qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
3,478,700 | <p>As you may know we can define the equation of a tangent line of a differentiable function at any point <span class="math-container">$a$</span> is given by:
<span class="math-container">$$y = f(a) + f'(a)(x-a)$$</span></p>
<blockquote>
<p>However how can I interpret this equation?
<span class="math-container">$$y = f(a) + f'(a)(x-a) + f''(a)(x-a)^2$$</span></p>
</blockquote>
<p>This would be very useful to me. Looks like a Taylor expansion at the point <span class="math-container">$a$</span>. However I can't see this geometrically.</p>
<p>If this doesn't have an answer, is there any geometric meaning to the third derivative of a function?</p>
| Lee Mosher | 26,501 | <p>The equation <span class="math-container">$y = f(a) + f'(a) (x-a)$</span> represents <strong>the only</strong> 1st degree polynomial function that has the same value and the same first derivative at <span class="math-container">$x=a$</span> as the function <span class="math-container">$f$</span>. Geometrically, this is the equation of the unique line that is the <strong>best fit</strong> or <strong>best approximation</strong> to the graph of <span class="math-container">$y=f(x)$</span> at the point <span class="math-container">$(a,f(a))$</span>.</p>
<p>The equation <span class="math-container">$y = f(a) + f'(a) (x-a) + \frac{f''(a)}{2} (x-a)^2$</span> represents the only 2nd degree polynomial function (quadratic function) that has the same value, the same first derivative, and the same second derivative at <span class="math-container">$x=a$</span> as the function <span class="math-container">$f$</span>. Geometrically, this is the equation of the unique parabola that is the best fit, or best approximation, to the graph of <span class="math-container">$y=f(x)$</span> at the point <span class="math-container">$(a,f(a))$</span>.</p>
<p>Want to guess what the equation <span class="math-container">$y = f(a) + f'(a) (x-a) + \frac{f''(a)}{2} (x-a)^2 + \frac{f'''(a)}{6}(x-a)^3$</span> represents?</p>
|
3,276,572 | <p>Let be <span class="math-container">$\lVert \cdot \rVert$</span> a matrix norm (submultiplicative).</p>
<p>Do we have for all matrices of determinant 1, the following lower bound:</p>
<p><span class="math-container">$$\lVert M \rVert \geq 1$$</span></p>
<p>I'm very confused and could not find any counterexample and I find this statement very fishy, I tried to experiment with:</p>
<p><span class="math-container">\begin{bmatrix}
1& x \\
0& 1
\end{bmatrix}</span></p>
<p>But, its Frobenius norm cannot be small enough.</p>
| Community | -1 | <p>If <span class="math-container">$\| \cdot \|$</span> is a matrix norm then <span class="math-container">$\rho(A) \leq \| A\|$</span> and <span class="math-container">$\rho(A) = 1$</span></p>
|
112,021 | <p>Let $n$ be a positive integer.
The $n$ by $n$ Fourier matrix may be defined as follows:</p>
<p>$$
F^{*} = (1/\sqrt{n}) (w^{(i-1)(j-1)})
$$</p>
<p>where </p>
<p>$$
w = e^{2 i \pi /n}
$$</p>
<p>is the complex $n$-th root of unity with smaller positive argument
and $*$ means transpose -conjugate.</p>
<p>It is well known that $F$ is diagonalizable with eigenvalues $1,-1,i,-i$</p>
<p>where $i^2 =-1.$</p>
<p>It is also known that $F$ has real eigenvectors:</p>
<p>COMMENT:
(I was unable to got this paper)</p>
<p>McClellan, James H.; Parks, Thomas W.
Eigenvalue and eigenvector decomposition of the discrete Fourier transform.
IEEE Trans. Audio Electroacoust. AU-20 (1972), no. 1, 66--74.
END of COMMENT</p>
<p>QUESTION:</p>
<p>There is some simple manner to get just one of these
real eigenvectors.</p>
<p>For example how to get a real vector with an odd number
$n=2k+1$ of coordinates and such that</p>
<p>$$
F(x) =x.
$$</p>
| Jeremy | 14,424 | <p>The vector $v = (1 - \sqrt{n}, 1, 1, 1, ...)$ is an eigenvector of $F(n)$ with an eigenvalue of -1 for all $n > 2$. To see this, note that the first row of $F(n)$ is $(\frac{1}{\sqrt{n}}, \frac{1}{\sqrt{n}}, \frac{1}{\sqrt{n}}, ...)$. From this is follows trivially that the first element of $F(n).v$ is $-1 + \sqrt{n}$. For all of the other rows of $F(n)$, we know that for a vector of all ones, the results of the row would sum to zero, since the row contains exactly the n roots of unity. Given this, and the fact the the first element of any row is again $\frac{1}{\sqrt{n}}$, if we were to multiply any row of $F(n)$ other than row 1, by the vector $(0,1,1,...)$, the total for a row is always $-\frac{1}{\sqrt{n}}$. Thus, for the vector $v$, we find for any row greater that zero, multiplying the row by $v$ results in $(1 - \sqrt{n})\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n}} = -1$</p>
|
4,114,180 | <p>The Theorem is as follows:</p>
<p>For any numbers x and y, the following statements are true:</p>
<ol>
<li><span class="math-container">$|x|<y$</span> if and only if <span class="math-container">$-y<x<y$</span></li>
<li><span class="math-container">$|x|\leq{y}$</span> if and only if <span class="math-container">$-y\leq{x}\leq{y}$</span></li>
<li><span class="math-container">$|x|\geq{y}$</span> if and only if either <span class="math-container">$x\leq{-y}$</span> or <span class="math-container">$x\geq{y}$</span></li>
<li><span class="math-container">$|x|>y$</span> if and only if either <span class="math-container">$x<{-y}$</span> or <span class="math-container">$x>{y}$</span></li>
</ol>
<p>Here's my progress so far,
<span class="math-container">$2|x|-3\geq{|x-1|}\\
\implies 2\left(|x|-\frac{3}{2}\right)\geq{|x-1|}\\
\implies |x|\geq{\frac{|x-1|}{2}+\frac{3}{2}}\\
\implies |x|\geq{\frac{|x-1|+3}{2}} \text{ can now use part 3.}\\
\implies x\leq{\frac{-|x-1|-3}{2}} \lor x\geq{\frac{|x-1|+3}{2}}\\$</span></p>
<p>I had this idea of replacing <span class="math-container">$|x-1|$</span> with 1 since it is the distance from 1 to x on the number line, but that is flawed since we could've done this initially and gotten the answer. I'm not sure where to go from here. Maybe I could've applied part 2 of the proof to the beginning of <span class="math-container">$|x-1|$</span> but then I'd arrive at <span class="math-container">$-(2|x|-3)\leq{|x-1|}\leq{2|x|-3}$</span> which doesn't help me at all. Any tips or hints?</p>
| Eric Towers | 123,905 | <p>You already have one absolute value isolated on one side of the inequality.</p>
<p><span class="math-container">$$ |x-1| \leq 2|x| - 3 \text{.} $$</span>
So apply part 2 to obtain
<span class="math-container">$$ -(2|x| - 3) \leq x-1 \leq 2|x| - 3 \text{.} $$</span></p>
<p>Work on <span class="math-container">$-(2|x| - 3) \leq x-1$</span> and <span class="math-container">$x-1 \leq 2|x| - 3$</span> separately, isolating the remaining absolute value in each. Now recall that <em>both</em> of these must be true at the same time and take the intersection of the results from each inequality.</p>
|
1,081,447 | <p>I'm talking about a Roulette wheel with $38$ equally probable outcomes. Someone mentioned that he guessed the correct number five times in a row, and said that this was surprising because the probability of this happening was $$\left(\frac{1}{38}\right)^5$$</p>
<p>This is true if you only play the game $5$ times. However, if you play it more than $5$ times there's a higher (should be much higher?) probability that you'll get $5$ in a row at <em>some point</em>. </p>
<p>I was thinking about <em>how</em> surprised this person should be at their streak of $m$ correct guesses given that they play $n$ games, each with probability $p$ of success. It makes intuitive sense that their surprise should be proportional to $1/q$ (or maybe $\log(1/q)$ since $1$ in a billion doesn't surprise you $10$ times more than $1$ in $100$ million), where $q$ is the probability that they get at least one streak of $m$ correct guesses at some point in their $n$ games. </p>
<p>So, with the Roulette example I was thinking about, $p=1/38$ and $m=5$. </p>
<p>I tried to find an explicit formula for $q$ in terms of $n$, and encountered some difficulty, because of the non-independence of "getting a streak in the first five tries" and "getting a streak in tries $2$ through $6$" (if the first is a failure, it's much more likely that the second will be too). </p>
<hr>
<p>In summary, two questions:</p>
<ol>
<li><p>How do I find the probability that you get $5$ correct guesses in a row at some point if you play $n$ games of Roulette?</p></li>
<li><p>More generally, what is the probability that you get $m$ successes at some point in a series of $n$ events, each with probability $p$ of success? </p></li>
</ol>
<p>The variables satisfy $\,\,\,m,n \in \mathbb{N}$, $\,\,\,m\leq n$, $\,\,\,p \in \mathbb{R}$, $\,\,\,0 \leq p \leq 1$.</p>
<hr>
<p>If we write the answer to the second question as a function $q(m,n,p)$, then we can say that $q$ should be increasing with $n$, decreasing with $m$, and increasing with $p$. It should equal $p^n$ when $m=n$ and should equal $1$ when $p=1$ and $0$ when $p=0$. </p>
<p>I feel as though this should be a basic probability problem, but I'm having trouble solving it. Maybe some kind of recursive approach would work? Given $q(n,m,p)$, I think I could write $q(n+1,m,p)$ using the probability that the last $m-1$ results are all successes ...</p>
| Samrat Mukhopadhyay | 83,973 | <p>To answer your general question, if the events are independent then the probability of getting <em>only</em> $m$ successes at a row <em>once</em> is $$(n-m+1)p^m(1-p)^{n-m}$$ This is because one can have success $m$ times at a row out of $n$ plays in $n-m+1$ ways and in each of these events have a probability $p^m$ of happening and the failures then happen in the rest of the $n-m$ cases with probability $(1-p)^{n-m}$. </p>
<p><strong>EDIT:</strong> As I understand now my previous answer does not correctly address the question. Now, let $P(m,n)$ denote the required probability which is of getting $m$ or more consecutive successes in a series of $n$ independent events. Then, basically we want the event that the <em>first</em> time $m>1$ consecutive successes occur. Now, let $R_k(m,n)$ denotes the probability that $m$ consecutive successes happen for the first time at $n\ge k> m$ th trail. Then we have $$R_k(m,n)=qR_{k-1}(m,n-1)+pqR_{k-2}(m,n-2)+p^2qR_{k-3}(m,n-3)+\cdots+p^{m-1}qR_{k-m}(m,n-m)$$ with $R_1(m,n)=\cdots=R_{m-1}(m,n)=0,R_{m}(m,n)=p^m$ where $q=1-p$. Now, the probability $R_{k}(m,n)$ should not depend upon the second argument as long as $k\le n$. Then calling only $R_{k}$, and defining $r=[R_1\cdots R_n]^T$, we have the equation $$R_k=qR_{k-1}+pqR_{k-2}+\cdots+qp^{m-1}R_{k-m}$$ which can be solved by solving the equation $$x^m-qx^{m-1}-\cdots-qp^{m-2}x-qp^{m-1}=0$$ to get $R_k=\sum_{i=1}^m a_i \lambda_i^k$ where $\lambda_i$ are the roots of the last equation and the coefficients $a_i$ can be obtained from the initial conditions. Then, we have $$P(n,m)=\sum_{k=n-m}^n R_k=\sum_{i=0}^m a_i \frac{\lambda_i^{n-m}(1-\lambda_i^{m+1})}{1-\lambda_i}$$</p>
|
1,231,095 | <p>How does one find $\mathcal{L}^{-1}\{\ln[\frac{s^2+a^2}{s^2+b^2}]\}$?</p>
<p>I've tried splitting it up into $\mathcal{L}^{-1}\{\ln(s^2+a^2)\}-\mathcal{L}^{-1}\{\ln(s^2+b^2)\}$. However, I can't think of any way to actually take the inverse transform of $\mathcal{L}^{-1}\{\ln(s^2+a^2)\}$.</p>
| doraemonpaul | 30,938 | <p>$\mathcal{L}^{-1}\left\{\ln\dfrac{s^2+a^2}{s^2+b^2}\right\}$</p>
<p>$=\mathcal{L}^{-1}\left\{\int_s^\infty\left(\dfrac{2s}{s^2+a^2}-\dfrac{2s}{s^2+b^2}\right)ds\right\}$</p>
<p>$=\dfrac{1}{t}\mathcal{L}^{-1}\left\{\dfrac{2s}{s^2+a^2}-\dfrac{2s}{s^2+b^2}\right\}$</p>
<p>$=\dfrac{2\cos at-2\cos bt}{t}$</p>
|
2,751,819 | <p>I need some help solving this.
I have tried:</p>
<p>$$
\begin{bmatrix}
a & b \\
c & d \\
\end{bmatrix}
=\frac{1}{\operatorname{det}A}\cdot \begin{bmatrix}
d & -b \\
-c & a \\
\end{bmatrix}$$
I ended up with $$a=\frac{d}{\operatorname{det}A},$$
and
$$d=\frac{a}{\operatorname{det}A}.$$
Then
$$\operatorname{tr}(A)=a+d=\frac{a+d}{\operatorname{det}A},$$
but I don't really think it works.</p>
| Ben Grossmann | 81,360 | <p>This problem is easier if you think about it in terms of the eigenvalues of $A$. Note that if $x$ is an eigenvector of $A$, we have $A^{-1}x = Ax$. What does this tell you about that eigenvalue? </p>
|
2,151,937 | <p>Let $A$ be an $m\times n$ real matrix, $x$ an $n\times 1$ vector and $b$ an $m\times 1$ vector. I want to compute
\begin{equation}
\dfrac{\partial }{\partial x} \Vert Ax+b\Vert^{2}.
\end{equation}
First, I expanded
\begin{equation}
\Vert Ax+b\Vert^{2}=(Ax+b)^{T}(Ax+b)=x^{T}A^{T}Ax+2x^{T}A^{T}b+b^{T}b
\end{equation}
then I computed
\begin{eqnarray}
\dfrac{\partial }{\partial x}(x^{T}A^{T}Ax+2x^{T}A^{T}b+b^{T}b)=A^{T}Ax+x^{T}A^{T}A+2A^{T}b
\end{eqnarray}
but I know the above is wrong since $A^{T}Ax$ and $x^{T}A^{T}A$ does not have the same dimention. Thanks for the help.</p>
| john316 | 418,508 | <p>Rather than expanding <em>first</em>, do the opposite. Define a new vector $$y=Ax+b$$ and write the function in terms of this new variable and the Frobenius product (which I'll denote by a colon). This approach reduces the visual "clutter". You can then expand the results <em>after</em> finding the derivative.</p>
<p>With the Frobenius product, finding the gradient is easy and fool-proof
$$\eqalign{
f &= \|y\|^2 = y:y \cr
\cr
df &= 2\,y:dy \cr
&= 2\,y:A\,dx \cr
&= 2\,A^Ty:dx\cr
\cr
\frac{\partial f}{\partial x} &= 2\,A^Ty \cr
&= 2\,A^T(Ax+b) \cr
\cr
}$$
The rules for rearranging the Frobenius product
$$\eqalign{
A:B &= B:A \cr
A:BC &= B^TA:C = AC^T:B\cr
}$$
can be derived from the familiar properties of the trace, since
$$A:B={\rm tr}(A^TB)$$</p>
|
2,779,429 | <blockquote>
<p>Evaluate $$\int \frac {dx}{\sin \frac x2\sqrt {\cos^3 \frac x2}}$$</p>
</blockquote>
<p>My try </p>
<p>Write $t=\frac x2$ and hence $dx=2dt$</p>
<p>To change the integral to $$\int \frac {\csc t dt}{\cos^{\frac 32} t}$$</p>
<p>Multiplying both bottom and top by $\csc t$ and then using $\csc^2 t=1+\cot^2 t$ in the numerator the problem simplifies to $$2\int (\sin t)(\cos ^{\frac {-3}{2}} t) dt+2\int \frac {\cot^3 t dt}{\sqrt {\cos t}}$$</p>
<p>Now the first integral is easy to go but I am not getting any idea for the second one. Any help would be very beneficial. New methods are also welcome. </p>
| Szeto | 512,032 | <p>Elaborating on answer of @JoseCarlosSantos:</p>
<p>By performing the substitution $t=u^2$,</p>
<p>$$\int\frac{dt}{(1-t^2)t\sqrt t}=\int\frac{2udu}{(1-u^4)(u^3)}=2\int\frac{du}{(1-u^4)u^2}$$</p>
<p>Performing partial fraction decomposition,</p>
<p>$$\frac1{(1-u^4)u^2}$$
$$=\frac1{u^2}+\frac{u^2}{1-u^4}$$
$$=\frac1{u^2}+\frac{u^2}{2(1+u^2)}+\frac{u^2}{2(1-u^2)}$$
$$= \frac1{u^2}+\frac{u^2+1-1}{2(1+u^2)}+\frac{u^2-1+1}{2(1-u^2)}$$
$$=\frac1{u^2}+\frac12-\frac{1}{2(1+u^2)}-\frac12+\frac{1}{2(1-u^2)}$$
$$= \frac1{u^2}-\frac{1}{2(1+u^2)}+\frac{1}{2(1-u^2)} $$
$$= \frac1{u^2}-\frac{1}{2(1+u^2)}+\frac{1}{4(1-u)}+\frac{1}{4(1+u)} $$</p>
<p>which can be integrated easily.</p>
|
3,152,021 | <p>I'm wondering if there are well known sorting techniques for the following problem.</p>
<p><strong>Problem</strong>:</p>
<p>Suppose you would like to sort a list of integer numbers <span class="math-container">$0, 1, 2, \ldots, d$</span>.
If one is only allowed to use swaps of adjacent positions the major part
of programmers would choose a bubble sort type algorithm.</p>
<p>The <span class="math-container">$d$</span> sized set of adjacent swaps <span class="math-container">$P^d_{\text{adj}}$</span> can be interpreted as the permutations, e.g. for <span class="math-container">$d=9$</span></p>
<p><span class="math-container">$P^9_{\text{adj}} = \{\\
\quad [1,0,2,3,4,5,6,7,8,9],\\
\quad [0,2,1,3,4,5,6,7,8,9],\\
\quad [0,1,3,2,4,5,6,7,8,9],\\
\quad [0,1,2,4,3,5,6,7,8,9],\\
\quad [0,1,2,3,5,4,6,7,8,9],\\
\quad [0,1,2,3,4,6,5,7,8,9],\\
\quad [0,1,2,3,4,5,7,6,8,9],\\
\quad [0,1,2,3,4,5,6,8,7,9],\\
\quad [0,1,2,3,4,5,6,7,9,8]\\
\}.$</span></p>
<p>Question: Are there sorting techniques for other sets of permutations?</p>
<p>I would like to use the following <span class="math-container">$1+d+d$</span> sized set, e.g. for <span class="math-container">$d=9$</span></p>
<p><span class="math-container">$P^9 = \{\\
\quad [0,1,2,3,4,5,6,7,9,8],\\
\quad\\
\quad [8,0,1,2,3,4,5,6,7,9],\\
\quad [8,1,0,2,3,4,5,6,7,9],\\
\quad [8,1,2,0,3,4,5,6,7,9],\\
\quad [8,1,2,3,0,4,5,6,7,9],\\
\quad [8,1,2,3,4,0,5,6,7,9],\\
\quad [8,1,2,3,4,5,0,6,7,9],\\
\quad [8,1,2,3,4,5,6,0,7,9],\\
\quad [8,1,2,3,4,5,6,7,0,9],\\
\quad [8,1,2,3,4,5,6,7,9,0],\\
\quad\\
\quad [9,0,1,2,3,4,5,6,7,8],\\
\quad [9,1,0,2,3,4,5,6,7,8],\\
\quad [9,1,2,0,3,4,5,6,7,8],\\
\quad [9,1,2,3,0,4,5,6,7,8],\\
\quad [9,1,2,3,4,0,5,6,7,8],\\
\quad [9,1,2,3,4,5,0,6,7,8],\\
\quad [9,1,2,3,4,5,6,0,7,8],\\
\quad [9,1,2,3,4,5,6,7,0,8],\\
\quad [9,1,2,3,4,5,6,7,8,0]\\
\}.$</span></p>
<p>For general <span class="math-container">$P^d$</span>, the first permutation swaps <span class="math-container">$d-1$</span> with <span class="math-container">$d$</span>.</p>
<p>The 1st subset of <span class="math-container">$d$</span> permutations start with <span class="math-container">$d-1$</span> and the
zero is moving to the right.</p>
<p>The 2nd subset of <span class="math-container">$d$</span> permutations start with <span class="math-container">$d$</span> and the
zero is moving to the right.</p>
<p>This set of permutations comes from an integral decomposition problem.</p>
<p>Additional question:<br>
Can the sorting be done in <span class="math-container">$d$</span> steps with the given set <span class="math-container">$P^d$</span>?</p>
<p><strong>Edit 2019-03-18</strong>:</p>
<p>@Jaap I will accept your answer if no other more sophisticated algorithm will be given.</p>
<p>Now that we have at least an algorithm which uses <span class="math-container">$\mathcal{O}(d^2)$</span> permutations<br>
I would like to tighten some constraints and give more information.</p>
<ul>
<li>Every permutation used will result in an inverse matrix multiplication
for my numerical analysis program, which will worsen the numerical error, thus I would like to have a better upper bound.</li>
<li>Instead of <span class="math-container">$P^d$</span> one is allowed to use <span class="math-container">$P^{-d} := \{ \pi^{-1} : \pi \in P^d\}$</span>. </li>
<li>Algorithms which recognize some/all of the permutations get a bonus.</li>
<li>Algorithms which use a specific permutation repeatedly do NOT get a bonus.</li>
<li>Algorithms which only work for odd permutations but have a
somehow nice twist get a bonus.</li>
</ul>
<p><strong>Edit 2019-03-21</strong>:</p>
<p>I wrote a small C++ program, which calculates the height of a shortest path tree (SPT) of the cayley graph <span class="math-container">$\Gamma = \Gamma(S_{d+1},P^d)$</span>
with the root node of the tree placed at the identity permutation.</p>
<p>This gives for <span class="math-container">$3 \le d \le 9$</span></p>
<ul>
<li><p><span class="math-container">$height(SPT(\Gamma)) = d$</span>.</p></li>
<li><p>If one removes the 1st subset we have<br>
<span class="math-container">$height(SPT(\Gamma)) = d$</span>.</p></li>
<li><p>If one removes the 2nd subset we have<br>
<span class="math-container">$height(SPT(\Gamma)) = d+1$</span>.</p></li>
</ul>
<p><strong>Edit 2019-04-04</strong>:</p>
<p>Numerical evidence suggests that for <span class="math-container">$d \ge 4$</span>
the only SPT being invariant under edge completion (EC)<br>
is the one coming from the subset <span class="math-container">$Q \subseteq P^d$</span>
consisting of the 2nd subset of permutations, i.e.</p>
<p><span class="math-container">$
Q = \{\\
\quad [9,0,1,2,3,4,5,6,7,8],\\
\quad [9,1,0,2,3,4,5,6,7,8],\\
\quad [9,1,2,0,3,4,5,6,7,8],\\
\quad [9,1,2,3,0,4,5,6,7,8],\\
\quad [9,1,2,3,4,0,5,6,7,8],\\
\quad [9,1,2,3,4,5,0,6,7,8],\\
\quad [9,1,2,3,4,5,6,0,7,8],\\
\quad [9,1,2,3,4,5,6,7,0,8],\\
\quad [9,1,2,3,4,5,6,7,8,0]\\
\}.$</span></p>
<p>Where edge completion (EC) means adding an edge<br>
iff two nodes of tree depth k and k+1
can be connected with a valid generator/permutation.</p>
<p><strong>Edit 2019-04-06</strong>: (Additional information)</p>
<p>We have an accepted answer now.</p>
<p>There are similar more difficult problems,<br>
where one has faster growing sets <span class="math-container">$P^{d,m}$</span> of valid permutations<br>
and one expects to do the sorting in <span class="math-container">$d-m$</span> steps. </p>
<p>Numerical evidence suggests there might be a relation to
<a href="http://oeis.org/A130477" rel="nofollow noreferrer">http://oeis.org/A130477</a> </p>
<p>where e.g. for <span class="math-container">$d=4$</span>, the row <span class="math-container">$1, 4, 15, 40, 60$</span> would tell us,<br>
there is 1 permutation of <span class="math-container">$S_{d+1}$</span> which can be sorted in <span class="math-container">$0$</span> steps.<br>
there are 4 permutations of <span class="math-container">$S_{d+1}$</span> which can be sorted in <span class="math-container">$1$</span> step.<br>
there are 15 permutations of <span class="math-container">$S_{d+1}$</span> which can be sorted in <span class="math-container">$2$</span> steps.<br>
there are 40 permutations of <span class="math-container">$S_{d+1}$</span> which can be sorted in <span class="math-container">$3$</span> steps.<br>
there are 60 permutations of <span class="math-container">$S_{d+1}$</span> which can be sorted in <span class="math-container">$4$</span> steps. </p>
<p><strong>Edit 2019-04-06</strong>: (Example of antkam's algorithm)</p>
<p>If one uses the "right-first" convention for multiplying two permutations,<br>
antkam's solution uses the inverted permutations of <span class="math-container">$Q$</span>.</p>
<p>As can be seen with the following example
<span class="math-container">$[0,\color{red}{4},3,1,2]
\rightarrow [2,\color{red}{4,0},3,1]
\rightarrow [1,\color{red}{2,4,0},3]
\rightarrow [3,\color{red}{1,2,4,0}]
\rightarrow [\color{red}{0,1,2,3,4}].
$</span></p>
<p>And as decomposition with inverted permutations from <span class="math-container">$Q$</span>:</p>
<p><span class="math-container">$
\begin{align}
&[0,\color{red}{4},3,1,2]\\
=&[2,\color{red}{4,0},3,1][2,1,3,4,0]\\
=&[1,\color{red}{2,4,0},3][1,2,3,4,0][2,1,3,4,0]\\
=&[3,\color{red}{1,2,4,0}][1,2,3,4,0][1,2,3,4,0][2,1,3,4,0]\\
=&[4,1,2,0,3]^{-1}[4,0,1,2,3]^{-1}[4,0,1,2,3]^{-1}[4,1,0,2,3]^{-1}.
\end{align}
$</span></p>
<p><strong>Edit 2019-04-08</strong>: (Example of improved algorithm)</p>
<p>There is an obvious improvement to antkam's algorithm,<br>
where one simply keeps the red run (ascending numbers) as large as possible.</p>
<p>Antkam's algorithm would give for <span class="math-container">$[0,2,3,4,1]$</span><br>
<span class="math-container">$[0,\color{red}{2},3,4,1]
\rightarrow [1,\color{red}{2,0},3,4]
\rightarrow [4,\color{red}{1,2,0},3]
\rightarrow [3,\color{red}{1,2,4,0}]
\rightarrow [\color{red}{0,1,2,3,4}].
$</span></p>
<p>Whereas the improved algorithm would give for <span class="math-container">$[0,2,3,4,1]$</span><br>
<span class="math-container">$[0,\color{red}{2,3,4},1]
\rightarrow [1,\color{red}{2,3,4,0}]
\rightarrow [\color{red}{0,1,2,3,4}].
$</span></p>
<p>It looks like the transition graph of the improved algorithm<br>
is isomorphic to the tree graph <span class="math-container">$SPT(\Gamma(S_{d+1},Q)$</span> discussed earlier. </p>
<p>I have added an image for <span class="math-container">$d=4$</span> with the tree graph,
<a href="https://i.stack.imgur.com/vKzrK.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vKzrK.png" alt="SPT(Gamma(S,Q))"></a>
where the root of the tree represents the identity permutation.</p>
| MachineLearner | 647,466 | <p>The eigenvalue equation is given by</p>
<p><span class="math-container">$$Av_i=\lambda_i v_i \quad \text{ for all } i=1,\ldots,n.$$</span></p>
<p>If you combine all equations for all eigenvalues you can rewrite the previous equations in a single matrix equation as </p>
<p><span class="math-container">$$A[v_1,\ldots,v_n]=[v_1,\ldots,v_n]\text{diag}[\lambda_1,\ldots,\lambda_n].$$</span></p>
<p>For simplicity assume that we have <span class="math-container">$n$</span> distinct eigenvalue and eigenvector pairs. If we introduce <span class="math-container">$V=[v_1,\ldots,v_n]$</span> and <span class="math-container">$\Lambda = \text{diag}[\lambda_1,\ldots,\lambda_n]$</span> solve for <span class="math-container">$\Lambda$</span> to obtain</p>
<p><span class="math-container">$$\Lambda = V^{-1}AV.$$</span> </p>
<p>Hence, by multiplying in this fashion with the eigenvector matrices we obtain a diagonal matrix with the eigenvalues on the diagonal. </p>
<p>You can think of this as a change of basis for the system. We first used an arbitrary coordinate system. After transforming the system we are using a system that has a coordinate axis aligned with the eigenvectors. </p>
|
1,463,419 | <p>A letter has come from exclusively LONDON or CLIFTON, but on the postmark only $2$ consecutive letters ''ON'' are found to be visible. What is the probability that the letter came from LONDON?</p>
<hr>
<p>This is a question of conditional probability. Let $A$ be the event that the letter has come from LONDON. Let $B$ be the event that consecutive letters ''ON'' are found to be visible. $A\cap B$ is the event that the letter has come from LONDON and consecutive letters ''ON'' are visible. We have to find $P(A\mid B)
=\frac{P(A\cap B)}{P(B)}$.</p>
<p>But then i am stuck. Please help me. Thanks.</p>
| joriki | 6,622 | <p>There is not enough information given to answer the question, since we don't know the prior probabilities of letters arriving from London or from Clifton. A reasonable assumption might be that letters are equally likely to come from any of the people living in those two places. London has a population of roughly $10$ million; the suburb Clifton of Bristol (assuming that that's the Clifton that is meant) has a population of roughly $10{,}000$. Thus the prior probabilities are roughly $0.999$ for arriving from London and $0.001$ for arriving from Clifton. </p>
<p>Regarding the letters on the postmark, again not enough information is given, since we don't know how these letters came to be visible and how likely they are to be from various parts of the words. A reasonable assumption is that a pair of consecutive letters was uniformly randomly selected from all such pairs. Then the probability of a letter from London showing "ON" would be $2/5$, and the probability of a letter from Clifton showing "ON" would be $1/6$.</p>
<p>Under these two assumptions, we can calculate the <em>a posteriori</em> probability of the letter having arrived from London as</p>
<p>$$
P(\text{London}\mid\text{ON})=\frac{P(\text{London}\cap\text{ON})}{P(\text{ON})}=\frac{0.999\cdot\frac25}{0.999\cdot\frac25+0.001\cdot\frac16}\approx0.9996\;.
$$</p>
<p>Thus we are now even more certain that the letter must have come from London than before.</p>
<p>If you want to solve this very badly posed problem the way it may have been intended, you may want to use the highly unrealistic prior probabilities $1/2$ instead.</p>
|
4,062,667 | <p>Let <span class="math-container">$A$</span> be a <span class="math-container">$n^{th}$</span> order square and skew-symmetric matrix, if <span class="math-container">$(E-A)$</span> is an invertible matrix show that
<span class="math-container">$(E+A)(E-A)^{-1}$</span> is an invertible matrix (where <span class="math-container">$E$</span> is an identity matrix)</p>
<p>This is a part of a question I have proved that <span class="math-container">$||(E-A)u|| \geq ||u||$</span> for any column vector <span class="math-container">$u \in \mathbb{R^n}$</span></p>
<p>If we can show that <span class="math-container">$(E+A)$</span> is invertible then the proof will be trivial. Any ideas would be appreciated.</p>
| user6725906 | 900,359 | <p>Hint: <span class="math-container">$\det(E+A)=0$</span> if and only if <span class="math-container">$A$</span> has an eigenvector with eigenvalue <span class="math-container">$-1$</span>. Assuming that for some <span class="math-container">$v\in \mathbb{R}^n\backslash\{0\}$</span> <span class="math-container">$$Av=-v,$$</span> try to find a contadiction with the assumption that <span class="math-container">$E-A$</span> is invertible. (Alternatively, in the same way you'll get a contradiction to the part of the question you already proved).</p>
|
11,073 | <p>I have three simple graphs in one Plot. Now I am trying to make a button for each graph so you can hide or show it in the plot. Until now I was just able to make a checkbox with the Manipulate function, but I don't now how to tell the checkbox that it should hide my graph when unchecked an display it when checked. </p>
<p>Here is what I was able to make so far, I know it looks simple but I also have many other mathematica documents which at the end just need this button.</p>
<p><img src="https://i.stack.imgur.com/QMqcr.png" alt="enter image description here"> </p>
| PlatoManiac | 240 | <p>As you mentioned that you wanted a GUI using <code>Button</code></p>
<pre><code>DynamicModule[{b = {True, True, True}, res,
funs = {0.5 x + 1, x, 2 x - 2}, pic},
fun[val_] := If[val == True, "Pressed", "DialogBox"];
res = Dynamic@
Row@{Button["f(x)", If[b[[1]] == True, b[[1]] = False, b[[1]] = True],
BaseStyle -> {"GenericButton", 20, Bold}, Appearance->fun[b[[1]]]],
Button["y(x)", If[b[[2]] == True, b[[2]] = False, b[[2]] = True],
BaseStyle -> {"GenericButton", 20, Bold},Appearance->fun[b[[2]]]],
Button["g(x)", If[b[[3]] == True, b[[3]] = False, b[[3]] = True],
BaseStyle -> {"GenericButton", 20, Bold},Appearance->fun[b[[3]]]]
};
pic = Column[{res,Dynamic@Plot[
Evaluate@Flatten@MapThread[If[#1 == True, #2, Null] &,{b,funs}],
{x, -1, 5},PlotRange -> {-1, 5}, AspectRatio ->1,PlotStyle->Thick,
Epilog -> {Text[If[b[[1]] == True, "f(x)", ""], {4.5, 2.7}],
Text[If[b[[2]] == True, "y=x", ""], {4.5, 4}],
Text[If[b[[3]] == True, "g(x)", ""], {3, 4.5}]},
ImageSize -> 500]},
Alignment -> Center, Frame -> True, Background -> {{LightRed, None}}
];
pic
]
</code></pre>
<p><img src="https://i.stack.imgur.com/yWpOC.png" alt="enter image description here"></p>
|
3,113,083 | <p>Why does every CNF formula for <span class="math-container">$(x_{1} \vee y_{1}) \wedge (x_{2} \vee y_{2})\wedge \ldots \wedge (x_{n} \vee y_{n})$</span> have at least <span class="math-container">$2^{n}$</span> terms?</p>
<p>This statement is on the Wikipedia page for DNF form here: <a href="https://en.wikipedia.org/wiki/Disjunctive_normal_form" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Disjunctive_normal_form</a></p>
<p>But, I don't understand why it's true. Can someone please clarify? As shown above, the CNF has <span class="math-container">$2n$</span> terms. I'm not sure why the DNF has <span class="math-container">$2^{n}$</span> at least terms though.</p>
| Bram28 | 256,001 | <p>If you think it is <span class="math-container">$(\frac{3}{6})^6$</span> because you assume these are all independent events ... then you are making a wrong assumption. For example, if the first can is put in the place of where the second can was, then the probability that the second can gets put in a place other than its initial or adjacent position is <span class="math-container">$\frac{3}{5}$</span>, rather than <span class="math-container">$\frac{3}{6}$</span></p>
<p>In general, the placement of cans effects the placement of other cans, making the outcomes of these events dependent on each other.</p>
|
2,860,360 | <p>It is a general question about simple examples of calculating class numbers in quadratic fields. Here are an excerpt from Frazer Jarvis' book <em>Algebraic Number Theory</em>:</p>
<p>"<em>Example 7.20</em> For $K=\mathbb{Q}(\sqrt[3]{2} )$, the discriminant is 108, and $r_{2}=1$. So the Minkowski bound is $\approx 2.940$. So every ideal is equivalent to one whose norm is at most 2. The only ideal of norm 1 is the full ring of integers, which is principal; the ideal $(2)=\mathcal{p}_{2}^{3}$, where $\mathcal{p}_{2}=(\sqrt[3]{2})$ is also principal. Thus every ideal is equivalent to a principal ideal, so the class group is trivial."</p>
<p>The question is why does it suffice to look at the principal ideal generated by 2?</p>
| Dhalsim | 579,015 | <p>The problem is that you cannot add probabilities when dealing with events that are not "disjoint". What you can do is multiply them, but <em>only</em> if the events are independent.</p>
<p>So, in your example, 6 people are given 300 attempts each to catch a pokemon each. First assume these are independent of each other (i.e. whatever one person does, that doesn't affect anybody else's chances of catching a pokemon). Now, the probability that <em>one particular person</em> catches <em>at least</em> 1 pokemon is about 70 % if each attempt has a 1/256 probability of success. </p>
<p>So say you want the probability that <em>each</em> person catches at least one pokemon. Then you can't say 6*70 %, but you can multiply 70% with itself 6 times. That gives you about 11 %. </p>
|
2,860,360 | <p>It is a general question about simple examples of calculating class numbers in quadratic fields. Here are an excerpt from Frazer Jarvis' book <em>Algebraic Number Theory</em>:</p>
<p>"<em>Example 7.20</em> For $K=\mathbb{Q}(\sqrt[3]{2} )$, the discriminant is 108, and $r_{2}=1$. So the Minkowski bound is $\approx 2.940$. So every ideal is equivalent to one whose norm is at most 2. The only ideal of norm 1 is the full ring of integers, which is principal; the ideal $(2)=\mathcal{p}_{2}^{3}$, where $\mathcal{p}_{2}=(\sqrt[3]{2})$ is also principal. Thus every ideal is equivalent to a principal ideal, so the class group is trivial."</p>
<p>The question is why does it suffice to look at the principal ideal generated by 2?</p>
| Alessandro Jeanteur | 579,032 | <p>I think that the easiest way for you to grasp where the 'missing' successes went is actually mostly in multiple successes.</p>
<p>If one event has 0.2 probability of success, twice the event has probabilities:</p>
<ul>
<li>0.64 -> no success at all (0.8*0.8)</li>
<li>0.04 -> two successes (0.2*0.2)</li>
<li>0.32 -> one success</li>
</ul>
<p>Therefore, relatively high probabilities and small number of trials means success rate is not at all being proportional to number of trials.</p>
<p>On the other hand, if you have a very low probability of success, then probability of at least one success is approximately proportional to number of trials.</p>
<p>The graph below maybe illustrates this better than my words, but since probability of success is very low (for instance, 0.4%) then probability of two or more successes is A LOT lower (0.0016%, 0.000064%, etc., so basically negligible) so if several experiments are repeated total success rates will closely match the number of times, on average that 'one' success occurs.</p>
<p>Probability for at least one success with probability of 0.004, for 1 to 1000 tries (straight approximation, actual curve):</p>
<p><a href="https://i.stack.imgur.com/mNi8n.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mNi8n.jpg" alt="Probability for at least one success with probability of 0.004, for 1 to 1000 tries (straight approximation, actual curve)"></a></p>
|
3,844,448 | <p>Find all values of <span class="math-container">$h$</span> such that rank(<span class="math-container">$A$</span>) = <span class="math-container">$2$</span>.</p>
<p><span class="math-container">$A$</span> = <span class="math-container">$\begin{bmatrix}
1 & h & -1\\
3 & -1 & 0\\
-4 & 1 & 3
\end{bmatrix} $</span></p>
<p>I used row transformations to get</p>
<p><span class="math-container">$A$</span> = <span class="math-container">$\begin{bmatrix}
1 & h & -1\\
0 & -1-3h & 3\\
0 & 1+4h & -1
\end{bmatrix} $</span></p>
<p>But how do I solve to get the rank? I know the general idea is that rank(<span class="math-container">$A$</span>) = <span class="math-container">$2$</span> when dim(col(<span class="math-container">$A$</span>)) = dim(row(<span class="math-container">$A$</span>)) = <span class="math-container">$2$</span></p>
| Fred | 380,717 | <p>It is easy to see that the first and the second row of <span class="math-container">$A$</span> are independent, hence</p>
<p><span class="math-container">$$rank(A) \ge 2.$$</span></p>
<p>Furthermore, since <span class="math-container">$\det(A)=-2-9h,$</span></p>
<p><span class="math-container">$$rank(A) =2 \iff \det(A)=0 \iff -2-9h=0.$$</span></p>
|
3,844,448 | <p>Find all values of <span class="math-container">$h$</span> such that rank(<span class="math-container">$A$</span>) = <span class="math-container">$2$</span>.</p>
<p><span class="math-container">$A$</span> = <span class="math-container">$\begin{bmatrix}
1 & h & -1\\
3 & -1 & 0\\
-4 & 1 & 3
\end{bmatrix} $</span></p>
<p>I used row transformations to get</p>
<p><span class="math-container">$A$</span> = <span class="math-container">$\begin{bmatrix}
1 & h & -1\\
0 & -1-3h & 3\\
0 & 1+4h & -1
\end{bmatrix} $</span></p>
<p>But how do I solve to get the rank? I know the general idea is that rank(<span class="math-container">$A$</span>) = <span class="math-container">$2$</span> when dim(col(<span class="math-container">$A$</span>)) = dim(row(<span class="math-container">$A$</span>)) = <span class="math-container">$2$</span></p>
| Aaron | 9,863 | <p>There are several ways to approach this problem. Here are two approaches that I will not use in this answer. (1) Rank is preserved by both row operations and column operations, and so we can do column operations to simpliffy the matrix further. (2) The matrix has rank 3 if and only if it is invertible, and we can use the determinant to test for invertibility. However, because the first and third columns (or second and third rows) are linearly independent, we must have rank at least 2. Therefore, we only need to find h where <span class="math-container">$\det(A)=0$</span>.</p>
<p>Let us observe that the first and third columns are linearly independent, and so we will have rank <span class="math-container">$2$</span> if and only if the second column is a linear combination of the first and third. Call the columns <span class="math-container">$v_1, v_2, v_3$</span>. Because the second coordinate of <span class="math-container">$v_3=0$</span>, if <span class="math-container">$v_2=av_1+bv_3$</span>, then looking at the second coordinate, that tells us <span class="math-container">$a=-1/3$</span>. Looking at the third coordinate, we get <span class="math-container">$(-4)/(-3)+3b=1$</span>, so <span class="math-container">$b=-1/9$</span>. Looking at the first coordinate, this gives <span class="math-container">$h=1/(-3)+(-1)/(-9)=-2/9$</span>. This is the only way that <span class="math-container">$v_2$</span> can be a linear combination of <span class="math-container">$v_1$</span> and <span class="math-container">$v_3$</span>.</p>
|
2,771,034 | <p>$\frac{a_n}{b_n} \rightarrow 1$ and $\sum_{n=1}^\infty b_n$ converges, can it be concluded that $\sum_{n=1}^\infty a_n$ converges?<br>
My attempt at an answer to this question: since $\sum_{n=1}^\infty b_n$ converges, $b_n \rightarrow 0$. Because of this, $a_n \rightarrow 0$ equally fast. However, I'm well aware that this does not imply that $\sum_{n=1}^\infty a_n$ converges. I'm stuck at that point, though, as I'm not sure what other conclusions can be drawn. Could anyone help me out?</p>
| Community | -1 | <p>I actually think the answer is no. Take $b_{n} = \frac{(-1)^{n}}{\sqrt{n}}$
and $a_{n} = \frac{(-1)^{n}}{\sqrt{n}} + \frac{1}{n}$. Then $\sum_{n}b_{n}$ converges and $\sum_{n}a_{n}$ diverges but $\frac{a_{n}}{b_{n}} \rightarrow 1$ as </p>
|
1,335,640 | <p>1) A disease has hit a city. The percentage of the population infected $t$ days after the disease arrives is approximated by $$p(t) = 12te^{\frac{-t}{7}} \qquad \mbox{for} \qquad0\leq t \leq 35.$$ </p>
<p>After how many days is the percentage of infected people a maximum? What is the maximum percent of the population infected?</p>
<p>The maximum percent of the population infected is ______ %</p>
<p>2) A container contains 12 diesel engines. The company chooses 5 engines at random and will not ship the container if any of the engines chosen are defective. Find the probability that a container will be shipped even though it contains 2 defectives if the sample size is 5.</p>
<p>For the first problem, the number of days at which the percentage is at maximum is 7. Clearly, if I substitute this to $p(t)$ I will get the maximum percentage. My problem is how did they get the answer of 7 days? How do I deal with this kind of problem? Is there a specific formula? I'm trying to figure it out but can't.</p>
<p>Also for the second problem I made use of the hypergeometric formula that is $$p(x) = \frac{\left[C(k,x) \cdot C(N-k, n-x)\right]}{C(N,n)}$$ where $N$ is the size of population, $k$ is the number of successes in the population, $x$ is the number of successes in the sample and $n$ is the sample size. I used this and I got a different answer. The answer should have been 0.318 but I got a different one. Please help. </p>
| Graham Kemp | 135,106 | <p>The gradient (or "slope") at a local maximum, minimum, or point of inflection, will equal zero.</p>
<p>We have $p(t) = 12 t e^{-t/7}$ , so then by differentiation (using the chain rule): $\frac{\mathrm d p(t)}{\mathrm d t} = 12 e^{-t/7} - \dfrac{12 t}{7} e^{-t/7}$.</p>
<p>Solving for $t$ when $\frac{\mathrm d p(t)}{\mathrm d t}=0$ gives us $t=7$. This exists within our interval $0\leq t \leq 35$, thus all we need do is check is that it is indeed a maximum.</p>
<p>$$p(7) \approx 30.9 \\ p(6) \approx 30.6 \\ p(8) \approx 30.62$$</p>
<hr>
<p>For the second problem the container will be shipped if the two defective containers are among the seven not inspected. By counting distinct ways to select places for defectives, the probability of this is: ${^{7}{\rm C}_{2}}\big/ {^{12}{\rm C}_{2}} = \frac{7\times 6}{12\times 11}$</p>
<p>Using the hypergeometric formula, this is $p(0) = \dfrac{{^{5}{\rm C}_{0}}\;{^{12-5}{\rm C}_{2-0}}}{ {^{12}{\rm C}_{2}}}$</p>
|
1,812,675 | <p>Is there a recurrence solution to $a_n=\frac{n}{a_{n-1}}$? I'm wondering if it could be done in the form of an alternating series partial to $n$ or as a trigonometric function.</p>
| peter.petrov | 116,591 | <p>For $n=2k+1$ odd you have: </p>
<p>$$a_{2k+1} = \frac{c . (2k+1)!!}{ (2k)!!}$$ </p>
<p>For $n=2k$ even you have: </p>
<p>$$a_{2k} = \frac{(2k)!!}{c . (2k-1)!!}$$ </p>
<p>where $c = a_1$. </p>
<p>This pattern can be seen by writing down the first 5-6 terms.<br>
It should be easy to prove it by induction. </p>
|
1,259,853 | <p>Why the derivative of $n^{1/n} = \sqrt[n]{n}$ is $n^{1/n} \left( \frac{1}{n^2} - \frac{\log(n)}{n^2}\right)$ (according to Maxima and other tools online)?</p>
<p>I have tried to applied the chain rule, but it comes something completely different:</p>
<p>$$\frac{1}{n} n^{\frac{1}{n} - 1} \cdot 1 = \frac{1}{n} n^\frac{1}{n}n^{-1} = \frac{1}{n^2} n^\frac{1}{n} = \frac{\sqrt[n]{n}}{n^2}$$</p>
<p>Sincerely, I am not seeing where that $\log$ and the rest of the stuff comes from. I have a more difficult problem that is similar and whose solution contains a $\log$ somewhere, but I am not seeing where it comes from.</p>
| E.H.E | 187,799 | <p>$$y=n^{1/n}$$
$$\log(y)=1/n\log(n)$$
$$\frac{y'}{y}=-\frac{1}{n^2}\log(n)+{1/n}(1/n)$$
multiply by $y$
$$y'=y(\frac{1}{n^2}-\frac{\log n}{n^2})$$</p>
|
894,476 | <p>I don't have a strong background in probability/statistics and I'm trying to understand the example at <a href="http://rationalwiki.org/wiki/Extraordinary_claims_require_extraordinary_evidence#Probability_theory" rel="nofollow">http://rationalwiki.org/wiki/Extraordinary_claims_require_extraordinary_evidence#Probability_theory</a></p>
<p>Is that the correct framework to explain the principle? Their $P(A)$ there seems arbitrary, and using other values and the Bayes' formula they employ one gets values for $P(A|B)$ greater than one. </p>
<p>I've been trying to work out a more general example in which the experiment consists of tossing $N$ times and the event $B$ would be guessing right $n$ times, but again it seems to me that there must be something limiting $P(A)$ or else one can get values of $P(A|B)$ higher than one.</p>
<p>But again, I don't really know how to make sense of the example in a completely rigorous way or even if that's the correct approach to illustrate the principle.</p>
<p>Thanks!</p>
| Abhishek Murali | 432,569 | <p>In my Applied Numerical Methods course, we are taught using Simpson's method also. Both works equally well. However, Trapezoidal rule works for any interval length as compared to Simpson's rule which needs even number of intervals.
So if we have say 5 values of h, we can apply trapezoidal rule for all those 5 values. But Simpson's method may not apply on certain h values. Those will have to be evaluated by Trapezoidal rule.
While coding these on any computer, it is easier then to use Trapezoidal rule since writing the algorithm would be easy. Hope this makes sense. </p>
|
38,439 | <p>I've mentioned before that I'm using this forum to expand my knowledge on things I know very little about. I've learnt integrals like everyone else: there is the Riemann integral, then the Lebesgue integral, and then we switch framework to manifolds, and we have that trick of using partitions of unity to define integrals.</p>
<p>This all seems very ad hoc, however. Not natural. I'm aware this is a pretty trivial question for a lot of you (which is why I'm asking it!), but what is the "correct" natural definition we should think of when we think of integrals?</p>
<p>I know there's some relation to a perfect pairing of homology and cohomology, somehow relating to Poincare duality (is that right?). And there's also something about chern classes? My geometry, as you can see, is pretty confused (being many years in my past).</p>
<p>If you can come up with a natural framework that doesn't have to do with the keywords I mentioned, that would also be very welcome.</p>
| G. Rodrigues | 2,562 | <p>My answer here is realy just a footnote to Paul Siegel's excellent answer, but it has become too long to fit in a comment box. Integrals are siamese brothers to measures; leaving them out seems rather perverse to me. Anyway, here is how I think of integrals. The objective here is to tackle the "categorical" part; the analytical viewpoint will necessarily obtrude. But bear with me a little, this is a somewhat long post, with a punchline at the end.</p>
<p>Fix a Boolean algebra $\Omega$. A map $\nu: \Omega\to V$ with values on a linear space $V$ is finitely additive if $\nu(E\cup F)= \nu(E) + \nu(F)$ for every disjoint $E, F$. Denote the linear space of such maps by $\mathbf{A}(\Omega, V)$.</p>
<p>Theorem 1: There is a linear space $\mathbf{S}(\Omega)$ and a finitely additive map $\chi:\Omega\to \mathbf{S}(\Omega)$ universal among all finitely additive maps.
proof: just follow the universal property and do the obvious thing (yeah, I suppose you can use the adjoint functor theorem but why would you?).</p>
<p>The universal property recast in terms of representability gives the natural isomorphism ($\mathbf{Vect}$ is the category of linear spaces)</p>
<p>$$\mathbf{A}(\Omega, V)\cong \mathbf{Vect}(\mathbf{S}(\Omega), V)$$</p>
<p>Before continuing, let me elucidate a little bit of the structure of $\mathbf{S}(\Omega)$.</p>
<p>Theorem 2: Let $f$ be a non-zero element of $\mathbf{S}(\Omega)$. Then there are non-zero scalars $k_n$ and non-zero, pairwise disjoint $E_n\in \Omega$ such that $f= \sum_n k_n\chi(E_n)$. Furthermore, if $\nu$ is a finitely additive map and $\widehat{\nu}$ the map induced on $\mathbf{S}(\Omega)$ by universality, then $\widehat{\nu}(f)= \sum_{n}k_n \nu(E_n)$.</p>
<p>To put it simply, $\mathbf{S}(\Omega)$ is the linear space of "simple functions on $\Omega$" and the map induced by universality is the integral. Now use theorem 2 to put a norm on $\mathbf{S}(\Omega)$:</p>
<p>$$\|\sum_n k_n\chi(E_n)\|= \max\{|k_n|\}$$</p>
<p>Denote the completion by $\mathbf{L}_{\infty}(\Omega)$. On the other hand, put on the linear subspace of the <em>bounded</em> finitely additive maps $\Omega\to B$ with $B$ a Banach space, the semivariation norm (which I am not going to define). Denote this space by $\mathbf{BA}(\Omega, B)$. Then:</p>
<p>Theorem 3: There is a bounded finitely additive $\chi:\Omega\to \mathbf{L}_{\infty}(\Omega)$ universal among all bounded finitely additive maps.</p>
<p>Once again, recasting the universal property in terms of representability, we have a natural isometric isomorphism ($\mathbf{Ban}$ is the category of Banach spaces and bounded linear maps).</p>
<p>$$\mathbf{BA}(\Omega, B)\cong \mathbf{Ban}(\mathbf{L}_{\infty}(\Omega), B)$$</p>
<p>It is illuminating to write down what does the naturality of the isomorphism implies: I will leave that as an exercise to the reader.</p>
<p>Note that $\mathbf{L}_{\infty}(\Omega)$ is a Banach algebra in a natural way (use theorem 2 or juggle the universal property around. Or "cheat" all the way up and use Stone duality) and that $\chi$ is <em>spectral</em> or <em>multiplicative</em>, that is, $\chi(E\cap F)= \chi(E)\chi(F)$. Theorem 3 can now be extended by saying that $\chi$ is universal among all spectral measures (with values in Banach algebras). This extension is trivial given theorem 3.</p>
<p>The case of $\mathbf{L}_{\infty}(\Omega)$ does not need the introduction of measures but of course, this is not so with $\mathbf{L}_{1}$. So fix a finitely additive, positive $\mu:\Omega\to \mathbb{R}$. For the sake of simplification I will assume $\mu$ non-degenerate, that is, $\mu(E)= 0$ implies $E= 0$ (otherwise, you will have to take some quotient along the way). A finitely additive $\nu:\Omega\to B$ with $B$ a Banach space is $\mu$-Lipschitz if there is a constant $C$ such that $\|\nu(E)\|\leq C\mu(E)$ for all $E$. The infimum of all the constants $C$ in the conditions of the inequality gives a norm and a normed space I will denote by $\mathbf{LA}(\Omega, \mu, B)$. On the other hand, endow $\mathbf{S}(\Omega)$ with the norm</p>
<p>$$\|\sum_n k_n\chi(E_n)\|= \sum_n |k_n|\mu(E_n)$$</p>
<p>and denote the completion by $\mathbf{L}_{1}(\Omega, \mu)$.</p>
<p>Theorem 4: There is a finitely additive, $\mu$-Lipschitz $\chi:\Omega\to \mathbf{L}_{1}(\Omega, \mu)$ universal among all such maps.</p>
<p>Before the conclusion let me address a few points.</p>
<ol>
<li><p>Measurable spaces are not needed. If you really want them, use Stone duality (that is, points count for nothing in measure theory so why not leave them out, heh?).</p></li>
<li><p>Finitely additive measures are really not that much more general than $\sigma$-additive ones. I will leave this cryptic comment as is, and just note that once again, Stone duality is the key here.</p></li>
<li><p>I am <em>not</em> advocating this approach to be used in teaching (unless your goal is to flunk and befuddle as many undergrads as humanly possible). For one, you need some functional analysis under the belt (Banach spaces, completions, semivariation, etc.). Intuition is very hard to come by as I have thrown away the measurable spaces without which THE most important example, Lebesgue measure (arguably, the core of a first measure theory course) cannot be constructed. The whole logic of the approach only makes sense after you have seen other instances of categorical thinking at work. I am sure you can think of other objections.</p></li>
</ol>
<p>How categorical is this approach? Certainly, the universal properties of the respective spaces are central to the whole business and at least, they make clear that some results are really just a consequence of abstract nonsense. In the words of P. Freyd, category theory is doing what it was invented for: to make the easy things really easy (or some such, my memory is lousy). For example, the Bochner vector integral is obtained simply by taking the projective tensor product. Fubini and Fubini-Tonelli on the equality of iterated integrals are other notable cases of categorical thinking at work. Now pepper with Stone duality and a few more tools (e.g. Hahn-Banach and the compact-Hausdorff monad) and you can get (a slight variation of) the Riesz representation theorem for compact Hausdorff spaces. Use the proper compactifications and generalize to wider classes of topological spaces. Or use Loomis-Sikorski to get Vitali-Hahn-Saks in one line (but this is really "cheating" as the crucial step in establishing Loomis-Sikorski is essentially the same as the one to establish Vitali-Hahn-Saks: a Baire category-theorem application). And a few more.</p>
<p>But once again, how categorical is this approach? Well, the argument is categorical enough to be generalized to symmetric monoidal closed categories. See R. Borger -- A categorical approach to integration, in the 23rd volume of TAC available online. For the modifications needed to internalize the arguments to a topos (and much more) see the delightful Phd thesis of Mathew Jackson "A Sheaf theoretic approach to measure theory" -- this is available online, just google for it. Oh, by the way, you can see (almost) everything I have explained above in volume 3 of D. Fremlin's measure theory 5-volume series, also available online.</p>
|
2,658,691 | <p>I have one last question regarding permutation. I understand the problem and the rule of product but this problem seems to be in a different format compared with the two questions I asked before.</p>
<p>A committee of eight is to form a round table discussion group. In how many ways
may they be seated if the 2 members are to be seated with each other?</p>
<p>If I understand it correctly, two persons are to choose from 6 spaces to sit,this would be 8*7=56 am I correct?</p>
| CostaZach | 517,147 | <p>Let $T\in L(X,Y)$ (linear and bounded, $X,Y$ normed spaces). We can show that $$T^{**}\circ J_{X}=J_{Y}\circ T,$$
with $J_X$ the natural embedding from $X$ to $X^{**}$ via $(J_X x)(x^*)=x^*(x)$.</p>
<p>Back to your case: If $E$ is reflexive, then $S=J_{E}^{-1}\circ T^*\circ J_{E}$. With this you have that $T=S^*$ and so $$\langle Te^*,e\rangle=\langle S^*e^*,e\rangle=\langle e^*,Se\rangle.$$</p>
|
3,408,846 | <p>This is an example in Serge Lang "Introduction to Linear Algebra", page 48. I try to multiply these two <span class="math-container">$2$</span>x<span class="math-container">$3$</span> and <span class="math-container">$3$</span>x<span class="math-container">$2$</span> matrices but fail to obtain the result as mentioned in the text.</p>
<p>I have:
<span class="math-container">$
\left( \begin{array}{cc}
2 & 1 & 5\\
1 & 3 & 2
\end{array} \right)
%
\left( \begin{array}{c}
3 & 4 \\
-1 & 2 \\
2 & 1
\end{array} \right)
=\left( \begin{array}{c}
15 & 30 \\
4 & 2
\end{array} \right)
$</span></p>
<p>Serge's result is, however:</p>
<p><span class="math-container">$
\left( \begin{array}{cc}
2 & 1 & 5\\
1 & 3 & 2
\end{array} \right)
%
\left( \begin{array}{c}
3 & 4 \\
-1 & 2 \\
2 & 1
\end{array} \right)
=\left( \begin{array}{c}
15 & 15 \\
4 & 2
\end{array} \right)
$</span></p>
<p>Where did I do wrong?</p>
| callculus42 | 144,421 | <p><strong>Hint:</strong></p>
<p><span class="math-container">$$\left(\begin{array}{}\color{blue }2&\color{red }1&\color{green }5\end{array} \right) \cdot \left(\begin{array}{}\color{blue }4\\ \color{red }2 \\ \color{green }1\end{array} \right)=\color{blue }{2\cdot 4}+\color{red }{1\cdot 2}+\color{green }{5\cdot 1}=\color{blue }8+\color{red }2+\color{green }5=15$$</span></p>
|
200,279 | <p>I'm trying to show that given a set $\{\mathbf{a}, \mathbf{b}\}$ of orthonormal vectors in a 2-dimensional vector space, I can construct the identity matrix by computing $aa^\dagger + bb^\dagger$. This should be straightforward but it's not working out. I get that my conditions for orthonormality are $$|a_1|^2+|a_2|^2 = 1,$$ $$|b_1|^2 + |b_2|^2 = 1,$$ $$a_1^*b_1+a_2^*b_2 = 0$$ but these don't directly lead me to the identity matrix. Where am I going wrong?</p>
| martini | 15,379 | <p>As $a$ and $b$ are orthonormal, they form a basis of your vector space $V$. We have
\[ (aa^* + bb^* )a = aa^* a + bb^* a = a|a|^2 + b0 = a
\]
and
\[(aa^* + bb^* )b = aa^* b + bb^* b = a0 + b|b|^2 = b
\]
So $(aa^* + bb^*)$ is the identity on a basis, hence on $V$, which gives $aa^* + bb^* = \mathrm{Id}$.</p>
<hr>
<p>$\def\abs#1{\left|#1\right|}$
More explicitly we first exploit $a_1^*b_1 + a_2^*b_2 = 0$, which gives
\[ \abs{a_1}^2\abs{b_1}^2 = \abs{a_1^*b_1}^2 = \abs{a_2^*b_2}^2 = \abs{a_2}^2\abs{b_2}^2 \]
Now $\abs{b_1}^2 = 1 - \abs{b_2}^2$ and $\abs{a_2}^2 = 1- \abs{a_1}^2$, that is
\[ \abs{a_1}^2 - \abs{a_1}^2\abs{b_2}^2 = \abs{b_2}^2 - \abs{a_1}^2\abs{b_2}^2\iff \abs{a_1} = \abs{b_2} \]
and therefore also $\abs{a_2} = \abs{b_1}$. </p>
<p>So we have
\begin{align*}
a_1^*a_1 + b_1^*b_1 &= \abs{a_1}^2 + \abs{b_1}^2\\
&= \abs{a_1}^2 + \abs{a_2}^2\\
&= 1,\\
a_2^*a_2 + b_2^*b_2 &= \abs{a_2}^2 + \abs{b_2}^2\\
&= \abs{b_1}^2 + \abs{b_2}^2\\
&= 1
\end{align*}
and hence the elements on the diagonal of $aa^* + bb^*$ are 1, as wished. It remains to conside the off-diagonal elements. We have
\begin{align*}
a_1(a_1^*a_2 + b_1^*b_2) &= \abs{a_1}^2a_2 + a_1b_1^*b_2\\
&= \abs{b_2}^2a_2 + a_1b_1^*b_2\\
&= b_2b_2^*a_2 + a_1b_1^*b_2\\
&= (b_1^*a_1 + b_2^*a_2)b_2\\
&= 0.
\end{align*}
So either $a_1 = 0$ or $a_1^*a_2 + b_1^*b_2 = 0$. In the letter case, we are done, in the former case, $a_2 \ne 0$ (as $\abs{a_2}^2 + \abs{a_1}^2 = 1$) and hence $a_2^* \ne 0$, but
\begin{align*}
a_2^*(a_1^*a_2 + b_1^*b_2) &= \abs{a_2}^2a_1^* + a_2^*b_1^*b_2\\
&= \abs{b_1}^2a_1^* + a_2^*b_1^*b_2\\
&= b_1^*b_1a_1^* + a_2^*b_1^*b_2\\
&= (a_1^*b_1 + a_2^*b_2)b_1^*\\
&= 0.
\end{align*}
So we are done in this case also.</p>
|
200,279 | <p>I'm trying to show that given a set $\{\mathbf{a}, \mathbf{b}\}$ of orthonormal vectors in a 2-dimensional vector space, I can construct the identity matrix by computing $aa^\dagger + bb^\dagger$. This should be straightforward but it's not working out. I get that my conditions for orthonormality are $$|a_1|^2+|a_2|^2 = 1,$$ $$|b_1|^2 + |b_2|^2 = 1,$$ $$a_1^*b_1+a_2^*b_2 = 0$$ but these don't directly lead me to the identity matrix. Where am I going wrong?</p>
| Owen Biesel | 41,747 | <p>Denote the two vectors by $\{\mathbf{a},\mathbf{b}\}$ instead. Then orthonormality is the condition that $\mathbf{a}^\dagger\mathbf{b}=\mathbf{a}^\dagger\mathbf{b}=0$, while $\mathbf{a}^\dagger\mathbf{a}= \mathbf{b}^\dagger\mathbf{b}=1$. Now $\mathbf{a}\mathbf{a}^\dagger + \mathbf{b}\mathbf{b}^\dagger$ is the block matrix product
$$\left(\begin{array}{cc}\mathbf{a} & \mathbf{b}\end{array}\right)\left(\begin{array}{c}\mathbf{a}^\dagger\\ \mathbf{b}^\dagger\end{array}\right),$$
where each factor is a $2\times 2$ matrix. This product is the identity matrix iff the two factors are inverses, which we can check by evaluating the product in the other order:
$$\left(\begin{array}{c}\mathbf{a}^\dagger\\ \mathbf{b}^\dagger\end{array}\right)\left(\begin{array}{cc}\mathbf{a} & \mathbf{b}\end{array}\right) = \left(\begin{array}{cc}\mathbf{a}^\dagger\mathbf{a} & \mathbf{a}^\dagger\mathbf{b}\\
\mathbf{b}^\dagger\mathbf{a} & \mathbf{b}^\dagger\mathbf{b}\end{array}\right)=\left(\begin{array}{cc}1 & 0\\
0 & 1\end{array}\right).$$
So $\left(\begin{array}{c}\mathbf{a}^\dagger\\ \mathbf{b}^\dagger\end{array}\right)$ and $\left(\begin{array}{cc}\mathbf{a} & \mathbf{b}\end{array}\right)$ are indeed inverses, and $$\left(\begin{array}{cc}\mathbf{a} & \mathbf{b}\end{array}\right)\left(\begin{array}{c}\mathbf{a}^\dagger\\ \mathbf{b}^\dagger\end{array}\right)=\mathbf{a}\mathbf{a}^\dagger + \mathbf{b}\mathbf{b}^\dagger$$ is the $2\times 2$ identity matrix.</p>
|
601,951 | <p><em><strong>2</strong> + <strong>5</strong> + <strong>8</strong> + . . . + <strong>(6n-1)</strong> = <strong>n(6n+1</strong>)</em></p>
<p>This is what I have so far. </p>
<p>The <strong>sum</strong> of <strong>(3j-1)</strong> from <strong>j=1</strong> to <em>something I`m not sure of</em>.</p>
| Grigory M | 152 | <p>LHS is the coefficient of $z^{2n}$ in
$$
(1+z)^m\frac{(1+\sqrt{1+z})^{2n+1}-(1-\sqrt{1+z})^{2n+1}}{2\sqrt{1+z}}.
$$
Such coefficient can be written as a residue,
$$
\operatorname{res}\left\{
(1+z)^m\frac{(1+\sqrt{1+z})^{2n+1}-(1-\sqrt{1+z})^{2n+1}}{2\sqrt{1+z}}\frac{dz}{z^{2n+1}}
\right\}.
$$
After the substitution $w=\sqrt{1+z}-1$ we get (using $dw=\frac{dz}{2\sqrt{1+z}}$)
$$
\operatorname{res}\left\{
(1+w)^{2m}\frac{(w+2)^{2n+1}-(-w)^{2n+1}}{(w(w+2))^{2n+1}}dw
\right\}
=
\operatorname{res}\left\{
\frac{(1+w)^{2m}}{w^{2n+1}}+
\frac{(1+w)^{2m}}{(2+w)^{2n+1}}
\right\}dw.
$$
The first summand gives the coefficient of $w^{2n}$ in $(1+w)^{2m}$ (i.e. RHS) and the second has zero residue at $w=0$. QED.</p>
|
3,347,264 | <p>I'm trying to find the distance between two points for a question in my textbook. The points are P(-2, 3) and Q(1, -3). Here's the working I have so far:</p>
<p>d = <span class="math-container">$\sqrt{(x_1 - x_0)^2 + (y_1 - y_0)^2}$</span></p>
<p>d = <span class="math-container">$\sqrt{(1 - (-2))^2 + (-3 - 3)^2}$</span></p>
<p>d = <span class="math-container">$\sqrt{3^2 + -6^2}$</span></p>
<p>d = <span class="math-container">$\sqrt -27$</span></p>
<p>This results in "Math ERROR" on my calculator.
Where am I going wrong?</p>
| Andreas | 317,854 | <p>Consider the inequality in the form</p>
<p><span class="math-container">$$\Big(\frac{x^n+y^n+(\frac{x+y}{2})^n}{x^{n-1}+y^{n-1}+(\frac{x+y}{2})^{n-1}}\Big)^n+\Big(\frac{x+y}{2}\Big)^n\leq x^n+y^n
$$</span>
Due to homegeneity, let <span class="math-container">$\frac{x+y}{2} = 1$</span>. Denote <span class="math-container">$\frac{x-y}{2} = a$</span>. Then we consider the equivalent question, w.l.o.g. with <span class="math-container">$0\le a \le 1$</span>:
<span class="math-container">$$
\Big(\frac{(1+a)^n+(1-a)^n+1}{(1+a)^{n-1} +(1-a)^{n-1} + 1 }\Big)^n \leq- 1+ (1+a)^n+(1-a)^n
$$</span></p>
<p>Note that for every convex function <span class="math-container">$f(x)$</span> and <span class="math-container">$p_i \ge 0$</span> the following is true (Jensen):</p>
<p><span class="math-container">$$
f\Big( \frac{p_1 x_1 + p_2 x_2 + p_3 x_3 }{p_1 + p_2 + p_3 }\Big) \le \frac{p_1 f(x_1) + p_2 f(x_2) + p_3 f(x_3) }{p_1 + p_2 + p_3 }
$$</span></p>
<p>Since <span class="math-container">$f(x) = x^n$</span> is a convex function, apply Jensen to the LHS with <span class="math-container">$p_1 = (1+a)^{n-1}$</span>, <span class="math-container">$p_2 = (1-a)^{n-1}$</span>, <span class="math-container">$p_3 = 1$</span>, and <span class="math-container">$x_1 = 1+a$</span>, <span class="math-container">$x_2 = 1-a$</span>, <span class="math-container">$x_3 = 1$</span>. Then it is sufficient to prove</p>
<p><span class="math-container">$$
\frac{(1+a)^{n-1} (1+a)^{n} + (1-a)^{n-1} (1-a)^{n} + 1}{(1+a)^{n-1} +(1-a)^{n-1} + 1 } \le - 1+ (1+a)^n+(1-a)^n
$$</span>
Clearing denominators gives
<span class="math-container">$$
g(a,n) = -(2 + (1+a)^n+(1-a)^n) + (1+a)^n(1+(1-a)^{n-1}) + (1-a)^n(1+(1+a)^{n-1}) \;\\
\; = 2 (1 - a^2)^{n-1} + a ((1+a)^{n-1} - (1-a)^{n-1} ) -2 \ge 0
$$</span>
But <span class="math-container">$g(a,n)$</span> is, for every <span class="math-container">$n \ge 2$</span>, a function which is monotonously increasing in <span class="math-container">$a$</span>. Namely,
<span class="math-container">$g(a=0,n) = 0$</span> and <span class="math-container">$g(a=1,n) = 2^{n-1} -2$</span>.</p>
<p>This can be proved by induction over <span class="math-container">$n$</span>. We have that the assumption holds for <span class="math-container">$n=2$</span> since <span class="math-container">$g(a,n=2) = 0$</span> <span class="math-container">$\forall a$</span>. Do the step <span class="math-container">$n \to n+1$</span>. We need to show
<span class="math-container">$$
g(a, n+1) = 2 (1 - a^2)^{n} + a ((1+a)^{n} - (1-a)^{n} ) -2 \ge 0
$$</span>
This can be rewritten
<span class="math-container">$$
g(a, n+1) = g(a, n) + a^2 [(1+a)^{n-2} + (1-a)^{n-2} - 2((1+a)(1-a))^{n-2}] \ge 0
$$</span></p>
<p>Since <span class="math-container">$g(a, n) \ge 0$</span> by induction hypothesis, this holds true if the second bracket is nonnegative. Use the AM-GM inequality for the first two terms in this bracket, giving the stronger condition
<span class="math-container">$2 ((1+a)(1-a))^{(n-2)/2} - 2((1+a)(1-a))^{n-2} \ge 0
$</span> or <span class="math-container">$(1-a^2)^{(n-2)/2} \le 1
$</span> which is obviously true.</p>
<p>This proves the claim. <span class="math-container">$\qquad \Box$</span></p>
<p><strong>Remark:</strong> This inequality is rather fine-tuned. Consider as a first factor <span class="math-container">$$\Big(\frac{x^n+y^n+z\cdot(\frac{x+y}{2})^n}{x^{n-1}+y^{n-1}+z\cdot(\frac{x+y}{2})^{n-1}}\Big)^n
$$</span>
and let <span class="math-container">$z$</span> increase from 0 to 1. It is easy to see that the increase of <span class="math-container">$z$</span> makes the term smaller. Choosing <span class="math-container">$z=1$</span> makes the term "just small enough" for the inequality to be "<span class="math-container">$\le$</span>". Indeed, for <span class="math-container">$z=0$</span>, this inverses to "<span class="math-container">$\ge$</span>", as <a href="https://math.stackexchange.com/questions/3304808/refinement-of-a-famous-inequality">this post</a> (with <a href="https://mathoverflow.net/questions/337457/prove-that-left-fracxn1xn-11-rightn-left-fracx12-rightn">this proof</a>) shows.</p>
<p>Numerically, one can solve for the particular value <span class="math-container">$z^*$</span> where equality occurs. It shows that for all <span class="math-container">$n$</span>, we have <span class="math-container">$\lim_{a \to 0} z^* = 0$</span>. The highest value for <span class="math-container">$z^*$</span> occurs for <span class="math-container">$n=2$</span> and <span class="math-container">$a=1$</span> with <span class="math-container">$z^* = \sqrt 3 - 1 \simeq 0.732$</span>. For <span class="math-container">$n \ge3$</span> and all <span class="math-container">$a$</span>, we have that <span class="math-container">$z^* < 0.423$</span>.</p>
|
3,347,264 | <p>I'm trying to find the distance between two points for a question in my textbook. The points are P(-2, 3) and Q(1, -3). Here's the working I have so far:</p>
<p>d = <span class="math-container">$\sqrt{(x_1 - x_0)^2 + (y_1 - y_0)^2}$</span></p>
<p>d = <span class="math-container">$\sqrt{(1 - (-2))^2 + (-3 - 3)^2}$</span></p>
<p>d = <span class="math-container">$\sqrt{3^2 + -6^2}$</span></p>
<p>d = <span class="math-container">$\sqrt -27$</span></p>
<p>This results in "Math ERROR" on my calculator.
Where am I going wrong?</p>
| Aforest | 337,161 | <p>Let's begin with your 2nd inequality with <span class="math-container">$x$</span> only.</p>
<p>Let <span class="math-container">$x = (1+p)/(1-p)$</span> and we have <span class="math-container">$x>0\iff p\in(-1,1)$</span>, then your inequality becomes</p>
<p><span class="math-container">$$\left(\frac{(1+p)^n+(1-p)^n+1}{(1+p)^{n-1}+(1-p)^{n-1}+1}\right)^n+1\le\!\!\!?\;(1+p)^n+(1-p)^n.$$</span></p>
<p>Let <span class="math-container">$a_n = (1+p)^n+(1-p)^n$</span>, by symmetry we can assume <span class="math-container">$p\in(0,1)$</span> (the case <span class="math-container">$p=0$</span> is trivial), then we have</p>
<p><span class="math-container">$$\left(\frac{a_n+1}{a_{n-1}+1}\right)^n+1\le\!\!\!?\;a_n.$$</span></p>
<p>Note that <span class="math-container">$\left(\frac{a_n+1}{a_{n-1}+1}\right)$</span> is increasing [1], we have</p>
<p><span class="math-container">$$\left(\frac{a_n+1}{a_{n-1}+1}\right)^n\le\left(\frac{a_n+1}{a_{n-1}+1}\right)\cdot\left(\frac{a_{n+1}+1}{a_{n}+1}\right)\cdots\left(\frac{a_{2n-1}+1}{a_{2n-2}+1}\right)=\left(\frac{a_{2n-1}+1}{a_{n-1}+1}\right).$$</span></p>
<p>Thus, your inequality can be proved if we have</p>
<p><span class="math-container">$$\frac{a_{2n-1}+1}{a_{n-1}+1}\le\!\!\!?\;a_n-1,$$</span></p>
<p>Note that <span class="math-container">$(a_n-1)(a_{n-1}+1)=a_{2n-1}+2(1-p^2)^{n-1}+a_n-a_{n-1}-1$</span>, so the inequality above becomes</p>
<p><span class="math-container">$$(1-p^2)^{n}+\frac12(a_{n+1}-a_n)\ge\!\!\!?\;1.$$</span></p>
<p>Let <span class="math-container">$(u_n)$</span> be the LHS, we have <span class="math-container">$u_0=u_1=1$</span>, and we can prove that <span class="math-container">$(u_n)$</span> is increasing [2], which finishes the proof.</p>
<p>PS. In fact, a stronger inequality has been proved. The following equalities can be useful:
<span class="math-container">$$a_{n+1}=a_n+pb_n,\quad b_{n+1}=b_n+pa_n,$$</span>
with <span class="math-container">$b_n=(1+p)^n-(1-p)^n$</span>.</p>
<hr />
<p>Update.</p>
<p>[1]
<span class="math-container">\begin{align}
\frac{a_{n+1}+1}{a_n+1}\ge\frac{a_{n}+1}{a_{n-1}+1} & \iff(a_{n+1}+1)(a_{n-1}+1)\ge(a_n+1)^2\\
&\impliedby a_{2n}+a_2(1-p^2)^{n-1}+1\ge a_{2n}+2(1-p^2)^n+1\tag{*}\\
&\iff (1+p)^2+(1-p)^2 \ge 2(1+p)(1-p)\\
&\iff (2p)^2\ge0.
\end{align}</span></p>
<p>(*): we have <span class="math-container">$a_{n+1}+a_{n-1}\ge2a_n$</span> since</p>
<p><span class="math-container">$$a_{n+1}+a_{n-1}\ge2a_n\iff b_{n}\ge b_{n-1}\iff a_{n-1}\ge 0.$$</span></p>
<p>[2]
<span class="math-container">\begin{align}
u_{n+1}\ge u_{n} & \iff \frac p2 b_{n+1}+(1-p^2)^{n+1}\ge\frac p2 b_{n}+(1-p^2)^{n}\\
& \iff\frac {p^2}2a_{n}\ge p^2(1-p^2)^{n}\\
& \iff \frac1{(1-p)^n}+\frac1{(1+p)^n}\ge2.\\
\end{align}</span></p>
<p>As mentioned by @Andreas, the last inequality is true since
<span class="math-container">$$\frac1{(1-p)^n}+\frac1{(1+p)^n}\ge \frac2{(1-p^2)^{n/2}}\ge2.$$</span></p>
|
1,435,196 | <p>In particular, I would like to know if the four velocity and the four acceleration are tensors. </p>
| Community | -1 | <p>"Tensors" are pretty general. Scalars are (rank 0) tensors. Vectors are (rank 1) tensors, including 4 vectors. So are matrices, etc.</p>
|
1,435,196 | <p>In particular, I would like to know if the four velocity and the four acceleration are tensors. </p>
| Mark Viola | 218,419 | <p>The four velocity and four acceleration are vectors in a space-time <a href="https://en.wikipedia.org/wiki/Minkowski_space" rel="nofollow">Minkowski space</a> and can be represented as a quadruplet $(u_0,u_1,u_2,u_3)$ and $(a_0,a_1,a_2,a_3)$ where the elements with subscripts $1-3$ represent the ordinary Euclidian vectors for velocity and acceleration, in proper time.</p>
<p>They are not tensors (of rank two), an example of which is the <a href="https://en.wikipedia.org/wiki/Electromagnetic_tensor" rel="nofollow">Electromagnetic Field tensor</a>.</p>
|
619,890 | <p>I have a question.There is a group of 5 men and a group of 7 women.With how many ways can each of the 5 men get married with one of the 7 women?</p>
| yig | 463,312 | <p>I don't know of a computationally more efficient way to project points onto the affine subspace (flat) than what you describe ($p + P_{\textit{null}(W)}( x - p )$).</p>
<p>Given ${W}$ with orthonormal columns, $P_{\textit{null}(W)} = {W} {W}^\top$
and ${I} - P_{\textit{null}(W)}$ are projection matrices onto the nullspace and row-space, respectively. If ${W}$'s columns are linearly independent but not orthonormal, $P_{\textit{null}(W)} = {W} ({W}^\top {W})^{-1} {W}^\top$. If W's columns aren't even linearly independent, I believe you can use the pseudoinverse instead of the inverse.</p>
<p>A couple of other useful properties of projection matrices: Given a projection matrix ${P}$, ${P}^\top={P}={P}{P}$. The eigenvalues of ${P}$ are either 0 or 1. If your linear subspace is defined implicitly by a basis orthogonal to it, it is probably computationally more efficient to get the needed projection matrix via $I - P$.</p>
|
984,558 | <p>Is it possible to find a representation of the infinitesimal generators of the special unitary group SU(3) that contains 4 by 4 matrices, by say taking a Kronecker product of its irreducible representation(s) with itself?</p>
<p>I know this is possible for SU(2), where one can express the three 4 by 4 matrices spanning the unit quaternion group in terms of some of the generators of SU(4), which are 4 by 4 matrices as well.</p>
<p>I am trying to do the same for SU(3). This question is motivated by the investigation of higher dimensional gauge theories. Thanks.</p>
| Jonathan Rayner | 90,675 | <p>As noted by the accepted answer and the comments, the answer is no.</p>
<p>Note, we can prove this without using Young Tableaux: we can check the possible dimensions as noted in Jyrki's comment above by directly using the <a href="https://en.wikipedia.org/wiki/Lie_algebra_representation#The_case_of_sl.283.2CC.29" rel="nofollow noreferrer">known dimension formula for <span class="math-container">$SU(3)$</span></a>:</p>
<blockquote>
<p>The irreducible representations are indexed by a pair of non-negative
integers <span class="math-container">$(m_1,m_2)$</span> and the dimension of the irreducible
representation is given by </p>
<p><span class="math-container">\begin{align}
\dim(V_{m_1,m_2}) = \frac{1}{2}(m_1+1)(m_2+1)(m_1+m_2+2)
\end{align}</span></p>
</blockquote>
<p>One checks that there is no pair of non-negative integers that gives <span class="math-container">$\dim(V_{m_1,m_2})=4$</span> by testing the first few (and only) possibilities directly.</p>
|
357,520 | <p>How can I show that
$$\lim_{a\to{0}}\frac{{\pi}a\ \coth{\mathrm{{\pi}a}-1}}{2a^2}=\frac{\pi^2}{6}$$
I think the limit is in $\frac{0}{0}$ form, so I am using L'Hospital's rule, and then I cannot solve further, Please Help.</p>
<p>Thanks!</p>
| Community | -1 | <p>Start with Marvis' approach, and evaluate $$\lim_{t\to{0}}\frac{t\ \coth(t)-1}{2t^2}=\lim_{t\to{0}}\frac{t\ \cosh(t)-\sinh(t)}{2t^2 \sinh(t)}$$</p>
<p>It is usually easier to work with sines and cosines. Apply l'Hopital rule once :</p>
<p>$$\lim_{t\to{0}}\frac{t\ \cosh(t)-\sinh(t)}{2t^2 \sinh(t)}=\lim_{t\to{0}}\frac{t\ \sinh(t)}{4 t \sinh(t) + 2t^2 \cosh(t)} $$</p>
<p>You can simplify $t$ in this last expression. Apply l'Hopital rule once more to find your result.</p>
|
14,508 | <p>Suppose that $f$ is a weight $k$ newform for $\Gamma_1(N)$ with attached $p$-adic Galois representation $\rho_f$. Denote by $\rho_{f,p}$ the restriction of $\rho_f$ to a decomposition group at $p$. When is $\rho_{f,p}$ semistable (as a representation of
$\mathrm{Gal}(\overline{\mathbf{Q}}_p/\mathbf{Q}_p)$?</p>
<p>To make things really concrete, I'm happy to assume that $k=2$ and that the $q$-expansion of $f$ lies in $\mathbf{Z}[[q]]$. </p>
<p>Certainly if $N$ is prime to $p$ then $\rho_{f,p}$ is in fact crystalline, while
if $p$ divides $N$ exactly once then $\rho_{f,p}$ is semistable (just thinking about the Shimura construction in weight 2 here, and the corresponding reduction properties of $X_1(N)$
over $\mathbf{Q}$ at $p$). For $N$ divisible by higher powers of $p$, we know that these representations are de Rham, hence potentially semistable. Can we say more? For example,
are there conditions on "numerical data" attached to $f$ (e.g. slope, $p$-adic valuation of $N$, etc.) which guarantee semistability or crystallinity over a specific
extension? Can we bound the degree and ramification of
the minimal extension over which $\rho_{f,p}$ becomes semistable in terms of numerical
data attached to $f$? Can it happen that $N$ is highly divisible by $p$ and yet $\rho_{f,p}$ is semistable over $\mathbf{Q}_p$?</p>
<p>I feel like there is probably a local-Langlands way of thinking about/ rephrasing this question, which may be of use... </p>
<p>As a possible example of the sort of thing I have in mind: if $N$ is divisible by $p$ and $f$ is ordinary at $p$ then $\rho_{f,p}$ becomes semistable over an abelian extension of
$\mathbf{Q}p$
and even becomes crystalline over such an extension provided that the Hecke eigenvalues
of $f$ for the action of $\mu_{p-1}\subseteq (\mathbf{Z}/N\mathbf{Z})^{\times}$ via the diamond operators
are not all 1.</p>
| Rob Harron | 1,021 | <p>Since <em>f</em> is potentially semi-stable, you can look at its attached filtered (φ, <em>N</em>, Gal(<em>L</em>/<strong>Q</strong><sub><em>p</em></sub>))-representation (where ρ<sub>f,p</sub> becomes semi-stable when restricted to <em>G</em><sub><em>L</em></sub>). If its <em>N</em> is zero, then it is potentially cristalline, otherwise it is not.</p>
<p>As for the ordinary case, I'm not sure what definition you're using. Under Greenberg's definition, an ordinary <em>p</em>-adic Galois representation is semi-stable (see Perrin-Riou's article in the Bures). Also, the Tate curve is ordinary at <em>p</em>, but not potentially cristalline (once something is semi-stable and non-cristalline, it can't be potentially cristalline).</p>
|
1,085,014 | <p>Let $G$ be a group. Prove that $g^2 = e$ for all $g \in G$, then $G$ is abelian. ($e$ is the identity element.)</p>
<p>My Solution: Let $a,b \in G$. Then $a(ab)b = a^2b^2 = e^2 =e$. Now I tried to reverse $ab$ in the brackets to get the same solution to show that the group is also commutative but I was not able to do so.</p>
| ILoveMath | 42,344 | <p>Let $a,b \in G$ be arbitrary elements. Notice:</p>
<p>$$abab = (ab)^2 = e = a^2 b^2 = aabb$$</p>
<p>Cancellation gives $ba = ab$.</p>
|
1,085,014 | <p>Let $G$ be a group. Prove that $g^2 = e$ for all $g \in G$, then $G$ is abelian. ($e$ is the identity element.)</p>
<p>My Solution: Let $a,b \in G$. Then $a(ab)b = a^2b^2 = e^2 =e$. Now I tried to reverse $ab$ in the brackets to get the same solution to show that the group is also commutative but I was not able to do so.</p>
| Leox | 97,339 | <p>we have that $ab ab= e$. Thus (multiply by $b$ two sides) $abab^2=b$ or $aba=b$. In the same way $aba^2=ba$ or $ab=ba.$</p>
|
1,575,764 | <p>$Q$ is a linear operator from $V \to V$ with $V$ being a finite dimensional complex inner-product-space. </p>
<p>Given: $Q^*=5Q$, $Q^*$ being the adjoint.</p>
<p>Show that $0$ is the only eigenvalue of $Q$. </p>
<p>I've been staring at this problem for quite a while now. I think the answer lies in the eigenvalues of $Q$ and $Q^*$ being the same, but I can't think of a way to prove $0$ is an eigenvalue let alone the only one. </p>
| Community | -1 | <p>Try doing this by direct computation. Suppose we have an eigenvalue $\lambda$ with (unit) eigenvector $v$. Then since we have a question about adjoints, it's natural that inner products will be used somewhere. Try considering $v$:</p>
<p>$$\langle Qv, v\rangle = \langle \lambda v, v \rangle = \lambda$$</p>
<p>On the other hand,</p>
<p>\begin{align*}
\langle Qv, v\rangle = \langle v, Q^* v\rangle = \langle v, 5Q v \rangle = 5\overline{\lambda}
\end{align*}</p>
<p>The result now follows.</p>
|
131,283 | <p>I came across a question which required us to find $\displaystyle\sum_{n=3}^{\infty}\frac{1}{n^5-5n^3+4n}$. I simplified it to $\displaystyle\sum_{n=3}^{\infty}\frac{1}{(n-2)(n-1)n(n+1)(n+2)}$ which simplifies to $\displaystyle\sum_{n=3}^{\infty}\frac{(n-3)!}{(n+2)!}$. I thought it might have something to do with partial fractions, but since I am relatively inexperienced with them I was unable to think of anything useful to do. I tried to check WolframAlpha and it gave $$\sum_{n=3}^{m}\frac{(n-3)!}{(n+2)!}=\frac{m^4+2m^3-m^2-2m-24}{96(m-1)m(m+1)(m+2)}$$
From this it is clear that as $m\rightarrow \infty$ the sum converges to $\frac{1}{96}$, however I have no idea how to get there. Any help would be greatly appreciated!</p>
| Prasad G | 25,314 | <p><code>(n−3)!/(n+2)! = 1/[(n+2)(n+1)n(n-1)(n-2)]</code> and you can easily solve by using fractional part.</p>
|
3,300,469 | <p>I have a problem counting all the possible ways of "pairing" two datasets of size n and m, including partial pairing. </p>
<p>Example:
Assume we have two sets <span class="math-container">$\{A,B\}$</span> and <span class="math-container">$\{1,2,3\}$</span>. My aim is to find all ways of pairing letters with numbers, including the consideration of situations, where only a fraction of possible pairs is actually present, including a case of "no pairing at all".</p>
<p>In this case I would get the following ways of pairing:<br>
<span class="math-container">$\{A,1\},\{B,2\},\{3\}$</span> (two pairs out of two concurrent pairs possible)<br>
<span class="math-container">$\{A,2\},\{B,1\},\{3\}$</span> (two pairs out of two concurrent pairs possible)<br>
<span class="math-container">$\{A,1\},\{B,3\},\{2\}$</span> (two pairs out of two concurrent pairs possible)<br>
<span class="math-container">$\{A,3\},\{B,1\},\{2\}$</span> (two pairs out of two concurrent pairs possible)<br>
<span class="math-container">$\{A,2\},\{B,3\},\{1\}$</span> (two pairs out of two concurrent pairs possible)<br>
<span class="math-container">$\{A,3\},\{B,1\},\{1\}$</span> (two pairs out of two concurrent pairs possible)<br>
<span class="math-container">$\{A,1\},\{B\},\{2\},\{3\}$</span> (one pair out of two coexisting pairs possible)<br>
<span class="math-container">$\{A,2\},\{B\},\{1\},\{3\}$</span> (one pair out of two coexisting pairs possible)<br>
<span class="math-container">$\{A,3\},\{B\},\{1\},\{2\}$</span> (one pair out of two coexisting pairs possible)<br>
<span class="math-container">$\{A\},\{B,1\},\{2\},\{3\}$</span> (one pair out of two coexisting pairs possible)<br>
<span class="math-container">$\{A\},\{B,2\},\{1\},\{3\}$</span> (one pair out of two coexisting pairs possible)<br>
<span class="math-container">$\{A\},\{B,3\},\{1\},\{2\}$</span> (one pair out of two coexisting pairs possible)<br>
<span class="math-container">$\{A\},\{B\},\{1\},\{2\},\{3\}$</span> (zero pairs out of two coexisting pairs possible) </p>
<p>How can I generalize it to bigger sets of size n and m?</p>
| Boyku | 567,523 | <p>there is a solution involving the exponential generating function. </p>
<p>The description using combinatorial species is <span class="math-container">$E(X)\cdot E(XY) \cdot E(Y)$</span> where:</p>
<p>X is the sort of letters</p>
<p>Y is the sort of digits</p>
<p>and the formula means that there is a set of unpaired letter, than a set of pairs, then a set of unpaired digits.</p>
<p>The exponential generating function is </p>
<p><span class="math-container">$$exp(x).exp(xy).exp(y)$$</span></p>
<p>By reading the coefficient of <span class="math-container">$$\frac {x^n}{n!} \frac {y^m}{m!}$$</span> one gets the solution.</p>
<p>For <span class="math-container">$n=2$</span> and <span class="math-container">$m=3$</span> the exponential series contains the term </p>
<p><span class="math-container">$$13 \frac {x^2}{2!} \frac {y^3}{3!}$$</span></p>
<p>so there are <span class="math-container">$13$</span> pairings.</p>
|
1,515,823 | <p>I am doing my research in Functional Analysis, especially in "Generalized inverse of Linear Maps".</p>
<p>I have come across Probability by studying only the methods or Distributions(like Binomial, poisson, normal,etc)</p>
<p>Now I wish to study the mathematical background and intuitive way of looking on it.</p>
<p>Can you suggest some probability text book which explains,</p>
<blockquote>
<p><span class="math-container">$\bullet$</span> Geometrical ideas of the Probability Distributions. May be using diagrams, graphs, etc.</p>
<p><span class="math-container">$\bullet$</span> Proofs for the distribution functions.</p>
<p><span class="math-container">$\bullet$</span> Problems with natural solutions and then generalizations
Like, for a binomial distribution(how they are giving the probability mass function).</p>
</blockquote>
| Jose Arnaldo Bebita Dris | 28,816 | <p>I would highly recommend <a href="http://rads.stackoverflow.com/amzn/click/032179477X" rel="nofollow">A First Course in Probability</a> by Sheldon Ross.</p>
<p>From the hyperlinked Amazon page:</p>
<blockquote>
<p><strong>A First Course in Probability</strong> features clear and intuitive explanations of the mathematics of probability theory, outstanding problem sets, and a variety of diverse examples and applications. This book is ideal for an upper-level undergraduate or graduate level introduction to probability for math, science, engineering and business students. It assumes a background in elementary calculus.</p>
</blockquote>
<p>I am recommending this book because I think it is suitable for self-study, as it is listed as a textbook for the Society of Actuaries' <a href="https://www.soa.org/education/exam-req/edu-exam-p-detail.aspx" rel="nofollow">Exam P</a>.</p>
<p>You can also try checking out the following books:</p>
<ol>
<li><p><a href="http://rads.stackoverflow.com/amzn/click/0495110817" rel="nofollow">Mathematical Statistics with Applications</a> by Dennis Wackerly, William Mendenhall and Richard L. Scheaffer</p></li>
<li><p><a href="http://rads.stackoverflow.com/amzn/click/0321923278" rel="nofollow">Probability and Statistical Inference</a> by Robert V. Hogg, Elliot Tanis and Dale Zimmerman</p></li>
<li><p><a href="http://rads.stackoverflow.com/amzn/click/1625424728" rel="nofollow">Probability and Statistics with Applications: A Problem Solving Text</a> by Leonard A. Asimow and Mark M. Maxwell</p></li>
</ol>
|
409,689 | <p>I have $(x_1, y_1), (x_2, y_2)$.</p>
<p>How do I find the point that's $d$ distance away from $(x_1, y_1)$ on a straight line to $(x_2, y_2)$?</p>
<p>I know I can get the length of the line with Pythagoras. I know if I drew a circle I could use the radius as distance and the point would be where the line and the circle intersect.</p>
<p>Could someone briefly explain each step to me please?</p>
<p>I don't understand <a href="https://math.stackexchange.com/questions/175896/finding-a-point-along-a-line-a-certain-distance-away-from-another-point">Finding a point along a line a certain distance away from another point!</a></p>
| John Douma | 69,810 | <p>If I can move one unit along a line I can move any distance along that line. We'll calculate how much we would have to add to each of $x_1$ and $y_1$ to move one unit along the line and then we'll multiply that by $d$ to get the answer.</p>
<p>Let $d^{\prime}$ be the distance between $(x_1,y_1)$ and $(x_2,y_2)$. It's value is $$d^{\prime}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$</p>
<p>If we move from $(x_1,y_1)$ to $(x_2,y_2)$ along the line connecting them we move $d^{\prime}$ units. We can represent this by the function</p>
<p>$$(x_1,y_1)\mapsto (x_1+(x_2-x_1),y_1+(y_2-y_1))$$</p>
<p>To move one unit along the same line, we divide the amount of the change by $d^{\prime}$ to get</p>
<p>$$(x_1,y_1)\mapsto (x_1+\frac{1}{d^{\prime}}(x_2-x_1),y_1+\frac{1}{d^{\prime}}(y_2-y_1))$$</p>
<p>Finally, to move a distance $d$ along the line we multiply the change by $d$ to get</p>
<p>$$(x_1,y_1)\mapsto (x_1+\frac{d}{d^{\prime}}(x_2-x_1),y_1+\frac{d}{d^{\prime}}(y_2-y_1))$$</p>
|
950,485 | <p>I have been trying to solve the following limit but am completely stuck.</p>
<p>$$\lim_{\alpha \rightarrow \infty} 1-\left( \frac{y+\alpha}{\alpha-1} \right)^{-\alpha}$$</p>
<p>I have tried inverting the ratio and came up with the following expression:</p>
<p>$$ 1 - \lim_{\alpha \rightarrow \infty} \left( 1-\frac{y+1}{y+\alpha}\right)^\alpha$$</p>
<p>Which roughly resembles the exponential function:</p>
<p>$$\lim_{\alpha \rightarrow \infty} \left( 1- \frac{x}{\alpha} \right)^\alpha = \exp(-x)$$</p>
<p>Except for the additive term in the denominator. Is there a u-substitution type trick to this?</p>
| Community | -1 | <p>$$\lim_{\alpha \rightarrow \infty} \left( 1-\frac{y+1}{y+\alpha}\right)^\alpha=\lim_{\alpha \rightarrow \infty} \dfrac{\left( 1-\frac{y+1}{y+\alpha}\right)^{y+\alpha}}{\left( 1-\frac{y+1}{y+\alpha}\right)^{y}}=\exp(-(y+1))$$</p>
|
3,203,100 | <p>This equation just came to my mind, I tried solving it but can't find any solution to this problem. Can anyone please tell what is the process to approach this problem? </p>
| Claude Leibovici | 82,404 | <p>Welcome to the world of <a href="https://en.wikipedia.org/wiki/Lambert_W_function" rel="nofollow noreferrer">Lambert function</a> !</p>
<p>The solution of <span class="math-container">$x=n^x$</span> is given by
<span class="math-container">$$x=-\frac{W(-\log (n))}{\log (n)}$$</span> and, in the real domain, <span class="math-container">$W(t)$</span> is defined if <span class="math-container">$t \geq -\frac 1e$</span> which implies <span class="math-container">$n\leq e^{-e}$</span>.</p>
|
3,653,979 | <p>Let A = <span class="math-container">$\begin{bmatrix}r_1 & r_2 & r_3 & r_4 & r_5\end{bmatrix}^T$</span> have rows <strong><span class="math-container">$r_1$</span></strong>, <strong><span class="math-container">$r_2$</span></strong>, <strong><span class="math-container">$r_3$</span></strong>, <strong><span class="math-container">$r_4$</span></strong>, <strong><span class="math-container">$r_5$</span></strong> <span class="math-container">$\in$</span> <span class="math-container">$\mathbb{R}^5$</span>. Assume det(A) = -3.</p>
<p>Compute </p>
<p>det <span class="math-container">$\begin{bmatrix}2r_1 + 3r_2 + 4r_3 + 4r_4\\ r_1 + 2r_2\\ r_2+3r_3\\r_3+4r_4\\r_1\end{bmatrix}$</span> and justify your answer.</p>
<p><em>(the part before this was the same thing but with this matrix)</em></p>
<p>det <span class="math-container">$\begin{bmatrix}5r_1 + 5r_2 + 5r_3 + 5r_4 + 5r_5\\ 4r_1 + 4r_2 + 4r_3 + 4r_4\\ r_1\\2r_1+2r_2\\3r_1+3r_2+3r_3\end{bmatrix}$</span> </p>
<p>and I think I figured this one out (you just Gauss-Jordan it and it ends up being equal to A so it's just -3)</p>
<p>But my main confusion is with the first matrix, because even with Gauss-Jordan it doesn't equal A so how can I even find the determinant because it's not a square matrix.</p>
<p>Any help is appreciated. Thanks in advance!</p>
| Shiv Tavker | 687,825 | <p><strong>Hint 1:</strong> What happens to new matrix if you do <span class="math-container">$R_1 \to R_1 - R_2 - R_3 -R_4 -R_5$</span> where <span class="math-container">$R$</span> represents the rows of the matrix in part (i). </p>
<p><strong>Hint 2:</strong> The answer for the determinant is zero.</p>
|
2,545,516 | <p>So I have to assess the convergence of $$\displaystyle\sum_{n=1}^{\infty}\sin\left(\displaystyle\frac{1}{\sqrt{n}}\right).$$</p>
<p>I'm told that it diverges, but can't really see why.</p>
<p>The divergence test doesn't really help, because
$\lim\limits_{x\to\infty}\displaystyle\frac{1}{\sqrt{n}}=0$, so</p>
<p>$\lim\limits_{x\to\infty}\sin\left(\displaystyle\frac{1}{\sqrt{n}}\right)=0$, which doesn't conclude its divergence.</p>
<p>I doubt the ratio test would be much of use in this situation.</p>
<p>I can't imagine using the integral comparison test, as I wouldn't know where to start with $\displaystyle\int_{1}^{\infty}\sin\left(\displaystyle\frac{1}{\sqrt{n}}\right) \mathrm dx$.</p>
| user284331 | 284,331 | <p>Use the inequality that $\sin x\geq\dfrac{2}{\pi}x$ for $x\in[0,\pi/2]$.</p>
|
3,705,539 | <p>I am trying to learn more about probability and came across an interesting question that I am stuck on and can no longer find online. There are 20 numbered balls and 10 bins. Someone is trying to assign the balls to the bins, but does it with replacement on accident.</p>
<p>So they did the following: Place a ball in bin 1, record it, then remove ball (with replacement remember). Place a ball in bin 2, record it, then remove ball. Place a ball in bin 3, record it, then remove ball. So for each bin, you have put in 1 ball. There are ten bins, therefore you do that process once for every bin. Once you have done that the experiment is over.</p>
<p>What is the probability exactly 1 ball was assigned to exactly 4 bins? What is the probability at least 2 bins received the same ball?</p>
<p>A) 1 Ball in 4 Bins:</p>
<p>We have <span class="math-container">${20 \choose 1}$</span> being the different ways we can choose the 1 ball that was assigned. Also, we have <span class="math-container">${19 \choose 6}$</span> being the different ways the other 19 balls can be picked for assignment. However, what is the sample size? Would it be <span class="math-container">$20^{10}$</span>? Thus the answer would be <span class="math-container">$\frac{{20 \choose 1}{19 \choose 6}}{20^{10}}$</span>.</p>
<p>B) Probability of at least 2 repeated can be represented as <span class="math-container">$1-P(\text{Zero Repeated})- P(\text{One Repeated})$</span>. So <span class="math-container">$P(0) = {20 \choose 10}/20^{10}$</span> and <span class="math-container">$P(1) = \frac{{20 \choose 1}{19 \choose 9}}{20^{10}}$</span>. Then we can plug and chug.</p>
<p>Are these right? Is this how to think about this type of problem?</p>
| user | 293,846 | <p>After clarification of the question it can be answered as following:</p>
<ol>
<li><em>What is the probability that exactly 1 ball was assigned to exactly 4 bins?</em></li>
</ol>
<p>We have <span class="math-container">$\binom{20}{1}$</span> ways to choose the "4-fold" ball and <span class="math-container">$\binom{10}{4}$</span> ways to choose the bins where it should go. The other 6 bins can be arbitrarily filled with remaining 19 balls. Hence the overall number of combinations is:
<span class="math-container">$$
\binom{20}{1}\binom{10}{4}19^6.
$$</span>
In this way we however double-count the cases where there are two balls each assigned to exactly 4 bins. There are <span class="math-container">$\binom{20}{2}$</span> such pairs and <span class="math-container">$\binom{10}4\binom{6}4$</span> ways to choose corresponding bins. The other 2 bins can be filled arbitrarily with remaining 18 balls. Bringing everything together the final result is:
<span class="math-container">$$
\frac{\binom{20}{1}\binom{10}{4}19^6-\binom{20}{2}\binom{10}{4}\binom{6}{4}18^2}{20^{10}}.
$$</span></p>
<ol start="2">
<li><em>What is the probability at least 2 bins received the same ball?</em></li>
</ol>
<p>The simplest way to answer this question is to use complementary probability of the event "all bins receive different balls":
<span class="math-container">$$
1-\frac{\frac{20!}{10!}}{20^{10}}.
$$</span>
Replacement of <span class="math-container">$\frac{20!}{10!}$</span> with <span class="math-container">$\binom{20}{10}$</span> would be wrong here, since after choosing <span class="math-container">$10$</span> balls out of <span class="math-container">$20$</span> there are still <span class="math-container">$10!$</span> ways to assign the balls to certain bins. </p>
|
2,209,438 | <p>I am trying to find this limit,</p>
<blockquote>
<p>$$\lim_{x \rightarrow 0} \frac{1}{x^4} \int_{\sin{x}}^{x} \arctan{t}dt$$</p>
</blockquote>
<p>Using the fundamental theorem of calculus, part 1,
$\arctan$ is a continuous function, so
$$F(x):=\int_0^x \arctan{t}dt$$
and I can change the limit to
$$\lim_{x \rightarrow 0} \frac{F(x)-F(\sin x)}{x^4}$$</p>
<p>I keep getting $+\infty$, but when I actually integrate $\arctan$ (integration by parts) and plot the function inside the limit, the graph tends to $-\infty$ as $x \rightarrow 0+$.</p>
<p>I tried using l'Hospital's rule, but the calculation gets tedious.</p>
<p>Can anyone give me hints?</p>
<p><strong>EDIT</strong></p>
<p>I kept thinking about the problem, and I thought of power series and solved it, returned to the site and found 3 great answers. Thank You!</p>
| Stefano | 387,021 | <p>Since $F(0) = 0$ and everything is smooth, you can apply de l'Hopital and get</p>
<p>$$\lim_{x \to 0}\frac{F(x)-F(\sin x)}{x^4} = \lim_{x \to 0}\frac{\arctan x - \cos x\arctan \sin x}{4x^3}.$$</p>
<p>This last limit can be evaluated using Taylor series:</p>
<p>$$\arctan x = x-\frac{x^3}{3}+O(x^5) $$
and</p>
<p>$$\cos x \arctan \sin x = x-x^3+O(x^5) $$</p>
<p>and the limit you are looking for is equal to</p>
<p>$$\lim_{x \to 0}\frac{x-\dfrac{x^3}{3}-x+x^3 + O(x^5)}{4x^3} = \frac{1}{6}. $$</p>
<p>The ''ugly'' Taylor expansion is obtained combining the Taylor expansions of $\sin$, $\cos$ and $\arctan$. Easier done than said.</p>
|
376,796 | <p>This is more of a pedagogical question rather than a strictly mathematical one, but I would like to find good ways to visually depict the notion of curvature. It would be preferable to have pictures which have a reasonably simple mathematical formalization and even better if there is a related diagram that explains torsion.</p>
<h2>One common picture</h2>
<p><a href="https://i.stack.imgur.com/bSiYsm.png" rel="noreferrer"><img src="https://i.stack.imgur.com/bSiYsm.png" alt="enter image description here" /></a></p>
<p>I've often used the above schematic to think about the Riemann curvature tensor
<span class="math-container">$$R(X,Y)Z = \nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z + \nabla_{[X,Y]} Z.$$</span></p>
<p>This diagram intuitively shows that the curvature involves the difference of covariant derivatives. However, it doesn't really explain why there is another term in the formula (i.e., <span class="math-container">$\nabla_{[X,Y]} Z$</span>). Also, it takes some work to translate the picture into a precise and correct mathematical formula.</p>
<p>One way to formalize this (suggested by Robert Bryant) is to consider a parallelogram with sides <span class="math-container">$\epsilon X$</span> and <span class="math-container">$\epsilon Y$</span> in <span class="math-container">$T_p M$</span>. Then the diagram depicts the parallel transport of <span class="math-container">$Z$</span> along the exponential of the sides of the parallelogram.
To understand the picture, you parallel transport the vector labelled <span class="math-container">$R(X,Y)Z$</span> back to <span class="math-container">$p$</span>, divide by <span class="math-container">$\epsilon^2$</span> and let <span class="math-container">$\epsilon$</span> go to <span class="math-container">$0$</span>.
This interpretation is conceptually simple, but has the disadvantage that the top and right hand sides of the parallelogram are not geodesics, so we cannot use this interpretation to draw a similar diagram for torsion.</p>
<p>There are other ways to formalize this diagram, and it would be interesting to hear other simple and correct explanations for this picture (or any variation of it).</p>
<h2>Another common picture</h2>
<p><a href="https://i.stack.imgur.com/MhGf1m.png" rel="noreferrer"><img src="https://i.stack.imgur.com/MhGf1m.png" alt="By Fred the Oyster, CC BY-SA 4.0, https://commons.wikimedia.org/w/index.php?curid=35124171" /></a></p>
<p>Another commonly used picture to explain curvature is a spherical triangle with two vertices on the equator and a third at a pole. This intuitively shows that curvature gives rise to holonomy, but also relies on the global geometry of the sphere. In other words, it doesn't really depict curvature as "local holonomy."</p>
| Mohammad Ghomi | 68,969 | <p>The best way I know to illustrate the notion of curvature is via Toponogov's theorem. We can compare any (geodesic) triangle in a Riemannian manifold <span class="math-container">$M$</span> with one with the same edge lengths in Euclidean plane <span class="math-container">$R^2$</span>. The (sectional) curvature of <span class="math-container">$M$</span> is positive (resp. negative) provided that all its triangles are fatter (resp. thinner) than the comparison triangle. More precisely, this means that the distance between each vertex and the midpoint of the opposite side is bigger (resp. smaller) than the corresponding distance in the comparison triangle.</p>
<p><a href="https://i.stack.imgur.com/yobXL.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/yobXL.jpg" alt="enter image description here" /></a></p>
|
9,918 | <p>I recently flagged as "rude or offensive" the comment </p>
<blockquote>
<p>As the tone should suggest, he’s a crank. It’s a hysterical screed with a few nuggets of fact surrounded by a great deal of nonsense. E.g., he may find that set theory ‘doesn’t make sense’, but a great many of us have no trouble making sense of it. – <a href="https://math.stackexchange.com/questions/356264#comment765805_356264">Brian M. Scott</a></p>
</blockquote>
<p>to the question</p>
<p><a href="https://math.stackexchange.com/questions/356264">Infinite sets don't exist!?</a></p>
<p>My complaint is very narrow: I am not arguing that the subject of the comment is right in his critique of infinite sets, only that the word "crank" is inappropriate in this context. In particular, none of the answers to the question provide evidence of crankdom, but rather disagreement over philosophy. If the comment had simply been </p>
<p>"He may find that set theory ‘doesn’t make sense’, but a great many of us have no trouble making sense of it."</p>
<p>I would have no objection.</p>
<p>I could also have flagged it as "not constructive/off topic" or as "too chatty". But to my mind it is the unwarranted rudeness that is the real problem here. More generally, I am quite upset that a productive mathematician is being called a crank. I don't know the subject of the comment personally, but I am familiar with his work and he is most definitely <em>not</em> a crank. </p>
<p>Comments like this reflect quite poorly on math.stackexchange and I hope not to see any more of them. What is the best way to encourage commenters to refrain from baseless accusations of crankdom? Is flagging the appropriate course of action?</p>
| Willie Wong | 1,543 | <p>I'm cleaning up <a href="https://math.stackexchange.com/questions/356264/infinite-sets-dont-exist">that comment thread</a>, since it veered a bit off topic. For the sake of transparency I'm preserving the comment thread below from screen caps. (To make sure that I don't miss any comments there are some over-laps in the three images, so some comments appear more than once. My apologies if this cause any inconvenience.) (Links given in the comments are reproduced under the screen capture, in case you want to check them out.)</p>
<p>Page 1: </p>
<p><img src="https://i.stack.imgur.com/uXyzu.png" alt="enter image description here"></p>
<p>Brian M. Scott's link to <a href="http://web.maths.unsw.edu.au/~norman/book.htm" rel="nofollow noreferrer">carry out his program for trigonometry</a></p>
<p>Page 2:</p>
<p><img src="https://i.stack.imgur.com/EcqcE.png" alt="enter image description here"></p>
<p>Brian M. Scott's link to <a href="http://en.wikipedia.org/wiki/Anatoly_Fomenko" rel="nofollow noreferrer">Anatoly Fomenko</a></p>
<p>Page 3:</p>
<p><img src="https://i.stack.imgur.com/zCfPY.png" alt="enter image description here"></p>
|
3,501,332 | <blockquote>
<p>A cryptoanalist, while trying to decipher a message, found that the
most frequent blocks were RH and NI, which must correspond to TH and
HE, which are the most common in the english language. Supposing the
text was codified using a 2x2 block cipher, what was the used matrix?</p>
</blockquote>
<p>I know that</p>
<p><span class="math-container">$\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}T&H\\H&E\end{bmatrix} =\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}19&7\\7&4\end{bmatrix} = \begin{bmatrix}R&N\\H&I\end{bmatrix}=\begin{bmatrix}17&13\\7&8\end{bmatrix} \pmod{26}$</span></p>
<p>I got as far as making a system of equations but that didn't get me anywhere. How do I solve this?</p>
| Siong Thye Goh | 306,553 | <p>Note that we have<span class="math-container">\begin{align}\begin{bmatrix} 19 & 7 \\ 7 & 4\end{bmatrix}^{-1} &= (19 \cdot 4 - 7^2)^{-1}\begin{bmatrix} 4 & -7 \\ -7 & 19\end{bmatrix}\\&=(-7\cdot 4 - 7^2)^{-1}\begin{bmatrix} 4 & -7 \\ -7 & -7\end{bmatrix} \\
&= (-77)^{-1}\begin{bmatrix} 4 & -7 \\ -7 & -7\end{bmatrix} \\
&= (-26
\cdot3+1)^{-1}\begin{bmatrix} 4 & -7 \\ -7 & -7\end{bmatrix}\\
&=\begin{bmatrix} 4 & -7 \\ -7 & -7\end{bmatrix}\end{align}</span></p>
<p>By post-multiplying this matrix to both sides, you should be able to recover your desired matrix.</p>
|
275,539 | <p>Kind of leading on from my other question, how would I solve for $i$? Or how would I check that it is possible to have such an $i$?</p>
<p>First I had to check for all $2^i$ and clearly this doesn't happen as all $2^i$ are even and so I will just get even $x's$ such that $2^i \equiv x \mod 28$. So the next one I go onto is $3$.</p>
<p>Now how do I go about doing this? </p>
| DonAntonio | 31,254 | <p>$$3^1=3\pmod{28}$$</p>
<p>$$3^2=9\pmod{28}$$</p>
<p>$$3^3=27=-1\pmod{28}$$</p>
<p>$$3^4=3\cdot3^3=-3=25\pmod{28}$$</p>
<p>$$3^5=3^2\cdot3^3=-9=19\pmod{28}$$</p>
<p>$$3^6=3^3\cdot3^3=1\pmod{28}\,\,\ldots\text{etc}$$</p>
|
200,903 | <p>My teacher was explaining quadratics in my class and it was a little bit unclear to me. The problem was <br> <br>
Suppose $at^2 + 5t + 4 > 0$, show that $a > 25/16$ . <br> <br></p>
<p>My teacher said that there are no solutions for this function when it is greater than $0$ and used $b^2-4ac \lt 0$, and this is the part that confused me. I understand why he used $b^2-4ac \lt 0$ but I cannot understand why there are no solutions by just looking at the function. Could someone explain this to me?</p>
| Ben | 27,132 | <p>The equation a$t^2$ + 5t + 4 = 0 may or may not have a solution. This depends on the value of <strong>a</strong> that you pick. Graphically, if this equation has a solution, the graph will touch and/or go through the x-axis. If there is not a solution, the graph will not have a solution. These x-intercepts are defined by the solutions: $$x = {-b \pm \sqrt{b^2-4ac}\over2a}$$ For these solutions to be real, $b^2-4ac$, the discriminant, must be greater than or equal to 0. If not, the solution will be irrational(sqrt of neg number).</p>
<p>Hope this helps!</p>
|
2,609,537 | <p>Is the following Proof Correct? In particular please comment on the correctness of the given formulas.</p>
<p><strong>Theorem.</strong> Given that $x$ is a real number, $x\neq 0$, and $x + \frac{1}{x}$ is an integer. For all $n\ge 1$, $x^n+\frac{1}{x^n}$ is an integer.</p>
<p><strong>Proof.</strong> We construct the proof by recourse to <em>Strong-Induction</em>. Assume for an arbitrary $n\in\mathbf{Z^+}$ that $x^k+\frac{1}{x^k}$ is an integer for any positive integer $k$ strictly less than $n$. Now Consider the following cases.</p>
<p><em>Case-1:</em> If $n$ is even then for some $l\in\mathbf{Z^+}$, $n = 2l$ thus the <em>Binomial-Theorem</em> implies that
$$(x+\frac{1}{x})^n - \sum_{j=1}^{l-1}\binom{n}{2j}\left(x^{2j}+\frac{1}{x^{2j}}\right)-\binom{n}{l} = x^n+\frac{1}{x^n}$$
from the inductive hypothesis we know that $x+\frac{1}{x}\in\mathbf{Z}$ which implies that $(x+\frac{1}{x})^n \in\mathbf{Z}$ in addition it also follows from the inductive hypothesis that $x^{2j}+\frac{1}{x^{2j}}\in\mathbf{Z}$ for $j\in\{1,2,3,...,l-1\}$ moreover we also that $\binom{n}{r}\in\mathbf{Z^+}$ is always a positive integer implying that $x^n+\frac{1}{x^n}$ is an integer.</p>
<p><em>Case-2:</em> If $n$ is odd then for some $l\in\mathbf{Z^+}$, $n = 2l+1$ thus the <em>Binomial-Theorem</em> implies that
$$(x+\frac{1}{x})^n - \sum_{j=1}^{l}\binom{n}{2j-1}\left(x^{2j-1}+\frac{1}{x^{2j-1}}\right) = x^n+\frac{1}{x^n}$$
and by using the same reasoning as in the previous case we can deduce that $x^n+\frac{1}{x^n}$ is an integer.</p>
| Matthew Leingang | 2,785 | <p>I get the gist of the argument and it looks correct to me. To verify it, though, I had to stare at the summations in the middle of the two displayed equations for a while.</p>
<p>This would be annoying to a grader (graders are <a href="http://www.jimpryor.net/teaching/guidelines/writing.html" rel="nofollow noreferrer">lazy, stupid, and mean</a>). So I suggest inserting your steps before those equations:</p>
<ol>
<li>Expand $\left(x + \frac{1}{x}\right)^{2l}$ with the Binomial Theorem</li>
<li>Visually collect the terms into multiples of $\left(x^{k} + \frac{1}{x^{k}}\right)$</li>
<li>Apply the inductive hypothesis.</li>
</ol>
<p>Basically I am asking for you to slow down at the point "The Binomial Theorem implies that" and be more explicit.</p>
|
4,495,950 | <blockquote>
<p>Why does <span class="math-container">$-\frac{1}{17-x}$</span> equal <span class="math-container">$\frac{1}{x-17}$</span>?</p>
</blockquote>
<p>Is there any simple computation to make this seem a little bit more intuitive? Right now, I cannot wrap my head around the fact that I can just switch signs of the term in the denominator.</p>
| JonathanZ supports MonicaC | 275,313 | <p>It comes from the following algebra facts:</p>
<p><span class="math-container">$$-\frac{a}{b}=\frac{-a}{b}=\frac{a}{-b}$$</span></p>
<p>and</p>
<p><span class="math-container">$$\begin{align}
-(c-d) &= (-c)-(-d)\\
&=-c+d\\
&=d-c
\end{align}$$</span></p>
<p>(For this second fact we first "distribute the negative sign", and then "minus a negative is plus".)</p>
<p>Are you familiar with those manipulations? The person who wrote the text you got this from just combined a few steps into one equation.</p>
<p>Generally, unless you're first being introduced to such algebra, it's not considered necessary to write down every step - it's like they're saying "well, we start over here, and end up over there", and expect that you can fill in the details. And after a few times where you write it out explicitly, you'll find you can "do it with just your eyes", as you are reading along.</p>
<hr />
<p>EDIT: (Since this is currently the accepted answer, I'll add another pattern that also sometimes shows up, and can be briefly puzzling, for the benefit of future readers.)</p>
<p>Similarly</p>
<p><span class="math-container">$$\frac{a}{b} = \frac{-a}{-b}$$</span></p>
<p>is pretty obvious written that way, but can be harder to recognize when it's used to rewrite</p>
<p><span class="math-container">$$\frac{1-x}{1-2x}$$</span>
as <span class="math-container">$$\frac{x-1}{2x-1}$$</span></p>
|
3,826,994 | <p>I would like to find <span class="math-container">$z$</span> which minimizes the below, when <span class="math-container">$x$</span> is held at a specific value.</p>
<p><span class="math-container">$f(x,z) =\sqrt{\sqrt{x^2 + z^2} - 0.25}$</span></p>
<p>For example; I would like to find the value of <span class="math-container">$z$</span> which minimizes the function when <span class="math-container">$x = 0.5$</span></p>
| Rezha Adrian Tanuharja | 751,970 | <p>Most of the time, polar form is better when dealing with complex numbers. Draw a circle with <span class="math-container">$2i$</span> and <span class="math-container">$6$</span> as the ends of its diameter. This diameter divide the circle into 2 parts, the lower half is the loci.</p>
<p>Hint: if <span class="math-container">$AB$</span> is the diameter of a circle and <span class="math-container">$P$</span> is on the circle, <span class="math-container">$\angle APB =\pm \frac{\pi}{2}$</span></p>
|
1,572,045 | <p>This is maybe a stupid question, but I want to find the roots of:</p>
<blockquote>
<p>$$2(x+2)(x-1)^3-3(x-1)^2(x+2)^2=0$$</p>
</blockquote>
<p>What that I did:</p>
<p>$$\underbrace{2(x+2)(x-1)(x-1)(x-1)}_{A}-\underbrace{3(x-1)(x-1)(x+2)(x+2)}_{B}=0$$</p>
<p>So the roots are when $A$ and $B$ are both zeros when $x=1$ and $x=-2$ </p>
<p>My questions:</p>
<p>$1)$ Is there an easy way to see that $x=-8$ is a root too?</p>
<p>$2)$ The degree of this polynomial is $4$, so I should have $4$ roots, and here I have only $3$</p>
| J.Gudal | 225,386 | <p>Well </p>
<p>$2(x+2)(x-1)^{3}-3(x-1)^{2}(x+2)^{2}=0 \Rightarrow (x-1)^{2}(x+2)[2(x-1)-3(x+2)]=0 \Rightarrow (x-1)^{2}(x+2)(-x-8)=0 $</p>
<p>From here it should be clear why $x=-8$ is also a root.</p>
<p>This is called <a href="https://www.mathsisfun.com/algebra/factoring.html" rel="nofollow">factoring</a>.</p>
|
1,307,085 | <p>How does one solve this equation?</p>
<blockquote>
<p>$$\cos {x}+\sin {x}-1=0$$</p>
</blockquote>
<p>I have no idea how to start it.</p>
<p>Can anyone give me some hints? Is there an identity for $\cos{x}+\sin{x}$?</p>
<p>Thanks in advance!</p>
| Eric R. Anschuetz | 162,230 | <p>As $\sin\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\left(\cos x+\sin x\right)$ by the angle addition formula we find that:
\begin{equation}
\begin{aligned}
\cos x+\sin x-1&=0\\
\implies\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)&=1\\
\implies\sin\left(x+\frac{\pi}{4}\right)&=\frac{\sqrt{2}}{2}\\
\implies x+\frac{\pi}{4}&=\frac{\pi}{4}+2\pi n,\frac{3\pi}{4}+2\pi n\\
\implies x&=2\pi n,\frac{\pi}{2}+2\pi n
\end{aligned}
\end{equation}
for $n\in\mathbb{Z}$.</p>
|
1,307,085 | <p>How does one solve this equation?</p>
<blockquote>
<p>$$\cos {x}+\sin {x}-1=0$$</p>
</blockquote>
<p>I have no idea how to start it.</p>
<p>Can anyone give me some hints? Is there an identity for $\cos{x}+\sin{x}$?</p>
<p>Thanks in advance!</p>
| Brian Tung | 224,454 | <p>I'll throw my hat in the ring to get a picture in edgewise. :-)</p>
<p><a href="https://i.stack.imgur.com/0dRHT.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0dRHT.png" alt="enter image description here"></a></p>
<p>I'll use $\theta$ for the angle, rather than $x$ (to avoid confusion with the $x$-coordinate). If we then let $x = \cos \theta, y = \sin \theta$, then the allowable results are along the red circle with $x^2+y^2 = 1$ (since $\cos^2 \theta + \sin^2 \theta = 1$). The solutions of the given equation are at the intersections of the blue line $x+y = 1$ with that red circle, yielding $(\cos \theta, \sin \theta) = (1, 0)$ and $(0, 1)$.</p>
<p>This in turn yields</p>
<p>$$
\theta = 2k\pi, \qquad \theta = 2k\pi + \frac{\pi}{2}
$$</p>
<p>for $k$ an integer.</p>
|
846,108 | <p>How do you solve this equation: $2x+8=6x-12$ by using the guess and check method?</p>
<p>I divide $2x+8$ and I get $4$ then I divide $6x-12$ and I get $-2$ but I don't know what to do next or is it wrong?</p>
| paw88789 | 147,810 | <p>One thing that you can get by 'guess-and-check' is how linear functions grow. </p>
<p>$$\begin{array}{r|r|r} x &2x+8 &6x-12\\ \hline 0&8&-12\\1&10&-6\\2&12&0 \\ \vdots&\vdots&\vdots \end{array}$$ Note that every time $x$ increases by $1$, $2x+8$ increases by $2$ and $6x-12$ increases by $6$. (do you see why this should be so?) That means whenever $x$ increases by $1$, $6x-12$ grows by $4$ units more than $2x+8$. Currently (at $x=2$), $2x+8$ is ahead by $12$. So if we increase $x$ by $3$, $6x-12$ will gain $3\cdot 4$, or $12$ more than $2x-8$, and so then the two expressions will be equal.</p>
<p>Thus the answer should be $x=5$.</p>
|
177,144 | <p>If G is a finite group, I understand that the category of RO(G)-graded spectra, when rationalized, becomes Quillen equivalent to the category of Mackey functors valued in chain complexes of rational vector spaces.</p>
<p>How does RO(G) act on the category of Mackey functors? For instance, if F = F(G/H) is a Mackey functor and V is a real representation of G, what is (S^V F)(G/H)?</p>
| Justin Noel | 8,818 | <p>Although I agree with Peter's comments, I believe I can add a few helpful comments of my own. First for $G$ finite, every rational Mackey functor is both injective and projective, so chain complexes are weakly equivalent to their homology. One can find this statement in appendix A of Greenlees-May 'Generalized Tate Cohomology.'</p>
<p>Given a representation sphere $S^V$ and a graded Mackey functor $M$ with associated Eilenberg-MacLane spectrum $HM$ we have an associated $G$-spectrum $S^V\wedge HM$ which represents the integer graded Bredon cohomology theory $$ X\mapsto H^{*-V}_{(-)}(X;M ).$$ Because rational equivariant cohomology theories are ordinary there is an equivalence $$S^V\wedge HM\simeq H(\pi_*^{(-)}(S^V\wedge HM)).$$ Hence the representation sphere takes a graded Mackey functor to another graded Mackey functor, namely the Bredon homology of $S^V$ with coefficients in $M$. </p>
<p>A similar argument shows that $H^{*+V}_{(-)}(X;M)$ is represented by $F(S^V,HM)\simeq D(S^V)\wedge HM$ which is again ordinary. Here $DX=F(X,S)$ is the equivariant Spanier-Whitehead dual of $X$. </p>
<p>If you like, after fixing $M$ you can extend this into a functor from the subcategory of the homotopy category spanned by the representation spheres, their Spanier-Whitehead duals, and their smash products to the category of graded Mackey functors. One can try to take a skeleton of this category to obtain an '$RO(G)$-graded' functor, but this involves keeping track of the automorphisms and solving a coherence problem. My understanding is that this coherence problem can be solved, but that it involves choices (which never seem to be specified).</p>
<p>As Peter pointed out the representation spheres are only distinguished by the fact that they are invertible, well understood, and play a pivotal role in the duality theory for $G$-manifolds. The canonical choice from a homotopical perspective would be to think of the Picard group of invertible objects instead.</p>
|
3,583,484 | <p>I am trying to work through a hw problem to show that the ring of gaussian integers <span class="math-container">$G=\{a+bi:a,b\in\mathbb{Z}\}$</span> is principal. To make it concrete I picked the ideal <span class="math-container">$I$</span> generated by <span class="math-container">$a=5$</span> and <span class="math-container">$b=3+4i$</span>. I believe <span class="math-container">$I=\{xa+yb:x,y\in G\}$</span> since this is closed under multiplication by gaussian integers and is an additive group and is contained in any ideal containing <span class="math-container">$a$</span> and <span class="math-container">$b$</span>. </p>
<p>An element in <span class="math-container">$I$</span> has squared modulus <span class="math-container">$(ax+by)(\overline{ax}+\overline{by})=25|x|^2+25|y|^2+5ax\overline{y}+5y\overline{ax}.$</span> So every element in <span class="math-container">$I$</span> has a squared modulus divisible by <span class="math-container">$5$</span>, so <span class="math-container">$I$</span> does not contain the units <span class="math-container">$\pm, 1\pm i$</span>.</p>
<p>So <span class="math-container">$I$</span> is a proper ideal generated by some gaussian integer <span class="math-container">$u+iv, u,v\in\mathbb{Z}$</span> with squared modulus divisible by 5. From <span class="math-container">$b=5=(x+iy)(u+iv), x,y\in\mathbb{Z}$</span>, setting real and imaginary parts equal, I get either <span class="math-container">$v=0$</span> or <span class="math-container">$-y(v^2+u^2)/v=5$</span>. This can only hold for <span class="math-container">$0\le v\le 5$</span> and working through the cases the only possibilities for <span class="math-container">$(u,v)$</span> one of <span class="math-container">$(1,0),(5,0),(0,0),(0,1),(1,2),(4,2),(0,5)$</span>. The only ones with squared modulus divisible by 5 are <span class="math-container">$(5,0),(1,2),(0,5)$</span>. But for none of these does <span class="math-container">$a=3+4i=(x'+iy')(u+iv)$</span> have a solution with <span class="math-container">$x,y\in \mathbb{Z}$</span>.</p>
| egreg | 62,967 | <p>You can note that
<span class="math-container">$$
(2+i)^2=4+4i-1=3+4i
$$</span>
and that <span class="math-container">$5=(2+i)(2-i)$</span>. So the greatest common divisor is <span class="math-container">$2+i$</span>, which is the generator of the ideal.</p>
|
3,583,484 | <p>I am trying to work through a hw problem to show that the ring of gaussian integers <span class="math-container">$G=\{a+bi:a,b\in\mathbb{Z}\}$</span> is principal. To make it concrete I picked the ideal <span class="math-container">$I$</span> generated by <span class="math-container">$a=5$</span> and <span class="math-container">$b=3+4i$</span>. I believe <span class="math-container">$I=\{xa+yb:x,y\in G\}$</span> since this is closed under multiplication by gaussian integers and is an additive group and is contained in any ideal containing <span class="math-container">$a$</span> and <span class="math-container">$b$</span>. </p>
<p>An element in <span class="math-container">$I$</span> has squared modulus <span class="math-container">$(ax+by)(\overline{ax}+\overline{by})=25|x|^2+25|y|^2+5ax\overline{y}+5y\overline{ax}.$</span> So every element in <span class="math-container">$I$</span> has a squared modulus divisible by <span class="math-container">$5$</span>, so <span class="math-container">$I$</span> does not contain the units <span class="math-container">$\pm, 1\pm i$</span>.</p>
<p>So <span class="math-container">$I$</span> is a proper ideal generated by some gaussian integer <span class="math-container">$u+iv, u,v\in\mathbb{Z}$</span> with squared modulus divisible by 5. From <span class="math-container">$b=5=(x+iy)(u+iv), x,y\in\mathbb{Z}$</span>, setting real and imaginary parts equal, I get either <span class="math-container">$v=0$</span> or <span class="math-container">$-y(v^2+u^2)/v=5$</span>. This can only hold for <span class="math-container">$0\le v\le 5$</span> and working through the cases the only possibilities for <span class="math-container">$(u,v)$</span> one of <span class="math-container">$(1,0),(5,0),(0,0),(0,1),(1,2),(4,2),(0,5)$</span>. The only ones with squared modulus divisible by 5 are <span class="math-container">$(5,0),(1,2),(0,5)$</span>. But for none of these does <span class="math-container">$a=3+4i=(x'+iy')(u+iv)$</span> have a solution with <span class="math-container">$x,y\in \mathbb{Z}$</span>.</p>
| Daniel Schepler | 337,888 | <p>Let us work through applying the Euclidean algorithm to find the gcd in <span class="math-container">$\mathbb{Z}[i]$</span> of <span class="math-container">$3+4i$</span> and <span class="math-container">$5$</span>. Now, <span class="math-container">$|3+4i|^2 = |5|^2 = 25$</span>, so neither element is really "smaller" than the other one. So, let us take the simpler quotient, and divide <span class="math-container">$3+4i$</span> by <span class="math-container">$5$</span>. In fact, <span class="math-container">$\frac{3+4i}{5} = \frac{3}{5} + \frac{4}{5} i$</span>, and the nearest Gaussian integer is <span class="math-container">$1 + i$</span>. So, the remainder is <span class="math-container">$(3+4i) - (1 + i) \cdot 5 = -2 - i$</span>; and <span class="math-container">$\gcd(3+4i, 5) = \gcd(5, -2 - i)$</span>.</p>
<p>Similarly, in this step, <span class="math-container">$\frac{5}{-2 - i} = -2 + i$</span>; thus, the remainder at this step has reached 0, and we conclude <span class="math-container">$\gcd(3+4i, 5) = -2 - i$</span>. Therefore, <span class="math-container">$-2 - i$</span> is a generator of the ideal.</p>
<p>(In fact, since the gcd is only defined up to multiplication by a unit of <span class="math-container">$\mathbb{Z}[i]$</span>, we could just as well have said that the gcd is <span class="math-container">$2 + i$</span>, or <span class="math-container">$1 - 2i$</span>, or <span class="math-container">$-1 + 2i$</span>. You could also apply some sort of normalization on the remainders at each step of the Euclidean algorithm if you wanted - for example, I tend to like to normalize to the region of the complex plane with argument in <span class="math-container">$(-\frac{\pi}{4}, \frac{\pi}{4}]$</span>.)</p>
|
2,592,007 | <blockquote>
<p>Let $V$ be a vector space and $W,U\subseteq V$ subspaces s.t $W\not \subseteq U$
$\dim(V)=5, \dim(W)=2, \dim(U)=4$ </p>
<p>Prove\Disprove: $\dim(U\cap W)=1$</p>
</blockquote>
<p>So I started with \begin{align} & \dim(W+U)=\dim(U)+\dim(W)-\dim(U\cap W) \\[10pt]
\iff & \dim(W+U)=4+2-\dim(U\cap W)\end{align}</p>
<p>Now in the following steps I am not sure</p>
<p>Because $W,U\subseteq V$ subspaces $\dim(W+U)\leq \dim(V)=5$ So $1\leq \dim(U\cap W)$</p>
<p>What can I conclude from $W\not \subseteq U$ ? How should I continue?</p>
| Dan | 500,478 | <p>From $\dim(W+U)\leq \dim(V)=5$ and using that $1\leq \dim(U\cap W)\leq\dim(W)=2$ we get two cases. If $\dim(U\cap W)=2=\dim(W)$, then $W\cap U=W$, as $W\cap U \subset W$, so $W\subset U$, which contradicts our initial assumption. So $\dim(U\cap W)=1$.</p>
|
1,829,030 | <p>The limit isn't too bad using l'hospital's rule, but I was wondering if there was a way to do it without l'hospital's. </p>
<p>Looking around the section limits without lhopital's, it seems usually evaluating without requires some clever factoring, while here the $\arctan$ seems to muck things up. </p>
<p>Here is the evaluation using l'hospital's in case someone visits with this question:
$$L=\lim_{x\rightarrow 0}(1+\arctan(\frac{x}{2}))^{\frac{2}{x}}\Rightarrow \log(L)=\lim_{x\rightarrow 0}\frac{2}{x}\log(1+\arctan(\frac{x}{2}))\\
\Rightarrow \log(L)=2\lim_{x\rightarrow 0}\frac{\log(1+\arctan(\frac{x}{2}))}{x}$$
and by l'hospital's
$$\log(L)=\lim_{x\rightarrow 0}\frac{1}{1+\arctan(\frac{x}{2})}*\frac{1}{1+(\frac{x}{2})^2}=1\\
\Rightarrow L=e$$</p>
| zhw. | 228,045 | <p>Let $f(x) = 1+\arctan (x/2).$ Apply $\ln $ to the expression of interest to get</p>
<p>$$\frac{\ln (f(x))}{x/2} = 2\frac{\ln f(x) - \ln f(0)}{x-0}.$$</p>
<p>As $x\to 0,$ the last expression $\to 2(\ln f)'(0)$ by definition of the derivative (no L'Hopital used). That's easy enough to compute. Exponentiate back for the original limit.</p>
|
872,889 | <p>What determinant is zero? What equation does this give for the plane?</p>
<p>I need some help here, am pretty stuck</p>
| Aritmo | 352,282 | <p>You can find plenty on information on those special continued fractions at
the following link (American Mathematical Monthly):</p>
<p><a href="https://www.facebook.com/AmerMathMonthly/photos/a.250425975006394.53155.241224542593204/1055084257873891/?type=3&theater" rel="nofollow noreferrer">https://www.facebook.com/AmerMathMonthly/photos/a.250425975006394.53155.241224542593204/1055084257873891/?type=3&theater</a></p>
<p>Also at:
<a href="https://youtu.be/mjSHenvXsEs" rel="nofollow noreferrer">https://youtu.be/mjSHenvXsEs</a></p>
|
2,024,997 | <blockquote>
<p>$$\lim_{x \rightarrow +\infty}\frac{\log_{1.1}x}{x}$$</p>
</blockquote>
<p>I can solve this easily by generating the graph with my calculator, but is there is a way to do this analytically?</p>
| hamam_Abdallah | 369,188 | <p>Let $g(x)=\ln(x)-\sqrt{x}$</p>
<p>we have $g(1)=0$ and</p>
<p>$$g'(x)=\frac{1}{2x}(2-\sqrt{x})$$</p>
<p>$g$ is decreasing at $[4,+\infty)$</p>
<p>thus</p>
<p>$$\forall x\geq 4 \;\;0<\frac{\log_{1.1}(x)}{x}\leq \frac{1}{\ln(1.1)\sqrt{x}}$$</p>
<p>and squeeze theorem.</p>
|
2,024,997 | <blockquote>
<p>$$\lim_{x \rightarrow +\infty}\frac{\log_{1.1}x}{x}$$</p>
</blockquote>
<p>I can solve this easily by generating the graph with my calculator, but is there is a way to do this analytically?</p>
| Mark Viola | 218,419 | <blockquote>
<p>I thought it might be instructive to present an approach that relies on elementary tools only. To that end, we begin with the following primer.</p>
<p><strong>PRIMER: BOUNDS FOR THE LOGARITHM FUNCTION</strong></p>
<p>In <a href="https://math.stackexchange.com/questions/1589429/how-to-prove-that-logxx-when-x1/1590263#1590263">THIS ANSWER</a>, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities</p>
<p><span class="math-container">$$\bbox[5px,border:2px solid #C0A000]{\frac{x}{x-1}\le \log(x)\le x-1} \tag 1$$</span></p>
<p>for <span class="math-container">$x<1$</span>.</p>
</blockquote>
<hr />
<p>Now, we note that <span class="math-container">$\log_{1.1}(x)=\frac{\log(x)}{\log(1.1)}$</span>. In addition, we note that for any number <span class="math-container">$a$</span>, we have <span class="math-container">$\log(x^a)=a\log(x)$</span>.</p>
<p>Using <span class="math-container">$(1)$</span>, we have for <span class="math-container">$a>0$</span>,</p>
<p><span class="math-container">$$\frac{x^{-1}-x^{-(a+1)}}{a}\le \frac{\log(x)}{x}\le \frac{x^{a-1}-x^{-1}}{a} \tag 2$$</span></p>
<p>Surely, <span class="math-container">$(2)$</span> is valid for <span class="math-container">$0<a<1$</span>. Then, applying the squeeze theorem to <span class="math-container">$(2)$</span> with <span class="math-container">$0<a<1$</span> (e.g., <span class="math-container">$a=1/2$</span>) yields the coveted limit</p>
<blockquote>
<p><span class="math-container">$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to \infty}\frac{\log_{1.1}(x)}{x}=0}$$</span></p>
</blockquote>
<p>and we are done!</p>
|
2,849,643 | <p>Consider the following recurrence problem:
\begin{align}
d_{i-1} &= 2\varphi_{i+1}+4\varphi_i + 8d_i-7d_{i+1} - F \left( \delta_{i,N} + \delta_{i,N+1} \right) \, , \\
\varphi_{i-1} &= -7\varphi_{i+1}-16\varphi_{i} + 24 \left( d_{i+1}-d_{i} \right) + F \left( \delta_{i,N} + \delta_{i,N+1} \right) \, ,
\end{align}
where $d_i, i \in \{2, \cdots , N+1\}$ represent displacements, $\varphi_i$ inclinations, and $F$ is a known force acting at the nodes $N$ and $N+1$.</p>
<p>We require by the system symmetry that $d_{N+1}=d_N = d_\mathrm{C}$ and $\varphi_N = -\varphi_{N+1}= \varphi_\mathrm{C}$, where $d_\mathrm{C}$ and $\varphi_\mathrm{C}$ are still to be determined from the boundary conditions:</p>
<p>$d_1 = 0$ (zero displacement) and $2\varphi_1+\varphi_2 = 3d_2$ (zero torque)</p>
<p>In order to proceed, i have tried to first determine $d_{N-1}$ and $\varphi_{N-1}$ from the system above, and then $d_{N-2}$ and $\varphi_{N-2}$, etc... in a recursive way and then try to find out the general term of the resulting sequences. </p>
<p>For the term $N-1$, we obtain
\begin{align}
d_{N-1} &= d_\mathrm{C}+2\varphi_\mathrm{C}-F \, \\
\varphi_{N-1} &= -9\varphi_\mathrm{C}+3F \, .
\end{align}</p>
<p>Analogously, we get for the term $N-2$
\begin{align}
d_{N-2} &= d_\mathrm{C}-18\varphi_\mathrm{C}+4F \, \\
\varphi_{N-2} &= 89\varphi_\mathrm{C}-24F \, .
\end{align}</p>
<p>I was wondering whether there is a particular way to figure out the general term of such sequence.</p>
<p>Any help or suggestions are most welcome.</p>
| Yuri Negometyanov | 297,350 | <p>$$\mathbf{\color{green}{The\ linear\ approach}}$$</p>
<p>As it follows from the comments, the issue task can be detalized in the form of
\begin{cases}
d_{i-1} = 2\varphi_{i+1}+4\varphi_i + 8d_i-7d_{i+1}\\[4pt]
\varphi_{i-1} = -7\varphi_{i+1}-16\varphi_{i} + 24 \left( d_{i+1}-d_{i} \right) \\[4pt]
i=2,3\dots N-1\\[4pt]
d_1 = 0\\[4pt]
2\varphi_1+\varphi_2 = 3d_2\\[4pt]
d_{N-1} = 2\varphi_{N+1}+4\varphi_N + 8d_N-7d_{N+1} - F\\[4pt]
\varphi_{N-1} = -7\varphi_{N+1}-16\varphi_{N} + 24 \left( d_{N+1}-d_{N}\right) + F\\[4pt]
d_{N} = 2\varphi_{N-1}+4\varphi_{N+1} + 8d_{N+1}-7d_{N-1} - F\\[4pt]
\varphi_{N} = -7\varphi_{N-1}-16\varphi_{N+1} + 24 \left( d_{N-1}-d_{N+1}\right) + F\\[4pt]
d_{N+1}=d_N\\[4pt]
\varphi_{N+1} = -\varphi_{N},\tag{I}
\end{cases}
or
\begin{cases}
d_{i-1} - 8d_i - 4\varphi_i + 7d_{i+1} - 2\varphi_{i+1} = 0\\[4pt]
\varphi_{i-1} + 24d_i + 16\varphi_{i} - 24d_{i+1} + 7\varphi_{i+1} = 0 \\[4pt]
i=2,3\dots N-1\\[4pt]
d_1 = 0\\[4pt]
2\varphi_1 - 3d_2 + \varphi_2 = 0\\[4pt]
d_{N-1} - d_N - 2\varphi_N = - F\\[4pt]
\varphi_{N-1} + 9\varphi_{N} = F.\tag{II}
\end{cases}
This gives the linear system
$$A\overrightarrow v=\overrightarrow f,\tag{III}$$
where
$$A=
\begin{pmatrix}
1&0&0&0&0&0&0&0&0&0&0&0&0&0&0&0\\
0&2&-3&1&0&0&0&0&0&0&0&0&0&0&0&0\\
1&0&-8&-4&7&-2&0&0&0&0&0&0&0&0&0&0\\
0&1&24&16&-24&7&0&0&0&0&0&0&0&0&0&0\\
0&0&1&0&-8&-4&7&-2&0&0&0&0&0&0&0&0\\
0&0&0&1&24&16&-24&7&0&0&0&0&0&0&0&0\\
0&0&0&0&1&0&-8&-4&7&-2&0&0&0&0&0&0\\
0&0&0&0&0&1&24&16&-24&7&0&0&0&0&0&0\\
&&&&&&&&&\dots&&&&&&\\
0&0&0&0&0&0&0&0&0&0&1&0&-8&-4&7&-2\\
0&0&0&0&0&0&0&0&0&0&0&1&24&16&-24&7\\
0&0&0&0&0&0&0&0&0&0&0&0&1&0&-1&-2\\
0&0&0&0&0&0&0&0&0&0&0&0&0&1&0&9\\
\end{pmatrix},$$
$$
\overrightarrow v=\begin{pmatrix}
d_1\\\varphi_1\\d_2\\\varphi_2\\\vdots\\d_{N-1}\\\varphi_{N-1}\\d_N\\\varphi_N
\end{pmatrix},
\overrightarrow f=\begin{pmatrix}
0\\0\\0\\0\\\vdots\\0\\0\\-F\\F
\end{pmatrix}.
$$
Methods for solving $n$-diagonal SLAE $(n=6)$ are well known.</p>
<p>$$\mathbf{\color{green}{Common\ solution}}$$</p>
<p>Let
\begin{cases}
d_{i-1} - 8d_i + 7d_{i+1} = 4\varphi_i + 2\varphi_{i+1}\\[4pt]
\varphi_{i-1} + 16\varphi_{i} + 7\varphi_{i+1} = -24d_i + 24d_{i+1}\\[4pt]
i=2,3\dots N-1\\[4pt]
d_1 = 0\\[4pt]
2\varphi_1 - 3d_2 + \varphi_2 = 0\\[4pt]
d_{N-1} - d_N - 2\varphi_N = - F\\[4pt]
\varphi_{N-1} + 9\varphi_{N} = F,\tag1
\end{cases}
Looking for the solution in the form of
$$d_i = Ap^{-i},\quad\varphi_i=Bq^{-i}.\tag2$$
we get the system
\begin{cases}
Ap^{-i-1}(p^2-8p+7) = Bq^{-i-1}(4q+2) \\
24Ap^{-i-1}(1-p) = Bq^{-i-1}(q^2+16q+7),\tag3
\end{cases}</p>
<p>which leads to the relation
$$\dfrac{d_i}{\varphi_i} = \frac pq\frac{4q+2}{p^2-8p+7} = \frac pq\frac{q^2+16q+7}{24(1-p)},\tag4$$
where
$$(p^2-8p+7)(q^2+16q+7) + 48(p-1)(2q+1)=0,$$
or
$$(p-1)\left((p-7)q^2+16(p-1)q + 7p-1\right) = 0,$$
with the solutions
$$\left[\begin{align}
&p=1\\
&p=7,\quad q=-\frac12\\[4pt]
&p=q=-5\pm2\sqrt6,
\end{align}\right.\tag5$$
wherein third and fourth solutions obtained, using additional condition $p=q$, which allows to hold the ratio $(4).$</p>
<p>Formulas $(5)$ and $(4)$ lead to the common solution in the form of
$$\begin{align}
&\binom{d_i}{\varphi_i} = C_1\binom{1}{0}+C_27^{-i}\binom{7}{-96}\\[4pt]
&+C_3(-5-2\sqrt6)^{i}\binom1{2\sqrt6}+C_4(-5+2\sqrt6)^{i}\binom1{-2\sqrt6},
\end{align}\tag6$$
(using the identity $(-5+2\sqrt6)(-5-2\sqrt6)=1$).</p>
<p>Coefficients $C_i$ can be defined from the boundary conditions.</p>
<p>$$\mathbf{\color{green}{Modified\ solution.}}$$</p>
<p>The previous model has a resonant solution $(p=1).$ To avoid this situation, should be used another basis.</p>
<p>Let
$$D_i=d_{i}-d_{i-1},\quad d_1=0,\tag7$$
then
\begin{cases}
-D_{i} + 7D_{i+1} = 4\varphi_i + 2\varphi_{i+1}\\[4pt]
\varphi_{i-1} + 16\varphi_{i} + 7\varphi_{i+1} = 24D_{i}\\[4pt]
i=2,3\dots N-1\\[4pt]
2\varphi_1 - 3D_2 + \varphi_2 = 0\\[4pt]
D_{N} + 2\varphi_N = + F\\[4pt]
\varphi_{N-1} + 9\varphi_{N} = F,\tag8
\end{cases}
Looking for the solution in the form of
$$D_i = Ap^{-i},\quad\varphi_i=Bp^{-i}.\tag9$$
we get the system
\begin{cases}
A(7-p) = B(4p+2) \\
24A = B(p^2+16p+7),\tag{10}
\end{cases}</p>
<p>which leads to the relation
$$\dfrac{D_i}{\varphi_i} = \frac{4p+2}{7-p},\tag{11}$$
where
$$(p-7)(p^2+16p+7)+96p+48=0,$$
$$(p - 1) (p^2 + 10 p + 1) = 0,$$
with the solutions
$$\left[\begin{align}
&p_1=1,\quad D_i=\varphi_i\\
&p_2=-5+2\sqrt6,\quad D_i=\dfrac{\sqrt6-2}2\varphi_i\\
&p_3=-5-2\sqrt6,\quad D_i=-\dfrac{\sqrt6+2}2\varphi_i.
\end{align}\right.\tag{12}$$</p>
<p>Formulas $(9)$ and $(12)$ lead to the common solution in the form of
$$\begin{align}
&\binom{D_i}{\varphi_i} = C_1\binom{1}{1}+C_2(-5-2\sqrt6)^{i}\binom{\sqrt6-2}2+C_3(-5+2\sqrt6)^{i}\binom{\sqrt6+2}{-2}
\end{align}\tag{13}$$
(using the identity $(-5+2\sqrt6)(-5-2\sqrt6)=1$).</p>
<p>Coefficients $C_i$ can be defined from the boundary conditions and then the values of $d_i$ can be calculafed using $(7).$</p>
<p>$$\mathbf{\color{green}{Analogies}}$$</p>
<p>Are known the formulas
$$1-\binom{n}{1}2^t+\binom{n}{2}n3^t+\dots+(-1)^n\binom{n}{n}n^t = 0,\quad t<n\tag{A1}$$
and
$$\Delta^nf(x) = \sum\limits_{m=0}^n(-1)^{n-m}f(x+m)=\sum\limits_{k=0}^{r}(-1)^k \binom{r}{k}\Delta r^{n+k}f(x).\tag{A2}$$
This leads to the some analogies between the linear recurrence relations and the linear DEs.</p>
<p>$\mathbf{Analogy 1.}$</p>
<p>The ODE
$$y'''+ay''+by'=R(x)\tag{A3}$$
does not contain the term with $cy$ and can be simplified using the substitution $Y=y'.$</p>
<p>This analogy is the hint to use the substitution $D=d_i-d_{i-1}.$</p>
<p>$\mathbf{Analogy 2.}$</p>
<p>Solution of the OOE $(A3)$ is the sum of the common solution of the homogenius ODE and the partial solution of the origin ODE.</p>
<p>The common solution can be defined using the substitution $y=e^{kx}.$</p>
<p>This analogy is the hint to use the approach with $D=q^n.$</p>
<p>Btw, the negative degree were used situatively and is not obligate.</p>
<p>$\mathbf{Analogy 3.}$</p>
<p>The partial solution of ODE $(3)$ with the polynomial RHS usually can be found as the same order polynomial with the unknown coefficients.</p>
<p>On the other hand, the ODE
$$y'''+ay''+by'= F\tag{A4}$$
already has the common solution $y=C$ (the resonant case), so the form $y_p=Dx$ is usually applied.</p>
<p>Hint is clear, but the analogy isn't. Maybe, the formulas $(1)$ are the second hint?</p>
<p>This is the main.</p>
|
635,077 | <p>$$\sin(a+b) = \sin(a) \cos(b) + \cos(a) \sin(b)$$</p>
<p>How can I prove this statement?</p>
| Edward ffitch | 26,243 | <p>$e^{i(a+b)}=\cos(a+b)+i\sin(a+b)$ by Euler's formula. But $e^{i(a+b)}=e^{ia}e^{ib}=(\cos(a)+i\sin(a))(\cos(b)+i\sin(b))= \cos(a)\cos(b)-\sin(a)\sin(b)+i(\sin(a)\cos(b) + \cos(a)\sin(b))$</p>
<p>So by comparing real and imaginary parts you obtain the trigonometric addition formulae for both $\sin$ and $\cos$.</p>
|
635,077 | <p>$$\sin(a+b) = \sin(a) \cos(b) + \cos(a) \sin(b)$$</p>
<p>How can I prove this statement?</p>
| dwarandae | 55,403 | <p>This is the way I learned it: a geometric proof like <a href="http://en.wikipedia.org/wiki/File%3aTrigSumFormula.svg" rel="nofollow">this</a> on Wikipedia.</p>
<p>The segment $OP$ has length $1$. We have then, $\sin(\alpha + \beta) = PB = PR + RB = \cos(\alpha) \sin(\beta) + \sin(\alpha) \cos(\beta)$.</p>
|
635,077 | <p>$$\sin(a+b) = \sin(a) \cos(b) + \cos(a) \sin(b)$$</p>
<p>How can I prove this statement?</p>
| Michael Hoppe | 93,935 | <p>This gem comes from E. Schmidt: Consider $f(x)=\sin(\alpha+\beta-x)\cos(x)+\cos(\alpha+\beta-x)\sin(x)$. Since $f'(x)=0$ we know that $f$ is constant, hence $f(0)=f(\beta)$.</p>
|
1,800,233 | <blockquote>
<p>If $n$ is composite and $\phi{(n)} | (n - 1)$ then prove that $n$ has at least four distinct prime factors.</p>
</blockquote>
<p><strong>Attempt:</strong></p>
<p>Since $n$ is not a prime, let's first take the case that $n$ is squarefree. Then $n = a_1 \cdot a_2 \cdots a_r$ where $a_i$ are the prime factors of $n$ listed in ascending order. Thus, $\dfrac{n-1}{\phi(n)} = \dfrac{a_1a_2\cdots a_r-1}{(a_1-1)(a_2-1)\cdots (a_n-1)}$. The denominator has a factor of $2^n$.</p>
<p>I am not sure how to continue from here.</p>
| user5713492 | 316,404 | <p>To eliminate $3$ prime factors is pretty simple because since $\phi(n)<n-1$ it implies that $\phi(n)\le\frac n2$. If $3$ isn't one of the factors the smallest $\frac{\phi(n)}n$ could be is $\frac45\frac67\frac{10}{11}=0.6234$. If $5$ isn't a factor, the smallest is $\frac23\frac67\frac{10}{11}=0.5165$. If both $3$ and $5$ are factors, $\frac23\frac45\frac{16}{17}=0.5020$ is too big, so we only have to test $\phi(105)=48$, $\phi(165)=80$, and $\phi(195)=96$ to eliminate all remaining possibilities.</p>
|
3,310,038 | <p>If <span class="math-container">$U$</span> is an open set of <span class="math-container">$\mathbb{R}^{m}$</span>, do we have that <span class="math-container">$U\times \mathbb{R}^{n-m}$</span> is an open set of <span class="math-container">$\mathbb{R}^{n}$</span>? </p>
<p>Here <span class="math-container">$\mathbb{R}^{n},\mathbb{R}^{m}$</span> both equip with the standard topology.</p>
| ayeayemaung | 249,040 | <p>Yes. Note <span class="math-container">$U \times \Bbb R^{n - m} = \pi^{-1}(U)$</span> where <span class="math-container">$\pi$</span> is the projection map
<span class="math-container">$$
\pi : \Bbb R^n \ni (x_1, \ldots, x_m, \ldots, x_n) \mapsto (x_1, \ldots, x_m) \in \Bbb R^m
$$</span> and as projection maps are continuous the inverse image of an open set is an open set.</p>
|
671,407 | <p>I have problem with equation: $4^x-3^x=1$. </p>
<p>So at once we can notice that $x=1$ is a solution to our equation. But is it the only solution to this problem? How to show that there aren't any other solutions? </p>
| IAmNoOne | 117,818 | <p>Prove it by contradiction. </p>
<p>Let $f(x) = 4^x - 3^x - 1$. This function is smooth and as you have shown, $x = 1$ is a root. By Rolle's theorem, suppose it had two other roots $f(a) = f(b) = 0$, then there exists $c$ in between $a$ and $b$ (where $a < 1 < b$) such that $f'(c) = 0$, but notice for $ x \geq 0$</p>
<p>$$f'(x) = 4^x \ln 4 - 3^x \ln 3 \geq 3^x (\ln 4 - \ln 3) > 0.$$</p>
<p>and for $x < 0$, we have </p>
<p>$$f'(x) = 4^x \ln 4 - 3^x \ln 3 \leq \ln 4 (4^x - 3^x) < 0.$$</p>
<p>Hence you can conclude there is only one root.</p>
|
671,407 | <p>I have problem with equation: $4^x-3^x=1$. </p>
<p>So at once we can notice that $x=1$ is a solution to our equation. But is it the only solution to this problem? How to show that there aren't any other solutions? </p>
| N. S. | 9,176 | <p>It is easy to see that there is no negative solution.</p>
<p>Assume by contradiction there exists a solution $x=a$ with $0 < a \neq 1$. </p>
<p>Let $g(y)=y^a$. By the mean value Theorem, there exists some $c \in (3,4)$ such that</p>
<p>$$\frac{g(4)-g(3)}{4-3} =g'(c)$$</p>
<p>Therefore
$$1= g'(c)=ac^{a-1}$$</p>
<p>But this is impossible: if $a>1$ then $a >1$ and $c^{a-1}>1$, while if $a<1$ then $0< a<1$ and $0< c^{a-1} <1$.</p>
|
708,604 | <blockquote>
<p>Let $f: \mathbb{R^4} \to \mathbb{R}$ be a linear transformation defined by $f(a,b,c,d)=a+b+c+d$. Find a basis for the $Im(f)$.</p>
</blockquote>
<p>So, $Im(f)=\{f(a,b,c,d) \in \mathbb{R}: (a,b,c,d) \in \mathbb{R^4} \}$.</p>
<p>Then $Im(f)=\{a+b+c+d \in \mathbb{R}: a,b,c,d \in \mathbb{R} \}=\mathbb{R}$.</p>
<p>I know that $\mathbb{R}$ is the set of the real numbers on the real line. That real line can be any axis of the plane, space or other higher dimension.</p>
<p>So, its correct to say that a basis for $\mathbb{R}$ can be any single vector of a canonical basis of any $n$ dimensional space?</p>
<p>For instance, can $\{(1,0,0)\}$ be a basis for $\mathbb{R}$? Or $\{(0,1,0)\}$? Thanks</p>
| mle | 66,744 | <p>if $\operatorname{im}(f)=\Bbb{R}^1$ then $\dim_\Bbb{R}(\operatorname{im}(f))=\dim_\Bbb{R}(\Bbb{R}^1)=1$, therefore a basis for the $\operatorname{im}(f)$ is a basis for $\Bbb{R}^1$</p>
<p><strong>Hint</strong>: let be $\{b_1,b_2,b_3,b_4\}$ a basis for $\Bbb{R}^4$, for example "<a href="http://de.wikipedia.org/wiki/Standardbasis#Standardbasis_in_den_Standardr.C3.A4umen" rel="nofollow">standard basis</a>", then $\operatorname{im}(f)=f(\Bbb{R}^4)=\operatorname{Span}(f(b_1),f(b_2),f(b_3),f(b_4))$</p>
|
279,520 | <p>So I have to find the integral of $$ \int \frac{\sin^{-1}(x)}{\sqrt{1+x}} \; dx$$</p>
<p>I think I have to do this using the integration by parts..so I will take $f = \sin^{-1}(x)$ and $ \sqrt {1+x}=g' $...what about now? </p>
| gt6989b | 16,192 | <p>Integrate by parts, letting $dv = (x+1)^{-1/2}$ and $u = \arcsin(x)$, then $du = \frac{1}{\sqrt{1-x^2}}$ and $v = \int (x+1)^{-1/2} dx = 2 \sqrt{x+1}$.</p>
<p>Use $\int u dv = uv - \int v du$ to reduce this to</p>
<p>$$
\int v du = \int 2 \sqrt{ \frac{x+1}{1-x^2} } dx
= 2 \int \frac{dx}{\sqrt{1-x}}
= -4 \sqrt{1-x}.
$$</p>
|
1,689,523 | <p>I need help with this Laplace question.
<span class="math-container">$$f(t) = e^{-t} \sin(t) $$</span></p>
<hr />
<p>Answer should be <span class="math-container">$\dfrac{1}{s^2 + 2s + 2}$</span></p>
<hr />
<p>What I'm currently doing is as follows:</p>
<p><span class="math-container">$u = \sin(t)\qquad$</span> <span class="math-container">$dv = e^{-(s+1)t}dt$</span></p>
<p><span class="math-container">$du = \cos(t)dt\qquad$</span> <span class="math-container">$v = \dfrac{e^{-(s+1)t}}{-(s+1)}$</span></p>
<p><span class="math-container">$\dfrac{-\sin(t) e^{-(s+1)t}}{-(s+1)} - \int\dfrac{ e^{-(s+1)t}\cos(t)}{ -(s+1)} dt$</span></p>
<p>But even if I solved the integral, I wouldn't get this (which is what I should, see picture).</p>
<blockquote>
<p><img src="https://i.stack.imgur.com/BhMOx.png" alt="enter image description here" /></p>
</blockquote>
| user873542 | 873,542 | <p>Here's the 'direct' proof:
<span class="math-container">$f(t)=e^{-t}\sin t.$</span>
<span class="math-container">\begin{eqnarray*}\mathcal{L}\{f(t)\}&=&\int_{0}^{+\infty}e^{-(s+1)t}\sin t\,dt~=~\lim_{\ell\to+\infty}\int_{0}^{\ell}e^{-(s+1)t}\sin t\,dt\\
&=&-\frac{1}{s+1}\lim_{\ell\to+\infty}\left[e^{-(s+1)t}\sin t\Big|_{0}^{\ell}-\int_{0}^{\ell}e^{-(s+1)t}\cos t\,dt\right]\\
&=&\frac{1}{s+1}\lim_{\ell\to+\infty}\int_{0}^{\ell}e^{-(s+1)t}\cos t\,dt,
\end{eqnarray*}</span><br />
since <span class="math-container">$$\lim_{\ell\to+\infty}e^{-(s+1)\ell}\sin\ell=0.$$</span> Integrating again by parts yields
<span class="math-container">\begin{eqnarray*}\mathcal{L}\{f(t)\}&=&-\frac{1}{(s+1)^2}\lim_{\ell\to+\infty}\left[e^{-(s+1)t}\cos t\Big|_{0}^{\ell}+\int_{0}^{\ell}e^{-(s+1)t}\sin t\,dt\right]\\
&=&\frac{1}{(s+1)^2}\left[1-\lim_{\ell\to+\infty}\int_{0}^{\ell}e^{-(s+1)t}\sin t\,dt\right]\\
&=&\frac{1}{(s+1)^2}(1-\mathcal{L}\{f(t)\})
\end{eqnarray*}</span><br />
and finally solving by <span class="math-container">$\mathcal{L}\{f(t)\}$</span>,
<span class="math-container">$$\boxed{\mathcal{L}\{f(t)\}=\frac{1}{(s+1)^2+1}=\frac{1}{s^2+2s+2}}~.$$</span></p>
|
510,080 | <p>This is not too obvious to me - what is the size of alternating group?</p>
<p>Following the hint in the comment, should it be $A_n = S_n/2$?</p>
<p>So I don't feel right up to here.....</p>
| Jared | 65,034 | <p>The map $\sigma:S_n\to\mathbb{Z}/2\mathbb{Z}$ defined by sending a permutation to $0$ if it has even parity, and $1$ if it has odd parity, is a group homomorphism. The kernel of this map is $A_n$, so by the first isomorphism theorem, we have $[S_n:A_n]=2$ for $n\ge 2$ (the map is not surjective for $n=1$). It follows that for $n\ge 2$ we have</p>
<p>$$|A_n|=\frac{|S_n|}{2}=\frac{n!}{2}$$</p>
|
106,126 | <blockquote>
<p><strong>Problem</strong> Prove that $n! > \sqrt{n^n}, n \geq 3$. </p>
</blockquote>
<p>I'm currently have two ideas in mind, one is to use induction on $n$, two is to find $\displaystyle\lim_{n\to\infty}\dfrac{n!}{\sqrt{n^n}}$. However, both methods don't seem to get close to the answer. I wonder is there another method to prove this problem that I'm not aware of? Any suggestion would be greatly appreciated.</p>
| Henry | 6,460 | <p>$(n!)^2 = (n \times 1) \times ((n-1)\times 2) \times \cdots \times (1 \times n) \gt n^n$</p>
<p>since $(n-1)\times 2 = 2n-2 \gt n$ iff $n \gt 2$. </p>
<p>Then take the square root. </p>
|
2,578,444 | <blockquote>
<p><span class="math-container">$\tan x> -\sqrt 3$</span></p>
</blockquote>
<p>How do I solve this inequality?</p>
<p>From the <a href="https://www.desmos.com/calculator/qb8bg1vbsf" rel="nofollow noreferrer">graph</a> it is evident that <span class="math-container">$\tan x>-\sqrt 3$</span> for <span class="math-container">$\left(\dfrac{2\pi}3 , \dfrac{5\pi} 3\right)$</span> <span class="math-container">$\forall x\in (0, 2\pi)$</span>.</p>
<p>Generalising this solution we get <span class="math-container">$\left(2n\pi +\dfrac{2\pi}3 , 2n\pi+\dfrac{5\pi} 3\right) \forall n \in \mathbb{Z}$</span> as the answer.</p>
<p>But the answer given is: <span class="math-container">$\left(n\pi - \dfrac \pi 3, n\pi + \dfrac \pi 2\right)$</span></p>
<p>Where have I gone wrong?</p>
| Martin Argerami | 22,857 | <p>You went wrong in reading the graph. The number $5\pi/3$ should have been $5\pi/2$. And the period is $\pi $ and not $2\pi $, which leads to the correct answer that you quoted: if you have $$\left( n\pi+\frac {5\pi}3,n\pi+\frac {5\pi}2\right), $$ Now,replacing $n $ with $n-2$ achieves the solution given.</p>
|
3,691,255 | <p>Pierre runs a game at a fair, where each player is guaranteed to win $10. </p>
<p>Players pay a certain amount each time they roll an unbiased die, and must keep rolling until a ‘6’ occurs. </p>
<p>When a ‘6’ occurs, Pierre gives the player $10 and the game concludes. </p>
<p>On average, Pierre wishes to make a profit of $2 per game. How much does he need to charge for each roll of the die?</p>
| Math Comorbidity | 731,254 | <p>The following question was answered by u/Alkalannar on reddit. </p>
<p>Answer: </p>
<p>Assumption not stated: This is a 6-sided die.</p>
<p>Consider the general n-sided die, and you want to roll max (or 1).</p>
<p>Expected income for the game is [Sum from k = 1 to infinity of xk(1 - 1/n)k-1(1/n)] = xn, where x is the price and n is the number of sides of the die.</p>
<p>So xn - 10 = 2, xn = 12, x = 12/n.</p>
<p>So when n = 6, x = 2.</p>
<p>And checking, [Sum from k = 1 to infinity of 2k(1 - 1/5)k-1(1/5)] = 12, which is what we want for expected income, so that expected profit is 2.</p>
|
4,278,505 | <p>I would like to clear up a confusion which might be trivial. In a proof the author proved <span class="math-container">$T = T'$</span> as following:</p>
<p>The author showed if <span class="math-container">$x \in T$</span> then <span class="math-container">$x \in T'$</span>, the next line is -</p>
<blockquote>
<p>... proving that <span class="math-container">$T\subseteq T'$</span>. As a result <span class="math-container">$T = T'$</span>.</p>
</blockquote>
<p>Reasoning:</p>
<p>If for any <span class="math-container">$x$</span>, it exists in both set <span class="math-container">$T$</span> and <span class="math-container">$T'$</span>, then we can directly write <span class="math-container">$T = T'$</span>, and if <span class="math-container">$T = T'$</span> by definition we can say <span class="math-container">$T \subset T'$</span>.</p>
<p>Full Context:
<a href="https://i.stack.imgur.com/eAuN2.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eAuN2.jpg" alt="enter image description here" /></a></p>
<p>Confusion:</p>
<p>Then why do author write first <span class="math-container">$T\subseteq T'$</span> then <span class="math-container">$T = T'$</span> (notice the sequence)?</p>
<p><span class="math-container">$T\subseteq T'$</span> means either <span class="math-container">$T\subset T'$</span> or <span class="math-container">$T= T'$</span>, it looks like at this point author does not know what should the case exactly but then based what he writes <span class="math-container">$T =T'$</span> in the next line?</p>
<p>If <span class="math-container">$T=T'$</span> is written for the reasoning given above then <span class="math-container">$T\subseteq T'$</span> is unnecessary, why do author write <span class="math-container">$T\subseteq T'$</span>?</p>
<p>What is the significance?</p>
<p>Another example is that often I found that we can say directly that <span class="math-container">$H_{i+1}= \rm Ker \; g$</span> but author(s) used <span class="math-container">$H_{i+1}\subseteq \rm Ker \; g$</span> (<a href="https://math.stackexchange.com/q/4256432/850314">see Question 1 in this post</a>), obviously <span class="math-container">$H_{i+1}= \rm Ker \; g$</span> does not contradict <span class="math-container">$H_{i+1}\subset \rm Ker \; g$</span> but why do author go for the weaker statement?</p>
| Anonymous M | 934,527 | <p>Based on the comments it seems the issue is that you need to return to the definition of <span class="math-container">$T'$</span> and make sure you understand it thoroughly. The author proved <span class="math-container">$T \subseteq T'$</span> and then immediately concluded <span class="math-container">$T = T'$</span> because it should so obvious to the reader that also <span class="math-container">$T'\subseteq T$</span> that it does not need to be written.</p>
<p>To be clear, you should not be discouraged if this is not obvious to you. Learning always takes time and I think all of us have at some point misunderstood some concept only to realize it later. Omitting an explicit mention of these types of "simple" statements in textbooks can often signal to the reader the level of understanding they probably need before progressing.</p>
|
787,926 | <p>I need some help to solve this integral:</p>
<p>$$\int_0^1 dy\int_0^{1-y} \cos \left(\frac{x-y}{x+y} \right) \mathrm dx$$</p>
<p>Thank you.</p>
| DaveBlackston | 149,057 | <p>Unless I am missing something, Julian's algorithm will simply check all numbers from 1, 2, 3, ... until it finds the 1000th 5-smooth number. This means it requires $O(u_n)$ time.</p>
<p>A more efficient algorithm is as follows. Let $N$ be large, and define $S_2$, $S_3$ and $S_5$ s follows.</p>
<p>$S_2 = \cup_{i=0}^{\infty} \{2^i\} \cap \{1, 2, \ldots, N\}$ </p>
<p>$S_3 = \cup_{i=0}^{\infty} \{3^i s | s \in S_2\} \cap \{1, 2, \ldots, N\}$</p>
<p>$S_5 = \cup_{i=0}^{\infty} \{5^i s | s \in S_3\} \cap \{1, 2, \ldots, N\}$</p>
<p>If $N$ is chosen large enough so that $|S_5|>1000$, then we can easily find $u_{1000}$. I believe (not a proof) that $|S_2| = O(\log{N})$ ,$|S_3|=O(\log^2 N)$, and $|S_5|=O(\log^3 N)$, which leads to a polynomial time algorithm for the problem, as long as we choose $N$ large enough. I also believe that we may choose $N \approx e^{n^\frac{1}{3}}$ in order to get $|S_5|>n$, but I have not proven this yet...</p>
|
942,030 | <p>Given a Hilbert space $\mathcal{H}$.</p>
<p>Consider spectral measures:
$$E:\mathcal{B}(\mathbb{C})\to\mathcal{B}(\mathcal{H}):\quad E(\mathbb{C})=1$$</p>
<p>Define its support:
$$\operatorname{supp}(E):=\bigg(\bigcup_{U=\mathring{U}:E(U)=0}U\bigg)^\complement=\bigcap_{C=\overline{C}:E(C)=1}C$$</p>
<p>By second countability:
$$E\bigg(\operatorname{supp}E^\complement\bigg)\varphi=E\left(\bigcup_{k=1}^\infty B_k'\right)\varphi=\sum_{k=1}^\infty E(B_k')\varphi=0$$</p>
<blockquote>
<p>But it may happen:
$$\Omega\subsetneq\operatorname{supp}E:\quad E(\Omega)=E(\operatorname{supp}E)=1$$</p>
</blockquote>
<p>What is an example?</p>
| Ri-Li | 152,715 | <p>You have $b(\theta) = \dot{\theta}^2$ .......(1)</p>
<p>Where $\dot{\theta}$ and $\ddot{\theta}$ are the first and second derivate respecting time.</p>
<p>In the above expression, you understand that $\dot{b}(\theta) = b'(\theta)\dot{\theta}$ because of the chain rule. Then $b'(\theta)\dot{\theta} = \dot{b}(\theta)= frac {d(b \theta}{dt}= \frac{d(\dot{\theta}^2)}{dt}$.[by 1]</p>
|
942,030 | <p>Given a Hilbert space $\mathcal{H}$.</p>
<p>Consider spectral measures:
$$E:\mathcal{B}(\mathbb{C})\to\mathcal{B}(\mathcal{H}):\quad E(\mathbb{C})=1$$</p>
<p>Define its support:
$$\operatorname{supp}(E):=\bigg(\bigcup_{U=\mathring{U}:E(U)=0}U\bigg)^\complement=\bigcap_{C=\overline{C}:E(C)=1}C$$</p>
<p>By second countability:
$$E\bigg(\operatorname{supp}E^\complement\bigg)\varphi=E\left(\bigcup_{k=1}^\infty B_k'\right)\varphi=\sum_{k=1}^\infty E(B_k')\varphi=0$$</p>
<blockquote>
<p>But it may happen:
$$\Omega\subsetneq\operatorname{supp}E:\quad E(\Omega)=E(\operatorname{supp}E)=1$$</p>
</blockquote>
<p>What is an example?</p>
| Chinny84 | 92,628 | <p>The chain rule is your friend
$$
\dfrac{d}{dt}b(\theta) = \dfrac{db}{d\theta}\cdot \dfrac{d\theta}{dt}
$$
but you also have $b=\dot{\theta}^2$</p>
|
3,045,899 | <p>In the lecture notes we have a fact:</p>
<blockquote>
<p>If <span class="math-container">$A$</span> has orthonormal columns then <span class="math-container">$||Ax||^2_2 = ||x||^2_2$</span></p>
</blockquote>
<p>Why is it the case? What properties of matrix-vector multiplication should I know to reason about this?</p>
<p>Thank you</p>
| littleO | 40,119 | <p>The fact that <span class="math-container">$A$</span> has orthonormal columns is expressed concisely by the statement that <span class="math-container">$A^T A = I$</span>. It follows from this fact that
<span class="math-container">\begin{align}
\| Ax \|^2 &= (Ax)^T Ax \\
&= x^T A^T A x \\
&= x^T x \\
&= \| x \|^2.
\end{align}</span></p>
<hr>
<p>Here's an alternative proof. Let <span class="math-container">$u_i$</span> be the <span class="math-container">$i$</span>th column of <span class="math-container">$A$</span> and let <span class="math-container">$x_i$</span> be the <span class="math-container">$i$</span>th component of a vector <span class="math-container">$x$</span>. If <span class="math-container">$y = Ax = \sum_i x_i u_i$</span>, then
<span class="math-container">\begin{align}
\|y\|^2&= \sum_i \| x_i u_i \|^2 \qquad \text{(by Pythagorean theorem)} \\
&= \sum_i x_i^2 \| u_i \|^2 \\
&= \sum_i x_i^2 \\
&= \| x \|^2.
\end{align}</span></p>
|
4,616,559 | <p>For each day we store a snapshot of data in a database. We want to balance the storage costs with the densitiy of snapshots.
The older a time frame is the fewer snapshots from this time frame we need.
For example: if we store 10 snapshots from last year then we would like to store only 1 snpashot from the time ten years ago.
The density of snapshots should be roughly 1/x where x is the age of the snapshot.
How can we determine when to delete a given snapshot?</p>
| Alexander Tissen | 1,139,101 | <p>This is the idea how to compute the date when to delete a snapshot based on the snapshot's date only.
First we need the ability to assign a number ( or an index ) to every possible date. So, let's choose a start date (for example 2020-01-01) and assign it the index 0.
The choice of the start date is arbitary and doesn't impact this solution. Is is only needed to be able to assign numbers to dates. The date after the start date gets the index 1. The date after that date gets the index 2 and so on. The dates before the start date will have negative indexes.</p>
<p>In the next step we assign a lifespan to every snapshot based on the index of its date. The lifespan states how many days a snapshot will survive (not be deleted) before it gets deleted. For example if the lifespan of the snapshot from 2020-01-01 is 10 then it will be deleted on 2020-01-11 (10 days after it was created).
For each date we can directly compute the date when the snapshot from this date will be deleted: date of the snapshot + lifespan( snapshot's date's index ).</p>
<p>The next step is to define the lifespan function. The lifespan function takes a number and computes the greatest divisor which is a power of 2. For example:
lifespan( 24 ) = 8 because 8 is the greatest power of 2 which divides 24 without rest.
lifespan( 25 ) = 1 because 1 is the greatest power of 2 which divides 25 wihout rest.
index 0 is an exception and needs special treatment because every power of 2 divides 0 without rest. So, let's just say lifespan( 0 ) = 1.</p>
<p>With this, given a snapshot and its creation date we can directly compute the date when it has to be deleted.</p>
<p>At any given time the surviving (not yet deleted) snapshots/dates will mostly have indexes which are powers of 2. I think that the densitiy of powers of 2 is distributed according to 1/x.</p>
|
659,256 | <p>This might be a silly question to some, but I need some help in this topic. <br />
Iota, denoted as <em>'i'</em> is equal to the principal root of -1.
Therefore, </p>
<p>$\iota^2 = -1$</p>
<p>When studying Modulus, I was wondering..</p>
<p>$|\iota| = ?$</p>
<p>A Google search revealed that the value is <strong>+1</strong>. This is because of the equation:</p>
<p>$z = x + y\iota$;
$Therefore, |z| = \sqrt(x^2 + y^2)$</p>
<p>Substituting the value of $0 + 1\iota$, we get our value. But how was this equation derived, and why would it hold true, since, if $\iota$ is +1, $\iota$ must be equal to either +1 or -1, making it a Real Number. My doubt is not analytical but intuitive. As it is an abstract concept, I
am having trouble understanding it. </p>
<p>Sorry for the lack of understanding of any fundamental concepts that render this question redundant. But, sadly, I do not know them, and would appreciate any help you gave me for understanding these.</p>
| heropup | 118,193 | <p>An axiomatic definition of $\mathbb C$ follows from its construction from $\mathbb R^2 = \{(x,y) : x \in \mathbb R, y \in \mathbb R\}$. An abbreviated treatment follows, and a rigorous treatment can be found in Walter Rudin's classic text <em>Principles of Mathematical Analysis</em>.</p>
<p>We <strong>define</strong> a complex number $z \in \mathbb C$ as an ordered pair $z = (a,b)$ of reals, that obey the following rules for addition and multiplication: for $z = (a,b)$ and $w = (c,d)$, we have $$\begin{align*} z \color{red}+ w &= (a,b) \color{red}+ (c,d) = (a \color{blue}+c, \; b\color{blue}+d), \\ z \color{red}\cdot w &= (a,b) \color{red}\cdot (c,d) = (a\color{blue}\cdot c \color{blue}- b\color{blue}\cdot d, \; a\color{blue}\cdot d \color{blue}+ b\color{blue}\cdot c), \end{align*}$$ where we have color coded the binary operations to emphasizes that the red symbols are operations in $\mathbb C$, and blue symbols are operations in $\mathbb R$. From this definition, it is easy to check that the field axioms are satisfied--that is to say, $(\mathbb C, \color{red}+, \color{red}\cdot )$ is a field, with additive identity $(0,0)$ and multiplicative identity $(1,0)$. We also can see that $\mathbb C$ contains a copy of $\mathbb R$ as a subfield: explicitly, $(a,0) = a$. Next, we can also see that $$(0,1)^2 = (0,1) \cdot (0,1) = (-1, 0) = -1,$$ so that the element $(0,1)$ has the property that its square equals $-1$. We <strong>define</strong> $$i = (0,1).$$ From this, we can now observe that for real numbers $a, b$, we can write $$(a,b) = (a,0) + (0,b) = a(1,0) + b(0,1) = a + bi.$$ Throughout this, we have not needed to appeal to square roots. Now, the <strong>magnitude</strong> of a complex number is again defined by its Euclidean norm in $\mathbb R^2$: the function $|\cdot| : \mathbb C \to \mathbb R^{\ge 0}$ is defined as $$|(a,b)| = \sqrt{a^2 + b^2}.$$ Then the fact that there are infinitely many such ordered pairs whose magnitude is $1$ is a direct consequence of elementary geometry in the real Euclidean plane.</p>
|
427,564 | <p>I'm supposed to give a 30 minutes math lecture tomorrow at my 3-grade daughter's class. Can you give me some ideas of mathemathical puzzles, riddles, facts etc. that would interest kids at this age?</p>
<p>I'll go first - Gauss' formula for the sum of an arithmetic sequence.</p>
| torbonde | 30,734 | <p>How about giving various proofs that $1+\cdots+n = n(n+1)/2$? There are at least a few different geometric, easily understandable ones (although I'm not sure about how easy, when it comes to 3rd-graders. You'll be the judge of that.)</p>
<p>I saw the one below recently, and it blew my mind. Using this, you will need to introduce the binomialcoefficient as well. But I guess they will be able to understand the idea of choosing two balls from $n$ balls, in different ways.</p>
<p><img src="https://i.stack.imgur.com/1FCaz.png" alt="Proof using the binomialcoefficient"></p>
|
962,242 | <p>$X \simeq Y$ reads as $X$ is <em>equivalent</em> to $Y$</p>
<p>If $X \simeq Y$, <strong>iff</strong> $X \leftrightarrow Y$ is a tautology.</p>
<p>Now given $X_1 \simeq X_2$, how do I prove,</p>
<ol>
<li>$\tilde X_1 \simeq \tilde X_2$</li>
<li>$X_1 \cap Y\simeq X_2 \cap Y$</li>
<li>$X_1 \cup Y\simeq X_2 \cup Y$</li>
<li>$X_1 \rightarrow Y\simeq X_2 \rightarrow Y$</li>
</ol>
<p>I know that I can write an algebraic proof by rewriting, $X \leftrightarrow Y$ as $( \tilde X \cup Y) \cap (\tilde Y \cup X)$, but is there a purely logical proof.</p>
<p><strong>Update</strong>: I figured the solution for $(1)$ and it is. Is that a right proof?</p>
<p>if $X \rightarrow Y$ is a tautology $(X \rightarrow Y)\cap (Y \rightarrow X)$ is a tautology.</p>
<p>And $X \rightarrow Y$ is same as $\tilde Y \rightarrow \tilde X$. Rewriting the equation in $(1)$ proves it.</p>
<p>The Axioms that are available are the basic axioms of Propositional Calculus</p>
<ol>
<li>Either $X$ or $\tilde X$ is true</li>
<li>If $X,Y$ is true, then so is $X\cap Y$ is true</li>
<li>If either $X$ or $Y$ is true, then $X\cup Y$ is true</li>
<li>If $X$ is not true or $Y$ is true, then $X\rightarrow Y$ is true</li>
</ol>
<p>There was a mistake and I had confused between $\cup$ and $\cap$. Apologies.
Thanks again</p>
| DSinghvi | 148,018 | <p>TRY TO PROVE AB<em>CD=AC</em>CB </p>
<p>THEN square it use AB^2=AC^2+BC^2 </p>
<p>To prove AB * CD=AC * CB </p>
<p>Equate the areas of triangle in terms of AC and CB and the other one in AB and CD </p>
<p>1/2 AC * CB=1/2 CD * AB</p>
|
962,242 | <p>$X \simeq Y$ reads as $X$ is <em>equivalent</em> to $Y$</p>
<p>If $X \simeq Y$, <strong>iff</strong> $X \leftrightarrow Y$ is a tautology.</p>
<p>Now given $X_1 \simeq X_2$, how do I prove,</p>
<ol>
<li>$\tilde X_1 \simeq \tilde X_2$</li>
<li>$X_1 \cap Y\simeq X_2 \cap Y$</li>
<li>$X_1 \cup Y\simeq X_2 \cup Y$</li>
<li>$X_1 \rightarrow Y\simeq X_2 \rightarrow Y$</li>
</ol>
<p>I know that I can write an algebraic proof by rewriting, $X \leftrightarrow Y$ as $( \tilde X \cup Y) \cap (\tilde Y \cup X)$, but is there a purely logical proof.</p>
<p><strong>Update</strong>: I figured the solution for $(1)$ and it is. Is that a right proof?</p>
<p>if $X \rightarrow Y$ is a tautology $(X \rightarrow Y)\cap (Y \rightarrow X)$ is a tautology.</p>
<p>And $X \rightarrow Y$ is same as $\tilde Y \rightarrow \tilde X$. Rewriting the equation in $(1)$ proves it.</p>
<p>The Axioms that are available are the basic axioms of Propositional Calculus</p>
<ol>
<li>Either $X$ or $\tilde X$ is true</li>
<li>If $X,Y$ is true, then so is $X\cap Y$ is true</li>
<li>If either $X$ or $Y$ is true, then $X\cup Y$ is true</li>
<li>If $X$ is not true or $Y$ is true, then $X\rightarrow Y$ is true</li>
</ol>
<p>There was a mistake and I had confused between $\cup$ and $\cap$. Apologies.
Thanks again</p>
| Irvan | 172,851 | <p>The area of the triangle can be expressed in two ways: $\displaystyle \frac{|AC| |BC|}{2}$ and $\displaystyle \frac{|AB| |CD|}{2}$. Thus, they must be equal:</p>
<p>$$\frac{|AC| |BC|}{2}=\frac{|AB| |CD|}{2}$$</p>
<p>Multiplying throughout by $2$ and squaring,</p>
<p>$$|AC|^2 |BC|^2 = |AB|^2 |CD|^2$$</p>
<p>$$|AC|^2 |BC|^2 = (|AC|^2 + |BC|^2) \cdot |CD|^2$$</p>
<p>$$\frac{1}{|CD|^2} = \frac{|AC|^2 + |BC|^2}{|AC|^2 |BC|^2}$$</p>
<p>$$\therefore \frac{1}{|CD|^2} = \frac{1}{|AC|^2} + \frac{1}{|BC|^2}$$</p>
<p>as required.</p>
|
4,206,286 | <p><span class="math-container">$$\int_0^1\int_0^\infty ye^{-xy}\sin x\,dx\,dy$$</span></p>
<p>How can I calculate out the value of this integral?</p>
<p>P.S. One easy way is to calculate this integral over <span class="math-container">$dy$</span> first, to get an integral form <span class="math-container">$\frac{1-e^{-x}(x+1)}{x^2}\sin x$</span>, if I calculated correctly, but I don't know any way to calculate out this value other than a hard work with contour integral. - so I wonder if there be a way other than this (=integrate over <span class="math-container">$dy$</span> and do contour integral)</p>
<p>(I tried some integration by parts and substitutions but it seems it does not work well, probably the <span class="math-container">$\infty$</span> at the second integral is crucial. I think this requires capturing a term in the integrand and converting to a further integral, and reverting the order of integral by Fubini's theorem.. but I'm not sure)</p>
| g.kov | 122,782 | <p><a href="https://i.stack.imgur.com/7UmqH.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7UmqH.png" alt="enter image description here" /></a></p>
<p>For the acute triangle <span class="math-container">$ABC$</span> with the circumcenter <span class="math-container">$O$</span></p>
<p><span class="math-container">\begin{align}
\angle CAB&=\alpha
,\quad
\angle ABC=\beta
,\quad
\angle BCA=\gamma
,\\
\angle COB&=2\alpha
,\quad
\angle AOC=2\beta
,\quad
\angle BOA=2\gamma
.
\end{align}</span></p>
<p>Triangles <span class="math-container">$COB$</span>, <span class="math-container">$AOC$</span> and <span class="math-container">$BOA$</span> are isosceles, hence</p>
<p><span class="math-container">\begin{align}
\angle COD&=\alpha
,\quad
\angle AOE=\beta
,\quad
\angle BOF=\gamma
\end{align}</span></p>
<p>and
<span class="math-container">\begin{align}
d_a=|OD|&=R\cos\alpha
,\\
d_b=|OE|&=R\cos\beta
,\\
d_c=|OF|&=R\cos\gamma
,\\
d_a+d_b+d_c&=
R(\cos\alpha+\cos\beta+\cos\gamma)
.
\end{align}</span></p>
<p>And it is well-known that for any triangle</p>
<p><span class="math-container">\begin{align}
\cos\alpha+\cos\beta+\cos\gamma&=\frac rR+1
,\\
\text{so, }\quad
d_a+d_b+d_c&=
R\cdot\left(\frac rR+1\right)
=r+R
.
\end{align}</span></p>
|
2,196,936 | <p>How to prove that $7p + 3^p -4$ is not a perfect square? </p>
<p>I calculated: $\left(\frac{7p+3^p-4}{p}\right) = \left(\frac{-1}{p}\right)$. So if $p \equiv 3 \mod 4$, the result is $-1$. So in that case, $7p+3^p -4$ can't be a square. But what about the case $p \equiv 1 \mod 4$? Any hints? Thanks in advance.</p>
| Shaun | 104,041 | <p>The squares modulo <span class="math-container">$4$</span> are</p>
<p><span class="math-container">$$(4k)^2 = 4(4k^2) \equiv 0 \pmod{4},$$</span></p>
<p><span class="math-container">$$(4k+1)^2 = 4(4k^2 + 2k) + 1 \equiv 1 \pmod{4},$$</span></p>
<p><span class="math-container">$$(4k + 2)^2 = 4(4k^2 + 4k + 1) \equiv 0 \pmod{4},$$</span></p>
<p>and</p>
<p><span class="math-container">$$(4k + 3)^2 = 4(4k^2 + 6k + 2) + 1 \equiv 1 \pmod{4},$$</span></p>
<p>but if <span class="math-container">$p = 4 \ell + 1$</span>, then
<span class="math-container">$$7p + 3^p - 4\equiv (-1)(1) + (-1)^{4 \ell}(-1)^1 \pmod{4} \equiv -2 \pmod{4},$$</span> which is equivalent to neither <span class="math-container">$0$</span> nor <span class="math-container">$1$</span> modulo <span class="math-container">$4$</span>, so <span class="math-container">$7p + 3^p - 4$</span> cannot be a square if <span class="math-container">$p = 4 \ell + 1$</span>.</p>
|
1,722,995 | <blockquote>
<blockquote>
<p>Question: Given the circle $x^2+y^2=25$ is inscribed in triangle $\triangle ABC$, where vertex $B$ lies on the first quadrant. Slope of $AB$ is $\sqrt 3$ and has a positive y-coordinate, and $|AB|=|AC|$. Find the equations for $AC$ and $BC$</p>
</blockquote>
</blockquote>
<p>I found out the equation for the straight line passing through $AB$: Let the line be $y=\sqrt 3 x+c$. Then</p>
<p>$3x^2+2\sqrt 3 cx+c^2+x^2=25$</p>
<p>$\Delta =0$ (discriminant)</p>
<p>$(2\sqrt 3 c)^2 - 4(4)(c^2-25)=0$</p>
<p>$c=10$</p>
<hr>
<p>However, I don't see any simple way to find out the equations of line for $AC$ and $BC$. While it seems like there is enough information, I have tried using similar triangles, etc, but I can't find out the coordinates of the vertices. Can anyone give me some hints? Thank you!</p>
| chenbai | 59,487 | <p><a href="https://i.stack.imgur.com/t9eXK.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/t9eXK.jpg" alt="enter image description here"></a></p>
<p>I will suggest to take $\alpha$ as parameter to write the equation. and you need to take care of the range of $\alpha$ .</p>
|
1,722,995 | <blockquote>
<blockquote>
<p>Question: Given the circle $x^2+y^2=25$ is inscribed in triangle $\triangle ABC$, where vertex $B$ lies on the first quadrant. Slope of $AB$ is $\sqrt 3$ and has a positive y-coordinate, and $|AB|=|AC|$. Find the equations for $AC$ and $BC$</p>
</blockquote>
</blockquote>
<p>I found out the equation for the straight line passing through $AB$: Let the line be $y=\sqrt 3 x+c$. Then</p>
<p>$3x^2+2\sqrt 3 cx+c^2+x^2=25$</p>
<p>$\Delta =0$ (discriminant)</p>
<p>$(2\sqrt 3 c)^2 - 4(4)(c^2-25)=0$</p>
<p>$c=10$</p>
<hr>
<p>However, I don't see any simple way to find out the equations of line for $AC$ and $BC$. While it seems like there is enough information, I have tried using similar triangles, etc, but I can't find out the coordinates of the vertices. Can anyone give me some hints? Thank you!</p>
| Steven Alexis Gregory | 75,410 | <p><span class="math-container">$AB=AC$</span> suggests to me that line <span class="math-container">$BC$</span> is a vertical line and point <span class="math-container">$A$</span> is on the negative x-axis. See the picture.</p>
<blockquote>
<p><span class="math-container">$A$</span> has to be on the line perpendicular to segmanet <span class="math-container">$OP$</span> at <span class="math-container">$P$</span>. Put <span class="math-container">$A$</span> in the second quadrant and <span class="math-container">$AB \lt AC$</span> Put <span class="math-container">$A$</span> in the third quadrant and <span class="math-container">$AB \gt AC$</span>.</p>
</blockquote>
<p>Let <span class="math-container">$P$</span> be the point where line <span class="math-container">$AB$</span> intersects the circle <span class="math-container">$x^2 + y^2 = 25$</span>. If the slope of line <span class="math-container">$AB$</span> is <span class="math-container">$\sqrt 3$</span>. Then the slope of line <span class="math-container">$OP$</span> must be <span class="math-container">$-\dfrac{1}{\sqrt 3}$</span>. So the line <span class="math-container">$OP$</span> is the line <span class="math-container">$y = -\dfrac{1}{\sqrt 3}x$</span>.</p>
<p>You said you just wanted a hint. So I will let you work out the rest.
<a href="https://i.stack.imgur.com/1Sbbi.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1Sbbi.jpg" alt="My solution" /></a></p>
|
278 | <p>If you take a look at our status in the <a href="http://area51.stackexchange.com/proposals/64216/mathematics-educators">area51</a>, all criteria seem to be satisfied (soon) but not the number of questions asked (which seems to be decreasing, actually). Do you think this is a problem for us? Is it something we should / can take care of?</p>
| Robert Cartaino | 2 | <p>This pattern is not unusual at all. Most sites go through a <em>honeymoon period</em> where users pile in from the excitement of a new launch with their questions at the ready. Then you experience a sharp drop in activity… followed by a pattern of slow, steady growth as you continue to compile content and the Google and network effects start to kick in. </p>
<p>This is all completely normal and expected.</p>
|
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