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 |
|---|---|---|---|---|
588,228 | <p>A,B,C are distinct digits of a three digit number such that </p>
<pre><code> A B C
B A C
+ C A B
____________
A B B C
</code></pre>
<p>Find the value of A+B+C.</p>
<p>a) 16 b) 17 c) 18 d) 19</p>
<p>I tried it out by using the digits 16 17 18 19 by breaking them in three numbers but due to so large number of ways of breaking I cannot help my cause.</p>
| md arif sarkar | 462,036 | <p>Let, $A=1, B=2, C=3$</p>
<p>The digit $A$ has $3$ times $=1\cdot 3=3$</p>
<p>The digit $B$ has $3$ times$=2\cdot 3=6$</p>
<p>The digit $C$ has $3$ times $=3\cdot 3=9$<br>
Now:</p>
<p>$A+B+C=3+6+9=18$</p>
|
1,515,645 | <p>The question is to find the values of a real number $\lambda$ for which the following equation is satisfied for all real values of $\alpha$ which are not integral multiples of $\pi/2$
$${\sin\lambda\alpha\over \sin\alpha}-{\cos\lambda\alpha\over \cos\alpha}=\lambda-1$$</p>
<p>All I could do was to guess some values that just came to mind by observation, like $-1,1,3$</p>
<p>What should be a more mathematical way to find all possible values of $\lambda$? </p>
<p>SOURCE: KVPY 2015 SB stream </p>
| fleablood | 280,126 | <p>Let <span class="math-container">$1111111111111111 \equiv k \pmod {17}$</span>.</p>
<p>Then <span class="math-container">$9999999999999999 \equiv 9k\pmod {17}$</span></p>
<p>And <span class="math-container">$10000000000000000 =10^{16} \equiv 9k + 1\pmod {17}$</span>.</p>
<p>ANd but Fermat's little theorem:</p>
<p><span class="math-container">$9k + 1 \equiv 10^{16} \equiv 1 \pmod{17}$</span></p>
<p>so <span class="math-container">$9k \equiv 0 \pmod 17$</span>.</p>
<p>Not <span class="math-container">$2\cdot 9 = 18 \equiv 1 \pmod {19}$</span> so <span class="math-container">$9^{-1} \equiv 2 \pmod {17}$</span> and we may do:</p>
<p>So <span class="math-container">$2\cdot 9k \equiv 2\cdot 0 \pmod {17}$</span></p>
<p><span class="math-container">$k \equiv 0 \pmod 7$</span>.</p>
<p>So <span class="math-container">$1111111111111111 \equiv 0 \pmod {17}$</span> and we are done. <span class="math-container">$17$</span> divides <span class="math-container">$1111111111111111$</span></p>
<p>Meanwhile if <span class="math-container">$11111111\equiv m \pmod{17}$</span> then</p>
<p><span class="math-container">$99999999 \equiv 9m\pmod {17}$</span></p>
<p><span class="math-container">$10^8 \equiv 9m + 1 \pmod {17}$</span>.</p>
<p>And be successive squaring <span class="math-container">$10^2 \equiv 100=6*17-2\equiv -2 \pmod {17}$</span>. <span class="math-container">$10^4\equiv (-2)^2\equiv 4\pmod{17}$</span> and <span class="math-container">$10^8 \equiv 4^2 =16\equiv -1 \pmod {17}$</span>.</p>
<p>SO <span class="math-container">$9m + 1 \equiv -1\pmod{17}$</span></p>
<p><span class="math-container">$9m \equiv -2 \pmod {17}$</span></p>
<p><span class="math-container">$m\equiv 2\cdot 9m \equiv -2\cdot 2 \equiv -4 \equiv 13 \pmod {17}$</span>.</p>
<p>And we are done <span class="math-container">$17$</span> does not divide <span class="math-container">$11111111$</span> and, indeed, we will have a remainder of <span class="math-container">$13$</span> if we attempt to.</p>
<p><span class="math-container">$11111098= 653594\times 17$</span> and <span class="math-container">$11111111=653594\times 17 + 13$</span>.</p>
|
259,431 | <p>In the book of Richard Hammack, I come accross with the following question:</p>
<blockquote>
<p>There are two different equivalence relations on the set $A = \{a,b\}$.
Describe them.</p>
</blockquote>
<p>OK, I found that the solution is,</p>
<p>$$R_1 = \{(a,a),(b,b),(a,b),(b,a)$$
and
$$R_2 = \{(a,a),(b,b)\}$$</p>
<p>Then I thought two more equivalence classes $R_3 = \{(a,a)\}$, $R_4 = \{(b,b)\}$. But when I looked the answer, I saw that, $R_1$ and $R_2$ are true but others are false. Why is that?</p>
| smackcrane | 12,039 | <p>$R_3 = \{(a, a)\}$ and $R_4 = \{(b, b)\}$ are not equivalence relations because they are not reflexive; $b$ must be related to itself.</p>
|
653,887 | <p>Why is the following set linearly independent for all x on ($-\infty$, $\infty$)?</p>
<p>$$\{1+x, 1-x, 1-3x\}$$</p>
<p>The Wronskian is $0$, but Wolfram Alpha says it is still linear independent? Why is this?</p>
<p>Thanks!</p>
<p><a href="http://www.wolframalpha.com/input/?i=Is+%7B1-x%2C+1%2Bx%2C+1-3x%7D+linearly+independent%3F" rel="nofollow">http://www.wolframalpha.com/input/?i=Is+%7B1-x%2C+1%2Bx%2C+1-3x%7D+linearly+independent%3F</a></p>
| Amzoti | 38,839 | <p>Hint:</p>
<p>Since the Wronskian is zero, no conclusion can be drawn about linear independence.</p>
<p>For linear independence, we want to go back to the basic definitions.</p>
<p>Can you proceed?</p>
|
1,600,266 | <p>I'm trying to work out sum of this series $$1 + \frac{2}{2} + \frac{3}{2^2} + \frac{4}{2^3} + \ldots$$</p>
<p>I know one method is to do substitutions and getting the series into a form of a known series. So far I've converted the series into $$ 1 + \frac{2}{x} + \frac{3}{x^2} + \frac{4}{x^3} + \ldots $$ where $x=2$ and I'm trying to get it into the form of the $\ln(1+x)$ series somehow. I have tried differentiating, integrating and nothing is working out. The closest I got is by inverting which gave me $ 1 + \frac{x}{2} + \frac{x^2}{3} + \frac{x^3}{4} \ldots $ </p>
<p>Now I'm just lost and have no idea what to do.</p>
<p>The other idea I had was converting it into $$ \Large{\sum_{n=1}^\infty{\frac{n}{2^{n-1}}}}, $$ but I have no idea how to do anything further to it.
How would you do this? Thnx.</p>
| Brian Tung | 224,454 | <p>Not sure what the book has in mind there. Perhaps there is a line missing? Is it possible to give a bit more context than you have so far?</p>
<p>Anyway, some remarks about divisibility for binary numbers: A binary number is divisible by $3$ (i.e., $11_2$) iff the number of even-indexed $1$'s and the number of odd-indexed $1$'s are equivalent modulo $3$. Thus, for instance, $57 = 111001_2$ is divisible by $3$, because there are two even-indexed $1$'s and two odd-indexed $1$'s. $2 \equiv 2 \pmod 3$ (trivially), so this number is divisible by $3$.</p>
<p>Note that this is not true of $29 = 11101_2$, which has one even-indexed $1$ and three odd-indexed $1$'s. This doesn't, however, show that $29$ is a prime, although obviously it happens to be. The first counterexample is $5^2 = 25 = 11001_2$, which has one even-indexed $1$ and two odd-indexed $1$'s, but is obviously composite.</p>
|
835,454 | <p>I was struggling with the following problem (from linear algebra):</p>
<blockquote>
<p>Let $V$ be the vector space of the $2 \times 2$ matrices with real coefficients. Consider the action of the group $SL_2(\mathbb{R})$ on $V$, namely for any matrix $T \in SL_2(\mathbb{R})$ and for every matrx $A \in V$ we define $\Phi_T(A)=T\cdot A\cdot T^{-1}$. Prove that for every $T\in SL_2(\mathbb{R})$, $\Phi_T$ is an endomorphism of $V$. Then starting from a basis of $V$, write the corresponding homomorphism $SL_2(\mathbb{R}) \to GL_4(\mathbb{R})$, then show that the image lies inside $SL_4(\mathbb{R})$</p>
</blockquote>
<p>First of all. Does anybody know the origin of this problem? Is there a book with standard solutions to this kind of problem? And is my solution any good? Apparently it is generalizable to many more dimensions, giving a nontrivial result on the determinant of such matrices.
And what does it happen when the trace is $0$? Can we show an homomorphism $SL_2(\mathbb{R}) \to SL_3(\mathbb{R})$ in the same way?</p>
| ThePunisher | 88,132 | <p>I have tried to show the actual homomorphism. I found the matrix in $GL_4(\mathbb{R})$ associated to the general matrix in $SL_2(\mathbb{R})$, namely:
$$ \left( \begin{matrix} a & b \\ c & d\end{matrix} \right) \quad \text{with} \ ad-bc\neq0 $$
And it should be the following:
$$ \left( \begin{matrix}ad & -ac & bd &-bc\\ -ab & a^2 & -b^2 & ab \\ cd & -c^2 & d^2 & -cd\\ -bc & ac & -bd & ad\end{matrix}\right) $$</p>
<p>But honestly I find it a tedious task to look at its determinant to show that it is always $1$.</p>
<p>I found an alternative solution by looking at the abstract algebra behind it. We can compose the homomorphism $SL_2(\mathbb{R}) \to GL_4(\mathbb{R})$ with the homomorphism given by the determinant $GL_4(\mathbb{R}) \to (\mathbb{R^*}, \cdot)$. We get a new homomorphism $SL_2(\mathbb{R}) \to (\mathbb{R^*}, \cdot)$. But I can prove that this homomorphism is trivial! <a href="http://math.berkeley.edu/~efuchs/250A/PSL2Rsimple.pdf" rel="nofollow">In this paper</a> I found a proof that the only normal subgroups of $SL_2(\mathbb{R})$ are $\{\pm 1\}$ and the trivial ones. Given that the kernel must be a normal subgroup and that the matrix $$\left( \begin{matrix} 0 & -1 \\ 1 & 0\end{matrix} \right)$$ must map to the identity since it has order $4$ we have that the whole group maps to the identity giving the thesis!</p>
|
2,278,120 | <p>In one of my algorithm courses, there is this: </p>
<blockquote>
<p>A subset $S$ of vertices in a directed graph $G$ is
strongly connected if for each pair of distinct
vertices ($v_i$, $v_j$) in $S$, $v_i$ is connected to $v_j$ and
$v_j$ is connected to $v_i$.</p>
</blockquote>
<p>And then the following example graph is given for this proposition: </p>
<p><a href="https://i.stack.imgur.com/m6PLe.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/m6PLe.jpg" alt="enter image description here"></a></p>
<p>Maybe i do not understand what that phrase means. What i think it means is this: A node, say $E$ can be a member of an ordered pair $(v_i, v_j)$ only once, ie. if $(E, A)$, then $\lnot (E, [someOtherNode]) $. But we clearly see here, that we have $(E,A)$ and $(E,D)$. </p>
<p>How should i correctly interpret this phrase. What does it mean exactly? Thanks.</p>
| Jose Arnaldo Bebita Dris | 28,816 | <p>I think "for each pair of distinct vertices $(v_i, v_j)$ in $S \times S$" means</p>
<blockquote>
<p>"for each $(v_i, v_j) \in S \times S$, $v_i \neq v_j$."</p>
</blockquote>
|
2,785,449 | <p>So, it is obvious that the empty set is a subset of every set, such that <strong>∀A:∅⊆A</strong>.</p>
<p>Therefore, if the null set is an element of some set A, such that <strong>A = {∅}</strong>, is the empty set considered a singleton element of A?</p>
<p>Thank you.</p>
| D. Brogan | 404,162 | <p>Yes, the set $\{\varnothing\}$ is a set whose only element is the empty set $\varnothing$.</p>
|
2,161,913 | <p>Consider the angle $\theta$ with $$0<\theta<\frac{\pi}{2}$$ Suppose that $$cos\theta=\frac{l}{n} \space and \space \space sin\theta=\frac{m}{n}$$
are rational numbers, with postive integers $l,m$ and $n$. Show that $l$ and $m$ cannot both odd integers.<br>
MY ATTEMPT: Assume that $l$ and $m$ are both odd integers. Then $cos^2\theta=\frac{l^2}{n^2}$ and $sin^2\theta=\frac{m^2}{n^2}$ then adding these equations and using the Pythagorean identity we obtain $$l^2+m^2=n^2$$
Notice we can conclude n is even, and we must find a contradiction. I am unsure where to go from here, any help is appreciated!</p>
| Claude Leibovici | 82,404 | <p><em>Much less elegant than RRL'answer.</em></p>
<p>Consider $$P_n=\left(\prod _{i=1}^n \sec \left(\frac{1}{i}\right)\right){}^{\frac{1}{n}}$$ $$\log(P_n)=\frac 1n \sum _{i=1}^n \log\left(\sec \left(\frac{1}{i}\right)\right)$$ Now, by Taylor for large values of $i$ $$\sec \left(\frac{1}{i}\right)=1+\frac{1}{2 i^2}+\frac{5}{24 i^4}+O\left(\frac{1}{i^6}\right)$$ $$\log\left(\sec \left(\frac{1}{i}\right)\right)=\frac{1}{2 i^2}+\frac{1}{12 i^4}+O\left(\frac{1}{i^6}\right)$$ $$\sum _{i=1}^n \log\left(\sec \left(\frac{1}{i}\right)\right)=\frac{H_n^{(2)}}{2}+\frac{H_n^{(4)}}{12}+\cdots$$ where appears harmonic numbers.</p>
<p>Using asymptotics $$\log(P_n)=\frac 1 n \left(\frac{90 \pi ^2+\pi ^4}{1080}-\frac{1}{2 n}+O\left(\frac{1}{n^2}\right)\right)$$ Taylor again $$P_n=e^{\log(P_n)}=1+\frac{90 \pi ^2+\pi ^4}{1080 n}+O\left(\frac{1}{n^2}\right)$$ which seems to be a quite reasonable approximation
$$\left(
\begin{array}{cccc}
n & P_n & \text{approximation} & \text{difference} \\
10 & 1.09393 & 1.09127 & 0.00266 \\
20 & 1.04713 & 1.04563 & 0.00149 \\
30 & 1.03145 & 1.03042 & 0.00103 \\
40 & 1.02360 & 1.02282 & 0.00078 \\
50 & 1.01889 & 1.01825 & 0.00063\\
60 & 1.01574 & 1.01521 & 0.00053 \\
70 & 1.01349 & 1.01304 & 0.00046 \\
80 & 1.01181 & 1.01141 & 0.00040 \\
90 & 1.01050 & 1.01014 & 0.00036 \\
100 & 1.00945 & 1.00913 & 0.00032
\end{array}
\right)$$</p>
|
4,037,206 | <p><a href="https://i.stack.imgur.com/VfORl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VfORl.png" alt="enter image description here" /></a></p>
<p>I have done the following for the median <br/>
<span class="math-container">$2(x-1) = .50$</span> <br/>
results in <span class="math-container">$x = 1.25$</span> <br/></p>
<p>For the quartiles, I have <span class="math-container">$q_{1}$</span>--> <span class="math-container">$2(x-1)=.25$</span> which results in <span class="math-container">$q_{1} = 1.125$</span> <br/>
for <span class="math-container">$q_{3}$</span>--> <span class="math-container">$2(x-1)=.75$</span> which results in <span class="math-container">$q_{3}=1.375$</span> and then subtract them to get <span class="math-container">$.25$</span>
Please correct me if I am wrong! thank you in advance</p>
| BruceET | 221,800 | <p><strong>An approach using a <a href="https://en.wikipedia.org/wiki/Beta_distribution" rel="nofollow noreferrer">beta distribution</a>.</strong> Let <span class="math-container">$Y \sim \mathsf{Beta}(2,1).$</span>
Then <span class="math-container">$X = Y+1.$</span> Results are the same as in @heropup's Answer (+1).</p>
<p>The quartiles of <span class="math-container">$Y$</span> are as follows (using R) are
<span class="math-container">$1/2$</span> for the lower quartile, <span class="math-container">$\sqrt{2}/2$</span> for the
median, and <span class="math-container">$\sqrt{3}/2$</span> for the upper quartile.
The IQR is <span class="math-container">$(\sqrt{3} - 1)/2.$</span>
Also, the lower
quartile for <span class="math-container">$X$</span> is <span class="math-container">$1.5$</span> and the IQR for <span class="math-container">$X$</span> is
the same as for <span class="math-container">$Y.$</span> [In R, <code>qbeta</code> is the quantile function (inverse CDF) of a beta distribution.]</p>
<pre><code>qbeta(c(1,2,3)/4, 2, 1)
[1] 0.5000000 0.7071068 0.8660254
sqrt(1:3)/2
[1] 0.5000000 0.7071068 0.8660254
</code></pre>
<p><em>Simulation:</em> By simulation of a million realizations of <span class="math-container">$X$</span> in R, we
can get good approximations to the quartiles, accurate
to about three significant digits--and a figure for
illustration.</p>
<pre><code>set.seed(223)
x = 1 + rbeta(10^6, 2, 1)
median(x); IQR(x)
[1] 1.706684 # aprx 1.7071
[1] 0.3666178 # aprx 0.8660
summary(x)
Min. 1st Qu. Median Mean 3rd Qu. Max.
1.001 1.499 1.707 1.666 1.866 2.000
hdr = "Simulated values of 1+BETA(2,1) with Quartiles"
hist(x, prob=T, br=30, col="skyblue2", main=hdr)
curve(dbeta(x-1, 2, 1), add=T, lwd=3, col="orange")
abline(v=quantile(x, c(1:3)/4), col="darkgreen", lwd=2)
</code></pre>
<p><a href="https://i.stack.imgur.com/icX2v.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/icX2v.png" alt="enter image description here" /></a></p>
|
1,609,446 | <p>I read the definition that $f$ is in $O(x^n)$ if $|f(x)|<C|x^n|$ for some $C$.</p>
<p>I'm struggling to understand how to check this. For example, supposedly $f(x) = 5x+3x^2$ is in $O(x)$ but not $O(x^2)$?</p>
<p>If I plot $f(x) = 5x + 3x^2$ and $g(x)=x$ I see that the first goes to infinity much quicker.</p>
<p><a href="https://i.stack.imgur.com/1H5Uv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1H5Uv.png" alt="enter image description here"></a></p>
<p>If I let $g(x) = Cx$, and plot $C=1,C=10, C=20, C=100$, it looks like it overcomes $f$ for $C>10$:
<a href="https://i.stack.imgur.com/owokf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/owokf.png" alt="enter image description here"></a></p>
<p>But, if you zoom out further, you can see that's not true:</p>
<p><a href="https://i.stack.imgur.com/ZSqk5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZSqk5.png" alt="enter image description here"></a></p>
<p>So, I know it doesn't matter what $C$ is, but how can I show that there exists a $C$ to make the definition hold so that I can tell if $f$ is in $O(x^n)$?</p>
<p>If you go out far enough, and $C$ is large enough can't I make either $f$ or $g$ as close to the y-axis as I want?</p>
| ml0105 | 135,298 | <p>The function $f(x) = 5x + 3x^{2} \not \in O(x)$, but it is in $O(x^{2})$. The way you show that $f(x) \in O(x^{2})$ is by picking a constant $C$ and appropriate constant $k$ such that $|f(x)| \leq Cx^{2}$ for all $x \geq k$.</p>
<p>Big-O can often be ascertained using limits. If the following condition holds:</p>
<p>$$L = \lim_{x \to \infty} \dfrac{f(x)}{g(x)} < \infty$$</p>
<p>Then we have that $f(x) \in O(g(x))$. The converse does not necessarily hold, as given in the comments below ($x*sin(x)$).</p>
<p>So consider:</p>
<p>$$\lim_{x \to \infty} \dfrac{5x + 3x^{2}}{x} = 5 + \lim_{x \to \infty} 3x = \infty$$</p>
<p>And:</p>
<p>$$\lim_{x \to \infty} \dfrac{5x + 3x^{2}}{x^{2}} = 0 + 3 = 3$$ </p>
<p>Thus, $f(x) \in O(x^{2})$ but not $O(x)$.</p>
<p>You can check out these links for more information on how to formulate Big-O proofs:
<a href="http://www.dreamincode.net/forums/topic/280815-introduction-to-proofs-induction-and-big-o/" rel="nofollow">http://www.dreamincode.net/forums/topic/280815-introduction-to-proofs-induction-and-big-o/</a>
<a href="http://www.dreamincode.net/forums/topic/321402-introduction-to-computational-complexity/" rel="nofollow">http://www.dreamincode.net/forums/topic/321402-introduction-to-computational-complexity/</a></p>
|
1,335,734 | <blockquote>
<p>A library leases its photocopier. One monthly bill was 750 dollars for
12,000 copies. Another month, the bill was $862.50 for 16,500 copies.
How much does the library pay for each copy?</p>
</blockquote>
<p>How should I start set up the equation for this problem?
How exactly do I find the linear equation giving the cost in terms of the number of copies used? </p>
<p>I am sorry if this is a stupid question, I am trying to review pre-calc and I forgot some techniques used to solve the problems. </p>
| TomMonTom | 237,366 | <p>Thank you Sky, I see now that upon one part of the chair rule becoming a complete polynomial with a coefficient it can then be distributed amongst the single variables of y<em>dy/dx and x respectively without having to foil as if it were (2x+2x)</em>(x^2+y^2).</p>
|
187,511 | <p>There are two questions:</p>
<ol>
<li><p>How to prove that in general </p>
<p>$[\hat{A}(\mathbb HP^m)]_{4m} = 0$</p>
<p>It is possible to verify it for low values of $m$.</p></li>
<li><p>How to prove that in general</p>
<p>$\left[\frac{\hat{A}(\mathbb HP^m)} { \hat{M}(\mathbb HP^m) }\right]_{4m} = 0$</p>
<p>where $\hat{M}(\mathbb HP^m)$ is the Mayer class defined by</p>
<p>$\hat{M}(V) = \prod _{i=1}^{s}\cosh \left( \frac{y_{{i}}}{2} \right)$</p>
<p>with</p>
<p>$p(V) =\prod _{i=1}^{s}(1+{y_{i}}^2)$.</p>
<p>It is possible to verify it for low values of $m$.</p></li>
</ol>
| Juan Ospina | 61,860 | <p>It is possible to obtain a more direct proof using other idea due to Professor Ebert.</p>
<p>It is well known that </p>
<p>$\hat{A} = \prod _ i{\frac {x_{{i}}/2}{\sinh \left( x_{{i}}/2 \right) }}= 1+\hat{A}_{1}+\hat{A}_{2}+\hat{A}_{3}+....$</p>
<p>where </p>
<p>$\hat{A}_{1}=-\frac{1}{24}p_{{1}}$</p>
<p>$\hat{A}_{2} = -{\frac {1}{1440}}\,p_{{2}}+{\frac {7}{5760}}\,{p_{{1}}}^{2}$</p>
<p>$\hat{A}_{3} = {\frac {11}{241920}}\,p_{{1}}p_{{2}}-{\frac {1}{60480}}\,p_{{3}}-{
\frac {31}{967680}}\,{p_{{1}}}^{3} $</p>
<p>From other side</p>
<p>$ {\frac {\hat{A}}{\hat{M}}}= \prod _ i{\frac {x_{{i}}}{\sinh \left( x_{{i}} \right) }} = 1+ [{\frac {\hat{A}}{\hat{M}}}]_{1}+ [{\frac {\hat{A}}{\hat{M}}}]_{2}+[{\frac {\hat{A}}{\hat{M}}}]_{3}+....$</p>
<p>where</p>
<p>$[{\frac {\hat{A}}{\hat{M}}}]_{1} = -\frac{1}{6}p_{{1}}$</p>
<p>$[{\frac {\hat{A}}{\hat{M}}}]_{2} = {\frac {7}{360}}\,{p_{{1}}}^{2}-{\frac {1}{90}}\,p_{{2}} $ </p>
<p>$[{\frac {\hat{A}}{\hat{M}}}]_{3} = {\frac {11}{3780}}\,p_{{1}}p_{{2}}-{\frac {31}{15120}}\,{p_{{1}}}^{3}-
{\frac {1}{945}}\,p_{{3}}
$</p>
<p>Then we have</p>
<p>$[{\frac {\hat{A}}{\hat{M}}}]_{1} = 4\hat{A}_{1}$</p>
<p>$[{\frac {\hat{A}}{\hat{M}}}]_{2} = 16\hat{A}_{2}$</p>
<p>$[{\frac {\hat{A}}{\hat{M}}}]_{3} = 64\hat{A}_{3}$</p>
<p>$[{\frac {\hat{A}}{\hat{M}}}]_{n} = 4^{n}\hat{A}_{n}$</p>
<p>From the last equation we obtain that given that </p>
<p>$[\hat{A}(\mathbb HP^m)]_{4m} = 0$</p>
<p>then</p>
<p>$[\frac{\hat{A}(\mathbb HP^m)} { \hat{M}(\mathbb HP^m) }]_{4m} = 0$.</p>
<p>Besides of this, the paragraph </p>
<p><img src="https://i.stack.imgur.com/iZshQ.jpg" alt="enter image description here"></p>
<p>extracted from <a href="http://geometrie.math.uni-potsdam.de/documents/baer/ellsymb.pdf" rel="nofollow noreferrer">Baer</a>, page 30; can be simplified to the integrality of</p>
<p>$2^{l+\frac{n}{2}}\int_{X}\hat{A}_{\frac{n}{4}}(TX)$.</p>
<p>As an application of the last result it is possible to prove that $CP^4$ cannot be immersed in $R^{11}$. Computing the last integral with $l=1$ and $n=8$ we obtain</p>
<p>$2^{5}\int_{CP^4}\hat{A}_{2}(TCP^4)= \frac{3}{4} $</p>
<p>which is not an integer.</p>
<p>In general for $CP^4$ the Mayer integral takes the form</p>
<p>$2^{l+4}\int_{CP^4}\hat{A}_{2}(TCP^4) = 3({2})^{l-3}$</p>
<p>Then this last integral indicates that $CP^4$ cannot be immersed in $R^{10}, R^{11}, R^{12}, R^{13}$. For $l\geq 3$ there is no topological obstruction to the existence of immersion of $CP^4$ in $R^{8+2l+1}$.</p>
<p>More in general, in the case of $CP^{2m}$ the Mayer integral is evaluated as</p>
<p>$2^{l+2m}\int_{CP^{2m}}\hat{A}_{m}(TCP^{2m}) = (-1)^m2^{l-2m}{2\,m\choose m}$.</p>
|
3,190,781 | <blockquote>
<p>Suppose the polynomial <span class="math-container">$P(x)$</span> with integer coefficients satisfies the following conditions:<br>
(A) If <span class="math-container">$P(x)$</span> is divided by <span class="math-container">$x^2 − 4x + 3$</span>, the remainder is <span class="math-container">$65x − 68$</span>.<br>
(B) If <span class="math-container">$P(x)$</span> is divided by <span class="math-container">$x^2 + 6x − 7$</span>, the remainder is <span class="math-container">$−5x + a$</span>.<br>
Then we know that <span class="math-container">$a =$</span>?</p>
</blockquote>
<p>I am struggling with this first question from the <a href="https://www.maa.org/sites/default/files/pdf/programs/JUEEDocument.pdf" rel="noreferrer">1990 Japanese University Entrance Examination</a>. Comments from the linked paper mention that this is a basic application of the "remainder theorem". I'm only familiar with the <a href="https://en.wikipedia.org/wiki/Polynomial_remainder_theorem" rel="noreferrer">polynomial remainder theorem</a> but I don't think that applies here since the remainders are polynomials. Do they mean the Chinese remainder theorem, applied to polynomials?</p>
<p>So for some <span class="math-container">$g(x)$</span> and <span class="math-container">$h(x)$</span> we have:
<span class="math-container">$$P(x) = g(x)(x^2-4x+3) + (65x-68),\\
P(x) = h(x)(x^2+6x-7) + (-5x+a),$$</span>
which looks to have more unknowns than equations. How should I proceed from here?</p>
| trancelocation | 467,003 | <p>You need only <span class="math-container">$P(1)$</span>, since</p>
<ul>
<li><span class="math-container">$x^2-4x+3 = (x-1)(x-3)$</span> and</li>
<li><span class="math-container">$x^2+6x-7 = (x-1)(x+7)$</span></li>
</ul>
<p>Hence,</p>
<ul>
<li><span class="math-container">$P(1) = 65-68 = -3$</span></li>
<li><span class="math-container">$P(1) = -5+a \Rightarrow a=2$</span></li>
</ul>
|
3,411,982 | <p>I know the answer using
Total-none of it is <span class="math-container">$6 = 900−648=252$</span></p>
<p>My doubt is that if we you it like</p>
<p>when unit digit is <span class="math-container">$6 = 9\cdot10\cdot1 = 90$</span></p>
<p>when tens place is <span class="math-container">$6 = 9\cdot1\cdot10 = 90$</span></p>
<p>when hundredth place is <span class="math-container">$6 = 1\cdot10\cdot10 = 100$</span></p>
<p>So total <span class="math-container">$= 90 + 90 + 100 = 280$</span></p>
<p>why my answer is not the same </p>
| Axion004 | 258,202 | <p>Suppose that 'three-digit' means <span class="math-container">$abc$</span>, where <span class="math-container">$a>0$</span>. </p>
<p>Now, we first count that there are <span class="math-container">$900$</span> of these numbers.</p>
<p>Of numbers without a <span class="math-container">$6$</span>, then it's <span class="math-container">$8\times 9 \times 9$</span>, since the first digit can be any of 1-5 or 7-9, and the rest 0-5 or 7-9. This gives <span class="math-container">$648$</span> numbers without a 6.</p>
<p>One then finds that there are <span class="math-container">$252 = 900-648$</span> numbers that contain at least one six (or any other specific non-zero digit).</p>
|
1,007,529 | <p>Suppose that the function $f:[-1,1]\rightarrow[-1,1]$ is continuous. Use the intermediate value theorem to prove that there exists a $c \in [-1,1]$ such that $f(c)=c^5$. You should carefully justify each of the hypothesis of the theorem. Question is to be done very formally.</p>
<p>First can I just say, shouldn't $c$ be $c \in (-1,1)$ because that is what the theorem has. So I am just going to assume that it is meant to say $c \in (-1,1)$. If this is not the case, then please let me know. Also this is what my thoughts are:</p>
<p>$f$ is continuous and is defined by $f(x)=x^5$ for $x\in[-1,1]$ so $f(-1)=-1<0$ and $f(1)=1$. </p>
<p>So $f(-1)$ is not equal to $f(1)$. And $f(-1) < 0 < f(1)$. By the intermediate value theorem there exists a point $c\in(-1,1)$ such that $f(c)=0$ that is $c^5=0$.</p>
<p>Is this correct?</p>
| Ben Grossmann | 81,360 | <p>To your first question, $c$ is not necessarily in $(-1,1)$. For example, what if $f(x) = x$?</p>
<p>Also, the question asks about an arbitrary function $f(x)$, so you shouldn't define $f(x) = x^5$.</p>
<p>Here's a hint: consider the function $g(x) = f(x) - x^5$. If $g(-1)$ or $g(1)$ is equal to $0$, the we're done. Otherwise, apply the intermediate value theorem.</p>
|
4,155,453 | <p>I am trying to evaluate <span class="math-container">$$\lim_{n\to \infty} \sum_{k=1}^\infty \frac{n}{n^2+k^2}.$$</span> Now I am aware that clearly <span class="math-container">$$\lim_{n\to \infty} \sum_{k=1}^n \frac{n}{n^2+k^2} = \int_0^1 \frac{1}{1+x^2}dx = \tan^{-1}(1) = \frac{\pi}{4},$$</span> but I do not know what to do if each sum is already sent to infinity. Im taking a limit of limits. I suppose I could rewrite my limit as <span class="math-container">$$\lim_{n\to \infty} \lim_{m\to \infty} \sum_{k=1}^m \frac{n}{n^2+k^2}?$$</span> But I am unaware if this is helpful at all. Any hints would be appreciated. Obviously, Wolfram calculates this as <span class="math-container">$\frac{\pi}{2}$</span> but I am unaware of the steps and logic to get there.</p>
| NoName | 649,394 | <p>Denote <span class="math-container">$\displaystyle \mathcal{S}(n) = \sum_{k \ge 1} \frac{n}{n^2+k^2}$</span>. We have</p>
<p><span class="math-container">\begin{aligned} \mathcal{S}(n) & = \sum_{k\geq 1}\frac{n}{ n^2+k^2} \\& = \frac{1}{2}\sum_{k\geq 1}\frac{2n }{ n^2+k^2} \\& = \frac{1}{2}\sum_{k\geq 1} \frac{d}{dn}\log\left(1+\frac{n^2}{k^2}\right) \\& = \frac{1}{2}\frac{d}{dn} \log \prod_{k\geq 1}\left(1+\frac{n^2}{k^2}\right) \\& = \frac{1}{2}\frac{d}{dn} \bigg[\log \pi n\prod_{k\geq 1}\left(1+\frac{n^2}{k^2}\right)- \log(\pi n)\bigg] \\& = \frac{1}{2}\frac{d}{dn} \log \pi n \prod_{k\geq 1}\left(1+\frac{n^2}{k^2}\right)-\frac{1}{2 n}\end{aligned}</span></p>
<p>Weierstrass factorization for hyperbolic sine is:</p>
<p><span class="math-container">$$\sinh \pi z=\pi z \prod_{k=1}^\infty\left(1+\frac{z^2}{k^2}\right)$$</span></p>
<p>Therefore <span class="math-container">\begin{aligned} \mathcal{S}(n) & = \frac{1}{2} \frac{d}{dn} \log \sinh \pi n -\frac{1}{2 n} \\&= \frac{1}{2}\pi \coth(n \pi)-\frac{1}{2 n } \end{aligned}</span></p>
<p>And finally taking the limit as <span class="math-container">$n \to \infty$</span></p>
<p><span class="math-container">\begin{aligned} \lim_{n \to \infty} \mathcal{S}(n) &= \lim_{n \to \infty} \bigg(\frac{1}{2}\pi \coth(n \pi)-\frac{1}{2 n }\bigg) \\& = \frac{\pi}{2}.\end{aligned}</span></p>
|
3,133,152 | <blockquote>
<p><span class="math-container">$g(x)=|\sin{x}-1|+|3-\cos{x}-\sin{x}|+2\sin{x}$</span></p>
<p>Answer: Above equality is simplified to <span class="math-container">$$1-\sin{x}+3-\cos{x}-\sin{x}+2\sin{x}=4-\cos{x}$$</span></p>
</blockquote>
<p><span class="math-container">$$-1 \le\sin{x}\le1$$</span></p>
<p>So , I know that <span class="math-container">$f(x)=|\sin{x}-1|$</span> will be equal to <span class="math-container">$1-\sin{x}$</span> when <span class="math-container">$-1 \le\sin{x}\lt1$</span>.</p>
<p>But what if <span class="math-container">$\sin{x}$</span> is exactly <span class="math-container">$1$</span>, then wouldn't the expression be <span class="math-container">$\sin{x}-1$</span>?</p>
<p>I always considered <span class="math-container">$|x|=x$</span>, whenever <span class="math-container">$x$</span> is <em>equal to</em> or greater than <span class="math-container">$0$</span>.</p>
| farruhota | 425,072 | <p>Yes, usually, the absolute value is defined as (see <a href="https://en.wikipedia.org/wiki/Absolute_value#Real_numbers" rel="nofollow noreferrer">Wikipedia</a>):
<span class="math-container">$$|x|=\begin{cases}\ \ \ x,\ x\ge 0\\ -x,\ x<0\end{cases}$$</span>
However, it can be adjusted (see the opening paragraph in the above article):
<span class="math-container">$$|x|=\begin{cases}\ \ \ x,\ x>0\\ -x,\ x<0 \\ \ \ \ 0, \ x=0\end{cases}$$</span>
So, using the adjusted form:
<span class="math-container">$$|\sin x-1|=\begin{cases}\ \ \ \ \sin x-1,\ \ \sin x-1>0\\ -(\sin x-1),\ \sin x-1<0 \\ \ \qquad \qquad \ 0, \ \ \sin x-1=0\end{cases} = \\
\begin{cases}\ \ \ \ \ \ \ \ \sin x-1,\ \ \sin x-1\not>0 \ (\emptyset)\\ \qquad \ 1-\sin x,\ \sin x-1<0 \\ \ 0=1-\sin x, \ \ \sin x-1=0\end{cases} = 1-\sin x, \sin x-1\le 0.
$$</span></p>
|
2,335,334 | <p>I'm trying to solve the following ODE:
$$3xy''+(3x+1)y'+y=0$$
So I started by assuming a solution(and its derivates) with the form:</p>
<p>$$
\tag{1}y = \sum_{n=0}^{\infty } a_{n} x^{n + r}$$</p>
<p>$$y' = \sum_{n=0}^{\infty } {(n+r)}a_{n} x^{n + r-1}$$</p>
<p>$$y'' = \sum_{n=0}^{\infty } {(n+r-1)(n+r)}a_{n} x^{n+r-2}$$
Replacing in original equation:
$$3x\sum_{n=0}^{\infty } {(n+r-1)(n+r)}a_{n} x^{n+r-2} + 3x\sum_{n=0}^{\infty } {(n+r)}a_{n} x^{n + r-1}+\sum_{n=0}^{\infty } {(n+r)}a_{n} x^{n + r-1} + \sum_{n=0}^{\infty } a_{n} x^{n + r} = 0$$</p>
<p>$$ 3\sum_{n=0}^{\infty } {(n+r-1)(n+r)}a_{n} x^{n+r-1} + \sum_{n=0}^{\infty } {(n+r)}a_{n} x^{n + r-1} + 3\sum_{n=0}^{\infty } {(n+r)}a_{n} x^{n + r} + \sum_{n=0}^{\infty } a_{n} x^{n + r} = 0 $$</p>
<p>To transform all $x^{n+r}$ into $x^{n+r-1}$ I've rewritten the equation as:</p>
<p>$$
3\sum_{n=0}^{\infty } {(n+r-1)(n+r)}a_{n} x^{n+r-1} +\sum_{n=0}^{\infty } {(n+r)}a_{n} x^{n + r-1} + 3\sum_{n=1}^{\infty } {(n+r-1)}a_{n-1} x^{n + r -1} + \sum_{n=1}^{\infty } a_{n-1} x^{n + r - 1} = 0 $$</p>
<p>In order to make all sums start from the $n=1$:</p>
<p>$$ \tag{2}
3r(r-1)a_{0}x^{r-1} + ra_{0}x^{r-1} + 3\sum_{n=1}^{\infty } {(n+r-1)(n+r)}a_{n} x^{n+r-1} + \sum_{n=1}^{\infty } {(n+r)}a_{n} x^{n + r-1} + 3\sum_{n=1}^{\infty } {(n+r-1)}a_{n-1} x^{n + r -1} + \sum_{n=1}^{\infty } a_{n-1} x^{n + r - 1} = 0
$$</p>
<p>If I understood correctly, the indicial equation should then be:
$$ 3r(r-1)a_{0}x^{r-1} + ra_{0}x^{r-1} = 0 $$</p>
<p>and since $x^{r-1} \neq 0 $ and $a_{0} \neq 0 $:</p>
<p>$$ 3r(r-1) + r = 0 $$
$$ r_{1} = 0 $$
$$ r_{2} = \frac{2}{3} $$</p>
<p>From $(2)$:</p>
<p>$$ [3(n+r)(n+r-1)+(n+r)]a_{n} + [3(n+r-1)+1]a_{n-1} = 0
$$
which seems to give for both solutions of $r$:
$$ a_{n} = \frac{-a_{n-1}}{n} $$
It seems to me that this is in terms of $a_{0}$:</p>
<p>$$ a_{n} = \frac{(-1)^{n}}{n!}a_{0} $$</p>
<p>So rewriting $(1)$ now would give me the solution and $y$ would be an infinite sum since all $a_{n} \neq 0$. And I know this can't be true because by deduction one can see that $y = e^{-x}$ is a solution to the ODE, which makes me think I made a mistake in some of the steps shown above, I'd be glad if anyone could show me where I made the mistake and give me some help in finding the correct general solution for this ODE.</p>
| Tom Himler | 457,359 | <p>So the most direct way of telling you what the answer is is this:</p>
<p>For each question you have 1/2 chance of getting the right answer.</p>
<p>Since there are four questions, each has a 1/2 chance of getting the right answer. So to find it, you do (1/2)x(1/2)x(1/2)x(1/2). Therefore, for this case, there is a 1/16 chance of getting the right answer or (1/(2^4)) chance. Or a 6.25% chance.</p>
|
3,579,457 | <p>Last Christmas, a kid in my neighborhood dressed in a pirate suit and knocked my door for "trick or treat". He explained to me that Dec 25, which is 25 written in base 10, is equal to 31 in base 8, which is Oct 31. So ingeniously, Halloween coincides with Christmas. </p>
<p>I really wonder which other two dates would possibly coincides with each other.</p>
<p>Can you help me find them? And explain what kind the two dates are. Your birthday is also appreciated.</p>
| Hagen von Eitzen | 39,174 | <p>The point in the original "date identity" is that the abbreviations Dec and Oct for the month names are also interpreted as abbreviations for the base, i.e., "decimal 25 = octal 31" becomes "Dec 25 = Oct 31". There is no month name abbreviated as Bin or Hex, for example, so binary and hexadecimal can be excluded. Indeed, at best we might be somewhat willing to accept Sep for septimal (base 7) and Nov for novemal(?) (base 9). However, these are far from common enough to be understood from such abbreviations (and I am not even sure about the correct adjective forms). </p>
<p>Nevertheless, in September, we have dates Sep 01, ..., Sep 06, Sep 10, ..., Sep 30 which can be seen as standing for the numbers one up to twenty-one.
In October, we have accordingly the numbers one up to twenty-five, in November the numbers one up to twenty-seven, and in December of course the numbers one up to thirty-one. Now you can pick any number that is common to at least two months and form such date equalities accordingly, e.g.,</p>
<blockquote>
<p>Sep 01 = Oct 01 = Nov 01 = Dec 01</p>
</blockquote>
<p>or</p>
<blockquote>
<p>Sep 10 = Oct 07 = Nov 07 = Dec 07</p>
</blockquote>
<p>or </p>
<blockquote>
<p>Sep 30 = Oct 25 = Nov 23 = Dec 21</p>
</blockquote>
<p>or </p>
<blockquote>
<p>Oct 26 = Nov 24 = Dec 22</p>
</blockquote>
<p>or</p>
<blockquote>
<p>Nov 30 = Dec 27</p>
</blockquote>
<p>and all cases in-between.</p>
|
245,880 | <p>A box contains 100 balls. Each ball has a number from 1 to 10. How many balls should I draw (ball is put back in box after drawing) to predict the number of balls for each number with 95% certainty.</p>
<p>A prediction is correct if for each of the 10 numbers, the number of balls with that number is correctly predicted.</p>
| Tunococ | 12,594 | <p>If I understand correctly, your problem is estimating the parameter of a Bernoulli distribution. I will attempt to find an approximate answer to the problem using CLT. (This is not an exact solution.)</p>
<p>Suppose we're interested in estimating the number of balls with number 1 written on them. Consider a draw a success if you draw a ball with 1 written on it. Then the actual rate of success is the number of balls with number 1 divided by 100. Let $p$ be this rate.</p>
<p>Let $X_n$ be the number of times 1 is drawn after $n$ drawings. Then CLT says that $\sqrt{n}\left(\frac{X_n}n - p\right)$ converges in distribution to a normal distribution with mean $0$ and variance $p(1 - p)$. A less formal statement is that the distribution of $\frac{X_n}n$ is roughly normal with mean $p$ and variance $\frac 1np(1 - p)$ for large $n$. So my approximation is that you estimate $p$ by $\hat p = \frac{X_n}n$, and also variance by $\hat p(1 - \hat p)$. Then you pick $n$ large enough such that the distribution $\mathcal{N}\left(0, \frac 1n\hat p(1 - \hat p)\right)$ has probability $0.95$ within the range $[-0.005, 0.005]$.</p>
|
245,880 | <p>A box contains 100 balls. Each ball has a number from 1 to 10. How many balls should I draw (ball is put back in box after drawing) to predict the number of balls for each number with 95% certainty.</p>
<p>A prediction is correct if for each of the 10 numbers, the number of balls with that number is correctly predicted.</p>
| Danica | 19,147 | <p>You're basically asking for confidence intervals on the probabilities of a <a href="http://en.wikipedia.org/wiki/Multinomial_distribution" rel="nofollow">multinomial distribution</a>.</p>
<p>Let $Q$ be the number of balls in the box (you said 100), of which there are distinct labels 1 to $m$ (you said 10). Let's say there are $N_i$ balls of type $i$, with $\sum_i N_i = Q$; we'll say $N = (N_1, \dots, N_m)$.</p>
<p>Say you draw $n$ balls and get a vector of counts for each type $x = (x_1, \dots, x_m)$. If $m = 3$ and you drew 10 with label 1, 2 with label 2, and 3 with label 3, you'd have $x = (10, 2, 3)$. $x$ is distributed according to $\text{Multinomial}(n, p_1, \dots, p_m)$, where $p_i = N_i / Q$.</p>
<p>After drawing $n$ balls, your maximum-likelihood prediction of the ball counts $\hat{N}$ is clearly $(Q x_1, \dots, Q x_m)$.</p>
<p>Now, you want to pick $n$ large enough such that your 95% confidence region on $N$ will contain only one integer for each component.</p>
<p>Confidence intervals on multinomials are actually somewhat complicated, because of the interactions between cells; <a href="http://tx.liberal.ntu.edu.tw/~purplewoo/literature/!Methodology/!Distribution_SampleSize/SimultConfIntervJSPI.pdf" rel="nofollow">here's</a> <a href="http://www.jstor.org/discover/10.2307/2288799?uid=3739864&uid=2129&uid=2&uid=70&uid=4&uid=3739256&sid=21101367407163" rel="nofollow">some</a> <a href="http://umr181.envt.fr/PubliLog/upload2/359.pdf" rel="nofollow">papers</a>. R has <a href="http://www.rforge.net/doc/packages/NCStats/gofCI.html" rel="nofollow">a function <code>gofCI</code> to do this</a> asymptotically, whose help page has some more references.</p>
|
77,466 | <p>There is a standard syllabus for a first graduate course in algebra. One teaches groups,
rings, fields, perhaps a little bit of Galois theory, perhaps a little bit of
category theory, perhaps a little bit of representation theory, all this a little bit superficially, to give an idea of the fundamental
algebraic structure to graduate students that will work in all parts of mathematics.</p>
<p>I have much more difficulties to see what to teach in a second, more advanced, course in algebra, whose student body is constituted of the grad students who like algebra, whatever they are eventually going to work in. Commutative algebra is excluded because in my department, as in many others, there is another course devoted to this specific subject.
But even so, there are so many loosely inter-related things (more category theory, more homological algebra, more representation theory, advanced theory of finite groups, study of classical groups, theory of groups defined by generators and relations, Brauer theory, etc.) one could think of that I find very difficult to
arbitrage between them. One is naturally pushed to give a course with no unity, which is not very pleasant. </p>
<p>Since the problem I experience has certainly been met by others, I'd like to know:
What did you or would you teach in such a course? What are the subjects that are
absolutely necessary to teach (if any)? How to give the course a backbone? What textbook to use?</p>
| David White | 11,540 | <p>As a graduate student in algebraic topology, but one who has taken many "second year" graduate courses in algebra, the one I think I would have enjoyed the most had it ever been offered (and the one which would have been most useful for me personally) would go something like this:</p>
<p>Textbook: <em><a href="http://books.google.com/books/about/An_introduction_to_homological_algebra.html?id=flm-dBXfZ_gC" rel="nofollow">An Introduction to Homological Algebra</a></em> - Charles A. Weibel</p>
<p>What to cover:</p>
<ul>
<li>Chain complexes and homology</li>
<li>Derived functors, Ext, and Tor</li>
<li>Spectral Sequences and/or homological dimension depending on which direction you want to go</li>
<li>Group Homology and Cohomology (I really enjoy Weibel's treatment of this)</li>
<li>Lie Algebra Homology and Cohomology (here you can bring in lots of related topics)</li>
<li>Last chapter and appendices on category theory and the derived category</li>
</ul>
<p>I agree with Richard Rast a bit that no one course can cover all the topics you like, but I think Weibel does a great job setting up the homology/cohomology framework using category theory and lots of homological algebra, applying this machinery to group cohomology and representation theory, and also bringing in classical groups. This seems to cover most of what you mention in your question.</p>
<p>A supplement I used when following this model on my own was <em>Representations and Cohomology</em> Parts I and II by D.J. Benson</p>
|
77,466 | <p>There is a standard syllabus for a first graduate course in algebra. One teaches groups,
rings, fields, perhaps a little bit of Galois theory, perhaps a little bit of
category theory, perhaps a little bit of representation theory, all this a little bit superficially, to give an idea of the fundamental
algebraic structure to graduate students that will work in all parts of mathematics.</p>
<p>I have much more difficulties to see what to teach in a second, more advanced, course in algebra, whose student body is constituted of the grad students who like algebra, whatever they are eventually going to work in. Commutative algebra is excluded because in my department, as in many others, there is another course devoted to this specific subject.
But even so, there are so many loosely inter-related things (more category theory, more homological algebra, more representation theory, advanced theory of finite groups, study of classical groups, theory of groups defined by generators and relations, Brauer theory, etc.) one could think of that I find very difficult to
arbitrage between them. One is naturally pushed to give a course with no unity, which is not very pleasant. </p>
<p>Since the problem I experience has certainly been met by others, I'd like to know:
What did you or would you teach in such a course? What are the subjects that are
absolutely necessary to teach (if any)? How to give the course a backbone? What textbook to use?</p>
| Joe Johnson | 10,206 | <p>When I took the second graduate algebra course we spent the first half on noncommutative ring theory. We covered topics like Jacobson radicals, artinian/noetherian rings, semi-simple algebras. We used Herstein and Jacobson's algebra books. The second half was all the ground work for complex representations of finite groups. No text was suggested, but it was a bit like "Linear Representations of Finite Groups" by Serre.</p>
|
1,016,585 | <p>I want to solve $$\int \frac{1}{\sqrt{x^2 - c}} dx\quad\quad\text{c is a constant}$$</p>
<p>How do I do this?</p>
<p>It looks like it is close to being an $\operatorname{arcsin}$?</p>
<hr>
<p>I would have thought I could just do:
$$\int \left(\sqrt{x^2 - c}\right)^{-\frac12}\, dx=\frac{2\sqrt{c+x^2}}{2x}\text{????}$$</p>
<p>But apparently not. </p>
| 5xum | 112,884 | <p><strong>HINT</strong>:
$$\frac{1}{\sqrt{x^2-c}} = \frac{1}{\sqrt{c}\sqrt{\frac{x^2}{c} - 1}} = \frac{1}{\sqrt{c}}\cdot \frac{1}{\sqrt{\left(\frac{x}{\sqrt{c}}\right)^2 - 1}}$$</p>
|
30,970 | <p>I try to filter sublists of a list which match a pattern. </p>
<pre><code>test = {{"String1", "a"}, {"String2", "b"}, {"String3",
"a"}, {"String4", "a"}};
</code></pre>
<p>The result should be:</p>
<pre><code>result = {{"String1", "a"}, {"String3", "a"}, {"String4", "a"}}
</code></pre>
<p>That means the first entry should be any String and the second should be "a".</p>
<p>I tried:</p>
<pre><code>Select[test, (# == {_, "a"}) &]
</code></pre>
<p>Which evaluates to {}. </p>
| halirutan | 187 | <p>If you use <code>Cases</code>, then you can give the pattern directly</p>
<pre><code>Cases[test, {_String, "a"}]
</code></pre>
<p>If you want to stick with <code>Select</code> you can do this as well but you have to transform your pattern match into a <em>test function</em></p>
<pre><code>Select[test, MatchQ[#, {_String, "a"}] &]
</code></pre>
|
1,782,796 | <p>Is there a particular name given to a matrix of <strong><em>m</em></strong> rows and <strong><em>n</em></strong> columns such that it must have one and only one 1 in each row and 0 elsewhere? For instance:</p>
<pre><code>0 1 0 0 0 0 0
0 0 0 0 1 0 0
0 1 0 0 0 0 0
0 0 0 0 0 0 1
0 0 1 0 0 0 0
</code></pre>
<p>Note: It may or may not be a square matrix.</p>
| L. Fermín | 523,546 | <p>Permutation Matrix
<a href="https://en.wikipedia.org/wiki/Permutation_matrix" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Permutation_matrix</a></p>
<p>It applies to squared matrices, but you can generalize for nonsquare matrices</p>
|
2,170,941 | <p>I know the derivative of position vector function $r(t)$ is velocity, and its unit vector is tangent vector, therefore, the derivative of tangent vector is the unit vector of acceleration and normal vector. <strong>Why isn't that the case?</strong></p>
| Daniel Robert-Nicoud | 60,713 | <p>By definition, acceleration is the variation of velocity in time, and thus the derivative of the velocity function.</p>
<p>Write $x(t)$ for the position, $v(t):=\dot x(t)$ for the velocity, and $a(t):=\dot v(t)$ for the acceleration. The direction vector is
$$d(t):=\frac{v(t)}{\|v(t)\|}\ .$$
If we differentiate it, we obtain
\begin{align}
\frac{d}{dt}d(t) =&\ \frac{\dot v(t)}{\|v(t)\|} -\frac{\langle\dot v(t),v(t)\rangle}{\|v(t)\|^3}v(t)\\
=&\ \frac{a(t)}{\|v(t)\|} -\frac{\langle a(t),v(t)\rangle}{\|v(t)\|^3}v(t)\ .
\end{align}
This is of course different from the normal vector in general.</p>
|
2,170,941 | <p>I know the derivative of position vector function $r(t)$ is velocity, and its unit vector is tangent vector, therefore, the derivative of tangent vector is the unit vector of acceleration and normal vector. <strong>Why isn't that the case?</strong></p>
| Doug M | 317,162 | <p>If speed $(\|r'(t)\|)$ is constant, then the acceleration must be normal to the direction of travel.</p>
<p>But if speed is variable then acceleration can be broken into a component parallel to the direction of travel and a component that is perpendicular to the direction of travel.</p>
|
2,211,461 | <p>Given that $f$ is an integrable function on $X$ and $\{E_k\}_{k=1}^\infty$ where each $E_k$ is a measurable set such that $\lim_{k\rightarrow \infty} \mu(E_k) = 0$</p>
<p>Can we show that $$\lim_{k\rightarrow \infty} \int_{E_k} fd\mu = 0$$ </p>
<p>I want to prove like this:
$$|\int_{E_k} fd\mu| \leq sup|f|\cdot \mu(E_k) \rightarrow 0 $$
The problem is when $|f| \rightarrow \infty$, I'm not sure if this is valid.</p>
<p>And if we remove the condition $f$ integrable and instead make f positive measurable, does the result still hold?</p>
| RiezFrechetKolmogorov | 390,584 | <p>Write integral over $E_k$ as integral of f*indicator ($E_k$) = $f_k$ , it should work</p>
|
1,381,017 | <p>Could you have me to find the ferivative of $$f(t)=\frac {1}{\rho}\log (1+\rho t)$$ with repsect to $t$?</p>
<p>And Is it
$$\lim_{t\to \infty} \frac {f'(t)}{t}=1?$$
<em>Update:</em>
Based on Hint of Surb:</p>
<p>$$f(t)'=\frac {1}{1+\rho t}$$</p>
<p>Then
$$\lim_{t\to \infty} \frac {f'(t)}{t}=0$$</p>
<p>Is it correct?</p>
| Peter | 82,961 | <p>Wolfram gives the following result :</p>
<p><a href="http://www.wolframalpha.com/input/?i=integral+sqrt%28sin%28x%29cos%28x%29%29" rel="nofollow">http://www.wolframalpha.com/input/?i=integral+sqrt%28sin%28x%29cos%28x%29%29</a></p>
<p>As you can see, there is no "nice" antiderivate to $\sqrt{sin(x)cos(x)}$</p>
|
27,896 | <p>How would I solve these differential equations? Thanks so much for the help!</p>
<p>$$P'_0(t) = \alpha P_1(t) - \beta P_0(t)$$
$$P'_1(t) = \beta P_0(t) - \alpha P_1(t)$$</p>
<p>We also know $P_0(t)+P_1(t)=1$</p>
| Community | -1 | <p>Note that from the equation you have $$P'_0(t) = \alpha P_1(t) - \beta P_0(t) = -P'_1(t)$$ which gives us $P'_0(t) + P'_1(t) = 0$ which gives us $P_0(t) + P_1(t) = c$. We are given that $c=1$. Use this now to eliminate one in terms of the other.</p>
<p>For instance, $P_1(t) = 1-P_0(t)$ and hence we get, $$P'_0(t) = \alpha (1-P_0(t)) - \beta P_0(t) \Rightarrow P'_0(t) = \alpha - (\alpha + \beta)P_0(t)$$</p>
<p>Let $Y_0(t) = e^{(\alpha + \beta)t}P_0(t) \Rightarrow Y'_0(t) = e^{(\alpha + \beta)t} \left[P'_0(t) + (\alpha + \beta) P_0(t) \right] = \alpha e^{(\alpha + \beta)t}$</p>
<p>Hence, $Y_0(t) = \frac{\alpha}{\alpha + \beta}e^{(\alpha + \beta)t} + k$ i.e. $$P_0(t) = \frac{\alpha}{\alpha + \beta} + k e^{-(\alpha+\beta)t}$$
$$P_1(t) = 1 - P_0(t) = \frac{\beta}{\alpha + \beta} - k e^{-(\alpha+\beta)t}$$</p>
|
800,802 | <p>Can someone give me a reference in which I can find the following result </p>
<p>Let $C$ be a curve, then $$g(C)=\frac{(n-1)(n-2)}{2}-s.$$</p>
<p>where $g=$ genus of $C,$ $n=$ degree of curve, $s=$ number of singular points.</p>
<p>Any help would be appreciated.</p>
| Marco | 149,606 | <p>You can try books on algebraic geometry, and look for the genus formula, or Plücker formulas. </p>
|
3,912,076 | <p>A particle of mass m slides down on the curve <span class="math-container">$z=1+\frac{x^2}{2}$</span> in the <span class="math-container">$xz$</span>-plane without friction under the action of constant gravity.Suppose <span class="math-container">$z$</span>-axis points vertically upward.Find the Lagrangian of given system.</p>
<p>The position vector of the particle is given by <span class="math-container">$\overline r=(x,0,1+\frac{x^2}{2})$</span>.
Then the velocity vector is given by
<span class="math-container">$ \dot {\overline r}=(\dot x,0,x\dot x)$</span>.
So the kinetic energy is given by
<span class="math-container">$T=\frac {m}{2}((\dot x)^2+x^2(\dot x)^2)$</span>. Now I am confused in potential energy. The potential energy is given by <span class="math-container">$V=mgz$</span> but what will be the sign of it? What does it mean that <span class="math-container">$z$</span>-axis points vertically upward?</p>
| Narasimham | 95,860 | <p>The Lagrangian is set up with Mechanics energy formulation and DE set up using Euler Lagrange Equation.</p>
<p>Angles <span class="math-container">$ \phi$</span> are tangent or normal rotations positive counterclockwise.</p>
<p>Parabola trough coordinates are taken wlog <span class="math-container">$ R=1, z_{min}=0 $</span>in parametric form:</p>
<p><span class="math-container">$$ (x,z)=( R \tan \phi, R/2 \tan^2 \phi)\tag1$$</span>
<span class="math-container">$$ ( \dot x, \dot z)= (R \sec^2 \phi\, \dot \phi,R \sec^2 \phi \tan \phi\;\dot\phi )\tag2$$</span></p>
<p>And the absolute value of velocity is</p>
<p><span class="math-container">$$R \sec^3 \phi \;\dot\phi \tag3$$</span></p>
<p><span class="math-container">$$ KE= \frac{m}{2}(R \sec^3 \phi \;\dot \phi)^2 \tag4 $$</span></p>
<p>PE is taken positive as shown and as usual <span class="math-container">$z>0$</span>. When PE increases KE decreases and vice-versa.</p>
<p><span class="math-container">$$ PE=mgz=mg \frac{R}{2}\dot\tan^2 \phi \tag5 $$</span></p>
<p><span class="math-container">$$\text{Lagrangian } = L =KE -PE $$</span></p>
<p><span class="math-container">$$ L= \dfrac{m}{2} \cdot (R^2 \sec^6 \phi\; \dot \phi ^2)- mg \dfrac{R}{2} \tan^2 \phi \tag6$$</span></p>
<p>Euler -Lagrange Equation ( after removing constants)</p>
<p><span class="math-container">$$ 6 \sec^5 \sec\phi \tan \phi\;{\dot{\phi}}^2-\frac{g}{R } 2 \tan {\phi} \sec^2\phi -\frac{d(2 \dot \phi)}{dt} \sec^6 \phi =0\tag7$$</span></p>
<p>Simplifying,</p>
<p><span class="math-container">$$ \dfrac{\ddot{\phi}}{\sin \phi \cos^2 \phi}+ \frac{g}{R}\cos \phi -3 \dot\phi^2=0 \tag8 $$</span></p>
<p>I have no closed form solution of the DE or any reference. So a numerical solution with the following boundary conditions is undertaken.</p>
<p><span class="math-container">$$( \phi_{max}= \frac{\pi}{3}, \dot\phi_{max}= 0, g=9.8, R=6) \tag9 $$</span></p>
<p>The parabolic trough profile and time oscillation of slope is plotted below:</p>
<p>The formula for shallow <span class="math-container">$\phi_{max}$</span> trough as a simple pendulum of constant <span class="math-container">$L$</span> for verification :</p>
<p><span class="math-container">$$ T= 2 \pi\sqrt{\dfrac{L}{g}} \tag {10} $$</span> is tallying alright.</p>
<p><a href="https://i.stack.imgur.com/75Tkf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/75Tkf.png" alt="enter image description here" /></a></p>
|
978,202 | <p>When I use $\sin x \sim x$ , the answer is $1$ , is the answer correct?</p>
| deepesh kumar singh | 185,074 | <p>in the limit $x \to \infty$, the numerator is bounded between $[-1,1]$ while the denominator approach infinity so the limit must be zero. </p>
|
3,787,576 | <p><span class="math-container">$$
\mbox{Prove}\quad
\int_{0}^{1}{\mathrm{d}x \over
\left(\,{x - 2}\,\right)\,
\sqrt[\Large 5]{\,x^{2}\,\left(\,{1 - x}\,\right)^{3}\,}\,}
=
-\,{2^{11/10}\,\pi \over \,\sqrt{\,{5 + \,\sqrt{\,{5}\,}}\,}\,}
$$</span></p>
<ul>
<li>Being honest I havent got a clue where to start. I dont think any obvious substitutions will help (<span class="math-container">$x \to 1-x, \frac{1}{x}, \sqrt{x},$</span> more).</li>
<li>The indefinite integral involves hypergeometric function so some miracle substitution has to work with the bounds I suspect.</li>
<li>Maybe gamma function is involved some how ??.</li>
</ul>
<p>If anyone has an idea and can provide help I would appreciate it.</p>
| Felix Marin | 85,343 | <p><span class="math-container">$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$</span>
<span class="math-container">$\ds{\bbox[10px,#ffd]{\int_{0}^{1}{\dd x \over \pars{x - 2}\root[\Large 5]{x^{2}\pars{1 - x}^{3}}}= -\,{2^{11/10}\,\pi \over \root{5 + \root{5}}}}:\ {\Large ?}}$</span>.</p>
<hr>
<span class="math-container">\begin{align}
&\bbox[10px,#ffd]{\int_{0}^{1}{\dd x \over \pars{x - 2}\root[\Large 5]{x^{2}\pars{1 - x}^{3}}}}
\,\,\,\stackrel{x\ \mapsto\ 1/x}{=}\,\,\,
\int_{\infty}^{1}{-\,\dd x/x^{2} \over \pars{1/x - 2}\pars{1/x}^{2/5}
\pars{1 - 1/x}^{3/5}}
\\[5mm] = &\
\int_{1}^{\infty}{\pars{x - 1}^{-3/5} \over 1 - 2x}\,\dd x
\,\,\,\stackrel{x + 1\ \mapsto\ x}{=}\,\,\,
-\,{1 \over 2}\int_{0}^{\infty}{x^{-3/5} \over x + 1/2}\,\dd x
\end{align}</span>
<hr>
Lets consider <span class="math-container">$\ds{\oint_{C}{z^{-3/5} \over z - 1/2}\,\dd z}$</span> where <span class="math-container">$\ds{C}$</span> is a key-hole contour which "takes care" of the principal branch of <span class="math-container">$\ds{\rule{0cm}{8mm}z^{-3/5}}$</span> ( with a branch-cut along
<span class="math-container">$\ds{\left(-\infty,0\right]}$</span> ). Namely,
<br><br>
<span class="math-container">\begin{align}
2\pi\ic\bracks{\pars{1 \over 2}^{-3/5}} & =
\int_{-\infty}^{0}{\pars{-x}^{-3/5}\expo{-3\pi\ic/5}
\over x - 1/2}\,\dd x + \int_{0}^{-\infty}{\pars{-x}^{-3/5}\expo{3\pi\ic/5}
\over x - 1/2}\,\dd x
\\[5mm] & =
-\expo{-3\pi\ic/5}\int_{0}^{\infty}{x^{-3/5} \over x + 1/2}\,\dd x +
\expo{3\pi\ic/5}\int_{0}^{\infty}{x^{-3/5} \over x + 1/2}\,\dd x
\\[5mm] & =
2\ic\sin\pars{3\pi \over 5}
\int_{0}^{\infty}{x^{-3/5} \over x + 1/2}\,\dd x
\\[5mm] \implies &
\bbox[10px,#ffd]{-\,{1 \over 2}\int_{0}^{\infty}{x^{-3/5} \over x + 1/2}\,\dd x =
- 2^{3/5}\pi\csc\pars{3\pi \over 5}} \approx -2.5034
\end{align}</span>
|
1,373,759 | <p>Given the natural number $n$,who is in the form $p^2 \cdot q^2$,with $p$,$q$ prime numbers.Also $φ(n)$ is given.Describe a fast algorithm(polynomial time) that calculates the $p$ and $q$.Apply your algorithm to calculate the $p$,$q$ when $n=34969$ and $φ(n)=29920$</p>
<p>I found this problem on a mathematical competition on cryptography.I tried to find a solution alone and in the internet and i didn't conclude anywhere, can we find a solution?</p>
| rpf | 256,834 | <p>Recall $\sqrt{gT} = \sqrt{g} * \sqrt{T}$.
As the book indicates, when calculating partial derivatives, we let only one variable change while the others are constant. </p>
<p>Say you're looking for $\frac{dn}{dT}$, then think of $n$ as </p>
<p>$n = constant * \sqrt{T}$. </p>
<p>Now calculating $\frac{dn}{dT}$ should be more clear. $\frac{dn}{d\sigma}$ follows the same method. </p>
|
3,669,266 | <p>The equation <span class="math-container">$(1-x)^n = x$</span> has a solution in <span class="math-container">$x' \in (0,1)$</span> and indeed the solution <span class="math-container">$x' \to 0$</span> as <span class="math-container">$n \to \infty$</span>. (Consider <span class="math-container">$f(x) = (1-x)^n$</span> noting that <span class="math-container">$f(0) = 1$</span> and <span class="math-container">$f(1) = 0$</span>. As <span class="math-container">$n$</span> increases, <span class="math-container">$f(x)$</span> is 'flat' in an increasingly-large neighbourhood around <span class="math-container">$x=1$</span>. Hence <span class="math-container">$f(x)$</span> crosses the line <span class="math-container">$y=x$</span> at a point that gets closer to <span class="math-container">$x=0$</span> as <span class="math-container">$n \to \infty$</span>.) </p>
<p><strong>What is the rate of convergence of <span class="math-container">$x' \to 0$</span> as <span class="math-container">$n \to \infty$</span>?</strong></p>
<p>What I have tried:</p>
<ul>
<li><p>I have obtained solutions for <span class="math-container">$x_n$</span> for <span class="math-container">$n = 1 \dots 17$</span> as shown below. (The ratio <span class="math-container">$x_{x+1}/x_n$</span> is increasing so it's not exponential decay (?))</p></li>
<li><p>I did a quick literature look for the bounds on the real-valued roots of a polynomial <a href="https://link.springer.com/chapter/10.1007/978-3-319-18275-9_10" rel="nofollow noreferrer">On Geometry of the Zeros of a Polynomial</a>. I was seeking an upper bound that decreases with <span class="math-container">$n$</span>. Unfortunately, the bounds in the literature (that I found) were all <span class="math-container">$1 + something$</span> which is unhelpful.</p></li>
</ul>
<p>Any suggestions most appreciated. Many thanks in advance.</p>
<p><a href="https://i.stack.imgur.com/nyuXD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nyuXD.png" alt="Solutions to <span class="math-container">$(1-x)^n = x$</span> obtained numerically"></a></p>
| Sangchul Lee | 9,340 | <p>Let <span class="math-container">$x_n$</span> be the unique solution of the equation <span class="math-container">$(1-x)^n=x$</span> on <span class="math-container">$(0, 1)$</span>.</p>
<ol>
<li><p>Let <span class="math-container">$f(x) = (1 - x)^n - x $</span>. Then by using the inequality <span class="math-container">$1-x \leq e^{-x}$</span>, we get
<span class="math-container">$$ f(0) = 1 \qquad\text{and}\qquad f\left(\frac{\log n}{n}\right) \leq e^{-\log n} - \frac{\log n}{n} = -\frac{\log(n/e)}{n}. $$</span>
So by the intermediate value theorem and the uniqueness of the zero, we conclude that
<span class="math-container">$$ 0 < x_n < \frac{\log n}{n} \qquad\text{for}\quad n \geq 3. $$</span></p></li>
<li><p>Write <span class="math-container">$y_n = \log x_n$</span> and <span class="math-container">$\epsilon_n=-\frac{\log(1-x_n)}{x_n}-1$</span>. Then
<span class="math-container">$$ n = \frac{\log x_n}{\log(1-x_n)} = -\frac{\log x_n}{x_n(1 + \epsilon_n)}. $$</span>
Rearranging this equation, we get
<span class="math-container">$$ -y_n e^{-y_n} = n(1 + \epsilon_n). $$</span>
This equation can be solved by using the <a href="https://en.wikipedia.org/wiki/Lambert_W_function" rel="nofollow noreferrer"><em>Lambert W-function</em></a> <span class="math-container">$W(x)$</span>. This is a well-studied special function defined as the inverse of <span class="math-container">$x \mapsto xe^{x}$</span>, Using this,
<span class="math-container">$$ y_n = -W(n(1 + \epsilon_n)). $$</span>
Then
<span class="math-container">$$ x_n = e^{y_n} = e^{-W(n(1 + \epsilon_n))} = \frac{W(n(1 + \epsilon_n))}{n(1 + \epsilon_n)} $$</span>
The previous step tells that <span class="math-container">$\epsilon_n = \mathcal{O}(\frac{\log n}{n})$</span>, and then a bit of effort shows that:
<span class="math-container">$$ x_n = \frac{W(n)}{n}\left(1+\mathcal{O}\left(\frac{\log n}{n}\right)\right). $$</span>
The following is a comparison between <span class="math-container">$x_n$</span> and <span class="math-container">$W(n)/n$</span>:</p>
<p><a href="https://i.stack.imgur.com/yxcg2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yxcg2.png" alt="Comparison"></a></p></li>
<li><p>If the use of a special function sounds less attractive, one may use the known bound
<span class="math-container">$$ W(x) = \log x - \log\log x + \mathcal{O}\left(\frac{\log\log x}{\log x}\right) $$</span>
to produce a bound involving only elementary functions:
<span class="math-container">$$ x_n = \frac{\log n - \log\log n}{n}\left(1 + \mathcal{O}\left(\frac{\log\log n}{(\log n)^2}\right) \right) $$</span></p></li>
</ol>
|
8,741 | <p>Here is a topic in the vein of <a href="https://mathoverflow.net/questions/1890/describe-a-topic-in-one-sentence" title="Describe a topic in one sentence"> Describe a topic in one sentence</a> and <a href="https://mathoverflow.net/questions/4994/fundamental-examples" > Fundamental examples </a> : imagine that you are trying to explain and justify a mathematical theory T to a skeptical mathematician who thinks T is just some sort of abstract nonsense for its own sake. The ideal solution consists
of a problem P which can be stated and understood without knowing anything about T, but which is difficult (or impossible, even better) to solve without T, and easier (or almost-trivial, even better) to solve with the help of T. What should be avoided is an example where T is "superimposed", e.g. when T is a model for some physical phenomenon, because there is always something arbitrary about the choice of a specific model. </p>
<p>A classical example is Galois theory for solving polynomial equations. </p>
<p>Any examples for homological algebra ? For Fourier analysis ? For category theory ?</p>
| José Figueroa-O'Farrill | 394 | <p>The <a href="http://en.wikipedia.org/wiki/Poincar%C3%A9_conjecture" rel="nofollow">Poincaré conjecture</a> is an example of a purely topological statement which apparently cannot be proved only by topological means.</p>
|
8,741 | <p>Here is a topic in the vein of <a href="https://mathoverflow.net/questions/1890/describe-a-topic-in-one-sentence" title="Describe a topic in one sentence"> Describe a topic in one sentence</a> and <a href="https://mathoverflow.net/questions/4994/fundamental-examples" > Fundamental examples </a> : imagine that you are trying to explain and justify a mathematical theory T to a skeptical mathematician who thinks T is just some sort of abstract nonsense for its own sake. The ideal solution consists
of a problem P which can be stated and understood without knowing anything about T, but which is difficult (or impossible, even better) to solve without T, and easier (or almost-trivial, even better) to solve with the help of T. What should be avoided is an example where T is "superimposed", e.g. when T is a model for some physical phenomenon, because there is always something arbitrary about the choice of a specific model. </p>
<p>A classical example is Galois theory for solving polynomial equations. </p>
<p>Any examples for homological algebra ? For Fourier analysis ? For category theory ?</p>
| José Figueroa-O'Farrill | 394 | <p>Many geometric problems cannot be solved without hard analysis. Perhaps the best known example is the <a href="http://en.wikipedia.org/wiki/Calabi_conjecture" rel="nofollow">Calabi Conjecture</a> proved by Yau.</p>
|
8,741 | <p>Here is a topic in the vein of <a href="https://mathoverflow.net/questions/1890/describe-a-topic-in-one-sentence" title="Describe a topic in one sentence"> Describe a topic in one sentence</a> and <a href="https://mathoverflow.net/questions/4994/fundamental-examples" > Fundamental examples </a> : imagine that you are trying to explain and justify a mathematical theory T to a skeptical mathematician who thinks T is just some sort of abstract nonsense for its own sake. The ideal solution consists
of a problem P which can be stated and understood without knowing anything about T, but which is difficult (or impossible, even better) to solve without T, and easier (or almost-trivial, even better) to solve with the help of T. What should be avoided is an example where T is "superimposed", e.g. when T is a model for some physical phenomenon, because there is always something arbitrary about the choice of a specific model. </p>
<p>A classical example is Galois theory for solving polynomial equations. </p>
<p>Any examples for homological algebra ? For Fourier analysis ? For category theory ?</p>
| Darsh Ranjan | 302 | <p>Mathematical logic can be motivated by other areas of math in at least two different ways: </p>
<p>[1] It allows you to formulate (and prove) results about the unsolvability of certain problems. These are obviously essential, since they tell you that you shouldn't spend too much time trying to solve those problems, which are often very natural problems. </p>
<p>For example, as a group theorist, you might often want to know whether two particular groups given by generators and relations are isomorphic. It would be nice to have some set of tools that allowed you to solve the problem mechanically, but no such tools exist (Novikov's Theorem). </p>
<p>Or, as a number theorist, you might wish for a set of tools allowing you to decide effectively whether a given polynomial equation has integer solutions. This is ruled out by the Davis-Putnam-Robinson-Matiyasevich Theorem. </p>
<p>[2] It can give you easier proofs of theorems in seemingly unrelated subfields. I hope somebody can provide/confirm examples here...for example, I think Gödel's Compactness Theorem gives some mileage in algebraic geometry (Nullstellensatz?), and nonstandard analysis can simplify a number of proofs (Tychonoff's Theorem?). (Although nonstandard analysis isn't exactly mathematical logic, the fact that proofs of standard results using nonstandard analysis can be trusted is a theorem of logic.) </p>
|
8,741 | <p>Here is a topic in the vein of <a href="https://mathoverflow.net/questions/1890/describe-a-topic-in-one-sentence" title="Describe a topic in one sentence"> Describe a topic in one sentence</a> and <a href="https://mathoverflow.net/questions/4994/fundamental-examples" > Fundamental examples </a> : imagine that you are trying to explain and justify a mathematical theory T to a skeptical mathematician who thinks T is just some sort of abstract nonsense for its own sake. The ideal solution consists
of a problem P which can be stated and understood without knowing anything about T, but which is difficult (or impossible, even better) to solve without T, and easier (or almost-trivial, even better) to solve with the help of T. What should be avoided is an example where T is "superimposed", e.g. when T is a model for some physical phenomenon, because there is always something arbitrary about the choice of a specific model. </p>
<p>A classical example is Galois theory for solving polynomial equations. </p>
<p>Any examples for homological algebra ? For Fourier analysis ? For category theory ?</p>
| Douglas Zare | 2,954 | <p><strong>Problem:</strong> You need to multiply large numbers, with $10^9$ digits (or take products of power series). Your computer doesn't have the ability to do $10^{18}$ calculations. </p>
<p><strong>Solution:</strong> Recognize multiplication of a power series as a convolution. Take a discrete Fourier transform of the digit sequences, multiply, and apply the inverse Fourier transform. Then perform the carries. This should take under $10^{12}$ calculations. The Fast Fourier Transform takes about $n \log n$ calculations.</p>
<p>The point is not that this is a fast algorithm or a clever trick. It's that you start out with a basic question about <em>integers</em> you can explain to someone comfortable with grade school math, and you end up dealing with complex or at least real numbers, characters of $\mathbb Z/n$, and properties of convolutions. </p>
|
1,072,427 | <p>Prove using contradiction that any prime number greater than $3$ is of the form $6n \pm 1$.</p>
<p>Thanks for any help</p>
| turkeyhundt | 115,823 | <p>Can you divide $(6n+2)$ by anything? What about $(6n+3)$?</p>
|
128,935 | <p>Hans Hahn is often credited with creating the modern theory of ordered algebraic systems with the publication of his paper <em>Über die nichtarchimedischen Grössensysteme</em> (Sitzungsberichte der Kaiserlichen Akademie der Wissenschaften, Wien, Mathematisch - Naturwissenschaftliche Klasse 116 (Abteilung IIa), 1907, pp. 601-655). Among the results established therein is Hahn’s Embedding Theorem, which is generally regarded to be the deepest result in the theory of ordered abelian groups. The following are two of its familiar formulations:</p>
<p>(i) Every ordered abelian group is isomorphic to a subgroup of a Hahn Group.</p>
<p>(ii) Every ordered abelian group G is isomorphic to a subgroup G’ of a Hahn Group, the latter of which is an Archimedean extension of G’.</p>
<p>(For definitions and modern proofs, see: A. H. Clifford [1954], <em>Note on Hahn’s theorem on ordered Abelian groups</em>, Proceedings of the American Mathematical Society, vol. 5, pp. 860–863; Laszlo Fuchs [1963], <em>Partially ordered algebraic systems</em>, Pergamon Press.)</p>
<p>Hahn’s proofs (and all subsequent proofs) of (i) and (ii) make use of the Axiom of Choice or some ZF-equivalent thereof. Moreover, while writing before the complete formulation of ZF (Foundation and Replacement had yet to be included), Hahn further maintained that he believed his embedding theorem could not be established without the well-ordering theorem, which had been established by Zermelo using Choice (and was subsequently shown to be equivalent in ZF to Choice). To my knowledge, this essentially amounts to the earliest conjecture that an algebraic result is equivalent (in ZF) to an assertion equivalent to the Axiom of Choice. Surprisingly, both Hahn’s use of Choice and his conjecture are overlooked in the well-known histories of the Axiom of Choice, including the excellent one by Gregory Moore. Apparently without knowledge of Hahn’s conjecture, D. Gluschankof (implicitly) asked if (i) is equivalent to the Axiom of Choice in ZF in his paper <em>The Hahn Representation Theorem for</em> ℓ-<em>Groups in ZFA</em>, (The Journal of Symbolic Logic, Vol. 65, No. 2 (Jun., 2000), pp. 519-52). However, Gluschankof did not answer the question and, unfortunately, died shortly after raising it. R. Downey and R. Solomon (in their paper <em>Reverse Mathematics, Archimedean Classes, and Hahn’s Theorem</em>) establish a countable version of Hahn’s theorem without using Choice, but their technique does not extend to the general case.</p>
<p>This leads to my two questions:</p>
<ol>
<li><p>Has anyone established or refuted Hahn’s Conjecture?</p></li>
<li><p>Assuming (as I suspect) the answer to 1 is “no”, is the status of Hahn’s Conjecture the longest standing open question in Set Theory?</p></li>
</ol>
<p><strong>Amendment (Response to request for references)</strong></p>
<p>Asaf: There are numerous proofs of Hahn’s Embedding Theorem in the literature besides the especially simple one due to Clifford. One proof is on pp. 56-60 of Laszlo Fuchs’s <em>Partially ordered algebraic systems</em> [1963] Pergamon Press. On page 60 of the just-said work there are also references to several other proofs including those of Clifford, Banaschewski, Gravett, Ribenboim and Conrad. Another proof, closely related to the one in Fuchs (including all preliminaries) can be found in Chapter 1 of Norman Alling’s <em>Foundations of Analysis over Surreal Number Fields</em>, North-Holland, 1987. Another very nice treatment, including all preliminaries, can be found in Chapter 1 of H. Garth Dales and W. H. Woodin’s <em>Super-Real Fields</em>, Oxford, 1996.There is also an interesting proof in Jean Esterle's <em>Remarques sur les théorèmes d'immersion de Hahn et Hausdorff et sur les corps de séries formelles</em>, Quarterly Journal of Mathematics 51 (2000), pp. 2011-2019. </p>
<p>For a now slightly dated history of Hahn’s Theorem, see my:</p>
<p><em>Hahn’s Über die nichtarchimedischen Grössensysteme and the Origins of the Modern Theory of Magnitudes and Numbers to Measure Them</em>, in <em>From Dedekind to Gödel: Essays on the Development of the Foundations of Mathematics</em>, edited by Jaakko Hintikka, Kluwer Academic Publishers, 1995, pp. 165-213. (A typed version of the paper can be downloaded from my website: <a href="http://www.ohio.edu/people/ehrlich/">http://www.ohio.edu/people/ehrlich/</a>)</p>
<p>Finally, I note that the earliest, but largely forgotten, altogether modern proof of Hahn’s theorem may be found on pp. 194-207 of Felix Hausdorff’s, <em>Grundzüge der Mengenlehre</em>, Leipzig [1914]. It was the lack of familiarity with Hausdorff’s proof and the need for a concise modern proof that led to the plethora of proofs in the 1950s. </p>
| Asaf Karagila | 7,206 | <p>I don't know the answer to (1), and would be glad to give it some thought later this week. Regardless to (1) the answer to (2) is semi-negative.</p>
<p>There are two conjectures which seem to be slightly older (although not by much) than this of Hahn, although both were not explicitly stated as conjectures but since they went without proof, and sometimes there were disputes over the truth values of these statements -- so I prefer to think about them as conjectures (and in modern terms, I believe that would be right, too).</p>
<ol>
<li><p>In 1905 Schoenflies asserted that the statement "There is no decreasing sequence of cardinals" implies the axiom of choice. This is still open, although in 1908 Zermelo rejected the claim.</p></li>
<li><p>In 1902 Beppo Levi introduced the Partition Principle stating that if $S$ is a partition of $A$ then $|S|\leq|A|$. Although he coined this principle to argue against its use by Bernstein, the latter rejected the criticism and claimed that this is one of the more important principles of set theory.</p>
<p>Of course all this was before Zermelo even introduced the axiom of choice in 1904. But in 1906 in an unpublished manuscript Russell claimed that AC is equivalent to PP, but this was without proof and the conjecture is still open to this very day.</p>
<p>More accurately, however, Russell proved that PP follows from another principle, and claimed the reverse implication holds as well (without proof), and in 1908 proved that the other principle is equivalent to the axiom of choice.</p></li>
</ol>
<p>So we have two conjectures from 1906, regarding the equivalence of two statements to the axiom of choice. Neither has been proven yet, and there has been very little progress (to my knowledge) in obtaining any concrete answer. My opinion is that we lack the proper tools to handle the complex structure of cardinals in models without choice. But I digress.</p>
<p>All the information I gave here is taken from the following paper:</p>
<blockquote>
<p>Bernhard Banaschewski, Gregory H. Moore, <strong>The dual Cantor-Bernstein theorem and the partition principle</strong>, <em>Notre Dame J. Formal Logic</em> <strong>31 (3)</strong>, (1990), 375–381.</p>
</blockquote>
<p>Along with the many hours that I have spent reading and searching results related to these topics (which made me rather certain that little progress has been made on these problems).</p>
<hr>
|
128,935 | <p>Hans Hahn is often credited with creating the modern theory of ordered algebraic systems with the publication of his paper <em>Über die nichtarchimedischen Grössensysteme</em> (Sitzungsberichte der Kaiserlichen Akademie der Wissenschaften, Wien, Mathematisch - Naturwissenschaftliche Klasse 116 (Abteilung IIa), 1907, pp. 601-655). Among the results established therein is Hahn’s Embedding Theorem, which is generally regarded to be the deepest result in the theory of ordered abelian groups. The following are two of its familiar formulations:</p>
<p>(i) Every ordered abelian group is isomorphic to a subgroup of a Hahn Group.</p>
<p>(ii) Every ordered abelian group G is isomorphic to a subgroup G’ of a Hahn Group, the latter of which is an Archimedean extension of G’.</p>
<p>(For definitions and modern proofs, see: A. H. Clifford [1954], <em>Note on Hahn’s theorem on ordered Abelian groups</em>, Proceedings of the American Mathematical Society, vol. 5, pp. 860–863; Laszlo Fuchs [1963], <em>Partially ordered algebraic systems</em>, Pergamon Press.)</p>
<p>Hahn’s proofs (and all subsequent proofs) of (i) and (ii) make use of the Axiom of Choice or some ZF-equivalent thereof. Moreover, while writing before the complete formulation of ZF (Foundation and Replacement had yet to be included), Hahn further maintained that he believed his embedding theorem could not be established without the well-ordering theorem, which had been established by Zermelo using Choice (and was subsequently shown to be equivalent in ZF to Choice). To my knowledge, this essentially amounts to the earliest conjecture that an algebraic result is equivalent (in ZF) to an assertion equivalent to the Axiom of Choice. Surprisingly, both Hahn’s use of Choice and his conjecture are overlooked in the well-known histories of the Axiom of Choice, including the excellent one by Gregory Moore. Apparently without knowledge of Hahn’s conjecture, D. Gluschankof (implicitly) asked if (i) is equivalent to the Axiom of Choice in ZF in his paper <em>The Hahn Representation Theorem for</em> ℓ-<em>Groups in ZFA</em>, (The Journal of Symbolic Logic, Vol. 65, No. 2 (Jun., 2000), pp. 519-52). However, Gluschankof did not answer the question and, unfortunately, died shortly after raising it. R. Downey and R. Solomon (in their paper <em>Reverse Mathematics, Archimedean Classes, and Hahn’s Theorem</em>) establish a countable version of Hahn’s theorem without using Choice, but their technique does not extend to the general case.</p>
<p>This leads to my two questions:</p>
<ol>
<li><p>Has anyone established or refuted Hahn’s Conjecture?</p></li>
<li><p>Assuming (as I suspect) the answer to 1 is “no”, is the status of Hahn’s Conjecture the longest standing open question in Set Theory?</p></li>
</ol>
<p><strong>Amendment (Response to request for references)</strong></p>
<p>Asaf: There are numerous proofs of Hahn’s Embedding Theorem in the literature besides the especially simple one due to Clifford. One proof is on pp. 56-60 of Laszlo Fuchs’s <em>Partially ordered algebraic systems</em> [1963] Pergamon Press. On page 60 of the just-said work there are also references to several other proofs including those of Clifford, Banaschewski, Gravett, Ribenboim and Conrad. Another proof, closely related to the one in Fuchs (including all preliminaries) can be found in Chapter 1 of Norman Alling’s <em>Foundations of Analysis over Surreal Number Fields</em>, North-Holland, 1987. Another very nice treatment, including all preliminaries, can be found in Chapter 1 of H. Garth Dales and W. H. Woodin’s <em>Super-Real Fields</em>, Oxford, 1996.There is also an interesting proof in Jean Esterle's <em>Remarques sur les théorèmes d'immersion de Hahn et Hausdorff et sur les corps de séries formelles</em>, Quarterly Journal of Mathematics 51 (2000), pp. 2011-2019. </p>
<p>For a now slightly dated history of Hahn’s Theorem, see my:</p>
<p><em>Hahn’s Über die nichtarchimedischen Grössensysteme and the Origins of the Modern Theory of Magnitudes and Numbers to Measure Them</em>, in <em>From Dedekind to Gödel: Essays on the Development of the Foundations of Mathematics</em>, edited by Jaakko Hintikka, Kluwer Academic Publishers, 1995, pp. 165-213. (A typed version of the paper can be downloaded from my website: <a href="http://www.ohio.edu/people/ehrlich/">http://www.ohio.edu/people/ehrlich/</a>)</p>
<p>Finally, I note that the earliest, but largely forgotten, altogether modern proof of Hahn’s theorem may be found on pp. 194-207 of Felix Hausdorff’s, <em>Grundzüge der Mengenlehre</em>, Leipzig [1914]. It was the lack of familiarity with Hausdorff’s proof and the need for a concise modern proof that led to the plethora of proofs in the 1950s. </p>
| Vladimir Kanovei | 34,207 | <p>Regarding </p>
<ol>
<li>In 1902 Beppo Levi introduced the Partition Principle stating that if S is a partition of A then |S|≤|A|. Although he coined this principle to argue against its use by Bernstein, the latter rejected the criticism and claimed that this is one of the more important principles of set theory. </li>
</ol>
<p>This is a trivial thm of ZFC, but definitely not provable in ZF+DC, in particular, in the case A = the reals and S = the Vitali partition R/Q, as Sierpinski established in 1920s that any injection R/Q\to R implies a non-measurable set of reals, whose existence is not provable in ZF+DC by Solovay. </p>
<p>Regarding the oldest unsolved concrete problem in set theory, this is most likely one of Hausdorff's questions on pantachies, namely, </p>
<p>is there a pantachy containing no $(\omega_1,\omega_1^\ast)$ gap
(H, Untersuchungen uber Ordnungstypen, V, 1907, May 05)</p>
<p>I underline that this is a <em>concrete</em> problem of existence of a certain mathematical object, rather than an abstract question on interrelations of different forms of AC spread over the whole set universe.</p>
<p>See more on this problem in my survey "Gaps in partially ordered sets" (12.1) in Hausdorff's Gesammelte Werke Band 1A, Springer 2013, 367-405, with additional references to Goedel (who rediscovered the problem with full ignorance of the Hausdorff's original formulation), Solovay, Kanamori. </p>
|
443,013 | <p>If a linear map $A:\mathbb{R}^m\rightarrow \mathbb{R}^n$ is injective, then there exists $c>0$ such that $|Ax|\geq c|x|$ for all $x\in\mathbb{R}^m$</p>
<p>Could someone give any solution or hint?</p>
<p>Thanks.</p>
| copper.hat | 27,978 | <p>Since $\ker A = \ker (A^T A)$, we see that $A$ is injective iff $A A^T$ is invertible.
It is easy to see that $A^T A$ is symmetric and positive semi-definite. Since $A^T A$ is invertible, we see that $A^T A$ is positive definite, hence $A^T A \ge \lambda I$, where $\lambda$ is the smallest eigenvalue of $A^TA$ (and $\lambda >0$).</p>
<p>Consequently we have $\|Ax\|^2 = \langle A x, Ax \rangle = \langle x, A^T Ax \rangle \ge \lambda \langle x, x \rangle = \lambda \|x\|^2$. Setting $c = \sqrt{\lambda}$ gives the desired result.</p>
|
942,738 | <p>First of all I must state that I am not a mathematician, so please correct me if I use wrong terminology.</p>
<p>I am building a web application which needs to calculate the rating for each entity based on both the quantity and score of the reviews for that entity.</p>
<p>In other words, I don't want to just calculate the rating based on the average score as that would make an entity with one hundred 9 score (review score can be from 0 to 10) reviews rate lower than an entity with only one 9.5 score review.</p>
<p>I need an algorithm to calculate rating and add rating "weight" to the final rating based on how many reviews the entity has, so that for instance in the above example the entity with 100 9 score reviews would get a rating that is higher than the entity with only one 9.5 score review. In other words, the final entity rating score will be based on the "relationship" between quality and quantity.</p>
<p>There is another important thing to note: an entity can not have a rating higher than 10, so the rating "weight" added by the quantity can not be linear.</p>
<p>In the algorithm we can use any data about the reviews/rating, that is individual review score, total number of reviews, sum of all reviews, number of good reviews (score 8 or higher) so far, etc, in each iteration of the rating calculation process.</p>
<p>Any kind of help or info regarding this would be appreciated.
Thank you.</p>
| Marc Bogaerts | 118,955 | <p>What you can do, for instance, is take the rate of reviews (w weighted mean), divide it by two (to reduce the scoring to a scale of <span class="math-container">$[0,5]$</span> and add this value to <span class="math-container">$5(1-e^{-q})$</span>. So the formula becomes <span class="math-container">$$\text{score}=5p/10+5(1-e^{-q/Q})$$</span> where <span class="math-container">$p$</span> is the review rating and <span class="math-container">$q$</span> is the number of ratings, and you chose for <span class="math-container">$Q$</span> an appropriate number that shows what importance you attach to the notion "quantity."</p>
<p>An example: An item has <span class="math-container">$3$</span> times a revision score of <span class="math-container">$6$</span> and <span class="math-container">$2$</span> times a revision score of <span class="math-container">$7$</span>.</p>
<p>Then <span class="math-container">$p=(3⋅6+2⋅7)/5=6.4$</span></p>
<p>If we take <span class="math-container">$Q=10$</span>, then <span class="math-container">$5(1-e^{-5/10})\approx 3.88$</span>, so the total score is <span class="math-container">$3⋅2+3⋅9=7.1$</span> rounded <span class="math-container">$7$</span>.</p>
<p>On the other hand, if somebody has <span class="math-container">$20$</span> scorings of <span class="math-container">$6$</span>, then <span class="math-container">$p=6$</span> and <span class="math-container">$5(1-e^{-20/10})\approx 4.58$</span>, so the final score is <span class="math-container">$3+4.6$</span> rounded giving <span class="math-container">$8$</span>.</p>
<p>The choice of <span class="math-container">$Q$</span> depends on what you call "few," "moderate," "many."</p>
<p>As a rule of thumb, consider a value <span class="math-container">$M$</span> that you consider "moderate", and take <span class="math-container">$Q=-M/\ln(1/2)\approx 1.44M$</span>.</p>
<p>So if you think <span class="math-container">$100$</span> is a moderate value, then take <span class="math-container">$Q=144$</span>.</p>
<p>Finally, you can also replace the equal weight on quantity and quality by a skewed one so that the final formula becomes:<span class="math-container">$$\text{score}=Pp+10(1-P)(1-e^{-q/Q}))$$</span> where <span class="math-container">$P\in [0,1]$</span> (in the original formula we had <span class="math-container">$P=0.5$</span>).</p>
|
3,808,111 | <p>Let <span class="math-container">$(X_n)_{n \geq 1}$</span> be a sequence of pairwise independent random variables such that :</p>
<p><span class="math-container">$$\sum_{n=1}^{\infty} n^{-1} P\left\{\max _{1 \leq m \leq n}\left|\sum_{k=1}^{m}\left(X_{k}-E X_{k}\right)\right|>\varepsilon n\right\}<\infty$$</span></p>
<p>show that <span class="math-container">$n^{-1} \sum_{k=1}^{n}\left(X_{k}-E X_{k}\right) \rightarrow 0$</span> almost surely.</p>
<p>I'm fairly certain Borel Cantelli Lemma for pairwise independent random variables is to be used here but I dont know how to get rid of the <span class="math-container">$n^{-1}$</span> inside the series.</p>
| Fawkes4494d3 | 260,674 | <p>Take <span class="math-container">$f(x)=x^4$</span>. Note that <span class="math-container">$f''(0)=0$</span> and <span class="math-container">$0$</span> is a point of global minima of <span class="math-container">$f$</span>.</p>
<p>The <a href="https://en.wikipedia.org/wiki/Derivative_test#Higher-order_derivative_test" rel="nofollow noreferrer">actual statement (given in this link)</a> which is most helpful for you to determine whether it is an inflection point or one of minima, maxima requires you to find as many derivatives as you can and then depends on the sign of the even-order derivative.</p>
<p>The method of classification of the critical point, given in the link works so far as a non-zero derivative is reached at the critical point <span class="math-container">$c$</span> you want to investigate. (A non-trivial function for which every derivative at <span class="math-container">$x=c_0$</span> is <span class="math-container">$0$</span>, which still achieves a global minima at <span class="math-container">$x=c_0$</span> is given by <span class="math-container">$$f(x)=\left\{\begin{matrix} \exp((x-c_0)^{-2}) & \text{if } x\ne c_0 \\ 0 & \text{if } x=c_0\end{matrix}\right\}$$</span> and this is a very useful/popular function with this property.)</p>
|
3,069,369 | <p>I know how to perform polynomial regression. But is there any method to use for estimating the degree of the polynomial that is best suited? Some kind of meta-regression.</p>
<p>With best suited I mean the grade that has the highest probability of being the true degree of the source for the data.</p>
<p>For example, if we look at this picture we can easily "see" that a polynomial of degree 4 would fit nicely:</p>
<p><a href="https://i.stack.imgur.com/ZOIbY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZOIbY.png" alt="enter image description here"></a></p>
<p>A more generalized question is if there is any method to determine if the source is polynomial at all or if it is exponential or something else.</p>
| Bill Dubuque | 242 | <p><strong>Hint</strong> <span class="math-container">$\ $</span> It is <span class="math-container">$\ (2x^4-7x^2+5) + (x^3-x)\, $</span> and both clearly have roots <span class="math-container">$\,x = \pm 1$</span> </p>
|
3,718,255 | <p>So i solved the first part of this question, that there's 588 ways for choosing 5 digit integers, but i can't solve "Find the number of integers in (i) which satisfies "if there's 8, it's followed by 6"
i tried to solve it using different cases:</p>
<p>Case 1) there's two 8s and two 6s => (3 * 3!)/2! - 1</p>
<p>Case 2) there's two 6s and 8 => 4 * 4!-3*3!</p>
<p>Case 3) 8 and 6 => 4!-3!</p>
<p>the sum was 104 but i don't think that's right, should i make another case for no 8s? or is there an easier way.</p>
| user-492177 | 492,177 | <p>If there is no <span class="math-container">$8$</span> then</p>
<p>Total <span class="math-container">$5$</span> digit distinct numbers=<span class="math-container">$\frac{5!}{2!}-\frac{4!}{2!}$</span>=<span class="math-container">$60-12=48$</span></p>
<p>If there is one <span class="math-container">$8$</span>, it must be followed by <span class="math-container">$6$</span>. We treat the pair as one. Notice that <span class="math-container">$8$</span> can't be unit place digit.So we can place <span class="math-container">$8$</span> in the first <span class="math-container">$4$</span> positions followed by <span class="math-container">$6$</span>.</p>
<p>Consider <span class="math-container">$8$</span> being not in the first position. Then for each of the other three position, we have <span class="math-container">$^4P_3-^3P_2=18$</span>(subtracting numbers starting with <span class="math-container">$0$</span>). So total distint <span class="math-container">$5$</span> digit numbers containing only one <span class="math-container">$8$</span> and not starting with <span class="math-container">$8=18×3=54$</span></p>
<p>Now if such a number starts with <span class="math-container">$8$</span>, then distinct numbers<span class="math-container">$=^4P_3=24$</span></p>
<p>So with one <span class="math-container">$8$</span>, there are <span class="math-container">$54+24=78$</span></p>
<p>If there are two <span class="math-container">$8s$</span> , there must be two <span class="math-container">$6s$</span> , then the two groups of <span class="math-container">$86$</span> are treated together.</p>
<p>Consider the remaining digit to be placed at the left, middle of the two groups and at the rightmost. Now there are 2 different choices for the leftmost, 3 for the middle, and 3 for the right.So with two <span class="math-container">$8$</span> , total numbers<span class="math-container">$=8$</span></p>
<p>Hence total numbers <span class="math-container">$=48+78+8=134$</span></p>
|
607,324 | <blockquote>
<p><span class="math-container">$d \in \mathbb{Z}$</span> is a square-free integer (<span class="math-container">$d \ne 1$</span>, and <span class="math-container">$d$</span> has no factors of the form <span class="math-container">$c^2$</span> except <span class="math-container">$c = \pm 1$</span>), and let <span class="math-container">$R=\mathbb{Z}[\sqrt{d}]= \{ a+b\sqrt{d} \mid a,b \in \mathbb{Z} \}$</span>. Prove that every nonzero prime ideal <span class="math-container">$P \subset R$</span> is a maximal ideal.</p>
</blockquote>
<p>I have a possible outline which I think is good enough to follow.</p>
<p>I think that we need to first prove that every ideal <span class="math-container">$I \subset R$</span> is finitely generated. </p>
<p>So if <span class="math-container">$I$</span> is non-zero, then <span class="math-container">$I \cap \mathbb{Z}$</span> is a non-zero ideal in <span class="math-container">$\mathbb{Z}$</span>. </p>
<p>Then I need to find <span class="math-container">$I \cap \mathbb{Z} = \{ xa \mid a \in \mathbb{Z} \}$</span> for some <span class="math-container">$x \in \mathbb{Z}$</span>. That way if I let <span class="math-container">$J$</span> be the set of all integers <span class="math-container">$b$</span> such that <span class="math-container">$a+b\sqrt{d} \in I$</span> for some <span class="math-container">$a\in \mathbb{Z}$</span>, then if there exists a integer <span class="math-container">$y$</span> such that <span class="math-container">$J=\{ yt \mid t\in \mathbb{Z} \}$</span>, then there must exist <span class="math-container">$s \in \mathbb{Z}$</span> such that <span class="math-container">$s+y\sqrt{d} \in I$</span>. </p>
<p>Then all I need to show is that <span class="math-container">$I = ( x,s+y\sqrt{d} )$</span>. </p>
<p>Now I need to derive that the factor ring <span class="math-container">$R / P$</span> is a finite ring without zero divisors, also finite, then since every finite integral domain is a field, every prime ideal <span class="math-container">$P \subset R$</span> is a maximal ideal, then I'll be done.</p>
| Brian Rushton | 51,970 | <p>Yes, you idea is right. When doing formal power series with matrices, you usually think of the constants as multiples of the identity.</p>
<p>One reason is that the ring of matrices is an $\mathbb{R}$ algebra, meaning that it contains a copy of the real numbers (i.e multiples of the identity) which commutes with all matrices. This allows us to think of the reals as a subset of the set of matrices.</p>
|
4,415,254 | <p>When is <span class="math-container">$n!>x^n$</span>? assuming that x is a fixed positive. I know we can take the log of both sides and use the following formula:</p>
<p><span class="math-container">$$n\log(x) = \log(x^n) < \log(n!) = \sum_{i = 1}^n\log(i)$$</span>.
But this still gives me difficulty trying to find a specific N that makes the right side larger. Is there another formula I should be using?</p>
| Ivan Kaznacheyeu | 955,514 | <p>If you need only estimate for <span class="math-container">$n$</span>, you can use following claim, which is simple to prove: <span class="math-container">$n! \geq (\sqrt{n})^n$</span>.</p>
<p><span class="math-container">$$(n!)^2=(1\cdot 2\cdot ...\cdot n)\cdot(n\cdot(n-1)\cdot...\cdot 1)
=(1\cdot n)\cdot(2\cdot(n-1))\cdot...\cdot(n\cdot 1)$$</span></p>
<p>Every product is of form</p>
<p><span class="math-container">$$k\cdot(n+1-k)=nk-k(k-1)=n+n(k-1)-k(k-1)=n+(n-k)(k-1)\geq n$$</span></p>
<p>There are exactly <span class="math-container">$n$</span> such products, then</p>
<p><span class="math-container">$$(n!)^2 \geq n^n \Rightarrow n! \geq (\sqrt{n})^n$$</span></p>
<p>So for <span class="math-container">$n > x^2$</span> one can be sure that <span class="math-container">$n! \geq (\sqrt{n})^n > x^n$</span>.</p>
|
4,113,744 | <p>The number of real values <span class="math-container">$(x,y)$</span> for which <span class="math-container">$$2^{x+1}+3^y=3^{ y+2}-2^x$$</span> is ?
I went like, <span class="math-container">$$2^{x+1}+2^x=3^{y+2}-3^y$$</span>
after solving which I got; <span class="math-container">$(x,y)=(3,1)$</span>
Is there any other process/ solution/ way to solve?</p>
| whatever124125 | 919,871 | <p><span class="math-container">$ 2^{x+1}+3^y=3^{y+2}-2^x \Leftrightarrow 2^{x+1}+2^x=3^{y+2}-3^y\Leftrightarrow 3\cdot 2^x= 8\cdot 3^y \Leftrightarrow 2^{x-3}=3^{y-1}$</span>, take <span class="math-container">$\log_3\implies
\log_32^{x-3}=\log_33^{y-1}\Leftrightarrow (x-3)\log_32=y-1 \implies y=(x-3)\log_32+1,~$</span>in<span class="math-container">$~\mathbb{R}$</span>.</p>
<p>Over the integers, we have only solution x=3, y=1.</p>
|
3,631,714 | <p>If <span class="math-container">$f: \mathbb{R}^n \to \mathbb{R} $</span> is a function on the Schwartz space <span class="math-container">$\mathcal{S} (\mathbb{R}^n) $</span> we define the Fourier transform of <span class="math-container">$f$</span> as the function
<span class="math-container">$$ \hat f(\xi)= \int_{\mathbb{R}^n} f(x)e^{2 \pi ix\xi} dx $$</span>
How can I prove that if <span class="math-container">$f \in L^1(\mathbb{R}^n)$</span>, then the same formula for <span class="math-container">$\hat f$</span> holds? I'd like to prove that that formula extends the Fourier transform on <span class="math-container">$L^1$</span>.</p>
| Amit Bendkhale | 440,224 | <p>It is the National Council of Educational Research and Training, that decides the syllabus, and publishes content of secondary and senior-secondary students.
You can check their site for E-Books <a href="http://ncert.nic.in/textbook/textbook.htm?lemh1=0-6" rel="nofollow noreferrer">http://ncert.nic.in/textbook/textbook.htm?lemh1=0-6</a>
These are mandatory books in high school and examinations are conducted based on them, yet interested students follow some other books as well.</p>
|
1,401,661 | <blockquote>
<p>Dice are cubes with pips (small dots) on their sides, representing
numbers 1 through 6. Two dice are considered the same if they can be
rotated and placed in such a way that they present matching numbers on
the top, bottom, left, right, front, and back sides.</p>
<p>Below is an example of two dice that can be rotated to show that they
are the same if the 2-pip and 4-pip sides are opposite and the 3-pip
and 5-pip sides are also opposite.
<a href="https://www.dropbox.com/s/6q56njm11hu3f36/Screenshot%202015-08-18%2012.02.11.png?dl=0" rel="nofollow">https://www.dropbox.com/s/6q56njm11hu3f36/Screenshot%202015-08-18%2012.02.11.png?dl=0</a></p>
<p>How many different dice exist? That is, how many ways can you make
distinct dice that cannot be rotated to show they are the same? Note:
This problem does not involve rolling the dice or the probability of
roll outcomes.</p>
</blockquote>
<p>I'm having trouble understanding exactly what is being asked in this question. I understand that I have to find how many different ways the dice can be placed to show that they are the same, but saying they cannot be rotated confuses me.</p>
<p>Could somebody make an attempt at rewording this? Or walking me through how to solve this?</p>
| wltrup | 232,040 | <p>I won't give you an actual answer, and this post is not a hint to solve the given question, but I will suggest a way that you can simplify the problem in order to understand better what's being asked.</p>
<p>(By the way, that's often a <em>very</em> useful approach, that is, simplifying the problem to get a better understanding of it)</p>
<p>Say you have a square and you assign numbers to the edges, in a clockwise manner, starting from the top edge. There are 4! = 24 different assignments you could make. Here are a few examples: 1234, 3241, 1324, 4123, 4321, and so on.</p>
<p>Now, note that the assignments 1234, 2341, 3412, and 4123 don't actually represent different squares because you could rotate them to match one another. You could put them into a grouping of their own. On the other hand, 1234 and 1324 are different assignments in the sense that no matter how you rotate either of the resulting squares, you won't be able to make them match. So, 1234 and 1324 must belong to different groupings.</p>
<p>What the question about the dice is asking is how many such groupings there are, but with a cube and its faces, instead of a square and its edges. Note that you can rotate a cube in more ways than you can rotate a square.</p>
<hr>
<p><strong>Edit</strong>: Based on the comment exchange below, here's the brute-force solution for the analogous problem with <em>triangles</em>. It's brute-force because I'm actually listing all the possible assignments and explicitly testing if they're pairwise equivalent or not. You will <em>not</em> want to do that in the dice problem. That's not how you want to solve that problem. I'm only brute-forcing the solution here to make it easier for you to understand what the problem is asking.</p>
<p>You can divide the 3! = 6 possible assignments into only 2 groups where every assignment in one group is equivalent to the other assignments in the <em>same</em> group but different from all assignments in the <em>other</em> group. The solution to "how many such groups are there?" is, then, 2. One is coloured blue, the other red in the figure below.</p>
<p><a href="https://i.stack.imgur.com/RkzCE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RkzCE.png" alt="enter image description here"></a></p>
<hr>
<p><strong>Hint #1</strong> In the case of 6-sided dice, if all the dice have opposite sides adding up to 7 (1 opposite to 6, 2 opposite to 5, 3 opposite to 4) then there are only <strong>2</strong> groups. Try to prove that result, then look at the case when opposite faces don't have to add up to 7. How many of those cases are there in total?</p>
<hr>
<p><strong>Hint #2</strong> So, if you fix the top and bottom numbers, you get 2 groups (that's the result of hint #1 above). In how many ways can you fix the top and bottom numbers?</p>
|
364,059 | <p>While reading up on "Glivenko Cantelli Theorem" from Probability Models by K.B Athreya, the author used 2 lemmae to prove it. One was called Scheffe's lemma, the other Polya's theorem.</p>
<p>Scheffe's Lemma is stated as follows:</p>
<p>Let $f_n, f$ be non negative $\mu$ integrable functions. If $f_n \to f$ a.e and $\int f_n d\mu \to \int fd\mu$, then
$$\int |f_n - f|d\mu \to 0$$</p>
<p>My proof is:</p>
<p>Let $g_n = |f_n -f|$. Now we have $g_n \to 0$ a.e. Now
$$0 \leq g_n = |f_n -f| \leq f + f_n$$
$$\Rightarrow \int g_n d\mu \leq \int fd\mu + \int f_nd\mu < \infty $$</p>
<p>Thus by Dominated convergence theorem,
$$\int g_nd\mu \to 0$$ QED.</p>
<p><strong>The Question</strong>:
My doubt is that in this proof, I have not used that $\int f_n d\mu \to \int fd\mu$, at least not explicitly. So is this condition superfluous?</p>
<p><strong>What I searched</strong>:
I searched for Scheffe on MSE, but got 3 results (none useful) and when I typed Scheffe's instead, I got a <a href="https://math.stackexchange.com/q/363271/35983">result</a> not belonging to these 3 which was actually on Scheffe's lemma (Though not helpful). It's strange (not the result but the search).</p>
<p>I'd appreciate any help/hints on this. Kindly ask me for clarifications if required.</p>
| Bombyx mori | 32,240 | <p>I don't think your "proof" works. Let $f_{n}=n\chi[0,\frac{1}{n}]$, $f=0$. Then $f_{n}\rightarrow f$ almost everywhere. But you cannot conclude that $$\int f_{n}d\mu\rightarrow 0$$ as it is constant $1$. As the commenter pointed out to apply dominated convergence theorem you need a function $g$ such that $|f_{n}|\le |g|$ and $g\in L^{1}(\Omega)$. </p>
|
4,014,144 | <p>Let <span class="math-container">$f(x) = \max(c_1^Tx, c_2^Tx, \dots, c_k^Tx)$</span>.
where <span class="math-container">$x, c_1, c_2, \dots, c_k \in \mathbb R^n$</span>.
What fast iterative methods are available for finding the (approximate) min of <span class="math-container">$f$</span> with the constraint <span class="math-container">$\lVert x \rVert_2 = 1$</span>?</p>
<p>Notes:</p>
<ol>
<li><span class="math-container">$f$</span> is convex and and non negative, that is, <span class="math-container">$c_i$</span> positively span <span class="math-container">$\mathbb R^n$</span></li>
<li>For my use case <span class="math-container">$k \gg n$</span> and (rougly) <span class="math-container">$3 \le n \le 50$</span> and <span class="math-container">$100 \le k \le 10000$</span>.
I tried <a href="https://en.wikipedia.org/wiki/Subgradient_method" rel="nofollow noreferrer">projected subgradients</a> (projected to the sphere) but it can be slow to converge and improving the initial guess doesn't seem to accelerate the method. From the literature, this seems to be equivalent to Riemannian manifold subgradient methods which use the exponential map. I haven't tried bundle methods yet but I am investigating them now. They seem like they might be too slow.</li>
</ol>
<p>I've also tried approximating <span class="math-container">$f$</span> with a smooth maximum <a href="https://en.wikipedia.org/wiki/LogSumExp" rel="nofollow noreferrer">LogSumExp</a> and doing gradient descent with projection. I also tried unconstrained gradient descent with a <a href="https://en.wikipedia.org/wiki/Penalty_method" rel="nofollow noreferrer">quadratic penalty</a>. The approximation is too inaccurate and the evaluation of exp is too slow for my use case. I'm not interested in <a href="https://en.wikipedia.org/wiki/Sequential_quadratic_programming" rel="nofollow noreferrer">SQP</a> because I suspect whatever method used to smooth (e.g. LogSumExp, hyperbolic) will be too costly/inaccurate to evaluate.</p>
<p>Edit:
I believe this problem is equivalent to a linear programming problem with quadratic <strong>equality</strong> constraints. I haven't been able to to find much literature on this type of problem. Perhaps I can use a <a href="https://en.wikipedia.org/wiki/Semidefinite_programming" rel="nofollow noreferrer">SDP</a> method but I'm not sure.</p>
| Alt | 123,685 | <p>Convert it to a quadratically constrained linear program by the following trick: Add an auxiliary variable <span class="math-container">$t$</span> such that: <span class="math-container">$\max(c_1^Tx, c_2^Tx, \dots, c_k^Tx)\leq t$</span>
Then, this will imply <span class="math-container">$c_i^Tx\leq t~~ \forall i$</span>.</p>
<p>Now your optimization program can be rewritten as:
<span class="math-container">$$
\text{minimize}_{x, t} ~~t\\
\quad\text{subject to}\\
\quad\|x\|_2= 1\\
\quad \quad \quad \quad \quad \quad \quad c_i^Tx\leq t~~ \forall i\in\{1, \ldots, k\} (*)
$$</span></p>
<p>Up to this point the optimization program is exactly the same as the original one. From a convex optimization perspective relaxing the norm constraint results in an efficiently solvable problem.</p>
<p>If you relax <span class="math-container">$\|x\|_2= 1$</span> to <span class="math-container">$\|x\|_2\leq 1$</span>, then the program is convex and can be efficiently solved using any convex optimization technic such as interior point method. The solution might satisfy the original constraint <span class="math-container">$\|x\|_2 = 1$</span> by itself. But there is no guarantee for that as that program is not convex unfortunately. However, you can perform Step two, and project back <span class="math-container">$x$</span> to <span class="math-container">$\|x\|_2 = 1$</span> by setting <span class="math-container">$x = x / \|x\|_2$</span></p>
<p>If the solution turns out to be zero, you should solve Problem (*) using iterative projection method, where in each iteration you project back to <span class="math-container">$\|x\|_2 = 1$</span>. This will give you an approximate solution to the non-convex problem you have.</p>
|
2,501,450 | <p>I was trying for a while to prove non-existence of theu following limit:</p>
<p>$$\lim_{n\to\infty}(-1)^n\frac{2^n+4n+6}{2^n(\sqrt[n]{5}-1)}$$</p>
<p>Unfortunately, with no results.</p>
<p>My hope was to show, that:</p>
<p>$$\lim_{n\to\infty}\frac{2^n+4n+6}{2^n(\sqrt[n]{5}-1)}\not=0$$</p>
<p>But showing that was harder than I thought.</p>
<p>Can anyone show me how to solve this problem?</p>
| Yiorgos S. Smyrlis | 57,021 | <p>Note first that
$$
\lim_{n\to\infty}n(\sqrt[n]{5}-1)=\ln 5,
$$
Hence
$$
(-1)^n\frac{2^n+4n+6}{2^n(\sqrt[n]{5}-1)}=n\cdot(-1)^n\cdot\frac{1+\frac{4n}{2^n}+\frac{6}{2^n}}{n(\sqrt[n]{5}-1)}.
$$
Clearly
$$
\frac{1+\frac{4n}{2^n}+\frac{6}{2^n}}{n(\sqrt[n]{5}-1)}\to \frac{1}{\ln 5}
$$
while
$$
n\cdot(-1)^n\,\,\,\text{diverges}.
$$
Hence, so does
$$
n\cdot(-1)^n\cdot\frac{1+\frac{4n}{2^n}+\frac{6}{2^n}}{n(\sqrt[n]{5}-1)}.
$$</p>
|
2,501,450 | <p>I was trying for a while to prove non-existence of theu following limit:</p>
<p>$$\lim_{n\to\infty}(-1)^n\frac{2^n+4n+6}{2^n(\sqrt[n]{5}-1)}$$</p>
<p>Unfortunately, with no results.</p>
<p>My hope was to show, that:</p>
<p>$$\lim_{n\to\infty}\frac{2^n+4n+6}{2^n(\sqrt[n]{5}-1)}\not=0$$</p>
<p>But showing that was harder than I thought.</p>
<p>Can anyone show me how to solve this problem?</p>
| stanly | 382,677 | <p>Try to show difference bwtween $\lim_{n \rightarrow \infty} \sup a_{n}$ and $\lim_{n \rightarrow \infty} \inf a_{n}$.</p>
|
235,425 | <p>Ok, here is what I have for the proof of this conjecture. Let me know if I'm on the right path? all input appreciated.</p>
<p>There exist integers $j$, $k$, and $m$, such that,
$b = aj $ and $ c = ajk.$ Then $c = ajk $ (substituting $aj$ for $b$)
let $m = jk$, then $c = ma, => a|c.$ </p>
| Mhenni Benghorbal | 35,472 | <p>Note that,</p>
<p>$$a|b \implies b=ma \,,$$</p>
<p>and </p>
<p>$$ b|c \implies c =nb \implies c = nma \implies c = q a \implies a|c \,,$$</p>
<p>for $m,n,q \in \mathbb{Z}$ and $q=nm.$ </p>
|
594,482 | <p>If a,b and c are elements of a ring, does the equation ax+b=c always have a solution x? If it does, must the solution be unique? </p>
| Boris Novikov | 62,565 | <p>It is enough to consider the equation $ax=b$.</p>
<p>If $b=0$ then the set of solutions coincides with the right annulator of $a$: $N_a=\{x|ax=0\}$. For arbitrary $b$ let $x_0$ is a solution of $ax=b$ (if exists!). Then the set of solutions coincides with the coset $N_a+x_0$.</p>
<p>This is all what one can say about solutions.</p>
|
591,014 | <p>I need help with this proof:</p>
<p>$f: X\rightarrow Y$</p>
<p>$C,D\subseteq Y$</p>
<p>$f^{-1}(C \cap D) = f^{-1}(C) \cap f^{-1}(D)$</p>
<p>Thanks.</p>
| angryavian | 43,949 | <p>To solve these kinds of questions, you want to show that $$f^{-1}(C\cap D) \subseteq f^{-1}(C) \cap f^{-1}(D)$$ and $$f^{-1}(C\cap D) \supseteq f^{-1}(C) \cap f^{-1}(D).$$</p>
<p>I will show $\subseteq$ here; you should try the other direction.</p>
<p>Suppose $x \in f^{-1} (C \cap D)$. This means $f(x) \in (C\cap D)$, which further implies $f(x) \in C$ AND $f(x) \in D$. Thus, $x \in f^{-1}(C)$ AND $x \in f^{-1}(D)$.</p>
|
690,729 | <p>Vancouver is 300 km away from Seattle. Two friends, one leaving from each city on a bike want to find a campsite between the two cities. One cyclist starts from vancouver at 25 km/h. His friend will start 2.0h later from seattle at 32 km/h. How far from Vancouver do the friends meet?</p>
<p>Attempt:</p>
<p>dA + dB = 300 </p>
<p>vA(t) + vB(t+2) = 300</p>
<p>25t + 32t + 2 = 300 </p>
<p>t = 4.14 s</p>
<p>now how do i find the distance travelled by each cyclist?</p>
<p>not sure if this is even the correct direction, but any help would be greatly appreciated! Thank you!!</p>
<p>btw- correct answer is 160 km</p>
| Ross Millikan | 1,827 | <p>Your direction is fine. It is good to define your variables, like $dA$ is the distance traveled by the one from Vancouver, etc. The vB should be multiplied by $(t-2)$ as the one from Seattle leaves two hours later, so that cyclist is traveling two hours less. You dropped multiplying the $2$ by vB, so it should become $64$ (with a minus sign). The unit on $t$ should be hr, not s. One cyclist traveled $vA(t)$ and the other $vB(t-2)$</p>
|
903,831 | <p>One may be curious why one wishes to convert a polynomial ring to a numerical ring. But as one of the most natural number system is integers, and many properties of rings can be easily understood in parallel to ring of integers, I think converting a polynomial ring to a numerical (i.e. integer) ring is useful.</p>
<p>What I mean by converting to a numerical ring is: in the standard ring of integers, $+$ and $\cdot$ are defined as in usual arithmetic. But is there universal way of converting any polynomial/monomial rings such that each object in the ring gets converted to an integer, and $+$ and $\cdot$ can be defined differently from standard integer $+$ and $\cdot$? This definition would be based on integer arithmetic, though. </p>
| lhf | 589 | <p>If you accept subrings of the complex numbers as numerical rings, then just take a homomorphism $\mathbb Z[X] \to \mathbb C$ with nontrivial kernel. This reduces to sending $X$ to an algebraic number.
A prime example is $X \mapsto i$, which gives you the numerical ring $\mathbb Z[i]$.</p>
<p>If you mean encoding the ring operations of $\mathbb Z[X]$ as integer operations, then this can be done by mapping $\sum_{k=0}^n a_k X^k$ to $\prod_{k=0}^n p_k^{a_k}$, where $p_k$ is the $k$-th prime. For other ideas along this line, see <a href="http://en.wikipedia.org/wiki/Godel_numbering_for_sequences" rel="nofollow">Gödel numbering for sequences</a>.</p>
|
2,873,309 | <p>Let's say each month a utility bill comes and it needs to be split among a variable number of roommates. Let's say each person occupied the house for a certain percentage of the month: person 1 100%, person 2 100%, person 3, 75%, person 4, 50%. How can I calculate how much each person has to pay? (The number of people could be greater than number of rooms because of people moving in and out during the month.) Let's say the utility bill is $150 and length of month is 28 days.</p>
<p>My algebraic and correct solution is to calculate utility per day: $5.357. Then I go through each day of the month and check number of people that lived in the house. I do 5.357/number of people in house at day d. Then I add up for each for person for each day.</p>
<pre><code><first 7 days>: 2 people occupied
<second 7 days>: 3 people occupied
<last 14 days>: 4 people occupied
This gives me a correct answer of:
-person 1: 50 = 18.75 + 12.5 + 18.75
-person 2: 50 = 18.75 + 12.5 + 18.75
-person 3: 31.25 = 12.5 + 18.75
-person 4: 18.75 = 18.75
</code></pre>
<p>How can I make this calculation without having to divide up the month day by day? My intuition is that there is a calculus solution to this. </p>
| Bram28 | 256,001 | <p>Another method (it's certainly simpler) is to just add up the occupancy rates for the people, and then figure out the percentage of the total bill that each person has to pay as their share.</p>
<p>For your example, we add up:</p>
<p>$$100+100+75+50=325$$</p>
<p>And now we can say that persons 1 and 2 are each responsible for $\frac{100}{325}$ part of the total bill, person 3 for $\frac{100}{325}$ mpart, and person 4 for $\frac{50}{325}$</p>
<p>That is, persons 1 and 2 pay: $\frac{100}{325}\cdot 150 \approx 46.15$</p>
<p>Person 3 pays $\frac{75}{325}\cdot 150 \approx 34.60$</p>
<p>Person 4 pays $\frac{75}{325}\cdot 150 \approx 23.10$</p>
<p>Now, these are different numbers than you got ... so ... which is the better (more 'fair'?) method? If it was just a matter of how one uses in terms of gas/electricity/water, etc, I think your method is better. However, utility bills typically also have some 'base rate' (typically called 'delivery charge') as well, just to cover the costs of the utility company setting up the infrastructure, maintaining and repairing pipes and wires, etc. Think of this as the cost to you for having the option and luxury to be able to use gas an electricity in the first place, before you actually get to use any of it.</p>
<p>So, look at how in your method persons 3 and 4 end up paying a bit less than in my method. So, are they getting stiffed by my my method? Well, you could argue that persons 3 and 4 enjoy the fact that there was this infrastructure just as much as persons 1 and 2: persons 3 and 4 know that they can come to the house and enjoy the fact that there ids gas, electricity, and running water.So sSure, they used less actual gas/water/electricity, but the pipes and wiring had to be there for them just as much ... and so maybe it's just proper that they pay a bit more ... </p>
<p>Indeed, how do you think about the rent? Say the rent for the whole house is $1000 and suppose the 4 people all have an equally nice and sized room. How would you split the rent? Probably you'd do something much closer to my method than yours ... in fact, you'd probably just split this 4-ways <em>regardless</em> of occupancy: and the reasoning is similar: they all have, at any time, a place to go to ... that is largely what they pay for. Well, something similar is going one with utilities, I would argue: at least partly you pay for the option of being able to use it; simply of it being there.</p>
<p>In fact, you could take my argument and make your life <em>more</em> complicated: split the bill between usage costs and 'delivery' costs, and then use your method to split up the usage costs, and simply evenly split the delivery costs.</p>
<p>In fact, with my method being 'in between' the individual-usage method of yours, and the simple 'even-split' method, my method could be seen as a kind of compromise ... and it's certainly fairly easy to calculate. In fact, as long as the differences in occupancy rates aren't too high, I would say don't sweat the few dollars more or less!</p>
|
2,897,374 | <p>So today in my Algebraic Topology class we were trying to construct a homotopy and at one point we needed to basically construct a (continuous) bijection between some intervals $[a, b]$ and $[c, d]$ sending $a \mapsto c$ and $b \mapsto d$. </p>
<p>My professor said that a bijection $\varphi : [a, b] \to [c, d]$ would be a linear function of the form $\varphi(x) = \alpha x + \beta$ and adding the contraints that $\varphi(a) = c$ and $\varphi(b) = d$ we get a linear system of equations $$\alpha \cdot a + \beta = c$$ $$\alpha \cdot b + \beta =d.$$</p>
<p>Solving this system we get $$\alpha = \frac{c-d}{a-b}$$ and $$\beta=\frac{ad-bc}{a-b}$$</p>
<p>so that $\varphi$ is given by $$\varphi(x) = \left(\frac{c-d}{a-b}\right)x + \frac{ad-bc}{a-b}$$</p>
<hr>
<p>Now I was not aware that we could construct a (continuous!) bijection between connected closed intervals of $\mathbb{R}$ so easily. My question is how exactly did my professor know that such a (continuous) bijection would be a linear function of the above form? I've never seen any theorem of the sort or something similar written in any textbook.</p>
| amsmath | 487,169 | <p>The projection that you are looking for is $Px = x - \frac 1{\|v\|^2}\langle x,v\rangle v$, i.e.,
$$
P = I - \frac{vv^T}{v^Tv},
$$
where $v = (1,1,\ldots,1)^T$.</p>
|
2,661,718 | <p>My basic understanding is that each time
I have 30% chance of winning the prize
so between 3 and 4 tries I should win it</p>
<p>cause .3 +.3 +.3 = 90% as I need to win only once with 3 try
.3+.3+.3+.3 = 120% with 4 try</p>
<p>I do remember a formula saying 1-(0.7)^3 = 65.7% chance but I dont remember what that number mean exactly</p>
<p>thanks for answering this noob question my math are way behind me now</p>
| user | 505,767 | <p>Let $y=x+1$ then</p>
<p>$$x > -1 \implies x^2 \ge (1+x) \{\log(1+x)\}^2$$</p>
<p>is equivalent to</p>
<p>$$y > 0 \implies (y-1)^2 \ge y \log^2 y$$</p>
<p>$$\iff -\frac{|y-1|}{\sqrt y}\le\log y\le\frac{|y-1|}{\sqrt y} \iff e^{-\frac{|y-1|}{\sqrt y}}\le y\le e^{\frac{|y-1|}{\sqrt y}}\iff 1\le ye^{\frac{|y-1|}{\sqrt y}} \le e^{2\frac{|y-1|}{\sqrt y}}$$</p>
<p>which is true, indeed for $1\le ye^{\frac{|y-1|}{\sqrt y}}$ </p>
<p>$$ye^{\frac{|y-1|}{\sqrt y}}\ge y\left(1+\frac{|y-1|}{\sqrt y}\right)=y+y\frac{|y-1|}{\sqrt y}\ge 1$$</p>
<p>$$\iff y\sqrt y+y|y-1|\ge \sqrt y\iff y|y-1|\ge\sqrt y(1-y)$$</p>
<p>and for $ye^{\frac{|y-1|}{\sqrt y}} \le e^{2\frac{|y-1|}{\sqrt y}} \iff e^{\frac{|y-1|}{\sqrt y}}\ge y $ note that</p>
<ul>
<li>for $0<y\le 1$ it is true since $\frac{|y-1|}{\sqrt y}\ge0$</li>
<li>for $y> 1$ </li>
</ul>
<p>$$\iff e^{\frac{y-1}{\sqrt y}}\ge 1+\frac{y-1}{\sqrt y}+\frac{(y-1)^2}{2y}+\frac{(y-1)^3}{6y\sqrt y}\ge y$$</p>
<p>$$\frac{y-1}{\sqrt y}+\frac{(y-1)^2}{2y}+\frac{(y-1)^3}{6y\sqrt y}\ge y-1$$</p>
<p>$$\frac{1}{\sqrt y}+\frac{(y-1)}{2y}+\frac{(y-1)^2}{6y\sqrt y}\ge 1$$</p>
<p>$$6y+3y\sqrt y-3\sqrt y +y^2-2y+1\ge 6y\sqrt y$$</p>
<p>$$y^2+4y+1\ge 3y\sqrt y$$</p>
<p>which is true indeed let $y=z^2>1$</p>
<p>$$y^2+4y+1\ge 3y\sqrt y\iff z^4-3z^3+4z^2+1\ge0$$</p>
<p>and </p>
<p>$$z^4-3z^3+4z^2+1\ge z^4-4z^3+4z^2+1=(z^2-2z)^2+1\ge0 \quad \square$$</p>
|
2,661,718 | <p>My basic understanding is that each time
I have 30% chance of winning the prize
so between 3 and 4 tries I should win it</p>
<p>cause .3 +.3 +.3 = 90% as I need to win only once with 3 try
.3+.3+.3+.3 = 120% with 4 try</p>
<p>I do remember a formula saying 1-(0.7)^3 = 65.7% chance but I dont remember what that number mean exactly</p>
<p>thanks for answering this noob question my math are way behind me now</p>
| Barry Cipra | 86,747 | <p>Let $x=u^2-1$ with $u\gt0$. The inequality to prove becomes $(u^2-1)^2\ge u^2(\log(u^2))^2$, or</p>
<p>$$\left| u-{1\over u}\right|\ge2|\log u|$$</p>
<p>Since each side is now invariant under $u\to1/u$, it suffices to prove the inequality for $u\ge1$, in which case we can remove the absolute value signs and write the inequality to prove as </p>
<p>$$f(u)=u-{1\over u}-2\log u\ge0$$</p>
<p>for $u\ge1$. But this is now easy: We see that $f(1)=0$ and</p>
<p>$$f'(u)=1+{1\over u^2}-{2\over u}=\left(1-{1\over u}\right)^2\ge0$$</p>
|
1,825,392 | <p>Let $f(z)$ be an entire function so that,</p>
<p>$$ \int \frac{|f(z)|}{1 + |z|^3} dA(z) < \infty$$</p>
<p>where the integral is taken over the entire complex plane. Show that $f$ is a constant.</p>
<p>I believe that the idea is to use the mean value property; that is:</p>
<p>$$f(z) = \frac{1}{\pi\delta^2}\int_{D(z, \delta)}f(w)dA(w)$$</p>
<p>and then do some manipulation to relate the two integrals. But I'm not sure otherwise how to proceed. Can anyone help? </p>
| Jonas Meyer | 1,424 | <p>You can use Cauchy's integral formula to show that $f^{(2)}=0$. Given $w\in \mathbb C$, take $r\geq\max\{1,2|w|\}$. Let $C_r$ be the circle with radius $r$ centered at $0$. Then $$f^{(2)}(w)=\frac{1}{\pi i}\int_{C_r}\frac{f(z)}{(z-w)^3}\,dz,$$</p>
<p>so $$|f^{(2)}(w)|\leq \frac1\pi \int_0^{2\pi}\frac{|f(re^{i\theta})|}{r^3-|w|^3}r\,d\theta\leq \frac3\pi\int_{0}^{2\pi}\frac{|f(re^{i\theta})|}{1+r^3}r\,d\theta.$$</p>
<p>The last inequality follows because $r\geq1$ and $r\geq2|w|$, so that</p>
<p>$$r^3-|w|^3\geq\frac78r^3>\frac13(r^3+r^3)\geq\frac13(1+r^3).$$</p>
<p>By integrating with respect to $r$ from $c=\max\{1,2|w|\}$ to $\infty$, $$\int_c^\infty |f^{(2)}(w)|\,dr\leq \frac3\pi\int_c^\infty\int_0^{2\pi}\frac{|f(re^{i\theta})|}{1+r^3}r\,d\theta\,dr=\frac3\pi\int_{|z|\geq c}\frac{|f(z)|}{1+|z|^3}\,dA(z)<\infty,$$</p>
<p>which implies $f^{(2)}(w)=0$, and $w$ was arbitrary.</p>
<p>Once you know $f^{(2)}=0$ you have $f(z) = az+b$ for constants $a$ and $b$, and $a$ must be zero for $\displaystyle{\int_\mathbb{C}\dfrac{|az+b|}{1+|z|^3}\,dA(z)}$ to converge. </p>
|
686,167 | <p>I'm working on an assignment where part of it is showing that $S_k=0$ for even $k$ and $S_k=1$ for odd $k$, where</p>
<blockquote>
<p>$$S_k:=\sum_{j=0}^{n}\cos(k\pi x_j)= \frac{1}{2}\sum_{j=0}^{n}(e^{ik\pi x_{j}}+e^{-ik\pi x_{j}}) $$</p>
<p>Here $x_j=j/(n+1)$.</p>
</blockquote>
<p>So, working through the algebra:</p>
<blockquote>
<p>$$\frac{1}{2}\sum_{j=0}^{n}(e^{ik\pi x_{j}} +e^{-ik\pi x_{j}}) =\dots
=\frac{1}{2}\cdot\frac{1-e^{ik\pi}}{1-e^{\frac{ik\pi}{n+1}}}+\frac{1}{2}\cdot\frac{1-e^{-ik\pi}}{1-e^{-\frac{ik\pi}{n+1}}}
$$</p>
</blockquote>
<p>Obviously $S_k=0$ for even $k$'s, since $e^{i\pi\cdot\text{even integer}}=1$. But when $k$ is odd we get $$\frac{1}{1-e^{\frac{ik\pi}{n+1}}}+\frac{1}{1-e^{-\frac{ik\pi}{n+1}}}$$
which isn't obviously one to me, at least. Wolfram alpha confirms it equals 1.</p>
<p>My question: How does one see that it equals 1?</p>
| Sandeep Thilakan | 124,957 | <p>Using trigonometric identities</p>
<p>\begin{align}
\cos (k \pi x_j) &= \frac{\sin (k \pi x_{j+1}) - \sin (k \pi x_{j-1})}{2 \sin (\frac{k \pi}{n+1})} \\
\sum_{j=0}^{n} \cos (k \pi x_j) &= 1+ \frac{1}{2 \sin (\frac{k \pi}{n+1})} \sum_{j=1}^{n} {\sin (k \pi x_{j+1}) - \sin (k \pi x_{j-1})} \\
\end{align}
In the summation on the RHS, the terms that remain are </p>
<p>$-\sin (\frac{k \pi}{n+1}) + \sin (\frac{k \pi n}{n+1}) + \sin (k \pi)$ which clearly equals $0$ when $k$ is even, and $-2 \sin(\frac{k \pi}{n+1})$ when $k$ is odd.</p>
|
1,219,851 | <p>If $n$ is a positive integer such that $2^n+n^2$ is a prime number , then is it true that $6|n-3$ ? Trivially $n$ cannot be even , so this leaves us only with the possibilities $n \equiv1,3,5( \mod 6) $ , but I cannot reduce the cases . Please help , Thanks in advance . </p>
| ajotatxe | 132,456 | <p>If $n\equiv1,5\pmod 6$ then $n^2\equiv 1\pmod 6$.</p>
<p>On the other hand, $2^n\equiv 2$ or $4\pmod 6$ depending on $n$ is odd or even. So if $n^2\equiv 1\pmod 6$, then
$$2^n+n^2\equiv 2+1\equiv 3\pmod 6$$
therefore $2^n+n^2$ is a multiple of three.</p>
<p>Notice the case $1^2+2^1$, that is the only exception to your statement.</p>
|
3,520,893 | <p>As in group theory, there is a concept of isomorphism between metric spaces called isometry.
Two metric spaces <span class="math-container">$X$</span> and <span class="math-container">$Y$</span> are isometric if there is a function that perserves the distance of two elements. That function is called isometry.</p>
<p>The thing is that properties of metric spaces (completeness, compactness, connectedness, etc.) are preserved under isometry. </p>
<p>So, thinking about the classification of finite simple groups, I was wondering if there is any classification of Metric Spaces up to isometry, or at least a specific category of Metric Spaces (like finite simple groups in Group Theory). Also, I am interested if there is a more general topological classification of Metric Spaces up to homeomorphism (isomorphism in topology).</p>
| Giuseppe Negro | 8,157 | <p>That's too ambitious, there is an enormous variety of metric spaces. But if you restrict to specific classes, then a lot has been done. I can think of two examples: Hilbert spaces and geodesic surfaces. These are completely classified. </p>
<p>Hilbert spaces are vector spaces equipped with a scalar product, which induces a norm, hence a distance and so a structure of metric space, which is required to be complete. Hilbert spaces are completely characterized up to isomorphism (a stronger form of isometry); in particular, any <strong>separable</strong> real Hilbert space is isomorphic to
<span class="math-container">$$
\ell^2:=\left\{\boldsymbol{x}=(x_1, x_2, x_3, \ldots)\ :\ x_j\in\mathbb R,\ \sum_{j=1}^\infty x_j^2<\infty\right\},$$</span>
where the scalar product is given by
<span class="math-container">$$
\langle \boldsymbol{x}, \boldsymbol{y}\rangle = \sum_{j=1}^\infty x_j y_j.$$</span>
(Non-separable Hilbert spaces are classified in terms of cardinality of their orthonormal bases. Also, the same result holds in the complex case, with obvious modifications). </p>
<p>Geodesic surfaces are something I know less. I will refer to <a href="http://perso.ens-lyon.fr/ghys/articles/disque-poincare.pdf" rel="nofollow noreferrer">this beautiful note</a> of Etienne Ghys, section 3. It is in French, but I am sure I have seen an English translation on the net.</p>
|
2,330,496 | <p>Players Ruby and Bob are given an undirected graph and a number $N$. First Ruby colors $N$ vertices red, then Bob colors $N$ vertices blue (they must be distinct from Ruby's choices). Afterward, all other points on the graph are given the color of whichever color they are closest to (shortest path) with ties left blank. The player with more of their color on the resulting graph wins.</p>
<p>Can the first player always win (or tie)?</p>
<p>Some context: This problem arose for me out of a mobile puzzle game. My knowledge in graph theory is pretty minimal, so I don't have much machinery to solve it (but I have spent a while with small graphs, always able to find an unbeatable strategy for the first player). My thoughts so far are to mark points as having an advantage in comparison to their neighbors, the maximum of which I'd guess provides a win with N=1. For higher N though the dynamics get much more challenging for me to express. There seems to be an aspect of even-spacing which I'm not sure how to formalize (perhaps it's picking vertices which minimize their shortest distance to any point).</p>
<p>Also if anyone has heard of a similar problem before (specifically related to coloring a graph based on the shortest path to a colored vertex) or has references I'd be happy to read them, but was unable to find much since I'm not certain what to search for.</p>
| Gregory J. Puleo | 183,812 | <p>I believe that the first player cannot always guarantee a tie, even when $N=1$. Some observations:</p>
<ul>
<li><p>Given a graph $G$ and a vertex subset $X \subset V(G)$, we can always modify $G$ so that the players are <em>forced</em> to choose a vertex in $X$: add a huge independent set $I$ to $G$ and make all the vertices in $I$ adjacent only to the vertices in $X$, so that if one player plays outside $X$ and the other plays inside it, then the player inside $X$ grabs all vertices in $I$ and wins even if the other player gets everything else.</p></li>
<li><p>This trick also lets us assign nonnegative integer weights to the edges, and calculate distances according to the weights: subdivide each edge an appropriate number of times, and choose our subset $X$ so that the internal vertices of the subdivided edges aren't feasible to play in. (Edit: As originally written this doesn't quite work, because the players in the modified graph would score points for the subdivided edges, but I think it can be made to work if you "blow up" the real vertices into large cliques or large independent sets, so that the value of getting one more "real vertex" far exceeds the value of the new vertices inside all the subdivided edges. This transformation doesn't <em>quite</em> preserve the rules of the game: in the transformed graph, Bob is essentially allowed to pick a vertex already chosen by Alice, so the transformed graph is friendlier to Bob than the original graph, but if the point is to come up with a graph where Bob can guarantee a win, then this is not a problem for us.)</p></li>
</ul>
<p>Once you've done these tricks, I think you can create a rock-paper-scissors scenario using the complete bipartite graph $K_{3,3}$. Say that the three vertices one partite set are $R,P,S$ and that the three vertices of the other set are $X,Y,Z$. The vertex $R$ has weights $1,2,3$ to $X,Y,Z$ respectively; $P$ has weights $3,1,2$, and $S$ has weights $2,3,1$. The players are only allowed to play in $\{R,P,S\}$, via the above tricks.</p>
<p>Now, if Alice picks $R$, then Bob picks $P$ and gets $Y,Z$ versus Alice's $X$. ($S$ is equidistant to both $R$ and $P$). If Alice picks $P$, Bob picks $S$ and gets $X,Z$ versus Alice's $Y$. If Alice picks $S$, Bob picks $R$ and gets $X,Y$ versus Alice's $Z$.</p>
|
1,083,141 | <p>According to Zeidler, 1995, in his book "Applied Functional Analysis: Application to Mathematical Physics".</p>
<p>Dirichlet problem is a problem to minimize $$F(u)=\frac{1}{2}\int_G(u')^2\ dx-\int_G fu\ dx,\qquad u=g \textrm{ in } bd(G).$$
Then, He says</p>
<blockquote>
<p>If $G$ is open, bounded, nonempty subset of $\mathbb{R}$ and given
continuous functions $f:G\rightarrow\mathbb{R}$ and
$g:bd(G)\rightarrow\mathbb{R}$. Then, for $u\in C^2(\overline{G})$:</p>
<p>$1.$ If $u$ is solution to Dirichlet problem then $u$ is also a
solution to generalized boundary value problem $$\int_G u'v'\ dx=\int_G fv\ dx,\qquad\forall v\in C_0^\infty(G)$$ $2.$ $u$ is a
solution to boundary value problem $$-u''=f,\qquad \textrm{in } G$$
$$u=g,\qquad \textrm{in }\partial G$$ if and only if $u$ is a solution
to generalized boundary value problem $$\int_G u'v'\ dx=\int_G fv\ dx,\qquad\forall v\in C_0^\infty(G)$$</p>
</blockquote>
<p>After that, he gives explanation that there is a Dirichlet problem which doesn't have solution (so Dirichlet Principle can't be justified). What I know about this, later on, is that the solution space needs to be complete so the Dirichlet Principle can be justified.</p>
<p>Therefore, the Dirichlet problem needs to be generalized by introducing Sobolev Space $W_2^1(G)$. The generalized Dirichlet Problem is to minimize the function $$\frac{1}{2}\int_G (\partial u)^2\ dx-\int_G fu\ dx, \qquad u-g\in \overset{\circ}{W^1_2}(G),$$ where $\partial u$ denotes weak derivatives of $u.$</p>
<p>The Dirichlet Principle is stated as below:</p>
<blockquote>
<p>If $G$ is open, bounded, nonempty subset of $\mathbb{R}$. Given two
functions $f\in L_2(G)$ and $g\in W_2^1(G)$. Then, these are true:</p>
<p>$1.$ The Generalized Dirichlet Problem has a unique solution $u\in W_2^1(G).$</p>
<p>$2.$ Solution to Generalized Dirichlet Problem is also a solution to
generalized boundary value problem $$\int_G \partial u \partial v \ dx=\int_G fv \ dx,\qquad \forall v\in \overset{\circ}{W^1_2}(G), \textrm{ and }u-g\in \overset{\circ}{W^1_2}(G)$$</p>
</blockquote>
<p>My questions are:</p>
<ol>
<li><p>Why the solution space $C^2(G)$ is generalized to $W_2^1(G)$? $C^2(G)$ is SECOND order continuously differentiable functions space while $W_2^1(G)$ is FIRST order weak derivable functions space.</p></li>
<li><p>Why the given function $g$ is generalized from only in $C(bd(G))$ to $W_2^1(G)$? The domain is extended from $bd(G)$ to $G$.</p></li>
<li><p>When proving the Dirichlet Principle, Zeidler, uses a theorem called Theorem in Quadratic Variational Problem, which only solved the minimizing problem and not the boundary condition, Why?</p></li>
</ol>
<p><strong>Note: Please help me. This is my undergraduate thesis topic and my undergraduate thesis defense seminar will be conducted on next Tuesday. Nobody familiar with this topic in my campus. So please, I'm desperately need an answer. Anything. Any explanation.</strong></p>
| Quickbeam2k1 | 63,384 | <p>I'll try to hint on answers to your first and second question.</p>
<p>Ad 1. As my comment suggested, the $C^2$ regularity of $u$ is needed to give sense to the term $-u''=f$. If you consider the generalized boundary value problem only first order derivatives appear. Essentially, this is the reason why only $W^{1,2}$ functions are considered in the second problem.</p>
<p>Ad 2. In the generalized setting one requires a definition of boundary values of $W^{1,2}$ functions. An easy way is to consider boundary data that are in some sense inherited by $W^{1,2}$-functions. Here the so-called trace theorem comes into play.</p>
|
1,360,910 | <p>My book (Introduction to Ring Theory, Paul Cohn) states this as a theorem and gives a proof. The book usually skips over trivial/easy proofs, so I don't really understand why this is in here.</p>
<p>Isn't the statement absolutely obvious for sets $A,B,C$ with $A\subseteq C$? You just need to draw a diagram and it becomes immediately obvious! <img src="https://i.stack.imgur.com/7YFcC.jpg" alt="enter image description here"></p>
<p>Is there a reason why it would not be obvious in modules?</p>
| rschwieb | 29,335 | <p>Judging from your diagram, it looks like (?) you are interpreting $+$ as set union. Certainly, if you are working with sets using union and intersection, your picture is accurate. </p>
<p>But using using the $+$ to join submodules of a module does not behave the same way. If $A$ and $B$ are submodules of a module, $A+B$ contains many more elements than just $A\cup B$ in general.</p>
<p>To summarize the situation, the property proved is that the lattice of submodules of a module (join $+$ and meet $\cap$) is a <a href="https://en.wikipedia.org/wiki/Modular_lattice" rel="nofollow noreferrer">modular lattice</a>. The lattice of subsets of a set, on the other hand (join $\cup$, meet $\cap$), is also modular, but even more strongly it is a <a href="https://en.wikipedia.org/wiki/Distributive_lattice" rel="nofollow noreferrer">distributive lattice</a>. In general, the lattice of submodules of a module does <em>not</em> have to be distributive.</p>
<p>For more about modularity, let me refer you to a question we had some time ago on the topic: <a href="https://math.stackexchange.com/q/142214/29335">Why are modular lattices important?</a></p>
|
1,576,007 | <p>I just completed an exam and was curious as to the following question:</p>
<blockquote>
<p>How many surjections are there from $\{1, 2, 3, ..., n\}$ to $\{1, 2\}$ if $n \geq 3$?</p>
</blockquote>
<p>My assumption was the following:</p>
<p>The first element of the domain has the possibility to map to either 1 or 2.</p>
<p>The second element then has the possibility to map to the opposite element of the codomain, in which the first element hasn't mapped to.</p>
<p>Now the functions are surjective, and then any remaining elements of the domain ($n-2$ of them) can then map to either of the 2 elements of the codomain.</p>
<p>My answer was $2^{n-1}$. Is this answer correct? Is there a fallacy in my thought process?</p>
| ultrainstinct | 177,777 | <p>Basically you can think of it as a binary string, $0$ means the element in the domain maps to the first element in the codomain, $1$ means the element maps to the second element in the codomain. There are $2^n$ ways to make a binary string of length $n$, but with the rules of surjectivity we need to take away the case where everything maps to the first element, and everything maps to the second element, so our final answer should be $2^n - 2$.</p>
|
2,604,660 | <p>In a box there are $16$ ice-creams: $6$ lemon flavor,$4$ mint flavor and $6$ strawberry flavor.When we extract two ice-creams,what's the probability of getting two different flavors,given that at least one is strawberry flavor.</p>
<p>That's my solution : </p>
<p>$P(A)$ = "Different flavors" = $\frac{\binom{16}{2}}{16!}$</p>
<p>$P(B^c)$ = "At least one is strawberry flavor" = $1 - \frac{\binom{10}{2}}{16!}$</p>
<p>So...We want $P(A|B^c)$ using conditional probability, where I go wrong?</p>
| true blue anil | 22,388 | <p>Assuming that each ice-cream has the same probability of being drawn,</p>
<p>you could compute it as the ratio of the ways of getting</p>
<p>$\text{ (one strawberry, one other) / (one strawberry, one other + two strawberries)}$</p>
<p>= $\dfrac{\binom61\binom{10}1}{\binom61\binom{10}1+\binom62}$ </p>
|
701,677 | <ol>
<li><p>If I know something about the representation theory of the general linear group $\mathrm{GL}_n(\mathbb C)$, what can I say about the representation theory of the unitary group $\mathrm U_n(\mathbb C)$? E.g. branching rules</p></li>
<li><p>Similarly for the groups $\mathrm{SO}_n(\mathbb C) \subset \mathrm O_n(\mathbb C)$?</p></li>
<li><p><strike>Is module over ${\mathbb C\mathrm{GL}_n(\mathbb C)}$ that is not semi-simple necessarily infinite-dimensional? How does it typically look like?</strike> See <a href="https://math.stackexchange.com/questions/707911/how-do-non-semisimple-modules-over-mathbb-c-mathrmgl-n-mathbb-c-look-like">new Question here</a></p></li>
<li><p><strike>If I have information about the restriction of representations of the general linear group, can I make any statements about the induction (by Frobenius reciprocity)? E.g. I know $$\mathrm{res}^{\mathrm{GL}_n}_{\mathrm{GL}_k\times \mathrm{GL}_{n-k}} V(\lambda)_n \cong \bigoplus_{\alpha, \beta} c_{\alpha, \beta}^\lambda V(\alpha)_k \otimes V(\beta)_{n-k}$$
where $V(\lambda)_n$ is the irreducible polynomial representations corresponding to a partition (or Young diagram) $\lambda$ of $\mathrm{GL}_n(\mathbb C)$ and $c^\lambda_{\alpha,\beta}$ are the Littlewood-Richardson numbers. Is it true that
$$ \mathrm{ind}_{\mathrm{GL}_k\times \mathrm{GL}_{n-k}}^{\mathrm{GL}_n} V(\alpha)_k \otimes V(\beta)_{n-k} \cong \bigoplus_\lambda c_{\alpha,\beta}^\lambda V(\lambda)_n ?$$
(I know it is not but true but it should be true up to being semi-simple.)</strike> See <a href="https://math.stackexchange.com/questions/707929/reciprocity-for-branching-rules-of-mathrmgl-n-mathbb-c">new question here</a></p></li>
</ol>
| Mauro ALLEGRANZA | 108,274 | <p>See the comment on the original question :</p>
<blockquote>
<p>if $H$ is invertible and the arguments of $H^{-1}$ are $f_A$ and $f_B$, then for $f$ to exists we need $f_A$ and $f_B$ to agree at $A \cap B$, that is $\forall x \in A \cap B$, $f_A(x) = f_B(x)$.</p>
</blockquote>
<p>But the definition of Brian Scott couple $<f,g>$ gives you , for $c \in A \cap B$ that :</p>
<blockquote>
<p>$f(c) = 1$ and $g(c) = 0$.</p>
</blockquote>
|
1,218,912 | <p>My problem is following:
$$\binom{n}{r} + \binom{n+1}{r+1} + \binom{n+2}{r+2} + \dots + \binom{n+M}{r+M}$$</p>
<p>how can we reduce it to a more short solution</p>
<p>Here $\dbinom{n}{r} = \dfrac{n!}{r! (n-r)!}$ and thus same as regular.Please help me in solving the above expression</p>
| Marko Riedel | 44,883 | <p>This question can be treated using basic complex variables, which is
an instructive exercise.</p>
<p><P>Suppose we seek to compute
$$\sum_{q=0}^M {n+q\choose r+q}$$
with $n\ge r.$</p>
<p><P>Introduce the integral representation
$${n+q\choose r+q}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n+q}}{z^{r+q+1}} \; dz.$$</p>
<p>This gives for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n}}{z^{r+1}}
\sum_{q=0}^M \frac{(1+z)^q}{z^q}\; dz.$$</p>
<p>This is
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n}}{z^{r+1}}
\frac{(1+z)^{M+1}/z^{M+1}-1}{(1+z)/z-1} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n}}{z^{r}}
\frac{(1+z)^{M+1}/z^{M+1}-1}{(1+z)-z} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n}}{z^{r}}
\left((1+z)^{M+1}/z^{M+1}-1\right) \; dz.$$</p>
<p>This has two pieces, the first is
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n+M+1}}{z^{r+M+1}} \; dz
= {n+M+1\choose r+M}$$
and the second is
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n}}{z^{r}} \; dz
= {n\choose r-1}.$$</p>
<p>Combine these two to get
$${n+M+1\choose r+M} - {n\choose r-1}.$$</p>
|
4,474,803 | <p>Let <span class="math-container">$f$</span> be a differentiate function and consider two operators
<span class="math-container">$$
A(f(x))=\int_0^1 \frac{ d}{dx}f(x t^\mu) dt,\\
B(f(x))=\mu xf(x)+\int_0^x f(t) dt,
$$</span>
where <span class="math-container">$\mu $</span> is a parameter.</p>
<p>I need to prove that</p>
<p><span class="math-container">$$
A \circ B =B \circ A = I
$$</span></p>
<p>where <span class="math-container">$I$</span> is the identity operator <span class="math-container">$I(f(x))=f(x).$</span></p>
<p>It is easy to prove it for powers <span class="math-container">$x^n$</span> and then for power series but it would be interesting to perform direct calculations and prove it for any function <span class="math-container">$f.$</span></p>
<p>But I was confused with calculation of <span class="math-container">$A(B(f(x))$</span>.
Any help?</p>
| Chris Eagle | 693,182 | <p>The answer to your general question is very much "no", because moving a quantifier past a connective can change the quantifier. For instance, <span class="math-container">$\neg \forall x \phi(x)$</span> is equivalent to <span class="math-container">$\exists x \neg\phi(x)$</span> and in the vast majority of cases is not equivalent to <span class="math-container">$\forall x \neg \phi(x)$</span>.</p>
|
966,267 | <p>Original PDE
$$T_t=\alpha T_{xx}$$
I need to solve this equation numerically and analytically and compared them. I've already done the numerical part. But I need to solve it analytically now. </p>
<p>Given the initial condition
$$T(x,0)=sin(\frac{\pi x}{L})$$
where $$L=1$$</p>
<p>I would like to find the exact solution of the heat equation.</p>
<p>I know what $$T(x,t)=\sum_{n=1}^{\infty}B_n sin(n\pi x)e^{-n^2\pi^2\alpha t}\\where\\B_n=2\int_0^1T(x,0)sin(n\pi x)dx$$</p>
<p>After evaluating this integral, I get the solution as</p>
<p>$$T(x,t)=\sum_{n=1}^{\infty}\frac{2sin(n\pi)}{(1-n^2)\pi} sin(n\pi x)e^{-n^2\pi^2\alpha t}$$</p>
<p>I think I've done something wrong here because $$n=1^{th}$$ term is not defined. Can someone point out my mistake if there is? Thank you!</p>
<p>Corrected</p>
<p>Bn is nonzero only at n=1. Evaluating the case for n=1, Bn=1</p>
<p>So the solution is </p>
<p>$$T(x,t)=sin(\pi x)e^{-\pi^2\alpha t}$$</p>
<p>Thanks to Leucippus and AlexZorn for the correction.</p>
| Leucippus | 148,155 | <p>Given the pde $u_{t} = \alpha u_{xx}$, $u(0,t) = u(L,t) = 0$ and $u(x,0) = \sin\left(\frac{\pi x}{L} \right)$ : the equation can be separated by use of $u(x,t) = F(t) G(t)$ and leads to the equations
\begin{align}
F' + \lambda^{2} \alpha F &= 0 \\
G'' + \lambda^{2} G &= 0
\end{align}
which have solutions
\begin{align}
F(t) &= e^{- \lambda^{2} \alpha t} \\
G(x) &= A \cos(\lambda x) + B \sin(\lambda x).
\end{align}
Now applying the boundary conditions it is seen that the general solution is
\begin{align}
u(x,t) = \sum_{n=1}^{\infty} B_{n} \, \sin\left( \frac{n \pi x}{L} \right) \, e^{- \frac{n^{2} \pi^{2} \alpha t}{L^{2}}}.
\end{align}
Applying the initial condition yields
\begin{align}
\sin\left(\frac{\pi x}{L} \right) = \sum_{n=1}^{\infty} B_{n} \, \sin\left( \frac{n \pi x}{L} \right).
\end{align}
This is a Fourier series for which the coefficients are given by
\begin{align}
B_{n} = \frac{2}{L} \, \int_{0}^{L} \sin\left(\frac{\pi x}{L} \right) \, \sin\left( \frac{n \pi x}{L} \right) \, dx.
\end{align}
For the case $n= 1$ it is seen that
\begin{align}
B_{1} = \frac{2}{L} \int_{0}^{L} \sin^{2}\left( \frac{\pi x}{L} \right) \, dx = 1
\end{align}
whereas for $n \geq 2$
\begin{align}
B_{n} = \frac{1}{2 \pi} \, \frac{\sin(n \pi) }{1 - n^{2}} = 0
\end{align}
since $n$ is an integer. The solution for $u$ is now
\begin{align}
u(x,t) = \sin\left( \frac{\pi x}{L} \right) \, e^{- \frac{\pi^{2} \alpha t}{L}}.
\end{align}</p>
|
252,679 | <p>I simply want to be able to assign the subscript from something like Subscript[x, 1] to a new variable. How should I do this?</p>
| Hans Olo | 61,267 | <p>If you have a variable</p>
<pre><code>y=Subscript[x, 1];
</code></pre>
<p>then to extract the index it suffices to do</p>
<pre><code>y[[2]]
</code></pre>
<p>which in this case returns 1 as expected.</p>
<p>For more than one indices, this works</p>
<pre><code>In[1]:= y=Subscript[x,1,2,3]
Out[1]= Subscript[x,1,2,3]
In[2]:= y[[2;;-1]]/.Subscript->List
Out[2]= {1,2,3}
</code></pre>
|
252,679 | <p>I simply want to be able to assign the subscript from something like Subscript[x, 1] to a new variable. How should I do this?</p>
| bbgodfrey | 1,063 | <p>For <code>y</code> with any number of subscripts ,</p>
<pre><code>Rest[List @@ y]
</code></pre>
<p>also works. For instance</p>
<pre><code>y = Subscript[x, 1, 2];
Rest[List @@ y]
(* {1, 2} *)
</code></pre>
<p>As a word of advice, use <code>Subscript</code> and the like for formatting expressions for display but not for symbolic computations, where it sometimes leads to problems.</p>
|
4,009,001 | <p>This problem was came to a Facebook post of mathigon ......the problem seems trickier than I expected...can you help me to find the distance??<a href="https://i.stack.imgur.com/CNQiX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CNQiX.png" alt="enter image description here" /></a></p>
| PTDS | 277,299 | <p><a href="https://i.stack.imgur.com/jmroO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jmroO.png" alt="enter image description here" /></a></p>
<p>Let <span class="math-container">$\angle CAB = \angle ABC = \angle BAD = \angle DBA = \theta$</span></p>
<p>Area of the <span class="math-container">$\bigtriangleup CAD = \frac{1}{2} 1^2 \sin 2 \theta = \frac{1}{2} \sin 2 \theta$</span></p>
<p>Area of the sector <span class="math-container">$CAD$</span> (consider the right side of the circle on the left) <span class="math-container">$ = \frac{1}{2} 1^2 2\theta = \frac{1}{2} (2 \theta)$</span></p>
<p>So the common area between the circles <span class="math-container">$ = 2 (\frac{1}{2} (2 \theta) - \frac{1}{2} \sin 2 \theta)$</span></p>
<p>However, it is given that the common area between the circles <span class="math-container">$= \frac{\pi}{2}$</span></p>
<p>Hence we need to solve the following equation for <span class="math-container">$\theta$</span></p>
<p><span class="math-container">$$2 \left(\frac{1}{2} \left(2 \theta \right) - \frac{1}{2} \sin 2 \theta \right) = \frac{\pi}{2}$$</span></p>
<p>Solving, <span class="math-container">$\theta = 1.15494$</span></p>
<p>Finally length of <span class="math-container">$AB$</span> is <span class="math-container">$2 \cos \theta = 0.80794684217$</span></p>
|
4,009,001 | <p>This problem was came to a Facebook post of mathigon ......the problem seems trickier than I expected...can you help me to find the distance??<a href="https://i.stack.imgur.com/CNQiX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CNQiX.png" alt="enter image description here" /></a></p>
| Quanto | 686,284 | <p>The area of the lens formed by two overlapping circle of radius <span class="math-container">$r$</span> and at distance <span class="math-container">$d$</span> apart is
given by <a href="https://mathworld.wolfram.com/Lens.html" rel="nofollow noreferrer">https://mathworld.wolfram.com/Lens.html</a></p>
<p><span class="math-container">$$Area= \pi r^2-2r^2 \tan^{-1}\frac d{\sqrt{4r^2 -d^2}}-\frac12d \sqrt{4r^2 -d^2}
$$</span>
Set <span class="math-container">$Area =\frac\pi2$</span> and <span class="math-container">$r=1$</span> to obtain numerically <span class="math-container">$d= 0.8079$</span>.</p>
|
2,668,779 | <p>The exercise consists of showing that the function $f(x,y)=x^4 + y^4$ has a global minimum and maximum under the constraint $x^4 + y^4 - 2xy = 2$.</p>
<p>In the solution to the exercise, it it follows that the constraint is compact if we can show that $\lim_{x^2 + y^2 \rightarrow \infty} x^4 + y^4 - 3xy - 2 \rightarrow \infty$.Why this is the case? </p>
<p>My intuition tells me that this is because the $x^4$ and $y^4$ terms dominates the other two terms when $x$ and $y$ gets large. This would then imply that $x$ and $y$ cannot get arbitrarily big without violating the constraint. Does this imply that if the limit of the constraint was $0$, that the domain would not be compact? Is my reasoning valid?</p>
<p>Many thanks,</p>
| Arian | 172,588 | <p>Take the set $V:=\{v\in\mathbb{R}^2: v=(x,0), v=(0,y), x,y\in\mathbb{R}\}$ Clearly if $v\in V$ then for any $\alpha\in\mathbb{R}$ we have $\alpha v\in V$, but $v_1+v_2\notin V$ where $v_1=(x,0),v_2=(0,y)$ for any such pairs and $x\neq0, y\neq 0$. </p>
|
2,668,779 | <p>The exercise consists of showing that the function $f(x,y)=x^4 + y^4$ has a global minimum and maximum under the constraint $x^4 + y^4 - 2xy = 2$.</p>
<p>In the solution to the exercise, it it follows that the constraint is compact if we can show that $\lim_{x^2 + y^2 \rightarrow \infty} x^4 + y^4 - 3xy - 2 \rightarrow \infty$.Why this is the case? </p>
<p>My intuition tells me that this is because the $x^4$ and $y^4$ terms dominates the other two terms when $x$ and $y$ gets large. This would then imply that $x$ and $y$ cannot get arbitrarily big without violating the constraint. Does this imply that if the limit of the constraint was $0$, that the domain would not be compact? Is my reasoning valid?</p>
<p>Many thanks,</p>
| John Coleman | 294,695 | <p>You could just take the empty set. Since it is empty it isn't a vector space, but it is vacuously closed under scalar multiplication.</p>
<p>Obviously this is a somewhat cheap example, but it does solve the problem as stated, and underscores that existential properties (e.g. the existence of a 0 vector) can't be derived from universally quantified implications.</p>
|
281,325 | <p>I am not very familiar with Stone-Čech compactification, but I would like to understand why the remainder $\beta\mathbb{R}\backslash\mathbb{R}$ has exactly two connected components.</p>
| Seirios | 36,434 | <p>I finally found something:</p>
<p>For convenience, let $X= \mathbb{R} \backslash (-1,1)$. </p>
<p>First, we show that $(\beta \mathbb{R})\backslash (-1,1)= \beta (\mathbb{R} \backslash (-1,1))$. Let $f_0 : X \to [0,1]$ be a continuous function. Obviously, $f_0$ can be extend to a continuous function $f : \mathbb{R} \to [0,1]$; then we extend $f$ to a continuous function $\tilde{f} : \beta \mathbb{R} \to [0,1]$ and we get by restriction a continuous function $\tilde{f}_0 : (\beta \mathbb{R}) \backslash (-1,1) \to [0,1]$. By the universal property of Stone-Čech compactification, we deduce that $\beta X= (\beta \mathbb{R}) \backslash (-1,1)$.</p>
<p>With the same argument, we show that for an closed interval $I \subset X$, $\beta I = \text{cl}_{\beta X}(I)$.</p>
<p>Let $F_1= \text{cl}_{\beta X} [1,+ \infty)$ and $F_2= \text{cl}_{\beta X} (-\infty,-1]$. Then, $F_1 \cup F_2 = \beta X$, $F_1 \cap F_1= \emptyset$ and $F_1$, $F_2$ are connected.</p>
<p>Indeed, there exists a continuous function $h : X \to \mathbb{R}$ such that $h_{|[1,+ \infty)} \equiv 0$ and $h_{|(-\infty,1]} \equiv 1$, so when we extend $h$ on $\beta X$ we get a continuous function $\tilde{h}$ such that $\tilde{h} \equiv 0$ on $F_1$ and $\tilde{h} \equiv 1$ on $F_2$; therefore $F_1 \cap F_2= \emptyset$. </p>
<p>Finally, $F_1$ and $F_2$ are connected as closures of connected sets.</p>
<p>So $(\beta \mathbb{R} ) \backslash (-1,1)$ has exactly two connected components; in fact, it is the same thing for $(\beta \mathbb{R}) \backslash (-n,n)$. Because $(\beta \mathbb{R}) \backslash \mathbb{R}= \bigcap\limits_{ n\geq 1 } (\beta \mathbb{R}) \backslash (-n,n)$, we deduce that $(\beta \mathbb{R}) \backslash \mathbb{R}$ has exactly two connected components (the intersection of a non-increasing sequence of connected compact sets is connected).</p>
|
4,574,446 | <p>This is question 13 on page 294 of Vector Calculus by Marsden and Tromba.</p>
<blockquote>
<p>Find the volume of the region determined by <span class="math-container">$x^2 + y^2 + z^2 \leq 10, z \geq 2.$</span></p>
</blockquote>
<p>I have attempted it as follows.</p>
<p>The region can be described by using spherical polar coordinates <span class="math-container">$(r,\theta,\phi)$</span> and we have <span class="math-container">$0 \leq r \leq \sqrt{10}$</span>, <span class="math-container">$0 \leq \theta \leq 2 \pi$</span>.</p>
<p>For the limits for <span class="math-container">$\phi$</span>, I think that we can find it by saying that <span class="math-container">$z=r \cos \phi = \sqrt{10} \cos \phi \geq 2$</span> using that <span class="math-container">$r = \sqrt{10}$</span> on the surface. This gives <span class="math-container">$0 \leq \phi \leq \cos^{-1}(2/\sqrt{10}).$</span> Hence I get</p>
<p><span class="math-container">\begin{align} \iiint_V dV &= \int_0^{2\pi} \int_0^{\cos^{-1}(2/\sqrt{10})}\int_0^{\sqrt{10}} r^2 \sin \phi drd\phi d\theta \\ &= 2\pi \int_0^{\cos^{-1}(2/\sqrt{10})} 10 \dfrac{\sqrt{3}}{3} \sin \phi d\phi \\&= 20\pi \dfrac{\sqrt{10}}{3} - 40\dfrac{\pi}{3}. \end{align}</span></p>
<p>However, the answer at the back of the book is</p>
<blockquote>
<p><span class="math-container">$20\pi \dfrac{\sqrt{10}}{3} - 52\dfrac{\pi}{3}$</span></p>
</blockquote>
<p>but I have been unable to identify the mistake. Please, could someone help me? Thank you very much.</p>
| Peter Phipps | 15,984 | <p>Let <span class="math-container">$n+2 = k^2+r$</span> where <span class="math-container">$0 \le r \lt 2k+1$</span>, so <span class="math-container">$k = \lfloor \sqrt{n+2}\rfloor$</span> is the largest integer less than <span class="math-container">$\sqrt{n+2}$</span>.</p>
<p>The requirement <span class="math-container">$k|n \rightarrow k|k^2+r-2 \rightarrow k| r-2 \rightarrow r = \lambda k+2$</span> for some integer <span class="math-container">$\lambda \ge 0$</span>.</p>
<p>The restriction <span class="math-container">$r \lt 2k+1$</span> means that <span class="math-container">$\lambda = 0$</span> or <span class="math-container">$1$</span>. <span class="math-container">$\lambda = 1$</span> leads to <span class="math-container">$n$</span> being even, so we are left with <span class="math-container">$\lambda = 0$</span>. So <span class="math-container">$r=2$</span> and <span class="math-container">$n = k^2$</span> as required.</p>
|
3,828,982 | <blockquote>
<p><strong>Statement</strong></p>
<p>If <span class="math-container">$U$</span> is open and <span class="math-container">$\text{int}(S)\neq\emptyset$</span> then <span class="math-container">$\text{int}(U\cap S)\neq\emptyset$</span> too when <span class="math-container">$U\cap S\neq\emptyset$</span> and when <span class="math-container">$S$</span> is path connected.</p>
</blockquote>
<p>We know that <span class="math-container">$\text{int}(U\cap S)=\text{int}(U)\cap\text{int}(S)=U\cap\text{int}(S)$</span> so that if <span class="math-container">$\text{int}(U\cap S)=\emptyset$</span> and <span class="math-container">$U\cap S\neq\emptyset$</span> then <span class="math-container">$(U\cap S)\subseteq\text{Bd}(S)$</span> but unfortunately I don't see in this any contradiction. So I don't be able to prove the statement. Could be that it is generally false? Anyway it seems to me that in the case where <span class="math-container">$U$</span> and <span class="math-container">$S$</span> are <strong>path connencted</strong> subset of <span class="math-container">$\Bbb R^n$</span> then the statement holds so I think that although it is gerenally false it could be true if we consider some particular case. If it is more difficul understand when the statement is generally true I ask to prove it when <span class="math-container">$U$</span> and <span class="math-container">$S$</span> are subset of <span class="math-container">$\Bbb R^n$</span>.
Finally I point out that the statement is generally false when <span class="math-container">$U$</span> or <span class="math-container">$S$</span> are not path connected: e.g. if you take <span class="math-container">$U=(0,1)$</span> and <span class="math-container">$S=\{\frac 1 2\}\cup(2,3)$</span> or <span class="math-container">$U=\big[(0,1)\cup(2,3)\big]$</span> and <span class="math-container">$S=[1,2]$</span> then the statement is clearly false!</p>
<p>So could someone help me, please?</p>
| Henno Brandsma | 4,280 | <p>Connectedness problems often degenerate in <span class="math-container">$\Bbb R$</span>, it's the plane where stuff gets interesting.</p>
<p><span class="math-container">$U=\{x\mid \|x\|< 1\}$</span> which is open, path-connected.</p>
<p><span class="math-container">$S= \{x \mid \|x\|\ge 1\} \cup \{(x,y):xy=0\}$</span>, the exterior plus axes, clearly <span class="math-container">$U \cap S \neq \emptyset$</span> and the interior of <span class="math-container">$S$</span> is <span class="math-container">$\{x\mid \|x\| > 1\} \neq \emptyset$</span>. <span class="math-container">$S$</span> is path-connected, and <span class="math-container">$S \cap U$</span> has empty interior (but is path-connected).</p>
|
3,025,375 | <p>What is<span class="math-container">$$\lim_{n→∞}n^3(\sqrt{n^2+\sqrt{n^4+1}}-n\sqrt2)?$$</span>So it is<span class="math-container">$$\lim_{n→∞}\frac{n^3(\sqrt{n^2+\sqrt{n^4+1}})^2-(n\sqrt{2})^2}{\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}}=\lim_{n→∞}\frac{n^3(n^2+\sqrt{n^4+1}-2n^2)}{\sqrt{n^2+\sqrt{n^4+1}}+n\sqrt{2}}.$$</span>
I do not know what to do next, because my resuts is <span class="math-container">$∞$</span> but the answer from book is <span class="math-container">$\dfrac{1}{4\sqrt{2}}$</span>.</p>
| lab bhattacharjee | 33,337 | <p>First replace <span class="math-container">$1/x=h$</span> to find </p>
<p><span class="math-container">$$L=\lim_{n→∞}n^3(\sqrt{n^2+\sqrt{n^4+1}}-n\sqrt2)=\lim_{h\to0^+}\dfrac{\sqrt{1+\sqrt{1+h^4}}-\sqrt2}{h^4}$$</span></p>
<p>Let <span class="math-container">$\sqrt{1+\sqrt{1+h^4}}=y+\sqrt2\implies1+\sqrt{1+h^4}=(\sqrt2+y)^2=2+2\sqrt2y+y^2$</span></p>
<p><span class="math-container">$$1+h^4=(1+y(2\sqrt2+y))^2=1+2y(2\sqrt2+y)+y^2(2\sqrt2+y)^2$$</span></p>
<p><span class="math-container">$$L=\lim_{y\to0}\dfrac{1+2y(2\sqrt2+y)+y^2(2\sqrt2+y)^2-1}y=?$$</span></p>
|
2,888,913 | <p>4 Boys & 4 Girls are to be seated in a line find number of ways , so that Boys & Girls are in alternate seats.</p>
<p>My approach:</p>
<p>If boys are seated in B$1$,B$2$,B$3$,B$4$ positions than at each gap between two consecutive boys a girl can sit so, there will be <strong>C$(5,4)$</strong> ways for girls and they can be arranged in <strong>C$(5,4)$</strong> *4! and boys too can be arranged in 4! so total number of ways are <strong>C$(5,4)$</strong> *4! *4!</p>
<p>But in textbook it's answer is 2*4!*4!</p>
<p><a href="https://i.stack.imgur.com/DsLnj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DsLnj.png" alt="enter image description here"></a> </p>
<p><strong>Please help me in finding my mistake and explain it too</strong></p>
<p>Please note: I am just a high school student . So,please don't close my question.</p>
| BallBoy | 512,865 | <p>One of the $C(5,4)$ ways you count to seat the girls fills gaps 1,3,4,5. But then boys B1 and B2 are seated next to each other, so boys and girls don't alternate. So there are not $C(5,4)$ ways to choose the spots for girls. </p>
|
2,888,913 | <p>4 Boys & 4 Girls are to be seated in a line find number of ways , so that Boys & Girls are in alternate seats.</p>
<p>My approach:</p>
<p>If boys are seated in B$1$,B$2$,B$3$,B$4$ positions than at each gap between two consecutive boys a girl can sit so, there will be <strong>C$(5,4)$</strong> ways for girls and they can be arranged in <strong>C$(5,4)$</strong> *4! and boys too can be arranged in 4! so total number of ways are <strong>C$(5,4)$</strong> *4! *4!</p>
<p>But in textbook it's answer is 2*4!*4!</p>
<p><a href="https://i.stack.imgur.com/DsLnj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DsLnj.png" alt="enter image description here"></a> </p>
<p><strong>Please help me in finding my mistake and explain it too</strong></p>
<p>Please note: I am just a high school student . So,please don't close my question.</p>
| WaveX | 323,744 | <p>Look at it this way, we have$$BGBGBGBG$$</p>
<p>We have $4$ ways for first boy, $4$ ways for first girl, $3$ ways for second boy, $3$ ways for second girl, and so on.</p>
<p>So we have $$4*4*3*3*2*2*1*1 = 4!*4!$$</p>
<p>But note that we could have also started with the pattern $$GBGBGBGB$$ so we need to double our previous answer, giving the desired result.</p>
<p><hr>
Edit: The original answer you found was number of ways only girls were not sitting next to each other. For your method to work, you need to complete an additional step where you subtract number of ways that girls aren't sitting next to each other but some of the boys are.</p>
<p>To find this number, note that girls must be on both ends of the line. So now we fill up our holes: $$GB\_ GB \_ GB\_ G$$
We have $3$ slots to choose for the last remaining boy. Therefore the number of ways girls are alternating but boys are touching is $C(3,1) = 3$ (and don't forget your permutations of these $3$ arrangements) So our new answer would be:</p>
<p>$$C(5,4) *4!*4! - C(3,1) *4!*4!$$
$$= 5*4!*4! - 3*4!*4!$$
$$=(5-3)*4!*4!$$
$$=2*4!*4!$$
Personally I feel this is a longer and slightly more complicated way to solve this problem, but it shows that your method does work if you include the extra step to correct it</p>
|
4,192,879 | <p>This question is about the <em>provided solution</em> to Question 11d from Chapter 8 of Spivak's Calculus. The relevant background material (i.e. 11a-11c) is as follows:</p>
<blockquote>
<p><span class="math-container">$(a)$</span> If <span class="math-container">$\{a_n\}$</span> is a sequence of positive terms such that <span class="math-container">$$a_{n+1}\leq a_n/2$$</span> then for every <span class="math-container">$\epsilon>0$</span> there exists an <span class="math-container">$n$</span> with <span class="math-container">$a_n<\epsilon$</span>.</p>
<p><span class="math-container">$(b)$</span> Let <span class="math-container">$P$</span> be a regular polygon inscribed in a circle. Let <span class="math-container">$P'$</span> be the regular polygon with twice the sides of <span class="math-container">$P$</span>. Let <span class="math-container">$A$</span> be the area of <span class="math-container">$P$</span>, <span class="math-container">$A'$</span> be the area of <span class="math-container">$P'$</span> and <span class="math-container">$C$</span> that of the circle. Prove</p>
<p><span class="math-container">$$(C-P')\leq \frac 1 2 (C-P)$$</span></p>
<p><span class="math-container">$(c)$</span> Prove there is a regular inscribed polygon with area as close as that to the circle.</p>
</blockquote>
<p>Part d reads as follows:</p>
<blockquote>
<p>Using the fact that the areas of two regular polygons with the same number of sides are in the same relation as the squares of their sides, prove the areas of two circles are in the same relation as the squares of their radii.</p>
</blockquote>
<p>Here is Spivak's solution (<strong>I will include an indexed dagger symbol <span class="math-container">$\dagger_i$</span> in places that I will subsequently ask questions about</strong>):</p>
<hr />
<p>Let <span class="math-container">$r_1$</span> and <span class="math-container">$r_2$</span> be the radii of the two circles <span class="math-container">$C_1$</span> and <span class="math-container">$C_2$</span>, and let <span class="math-container">$A_i$</span> be the area of the region bounded by <span class="math-container">$C_i$</span>. We know that there are numbers <span class="math-container">$\delta_1, \delta_2 \gt 0$</span> such that</p>
<p><span class="math-container">$$\Big\lvert \frac{A_1}{A_2}-\frac{B_1}{B_2} \Big\rvert \lt \epsilon \quad \dagger_1$$</span></p>
<p>for any numbers <span class="math-container">$B_1, B_2$</span> with <span class="math-container">$|A_i-B_i| \lt \delta_i \quad \dagger_2$</span>. By part (c), there are numbers <span class="math-container">$n_i$</span> such that the area of a regular polygon, with <span class="math-container">$n_i$</span> sides, inscribed in <span class="math-container">$C_i$</span> differs from <span class="math-container">$A_i$</span> by less than <span class="math-container">$\delta_i$</span>. Let <span class="math-container">$P_i$</span> be the area of a regular polygon inscribed in <span class="math-container">$C_i$</span> with <span class="math-container">$\max(n_1,n_2)$</span> sides. Then</p>
<p><span class="math-container">$$\Big\lvert \frac{A_1}{A_2}-\frac{P_1}{P_2} \Big\rvert \lt \epsilon$$</span></p>
<p><span class="math-container">$$\Big\lvert \frac{A_1}{A_2}-\frac{r_1^{\ 2}}{r_2^{\ 2}} \Big\rvert \lt \epsilon$$</span></p>
<p>Since this is true for each <span class="math-container">$\epsilon \gt 0$</span>, it follows that <span class="math-container">$A_1/A_2=r_1^{\ 2}/r_2^{\ 2}$</span></p>
<hr />
<p>Referencing <span class="math-container">$\dagger_1$</span> and <span class="math-container">$\dagger_2$</span>, which I view as being directly connected, how is the equation <span class="math-container">$\Big\lvert \frac{A_1}{A_2}-\frac{B_1}{B_2} \Big\rvert \lt \epsilon$</span>, where <span class="math-container">$\epsilon$</span> is <em>arbitrary</em>, derived from the two equations: (1) <span class="math-container">$|A_1-B_1| \lt \delta_1$</span> and <span class="math-container">$|A_2-B_2| \lt \delta_2$</span>. Further, where are the statements <span class="math-container">$|A_1-B_1| \lt \delta_1$</span> and <span class="math-container">$|A_2-B_2| \lt \delta_2$</span> coming from? (If <span class="math-container">$B_1$</span> and <span class="math-container">$B_2$</span> are arbitrary numbers, and <span class="math-container">$\delta_1$</span> and <span class="math-container">$\delta_2$</span> are constructed numbers, I am having difficulties seeing how the <span class="math-container">$\delta$</span>'s can be related to an arbitrary <span class="math-container">$\epsilon$</span>)</p>
| Ben | 754,927 | <p>Recall the lemmas used in the proofs concerning limits for products, and quotients:</p>
<blockquote>
<p>(2) If
<span class="math-container">$$|x - x_0| < \min\left(1,\frac{\varepsilon}{2(|y_0|+ 1)}\right) \text{ and } |y - y_0| < \frac{\varepsilon}{2(|x_0|+ 1)},$$</span>
then
<span class="math-container">$$ |xy - x_0y_0| < \varepsilon.$$</span></p>
<p>(3) If <span class="math-container">$y_0 \neq 0$</span> and
<span class="math-container">$$|y-y_0| < \min\left(\frac{|y_0|}{2}, \frac{\varepsilon|y_0|^2}{2}\right),$$</span>
then <span class="math-container">$y \neq 0$</span> and
<span class="math-container">$$\left|\frac{1}{y} - \frac{1}{y_0}\right| < \varepsilon.$$</span></p>
</blockquote>
<p>Informally, given numbers <span class="math-container">$x_0$</span> and <span class="math-container">$y_0$</span> with <span class="math-container">$y_0 \neq 0$</span>, we can make <span class="math-container">$\frac{x}{y}$</span> arbitrarily close to <span class="math-container">$\frac{x_0}{y_0}$</span> by restricting <span class="math-container">$x$</span> and <span class="math-container">$y$</span> to be sufficiently close to <span class="math-container">$x_0$</span> and <span class="math-container">$y_0$</span>.</p>
<p>Spivak applies this same argument here. Let <span class="math-container">$n$</span> be the number of sides of the regular polygons inscribed in the circles <span class="math-container">$C_1$</span> and <span class="math-container">$C_2$</span>. (The polygons each have <span class="math-container">$n$</span> sides. They are both regular <span class="math-container">$n$</span>-gons). <span class="math-container">$P_1$</span> and <span class="math-container">$P_2$</span>, the areas of these polygons can be made arbitrarily close to the areas of their inscribing circles <span class="math-container">$A_1$</span> and <span class="math-container">$A_2$</span> by choosing <span class="math-container">$n$</span> to be large enough.</p>
<p>Therefore, we can make the ratio <span class="math-container">$\frac{P_1}{P_2}$</span> arbitrarily close to <span class="math-container">$\frac{A_1}{A_2}$</span>.</p>
<p>Spivak then notes that for either polygon, its area is proportional to its incribing circle's radius squared, that is
<span class="math-container">$$\frac{P_1}{P_2} = \frac{{r_1}^2}{{r_2}^2},$$</span>
where <span class="math-container">$r_i$</span> is the radius of circle <span class="math-container">$C_i$</span>.</p>
<p>As we chose bigger and bigger <span class="math-container">$n$</span>, <span class="math-container">$\frac{P_1}{P_2}$</span> remains constant and equal to <span class="math-container">$\frac{{r_1}^2}{{r_2}^2}$</span>. And yet, we know by choosing the right <span class="math-container">$n$</span> we can get arbitrarily close to <span class="math-container">$\frac{A_1}{A_2}$</span>. The only way this is possible is if
<span class="math-container">$$\frac{A_1}{A_2} = \frac{P_1}{P_2} = \frac{{r_1}^2}{{r_2}^2}.$$</span></p>
<p>That is, the polygon areas get closer and closer to the circle areas, but the ratios remain constant</p>
<p>Finally, note that this is one of those problems that relies a bit on "outside" facts that aren't quite rigorously justified here. That the area of the polygons is proportional to <span class="math-container">$r^2$</span> is hopefully not too hard to swallow.</p>
<p>You might find the following geometric argument mildly convincing:</p>
<p>Around the outside of each circle <span class="math-container">$C_i$</span> let us construct a square with sides of length <span class="math-container">$2r_i$</span> (The circles are themselves inscribed within the squares.)</p>
<p>Now, this argument hinges on the expectation that the polygons, since they each have <span class="math-container">$n$</span> sides should "punch out" the same proportion of the areas of their inscribing squares, that is,
<span class="math-container">$$P_i = k \cdot S_i,$$</span>
where <span class="math-container">$S_i$</span> is the area of the square and <span class="math-container">$k$</span> is some constant. (To be more precise, we could write <span class="math-container">$k_n$</span>. <span class="math-container">$k$</span> will be different depending on the number of sides, but this doesn't matter for what follows.)</p>
<p>If we accept this, we have
<span class="math-container">\begin{align}
\frac{P_1}{P_2} & = \frac{k\cdot S_1}{k\cdot S_2} \\
& = \frac{S_1}{S_2} \\
& = \frac{2r_1 \cdot 2r_1}{2r_2 \cdot 2r_2} \\
& =\frac{{r_1}^2}{{r_2}^2}.
\end{align}</span></p>
<p>(IIRC you can avoid this hand wavy "proportional punch out" talk and get the same result on somewhat firmer ground by instead using triangle areas to construct the areas of <span class="math-container">$n$</span>-gons and so forth.)</p>
<p>This problem provides a cool result, but we're not yet in a position to 100% prove it all with absolute rigor. The concept of "area" hasn't been formally introduced, much less justification for the area formulas for triangles or other polygons.</p>
|
2,738,023 | <p>We know we can check if two vectors are 'orthogonal' by doing an inner product.</p>
<p>$a*b=0$</p>
<p>tells us that these two vectors are orthogonal</p>
<p>here comes the question:</p>
<p>if there a way to compute if they are 'parallel'? i.e., they are pointing at the same direction. </p>
| Michael Hoppe | 93,935 | <p>Two vectors are parallel iff the absolute value of their dot product equals the product of their lengths.</p>
<p>Iff their dot product equals the product of their lengths, then they “point in the same direction”.</p>
|
2,064,403 | <p>Example. Let $V$ be a finite dim vector space with two different bases</p>
<p>$S = \{ u_1,u_2 \} = \{ (1,2),(3,5) \}$ and $S' = \{ v_1, v_2 \} = \{ (1,-1), (1,-2) \} $</p>
<p>You can check that $v_1 = -8u_1 + 3u_2$ and $v_2 = -11u_1+4u_2$ and $P = \begin{bmatrix}
-8 &-11 \\
3& 4
\end{bmatrix}$ is the change of basis where the columns are the coords.</p>
<p>But why can't we define change of basis by their rows? Since it works nicely that</p>
<p>$\begin{bmatrix}
v_1 \\
v_2
\end{bmatrix} = \begin{bmatrix}
-8 &3 \\
-11& 4
\end{bmatrix}\begin{bmatrix}
u_1 \\
u_2
\end{bmatrix}$</p>
<p>So to move from the old basis $S$, you apply the matrix $\begin{bmatrix}
-8 &3 \\
-11& 4
\end{bmatrix}$ to get a new basis. Why do we have to transpose? If you transpose, how do you even use this change of basis? Why can't I use this definition of change of basis.</p>
| levap | 32,262 | <p>This is a good question. It might be comforting to know that there is always some arbitrary choices involved in the whole issue of representing vectors of an abstract vector space as an array of numbers, linear maps as a double array of numbers (matrices) and what one means by changing a basis. The most important thing is to set up a notation that minimizes the number of arbitrary choices and is self consistent. </p>
<p>Let me try and explain the motivation behind the most popular notation and then reconsider your example. Fix a vector space $V$ and let $\mathcal{B} = (v_1, \dots, v_n)$ be some basis of $V$. The basis $\mathcal{B}$ allows us to identify a vector $v \in V$ with a list of scalars by representing $v$ (uniquely) as $v = a_1 v_1 + \dots + v_n a_n$ and identifying $v$ with the list $(a_1,\dots,a_n)$ which is called the coordinates of $v$ with respect to $\mathcal{B}$. The <strong>convention</strong> is that we treat this list as a <strong>column</strong> vector and write</p>
<p>$$ [v]_{\mathcal{B}} := \begin{pmatrix} a_1 \\ \vdots \\ a_n \end{pmatrix}. $$</p>
<p>Given a linear map $T \colon V \rightarrow W$, a basis $\mathcal{B} = (v_1,\dots,v_n)$ of $V$ and a basis $\mathcal{C} = (w_1,\dots,w_m)$ of $W$, we can represent each vector $T(v_i)$ as a linear combination $T(v_j) = \sum_{i=1}^m a_{ij} w_i$. The <strong>convention</strong> is that we treat the array $A = (a_{ij})$ as a double array (which we call a matrix) for which $i$ is the row index and $j$ is the column index. The matrix $A \in M_{m \times n}(\mathbb{F})$ is denoted by $A = [T]^{\mathcal{B}}_{\mathcal{C}}$ and is called the matrix representing $T$ with respect to the basis $\mathcal{B}$ (of the domain) and the basis $\mathcal{C}$ (of the codomain). This convention has the slightly annoying feature (especially to beginners) that a linear map from an $n$-dimensional space to an $m$-dimensional space is represented by an $m \times n$ matrix (so the dimensions are "reversed") but its most important advantage is that it identifies matrix multiplication and composition/evaluation. Namely, we have the following formulas:</p>
<p>$$ [T(v)]_{\mathcal{C}} = [T]^{\mathcal{B}}_{\mathcal{C}} \cdot [v]_{\mathcal{B}}, \,\,\, [T \circ S]^{\mathcal{B}}_{\mathcal{D}} = [T]^{\mathcal{C}}_{\mathcal{D}} \cdot [S]_{\mathcal{C}}^{\mathcal{B}} $$</p>
<p>where $\cdot$ is matrix multiplication.</p>
<p>Finally, let us discuss the change of basis matrices. If $\mathcal{B} = (u_1,\dots,u_n)$ and $\mathcal{B}' = (v_1,\dots,v_n)$ are two bases of $V$, the change of basis matrix "from $\mathcal{B}'$ to $\mathcal{B}$" is the matrix $P = [\operatorname{id}]_{\mathcal{B}}^{\mathcal{B'}}$ where $\operatorname{id} \colon V \rightarrow V$ is the identity transformation. Using the properties above, we see that we have</p>
<p>$$ P[v]_{\mathcal{B}'} = [\operatorname{id}]_{\mathcal{B}}^{\mathcal{B'}} [v]_{\mathcal{B}'} = [v]_{\mathcal{B}}. $$</p>
<p>Thus, given the <strong>coordinates</strong> a vector $v \in V$ in the "new basis" $\mathcal{B}'$, the matrix $P$ allows us to compute the coordinates of $v$ in the "old basis" $\mathcal{B}$ by performing matrix multiplication. The decision which basis to call "the old" and which "the new" is not entirely standard and depends on whether you prefer to change basis vectors or coordinates. In physics, this is related to the "passive v.s active" point of view of linear transformations.</p>
<hr>
<p>Finally, let me reconsider your example. We have $V = \mathbb{R}^2$, $\mathcal{B} = (u_1 = (1,2),u_2 = (3,5))$ and $\mathcal{B}' = (v_1 = (1,-1), v_2 = (1,2))$. When representing elements in the basis $\mathcal{B}$, I'll use the letter $a$ and when writing elements in the basis $\mathcal{B}'$ I'll use the letter $b$ for the coefficients. That is,</p>
<p>$$ v = a_1 u_1 + a_2 u_2 = b_1 v_1 + b_2 v_2. $$</p>
<p>The matrix $P$ has the feature that</p>
<p>$$ P \begin{pmatrix} b_1 \\ b_2 \end{pmatrix} = \begin{pmatrix} a_1 \\ a_2 \end{pmatrix} $$</p>
<p>and so it tells you how to transform the coordinates of an arbitrary vector in the basis $\mathcal{B}'$ to its coordinates in the basis $\mathcal{B}$. For example, if </p>
<p>$$ v = 1 \cdot v_1 + 1 \cdot v_2 = 1 \cdot (-8u_1 + 3u_2) + 1 \cdot (-11u_1 + 4u_2) = -19 u_1 + 7 u_2 $$</p>
<p>we have</p>
<p>$$ [v]_{\mathcal{B}} = \begin{pmatrix} -19 \\ 7 \end{pmatrix}, \,\,\, [v]_{\mathcal{B}'} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}, \\
\begin{pmatrix} -8 && -11 \\ 3 && 4 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} -8 && -11 \\ 3 && 4 \end{pmatrix} [v]_{\mathcal{B}'} = \begin{pmatrix} -19 \\ 7 \end{pmatrix} = [v]_{\mathcal{B}}. $$</p>
|
21,257 | <p>I am interested in examples where the <a href="http://en.wikipedia.org/wiki/Shooting_method" rel="nofollow">Shooting Method</a> has been used to find solutions to systems of ordinary differential equations that are either </p>
<ul>
<li>reasonably large systems, or </li>
<li>the search algorithm in the shooting parameters is somewhat prohibitive because of the nature of the solutions, or</li>
<li>both of the above.</li>
</ul>
<p>Any references, descriptions, recent progress, folklore, in the ballpark would be of interest. Feel free to interpret "reasonably large" subjectively if necessary. </p>
| Jitse Niesen | 2,610 | <p>Another shameless plug ... Coworkers and I used the Evans function formalism, which is a variant of the shooting method to deal with unstable directions (probably the same problem as mentioned by yfarjoun), on a boundary value problem of the form $y'(t) = (\lambda A_1 + A_2(t)) y(t)$ with $y$ specified as $t \to \pm \infty$. This is very similar to a Sturm-Liouville problem except that the differential operator is not self-adjoint. The application we're interested in is to do stability analysis of a travelling wave of a 2d reaction-diffusion equation. The main problem is that $y(t)$ is a fairly big vector with up to about 200 entries. </p>
<p>For details, please ask or see the paper at <a href="http://arxiv.org/abs/0805.1706" rel="nofollow">http://arxiv.org/abs/0805.1706</a> and references therein.</p>
|
3,579,739 | <p>Its also mentioned that this question is an example of <strong>Sampling without replacement</strong></p>
<p>My question is , by the general method of how I do these kind of problems , I would assume that there are two possibilities</p>
<ol>
<li><p>You chose a defective bottle and then a non defective which will have the probability of </p>
<pre><code> 1/5 X 4/4 = 4/20
OR
</code></pre></li>
<li><p>Choosing a non defective one first and then a defective one which will have the probability of</p>
<pre><code>4/5 x 1/4 =4/20
</code></pre></li>
</ol>
<p>Total probability is 8/20= 2/5 which doesn't make sense at all, since there is just one bottle to begin with
But the problem is I cant find a fault in my logic. Though I feel both the options are redundant , isn't both two different ways of selecting the pair?. Or is the fact that the phrase "one after the other" is absent a reason on why this might be a wrong approach?</p>
<pre><code> Thank you in advance
</code></pre>
| Jaap Scherphuis | 362,967 | <p>Here is another viewpoint.</p>
<p>Suppose you don't know there is a flawed bottle. You choose two of the five bottles. You are then told that one of the five bottles contains a crack. It is equally likely to be any of the five bottles, so each bottle has a probability of <span class="math-container">$\frac{1}{5}$</span> of being the cracked one.</p>
<p>You hold two bottles, so the probability of either one of them being the cracked one is <span class="math-container">$\frac{1}{5}+\frac{1}{5}=\frac{2}{5}$</span>. Note that you can add these two probabilities because they are mutually exclusive events - they cannot both be cracked as there is only one cracked bottle.</p>
<p>Edit: I see now that JMP already briefly mentioned this at the end of his answer.</p>
|
3,287,067 | <p>I'm reading a solution to the following exercise:</p>
<p>"Assume that <span class="math-container">$\lim_{x\to c}f\left(x\right)=L$</span>, where <span class="math-container">$L\ne0$</span>, and assume <span class="math-container">$\lim_{x\to c}g\left(x\right)=0.$</span> Show that <span class="math-container">$\lim_{x\to c}\left|\frac{f\left(x\right)}{g\left(x\right)}\right|=\infty.$</span>" </p>
<p>And at some point in the proof the following step appears:</p>
<p>"Choose <span class="math-container">$\delta_1$</span> so that <span class="math-container">$0<\left|x-c\right|<\delta _1$</span> implies <span class="math-container">$\left|f\left(x\right)-L\right|<\frac{|L|}{2}$</span>. <strong>Then we have <span class="math-container">$\left|f\left(x\right)\right|\ge\frac{\left|L\right|}{2}$</span></strong>."</p>
<p>It's precisely the implication in bold that I'm struggling to understand. How does the writer go from <span class="math-container">$\left|f\left(x\right)-L\right|<\frac{\left|L\right|}{2}$</span> to <span class="math-container">$\left|f\left(x\right)\right|\ge\frac{\left|L\right|}{2}$</span>? </p>
<p>I'm probably failing to see something that may be very clear, but I've been attempting unsuccessfully to reach the conclusion algebraically long enough, and can't quite see why it is true either! </p>
<p>Here is the rest of the solution, if necessary. </p>
<p>"Let <span class="math-container">$M>0\ $</span> be arbitrary. [...]. Because <span class="math-container">$\lim_{x\to c}g\left(x\right)=0$</span>, we can choose <span class="math-container">$\delta_2$</span> such that <span class="math-container">$\left|g\left(x\right)\right|<\frac{\left|L\right|}{2M}\ $</span>provided <span class="math-container">$0<\left|x-c\right|<\delta_2$</span>. </p>
<p>Let <span class="math-container">$\delta=\min\left\{\delta_1,\delta_2\right\}.\ $</span> Then we have </p>
<p><span class="math-container">$\left|\frac{f\left(x\right)}{g\left(x\right)}\right|\ge\left|\frac{\frac{\left|L\right|}{2}}{\frac{\left|L\right|}{2M}}\right|=M$</span> provided <span class="math-container">$0<\left|x-c\right|<\delta$</span>, as desired." </p>
| J.G. | 56,861 | <p>In the spirit of the cosine example, take <span class="math-container">$f(x)=g\left(|x|-\frac12 a\right)$</span> for odd <span class="math-container">$g$</span> with <span class="math-container">$g\left(\frac12a\right)\ne0$</span>.</p>
|
3,930,466 | <p>I want to prove by induction that</p>
<p><span class="math-container">$$\left(\tfrac{1}{n}\sum_{k=1}^{n} x_k\right)^2\leq\tfrac{1}{n}\sum_{k=1}^{n} x_k^2$$</span></p>
<p>But I have no idea how. For n=1 it is trivial that
<span class="math-container">$$\left(\tfrac{1}{1}\sum_{k=1}^{1} x_k\right)^2\leq\tfrac{1}{1}\sum_{k=1}^{1} x_k^2=x_k^2$$</span></p>
<p>But if i want to conclude n+1 from n I get</p>
<p><span class="math-container">$$\left(\tfrac{1}{n+1}\sum_{k=1}^{n+1} x_k\right)^2\leq\tfrac{1}{n+1}\sum_{k=1}^{n+1} x_k^2$$</span></p>
<p>And now I cannot seperate <span class="math-container">$\tfrac{1}{n+1}$</span> in order to substitute the rest with the induction hypothesis.</p>
<p>Maybe one needs to use a different version of induction? Thanks for any help.</p>
| G Cab | 317,234 | <p>Hint:</p>
<p>you can resolve the <span class="math-container">$1/n+1$</span> hurdle by considering <em>weighted averages</em> of previous <span class="math-container">$n$</span> points and new added one.</p>
<p>Then consider that <span class="math-container">$f(x)=x^2$</span> is convex</p>
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.