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 |
|---|---|---|---|---|
3,276,332 | <p>I'm studying for my exam in discrete mathematics and found the following problem on last years exam:</p>
<p>Find a closed formula without using induction for <span class="math-container">$\sum_{k=0}^n k^3$</span>.</p>
<p>I tried it by finding the Generating Function first:</p>
<p><span class="math-container">$F(x) = F_0 + \sum_{k=1}^nF_nx^n = \sum_{k=1}^n (F_{n-1}+n^3)x^n = \sum_{k=1}^n F_{n-1}x^n + n^3x^n = \sum_{k=0}^n F_n x^{n+1} + \sum_{k=1}^n n^3x^n = xF(x) + \sum_{k=1}^nn^3x^n$</span></p>
<p>The problem seems to be, that I lack an actual recursive definition of <span class="math-container">$\sum_{k=0}^n k^3$</span> which is, as far as I know, needed to find a generating function. Above, I pretty much used, that <span class="math-container">$X_n = X_{n-1}+n^3$</span>, but obviously, that<span class="math-container">`</span>s not enough. Because a recursive definition was always given in our lectures, I don't now other possibilities to solve this, except for finding the Generating Function with help of recursive definitions.</p>
| Bernard | 202,857 | <p><em>With a telescoping sum</em>: note that
<span class="math-container">$$(k+1)^4-k^4= 4k^3+6k^2+4k+1.$$</span>
Write this equation for <span class="math-container">$k=1, 2,\dots, n$</span> and add them. You'll need to know the sums <span class="math-container">$1+2+\dots +n$</span> and <span class="math-container">$1^2+2^2+\dots +n^2$</span>, which are standard.</p>
|
3,276,332 | <p>I'm studying for my exam in discrete mathematics and found the following problem on last years exam:</p>
<p>Find a closed formula without using induction for <span class="math-container">$\sum_{k=0}^n k^3$</span>.</p>
<p>I tried it by finding the Generating Function first:</p>
<p><span class="math-container">$F(x) = F_0 + \sum_{k=1}^nF_nx^n = \sum_{k=1}^n (F_{n-1}+n^3)x^n = \sum_{k=1}^n F_{n-1}x^n + n^3x^n = \sum_{k=0}^n F_n x^{n+1} + \sum_{k=1}^n n^3x^n = xF(x) + \sum_{k=1}^nn^3x^n$</span></p>
<p>The problem seems to be, that I lack an actual recursive definition of <span class="math-container">$\sum_{k=0}^n k^3$</span> which is, as far as I know, needed to find a generating function. Above, I pretty much used, that <span class="math-container">$X_n = X_{n-1}+n^3$</span>, but obviously, that<span class="math-container">`</span>s not enough. Because a recursive definition was always given in our lectures, I don't now other possibilities to solve this, except for finding the Generating Function with help of recursive definitions.</p>
| Robert Z | 299,698 | <p>Let <span class="math-container">$S_n^{(r)}=\sum_{k=1}^n k^r$</span>. Note that
<span class="math-container">$$(k+1)^4-k^4=4k^3+6k^2+4k+1$$</span>
and after summing for <span class="math-container">$k=1,\dots,n$</span>, (the sum on the left is telescopic) we get
<span class="math-container">$$(n+1)^4-1=4 S_n^{(3)}+6S_n^{(2)}+4S_n^{(1)}+n.$$</span>
In a similar way,
<span class="math-container">$$(k+1)^3-k^3=3k^2+3k+1\implies (n+1)^3-1=3S_n^{(2)}+3S_n^{(1)}+n$$</span>
and
<span class="math-container">$$(k+1)^2-k^2=2k+1\implies (n+1)^2-1=2S_n^{(1)}+n.$$</span>
Now starting from the last one and going back we can obtain <span class="math-container">$S_n^{(1)}$</span>, <span class="math-container">$S_n^{(2)}$</span>, and finally <span class="math-container">$S_n^{(3)}$</span>.</p>
|
2,867,907 | <p>Is the function </p>
<p>$$ f\left(\frac{x}{\epsilon}\right)=\exp\left(\frac{2\pi ix}{\epsilon}\right)=g(x) $$</p>
<p>equivalent to zero ? in the limit $ \epsilon \to 0 $ ?</p>
<p>If I take the derivative $$\frac{ g(x+\epsilon)-g(x)}{\epsilon} $$ is $0$ because the function $g(x)$ has a period 'epsilon'</p>
<p>Also if I take the integral of $g(x)$ is $0$ almost everywhere</p>
<p>So is this function equivalent to $0$ ??</p>
| Community | -1 | <p>You cannot take the derivative with the same $\epsilon$ as that in the function.</p>
<p>$$\lim_{\epsilon\to0}\frac{e^{2i\pi\epsilon/\epsilon}-1}\epsilon\ne
\lim_{\epsilon\to0}\lim_{\eta\to0}\frac{e^{2i\pi \eta/\epsilon}-1}\eta.$$</p>
|
3,649,221 | <blockquote>
<p>Suppose <span class="math-container">$\{f_n\}$</span> is an equicontinuous sequence of functions defined on <span class="math-container">$[0,1]$</span> and <span class="math-container">$\{f_n(r)\}$</span> converges <span class="math-container">$∀r ∈ \mathbb{Q} \cap [0, 1]$</span>. Prove that <span class="math-container">$\{f_n\}$</span> converges uniformly on
<span class="math-container">$[0, 1]$</span>.</p>
</blockquote>
<p>Since I know that <span class="math-container">$\mathbb{Q} \cap [0, 1]$</span> is not compact, I am a bit stuck on my proof.</p>
<p>So far I have:</p>
<p>Let <span class="math-container">$f_n \to f$</span> pointwise on <span class="math-container">$\mathbb{Q} \cap [0, 1]$</span> </p>
<p>Since <span class="math-container">$\{f_n\}_n$</span> is equicontinuous and point-wise bounded (it’s pointwise convergent, so in particular), there exists a subsequence <span class="math-container">$\{f_{n_k}\}_k$</span> such that <span class="math-container">$f_{n_k} \to f$</span> uniformly. </p>
<p>Since each <span class="math-container">$f_n$</span> is continuous, <span class="math-container">$f$</span> is then continuous.</p>
<p>Now take <span class="math-container">$\varepsilon > 0$</span>. Using equicontinuity of <span class="math-container">$\{f_n\}_n$</span>, we find <span class="math-container">$\delta_1 > 0$</span> such that if <span class="math-container">$d(x, y) < δ_1$</span>, <span class="math-container">$x, y \in K$</span>, then
<span class="math-container">$|f_n(x) − f_n(y)| < \varepsilon/3$</span> for all <span class="math-container">$n ∈ \mathbb{Z}^+$</span>. </p>
<p>Using continuity of <span class="math-container">$f$</span>, for each <span class="math-container">$x \in K$</span>, let <span class="math-container">$\delta_2 = \delta_2(x) > 0$</span> be such that if <span class="math-container">$|x − y| < \delta_2(x)$</span>, <span class="math-container">$y \in \mathbb{Q} \cap [0, 1]$</span>, then <span class="math-container">$|f(x) − f(y)| < \varepsilon/3$</span>. For <span class="math-container">$x \in \mathbb{Q} \cap [0, 1]$</span>, let <span class="math-container">$\delta(x) = \min(\delta_1, \delta_2(x)) > 0$</span></p>
<p>I am not sure how to continue nor am I too sure I am on the right path.</p>
| Danny Pak-Keung Chan | 374,270 | <p>Firstly, we show that <span class="math-container">$\lim_{n\rightarrow\infty}f_{n}(x)$</span> exists
for all <span class="math-container">$x\in[0,1]$</span>. Let <span class="math-container">$x\in[0,1]$</span>. Let <span class="math-container">$\varepsilon>0$</span> be arbitrary.
By equicontinuity, there exists <span class="math-container">$\delta>0$</span> such that <span class="math-container">$|f_{n}(x)-f_{n}(y)|<\varepsilon$</span>
whenenver <span class="math-container">$n\in\mathbb{N}$</span> and <span class="math-container">$y\in[0,1]\cap(x-\delta,x+\delta)$</span>.
By density of <span class="math-container">$\mathbb{Q}$</span>, there exists <span class="math-container">$r\in[0,1]\cap(x-\delta,x+\delta)$</span>.
Choose <span class="math-container">$N$</span> such that <span class="math-container">$|f_{n+p}(r)-f_{n}(r)|<\varepsilon$</span> whenever
<span class="math-container">$n\geq N$</span> and <span class="math-container">$p\in\mathbb{N}$</span> (this is possible because <span class="math-container">$\{f_{n}(r)\}_{n}$</span>
is convergent). For any <span class="math-container">$n\geq N$</span> and <span class="math-container">$p\in\mathbb{N}$</span>, we have
<span class="math-container">\begin{eqnarray*}
& & |f_{n+p}(x)-f_{n}(x)|\\
& \leq & |f_{n+p}(x)-f_{n+p}(r)|+|f_{n+p}(r)-f_{n}(r)|+|f_{n}(r)-f_{n}(x)|\\
& < & 3\varepsilon.
\end{eqnarray*}</span>
This shows that <span class="math-container">$\{f_{n}(x)\}$</span> is a Cauchy sequence and hence it
converges.</p>
<p>Next, we show that <span class="math-container">$\{f_{n}(x)\}$</span> converges uniformly in <span class="math-container">$x$</span>. Let
<span class="math-container">$\varepsilon>0$</span> be arbitrary. By equicontinuity, for each <span class="math-container">$x\in[0,1]$</span>,
there exists <span class="math-container">$\delta_{x}>0$</span> such that <span class="math-container">$|f_{n}(x)-f_{n}(y)|<\varepsilon$</span>
whenever <span class="math-container">$y\in[0,1]\cap(x-\delta_{x},x+\delta_{x})$</span> and <span class="math-container">$n\in\mathbb{N}$</span>.
Note that <span class="math-container">$\{(x-\delta_{x},x+\delta_{x})\mid x\in[0,1]\}$</span> is an open
convering for the compact set <span class="math-container">$[0,1]$</span>, so it has a finite subcover
<span class="math-container">$\{(x_{i}-\delta_{x_{i}},x_{i}+\delta_{x_{i}})\mid i=1,\ldots,K\}$</span>.
Choose <span class="math-container">$N$</span> such that <span class="math-container">$|f_{n+p}(x_{i})-f_{n}(x_{i})|<\varepsilon$</span>
whenever <span class="math-container">$n\geq N$</span>, <span class="math-container">$p\in\mathbb{N}$</span>, and <span class="math-container">$i=1,2,\ldots,K$</span>. Now,
let <span class="math-container">$x\in[0,1]$</span>, <span class="math-container">$n\geq N$</span>, and <span class="math-container">$p\in\mathbb{N}$</span> be arbitrary. Choose <span class="math-container">$i$</span> such that <span class="math-container">$x\in (x_{i}-\delta_{x_{i}},x_{i}+\delta_{x_{i}})$</span>.
We have
<span class="math-container">\begin{eqnarray*}
& & |f_{n+p}(x)-f_{n}(x)|\\
& \leq & |f_{n+p}(x)-f_{n+p}(x_{i})|+|f_{n+p}(x_{i})-f_{n}(x_{i})|+|f_{n}(x_{i})-f_{n}(x)|\\
& < & 3\varepsilon.
\end{eqnarray*}</span>
This shows that <span class="math-container">$\{f_{n}(x)\}$</span> is uniformly Cauchy in <span class="math-container">$x$</span> and hence <span class="math-container">$\{f_n\}$</span>
converges uniformly.</p>
|
3,484,293 | <p>In the <span class="math-container">$xy$</span> - plane, the point of intersection of two functions <span class="math-container">$f(x) = x^2$</span> and <span class="math-container">$g(x) = x + 2$</span> lies in which quadrant/s ?</p>
<p>I have no idea how to begin with this question.</p>
| The Demonix _ Hermit | 704,739 | <p><strong>Hint :</strong> We know that <span class="math-container">$f(x) = x^2 \ge 0$</span> for all <span class="math-container">$x$</span> . In which <em>quadrants</em> would this function lie ?</p>
<p>Also <span class="math-container">$f(x) = x+2$</span> is defined in all quadrants except <span class="math-container">$4^{\text{th}}$</span> quadrant.</p>
<p>Can you now find the intersection of both the functions ?</p>
|
2,773,515 | <p>Given $X_1 \sim \exp(\lambda_1)$ and $X_2 \sim \exp(\lambda_2)$, and that they are independent, how can I calculate the probability density function of $X_1+X_2$? </p>
<hr>
<p>I tried to define $Z=X_1+X_2$ and then: $f_Z(z)=\int_{-\infty}^\infty f_{Z,X_1}(z,x) \, dx = \int_0^\infty f_{Z,X_1}(z,x) \, dx$.<br>
And I don't know how to continue from this point.</p>
| drhab | 75,923 | <p>$$f_Z(z)=\int f_{X_1}(x)f_{X_2}(z-x)dx$$</p>
<p>Note that in your case the RHS has integrand $0$ if $z\leq0$ so that $f_Z(z)=0$ if $z\leq0$. </p>
<p>For $z>0$ we have:$$f_Z(z)=\int f_{X_1}(x)f_{X_2}(z-x)dx=\int_0^{z}f_{X_1}(x)f_{X_2}(z-x)dx$$</p>
<p>Work this out yourself.</p>
|
4,410,917 | <p>A student is looking for his teacher. There is a 4/5 chance that the teacher is in one of 8 rooms, and he has no specific room preferences. Student checked 7 of the rooms, but the teacher wasn't in any of them. What's the probability that he is in one of the 8 rooms?</p>
<p>I tried dividing the P(4/5) by 8 and getting probability of teacher being in any one room of 0.1, and then subtracting 0.1*7 from 1 to get 0.3 - probability that he is in the last room. However that's not the right answer.</p>
| drhab | 75,923 | <p><strong>Hint:</strong></p>
<p>Number the rooms with <span class="math-container">$i=1,2,\dots,8$</span>.</p>
<p>Let <span class="math-container">$E_i$</span> denote the event that the teacher is in room <span class="math-container">$i$</span>.</p>
<p>Now find:<span class="math-container">$$P(E_8|E_1^c\cap\cdots\cap E_7^c)$$</span>
Beware that the events are mutually exclusive with:<span class="math-container">$$P(E_1\cup\cdots\cup E_8)=\frac45\text{ and }P(E_1)=\cdots=P(E_8)$$</span></p>
|
753,553 | <p>Let $R$ be a commutative ring and let $0 \to L \to M \to N \to0$ be an exact sequence of $R$-modules. Prove that if $L$ and $N$ are noetherian, then $M$ is noetherian. I tried considering the pre image of the map $L \to M$ and the image of the map $M \to N$ as they are submodules of $L$ and $N$ respectively, but I couldn't make headway.</p>
| jwsiegel | 143,003 | <p>Let $S_1\subset S_2 \subset S_3 \subset ... $ be an ascending chain of submodules
in $M$. </p>
<p>Now consider the chains $S_1\cap L \subset S_2\cap L \subset ... $
and $f(S_1)\subset f(S_2)\subset ... $ in $L$ and $N$, respectively
(here $f:M\rightarrow N$ is the second map in the exact sequence).</p>
<p>As $L$ and $N$ are noetherian, both of these chains stabilize.
Now you must show that if $S_i\subset S_j$,
$S_i\cap L = S_j\cap L$ and $f(S_i) = f(S_j)$, then $S_i = S_j$.</p>
|
2,378,577 | <p>How do I prove or disprove that for a rational number x and an irrational number y, $\ x^y\ $ is irrational?</p>
| Siong Thye Goh | 306,553 | <p>$2^{\log_2 3} =3$ is rational.</p>
<p>Check that $\log_2 3$ is irrational.</p>
<p>Suppose it is rational. </p>
<p>$$\log_2 3 = \frac{a}{b}$$ where $gcd(a,b)=1$.</p>
<p>$$3^b=2^a$$
which is a contradition.</p>
|
32,088 | <h2>Motivation</h2>
<p>One of the methods for strictly extending a theory <span class="math-container">$T$</span> (which is axiomatizable and consistent, and includes enough arithmetic) is adding the sentence expressing the consistency of <span class="math-container">$T$</span> ( <span class="math-container">$Con(T)$</span> ) to <span class="math-container">$T$</span>. But this extension ( <span class="math-container">$T+Con(T)$</span> ) looks very artificial from the mathematical viewpoint, i.e. does not seem to have any mathematically interesting new consequences, and therefore is probably of no interest to a typical mathematician.</p>
<hr />
<p>I would like to know if there is a <em>natural</em> theory (like PA, ZFC, ... ) which by adding the consistency statement we can prove new <em>mathematically interesting</em> statements.
I don't have a definition for what is a natural theory or a mathematically interesting statement, but a theory artificially build for the sole purpose of this question would not be natural, and a purely metamathematical statement (like consistency of <span class="math-container">$T$</span>, or a statement depending on the encoding of <span class="math-container">$T$</span> or its language, or ...) would not count as a mathematically interesting statement.</p>
<p>Questions:</p>
<blockquote>
<ol>
<li><p>Is there a natural theory <span class="math-container">$T$</span> and an mathematically interesting statement <span class="math-container">$\varphi$</span>, such that it is <strong>not known</strong> that <span class="math-container">$T \vdash \varphi$</span>, but <span class="math-container">$T + Con(T) \vdash \varphi$</span>?</p>
</li>
<li><p>Is there a natural theory <span class="math-container">$T$</span> and an interesting mathematical statement <span class="math-container">$\varphi$</span>, such that <span class="math-container">$T \nvdash \varphi$</span> but <span class="math-container">$T + Con(T) \vdash \varphi$</span>?</p>
</li>
</ol>
</blockquote>
| François G. Dorais | 2,000 | <p>The <a href="http://en.wikipedia.org/wiki/Paris%E2%80%93Harrington_theorem" rel="nofollow">Paris–Harrington Theorem</a> is equivalent (over IΣ<sub>1</sub>) to Con(PA + Tr(Π<sub>1</sub>)), where Tr(Π<sub>1</sub>) is the set of all true Π<sub>1</sub> sentences in the language of arithmetic.</p>
<p>For clarity, the <em>1-consistency of PA</em>, i.e. what is meant by Con(PA + Tr(Π<sub>1</sub>)), is the following statement:</p>
<blockquote>
<p>If φ is a (Gödel code for a) true Π<sub>1</sub> sentence, then PA ⊬ ¬φ.</p>
</blockquote>
<p>Or the dual reflection principle:</p>
<blockquote>
<p>If φ is a (Gödel code for a) Σ<sub>1</sub> sentence and PA ⊦ φ, then φ is true.</p>
</blockquote>
<p>Note that IΣ<sub>1</sub> proves the existence of Σ<sub>1</sub> truth predicates, so the above is expressible as a Π<sub>2</sub> statement.</p>
|
1,791,631 | <p>The following is stated on <a href="https://en.wikipedia.org/wiki/Constructible_universe#L_and_large_cardinals" rel="nofollow">Wikipedia</a> for <a href="https://en.wikipedia.org/wiki/Mahlo_cardinal" rel="nofollow">Mahlo cardinals</a>. Unfortunately, it's not sourced. Where can I find details? I wasn't able to google any articles dealing with Mahlo cardinals in $L$.</p>
<blockquote>
<p>Since $On⊂L⊆V$, properties of ordinals that depend on the absence of a function or other structure (i.e. $\Pi_1^{ZF}$ formulas) are preserved when going down from $V$ to $L$. Hence initial ordinals of cardinals remain initial in L. Regular ordinals remain regular in $L$. Weak limit cardinals become strong limit cardinals in $L$ because the generalized continuum hypothesis holds in $L$. Weakly inaccessible cardinals become strongly inaccessible. Weakly Mahlo cardinals become strongly Mahlo. And more generally, any large cardinal property weaker than 0# (see the list of large cardinal properties) will be retained in $L$.</p>
</blockquote>
| DanielWainfleet | 254,665 | <p>Let $C(k)$ be the set of club subsets of $k.$ Let $R(k)=\{l\in k: l=cf(l)\}.$ Observe that for any set $S\subset On,$ if $S\in L$ then </p>
<p>(1) $\forall a\in On \;[\;\{b\cap S :b\in a\} \in L\;], \; \text {and}$</p>
<p>(2) $\forall a\in On\;[\;a=\cup (a\cap S)\iff L\Vdash (a= \cup (a\cap S)\;].$</p>
<p>(3) Also observe that $\forall a\in On\;[a=|a|\implies L\Vdash a=|a|.)]$ </p>
<p>Let $C(k)$ be the set of club subsets of $k.$ From (1) and (2) we have $ C(k)\supset (C(k))^L.$ Let $R(k)=\{l<k: l=|l|\}.$ From (3), we have $R(k)\subset (R(k))^L.$ </p>
<p>So for any $c\in (C(k))^L$ we have $c\in C(k), $ so $\emptyset\ne c\cap R(k)\subset c\cap (R(k))^L.$</p>
<p>Remark. If $k=|k|>\omega$ and $R(k)$ is stationary in $k,$ then $k$ must be weakly inaccessible. Obviously $k$ can't be a successor cardinal. If $k$ is singular then </p>
<p>(i): If $k>cf(k)=\omega $ let $f(n):\omega \to k$ be a co-final strictly increasing map with $f(0)=\omega.$ Let $g(n)=f(n)+1$ for $n\in \omega.$ Then $\{g(n):n\in \omega\}$ is club in $k$ and contains no cardinals.</p>
<p>(ii): If $k>cf (k)=l>\omega,$ let $S\subset k$ with $|S|=l$ and $\cup S=k.$ Let $C=\{a\in k: l<a=\cup (a\cap S)\}.$ It is easy to show that $C$ is club in $k.$ For $a\in C$ we have $cf (a)\leq |a\cap S|\leq l<a.$ So $C$ has no regular members.</p>
|
1,676,347 | <p>Suppose that $V$ and $W$ are vector spaces, $g:V\rightarrow W$ a linear map. Show that g is surjective $\iff$ For any vector space $U$ the map $g^{\ast}:Hom(W,U) \rightarrow Hom(V,U)$ defined by $g^{\ast}(f) = f\circ g$ is injective.</p>
<p>I've failed too much trying to solve this problem. Any hint could be useful, thank you.</p>
| DonAntonio | 31,254 | <p>Suppose $\;g\;$ is surjective and assume $\;f\in\ker g^*\;,\;\;g^*:\text{Hom},(W,U)\to\text{Hom}\,(V,U)\;\implies$</p>
<p>$$g^*(f)=0\implies\color{red}{\forall v\in V}\;,\;\;f\circ g(v)=f(gv)=0\implies g(v)\in \ker f$$</p>
<p>But since $\;g\;$ is surjective, any $\;w\in W\;$ is $\;f(v_w)\;$ , for some $\;v_w\in V\;$, so for any $\;w\in W\;$ we get</p>
<p>$$fw=f(gv_w)=0\implies f=0\implies \ker g^*=0\implies g^*\;\;\text{injective}$$</p>
<p>Perhaps you can do now the other direction by yourself. If you've any problem I shall be back later to try to help.</p>
|
1,555,429 | <p>Hi I am trying to solve the sum of the series of this problem:</p>
<p>$$
11 + 2 + \frac 4 {11} + \frac 8 {121} + \cdots
$$</p>
<p>I know its a geometric series, but I cannot find the pattern around this. </p>
| Michael Hardy | 11,667 | <p><b>Q:</b> What does it mean to say a series is geometric?</p>
<p><b>A:</b> It means there is a common ratio.<br>${}\qquad{}$I.e. the number by which you multiply each term to get the next is the <b>same</b> in every case.</p>
<p>What do you multiply $11$ by to get $2$? You multiply it by $\dfrac 2 {11}$.</p>
<p>What do you multiply <b>that</b> by to get $\dfrac 4 {11}$? You multiply it by $\dfrac 2 {11}$.</p>
<p>What do you multiply <b>that</b> by to get $\dfrac 8 {121}$? You multiply it by $\dfrac 2 {11}$.</p>
<p>The fact that the thing by which you multiply is always the <b>same</b> thing is what the word "common" means.</p>
<p>The common ratio is $r=\dfrac 2 {11}$.</p>
<p>Remember that the sum of a geometric series is $\dfrac a {1-r}$, where $a$ is the first term (provided the common ratio is between $\pm1$).</p>
|
2,505,757 | <p>Today, I was trying to prove <a href="https://math.stackexchange.com/questions/2505714/showing-cantor-set-is-uncountable">Cantor set is uncountable</a> and I completed it just a while ago.</p>
<p>So, I know that the end-points of each $A_n$ are elements of $C$ and those end-points are rational numbers. But since $C$ is uncountable, $C$ must contain uncountable numbers of irrational numbers. Then, is their a way to prove that, a specific irrational number (say $1/\sqrt2$ or $1/4\pi$) belongs to the set $C$ or not?</p>
<p>(Description of notation can be found in the link given above or <a href="https://math.stackexchange.com/questions/2505714/showing-cantor-set-is-uncountable">here</a>)</p>
<p>Lets say,</p>
<blockquote>
<p>Prove or disprove that $1/\sqrt2\in$ Cantor set on $[0,1]$.</p>
</blockquote>
<p>Can we do that? Or is their a way to solve such problem?</p>
| Eric Wofsey | 86,856 | <p>No Hausdorff space with more than one point is irreducible. Indeed, if $X$ is a Hausdorff space with two different points $x,y\in X$, there are disjoint open sets $U$ and $V$ with $x\in U$ and $y\in V$. The complements of $U$ and $V$ are then closed sets whose union is $X$ and neither is all of $X$, so $X$ is not irreducible.</p>
<p>So to show a variety is not irreducible in the Euclidean topology, all you have to do is show it has more than one point, since the Euclidean topology is always Hausdorff. I'll leave it to you to find two different points on your curve.</p>
|
956,256 | <p>If $a_n \ge 0 $ for all n, prove that $\sum_{n=1}^\infty a_n$ converges if and only if $\sum_{n=1}^\infty {a_n\over 1+a_n}$ converges. </p>
<p>Here is my attempt!</p>
<p>=> Suppose that $\epsilon \ge 0$ is given and $a_n$ converges, then for all $N\le n \le m$$$\sum_{k=n+1}^m a_k \lt \epsilon.$$
let $b_n={a_n\over 1+a_n}$ and Notice that $b_n \lt a_n$ and </p>
<p>Then, applying the comparison test $$|\sum_{k=n+1}^m b_k| \le \sum_{k=n+1}^m |b_k |\le\sum_{k=n+1}^m a_k\lt\epsilon$$Hence,${a_n\over 1+a_n}$ converges.</p>
<p>let me know if it is worng and I don't know how to start for the other direction. Any help is appreciated!</p>
| Leucippus | 148,155 | <p>Given $s(t) = 32 + 112 t - 16 t^{2}$ then $v(t)$, being the derivative of $s(t)$, is $v(t) = 112 - 32 t$. If $v(t) = 0$ then $t = 7/2$. Now, $s(7/2) = 32 + 56 \cdot 7 - 4 \cdot 49 = 228$. </p>
<p>To find the velocity at impact solve for $s(t) = 0$. This yields $16 t^{2} - 112 t - 32 = 0$, or $t^{2} - 7 t - 2 = 0$. Solving this equation leads to the two possible values
\begin{align}
t = \frac{7}{2} \pm \frac{\sqrt{49+8}}{2} = \frac{7}{2} \pm \frac{\sqrt{57}}{2}.
\end{align}
The negative component is negative and is to be tossed out of consideration leaving $2 t_{I} = 7 + \sqrt{57}$. Now,
\begin{align}
v(t_{I}) = 112 - 32 t_{I} = 112 - 16 \cdot 7 - 16 \cdot \sqrt{57} = - 16 \sqrt{57} = -120.79735\cdots .
\end{align}</p>
|
3,715,987 | <p>The domain is: <span class="math-container">$\forall x \in \mathbb{R}\smallsetminus\{-1\}$</span></p>
<p>The range is: first we find the inverse of <span class="math-container">$f$</span>:
<span class="math-container">$$x=\frac{y+2}{y^2+2y+1} $$</span>
<span class="math-container">$$x\cdot(y+1)^2-1=y+2$$</span>
<span class="math-container">$$x\cdot(y+1)^2-y=3 $$</span>
<span class="math-container">$$y\left(\frac{(y+1)^2}{y}-\frac{1}{x}\right)=\frac{3}{x} $$</span>
I can't find the inverse... my idea is to find the domain of the inverse, and that would then be the range of the function. How to show otherwise what is the range here?</p>
| user0102 | 322,814 | <p>Notice that
<span class="math-container">\begin{align*}
f(x) = \frac{x+2}{x^{2} + 2x + 1} = \frac{(x+1) + 1}{(x+1)^{2}} = \frac{1}{x+1} + \frac{1}{(x+1)^{2}}
\end{align*}</span></p>
<p>If we set <span class="math-container">$y = 1/(x+1)$</span> and consider <span class="math-container">$c\in\textbf{R}$</span>, we arrive at the following equation
<span class="math-container">\begin{align*}
y + y^{2} = c \Longleftrightarrow y^{2} + y - c = 0\Longleftrightarrow y = \frac{-1\pm\sqrt{1+4c}}{2}
\end{align*}</span>
which has real roots if and only if <span class="math-container">$c\geq-1/4$</span>.</p>
<p>So the answer is <span class="math-container">$\text{Im}(f) = [-1/4,\infty)$</span>.</p>
|
312,238 | <p>Reading my textbook, I came across exercises for nested quantifiers.</p>
<p>The question: Let $L(x, y)$ be the statement “$x$ loves $y$,” where the domain for both $x$ and $y$ consists of all people in the world.
Use quantifiers to express each of these statements.</p>
<p>i) Everyone loves himself or herself.</p>
<p>Textbook answer:
$$
\forall xL(x, x)
$$</p>
<p>Is this equivalent to my answer? :
$$
\forall x\forall y((x=y)\to L(x,y))
$$</p>
| Petr | 37,490 | <p>They're equivalent.</p>
<p>Let's start with $(\forall x)(\forall y) ((x=y)\to L(x,y))$:</p>
<ol>
<li>Apply <a href="https://en.wikipedia.org/wiki/Universal_instantiation" rel="noreferrer">instantiation</a> twice and set $x/x$ and $x/y$, getting $(x=x)\to L(x,x)$.</li>
<li>$x=x$ is an axiom of <a href="https://en.wikipedia.org/wiki/First-order_logic_with_equality#Equality_and_its_axioms" rel="noreferrer">reflexivity of equality</a>.</li>
<li>From these two infer $L(x,x)$ (<a href="https://en.wikipedia.org/wiki/Modus_ponens" rel="noreferrer">modus ponens</a>).</li>
<li>Apply <a href="https://en.wikipedia.org/wiki/Universal_generalization" rel="noreferrer">generalization</a> to get $(\forall x) L(x,x)$.</li>
</ol>
<p>And the other way around: Let's start with $(\forall x) L(x,x)$.</p>
<ol>
<li>Apply instantiation to get $L(x,x)$.</li>
<li>Use <a href="https://en.wikipedia.org/wiki/First-order_logic_with_equality#Equality_and_its_axioms" rel="noreferrer">substitution for formulas</a> axiom to get $(x = y) \to (L(x,x) \to L(x,y))$.</li>
<li>Exchange the antecendents to get $L(x,x) \to ((x = y) \to L(x,y))$ (we can do this because $\phi\to(\psi\to\rho) \equiv \psi\to(\phi\to\rho)$ is a tautology).</li>
<li>Apply modus ponens on 3. and 1. to get $(x = y) \to L(x,y)$.</li>
<li>Use generalization twice to get $(\forall x)(\forall y) (x = y) \to L(x,y)$</li>
</ol>
|
1,119,010 | <p>Write down the assumptions in a form of clauses and give a resolution proof that the proposition
$$\Big((p \rightarrow q) \land ( q \rightarrow r) \land p \Big) \rightarrow r$$
is a tautology.</p>
| Rob Arthan | 23,171 | <p>The negation of the goal has the following conjunctive normal form:
$$
\lnot(((p \rightarrow q) \land ( q \rightarrow r) \land p) \rightarrow r) \equiv
(\lnot p \lor q) \land (\lnot q \lor r) \land p \land \lnot r
$$
I.e., you have four clauses:
$$
\begin{array}{cl}
A:& \{\lnot p, q\}\\
B:& \{\lnot q, r\}\\
C:& \{p\}\\
D:& \{\lnot r\}
\end{array}
$$
$A$, $B$ and $C$ give the "assumptions in clause form" that your question is asking for.
The rest of the resolution proof goes like this:
$$
\begin{array}{clr}
E:& \{\lnot q\} & [B, D] \\
F:& \{\lnot p\} & [A, E] \\
G:& \{\mathsf{false}\} & [C, F]
\end{array}
$$
where the letters in square brackets tell you which earlier clauses to resolve to get the new clause. We have derived $\mathsf{false}$ from the negation of the goal by putting it into clause form and applying resolution inference steps. That constitutes a resolution proof of the goal.</p>
|
3,971,833 | <p>If there is an <span class="math-container">$n$</span> by <span class="math-container">$n$</span> matrix where each element is either 1 or -1, how many unique matrices are there such that each row and each column multiplies to 1?</p>
<p>I solved for the trivial case of <span class="math-container">$n = 2$</span>, which is 2. However, for larger <span class="math-container">$n$</span>, I'm not sure how to find a systematic way to count. I have tried using the fact that you can transform an existing correct matrix into another correct matrix by selecting 4 points that form a rectangle to all become negative, but I am not sure how I could count these in an organized way.</p>
| Calvin Lin | 54,563 | <p>Consider any <span class="math-container">$(n-1) \times (n-1) $</span> minor.<br />
This could be filled with any option of <span class="math-container">$ \pm 1$</span>.</p>
<p>Claim: The unfilled cells are uniquely determined by the column / row that they are in, and there is no conflict.</p>
<p>Proof: Do it yourself.</p>
<p>Corollary: There are <span class="math-container">$ 2^{(n-1)^2 } $</span> ways to fill up the matrix.</p>
|
3,459,205 | <p>Explain why <span class="math-container">$\arccos(t)=\arcsin(\sqrt{{1}-{t^2}})$</span> when <span class="math-container">$0<t≤1$</span>. </p>
<p>I tried researching online, couldn't find anything related to this question though. Know this equation is correct and make sense, just don't know how to explain it using algebra only, solving for the left side of the equation.</p>
| Semiclassical | 137,524 | <p>Here is a calculus proof: Given the integral definitions of <span class="math-container">$\arcsin$</span> and <span class="math-container">$\arcsin$</span>, which respectively are </p>
<p><span class="math-container">$$\arcsin t=\int_0^t \frac{dz}{\sqrt{1-z^2}},\\ \arccos t=\int_t^1 \frac{dz}{\sqrt{1-z^2}}$$</span>
for <span class="math-container">$-1\leq t\leq 1$</span>, the substitution <span class="math-container">$w=\sqrt{1-z^2}$</span> yields (note that <span class="math-container">$w\,dw=-z\,dz$</span>)</p>
<p><span class="math-container">$$\arccos t=\int_t^1\frac{dz}{\sqrt{1-z^2}} = \int_{\sqrt{1-t^2}}^0 \frac{(-w/z)}{w}dw=\int_0^{\sqrt{1-t^2}}\frac{dw}{\sqrt{1-w^2}}=\arcsin(\sqrt{1-t^2})$$</span> as desired.</p>
|
3,622,508 | <p>I’m not sure exactly about the conditions needed for a subset <span class="math-container">$S$</span> to localise a ring <span class="math-container">$R$</span>. I know <span class="math-container">$S$</span> has to be multiplicative. But does <span class="math-container">$S$</span> also have to be a subset of the non-zero divisors of <span class="math-container">$R$</span> or does it have to be a subset of the group of units of <span class="math-container">$R$</span>?</p>
<p>I can’t find a clear answer. </p>
| James | 506,916 | <p>Localisation is defined for any commutative ring with identity <span class="math-container">$R$</span>, and for any multiplicatively closed subset <span class="math-container">$S \subset R$</span>: <span class="math-container">$1 \in S$</span> and <span class="math-container">$s,t \in S$</span> implies <span class="math-container">$st \in S$</span>.</p>
<p>In fact the inclusion homomorphism <span class="math-container">$R \rightarrow S^{-1}R$</span> is injective if and only if <span class="math-container">$S$</span> contains no zero divisiors, and <span class="math-container">$S^{-1}R = 0$</span> if and only if <span class="math-container">$S$</span> contains <span class="math-container">$0$</span>.</p>
|
484,313 | <p>I am taking linear algebra and none of this stuff is expained. I found this helpful link <a href="http://www.math.ucla.edu/~pskoufra/M115A-Notation.pdf" rel="nofollow">http://www.math.ucla.edu/~pskoufra/M115A-Notation.pdf</a></p>
<p>but it is missing a lot of what I need to know. Just right now though what does v and ^ mean in the context of linear algebra and set stuff? It is not defined anywhere in my book and it is exceptionally frustrating trying to read this...stuff. Also what does something like upside down A$ x(x \epsilon A _> x \epsilon B)$ mean?</p>
<p>More context </p>
<p>$ x(x \epsilon A _> x \epsilon B) $^ 4 \exists x \in B $ ^ $x $ \not\in A)$</p>
| Adriano | 76,987 | <p><a href="http://en.wikipedia.org/wiki/List_of_logic_symbols" rel="nofollow">Wikipedia</a> has a list of logic symbols. The expression:</p>
<blockquote>
<p>$\forall x ~ [(x \in A \implies x \in B) ~~~\land~~~ (\exists x \in B ~\text{ s.t. }~ x \in A)]$</p>
</blockquote>
<p>can be interpreted to mean:</p>
<blockquote>
<p>For all $x$, both of the following claims hold true:</p>
<ul>
<li>If $x$ is in $A$, then we also know that $x$ is in $B$.</li>
<li>There exists some $x$ in $B$ such that $x$ is also in $A$.</li>
</ul>
</blockquote>
|
2,648,549 | <p>Let $\tau_{ij}$ be a transposition if degree n. What does it mean when one says that $\tau_{ij}=\tau_{ji}$? Thanks in advance!</p>
| projectilemotion | 323,432 | <p>It is possible that you have not learnt that the trace of a matrix is the sum of its eigenvalues and the determinant of a matrix is the product of its eigenvalues. Hence, I provide an alternative approach:</p>
<hr>
<p>Note that the following statements are equivalent:</p>
<ul>
<li>$\lambda$ is an eigenvalue of $P$</li>
<li>$(P-\lambda I)\mathbf{x}=\mathbf{0}$ has a nontrivial solution</li>
<li>$N(P-\lambda I)\neq \{\mathbf{0}\}$</li>
<li>$(P-\lambda I)$ is singular</li>
<li>$\det(P-\lambda I)=0$</li>
</ul>
<p>Hence, all we need to do is check that one of these statements satisfy, then all the others satisfy. I will assume you have correctly shown that $\lambda_1=1$ is an eigenvalue of $P$. I will prove that $\lambda_2=1-p-q$ is an eigenvalue of $P$.</p>
<hr>
<p>The easiest condition to check, in my opinion is if $\det(P-\lambda_2 I)=0$. Note that:
$$\det(P-\lambda_2 I)=\begin{vmatrix} 1-p-(1-p-q) & p \\ q & 1-q-(1-p-q) \end{vmatrix}=\begin{vmatrix} q & p \\ q & p \end{vmatrix}=0$$
Where in the last step, we have used the fact that if a $n\times n$ matrix has two identical rows or columns, its determinant is zero. Thus, $1-p-q$ is an eigenvalue of $P$.
$$\tag*{$\square$}$$</p>
|
2,796,694 | <p>So for my latest physics homework question, I had to derive an equation for the terminal velocity of a ball falling in some gravitational field assuming that the air resistance force was equal to some constant <em>c</em> multiplied by $v^2.$ <br> So first I started with the differntial equation: <br>
$\frac{dv}{dt}=-mg-cv^2$
<br>
Rearranging to get:
<br>
$\frac{dv}{dt}=-\left(g+\frac{cv^2}{m}\right)$
<br>
From here I tried solving it and ended up with: <br>
$\frac{\sqrt{m}}{\sqrt{c}\sqrt{g}}\arctan \left(\frac{\sqrt{c}v}{\sqrt{g}\sqrt{m}}\right)+C=-t$
<br>
I rearranged this to get:
$v\left(t\right)=\left(\frac{\sqrt{g}\sqrt{m}\tan \left(\frac{\left(-C\sqrt{c}\sqrt{g}-\sqrt{c}\sqrt{g}t\right)}{\sqrt{m}}\right)}{\sqrt{c}}\right)$ <br>
In order to calculate the terminal velocity I took the limit as t approaches infinity:<br>
$\lim _{t\to \infty }\left(\frac{\sqrt{g}\sqrt{m}\tan \:\left(\frac{\left(-C\sqrt{c}\sqrt{g}-\sqrt{c}\sqrt{g}t\right)}{\sqrt{m}}\right)}{\sqrt{c}}\right)$ <br>
This reduces to:
$\frac{\sqrt{g}\sqrt{m}\tan \left(\infty \right)}{\sqrt{c}}$ <br>
The problem with this is that tan $(\infty)$ is indefinite. <br>
Where did I go wrong? Could someone please help properly solve this equation.
<br>
Cheers, Gabriel.</p>
| Community | -1 | <p>In order to determine the teriminal velocity, set $m\frac{dv}{dt}=-mg+cv^2=0$, which implies that $v_t=\sqrt{\frac{mg}{c}}$. The differential equation itself can be solved as follows. Since we know $v_t$, we can rewrite the orign differential equation as $\frac{dv}{dt}=g(1-\frac{v^2}{v_t^2})$ with boundary conditions $v(t_0)=v_0$. Then, we can solve this differential equation by integration over both sides.
$$
\int_{t_0}^t dt'=\int_{v_0}^{v(t)}\frac{dv'}{g(1-\frac{v'^2}{v_t^2})}
$$
Let us write $\tau=\frac{v_t}{g}=\sqrt{\frac{m}{cg}}$. Then
$$t-t_0=\tau(\tanh^{-1}\frac{v}{v_t}-\tanh^{-1}\frac{v_0}{v_t})$$
Solving for $v$, we find that
$$
v=v_t\tanh(\frac{t-t_0}{\tau}-\tanh^{-1}\frac{v_0}{v_t})
$$</p>
<p>If the body is released at rest at $t_0=0$, $$v=v_t\tanh\frac{t}{\tau}$$</p>
|
2,405,505 | <p>How to prove that the infinite product $\prod_{n=1}^{+\infty} \left(1-\frac{1}{2n^2}\right)$ is positive ?</p>
<p>Thanks</p>
| Felix Marin | 85,343 | <p>$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$</p>
<blockquote>
<p>With $\ds{N \in \mathbb{N}_{\geq 1}}$:</p>
</blockquote>
<p>\begin{align}
\prod_{n = 1}^{N}\pars{1 - {1 \over 2n^{2}}} & =
\prod_{n = 1}^{N}{2\pars{n - \root{2}/2}\pars{n + \root{2}/2} \over 2n^{2}} =
{\pars{1 - \root{2}/2}^{\overline{N}}\pars{1 + \root{2}/2}^{\overline{N}} \over \pars{N!}^{2}}
\\[5mm] & =
{\Gamma\pars{N + 1 - \root{2}/2} \over \Gamma\pars{1 - \root{2}/2}N!}\,
{\Gamma\pars{N + 1 + \root{2}/2} \over \Gamma\pars{1 + \root{2}/2}N!}
\\[5mm] & =
{1 \over \Gamma\pars{1 - \root{2}/2}\Gamma\pars{\root{2}/2}\root{2}/2}\,
{\pars{N - \root{2}/2}! \over N!}\,{\pars{N + \root{2}/2}! \over N!}
\\[5mm] & =
{\root{2} \over \pi/\sin\pars{\pi\root{2}/2}}\,
{\pars{N - \root{2}/2}! \over N!}\,{\pars{N + \root{2}/2}! \over N!}
\end{align}
<hr>Note that
\begin{align}
{\pars{N + \alpha}! \over N!} &
\,\,\,\stackrel{\mrm{as}\ N\ \to\ \infty}{\sim}\,\,\,
{\root{2\pi}\pars{N + \alpha}^{N + \alpha + 1/2}\expo{-\pars{N + \alpha}} \over
{\root{2\pi}N^{N + 1/2}\expo{-N}}} =
{N^{N + \alpha + 1/2}\pars{1 + \alpha/N}^{N + \alpha + 1/2}\expo{-\alpha} \over
N^{N + 1/2}}
\\[5mm] & \,\,\,\stackrel{\mrm{as}\ N\ \to\ \infty}{\sim}\,\,\,
N^{\alpha}
\end{align}</p>
<blockquote>
<p>such that</p>
</blockquote>
<p>\begin{align}
\prod_{n = 1}^{\infty}\pars{1 - {1 \over 2n^{2}}} & =
\bbx{{\root{2}\sin\pars{\sqrt{2}\,\pi/2} \over \pi}} \approx 0.3582
\end{align}</p>
|
4,008,987 | <p>I am reading a math book and in it, it says, "Let <span class="math-container">$V$</span> be the set of all functions <span class="math-container">$f: \mathbb{Z^n_2} \rightarrow \mathbb{R}.$</span> I know that <span class="math-container">$\mathbb{Z^n_2}$</span> is just the cyclic group of order <span class="math-container">$2$</span> taken to the <span class="math-container">$n$</span>th power, but I don't get what the actual function means.</p>
<p>How would such a function, <span class="math-container">$f$</span>, work? Could someone please give an example?</p>
| mrtaurho | 537,079 | <p>A function is by definition a mapping between sets. So we have to look at the underlying sets. For simplicity, we write <span class="math-container">$\Bbb Z_2=\{0,1\}$</span>. Then <span class="math-container">$\Bbb Z_2^n=\{0,1\}^n$</span>. A map <span class="math-container">$f\colon\{0,1\}^n\to\Bbb R$</span> now is an assigment which takes an element of the form <span class="math-container">$(a_1,\dots,a_n)$</span> and maps it to some real number <span class="math-container">$r$</span> depending on all <span class="math-container">$a_i\in\{0,1\}$</span>. For example, we can trivially define <span class="math-container">$f(a_1,\dots,a_n)=\pi$</span> for <em>all</em> <span class="math-container">$n$</span>-tupel <span class="math-container">$(a_1,\dots,a_n)\in\Bbb Z_2^n$</span>. Formally, we write</p>
<p><span class="math-container">$$\Bbb R^{\Bbb Z_2^n}=\{f\colon\{0,1\}^n\to\Bbb R\}$$</span></p>
<p>where we do not enforce any further conditions on <span class="math-container">$f\in\Bbb R^{\Bbb Z_2^n}$</span> (like, being a group homomorphism, being linear, etc. which we <em>could</em> ask for; but then we would get a notably different result).</p>
<hr />
<p>Naturally, this set is equipped with a <span class="math-container">$\Bbb R$</span>-vector space structure.</p>
<p>Indeed, define the <em>pointwise sum</em> of two elements <span class="math-container">$f,g\in\Bbb R^{\Bbb Z_2^n}$</span> by</p>
<p><span class="math-container">$$(f+g)(x):=f(x)+g(x)\quad\text{where}\quad x\in\Bbb Z_2^n\,.$$</span></p>
<p>This is again just a function with domain <span class="math-container">$\Bbb Z_2^n$</span> taking values in <span class="math-container">$\Bbb R$</span> , hence an element of <span class="math-container">$\Bbb R^{\Bbb Z_2^n}$</span>. We can define a trivial function sending everything to <span class="math-container">$0\in\Bbb R$</span> as neutral element and <span class="math-container">$g(x)=-f(x)$</span> is an additive inverse. Thus, this set has an (abelian) group structure inherited from <span class="math-container">$\Bbb R$</span> as one can check.</p>
<p>Also, as the functions are valued in <span class="math-container">$\Bbb R$</span>, we can define a new function <span class="math-container">$(\lambda\cdot f)(x)=\lambda\cdot f(x)$</span> for every <span class="math-container">$\lambda\in\Bbb R$</span>. One can check that the usual compability axioms hold making this set into a <span class="math-container">$\Bbb R$</span>-vector space.</p>
<hr />
<p>The functions <span class="math-container">$f_u(v)=\delta_{uv}$</span> (where <span class="math-container">$u,v\in\Bbb Z_2^n$</span>) form a basis for <span class="math-container">$V=\Bbb R^{\Bbb Z_2^n}$</span> as <span class="math-container">$\Bbb R$</span>-vector space as a function is uniquely and completely determined on its values at any point in its domain. The domain happens to be <span class="math-container">$\Bbb Z_2^n=\{0,1\}^n$</span> which has <span class="math-container">$2^n$</span> elements <span class="math-container">$u$</span>. Every one of them gives us a unique function <span class="math-container">$f_u$</span> as defined earlier and we can decompose a function <span class="math-container">$f\in\Bbb R^{\Bbb Z_2^n}$</span> into how it acts on every element, giving us the needed linear combination of the form
<span class="math-container">$$f=\sum_{u\in\Bbb Z_2^n} a_uf_u,\quad a_u\in\Bbb R$$</span></p>
|
403,184 | <p>A (non-mathematical) friend recently asked me the following question:</p>
<blockquote>
<p>Does the golden ratio play any role in contemporary mathematics?</p>
</blockquote>
<p>I immediately replied that I never come across any mention of the golden ratio in my daily work, and would guess that this is the same for almost every other mathematician working today.</p>
<p>I then began to wonder if I were totally correct in this statement . . . which has led me to ask this question.</p>
<p>My apologies is this question is unsuitable for Mathoverflow. If it is, then please feel free to close it.</p>
| ho boon suan | 118,745 | <p>Here's a chess problem. If one greedily places queens on an infinite <span class="math-container">$\mathbf{N}\times\mathbf{N}$</span> chessboard, column by column, such that at each step no two queens may attack one another, then one obtains a permutation of <span class="math-container">$\mathbf{N}$</span> (<a href="https://oeis.org/A065188" rel="noreferrer">OEIS A065188</a>): 1, 3, 5, 2, 4, 9, 11, 13, 15, 6, ..., and the queens appear to form two lines, of slopes <span class="math-container">$\phi$</span> and <span class="math-container">$1/\phi$</span>.
<a href="https://i.stack.imgur.com/iQyeP.png" rel="noreferrer"><img src="https://i.stack.imgur.com/iQyeP.png" alt="N. J. A. Sloane, Scatterplot of A065188(n)" /></a></p>
<p>While this has not been proven (it is conjectured that every term is either <span class="math-container">$n\phi+O(1)$</span> or <span class="math-container">$n/\phi+O(1)$</span>), a similar result for queens greedily placed along a square spiral on a <span class="math-container">$\mathbf{Z}\times\mathbf{Z}$</span> board was established in F. M. Dekking, J. Shallit, and N. J. A. Sloane, <a href="https://www.combinatorics.org/ojs/index.php/eljc/article/view/v27i1p52" rel="noreferrer"><em>Queens in Exile: Non-attacking Queens on Infinite Chess Boards</em></a> (2020), Electronic Journal of Combinatorics <strong>27</strong>, #P1.52. The queens there lie on lines with slopes related to the <em>Tribonacci constant</em> (<span class="math-container">$\approx1.83929$</span>), which is the real root of <span class="math-container">$x^3-x^2-x-1$</span>.</p>
|
403,184 | <p>A (non-mathematical) friend recently asked me the following question:</p>
<blockquote>
<p>Does the golden ratio play any role in contemporary mathematics?</p>
</blockquote>
<p>I immediately replied that I never come across any mention of the golden ratio in my daily work, and would guess that this is the same for almost every other mathematician working today.</p>
<p>I then began to wonder if I were totally correct in this statement . . . which has led me to ask this question.</p>
<p>My apologies is this question is unsuitable for Mathoverflow. If it is, then please feel free to close it.</p>
| László Kozma | 7,368 | <p>There are several exponential-time algorithms (in theoretical computer science, arguably a subfield of mathematics), whose running time can be bounded by an expression of the form <span class="math-container">${n-k \choose k}$</span> where <span class="math-container">$k$</span> is an integer between <span class="math-container">$0$</span> and <span class="math-container">$n$</span>. Plugging in the <span class="math-container">$k$</span> that maximizes this term, we get an upper bound <span class="math-container">$\sim \phi^n$</span> where <span class="math-container">$\phi = \frac{1+\sqrt{5}}{2}$</span>, the golden ratio. This is often in contrast to the naive approach whose running time would be <span class="math-container">$\sim 2^n$</span>.</p>
<p>See for example in Section 3.4 of the book: Exact Exponential Algorithms, by Fomin and Kratsch:
<a href="https://folk.uib.no/nmiff/BookEA/index.html" rel="noreferrer">https://folk.uib.no/nmiff/BookEA/index.html</a></p>
|
403,184 | <p>A (non-mathematical) friend recently asked me the following question:</p>
<blockquote>
<p>Does the golden ratio play any role in contemporary mathematics?</p>
</blockquote>
<p>I immediately replied that I never come across any mention of the golden ratio in my daily work, and would guess that this is the same for almost every other mathematician working today.</p>
<p>I then began to wonder if I were totally correct in this statement . . . which has led me to ask this question.</p>
<p>My apologies is this question is unsuitable for Mathoverflow. If it is, then please feel free to close it.</p>
| coudy | 6,129 | <p>The golden L is a translation surface that has received much attention recently in the theory of billiard dynamics. It is built out of a <span class="math-container">$1 \times 1$</span> square with <span class="math-container">$1 \times \phi$</span> and <span class="math-container">$\phi\times1$</span> rectangles glued to it along their length-<span class="math-container">$1$</span> sides. See for example the preprint <a href="https://arxiv.org/abs/1810.11310" rel="nofollow noreferrer">Davis and Lelievre - Periodic paths on the pentagon, double pentagon and golden L</a>.</p>
|
403,184 | <p>A (non-mathematical) friend recently asked me the following question:</p>
<blockquote>
<p>Does the golden ratio play any role in contemporary mathematics?</p>
</blockquote>
<p>I immediately replied that I never come across any mention of the golden ratio in my daily work, and would guess that this is the same for almost every other mathematician working today.</p>
<p>I then began to wonder if I were totally correct in this statement . . . which has led me to ask this question.</p>
<p>My apologies is this question is unsuitable for Mathoverflow. If it is, then please feel free to close it.</p>
| James Propp | 3,621 | <p>In the active field of research called symbolic dynamics (for a nice overview see <a href="https://www.southalabama.edu/mathstat/personal_pages/williams/wilshort.pdf" rel="nofollow noreferrer">https://www.southalabama.edu/mathstat/personal_pages/williams/wilshort.pdf</a> by Susan Williams), the canonical example of a subshift of finite type that is not isomorphic to a full shift is the “golden mean subshift”, so called because its entropy is the log of the golden ratio.</p>
|
2,394,815 | <p>Let $A$ be a (not necessarily finitely generated) abelian group where all elements have order 1, 2, or 4. Does it follow that $A$ can be written as a direct sum $(\bigoplus _\alpha \mathbb Z/4) \oplus (\bigoplus_\beta \mathbb Z/2)$?</p>
| Eric Wofsey | 86,856 | <p>Yes. Note that such an abelian group is the same thing as a module over the ring <span class="math-container">$\mathbb{Z}/4$</span>. Note also that any direct sum of copies of <span class="math-container">$\mathbb{Z}/4$</span> is injective as a module over <span class="math-container">$\mathbb{Z}/4$</span> (this is easy to check by Baer's criterion, since the only nontrivial proper ideal in <span class="math-container">$\mathbb{Z}/4$</span> is <span class="math-container">$(2)$</span>). Now let <span class="math-container">$A$</span> be a <span class="math-container">$\mathbb{Z}/4$</span>-module. If <span class="math-container">$A$</span> has an element of order <span class="math-container">$4$</span>, that element generates a submodule isomorphic to <span class="math-container">$\mathbb{Z}/4$</span>. Since <span class="math-container">$\mathbb{Z}/4$</span> is an injective module, <span class="math-container">$A$</span> splits as a direct sum <span class="math-container">$\mathbb{Z}/4\oplus A'$</span> for some submodule <span class="math-container">$A'\subset A$</span>. If <span class="math-container">$A'$</span> has an element of order <span class="math-container">$4$</span>, we can split off a direct summand of <span class="math-container">$\mathbb{Z}/4$</span> from it, and so on.</p>
<p>Repeating this process by transfinite induction until there are no elements of order <span class="math-container">$4$</span> left, we can write <span class="math-container">$A$</span> as a direct sum <span class="math-container">$B\oplus C$</span> where <span class="math-container">$B$</span> is a direct sum of copies of <span class="math-container">$\mathbb{Z}/4$</span> and <span class="math-container">$C$</span> has no elements of order <span class="math-container">$4$</span>. But then every element of <span class="math-container">$C$</span> has order <span class="math-container">$1$</span> or <span class="math-container">$2$</span>, so <span class="math-container">$C$</span> is a <span class="math-container">$\mathbb{Z}/2$</span>-vector space. Thus <span class="math-container">$C$</span> is a direct sum of copies of <span class="math-container">$\mathbb{Z}/2$</span>, and <span class="math-container">$A=B\oplus C$</span> is the direct sum decomposition you ask for.</p>
|
2,394,815 | <p>Let $A$ be a (not necessarily finitely generated) abelian group where all elements have order 1, 2, or 4. Does it follow that $A$ can be written as a direct sum $(\bigoplus _\alpha \mathbb Z/4) \oplus (\bigoplus_\beta \mathbb Z/2)$?</p>
| Steve D | 265,452 | <p>What I said in the comments is true, and goes by the name</p>
<p><strong>Prüfer's First Theorem</strong>: An abelian $p$-group $G$ with bounded exponent (an integer $k$ such that $g^k=1$ for all $g\in G$) is a direct sum of cyclic subgroups.</p>
<p>The proof is by induction on $k=p^e$, the base case $e=1$ being the vector-space case.</p>
<p>For the inductive step, write $pG=\oplus_\alpha\langle g_\alpha\rangle$, and choose $h_\alpha$ in $G$ with $ph_\alpha=g_\alpha$. Then the $h_\alpha$ generate a subgroup of $G$ (call it $H$) that is a direct sum of $\langle h_\alpha\rangle$. Let $L$ be a subgroup of $G$, maximal with respect to having trivial intersection with $H$. Then $L$ is also a direct sum of cyclic subgroups (by the vector-space case), and you can show $G=H\oplus L$.</p>
<p>Reference: <em>Fundamentals of the Theory of Groups</em>, by Kargapolov and Merzljakov, $\S$10.</p>
|
2,556,339 | <p>This is the function $f(x)$$=\frac{1}{\sqrt{3x-2}}$ .
I wrote that $$\lim_{h\to 0}\frac{\frac{\sqrt{3x+3h-2}}{3x+3h-2}-\frac{\sqrt{3x-2}}{3x-2}}{h}.$$
I am not able to continue further.</p>
| Jack | 510,857 | <p>Let $a = \sqrt{3x+h-2}$ and $b = \sqrt{3x-2}$, then $a^2 - b^2 = h$ which means $h \to 0$ as $a \to b$. So we've:</p>
<p>$$ \lim_{a \to b} \frac{1/a-1/b}{a^2-b^2} = \lim_{a \to b} \frac{b-a}{(ab)(a+b)(a-b)} = - \lim_{a \to b} \frac{1}{ab(a+b)} = -\frac{1}{2 b^3} = -\frac{1}{2\sqrt{(3x-2)^3}.} $$</p>
|
3,407,852 | <p>A space X is said to be h-homogeneous if
every non-empty clopen subset of <span class="math-container">$X$</span> is homeomorphic to <span class="math-container">$X.$</span></p>
<p>Is the space <span class="math-container">$L = 2^{\mathbb N} - \{p\}$</span> for <span class="math-container">$p \in 2^{\mathbb N}$</span> h-homogeneous?</p>
| Ross Millikan | 1,827 | <p>Compute the number of squares, cubes, and fifth powers. Now note that you have counted the sixth, tenth, and fifteenth powers twice each, so subtract them. You have counted the thirtieth powers three times and subtracted them three times, so add them once more.</p>
|
2,601,601 | <p>Consider the complex matrix $$A=\begin{pmatrix}i+1&2\\2&1\end{pmatrix}$$ and the linear map $$f:M(2,\mathbb{C})\to M(2,\mathbb{C}),\qquad X\mapsto XA-AX.$$</p>
<p>I want to find a basis of $\ker f$.</p>
<p>I already know the canonical basis $\{E_{11},E_{12},E_{21},E_{22}\}$ and computed $$f(E_{11})=\begin{pmatrix}0&2\\-2&0\end{pmatrix},f(E_{12})=\begin{pmatrix}2&0\\0&-2\end{pmatrix},f(E_{21})=\begin{pmatrix}-2&0\\0&2\end{pmatrix},f(E_{22})=\begin{pmatrix}0&-2\\2&0\end{pmatrix}$$</p>
<p>Does this help to find the basis?</p>
| Cameron Buie | 28,900 | <p>It does. It means that for an arbitrary matrix $$X=\begin{pmatrix}a&b\\c&d\end{pmatrix},$$ we have $$f(X)=af(E_{11})+bf(E_{12})+cf(E_{21})+df(E_{22}),$$ or $$f(X)=(a-d)f(E_{11})+(b-c)f(E_{12}).$$ Thus, we have $f(X)$ is the zero matrix if and only if...what?</p>
|
3,523,213 | <p>I've read some simple explanations of Cantor's diagonal method.</p>
<p>It seems to be:</p>
<pre><code>1) Changing the i-th value in a row.
2) Do the same to the next row with the (i+1)th element.
3) Now you get an element not in any other row. So add it to list.
4) This process never ends.
</code></pre>
<p>This looks very like induction since it uses the (n+1) trick.</p>
<p>However, induction only works for finite numbers.</p>
<p>And the row lengths are not finite.</p>
<p>So how did Cantor get around this?</p>
| Vsotvep | 176,025 | <p>Cantor's diagonal proof is <strong>not</strong> infinite in nature, and neither is a proof by induction an infinite proof. </p>
<hr>
<p>For Cantor's diagonal proof (I'll assume the variant where we show the set of reals between <span class="math-container">$0$</span> and <span class="math-container">$1$</span> is uncountable), we have the following claims:</p>
<ul>
<li>Every real number (between <span class="math-container">$0$</span> and <span class="math-container">$1$</span>) has a representation as an infinite string <span class="math-container">$0.d_0d_1d_2d_3\cdots$</span>, where each <span class="math-container">$d_i$</span> is a digit, i.e. <span class="math-container">$0$</span>, <span class="math-container">$1$</span>, <span class="math-container">$2$</span>, ... or <span class="math-container">$9$</span>.</li>
<li>If the real numbers were countable, then we can make a function such that each real is assigned some natural number <span class="math-container">$n\in\Bbb N$</span> and vice versa, also known as a countable list of real numbers.</li>
<li>It is possible to define a real that is not assigned any natural number: we define the <span class="math-container">$i$</span>th digit of our new real to be <span class="math-container">$5$</span> if the <span class="math-container">$i$</span>th digit of the real number associated with the natural number <span class="math-container">$i$</span> is unequal to <span class="math-container">$5$</span>, and otherwise we define the <span class="math-container">$i$</span>th digit to be <span class="math-container">$6$</span>. Note that we define this real all at once. We do not need to know the construction of earlier steps, so we do not use induction.</li>
<li>This new real is not part of our list, since for any number of the list, there is some digit on which it disagrees.</li>
</ul>
<p>None of these steps need you to do an infinite process. The only infinite thing we do is that we work with infinite sets of elements. </p>
<hr>
<p>Induction in itself (which is unrelated to Cantor's diagonal argument) is also not an infinite process. To show that a property <span class="math-container">$\phi$</span> holds for all natural numbers, we prove that <span class="math-container">$\phi$</span> holds for <span class="math-container">$0$</span>, and that if <span class="math-container">$\phi$</span> holds for <span class="math-container">$n$</span>, then it holds for <span class="math-container">$n+1$</span>. These are two finite proofs.</p>
<p>Then, to show that the property holds for <em>every</em> natural number, we pick an arbitrary number <span class="math-container">$m$</span>. If we can show the property holds for <span class="math-container">$m-1$</span>, then it must hold for <span class="math-container">$m$</span>, so we try to show it holds for <span class="math-container">$m-1$</span>. If it holds for <span class="math-container">$m-2$</span>, then it must hold for <span class="math-container">$m-1$</span>, so we try to show it holds for <span class="math-container">$m-2$</span>. Et cetera, until we reach <span class="math-container">$m-m=0$</span>. This is the base case, which we have already proved.</p>
<p>Therefore, we can show that the property holds for all natural numbers, since for any number, we only have to work back a finite number of steps to get to the base case. </p>
<hr>
<p>As a sidenote, induction can be used on any <a href="https://en.wikipedia.org/wiki/Well-founded_relation" rel="nofollow noreferrer">well-founded class</a>, regardless of it being infinite or finite. </p>
<p>The trick is not that you show some property of interest for every element individually (which does indeed cost an infinite number of steps), but that you show that the property of an individual depends only on its predecessors having the property. You show "if all predecessors have the property, then this individual has the property". </p>
<p>If we then want to prove that an individual does not have the property, you would need to be able to find a predecessor that does not have the property.</p>
<p>In a well-founded set, there are no infinitely descending chains, so for any predecessor that is not the base case, we must be able to find a predecessor of this predecessor that does not have the property. We only can do this finitely many times, since a well-founded set has no infinitely descending chains. </p>
<p>It follows that eventually we will reach the base case, and for the base case we (presumably) have already proved that it has the property. Therefore our finite chain of predecessors must all have the property, which means our original element of interest must have the property.</p>
|
2,332,419 | <p>What's the angle between the two pointers of the clock when time is 15:15? The answer I heard was 7.5 and i really cannot understand it. Can someone help? Is it true, and why?</p>
| Steven Alexis Gregory | 75,410 | <p>Lets assume we have a 12 hour clock and let 1200 hrs be 0 degrees and lets measure angles clockwise from there. At 1500 hrs, the hour hand is exactly on the three, that is at 90 degrees.</p>
<p>It takes the hour hand 12 hours to travel 360 degrees. That comes to
$\dfrac{360^\circ}{12 \text{hrs}} = 30 \frac{\text{degrees}}{\text{hr}}$</p>
<p>So, at 1515 hrs, $\dfrac 14$ hr later, the minute hand will be at 90 degrees and the the hour hand will have moved
$\left(\dfrac 14 \text{hr} \right)
\left(30 \frac{\text{degrees}}{\text{hr}}\right) = 7.5 \, \text{degrees}$.</p>
<p>So the angle between the two hands will be 7.5 degrees.</p>
|
3,745,551 | <p>I often see people say that if you have 2 IID gaussian RVs, say <span class="math-container">$X \sim \mathcal{N}(\mu_x, \sigma_x^2)$</span> and <span class="math-container">$Y \sim \mathcal{N}(\mu_y, \sigma_y^2)$</span>, then the distribution of their sum is <span class="math-container">$\mathcal{N}(\mu_x + \mu_y, \sigma_x^2 + \sigma_y^2)$</span>.</p>
<p>This is only true when <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> have the same units right, otherwise you can't even sum them to begin with without standardization?</p>
<p>e.g., if <span class="math-container">$X$</span> was some measure of distance in meters and <span class="math-container">$Y$</span> was some measure of velocity in <span class="math-container">$\frac{meters}{second}$</span>, then you can't simply just add their means and variances together. That wouldn't make sense. You'd have to standardize them first so they're both unitless before you can do the above.</p>
| Kavi Rama Murthy | 142,385 | <p>If <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> are independent , <span class="math-container">$X \sim N(\mu_X,\sigma_X^{2})$</span> and <span class="math-container">$Y \sim N(\mu_Y,\sigma_Y^{2})$</span> then <span class="math-container">$X+Y \sim N(\mu_X+\mu_Y,\sigma_X^{2}+\sigma_Y^{2})$</span>. [IID would force <span class="math-container">$\mu_X=\mu_Y, \sigma_X=\sigma_Y$</span> which is not required here. Independence is enough]. There is no need to standardize the random variables for this.</p>
<p>[There are many ways of proving this and one way is to use characteristic functions. Use the fact that <span class="math-container">$Ee^{it(X+Y)}=Ee^{itX}Ee^{itY}$</span> and you would be able to supply a proof].</p>
|
253,271 | <p>I recently found that <code>\[Gradient]</code> and <code>\[InlinePart]</code> both expand (contract) to special symbols in MMA.</p>
<p>So far as I can tell (see <a href="https://mathematica.stackexchange.com/questions/134506/inlinepart-what-is-it-and-what-happened-to-it">InlinePart. What is it and what happened to it?</a>) they have no built in meaning. Further, I can't find any Wolfram documentation of them, even in <a href="https://reference.wolfram.com/language/guide/ListingOfNamedCharacters.html" rel="nofollow noreferrer">https://reference.wolfram.com/language/guide/ListingOfNamedCharacters.html</a>.</p>
<p>Is there some introspection that can list all special symbols that expand like <code>\[stuff]</code>? The same goes for symbols entered as <code>\[AliasDelimiter]stuff\[AliasDelimiter]</code>.</p>
<hr />
<p>Here's what the symbols look like</p>
<p><a href="https://i.stack.imgur.com/ZCKJZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZCKJZ.png" alt="symbols" /></a></p>
<hr />
<p>OK the second part of the question concerns <code>InputAlias</code>s, which don't seem to be collected in any one place, but are sprinkled throughout the filesystem, for instance, in Mathematica/SystemFiles/FrontEnd/TextResources/CommonFrontEndInit.tr</p>
<p>This question, then, amounts to basically lamenting that no method so far lists <em>all</em> of the existing glyphs. So far there's</p>
<ul>
<li>the Wolfram ListingOfNamedCharacters, missing many</li>
<li>UnicodeCharacters.tr, missing a bit fewer glyphs</li>
<li>Brute forcing with <code>"PrintableASCII"</code> export of every char code, missing only a couple</li>
</ul>
| Syed | 81,355 | <p>The list of keyboard shortcuts for the Wolfram language and System is <a href="https://reference.wolfram.com/language/tutorial/KeyboardShortcutListing.html" rel="nofollow noreferrer">here</a>.</p>
<p>The list of named characters for the Wolfram language and System is <a href="https://reference.wolfram.com/language/guide/ListingOfNamedCharacters.html" rel="nofollow noreferrer">here</a>.</p>
<p>For system names, try: <code>Names["System`*"]</code> (* edit suggested by Adam Thanks *)</p>
<p>---------- edit---- More digging for programmatic retrieval of special characters.</p>
<p><a href="https://mathematica.stackexchange.com/questions/104299/how-could-get-all-escape-characters-in-mathematica">@LouisB</a> has a solution here.</p>
<p><a href="https://mathematica.stackexchange.com/questions/7610/list-of-mathematica-glyphs">@george2017</a> posted a solution here 9 years ago.</p>
<p>With some effort:</p>
<p><a href="https://i.stack.imgur.com/dS97t.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dS97t.png" alt="enter image description here" /></a></p>
<p>I hope this answers your question.</p>
|
393,467 | <p>I am looking for a proof that:</p>
<p>if <span class="math-container">$A_{11}A_{12}...A_{1n}$</span>; <span class="math-container">$A_{21}A_{22}...A_{2n}$</span>; <span class="math-container">$\cdots$</span>; <span class="math-container">$A_{i1}A_{i2}...A_{in}$</span>; <span class="math-container">$\cdots$</span>; <span class="math-container">$A_{m1}A_{m2}...A_{mn}$</span> are <span class="math-container">$m$</span> oriented regular polygons (<span class="math-container">$n$</span>-gons), where <span class="math-container">$n=2k$</span>, then<span class="math-container">$\DeclareMathOperator\Area{Area}$</span>
<span class="math-container">$$
\begin{align*}
& \Area(A_{11}A_{21}...A_{m1})+\Area(A_{1\;k+1}A_{2\;k+1}...A_{m\;k+1})\\ =\ & \Area(A_{12}A_{22}...A_{m2})+\Area(A_{1\;k+2}A_{2\;k+2}...A_{m\;k+2})\\=\ & \cdots\\ =\ & \Area(A_{1i}A_{2i}...A_{mi})+\Area(A_{1\;k+i}A_{2\;k+i}...A_{m\;k+i})\\ =\ & \cdots
\end{align*}
$$</span></p>
<p><strong>Reference:</strong></p>
<ul>
<li><p><a href="http://www.xente.mundo-r.com/ilarrosa/GeoGebra/AreasIg_Npoligonos.html" rel="nofollow noreferrer">Areas que suman lo mismo</a></p>
</li>
<li><p>A case <span class="math-container">$m=4$</span> and <span class="math-container">$n=4$</span>, I posed in <a href="https://math.stackexchange.com/questions/1447773/two-conjectures-of-four-squares">here</a> is near six year ago. But no have a proof.</p>
</li>
</ul>
| Algernon | 23,297 | <p>To complement <a href="https://mathoverflow.net/a/393477/23297">Iiro Ullin</a>'s answer, we have the following inequality in one direction:</p>
<p><strong>Lemma</strong> (<a href="https://doi.org/10.1109/TIT.2005.864431" rel="nofollow noreferrer" title="I. Csiszar and Z. Talata,Context tree estimation for not necessarily finite memory processes, via BIC and MDL,">László Györfi</a>). If <span class="math-container">$p$</span> and <span class="math-container">$q$</span> are probability densities both supported on a bounded interval <span class="math-container">$I$</span>, then we have
<span class="math-container">$$D_{\textrm{KL}}(p,q)\leq\frac{1}{\inf_{x\in I}q(x)}\|p-q\|_2^2$$</span></p>
<p><em>Proof</em>.
<span class="math-container">\begin{align}
D_{\textrm{KL}}(p,q) &= \int_I p(x)\log\frac{p(x)}{q(x)}\mathrm{d}x \\
&\leq \int_I p(x)\left(\frac{p(x)}{q(x)}-1\right)\mathrm{d}x \\
&= \int_I \frac{\big(p(x)-q(x)\big)^2}{q(x)}\mathrm{d}x \;,
\end{align}</span>
from which the claim follows. <span class="math-container">$\square$</span></p>
|
393,467 | <p>I am looking for a proof that:</p>
<p>if <span class="math-container">$A_{11}A_{12}...A_{1n}$</span>; <span class="math-container">$A_{21}A_{22}...A_{2n}$</span>; <span class="math-container">$\cdots$</span>; <span class="math-container">$A_{i1}A_{i2}...A_{in}$</span>; <span class="math-container">$\cdots$</span>; <span class="math-container">$A_{m1}A_{m2}...A_{mn}$</span> are <span class="math-container">$m$</span> oriented regular polygons (<span class="math-container">$n$</span>-gons), where <span class="math-container">$n=2k$</span>, then<span class="math-container">$\DeclareMathOperator\Area{Area}$</span>
<span class="math-container">$$
\begin{align*}
& \Area(A_{11}A_{21}...A_{m1})+\Area(A_{1\;k+1}A_{2\;k+1}...A_{m\;k+1})\\ =\ & \Area(A_{12}A_{22}...A_{m2})+\Area(A_{1\;k+2}A_{2\;k+2}...A_{m\;k+2})\\=\ & \cdots\\ =\ & \Area(A_{1i}A_{2i}...A_{mi})+\Area(A_{1\;k+i}A_{2\;k+i}...A_{m\;k+i})\\ =\ & \cdots
\end{align*}
$$</span></p>
<p><strong>Reference:</strong></p>
<ul>
<li><p><a href="http://www.xente.mundo-r.com/ilarrosa/GeoGebra/AreasIg_Npoligonos.html" rel="nofollow noreferrer">Areas que suman lo mismo</a></p>
</li>
<li><p>A case <span class="math-container">$m=4$</span> and <span class="math-container">$n=4$</span>, I posed in <a href="https://math.stackexchange.com/questions/1447773/two-conjectures-of-four-squares">here</a> is near six year ago. But no have a proof.</p>
</li>
</ul>
| Mark | 101,207 | <p>This is probably obvious, but just wanted to explicitly mention the following.
It is perhaps more natural to look at the modified <span class="math-container">$L_2$</span> norm <span class="math-container">$L_2'(p,q) = L_2(\sqrt{p}, \sqrt{q})$</span>. Note that probability distributions (which are a subset of unit vectors for the <span class="math-container">$L_1$</span> norm) are unit vectors for this modified norm (but need not be unit vectors for the <span class="math-container">$L_2$</span> norm).
Note that <span class="math-container">$L_2'$</span> shares many properties with <span class="math-container">$L_2$</span>, for example despite the additional <span class="math-container">$\sqrt{\cdot}$</span> terms it is still a metric.</p>
<p>Anyway, this modified norm is precisely the Hellinger distance, typically notated <span class="math-container">$H(p,q) := L_2(\sqrt{p}, \sqrt{q})$</span>.
The inequality <span class="math-container">$H(p,q)^2 \leq D_{KL}(p,q)$</span> is then known, and can be seen as a refinement of Pinsker's inequality (as <span class="math-container">$\delta_{TV}(p,q) \leq H(p,q)$</span> is also known --- I might be missing some small constant factors in some definitions here).</p>
|
1,252,167 | <p>I'm trying to understand what a vector of functions is, from trying to understand how to solve linear homogeneous differential equations. </p>
<p>It seems that functions can be manipulated as vectors as long as they are not interpreted as having real values.<br>
Suppose the solution space of a linear homogeneous diff equation is spanned by $\cos(x)$, $\sin(x)$, then the solution is $a \sin(x) + b \cos(x)$, and it's a vector.</p>
<p>But, if $y$ is a vector $y = a \sin(x) + b \cos(x)$, then how is it that for any value of $x$, $y$ is always a scalar value?</p>
<p>If $x$ is a set of values rather than a symbol then how can $y$ remain a vector if, for each element of $x$, $y$ is scalar?</p>
| triple_sec | 87,778 | <p>I'm not sure if it helps you with your original question, but let me try proposing an alternative concept capturing the intuitive notion of irredundancy: for each point, take the “smallest” open set in the cover containing that point.</p>
<p>Formally, let $\{U_{\lambda}\}_{\lambda\in\Lambda}$ be an open cover of the topological space $X$, where $\Lambda$ is a non-empty index set. Now endow $\Lambda$ with a well-ordering $\succsim$. For each point $x\in X$, let $\lambda_x$ be the least element of $\{\lambda\in\Lambda\,|\,x\in U_{\lambda}\}$
according to the well-ordering. Then, $\{U_{\lambda_x}\}_{x\in X}$ is a refined subcover of $\{U_{\lambda}\}_{\lambda\in\Lambda}$.</p>
|
1,824,966 | <p>Ok, I was asked this strange question that I can't seem to grasp the concept of..</p>
<blockquote>
<p>Let $T$ be a linear transformation such that:
$$T \langle1,-1\rangle = \langle 0,3\rangle \\
T \langle2, 3\rangle = \langle 5,1\rangle $$
Find $T$.</p>
</blockquote>
<p>Is there suppose to be a function out of this? A matrix of some kind? Maybe both? If so, what is it?</p>
| Mnifldz | 210,719 | <p>Yes, $T$ is a matrix of some kind. You can tell that it is $2\times 2$ since both its inputs and outputs are vectors of length $2$. My recommendation for solving this is the following. Suppose $T$ has matrix representation</p>
<p>$$
T \;\; =\;\; \left [ \begin{array}{cc}
a & b \\
c & d \\
\end{array} \right ].
$$</p>
<p>Start with the first equation which you can write as </p>
<p>$$
\left [ \begin{array}{cc}
a & b \\
c & d \\
\end{array} \right ] \left [ \begin{array}{c}
1 \\
-1\\
\end{array} \right ] \;\; =\;\; \left [ \begin{array}{c}
0 \\
3\\
\end{array} \right ].
$$</p>
<p>You can find the values of $a,b,c,d$ by working through these equations. </p>
|
3,663,526 | <p><span class="math-container">$F(x)=\begin{cases} x^3+5, & x\ge 1\\ x^3+2, & 0\leq x<1\\ x^3, & x<0 \end{cases}$</span></p>
<p>Let <span class="math-container">$\mu_{F}$</span> be the Lebesgue-Stieltjes measure associated with <span class="math-container">$F$</span>. Find the lebesgue decomposition of <span class="math-container">$\mu_{F}$</span> with respect to the Lebesgue measure <span class="math-container">$m$</span>.</p>
<p><span class="math-container">$\frac{dF(x)}{dx}=3x^2$</span>. It has jump of height 2 at <span class="math-container">$x=0$</span> and has jump of height 3 at <span class="math-container">$x=1$</span>. </p>
<p>My answer is <span class="math-container">$\mu_{F}=\mu_{a}(E)+\mu_{d}(E)=\int_{E} 3x^2dm(x)+2\delta_{0}+3\delta_{1}$</span>, where <span class="math-container">$\mu_{a}<<m$</span>, and <span class="math-container">$\mu_{a}\perp\mu_{d}$</span>.</p>
<p>I am asking because I do not quite understand the concept of Lebesgue decomposition. Are there any flaws in my analysis? I asked similar questions here before, but no one answers.</p>
| Ng Tze Beng | 866,955 | <p>F is an increasing function, so you have the Lebesgue decomposition for the positive Lebesgue Stieltjes measure. See for example Theorem 8 of <em>Complex Measure, Dual Space of L
p Space, Radon-Nikodym Theorem and Riesz Representation Theorems</em>. (<a href="https://my-calculus-web.firebaseapp.com/MA3110/Complex%20Measure_Dual_space_Riesz_Thm.pdf" rel="nofollow noreferrer">https://my-calculus-web.firebaseapp.com/MA3110/Complex%20Measure_Dual_space_Riesz_Thm.pdf</a>). The proof there is a general existence proof. For Lebesgue Stieltjes measure, specifically we have a more detail decomposition: for the function F here, we can write it as a sum of an increasing continuous function <i>x</i><sup>3</sup> and an obvious Saltus function. Note that <i>x</i><sup>3</sup> is not absolutely continuous on the set of real numbers but is locally absolutely continuous. Taking the domain as a bounded interval, from Theorem 19 of <em>Lebesgue Stieltjes Measure, de La Vallée Poussin’s Decomposition,
Change of Variable, Integration by Parts for Lebesgue Stieltjes Integrals</em> (<a href="https://my-calculus-web.firebaseapp.com/MA3110/Lebesgue%20Stieltjes%20Measure%20and%20de%20La%20Vallee%20Poussin%20Decomposition.pdf" rel="nofollow noreferrer">https://my-calculus-web.firebaseapp.com/MA3110/Lebesgue%20Stieltjes%20Measure%20and%20de%20La%20Vallee%20Poussin%20Decomposition.pdf</a>), appropriate interpreted, we have a Lebesgue decomposition on any bounded interval. By Passing the domain to the whole of the reals we have the Lebesgue decomposition. The explicit form of the decomposition is given by Theorem 22 of <em>Lebesgue Stieltjes Measure, de La Vallée Poussin’s Decomposition, Change of Variable, Integration by Parts for Lebesgue Stieltjes Integrals</em>, where we can ignore the contributions from the points having infinite derivative, since there are none. Note that the absolute part is given by the integral of the derivative of <i>x</i><sup>3</sup>. The contribution from the saltus part of the function F is given by the sum of the jumps at 0 and 1 which is the same as the sum of the jumps at all points of a Borel set and so are the sum of the function given by you, 2 times the characteristic function of {0} plus 3 times the characteristic function at {1}. Note that Theorem 19 of <em>Lebesgue Stieltjes Measure, de La Vallée Poussin’s Decomposition,
Change of Variable, Integration by Parts for Lebesgue Stieltjes Integrals</em> says that these two parts are mutually singular. Proposition 12 of <em>Lebesgue Stieltjes Measure, de La Vallée Poussin’s Decomposition,Change of Variable, Integration by Parts for Lebesgue Stieltjes Integrals</em> affirms that the absolute part of the Lebesgue Stieltjes measure of F, is absolutely continuous with respect to the Lebesgue measure on the reals since <i>x</i><sup>3</sup> is a locally absolutely continuous function on the reals. Hope the references cited here would be useful.</p>
|
85,717 | <p>Nowadays we can associate to a topological space $X$ a category called the fundamental (or Poincare) $\infty$-groupoid given by taking $Sing(X)$.</p>
<p>There are many different categories that one can associate to a space $X$. For example, one could build the small category whose object set is the set of points with only the identity morphisms from a point to itself. It is claimed that the classifying space of this category returns the space: $BX=X$</p>
<p>The inspiration for these examples comes from three primary sources: Graeme Segal's famous 1968 paper <em>Classifying Spaces and Spectral Sequences</em>, Raoul Bott's Mexico notes (taken by Lawrence Conlon) <em>Lectures on characteristic classes and foliations</em>, and a 1995 pre-print called <em>Morse Theory and Classifying Spaces</em> by Ralph Cohen, G. Segal and John Jones. </p>
<p>In each of these papers there is a notion of a topological category. It is not just a category enriched in <strong>Top</strong>, since the set of objects can have non-discrete topology. Here is the definition that I can gleam from these articles:</p>
<p>A <strong>topological category</strong> consists of a pair of spaces $(Obj,Mor)$ with four continuous structure maps:</p>
<ul>
<li>$i:Obj\to Mor$, which sends an object to the identity morphism</li>
<li>$s:Mor\to Obj$, which gives the source of an arrow</li>
<li>$t:Mor\to Obj$, which gives the target of an arrow</li>
<li>$\circ:Mor\times_{t,s}Mor\to Mor$, which is composition.</li>
</ul>
<p>Were $i$ is a section of both $s$ and $t$, and all the axioms of a small category hold.</p>
<p><strong>Is the appropriate modern terminology to describe this a <a href="http://ncatlab.org/nlab/show/Segal+space" rel="nofollow noreferrer">Segal Space</a>? What would Lurie call it?</strong> Based on reading <a href="https://mathoverflow.net/questions/29728/a-model-category-of-segal-spaces">Chris Schommer-Pries MO post</a> and elsewhere this seems to be true. Would the modern definition of the above be a Segal Space where the Segal maps are identities? Also, why do we demand that the topology on objects be discrete for Segal Categories? <strong>Is there something wrong with allowing the object sets to have topologies?</strong></p>
| Peter May | 14,447 | <p>Could I ask young people to use precise language? Calling a Kan complex an $\infty$-groupoid
and asking what kind of category it is just jars. It feels so pointless (I'm toning down the language I'm tempted to use). As Tony pointed out, a topological category in the proposed sense is just a category internal to topological spaces. The notion of internal category is so familiar and elementary that it must long antecede any reference made in the question. (It seemed an old notion when I was using it in the early 1970s). A Segal space (original version) is a covariant functor from the category $\mathcal{F}$ of finite based spaces (the opposite of Segal's category $\Gamma$) to the category of based spaces. It is not a kind of category. Similarly, a Segal category is a functor from $\mathcal{F}$ to Cat. There is a forgetful functor from Segal spaces to simplicial spaces (simplicial objects in spaces). Parenthetically, the terms topological category and simplicial category are both ambiguous since, without clarification, they could mean either categories enriched in spaces or in simplicial sets, or they could mean categories internal to spaces or to simplicial sets. A category internal to simplicial sets is the same notion as a simplicial object in Cat, so the consistent meaning would be the internal one, but the more standard usage is that a simplicial category is a simplicially enriched category.</p>
|
174,573 | <p>I am generating a random matrix $F$ and then plotting norm of the matrices $E+Ft$, representing them as a table. But I also want to print the matrix $F$ near every plot. If I remove the semicolon from <code>F // MatrixForm;</code> in the code below I get some errors.</p>
<p>My code:</p>
<pre><code>Dimension := 3;
Iterations := 2;
Table[
M = RandomComplex[{0, 20}, {Dimension, Dimension}];
F = UpperTriangularize[M, 1];
F // MatrixForm;
Plot[
Norm[
Inverse[IdentityMatrix[Dimension] + t*F]
],
{t, 0, 1}
],
{n, 1, Iterations}
]
</code></pre>
| kglr | 125 | <p><strong>Update:</strong> <code>f4</code> and <code>f5</code> should work as is in version 8. Variants of <code>f1</code>, <code>f2</code> and <code>f3</code> that avoid the functions and forms that are only available in later versions are: </p>
<pre><code>ClearAll[f1b, f2b, f3b]
f1b[l_List, {p__}] := Boole[MatchQ[#, {p, ___}] & /@ l]
f2b[l_List, p_List] := Boole[MatchQ[#[[;; Min[Length@#, Length@p]]], p] & /@ l]
f3b[l_List, p_List] := Boole[Equal[#[[;; Min[Length@#, Length@p]]], p] & /@ l]
b = {{0}, {1, 0, 1, 1, 1, 1, 1}, {0, 0, 0}, {0, 0, 1}, {0, 1, 0}, {0, 1, 1},
{1, 0, 0}, {1, 0, 1}, {1, 1, 0}, {1, 1, 1}};
f1b[b, {0}]
</code></pre>
<blockquote>
<p>{1, 0, 1, 1, 1, 1, 0, 0, 0, 0}</p>
</blockquote>
<pre><code>f1b[b, {1, 0}]
</code></pre>
<blockquote>
<p>{0, 1, 0, 0, 0, 0, 1, 1, 0, 0}</p>
</blockquote>
<pre><code>Equal @@ (#[b, {0}] & /@ {f1b, f2b, f3b})
</code></pre>
<blockquote>
<p>True</p>
</blockquote>
<p><strong>Original answer:</strong></p>
<pre><code>ClearAll[f1, f2, f3, f4, f5]
f1[l_List, {p__}] := Boole[MatchQ[{p, ___}] /@ l]
f2[l_List, p_List] := Boole[MatchQ[p] /@ l[[All, ;; UpTo@Length@p]]]
f3[l_List, p_List] := Boole[EqualTo[p] /@ l[[All, ;; UpTo @ Length@p]]]
f4[a_, {p__}] := Block[{f}, f[{p, ___}] = 1; f[_] := 0; f /@ a];
f5[a_, {p__}] := Replace[a, {{p, ___} :> 1, {__} :> 0}, 1]
f1[a, {0}]
</code></pre>
<blockquote>
<p>{1, 1, 1, 1, 0, 0, 0, 0}</p>
</blockquote>
<pre><code>f1[a, {1, 0}]
</code></pre>
<blockquote>
<p>{0, 0, 0, 0, 1, 1, 0, 0}</p>
</blockquote>
<pre><code> f5[a, {0}] == f4[a, {0}] == f3[a, {0}] == f2[a, {0}] == f1[a, {0}]
</code></pre>
<blockquote>
<p>True</p>
</blockquote>
<pre><code>f3[a, {1, 0}] == f2[a, {1, 0}] == f1[a, {1, 0}]
</code></pre>
<blockquote>
<p>True</p>
</blockquote>
<pre><code>b= {{0}, {1, 0, 1, 1, 1, 1, 1}, {0, 0, 0}, {0, 0, 1}, {0, 1, 0},
{0, 1, 1}, {1, 0, 0}, {1, 0, 1}, {1,1, 0}, {1, 1, 1}} ;
f1[b, {0}]
</code></pre>
<blockquote>
<p>{1, 0, 1, 1, 1, 1, 0, 0, 0, 0} </p>
</blockquote>
<pre><code>f1[b, {0}] == f2[b, {0}] == f3[b, {0}] == f4[b, {0}] == f5[b, {0}]
</code></pre>
<blockquote>
<p>True</p>
</blockquote>
<pre><code>f1[b, {1,0}]
</code></pre>
<blockquote>
<p>{0, 1, 0, 0, 0, 0, 1, 1, 0, 0}</p>
</blockquote>
<pre><code>f1[b, {1, 0}] == f2[b, {1, 0}] == f4[b, {1, 0}] == f5[b, {1, 0}]
</code></pre>
<blockquote>
<p>True</p>
</blockquote>
|
3,928,937 | <p>Determine all the solutions of the congruence<br />
<span class="math-container">$x^{85} ≡ 25 \pmod{31}$</span><br />
using index function in base <span class="math-container">$3$</span> module <span class="math-container">$31$</span>.<br />
It is clear to me that <span class="math-container">$3$</span> is primitive root module <span class="math-container">$31$</span>, but how do I use this information in the solution?</p>
| J. W. Tanner | 615,567 | <p><span class="math-container">$3^22^2\equiv5 $</span> and <span class="math-container">$2^2\equiv(-3)^3$</span>, so <span class="math-container">$3^5\equiv -5$</span>, and <span class="math-container">$3^{10}\equiv25\pmod{31}$</span>.</p>
<p>Let <span class="math-container">$x=3^y$</span>, so you're asking for <span class="math-container">$(3^y)^{85}\equiv3^{10}\pmod{31},$</span> which means <span class="math-container">$85y\equiv10\pmod{30}$</span></p>
<p><span class="math-container">$\iff 17y\equiv2\bmod6\iff y\equiv4\bmod6\iff y\equiv 4, 10, 16, 22, $</span> or <span class="math-container">$28\pmod{30}$</span>.</p>
<p>Now do you see how <span class="math-container">$x^{85}\equiv25\pmod{31}$</span> can be solved using indices with base <span class="math-container">$3$</span>?</p>
|
3,815,990 | <p>The following question was asked in End Term exam of real analysis and I was clueless on how it can be approached .</p>
<p>So, I am asking for guidence here.</p>
<blockquote>
<p>Question: Let <span class="math-container">$\phi :[0,1] \to \mathbb{R}$</span> be a continuous function such that <span class="math-container">$\int_{0}^{1}\phi(t)e^{-at} dt=0$</span> for every <span class="math-container">$ a \in\mathbb{R}_{+}$</span> . Show that for every non-negative integer n , <span class="math-container">$\int_{0}^{1}\phi(t) {t}^n dt=0$</span> .</p>
</blockquote>
<p>I am sorry but for this question I would not be able to show how I tried to proceed as I am absolutely clueless on how this problem should be approached .</p>
<p>Kindly tell how should I procced.</p>
<p>Thanks!!</p>
| TheSilverDoe | 594,484 | <p><strong>Hint :</strong> Show, using the appropriate version of DCT, that the function
<span class="math-container">$$F : a \mapsto \int_0^1 \phi(t)e^{-at} dt$$</span></p>
<p>is <span class="math-container">$\mathcal{C}^{\infty}$</span>, and that for all <span class="math-container">$n \in \mathbb{N}$</span>,
<span class="math-container">$$F^{(n)}(a)= \int_0^1 \phi(t)(-t)^ne^{-at} dt$$</span></p>
<p>Then evaluate in <span class="math-container">$a=0$</span>.</p>
<hr />
<p><strong>Addendum : detailed solution :</strong>
Let <span class="math-container">$K>0$</span> such that for every <span class="math-container">$t \in [0,1]$</span>,
<span class="math-container">$$|\phi(t)|\leq K$$</span></p>
<p>(such a <span class="math-container">$K$</span> exists since <span class="math-container">$\phi$</span> is continuous on the segment <span class="math-container">$[0,1]$</span>).</p>
<p>Let <span class="math-container">$f : \mathbb{R}_+ \times [0,1] \rightarrow \mathbb{R}$</span> be the function defined for every <span class="math-container">$a \in \mathbb{R}_+$</span> and <span class="math-container">$t \in [0,1]$</span> by
<span class="math-container">$$f(a,t)=\phi(t)e^{-at}$$</span></p>
<p>Then, you have the following facts :</p>
<p><span class="math-container">$\bullet \ $</span> for every <span class="math-container">$a \in \mathbb{R}_+$</span>, the function <span class="math-container">$t \mapsto f(a,t)$</span> is measurable.</p>
<p><span class="math-container">$\bullet \ $</span> for every <span class="math-container">$t \in [0,1]$</span>, the function <span class="math-container">$a \mapsto f(a,t)$</span> is <span class="math-container">$\mathcal{C}^{\infty}$</span>.</p>
<p><span class="math-container">$\bullet\ $</span> for every <span class="math-container">$n \in \mathbb{N}$</span>, and every <span class="math-container">$(a,t) \in \mathbb{R}_+ \times [0,1]$</span>, one has
<span class="math-container">$$\left|\frac{\partial^n f}{\partial a^n}(a,t)\right| = \left|\phi(t)(-t)^ne^{-at} \right| \leq K t^n$$</span></p>
<p>which is integrable w.r.t. <span class="math-container">$t$</span> over <span class="math-container">$[0,1]$</span>.</p>
<p>Hence, applying <a href="https://en.wikipedia.org/wiki/Leibniz_integral_rule#Measure_theory_statement" rel="nofollow noreferrer">Leibniz integral rule</a> (which is just a particular case of the DCT), you get that the function
<span class="math-container">$$F : a \mapsto \int_0^1 \phi(t)e^{-at} dt$$</span> is <span class="math-container">$\mathcal{C}^{\infty}$</span> <em>and</em> that for every <span class="math-container">$n \in \mathbb{N}$</span>, and <span class="math-container">$a \in \mathbb{R}_+$</span>,
<span class="math-container">$$F^{(n)}(a) = \int_0^1 \frac{\partial^n f}{\partial a^n}(a,t) dt = \int_0^1 \phi(t)(-t)^ne^{-at} dt$$</span></p>
<p>In particular, you get
<span class="math-container">$$F^{(n)}(0) = \int_0^1 \phi(t)(-t)^n dt$$</span></p>
<p>Now, if <span class="math-container">$F \equiv 0$</span>, then obviously <span class="math-container">$F^{(n)}(0)=0$</span>, and you are done.</p>
|
3,815,990 | <p>The following question was asked in End Term exam of real analysis and I was clueless on how it can be approached .</p>
<p>So, I am asking for guidence here.</p>
<blockquote>
<p>Question: Let <span class="math-container">$\phi :[0,1] \to \mathbb{R}$</span> be a continuous function such that <span class="math-container">$\int_{0}^{1}\phi(t)e^{-at} dt=0$</span> for every <span class="math-container">$ a \in\mathbb{R}_{+}$</span> . Show that for every non-negative integer n , <span class="math-container">$\int_{0}^{1}\phi(t) {t}^n dt=0$</span> .</p>
</blockquote>
<p>I am sorry but for this question I would not be able to show how I tried to proceed as I am absolutely clueless on how this problem should be approached .</p>
<p>Kindly tell how should I procced.</p>
<p>Thanks!!</p>
| zhw. | 228,045 | <p>If you are familiar with the Stone-Weierstrass theorem: Let <span class="math-container">$\mathcal A$</span> be the algebra of functions of the form <span class="math-container">$\sum_{k=1}^{n}a_ke^{-kt}.$</span> By S-W, <span class="math-container">$\mathcal A$</span> is dense in <span class="math-container">$C[0,1].$</span> Thus there is a sequence <span class="math-container">$f_m$</span> in <span class="math-container">$\mathcal A$</span> that converges uniformly to <span class="math-container">$\phi$</span> on <span class="math-container">$[0,1].$</span></p>
<p>Now our assumption on <span class="math-container">$\phi$</span> implies <span class="math-container">$\int_0^1\phi(t)f_m(t)\,dt =0$</span> for all <span class="math-container">$m.$</span> So as <span class="math-container">$m\to \infty,$</span></p>
<p><span class="math-container">$$0=\int_0^1\phi(t)f_m(t)\,dt \to \int_0^1\phi(t)^2\,dt.$$</span>
It follows that <span class="math-container">$\int_0^1\phi(t)^2\,dt =0.$</span> From this we conclude <span class="math-container">$\phi \equiv 0,$</span> which implies the desired conclusion (and a lot more).</p>
|
787,558 | <p>we have $ad-bc >1$ is it true that at least one of $a,b,c,d$ is not divisible by $ad-bc$ ?
Thanks in advance.</p>
<p><strong>Example:</strong>
$a=2$ , $b = 1$, $c = 2$, $d = 2$, $ad-bc = 2$ </p>
<p>so $b$ is not divisible by $ad-bc$</p>
| Magdiragdag | 35,584 | <p>Yes, that is true. </p>
<p><strong>Hint.</strong> Consider a common divisor $q$ of $a, b, c, d$. What does that mean for $q^2$ and $ad - bc$? Now assume that $ad - bc$ is a common divisor of $a, b, c, d$ and apply this to $ad - bc$ itself. This is going to give a contradiction with the assumption that $ad - bc > 1$.</p>
|
379,194 | <p>Let $X$ be a topological space and $X^*$ be its supspace. It is stated in my textbook that if $c(A)$ represents the closure of set $A$ in $X$, then $c(A) \bigcap X^*$ is closed in $X^*$. </p>
<p>A closed set is one which contains all its limit points, and a limit point of a set is a point such that every open set containing it contains a different point from the aforementioned set. </p>
<p>Let $l$ is an external limit point of set $A$. If there is an open set containing $l$, it has to contain a point in $A$- let's call it $p$. Let the open set containing $p$ and $l$ not contain any other point in $A$. I don't see why that should be a problem at all. Let the subspace $X^*$ contain $l$, but not $p$. </p>
<p>$c(A) \bigcap X^*$ will contain $l$, but $l$ will be not a limit point of $A$, as there is an open set containing $l$ and no point in $A \bigcap X^*$ ($p$ is not there in $X^*$). How is $A \bigcap X^*$ closed in $X^*$ then?</p>
| Jared | 65,034 | <p>Since the problem is a power series, we must first write it in summation notation, and then differentiate term by term within its radius of convergence. Note we can write the power series as
$$\sum_{n=0}^{\infty}\frac{(x-1)^{2n+1}}{8n+4}$$
Now, we apply the ratio test
$$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim_{n\to\infty}\left|\frac{(x-1)^{2(n+1)+1}}{8(n+1)+4}\frac{8n+4}{(x-1)^{2n+1}}\right|=(x-1)^2<1\Longrightarrow 0<x<2$$
Now we can actually show that the interval of convergence is $0\le x<2$, but differentiating term by term is only valid in the interior of the interval of convergence. What does it mean to 'differentiate term by term?' It means for $0<x<2$, the following equality holds
$$\frac{d}{dx}\sum_{n=0}^{\infty}\frac{(x-1)^{2n+1}}{8n+4}=\sum_{n=0}^{\infty}\frac{d}{dx}\frac{(x-1)^{2n+1}}{8n+4}$$</p>
<p>Now all you need to do is compute the derivative on the inside of the sum (treating $n$ as a constant).</p>
|
361,862 | <p>I would like you to expose and explain briefly some examples of theorems having some hypothesis that are (as far as we know) actually necessary in their proofs but whose uses in the arguments are extremely subtle and difficult to note at a first sight. I am looking for hypothesis or conditions that appear to be almost absent from the proof but which are actually hidden behind some really abstract or technical argument. It would be even more interesting if this unnoticed hypothesis was not noted at first but later it had to be added in another paper or publication not because the proof of the theorem were wrong but because the author did not notice that this or that condition was actually playing a role behind the scene and needed to be added. And, finally, an extra point if this hidden hypothesis led to some important development or advance in the area around the theorem in the sense that it opened new questions or new paths of research. This question might be related with this <a href="https://mathoverflow.net/questions/352249/nontrivially-fillable-gaps-in-published-proofs-of-major-theorems">other</a> but notice that it is not the same as I am speaking about subtleties in proof that were not exactly incorrect but incomplete in the sense of not mentioning that some object or result had to be use maybe in a highly tangential way.</p>
<p>In order to put some order in the possible answers and make this post useful for other people I would like you to give references and to at least explain the subtleties that helps the hypothesis to hide at a first sight, expose how they relate with the actual proof or method of proving, and tell the main steps that were made by the community until this hidden condition was found, i.e., you can write in fact a short history about the evolution of our understanding of the subtleties and nuances surrounding the result you want to mention.</p>
<p>A very well known and classic example of this phenomenon is the full theory of classical greek geometry that although correctly developed in the famous work of Euclides was later found to be incompletely axiomatized as there were some axioms that Euclides uses but <a href="https://en.wikipedia.org/wiki/Euclidean_geometry#cite_note-6" rel="noreferrer">he did not mention</a> as such mainly because these manipulations are so highly intuitive that were not easy to recognize that they were being used in an argument. Happily, a better understanding of these axioms and their internal respective logical relations through a period of long study and research lasting for millennia led to the realization that these axioms were not explicitly mention but necessary and to the development of new kinds of geometry and different geometrical worlds.</p>
<p>Maybe this one is (because of being the most classic and expanded through so many centuries and pages of research) the most known, important and famous example of the phenomena I am looking for. However, I am also interested in other small and more humble examples of this phenomena appearing and happening in some more recent papers, theorems, lemmas and results in general.</p>
<p>Note: I vote for doing this community wiki as it seems that this is the best way of dealing with this kind of questions.</p>
| Timothy Chow | 3,106 | <p>The convergence conditions for the Fourier series of a function <span class="math-container">$f:S^1 \to \mathbb{R}$</span> are a good example. The investigation of convergence conditions for Fourier series was a major motivation for Cantor's set theory and Lebesgue's measure theory. Depending on what kind of convergence you want, the conditions can be very subtle. For example, if you want the Fourier series of a continuous function to converge pointwise everywhere, then I don't think that there is any nice set of necessary and sufficient conditions known. Various sufficient conditions are known, e.g., the <a href="https://en.wikipedia.org/wiki/Dirichlet_conditions" rel="noreferrer">Dirichlet conditions</a>, which are fairly subtle.</p>
<p>Nowadays, I think it is generally considered that asking for convergence everywhere is the "wrong question"; one should ask for convergence <i>almost</i> everywhere. Then the most famous theorem is Carleson's theorem that the Fourier series of a function in <span class="math-container">$L^2$</span> converges almost everywhere. The hypothesis here is easy to state, but the way that the hypothesis is used is subtle. There are various proofs known now but none of them is easy. Note for example that Kolmogorov's first paper gave an example of a function in <span class="math-container">$L^1$</span> whose Fourier series diverges almost everywhere.</p>
|
361,862 | <p>I would like you to expose and explain briefly some examples of theorems having some hypothesis that are (as far as we know) actually necessary in their proofs but whose uses in the arguments are extremely subtle and difficult to note at a first sight. I am looking for hypothesis or conditions that appear to be almost absent from the proof but which are actually hidden behind some really abstract or technical argument. It would be even more interesting if this unnoticed hypothesis was not noted at first but later it had to be added in another paper or publication not because the proof of the theorem were wrong but because the author did not notice that this or that condition was actually playing a role behind the scene and needed to be added. And, finally, an extra point if this hidden hypothesis led to some important development or advance in the area around the theorem in the sense that it opened new questions or new paths of research. This question might be related with this <a href="https://mathoverflow.net/questions/352249/nontrivially-fillable-gaps-in-published-proofs-of-major-theorems">other</a> but notice that it is not the same as I am speaking about subtleties in proof that were not exactly incorrect but incomplete in the sense of not mentioning that some object or result had to be use maybe in a highly tangential way.</p>
<p>In order to put some order in the possible answers and make this post useful for other people I would like you to give references and to at least explain the subtleties that helps the hypothesis to hide at a first sight, expose how they relate with the actual proof or method of proving, and tell the main steps that were made by the community until this hidden condition was found, i.e., you can write in fact a short history about the evolution of our understanding of the subtleties and nuances surrounding the result you want to mention.</p>
<p>A very well known and classic example of this phenomenon is the full theory of classical greek geometry that although correctly developed in the famous work of Euclides was later found to be incompletely axiomatized as there were some axioms that Euclides uses but <a href="https://en.wikipedia.org/wiki/Euclidean_geometry#cite_note-6" rel="noreferrer">he did not mention</a> as such mainly because these manipulations are so highly intuitive that were not easy to recognize that they were being used in an argument. Happily, a better understanding of these axioms and their internal respective logical relations through a period of long study and research lasting for millennia led to the realization that these axioms were not explicitly mention but necessary and to the development of new kinds of geometry and different geometrical worlds.</p>
<p>Maybe this one is (because of being the most classic and expanded through so many centuries and pages of research) the most known, important and famous example of the phenomena I am looking for. However, I am also interested in other small and more humble examples of this phenomena appearing and happening in some more recent papers, theorems, lemmas and results in general.</p>
<p>Note: I vote for doing this community wiki as it seems that this is the best way of dealing with this kind of questions.</p>
| Timothy Chow | 3,106 | <p><a href="https://en.wikipedia.org/wiki/G%C3%B6del%27s_ontological_proof" rel="noreferrer">Gödel's ontological proof</a> requires a subtle assumption that if <span class="math-container">$\varphi$</span> is an <i>essential property</i> of <span class="math-container">$x$</span> then <span class="math-container">$x$</span> must possess <span class="math-container">$\varphi$</span>. The first time that Gödel showed his proof to anyone was in 1970, when he provided Dana Scott with a terse writeup. In this writeup, Gödel's definition of the essence of <span class="math-container">$x$</span> omitted the conjunct <span class="math-container">$\varphi(x)$</span>, thereby omitting the subtle assumption.</p>
<p>Many years later, Christoph Benzmüller and Bruno Woltzenlogel Paleo discovered, using a computer proof assistant, that omitting the subtle assumption not only creates a gap in the proof, but leads to a contradiction. See their 2016 paper, <a href="https://dl.acm.org/doi/abs/10.5555/3060621.3060751" rel="noreferrer">The Inconsistency in Gödel’s Ontological Argument: A Success Story for AI in Metaphysics</a>.</p>
<p>To be fair to Gödel, it seems likely that he knew the assumption was needed, since older private notes of his include the assumption. Also, when Dana Scott presented the proof to others, he included the assumption without indicating that he was doing anything more than reporting what Gödel had told him. (For further information, see page 392 of <a href="https://monoskop.org/images/a/aa/Kurt_G%C3%B6del_Collected_Works_Volume_III_1995.pdf" rel="noreferrer">Volume III of Gödel's Collected Works</a>.) Nevertheless, it was subtle enough that Gödel did omit it on at least once crucial occasion (even if by accident), and the fact that its omission leads to a contradiction seems to have gone unnoticed for over 40 years.</p>
|
4,651,047 | <p>I've recently come across an interesting integral, which is of the form:</p>
<p><span class="math-container">$$\int_0^1\arctan(x)\log\left(\frac{1-x}{1+x}\right)\mathrm{d}x$$</span></p>
<p>To start, I expanded the arctangent into its series expansion, then utilized the Weierstraß substitution in order to remove the fractional term from the logarithm:</p>
<p><span class="math-container">$$t = \frac{1-x}{1+x}$$</span></p>
<p>Finally, I'm left with this integral:</p>
<p><span class="math-container">$$2 \sum_{k \geq 0} \frac{(-1)^k}{2k+1} \int_0^1 \frac{(1-t)^{2k+1}}{(1+t)^{2k+3}} \log(t)\mathrm{d}t$$</span></p>
<p>Which looks an <strong>awful</strong> lot like the beta function, namely:</p>
<p><span class="math-container">$$B(x, y) = (1-a)^y \int_0^1 \frac{(1-t)^{x-1} t^{y-1}}{(1-at)^{x+y}} \mathrm{d}t, \quad a \leq 1$$</span></p>
<p>For the following values, the integrals are <strong>nearly</strong> identical:</p>
<p><span class="math-container">$$a=-1,$$</span>
<span class="math-container">$$x=2k+2,$$</span>
<span class="math-container">$$y=1$$</span></p>
<p>However, this is the bit where I fail to make progress.</p>
<p>I see that the integrals are clearly just off by that logarithm, but I cannot find a relation between them in order to progress with this integral.</p>
<p>I've tried differentiating with respect to the parameter <span class="math-container">$y$</span> in order to bring in that logarithm, but that obviously doesn't do much - as the parameter also lies in the denominator and causes unwanted trouble.</p>
<p>I've also tried constructing integrals which are similar to this one, but only have the parameter <span class="math-container">$y$</span> in the numerator; however, I haven't been able to make much progress doing that either. These integrals end up looking nothing like the beta function.</p>
| Quanto | 686,284 | <p>Integrate by parts with <span class="math-container">$dx=d(x-1)$</span></p>
<p><span class="math-container">\begin{align}
&\int_0^1\tan^{-1}x\ln\frac{1-x}{1+x}\ {dx}\\
\overset{ibp}= & \int_0^1 \frac{(1-x)\ln \frac{1-x}{1+x}}{1+x^2}\ \overset{\frac{1-x}{1+x}\to x}{dx}
-2\int_0^1 \frac{\tan^{-1}x}{1+x}dx\\
=& \ \int_0^1 \frac{\ln x}{1+x^2}dx -\frac34 \int_0^1\frac{\ln x}{1+x}dx
-2\int_0^1 \frac{\tan^{-1}x}{1+x}dx\\
=& -G-\frac34 \left(-\frac{\pi^2}{12}\right)-2\cdot \frac\pi8\ln2
=-G +\frac{\pi^2}{16} -\frac\pi4\ln2
\end{align}</span>
where <span class="math-container">$\int_0^1 \frac{\ln x}{1+x^2}dx=-G$</span> and <a href="https://math.stackexchange.com/a/3826136/686284"><span class="math-container">$\int_0^1\frac{\ln x}{1+x}dx= -\frac{\pi^2}{12} $</span></a></p>
|
4,651,047 | <p>I've recently come across an interesting integral, which is of the form:</p>
<p><span class="math-container">$$\int_0^1\arctan(x)\log\left(\frac{1-x}{1+x}\right)\mathrm{d}x$$</span></p>
<p>To start, I expanded the arctangent into its series expansion, then utilized the Weierstraß substitution in order to remove the fractional term from the logarithm:</p>
<p><span class="math-container">$$t = \frac{1-x}{1+x}$$</span></p>
<p>Finally, I'm left with this integral:</p>
<p><span class="math-container">$$2 \sum_{k \geq 0} \frac{(-1)^k}{2k+1} \int_0^1 \frac{(1-t)^{2k+1}}{(1+t)^{2k+3}} \log(t)\mathrm{d}t$$</span></p>
<p>Which looks an <strong>awful</strong> lot like the beta function, namely:</p>
<p><span class="math-container">$$B(x, y) = (1-a)^y \int_0^1 \frac{(1-t)^{x-1} t^{y-1}}{(1-at)^{x+y}} \mathrm{d}t, \quad a \leq 1$$</span></p>
<p>For the following values, the integrals are <strong>nearly</strong> identical:</p>
<p><span class="math-container">$$a=-1,$$</span>
<span class="math-container">$$x=2k+2,$$</span>
<span class="math-container">$$y=1$$</span></p>
<p>However, this is the bit where I fail to make progress.</p>
<p>I see that the integrals are clearly just off by that logarithm, but I cannot find a relation between them in order to progress with this integral.</p>
<p>I've tried differentiating with respect to the parameter <span class="math-container">$y$</span> in order to bring in that logarithm, but that obviously doesn't do much - as the parameter also lies in the denominator and causes unwanted trouble.</p>
<p>I've also tried constructing integrals which are similar to this one, but only have the parameter <span class="math-container">$y$</span> in the numerator; however, I haven't been able to make much progress doing that either. These integrals end up looking nothing like the beta function.</p>
| xpaul | 66,420 | <p>Noting
<span class="math-container">$$ d\bigg[(x-1)\ln\bigg(\frac{1-x}{1+x}\bigg)-2\ln(1+x)\bigg]=\ln\bigg(\frac{1-x}{1+x}\bigg) $$</span>
one has, by IBP,
<span class="math-container">\begin{eqnarray}
&&\int_0^1\tan^{-1}x\ln\bigg(\frac{1-x}{1+x}\bigg)\ {dx}\\
&=&-\int_0^1\bigg[(x-1)\ln\bigg(\frac{1-x}{1+x}\bigg)-2\ln(1+x)\bigg]\frac{1}{1+x^2}dx\\
&\overset{x\to\frac{1-x}{1+x}}{=}&\int_0^1\bigg(\frac{\ln x}{1+x^2}+\frac{x\ln x}{1+x^2}-\frac{\ln x}{1+x}+\frac{2\ln(\frac2{1+x})}{1+x^2}\bigg)dx\\
&=&-G-\frac{\pi^2}{48}+\frac{\pi^2}{12}+\frac14\pi\ln2\\
&=&-G+\frac{\pi^2}{16}+\frac14\pi\ln2.
\end{eqnarray}</span>
Update: Let
<span class="math-container">\begin{eqnarray}
I&=&\int_0^1\frac{\ln(1+x)}{1+x^2}\\
&=&\int_0^1\int_0^1\frac{x}{(1+tx)(1+x^2)}dtdx\\
&=&\int_0^1\int_0^1\frac{x}{(1+tx)(1+x^2)}dxdt\\
&=&\int_0^1\bigg(\frac\pi4\frac{t}{1+t^2}+\frac{2\ln2}{1+x^2}-\frac{1+t}{1+t^2}\bigg)dt\\
&=&\frac14\pi\ln2-I
\end{eqnarray}</span>
and hence
<span class="math-container">$$ I=\frac18\pi\ln2. $$</span>
<span class="math-container">\begin{eqnarray}
&&\int_0^1\frac{2\ln(\frac2{1+x})}{1+x^2}\\
&=&\frac{\pi}2\ln2-2I=\frac{\pi}4\ln2.
\end{eqnarray}</span></p>
|
1,041,177 | <p>Proof that if $p$ is a prime odd and $k$ is a integer such that $1≤k≤p-1$ , then the binomial coefficient</p>
<p>$$\displaystyle \binom{p-1}{k}\equiv (-1)^k \mod p$$</p>
<p>This exercise was on a test and I could not do!!</p>
| erfan soheil | 195,909 | <p>You can prove it by induction on $k$.
If $ k=1$ $\to$ $\displaystyle \binom{p-1}{k} = p-1$ that $p-1 \equiv -1 \mod p$.
For $k= n +1$ use this
$\displaystyle \binom{m}{n+1} =\displaystyle \binom{m}{n} +\displaystyle \binom{m}{n+1}$</p>
|
2,311,979 | <p>Let $A = (a_{i,j})_{n\times n}$ and $B = (b_{i,j})_{n\times n}$</p>
<p>$(AB) = (c_{i,j})_{n\times n}$, where $c_{i,j} = \sum_{k=1}^n a_{i,k} b_{k,j}$, so</p>
<p>$(AB)^T = (c_{j,i})$, where $c_{j,i} = \sum_{k=1}^n a_{j,k}b_{k,i} $, and
$B^T = b_{j,i}$ and $A^T = a_{j,i}$, so </p>
<p>$B^T A^T = d_{j,i}$ where $d_{j,i} = \sum_{k=1}^n b_{j,k} a_{k,i}$, but this mean that $(AB)^T \not = (B^T A^T)$, so where is the problem in this derivation ?</p>
<p>Edit: To be clear, lets be more precise;
Let $A = (a_{x,y})_{p\times n}$ and $B = (b_{z,t})_{n\times q}$</p>
<p>So, $A^T_{n\times p} = (a_{y,x})$ and $B^T_{q\times n} = (b_{t,z})$, which implies</p>
<p>$$(B^T A^T)_{i,j}^{q \times p} = \sum_{k=1}^n b_{i,k} a_{k,j},$$ and</p>
<p>$(AB)_{c,d}^{p\times q} = \sum_{k=1}^n a_{c,k} b_{k,d}$, which implies
$$((AB)^T)_{d,c}^{q\times p} = \sum_{k=1}^n = a_{d,k} b_{k,c}.$$
Since $i,d \in \{1,...,q\}$ and $j,c \in \{1,...,p\}$,
$$((AB)^T)_{d,c}^{q\times p} = \sum_{k=1}^n = a_{d,k} b_{k,c} = = \sum_{k=1}^n = a_{i,k} b_{k,j},$$
which again concludes that $(AB)^T \not = (B^T A^T)$.</p>
| 5xum | 112,884 | <p>OK, your notations confusing. It's very hard to understand what you mean by $(c_{j,i})$, so let's start over.</p>
<hr>
<p>Let's call $AB = C$, and let's call $B^TA^T=D$. What we want to prove is $C^T=D$.</p>
<p>First of all, let's say that the element in $C$'s $i$-th row and $j$-th column is $c_{ij}$. Then you know that $$c_{ij} = \sum_{i=1}^n a_{ik}b_{kj}.$$</p>
<p>Now, let's say that the element in $D$'s $i$-th row and $j$-th column is $d_{ij}$. Then we know that $$d_{ij} = \sum_{k=1}^n b'_{ik}a'_{kj}$$</p>
<p>Where $a'_{ik}$ is the element of $A^T$ and $b'_{kj}$ of $B^T$.</p>
<p>Next, since that's what transposition is, we know that $a'_{ik} = a_{ki}$ which means that $$d_{ij} = \sum_{k=1}^n b_{ki}a_{jk}$$</p>
<hr>
<p>Now you have to show that $d_{ij} = c_{ji}$ for all $i,j$. Since you know what $c_{ij}$ is equal to, you know that $c_{ji}$ is equal (if you replace every $i$ with $j$ and every $j$ with $i$) to</p>
<p>$$c_{ji} = \sum_{i=1}^n a_{jk}b_{ki}$$
which is exactly the same as $d_{ij}$ and you are done.</p>
<hr>
<p>If you don't like the fact that we switch $i$ and $j$, you can also introduce new variables to make it more clear:</p>
<p>We want to prove that $C^T=D$ which means we want to prove that $c_{qp} = d_{pq}$ for all $q,p$ (I am using different indices to avoid confusion). Since we know what $c_{ij}$ is <em>for any value $i,j$</em>, we can now substitute $i=p$ and $j=q$ to get</p>
<p>$$c_{qp} = \sum_{k=1}^n a_{qk} b_{kp}$$</p>
<p>and</p>
<p>$$d_{pq} = \sum_{k=1}^n b_{kp} a_{qk}.$$</p>
<p>Now it's easy to see that $c_{qp}=d_{pq}$.</p>
<hr>
<p><strong>AFTER EDIT</strong>:</p>
<hr>
<p>After your edit, the mistake comes in the line</p>
<blockquote>
<p>$$(B^T A^T)_{i,j}^{q \times p} = \sum_{k=1}^n b_{i,k} a_{k,j}$$</p>
</blockquote>
<p>because that is not true. In fact, $$\sum_{k=1}^n b_{i,k} a_{k,j}$$ is equal to $$(BA)_{i,j}.$$</p>
<p>Since you want $(B^TA^T)_{i,j}$, you want </p>
<p>$$(B^T A^T)_{i,j}^{q \times p} = \sum_{k=1}^n b'_{i,k} a'_{k,j}$$</p>
<p>where $b'_{i,k} = (B^T)_{i,k}$ and $a'_{i,k} = (A^T)_{k,j}$.</p>
<p>Then, you use the fact that $$b'_{i,k} = b_{k,i}$$ to get the correct result, which is</p>
<p>$$(B^T A^T)_{i,j}^{q \times p} = \sum_{k=1}^n b_{k,i} a_{j,k}$$</p>
|
1,463,881 | <blockquote>
<p>By considering $\sum_{r=1}^n z^{2r-1}$ where z= $\cos\theta + i\sin\theta$, show that if $\sin\theta$ $\neq$ 0, $$\sum_{r=1}^n \sin(2r-1)\theta=\frac{\sin^2n\theta}{\sin\theta}$$</p>
</blockquote>
<p>I couldn't solve this at first but with some hints some of you gave, I was able to come up with my own solution. Here it is:</p>
<p>First, we want to consider what is given in the question,
$$\begin{align}
\sum_{r=1}^n z^{2r-1} & =z+z^3+z^5...\\
& = z + z(z^2)+z^3(z^2)+...\\
\end{align}$$
In Geometric Progression, sum is given by
$$\sum_{r=1}^n z^{2r-1}=S_n = \frac{a (1-r^n)}{1-r}=\frac{z(1-(z^2)^n)}{1-z^2} = \frac {z-z^{2n+1}}{1-z^2}=\frac {1-z^{2n}}{z^{-1}-z} $$
Now, substitute $z = \cos \theta +i \sin \theta$</p>
<p>$$\begin{align}
\sum_{r=1}^n z^{2r-1} & =\frac{1-(\cos (2n\theta) + i \sin(2n\theta))}{\cos\theta - i\sin\theta-(\cos\theta+i\sin\theta)} \\
& = \frac{1-\cos (2n\theta) - i \sin(2n\theta)}{-2i\sin\theta} \cdot \frac{(i\sin\theta)}{(i\sin\theta)}\\
& = \frac{i\sin\theta-i\sin\theta \cos (2n\theta) + \sin\theta \sin(2n\theta)}{2\sin^2\theta} \\
\end{align}$$
Equating imaginary parts,
$$\sum_{r=1}^n \sin(2r-1)\theta = \frac{\sin\theta (1 - \cos(2n\theta))}{2\sin^2\theta} = \frac{\sin^2n\theta}{2\sin\theta} $$</p>
| Jack D'Aurizio | 44,121 | <p>The number of solutions of
$$ x_1+x_2+x_3+x_4+x_5= 5 $$
with the constraints $x_1\in[0,3],x_2\in[0,2]$ and $x_3,x_4,x_5\geq 0$ is given by the coefficient of $z^5$ in the product:
$$ (1+z+z^2+z^3)(1+z+z^2)(1+z+z^2+\ldots)(1+z+z^2+\ldots)(1+z+z^2+\ldots)$$
i.e. by the coefficient of $z^5$ in the Taylor series around $z=0$ of
$$ f(z) = \frac{(1-z^4)(1-z^3)}{(1-z)^5} = (1-z^3-z^4+z^7)\cdot\sum_{n\geq 0}\binom{n+4}{4}z^n $$
i.e. by:
$$ \binom{5+4}{4}-\binom{2+4}{4}-\binom{1+4}{4}=\color{red}{106}.$$</p>
|
4,330,755 | <p>Given a convex pentagon <span class="math-container">$ABCDE$</span>, there is a unique ellipse with center <span class="math-container">$F$</span> that can be inscribed in it as shown in the image below. I've written a small program to find this ellipse, and had to numerically (i.e. by iterations) solve five quadratic equations in the center <span class="math-container">$F$</span> coordinates and the entries of the inverse of the <span class="math-container">$Q$</span> matrix, such that the equation of the inscribed ellipse is</p>
<p><span class="math-container">$(r - F)^T Q (r - F) = 1 $</span></p>
<p>My question is: Is it possible to determine the coordinates of the center in closed form, from the given coordinates of the vertices of the pentagon ?</p>
<p><a href="https://i.stack.imgur.com/I1qFM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/I1qFM.png" alt="Ellipse Inscribed in Pentagon" /></a></p>
| achille hui | 59,379 | <p><strong>Disclaimer</strong> - this is not an independent answer!</p>
<blockquote>
<p>I have transformed the geometric construction in Intelligenti pauca's excellent answer into algebra using homogeneous coordinates. From that,
I observed the center of inellipse can be expressed as a weighted sum of vertices (<span class="math-container">$*5$</span>) and the weights can be computed from the areas of various triangles associated with the pentagon.</p>
<p>In the answer below, I have thrown away the uninspiring algebra and proved the most "unexpected" relation (<span class="math-container">$*4$</span>) by more elementary means (trigonometry).</p>
</blockquote>
<ul>
<li>For any <span class="math-container">$P = (x_p,y_p), Q = (x_q,y_q)$</span> in the plane, let <span class="math-container">$$[PQ] \stackrel{def}{=} x_py_q - x_qy_p$$</span> Please note that <span class="math-container">$[PQ] = -[QP]$</span> and geometrically, it is twice the signed area of a triangle with vertices at <span class="math-container">$P,Q$</span> and origin <span class="math-container">$O$</span>.</li>
<li>For any <span class="math-container">$P, Q, R$</span> in the plane, let
<span class="math-container">$$[PQR] \stackrel{def}{=} [PQ] + [QR] + [RQ]$$</span>
Please note that <span class="math-container">$[PQR] = [QRP] = [RPQ]$</span> and geometrically, it is twice the signed area of a triangle with vertices at <span class="math-container">$P,Q,R$</span>.</li>
</ul>
<p>Let <span class="math-container">$A,B,C,D,E$</span> be the vertices of a convex pentagon (ordered counterclockwisely). Let <span class="math-container">$G$</span> be its center of mass and <span class="math-container">$K$</span> be the center of its in-ellipse.</p>
<p>Pick any point <span class="math-container">$P$</span> within the pentagon. Cut the pentagon at <span class="math-container">$P$</span> to form 5 triangles. We can compute <span class="math-container">$G$</span> as the area weighted average of the centroid of the <span class="math-container">$5$</span> triangles:</p>
<p><span class="math-container">$$3G = \frac{\sum_{cyc} [ABP] (A+B+P)}{\sum_{cyc}[ABP]} \tag{*1}$$</span>
RHS of <span class="math-container">$(*1)$</span> can be viewed as a rational function in coordinates of <span class="math-container">$P$</span>. Since this is valid for all points within the pentagon, it is valid for all points in the plane. In particular, if we substitute <span class="math-container">$P$</span> by origin <span class="math-container">$O$</span> in <span class="math-container">$(*1)$</span>, we obtain
<span class="math-container">$$3G = \frac{\sum_{cyc}[AB](A+B)}{\sum_{cyc}[AB]}\tag{*2}$$</span>
Substitute <span class="math-container">$P$</span> by <span class="math-container">$K$</span> in <span class="math-container">$(*1)$</span> and combine with <span class="math-container">$(*2)$</span>, we can express <span class="math-container">$K$</span> as a weighted sum of vertices:</p>
<p><span class="math-container">$$K = \frac{\sum_{cyc}[AB](A+B)}{\sum_{cyc}[AB]} - \frac{\sum_{cyc} [ABK] (A+B)}{\sum_{cyc}[ABK]}\tag{*3}$$</span></p>
<p>It turns out the ratios among <span class="math-container">$[ABK], [BCK], \ldots$</span> can be computed from the areas of various triangles associated with the pentagon. More precisely, let</p>
<p><span class="math-container">$$
\lambda_A = [ABE][ACD],\;\;
\lambda_B = [BCA][BDE], \ldots $$</span>
(definition of <span class="math-container">$\lambda_C,\lambda_D,\lambda_E$</span> can be derived from that of <span class="math-container">$\lambda_A$</span> by cyclic permuting the vertices). We have</p>
<p><span class="math-container">$$\frac{[ABK]}{\lambda_A+\lambda_B}
= \frac{[BCK]}{\lambda_B+\lambda_C}
= \frac{[CDK]}{\lambda_C+\lambda_D}
= \frac{[DEK]}{\lambda_D+\lambda_E}
= \frac{[EAK]}{\lambda_E+\lambda_A}\tag{*4}$$</span></p>
<p>Combine these with <span class="math-container">$(*3)$</span>, we obtain a formula we seek.</p>
<p><span class="math-container">$$
\bbox[border: 1px solid blue;padding: 1em;]{
K = \frac{\sum_{cyc}[AB](A+B)}{\sum_{cyc}[AB]} - \frac{\sum_{cyc} (\lambda_A + \lambda_B) (A+B)}{2\sum_{cyc}\lambda_A}}\tag{*5}$$</span></p>
<p>What's remain is to justify <span class="math-container">$(*4)$</span>. Since under affine transform, the ratio of areas remains invariant. It suffices to verify <span class="math-container">$(*4)$</span> when the inellipse is the unit circle centered at origin and <span class="math-container">$A$</span> lies on the +ve <span class="math-container">$x$</span>-axis.</p>
<p>Let <span class="math-container">$A_1,B_1,C_1,D_1,E_1$</span> be points on the inellipse touching the pentagon edges <span class="math-container">$CD$</span>, <span class="math-container">$DE$</span>, <span class="math-container">$EA$</span>, <span class="math-container">$AB$</span>, <span class="math-container">$BC$</span> respectively. Let
<span class="math-container">$$2\alpha = \angle C_1OD_1, 2\beta = \angle D_1OE_1, 2\gamma = \angle E_1OA_1, 2\delta = \angle A_1OB_1, 2\epsilon = \angle B_1OC_1$$</span></p>
<p>It is not hard to see the vertices are located at
<span class="math-container">$$
\begin{align}
A &= \frac{1}{\cos\alpha}(1, 0)\\
B &= \frac{1}{\cos\beta}(\cos(\alpha+\beta),\sin(\alpha+\beta))\\
C &= \frac{1}{\cos\gamma}(\cos(\alpha+2\beta+\gamma),\sin(\alpha+2\beta+\gamma))\\
D &= \frac{1}{\cos\delta}(\cos(\alpha+2\epsilon+\delta),-\sin(\alpha+2\epsilon+\delta))\\
E &= \frac{1}{\cos\epsilon}(\cos(\alpha+\epsilon),-\sin(\alpha+\epsilon))
\end{align}$$</span>
With this, it is not hard to verify</p>
<p><span class="math-container">$$[AB] = \tan\alpha + \tan\beta, [BC] = \tan\beta + \tan\gamma,\ldots$$</span></p>
<p>Furthermore,
<span class="math-container">$$
\begin{align}[ABE]
&= (\tan\alpha + \tan\beta)(\tan\alpha + \tan\epsilon)\sin(2\alpha)\\
&= 2\frac{\sin(\alpha+\beta)\sin(\alpha+\epsilon)}{\cos\alpha\cos\beta\cos\epsilon}\sin\alpha\\
[ACD]
&= (\tan\gamma + \tan\delta)\left(1 - \frac{\cos(\alpha+2\beta+2\gamma)}{\cos\alpha}\right)\\
&= 2\frac{\sin(\gamma+\delta)}{\cos\alpha\cos\gamma\cos\delta}
(\sin(\alpha+\beta+\gamma)\sin(\beta+\gamma)\\
&= 2\frac{\sin(\gamma+\delta)\sin(\delta+\epsilon)\sin(\beta+\gamma)}{\cos\alpha\cos\gamma\cos\delta}
\end{align}$$</span>
This leads to
<span class="math-container">$$\lambda_A = [ABE][ACD] = 4\tan\alpha \prod_{cyc}\frac{\sin(\alpha+\beta)}{\cos\alpha}$$</span>
There are similar formulas for <span class="math-container">$\lambda_B,\lambda_C$</span> and at the end, we get</p>
<p><span class="math-container">$$\frac{\lambda_A + \lambda_B}{[AB]} = \frac{\lambda_A + \lambda_B}{\tan\alpha + \tan\beta} = 4\prod_{cyc}\frac{\sin(\alpha+\beta)}{\cos\alpha}$$</span>
Notice RHS is invariant under cyclic permutation of vertices. This justifies <span class="math-container">$(*4)$</span> in this special case and hence in general.</p>
|
149,049 | <p>Suppose you have a list of intervals (or tuples), such as:</p>
<pre><code>intervals = {{3,7}, {17,43}, {64,70}};
</code></pre>
<p>And you wanted to know the intervals of all numbers not included above, e.g.:</p>
<pre><code>myRange = 100;
numbersNotUed[myRange,intervales]
(*out: {{1,2},{8,16},{44,63},{71,100}}*)
</code></pre>
<p>What would be the most efficient way to approach this?</p>
<p><em>Mathematica</em> currently supports <code>IntervalIntersection</code> but not <code>IntervalComplement</code>.</p>
| kglr | 125 | <pre><code>f1 = Partition[Flatten[{0, Select[Flatten@#, Function[x, x < #2]], #2 + 1}],
2, 2, {1, -1}, {}, ({1, -1} + {##}) /. {1, 0} -> Nothing &]&;
f1[{{3, 7}, {17, 43}, {64, 70}}, 80]
</code></pre>
<blockquote>
<p>{{1, 2}, {8, 16}, {44, 63}, {71, 80}}</p>
</blockquote>
<p>Also</p>
<pre><code>f2 = BlockMap[({1, -1} + #) /. {1, 0} -> Nothing &,
Flatten[{0, Select[Flatten@#, Function[x, x < #2]], #2 + 1}], 2] &;
f2[{{3, 7}, {17, 43}, {64, 70}}, 80]
</code></pre>
<blockquote>
<p>{{1, 2}, {8, 16}, {44, 63}, {71, 80}}</p>
</blockquote>
<p><strong>Note:</strong> In versions before 10.2 you can use <code>Developer`PartitionMap</code> in place of <a href="http://reference.wolfram.com/language/ref/BlockMap.html" rel="nofollow noreferrer"><code>BlockMap</code></a> above. (thanks: @CarlWoll)</p>
|
86,202 | <p>Let $\mathcal{L},\mathcal{U}$ be invertible sheaves over a
noetherian scheme $X$, where $X$ is of finite type over a noetherian
ring $A$. If $\mathcal{L}$ is very ample, and $\mathcal{U}$ is
generated by global sections, then $\mathcal{L} \otimes \mathcal{U}$
is very ample.</p>
<p>Since $\mathcal{L}$ is very ample, there exists $n$, s.t. $i:
X\mapsto \mathbb{P}^n$ is an immersion with $\mathcal{L}=
i^*\mathcal{O}(1)$, and since $\mathcal{U}$ is generated by global
sections, one can construct $j:X \to \mathbb{P}^m$ with
$j^*\mathcal{O}(1) = \mathcal{U}$. From this I can construct the
following morphism:</p>
<p>$$
h: X \xrightarrow{\Delta} X\times X \xrightarrow{i\times j}
\mathbb{P}^n \times \mathbb{P}^m \xrightarrow{ \operatorname{segre \
embedding}} \mathbb{P}^N $$</p>
<p>I can prove $\mathcal{L}\otimes \mathcal{U } \cong
h^*\mathcal{O}(1)$, and the segre embedding is a closed immersion.
But I don't know whether the map $(i\times j) \circ \Delta$ is an
immersion, which is suspicious to be such, especially for the
$\Delta$.</p>
| KReiser | 21,412 | <p>From the comments, to clear this from the unanswered queue: </p>
<blockquote>
<p>Let <span class="math-container">$\pi:\Bbb P^m\to \operatorname{Spec} A$</span> be the projection. Then, <span class="math-container">$(\text{id}\times\pi)\circ (i\times j) \circ \Delta = i$</span> is an immersion, while <span class="math-container">$\text{id}\times \pi$</span> is separated (by Hartshorne's Corollary II.4.6(c)), so that Hartshorne's Exercise II.4.8 (applied to <span class="math-container">$\mathcal{P} = \text{"being an immersion"}$</span>) yields that <span class="math-container">$(i\times j)\circ \Delta$</span> is an immersion. </p>
</blockquote>
<p>This argument is sourced to Liu's Algebraic Geometry and Arithmetic Curves, Chapter 5, Excercise 1.28. This comment was originally written by darij grinberg (2011-11-29), and at the time of this posting was comment-upvoted at +7 as well as confirmed by the original asker in a subsequent comment.</p>
|
844,700 | <p>I am looking for a calculator which can calculate functions like $f(x) = x+2$
at $x=a$ etc; but I am unable to do so. Can you recommend any online calculator?</p>
| murkle | 177,867 | <p>GeoGebra works well for this, eg
<a href="http://web.geogebra.org/?command=f(x)=x%2B2;a=2;f(a)" rel="nofollow">http://web.geogebra.org/?command=f(x)=x+2;a=2;f(a)</a></p>
|
2,623,560 | <blockquote>
<p>Decide if $\mathbb Z[i]/\langle i\rangle$ and $\mathbb Z$ are isomorphic, if $\mathbb Z[i]/\langle i+1\rangle$ and $\mathbb Z_2$ are isomorphic</p>
</blockquote>
<p>I know that in the first case if there exist such homomorphism then $f(i)=0$ (and in the second case $f(i+1)=0$), but I don't know exactly how to prove it.</p>
| egreg | 62,967 | <p>It seems you are in the context of rings.</p>
<p>Note that $i$ is invertible in $\mathbb{Z}[i]$, so the ideal it generates is the whole ring. Hence the quotient $\mathbb{Z}[i]/\langle i\rangle$ is the trivial ring.</p>
<p>For the second part, consider $a+bi=a-b+b(1+i)$ and the ring homomorphism
$$
\varphi\colon\mathbb{Z}[i]\to\mathbb{Z}/2\mathbb{Z}
$$
defined by $\varphi(a+bi)=[a-b]$ where $[x]$ denotes the residue class of $x$ modulo $2$ (prove it is a ring homomorphism and that its kernel is $\langle 1+i\rangle$).</p>
|
834,949 | <p>I have this HW where I have to calculate the $74$th derivative of $f(x)=\ln(1+x)\arctan(x)$ at $x=0$.
And it made me think, maybe I can say (about $\arctan(x)$ at $x=0$) that there is no limit for the second derivative, therefore, there are no derivatives of degree grater then $2$.
Am I right?</p>
| André Nicolas | 6,312 | <p>To get an <em>expression</em> for the $n$-th derivative, you can use the <a href="http://en.wikipedia.org/wiki/General_Leibniz_rule" rel="nofollow">Leibniz Rule</a>
$$(fg)^{(n)}=\sum_{k=0}^n \binom{n}{k}f^{(k)}g^{(n-k)}.$$</p>
<p>Here $h^{(i)}$ denotes the $i$-th derivative of $h$. </p>
|
1,554 | <p>Suppose you have an incomplete Riemannian manifold with bounded sectional curvature such that its completion as a metric space is the manifold plus one additional point. Does the Riemannian manifold structure extend across the point singularity?</p>
<p>(Penny Smith and I wrote a paper on this many years ago, but we had to assume that no arbitrarily short closed geodesics existed in a neighborhood of the singularity. I was never able to figure out how to get rid of this assumption and still would like someone better at Riemannian geometry than me to explain how. Or show me a counterexample.)</p>
<p>EDIT: For simplicity, assume that the dimension of the manifold is greater than 2 and that in any neighborhood of the singularity, there exists a smaller punctured neighborhood of the singularity that is simply connected. In dimension 2, you have to replace this assumption by an appropriate holonomy condition. </p>
<p>EDIT 2: Let's make the assumption above simpler and clearer. Assume dimension greater than 2 and that for any r > 0, there exists 0 < r' < r, such that the punctured geodesic ball B(p,r'){p} is simply connected, where p is the singular point. The precludes the possibility of an orbifold singularity.</p>
<p>ADDITIONAL COMMENT: My approach to this was to construct a differentiable family of geodesic rays emanating from the singularity. Once I have this, then it is straightforward using Jacobi fields to show that this family must be naturally isomorphic to the standard unit sphere. Then using what Jost and Karcher call "almost linear coordinates", it is easy to construct a C^1 co-ordinate chart on a neighborhood of the singularity. (Read the paper. Nothing in it is hard.)</p>
<p>But I was unable to build this family of geodesics without the "no small geodesic loop" assumption. To me this is an overly strong assumption that is essentially equivalent to assuming in advance that that differentiable family of geodesics exists. So I find our result to be totally unsatisfying. I don't see why this assumption should be necessary, and I still believe there should be an easy way to show this. Or there should be a counterexample.</p>
<p>I have to say, however, that I am pretty sure that I did consult one or two pretty distinguished Riemannian geometers and they were not able to provide any useful insight into this.</p>
| Anton Petrunin | 1,441 | <p>Once we considered a similar problem but around infinity,
try to look in our paper "Asymptotical flatness and cone structure at infinity".</p>
<p>Let us denote by $r$ the distance to the singular point.
If dimensions $\not= 4$ then the same method shows that at singular point we have Euclidean tangent cone even if curvature is "much less" than $r^{-2}$ (say if $K=O(\tfrac{1}{r^{2-\varepsilon}})$ for some $\varepsilon>0$, but one can make it bit weaker).</p>
<p>In dimension 4 there might be some funny examples: Your singular point has tangent space $\mathbb R^3$,
the $r$-spheres around this point are Berger spheres, so its curvature is very much like curvature of $r\cdot(S^2\times \mathbb R)$, the size of Hopf fibers goes to $0$ very fast.
However if you know that dimension of the tangent space is $4$ then it has to be Euclidean.</p>
<p>All this can happen if curvature grows slowly.
If it is bounded then one can extend Bishop--Gromov type inequality for balls around singular point.
It implies that the dimension of the tangent space is $4$. That will finish the proof.</p>
|
4,369,232 | <p>I have the following problem:</p>
<blockquote>
<p>Let <span class="math-container">$\{(M_i,\tau_i)\}_{i\in I}$</span> be nonempty topological spaces where <span class="math-container">$I$</span> is arbitrary but non empty. Let <span class="math-container">$M=\prod_{i\in I} M_i$</span>. Let <span class="math-container">$F$</span> be a filter on <span class="math-container">$M$</span> and denote by</p>
<p><span class="math-container">$$F_i=(pr_i)_*F=\{B\subseteq M_i: \exists\,\,A\in F\,\,s.t.\,\,pr_i(A)\subseteq B\}$$</span> the corresponding image filter on each componente <span class="math-container">$M_i$</span>. Show that <span class="math-container">$F$</span> converges to <span class="math-container">$p\in M$</span> iff <span class="math-container">$F_i$</span> converges to <span class="math-container">$p_i=pr_i(p)$</span> for all <span class="math-container">$i\in I$</span>.</p>
</blockquote>
<p>Then <span class="math-container">$\Rightarrow$</span> implication worked perfectly by using some theorem from the lecture. But I somehow struggled a bit with the other implication. In the solution they did something with subbasis ect but I didn't understand that. Therefore I wandet do do it different. My Idea was the following:</p>
<p><span class="math-container">$\Leftarrow$</span> Let us assume tat <span class="math-container">$F_i$</span> converges to <span class="math-container">$p_i$</span>. We need to show that <span class="math-container">$F$</span> converges to <span class="math-container">$p$</span>, i.e. we need to show that <span class="math-container">$\mathfrak{U}(p)\subset F$</span> where <span class="math-container">$\mathfrak{U}(p)$</span> is the neighbourhood filter of <span class="math-container">$p$</span>. Now let us take any basis open set <span class="math-container">$B$</span> in M. <span class="math-container">$$B=\bigcap_{i\in I} pr_i^{-1}(O_i)$$</span>where <span class="math-container">$I$</span> is finite and <span class="math-container">$p_i\in O_i$</span> are open sets in <span class="math-container">$M_i$</span>. We see that it is enought to show that <span class="math-container">$pr_i^{-1}(O_i)\in F$</span>. To do so let us use that <span class="math-container">$F_i\rightarrow p_i\Leftrightarrow \mathfrak{U}(p_i)\subset F_i$</span>. But then <span class="math-container">$O_i\in (pr_i)_*F$</span> which implies that there is some <span class="math-container">$A_i\in F$</span> such that <span class="math-container">$pr_i(A_i)\subset O_i$</span>. This implies that <span class="math-container">$A_i\subset pr_i^{-1}(O_i)\in F$</span>. Then since the intersection is finite also <span class="math-container">$B\in F$</span>.</p>
<p>Now what I don't see is why then <span class="math-container">$\mathfrak{U}(p)\subset F$</span>.</p>
<p>I'm really not sure if this workes therefore it would be nice if someone could take a look at it. Thanks a lot</p>
| Nicolás Vilches | 413,494 | <p>You have to be a bit careful with the details here. When you write <span class="math-container">$U=\prod_{i \in I} O_i$</span>, it is not true that every open set of <span class="math-container">$M$</span> is of this form. Rather, you can show that given a neighborhood <span class="math-container">$p \in U$</span>, there are some open sets <span class="math-container">$p_i \in O_i \subseteq M_i$</span>, <strong>all but finitely many equal to <span class="math-container">$M_i$</span></strong>, such that <span class="math-container">$U \supseteq \prod_{i \in I} O_i$</span>. (This is essentially the claim that the <span class="math-container">$\mathrm{pr}_i^{-1}(O_i)$</span> form a subbasis of the product topology on <span class="math-container">$M$</span>.) Now, the convergence tells you that <span class="math-container">$O_i \in F_i$</span>, which implies that some set <span class="math-container">$B_i \subseteq M$</span> has <span class="math-container">$\mathrm{pr}(B_i)=O_i$</span>. Here you can asusme that <span class="math-container">$B_i=\mathrm{pr}_i^{-1}(O_i)$</span> by the monotonicity of the filter, and then this intersection (which is finite!) tells you that <span class="math-container">$U \in F$</span>.</p>
|
456,583 | <p>I was searching for a Latex symbol that indicates $A \Rightarrow B$ and $A \not\Leftarrow B$ ($B$ if not only if $A$, $B$ ifnf $A$). I thought of using $A \Leftrightarrow B$ with the left arrow tick <code><</code> crossed out. Since I did not find such a symbol:</p>
<p>Is there a Latex symbol for this?</p>
<p>How common or understandable is this symbol?</p>
<p>If it isn't common: How easily is it confused with the symbol $\not\Leftrightarrow$?</p>
<hr>
<p>Update: </p>
<p>I need it for a sequence $A$ ifnf $B$ ifnf $C$ ifnf $D$, which I find more understandable than $A \Leftarrow B \Leftarrow C \Leftarrow D$ and $A \not\Rightarrow B \not\Rightarrow C \not\Rightarrow D$.</p>
<p>Of course I will prove both directions.</p>
| Emily | 31,475 | <p>It is doubtful that a symbol exists; I do not believe it is common usage.</p>
<p>Note that your situation is equivalent to "$A$ implies $B$, but $B$ does not imply $A$". There are many, many situations in mathematics when this is the case. For instance:</p>
<blockquote>
<p>Independent random variables have zero correleation coefficient, but a zero correlation coefficient does not imply that the random variables are independent.</p>
</blockquote>
<p>In fact, the distinction that $A \implies B$ does not imply that $B \implies A$ is so important that it is almost always best addressed with more than a basic symbolic representation. The reader demands to know why the converse does not hold! Examples of situations where the converse does not hold are almost always useful.</p>
|
34,487 | <p>A few years ago Lance Fortnow listed his favorite theorems in complexity theory:
<a href="http://blog.computationalcomplexity.org/2005/12/favorite-theorems-first-decade-recap.html" rel="nofollow">(1965-1974)</a>
<a href="http://blog.computationalcomplexity.org/2006/12/favorite-theorems-second-decade-recap.html" rel="nofollow">(1975-1984)</a>
<a href="http://eccc.hpi-web.de/eccc-reports/1994/TR94-021/index.html" rel="nofollow">(1985-1994)</a>
<a href="http://blog.computationalcomplexity.org/2004/12/favorite-theorems-recap.html" rel="nofollow">(1995-2004)</a>
But he restricted himself (check the third one) and his last post is now 6 years old. An updated and more comprehensive list can be helpful.</p>
<blockquote>
<p>What are the most important results (and papers) in complexity theory that every one should know? What are your favorites?</p>
</blockquote>
| Marcos Villagra | 7,692 | <p>I think you should add as a recent result the proof for QIP=IP=PSPACE</p>
|
704,921 | <p>This is the question:
$$
\frac{(2^{3n+4})(8^{2n})(4^{n+1})}{(2^{n+5})(4^{8+n})} = 2
$$
I've tried several times but I can't get the answer by working out.I know $n =2$, can someone please give me some guidance? Usually I turn all the bases to 2, and then work with the powers, but I probaby make the same mistake every time, unfortunately I don't know what that is. Thank you in advance.</p>
<p><em>EDIT</em></p>
<p>This is what I simplified it to in the beginning of every attempt.</p>
<p>$$
\frac{(2^{3n} *16)(2^{6n})(2^{2n}*4)}{(2^{n}*32)(2^{n}*2^{16})} = 2
$$</p>
<p>Therefore</p>
<p>$$
\frac{64(2^{3n+6n+2n})}{(2^{16}*32)(2^{2n})} = 2
$$
<br> I simplified further:
$$
\frac{2^{11n}}{32768(2^{2n})} =2
$$
<br>
$$
2^{11n} = (2^{2n+1})*32768 \\
$$
$$
\frac{2^{11n}}{2^{2n+1}} = 32768
$$</p>
<p>$$
\frac{2^{11n}}{2*2^{2n}} = 32768
$$</p>
<p>And this is the furthest I get, what do i do now?</p>
| lab bhattacharjee | 33,337 | <p>HINT:</p>
<p>Use $$a^{mn}=(a^m)^n$$ and $$a^m\cdot a^n=a^{m+n}\text{ and }\frac{a^m}{a^n}=a^{m-n}$$</p>
<p>For example, $\displaystyle4^{8+n}=(2^2)^{(8+n)}=2^{2(8+n)}$</p>
<p>Finally $a^x=a^y\implies x=y$ if real $a\ne\pm1$ </p>
<p><strong>Reference</strong> : <a href="http://www.proofwiki.org/wiki/Exponent_Combination_Laws" rel="nofollow">Exponent Combination Laws</a></p>
|
69,476 | <p>Hello everybody !</p>
<p>I was reading a book on geometry which taught me that one could compute the volume of a simplex through the determinant of a matrix, and I thought (I'm becoming a worse computer scientist each day) that if the result is exact this may not be the computationally fastest way possible to do it.</p>
<p>Hence, the following problem : if you are given a polynomial in one (or many) variables $\alpha_1 x^1 + \dots + \alpha_n x^n$, what is the cheapest way (in terms of operations) to evaluate it ?</p>
<p>Indeed, if you know that your polynomial is $(x-1)^{1024}$, you can do much, much better than computing all the different powers of $x$ and multiply them by their corresponding factor.</p>
<p>However, this is not a problem of factorization, as knowing that the polynomial is equal to $(x-1)^{1024} + (x-2)^{1023}$ is also much better than the naive evaluation.</p>
<p>Of course, multiplication and addition all have different costs on a computer, but I would be quite glad to understand how to minimize the "total number of operations" (additions + multiplications) for a start ! I had no idea how to look for the corresponding litterature, and so I am asking for your help on this one :-)</p>
<p>Thank you !</p>
<p>Nathann</p>
<p>P.S. : <em>I am actually looking for a way, given a polynomial, to obtain a sequence of addition/multiplication that would be optimal to evaluate it. This sequence would of course only work for <strong>THIS</strong> polynomial and no other. It may involve working for hours to find out the optimal sequence corresponding to this polynomial, so that it may be evaluated many times cheaply later on.</em></p>
| Igor Rivin | 11,142 | <p>The simplest version of this question is: what is the quickest way to evaluate $x^n?$ For $n = 2^k,$ $k$ repeated squarings is obviously best, but for more complicated $n$ I believe that finding the optimum is very hard -- see Knuth, vol 2 for (much) more on these so-called "multiplication trees".</p>
|
823,928 | <p>Prove that all of the rings, which mediate between principal ideal ring $K$ and the field of fractions $Q$, are the principal ideal ring.</p>
| John Machacek | 155,418 | <p>$K = \mathbb{Z}$</p>
<p>$R = \{\frac{a}{2^n} : a \in \mathbb{Z}, n \in \mathbb{N}\}$</p>
<p>$Q = \mathbb{Q}$</p>
<p>If I understand your question here is an counter example.</p>
|
31,308 | <p>Apologies if my question is poorly phrased. I'm a computer scientist trying to teach myself about generalized functions. (Simple explanations are preferred. -- Thanks.)</p>
<p>One of the references I'm studying states that the space of Schwartz test functions of rapid decrease is the set of infinitely differentiable functions: $\varphi: \mathbb{R} \rightarrow \mathbb{R}$ such that for all natural numbers $n$ and $r$,</p>
<p>$\lim_{x\rightarrow\pm\infty} |x^n \varphi^{(r)}(x)|$</p>
<p>What I would like to know is why is necessary or important for test functions to decay rapidly in this manner? i.e. faster than powers of polynomials. I'd appreciate an explanation of the intuition behind this statement and if possible a simple example.</p>
<p>Thanks.</p>
<p>EDIT: the OP is actually interested in a particular 1994 paper on "Spatial Statistics" by
Kent and Mardia,
1994 Link between kriging and thin plate splines (with J. T. Kent). In Probability, Statistics and Optimization (F. P. Kelly ed.). Wiley, New York, pp 325-339.</p>
<p>Both are in Statistics at Leeds,</p>
<p><a href="http://www.amsta.leeds.ac.uk/~sta6kvm/" rel="nofollow">http://www.amsta.leeds.ac.uk/~sta6kvm/</a> </p>
<p><a href="http://www.maths.leeds.ac.uk/~john/" rel="nofollow">http://www.maths.leeds.ac.uk/~john/</a> </p>
<p><a href="http://www.amsta.leeds.ac.uk/~sta6kvm/SpatialStatistics.html" rel="nofollow">http://www.amsta.leeds.ac.uk/~sta6kvm/SpatialStatistics.html</a> </p>
<p>Scanned article:
<a href="http://www.gigasize.com/get.php?d=90wl2lgf49c" rel="nofollow">http://www.gigasize.com/get.php?d=90wl2lgf49c</a> </p>
<p>FROM THE OP: Here is motivation for my question: I'm trying to understand a paper that replaces an integral $$\int f(\omega) d\omega$$ with $$\int \frac{|\omega|^{2p + 2}}{ (1 + |\omega|^2)^{p+1}} \; f(\omega) \; d\omega$$ where $p \ge 0$ ($p = -1$ yields to the unintegrable expression) because $f(\omega)$ contains a singularity at the origin i.e. is of the form $\frac{1}{\omega^2}.$ </p>
<p>LATER, ALSO FROM THE OP:
I understand some parts of the paper but not all of it. For example, I am unable to justify the equations (2.5) and (2.7). Why do they take these forms and not some other form?</p>
| Olumide | 7,486 | <p>I believe I now have the answer to the question. The power of $\omega$ appear from the taylor expansion of $e^{i\omega.t_j}$ (in section 2.3 of Kent and Mardia's paper)</p>
<p>Thanks.</p>
<p>(Apologies for the seeming bit of self promotion, but I've tagged this as the correct answer.)</p>
|
2,280,243 | <blockquote>
<p>A tribonacci sequence is a sequence of numbers such that each term from the fourth onward is the sum of the previous three terms. The first three terms in a tribonacci sequence are called its <em>seeds</em> For example, if the three seeds of a tribonacci sequence are $1,2$,and $3$, it's 4th terms is $6$<br>
($1+2+3$),then $11(2+3+6)$.</p>
</blockquote>
<p>Find the smallest 5 digit term in a tribonacci sequence if the seeds are $6,19,22$</p>
<p>I'm having trouble with this. I don't know where to start. The formula for the tribonacci sequence in relation to its seeds is $$u_{n+3} = u_{n} + u_{n+1} + u_{n+2}$$
This tribonacci formula holds for all integer $n$. But that's all I know how to work out. And just if it helps, the next few numbers in the sequence mentioned in the question are $47,88,157,292$. Is there some shortcut to it, because I need to show some working out and having two pages full of addition doesn't sound very easy to mark, does it?</p>
| Doug M | 317,176 | <p>I would consider the matrix <span class="math-container">$T = \begin{bmatrix} 1&1&1\\1&0&0\\0&1&0\end{bmatrix}$</span>.</p>
<p>And note that when applied to any sequence <span class="math-container">$$S = \begin{bmatrix} s_3\\s_2\\s_1\end{bmatrix}$$</span></p>
<p><span class="math-container">$TS = \begin{bmatrix}s_3 + s_2 + s_1 \\s_3\\s_2\end{bmatrix}$</span></p>
<p>And <span class="math-container">$T^{n}S = \begin{bmatrix} s_{n+3}\\s_{n+2}\\s_{n+1}\end{bmatrix}$</span></p>
<p>generating a "tribbonicci" sequence.</p>
<p>The largest eigenvalue of <span class="math-container">$T$</span> will tell you how fast the sequence grows.</p>
<p><span class="math-container">$$s_{n+a} \approx 1.84^n s_a$$</span></p>
<p><span class="math-container">$$\Bigl\lfloor\frac {\log\left(\frac {10,000}{22}\right)}{\log 1.84}\Bigr\rfloor = 10
\Bigl\lfloor\frac {\log\left(\frac {10,000}{19}\right)}{\log 1.84}\Bigr\rfloor = 10$$</span></p>
<p>By one seed, it is ten steps beyond the 2nd seed and by the other it is 10 steps beyond the 13th seed, making either the 13th or the 14th member of the sequence as the target.</p>
|
2,927,079 | <p><a href="https://i.stack.imgur.com/ih7X2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ih7X2.png" alt="enter image description here"></a></p>
<p>In the second paragraph, Munkres assumes that there exists a separation of <span class="math-container">$Y$</span> (in the sense he defined in Lemma 23.1) and proves that <span class="math-container">$Y$</span> is not connected. </p>
<p>So I would think the first paragraph should prove the converse statement: if <span class="math-container">$Y$</span> is not connected, then there is a separation. But in the first paragraph Munkres says "Suppose <span class="math-container">$A$</span> and <span class="math-container">$B$</span> form a separation of <span class="math-container">$Y$</span>". Is it true that he does NOT mean the separation that he defined in the statement of Lemma 23.1? Does he mean "Suppose <span class="math-container">$Y$</span> is not connected"? If so, the next sentence that claims that <span class="math-container">$A$</span> and <span class="math-container">$B$</span> are open in <span class="math-container">$Y$</span> makes sense, but otherwise (if <span class="math-container">$A$</span> and <span class="math-container">$B$</span> form a separation in the sense he defined in Lemma 23.1) it doesn't.</p>
| Steve Kass | 60,500 | <p>I agree that this is unclear. It looks to me as though Munkres means “Suppose <span class="math-container">$Y$</span> is not connected.”</p>
<p>The first use of “separation” in the proof seems to be his way to say that <span class="math-container">$A$</span> and <span class="math-container">$B$</span> form a witness to the non-connectedness of <span class="math-container">$Y$</span>: they partition <span class="math-container">$Y$</span> into two disjoint open (in <span class="math-container">$Y$</span>) sets, and not that they form a separation as defined at the beginning of the lemma.</p>
<p>Whether that’s carelessness or perhaps because he uses the term “separation” to mean something else in his definition of connected, I don’t know. (Additionally suggestive of some carelessness, the statement of the lemma is a one-way implication, but he seems to prove an if-and-only-if proposition, and some writers might argue that beginning a lemma statement with an “If <span class="math-container">$P$</span>, <span class="math-container">$Q$</span>” sentence that is intended as a definition, and not the sentence to be proved, is also a problem.)</p>
|
1,651,427 | <blockquote>
<p>Let $f$ be a bounded function on $[0,1]$. Assume that for any $x\in[0,1)$, $f(x+)$ exists. Define $g(x)=f(x+)$, $x\in [0,1)$, and $g(1)=f(1)$. Is $g(x)$ right continuous? </p>
</blockquote>
<p>Prove it or give me a counterexample.</p>
<p>My ideas:</p>
<p>$(1)$If $f$ is of bounded variation, then $g$ must be right continuous.</p>
<p>$(2)$If the continuous points of $f$ are dense in $[0,1]$, then $g$ must be right continuous.</p>
<p>But I can not find a counterexample. Please help me. Thanks!</p>
| Qiyu Wen | 310,935 | <p>Given any $\epsilon > 0$ and any $x \in [0,1)$, pick $\delta > 0$ such that $|f(y) - f(x+)| < \epsilon/2$ for all $y \in (x,x+\delta)$. For any $y \in (x,x+\delta)$, let $\{x_n\}$ and $\{y_n\}$ be two sequences approaching $x$ and $y$ respectively from the right. Then there exists an integer $N$ such that $x_n$ and $y_n$ lie in $(x,x+\delta)$ for all $n>N$. For these $n$,
$$
|f(x_n)-f(y_n)| \leq |f(x_n)-f(x+)|+|f(y_n)-f(x+)| < \epsilon.
$$
By definition of $g$, we have
$$
|g(x)-g(y)| = \left|\lim_{n\rightarrow \infty}f(x_n) - \lim_{n\rightarrow \infty}f(y_n)\right| = \lim_{n\rightarrow \infty} |f(x_n)-f(y_n)| \leq \epsilon
$$
for all $y \in (x,x+\delta)$. Hence $g$ is right-continuous. Note that we do not need $f$ to be bounded.</p>
|
1,651,427 | <blockquote>
<p>Let $f$ be a bounded function on $[0,1]$. Assume that for any $x\in[0,1)$, $f(x+)$ exists. Define $g(x)=f(x+)$, $x\in [0,1)$, and $g(1)=f(1)$. Is $g(x)$ right continuous? </p>
</blockquote>
<p>Prove it or give me a counterexample.</p>
<p>My ideas:</p>
<p>$(1)$If $f$ is of bounded variation, then $g$ must be right continuous.</p>
<p>$(2)$If the continuous points of $f$ are dense in $[0,1]$, then $g$ must be right continuous.</p>
<p>But I can not find a counterexample. Please help me. Thanks!</p>
| Phillip Hamilton | 312,810 | <p>We are assuming that $f(x+)$ exists. Its definition is:</p>
<p>$f(x+) = q$ where $q\in [0,1)$, if $f(t_n) \to q$ as $n \to \infty$ for all {$t_n$} in $[x,1)$ s.t. $t_n \to x$</p>
<p>So try setting $g(x) = f(x+)$, then think about {$t_n$}</p>
|
194,218 | <blockquote>
<p>Let A, B be two sets. Prove that <span class="math-container">$A \subset B \iff A \cup B = B$</span></p>
</blockquote>
<p>I'm thinking of using disjunctive syllogism by showing that <span class="math-container">$\neg \forall Y(Y \in A).$</span> However, I'm not sure how the proving steps should proceed such that it leads me to that premise.</p>
<p>Edit: Thanks for the input. FYI, I need to prove this using predicate logic.</p>
| Community | -1 | <p>Let $A\subset B$. Since $B\subset B$, we have $A\cup B\subset B$. Clearly, $B\subset A\cup B$. Hence $A\cup B=B$.</p>
<p>Let $A\not\subset B$. Then there is some $x\in A$ with $x\not\in B$. Clearly, $x\in A\cup B$. Hence $A\cup B\neq B$.</p>
|
194,218 | <blockquote>
<p>Let A, B be two sets. Prove that <span class="math-container">$A \subset B \iff A \cup B = B$</span></p>
</blockquote>
<p>I'm thinking of using disjunctive syllogism by showing that <span class="math-container">$\neg \forall Y(Y \in A).$</span> However, I'm not sure how the proving steps should proceed such that it leads me to that premise.</p>
<p>Edit: Thanks for the input. FYI, I need to prove this using predicate logic.</p>
| Iuli | 33,954 | <p>For the first implication a draw can help us:
<img src="https://i.stack.imgur.com/ySAJW.png" alt="enter image description here"></p>
<p>and now it is obvious. </p>
<p>Now conversely, we have: </p>
<p>$A \cup B =B \Rightarrow$ $A \cup B \subset B \tag{1}$ and $B \subset A \cup B\tag{2}.$ We need only the relation $(1)$ which help us to conclude that: $A \subset B.$</p>
|
194,218 | <blockquote>
<p>Let A, B be two sets. Prove that <span class="math-container">$A \subset B \iff A \cup B = B$</span></p>
</blockquote>
<p>I'm thinking of using disjunctive syllogism by showing that <span class="math-container">$\neg \forall Y(Y \in A).$</span> However, I'm not sure how the proving steps should proceed such that it leads me to that premise.</p>
<p>Edit: Thanks for the input. FYI, I need to prove this using predicate logic.</p>
| Asaf Karagila | 622 | <p>Taking the "longer" road. Let us review the definitions:</p>
<ol>
<li>$A\subseteq B$ if and only if <em>for every $x\in A$, $x\in B$</em>.</li>
<li>$x\in A\cup B$ if and only if $x\in A$ <strong>or</strong> $x\in B$.</li>
<li>$A=B$ if and only if $A\subseteq B$ <strong>and</strong> $B\subseteq A$.</li>
<li>$P\iff Q$ means that if we assume that $P$ holds, then $Q$ must hold; and vice versa.</li>
</ol>
<hr>
<p>Now assume $A\subseteq B$. This means that for every $x\in A$ we have $x\in B$. We want to show that $A\cup B=B$. </p>
<ul>
<li>So we take $x\in B$, then $x\in A$ or $x\in B$, and therefore $x\in A\cup B$.</li>
<li><p>Now take $x\in A\cup B$, we want to show that $x\in B$. By definition either $x\in A$ or $x\in B$.</p>
<ul>
<li>If $x\in B$ we are done.</li>
<li>If $x\in A$ then by the assumption that $A\subseteq B$ we have that $x\in B$.</li>
</ul>
<p>Either way we have that $x\in B$.</p></li>
</ul>
<p>We have shown that <strong>if</strong> $A\subseteq B$ <strong>then</strong> $A\cup B\subseteq B$ and $B\subseteq A\cup B$, which is by fact number $3$ to say $A\cup B=B$.</p>
<hr>
<p>Now we need to assume that $A\cup B=B$, and deduce that $A\subseteq B$. So we need to show that if $x\in A$ <em>then</em> $x\in B$.</p>
<p>Take $x\in A$ to be an arbitrary element. Because $x\in A$ we have that $x\in A$ or $x\in B$, and therefore $x\in A\cup B$. The assumption was, however, that $A\cup B=B$ and therefore we have that $x\in B$ as wanted.</p>
|
194,218 | <blockquote>
<p>Let A, B be two sets. Prove that <span class="math-container">$A \subset B \iff A \cup B = B$</span></p>
</blockquote>
<p>I'm thinking of using disjunctive syllogism by showing that <span class="math-container">$\neg \forall Y(Y \in A).$</span> However, I'm not sure how the proving steps should proceed such that it leads me to that premise.</p>
<p>Edit: Thanks for the input. FYI, I need to prove this using predicate logic.</p>
| A.P. Phaneendra kumar | 811,292 | <p>Let <span class="math-container">$A\subseteq B$</span>.<br>
<strong>Claim:</strong> We need to prove <span class="math-container">$A\cup B\subseteq B$</span> and <span class="math-container">$B\subseteq A\cup B$</span></p>
<p>Let <span class="math-container">$x\in A\cup B$</span>. Then <span class="math-container">$x\in A$</span> or <span class="math-container">$x\in B$</span>. But <span class="math-container">$A\subseteq B$</span>. Hence <span class="math-container">$x\in B$</span>. Therefore <span class="math-container">$A\cup B\subseteq B$</span>.</p>
<p>Now conversely <span class="math-container">$x\in B$</span>. Then <span class="math-container">$x\in A\cup B$</span>. Therefore <span class="math-container">$B\subseteq A\cup B$</span>. Thus <span class="math-container">$A\cup B=B$</span>.</p>
<p>Consider <span class="math-container">$A\cup B=B$</span>. <br>
<strong>Claim:</strong> <span class="math-container">$A\subseteq B$</span>
Let <span class="math-container">$x\in A$</span>. Then <span class="math-container">$x\in A\cup B=B$</span>. Therefore <span class="math-container">$A\subseteq B$</span>.</p>
|
113,963 | <p>While the common approach to algebraic groups is via representable functors, it seems that there is no such for differential algebraic groups (defined by differential polynomials). Neither the book by E. Kolchin, nor the texts by Ph. J. Cassidy contain anything like this — they work only with the groups of points over differential fields (and, naturally, don't say the words "group of points").</p>
<p>Concerning difference algebraic groups, i.e. defined by polynomials with some fixed endomorphism (also, I don't like the ambiguity with the notion of "difference algebraic equation"), there is no systematic treatment at all, although some of these groups are intensively studied (twisted groups of Lie type as an example).</p>
<p>So the question is: is there really no modern (scheme-like) exposition of the subject? If so, why?</p>
| Peter | 69,733 | <p>For difference algebraic groups, I think the paper</p>
<p>Michael Wibmer: <a href="http://arxiv.org/abs/1405.6603" rel="nofollow">Affine difference algebraic groups</a></p>
<p>is what you asked for.</p>
|
150,180 | <p>I try to read Gross's paper on Heegner points and it seems ambiguous for me on some points:</p>
<p>Gross (page 87) said that $Y=Y_{0}(N)$ is the open modular curve over $\mathbb{Q}$ which classifies ordered pairs $(E,E^{'})$ of elliptic curves together with cyclic isogeny $E\rightarrow E^{'}$ of degree $N$. Gross uses on some steps the cyclic isogeny between two elliptic curves over $\mathbb{C}$. One of the books that I have read to understand the theory of modular curves is "A first course in Modular forms, written by Fred Diamond and Jerry Shurman". </p>
<p>Theorem 1.5.1.(page 38)</p>
<p>Let $N$ be a positive integer. </p>
<p>(a) The moduli space for $\Gamma_{0}(N)$ is
$$S_{0}(N)=\{[E_{\tau},\langle1/N+\Lambda_{\tau}\rangle]:\tau\in H\},$$
with $H$ is the upper half plan. Two points $[E_{\tau},\langle1/N+\Lambda_{\tau}\rangle]$ and $[E_{\tau^{'}},\langle1/N+\Lambda_{\tau^{'}}\rangle]$ are equal if and only if $\Gamma_{0}(N)\tau=\Gamma_{0}(N)\tau^{'}$. Thus there is a bijection
$$\psi:S_{0}(N)\rightarrow Y_{0}(N), [\mathbb{C}/\Lambda_{\tau},\langle1/N+\Lambda_{\tau}\rangle]\mapsto \Gamma_{0}(N)\tau.$$</p>
<p>So, how can we make the link between the equivalence classes of the enhanced elliptic curves for $\Gamma_{0}(N)$ (= the equivalence classes of $(E,C)$ where $E$ is a complex elliptic curve and $C$ is a cyclic subgroup of $E$ of order $N$) defined in Diamond/Shurman's book and the cyclic isogenies $E\rightarrow E^{'}$ of degree $N$ used by Gross.</p>
<p>I also ask if there is any other paper which explains the theory of Heegner points explicitly? </p>
<p>I have look at Darmon's note and Gross-Zagier paper "Heegner points and derivatives of L-series" and it seems that the both were influenced by Gross's paper! Is there any other paper which explains Heegner points explecitely and independently of Gross's paper?</p>
<p>(I keep this post open for any further question about Gross's paper and I apologise for any mistakes in my English.)</p>
<p>Thank you.</p>
| Qiaochu Yuan | 290 | <blockquote>
<p>(obviously, in the affine case this question translates into: can the category of (finitely generated) modules be defined via the category of projective modules (of finite rank)?)</p>
</blockquote>
<p>Yes (you're assuming Noetherian here, right?). We will need to combine two observations. Let $A$ be an $\text{Ab}$-enriched category and let $\text{Mod}(A)$ be the category of additive functors $A^{op} \to \text{Ab}$ (generalizing either right modules or presheaves according to taste). The <a href="http://ncatlab.org/nlab/show/Cauchy+complete+category" rel="nofollow noreferrer">Cauchy completion</a> $\hat{A}$ of $A$ is the category obtained from $A$ by first formally adjoining biproducts and then splitting all idempotents. When $A$ has one object with endomorphism ring $R$ then $\hat{A}$ is the category of finitely generated projective right $R$-modules. </p>
<blockquote>
<p><strong>Observation 1:</strong> The natural restriction map $\text{Mod}(\hat{A}) \to \text{Mod}(A)$ is an equivalence.</p>
</blockquote>
<p>(This is the easy direction of <a href="https://mathoverflow.net/questions/149331/homotopy-of-quivers/149368#149368">Morita theory for $\text{Ab}$-enriched categories</a>.) </p>
<p>In particular, the category of right modules over (finitely generated projective right $R$-modules) is $\text{Mod}(R)$. </p>
<p>To prove this it suffices to check that a right module $F : A^{op} \to \text{Ab}$ uniquely extends both to formal biproducts and to split idempotents, or in other words that biproducts and split idempotents are both <a href="http://ncatlab.org/nlab/show/absolute+colimit" rel="nofollow noreferrer">absolute colimits</a> for $\text{Ab}$-enriched functors. Existence follows from the fact that $\text{Ab}$ has all biproducts and all idempotents split in it. For uniqueness the essential point is that both being a biproduct and being a split idempotent are defined by equations among morphisms and that equations among morphisms are always preserved; see <a href="http://qchu.wordpress.com/2012/09/14/a-meditation-on-semiadditive-categories/" rel="nofollow noreferrer">this blog post</a> for more details in the case of biproducts, and the case of split idempotents is even easier (these are already absolute colimits for ordinary functors). </p>
<p>So starting from finitely generated projective modules / vector bundles, we've recovered all modules / quasicoherent sheaves. But we wanted to recover just the finitely generated modules. There are several options from here; for example, there is a canonical inclusion $\hat{A} \to \text{Mod}(A)$ and we can take the smallest abelian subcategory containing its image. Perhaps the most categorical answer is the following. </p>
<blockquote>
<p><strong>Observation 2:</strong> Let $R$ be a ring. A module $M \in \text{Mod}(R)$ is finitely generated iff $\text{Hom}(M, -)$ preserves filtered colimits where all of the maps in the diagram are monomorphisms.</p>
</blockquote>
<p>(In particular, being finitely generated is a Morita invariant property: it does not depend on the choice of ring $R$.) For a proof of a closely related fact see <a href="https://math.stackexchange.com/questions/111472/reference-request-compact-objects-in-r-mod-are-precisely-the-finitely-presented">this math.SE answer</a>. </p>
<hr>
<p><strong>Edit:</strong> Actually there is an easier construction. If $A = \hat{R}$ is the category of finitely generated projective right $R$-modules then we can just consider "finitely generated (right) $A$-modules," namely those which are a quotient of a finite direct sum of representables. This reproduces the usual notion of finite generation. </p>
|
3,599,893 | <p>I had this idea to build a model of Earth in Minecraft. In this game, everything is built on a 2D plane of infinite length and width. But, I wanted to make a world such that someone exploring it could think that they could possibly be walking on a very large sphere. (Stretching or shrinking of different places is OK.) </p>
<p>What I first thought about doing was building a finite rectangular model of the world as like a mercator projection, and tessellating this model infinitely throughout the plane. </p>
<p><a href="https://i.stack.imgur.com/bzdjA.png" rel="noreferrer"><img src="https://i.stack.imgur.com/bzdjA.png" alt="enter image description here"></a></p>
<p>Someone starting in the US could swim eastwards in a straight line across the Atlantic, walk across Africa and Asia, continue through the Pacific and return to the US. This would certainly create a sense of 3D-ness. However, if you travel north from the North Pole, you would wind up immediately at the South Pole. That wouldn't be right.</p>
<p>After thinking about it, I hypothesized that an explorer of this model might conclude that they were walking on a donut-shaped world, since that would be the shape of a map where the left was looped around to the right (making a cylinder), and then the top was looped to the bottom. For some reason, by simply tessellating the map, I was creating a hole in the world.</p>
<p>Anyway, to solve this issue, I thought about where one ends up after travelling north from various parts of the world. Going north from Canada, and continuing to go in that direction, you end up in Russia and you face south. The opposite is true as well: going north from Russia, you end up in Canada pointing south. Thus, I started to modify the tessellation to properly connect opposing parts of Earth at the poles. </p>
<p>When going north of a map of Earth, the next (duplicate) map would have to be rotated 180 degrees to reflect the fact that one facing south after traversing the north pole. This was OK. However, to properly connect everything, the map also had to be <em>flipped</em> about the vertical axis. On a globe, if Alice starts east of Bob and they together walk North and cross the North Pole, Alice still remains east of Bob. So, going north from a map, the next map must be flipped to preserve the east/west directions that would have been otherwise rotated into the wrong direction.</p>
<p><a href="https://i.stack.imgur.com/U5n9t.png" rel="noreferrer"><img src="https://i.stack.imgur.com/U5n9t.png" alt="enter image description here"></a></p>
<p>Now the situation is hopeless. After an explorer walks across the North Pole in this Minecraft world, he finds himself in a mirrored world. If the world were completely flat, it would feel as if walking North will take you from the outside of a 3D object to its inside.</p>
<p>Although I now think that it is impossible to trick an explorer walking on infinite plane into thinking he is on a sphere-like world, a part of me remains unconvinced. Is it really impossible? Also, how come a naive tessellation introduces a hole? And finally, if an explorer were to roam the world where crossing a pole flips everything, what would he conclude the shape of the world to be?</p>
| James K | 92,207 | <p>Although you can't make a sphere from a plane, there are map projections that tessellate "naturally" (and place the tricky singular points in the ocean where people tend not to notice them). You can't, for topological reasons, avoid the points at the corners, but this kind of map does avoid some of the problems of mirroring and is continuous except at those corner points.</p>
<p>Most well known is the "Peirce quincuncial" projection. Wikipedia has an image showing the projection <a href="https://i.stack.imgur.com/DDXEQ.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/DDXEQ.jpg" alt="quincuncial projection"></a></p>
<p><sup><a href="https://commons.wikimedia.org/w/index.php?curid=20378696" rel="noreferrer">Image by Strebe - Own work, CC BY-SA 3.0</a></sup></p>
|
2,468,067 | <p>Can we say that that if $f(x)$ and $f^{-1}(x)$ intersect, then at least one point of intersection will lie on $y=x$? </p>
<p>Also there are many function e.g. $f(x)=1-x^3$ where point of intersection exists outside $y=x$ There will be $5$(odd) point of intersection of $f(x)=1-x^3$ and $f^{-1}(x)=(1-x)^{1/3}$ out of which one lie on $y=x$. Will there exist a function in which there will be even number of point of intersection but odd number of point of intersection will lie outside $y=x$? </p>
<p>Also it is clear that in case of strictly increasing continuous function , point of intersection if exists will lie on $y=x$ but will also be true for
strictly increasing discontinuous function?</p>
| hmakholm left over Monica | 14,366 | <p>if the graphs of $f$ and $f^{-1}$ intersect at $(x,y)$, then they will also intersect at $(y,x)$.</p>
<p>If the function is additionally assumed to be <em>continuous</em>, then -- since $(x,y)$ and $(y,x)$ are on different sides of the diagonal $x=y$ -- the function graph must cross the line $x=y$. And such a point is necessarily also a crossing of $f$ and $f^{-1}$.</p>
|
3,891,749 | <p>I got a pretty good idea for the proof but it feels like it's missing some details.</p>
<p>Proof:
<span class="math-container">$$X \times Y \Leftrightarrow a\in X \land b \in Y \Leftrightarrow a\in X \land b\in Z$$</span></p>
<p>Since <span class="math-container">$b\in Y \Leftrightarrow b\in Z$</span>, then <span class="math-container">$Y = Z \blacksquare$</span></p>
<p>I feel like the jump to <span class="math-container">$b\in Y \Leftrightarrow b\in Z$</span> is flawed is some way, missing some details. Can anyone help me justify it more or tell me that it's good?</p>
| Hagen von Eitzen | 39,174 | <p>You do not clearly state where and how you use <span class="math-container">$X\ne \emptyset$</span>, and you should start from <span class="math-container">$Y$</span> or <span class="math-container">$Z$</span> and not the products.</p>
<p>Let <span class="math-container">$y\in Y$</span> be arbitrary. As <span class="math-container">$X\ne\emptyset$</span>, there exists some <span class="math-container">$x\in X$</span>.
Then <span class="math-container">$(x,y)\in X\times Y=X\times Z$</span>, i.e., <span class="math-container">$(x,y)=(u,v)$</span> for some <span class="math-container">$u\in X$</span>, <span class="math-container">$v\in Z$</span>. By definition of ordered pairs, this means (<span class="math-container">$x=u$</span> and) <span class="math-container">$y=v$</span>. In particular, <span class="math-container">$y\in Z$</span>. We conclude <span class="math-container">$Y\subseteq Z$</span>.
The other direction, <span class="math-container">$Z\subseteq Y$</span> , follows the same way,</p>
|
1,032,650 | <p><img src="https://i.stack.imgur.com/GVk1i.png" alt="enter image description here"></p>
<p>Here, ABCD is a rectangle, and BC = 3 cm. An Equilateral triangle XYZ is inscribed inside the rectangle as shown in the figure where YE = 2 cm. YE is perpendicular to DC. Calculate the length of the side of the equilateral triangle XYZ.</p>
| Jack D'Aurizio | 44,121 | <p>Consider the reference system with the origin in $Z$ in which $DC$ is the real axis, and let $EZ=a$. Then we have $Y=a+2i$ and:</p>
<p>$$e^{\pi i/6}(a+2i) = X $$
so:
$$\Im \left[\left(\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)\cdot\left(a+2i\right)\right]=3, $$
or:
$$ 1+\frac{\sqrt{3}}{2}a = 3 $$
so $a = \frac{4}{\sqrt{3}}$, and by the Pythagorean theorem:
$$ ZY^2 = a^2 + 4 = \frac{16}{3}+4 = \frac{28}{3} $$
so the side length is $2\sqrt{\frac{7}{3}}$.</p>
|
3,242,363 | <blockquote>
<p>Why does this function, <span class="math-container">$$\tan\left(x ^ {1/x}\right)$$</span>
have a maximum value at <span class="math-container">$x=e$</span>?</p>
</blockquote>
<p><a href="https://i.stack.imgur.com/pqE0Q.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pqE0Q.png" alt="Graph"></a></p>
| Acccumulation | 476,070 | <p>The general term for this is "simultaneous equations". In this case, as Ovi notes, they are linear equations. You can write is as</p>
<p><span class="math-container">$$1a+1b+-y=-x$$</span>
<span class="math-container">$$1a-.029b=.029x+.3$$</span>
<span class="math-container">$$.015a-1b=-.015x$$</span></p>
<p>In matrix form, that's</p>
<p><span class="math-container">$\begin{bmatrix}1&1&1\\1&-.029&-1\\ .015&.015&0\end{bmatrix}\begin{bmatrix}a\\b\\y \end{bmatrix}=\begin{bmatrix}-x\\.029x+.3\\-.015x \end{bmatrix}$</span></p>
<p>If you find the inverse of that matrix on the left (a web search of "matrix calculator" should get you a page that can give you the inverse), and then do a matrix multiplication of the inverse and <span class="math-container">$\begin{bmatrix}-x\\.029x+.3\\-.015x \end{bmatrix}$</span> will get you <span class="math-container">$a$</span>, <span class="math-container">$b$</span>, and <span class="math-container">$y$</span>.</p>
|
554,025 | <p>I have a question as such:</p>
<blockquote>
<p>Class A has 45 students in it, and class B has 30 students in it. In class A, every student attends any particular lecture with probability 0.7 independent of the other students. For class B, two thirds of lectures are attended by everyone, with probability 1/3 that a student is missing.</p>
<p>Suppose you are looking for Class A, but the doors are not labelled. You open one of the two doors at random, and see 30 students. What is the probability that you opened the right door?</p>
</blockquote>
<p>My work so far:</p>
<p>I worked out the expected value of the number of students in each class: for class <span class="math-container">$A$</span>, it is <span class="math-container">$31.5$</span>, and for class <span class="math-container">$B$</span>, it is <span class="math-container">$29\frac23$</span>. Given this, it is likelier that I have opened the wrong door - but I don't know how to work out the probabilities to arrive at a precise numerical statement. Can anyone help me out?</p>
| N. S. | 9,176 | <p>Pick any five vertices. Then, as $K_5$ is not planar, there is a pair of edges in the graph generated by these 5 vertices which cross. </p>
<p>Thus, we get at least one crossing for every combination of $5$ vertices. In total, our count is at least $\binom{n}{5}$.</p>
<p>But, we might had counted the same crossing multiple times. Look at a crossing: lets say $v_iv_j$ and $v_kv_l$.We counted this crossing at most $n-4$ times: we might have counted it in the quintle $v_i v_j v_k v_l v_s$ where $1 \leq s \leq n$ and $s \neq i, s \neq j , s\neq l s\neq m$. </p>
<p>If $k$ is the number of crossings, and we counted each crossing at most $n-4$ times, our count is at most $k(n-4)$. Thus</p>
<p>$$k(n-4) \geq \binom{n}{5}\,.$$</p>
<p>Thus the number of crossings is at least </p>
<p>$$\frac{\binom{n}{5}}{n-4}=\frac{1}{5}\binom{n}{4}$$</p>
|
2,828,487 | <p>If $\mathcal{R}$ is a von Neumann algebra acting on Hilbert space $H$, and $v \in H$ is a cyclical and separating vector for $\mathcal{R}$ (hence also for its commutant $\mathcal{R}'$), and $P \in \mathcal{R}, Q \in \mathcal{R}'$ are nonzero projections, can we have $PQv = 0$?</p>
<p>[note i had briefly edited this to a reformulated version of the question, but am rolling it back to align with the answer below.]</p>
| José Carlos Santos | 446,262 | <p>As far as textbooks are concerned, I learned Real Analysis mainly from Michael Spivak's <em>Calculus</em>. My contact with Walter Rudin's textbooks started only after graduation, but I also have a very high opinion about them.</p>
|
2,828,487 | <p>If $\mathcal{R}$ is a von Neumann algebra acting on Hilbert space $H$, and $v \in H$ is a cyclical and separating vector for $\mathcal{R}$ (hence also for its commutant $\mathcal{R}'$), and $P \in \mathcal{R}, Q \in \mathcal{R}'$ are nonzero projections, can we have $PQv = 0$?</p>
<p>[note i had briefly edited this to a reformulated version of the question, but am rolling it back to align with the answer below.]</p>
| Thomas | 26,188 | <p><strong>People will have different experiences of doing research.</strong> </p>
<p>Typically in a graduate program in the US one starts with taking general classes. After covering some basic classes (often in algebra, topology, and analysis (and maybe others)) you might start to take more specialized classes. At this point you might also start working with a specific professor on a project. It is really dependent on the project and you how "fast" you get to do actual research. Some professors will have students start almost immediately with smaller projects. Others will insist that you cover certain background before giving you a project to work on.</p>
<p>So, it is really difficult to say how long to "get to a serious level". My opinion is that it all is serious. Doing research now I often find my self consulting graduate (or even undergraduate) level textbooks. Often I have questions that might be answered in a graduate level class! In general I would say to <strong>not worry so much about whether the level is "serious".</strong></p>
<p>As for books, I used both Rudin and Royden. </p>
<p>The best way to know about summer programs is to ask faculty in the math department. They often have an undergraduate advisor. You can also ask your current instructors if they know anything. At least you would probably need to seek our letters of recommendation from past instructors. You institution might even have funding for you. You can also look online for programs. I always get excited when a student approaches me about wanting to do a summer program!</p>
|
3,070,258 | <p>In a (Partial Differential Equations / Laplace Equation) , I try to solving a problem of Laplace eq. by using separation of variables method.</p>
<p>I usually using the rule : if <span class="math-container">$e^{2 \sqrt{k} b} = 1$</span>, then I have: <span class="math-container">$2\sqrt{k} b = 2ni\pi$</span>. </p>
<p>Now in my problem I have : <span class="math-container">$e^{2 \sqrt{k}\pi} = 1$</span>
Can I use the same rule which lead to cancel the <span class="math-container">$\pi$</span> ?</p>
| Rhys Hughes | 487,658 | <p>You're using Euler's formula for complex numbers</p>
<p><span class="math-container">$$e^{i\theta}=\cos(\theta)+i\sin(\theta)$$</span></p>
<p>We know that <span class="math-container">$\cos(2n\pi)=1, \sin(2n\pi)=0$</span>, for <span class="math-container">$n\in \Bbb Z$</span> so:</p>
<p><span class="math-container">$$n\in \Bbb Z \implies e^{2ni\pi}=1$$</span></p>
<p>You have, therefore, in the first part of your question:</p>
<p><span class="math-container">$$2b\sqrt k =2ni\pi\to b\sqrt k=ni\pi\to k=-(\frac{n\pi}{b})^2$$</span></p>
<p>This CAN be applied to part b as well:</p>
<p><span class="math-container">$$2\pi\sqrt k=2ni\pi\to\sqrt k=ni\to k=-n^2$$</span></p>
|
930,611 | <blockquote>
<p>Find the maximal value of the function for $a=24.3$, $b=41.5$:
$$f(x,y)=xy\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}$$</p>
</blockquote>
<p>Using the second derivative test for partial derivatives, I find the critical point in terms of $a$ and $b$ by taking partial derivatives of $x$ and $y$ and equating them to $0$. </p>
<p>$$f_y=0$$
$$f_x=0$$</p>
<p>Getting</p>
<p>$$y \left ( 1-\frac{x^2}{a^2}-\frac{y^2}{b^2} \right ) -\frac{yx^2}{a^2}=0$$</p>
<p>and</p>
<p>$$x \left ( 1-\frac{x^2}{a^2}-\frac{y^2}{b^2} \right ) - \frac{xy^2}{b^2}=0$$</p>
<p>Then i combine both equations together to get</p>
<p>$$\left ( 1-\frac{y^2}{b^2} \right ) =\frac{2x^2}{a^2}$$</p>
<p>and</p>
<p>$$\left ( 1-\frac{x^2}{a^2} \right ) =\frac{2y^2}{b^2}$$</p>
<p>Solving both equations to get the critical points in terms of a and b. I got</p>
<p>$$(0,0)$$<br>
$$\left(\frac{a}{\sqrt{3}},\frac{b}{\sqrt{3}}\right)$$</p>
<p>Hence to get the maximum value I substitute the second critical point back into the original function. And I let $a=24.3$, $b=41.5$ to get the maximal point. However, I don't seem to get the right answer.</p>
<p>Is my method correct?</p>
| Macavity | 58,320 | <p>Another way is to consider equivalently the maximum of
$$\frac{f^2}{a^2b^2}=\frac{x^2}{a^2} \frac{y^2}{b^2} \left(1-\frac{x^2}{a^2} -\frac{y^2}{b^2} \right)$$
which is the product of three positive terms with constant sum, so each term must be equal to (in this case) $\frac13$ at maximum.</p>
|
1,936,043 | <p>I would like to prove that the sequence $n^{(-1)^{n}}$ is divergent. </p>
<p>My thoughts: I know $(-1)^n$ is divergent, so $n$ to the power of a divergent sequence is still divergent? I am not sure how to give a proper proof, pls help!</p>
| Bernard | 202,857 | <p>Set $u_n=n^{(-1)^n}$? Explicitly, $u_{2n}=2n$, $u_{2n+1}=\dfrac1{2n+1}$.</p>
<p>If the sequence were convergent, all subsequences would converge to the same limit. However we see the subsequence of odd terms converges to $0$, whereas the subsequence of even terms tends to $+\infty$.</p>
|
2,011,181 | <blockquote>
<p><strong>Question:</strong> Find the area of the shaded region given $EB=2,CD=3,BC=10$ and $\angle EBC=\angle BCD=90^{\circ}$.</p>
</blockquote>
<p><a href="https://i.stack.imgur.com/BFf2h.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BFf2h.jpg" alt="Diagram"></a></p>
<p>I first dropped an altitude from $A$ to $BC$ forming two cases of similar triangles. Let the point where the altitude meets $BC$ be $X$. Thus, we have$$\triangle BAX\sim\triangle BDC\\\triangle CAX\sim\triangle CEB$$
Using the proportions, we get$$\frac {BA}{BD}=\frac {AX}{CD}=\frac {BX}{BC}\\\frac {CA}{CE}=\frac {AX}{EB}=\frac {CX}{CB}$$
But I'm not too sure what to do next from here. I feel like I'm <em>very</em> close, but I just can't figure out $AX$.</p>
| kotomord | 382,886 | <p>Coordinates:</p>
<p>B (0,0)</p>
<p>E(0,2)</p>
<p>C (10, 0)</p>
<p>D( 10, 3)</p>
<p>BD 10y - 3x = 0</p>
<p>EC x+5y = 10</p>
<p>Find y coordinate of A:
10y - 3x = 0, 3x+15y = 30 => 25y = 30 => y = 6/5</p>
<p>Area size is 10*(6/5)/2 = 6;</p>
|
451,063 | <p>Alright so I am having the following issue: I want to figure out how to find the fourier coefficients of the following function:
$$D(X)=\frac {a'(x)} {1+a'(x)^2}$$</p>
<p>Where $a(x)$ is an arbitrary function. I already have a model for finding the fourier coefficients for $a(x)$ and $a'(x)$:</p>
<pre><code>fc = fft(a) / Nfft;
fc = fftshift(fc); % fft of a(x)
fc = conj(fc); % sign correction
aprimec = -i * [0:Dim2-1] .* fc; % fc of derivative (definition)
</code></pre>
<p>The equation I am given to use is:
$$f_m=\frac 1 N \sum^Nf_ie^{+im2\pi x}$$</p>
<p>Which confuses me because of the $f_i$. So does any one have any suggestions? </p>
<p>Additionally, I do not know how to define d</p>
<pre><code>d = diff(a)/(1+diff(a)^2);
</code></pre>
<p>I do not think that this would work because doesn't <code>diff(x)</code> just take the difference between two consecutive components in the vector?</p>
<p>I would greatly appreciate any help. Thanks!</p>
| AnonSubmitter85 | 33,383 | <p>The DFT is isomorphic, so if you have the DFT coefficients for $a'$, then all you need to do to get $a'$ is apply the inverse DFT:</p>
<pre><code>D = ifft(aprimec) ./ (1 + ifft(aprimec).^2);
</code></pre>
|
4,378,215 | <p>Is there are a big-O notation for a function <span class="math-container">$f(x)$</span> that tends <span class="math-container">$\infty$</span> at an arbitrarily small rate?</p>
<p>Obviously, the expression <span class="math-container">$\mathcal{O(1)}<f(x)<\mathcal{O}(\sqrt{\log(x)})$</span> is not good enough (I just randomly choose <span class="math-container">$\sqrt{\log(x)}$</span>. What is a smart expression that I can replace <span class="math-container">$\sqrt{\log(x)}$</span> with?</p>
| Thomas Lesgourgues | 601,841 | <p>You want to use the notation <span class="math-container">$f(x)=\omega(1)$</span>.</p>
<p>The notation <span class="math-container">$f = \omega(g)$</span>, <span class="math-container">$f$</span> dominate <span class="math-container">$g$</span>, is defined as
<span class="math-container">$$\forall k > 0 \, \exists n_0 \, \forall n > n_0 \colon |f(n)| > k\cdot g(n)$$</span>
or
<span class="math-container">$$\lim_{n \to \infty} \frac{\left|f(n)\right|}{g(n)} = \infty$$</span></p>
|
4,378,215 | <p>Is there are a big-O notation for a function <span class="math-container">$f(x)$</span> that tends <span class="math-container">$\infty$</span> at an arbitrarily small rate?</p>
<p>Obviously, the expression <span class="math-container">$\mathcal{O(1)}<f(x)<\mathcal{O}(\sqrt{\log(x)})$</span> is not good enough (I just randomly choose <span class="math-container">$\sqrt{\log(x)}$</span>. What is a smart expression that I can replace <span class="math-container">$\sqrt{\log(x)}$</span> with?</p>
| Arthur | 15,500 | <p>There isn't a good way to do this with big-<span class="math-container">$O$</span> notation. No matter what expression you put into it, you can take a logarithm or a square root to find a significantly smaller function that still goes to infinity.</p>
<p>But there are other (albeit less common) notations for similar notions. In your case, you want <span class="math-container">$$f(x)\in \omega(1)$$</span>This says that <span class="math-container">$f$</span> dominates <span class="math-container">$1$</span>. Which is to say, for any multiple of <span class="math-container">$1$</span>, <span class="math-container">$f$</span> is eventually bigger.</p>
|
19,356 | <p>So I was wondering: are there any general differences in the nature of "what every mathematician should know" over the last 50-60 years? I'm not just talking of small changes where new results are added on to old ones, but fundamental shifts in the nature of the knowledge and skills that people are expected to acquire during or before graduate school.</p>
<p>To give an example (which others may disagree with), one secular (here, secular means "trend over time") change seems to be that mathematicians today are expected to feel a lot more comfortable with picking up a new abstraction, or a new abstract formulation of an existing idea, even if the process of abstraction lies outside that person's domain of expertise. For example, even somebody who knows little of category theory would not be expected to bolt if confronted with an interpretation of a subject in his/her field in terms of some new categories, replete with objects, morphisms, functors, and natural transformations. Similarly, people would not blink much at a new algebraic structure that behaves like groups or rings but is a little different.</p>
<p>My sense would be that the expectations and abilities in this regard have improved over the last 50-60 years, partly because of the development of "abstract nonsense" subjects including category theory, first-order logic, model theory, universal algebra etc., and partly because of the increasing level of abstraction and the need for connecting frameworks and ideas even in the rest of mathematics. I don't really know much about how mathematics was taught thirty years ago, but I surmised the above by comparing highly accomplished professional mathematicians who probably went to graduate school thirty years ago against today's graduate students.</p>
<p>Some other guesses:</p>
<ol>
<li>Today, people are expected to have a lot more of a quick idea of a larger number of subjects, and less of an in-depth understanding of "Big Proofs" in areas outside their subdomain of expertise. Basically, the Great Books or Great Proofs approach to learning may be declining. The rapid increase in availability of books, journals, and information via the Internet (along with the existence of tools such as Math Overflow) may be making it more profitable to know a bit of everything rather than master big theorems outside one's area of specialization.</li>
<li>Also, probably a thorough grasp of multiple languages may be becoming less necessary, particularly for people who are using English as their primary research language. Two reasons: first, a lot of materials earlier available only in non-English languages are now available as English translations, and second, translation tools are much more widely available and easy-to-use, reducing the gains from mastery of multiple languages.</li>
</ol>
<p>These are all just conjectures. Contradictory information and ideas about other possible secular trends would be much appreciated.</p>
<p>NOTE: This might be too soft for Math Overflow! Moderators, please feel free to close it if so.</p>
| Steve Huntsman | 1,847 | <p>Practically, mathematicians today should know the rudiments of programming in at least one language (Mathematica and MATLAB count). They should know the basics of probability and linear algebra. They should know these three things because if they get jobs outside of academia they will generally be expected to use at least two of these three, and probably all of them.</p>
<p>Mathematicians should know how to use the internet and how to learn there. They need not recall many formulas, as the convenience of having them at one's fingertips can be "outsourced" to the internet. By the same token, they need not even recall most of what they have learned, but instead should be able to refresh their memory quickly.</p>
<p>Classical and complex analysis have clearly (I think) become less important to command in detail. Combinatorics and algebra have become more so. This is because of computers, and the interplay between mathematics and technology more generally.</p>
|
19,356 | <p>So I was wondering: are there any general differences in the nature of "what every mathematician should know" over the last 50-60 years? I'm not just talking of small changes where new results are added on to old ones, but fundamental shifts in the nature of the knowledge and skills that people are expected to acquire during or before graduate school.</p>
<p>To give an example (which others may disagree with), one secular (here, secular means "trend over time") change seems to be that mathematicians today are expected to feel a lot more comfortable with picking up a new abstraction, or a new abstract formulation of an existing idea, even if the process of abstraction lies outside that person's domain of expertise. For example, even somebody who knows little of category theory would not be expected to bolt if confronted with an interpretation of a subject in his/her field in terms of some new categories, replete with objects, morphisms, functors, and natural transformations. Similarly, people would not blink much at a new algebraic structure that behaves like groups or rings but is a little different.</p>
<p>My sense would be that the expectations and abilities in this regard have improved over the last 50-60 years, partly because of the development of "abstract nonsense" subjects including category theory, first-order logic, model theory, universal algebra etc., and partly because of the increasing level of abstraction and the need for connecting frameworks and ideas even in the rest of mathematics. I don't really know much about how mathematics was taught thirty years ago, but I surmised the above by comparing highly accomplished professional mathematicians who probably went to graduate school thirty years ago against today's graduate students.</p>
<p>Some other guesses:</p>
<ol>
<li>Today, people are expected to have a lot more of a quick idea of a larger number of subjects, and less of an in-depth understanding of "Big Proofs" in areas outside their subdomain of expertise. Basically, the Great Books or Great Proofs approach to learning may be declining. The rapid increase in availability of books, journals, and information via the Internet (along with the existence of tools such as Math Overflow) may be making it more profitable to know a bit of everything rather than master big theorems outside one's area of specialization.</li>
<li>Also, probably a thorough grasp of multiple languages may be becoming less necessary, particularly for people who are using English as their primary research language. Two reasons: first, a lot of materials earlier available only in non-English languages are now available as English translations, and second, translation tools are much more widely available and easy-to-use, reducing the gains from mastery of multiple languages.</li>
</ol>
<p>These are all just conjectures. Contradictory information and ideas about other possible secular trends would be much appreciated.</p>
<p>NOTE: This might be too soft for Math Overflow! Moderators, please feel free to close it if so.</p>
| Joel David Hamkins | 1,946 | <p>As mathematics grows and diversifies beyond belief, surely the collection of topics that <em>every</em> mathematician must know is shrinking fast. One can carry out serious mathematical research in one area while knowing very little of another, even when many mathematicians regard that other area as fundamentally important. Thus, the assumption in the question that there is anything substantial in the list of topics that ALL mathematicians must know seems to me unwarranted. Of course, the interdisciplinary work that connects widely separated research areas is often very important (as well as difficult), but a lot of progress is also made within the various specialities without interacting with other areas. But for someone to to insist that every mathematician must know category theory, say, or homology, seems to exhibit just as narrow a conception of mathematics as to insist that every mathematician must know how to program. There have been profound mathematical advances in subjects requiring none of that knowledge. All other things being equal, of course, a mathematician would be better off knowing some category theory or logic or homology or programming, but in practice, all other things are not equal, since we must all choose how best to spend our time, choosing the topics that seem most relevant to the research we seek to undertake. </p>
<p>Ultimately, we need all kinds of mathematicians: some who are deeply specialized, some who know various areas to build the bridges that can connect diverse subjects, some who know how to communicate ideas from one area to another, and others who know how to communicate the deep ideas of one area to the future specialists in that area, or to the public. Perhaps the intersection of the knowledge of all these people is rather smaller than one might think, and this isn't necessarily a problem.</p>
<p>Contemporary mathematical research is indeed a big tent, as Charlie Frohman said in the comments.</p>
|
19,356 | <p>So I was wondering: are there any general differences in the nature of "what every mathematician should know" over the last 50-60 years? I'm not just talking of small changes where new results are added on to old ones, but fundamental shifts in the nature of the knowledge and skills that people are expected to acquire during or before graduate school.</p>
<p>To give an example (which others may disagree with), one secular (here, secular means "trend over time") change seems to be that mathematicians today are expected to feel a lot more comfortable with picking up a new abstraction, or a new abstract formulation of an existing idea, even if the process of abstraction lies outside that person's domain of expertise. For example, even somebody who knows little of category theory would not be expected to bolt if confronted with an interpretation of a subject in his/her field in terms of some new categories, replete with objects, morphisms, functors, and natural transformations. Similarly, people would not blink much at a new algebraic structure that behaves like groups or rings but is a little different.</p>
<p>My sense would be that the expectations and abilities in this regard have improved over the last 50-60 years, partly because of the development of "abstract nonsense" subjects including category theory, first-order logic, model theory, universal algebra etc., and partly because of the increasing level of abstraction and the need for connecting frameworks and ideas even in the rest of mathematics. I don't really know much about how mathematics was taught thirty years ago, but I surmised the above by comparing highly accomplished professional mathematicians who probably went to graduate school thirty years ago against today's graduate students.</p>
<p>Some other guesses:</p>
<ol>
<li>Today, people are expected to have a lot more of a quick idea of a larger number of subjects, and less of an in-depth understanding of "Big Proofs" in areas outside their subdomain of expertise. Basically, the Great Books or Great Proofs approach to learning may be declining. The rapid increase in availability of books, journals, and information via the Internet (along with the existence of tools such as Math Overflow) may be making it more profitable to know a bit of everything rather than master big theorems outside one's area of specialization.</li>
<li>Also, probably a thorough grasp of multiple languages may be becoming less necessary, particularly for people who are using English as their primary research language. Two reasons: first, a lot of materials earlier available only in non-English languages are now available as English translations, and second, translation tools are much more widely available and easy-to-use, reducing the gains from mastery of multiple languages.</li>
</ol>
<p>These are all just conjectures. Contradictory information and ideas about other possible secular trends would be much appreciated.</p>
<p>NOTE: This might be too soft for Math Overflow! Moderators, please feel free to close it if so.</p>
| Kevin O'Bryant | 935 | <p>One thing I'm sure we'll <em>all</em> agree on: every mathematician should know some flavor of TeX!</p>
|
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