen, I know that a set of integers mod 3 will create three equivalence classes...namely, the integers with remainder 0, 1, and 2 when divided by 3. </p>
<p><strong>3rd question</strong>:
However, I do not really understand, mechanistically, how we "separate" these ordered pairs. All of the ordered pairs are initially grouped together. How do we decide, from this initial <span class="math-container">$R_1$</span> set, which ordered pairs belong to which equivalence class? Obviously, if you know how mod 3 works, you could sort of intuit that 1 and 4 go together because </p>
<p><span class="math-container">$1 mod (3) = 4 mod (3)$</span></p>
<p>...however, if I knew nothing about how <span class="math-container">$mod (3)$</span> worked, how would I know how to make the appropriate partitions? </p>
| Hagen von Eitzen | 39,174 | <p><strong>1.</strong> No. Without <span class="math-container">$(6,6)$</span> the relation would not be reflexive because <span class="math-container">$6\in S$</span>. If you want to argue with <span class="math-container">$(6,3),(3,6)\in R$</span>, you would show that <span class="math-container">$R$</span> is not transitive if <span class="math-container">$(6,6)$</span> is missing.</p>
<p><strong>2.</strong> No, I'd rather say that the set is <em>too big</em> to see at once that all instances of transitivity are fulfilled. In principle (i.e., arguing only with the concrete elements of <span class="math-container">$R$</span>, not algebraically), you'd have to check every pair of pairs in <span class="math-container">$R$</span>, and if they are of the form <span class="math-container">$(a,b)$</span>, <span class="math-container">$(b,c)$</span> verify that <span class="math-container">$(a,c)\in R$</span>. That's a lot to check if the relation is uncomfortably big. </p>
<p><strong>3.</strong> The equivalence classes do not consist of pairs <span class="math-container">$\in R$</span>, but of elements <span class="math-container">$\in S$</span>. We can build the classes successivley: What is the eequivalence class of <span class="math-container">$1\in S$</span>? From <span class="math-container">$(1,4)\in S$</span>, we see that <span class="math-container">$4$</span> is also in the equivalence class of <span class="math-container">$1$</span>. As we do not find any additional relations involving <span class="math-container">$1$</span> and/or <span class="math-container">$4$</span> (apart from the corresponding reflexive and symmetric pairs) we conclude that the equivalence class of <span class="math-container">$1$</span> (as well as of <span class="math-container">$4$</span>) is <span class="math-container">$\{1,4\}$</span>. Similarly, we find the other equivalence classes <span class="math-container">$\{2,5\}$</span> and <span class="math-container">$\{3,6\}$</span>. To emphasize that the classes are not pairs: Had <span class="math-container">$S$</span> been bigger, the equivalence class of <span class="math-container">$1$</span> might have been <span class="math-container">$\{1,4,7\}$</span>.</p>
|
1,615,177 | <p>The primes $p$ are, of course, in one-to-one correspondence with the squares of primes $p^2$. But is there any interval $a < x < b$ possible where the primes thin out so much, that it contains more squares of primes than primes?</p>
| Ross Millikan | 1,827 | <p>The interval $23 \lt x \lt 29$ is one such</p>
|
344,725 | <p>in $\Delta ABC$,and </p>
<p>$$\dfrac{\sin{(\dfrac{B}{2}+C)}}{\sin^2{B}}=\dfrac{\sin{(\dfrac{C}{2}+B)}}{\sin^2{C}}$$</p>
<p>prove that $B=C$</p>
<p>I think $\sin{(\dfrac{B}{2}+C)}\sin^2{C}=\sin{(\dfrac{C}{2}+B)}\sin^2{B}$</p>
<p>then
$$\sin{(\dfrac{B}{2})}\cos{C}\sin^2{C}+\cos{\dfrac{B}{2}}\sin^3C=\sin{\dfrac{C}{2}}\cos{B}\sin^2B+\cos{\dfrac{C}{2}}\sin^3B$$</p>
<p>so
$$(\sin{\dfrac{B}{2}}-\sin{\dfrac{C}{2}})f(B,C)=0$$</p>
<p>my question:How can prove $f(B,C)\neq 0$ ？</p>
| Inceptio | 63,477 | <p><img src="https://i.stack.imgur.com/tJd6e.png" alt="enter image description here"></p>
<p>Its just kind of a hint, I hope its of some help.</p>
<p>Using sine-rule on $\triangle ABC$, you get :</p>
<p>$$\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}= 2R$$</p>
<p>$$\dfrac{\sin B}{\sin C}= \dfrac{b}{c} \implies \dfrac{\sin^2 B}{\sin^2 C}= \dfrac{b^2}{c^2}$$</p>
<p>Use sine-rule for $\triangle ADB$ and $\triangle AEC$</p>
<p>$$\dfrac{c}{\sin (\frac{B}{2}+C)}=\dfrac{AD}{\sin \frac{B}{2}}=\dfrac{BD}{\sin A}= 2R'$$</p>
<p>$$\dfrac{b}{\sin (\frac{C}{2}+B)}=\dfrac{AE}{\sin \frac{C}{2}}=\dfrac{EC}{\sin A}= 2R''$$</p>
<p>$\sin A=( \dfrac{c}{\sin (\frac{B}{2}+C) \times BD} )^{-1}$</p>
<p>$\sin A= (\dfrac{b}{\sin (\frac{C}{2}+B ) \times EC})^{-1}$</p>
<p>Equate them:</p>
<p>$( \dfrac{c}{\sin (\frac{B}{2}+C) \times BD} )=(\dfrac{b}{\sin (\frac{C}{2}+B ) \times EC})$</p>
<p>$\dfrac{c \times EC}{b \times BD} = \dfrac{\sin (\dfrac{B}{2}+C)}{\sin (\dfrac{C}{2}+B)}$</p>
<p>$\dfrac{c \times EC}{b \times BD}=\dfrac{b^2}{c^2} \implies \dfrac{EC}{BD}=\dfrac{b^3}{c^3}$</p>
<p>In $\triangle BDC$ and $\triangle BEC$</p>
<p>$\dfrac{a}{\sin (\dfrac{B}{2}+C)}=\dfrac{BD}{\sin C}$ .....$1$</p>
<p>$\dfrac{a}{\sin (\dfrac{C}{2}+B)}=\dfrac{EC}{\sin B}$......$2$</p>
<p>Dividing ($1$) and ($2$), you get:</p>
<p>$\dfrac{\sin (\dfrac{C}{2}+B)}{\sin (\dfrac{B}{2}+C)}=\dfrac{BD \cdot \sin B}{EC \cdot \sin C}$</p>
<p>I couldn't get the conclusion right. Maybe this kinda approach is useful.:)</p>
|
344,725 | <p>in $\Delta ABC$,and </p>
<p>$$\dfrac{\sin{(\dfrac{B}{2}+C)}}{\sin^2{B}}=\dfrac{\sin{(\dfrac{C}{2}+B)}}{\sin^2{C}}$$</p>
<p>prove that $B=C$</p>
<p>I think $\sin{(\dfrac{B}{2}+C)}\sin^2{C}=\sin{(\dfrac{C}{2}+B)}\sin^2{B}$</p>
<p>then
$$\sin{(\dfrac{B}{2})}\cos{C}\sin^2{C}+\cos{\dfrac{B}{2}}\sin^3C=\sin{\dfrac{C}{2}}\cos{B}\sin^2B+\cos{\dfrac{C}{2}}\sin^3B$$</p>
<p>so
$$(\sin{\dfrac{B}{2}}-\sin{\dfrac{C}{2}})f(B,C)=0$$</p>
<p>my question:How can prove $f(B,C)\neq 0$ ？</p>
| Jeppe Stig Nielsen | 70,134 | <p>The other answers make one realize there are many counter-examples, so we can compose an answer as follows.</p>
<p>We shal prove that there exists a right-angled triangle $ABC$ which is not isosceles, but for which
$$\frac{\sin\left( \frac{B}{2} + C \right)}{(\sin B)^2}
-
\frac{\sin\left( \frac{C}{2} + B \right)}{(\sin C)^2}
\quad\quad\quad(*)$$
is zero. So for the entire post, <strong>fix $C=90^\circ$,</strong> and consider $60^\circ \le B \le 75^\circ$ (and $A$ is the complement which makes the triangle Euclidean, of course).</p>
<p>In this range, clearly $A<B<C$, so the triangles we consider are all scalene.</p>
<p>Direct calculation (exact or by floating-point) shows that for $B=60^\circ$ the quantity $(*)$ above is positive, but for $B=75^\circ$ it is negative. So by continuity there exists a value of $B$ in $\left[ 60^\circ , 75^\circ \right]$ which makes $(*)$ zero. So we have a counter-example as promised.</p>
|
2,293,600 | <p>How to calculate X $\cap$ $\{X\}$ for finite sets to develop an intuition for intersections?</p>
<p>If $X$ = $\{$1,2,3$\}$, then what is $X$ $\cap$ $\{X\}$? </p>
| Jay Zha | 379,853 | <p>For your example, it is $\emptyset$, because none of elements of $X$ is in $\{X \}$, and none of element me of $\{X\}$ is in $X$.</p>
<p>For general case, one axiom of set theory is that $A \notin A$ for any set (see <a href="https://math.stackexchange.com/questions/1046863/how-can-a-set-contain-itself">this post</a>), which means $\{A\}$ does not have any element in $A$, and thus they intersection is $\emptyset$.</p>
|
3,917,601 | <p>Let <span class="math-container">$(X,\mathcal{M},\mu)$</span> be a measure space. Suppose <span class="math-container">$E_n\in \mathcal{M}$</span> such that <span class="math-container">$$\sum_{n=1}^\infty \mu(E_n) < \infty$$</span> show <span class="math-container">$\mu(\lim\sup_{n\to\infty} E_n) = 0.$</span></p>
<p>Also, how can I prove or give a counterexample of the conclusion if the hypothesis is replaced with <span class="math-container">$$\sum_{n=1}^\infty \mu(E_n)^2 < \infty$$</span></p>
<p>Can I prove this by Fatou's Lemma? Thank you</p>
| Stella Biderman | 123,230 | <p>It is a very general and very useful theorem of calculus that if <span class="math-container">$\sum a_i < \infty$</span> then it must be that <span class="math-container">$\lim\sup a_i = 0$</span>. To get your result from this theorem, simply apply subadditivity.</p>
<p>For the counter example, can you think of a series such that <span class="math-container">$\sum a_i^2$</span> converges
and <span class="math-container">$\sum a_i$</span> diverges?</p>
|
3,917,601 | <p>Let <span class="math-container">$(X,\mathcal{M},\mu)$</span> be a measure space. Suppose <span class="math-container">$E_n\in \mathcal{M}$</span> such that <span class="math-container">$$\sum_{n=1}^\infty \mu(E_n) < \infty$$</span> show <span class="math-container">$\mu(\lim\sup_{n\to\infty} E_n) = 0.$</span></p>
<p>Also, how can I prove or give a counterexample of the conclusion if the hypothesis is replaced with <span class="math-container">$$\sum_{n=1}^\infty \mu(E_n)^2 < \infty$$</span></p>
<p>Can I prove this by Fatou's Lemma? Thank you</p>
| Sam Wong | 507,382 | <p>As Bernard comments, the first part is actually the statement of Borel-Cantelli lemma.</p>
<p>For the second part, as Stella mentions, we may want to find a series which is not convergent but the series consists of square of its terms is convergent. The series <span class="math-container">$\sum_{n=1}^\infty \frac{1}{n}$</span> will satisfy this condition.</p>
<p>So we need to construct a sequence of sets <span class="math-container">$\{E_n\}$</span> s.t. <span class="math-container">$\mu(E_n)=\frac{1}{n}, \forall n\ge1.$</span> Also, we want that <span class="math-container">$\mu(\lim\sup_{n\to\infty} E_n) \neq 0.$</span></p>
<p>To construct a concrete example, we may need a proposition.</p>
<p><strong>Proposition: For every real number x, there exists a sequence <span class="math-container">$x_1,x_2,x_3,...$</span> of integers such that <span class="math-container">$$x=x_1+\frac{x_2}{2!}+\frac{x_3}{3!}+\cdot\cdot\cdot \,,$$</span> where <span class="math-container">$x_1$</span> can be any integer, but for <span class="math-container">$n\ge2,x_n\in\{0,1,...,n-1\}.$</span> Furthermore, if we require that the partial sums be strictly smaller than x, then such a representation is unique.</strong></p>
<p>A proof can be found here: <a href="https://blogs.ams.org/mathgradblog/2017/09/20/real-numbers-base-factorials-by-product/" rel="nofollow noreferrer">https://blogs.ams.org/mathgradblog/2017/09/20/real-numbers-base-factorials-by-product/</a></p>
<p>From the proposition and its proof, it's easy to see that, if <span class="math-container">$x\in(0,1)$</span>, being an irrational number, then we can find an unique sequence <span class="math-container">$x_1,x_2,x_3,...$</span> s.t. <span class="math-container">$x_1=0$</span> and <span class="math-container">$$x=\frac{x_2}{2!}+\frac{x_3}{3!}+\cdot\cdot\cdot \,,$$</span> with <span class="math-container">$x_n\in\{0,1,...,n-1\},\forall n\ge2.$</span></p>
<p>Also note that, if <span class="math-container">$x$</span> is an irrational number, then the corresponding series will consist of infinitely many terms. (N.B. This is crucial to our construction, which makes sure the construction process will not terminate in finite steps.)</p>
<p>Our example will be given in the language of probability space, but in essence, it can be easily converted into the language of general measure theory.</p>
<p>We define the event <span class="math-container">$E_1=\{\text{pick a random irrational number $x$ in $(0,1)$, $x_1$ is $0$}. \}$</span>. Then it's clear that <span class="math-container">$P(E_1)=1.$</span></p>
<p>For each <span class="math-container">$n\ge 2,$</span> define event <span class="math-container">$E_n=\{\text{pick a random irrational number $x$ in $(0,1)$, $x_n$ is $0$}. \}$</span>. Then it's also clear that <span class="math-container">$P(E_n)=\frac{1}{n}.$</span></p>
<p>Note that these events are <strong>mutually independent</strong>.</p>
<p>By the second Borel-Cantelli lemma(c.f. <a href="https://en.wikipedia.org/wiki/Borel%E2%80%93Cantelli_lemma#Converse_result" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Borel–Cantelli_lemma#Converse_result</a>), we conclude that <span class="math-container">$P(\lim\sup_{n\to\infty} E_n)=1\neq 0.$</span></p>
<p>So, even though <span class="math-container">$\sum_{n=1}^\infty \mu(E_n)^2 < \infty$</span>, we may still have <span class="math-container">$\mu(\lim\sup_{n\to\infty} E_n)=1\neq 0.$</span></p>
<p><span class="math-container">$$\tag*{$\blacksquare$}$$</span></p>
|
3,964,910 | <p>Let <span class="math-container">$E$</span> be a metric space, <span class="math-container">$(\mu_n)_{n\in\mathbb N}$</span> be a sequence of finite nonnegative measures on <span class="math-container">$\mathcal B(E)$</span> and <span class="math-container">$\mu$</span> be a probability measure on <span class="math-container">$\mathcal B(E)$</span> s.t. <span class="math-container">$$\frac{\mu_n}{\mu_n(E)}\xrightarrow{n\to\infty}\mu\tag1$$</span> with respect to the to the topology of <a href="https://en.wikipedia.org/wiki/Convergence_of_measures#Weak_convergence_of_measures" rel="nofollow noreferrer">weak convergence of measures</a>.</p>
<p>Assuming that <span class="math-container">$c:=\sup_{n\in\mathbb N}\mu_n(E)<\infty$</span>, are we able to deduce that the nonnormalized sequence <span class="math-container">$(\mu_n)_{n\in\mathbb N}$</span> is convergent as well?</p>
<p>If not, does the other direction of this assertion hold, i.e. can we infer that the normalized sequence (i.e. the left-hand side of <span class="math-container">$(1)$</span>) is convergent from knowing that the nonnormalized sequence is convergent? It's easy to see that <span class="math-container">$(\mu_n(E))_{n\in\mathbb N}$</span> is convergent in that case (since the function constantly equal to <span class="math-container">$1$</span> is bounded and continuous). So, this should be possible to obtain.</p>
<p><em>Remark</em>: Without assuming <span class="math-container">$c<\infty$</span>, the first implication is clearly wrong. We simply can consider any finite nonnegative measure <span class="math-container">$\nu$</span> on <span class="math-container">$\mathcal B(E)$</span> and set <span class="math-container">$\nu_n:=n\nu$</span> for <span class="math-container">$n\in\mathbb N$</span>.</p>
| Botnakov N. | 452,350 | <p>Put <span class="math-container">$E = $</span> real line. Consider <span class="math-container">$\mu_{2n} = 2\delta_{ \{0 \} }$</span> and <span class="math-container">$\mu_{2n-1} = \delta_{ \{0 \} }$</span>, where <span class="math-container">$\delta_{ \{x \} }$</span> is a Dirac measure.</p>
<p>So the answer is negative.</p>
<p>The opposite is true if <span class="math-container">$lim_n \mu_n (E)$</span> (it exists, as you noticed) is not equal to <span class="math-container">$0$</span>.</p>
<p>But if <span class="math-container">$lim_n \mu_n (E) = 0$</span> we may have smth. like that: <span class="math-container">$\mu_{n} = \frac{(-1)^n + 1}{n}\delta_{ \{0 \} }$</span>, that is counterexample, because normalized version of <span class="math-container">$\mu_{2n+1}$</span> doesn't exist.</p>
<p><strong>Addition.</strong> Suppose that normalized sequence of measures converges to some measure <span class="math-container">$\nu$</span> and that <span class="math-container">$\mu_n(E) \to a \ne 0$</span>, (hence <span class="math-container">$c < \infty$</span> and <span class="math-container">$a = \nu(E)$</span>).
<strong>Statement:</strong> <span class="math-container">$\mu_n \to a \nu$</span> and <span class="math-container">$\nu$</span> is normalized (i.e. a probability measure).</p>
<p>Indeed, as <span class="math-container">$\lim_n \mu_n(E) \ne 0$</span> we have <span class="math-container">$\mu_n(E) \ne 0$</span> for all <span class="math-container">$n\ge n_0$</span>. Thus <span class="math-container">$$\mu_n = \mu_n(E) \cdot \frac{\mu_n}{\mu_n(E)}.$$</span> Put <span class="math-container">$a_n = \mu_n(E)$</span>, <span class="math-container">$\nu_n = \frac{\mu_n}{\mu_n(E)}$</span>. We want to show that <span class="math-container">$a_n \nu_n \to a \nu$</span>. We know that <span class="math-container">$\nu_n \to \nu$</span> and <span class="math-container">$a_n \to a \ne 0$</span>.
By definition <span class="math-container">$\int f(x) d\nu_n \to \int f(x) d\nu$</span> for all bounded continuous <span class="math-container">$f$</span>. Thus <span class="math-container">$$\int f(x) d(a_n \nu_n) = a_n \int f(x) d\nu_n \to a \int f(x) d\nu = \int f(x) d(a \nu)$$</span> and hence <span class="math-container">$a_n \nu_n \to a \nu$</span>.</p>
<p>Further, <span class="math-container">$\nu(E) = \lim_n \nu_n(E) = 1$</span>.</p>
<p>Moreover, under the condition <span class="math-container">$\lim_n \mu_n(E) \ne 0$</span> convergence of <span class="math-container">$\nu_n$</span> and convergence of <span class="math-container">$\mu_n$</span> are equivalent (because we may multiply and devide convergent sequence of measures by <span class="math-container">$ \mu_n(E)$</span>).</p>
<p>Conclusion: if normalized measures converge then <span class="math-container">$\mu_n$</span> may not converge. If <span class="math-container">$\mu_n$</span> converges then in general case it doesn't mean that normalized measures converge (al least because normalized measures may not exist). But if <span class="math-container">$\mu_n$</span> converge and there is an additional condition <span class="math-container">$lim_n \mu_n(E) \ne 0$</span> then normalized measures converge.</p>
|
1,557,165 | <p>Prove that
$$\int_1^\infty\frac{e^x}{x (e^x+1)}dx$$
does not converge.</p>
<p>How can I do that? I thought about turning it into the form of $\int_b^\infty\frac{dx}{x^a}$, but I find no easy way to get rid of the $e^x$.</p>
| Kay K. | 292,333 | <p>\begin{align}
&\int_1^{\infty}{e^x\over x(e^x+1)}dx,\quad e^x+1\mapsto y\\
&=\int_{e+1}^{\infty}{1\over y\ln (y-1) }dy>\int_{e+1}^{\infty}{1\over y\ln y }dy=\left[\ln (\ln y)\right]_{e+1}^\infty=\infty
\end{align}</p>
|
3,371,104 | <p>How could I show <span class="math-container">$$\int_{\mathbb{R}}\dfrac{1}{\sqrt{1+t^{2}}}dt=\infty?$$</span> </p>
<p>I tried to use comparison test so that <span class="math-container">$$\dfrac{1}{\sqrt{1+t^{2}}}\geq \dfrac{C}{t},$$</span> for some <span class="math-container">$C$</span>, and we can use the fact that <span class="math-container">$$\int_{\mathbb{R}}\dfrac{1}{t}dt=\infty$$</span> to conclude, but this requires me to find a <span class="math-container">$C$</span> such that <span class="math-container">$$1+t^{2}\leq\dfrac{1}{C^{2}}t^{2},$$</span> which is <span class="math-container">$$\dfrac{1-C^{2}}{C^{2}}t^{2}\geq 1$$</span> for all <span class="math-container">$t$</span>.</p>
<p>How could I find such value?</p>
<p>Thank you!</p>
| angryavian | 43,949 | <p>Since the integrand is positive, it suffices to show the tail <span class="math-container">$\int_a^\infty \frac{1}{\sqrt{1+t^2}} \, dt$</span> diverges for some <span class="math-container">$a$</span>.
Then you can note that <span class="math-container">$\sqrt{1+t^2} \le 2t$</span> for <span class="math-container">$t \ge 1$</span>, for instance.</p>
|
3,371,104 | <p>How could I show <span class="math-container">$$\int_{\mathbb{R}}\dfrac{1}{\sqrt{1+t^{2}}}dt=\infty?$$</span> </p>
<p>I tried to use comparison test so that <span class="math-container">$$\dfrac{1}{\sqrt{1+t^{2}}}\geq \dfrac{C}{t},$$</span> for some <span class="math-container">$C$</span>, and we can use the fact that <span class="math-container">$$\int_{\mathbb{R}}\dfrac{1}{t}dt=\infty$$</span> to conclude, but this requires me to find a <span class="math-container">$C$</span> such that <span class="math-container">$$1+t^{2}\leq\dfrac{1}{C^{2}}t^{2},$$</span> which is <span class="math-container">$$\dfrac{1-C^{2}}{C^{2}}t^{2}\geq 1$$</span> for all <span class="math-container">$t$</span>.</p>
<p>How could I find such value?</p>
<p>Thank you!</p>
| Barry Cipra | 86,747 | <p>From <span class="math-container">$\sqrt{1+t^2}\gt|t|$</span> we have</p>
<p><span class="math-container">$$\int_\mathbb{R}{1\over\sqrt{1+t^2}}\,dt\gt\int_0^\infty{t\over1+t^2}\,dt={1\over2}\ln(1+t^2)\big|_0^\infty=\infty$$</span></p>
|
14,391 | <p>I am currently tutoring a student who is really lacking in math. At first I thought she was just resistant; perhaps she thought that her teacher had no choice but to pass her. What truly stunned me is that she is going to tenth-grade and has very little common sense with numbers.</p>
<p>I once asked her what is 3 minus 1/2, and she could not answer it. So, I drew three circles in front of her and shaded out one half. Now I asked how many circles are left, and she still could not answer it .... </p>
<p>An episode like this is not a rarity with her. Moreover, too often she would give responses that even a third-grader would know were nonsense. (What's half-way between 19.5 and 20? She said 9.) According to her mom, she is fine with her other classes. That's why I am wondering if a person can have issues with learning math specifically. </p>
| Dan Fox | 672 | <p>The inabilities described in the question don't strike me as either particularly unusual nor as clear signs of a neurological learning disability. They strike me as more likely the passivity and mental shutdown that one frequently encounters in students who, whether consciously or unconsciously, are hiding a sense of incapacity, inability, confusion, etc. </p>
<p>Many students learn arithmetic operationally, as a series of rules for obtaining an answer, without internalizing the significance of the rules or their outcomes. Such a student neither has the habit of associating some mental model (e.g. the circles) with an arithmetic problem nor in many cases is able to relate such a model, posed by someone else, with the problem at hand. </p>
<p>Often students are not asked to consider whether their answers are reasonable (in some qualitative sense). Often they are rewarded (in grading) for performing certain operations correctly, whether or not the outcome is reasonable (with respect to extra-mathematical considerations). So a student who does not think much about what halfway means might easily obtain an answer that is less than either of the averaged numbers. That's not a sign of a learning disability, it's a sign of someone who hasn't learned how to connect arithmetic operations with common sense (in any case, common sense is mostly learned).</p>
<p>As an innumerate student progresses through the educational system, the psychological need to hide inadequacies becomes stronger and stronger, and the possibilities to correct basic misunderstandings become rarer and more remote. The algebra teacher often won't have time, energy, or inclination to teach arithmetic to a student who supposedly has passed several courses that require knowing arithmetic. </p>
<p>It's an extreme case, but once I taught a fourth year university student who was functionally illiterate (how could such a situation occur? This was in the US, and he was the starting tailback on the football team) and largely innumerate. He was adept at coaxing hints from the teacher, and at concealing his ignorance (of which he was fully and quite self-consciously aware). It took considerable care and effort to develop with him the confidence necessary to deal with the problem openly and honestly. </p>
<p>My own suspicion is that far more students have difficulties like this than most teachers care to admit. However, to detect such problems, one needs to interact with such a student personally, individually, and with the confidence sufficient to break down the barriers that prevent honest realization of the psychological obstacles (and the institutional context often makes that difficult or impossible). At the university one experiences that most students who struggle with calculus do so because they don't know basic algebra, even basic arithmetic. (A typical example is that students know that the logarithm and exponential satisfy functional identities, but can't remember if it is $f(xy) = f(x) + f(y)$ or $f(xy) = f(x)f(y)$ of $f(x + y) = f(x)f(y)$ and so guess.) Surely for students struggling in high school mathematics the common difficulty is this sort of inability to relate basic arithmetic to something other than a chain of operations aimed at getting points on an exam.</p>
|
1,507,298 | <p>$f$ and $g$ are both holomorphic functions on some domain $\Omega_1, \Omega_2$. Is $\frac f g$ going to be holomorphic on $\Omega_1\cap\Omega_2-\{z\mid g(z)=0\}$? What's the proof?</p>
| Nikita Evseev | 23,566 | <ol>
<li><p>$g(z)\ne 0$ for $z\in \Omega_1\cap\Omega_2$.</p></li>
<li><p>Complex function is holomorphic in $\Omega$ if there exist the derivative in each point of the domain. Then in usual manner one could prove that there exist derivative $\left(\frac{f}{g}\right)'(z)$ for all $z\in \Omega_1\cap\Omega_2$. Actually we should to check that
$$
\left(\frac{f}{g}\right)'(z) = \frac{f'(z)g(z)-f(z)g'(z)}{g^2(z)}.
$$</p></li>
</ol>
<p>The proof is the same as in <a href="https://en.wikipedia.org/wiki/Quotient_rule#Proof" rel="nofollow">real case</a> (but before they prove <a href="https://proofwiki.org/wiki/Product_Rule_for_Derivatives#Proof" rel="nofollow">product rule</a>).</p>
|
1,507,298 | <p>$f$ and $g$ are both holomorphic functions on some domain $\Omega_1, \Omega_2$. Is $\frac f g$ going to be holomorphic on $\Omega_1\cap\Omega_2-\{z\mid g(z)=0\}$? What's the proof?</p>
| Michael Hardy | 11,667 | <p>\begin{align*}
& \lim_{h\to0} \frac{\frac{f(x+h)}{g(x+h)} - \frac{f(x)}{g(x)}} h \\[10pt]
= {} & \lim_{h\to0} \frac{ f(x+h)g(x) - f(x)g(x+h) }{h g(x+h) g(x)}\\[10pt]
= {} & \lim_{h\to0} \frac{ \Big( f(x+h)g(x) - f(x)g(x) \Big) - \Big( f(x)g(x+h) - f(x)g(x) \Big)}{hg(x)g(x+h)} \\[10pt]
= {} & \lim_{h\to0} \left( \frac 1 {g(x)g(x+h)} \left( g(x) \frac{f(x+h)-f(x)} h - f(x) \frac{g(x+h)-g(x)} h \right) \right) \\[15pt]
= {} & \frac{g(x)\left(\lim_{h\to0} \frac{f(x+h)-f(x)} h\right) - f(x) \left( \lim_{h\to0}\frac{g(x+h)-g(x)} h \right) }{\lim_{h\to0} g(x)g(x+h)} \tag 1 \\[15pt]
= {} & \frac{g(x)f'(x) - f(x) g'(x)}{g(x)^2}
\end{align*}</p>
<p>In $(1)$ we used an assumption that $f$ and $g$ are differentiable and a theorem that says differentiable functions are continuous.</p>
|
340,886 | <p>Suppose $x=(x_1,x_2),y = (y_1,y_2) \in \mathbb{R}^2$. I noticed that
\begin{align*}
\|x\|^2 \|y\|^2 - \langle x,y \rangle^2 &=
x_1^2y_1^2 + x_1^2 y_2^2 + x_2^2 y_1^2 + x_2^2 y_2 ^2 - (x_1^2 y_1^2 + 2 x_1 y_1 x_2 y_2 + x_2^2 y_2^2) \\
&=(x_1 y_2)^2 - 2x_1 y_2 x_2 y_1 + (x_2 y_2)^2 \\
&=(x_1 y_2 - x_2 y_1)^2
\end{align*}
which proves the CSB inequality in dimension two. This begs the question:</p>
<blockquote>
<p>If $x = (x_1,\ldots,x_n),y=(y_1,\ldots,y_n) \in \mathbb{R}^n$, is there a polynomial $p \in \mathbb{R}[x_1,\ldots,x_n;y_1,\ldots,y_n]$ such that $ \|x\|^2 \|y\|^2 - \langle x,y \rangle^2 = p^2$?</p>
</blockquote>
| urandon | 68,449 | <p>CSB inquality can be generalized: Gramian's matrix determinant is equal or greater than a zero. And equality means linear dependence of system of vectors. In case of two vectors $u$ and $v$, we have:</p>
<p>$$ 0 \leqslant
\begin{vmatrix}
\langle u,u \rangle & \langle u,v \rangle \\
\langle v,u \rangle & \langle v,v \rangle \\
\end{vmatrix}
= \|u\|^2 \cdot \|v\|^2 - \langle u,v \rangle^2 $$</p>
<p>So, it's another proof of CSB.</p>
|
231,887 | <p>I'm learning to do proofs, and I'm a bit stuck on this one.
The question asks to prove for any positive integer $k \ne 0$, $\gcd(k, k+1) = 1$.</p>
<p>First I tried: $\gcd(k,k+1) = 1 = kx + (k+1)y$ : But I couldn't get anywhere.</p>
<p>Then I tried assuming that $\gcd(k,k+1) \ne 1$ , therefore $k$ and $k+1$ are not relatively prime, i.e. they have a common divisor $d$ s.t. $d \mid k$ and $d \mid k+1$ $\implies$ $d \mid 2k + 1$</p>
<p>Actually, it feels obvious that two integers next to each other, $k$ and $k+1$, could not have a common divisor. I don't know, any help would be greatly appreciated.</p>
| i. m. soloveichik | 32,940 | <p>Let $d$ be the $gcd(k, k+1)$ then $k=rd$, $k+1=sd$, so $1=(s-r)d$, so $d |1$.</p>
|
1,534,675 | <p>Since differential quantities are defined as any variable /function tending to zero ($\lim_{x\to0} x= dx$). This is basically the smallest value that we can imagine. Doesn't this mean that there is only one smallest value that ew can imagine? Doesn't the existence of another differential quantity, say $dl$ also mean that $dl$ can be smaller than dx?
For example,<a href="https://i.stack.imgur.com/Y0VJH.png" rel="nofollow noreferrer">right angled triangle</a></p>
<p>Means that $dl>dh$ ? Or do the differential quantities differ by a differential differential amount that is negligible in first order integrations?</p>
<p>EDIT: It seems I was mistaken in my understanding of limits and differentials. I am not taking down this question as it may be useful to others with the same confusion.</p>
| Cameron Buie | 28,900 | <p>There seems to be a(t least one) serious (but rather common) misunderstanding.</p>
<p>When you say things like "the smallest value we can imagine," what do you mean? A natural answer (assuming that, by "smallest," you mean "closest to <span class="math-container">$0$</span>") is <span class="math-container">$0$</span>. However, context leads me to believe that you are referring to some nonzero value. Now, here's the problem: if <span class="math-container">$v$</span> is a non-zero value, then so is <span class="math-container">$\frac12v,$</span> and the latter is <strong>strictly smaller</strong>! Hence, there is no such thing as "the smallest nonzero value we can imagine."</p>
<hr />
<p><strong>Added</strong>: It seems you're also misunderstanding what a limit is. Again, this is a very common mistake to make! Within the last two weeks, I was explaining this to another user. Unfortunately, the question (and my answer to it) have since been deleted, so I can't link to it. Alas!</p>
<p>Let me first give you a few verbal definitions. After each verbal definition, I will give the common rigorous definition and connect the two.</p>
<p>We will be supposing hereinafter that <span class="math-container">$f$</span> is a function from <span class="math-container">$E$</span> to <span class="math-container">$\Bbb R$</span> for some subset <span class="math-container">$E$</span> of <span class="math-container">$\Bbb R.$</span></p>
<p>We will also be assuming that <span class="math-container">$x_0$</span> is a <em>limit point</em> of <span class="math-container">$E$</span> in <span class="math-container">$\Bbb R.$</span> Put into words, this means that <span class="math-container">$x_0$</span> is a point of <span class="math-container">$\Bbb R,$</span> and no matter how close we get to <span class="math-container">$x_0,$</span> we can always find a point of <span class="math-container">$E$</span> that is closer to (and yet distinct from) <span class="math-container">$x_0.$</span> Put rigorously: given any <span class="math-container">$\delta>0,$</span> there is some <span class="math-container">$x\in E$</span> such that <span class="math-container">$0<|x-x_0|<\delta.$</span> Here, <span class="math-container">$\delta$</span> tells us how "close" we need to get to <span class="math-container">$x_0$</span> and <span class="math-container">$|x-x_0|$</span> is the distance from <span class="math-container">$x$</span> to <span class="math-container">$x_0.$</span> So, <span class="math-container">$|x-x_0|<\delta$</span> says that <span class="math-container">$x$</span> is "even closer" than our required distance <span class="math-container">$\delta,$</span> while <span class="math-container">$0<|x-x_0|$</span> tells us that <span class="math-container">$x$</span> and <span class="math-container">$x_0$</span> are distinct. Note that <span class="math-container">$x_0$</span> may or may not be an element of <span class="math-container">$E.$</span> For example, <span class="math-container">$0$</span> is a limit point of the set of non-zero real numbers (of which it is <em>not</em> an element) and of the set of nonnegative real numbers (of which it <em>is</em> an element).</p>
<p>Now, we say that <span class="math-container">$L$</span> is a* <em>limit of</em> <span class="math-container">$f(x)$</span> <em>as</em> <span class="math-container">$x$</span> <em>approaches</em> <span class="math-container">$x_0$</span> if we can make sure that <span class="math-container">$f(x)$</span> is as close as we like to <span class="math-container">$L,$</span> simply by making sure that <span class="math-container">$x$</span> sufficiently close to (but not equal to) <span class="math-container">$x_0.$</span> Put rigorously: given any <span class="math-container">$\epsilon>0,$</span> there is some <span class="math-container">$\delta>0$</span> such that for any <span class="math-container">$x$</span> in <span class="math-container">$E$</span> with <span class="math-container">$0<|x-x_0|<\delta,$</span> we have that <span class="math-container">$|f(x)-L|<\epsilon.$</span> Here, <span class="math-container">$\epsilon$</span> tells us the desired "closeness" level between <span class="math-container">$f(x)$</span> and <span class="math-container">$L,$</span> so <span class="math-container">$|f(x)-L|<\epsilon$</span> says that <span class="math-container">$f(x)$</span> is as close to <span class="math-container">$L$</span> as we want it to be. Again, <span class="math-container">$0<|x-x_0|$</span> tells us that <span class="math-container">$x$</span> and <span class="math-container">$x_0$</span> are distinct,. Also, <span class="math-container">$\delta$</span> shows us the sufficient "closeness" level between <span class="math-container">$x$</span> and <span class="math-container">$x_0,$</span> so that <span class="math-container">$|x-x_0|<\delta$</span>. shows that <span class="math-container">$x$</span> is sufficiently close to <span class="math-container">$x_0$</span> for our purposes. Depending on how close we want <span class="math-container">$f(x)$</span> and <span class="math-container">$L$</span> to be (that is, depending on how small <span class="math-container">$\epsilon$</span> is), we may need to change our requirement for how close <span class="math-container">$x$</span> and <span class="math-container">$x_0$</span> must be (that is, our <span class="math-container">$\delta$</span> may change).</p>
<blockquote>
<p>*It turns out that, if there is a limit of <span class="math-container">$f(x)$</span> as <span class="math-container">$x$</span> approaches <span class="math-container">$x_0,$</span> then it is unique, and we denote it by <span class="math-container">$\lim\limits_{x\to x_0}f(x).$</span></p>
</blockquote>
<p>So, what's the point? Well, let's consider what <span class="math-container">$\lim\limits_{x\to 0}x$</span> means, if anything.</p>
<p>The function <span class="math-container">$f(x):=x$</span> is readily defined on all of <span class="math-container">$\Bbb R$</span>--that is, we have <span class="math-container">$E=\Bbb R$</span> in this case--and <span class="math-container">$x_0=0$</span> is readily a limit point of <span class="math-container">$\Bbb R.$</span> We can prove this by adapting the argument from my initial answer (before the "Added" part).</p>
<p>Now, for we want to know if there is some <span class="math-container">$L$</span> such that, given any <span class="math-container">$\epsilon>0,$</span> we can pick <span class="math-container">$\delta>0$</span> such that if <span class="math-container">$0<|x-x_0|<\delta,$</span> then <span class="math-container">$|f(x)-L|<\epsilon.$</span> Translating this into our particular situation, we want to know if there is some <span class="math-container">$L$</span> such that, given any <span class="math-container">$\epsilon>0,$</span> we can pick <span class="math-container">$\delta>0$</span> such that if <span class="math-container">$0<|x-0|<\delta,$</span> then <span class="math-container">$|x-L|<\epsilon.$</span> Looking at it that way, it should be clear that if we put <span class="math-container">$L=0$</span> and <span class="math-container">$\delta=\epsilon,$</span> then this works out just fine. The upshot is this: <span class="math-container">$$\lim\limits_{x\to 0}x=0.$$</span></p>
<p>This may come as a shock. After all, didn't we require <span class="math-container">$x$</span> to stay away from <span class="math-container">$0$</span> in our definition of limit? Well, yes, <em>in part</em>. We required our <em>domain</em> values to be distinct from <span class="math-container">$0$</span>. Consequently, our range values were also required to be distinct from <span class="math-container">$0.$</span> However, the <em>limit</em> was not restricted! Not only does <span class="math-container">$0$</span> fit the definition of the desired limit, it is <em>the only number that does</em>!</p>
<hr />
<p><strong>Added</strong>: Let me go through a proof of limit uniqueness in a less formal way that will hopefully be easier to understand.</p>
<blockquote>
<p><strong>Claim</strong>: If <span class="math-container">$L$</span> is a limit of <span class="math-container">$f(x)$</span> as <span class="math-container">$x$</span> approaches <span class="math-container">$x_0,$</span> then it is the only number that is. Hence, <span class="math-container">$\lim\limits_{x\to x_0}f(x)$</span> is well-defined.</p>
<p><strong>Heuristic Proof</strong>: Take <span class="math-container">$L'$</span> to be any number not equal to <span class="math-container">$L,$</span> so that the distance between <span class="math-container">$L$</span> and <span class="math-container">$L'$</span> is positive. Call this distance <span class="math-container">$d,$</span> and note that <span class="math-container">$\frac12d$</span> is positive. Hence, we have by definition of limit that, for <span class="math-container">$x$</span> sufficiently close (but not equal to) <span class="math-container">$x_0,$</span> <span class="math-container">$f(x)$</span> is within <span class="math-container">$\frac12d$</span> of <span class="math-container">$L.$</span> Consequently, for such <span class="math-container">$x,$</span> the distance from <span class="math-container">$f(x)$</span> to <span class="math-container">$L$</span> is less than half of the distance from <span class="math-container">$L'$</span> to <span class="math-container">$L.$</span> Hence, for such <span class="math-container">$x,$</span> we have that <span class="math-container">$f(x)$</span> is <em>more than</em> <span class="math-container">$\frac12d$</span> away from <span class="math-container">$L',$</span> and less than <span class="math-container">$\frac32d$</span> away from <span class="math-container">$L'$</span> (though the latter isn't important). In order to have <span class="math-container">$L'$</span> as a limit of <span class="math-container">$f(x)$</span> as <span class="math-container">$x$</span> approaches <span class="math-container">$x_0,$</span> though, we would need to be able to make <span class="math-container">$f(x)$</span> be <em>less than</em> <span class="math-container">$\frac12d$</span> away from <span class="math-container">$L'.$</span> Having showed this to be impossible, we find that for any <span class="math-container">$L'\ne L,$</span> <span class="math-container">$L'$</span> is not a limit of <span class="math-container">$f(x)$</span> as <span class="math-container">$x$</span> approaches <span class="math-container">$x_0.$</span></p>
</blockquote>
<p>The approach above is basically a more general (and direct) version of the formalistic and particular proof I gave in the comments. There, I used proof by contradiction, and "<span class="math-container">$L$</span>" there corresponds to "<span class="math-container">$L'$</span>" here. From this claim, it follows that <span class="math-container">$\lim\limits_{x\to 0}x=0,$</span> in particular.</p>
|
644,494 | <p><strong>Question</strong></p>
<blockquote>
<p>If for some real number $a$, $\lim_{x\to 0}\frac{\sin 2x + a\sin x}{x^3}$ exists, then the limit is equal to:</p>
</blockquote>
<p>Here what i have done</p>
<p>since it is of $0/0$ form applying L' Hospital's rule$$\implies\lim_{x\to0}\frac{\sin 2x + a \sin x}{x^3} = \frac{2\cos 2x + a\cos x}{3x^2}$$ now what to do i am stuck here please help</p>
<p>Thanks</p>
<p>Akash</p>
| 2'5 9'2 | 11,123 | <p>Using no L'Hospital's Rule or Maclaurin series, but instead trig identities and that $\lim_{x\to0}\frac{\sin(x)}{x}=1$:</p>
<p>$$\begin{align}
\lim_{x\to0}\frac{\sin(2x)+a\sin(x)}{x^3}&=\lim_{x\to0}\frac{2\sin(x)\cos(x)+a\sin(x)}{x^3}\\
&=\lim_{x\to0}\frac{\sin{x}}{x}\frac{2\cos(x)+a}{x^2}\\
&=\lim_{x\to0}\frac{2\cos(x)+a}{x^2}\\
\end{align}$$</p>
<p>The denominator of this approaches $0$, so the only way the limit can exist is if the numerator also approaches $0$. So $a$ would have to be $-2$.</p>
<p>$$\begin{align}
\lim_{x\to0}\frac{2\cos(x)-2}{x^2}&=2\lim_{x\to0}\frac{\cos(x)-1}{x^2}\\
&=2\lim_{x\to0}\frac{\cos(x)-1}{x^2}\frac{\cos(x)+1}{\cos(x)+1}\\
&=2\lim_{x\to0}\frac{\cos^2(x)-1}{x^2(\cos(x)+1)}\\
&=2\lim_{x\to0}\frac{-\sin^2(x)}{x^2(\cos(x)+1)}\\
&=-2\lim_{x\to0}\frac{\sin^2(x)}{x^2}\frac{1}{\cos(x)+1}\\
%&=-2\lim_{x\to0}\frac{1}{\cos(x)+1}\\
&=\ldots
\end{align}$$</p>
|
1,221,221 | <p>Suppose $f:\mathbb{R}\rightarrow \mathbb{R}$ is continuous and has a left derivative, $f^-$, everywhere in a neighborhood of $x.$ Suppose $f^-$ is continuous at $x.$ Does this imply that $f$ is differentiable at $x$?</p>
| Michael Albanese | 39,599 | <p>Not necessarily. Consider the series $\displaystyle\sum_{n=1}^{\infty}a_n$ where $a_n = 1$ for every $n$.</p>
|
2,050,867 | <p>I would like to prove for all $x, y \in \mathbb{R}$ that $\dfrac{e^{x}+e^{y}}{2} \geq e^{\frac{x+y}{2}}$. My idea, is to show that $f(x,y) \ge 0$, it means that $(0,0)$ is the minimum of $f(x,y)$. So, I compute the equation: $\nabla f(x,y)=\begin{pmatrix}0 \\0 \end{pmatrix}$. I find that the solutions are $x=y$. <strong>My first question is may I choose $\textbf{x=y=0}$?</strong> After I have made this assumption, I compute the eigenvalues of $\nabla^2f(x,y)_{(0,0)}$ and got $\lambda_1=0$ and $\lambda_2=\dfrac{1}{2}$. Thus, I can't conclude anything about $(0,0)$ from this point since one of the eigenvalues is zero. <strong>Do you have any idea about what can I do further using this method or a different path to prove it?</strong> Thank you.</p>
| Community | -1 | <p>Setting $e^x=u,e^y=v$, this is just</p>
<p>$$\frac{u+v}2\ge\sqrt{uv}.$$</p>
<hr>
<p>Dividing by $e^x$ and setting $t:=y-x$,</p>
<p>$$\frac{1+e^t}2\ge e^{t/2}.$$</p>
<p>$$1+\frac t2+\frac{t^2}{2\,2!}+\frac{t^3}{2\,3!}+\cdots\ge1+\frac t2+\frac{t^2}{2^22!}+\frac{t^3}{2^33!}+\cdots$$</p>
|
4,155,718 | <p>I'm given <span class="math-container">$n$</span> points <span class="math-container">$(p_1, p_2, \ldots, p_n)$</span>, lying on the boundary of a polygon and constituting this polygon (not necessarily convex), whereby those points are given me in clockwise order and I want to compute the convex hull of this polygon, i.e. determine the set of points <span class="math-container">$S \subseteq \{p_1, p_2, \ldots, p_n\}$</span> s.t. the points in <span class="math-container">$S$</span> shape the convex hull of the polygon. The tricky part is that my runtime constraint is <span class="math-container">$O(n)$</span>, which is why I'm very stuck on this task. One should output the points of the convex hull in counterclockwise order.</p>
| TaChu | 933,577 | <p>One algorithm that does the above is Lee's algorithm, explained in this paper
<a href="https://link.springer.com/content/pdf/10.1007/BF00993195.pdf" rel="nofollow noreferrer">https://link.springer.com/content/pdf/10.1007/BF00993195.pdf</a></p>
|
27,904 | <blockquote>
<p>If $f(z) = (g(z),h(z))$ is continuous then $g$ and $h$ are as well.</p>
</blockquote>
<p>The converse is easy for me to prove, but I'm not seeing how to prove it using the terminology of open sets and not metric spaces.</p>
| Arturo Magidin | 742 | <p>So, $g\colon X\to Y$ and $h\colon X\to Z$ are maps, and you let $f\colon X\to Y\times Z$, endowing $Y\times Z$ with the product topology. You know $f$ is continuous, and you want to show that $g$ and $h$ are continuous.</p>
<p>The simplest way does not use open sets at all: it just notes that the projections $\pi_Y\colon Y\times Z\to Y$ and $\pi_Z\colon Y\times Z\to Z$ are continuous, and composition of continuous functions is continuous. What is $\pi_Y\circ f$ and what is $\pi_Z\circ g$?</p>
<p>If you absolutely, definitely, <em>positively</em> must use open sets, note that if $A\subseteq Y$ and $B\subseteq Z$, then
\begin{align*}
x\in f^{-1}(A\times B) &\Longleftrightarrow g(x)\in A\text{ and }h(x)\in B\\
&\Longleftrightarrow x\in g^{-1}(A)\text{ and }x\in h^{-1}(B)\\
&\Longleftrightarrow x\in g^{-1}(A)\cap h^{-1}(B).
\end{align*}
So, let $\mathcal{U}\subset Y$ be an open subset of $Y$. We know that $\mathcal{U}\times Z$ is open, sos= $f^{-1}(\mathcal{U}\times Z)$ is open. What is it?</p>
<p>And let $\mathcal{O}\subseteq Z$ be an open subset of $Z$; then $Y\times\mathcal{O}$ is open, so $f^{-1}(Y\times\mathcal{O})$ is open. What is it?</p>
|
4,218 | <p>I could imagine a system of categorizing the questions that would work alongside the current tagging system. If you select the "homework" tag (or some special tag or option), it would give you the option to specify which textbook problem your question pertains to in terms of title/chapter/section/problem number. Maybe the site could present a list of textbooks that have already had one solution entered and the user could navigate through a hierarchical tree of problems.</p>
<p>Now the users would be able to search or browse by textbook title, go to the chapter/section/problem and see if there is a solution present from someone else who has asked a question about it before.</p>
<p>This feature would make the site a lot more organized and eliminate redundant questions, or at least make it possible to find all the questions related to a specific problem in a textbook by providing explicit links. Math Stackexchange could eventually turn into the authoritative source of solutions for textbook problems. I think it would make the site a little less intimidating, too, if it was easier to use without having to ask questions all the time. Searching for mathematical symbols is pretty hard to do.</p>
| Dan Brumleve | 1,284 | <p>I upvoted because I don't see why we shouldn't allow copies of common problems (and index them by metadata). I think the critical issue here is "cheating", which nobody supports, but there is a lack of consensus as to what extent we should facilitate or tolerate it.</p>
<p>I am in favor of such an index, because it would be an aid to using the site, but I don't pretend to fully understand the concerns of academic users.</p>
|
4,218 | <p>I could imagine a system of categorizing the questions that would work alongside the current tagging system. If you select the "homework" tag (or some special tag or option), it would give you the option to specify which textbook problem your question pertains to in terms of title/chapter/section/problem number. Maybe the site could present a list of textbooks that have already had one solution entered and the user could navigate through a hierarchical tree of problems.</p>
<p>Now the users would be able to search or browse by textbook title, go to the chapter/section/problem and see if there is a solution present from someone else who has asked a question about it before.</p>
<p>This feature would make the site a lot more organized and eliminate redundant questions, or at least make it possible to find all the questions related to a specific problem in a textbook by providing explicit links. Math Stackexchange could eventually turn into the authoritative source of solutions for textbook problems. I think it would make the site a little less intimidating, too, if it was easier to use without having to ask questions all the time. Searching for mathematical symbols is pretty hard to do.</p>
| zyx | 14,120 | <p>I am curious to know if the anti-homework bloc that has downvoted this question <em>objects</em> to askers of exercises stating the book, problem number, and complete original wording of the problem. </p>
<p>Search engines would then locate the problem and solution for anyone who wants it. MSE's page ranking will put the question/answer at the top of the results.</p>
|
1,252,591 | <p>How many odd numbers can be formed using digits $0,4,5,7$.
I am getting answer $12$ but the actual answer is $14$. </p>
| Paolo Leonetti | 45,736 | <p>Suppose the number has to be of $n$ digits. Then the last digit has to be $5$ or $7$. Hence the solution is
$$
4^{n-1} \cdot 2=2^{2n-1}.
$$
If the number has at most $n$ digits then the answer is
$$
\sum_{i=1}^n2^{2i-1}=2\sum_{i=0}^{n-1}4^i=\frac{2}{3}(4^n-1).
$$</p>
<hr>
<p>Probably the correct interpretation:</p>
<p>We have to use exactly one each digit $0,4,5,7$. Then the last digit can be $5$ or $7$. The other can be chosen in $3!$ ways. Therefore the answer $2\cdot 3!=12$.</p>
<p>[Moreover, if we add the number has to have exactly $4$ digits (i.e. it doesn't start with $0$), then the answer would be $8$]</p>
<p>Ps. I still don't get why it could be $14$ :P</p>
|
238,577 | <p>The following working program uses
<a href="https://mathematica.stackexchange.com/questions/58560/graph-and-markov-chain">Graph and Markov Chain</a></p>
<pre><code>P = {{1/2, 1/2, 0, 0}, {1/2, 1/2, 0, 0}, {1/4, 1/4, 1/4, 1/4}, {0, 0,
0, 1}}; proc = DiscreteMarkovProcess[3, P];
Graph[proc, GraphStyle -> "DiagramBlue",
EdgeLabels ->
With[{sm = MarkovProcessProperties[proc, "TransitionMatrix"]},
Flatten@Table[DirectedEdge[i, j] -> sm[[i, j]], {i, 2}, {j, 2}]]]
sm = MarkovProcessProperties[proc, "TransitionMatrix"]
sm == P
</code></pre>
<p>Since I couldn't make it work for larger matrices, I clarified in the last two lines that sm is just P. But, if I try to replace sm by P in the first part, all hell breaks loose. So, I tried copy paste changing just P to a larger matrix, but this does not work. Why?</p>
<pre><code>P = {{0, 1/4, 1/2, 1/4, 0, 0}, {0, 1, 0, 0, 0, 0}, {0, 0, 1/3, 0, 2/3,
0}, {0, 0, 0, 0, 0, 1},
{0, 0, 1/4, 0, 3/4, 0}, {1/4, 0, 0, 0, 3/4, 0}};
P // MatrixForm
proc = DiscreteMarkovProcess[1, P];
Graph[proc,
EdgeLabels ->
With[{sm = MarkovProcessProperties[proc, "TransitionMatrix"]},
Flatten@Table[DirectedEdge[i, j] -> sm[[i, j]], {i, 6}, {j, 6}]]]
</code></pre>
| kglr | 125 | <pre><code>P2 = {{0, 1/4, 1/2, 1/4, 0, 0}, {0, 1, 0, 0, 0, 0}, {0, 0, 1/3, 0, 2/3, 0},
{0, 0, 0, 0, 0, 1}, {0, 0, 1/4, 0, 3/4, 0}, {1/4, 0, 0, 0, 3/4, 0}};
proc2 = DiscreteMarkovProcess[1, P2];
tm2 = MarkovProcessProperties[proc2, "TransitionMatrix"];
</code></pre>
<p>You can specify the edge labels using patterns:</p>
<pre><code>Graph[proc2, EdgeLabels -> {DirectedEdge[i_, j_] :> tm2[[i, j]]}]
</code></pre>
<p><a href="https://i.stack.imgur.com/DidAn.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DidAn.png" alt="enter image description here" /></a></p>
<p><em><strong>Why your code does not work:</strong></em></p>
<pre><code>EdgeList @ Graph[proc2]
</code></pre>
<p><a href="https://i.stack.imgur.com/TtIBg.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TtIBg.png" alt="enter image description here" /></a></p>
<p>and the spec</p>
<pre><code>With[{sm = MarkovProcessProperties[proc2, "TransitionMatrix"]},
Flatten @ Table[DirectedEdge[i, j] -> sm[[i, j]], {i, 6}, {j, 6}]]
</code></pre>
<p>refers to non-existent edges (e.g, <code>1 -> 5</code> or <code>6 ->6</code>).</p>
<p>Using a simpler example: Compare</p>
<pre><code>Graph[{1 -> 2, 2 -> 3}, EdgeLabels -> {DirectedEdge[1, 2] -> 100}]
</code></pre>
<p><a href="https://i.stack.imgur.com/hshG7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hshG7.png" alt="enter image description here" /></a></p>
<p>with</p>
<pre><code>Graph[{1 -> 2, 2 -> 3},
EdgeLabels -> {DirectedEdge[1, 2] -> 100, DirectedEdge[2, 2] -> 50}]
</code></pre>
<blockquote>
<p><a href="https://i.stack.imgur.com/yNay4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yNay4.png" alt="enter image description here" /></a></p>
</blockquote>
|
288,974 | <p>Alright this maybe really funny but I want to know why is this wrong. We often come across identities which we prove by multiplying both the sides of the identity by a certain entity but why don't we multiply it by $0$. That way every identity will be proved in one single line. That is so stupid. I mean, by that way we may also say that $1=2=3$. I know it is wrong. But why? I mean if we can multiply both the sides by $2$ then why not by $0$. For example, consider the following trigonometric identity :</p>
<p>Prove the identity : $\sin^2 \theta = \tan^2 \theta \cos ^2 \theta$</p>
<h2>Usual way</h2>
<p>To Prove : $\sin^2 \theta = \tan^2 \theta \cos ^2 \theta $</p>
<p>$\displaystyle \implies {\sin^2 \theta \over \cos^2 \theta } = {\tan^2 \theta \cos ^2 \theta \over \cos^2 \theta}$ (multiplying both the sides by $\displaystyle 1 \over \cos^2 \theta$)</p>
<p>$\implies \tan ^2 \theta = \tan^2\theta$</p>
<p>$\implies LHS=RHS$</p>
<p>$\therefore proved$</p>
<h2>Funny way</h2>
<p>To Prove : $\sin^2 \theta = \tan^2 \theta \cos ^2 \theta $</p>
<p>$\displaystyle \implies {\sin^2 \times 0} = {\tan^2 \theta \cos ^2 \theta \times 0}$ (multiplying both the sides by $0$)</p>
<p>$\implies 0 = 0$</p>
<p>$\therefore proved$</p>
<p>Please explain why is this wrong.</p>
| alexis | 25,773 | <p>Here's how you see that <strong>something</strong> is wrong with your argument:</p>
<ol>
<li><p>To prove: <span class="math-container">$1 = 2$</span></p>
</li>
<li><p>Multiply both sides by zero:</p>
<p><span class="math-container">$ \Rightarrow 1 \times 0 = 2 \times 0$</span> (multiplying both sides by <span class="math-container">$0$</span>)</p>
</li>
<li><p><span class="math-container">$ \Rightarrow 0 = 0 $</span></p>
</li>
</ol>
<p>Conclusion: This is useless as a proof method. Multiplying <em>anything</em> by zero gives you zero. So, although both sides of your equation become zero when you do this, it doesn't mean that they were equal at the start.</p>
<p>Arithmetic proofs of (in)equality are only valid if each step guarantees that the results could <em>only</em> be equal if the inputs were equal. Essentially, the steps have to work backwards as well as forwards. In this case, the reverse step would amount to dividing by zero.</p>
|
288,974 | <p>Alright this maybe really funny but I want to know why is this wrong. We often come across identities which we prove by multiplying both the sides of the identity by a certain entity but why don't we multiply it by $0$. That way every identity will be proved in one single line. That is so stupid. I mean, by that way we may also say that $1=2=3$. I know it is wrong. But why? I mean if we can multiply both the sides by $2$ then why not by $0$. For example, consider the following trigonometric identity :</p>
<p>Prove the identity : $\sin^2 \theta = \tan^2 \theta \cos ^2 \theta$</p>
<h2>Usual way</h2>
<p>To Prove : $\sin^2 \theta = \tan^2 \theta \cos ^2 \theta $</p>
<p>$\displaystyle \implies {\sin^2 \theta \over \cos^2 \theta } = {\tan^2 \theta \cos ^2 \theta \over \cos^2 \theta}$ (multiplying both the sides by $\displaystyle 1 \over \cos^2 \theta$)</p>
<p>$\implies \tan ^2 \theta = \tan^2\theta$</p>
<p>$\implies LHS=RHS$</p>
<p>$\therefore proved$</p>
<h2>Funny way</h2>
<p>To Prove : $\sin^2 \theta = \tan^2 \theta \cos ^2 \theta $</p>
<p>$\displaystyle \implies {\sin^2 \times 0} = {\tan^2 \theta \cos ^2 \theta \times 0}$ (multiplying both the sides by $0$)</p>
<p>$\implies 0 = 0$</p>
<p>$\therefore proved$</p>
<p>Please explain why is this wrong.</p>
| Community | -1 | <p>There is no problem with multiplying by zero, this is quite legal.</p>
<p>Consider your text:</p>
<p>"To Prove : $\sin^2 \theta = \tan^2 \theta \cos ^2 \theta $</p>
<p>$\displaystyle \implies {\sin^2 \times 0} = {\tan^2 \theta \cos ^2 \theta \times 0}$ (multiplying both the sides by $0$)</p>
<p>$\implies 0 = 0$".</p>
<p>This is perfectly right, you proved that $0=0$. Congrats!</p>
|
3,110,660 | <p>Let <span class="math-container">$f:\mathbb{R}^n\to \mathbb{R}^n$</span> a function of class <span class="math-container">$C^1$</span> such that <span class="math-container">$Df(x)$</span> is invertible for all <span class="math-container">$x\in \mathbb{R}^n$</span>. Show that <span class="math-container">$\{x\in \mathbb{R}^n:f(x)=0\}$</span> is a countable set.</p>
<p>My attemp:</p>
<p>I prove that, if <span class="math-container">$a\in \mathbb{R}^n$</span>, then there exists <span class="math-container">$\delta>0$</span> such that, if <span class="math-container">$x\in B_{\delta}(a)$</span> then <span class="math-container">$f(x)\neq f(a)$</span>.</p>
| Myunghyun Song | 609,441 | <p>This is a supplement of @Mindlack's comment. Let us consider for each <span class="math-container">$n\ge 1$</span>,
<span class="math-container">$$
\mathcal C_n =\{x\in\Bbb R^n: f(x) = 0,\ \|x\|\le n\}.
$$</span> By inverse mapping theorem, for each <span class="math-container">$x\in \mathcal C_n$</span>, there exists <span class="math-container">$\delta_x>0$</span> such that <span class="math-container">$$
B(x,\delta_x)\cap \mathcal{C}_n =\{x\}.
$$</span> Since <span class="math-container">$\mathcal C_n$</span> is compact and <span class="math-container">$\{B(x,\delta_x)\cap \mathcal C_n:x\in\mathcal C_n\}$</span> forms an open cover of <span class="math-container">$\mathcal C_n$</span>, there exists a finite subcover <span class="math-container">$\{B(x_i,\delta_{x_i})\cap \mathcal C_n:i=1,2,\ldots, N\}$</span> such that
<span class="math-container">$$
\mathcal C_n = \bigcup_{i=1}^N\ \big(B(x_i,\delta_{x_i})\cap \mathcal C_n \big)=\{x_1,\ldots,x_N\}.
$$</span> This shows for each <span class="math-container">$n\ge 1$</span>, <span class="math-container">$\mathcal{C}_n$</span> is finite, hence
<span class="math-container">$$
\{x\in\Bbb R^n : f(x) = 0\} =\bigcup_{n\ge 1} \mathcal C_n
$$</span> is at most countable.</p>
|
1,897,849 | <p>I'm reading <strong>Awodey's</strong>: Category Theory. I harbor a little confusion: When we speak about a category, say: $\mathbf{Set}$. In the book, usually he talks about this category but there is no notion of quantity of sets. How many sets are actually there?</p>
<p>Another thing that is getting me confused is this: Suppose that I have this category with $n$ sets, I must have the arrows, the composition, the identity, etc. But for these sets, should I assume that functions can be <em>constructed</em> from one set to every other set? That is: For a category with $3$ objects: $A,B,C$, should I assume that I have the following functions?</p>
<p>$$f_1:A\to B\\f_2:A\to C\\f_3:B\to C\\f_4:C\to B\\f_5:C\to A\\f_6:B\to A\\$$</p>
<p>I'm not sure if it's about <em>what I have</em> or <em>what could be made</em>. </p>
| Ricardo J Rademacher | 193,025 | <blockquote>
<p>How many sets are actually there?</p>
</blockquote>
<p>The powerset of all sets P(SET) in the reals is infinite. Hence, from an infinite number of real numbers you can create an infinite number of sets. This is not true of all powersets P(_), but true of SET.</p>
<blockquote>
<p>Suppose that I have this category with nn sets, I must have the arrows, the composition, the identity, etc. But for these sets, should I assume that functions can be constructed from one set to every other set?</p>
</blockquote>
<p>P(n), the power set of all sets created up to "n" will give you all possible combinations. Remember, SET doesn't have arrows (no structure beyond associativity and commutivity of elements) except from powerset to powerset.</p>
<blockquote>
<p>That is: For a category with 3 objects: A,B,CA,B,C, should I assume that I have the following functions?</p>
</blockquote>
<p>Absolutely not; you should never "assume" any functions in CT. The only categorical assumption you are allowed of sorts is that if you assume you are dealing with a category, then you can assume that composition and associativity apply. </p>
<p>Instead, the powerset of 3 objects is:</p>
<p>$$P(3) = (a,b,c), (b,c,a), (c,a,b), (b,a,c), (a,c,b), (c,b,a), (a,c,b), (c,b,a), (b,a,c)$$</p>
<p>Thus the only arrows possible are those that take you from one set to the other:
eg. between (a,b,c) and (b,c,a) you have the following:</p>
<p>$$ f1:a \rightarrow b$$ </p>
<p>$$f2: b\rightarrow c$$</p>
<p>$$f3: c\rightarrow a $$</p>
|
3,142,339 | <p>Let <span class="math-container">$p$</span> be a real number. I am looking for all <span class="math-container">$(x,y)$</span> such that <span class="math-container">$\ln[e^{x}+e^{y}]=px+(1-p)y$</span>. My effort:</p>
<p>Take exponent of both sides to obtain <span class="math-container">$e^{x}+e^{y}=e^{px}e^{(1-p)y}$</span> and then let <span class="math-container">$X=e^{x}, Y=e^{y}$</span>, so that <span class="math-container">$X+Y=X^{p}Y^{1-p}$</span>. How can I proceed from here?</p>
| user21820 | 21,820 | <p>I shall answer the general question completely. The continued fraction <span class="math-container">$[x;x,x,...]$</span> converges for any <span class="math-container">$x ∈ \mathbb{C} ∖ i(-2,2)$</span> <span class="math-container">$= \{ z : z ∈ \mathbb{C} ∧ z ∉ \{ ir : r ∈ (-2,2) \} \}$</span>, and here is a proof sketch.</p>
<p>If <span class="math-container">$[x;x,x,...]$</span> converges to <span class="math-container">$c$</span>, then <span class="math-container">$c = x + 1/c$</span> and hence <span class="math-container">$c$</span> is a root of the quadratic <span class="math-container">$( t ↦ t^2 - x t - 1 )$</span>.</p>
<p>Let <span class="math-container">$r,s$</span> be the roots of the quadratic <span class="math-container">$( t ↦ t^2 - x t - 1 )$</span> and so <span class="math-container">$r + s = x$</span> and <span class="math-container">$r s = -1$</span>.</p>
<p>Let the sequence of approximants be <span class="math-container">$(a_n)$</span> where <span class="math-container">$a_1 = x$</span> (and the sequence stops if it becomes <span class="math-container">$0$</span>).</p>
<p>Let <span class="math-container">$b_0 = 1$</span> and <span class="math-container">$b_n = a_n b_{n-1}$</span> for each <span class="math-container">$n$</span> such that <span class="math-container">$a_n$</span> is defined, giving <span class="math-container">$a_n = \frac{b_n}{b_{n-1}}$</span>.</p>
<p>Given any <span class="math-container">$n ∈ \mathbb{N}^+$</span> such that <span class="math-container">$a_n ≠ 0$</span>:</p>
<p>  <span class="math-container">$b_{n+1} = ( x + \frac1{a_n} ) b_n = x b_n + b_{n-1}$</span>.</p>
<p>  Thus <span class="math-container">$b_{n+1} - r b_n = s ( b_n - r b_{n-1} ) = s^n ( b_1 - r b_0 ) = (x-r) s^n = s^{n+1}$</span>.</p>
<p>  Thus <span class="math-container">$b_n - r^n b_0 = \sum_{k=1}^n r^{n-k} s^k$</span> and hence <span class="math-container">$b_n = \sum_{k=0}^n r^{n-k} s^k$</span>.</p>
<p>  If <span class="math-container">$r ≠ s$</span>, then <span class="math-container">$b_n = {\large \frac{ r^{n+1} - s^{n+1} }{r-s} }$</span> and hence <span class="math-container">$a_n = {\large \frac{ r^{n+1} - s^{n+1} }{ r^n - s^n } }$</span>.</p>
<p>  If <span class="math-container">$r = s$</span>, then <span class="math-container">$b_n = (n+1) r^n$</span> and hence <span class="math-container">$a_n = \frac{n+1}{n} r$</span>.</p>
<p>If <span class="math-container">$x ∉ i[-2,2]$</span>,</p>
<p>  <span class="math-container">$|r| ≠ 1$</span> otherwise <span class="math-container">$x = r + s = r - \frac{1}{r} = r - r^* = 2i·Im(r) ∈ i[-2,2]$</span>.</p>
<p>  Permute <span class="math-container">$r$</span>,<span class="math-container">$s$</span> such that <span class="math-container">$|r| > 1 > |s|$</span>, possible since <span class="math-container">$|r| · |s| = |rs| = 1$</span>.</p>
<p>  Then by induction we get <span class="math-container">$a_n = {\large \frac{ r^{n+1} - s^{n+1} }{ r^n - s^n } } ≠ 0$</span> for every <span class="math-container">$n ∈ \mathbb{N}^+$</span>.</p>
<p>  Thus <span class="math-container">$a_n = r + {\large \frac{ (r-s) s^n } { r^n - s^n } } = r + {\large \frac{r-s}{ (\frac{r}{s})^n - 1 } } \to r$</span> as <span class="math-container">$n \overset{∈\mathbb{N}}\to \infty$</span>.</p>
<p>If <span class="math-container">$x ∈ \{2i,-2i\}$</span>,</p>
<p>  <span class="math-container">$r = s ∈ \{i,-i\}$</span> because <span class="math-container">$(r-s)^2 = (r+s)^2 - 4rs = 0$</span>.</p>
<p>  Then by induction we get <span class="math-container">$a_n = \frac{n+1}{n} r ≠ 0$</span> for every <span class="math-container">$n ∈ \mathbb{N}^+$</span>.</p>
<p>  Thus <span class="math-container">$a_n \to r$</span> as <span class="math-container">$n \overset{∈\mathbb{N}}\to \infty$</span>.</p>
<p>If <span class="math-container">$x ∈ i(-2,2)$</span>,</p>
<p>  By induction we get <span class="math-container">$a_n ∈ i\mathbb{R}$</span> whenever <span class="math-container">$a_n$</span> is defined.</p>
<p>  If <span class="math-container">$a_n \to c$</span> as <span class="math-container">$n \to \infty$</span>:</p>
<p>    <span class="math-container">$c ∈ i\mathbb{R}$</span> since <span class="math-container">$i\mathbb{R}$</span> is closed.</p>
<p>    But <span class="math-container">$c ∈ \{r,s\} = {\large \frac{ x \pm \sqrt{x^2+4} }{2} }$</span> and hence <span class="math-container">$c ∉ i\mathbb{R}$</span> since <span class="math-container">$x^2+4 > 0$</span>.</p>
<p>    Contradiction.</p>
<p>  Therefore <span class="math-container">$( a_n )$</span> either terminates in a <span class="math-container">$0$</span> or does not converge.</p>
<hr>
<p>It is worth <strong>emphasizing</strong> that one cannot assume that the continued fraction converges. <strong>If</strong> it converges then its limit must be one of the roots of the quadratic (and the above proof shows us explicitly which one). But it <strong>may</strong> be that it does not converge in the first place, such as for <span class="math-container">$x = i$</span>.</p>
|
285,167 | <p>This problem is taken from the Kosovo Mathematical Olympiad for Grade-$ 10 $ students.</p>
<p>Let $ a $, $ b $ and $ c $ be the lengths of the edges of a given triangle.</p>
<p>How can one prove the following inequality?</p>
<blockquote>
<p>$$
\left| \frac{a}{b} + \frac{b}{c} + \frac{c}{a} - \frac{a}{c} - \frac{b}{a} - \frac{c}{b} \right| < 1.
$$</p>
</blockquote>
| mathemagician | 49,176 | <p>You can write the question as
$|\frac{a^2c+b^2a+c^2b-b^2c-c^2a-a^2b}{abc}|<1$. This means you would like to show that
$-1<\frac{a^2c+b^2a+c^2b-b^2c-c^2a-a^2b}{abc}<1$. Since $a,b,c$ are the sides of a triangle, you have that $a,b,c>0$. This means you want to show $-abc<a^2c+b^2a+c^2b-b^2c-c^2a-a^2b<abc$. We write the inside term more compactly, we are trying to show $-abc<ac(a-c)+ab(b-a)+bc(c-b)<abc.$ Because the expression we are trying to show is cyclic, without loss of generality, we assume $a\geq b\geq c >0$. Since they are the sides of a triangle, you have that $a-b<c$, $b-c<a$ and $a-c<b$ (sum of two sides exceeds the third one). Therefore, this gives you $ac(a-c)<abc$. Also $a\geq b\geq c>0$ give you $ab(b-a)\leq 0$ and $bc(c-b)\leq 0$. Adding the three inequalities gives $ac(a-c)+ab(b-a)+bc(c-b)<abc$. </p>
<p>Now since $b-c<a$ you have that $bc(b-c)<abc$. Call this (1). Since $b+c > a$ and since $b-c\geq 0$, it follows that $(b+c)(b-c)\geq a(b-c)$, which means $b^2-c^2-ab+ac\geq 0$ implying $b(a-b)+c(c-a)\leq 0$. This means that $ab(a-b)+ac(c-a)\leq 0$. Call this (2).
Adding (1) and (2) side-by-side gives $ab(a-b)+ac(c-a)+bc(b-c)< abc$ which is the same as $-abc<ab(b-a)+ac(a-c)+bc(c-b)$, which is what we wanted to show. </p>
|
285,167 | <p>This problem is taken from the Kosovo Mathematical Olympiad for Grade-$ 10 $ students.</p>
<p>Let $ a $, $ b $ and $ c $ be the lengths of the edges of a given triangle.</p>
<p>How can one prove the following inequality?</p>
<blockquote>
<p>$$
\left| \frac{a}{b} + \frac{b}{c} + \frac{c}{a} - \frac{a}{c} - \frac{b}{a} - \frac{c}{b} \right| < 1.
$$</p>
</blockquote>
| Haskell Curry | 39,362 | <p>The inequality can be proven very cleanly using a powerful technique known as the <a href="http://mblog1024.wordpress.com/2011/02/14/ravi-substitution-explained/" rel="noreferrer">Ravi Substitution</a>. Before introducing this technique, let us first give a definition.</p>
<blockquote>
<p><strong>Definition</strong> An ordered triple $ (a,b,c) $ is said to be a <strong>triangular triple</strong> if and only if $ (a,b,c) \in \mathbb{R}_{+}^{3} $ and there exists a triangle whose edges have lengths $ a $, $ b $ and $ c $.</p>
</blockquote>
<p>The following result says that triangular triples have a particularly nice form.</p>
<blockquote>
<p><strong>Theorem 1</strong> An ordered triple $ (a,b,c) $ is a triangular triple if and only if there exists $ (x,y,z) \in \mathbb{R}_{+}^{3} $ such that
$$
a = x + y, \quad b = y + z, \quad c = z + x.
$$
Furthermore, $ x $, $ y $ and $ z $ are unique.</p>
</blockquote>
<p>The Ravi Substitution is then the act of making the substitutions
$$
a \to x + y, \quad b \to y + z, \quad c \to z + x.
$$</p>
<hr>
<p>We are now ready to prove the given inequality, so let $ (a,b,c) $ be a triangular triple. As
$$
\left| \frac{a}{b} + \frac{b}{c} + \frac{c}{a} - \frac{a}{c} - \frac{b}{a} - \frac{c}{b} \right| < 1 \iff \left| a^{2} c + b^{2} a + c^{2} b - a^{2} b - b^{2} c - c^{2} a \right| < abc,
$$
it suffices to prove the inequality on the right.</p>
<p>Applying the Ravi Substitution, we obtain (after some algebraic manipulation, which I shall leave as an exercise)
$$
a^{2} c + b^{2} a + c^{2} b - a^{2} b - b^{2} c - c^{2} a = -(x - y)(y - z)(z - x).
$$
Hence,
\begin{align}
&\left| a^{2} c + b^{2} a + c^{2} b - a^{2} b - b^{2} c - c^{2} a \right| \\
= &|-(x - y)(y - z)(z - x)| \\
= &|(x - y)(y - z)(z - x)| \\
= &|x - y||y - z||z - x| \\
< &(|x| + |y|)(|y| + |z|)(|z| + |x|) \quad (\text{By the Triangle Inequality.}) \\
= &(x + y)(y + z)(z + x) \quad (\text{As $ x,y,z > 0 $.}) \\
= &abc.
\end{align}</p>
<hr>
<p><strong>Further Notes on the Ravi Substitution</strong></p>
<p>Given $ S \subseteq \mathbb{R}^{3} $, we call $ S $ a <strong>positive cone</strong> in $ \mathbb{R}^{3} $ if and only if $ S $ is closed under</p>
<ul>
<li><p>addition, i.e., $ \mathbf{x},\mathbf{y} \in S \implies \mathbf{x} + \mathbf{y} \in S $, and</p></li>
<li><p>scalar multiplication by a positive real number, i.e., $ \mathbf{x} \in S, \lambda \in \mathbb{R}_{+} \implies \lambda \cdot \mathbf{x} \in S $.</p></li>
</ul>
<p>Let $ \Delta $ denote the set of all triangular triples. Using the Triangle Inequality, it is easy to show that $ \Delta $ is a positive cone in $ \mathbb{R}^{3} $. Clearly, $ \mathbb{R}_{+}^{3} $ is also a positive cone in $ \mathbb{R}^{3} $. Next, define a mapping $ \mathcal{R}: \mathbb{R}_{+}^{3} \to \mathbb{R}^{3} $ as follows:
$$
\forall (x,y,z) \in \mathbb{R}_{+}^{3}: \quad \mathcal{R}(x,y,z) \stackrel{\text{def}}{=} (x + y,y + z,z + x).
$$</p>
<blockquote>
<p><strong>Theorem 2</strong> The mapping $ \mathcal{R} $ is a bijective positively linear transformation from $ \mathbb{R}_{+}^{3} $ to $ \Delta $, i.e., $ \mathcal{R}: \mathbb{R}_{+}^{3} \to \Delta $ is a bijection and
$$
\forall \mathbf{x},\mathbf{y} \in \mathbb{R}_{+}^{3}, ~ \forall \lambda \in \mathbb{R}_{+}: \quad \mathcal{R}(\lambda \cdot \mathbf{x} + \mathbf{y}) = \lambda \cdot \mathcal{R}(\mathbf{x}) + \mathcal{R}(\mathbf{y}).
$$
We call $ \mathcal{R} $ the <strong>Ravi Substitution mapping</strong>.</p>
</blockquote>
<p>Theorem 2 thus says that the Ravi Substitution mapping $ \mathcal{R} $ is an isomorphism of positive cones. Theorem 1 is an immediate consequence.</p>
<p><em>Proof of Theorem 2</em>: Let $ (x,y,z) \in \mathbb{R}_{+}^{3} $, so $ \mathcal{R}(x,y,z) = (x + y,y + z,z + x) $. As
\begin{align}
(x + y) + (y + z) &= x + 2y + z > x + z, \\
(x + y) + (z + x) &= 2x + y + z > y + z, \quad \text{and} \\
(y + z) + (z + x) &= x + y + 2z > x + y,
\end{align}
there exists a triangle whose edges have lengths $ x + y $, $ y + z $ and $ z + x $. Hence, $ \text{Range}(\mathcal{R}) \subseteq \Delta $.</p>
<p>If $ (a,b,c) \in \Delta $, then by the Triangle Inequality, we have
$$
\left( \frac{a + c - b}{2},\frac{b + a - c}{2},\frac{c + b - a}{2} \right) \in \mathbb{R}_{+}^{3}.
$$
Hence, we can define a mapping $ \mathcal{S}: \Delta \to \mathbb{R}_{+}^{3} $ as follows:
$$
\forall (a,b,c) \in \Delta: \quad \mathcal{S}(a,b,c) \stackrel{\text{def}}{=} \left( \frac{a + c - b}{2},\frac{b + a - c}{2},\frac{c + b - a}{2} \right).
$$
As $ \mathcal{R} \circ \mathcal{S} = \text{id}_{\Delta} $ and $ \mathcal{S} \circ \mathcal{R} = \text{id}_{\mathbb{R}_{+}^{3}} $, we deduce that $ \mathcal{R}: \mathbb{R}_{+}^{3} \to \Delta $ is a bijection.</p>
<p>The proof that $ \mathcal{R} $ is a positively linear transformation is not difficult at all, so we leave it to the reader. $ \quad \spadesuit $</p>
<p>Suppose now that we have a function $ F: \Delta \to \mathbb{R} $ and are asked to prove
$$
\forall (a,b,c) \in \Delta: \quad F(a,b,c) \geq 0.
$$
By Theorem 2, this is equivalent to proving
$$
\forall (x,y,z) \in \mathbb{R}_{+}^{3}: \quad F(\mathcal{R}(x,y,z)) = F(x + y,y + z,z + x) \geq 0.
$$
The reason why the second statement may be easier to prove than the first one is that $ \mathbb{R}_{+}^{3} $ is easier to handle than $ \Delta $. (Do you find it easy to visualize $ \Delta $ in $ 3 $-space? I do not.) Many inequalities involving three variables, such as the three-variable version of the AM-GM Inequality, are tailor-made for $ \mathbb{R}_{+}^{3} $, so applying the Ravi Substitution mapping usually simplifies matters considerably as we may then use these inequalities directly.</p>
|
934,660 | <p>Prove that for $ n \geq 2$, n has at least one prime factor.</p>
<p>I'm trying to use induction. For n = 2, 2 = 1 x 2. For n > 2, n = n x 1, where 1 is a prime factor. Is this sufficient to prove the result? I feel like I may be mistaken here.</p>
| André Nicolas | 6,312 | <p>For a formal proof, we use <em>strong induction</em>. Suppose that for all integers $k$, with $2\le k\lt n$, the number $k$ has at least one prime factor. We show that $n$ has at least one prime factor.</p>
<p>If $n$ is prime, there is nothing to prove. If $n$ is not prime, by definition there exist integers $a$ and $b$, with $2\le a\lt n$ and $2\le b\lt n$, such that $ab=n$.</p>
<p>By the induction assumption, $a$ has a prime factor $p$. But then $p$ is a prime factor of $n$. </p>
|
934,660 | <p>Prove that for $ n \geq 2$, n has at least one prime factor.</p>
<p>I'm trying to use induction. For n = 2, 2 = 1 x 2. For n > 2, n = n x 1, where 1 is a prime factor. Is this sufficient to prove the result? I feel like I may be mistaken here.</p>
| selva kumar | 146,563 | <p>Let n be a number greater than 2. It can be proved in following way. </p>
<p>Case 1:If n is a prime number and $n \geq 2 $ , then the number n can be factorized like $n= 1 * n $, where n is the only prime factor. </p>
<p>Case 2:If n is not a prime number and $n \geq 2 $, then the number n may be factorized in the following manner
$n = p_1^m * p_2^r *...p_i^a$ where $p_i$ denotes the prime factor and i denotes the number of prime numbers involved in its factorization such that $i > 1$. </p>
<p>If $i=1$ , it falls in the case 1.</p>
<p>Hence proved. </p>
|
4,491,251 | <p>Per the question title, what's the easiest way to evaluate the following?
<span class="math-container">$$\int_0^{\pi/6}\sec x\,dx$$</span></p>
<p>You can do something like computing the derivatives of <span class="math-container">$\sec x$</span> and <span class="math-container">$\tan x$</span>, adding them up, computing the derivative of the logarithm of the absolute value of the sum of <span class="math-container">$\sec x$</span> and <span class="math-container">$\tan x$</span>, and then completing the integration-by-parts, getting the final answer of <span class="math-container">$\ln(\sqrt{3})$</span>.</p>
<p>But that feels like pulling something out of thin air.</p>
<p>I'm wondering if there's an easier way to compute the integral.</p>
| mr_e_man | 472,818 | <p><span class="math-container">$$\int_0\frac{1}{1-u^2}du=\text{artanh}(u)$$</span>
<span class="math-container">$$\int_0\frac{1}{\cos x}dx=\int_0\frac{1}{1-\sin^2 x}\cos x\,dx=\text{artanh}(\sin x)$$</span>
<span class="math-container">$$\int_0^{\pi/6}\frac{1}{\cos x}dx=\text{artanh}(\sin(\pi/6))=\text{artanh}(1/2)$$</span></p>
|
188,947 | <p>I have a rather complex looking plot which is a combination of graphics objects, generated by </p>
<pre><code>data = Import["o2ld.csv"];
data2 = Import["stemld.csv"];
a1 = ListContourPlot[data, Contours -> 25, Axes -> False,
PlotRangePadding -> 0, Frame -> False,
ColorFunction -> "DarkRainbow", PlotRange -> {0.001, 100}];
a2 = ListPlot3D[data2, ClippingStyle -> None, Mesh -> None,
ColorFunction -> "TemperatureMap", PlotRange -> {0.001, 1},
PlotStyle -> Opacity[0.33]];
level = -0.01; gr =
Graphics3D[{Texture[a1], EdgeForm[],
Polygon[{{0, 0, level}, {200, 0, level}, {200, 200, level},
{0, 200, level}},
VertexTextureCoordinates -> {{0, 0}, {1, 0}, {1, 1}, {0,
1}}]}, Lighting -> "Neutral"];
out = Show[a2, gr, PlotRange -> All, BoxRatios -> {1, 1, 1},
BoxStyle -> Directive[Dashed, Black, Thin],
ViewPoint -> {-0.35, -2, 1.5},
AxesLabel -> {"x", "y", "Proportion"},
LabelStyle -> Directive[Blue, Bold],
Ticks -> {{{0, 0}, {40, 0.5}, {80, 1}, {120, 1.5}, {160, 2},
{200, 2.5}}, {{0, 0}, {40, 0.5}, {80, 1}, {120, 1.5}, {160,
2}, {200, 2.5}}, {0, 0.5, 1}}, ImageSize -> Large]
</code></pre>
<p>Which produces graphics like;
<a href="https://i.stack.imgur.com/WPzT8.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WPzT8.png" alt="enter image description here"></a></p>
<p>What I would like to do is add a bar legend to the right of the right, based on the values from data (a1). I can create the precise bar easily enough with </p>
<pre><code> vr = BarLegend[{"DarkRainbow", {0, Max[data]}}, 25];
</code></pre>
<p>But as this is not a graphics object, I cannot get it to display in with Show. Nor can I seem to get it working with epilog or inset. Does anyone have any idea how to include the bar to the right? I can included csv files for ease of recreation if required, downloadable in a RAR <a href="http://s000.tinyupload.com/download.php?file_id=89424804368267636644&t=8942480436826763664489404" rel="nofollow noreferrer">here</a>.</p>
| DRG | 6,962 | <p>Following on from the excellent answer by Robert Jacobson, I eventually found it's possible too to use placed with the graphics object in A2, as laid out below. </p>
<pre><code>data = Import["o2ld.csv"];
data2 = Import["stemld.csv"];
a1 = ListContourPlot[data, Contours -> 25, Axes -> False,
PlotRangePadding -> 0, Frame -> False,
ColorFunction -> "DarkRainbow", PlotRange -> {0.001, 100}];
a2 = Legended[
ListPlot3D[data2, ClippingStyle -> None, Mesh -> None,
ColorFunction -> "TemperatureMap", PlotRange -> {0.001, 1},
PlotStyle -> Opacity[0.33], ImageSize -> 500],
Placed[BarLegend[{"DarkRainbow", {0, Max[data]}}, 25,
LegendLayout -> "Row", LegendMarkerSize -> 500,
LabelStyle -> {FontSize -> 16}], Below]];
level = -0.01; gr =
Graphics3D[{Texture[a1], EdgeForm[],
Polygon[{{0, 0, level}, {200, 0, level}, {200, 200, level}, {0,
200, level}},
VertexTextureCoordinates -> {{0, 0}, {1, 0}, {1, 1}, {0, 1}}]},
Lighting -> "Neutral"];
vr = BarLegend[{"DarkRainbow", {0, Max[data]}}, 25];
out = Show[a2, gr, PlotRange -> All, BoxRatios -> {1, 1, 1},
BoxStyle -> Directive[Dashed, Black, Thin],
ViewPoint -> {-0.35, -2, 1.5},
AxesLabel -> {" mm", " mm", "Ratio"},
LabelStyle -> {Directive[Black], FontSize -> 15},
PlotLabel ->
Style["(a) (i) Spatial map of stem divisions", FontSize -> 20],
Ticks -> {{{0, 0}, {40, 0.5}, {80, 1}, {120, 1.5}, {160, 2}, {200,
2.5}}, {{0, 0}, {40, 0.5}, {80, 1}, {120, 1.5}, {160, 2}, {200,
2.5}}, {0, 0.5, 1}}, ImageSize -> 700];
</code></pre>
<p>This yields the image below (with minor stylistic differences to original, just for clarity, but idea should be there). So this is another option anyway! </p>
<p><a href="https://i.stack.imgur.com/pmKmI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pmKmI.png" alt="enter image description here"></a></p>
|
293,047 | <p>When I am reading through higher Set Theory books I am frequently met with statements such as '$V$ is a model of ZFC' or '$L$ is a model of ZFC' where $V$ is the Von Neumann Universe, and $L$ the Constructible Universe. For instance, in Jech's 'Set Theory' pg 176, in order to prove the consistency of the Axiom of Choice with ZF, he constructs $L$ and shows that it models the ZF axioms plus AC. </p>
<p>However isn't this strictly inaccurate as $V$ and $L$ are proper classes? For instance, by this very method we might as well take it as a $Theorem$ in ZFC that ZFC is consistent since $V$ models ZFC. However this is obviously impossible as ZFC cannot prove its own consistency. I highly doubt that Jech would make a mistake in such classic textbook, so I must be missing something.</p>
<p>How could we, for instance, show Con(ZF) $\implies$ Con(ZF + AC) without invoking the use of proper classes? I imagine, for instance, that we would start with some (set sized) model $M$ of ZFC and apply some sort of 'constructible universe' construction to $M$. </p>
| Justina Colmena | 121,834 | <p>The class $V$ of all sets is not a model of $ZFC$, because it is a proper class, not a set.</p>
<p>A model of $ZFC$ is a <strong>set</strong> (or <em>small class</em>) $U\in V$ which satisfies all the axioms of $ZFC$ when these axioms are restricted or relativized to $U$, even though $U$ does not include all the sets of the $ZFC$ universe.</p>
<p>If a model $U$ of $ZFC$ exists within the universe $V$ of $ZFC$, then its cardinality is "inaccessible" with respect to the universe $V$. Conversely, if an inaccessible cardinal exists in the universe $V$ of $ZFC$, then a small class, or set $U\in V$ exists, which is a model of $ZFC$.</p>
<p>The existence of a model $U$ within the universe $V$ of $ZFC$ implies that $ZFC$ is consistent. $ZFC$ would be <em>inconsistent</em> if its axioms could prove the existence of a model of itself within itself.</p>
<p>I fear that the meta-theorems, meta-theory, and other meta-language are unnecesssary and ill-defined, unless in the universe $V$ of $ZFC$ we are speaking of the properties of a possible model $U, U\in V, U\subsetneqq V$ where the original axioms of $ZFC$ have been restricted to a $U$.</p>
|
1,480,671 | <p>How to prove $\int_{0}^{\infty}{h(t)\mathbb{E}(I(X>t))dt}=\mathbb{E}(\int_{0}^{\infty}{h(t)I(X>t)dt})$.
Can I treat $h(t)$ as a constant respect to $X$? Then, directly get the result?</p>
<p>The point is I do not understand what $\mathbb{E}(\int_{0}^{\infty}{h(t)I(X>t)dt})$ is.</p>
| jdods | 212,426 | <p>Ok, it looks better now, and I think all requirements for switching the order of integration are satisfied. It seems you are assuming finite expectation of $X$. To get at your last line/question: $X$ is a random variable and thus $I(X>t)$ is a random variable (for each fixed $t$). It is either $1$ or $0$, and depends (randomly) on the value $X$ takes. Thus $\int_0^\infty h(t) I(X>t)dt$ is a random variable. The function $h(t) I(X>t)$ is going to be set to zero over a certain interval depending on what value $X$ takes, thus we don't know the value of the integral with certainty.</p>
|
2,566,193 | <p>Let's say we draw $49$ numbers from $1,\ldots,100$ without returning them back. Then we use the arithmetic mean from the sample.
$$M=\frac{1}{49}\sum_{i=1}^{49}X_i$$</p>
<p>They gave me the hint that $M$ is roughly normal distributed despite the dependencies in the draws. Now I have to determine an symmetric intervall $J$ around the mean with $\mathbb{P}(M\in J)\approx0.95$</p>
<p>My idea was to use the variance and covariance to solve this problem.</p>
<p>$$Var\left(\frac{1}{49}\left(X_1+\ldots+X_{49}\right)\right)$$</p>
<p>$$=\frac{1}{49^2}\left(49\cdot Var(X_1)+49\cdot48Cov(X_1,X_2\right))$$</p>
<p>But I'm not sure how this works</p>
| José Carlos Santos | 446,262 | <p>You know that $(A+3I)(A+I)=0$. Hence, the eigenvalues are a subset of $\{-1,-3\}$. Therefore, the possible values of $\operatorname{tr}A$ (taking into account that it is a $3\times3$ matrix) are $-3$, $-5$, $-7$, and $-9$.</p>
|
2,964,359 | <blockquote>
<p>Let <span class="math-container">$(X, d)$</span> be a metric space with no isolated points, and let <span class="math-container">$A$</span>
be a relatively discrete subset of <span class="math-container">$X$</span>. Prove that <span class="math-container">$A$</span> is nowhere
dense in <span class="math-container">$X$</span>.</p>
</blockquote>
<p><strong>relatively discrete subset of</strong> <span class="math-container">$X$</span>:= A subset <span class="math-container">$A$</span> of a topological space <span class="math-container">$(X,\mathscr T)$</span> is relatively discrete provided that for each <span class="math-container">$a\in A$</span>, there exists <span class="math-container">$U\in \mathscr T$</span> such that <span class="math-container">$U \cap A=\{a\}$</span>.</p>
<p>My aim is to prove <span class="math-container">$int(\overline{A})=\emptyset$</span>. Let if possible <span class="math-container">$int(\overline{A})\neq \emptyset$</span>. Let <span class="math-container">$x\in int(\overline{A})$</span>. which implies there exists <span class="math-container">$B_d(x,\epsilon)\subset \overline A=A\cup A'$</span>.</p>
<p>How do I complete the proof?</p>
<p>What if metric space is replaced by arbitrary topological space, will the result still hold?</p>
| Henno Brandsma | 4,280 | <p>Suppose <span class="math-container">$X$</span> is a crowded <span class="math-container">$T_1$</span> space and <span class="math-container">$D$</span> is relatively discrete.</p>
<p>Suppose (for a contradiction) that there is some non-empty open set <span class="math-container">$U \subseteq \overline{D}$</span></p>
<p>In particular, there is some <span class="math-container">$d \in D \cap U$</span> (being in the closure of <span class="math-container">$D$</span> means every neighbourhood intersects <span class="math-container">$D$</span>) and as <span class="math-container">$D$</span> is relatively discrete, <span class="math-container">$\{d\}$</span> is open in <span class="math-container">$D$</span>, so there is an open set <span class="math-container">$U_d$</span> of <span class="math-container">$X$</span> such that <span class="math-container">$U_d \cap D = \{d\}$</span>.</p>
<p>Now I claim that <span class="math-container">$U \cap U_d = \{d\}$</span>:</p>
<p>The right to left inclusion is clear, as both open sets contain <span class="math-container">$d$</span> and if <span class="math-container">$x \neq d$</span> existed in <span class="math-container">$U \cap U_d$</span>, by <span class="math-container">$T_1$</span>-ness of <span class="math-container">$X$</span> it follows that <span class="math-container">$U \cap U_d \cap (X\setminus\{d\})$</span> is an open set containing <span class="math-container">$x$</span> that misses <span class="math-container">$D$</span> entirely (clearly, as <span class="math-container">$U_d \cap D = \{d\}$</span> and <span class="math-container">$(X\setminus \{d\}) \cap \{d\} = \emptyset$</span>) but <span class="math-container">$x \in U \subseteq \overline{D}$</span>, so this cannot happen. This shows that indeed <span class="math-container">$U \cap U_d = \{d\}$</span>, making <span class="math-container">$\{d\}$</span> open, but this contradicts in turn that <span class="math-container">$X$</span> is crowded (has no isolated points)! </p>
<p>So no such <span class="math-container">$U$</span> can exist and <span class="math-container">$\operatorname{int}(\overline{A}) = \emptyset$</span>. </p>
|
444,486 | <p>I am teaching myself real analysis, and in this particular set of lecture notes, the <a href="http://www.math.louisville.edu/~lee/RealAnalysis/IntroRealAnal-ch01.pdf" rel="nofollow">introductory chapter on set theory</a> when explaining that not all sets are countable, states as follows:</p>
<blockquote>
<p>If $S$ is a set, $\operatorname{card}(S) < \operatorname{card}(\mathcal{P}(S))$.</p>
</blockquote>
<p>Can anyone tell me what this means? It is theorem 1.5.2 found on page 13 of the article.</p>
| Umberto P. | 67,536 | <p>If $X = \mathbb C$ (a one-dimensional vector space over the scalar field $\mathbb C$), [its] balanced sets are $\mathbb C$, the empty set $\emptyset$, and every circular disc (open or closed) centered at $0$. If $X = \mathbb R^2$ (a two-dimensional vector space over the scalar field $\mathbb R$), there are many more balanced sets; any line segment with midpoint at $(0,0)$ will do. The point is that, in spite of the well-known and obvious identification of $\mathbb C$ with $\mathbb R^2$, these two are entirely different as far as their vector space structure is concerned.</p>
<p>-W. Rudin (1973)</p>
|
444,486 | <p>I am teaching myself real analysis, and in this particular set of lecture notes, the <a href="http://www.math.louisville.edu/~lee/RealAnalysis/IntroRealAnal-ch01.pdf" rel="nofollow">introductory chapter on set theory</a> when explaining that not all sets are countable, states as follows:</p>
<blockquote>
<p>If $S$ is a set, $\operatorname{card}(S) < \operatorname{card}(\mathcal{P}(S))$.</p>
</blockquote>
<p>Can anyone tell me what this means? It is theorem 1.5.2 found on page 13 of the article.</p>
| rurouniwallace | 35,878 | <p>I'll explain this more from an electrical engineer's perspective (which I am) than a mathematician's perspective (which I'm not).</p>
<p>The complex plane has several useful properties which arise due to Euler's identity:</p>
<p>$$Ae^{i\theta}=A(\cos(\theta)+i\sin(\theta))$$</p>
<p>Unlike points in the real plane $\mathbb{R}^2$, complex numbers can be added, subtracted, multiplied, and divided. Multiplication and division have a useful meaning that comes about due to Euler's identity:</p>
<p>$$Ae^{i\theta_1}\cdot{Be^{i\theta_2}}=ABe^{i(\theta_1+\theta_2)}$$</p>
<p>$$Ae^{i\theta_1}/{Be^{i\theta_2}}=\frac{A}{B}e^{i(\theta_1-\theta_2)}$$</p>
<p>In other words, multiplying two numbers in the complex plane does two things: multiplies their absolute values, and adds together the angle that they make with the real number line. This makes calculating with phasors a simple matter of arithmetic.</p>
<p>As others have stated,addition, subtraction, multiplication, and division can simply be defined likewise on $\mathbb{R}^2$, but it makes more sense to use the complex plane, because this is a property that comes about naturally due to the definition of imaginary numbers: $i^2=-1$.</p>
|
1,054,595 | <p>I have been thinking about this problem for a while and I still can't come up with a solution. Could you please point me in a direction? Here's the problem.</p>
<pre><code>Let A, B be two 2x2 matrices, A = a b . A and B belong to M2(C). A*B - B*A = A.
c d
Prove that for every n >= 2:
A^n * B - B * A^n = n*A^n
</code></pre>
<p>Any kind of help is appreciated. Thanks!</p>
| Community | -1 | <p>Assume that $A=[a_{i,j}]$ satisfies $||A||_2\leq \epsilon$. Then (with the Frobenius norm) $||A||_F\leq \sqrt{n}||A||_2\leq \sqrt{n}\epsilon$. Then, for every $(i,j)$, $|a_{i,j}|\leq \sqrt{n}\epsilon$. (the Dumpster's result is false -he forgot the square root-). </p>
|
1,671,357 | <p>I'm trying to solve a minimization problem whose purpose is to optimize a matrix whose square is close to another given matrix. But I can't find an effective tool to solve it.</p>
<p>Here is my problem:</p>
<blockquote>
<p>Assume we have an unknown Q with parameter $q11, q12,q14,q21,q22,q23,q32,q33,q34,q41,q43,q44$, and a given matrix G, that is,
$Q=\begin{pmatrix}
q11&q12 &0 &q14 \\q21&q22& q23&0\\ 0&q32& q33&q34\\ q41&0& q43&q44\\
\end{pmatrix} $, $G=\begin{pmatrix}
0.48&0.24 &0.16 &0.12 \\ 0.48&0.24 &0.16 &0.12\\0.48&0.24 &0.16 &0.12\\0.48&0.24 &0.16 &0.12
\end{pmatrix} $,</p>
<p>The problem is how to find the values of $q11, q12,q14,q21,q22,q23,q32,q33,q34,q41,q43,q44$ such that the square of $Q$ is very close to matrix $G$.</p>
</blockquote>
<p>I choose to minimize the Frobenius norm of their difference, that is, </p>
<blockquote>
<p>$ Q* ={\arg\min}_{Q} \| Q^2-G\|_F$</p>
<p>s.t. $0\leq q11, q12,q14,q21,q22,q23,q32,q33,q34,q41,q43,q44 \leq 1$,$\quad$<br>
$\quad$ $q11+q12+q14=1$,
$\quad$ $q21+q22+q23=1$,
$\quad$ $q32+q33+q34=1$,
$\quad$ $q41+q43+q44=1$.</p>
</blockquote>
<p>During those days, I am frustrated to find a effective tool to execute the above optimization algorithm, can someone help me to realize it?</p>
| Set | 26,920 | <p>Minimizing $||Q^2-G||_F^2$ is equivalent to minimizing an SOS (sums of squares) polynomial whose variables are the non-zero entries of $Q$. </p>
<p>Since your constraints can also be written as polynomial constraints, this should be possible using the MOSEK.Polyopt.jl package for Julia. <a href="http://docs.mosek.com/slides/2015/ismp-2015-dahl-polyopt.pdf" rel="nofollow">Here</a> is a link to a powerpoint tutorial on how to use the package and an explanation of the technique used to solve these types of problems.</p>
<p>I haven't used this package, but I think it's based on techniques from the field of convex algebraic geometry, which adapts semidefinite programming to polynomial optimization.</p>
<p><em>Postscript</em>: I see you tagged MATLAB, and it looks like there should be a similar package for MATLAB so if you don't have Julia you could check that one out instead</p>
|
189,069 | <p>The Survival Probability for a walker starting at the origin is defined as the probability that the walker stays positive through n steps. Thanks to the Sparre-Andersen Theorem I know this PDF is given by</p>
<pre><code>Plot[Binomial[2 n, n]*2^(-2 n), {n, 0, 100}]
</code></pre>
<p>However, I want to validate this empirically. </p>
<p>My attempt to validate this for <code>n=100</code>:</p>
<pre><code>FoldList[
If[#2 < 0, 0, #1 + #2] &,
Prepend[Accumulate[RandomVariate[NormalDistribution[0, 1], 100]], 0]]
</code></pre>
<p>I want<code>FoldList</code> to stop if <code>#2 < 0</code> evaluates to <code>True</code>, not just substitute in 0. </p>
| JimB | 19,758 | <p>How about the following brute force approach:</p>
<pre><code>n = 100;
SeedRandom[12345];
nsim = 1000000;
Total[Table[If[Min[Accumulate[RandomVariate[NormalDistribution[0, 1], n]]] <= 0, 0, 1],
{i, nsim}]]/nsim // N
(* 0.056092 *)
Binomial[2 n, n] 2^(-2 n) // N
(* 0.0563485 *)
</code></pre>
<p>To get all of the values from 1 to 100 "simultaneously"...</p>
<pre><code>SeedRandom[12345];
nsim = 100000;
n = 100;
z = ConstantArray[0, n];
Do[
x = Accumulate[RandomVariate[NormalDistribution[0, 1], n]];
i = Flatten[Position[x, _?NonPositive]];
If[Length[i] > 0, If[i[[1]] > 1, z[[1 ;; i[[1]] - 1]] = z[[1 ;; i[[1]] - 1]] + 1], z = z + 1],
{j, nsim}]
z = z/nsim;
ListPlot[{z, Table[Binomial[2 j, j] 2^(-2 j), {j, n}]}, PlotRange -> All, ImageSize -> Large]
</code></pre>
<p><a href="https://i.stack.imgur.com/nN3LT.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nN3LT.png" alt="Simulation and exact formula"></a></p>
|
1,488,388 | <p><strong>The Statement of the Problem:</strong></p>
<p>Let $G$ be a finite abelian group. Let $w$ be the product of all the elements in $G$. Prove that $w^2 = 1$.</p>
<p><strong>Where I Am:</strong></p>
<p>Well, I know that the commutator subgroup of $G$, call it $G'$, is simply the identity element, i.e. $1$. But, can I conclude from this that $\forall g \in G, g=g^{-1}$, i.e., $\forall g \in G, g^2 = gg^{-1} = 1$, which is our desired result? That just seems... strange. But, it kind of makes sense. After all, each element in $G$ has an associated inverse element (because it's a group), and because it's abelian, we can always position an element next to its inverse, i.e.</p>
<p>$$ w^2 = (g_1g_1^{-1}g_2g_2^{-1}g_3g_3^{-1}\cdot \cdot \cdot g_ng_n^{-1})^2 = (1\cdot 1\cdot 1\cdot \cdot \cdot 1)^2=1.$$</p>
<p>Is that all there is to it? Actually, looking at it now, I don't even need to mention the commutator subgroup, do I...</p>
| Oiler | 270,500 | <p>Let $G$ be finite group and let $I =\{g \in G \mid g^2 = 1 \}$. For notational purposes, write $I = \{ i_{1}, i_{2}, \dots, i_{k} \}$ and $G \smallsetminus I = \{g_{1}, g_{2}, \dots , g_{l} \}$ Then since $G$ is abelian we have that
\begin{eqnarray*}
w^2 & = & \left( \prod_{i \in I} i^2 \right) \left( \prod_{g \in G \smallsetminus I} g^2 \right) \\
& = & [i_{1}^2i_{2}^2 \cdots i_{k}^2][g_{1}^2g_{2}^2 \cdots g_{l}^2].
\end{eqnarray*}
Now, since $i_{j}^2 = 1$ for all $i_{j} \in I$ by definition, the left product is simply equal to $\text{id}_{G}$. For the second product, since each $g_{i} \in G \smallsetminus I$ is not self inverse by definition and $G$ is abelian, we may pair up each element with its inverse to give the identity. Hence we have that $w^2 = 1$. </p>
|
1,773,375 | <p>I am having a problem with solving this equation.</p>
<p>I've tried different ways but nothing works.</p>
<p>$$y^2\frac{dy}{dx}+2xy=e^y$$</p>
| doraemonpaul | 30,938 | <p>$y^2\dfrac{dy}{dx}+2xy=e^y$</p>
<p>$y^2\dfrac{dy}{dx}=e^y-2yx$</p>
<p>$(e^y-2yx)\dfrac{dx}{dy}=y^2$</p>
<p>Let $u=\dfrac{e^y}{2y}-x$ ,</p>
<p>Then $x=\dfrac{e^y}{2y}-u$</p>
<p>$\dfrac{dx}{dy}=\dfrac{(y-1)e^y}{2y^2}-\dfrac{du}{dy}$</p>
<p>$\therefore2yu\left(\dfrac{(y-1)e^y}{2y^2}-\dfrac{du}{dy}\right)=y^2$</p>
<p>$\dfrac{(y-1)e^yu}{2y^2}-u\dfrac{du}{dy}=\dfrac{y}{2}$</p>
<p>$u\dfrac{du}{dy}=\dfrac{(y-1)e^yu}{2y^2}-\dfrac{y}{2}$</p>
<p>This belongs to an Abel equation of the second kind.</p>
<p>Let $t=\dfrac{e^y}{2y}$ ,</p>
<p>Then $y=-W\left(-\dfrac{1}{2t}\right)$</p>
<p>$\dfrac{du}{dy}=\dfrac{du}{dt}\dfrac{dt}{dy}=\dfrac{(y-1)e^y}{2y^2}\dfrac{du}{dt}$</p>
<p>$\therefore\dfrac{(y-1)e^yu}{2y^2}\dfrac{du}{dt}=\dfrac{(y-1)e^yu}{2y^2}-\dfrac{y}{2}$</p>
<p>$u\dfrac{du}{dt}=u-\dfrac{y^3}{(y-1)e^y}$</p>
<p>$u\dfrac{du}{dt}-u=\dfrac{\left(W\left(-\dfrac{1}{2t}\right)\right)^2}{2t\left(W\left(-\dfrac{1}{2t}\right)+1\right)}$</p>
<p>This belongs to an Abel equation of the second kind in the canonical form.</p>
<p>Please follow the method in <a href="https://arxiv.org/ftp/arxiv/papers/1503/1503.05929.pdf" rel="nofollow">https://arxiv.org/ftp/arxiv/papers/1503/1503.05929.pdf</a> or in <a href="http://www.iaeng.org/IJAM/issues_v43/issue_3/IJAM_43_3_01.pdf" rel="nofollow">http://www.iaeng.org/IJAM/issues_v43/issue_3/IJAM_43_3_01.pdf</a></p>
|
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