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 |
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3,506,091 | <p>Solve <span class="math-container">$2^m=7n^2+1$</span> with <span class="math-container">$(m,n)\in \mathbb{N}^2$</span></p>
<p>Here is what I did:
First try, I have seen first that the obvious solutions are <span class="math-container">$n=1$</span> and <span class="math-container">$m=3$</span> , and <span class="math-container">$n=3$</span> and <span class="math-container">$m=6$</span>, then I proved by simple congruences that <span class="math-container">$m$</span> must be divisible by <span class="math-container">$3$</span> so <span class="math-container">$m=3k$</span>, If we add <span class="math-container">$27$</span> to the equation we will have <span class="math-container">$2^{3k}+3^3=7(n^2+2^2)$</span>, but unfortunately I tried to do something with Legendre symbol or the multiplicative order but I found nothing interesting.</p>
<p>Second try,I let <span class="math-container">$n=2k+1$</span> then I worked in <span class="math-container">$\mathbb{Z}\left[ \frac{-1+\sqrt{-7}}{2} \right] $</span> and the equation becomes <span class="math-container">$7\times 2^{m-2}=\left( 7k+4+\frac{-1+\sqrt{-7}}{2} \right) \left( 7k+3-\frac{-1+\sqrt{-7}}{2} \right) $</span> but I didn't find something interesting because the two factors are not coprime.</p>
| Mastrem | 253,433 | <p><strong>tl;dr</strong> The only solutions are <span class="math-container">$(m,n)=(6,3)$</span> and <span class="math-container">$(m,n)=(3,1)$</span>.</p>
<p>This answer is rather long and has been split in two parts. In part one, we show that, with the exception of <span class="math-container">$(6,3)$</span> we must have <span class="math-container">$m$</span> a power of <span class="math-container">$3$</span> for all solutions <span class="math-container">$(m,n)$</span>.</p>
<p>In part <span class="math-container">$2$</span>, we show that <span class="math-container">$(r,n)=(1,1)$</span> is the only solution to
<span class="math-container">$$2^{3^r}=7n^2+1.$$</span></p>
<h1>Part 1</h1>
<p>Let <span class="math-container">$(m,n)$</span> be any solution to the equation
<span class="math-container">$$2^m=7n^2+1.$$</span>
First, suppose that <span class="math-container">$m$</span> is even. Write <span class="math-container">$m=2k$</span> and note that the equation is equivalent with
<span class="math-container">$$(2^k+1)(2^k-1)=7n^2.$$</span>
Since <span class="math-container">$2^k+1$</span> and <span class="math-container">$2^k-1$</span> are both odd and just two apart, they're coprime. This means that one of these factors is a square and the other is seven times a square. Note that <span class="math-container">$2^k+1$</span> is congruent to one of <span class="math-container">$2$</span>, <span class="math-container">$3$</span> and <span class="math-container">$5$</span> modulo <span class="math-container">$7$</span>. Therefore, it must be a square. Write <span class="math-container">$2^k+1=l^2$</span>, then
<span class="math-container">$$2^k = (l+1)(l-1),$$</span>
so both <span class="math-container">$l-1$</span> and <span class="math-container">$l+1$</span> are powers of <span class="math-container">$2$</span>. Clearly, the only solution is <span class="math-container">$k=l=3$</span>, which corresponds to the solution <span class="math-container">$(m,n)=(6,3)$</span> of the original equation.</p>
<hr />
<p>Otherwise, <span class="math-container">$m$</span> is odd. Write <span class="math-container">$m=3^rs$</span> with <span class="math-container">$s$</span> odd and not divisible by <span class="math-container">$3$</span>. Note that
<span class="math-container">$$7n^2=2^{3^rs}-1=(2^s-1)\sum_{i=0}^{3^r-1}2^{si}.$$</span>
Let <span class="math-container">$p$</span> be any prime dividing <span class="math-container">$2^s-1$</span>. We have <span class="math-container">$2^{si}\equiv 1\pmod p$</span> for all <span class="math-container">$i$</span>, meaning that
<span class="math-container">$$\sum_{i=0}^{3^r-1}2^{si}\equiv 3^r\pmod p.$$</span>
Because <span class="math-container">$s$</span> is odd, <span class="math-container">$p\neq 3$</span>. We conclude that <span class="math-container">$2^s-1$</span> and <span class="math-container">$\prod_{i=0}^{3^r-1}2^{si}$</span> are coprime. Therefore, <span class="math-container">$2^s-1$</span> is either a square or seven times a square. Since <span class="math-container">$3\nmid s$</span>, it cannot not be the latter. Because <span class="math-container">$2^s-1$</span> is an odd square, it must be congruent to <span class="math-container">$1$</span> modulo <span class="math-container">$8$</span>, so <span class="math-container">$s<3$</span>. Because <span class="math-container">$s$</span> is odd, it must be <span class="math-container">$1$</span> and <span class="math-container">$m=3^r$</span>.</p>
<h1>Part 2</h1>
<p>The key result here is:</p>
<p><strong>Lemma:</strong> Let <span class="math-container">$a\in\mathbb{Z}_{\ge 1}$</span> be given with <span class="math-container">$a\pmod 3\in \{0,2\}$</span>. All elements of the sequence
<span class="math-container">$$\left\{1+a^{2\cdot 3^n}+a^{3^n}\right\}_{n\in\mathbb{Z}_{\ge 0}}$$</span>
are pairwise coprime.</p>
<p><strong>Proof:</strong> Let <span class="math-container">$k,l$</span> be non-negative integers, <span class="math-container">$k>l$</span>. We have the polynomial identities
<span class="math-container">$$
\begin{align*}
X^{3^k}-1 &= \left(X^{3^{l+1}}-1\right)\sum_{i=0}^{3^{k-l-1}-1}X^{3^{l+1}i}\\
&= \left(1+X^{2\cdot 3^l}+X^{3^l}\right)\left(X^{3^l}-1\right)\sum_{i=0}^{3^{k-l-1}-1}X^{3^{l+1}i}
\end{align*}
$$</span>
and
<span class="math-container">$$1+X^{2\cdot 3^k}+X^{3^k}=X^{3^k}\left(X^{3^k}+1\right)+1.$$</span>
It follows that
<span class="math-container">$$1+X^{2\cdot 3^k}+X^{3^k}=q\left(1+X^{2\cdot 3^l}+X^{3^l}\right)+3,$$</span>
where
<span class="math-container">$$q=\left(1+X^{2\cdot 3^l}+X^{3^l}\right)(X^{3^l}-1)^2\left[\sum_{i=0}^{3^{k-l-1}-1}X^{3^{l+1}i}\right]^2+3(X^{3^l}-1)\sum_{i=0}^{3^{k-l-1}-1}X^{3^{l+1}i}$$</span>
has integer coefficients. Evaluating in <span class="math-container">$a$</span> yields
<span class="math-container">$$1+a^{2\cdot 3^{k}}+a^{3^k}=q(a)\left(1+a^{2\cdot 3^l}+a^{3^l}\right)+3,$$</span>
where <span class="math-container">$q(a)$</span> is an integer. By Fermat's little theorem,
<span class="math-container">$$1+a^{2\cdot 3^{k}}+a^{3^k}\equiv 1+a^{2\cdot 3^{l}}+a^{3^l}\equiv 2a+1\not\equiv 0\pmod 3.$$</span>
Hence, <span class="math-container">$1+a^{2\cdot 3^k}+a^{3^k}$</span> and <span class="math-container">$1+a^{2\cdot 3^l}+a^{3^l}$</span> are coprime. Q.E.D.</p>
<hr />
<p>Consider any solution <span class="math-container">$(r,n)$</span> to
<span class="math-container">$$2^{3^r}=7n^2+1.$$</span>
Note that this is equivalent with
<span class="math-container">$$
\begin{align*}
7n^2 &= 2^{3^r}-1\\
&= \prod_{i=0}^{r-1}\frac{2^{3^{i+1}}-1}{2^{3^i}-1}\\
&= \prod_{i=0}^{r-1}\left(1+2^{2\cdot 3^i}+2^{3^i}\right)\\
&= 7\prod_{i=1}^{r-1}\left(1+2^{2\cdot 3^i}+2^{3^i}\right).
\end{align*}
$$</span>
By the lemma, the factors in the final product must be pairwise coprime, so they must be squares. However, the factor given by <span class="math-container">$i=1$</span> is
<span class="math-container">$$1+2^{2\cdot 3}+2^3=1+64+8=73,$$</span>
which isn't a square. Therefore, <span class="math-container">$r\le 1$</span>. We find that <span class="math-container">$(r,n)=(1,1)$</span> is the only solution.</p>
|
3,506,091 | <p>Solve <span class="math-container">$2^m=7n^2+1$</span> with <span class="math-container">$(m,n)\in \mathbb{N}^2$</span></p>
<p>Here is what I did:
First try, I have seen first that the obvious solutions are <span class="math-container">$n=1$</span> and <span class="math-container">$m=3$</span> , and <span class="math-container">$n=3$</span> and <span class="math-container">$m=6$</span>, then I proved by simple congruences that <span class="math-container">$m$</span> must be divisible by <span class="math-container">$3$</span> so <span class="math-container">$m=3k$</span>, If we add <span class="math-container">$27$</span> to the equation we will have <span class="math-container">$2^{3k}+3^3=7(n^2+2^2)$</span>, but unfortunately I tried to do something with Legendre symbol or the multiplicative order but I found nothing interesting.</p>
<p>Second try,I let <span class="math-container">$n=2k+1$</span> then I worked in <span class="math-container">$\mathbb{Z}\left[ \frac{-1+\sqrt{-7}}{2} \right] $</span> and the equation becomes <span class="math-container">$7\times 2^{m-2}=\left( 7k+4+\frac{-1+\sqrt{-7}}{2} \right) \left( 7k+3-\frac{-1+\sqrt{-7}}{2} \right) $</span> but I didn't find something interesting because the two factors are not coprime.</p>
| Community | -1 | <p>HINT</p>
<p><span class="math-container">$2^m\equiv 1$</span> mod <span class="math-container">$7$</span> and so <span class="math-container">$m=3k$</span>. For <span class="math-container">$n>0$</span>, we now have
<span class="math-container">$$2^k-1=au^2, 2^{2k}+2^k+1=bv^2$$</span>
where either <span class="math-container">$\{a,b\}=\{1,7\}$</span> or <span class="math-container">$\{a,b\}=\{3,21\}.$</span></p>
<p>Each of the four possibilities gives an elliptic curve <span class="math-container">$$bv^2=3+3au^2+a^2u^4.$$</span></p>
<p>Of these, the case <span class="math-container">$a=7,b=1$</span> is impossible modulo <span class="math-container">$7$</span>. </p>
|
885,450 | <blockquote>
<p>After covering a distance of 30Km with a uniform speed, there got some
defect in train engine and therefore its speed is reduced to 4/5 of
its original speed. Consequently, the train reaches its destination 45
minutes late. If it had happened after covering 18Km of distance, the
train would have reached 9 minutes earlier>Find the speed of the train
and the distance of the journey.</p>
</blockquote>
<p>This is my question. .</p>
<p>now after letting the original speed of train be $x Km/Hr$ and the time taken be y Hr, threfore
distance = $xy$</p>
<p>CASE I:</p>
<p>speed = $x-4x/5$ Km/hr</p>
<p>time = $y + 45/60$</p>
<p>$xy=60xy/300 +45x/300 => +300xy-60xy = 45x => 240xy = 45x$ .............[i]</p>
<p>similarly in CASE II we get the equation:</p>
<p>$240xy = -9x$.........[ii]</p>
<p>but after solving these two equations the answer is coming to 0 which is wrong please tell me the correct solution.</p>
<p>thanks</p>
<p>(fast please)</p>
| Start wearing purple | 73,025 | <p>The logarithm of the expression under the limit can be rewritten as
$$\sum_{k=0}^{2^{2^n-1}}\ln\left(1+\frac{1}{2^{2^n}+2k}\right)=\sum_{k=0}^{2^{2^n-1}}\frac{1}{2^{2^n}+2k}+O\left(2^{-2^n}\right).$$
Denoting $N=2^{2^n-1}$, it is easy to see that the limit of the logarithm can be computed as the limit of a Riemann sum:
$$\frac{1}{N}\sum_{k=0}^N\frac{1}{2\left(1+\frac{k}{N}\right)}\stackrel{N\rightarrow\infty} \longrightarrow \frac12\int_0^1\frac{dx}{1+x}=\ln\sqrt2.$$</p>
|
105,040 | <p>This question in stackExchange remained unanswered. </p>
<p>Let $\mathbb F$ be a finite field. Denote by $M_n(\mathbb F)$ the set of matrices of order $n$ over $\mathbb F$ . For a matrix $A∈M_n(\mathbb F)$ what is the cardinality of $C_{M_n(\mathbb F)} (A)$ , the centralizer of $A$ in $M_n(\mathbb F)$? There are papers about it? </p>
| Alireza Abdollahi | 19,075 | <p>Let me add some cases in which one has a clear answer:</p>
<p>[R.A. Horn, C.R. Johnson, Topics in Matrix Analysis, Cambridge University Press, Cambridge, 1991., Corollary 4.4.18]. Let $F$ be a field and $n$ is a natural number. If
$A\in M_n(F)$ is a cyclic matrix, then $C_{M_n(F)}(A)$ is the set of all matrices which are
polynomial in $A$ with coefficients in $F$.</p>
<p>Recall that a cyclic matrix in $M_n(F)$ is a matrix whose minimal and characteristic polynomials are the same.</p>
<p>[Lemma 3 of S. Akbari et al. / Linear Algebra and its Applications 390 (2004) 345–355] Let $F$ be a field and $n\geq 2$. If $A$ is a non-scalar matrix in $M_n(F)$ and
$C_{M_n(F)}(A)$ has maximum dimension over $F$, then $\dim_F C_{M_n(F)}(A) = n^2 − 2n + 2$
and $A$ is similar to either $aI_1 \oplus bI_{n−1}$ or $aI_n + bE_{12}$, for some $a, b \in F$. </p>
<p>See for the notation the latter mentioned paper.</p>
<p>I suggest you to look for papers on the commuting graphs of rings, you may find some other cases which are treated in the proofs.
One paper is quoted above and the another is
S. Akbari, P. Raja / Linear Algebra and its Applications 416 (2006) 1038–1047</p>
|
105,040 | <p>This question in stackExchange remained unanswered. </p>
<p>Let $\mathbb F$ be a finite field. Denote by $M_n(\mathbb F)$ the set of matrices of order $n$ over $\mathbb F$ . For a matrix $A∈M_n(\mathbb F)$ what is the cardinality of $C_{M_n(\mathbb F)} (A)$ , the centralizer of $A$ in $M_n(\mathbb F)$? There are papers about it? </p>
| Amritanshu Prasad | 9,672 | <p>Treat $F^n$ as an $F[t]$-module $M^A$, where $t$ acts by the matrix $A$. Then the centralizer can be thought of as $\mathrm{End}_{F[t]} M^A$. Now, $M^A$ has a primary decomposition</p>
<p>$ M^A = \bigoplus_{p \in \mathrm{Irr}(F[t])} M_p$</p>
<p>where $M_p$ consists of vectors in $M^A$ which are annihilated by some power of $p(A)$. Likewise, </p>
<p>$C_{M_n(F)}(A)= \mathrm{End}_{F[t]} M^A = \bigoplus_p \mathrm{End}_{F[t]} M_p$</p>
<p>So the problem is reduced to the primary case, where the characteristic polynomial of $A$ is a power of some irreducible polynomial $p$.</p>
<p>Now, there exists a unique partition $\lambda=(\lambda_1,\dotsc,\lambda_l)$ such that</p>
<p>$M_p = \bigoplus_{i=1}^l F[t]/(p(t))^{\lambda_i}$.</p>
<p>As a vector space (and even as an $F[t]$-module), the endomorphism algebra of this module is the sum</p>
<p>$\bigoplus_{i,j} \mathrm{Hom}_{F[t]} (F[t]/(p(t))^{\lambda_i},F[t]/(p(t))^{\lambda_j})$.</p>
<p>The $(i,j)$th summand has dimension $(\deg p)\min\{\lambda_i,\lambda_j\}$. Therefore, the endomorphism algebra of this primary part is of dimension</p>
<p>$ (\deg p)\sum_{i,j} \min\{\lambda_i,\lambda_j\}$</p>
<p>To get the centralizer of the original matrix, you would add these numbers over all primary parts.
Finally, raising $q$ to this number is the cardinality that you want.</p>
<p>These centralizers are discussed in great detail in Pooja Singla's PhD thesis <a href="http://www.hbni.ac.in/phdthesis/allthesis/MATH10200604007_PSingla.pdf" rel="nofollow">http://www.hbni.ac.in/phdthesis/allthesis/MATH10200604007_PSingla.pdf</a>
and a related paper in J. Algebra 2010 (available on the arXiv at <a href="http://arxiv.org/abs/1001.5304v1" rel="nofollow">http://arxiv.org/abs/1001.5304v1</a>).</p>
|
105,040 | <p>This question in stackExchange remained unanswered. </p>
<p>Let $\mathbb F$ be a finite field. Denote by $M_n(\mathbb F)$ the set of matrices of order $n$ over $\mathbb F$ . For a matrix $A∈M_n(\mathbb F)$ what is the cardinality of $C_{M_n(\mathbb F)} (A)$ , the centralizer of $A$ in $M_n(\mathbb F)$? There are papers about it? </p>
| Victor Miller | 2,784 | <p>I've recently come across this question about the size of the centralizer (I assume that you mean the subgroup of $\text{GL}_n(F)$ which commute with the matrix). It seems to be hard to find. I wonder why it's been omitted from every Algebra book (even Bourbaki!) that I've consulted. There is a derivation of it in Dickson's "Linear Groups with an Exposition of the Galois Field Theory" p. 235 (Dover edition) dating from 1900. A more modern way of seeing this is to let $R = F[x]$, and construct the $R$-module on a vector space $V$ of dimension $n$ (if $A$ is $n \times n$) by having $x v = A v$. Then the centralizer is $\text{Aut}_R(V)$. In MacDonald's book "Symmetric Functions and Hall Polynomials" p. 87 (1979 edition) he derives the formula in terms of the theory of finitely generated modules over a commutative discrete valuation ring. This, incidentally, also gives the formula for the number of automorphisms of a finite abelian group.</p>
|
300,745 | <p>If a function is uniformly continuous in $(a,b)$ can I say that its image is bounded?</p>
<p>($a$ and $b$ being finite numbers).</p>
<p>I tried proving and disproving it. Couldn't find an example for a non-bounded image. </p>
<p>Is there any basic proof or counter example for any of the cases?</p>
<p>Thanks a million!</p>
| N. S. | 9,176 | <p>Let $\delta$ be so that $|x-y|< \delta \Rightarrow |f(x)-f(y)|<1$. Now prove that there exists a finite set $F$ so that </p>
<p>$$(a,b) \subset \cup_{z \in F} (z-\frac{\delta}{2}, z+\frac{\delta}{2} ) \,.$$</p>
<p>Now, what can you say about $|f(x)|$ and $\max\{f(z)|z \in F \}+1\,.$?</p>
<p>P.S. The existence of the finite set is really the fact that $(0,1)$ is precompact in the topology of $\mathbb R$. So this solution is not really different than Ittay's...</p>
|
741,436 | <p>I get stuck at the following question:</p>
<p>Consider the matrix<br>
$$A=\begin{bmatrix}
0 & 2 & 0 \\
1 & 1 & -1 \\
-1 & 1 & 1\\
\end{bmatrix}$$</p>
<p>Find $A^{1000}$ by using the Cayley-Hamilton theorem.</p>
<p>I find the characteristic polynomial by $P(A) = -A^{3} + 2A^2 = 0$ (by Cayley-Hamilton) but I don't see how to find $A^{1000}$ by this characteristic polynomial.</p>
| Potato | 18,240 | <p>Your formula tells you, after you multiply through by $A^{997}$, that
$$A^{1000}=2A^{999}.$$
Similarly,
$$2A^{999}=4A^{998}.$$</p>
<p>This process can be repeated to find $A^{1000}$ in terms of $A^2$, which you can then compute. </p>
|
741,436 | <p>I get stuck at the following question:</p>
<p>Consider the matrix<br>
$$A=\begin{bmatrix}
0 & 2 & 0 \\
1 & 1 & -1 \\
-1 & 1 & 1\\
\end{bmatrix}$$</p>
<p>Find $A^{1000}$ by using the Cayley-Hamilton theorem.</p>
<p>I find the characteristic polynomial by $P(A) = -A^{3} + 2A^2 = 0$ (by Cayley-Hamilton) but I don't see how to find $A^{1000}$ by this characteristic polynomial.</p>
| Oria Gruber | 76,802 | <p>There is another way of approaching this.</p>
<p>You could divide $x^{1000}$ by the characteristic polynomial:</p>
<p>$x^{1000} = (-x^3+2x^2)Q+R$ where $R$ is a polynomial of degree less than 3 with unknown coefficients.</p>
<p>write down $R=ax^2+bx+c$ and evaluate $R$ at the roots of the characteristic polynomial.</p>
<p>Meaning, write down $\lambda^{1000}=a\lambda ^2+b\lambda+c$</p>
<p>and </p>
<p>$\xi^{1000} = a\xi ^2+b\xi+c$</p>
<p>and</p>
<p>$\rho^{1000} = a\rho ^2+b\rho+c$</p>
<p>where $\lambda$ and $\xi$ and $\rho$ are roots of the characteristic polynomial. as you can see, $Q$ wont matter because it is multiplied by zero.</p>
<p>Do this to find the coeffiecents of the remainder, $R$.</p>
<p>after you have done that, insert $x=A$ to get $A^{1000}=aA^2+bA+c$ with the coeffiecents $a,b,c$ that you found.</p>
<p>Edit: The problem here, is that you have a double root, so you need to use the derivative as well.</p>
<p>Full answer:</p>
<p>divide $x^{1000}$ by $(-x^3+2x^2)$ to get:</p>
<p>$x^{1000} = (-x^3+2x^2)Q+ax^2+bx+c$ where $Q$ is some polynomial unknown to us.</p>
<p>the roots of the char. polynomial are $0,2$. put $x=0$ to get:</p>
<p>$0^{1000}=0=0*Q+c=c$ so $c=0$.</p>
<p>now derive $x^{1000} = (-x^3+2x^2)Q+ax^2+bx$ to get:</p>
<p>$1000x^{999}=(-3x^2+4x)Q+Q'(-x^3+2x^2)+2ax+b$ and insert $x=0$ again t oget:</p>
<p>$1000*0^{999} = 0 =b$ meaning $b=0$.</p>
<p>Now back to our original formula with $b=c=0$:</p>
<p>$x^{1000} = (-x^3+2x^2)Q+ax^2$</p>
<p>Insert $x=2$ to get:</p>
<p>$2^{1000} = 4a$ meaning $a=2^{998}$.</p>
<p>Now our original formula looks like $x^{1000} = (-x^3+2x^2)Q+2^{998}x^2$</p>
<p>Inserts $x=A$ to get:</p>
<p>$A^{1000} = 2^{998}A^2$</p>
|
741,436 | <p>I get stuck at the following question:</p>
<p>Consider the matrix<br>
$$A=\begin{bmatrix}
0 & 2 & 0 \\
1 & 1 & -1 \\
-1 & 1 & 1\\
\end{bmatrix}$$</p>
<p>Find $A^{1000}$ by using the Cayley-Hamilton theorem.</p>
<p>I find the characteristic polynomial by $P(A) = -A^{3} + 2A^2 = 0$ (by Cayley-Hamilton) but I don't see how to find $A^{1000}$ by this characteristic polynomial.</p>
| Guy Fsone | 385,707 | <p>Set $X_n=A^n$ since, $A^3 =2A^2$
then, we have $$X_{n+3} = A^{n+3} = 2A^2A^n = 2X_{n+2}$$
Hence, $(X_n)_n$ is geometric and
$$X_n =2^{n-2}X_2\Longleftrightarrow A^n =2^{n-2} A^2$$
that is for $n=1000$ we get
$$A^{1000} = 2^{998}A^2$$</p>
|
1,250,020 | <blockquote>
<p>Find $\int \frac{1+\sin x \cos x}{1-5\sin^2 x}dx$</p>
</blockquote>
<p>I used a bit of trig identities to get: $\int \frac {2+\sin (2x)}{-4+\cos(2x)}dx$ and using the substitution: $t= \tan (2x)$ I got to a long partial fractions calculation which doesn't seem right.</p>
<p>Any hints on how to do it please?</p>
| Américo Tavares | 752 | <p>The denominator of your second integral should be $-3+5\cos 2x$, because
from the identities $\sin x\cos x=\frac{\sin 2x}{2}$ and $\sin ^{2}x=\frac{
1-\cos 2x}{2}$ we obtain</p>
<p>\begin{equation*}
\frac{1+\sin x\cos x}{1-5\sin ^{2}x}=\frac{2+\sin 2x}{-3+5\cos 2x}.
\end{equation*}</p>
<p>To evaluate </p>
<p>\begin{equation*}
\int \frac{2+\sin 2x}{-3+5\cos 2x}dx
\end{equation*}</p>
<p>we can use the standard half-angle substitution $t=\tan x$. Since $dt=\left(
t^{2}+1\right) dx$, $\sin 2x=\frac{2t}{t^{2}+1}$ and $\cos 2x=\frac{1-t^{2}}{
t^{2}+1}$, we thus have</p>
<p>\begin{eqnarray*}\int \frac{2+\sin 2x}{-3+5\cos 2x}dx
=\int \frac{t^{2}+t+1}{-\left( 2t-1\right) \left( 2t+1\right) \left(
t^{2}+1\right) }dt
\end{eqnarray*}</p>
<p>Now, using partial fractions decomposition, we obtain</p>
<p>\begin{equation*}
\frac{t^{2}+t+1}{-\left( 2t-1\right) \left( 2t+1\right) \left(
t^{2}+1\right) }=-\frac{7/20}{t-1/2}+\frac{3/20}{t+1/2}+\frac{t/5}{t^{2}+1}.
\end{equation*}</p>
<p>As such, </p>
<p>\begin{eqnarray*}
\int \frac{2+\sin 2x}{-3+5\cos 2x}dx
&=&-\frac{7}{20}\ln \left| \tan x-\frac{1}{2}\right| +\frac{3}{20}\ln \left|
\tan x+\frac{1}{2}\right| \\&&+\frac{1}{10}\ln \left| \tan ^{2}x+1\right| +C.
\end{eqnarray*}</p>
|
131,179 | <p>Consider the assumptions</p>
<pre><code>$Assumptions = {Element[a,Reals], Element[z,Complexes]}
</code></pre>
<p>I'm looking for a test, to be applied on <code>a</code> and <code>z</code>, that gives <code>True</code> if the argument is a complex number such as <code>a</code> and <code>False</code> if it's real such as <code>z</code>.</p>
<p>The aim is to use this test in a replacement in this way</p>
<pre><code>set = {a, z, x};
set /. (x_ :> img /; test[x])
(* {a, img, x} *)
</code></pre>
<p>An example is</p>
<pre><code>set /. (x_ :> img /; Simplify[NotElement[x, Reals] && Element[x,Complexes]] === (NotElement[x, Reals]))
</code></pre>
<p>It is based on the fact that </p>
<pre><code>Simplify[ Element[z,Reals] ]
</code></pre>
<p>remain unevaluated.</p>
<p>Is there another possible test that doesn't rest on this (and, possibly, simpler and without <code>If</code> and similar)?</p>
| Edmund | 19,542 | <p>The following uses <a href="http://reference.wolfram.com/language/ref/$Assumptions.html" rel="nofollow noreferrer"><code>$Assumptions</code></a> to <em>loosely</em> check if a symbol has a definition using a particular domain.</p>
<pre><code>ClearAll[symbolDomainQ];
SetAttributes[symbolDomainQ, HoldFirst];
symbolDomainQ[s_Symbol, domain_] :=
Or @@ Nor @@@ (Through@{FreeQ[s], FreeQ[domain]}[#] & /@ $Assumptions)
</code></pre>
<p>With</p>
<pre><code>$Assumptions = {a ∈ Reals, z ∈ Complexes, m ∈ Matrices[{4, 4}, Reals, Symmetric[{1, 2}]]}
</code></pre>
<p>Then</p>
<pre><code>{symbolDomainQ[a, Reals], symbolDomainQ[z, Complexes], symbolDomainQ[m, Reals]}
</code></pre>
<blockquote>
<pre><code>{True, True, True}
</code></pre>
</blockquote>
<pre><code>symbolDomainQ[m, Matrices]
</code></pre>
<blockquote>
<pre><code>True
</code></pre>
</blockquote>
<pre><code>symbolDomainQ[a, Complexes]
</code></pre>
<blockquote>
<pre><code>False
</code></pre>
</blockquote>
<p>This only works if the symbols have not been assigned values. When they are assigned values <code>$Assumptions</code> changes so that it is not searchable for that symbol.</p>
<pre><code>a = 1;
$Assumptions
a=.
</code></pre>
<blockquote>
<pre><code>{True, z ∈ Complexes, m ∈ Matrices[{4, 4}, Reals, Symmetric[{1, 2}]]}
</code></pre>
</blockquote>
<p>It would be nice to have a built-in function that could check against the assumptions and work with or without a symbol having a value. Perhaps you should make a suggestion to WRI.</p>
<p>Hope this helps.</p>
|
11,244 | <p>In order to evaluate new educational material the contentment of students with this material is often measured. However, just because a student is contented doesn't mean that he/she has actually learned something. Is there any research investigating the correlation between students contentment and the educational quality of the presented material?</p>
| Michael Hardy | 205 | <p>Vast numbers of students are trained from early childhood to engage in a ruthless competition for grades. When such students are required to take math courses, if they're not interested in understanding math, the course degenerates into one in which getting good grades depends only on believing and obeying and working hard, rather than on understanding. Such students learn that mathematics consists of meaningless algorithms. That is a lie. They are <em>content</em> when they excel in the competition for grades. That differs from understanding.</p>
<p>But what is the correlation? Has a study been done? I wonder if there is anyone with expertise either in education or in mathematics who is competent to conduct such a study.</p>
|
159,026 | <p>It is known (and easy) to prove that if $T: H\longrightarrow H $ is compact, where $H$ is a Hilbert space, then for any orthonormal basis $ e_n $ we have $||Te_n||\longrightarrow 0$.</p>
<p>My question is the following:
Let $P$ be a orthogonal projection on $H$. Let $ e_n $ be a fixed orthonormal basis such that $$||Pe_n||\longrightarrow 0$$ Does this guarantee that $P$ is compact, i.e. finite rank?</p>
| Bill Johnson | 2,554 | <p>No. Consider $H = (\sum_n \ell_2^{2^n})_2$ with the unit vector basis. In $\ell_2^{2^n}$, let $x_n$ be the sum of the unit vector basis. For the subspace take the closed linear span of $(x_n)$.</p>
|
159,026 | <p>It is known (and easy) to prove that if $T: H\longrightarrow H $ is compact, where $H$ is a Hilbert space, then for any orthonormal basis $ e_n $ we have $||Te_n||\longrightarrow 0$.</p>
<p>My question is the following:
Let $P$ be a orthogonal projection on $H$. Let $ e_n $ be a fixed orthonormal basis such that $$||Pe_n||\longrightarrow 0$$ Does this guarantee that $P$ is compact, i.e. finite rank?</p>
| Alex Degtyarev | 44,953 | <p>No. Let the basis be $e_{1,1}, e_{2,1}, e_{2,2},\ldots e_{n,1},\ldots,e_{n,n},\ldots$, in this order, i.e., split it into groups of growing size. Then the projector to the subspace generated by $e_{1,1},\frac12(e_{2,1}+e_{2,2}),\ldots,\frac1n(e_{n,1}+\ldots+e_{n,n}),\ldots$ is a counterexample.</p>
|
1,227,609 | <p>Let $X,Y,Z$ be finite sets, and consider probability distributions $p$ over $X\times Y\times Z$. If we know the marginals of $p$ over all the pairs $X\times Y$, $X\times Z$ and $Y\times Z$, is that enough to pin down $p$ uniquely?</p>
| snar | 24,723 | <p>No, marginals never give you the joint distribution. For a simpler case, take $p$ uniformly distributed on $[0,1]^2$ and $q$ uniformly distributed on the diagonal line segment $\{(x,y) : x = y, 0 \leq x \leq 1\} \subset \mathbb{R}^2$. Both have both marginals being uniform on $[0,1]$ but $p$ and $q$ are distinct.</p>
|
419,176 | <p>Let $f:\left\{x\in\mathbb{R}^n\vert\parallel x\parallel<1\right\}\rightarrow\mathbb{R}$ be an one-to-one bounded continuous function.<br>
I want to construct such $f$ which is not uniformly continuous.<br><br>
In this case, I thought I can construct $f$ with a restriction $n=2$.<br>
But I'm confused because $f$ is bounded so I can't use functions like $\frac{1}{x}$ on $(0,1)$.<br>
To top that off, $f$ is even one-to-one so I gave up and now I'm writing this to get some help from you who is smarter than me.<br><br>
Please give me some help to solve this problem.<br>
Thanks.</p>
| Mher | 80,548 | <p>You can't construct such function. You can prove that $f$ has the limit on the boundary of the closed ball and $f$ will be uniformly continuous.</p>
<p>Let $U_n$ be the open unit ball in $\mathbb{R}^n$. Consider $f$ on the interval $[0,x)\subset U_n$ that connects the center of the ball and the point $x$ from the edge of the ball (i.e. $\|x\|=1$). Now we consider $f$ only $[0,x)$ and using that we can consider $f$ as a function of one variable: $f(y)=f(tx), t\in[0,1)$. Since $f$ is continuous on $[0,x)$ and one-to-one then it is monotonic and hence has the limit at $x$.</p>
<p>Actually, I must take any curve $C$, that connects the center of the ball and the point $x$, instead of $[0,x)$. But that case is proven similarly.</p>
|
419,176 | <p>Let $f:\left\{x\in\mathbb{R}^n\vert\parallel x\parallel<1\right\}\rightarrow\mathbb{R}$ be an one-to-one bounded continuous function.<br>
I want to construct such $f$ which is not uniformly continuous.<br><br>
In this case, I thought I can construct $f$ with a restriction $n=2$.<br>
But I'm confused because $f$ is bounded so I can't use functions like $\frac{1}{x}$ on $(0,1)$.<br>
To top that off, $f$ is even one-to-one so I gave up and now I'm writing this to get some help from you who is smarter than me.<br><br>
Please give me some help to solve this problem.<br>
Thanks.</p>
| Etienne | 80,469 | <p>For $n=1$, as pointed out by Vishal, there is no such function $f$.</p>
<p>For $n\geq 2$, such an $f$ doesn't exist either, but for a "trivial" reason: there is no continuous and one-to-one function at all from the open unit ball $B\subset\mathbb R^n$ into $\mathbb R$. One can see this as follows. Assume that $f:B\to\mathbb R$ is continuous and $1$-$1$. Then $I=f(B)$ is a nontrivial interval of $\mathbb R$ because $B$ is connected and $f$ is continuous and non-constant. Take any point $a\in B$ such that $f(a)$ is an interior point of $I$. Then $f(B\setminus\{ a\})$ has to be connected because $B\setminus\{ a\}$ is connected ($n\geq 2$) and $f$ is continuous. But $f(B\setminus\{ a\})$ is equal to $I\setminus \{ f(a)\}$ because $f$ is $1$-$1$, and this is not a connected set. So we have a contradiction.</p>
|
2,634,277 | <p>I am working on some development formulas for surfaces and as a byproduct of abstract theory i get that:
$$\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\frac{1+\sin^2\theta}{(\cos^4\theta+(\gamma\cos^2\theta-\sin\theta)^2)^\frac{3}{4}}d\theta$$
is independent on the parameter $\gamma\in\mathbb{R}$. I thought that there was something wrong with my calculations but actually turns out that using Mathematica that the value is somewhat near $5,24412$ independently on the $\gamma$ I plug in the calculation of the integral. Is there any way to verify that actually this is a constant by direct computations, complex analysis, or at least is this kind of integrals studied?</p>
<p>Edit:obviously differentiating in the integral does not help much</p>
| JanG | 266,041 | <p>Put
\begin{equation*}
I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\dfrac{1+\sin^2\theta}{(\cos^4\theta +(\gamma\cos^2\theta-\sin \theta)^2)^{\frac{3}{4}}}\, d\theta
\end{equation*}
If $x = \dfrac{\sin\theta}{\cos^2\theta}$, $\, y = \gamma-x$ and $y = \sqrt{z}$ then
\begin{equation*}
dx = \dfrac{\cos^2\theta+2\sin^2\theta}{\cos^3\theta}\, d\theta
\end{equation*}
and
\begin{gather*}
I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\dfrac{\cos^2\theta+2\sin^2\theta}{\cos^3\theta\left(1 +\left(\gamma-\frac{\sin \theta}{\cos^2\theta}\right)^2\right)^{\frac{3}{4}}}\, d\theta = \int_{-\infty}^{\infty}\dfrac{1}{\left(1 +\left(\gamma- x\right)^2\right)^{\frac{3}{4}}}\, dx = \int_{-\infty}^{\infty}\dfrac{1}{\left(1 +y^2\right)^{\frac{3}{4}}}\, dy = \\[2ex]
\int_{0}^{\infty}\dfrac{z^{\frac{1}{2}-1}}{(1+z)^{\frac{1}{2}+\frac{1}{4}}}\, dz = {\rm B}\left(\frac{1}{4},\frac{1}{2}\right) \approx 5.244115109
\end{gather*}
where ${\rm B}$ is the Beta function.</p>
|
520,046 | <blockquote>
<p>Find the smallest natural number that leaves residues $5,4,3,$ and $2$ when divided respectively by the numbers $6,5,4,$ and $3$.</p>
</blockquote>
<p>I tried
$$x\equiv5\pmod6\\x\equiv4\pmod5\\x\equiv3\pmod4\\x\equiv2\pmod3$$What $x$ value?</p>
| wendy.krieger | 78,024 | <p>The number which leaves (5, 4, 3, 2) mod (6, 5, 4, 3) is one less than the one that leaves a residue of (0, 0, 0, 0), mod (6, 5, 4, 3). So one finds $n=\operatorname{lcm}(6,5,4,3)-1$ to get 59, which is the desired answer.</p>
|
2,594,669 | <p>Given the Pythagoras Theorem: <strong>a² + b² = c²</strong></p>
<p>Is there a way to get the value of <strong>b</strong> when we only have a value for <strong>a</strong> and the angle <strong>α</strong>?</p>
<p>To be frank, I have no clue about that, what I want isn't the angle of <strong>β</strong> but the length of <strong>b</strong> (Opposite).</p>
<p><a href="https://i.stack.imgur.com/0jRad.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0jRad.png" alt="enter image description here"></a></p>
| ArsenBerk | 505,611 | <p>Pythagoras Theorem is a theorem about right-angle triangles so if we know one of the angles (except $90^{\circ}$, which is not the case in here), then we know all of the angles and knowing one of the sides of the triangle is enough for finding other two.</p>
|
3,471,292 | <p>I need to find the value of the series <span class="math-container">$\sum_{n=0}^{\infty}\frac{(n+1)x^n}{n!}$</span>.I've computed its radius of convergence which comes out to be zero.</p>
<p>I'm not getting how to make adjustments in the general terms of the series to get the desired result...</p>
| lab bhattacharjee | 33,337 | <p>Hint</p>
<p><span class="math-container">$$\dfrac{(n+1)x^n}{n!}=x\cdot\dfrac{x^{n-1}}{(n-1)!}+\dfrac{x^n}{n!}$$</span></p>
<p>Now <span class="math-container">$$e^y=\sum_{r=0}^\infty\dfrac{y^r}{r!}$$</span></p>
|
3,471,292 | <p>I need to find the value of the series <span class="math-container">$\sum_{n=0}^{\infty}\frac{(n+1)x^n}{n!}$</span>.I've computed its radius of convergence which comes out to be zero.</p>
<p>I'm not getting how to make adjustments in the general terms of the series to get the desired result...</p>
| Marios Gretsas | 359,315 | <p>Let <span class="math-container">$a_n$</span> the sequence in the series.</p>
<p>Now <span class="math-container">$a_n=\frac{z^n}{(n-1)!}+\frac{z^n}{n!}$</span></p>
<p>Now <span class="math-container">$\sum_{n=0}^{\infty}\frac{nz^n}{n!}=\sum_{n=1}^{\infty}\frac{z^n}{(n-1)!}=z\sum_{k=0}^{\infty}\frac{z^k}{k!}=ze^z$</span></p>
<p>Also <span class="math-container">$\sum_{k=0}^{\infty}\frac{z^k}{k!}=e^z$</span></p>
<p>So the whole sum is <span class="math-container">$(z+1)e^z$</span></p>
|
228,889 | <p><em>[Attention! This question requires some reading and it's answer probably is in form of a "soft-answer", i.e. it can't be translated into a hard mathematical proposition. (I hope I haven't scared away all readers with this.)]</em></p>
<p>Consider the following three examples:</p>
<p>1) <em>[If this example seems too technical just skip it - it isn't that important for the idea I want to convey.]</em> The set $T$ of all terms (of a functional structure) is the set $$T=\bigcap_{O\text{ closed under concatenation with function symbols}}_{O\supseteq X} O,$$where $X$ is the (countable) of all variables and "closed under concatenation with function symbol" means: If $f$ is some function symbol of arity $n$ and $x_1,\ldots,x_n\in X$, then $fx_1\ldots x_n\in T$. The above $T$ is the <em>smallest</em> set such that it contains the variables and is closed under concatenation with function symbols.</p>
<p>2) The smallest subgroup $G$ of a group $X$, containing a set $A\subseteq X$, is the set $$G:=\bigcap_{O\ \text{ is a subgroup of $X$}}_{O\supseteq A} O.$$</p>
<p>3) The smallest $\sigma$-algebra $\mathcal{A}$ on a set $X$ containing a set $A\subseteq X$ is the set $$\mathcal{A}:=\bigcap_{O\ \text{ is a $\sigma$-algebra on $X$}}_{O\supseteq A} O.$$</p>
<p>4) The set $$C:=\bigcap_{O\ \text{ is open in $X$}}_{O\supseteq A} O,$$ where $X$ is a metric space and $A\subseteq X$ is arbitrary. </p>
<p>Now here's the thing: The sets $T$, $G$, and $\mathcal{A}$, from examples 1),2) and 3) are also <strong>closed under the closing condition defining them</strong>, i.e. $T$ is also closed under concatenation with function symbols, $G$ is also a group and $\mathcal{A}$ is also a $\sigma$-algebra; for $G$ and $\mathcal{A}$ this is already implied by their name (the smallest <em>subgroup</em>, the smallest $\sigma$*-algebra*), which was the reason I also gave $T$ as an example, where it's name doesn't already imply it's closure under it's defining closure operations. <strong>But</strong> the set $C$ from example 4) need not be open, if for example $\mathbb{R}=X$ and $A=[0,1]$ (maybe there are nontrivial metric space, where it <em>is</em> open for nontrivial sets $A$, but I didn't want to waste time checking that). That is, $C$ isn't closed (no pun intended) under the closing condition I used to define it; or differently said: There isn't a smallest open set containing $A$.</p>
<p>Notice that the closure conditions in 1) and 2) have a more algebraic character, since we close under some algebraic operations, where the closure condition in 3) as a more "set-theoretical-topology"-type character, since we close under set-theoretic operations. Nonetheless in all cases the outcome is again "closed". </p>
<p>My question is: How do generally (abstractly) "closing conditions" $\mathscr{C}$ have to look like such that the set $$ S:=\bigcap_{O\ \text{ is closed under $\mathscr{C}$}}_{O\supseteq A} O$$ is itself closed in $X$ under $\mathscr{C}$, where $A\subseteq X$ is an arbitrary set ? Differently said: How do generally (abstractly) "closing conditions" have to look like such that there is a smallest set being closed under these conditions, containing some arbitrary fixed set.</p>
| user642796 | 8,348 | <p>Take a <a href="http://en.wikipedia.org/wiki/Vitali_set">Vitali set</a> in the unit interval $[0,1)$, and call it $V$. (Define an equivalence relation on $[0,1)$ by $x \sim y$ iff $x - y \in \mathbb{Q}$, and let $V$ be obtained from the Axiom of Choice by choosing exactly one element from every $\sim$-equivalence class.)</p>
<p>By $Q$ I will denote $\mathbb{Q} \cap [0,1)$. For each $q \in Q$ let $V_q$ denote the modulo $1$ shift of $V$ by $q$: $$V_q = \{ x + q : x \in V, x < 1-q \} \cup \{ x + q - 1 : x \in V , 1-q \leq x \}.$$ Note the following facts:</p>
<ul>
<li>$V_q \cap V_p = \emptyset$ if $p \neq q$.</li>
<li>$\bigcup_{q \in Q} V_q = [0,1)$.</li>
</ul>
<p>(Both of the above follow from the fact that $V$ contains exactly one element of every $\sim$-equivalence class.)</p>
<p>By the second fact above we have that $$\mu^* \left( \textstyle{\bigcup_{q \in Q}} V_q \right) = 1.$$
On the other hand, by the translation invariance of $\mu^*$ is follows that $\mu^* ( V_q ) = \mu^* ( V )$ for all $q$, and since $V$ is not Lebesgue measurable it follows that $\mu^* ( V ) > 0$, and therefore $$\sum_{q \in Q} \mu^* ( V_k ) = + \infty.$$</p>
|
2,361,920 | <p>The question is as follows:
<a href="https://i.stack.imgur.com/RHPwG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RHPwG.png" alt="enter image description here"></a> </p>
<p>I can not solve this question so I am asking what exactly in the probability theory that I must revise so I could solve it, could anyone give me advice please?</p>
| Graham Kemp | 135,106 | <p>There are three colours. You are to draw two socks, and may as well place them in front of you. For each of the colours, what is the probability that the sock on your left is that colour and, <em>given that</em>, that the sock on the right is so too?</p>
<p>$$\def\blank{{\underline{\quad}}} \blank\times\blank~~+~~\blank\times\blank~~+~~\blank\times\blank $$</p>
<p>Fill in the blanks.</p>
|
22 | <p>By matrix-defined, I mean</p>
<p>$$\left<a,b,c\right>\times\left<d,e,f\right> = \left|
\begin{array}{ccc}
i & j & k\\
a & b & c\\
d & e & f
\end{array}
\right|$$</p>
<p>...instead of the definition of the product of the magnitudes multiplied by the sign of their angle, in the direction orthogonal)</p>
<p>If I try cross producting two vectors with no $k$ component, I get one with only $k$, which is expected. But why?</p>
<p>As has been pointed out, I am asking why the algebraic definition lines up with the geometric definition.</p>
| Isaac | 72 | <p>Note that if you replace $i$, $j$, and $k$ with $m$, $n$, and $p$, the determinant becomes the dot-product of the vector $(m, n, p)$ with the cross-product of the two original vectors. If $(m, n, p) = (a, b, c)$ or $(m, n, p) = (d, e, f)$, the determinant is zero (any matrix with two identical rows has determinant zero), so the dot product of $(a, b, c)$ or $(d, e, f)$ with the cross-product is zero. Hence, $(a, b, c)$ and $(d, e, f)$ are orthogonal to their cross-product.</p>
|
2,076 | <p>I'm attempting for the first time to create a map within <em>Mathematica</em>. In particular, I would like to take an output of points and plot them according to their lat/long values over a geographic map. I have a series of latitude/longitude values like so:</p>
<pre><code> {{32.6123, -117.041}, {40.6973, -111.9}, {34.0276, -118.046},
{40.8231, -111.986}, {34.0446, -117.94}, {33.7389, -118.024},
{34.122, -118.088}, {37.3881, -122.252}, {44.9325, -122.966},
{32.6029, -117.154}, {44.7165, -123.062}, {37.8475, -122.47},
{32.6833, -117.098}, {44.4881, -122.797}, {37.5687, -122.254},
{45.1645, -122.788}, {47.6077, -122.692}, {44.5727, -122.65},
{42.3155, -82.9408}, {42.6438, -73.6451}, {48.0426, -122.092},
{48.5371, -122.09}, {45.4599, -122.618}, {48.4816, -122.659},
{42.3398, -70.9843}}
</code></pre>
<p>I've tried finding documentation on how I would proceed but I cannot find anything that doesn't assume a certain level of introduction to geospatial data. Does anyone know of a good resource online or is there a simple explanation one can supply here? </p>
| Stefan | 2,448 | <p>I am a big fan of the <code>Texture</code> functionality...
(I'd like to point out, that this is all inspired by graphics/visualisation genius <a href="https://mathematica.stackexchange.com/users/820/yu-sung-chang">Yu-Sung Chang</a>)</p>
<pre><code>nightEarth = SphericalPlot3D[1, {u, 0, Pi}, {v, 0, 2 Pi},
PlotPoints -> 50, MaxRecursion -> 0,
Mesh -> None, TextureCoordinateFunction -> ({#5, 1 - #4} &),
PlotStyle -> Directive[
Texture[Import["http://eoimages.gsfc.nasa.gov/images/imagerecords/55000/55167/earth_lights_lrg.jpg"]],
Specularity[White, 50]],
Lighting -> "Neutral", Boxed -> False, Axes -> False, Background -> Gray]
</code></pre>
<p><img src="https://i.stack.imgur.com/wZZsK.jpg" alt="enter image description here"></p>
<p>Well this is already gospel:</p>
<pre><code>SC[{lat_, lon_}] := {Cos[(lon + 180) °] Cos[lat °],
Sin[(lon + 180) °] Cos[lat °], Sin[lat °]};
</code></pre>
<p>Your coordinates:</p>
<pre><code>centers = {{32.6123, -117.041}, {40.6973, -111.9}, {34.0276, \
-118.046}, {40.8231, -111.986}, {34.0446, -117.94}, {33.7389, \
-118.024}, {34.122, -118.088}, {37.3881, -122.252}, {44.9325, \
-122.966}, {32.6029, -117.154}, {44.7165, -123.062}, {37.8475, \
-122.47}, {32.6833, -117.098}, {44.4881, -122.797}, {37.5687, \
-122.254}, {45.1645, -122.788}, {47.6077, -122.692}, {44.5727, \
-122.65}, {42.3155, -82.9408}, {42.6438, -73.6451}, {48.0426, \
-122.092}, {48.5371, -122.09}, {45.4599, -122.618}, {48.4816, \
-122.659}, {42.3398, -70.9843}};
</code></pre>
<p>and now:</p>
<pre><code>Show[nightEarth, Graphics3D[{Opacity[.9], {Red, PointSize[Medium],
Point[SC[#]] & /@ centers}}, Boxed -> False,
SphericalRegion -> True, ViewAngle -> .3], ImageSize -> Large]
</code></pre>
<p><img src="https://i.stack.imgur.com/RfAzy.jpg" alt="enter image description here"></p>
|
93,499 | <p>Let $P$ be a normal sylow $p$-subgroup of a finite group $G$.</p>
<p>Since $P$ is normal it is the unique sylow $p$-subgroup.</p>
<p>I would like to say if $\phi$ is an automorphism then $\phi(P)$ is also a sylow $p$-subgroup. Then uniqueness would finish the proof. But is that true?</p>
<p>Does an automorphism of a group always send subgroups to subgroups of the same order?</p>
| Santropedro | 83,420 | <p>Take any automorphism $f$, then $\#f(P)=\#P$ (because in particular $f$ is a bijection), and since also the image of subgroups (in this case $P$) is a subgroup, you conclude $f(P)$ is a p-sylow, and since you remarked:</p>
<blockquote>
<p>Since $P$ is normal it is the unique Sylow $p$ subgroup.</p>
</blockquote>
<p>That means $f(P)=P$, </p>
|
2,806,432 | <p>Let $(\mathbb{R}^N,\tau)$ a topological space, where $\tau$ is the usual topology.
Let $A\subset\mathbb{R}^N$ a compact. If $(A_n)_n$ is a family of open such that
\begin{equation}
\bigcup_nA_n\supset A,
\end{equation}
then, from compact definition
\begin{equation}
\bigcup_{i=1}^{k}A_i\supset A
\end{equation}
Now, if I find a family of open $(B_n)_n$ such that
\begin{equation}
cl\bigg(\bigcup_n B_n\bigg)\supset A,
\end{equation}
can I say that
\begin{equation}
cl\bigg(\bigcup_{i=1}^kB_i\bigg)\supset A\quad?
\end{equation}
Thanks for the attention!</p>
| drhab | 75,923 | <p>The values $f(x)$ where $f$ denotes a PDF cannot be interpreted as probabilities (as you seem to think). Note for instance that they can take values that exceed $1$.</p>
<p>That is the mistake you made.</p>
<p>We have the equalities $H+T=50$ and $H+8T=225$ here leading to $H=25=T$, so the question can be rephrased as:</p>
<p>"By $50$ throws of a fair coin what is the probability on $25$ tails?"</p>
<p>This is the route taken by your teacher.</p>
|
1,369,409 | <p>I have a bit of an advanced combination problem that has left me stumped for a few days. Essentially my question is if you have n sets of items, and you can select a different number of items from each set, how do you compute the combinations without first creating new sets.</p>
<p>An example in pictures:
I have three sets:
<br>Set A has elements: <br>
Dog<br>
Cat<br>
Rhino<br><br>
Set B has elements:
<br>Pig
<br>Horse
<br>Cow
<br><br>Set C has elements:
<br>Lizard
<br>Snake
<br>Crocodile
<br>Alligator
<br><br>
And now I would like to compute all of the combinations with the criteria that 2 elements be selected from set a, 1 element is selected from set
B, and 2 elements are selected from set C.
<br><br>
The end result would contain all the unique combinations with those specifications.<br>
A current way I am using is to take Set A and turn Set A into all of the combinations of Set A that has 2 elements and storing it in a different set, Set D, then doing the same for Set C, storing in Set E, and then selecting one item from each Set B, Set D, and Set E to get all the unique combinations but I was wondering if there was a better solution.</p>
<p><br>EDIT: By compute I am referring to generating a list of all the possible sets, NOT figure out the count or number of items. That being said, this implies that each set must be unique (Dog, Cat, Pig, Lizard, Snake) is the same as (Cat, Dog, Pig, Snake, Lizard). </p>
| Mark | 169,199 | <p>This seems like a good approach. Combinations from A are {Dog, Cat}, {Dog, Rhino}, and {Cat Rhino}. There are 3 combinations for set B, and 6 combinations for set C. Thus you will have $3*3*6=54$ different sets that meet your criteria.</p>
|
1,421,740 | <p>Let $90^a=2$ and $90^b=5$, Evaluate </p>
<h1>$45^\frac {1-a-b}{2-2a}$ </h1>
<p>I know that the answer is 3 when I used logarithm, but I need to show to a student how to evaluate this without involving logarithm. Also, no calculators.</p>
| lab bhattacharjee | 33,337 | <p>$$2=90^a=(2\cdot5\cdot3^2)^a\iff5^a3^{2a}=2^{1-a}$$</p>
<p>and similarly, $$5=90^b\iff5^{1-b}3^{-2b}=2^b$$</p>
<p>Equating the powers of $2,$
$$\implies(5^a3^{2a})^b=(5^{1-b}3^{-2b})^{1-a}\iff3^{2b}=5^{1-a-b}$$</p>
<p>$$45^{1-a-b}=(3^2\cdot5)^{1-a-b}=3^{2-2(a+b)}\cdot5^{1-a-b}=3^{2-2(a+b)}\cdot3^{2b}$$
$$\implies45^{1-a-b}=3^{2(1-a)}$$</p>
<p>$\implies$ one of the values of $$45^{\frac{1-a-b}{2(1-a)}}$$ is $3$</p>
|
1,044,009 | <p>I have two numbers $N$ and $M$.
I efficiently want to calculate how many pairs of $a$,$b$ are there such that $1 \leq a \leq N$ and $1 \leq b \leq M$ and $ab$ is a perfect square.</p>
<p>I know the obvious $N*M$ algorithm to compute this. But i want something better than that. I think it can be done in a better time may $\operatorname{O}(m+n)$ or something like that but calculating new pairs directly from previous pairs rather than iterating over all $a$ and $b$.
Thanks for any help in advance. A pseudo code will be more helpful.</p>
| Mike Pierce | 167,197 | <p><strong>Step 1:</strong></p>
<p>Write a function $f$ that takes a perfect square $x$ and returns a list of all (or just the number of) possible pairs $(a,b)$ such that $ab = x$.</p>
<p><strong>Step 2:</strong></p>
<p>Have the function $f$ iterate over the squares of the integers $1$ to $n$, where
$n = \lfloor\sqrt{M*N}\rfloor$ and count the results as you go.</p>
|
1,736,098 | <p>Wrote some Python code to verify if my Vectors are parallel and/or orthogonal. Parallel seems to be alright, orthogonal however misses out in one case. I thought that if the dotproduct of two vectors == 0, the vectors are orthogonal? Can someone tell me what's wrong with my code?</p>
<pre><code>def isParallel(self,v):
try:
print self.findTheta(v,1)
if (self.findTheta(v,1) == 0 or self.findTheta(v,1) == 180):
return True
except ZeroDivisionError:
return True
return False
def isOrthogonal(self,v):
print self.dotProduct(v)
if self.dotProduct(v) == 0:
return True
return False
def dotProduct(self,v):
dotproduct = sum([self.coordinates[i]*v.coordinates[i] for i in range(len(self.coordinates))])
return dotproduct
</code></pre>
| paulmelnikow | 640,314 | <p>If you're working with 3D vectors, you can do this concisely using the toolbelt <a href="https://github.com/lace/vg" rel="nofollow noreferrer">vg</a>. It's a light layer on top of numpy and it supports single values and stacked vectors.</p>
<pre><code>import numpy as np
import vg
v1 = np.array([1.0, 2.0, 3.0])
v2 = np.array([-2.0, -4.0, -6.0])
vg.almost_collinear(v1, v2)
# True
</code></pre>
<p>I created the library at my last startup, where it was motivated by uses like this: simple ideas which are verbose or opaque in NumPy.</p>
|
381,566 | <p>I know practically nothing about fractional calculus so I apologize in advance if the following is a silly question. I already tried on math.stackexchange.</p>
<p>I just wanted to ask if there is a notion of fractional derivative that is linear and satisfy the following property <span class="math-container">$D^u((f)^n) = \alpha D^u(f)f^{(n-1)}$</span> where <span class="math-container">$\alpha$</span> is a scalar. In the case of standard derivatives we would have <span class="math-container">$\alpha = n$</span>.</p>
<p>Thank you very much.</p>
| Tom Copeland | 12,178 | <p>The generalized Leibniz formula applicable to the classic fractional integroderivative is</p>
<p><span class="math-container">$$ D^{\omega}\; f(x)g(x) = \sum_{n \geq 0} \binom{\omega}{n} [D^{\omega-n}f(x)]D^ng(x)=(D_L+D_R)^{\omega} g(x)f(x),$$</span></p>
<p>where <span class="math-container">$D_L$</span> acts on the function on the left of the product and <span class="math-container">$D_R$</span> on the right function. See, e.g., <a href="https://www.researchgate.net/publication/275043277_Leibniz_Rules_and_Integral_Analogues_for_Fractional_Derivatives_Via_a_New_Transformation_Formula" rel="nofollow noreferrer">Leibniz rules and integral analogues for fractional derivatives via
a new transformation formula</a> by Fugere, Gaboury, and Tremblay.</p>
<p>This generalized Leibniz rule applies to the fractional integroderivative satisfying the sensible axioms given by Pincherle described in "The Role of Salvatore Pincherle in the Development of Fractional Calculus" by Francesco Mainardi and Gianni Pagnini--those satisfied by the usual derivative raised to integral powers, negative or positive. Reps of this op are presented in <a href="https://math.stackexchange.com/questions/125343/lie-group-heuristics-for-a-raising-operator-for-1n-fracdnd-betan-fra">this MSE-Q</a> and can be used to define the confluent (see <a href="https://mathoverflow.net/questions/107159/pochhammer-symbol-of-a-differential-and-hypergeometric-polynomials/107191#107191">this MO-Q</a>) and regular hypergeometric fcts.</p>
<p>These reps of <span class="math-container">$D^{\omega}$</span> are at the heart of the definitions of the Euler gamma and beta functions via integrals, generalizations of the integral factorials and integral binomial coefficients (see my reply to/refs in <a href="https://mathoverflow.net/questions/353282/an-analytic-continuation-of-power-series-coefficients/379477#379477">this MO-Q</a>), which most researchers use frequently in their math endeavors--contrary to some opinions expressed on MO. See an example of the half-derivative in <a href="https://mathoverflow.net/questions/153542/geometric-interpretation-of-the-half-derivative">this MO-Q</a> (which many users apparently confuse with some pseudo-differential operator defined by the Fourier transform).</p>
|
876,209 | <p>I am confused on how to write a formal proof for sum notations. How would I write a formal proof for this example?</p>
<p>Prove that $$\sum\limits_{k = 0}^\infty\frac{2}{3^k} = 3.$$ Prove that for any $\alpha \in \{0, 2\}^\mathbb{N}$ that $$0 \le \sum\limits_{k = 0}^\infty\frac{\alpha(k)}{3^k} \le 3.$$</p>
| Joel | 85,072 | <p>It all really depends on how many theorems you have at your disposal. Since you have $\sum_{k=0}^\infty \frac{2}{3^k} = 3$ already, you know by the comparison theorem that $$\sum_{k=0}^\infty \frac{\alpha(k)}{3^k} \le \sum_{k=0}^\infty \frac{2}{3^k} = 3$$ and it is positive since all of the terms are positive.</p>
<p>If you can't use the comparison theorem, then you could notice that $$0 \le \sum_{k=0}^N \frac{\alpha(k)}{3^k} \le \sum_{k=0}^N \frac{2}{3^k} < 3$$</p>
<p>Since $$ \sum_{k=0}^N \frac{\alpha(k)}{3^k}$$ is a monotonic increasing sequence that is bounded above, it converges. At the same time we have shown that its limit is less than $$\sum_{k=0}^\infty \frac{2}{3^k} = 3.$$</p>
<hr>
<p>Finally to prove the convergence of the first series notice that the partial sums satisfies: $$\sum_{k=0}^N \frac{2}{3^k} = 2 \cdot \frac{1-(1/3)^{N+1}}{1-(1/3)}$$</p>
<p>In the limit, the term $(1/3)^N \to 0$, which means the series converges to $$\frac{2}{1-(1/3)} = 3$$</p>
|
3,525,488 | <p>So I have the polar curve </p>
<p><span class="math-container">$r=\sqrt{|\sin(n\theta)|}$</span></p>
<p>Which I am trying to evaluate between <span class="math-container">$0$</span> and <span class="math-container">$2\pi$</span>. By smashing it into wolfram it returns a constant value 4 for any <span class="math-container">$n$</span>.</p>
<p>I tried calculating it manually (I suspect my calculation might be wrong), but I arrived at </p>
<p><span class="math-container">$$\textrm{I}=\frac{1}{2}\int_{0}^{2\pi}|\sin(n\theta)|= \left[\text{sgn}(\sin(nx))\frac{\cos(nx)}{n}\right]_{0}^{2\pi} $$</span> Which when looking at it can't be evaluated at <span class="math-container">$0$</span> or <span class="math-container">$2\pi$</span> with taking the limits. I suspect that whenever <span class="math-container">$n$</span> increases, the curve becomes "tighter", which could explain why in the integral stays permanently at 4, but I can't come up with a sound argument for it, so if someone could give me a pointer, that would be greatly appreciated. </p>
| Claude Leibovici | 82,404 | <p>The constant value of <span class="math-container">$4$</span> is normal
<span class="math-container">$$\int_{0}^{2\pi}|\sin(t)|\,dt=\int_{0}^{\pi}\sin(t)\,dt-\int_{\pi}^{2\pi}\sin(t)\,dt=2+2$$</span>
<span class="math-container">$$\int_{0}^{2\pi}|\sin(2t)|\,dt=\int_{0}^{\frac\pi 2}\sin(2t)\,dt-\int_{\frac\pi 2}^{\pi}\sin(2t)\,dt+\int_{\pi}^{\frac{3\pi} 2}\sin(2t)\,dt-\int_{\frac{3\pi} 2}^{2\pi}\sin(2t)\,dt=1+1+1+1$$</span> Just continue.</p>
|
1,781,117 | <h1>The question</h1>
<p>Prove that:
$$\prod_{n=2}^∞ \left( 1 - \frac{1}{n^4} \right) = \frac{e^π - e^{-π}}{8π}$$</p>
<hr>
<h2>What I've tried</h2>
<p>Knowing that:
$$\sin(πz) = πz \prod_{n=1}^∞ \left( 1 - \frac{z^2}{n^2} \right)$$
evaluating at $z=i$ gives
$$ \frac{e^π - e^{-π}}{2i} = \sin(πi) = πi \prod_{n=1}^∞ \left( 1 + \frac{1}{n^2} \right)$$
so:
$$ \prod_{n=1}^∞ \left( 1 + \frac{1}{n^2} \right) = \frac{e^π - e^{-π}}{2π}$$</p>
<p>I'm stucked up and don't know how to continue, any help?</p>
| Jorge Martín Pérez | 308,745 | <p>I'll reproduce the answer @C.Dubussy have just deleted:
$$ \prod_{n=2} \left( 1 - \frac{1}{n^4} \right) = \prod_{n=2}^∞ \left( 1 + \frac{1}{n^2}\right) \prod_{n=2}^∞ \left( 1 - \frac{1}{n^2}\right) = \frac{\sin{iπ}}{iπ} \prod_{n=2}^∞ \frac{n-1}{n} \prod_{n=2}^∞ \frac{n+1}{n} $$</p>
<p>And because the last product gives $\frac{1}{2}$, we have it!</p>
|
1,781,117 | <h1>The question</h1>
<p>Prove that:
$$\prod_{n=2}^∞ \left( 1 - \frac{1}{n^4} \right) = \frac{e^π - e^{-π}}{8π}$$</p>
<hr>
<h2>What I've tried</h2>
<p>Knowing that:
$$\sin(πz) = πz \prod_{n=1}^∞ \left( 1 - \frac{z^2}{n^2} \right)$$
evaluating at $z=i$ gives
$$ \frac{e^π - e^{-π}}{2i} = \sin(πi) = πi \prod_{n=1}^∞ \left( 1 + \frac{1}{n^2} \right)$$
so:
$$ \prod_{n=1}^∞ \left( 1 + \frac{1}{n^2} \right) = \frac{e^π - e^{-π}}{2π}$$</p>
<p>I'm stucked up and don't know how to continue, any help?</p>
| Felix Marin | 85,343 | <p>$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$</p>
<blockquote>
<p>With $\ds{N \in \mathbb{N}_{\ \geq\ 2}}$:</p>
</blockquote>
<p>\begin{align}
\prod_{n = 2}^{N}\pars{1 - {1 \over n^{4}}} & =
\prod_{n = 2}^{N}
{\pars{n - 1}\pars{n + 1}\pars{n - \ic}\pars{n + \ic} \over n^{4}}
\\[5mm] & =
{\pars{N - 1}!\bracks{\pars{N + 1}!/2} \over \pars{N!}^{4}}\,
\verts{\pars{2 + \ic}^{\overline{N - 1}}}^{2} =
{1 \over 2}\,{N + 1 \over N}\,\verts{{1 \over N!}\,{\Gamma\pars{N + 1 + \ic} \over \Gamma\pars{2 + i}}}^{\,2}
\\[5mm] & =
{1 \over 2}\,{N + 1 \over N}\,\verts{{1 \over 1 + \ic}
\,{\pars{N + \ic}! \over N!}}^{2}\,{1 \over \verts{\Gamma\pars{1 + i}}^{2}}
\end{align}</p>
<blockquote>
<p>With <a href="http://people.math.sfu.ca/~cbm/aands/page_256.htm" rel="nofollow noreferrer">A & S Table $\ds{\mathbf{\color{#000}{6.1.31}}}$ identity</a>
$\ds{\verts{\Gamma\pars{1 + \ic y}}^{\,2} = {\pi y \over \sinh\pars{\pi y}}}$ and the <em>Stirling Asymptotic Expansion</em>:</p>
</blockquote>
<p>\begin{align}
\prod_{n = 2}^{N}\pars{1 - {1 \over n^{4}}} &
\stackrel{\mrm{as}\ N\ \to\ \infty}{\sim}\,\,\,
{1 \over 4}\verts{\root{2\pi}\pars{N + \ic}^{N + \ic + 1/2}\expo{-N - \ic} \over \root{2\pi}N^{N + 1/2}\expo{-N}}^{2}\,{1 \over \pi/\sinh\pars{\pi}}
\\[5mm] & =
{\expo{\pi} - \expo{-\pi} \over 8\pi}
\,\verts{N^{\ic}\pars{1 + {\ic \over N}}^{N + \ic + 1/2}}^{2} =
{\expo{\pi} - \expo{-\pi} \over 8\pi}
\,\verts{\exp\pars{\ic\ln\pars{N}}\expo{\ic}}^{2}
\\[5mm] & \stackrel{\mrm{as}\ N\ \to\ \infty}{\to}\,\,\,
\bbx{{\expo{\pi} - \expo{-\pi} \over 8\pi}}
\end{align}</p>
|
2,077,694 | <p>How to find $A = M^{A}_{B}$ in linear transformation $F = \mathbb{P_{2}} \rightarrow \mathbb{R^{2}} $, where $ F(p(t)) = \begin{pmatrix} p(0) \\ P(1) \end{pmatrix},$
$ A = \{1,t,t^{2}\},$
$B=\left \{ \begin{pmatrix} 1\\ 0\end{pmatrix},\begin{pmatrix} 0\\ 1\end{pmatrix} \right \}$?</p>
| tomasz | 30,222 | <p><strong>Hint</strong>: If $d$ is a proper divisor of $n$, then $dk$ is a proper divisor of $nk$.</p>
|
9,484 | <p>Let <span class="math-container">$F(k,n)$</span> be the number of permutations of an n-element set that fix exactly <span class="math-container">$k$</span> elements.</p>
<p>We know:</p>
<ol>
<li><p><span class="math-container">$F(n,n) = 1$</span></p>
</li>
<li><p><span class="math-container">$F(n-1,n) = 0$</span></p>
</li>
<li><p><span class="math-container">$F(n-2,n) = \binom {n} {2}$</span></p>
<p>...</p>
</li>
<li><p><span class="math-container">$F(0,n) = n! \cdot \sum_{k=0}^n \frac {(-1)^k}{k!}$</span> (the subfactorial)</p>
</li>
</ol>
<p>The summation formula is obviously</p>
<p><span class="math-container">$\displaystyle\sum_{k=0}^n F(k,n) = n!$</span></p>
<p>A recursive definition of <span class="math-container">$F(k,n)$</span> is (my claim):</p>
<p><span class="math-container">$$F(k,n) = \binom {n} {k} \cdot \Big( k! - \displaystyle\sum_{i=0}^{k-1} F(i,k) \Big)$$</span></p>
<p>Question 1: Is there a common name for the "generalized factorial" <span class="math-container">$F(k,n)$</span>?</p>
<p>Question 2: Does anyone know a closed form for <span class="math-container">$F(k,n)$</span> or have an idea how to get it from the recursive definition? (generating function?)</p>
| Reid Barton | 126,667 | <p>A permutation of {1, ..., n} with k fixed points is determined by choosing which k elements of {1, ..., n} it fixes and choosing a derangement of the remaining n-k elements. So,</p>
<p>$F(k, n) = {n \choose k} F(0, n-k)$.</p>
<p>(This formula is also on the page Michael Lugo linked to.) You have already given one formula for the number of derangements on n letters. Another one is F(0, n) = the nearest integer to n!/e.</p>
|
1,821,849 | <p>Let $K/F$ be a field extension and $L_1,L_2$ subfields of $K$ such that $L_1$ and $L_2$ have finite degree over $F$. </p>
<p>Does $L_1 \cong L_2$ imply $[L_1 : F ]=[L_2 : F]$? Obviously, if the isomorphism fixes $K$ (which isn't always necessarily true) the result holds. The result even holds if $F$ is of finite degree over its prime field. </p>
<p>When trying the usual proof, we get that $[L_1 : F] = [L_1^{\theta} : F^{\theta} ]=[L_2 : F^{\theta}]$ when $\theta \,\colon L_1 \rightarrow L_2$ is the given isomorphism. But I don't see how to relate $F^{\theta}$ to $F$ in an easy way. Any help or counterexample?</p>
<p>Many thanks in advance.</p>
| Jyrki Lahtonen | 11,619 | <p>No.</p>
<p>A simple counterexample is $K=\Bbb{Q}(\pi)$, $F=\Bbb{Q}(\pi^6)$,
$L_1=\Bbb{Q}(\pi^2)$ and $L_2=\Bbb{Q}(\pi^3)$. Then</p>
<ol>
<li>All the fields $K,F,L_1,L_2$ are simple transcendental extensions of $\Bbb{Q}$ and thus they are all isomorphic to each other.</li>
<li>$[K:F]=6$, $[L_1:F]=3$, $[L_2:F]=2$.</li>
</ol>
|
346,950 | <p>The equation $$3\sin^2 x - 3\cos x -6\sin x + 2\sin 2x + 3=0$$ has a solution $x = 0$. That is mean it has a factor $\cos x - 1$. I tried write the given equation has the form
$$(\cos x - 1)P(x)=0.$$ I am looking for the factor $P(x)$. How to do that?</p>
| Ron Gordon | 53,268 | <p>It looks like the simplest way to factor out the zero-ing function is to use a half angle. For example, I rewrite your equation as</p>
<p>$$\sin{\frac{x}{2}} \left [ 2 \cos{\frac{x}{2}} ( 3 \sin{x} + 4 \cos{x} - 6) + 6 \sin{\frac{x}{2}} \right ] = 0$$</p>
<p>Note that </p>
<p>$$\sin{\frac{x}{2}} = \sqrt{\frac{1-\cos{x}}{2}}$$</p>
<p>so the $1-\cos{x}$ factor you wanted is there, in a way.</p>
|
346,950 | <p>The equation $$3\sin^2 x - 3\cos x -6\sin x + 2\sin 2x + 3=0$$ has a solution $x = 0$. That is mean it has a factor $\cos x - 1$. I tried write the given equation has the form
$$(\cos x - 1)P(x)=0.$$ I am looking for the factor $P(x)$. How to do that?</p>
| lab bhattacharjee | 33,337 | <p>$$\implies 3\left(\sin^2x-2\sin x+1\right)+2\sin2x-3\cos x=0$$</p>
<p>$$\implies 3\left(1-\sin x\right)^2+\cos x(4\sin x-3)=0$$</p>
<p>$$\text{We know,}\sin x=\frac{2t}{1+t^2},\cos2x=\frac{1-t^2}{1+t^2}\text{ where }t=\tan\frac x2$$</p>
<p>So, $1-\sin x=\frac{(1-t)^2}{1+t^2}, 4\sin x-3=4\frac{2t}{1+t^2}-3=-\frac{3t^2-8t+3}{1+t^2}$</p>
<p>$$\implies 3\left(\frac{(1-t)^2}{1+t^2}\right)^2-\frac{(1-t^2)}{1+t^2}\frac{(3t^2-8t+3)}{1+t^2}=0$$</p>
<p>If $t$ is finite, $$3(1-t)^4-(1-t)(1+t)(3t^2-8t+3)=0$$ </p>
<p>$$\text{Or, }(1-t)\{3(1-t)^3-(1+t)(3t^2-8t+3)\}=0$$</p>
<p>If $t=1, \tan\frac x2=1\implies \frac x2=n\pi+\frac\pi4\implies x=2n\pi+\frac\pi2$ where $n$ is any integer .</p>
<p>Else $$2t(3t^2-7t+2)=0$$</p>
<p>If $t=0,t=2m\pi+0$ where $m$ is any integer (like the previous case)</p>
<p>Else $3t^2-7t+2=0$ (Solve the Quadratic Equation and find $x$ like the previous cases )</p>
<p>Finally, if $t$ is $+\infty,\frac x2=r\pi+\frac\pi2, x=2r\pi+\pi=(2r+1)\pi$ where $r$ is any integer</p>
<p>if $t$ is $-\infty,\frac x2=s\pi-\frac\pi2, x=2s\pi-\pi=(2s-1)\pi$ where $s$ is any integer</p>
<p>In the last two case $x$ is an odd multiple of $\pi$ which does <strong>not</strong> satisfy the given equation </p>
|
138,658 | <p>Suppose $X$ is a topological space, and $\mu$ is a Borel measure on $X$. Also suppose we have an $n$-dimensional vector bundle $E \to X$, with an inner product $\langle \cdot,\cdot \rangle_x$ on the fibre $E_x$ for all $x \in X$, in such a way that each $E_x$ is complete and such that there exists a vector bundle trivialisation which is compatible with the fibrewise inner products. The inner product $\langle \cdot, \cdot \rangle_x$ determines a norm $||\cdot||_x$ on $E_x$.</p>
<p>Say that a (not necessarily continuous) section $\sigma \colon X \to E$ is <em>measurable</em> if its restriction to each trivialising open set $U \subset X$ is given by a measurable function $U \to \mathbb{F}^n$ (here $\mathbb{F}$ is either $\mathbb{R}$ or $\mathbb{C}$, depending on how you feel). Denote the set of measurable sections (understood as being defined up to measure zero) by $\Gamma(E)$.</p>
<p>Given a section $\sigma \in \Gamma(E)$ and a number $p \in (0,\infty)$, we can define
$$||\sigma||_p := \left( \int_X ||\sigma(x)||_x^p \; d\mu(x) \right)^{1/p},$$
and we can then define the corresponding $E$-valued Lebesgue space $L^p(X;E)$ in the obvious way.</p>
<blockquote>
<p><strong>Question:</strong> do we have the usual duality relations for Lebesgue spaces, i.e. $(L^p(X;E))^* \cong L^{p^\prime}(X;E)$ where $1 = \frac{1}{p} + \frac{1}{p^\prime}$?</p>
</blockquote>
<p>As one would expect, there is a kind of Hölder inequality: if $\sigma \in L^p(X;E)$ and $\tau \in L^{p^\prime}(X;E)$, then the function $\langle \sigma,\tau \rangle$ on $X$, given by
$$\langle \sigma,\tau \rangle(x) := \int_X \langle \sigma(x), \tau(x) \rangle_x \; d\mu(x),$$
satisfies $|| \langle \sigma,\tau \rangle||_{L^1(X)} \leq ||\sigma||_p ||\tau||_{p^\prime}$. It follows that the pairing $\langle \cdot,\cdot \rangle$ can be used to isometrically embed $L^{p^\prime}(X;E)$ into $(L^p(X;E))^*$.</p>
<p>However, I haven't been able to prove the reverse containment - that each functional $\phi \in (L^p(X;E))^*$ is given by pairing with an element of $L^{p^\prime}(X;E)$ - without additional assumptions, such as the existence of a finite trivialising cover for $E$ with uniform norm control (for example, when $X$ is compact). In this case, one can recover the result from the corresponding result for trivial bundles - which is essentially the case of vector-valued Lebesgue spaces - but constants appear which depend on the cardinality of a trivialising cover, which is somewhat unexpected.</p>
<p>Has this been explicitly proven anywhere? Is it even true in general? (I'll be very surprised if it isn't)</p>
| TaQ | 12,643 | <p>At least if $X$ is $\sigma$-compact and if compact sets have finite measure and if the local vector bundle trivializations are homeomorphisms, I would expect the representation of $(L^p(E))^*$ as $L^{p'}(E)$ to hold. The proof would proceed by first dividing $X$ into countably many disjoint relatively compact sets $B_i$ over which the bundle is trivial. Given $u$ in $(L^p(E))^*$, first consider the functionals $u_i:z\mapsto u(\bar z_i)$ on $L^p(E_{\,|B_i})$ where $\bar z_i$ denotes the zero extension. Then you get the representation of $u_i$ by some $y_i$ in $L^{p'}(E_{\,|B_i})$ . Patch these together to get a measurable section $y$ of $E$ such that $u(x)=\int_X\langle y(t),x(t)\rangle_t\,d\mu(t)$ holds for $x$ in $L^p(E)$. Then you can show $\|y\|_{p'}\le\|u\|$ in the same way as this is done in the case of real or complex valued functions just by considering the inequality $|u(x)|\le\|u\|\,\|x\|_p$ for $x$ chosen so that it gets the form $\|y\|_{p'}^{\,p'}\le\|u\|\,\|y\|_{p'}^{\,\frac{p'}p}$ which directly gives the required inequality. Note here that you can get $\sum_{j=1}^n(\eta_j\xi_j)=(\sum_{j=1}^n\eta_j^2)^{\frac{p'}2}$ by choosing $\xi_j=\dfrac{\eta_j}{(\sum_{j=1}^n\eta_j^2)^{1-\frac{p'}2}}\ $.</p>
|
1,784,679 | <p>if $p,q,r$ are three positive integers prove that</p>
<p>$$LCM(p,q,r)=\frac{pqr \times HCF(p,q,r)}{HCF(p,q) \times HCF(q,r) \times HCF(r,p)}$$</p>
<p>I tried in this way;</p>
<p>Let $HCF(p,q)=x$ hence $p=xm$ and $q=xn$ where $m$ and $n$ are relatively prime.</p>
<p>similarly let $HCF(q,r)=y$ hence $q=ym_1$ and $r=yn_1$ where $m_1$ and $n_1$ are Relatively prime.</p>
<p>Alo let $HCF(r,p)=z$ hence $r=zm_2$ and $p=zn_2$</p>
<p>we have $$p=xm=zn_2$$</p>
<p>$$q=xn=ym_1$$and</p>
<p>$$r=yn_1=zm_2$$</p>
<p>can i have any hint to proceed?</p>
| sTertooy | 336,630 | <p>I decided to write my comment as an answer. Rather than start with naming $HCF(p,q)$, $HCF(q,r)$ and $HCF(r,p)$, start with $HCF(p,q,r)$. So let's call $HCF(p,q,r) = h$.</p>
<p>Next, write $HCF(p,q) = xh$, $HCF(q,r) = yh$ and $HCF(r,p) = zh$. It should be clear why we can assume the factor $h$ appears in all three, but you also know that $x,y,z$ are relatively prime. (Why?)</p>
<p>Thus, you can write $p = p'xzh$ for some $p'$, and similarly $q = q'xyh$ and $r = r'yzh$ (again, why?). What do you get when you plug those into your equation?</p>
|
4,377,932 | <p><span class="math-container">$$3000 = ( p6300 + (1-p)2200 ) / 1.06
\\
3000 = p6300 + 2200 - p2200 / 1.06
\\
3000 = p4100 + 2200 / 1.06
\\
800 = p4100 / 1.06
\\
p4100 = 800\times1.06
\\
p4100 = 848
\\
p = 0.2068$$</span></p>
<p>The correct answer is <span class="math-container">$p=0.239.$</span>
What mistake(s) have I made?</p>
| Magdiragdag | 35,584 | <p>As written, the first line says that <span class="math-container">$(1 - p) 2200$</span> has to be divided by <span class="math-container">$1.06$</span>. However, you're messing that up in several ways in the subsequent steps.</p>
<p>Edit: in the comments the OP said the correct answer is 0.239. And, also mentioned in the comments, that indicates that the original question should be
<span class="math-container">$3000 = (p 6300 + (1 - p) 2200) / 1.06.$</span></p>
|
4,377,932 | <p><span class="math-container">$$3000 = ( p6300 + (1-p)2200 ) / 1.06
\\
3000 = p6300 + 2200 - p2200 / 1.06
\\
3000 = p4100 + 2200 / 1.06
\\
800 = p4100 / 1.06
\\
p4100 = 800\times1.06
\\
p4100 = 848
\\
p = 0.2068$$</span></p>
<p>The correct answer is <span class="math-container">$p=0.239.$</span>
What mistake(s) have I made?</p>
| ryang | 21,813 | <blockquote>
<p>I'm not good at solving an equation with brackets. Could you show it in a simple way, with steps, with the equation and values from my initial post?</p>
</blockquote>
<p>Click <a href="https://www.symbolab.com/solver/step-by-step/3000%20%3D%20%5Cleft(p6300%20%2B%20%5Cleft(1-p%5Cright)2200%5Cright)%20%2F%201.06?or=input" rel="nofollow noreferrer">here</a> for the step-by-step working with the correct algebraic manipulations.</p>
<p>Regarding your suggested working, note that when you have parentheses in an arithmetic expression, first perform the operations within the innermost pair then keep working outwards. Also, note that:</p>
<ol>
<li><span class="math-container">\begin{align}(a+b)c/d&=\frac{ac+bc}d\\\\&\ne ac+\frac{bc}d\end{align}</span>
<span class="math-container">\begin{align}(3+5)2/7&=\frac{6+10}7\\\\&\ne 6+\frac{10}7\end{align}</span></li>
<li><span class="math-container">\begin{align}a+ b/c &= a+ \frac bc \\&\ne \frac{a+b}c\end{align}</span>
<span class="math-container">\begin{align}2+ 3/7 &= 2+ \frac 37 \\&\ne \frac{5}7\end{align}</span></li>
</ol>
|
3,313,603 | <p>I am assigned with a question which states the rate of a microbial growth is exponential at a rate of (15/100) per hour. where y(0)=500, how many will there be in 15 hours?</p>
<p>I know this question is generally modelled as: </p>
<p><span class="math-container">$y=y_0*e^{kt}$</span></p>
<p>However, my solution ended up being modelled as :</p>
<p><span class="math-container">$y=e^{kt}*e^{c}$</span> via <span class="math-container">$y'=ky$</span></p>
<p>The resulting equation was:</p>
<p><span class="math-container">$y=e^{kt}*e^{ln500}$</span></p>
<p>I ended up getting <span class="math-container">$y(15)=4743.86$</span> which is the same answer for both methods.</p>
<p>I'm wondering how the general equation was modelled, and if someone could explain how I could tidy up my equations.</p>
<p>Thanks. </p>
| Abdulrahman Albishri | 1,024,527 | <p>The general equation can be modeled in the following way:</p>
<p>If we call the bacteria <span class="math-container">$B$</span>, and time (obviously) <span class="math-container">$t$</span>, then <span class="math-container">$B$</span> is clearly a function of time <span class="math-container">$B(t)$</span>. Notice that <span class="math-container">$B$</span> is proportional with the rate of change.</p>
<p><span class="math-container">$$\frac{dB}{dt} \propto B
\quad\implies\quad \frac{dB}{dt} = kB,$$</span></p>
<p>Where <span class="math-container">$k$</span> is the proportionality constant. As we can see, this is just a separable differential equation, and to solve this we separate the variables and integrate in the following way:</p>
<p><span class="math-container">$$\left(\frac{dB}{dt}\right)\frac{1}{B} \,=\, (kB)\frac{1}{B} \;\implies\; \frac{1}{B}\frac{dB}{dt}dt \,=\, kdt$$</span> <span class="math-container">$$\implies\; \int \frac{1}{B}dB \,=\, \int k\,dt \;\implies\; \ln|B|+C_1 \,=\, k(t+C_2) $$</span></p>
<p><span class="math-container">$$\implies\; \ln|B| \,=\, kt+(kC_2-C_1),\quad \text{and letting}\quad kC_2-C_1=C_3,$$</span> <span class="math-container">$$\implies\; \ln|B|=kt+C_3.$$</span> <span class="math-container">$$\because B\geq 0\quad\quad (\text{you can't have negative bacteria}) $$</span> <span class="math-container">$$\therefore \ln(B)=kt+C_3
\;\implies\; e^{\ln(B)}=e^{kt+C_3} $$</span> <span class="math-container">$$\implies B=e^{kt}e^{C_3},\quad \text{and letting}\quad e^{C_3}=C,$$</span> <span class="math-container">$$\therefore B=Ce^{kt} \quad\quad (\text{exponential growth})$$</span></p>
<p><strong>B(t) plot:</strong></p>
<p><img src="https://i.stack.imgur.com/yQVff.png" alt="" /></p>
<p>Note that <span class="math-container">$C$</span>, and <span class="math-container">$k$</span> are determined by the initial conditions. I'm guessing there can be some confusion about how <span class="math-container">$e^{C_3}$</span> can be equated to <span class="math-container">$C$</span>, but remember that <span class="math-container">$e$</span>, <span class="math-container">$C_1$</span>, <span class="math-container">$C_2$</span>, <span class="math-container">$C_3$</span>, and <span class="math-container">$C$</span> are all constants, therefore, it wouldn't make a difference if we equate <span class="math-container">$e^{C_3}$</span> with <span class="math-container">$C$</span>.</p>
<p>I hope i did answer your question.</p>
|
3,405,622 | <p>Let <span class="math-container">$\int_{0}^{1}fg \text{ }d\mathbb{P}=0$</span>, for all <span class="math-container">$f$</span> <span class="math-container">$\in$</span> <span class="math-container">$L^{\infty}([0,1],\mathbb{P})$</span> and <span class="math-container">$g$</span> be a fixed function in <span class="math-container">$L^{1}([0,1],\mathbb{P})$</span> and where <span class="math-container">$\mathbb{P}$</span> is a probability measure in <span class="math-container">$[0,1]$</span>. Is it true <span class="math-container">$g=0$</span> almost everywhere?
Attempt: I was trying to use dominated convergence theorem to show <span class="math-container">$\int|g|^{2}\text{ }d\mathbb{P}=0$</span>, just want to find <span class="math-container">$f_{n}\rightarrow \overline{g}$</span> bounded sequence, that I can do using density of <span class="math-container">$L^{\infty}$</span> in <span class="math-container">$L^{1}$</span>, but the problem is <span class="math-container">$f_{n}$</span> need not dominated by <span class="math-container">$L^{1}$</span> function for applying DCT.</p>
| Kavi Rama Murthy | 142,385 | <p>Yes. Take <span class="math-container">$f=I_{g>0}$</span> to get <span class="math-container">$\int_{g>0} gdP=0$</span>. Similarly <span class="math-container">$\int_{g<0} gdP=0$</span>. These give <span class="math-container">$\int |g|dP=0$</span> so <span class="math-container">$g=0$</span> almost everywhere w.r.t. <span class="math-container">$P$</span>. </p>
|
2,857,769 | <blockquote>
<p>Find $t$ such that $$\lim_{n\to\infty} \frac {\left(\sum_{r=1}^n r^4\right)\cdot\left(\sum_{r=1}^n r^5\right)}{\left(\sum_{r=1}^n r^t\right)\cdot\left(\sum_{r=1}^n r^{9-t}\right)}=\frac 45.$$</p>
</blockquote>
<p>At first sight this question scared the hell out of me. I tried using the general known formulas like $$\sum_{r=1}^n r^4=\frac {n(n+1)(2n+1)(3n^2+3n-1)}{6}$$ and $$\sum_{r=1}^n r^5=\frac {n^2(n+1)^2(2n^2+2n-1)}{12}.$$</p>
<p>But the denominator portion really doesn't go with it. I tried to write it in form of integrals. I also searched the internet for some information but it dealt higher level calculus relating the harmonic functions, Bernoulli numbers and the Riemann zeta function. I read about it but couldn't get much out of it. </p>
<p>Any help would be greatly appreciated. Thanks. </p>
| Oleg567 | 47,993 | <p>The skeleton:</p>
<p>First thought is to apply integrals:</p>
<p>$$
\dfrac{\int\limits_1^n x^4 dx \int\limits_1^n x^5 dx}{\int\limits_1^n x^t dx \int\limits_1^n x^{9-t} dx} \sim
\dfrac{\dfrac{n^5}{5} \cdot \dfrac{n^6}{6}}{\dfrac{n^{t+1}}{t+1} \cdot \dfrac{n^{9-t+1}}{9-t+1}}
$$</p>
<p>Then we'll have fraction
$$
\dfrac{(9-t+1)(t+1)}{30}=\dfrac{4}{5} = \dfrac{24}{30}
$$</p>
<p>Then quadratic equation
$$(9-t+1)(t+1) = 24$$
So $t=2$, $t=7$ should work.</p>
|
2,512,556 | <p>What would be the solution of $ y''+y=\cos (ax) \ $ if $ \ a \to 1 \ $. </p>
<p><strong>Answer:</strong></p>
<p>I have found the complementary function $ y_c \ $ </p>
<p>$ y_c(x)=A \cos x+B \sin x \ $</p>
<p>But How can I find the particular integral if $ a \to 1 \ $ </p>
| Ian | 83,396 | <p>For simplicity let's just solve the complex exponential problem $z''+z=e^{iax}$ (your solution is the real part of this). When $a \neq 1$, the particular solution to this is of the form $Ae^{iax}$ (where $A$ is in general complex). When $a=1$ this is no longer a particular solution, because it is actually a homogeneous solution. Instead the particular solution looks like $Axe^{ix}$ in that case. Physically, this linearly growing amplitude arises due to a resonance effect (driving the system at its natural frequency).</p>
<p>You mention $a \to 1$, which makes me think you might also want to understand how the $a \neq 1$ case approaches the $a=1$ case. This is something that confuses students a lot, because it seems like the situation is "discontinuous" when it shouldn't be. </p>
<p>To understand how the two cases come together, you really need to pin down some initial conditions. For example, $z''+z=e^{iax},z(0)=1,z'(0)=0$ has solution $\frac{a^2 \cos(x)-e^{iax}+ia\sin(x)}{a^2-1}$. Computing the limit of this as $a \to 1$ at fixed $x$ requires L'Hospital's rule. Doing that gives the factor of $x$ when you differentiate $e^{iax}$ with respect to $a$.</p>
|
3,500,799 | <p>what is the dimension of the vector space spanned by the set of vectors <span class="math-container">$(a,b,c) $</span>where <span class="math-container">$a^2+b^2=c$</span>?</p>
| Lukas Rollier | 737,665 | <p>It is a subspace of <span class="math-container">$\mathbb{R}^3$</span> containing the linearly independent vectors <span class="math-container">$(1,0,1)$</span>, <span class="math-container">$(0,1,1)$</span>, and <span class="math-container">$(1,2,5)$</span>. Hence, it must be <span class="math-container">$\mathbb{R}^3$</span> in its entirety. The dimension is 3.</p>
|
12,102 | <p>For local field, the reciprocity map establishes almost an isomorphism from the multiplicative group to the Abelian Absolute Galois group. (In global case the relationship is almost as nice). It is tempted to think that there can be no such nice accident. </p>
<p>Do we know any explanation which suggest that there "should be" a relationship between the multiplicative group and the Galois group?</p>
<p>Actually, my current belief is that the reciprocity map is half accidental. I think that there is a natural extension where we can define a natural action of the multiplicative group. In the local case this is the Lubin-Tate extension (a generalization of cyclotomic extension). The fact that this Lubin-Tate extension is the Abelian Absolute Galois group is an accident. </p>
<p>Do we know something that might support/ reject this view?</p>
<p>Please feel free to edit the question into a form that you think might be better.</p>
| Johnson Jia | 3,247 | <p>The reciprocity map in the local case can be motivated by considering the unramified case. Let me try to explain this. For simplicity, let us consider the case of a finite abelian unramified extension $K_n/\bf Q_p$ of degree $n$. Denote by $k_n$ the residue field of $K_n$. The unramified condition gives rise to the isomorphism
$$Gal(K_n/ {\bf Q_p}) \simeq Gal(k_n / {\bf F_p}) \simeq {\bf Z}/n.$$</p>
<p>Now we like to parametrize these abelian unramified extensions {$ K_n$} of $\bf Q_p$ using information from $\bf Q_p$. However, $\bf Q_p$ is not finitely generated as a module over $\bf Z_p$, let along over $\bf Z$. $\bf Q_p^\times$, on the other hand, decomposes as
$$\bf Q_p^\times \simeq p^{\bf Z} \times {\bf Z}_p^\times \simeq \bf Z \times \mu_{p-1} \times Z_p,$$
which, if anything, at least contains a copy of $\bf Z$ (depending on the choice of a uniformizer, say $p$).</p>
<p>It then seems somewhat natural to consider the map $${\bf Q_p^\times} \xrightarrow{Art} Gal({\bf Q^{ab, un}_p}/{\bf Q_p}) \qquad p \mapsto Frob$$</p>
<p>where $Frob$ is a choice of a topological generator for $Gal({\bf Q_p^{ab, un}}/{\bf Q_p})$, the Galois group of the maximal abelian unramified extension ${\bf Q_p^{ab, un}}$ of $\bf Q_p$, which is naturally isomorphic to $Gal(\bar{\bf F_p} / \bf F_p)$ (which is itself non-canonically isomorphic to $\hat {\bf Z}$).</p>
<p>Now compose the Artin map $Art$ with the restriction map $Gal({\bf Q_p^{ab, un}}/{\bf Q_p}) \to Gal(K_n/ {\bf Q_p})$, we obtain a map
$${\bf Q_p}^\times \xrightarrow{Art_n} Gal(K_n/ {\bf Q_p})$$ whose kernel is $$p^{n {\bf Z}} \times \bf Z_p^\times,$$ which coincidentally is also the image of the norm of $K_n^\times$ in $\bf Q_p^\times$.</p>
<p>This also sheds some light on the global situation, where you have Frobenius at all but finitely many primes. Don't quote me on this, but I recall the Artin reciprocity map is uniquely determined by its action on the Frobenii (I assume a Chebatorev density argument will show this, and you can prove Chebatorev Density Theorem independent of class field theory if I remember correctly.)</p>
<p>Lastly, we saw that the (local) reciprocity map depends on the choice of a Frobenius as well as that of a uniformizer.</p>
|
3,995,046 | <p>Refer to <a href="https://oeis.org/A340800" rel="nofollow noreferrer">https://oeis.org/A340800</a> to notice that the number of primes between two primes having the same last digit is increasing as the primes themselves increase. Is there an explanation for this? How can the size of primes have any influence on the last digit of the following primes?</p>
| Rhys Hughes | 487,658 | <p>Let <span class="math-container">$$\mathfrak P(x,n)=\Bbb P(\exists m\leq n \in \Bbb N : x+10m \text{ is prime} | x \text{ is prime})$$</span></p>
<p>The relative density of primes decreases as numbers get larger. Because of this, <span class="math-container">$\mathfrak P$</span> is very high for small <span class="math-container">$x$</span> and generally decreasing in <span class="math-container">$x$</span>. At the same time, it is increasing in <span class="math-container">$n$</span> (Every time you add another potential number that might be prime, the probability increases).</p>
<p>One can see from this exactly what the OEIS results show, that if a prime is small, the next prime with the same last digit is very likely not much bigger, and for a large prime, you'll likely need a larger <span class="math-container">$n$</span> to satisfy, which implies the next prime will be further away.</p>
|
3,995,046 | <p>Refer to <a href="https://oeis.org/A340800" rel="nofollow noreferrer">https://oeis.org/A340800</a> to notice that the number of primes between two primes having the same last digit is increasing as the primes themselves increase. Is there an explanation for this? How can the size of primes have any influence on the last digit of the following primes?</p>
| Robert Israel | 8,508 | <p>Of course the general trend of A340800 must be increasing (although it is not monotonic). For any <span class="math-container">$M$</span>, the maximum of <span class="math-container">$A072971(k)$</span> for <span class="math-container">$k \le M$</span> is some finite
value <span class="math-container">$N$</span>, and so if <span class="math-container">$n > N$</span> and <span class="math-container">$A340800(n)$</span> exists at all it must be <span class="math-container">$prime(k)$</span> for some <span class="math-container">$k > M$</span>. The existence of <span class="math-container">$A340800(n)$</span> follows from the prime <span class="math-container">$k$</span>-tuples conjecture.</p>
|
68,803 | <p>I am trying to understand how all the players in the title relate, but with all the grading shifts,and difficult isomorphisms involved in the subject I am having a hard time being sure that I have the picture right. I am going to write what I think is true, and if someone would confirm or deny it, that would be really nice. </p>
<p>The basic jumping off point is that if N is a simply connected manifold, symplectic cohomology of the cotangent disk bundle $D^*(N)$, symplectic cohomology $SH^*(D^*(N))$ as defined in Seidel's "A Biased View of Symplectic Cohomology" is naturally identified with isomorphic to $H_{n-*}(LN)$. And symplectic homology is isomorphic to $H^*(LN)$, these are the Hochschild cohomologies and homologies respectively of the algebra $C^*(N)$. </p>
<p>The reason is that the zero section generates the compact Fukaya category $Fuk^{cpt}(D^*(N))$ and $End(N_0,N_0) \cong C^*(N)$. There is expected to be a geometric Seidel map from $SH^*(D^*(N)) \to HH^*(Fuk^{cpt}(D^*(N))$ basically one considers cylinders in with a puncture which satisfy a deformed d-bar equation and which are asymptotic to periodic orbits of the Hamiltonian vector field and whose boundary lies on the zero section and is able to deform the compositions in the category to first order.</p>
<p>Question 1) Has anybody checked in this example that Seidel's map is an isomorphism?</p>
<p>Now we take equivariant versions of this, $SH^*_{eq}$ which should be identified with cyclic cohomology $CC^*(C^*(N))$ and we need to consider. Now we move to (linearized) contact homology, which Bourgeois and Oancea claim can be identified with $H_{eq}(LN,N)$(I give up on the gradings at this point :)) where N is included as the constant loops. Reasonable enough, since those are somehow the generators missing from contact homology. However, they also seem to be making an identification that $SH_* \cong H_*(LN)$ and not the cohomology... I get that it's not really a huge deal as vector spaces to identify a vector space and it's dual, but it adds to the confusion below.</p>
<p>With contact homology one can try a similar map to the Seidel map, namely work in the symplectic completion $T^*(N)$ and consider holomorphic disks with a puncture, boundary on the zero section, and as asymptotic now to a Reeb orbit as $|\rho|\to \infty$ `($|\rho|$ is some norm on $T^*(N)$). This gives a map from $CH_{*}\to CC^{*} $ which is mysterious because...(or maybe this is a map from "contact cohomology", I'm getting confused). Edit: A reference for this map is in a paper by Xiaojun Chen called "Lie Bialgebras and cyclic homology of A(\infty structures) in topology" </p>
<p>Question 2) We are supposed to somehow map $H_{*,eq}(LN,N) \to H_{*,eq}(LN)$. I'm not great with topology but this is a strange map, since it seems to go the wrong way. On the other hand one should expect it to be interesting by analogy with the Seidel map. What is this map at least conjecturally supposed to be? The only thing I could think of is the kernel of the map: $C_*(LN) \to C_*(N)$ induced by the map $LN \to N$ but this map doesn't even exist equivariantly.</p>
| Tim Perutz | 2,356 | <p><b>Some blah on symplectic homology vs. cohomology.</b> There's an invariant $SH(M)$ of Liouville domains $M$ which some people call symplectic homology and some symplectic cohomology. This is the direct limit of Hamiltonian Floer groups associated with functions of increasing eventual slope. The dual theory has two rather unpleasant features: it involves inverse limits, hence one must worry about $\varprojlim^1$-terms; and in general it's not countably generated. It's not often used.</p>
<p>Why the confusion about terminology? Well, depending on your convention for the sign of the symplectic action functional, you may regard this as Morse homology or compactly supported Morse cohomology of this function. From the perspective of Lagrangian Floer cohomology, consistency demands that one calls it symplectic cohomology; I do. However, symplectic field theorists (including Bourgeois-Oancea, I think) prefer the contrary convention.</p>
<p><b>Blah about grading.</b> The integer grading on $SH^\ast(M)$ is defined when $c_1=0$, and is canonical when $H^1(M)=0$. One has Viterbo's map $H^\ast(M)\to SH^\ast(M)$, and one convention makes this preserve degree (I'll take that option), while another makes it shift degree by the complex dimension $n$ of $M$.</p>
<p><b>Seidel's map for cotangent bundles.</b>
(Edited: my first version was not correct.) There are actually <i>two</i> versions of Seidel's open-closed string map, derived from the same moduli spaces:
$$ \kappa: SH^\ast(M) \to HH^\ast(F(M),F(M))$$
and
$$ \lambda : HH_\ast(F(M),F(M)) \to SH^{n+\ast}(M). $$
Here $F(M)$ is the Fukaya category of exact, compact Lagrangians. Moreover, there are extensions of these maps to the wrapped Fukaya category $W(M)$. </p>
<p>The absence of a dualisation - hence the connection of symplectic cohomology to both Hochschild homology and cohomology - looks strange. Mohammed Abouzaid points out in his comment below that this is a manifestation of a self-duality property for the wrapped category. He shows in <a href="http://arxiv.org/abs/1003.4449" rel="noreferrer">http://arxiv.org/abs/1003.4449</a> that for $M=T^\ast L$ and the wrapped category, the map $\lambda$ is an isomorphism. </p>
<p><b>Cyclic version.</b> My expectations are slightly different from those stated in the question. I'd guess that $\lambda$ extends to a map from cyclic homology to circle-equivariant symplectic cohomology,
$$ HC_\ast(F(M)) \to SH^{n+\ast}_{S^1}(M) $$
and that this should be an isomorphism when $\lambda$ is. </p>
<p>For cotangent bundles $T^\ast L$ of simply connected, spin manifolds, one has $SH^\ast_{S^1}(M) \cong H_{n-*}^{S^1}(\mathcal{L}L)$. </p>
<p>Linearised contact (co)homology is, according to Bourgeois-Oancea (if I have it right), the mapping cone of the (cochain level) Viterbo map $H^\ast(M; H^\ast_{S^1}(pt.)) \to SH^\ast_{S^1}(M)$. For cotangent bundles as before, Viterbo's map should be identifiable with the map induced by the equivariant inclusion of constant loops: $H^\ast(T^\ast L)[u] = H^\ast (L)[u] = H_{n-\ast}(L)[u] \to H_{n-\ast}^{S^1}(\mathcal{L}L)$. </p>
|
2,804,716 | <p>Given this Maclaurin series:</p>
<p>$$f(x)=\sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!}$$</p>
<p>And the following Catenary curve, assuming that $a=1$:</p>
<p>$$g(x)=\frac{a(e^\frac{x}{a}+e^{-\frac{x}{a}})}{2}$$</p>
<p>Why does $f(x)=g(x)$ seem to hold true (at least when graphed)?</p>
<p>I'm looking for a purely algebraic reason here as to why these two are equal, ideally in terms that are at or around a high-school calculus level (where I'm at currently).</p>
<p>If I am mistaken, and these two are not equal to each other, an explanation of why that is would be great too. </p>
| cansomeonehelpmeout | 413,677 | <p>Remember that $$e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$$</p>
<p>Thus $$\frac{1}{2}(e^\frac{x}{a}+e^\frac{-x}{a})=\frac{1}{2}\sum_{n=0}^{\infty}\left [\left (\frac{x}{a}\right )^n\frac{1}{n!}+\left (\frac{-x}{a}\right )^n\frac{1}{n!}\right ]$$</p>
<p>This can be simplified to $$\frac{1}{2}\sum_{n=0}^\infty \left (\frac{1+(-1)^n}{a^n n!}\right )x^n$$</p>
<p>When $n$ is odd, $1+(-1)^n=0$, so we can replace $n$ by $2n$ everywhere to get $$\frac{1}{2}\sum_{n=0}^\infty \left [\frac{1+1}{a^{2n} (2n)!}x^{2n}\right ]=\sum_{n=0}^\infty\frac{x^{2n}}{a^{2n}(2n!)}$$</p>
<p>When $a=1$, you get the series $$\sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!}$$</p>
|
1,641,255 | <p>I am having difficulty find the centroid of the region that is bound by the surfaces $x^2+y^2+z^2-2az=0$ and $3x^2+3y^2-z^2=0$ (lying above $xy$ plane, consider the inner region). I know the first surface is a sphere, while the second is an infinite cone.</p>
<p>I just dont know how to even approach it as I dont know how to even visualize these surfaces.</p>
<p>Can anyone help?</p>
<p>My only thoughts would be to probably use spherical coordinates.</p>
<p>My possible thoughts( I have no clue if this will work, just some thoughts)</p>
<p>if we had</p>
<p>$$\rho^{2}=x^2+y^2+z^2$$</p>
<p>$$x=\rho\sin(\psi)\cos(\theta)$$</p>
<p>$$y=\rho\sin(\psi)\sin(\theta)$$</p>
<p>$$z=\rho\sin(\psi)$$</p>
<p>We could do some rearranging </p>
<p>$\rho=2acos(\psi)$</p>
<p>Since we are only considering above xy plane, maybe then $\theta$ would only run though fourth quadrant?</p>
<p>In regard to the centroid aspect, I know my answer will need to be of the form $C=(X_{c},Y_{c},Z_{c})$ where each component is obtained by evaluating the respective integral and dividing by the total mass.</p>
<p>But I am not sure, so looking for help.</p>
<p>Update: A suggestion was made to use cylindrical coordinates instead.</p>
<p>If I were to try this, I would have</p>
<p>$$x=r\cos\theta$$</p>
<p>$$y=r\sin\theta$$</p>
<p>$$x^2+y^2=r^2$$</p>
<p>$$z=z$$</p>
<p>And from rearranging I would have</p>
<p>$r^2+(z-a)^2=a^2$
and
$3r^2=z^2$</p>
<p>$$r=\pm \sqrt{\frac{z^2}{3}}$$</p>
<p>But I am still left confused on how to tie it all together. I am still wondering this even after all this time. I really would like some advice and help. Anyone? I am tried all I know and tried the current answer, but I am making no progress.</p>
| user84413 | 84,413 | <p>By symmetry, you have that $\overline{x}=\overline{y}=0$; so you just need to find $\overline{z}$.</p>
<p>If you use spherical coordinates, you have </p>
<p>$\hspace{.5 in}\displaystyle\overline{z}=\frac{\int z\;dV}{\int 1 \;dV}$ where</p>
<p>$\;\;\displaystyle\int z\;dV=\int_0^{2\pi}\int_0^{\frac{\pi}{6}}\int_0^{2a\cos\psi}(\rho\cos\psi)(\rho^2\sin\psi)\;d\rho d\psi d\theta$ and</p>
<p>$\;\;\displaystyle\int 1\;dV=\int_0^{2\pi}\int_0^{\frac{\pi}{6}}\int_0^{2a\cos\psi}(\rho^2\sin\psi)\;d\rho d\psi d\theta$. </p>
<hr>
<p>explanation: </p>
<p>A ray through the solid emanating from the origin exits along the surface of the sphere, which is given by $\rho=2a\cos\psi$.</p>
<p>The maximum value of $\psi$ is on the cone, where $\psi=\frac{\pi}{6}$ since $\displaystyle\tan\psi=\frac{r}{z}=\frac{\frac{z}{\sqrt{3}}}{z}=\frac{1}{\sqrt{3}}$. </p>
|
1,641,255 | <p>I am having difficulty find the centroid of the region that is bound by the surfaces $x^2+y^2+z^2-2az=0$ and $3x^2+3y^2-z^2=0$ (lying above $xy$ plane, consider the inner region). I know the first surface is a sphere, while the second is an infinite cone.</p>
<p>I just dont know how to even approach it as I dont know how to even visualize these surfaces.</p>
<p>Can anyone help?</p>
<p>My only thoughts would be to probably use spherical coordinates.</p>
<p>My possible thoughts( I have no clue if this will work, just some thoughts)</p>
<p>if we had</p>
<p>$$\rho^{2}=x^2+y^2+z^2$$</p>
<p>$$x=\rho\sin(\psi)\cos(\theta)$$</p>
<p>$$y=\rho\sin(\psi)\sin(\theta)$$</p>
<p>$$z=\rho\sin(\psi)$$</p>
<p>We could do some rearranging </p>
<p>$\rho=2acos(\psi)$</p>
<p>Since we are only considering above xy plane, maybe then $\theta$ would only run though fourth quadrant?</p>
<p>In regard to the centroid aspect, I know my answer will need to be of the form $C=(X_{c},Y_{c},Z_{c})$ where each component is obtained by evaluating the respective integral and dividing by the total mass.</p>
<p>But I am not sure, so looking for help.</p>
<p>Update: A suggestion was made to use cylindrical coordinates instead.</p>
<p>If I were to try this, I would have</p>
<p>$$x=r\cos\theta$$</p>
<p>$$y=r\sin\theta$$</p>
<p>$$x^2+y^2=r^2$$</p>
<p>$$z=z$$</p>
<p>And from rearranging I would have</p>
<p>$r^2+(z-a)^2=a^2$
and
$3r^2=z^2$</p>
<p>$$r=\pm \sqrt{\frac{z^2}{3}}$$</p>
<p>But I am still left confused on how to tie it all together. I am still wondering this even after all this time. I really would like some advice and help. Anyone? I am tried all I know and tried the current answer, but I am making no progress.</p>
| Community | -1 | <p>We can solve for $a=1$ and in the end multiply by $a$.</p>
<p>As said, in cylindrical coordinates</p>
<p>$$r^2+(z-1)^2\le 1,\\
3r^2\le z^2.$$</p>
<p>The intersection of the two surfaces is a circle in the plane $z=\frac32$ (by eliminating $r^2$).</p>
<p>We can decompose the domain in horizontal slices, which are disks of section $\pi r^2$, and split for the cone and the sphere. Then</p>
<p>$$\bar z=a\frac{\int_0^{3/2}z\frac{z^2}3dz+\int_{3/2}^2z(2z-z^2)dz}{\int_0^{3/2}\frac{z^2}3dz+\int_{3/2}^2(2z-z^2)dz}.$$</p>
<p>The integrands are polynomials, so the computation is straightforward.</p>
|
2,963,587 | <p>I'm working on a relatively low-level math project, and for one part of it I need to find to a function that returns how many many configurations are reachable within n moves. from the solved state.</p>
<p>Because there are 18 moves ( using the double moves metric ), one form of the function could be <span class="math-container">$\sum_{k=1}^{i} 18^{k} $</span>, since it would technically be the sum of all permutations reachable by the amount of moves 0,1,...,n. However, what would a more optimised function, which takes into account factors like inverses, cube symmetry, ...etc look like?</p>
| Mark S | 447,928 | <p><strong>Too long for a comment, making CW</strong></p>
<p>Alexander Chervov recently asked a very similar <a href="https://mathoverflow.net/questions/322877/number-of-positions-of-rubiks-cube-grows-with-multiplier-13-with-the-distance">question</a> on MO. He noted that the <a href="http://cube20.org" rel="nofollow noreferrer">http://cube20.org</a> data, mentioned by Henning, grows as <span class="math-container">$13^n$</span>, slower than the <span class="math-container">$18^n$</span> bound proposed by Mychil and corresponding to a free group. He wondered why it's dissimilar from an abelian group such as <span class="math-container">$\mathbb{Z}/2\mathbb{Z}^n$</span>, which looks like a Galton board and grows quadratically.</p>
<p>In his answer, Derek Holt noted that under the half-turn metric, moving the same face twice in a row does not lead to a new position, thus reducing the growth from <span class="math-container">$18$</span> to <span class="math-container">$15$</span>. He developed his reasoning further to consider the presentation of another infinitude group having a subset of relations of the Rubik's Cube group. Holt's presentation was </p>
<p><span class="math-container">$$\langle a_i, b_i\ (1 \le i \le 6) \mid a_i^4= 1,a_i^2=b_i\ (1 \le i\le 6), a_1a_4=a_4a_1,a_2a_5=a_5a_2,a_3a_6=a_6a_3 \rangle$$</span></p>
<p>With Singmaster notation we can have <span class="math-container">$F=a_1,\:F'=a_1^{-1},\:F^2=b_1,\:B=a_4,\vdots$</span></p>
<p>This group is infinite and automatic. It is amenable to study with the <a href="https://en.wikipedia.org/wiki/Knuth%E2%80%93Bendix_completion_algorithm" rel="nofollow noreferrer">Knuth-Bendix completion algorithm</a>, which is similar to Buchberger's algorithm.</p>
<p>Holt goes on to find a Taylor series expansion of the growth function for his infinite group. The coefficients begin to deviate from the cube20 data after the 4th move, which he takes to mean that the Rubik's Cube group has relators of length <span class="math-container">$8$</span> that cause the deviation (and may be the source of the finiteness of the Rubik's Cube group), although he was unable to find the relation. </p>
<p>I didn't know anything about the Knuth-Bendix completion algorithm prior to Holt's answer. As I envision it, Holt has managed to find the above infinite group with only a small number of relations, and show that it is in a certain sense "tangential" to the finite Rubik's Cube group.</p>
|
1,783,458 | <blockquote>
<p>Prove that the equation
$$z^n + z + 1=0 \ z \in \mathbb{C}, n \in \mathbb{N} \tag1$$
has a solution $z$ with $|z|=1$ iff $n=3k +2, k \in \mathbb{N} $.</p>
</blockquote>
<hr>
<p>One implication is simple: if there is $z \in \mathbb{C}, |z|=1$ solution for (1) then $z=cos \alpha + i \cdot sin\alpha$ and $|z + 1|=1$. It follows $cos\alpha=-\frac 1 2$ etc.</p>
<p>The other implication is the one I failed to prove.</p>
| David | 119,775 | <p>You have to prove that if $n=3k+2$, then the equation has a solution $z$ with $|z|=1$.</p>
<p>So, suppose $n=3k+2$ and take $z=\frac12(-1\pm i\sqrt3)$, which is the solution you found for the other direction. Certainly it is true that $|z|=1$, and by substituting and simplifying you can check that $z^n+z+1=0$.</p>
|
366,096 | <p>Let's consider $J\subset \mathbb R^2$ such that J is convex and such that it's boundary it's a curve $\gamma$. Let's suppose that $\gamma$ is anti-clockwise oriented, let's consider it signed curvature $k_s$. I want to prove the intuitive following fact:</p>
<p>$$
\int\limits_\alpha {k_s } \left( s \right)ds \geqslant 0
$$</p>
<p>For every sub-curve $\alpha \subset \gamma $.</p>
<p>And then prove that $k_s(s) \ge 0$</p>
<p>I have no idea how to attack this problem, intuitively I can see the result.</p>
| Robert Israel | 8,508 | <p>If $s$ is arc-length, $T(s)$ is the unit tangent vector and $N(s)$ the counterclockwise unit normal, $\dfrac{d}{ds} T(s) = k(s) N(s)$. It's convenient to consider the plane as the complex plane, so $T(s) = e^{i\theta(s)}$ and $N(s) = i e^{i \theta(s)}$. Then we have
$\dfrac{d\theta}{ds} = k(s)$. Now you want to show that $\theta(s)$ is nondecreasing...</p>
|
2,663,303 | <blockquote>
<p>Let $G$ be finite. Suppose that $\left\vert \{x\in G\mid x^n =1\}\right\vert \le n$ for all $n\in \mathbb{N}$. Then $G$ is cyclic.</p>
</blockquote>
<p>What I have attempted was the fact that every element is contained in a maximal subgroup following that <a href="https://groupprops.subwiki.org/wiki/Cyclic_iff_not_a_union_of_proper_subgroups" rel="nofollow noreferrer">cyclic iff not a union of cyclic subgroups</a>, <a href="https://math.stackexchange.com/questions/2060998/count-elements-in-a-cyclic-group-of-given-order">the order of elements of a cyclic group</a>, and Sylow-$p$ subgroups......But none of them seems helpful. </p>
<p>: ) It’s very kind of you to give me some hints to push me further. Thanks!</p>
| Ri-Li | 152,715 | <p>I have another idea, please have a look.</p>
<p>For each prime $p\mid |G|$ we have $H$, a Sylow $p$-subgroup(uniqueness comes from the given condition) s.t. $|H|=p^n$(say). Then $G=H_1 \oplus \cdots \oplus H_n$. Now each $H_i$ has at least one normal subgroup of order $p^a$ s.t. $a\mid n_i$ for all $0 \leq a \leq n_i$.</p>
<blockquote>
<p>Claim: Each $H_i$ is cyclic.</p>
</blockquote>
<p>Suppose $H_1$ is not cyclic then $\exists P_1\neq P_2$ subgroups of $H_1$ s.t. $|P_1|=|P_2|=p^b$ for some $0\leq b\leq n_1$ then the given condition would break.</p>
|
445 | <p>Under what circumstances should a question be made community wiki?</p>
<p>Probably any question asking for a list of something (e.g. <a href="https://math.stackexchange.com/questions/81/list-of-interesting-math-blogs">1</a>) must be CW. What else? What about questions asking for a list of applications of something (like, say, <a href="https://math.stackexchange.com/questions/804/applications-for-the-class-equation-from-group-theory">3</a>) or questions like <a href="https://math.stackexchange.com/questions/1446/interesting-properties-of-ternary-relations">2</a>? Should all soft-questions be made community wiki (and how we should define soft-question, in that case)?</p>
| Larry Wang | 73 | <p>The only useful effect I see in making a question community wiki is to force answers to be community wiki: that is, answers should be more easily editable, or should not grant reputation. So when do we want this? </p>
<p>Polls or lists, <a href="https://math.stackexchange.com/questions/329/best-ever-book-on-number-theory">explicit</a> or <a href="https://math.stackexchange.com/questions/3/list-of-interesting-math-podcasts">implicit</a>. Voting up or down is not a measure of how complete or correct the answer is, because those metrics do not apply to these questions. Instead, voting is used to express agreement or disagreement with a recommendation or opinion. </p>
<p>Questions asking for a comprehensive list of X. This would allow the top rated/accepted answer to contain a complete answer, edited whenever new answers come. I don't think there have been any like this yet, but some could eventually get converted to this format when too many answers get posted. <a href="https://stackoverflow.com/questions/194812/list-of-freely-available-programming-books">An example</a>. </p>
<p>Other possible situations:<br>
Questions whose answers reveal little about the answerer's mathematical ability.<br>
<a href="http://en.wikipedia.org/wiki/Breaching_experiment" rel="nofollow noreferrer">Breaching experiments</a> such as <a href="https://math.stackexchange.com/questions/459/solved-the-riemann-hypothesis-closed">this</a> (although we shouldn't have many of those anymore)<br>
Known open questions (these should probably go to mathoverflow or be closed instead, unless some really good answers get posted)</p>
|
990,340 | <p>I have an arguement with my friends on a probability question.</p>
<p><strong>Question</strong>: There are lots of stone balls in a big barrel A, where 60% are black and 40% are white, black and white ones are identical, except the color.</p>
<p>First, John, blindfolded, takes 110 balls into a bowl B; afterwards, Jane, blindfolded also, from bowl B takes 10 balls into cup C -- and find all 10 balls in C are white.</p>
<p>Now, what's the expectation of black balls in bowl B?</p>
<p>It seems there are 3 answers</p>
<p><strong>Answer 1</strong>
My friend thinks the probability of black stones in bowl B is still 60%, or 60% * 100 = 60 black balls expected in bowl B.</p>
<p><strong>Answer 2</strong>
However, I think 60% is just prior probability; with 10 balls all white from bowl B, we shall calculate the posterior probability. Denote $B_k$ as the black ball numbers in bowl B, and $C_{w}=10$ as the event that 10 balls in cup C are all white.</p>
<p>$$E(B_B | C_w=10)
= \sum_{x=10}^{110}\left[x P(B_k = x | C_w=10)\right] = \sum_{x=10}^{110} \left[x \frac{P(B_k = x \text{ and } C_w=10)}{P(C_w=10)}\right] = \sum_{x=10}^{110} \left[x \frac{P(B_k = x) P( C_w=10 | B_k=x)}{P(C_w=10)}\right]
$$</p>
<p>, where $$P(C_w=10) = \sum_{x=10}^{110} \left[ P(B_k = x) P(C_w = 10 | B_k=x) ) \right] $$</p>
<p>, and according to binomial distribution $$P(B_k=x) = {110 \choose x} 0.6^x 0.4^{110-x} $$ , and $$P(C_w = 10 | B_k=x) = \frac{1}{(110-x)(110-x-1)\cdots (110-x-9)}$$ </p>
<p><strong>Answer 3</strong>
This is from Stefanos below:
You can ignore the step that John takes 110 balls into a bowl B, this does not affect the answer. The expected percentages in bowl $B$ are again $60\%$ and $40\%$ percent, i.e. 66 black balls and 44 white balls. Now, after Jane has drawn 10 white balls obviously the posterior probability changes, since the expected number of balls is now 66 and 34. So, you are correct.</p>
<p>Which answer is right?</p>
<p>I sort of don't agree with Stefanos that, the black ball distribution from bowl B could vary a lot from barrel A, as the sampling distribution could be different from the universe distribution.</p>
<p>in other words, if Janes draws a lot balls and all are white, e.g. 50 balls are all white, I fancy it's reasonable to suspect that bowl B does not have a 60%-40% black-white distribution.</p>
| G Tony Jacobs | 92,129 | <p>The exponents are all multiples of 17. Thus, we can order the following:</p>
<p>$7<2^3<3^2$</p>
<p>Raising each to the $17$th power, we obtain:</p>
<p>$7^{17} < 2^{51} < 3^{34}$.</p>
<p>The inequality is preserved, because $x\mapsto x^{17}$ is a strictly increasing function.</p>
|
990,340 | <p>I have an arguement with my friends on a probability question.</p>
<p><strong>Question</strong>: There are lots of stone balls in a big barrel A, where 60% are black and 40% are white, black and white ones are identical, except the color.</p>
<p>First, John, blindfolded, takes 110 balls into a bowl B; afterwards, Jane, blindfolded also, from bowl B takes 10 balls into cup C -- and find all 10 balls in C are white.</p>
<p>Now, what's the expectation of black balls in bowl B?</p>
<p>It seems there are 3 answers</p>
<p><strong>Answer 1</strong>
My friend thinks the probability of black stones in bowl B is still 60%, or 60% * 100 = 60 black balls expected in bowl B.</p>
<p><strong>Answer 2</strong>
However, I think 60% is just prior probability; with 10 balls all white from bowl B, we shall calculate the posterior probability. Denote $B_k$ as the black ball numbers in bowl B, and $C_{w}=10$ as the event that 10 balls in cup C are all white.</p>
<p>$$E(B_B | C_w=10)
= \sum_{x=10}^{110}\left[x P(B_k = x | C_w=10)\right] = \sum_{x=10}^{110} \left[x \frac{P(B_k = x \text{ and } C_w=10)}{P(C_w=10)}\right] = \sum_{x=10}^{110} \left[x \frac{P(B_k = x) P( C_w=10 | B_k=x)}{P(C_w=10)}\right]
$$</p>
<p>, where $$P(C_w=10) = \sum_{x=10}^{110} \left[ P(B_k = x) P(C_w = 10 | B_k=x) ) \right] $$</p>
<p>, and according to binomial distribution $$P(B_k=x) = {110 \choose x} 0.6^x 0.4^{110-x} $$ , and $$P(C_w = 10 | B_k=x) = \frac{1}{(110-x)(110-x-1)\cdots (110-x-9)}$$ </p>
<p><strong>Answer 3</strong>
This is from Stefanos below:
You can ignore the step that John takes 110 balls into a bowl B, this does not affect the answer. The expected percentages in bowl $B$ are again $60\%$ and $40\%$ percent, i.e. 66 black balls and 44 white balls. Now, after Jane has drawn 10 white balls obviously the posterior probability changes, since the expected number of balls is now 66 and 34. So, you are correct.</p>
<p>Which answer is right?</p>
<p>I sort of don't agree with Stefanos that, the black ball distribution from bowl B could vary a lot from barrel A, as the sampling distribution could be different from the universe distribution.</p>
<p>in other words, if Janes draws a lot balls and all are white, e.g. 50 balls are all white, I fancy it's reasonable to suspect that bowl B does not have a 60%-40% black-white distribution.</p>
| Yola | 31,462 | <p>You have $7^{17}, (2^3)^{17}, (3^2)^{17}$</p>
|
94,525 | <p>I am trying to solve the equation
$$z^n = 1.$$</p>
<p>Taking $\log$ on both sides I get $n\log(z) = \log(1) = 0$.</p>
<p>$\implies$ $n = 0$ or $\log(z) = 0$</p>
<p>$\implies$ $n = 0$ or $z = 1$.</p>
<p>But I clearly missed out $(-1)^{\text{even numbers}}$ which is equal to $1$.</p>
<p>How do I solve this equation algebraically?</p>
| Gigili | 181,853 | <p>To solve it algebraically, I'd say:</p>
<ul>
<li>For even values of $n$, $z=1$ or $z=-1$ are the solutions.</li>
<li>For odd values of $n$, $z=1$ is the answer.</li>
<li>And if $n=0$, $\forall z \in \mathbb{R}$ would be valid as an answer.</li>
</ul>
|
1,114 | <p>Or more specifically, why do people get so excited about them? And what's your favorite easy example of one, which illustrates why I should care (and is not a group)?</p>
| Andrew Stacey | 45 | <p>Any vector bundle is a groupoid: you can add and subtract vectors only if they are in the same fibre. Similarly, if you take a vector bundle E → M (or some other fibre bundle) then consider the automorphism bundle Aut(E) → M where a point in Aut(E) above p ∈ M is an automorphism of E<sub>p</sub>. This is a groupoid since these automorphisms can only be composed if they lie in the same fibre.</p>
<p>These are <strong>discrete</strong> groupoids in the sense that there are no morphisms between distinct objects (aka points in the base space). However, they are not discrete topologically as they clearly have topologies! (Which, incidentally, shows that you should be careful when using the statement about groupoids having no homotopy above degree 2: this is a statement about groupoids in Set but groupoids exist enriched in other categories where they can have lots of interesting homotopy). To get a more general groupoid, you can consider the bundle Iso(E) → M × M where a point in Iso(E) above (p,q) ∈ M × M is an isomorphism from E<sub>p</sub> to E<sub>q</sub>.</p>
<p>One reason for liking groupoids is that they allow you to talk about quotients by group actions without actually having to take the quotient. That's useful because some categories don't have quotients - such as the category of smooth manifolds. So when you have a group G acting on a manifold M you can try to take the quotient M/G but that is quite often not a manifold. So instead you can take the groupoid with objects M and morphisms G × M, where a morphism (g,m) has source m and target gm. Even when the quotient exists, or when you've extended the category to include quotients, this can be much better behaved than the corresponding quotient.</p>
|
1,114 | <p>Or more specifically, why do people get so excited about them? And what's your favorite easy example of one, which illustrates why I should care (and is not a group)?</p>
| javier | 914 | <p>Beyond all the categorical and bundle-like examples already given, you can easily understand groupoids as generalizations of groups in a purely geometrical sense.</p>
<p>If you think of groups as the sets of <em>symmetries</em> of certain geometrical objects, then groupoids are <strong>local</strong> symmetries of geometrical objects. My favorite example of this consists on taking a manifold M and defining a groupoid G as the set of all the local diffeomorphisms f:U-->V where U and V are open sets of M, with multiplication given by composition of maps (whenever it makes sense). </p>
|
1,114 | <p>Or more specifically, why do people get so excited about them? And what's your favorite easy example of one, which illustrates why I should care (and is not a group)?</p>
| Sean Rostami | 2,047 | <p>Lots of things are groupoids, but many are not groups. There is a theory of groupoids, and if you don't acknowledge groupoids, they won't let you use their theory =)</p>
<p>The fundamental group(oid) example is really good. What happens if you want to do Van Kampen's theorem on a pair of sets whose intersection is <em>not</em> connected? There's an answer but you have to use fundamental groupoids. </p>
<p>Also, categories are sometimes useful for isolating data. For example, you can replace the usual notion of a local system with a "representation of the fundamental groupoid" and doing so lets you think of a local system the way you already wanted to in your heart: as a collection of operators on fibers coming from "going around the bad points".</p>
<p>Here's a stupid riddle for everyone: what's another word for a "monoidoid"?</p>
|
1,114 | <p>Or more specifically, why do people get so excited about them? And what's your favorite easy example of one, which illustrates why I should care (and is not a group)?</p>
| Maik Köster | 41,401 | <p>One reason why people get existed about groupoids which has not been mentioned so far: The category of reduced smooth orbifolds is isomorphic to a category of proper effective etale Lie groupoids. For some investigations on orbifolds, working with their local charts can become very clumpsy. Sometimes, the corresponding question on groupoids has an elegant solution.</p>
|
3,845,968 | <p><a href="https://math.stackexchange.com/q/29666/717872">There</a>
<a href="https://math.stackexchange.com/q/11/717872">are</a>
<a href="https://math.stackexchange.com/q/363977/717872">tens</a>
<a href="https://math.stackexchange.com/q/3339682/717872">of</a>
<a href="https://math.stackexchange.com/q/2005492/717872">posts</a>
<a href="https://math.stackexchange.com/q/281492/717872">already</a>
<a href="https://math.stackexchange.com/q/1291388/717872">on</a>
<a href="https://math.stackexchange.com/q/1600940/717872">this</a>
site about whether <span class="math-container">$0.\overline{9} = 1$</span>.</p>
<p>This is something that intrigues me, and I have a question about this, including a "proof" which I have found myself.</p>
<h3>Question:</h3>
<p><a href="https://math.stackexchange.com/questions/281492/about-0-999-1#comment614377_281492">This comment</a> says that</p>
<blockquote>
<p>you need a <em>terminating</em> decimal to get something less than 1.</p>
</blockquote>
<p>If so, does it mean that a non-terminating decimal (e.g. <span class="math-container">$0.\overline{9}$</span>) is <span class="math-container">$\ge 1$</span>?</p>
<p>So is <span class="math-container">$\frac{1}{3}$</span> (<span class="math-container">$0.\overline{3}$</span>) also <span class="math-container">$\ge 1$</span>? It is non-terminating, but you can subtract <span class="math-container">$\frac{1}{3}$</span> from <span class="math-container">$1$</span> to get <span class="math-container">$\frac{2}{3} = 0.\overline{6}$</span>, which is another non-terminating decimal. How do those mechanics work?</p>
<hr />
<p><strong>Theorem:</strong> <span class="math-container">$0.99999... = 1(.00000... = 1)$</span></p>
<p><strong>Proof:</strong></p>
<p><span class="math-container">\begin{align}
\frac{1}{9} &= 0.11111... \\
\frac{2}{9} &= 0.22222... \\
\frac{3}{9} &= 0.33333... \\
\frac{4}{9} &= 0.44444... \\
\frac{5}{9} &= 0.55555... \\
\frac{6}{9} &= 0.66666... \\
\frac{7}{9} &= 0.77777... \\
\frac{8}{9} &= 0.88888... \\
\therefore \frac{9}{9} &= 0.99999... \\
&= 1
\end{align}</span></p>
<p>Is the above proof correct? I came up with it myself before I decided to ask this question, but I do not know if it is mathematically valid.</p>
| Ennar | 122,131 | <p>What you wrote is simply
<span class="math-container">$$\frac 19 = 0.\bar 1 \implies 1 = \frac 99 = 0.\bar 9$$</span>
which is correct.</p>
<p>However, you didn't write anything to prove that <span class="math-container">$\frac 19 = 0.\bar 1$</span>, so I wouldn't count this as a proof.</p>
<p>Any proof really needs to use that decimal number system is using geometric series to represent numbers that they converge to. You can use formula for the sum of geometric series, like Yves Daoust did, or you can use the same technique that's used in the proof of the formula directly:</p>
<p><span class="math-container">$$x = 0.9999\ldots \implies 10 x = 9.9999\ldots = 9 + x \implies x = 1.$$</span></p>
<p>The above is just a notational shortcut for the following:
<span class="math-container">\begin{align}
10\cdot\lim_{n\to\infty} \left(\frac 9{10} + \frac 9{10^2} + \ldots + \frac 9{10^n}\right)
&= \lim_{n\to\infty} 10\left(\frac 9{10} + \frac 9{10^2} + \ldots + \frac 9{10^n}\right)\\
&= 9 + \lim_{n\to\infty} \left(\frac 9{10} + \frac 9{10^2} + \ldots + \frac 9{10^{n-1}}\right).
\end{align}</span></p>
|
2,204,812 | <p>The solution of the differential equation $\frac{dy}{dx}-xtan(y-x)=1$ will be?</p>
<p>For solving such problems first we should see if the equation is in variable seperable form or not. Obviously here it is not. So I tried to see if it can be made to variable seperable by substitution, but substituting $y-x=z$ would not give the answer as there is one $x$ remaining outside that $tan(y-x)$. Also it is not a homogenous nor is getting converted into homogenous form so that I could substitute $y=vx$ or $x=vy$. So which method should I use here? I am getting wrong answer everytime, please help in dealing with this. Atleast provide me a hint.</p>
| projectilemotion | 323,432 | <p>Your decision to substitute $z=y-x$ is a good one.</p>
<p>Differentiating with respect to $x$ gives:
$$\frac{dz}{dx}=1\cdot \frac{dy}{dx}-1$$
Therefore, rearranging gives:
$$\frac{dy}{dx}=\frac{dz}{dx}+1$$
Substituting gives a separable ODE:
$$\frac{dz}{dx}+1-x\tan{z}=1$$
$$\frac{dz}{dx}=x\tan{z}$$
Can you continue?</p>
|
3,932,803 | <p>I need to prove <span class="math-container">$$ \lim_{x\rightarrow\ 0}\frac{x^2-8}{{x-8}} =1 $$</span> using epsilon-delta definition.
I know I need to show that for every <span class="math-container">$\epsilon >0$</span> there exist a <span class="math-container">$\delta >0$</span> such that if <span class="math-container">$|x|<\delta$</span>, then <span class="math-container">$| {\frac{x^2-8}{x-8}}-1|<\epsilon$</span></p>
<p>WHat I did:</p>
<p><span class="math-container">$|{\frac{x^2-8}{x-8}}-1|=|{\frac{x^2-x}{x-8}}|$</span>, but Im having troubles now deciding what my delta should be.</p>
<p>I would really appreciate if someone could give me an explanation on how should I choose delta and why.
Thanks.</p>
| 5xum | 112,884 | <p><strong>Hint</strong>:</p>
<p><span class="math-container">$$\left|\frac{x^2-x}{x-8}\right| = |x|\cdot \left|\frac{x-1}{x-8}\right|.$$</span></p>
<p>Now, if <span class="math-container">$\delta<1$</span>, then you can further estimate that:</p>
<ol>
<li><span class="math-container">$|x-1| < 3$</span></li>
<li><span class="math-container">$|x-8| > 6$</span>.</li>
</ol>
|
3,932,803 | <p>I need to prove <span class="math-container">$$ \lim_{x\rightarrow\ 0}\frac{x^2-8}{{x-8}} =1 $$</span> using epsilon-delta definition.
I know I need to show that for every <span class="math-container">$\epsilon >0$</span> there exist a <span class="math-container">$\delta >0$</span> such that if <span class="math-container">$|x|<\delta$</span>, then <span class="math-container">$| {\frac{x^2-8}{x-8}}-1|<\epsilon$</span></p>
<p>WHat I did:</p>
<p><span class="math-container">$|{\frac{x^2-8}{x-8}}-1|=|{\frac{x^2-x}{x-8}}|$</span>, but Im having troubles now deciding what my delta should be.</p>
<p>I would really appreciate if someone could give me an explanation on how should I choose delta and why.
Thanks.</p>
| Neat Math | 843,178 | <p>There are many ways to get an upper bound for <span class="math-container">$\left| \frac{x-1}{x-8} \right|$</span> when <span class="math-container">$x$</span> is sufficiently small. A couple of things you can do to make this easier: 1) Use the triangle inequality, as one of the answerers did; 2) separate a constant so that you only need to deal with one term involving <span class="math-container">$x$</span>. So if <span class="math-container">$|x| < 1$</span> then</p>
<p><span class="math-container">$$\left| \frac{x-1}{x-8} \right|=\left| 1+\frac{7}{x-8} \right| \le 1+\frac{7}{8-x} \le 1+\frac{7}{8-1}=2.$$</span></p>
<p>Then you have <span class="math-container">$\delta = \min \left(1, \frac{\varepsilon}{2}\right).$</span></p>
|
2,002,385 | <blockquote>
<p>Prove $|e^{i\theta_1}-e^{i\theta_2}|\geq|e^{i\theta_1/2}-e^{i\theta_2/2}|$
where $ \theta_1, \theta_2 \in (0,\pi]$.</p>
</blockquote>
<p>Even though geometrically it is an obvious fact, somehow I couldn't prove it elegant way (it's really frustrating), and I'm sure some of you guys know how to prove it in two or three lines. </p>
<p>That's pretty much it.</p>
| Jan Eerland | 226,665 | <p>When $\theta_1\space\wedge\space\theta_2\in\mathbb{R}$:</p>
<ol>
<li>$$e^{\theta_1i}-e^{\theta_2i}=\cos\left(\theta_1\right)+\sin\left(\theta_1\right)i-\left(\cos\left(\theta_2\right)+\sin\left(\theta_2\right)i\right)=$$
$$\cos\left(\theta_1\right)-\cos\left(\theta_2\right)+\left(\sin\left(\theta_1\right)-\sin\left(\theta_2\right)\right)i$$</li>
<li>$$e^{\frac{\theta_1i}{2}}-e^{\frac{\theta_2i}{2}}=\cos\left(\frac{\theta_1}{2}\right)+\sin\left(\frac{\theta_1}{2}\right)i-\left(\cos\left(\frac{\theta_2}{2}\right)+\sin\left(\frac{\theta_2}{2}\right)i\right)=$$
$$\cos\left(\frac{\theta_1}{2}\right)-\cos\left(\frac{\theta_2}{2}\right)+\left(\sin\left(\frac{\theta_1}{2}\right)-\sin\left(\frac{\theta_2}{2}\right)\right)i$$</li>
</ol>
<p>So:</p>
<ol>
<li>$$\left|e^{\theta_1i}-e^{\theta_2i}\right|=\sqrt{\left(\cos\left(\theta_1\right)-\cos\left(\theta_2\right)\right)^2+\left(\sin\left(\theta_1\right)-\sin\left(\theta_2\right)\right)^2}=\sqrt{2\left(1-\cos\left(\theta_1-\theta_2\right)\right)}$$</li>
<li>$$\left|e^{\frac{\theta_1i}{2}}-e^{\frac{\theta_2i}{2}}\right|=\sqrt{\left(\cos\left(\frac{\theta_1}{2}\right)-\cos\left(\frac{\theta_2}{2}\right)\right)^2+\left(\sin\left(\frac{\theta_1}{2}\right)-\sin\left(\frac{\theta_2}{2}\right)\right)^2}=$$
$$2\sqrt{\sin^2\left(\frac{\theta_1-\theta_2}{4}\right)}$$</li>
</ol>
|
780,611 | <p>The differential equation in question is a FODE,</p>
<p>$$
\frac{dy}{dt} = -\frac{a^2\sqrt{2g}}{\sqrt{(R+y)(R-y)}}
$$</p>
<p>Upon first inspection, this is separable, but I don't know how to proceed from there.</p>
<p>Thanks.</p>
| Chinny84 | 92,628 | <p>$$
\frac{dy}{dt} = -\frac{a^2\sqrt{2g}}{\sqrt{(R+y)(R-y)}}
$$
is indeed seperable
$$
\int\sqrt{R^{2}-y^{2}}dy = -a^{2}\sqrt{2g}t + \lambda
$$
using $y=R\cos\theta$ leads to
$$
\int R\sin\theta d\left(R\cos\theta\right) = -\int R^2\sin^{2}\theta d\theta = -a^{2}\sqrt{2g}t + \lambda
$$
Do you need this to be completed?</p>
<p>$\textbf{Edit:}$
The question was actually meant to be
$$
\frac{dy}{dt} = -\frac{a^2\sqrt{2g}}{\sqrt{(R+y)}(R-y)}
$$
This is not too bad (it is actually easier), and is instead separated as
$$
\int\sqrt{R+y}(R-y)dy = -a^{2}\sqrt{2g}t + \lambda
$$
with the substitution $v=R+y$
$$
\int v^{1/2}\left(2R-v\right)dv = -a^{2}\sqrt{2g}t + \lambda
$$</p>
<p>This is all assuming $R=$constant.</p>
|
2,930,292 | <p>I'm currently learning the unit circle definition of trigonometry. I have seen a graphical representation of all the trig functions at <a href="https://www.khanacademy.org/math/trigonometry/unit-circle-trig-func/unit-circle-definition-of-trig-functions/a/trig-unit-circle-review" rel="nofollow noreferrer">khan academy</a>. </p>
<p><img src="https://i.stack.imgur.com/zp5WB.png" alt="enter image description here"></p>
<p>I understand how to calculate all the trig functions and what they represent. Graphically, I only understand why sin and cos is drawn the way it is. I'm having trouble understand why tangent, cotangent, secant, cosecant are drawn the way they are. </p>
<p>Can someone please provide me with some intuitions.</p>
| Mohammad Riazi-Kermani | 514,496 | <p>It would be very helpful if you connected the center of the unit circle to the point of tangency.</p>
<p>Notice that this segment has length one and is perpendicular to the tangent line.</p>
<p>Now there are many similar right triangles to be found and the proprtionality relations gives you the trig functions as indicated in the graph.</p>
<p>It is very time consuming but the result is interesting.</p>
|
265,949 | <p>Consider the set of all $n \times n$ matrices with real entries as the space $ \mathbb{R^{n^2}}$ .
Which of the following sets are compact?</p>
<ol>
<li>The set of all orthogonal matrices.</li>
<li>The set of all matrices with determinant equal to unity.</li>
<li><p>The set of all invertible matrices.</p>
<p>I am stuck on this problem. Can anyone help me please...............</p>
<p>how can we check closedness and boundedness of matrices???</p></li>
</ol>
| froggie | 23,685 | <p>To check that a set of $n\times n$ matrices $\subset \mathbb{R}^{n^2}$ is compact, it is probably easiest to check that it is closed and bounded. </p>
<p>I'm going to give an example that is not any of the three you have in the problem, but will hopefully help you with them. Let $S$ be the collection of $n\times n$ symmetric matrices. We can then ask: is $S$ closed? Is $S$ bounded? Is $S$ compact?</p>
<p>I claim that $S$ is in fact closed. To see this, suppose that $M_n$ is a sequence of elements in $S$, i.e., a sequence of symmetric matrices, and suppose $M_n$ converges to a matrix $M$. To prove that $S$ is closed, we must show that $M$ is also a symmetric matrix. Write $M_n = (a_{ij}^{(n)})$, and $M = (a_{ij})$. The fact that $M_n\to M$ says exactly that $a_{ij}^{(n)}\to a_{ij}$ as $n\to \infty$ for each $i$ and $j$. But $M_n$ is symmetric, so $a_{ij}^{(n)} = a_{ji}^{(n)}$, so we get $$a_{ij} = \lim_n a_{ij}^{(n)} = \lim_n a_{ji}^{(n)} = a_{ji}.$$ This proves $M$ is symmetric, and that $S$ is closed.</p>
<p>I claim next that $S$ is not bounded. $S$ is bounded if and only if the entries of all the matrices in $S$ are bounded uniformly, so to prove my claim I need only find elements of $S$ with arbitrarily large entries. For each $c\in \mathbb{R}$, the matrix $M$ all of whose entries are $c$ lies in $S$, proving that the matrices in $S$ do not have uniformly bounded entries. This proves $S$ is not bounded.</p>
<p>Now is $S$ compact? $S$ is compact if and only if it is closed and bounded. Since it is not bounded, it is not compact. I hope this helps.</p>
|
514,922 | <p>I need to prove the following affirmation: If $ \lim x_{2n} = a $ and $ \lim x_{2n-1} = a $, prove that $\lim x_n = a $ (in $ \mathbb{R} $ )</p>
<p>It is a simple proof but I am having problems how to write it. I'm not sure it is the right way to write, for example, that the limit of $(x_{2n})$ converges to a:</p>
<p>$ \forall \epsilon > 0 \; \exists n_1 \in \mathbb{N} $ such that if $n \in \mathbb{N}$ and $n \geq n_1$, then (*) $ | x_{2n} - a| < \epsilon $</p>
<p>About the (*) step, is that correct to write $ x_{2n} $? Or should I use another notation?</p>
<p>Thanks for the help! </p>
| Hagen von Eitzen | 39,174 | <p>You can do so. As a matter of fact,you consider the sequence $(y_n)$ given by $y_n=x_{2n}$ and could write - as you may be more accustomed to - that $|y_n-a|<\epsilon$ for all $n>n_1$. But as $y_n=x_{2n}$ you really get back what you (correctly) wrote: $|x_{2n}-a|<\epsilon$.</p>
|
2,918,091 | <p>Suppose I want to find the locus of the point $z$ satisfying $|z+1| = |z-1|$</p>
<p>Let $z = x+iy$</p>
<p>$\Rightarrow \sqrt{(x+1)^2 + y^2} = \sqrt{(x-1)^2 + y^2}$ <br/>
$\Rightarrow (x+1)^2 = (x-1)^2$ <br/>
$\Rightarrow x+1 = x-1$ <br/>
$\Rightarrow 1= -1$ <br/>
$\Rightarrow$ Loucus does not exist</p>
<p>Is my approach incorrect? The answer I was given was that the y-axis describes the locus.</p>
<p>Any help would be appreciated.</p>
| Fabio Lucchini | 54,738 | <p>When you removed squares, you have also to consider $x+1=1-x $ which gives $x=0$.</p>
<p>Alternatively, this is the locus of the points equidistant from $\pm 1$, that's the axis of the segment $[-1,1] $.</p>
|
140,134 | <p>Does anyone know whether there is any geometric applications of the Iwaniec's conjecture on $ l^p $ bound of Beurling Alfhors transform (or the complex Hilbert transform). One application could have been was Behrling's conjecture (solved). Is there is any thing else that might be of geometric significance.
For p=2, it is known that $ l^p $ bound is 1 and it coincides with the conjecture.It is already known to be $ l^p $ bounded by Calderon, Zygmund and even more. Is the exact bound is of any geometric significance.</p>
| Fabrice Baudoin | 48,356 | <p>Iwaniec conjecture is closely connected to the important Morrey's conjecture on the relationship between rank one convexity and quasiconvexity.</p>
<p>You will find a nice discussion in the Section 5 of the following survey by Banuelos</p>
<p><a href="http://arxiv.org/pdf/1012.4850v2.pdf" rel="nofollow">http://arxiv.org/pdf/1012.4850v2.pdf</a></p>
|
2,125,206 | <ol>
<li><p>Let $W$ be the region bounded by the planes
$x = 0$, $y = 0$, $z = 0$, $x + y = 1$, and $z = x + y$.</p></li>
<li><p>$(x^2 + y^2 + z^2)\, \mathrm dx\, \mathrm dy\, \mathrm dz$; $W$ is the region bounded by $x + y + z = a$ (where $a > 0$), $x = 0$, $y = 0$, and $z = 0$.</p></li>
</ol>
<p>$x,y,z$ being $0$ is throwing me off because I'm not sure how to graph it and get a bounded area; it seems like it would be infinite to me. What would the bounds for the triple integrals be?</p>
| Doug M | 317,162 | <p>1) give you nothing to integrate. But we can still find the limits</p>
<p>$z = 0$ lower limit for $z$</p>
<p>$z = x+y$ upper limit for $z$</p>
<p>$y = 0$ lower limit for $y$</p>
<p>$y = 1-x$ upper limit for $y$</p>
<p>$x = 0$ lower limit for $x$</p>
<p>$x = 1$ upper limit for $x$</p>
<p>For the last one, because that is where $y = 0$ intersects $x+y = 1$</p>
<p>2)</p>
<p>$\int_0^a\int_0^{a-x}\int_0^{a-x-y} x^2 + y^2 + z^2 \;dz\;dy\;dx$</p>
|
447,484 | <p>I am just learning about differential forms, and I had a question about employing Green's theorem to calculate area. Generalized Stokes' theorem says that $\int_{\partial D}\omega=\int_D d\omega$. Let's say $D$ is a region in $\mathbb{R}^2$. The familiar formula to calculate area is $\iint_D 1 dxdy = \frac{1}{2}\int_{\partial D}x\,dy - y\,dx$, and indeed, $d(x\,dy - y\,dx)=2\,dx\,dy$. But why aren't we allowed to simply use $\int_{\partial D}x\,dy$? Doesn't $d(x\,dy)=d(x)dy = (1\,dx + 0\,dy)dy = dx\,dy$?</p>
| Steven Gubkin | 34,287 | <p>I would like to point out the integrating $xdy$ to get area has a natural geometric interpretation: you are summing the areas of small horizontal rectangles. The sign of these areas is determined by whether you are moving up or down, and the sign of x. Draw a picture of a wild blob, intersect it with a horizontal line, and see how at each intersection point you will get a rectangle. The areas of these rectangles will cancel out when you are not inside the blob.</p>
<p>Here is my blob:</p>
<p><img src="https://i.stack.imgur.com/ho4TG.png" alt="enter image description here"></p>
<p>if I am integrating $xdy$ around this closed curve in the indicated direction, then I am going to move incrementally along the curve a little step at a time, and keep a running sum of $x$ times the small vertical distance I just traveled. Moving up is a positive distance, moving down is a negative distance. That is just the signed area of a little rectangle whose length is my current position to the $y$ -axis, and whose height is the small height I just traversed.</p>
<p>I will denote positive areas as green, and negative areas as red.</p>
<p><img src="https://i.stack.imgur.com/txHJG.png" alt="enter image description here"></p>
<p>This point is positive since I am moving up, and $x$ is positive. Some time later I come back to the same height, but at a different $x$ value.</p>
<p><img src="https://i.stack.imgur.com/I1sLa.png" alt="enter image description here"></p>
<p>This time the area is negative because I am moving down, and $x$ is positive.</p>
<p>Later on I am again at the same $y$ value:</p>
<p><img src="https://i.stack.imgur.com/k3lqS.png" alt="enter image description here"></p>
<p>moving up, x pos</p>
<p><img src="https://i.stack.imgur.com/bWrlR.png" alt="enter image description here"></p>
<p>moving down, x negative!</p>
<p>Putting these all together, I have something like this:</p>
<p><img src="https://i.stack.imgur.com/4awmD.png" alt="enter image description here"></p>
<p>Now the red and part of the first green rectangle cancel, and you can see that I only have the area of the horizontal strip which is inside the blob: the orientations have automatically recorded the difference between inside and outside!</p>
<p><img src="https://i.stack.imgur.com/ioQSu.png" alt="enter image description here"></p>
<p>Hopefully this application of Green's theorem is not so mysterious any more.</p>
|
108,372 | <p>Given a map $\psi: S\rightarrow S,$ for $S$ a closed surface, is there any algorithm to compute its translation distance in the curve complex? I should say that I mostly care about checking that the translation distance is/is not very small. That is, if the algorithm can pick among the possibilities: translation distance is 0, 1, 2, 3, many, then I am happy...</p>
<p>I know there are algorithms for computing distances IN the curve complex, but this is not quite the same...</p>
| Ian Agol | 1,345 | <p>I don't know an algorithm, but here's a possible approach. As Richard and Lee have observed, one may assume that <span class="math-container">$\psi$</span> is pseudo-Anosov. In that case, the mapping torus <span class="math-container">$T_\psi$</span> is a hyperbolic 3-manifold fibering over <span class="math-container">$S^1$</span>, with fiber <span class="math-container">$S$</span>. There is a short exact sequence <span class="math-container">$\pi_1(S)\to\pi_1(T_\psi)\to \mathbb{Z}$</span>.</p>
<p>Here's a characterization of the translation length on the curve complex in terms of the topology of <span class="math-container">$T_\psi$</span>. The fiber <span class="math-container">$S$</span> represents a homology class <span class="math-container">$[S]\in H_2(T_\psi)$</span>.
Let <span class="math-container">$\Sigma \looparrowright T_\psi$</span> be an immersed connected surface, such that <span class="math-container">$[\Sigma]=k[S]$</span> and such that <span class="math-container">$\chi(\Sigma)=k\chi(S)$</span>. Moreover, assume that the composite map <span class="math-container">$\pi_1(\Sigma)\to \pi_1(T_\psi) \to \mathbb{Z}$</span> is non-trivial, so that <span class="math-container">$\Sigma$</span> is not homotopic to a finite-sheeted cover of <span class="math-container">$S$</span>. Let <span class="math-container">$K(\psi)$</span> be the minimal such <span class="math-container">$k$</span>, and let <span class="math-container">$D(\psi)$</span> be the minimal such <span class="math-container">$k$</span> so that the surface has only double curves of intersection. Clearly <span class="math-container">$K(\psi)\leq D(\psi)$</span>.</p>
<p><strong>Claim:</strong> The curve complex translation distance of <span class="math-container">$\psi$</span> is <span class="math-container">$=D(\psi)$</span>.</p>
<p>One direction: Let <span class="math-container">$k$</span> be the translation length of <span class="math-container">$\psi$</span>. There exists a sequence of non-separating curves <span class="math-container">$c_1,c_2,\ldots, c_k \subset S$</span>, such that <span class="math-container">$\psi(c_1)=c_k$</span>, and <span class="math-container">$c_i \cap c_{i+1}=\emptyset$</span>. One creates a surface <span class="math-container">$\Sigma\subset T_\psi$</span> by taking <span class="math-container">$k$</span> copies of <span class="math-container">$S$</span>, <span class="math-container">$S_1 \sqcup \cdots \sqcup S_k \subset T_\psi$</span> in circular order. Cut out annular neighborhoods of <span class="math-container">$c_i, c_{i+1}$</span> inside <span class="math-container">$S_i$</span>, and insert cross annuli between <span class="math-container">$S_{i-1}$</span> and <span class="math-container">$S_i$</span> (taking indices <span class="math-container">$\mod k$</span>) between the 4 copies of <span class="math-container">$c_i$</span>.
This construction generalizes a construction of <a href="https://doi.org/10.1007/BF01231534" rel="nofollow noreferrer">Cooper-Long-Reid</a>. One can see that the resulting surface has the properties above.</p>
<p>Conversely, if one has such an immersed surface with only double curves, one may cut and paste the self-intersection curves to get <span class="math-container">$k$</span> parallel copies of <span class="math-container">$S$</span>. The cross cut curves gives a sequence of closed curves in <span class="math-container">$S$</span>, which one can prove using the homology condition forms a closed loop in the curve complex <span class="math-container">$\mod \psi$</span>.</p>
<p>I don't know yet how to make this criterion into an algorithm. I think there is an algorithm to compute <span class="math-container">$K(\psi)$</span>. For a given genus <span class="math-container">$g$</span>, <a href="http://www.sciencedirect.com/science/article/pii/0040938394000557" rel="nofollow noreferrer">Canary proved</a> that there are only finitely many homotopy classes of immersed surfaces of genus <span class="math-container">$g$</span>. I think this proof can be made effective, and should give one a method to compute <span class="math-container">$K(\psi)$</span>. This would at least give an algorithmic lower bound, since <span class="math-container">$K(\psi)\leq D(\psi)$</span>. Also, there is a constant <span class="math-container">$0< c_S <1$</span>
such that <span class="math-container">$D(\psi)\leq c_S K(\psi)$</span> (this may be proved using hyperbolic geometry techniques).</p>
<p>One could try to algorithmically to construct all surfaces realizing <span class="math-container">$K(\psi)$</span>, and then try to homotope them to have only double curves of intersection, e.g. using normal surfaces. However, there is a result of <a href="http://www.ams.org/mathscinet-getitem?mr=899054" rel="nofollow noreferrer">Gulliver-Scott</a> that an immersed surface with only double curves of intersection might have a minimal area representative which has triple points. So I don't know yet how to make an algorithm by computing <span class="math-container">$D(\psi)$</span> using this approach.</p>
|
108,372 | <p>Given a map $\psi: S\rightarrow S,$ for $S$ a closed surface, is there any algorithm to compute its translation distance in the curve complex? I should say that I mostly care about checking that the translation distance is/is not very small. That is, if the algorithm can pick among the possibilities: translation distance is 0, 1, 2, 3, many, then I am happy...</p>
<p>I know there are algorithms for computing distances IN the curve complex, but this is not quite the same...</p>
| Mark Bell | 3,121 | <p>Recently, <a href="https://arxiv.org/abs/1609.09392" rel="nofollow noreferrer">Richard Webb and myself gave a polynomial-time algorithm</a> for computing the asymptotic translation length of a mapping class
$$ \ell(h) = \lim_{n \to \infty} d(x, h^n(x)). $$
This appears as Algorithm 6 of the paper and relies on being able to compute geodesics in the curve complex. Assuming this, the key observation is that a midpoint $c'$ of a curve $c$ and its image under a large enough power of $h$ lies close to the tight axis of $h$. Hence $d(c', h^n(c')) \approx n \ell(h)$. Therefore, since $\ell(h)$ is a rational number with bounded denominator, if we take $n$ large enough then
$$ \ell(h) = \frac{1}{n} d(c', h^n(c')). $$</p>
<p>If you fix a finite generating set $\langle X \rangle = \textrm{Mod}(S)$ then the running time of this algorithm is a polynomial function of the word length of the mapping class $|h|_X$ since this is the running time of our algorithm for computing a geodesic from $c$ to $h(c)$ (Algorithm 4).</p>
|
141,823 | <p>I am thinking about the simplest version of Hensel's lemma. Fix a prime $p$. Let $f(x)\in \mathbf{Z}[x]$ be a polynomial. Assume there exists $a_0\in \mathbf{F}_p$ such that $f(a_0)=0\mod p$, and $f'(a_0)\neq 0\mod p$. Then there exists a unique lift $a_n\in \mathbf{Z}/p^{n+1}\mathbf{Z}$ for every $n$. I know there is an elementary proof. However, I want to prove it by using standard deformation theory. It is simply a problem about extending the section $a_0$ order by order. Let $X$ be the scheme defined by $f(x)$ over $k=\mathbf{F}_p$. $I$ is the $k$-module $p\mathbf{Z}/p^2$. If I am correct, the obstruction class is in $Ext^1(a_0^*L_{X/k},I)$, and if it vanishes, the extension is classified by $Ext^0(a_0^*L_{X/k},I)$. Is there a proof of Hensel's lemma along this line?</p>
<p>I know this is like using a big machine to solve a simple problem. However, I really want to understand why $f'(a_0)\neq 0 \mod p$ implies that the obstruction class vanishes. </p>
| Daniel Litt | 6,950 | <p>There is indeed a proof along these lines. Suppose one has a polynomial $f$ over $\mathbb{Z}_p$; I'll use $f$ to refer to its reductions mod $p^k$ as well. You have a diagram:</p>
<p>$$\text{Spec}((\mathbb{Z}/p^k)[t]/f(t))\to~\text{Spec}((\mathbb{Z}/p^{k+1})[t]/f(t))~~~~~~~~~~~~~$$
$$\downarrow\uparrow~~~~~~~~~~~~~~~~~~~~~~\downarrow$$
$$\text{Spec}(\mathbb{Z}/p^k)\to \text{Spec}(\mathbb{Z}/p^{k+1})$$</p>
<p>The vertical arrow on the left is a section $s$, e.g. a choice of root of $f$ in $\mathbb{Z}/p^k$. It is this section $s$ that we would like to show lifts uniquely. To do so, it suffices to show that the section $s: \text{Spec}(\mathbb{Z}/p^k)\to \text{Spec}((\mathbb{Z}/p^k)[t]/f(t))$ is etale, because this implies the associated $L_s=\Omega^1_s=0$; as you observe, $\Omega^1_s$ controls deformations and thus $\Omega^1_s=0$ implies deformations are unobstructed and unique.</p>
<p>In fact, we may see that $s$ is the inclusion of a connected component $X_k$ of $\text{Spec}((\mathbb{Z}/p^k)[t]/f(t))$ (which implies $\Omega^1_s=0$); to do so, it suffices to show that the map $\pi: X_k\to \mathbb{Z}/p^k$ is an isomorphism, as $s$ is a left inverse to this map. For $k=1$ this follows from the Chinese remainder theorem (using that $f'(a_0)\not=0$ implies $a_0$ is not a double root); in general, for $k>1$, this follows from $k=1$ case plus Nakayama.</p>
<p>Of course, this argument also shows that $\pi: X_{k+1}\to \text{Spec}(\mathbb{Z}/p^{k+1})$ is an isomorphism, whence its inverse is the required section. So deformation theory is strictly optional.</p>
|
1,168,968 | <p>So I'm doing some cryptography assignment and I'm dealing with a modular arithmetic in hexadecimal. Basically I have the values for $n$ and the remainder $x$, but I need to find the original number $m$, e.g.</p>
<p>$$m \mod 0x6e678181e5be3ef34ca7 = 0x3a22341b02ad1d53117b.$$</p>
<p>I just need a formula to calculate $m$.</p>
<p>Edit: ok, let's put it this way, $x = K^e \mod n$, I know the values for $x$, $e$ and $n$. Does that help?</p>
<p>Ok, I realized I was approaching the problem in a wrong way, basically I had the RSA public key and I should have used RSA problem to decrypt the file without having the private key. Sorry for the stupid question.</p>
| kryomaxim | 212,743 | <p>Derivative of the numerator: $(5+ \sqrt{x^2+5})'= (\sqrt{x^2+5})'= \frac{(x^2+5)'}{2}\sqrt{x^2+5}^{-1} = \frac{2x}{2 \sqrt{x^2+5}} = \frac{1}{\sqrt{1+\frac{5}{x^2}}}$.</p>
<p>Derivative of the denominator: $(x-6)'=1$. </p>
<p>L'Hospital's rule is possible.</p>
|
2,377,946 | <blockquote>
<p>The integral is:
$$\int_0^a \frac{x^4}{(x^2+a^2)^4}dx$$</p>
</blockquote>
<p>I used an approach that involved substitution of x by $a\tan\theta$. No luck :\ . Help?</p>
| Robert Z | 299,698 | <p>Let $x=at$ and by <a href="https://en.wikipedia.org/wiki/Partial_fraction_decomposition" rel="nofollow noreferrer">Partial-Fraction Decomposition</a> we get
\begin{align*}\int_0^a \frac{x^4}{(x^2+a^2)^4}dx&=\frac{1}{a^3}\int_0^1 \frac{t^4}{(t^2+1)^4}dt\\
&=\frac{1}{a^3}\int_0^1\left(\frac{1}{(t^2+1)^2}-\frac{2}{(t^2+1)^3}+\frac{1}{(t^2+1)^4}\right)\,dt
\\&=\frac{1}{16a^3}\left[\arctan(t)
+\frac{t}{t^2+1}-\frac{(14/3)t}{(t^2+1)^2}+\frac{(8/3)t}{(t^2+1)^3}\right]_0^1\\&=\frac{3\pi-4}{192a^3}.
\end{align*}
P.S. Note that for $p\geq 2$,
$$(2p-2)\int\frac{dt}{(t^2+1)^p}=\frac{t}{(t^2+1)^{p-1}}
+(2p-3)\int\frac{dt}{(t^2+1)^{p-1}}.$$</p>
|
1,956,897 | <p>I have the question "Find the centre and radius of the circle whose equation is $x^2+y^2+x+3y-2=0$"</p>
<p>So I've worked out the centre to be $(-1/2, -3/2)$ which when I checked the solutions is correct however I got $9/2$ for the radius and in the solutions the radius should be $\frac{3\sqrt{2}}{2}$.</p>
<p>Could you explain how this is achieved for the radius? </p>
| Theorem | 346,898 | <p>You got $R^2$! you need $R$. So just take the square root.</p>
|
1,956,897 | <p>I have the question "Find the centre and radius of the circle whose equation is $x^2+y^2+x+3y-2=0$"</p>
<p>So I've worked out the centre to be $(-1/2, -3/2)$ which when I checked the solutions is correct however I got $9/2$ for the radius and in the solutions the radius should be $\frac{3\sqrt{2}}{2}$.</p>
<p>Could you explain how this is achieved for the radius? </p>
| Stefan | 375,579 | <p>$x^2+y^2+x+3y-2=0 \Leftrightarrow (x+0.5)^2+(y+1.5)^2=4.5$</p>
<p>i suppose this is what you did.
The common formula for circles is $(x-a)^2+(y-b)^2=r^2$ so you just need to take the root of $\frac{9}{2}$.</p>
<p>$\sqrt{\frac{9}{2}}=\frac{3}{\sqrt{2}}=3\frac{\sqrt{2}}{2}$</p>
|
798,307 | <p>I have a linear functional from the space of nxn matrices over a field F. The functional satisfies $f(A) = f(PAP^{-1})$ for all invertible $P$ and $A$ an nxn matrix. I'm trying to show that $f(A) = \lambda tr(A)$ for some constant $\lambda$.</p>
<p>So far I have:</p>
<ul>
<li>The linear functionals have basis $f_{E_{i,j}}(A) = tr({E_{i,j}}^t A)$ where $E_{i,j}$ is zero everywhere except in the $i,j$ position.</li>
<li>If $PAP^{-1} = A$ for all invertible $P$, then A is a multiple of the identity matrix.</li>
</ul>
<p>Thanks for any help </p>
| EPS | 133,563 | <p>From the given condition it is clear that $f$ satisfies $f(AB)=f(BA)$ for any two square matrices $A$ and $B$. Next recall that the kernel of the trace function on $M_n(\mathbb{R})$ is $n^2-1$ dimensional. Thus any square matrix $X$ can be written in the form $X=Z+cA$ where $Z$ is a matrix with zero trace and $A$ is any fixed matrix with nonzero trace. Hence
$$f(X)=cf(A).$$
This shows that $f$ depends only on the value of $c$. On the other hand tr $X=c\text{tr }A$ and hence $f$ has to be a multiple of the trace function.</p>
|
3,306,089 | <p>I came across this meme today:</p>
<p><a href="https://i.stack.imgur.com/RfJoJ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RfJoJ.jpg" alt="enter image description here"></a></p>
<p>The counterproof is very trivial, but I see no one disproves it. Some even say that the meme might be true. Well, <span class="math-container">$\pi$</span> cannot contain itself.</p>
<p>Well, everything means <span class="math-container">$\pi$</span> might contain <span class="math-container">$\pi$</span> somewhere in it. Say it starts going <span class="math-container">$\pi=3.1415...31415...$</span> again on the <span class="math-container">$p$</span> digit. Then it will have to do the same at the <span class="math-container">$2p$</span> digit, since the "nested <span class="math-container">$\pi$</span>" also contains another <span class="math-container">$\pi$</span> in it. <span class="math-container">$\pi$</span> then will be rational, which is wrong. Thus <span class="math-container">$\pi$</span> does not contain all possible combination.</p>
<p>Is this proof correct? I'm not a mathematician so I'm afraid I make silly mistakes.</p>
| Ross Millikan | 1,827 | <p>I am sure it has <span class="math-container">$31415$</span> again in the decimal expansion, but why should it continue <span class="math-container">$926535$</span> after that? Sometimes it will, but it will eventually diverge from the decimals at the start. You have not made any argument that when you see <span class="math-container">$31415$</span> it should repeat from there and in fact it will not. </p>
<p>You are correct that <span class="math-container">$\pi$</span> cannot contain itself. The <a href="https://math.stackexchange.com/questions/216343/does-pi-contain-all-possible-number-combinations?rq=1">claim, not known to be true</a>, is that <span class="math-container">$\pi$</span> contains all finite sequences of digits.</p>
|
2,401,281 | <blockquote>
<p>Show that for $\{a,b,c\}\subset\Bbb Z$ if $a+b+c=0$ then $2(a^4 + b^4+ c^4)$ is a perfect square. </p>
</blockquote>
<p>This question is from a math olympiad contest. </p>
<p>I started developing the expression $(a^2+b^2+c^2)^2=a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2$ but was not able to find any useful direction after that.</p>
<p><strong>Note</strong>: After getting 6 answers here, another user pointed out other question in the site with similar but not identical content (see above), but the 7 answers presented include more comprehensive approaches to similar problems (e.g. newton identities and other methods) that I found more useful, as compared with the 3 answers provided to the other question. </p>
| ajotatxe | 132,456 | <p>$$c^4=(-a-b)^4=(a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4$$</p>
<p>Therefore,</p>
<p>$$2(a^4+b^4+c^4)=4(a^4+2a^3b+3a^2b^2+2ab^3+b^4)$$</p>
<p>Now compute $(a^2+ab+b^2)^2$.</p>
|
2,401,281 | <blockquote>
<p>Show that for $\{a,b,c\}\subset\Bbb Z$ if $a+b+c=0$ then $2(a^4 + b^4+ c^4)$ is a perfect square. </p>
</blockquote>
<p>This question is from a math olympiad contest. </p>
<p>I started developing the expression $(a^2+b^2+c^2)^2=a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2$ but was not able to find any useful direction after that.</p>
<p><strong>Note</strong>: After getting 6 answers here, another user pointed out other question in the site with similar but not identical content (see above), but the 7 answers presented include more comprehensive approaches to similar problems (e.g. newton identities and other methods) that I found more useful, as compared with the 3 answers provided to the other question. </p>
| lhf | 589 | <p>We can simplify the algebra by noticing that the expressions are homogenous:</p>
<p>$
2(a^4+b^4+c^4)
\\\quad=2(a^4+b^4+(a+b)^4)
\\\quad=2b^4(x^4+1+(x+1)^4),
\quad x=a/b
$</p>
<p>$
2(x^4+1+(x+1)^4)
\\\quad=4 x^4 + 8 x^3 + 12 x^2 + 8 x + 4
\\\quad=4 (x^2 + x)^2 + 8 (x^2 + x) + 4
\\\quad=4(y^2+2y+1),
\quad y=x^2+x
\\\quad=4(y+1)^2
\\\quad=(2(y+1))^2
\\\quad=(2(x^2+x+1))^2
$</p>
<p>$
2(a^4+b^4+c^4)
\\\quad=b^4(2(x^2+x+1))^2
\\\quad=(2b^2(x^2+x+1))^2
\\\quad=(2(a^2+ab+b^2)^2
$</p>
|
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.youtube.com/watch?v=0z1fIsUNhO4" rel="nofollow noreferrer">Moebius transformation revealed</a></li>
</ol>
<p>Do you know of any other good math videos? Share.</p>
| Matt Noonan | 2,510 | <p>Along the sphere eversion lines, there is also the energy-minimizing sphere eversion constructed by Rob Kusner. I think there is a video of it at <a href="https://www.youtube.com/watch?v=I6cgca4Mmcc" rel="nofollow noreferrer">https://www.youtube.com/watch?v=I6cgca4Mmcc</a>, though it isn't labelled as such.</p>
<p>Rob also has written <a href="http://www.gang.umass.edu/%7Ekusner/other/minimax.pdf" rel="nofollow noreferrer">a paper</a> about the history of the minimax eversion.</p>
|
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.youtube.com/watch?v=0z1fIsUNhO4" rel="nofollow noreferrer">Moebius transformation revealed</a></li>
</ol>
<p>Do you know of any other good math videos? Share.</p>
| Matt | 4,265 | <p>MIT's OpenCourseWare has a few math courses up:</p>
<p><a href="http://ocw.mit.edu/OcwWeb/web/courses/av/index.htm#Mathematics">http://ocw.mit.edu/OcwWeb/web/courses/av/index.htm#Mathematics</a></p>
|
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.youtube.com/watch?v=0z1fIsUNhO4" rel="nofollow noreferrer">Moebius transformation revealed</a></li>
</ol>
<p>Do you know of any other good math videos? Share.</p>
| Jan Weidner | 2,837 | <p>A nice introduction to representation theory of compact lie groups, sl2(R) and other topics:
<a href="http://www.math.utah.edu/vigre/minicourses/sl2/schedule.html" rel="nofollow">http://www.math.utah.edu/vigre/minicourses/sl2/schedule.html</a></p>
|
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.youtube.com/watch?v=0z1fIsUNhO4" rel="nofollow noreferrer">Moebius transformation revealed</a></li>
</ol>
<p>Do you know of any other good math videos? Share.</p>
| Anweshi | 2,938 | <p>I am surprised that nobody mentioned the four-week <a href="http://www.uni-math.gwdg.de/aufzeichnungen/SummerSchool/">workshop at Göttingen</a> on arithmetic geometry in 2006 summer. Almost all of the videos are still available. Wonderful videos.</p>
|
1,714 | <p>I know of two good mathematics videos available online, namely:</p>
<ol>
<li>Sphere inside out (<a href="https://www.youtube.com/watch?v=BVVfs4zKrgk" rel="nofollow noreferrer">part I</a> and <a href="https://www.youtube.com/watch?v=x7d13SgqUXg" rel="nofollow noreferrer">part II</a>)</li>
<li><a href="https://www.youtube.com/watch?v=0z1fIsUNhO4" rel="nofollow noreferrer">Moebius transformation revealed</a></li>
</ol>
<p>Do you know of any other good math videos? Share.</p>
| T.B. | 12,312 | <p>Ken Ribet's introductory lecture on Serre's modularity conjecture. Useful and quite easy to follow and understand. <a href="http://fora.tv/2007/10/25/Kenneth_Ribet_Serre_s_Modularity_Conjecture" rel="nofollow">http://fora.tv/2007/10/25/Kenneth_Ribet_Serre_s_Modularity_Conjecture</a></p>
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