Q stringlengths 18 13.7k | A stringlengths 1 16.1k | meta dict |
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I'm confused how to get this result using a different definition of an adjoint operator. In the problem we're given that $ \hat{A}|0 \rangle = |0\rangle -2i|1 \rangle$ and that $ \hat{A}|1 \rangle = 2|0 \rangle -i |1 \rangle$.
We want to know what $\hat{A}^{\dagger}|0\rangle, \hat{A}^{\dagger}|1\rangle$ are using the definition of an adjoint operator,
$$\langle \psi \mid A^{\dagger}\phi \rangle = \langle \phi\mid A\psi \rangle^* $$
When I try to use this definition I get lost, but I have a value for A via my own methods on the information in sentence 1. How would I use the definition of adjoint to get the same answer as below?
I got, $$\hat{A} = \left(\begin{matrix} -i & 2i \\ 2 & 1\end{matrix}\right), \hat{A}^{\dagger} = \left(\begin{matrix} 1 & -2i \\ -2 & -i\end{matrix}\right) $$
| Rewriting the definition so it can be used in our case as follows
$$
\langle \psi |A^{\dagger}\phi \rangle = \langle \phi|A\psi \rangle^*
\iff
\langle \psi | A^{\dagger}|\phi \rangle = \langle \phi| A|\psi \rangle^*
$$
using the definition you have to find the matrix elements $A^\dagger_{j,k}$ of the operator $\hat A^\dagger$
$$
A^\dagger_{j,k} = \langle j | A^{\dagger}|k \rangle = \langle k| A|j \rangle^*
$$
Since we have $\hat{A}|0 \rangle = |0\rangle -2i|1 \rangle$ and $\hat{A}|1 \rangle = 2|0 \rangle -i |1 \rangle$ you can find, for example
$$
A^\dagger_{0,1} = \langle 0 | A^{\dagger}|1 \rangle = \langle 1| A|0 \rangle^* = (-2i)^* = 2i
$$
and the matrix of $\hat A^\dagger$ is
$$
\hat{A}^{\dagger} = \left(\begin{matrix} 1 & 2i \\ 2 & i\end{matrix}\right)
$$
The operator in the dyadic form
$$
\hat A^\dagger = |0\rangle\langle0| + 2i|0\rangle\langle1| + 2|1\rangle\langle0| + i|1\rangle\langle1|
$$
Now you can find $\hat{A}^{\dagger}|0\rangle , \hat{A}^{\dagger}|1\rangle$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find a number divisible by 7 by erasing digits from a seven digit number Let $x$ be a seven-digit number. Prove that only by erasing digits from the left side or the right side (or both) of the number, you can remain with a number that is divisible by $7$.
I assume the Pigeonhole Principal is useful here, however I'm having trouble finding its application here. I would like to get a hint (or the full proof with a hint beforehand). Thanks
| Let our number be $n=\overline {a_7a_6a_5a_4a_3a_2a_1}$
Now consider the $7$ numbers $n_i=\overline {a_i\cdots a_1}\pmod 7$.
As there are $7$ of these we either hit every reside class $\mod 7$ or we hit one twice. If we hit all the classes we must hit $0$. If we hit some class twice, say $n_i\equiv n_j\pmod7$ for some $i<j$ then it is easy to see that $\overline {a_j\cdots a_{i+1}}\equiv 0 \pmod 7$ and we are done.
Example: take $n=1248123$. Then, $\pmod 7$ we get $$\{n_1,\cdots, n_7\}\equiv \{3,2,4,3,5,1,2\}\pmod 7$$
We see that $n_1\equiv n_4\pmod 7$ so $$8123\equiv 3\pmod 7\implies 7\,|\,8120\implies 7\,|\,812$$
Or $$n_7\equiv n_2\pmod 7\implies 7\,|\,12481$$
Here, of course, we are relying on the fact that $\gcd(7,10)=1$. The claim generalizes from $7$ to any $m$ prime to $10$ using the same argument.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove Segner's Recurrence Relation $C_{n+1} = \sum\limits_{i=0}^n C_i C_{n-i}$ on Catalan Numbers $C_n = \frac{1}{n+1} \binom{2n}{n}$
Prove Segner's Recurrence Relation $C_{n+1} = \sum\limits_{i=0}^n C_i C_{n-i}$ on Catalan Numbers $C_n = \frac{1}{n+1} \binom{2n}{n}$
Plugging in the Catalan Equation, we want to prove:
\begin{align*}
\frac{1}{n+2} \binom{2(n+1)}{n+1} &= \sum\limits_{i=0}^n \frac{1}{i+1} \binom{2i}{i} \frac{1}{n-i+1} \binom{2(n-i)}{n-i} \\
\end{align*}
Expanding the binomial coefficients to factorials (again this is a formula we want to prove, not a result we have demonstrated):
\begin{align*}
\frac{1}{n+2} \frac{(2n+2)!}{(n+1)!(n+1)!} &= \sum\limits_{i=0}^n \frac{1}{i+1} \frac{(2i)!}{i!i!} \frac{1}{n-i+1} \frac{(2n-2i)!}{(n-i)!(n-i)!} \\
\end{align*}
How would one go from here?
| In all honesty, I’d prove it combinatorially, e.g., by showing that there are $\frac1{n+1}\binom{2n}n$ well-formed strings of $n$ pairs of parentheses and that the number of such strings satisfies the recurrence. (Using generating functions is also a reasonable approach.)
You said that you’re familiar with the first fact. As for the second, if $s$ is such a string, split it into two substrings at the first point at which the initial segment is well-formed. Say that you have $\ell$ pairs in that initial segment and $n−\ell$ in the final segment. There are $C_{n−\ell}$ possibilities for the final segment. Try to see why there are $C_{\ell−1}$ possibilities for the initial segment; use the fact that it’s the minimal well-formed prefix.
| {
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Evaluating $\lim_{x\to0}\left(\frac{\sin x}x\right)^\frac{1}{1-\cos x}$ I was trying to evaluate
$$\lim_{x\to0}\left(\frac{\sin x}x\right)^\frac{1}{1-\cos x}$$
I have tried taking natural logarithm first:
$\lim_{x\to0}\frac{\ln(\sin x)-\ln x}{1-\cos x}=\lim_{x\to0}\frac{\frac{\cos x}{\sin x}-\frac{1}x}{\sin x}\quad\quad\text{(L'Hopital Rule)}\\
=\lim_{x\to0}\frac{\cos x}{x\sin x}-\frac{1}{x^2}\quad\quad(\lim_{x\to0}\frac{\sin x}x=1)\\
=\lim_{x\to0}\frac{\cos x-1}{x^2}\quad\quad\quad(\lim_{x\to0}\frac{\sin x}x=1)$
and after this I eventually have the limit equaling $-\frac{1}2$, which means that the original limit should be $\frac{1}{\sqrt{e}}$.
However, I graphed $f(x)=\left(\frac{\sin x}x\right)^\frac{1}{1-\cos x}$ on Desmos, and it turned out that the limit is approximately $0.7165313$, or $\frac{1}{\sqrt[3]{e}}$.
Therefore I think there's something wrong in my approach, but I couldn't find it. Any suggestions?
| $$\lim_{x\to0}\frac{\frac{\cos x}{\sin x}-\frac{1}x}{\sin x}=\lim_{x\to0}\frac{(x-\frac{x^3}{2!}+\frac{x^5}{4!}+\cdots)-(x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots)}{x^3}=\frac{1}{6}-\frac{1}{2}=-\frac{1}{3}$$
| {
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What is the limiting sum of $\frac{1}{1\left(3\right)}+\frac{1}{3\left(5\right)}+\frac{1}{5\left(7\right)}+···+\frac{1}{(2n-1)(2n+1)}=\frac{n}{2n+1}$ I know $a$ (first term) is $\frac{1}{3}$, but I'm not sure what the common ratio would be from this. I assume it is the $\frac{1}{(2n-1)(2n+1)}$, but this is not a number. Any hints?
| $$S=\frac12\big(\frac{1}{2-1}-\frac{1}{2+1}+\frac{1}{4-1}-\frac{1}{4+1}+ \frac{1}{6-1}-\frac{1}{6+1}+\cdot\cdot\cdot+\frac{1}{2n-1}-\frac{1}{2n+1}\big) $$
Cancelling similar terms we finally get:
$$S= \frac12\big(1-\frac{1}{2n+1}\big)=\frac12\times\frac{2n}{2n+1}$$
$$\lim _{n→∞ } S=\frac{1}{2}$$
| {
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surjectivity and injectivity of a linear transformation I have the following question here.
A certain linear transformation $\varphi:P_3 \rightarrow\mathbb{R}^2$
maps the polynomial $x^3+3x^2+5x+7$ to $\begin{pmatrix}0 \\ 0\\
\end{pmatrix}$ and $x^3+2x^2+3x+4$ to $\begin{pmatrix}1 \\ 0\\\end{pmatrix}$. What can we conclude aboout $\varphi$?
a) It is injective and surjective.
b) It is injective but not surjective.
c) It is surjective but not injective.
d) It is neither injective nor surjective.
e) It is impossible to decide whether it is surjective, but we know it is not injective.
I'm tempted to say neither. For the transformation to be surjective, $\ker(\varphi)$ must be the zero polynomial but I can't really say that's the case here. I can't even conclude if the transformation is $1-1$ so I don't think it's injective.
For surjectivity, the codomain of the linear transformation (The image) has to describe all of $\mathbb{R}^2$ but I don't think I can really conclude this either.
So as such, my answer choice is D but I don't think this is right.
Can someone offer some guidance?
EDIT: I changed $\mathbb{R}^3$ to $\mathbb{R}^2$.
| Firstly, I think you mean in your first paragraph that for the transformation to be injective, $\operatorname{ker} \varphi$ must be the zero polynomial. This is a true statement, so the fact that $\varphi$ maps $x^3+3x^2+5x+7$ to $(0,0)$ tells us that $\varphi$ is not injective.
Secondly, for surjectivity, the image of the linear transformation has to be all of $\mathbb R^2$. The only information we have here is that $\varphi$ maps $x^3+2x^2+3x+4$ to $(1,0)$, so your reasoning is correct that we can't necessarily conclude that $\varphi$ is surjective.
Now can you conclude?
| {
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Number of $B\subset A$ with $s(B)$ divisible by $n$ I recently saw this IMO $1995$ problem:
How many subsets of $\{1,2,...,2p\}$ are there, with $p$ elements, such that the sum of the elements is divisible by $p$, given that $p$ is a prime, $p\geq 3$.
I solved this using the classical (well not really, but not unheard of) method of considering $a_i$ the number of subsets with $p$ elements whose sum is $\equiv i\pmod{p}$ and then constructing the following polynomial:
$$\sum_{i=0}^{p-1}a_i\cdot\epsilon^i$$
Where $\epsilon$ is the $p^{th}$ primitive root of unity $\big($i.e. $\epsilon=\cos{\frac{2\pi}{p}+i\cdot\sin{\frac{2\pi}{p}}}\big)$, and then using this lemma:
If $\epsilon$ is the $p^{th}$ primitive root of unity, $p\geq 3$ and $p$ is a prime, then: $$\sum_{i=0}^{p-1} a_i\cdot\epsilon^i=0\Leftrightarrow a_0=a_1=...=a_{p-1}$$
And a bit of interpretation, I get that there are $$2+\frac{1}{p}\bigg(\binom{2p}{p}-2\bigg)$$
such subsets. This can be easily generalized in many ways $\big($for example to count all subsets, or count subsets of $\{1,2,..,k\cdot p\}\big)$, as long as $p$ is a prime. However, what should we do with this problem?
How many subsets of $\{1,2,...,an\}$ are there, such that the sum of the elements is divisible by $n$, where $n$ is an arbirary positive integer.
Thank you!
| Here is a calculation using something similar to the polynomial you consider. Set $\epsilon = \cos(2\pi /n)+i\sin(2\pi /n)$. For every integer $k\geq 1$, there is a polynomial factorization
$$
\prod_{j=1}^{an} \left(x-\epsilon^{jk}\right) = \left(x^{n/(n,k)}-1\right)^{a(n,k)}.
$$
Also, we have
$$
\sum_{j=1}^{n}\epsilon^{jb}=\begin{cases}n&:n|b,\\0&n\nmid b.\end{cases}
$$
So, the number of subsets $B\subseteq \{1,\ldots,an\}$ with sum divisible by $n$ is
$$
\begin{align*}
\frac{1}{n}\sum_{B\subseteq\{1,\ldots,an\}}\sum_{j=0}^{n-1}\epsilon^{js(B)}&=\frac{1}{n}\sum_{j=1}^{n}\prod_{k=1}^{an}\left(1+\epsilon^{jk}\right)\\
&=\lim_{x\to 1}\frac{1}{n}\sum_{j=1}^{n}\prod_{k=1}^{an}\left(x+\epsilon^{jk}\right)\\
&=\lim_{x\to 1}\frac{1}{n}\sum_{j=1}^{n}\prod_{k=1}^{an}\left(\frac{x^2-\epsilon^{2jk}}{x-\epsilon^{jk}}\right)\\
&=\frac{1}{n}\sum_{j=1}^{n} \lim_{x\to 1}\frac{\left(x^{2n/(n,2j)}-1\right)^{a(n,2j)}}{\left(x^{n/(n,j)}-1\right)^{a(n,j)}}
\end{align*}.
$$
The $j$-th term in the sum is $0$ if $(n,2j)>(n,j)$ (equivalently, $n/(n,j)$ is even), and is $2^{a(n,j)}$ if $(n,2j)=(n,j)$ (equivalently, $n/(n,j)$ is odd). So, if we write $n=2^km$ with $m$ odd, the number of subsets in question is
$$
\begin{align*}
\frac{1}{n}\sum_{\substack{j=1\\n/(n,j)\text{ odd}}}^{n} 2^{a(n,j)}=\frac{1}{n}\sum_{j=1}^m 2^{a2^k(m,j)}=\frac{1}{n}\sum_{d|m}\varphi(m/d)2^{2^kad}.
\end{align*}
$$
I don't know if this sum can be further simplified.
| {
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How to solve this question of picking balls from bins, question is probability of ball number vs selection number? I'm a student and always get confused with different combinations of balls in bins problem. I've encountered this question today and not able to solve this. Hope someone can give me some pointers, thanks! If there are any related materials I need to refresh, that would be helpful as well.
You have been given 3 balls which are numbered as 1, 2 and 3 and they are placed inside a bag. You randomly select 3 balls, one ball at a time. Considering your selection, you win if at least one selection number matches with the number on the ball. So for example if you get ball with number 1 on your first draw you call this as win. The question is what are the chances of winning if you do this with replacement versus without replacement.
| I like to simulate such problems as a double check on the
combinatorial logic. [@MathLover's Answer (+1), which you might consider Accepting, and @JMoravitz' Comment have
put you on the right track.]
# without replacement
set.seed(2020)
win.wo = replicate(10^6, sum(sample(1:3, 3)==1:3))
mean(win.wo > 0)
[1] 0.666828 # aprx 1 - (2/3)(1/2) = 2/3
# with replacement
win.wr = replicate(10^6, sum(sample(1:3, 3, rep=T)==1:3))
mean(win.wr > 0)
[1] 0.702987 # aprx 1 = (2/3)^2 = 19/27
1 - dbinom(0, 3, 1/3)
[1] 0.7037037 # Exact binomial computation
Notes: win.w0 and win.wr are vectors of length 1,000,000 with numbers of wins; we want one or more. Logical vector
(win.wr > 0) has a million TRUEs and FALSEs; its mean
is the proportion of its TRUEs. Similarly, for win.wr.
With a million iterations of the game we can expect 2 or 3
place accuracy for the probabilities.
'sample' does the sampling. Examples:
sample(1:3, 3)
[1] 2 3 1
sample(1:3, 3)
[1] 1 2 3
s = sample(1:3, 3)
[1] 2 1 3
s == 1:3
[1] FALSE FALSE TRUE
sum(s == 1:3)
[1] 1
sample(1:3, 3, rep=T)
[1] 1 2 2
sample(1:3, 3, rep=T)
[1] 3 3 3
sample(1:3, 3, rep=T)
[1] 2 3 2
sample(1:3, 3, rep=T)
[1] 1 3 2
| {
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If we assume $f$ is Riemann integrable and the Fourier series of $f$ is pointwise convergent, do we have... If we assume $f$ is Riemann integrable and the Fourier series of $f$ is pointwise convergent, do we have
$f(x)=\sum_{n=-\infty}^{n=+\infty} \hat {f}(n) e^{2\pi i nx/L}$?
Could someone recommend a source that answers this question?
| The Cesaro means of the Fourier series converges to $f$ everywhere that $f$ is continuous, which is almost everywhere for a Riemann integrable function. So, if the Fourier series converges at a point of continuity of $f$, then it must converge to $f$. So under your assumptions, that would mean that $f$ equals the Fourier series at least at every point of continuity of $f$, which is almost everywhere.
| {
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Can a pure gradient be made equal to an arbitrary function? This question must have a really simple answer that I'm not seeing, since I'm reading it just stated in an undergrad textbook without any proof. It's about fluid dynamics, but the physics of it is actually irrelevant.
It's stated that the following equation
$$\nabla p + \rho\nabla\phi = 0$$
where $p$, $\rho$ and $\phi$ are smooth scalar functions on $\mathbb R^3$, has no solution if $\rho$ is not a constant, since the pressure term $p$ is a pure gradient, and the second term is not.
If $\rho$ is constant then of course there is the trivial solution $p=-\rho\phi + C$ for a constant $C$.
Can someone explain why it cannot have a solution if $\rho$ is not constant? I have some intuition about it, but can't see the statement to be trivial for any arbitrary functions $\rho$ and $\phi$.
| Just to be clear, extending an aspect of @QiaochuYuan's answer: there is the very basic equality-of-mixed-partials criterion, that if $F(x,y,z)=(f_x,f_y,f_z)$ for some scalar-valued $f$, then (because under mild hypotheses $(f_x)_y=(f_y)_x)$, etc.), the partial with respect to $y$ of the first component of $F$ must be equal to the partial with respect to $x$ of the second component of $F$, and so on.
In your set-up, this gives a necessary condition on $\rho$.
In fact, locally this is also a sufficient condition, but some of the interesting cases are definitely not local...
| {
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Derivative of $\left | x-\left \lfloor x+1 \right \rfloor \right |$ at $x = 1.5$? Q: If $f(x)=\left | x-\left \lfloor x+1 \right \rfloor \right |$, where $\left \lfloor x \right \rfloor$ denotes the greatest integer less than or equal to x and $\left | x \right |$ denotes the absolute value of x, then $f'(1.5)$ =
I am not quite sure how the derivative of floor function and absolute function. I did some research and found out that the derivative of an absolute function is $\frac{\left \lfloor x \right \rfloor}{x}$. But I am stuck with the floor function.
What concept should I be aware of to solve this question?
| For $x \in (1,2)$ we have $f(x)=\left | x-\left \lfloor x+1 \right \rfloor \right |=2-x$, so $f'(1.5)=-1$.
Same will be on any $(k,k+1)$ interval for $k \in \mathbb{Z}$.
| {
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A problem about how many evening we can meet all friends There are 30 friends and everyone lives in their own apartment. Everybody wants to
visit everybody else's apartment. Every evening a person can either stay home for the
whole evening, or visit as many friends as she likes. (Of course, if she visits somebody
who is not at home, we don't count it as a visit.) Prove that for each person to visit all
the others:
(1) 4 evenings are not enough;
(2) 5 evenings are also not enough;
(3) but 10 evenings is enough;
(4) and even 7 evenings is enough.
If the "visit" is a single direction, I can say in five days, each person can visit all other persons. However the problem seems to be two direction? how can we do?
| I have worked out the problem . Actually it is very simple. We can let S be the days we consider. S_i be the set of visit days for each i. The problems say that S_i and S_j can not be the subset of the other. For 7 evenings, since 7 choose 3 equal 35, hence we can construct 30 different 3-tuple set.
| {
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Should a partial order and total order be related in the context of the "condition" specified in the relation? I think using examples would convey my question the fastest.
Let there be a set $A = \{ 1,2,3,5,10 \}$. Let $R$ be relation such that $a | b$.
Note that in this case $a | b$ is what I mean by "condition" which I mentioned in the question
Thus $R= \{ (1,2), (1,3), (1,5), (1,10), (2,10), (5,10) \}$
One possible total order $T$ would be $T= \{ (1,2), (1,10) \}$ since $1,2,10$ are comparable,ie ($1$⪯$2$⪯$10$) and is partial order.
Both $T$ and $R$ are related by the same condition, $a | b$... right?
But then my school showed that another possible total order $T$ to be $1 ≺ 2 ≺ 3 ≺ 5 ≺ 10$ ie $T= \{ (1,2), (1,3),...(2,3),(2,5)... (2,10), (5,10) \}$
But how is $2|3$ or $2|5$? Is the condition no longer $a | b$?
My question essentially boils down to the title. So can the partial and total order not have the same "condition" at all?
| You're right: the condition (that is to say, the relation itself) is different in the second case: it's simply the usual $\le$.
Nothing prevents us to consider two different relations on the same set. Nevertheless the chain $1,2,3,5,10$ is not a chain with respect to the first relation.
Note that though these are two different relations, they have a connection, namely the first one is included in the second one, meaning that $a|b\implies a\le b$ (at least for positive integers).
| {
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Proving that cardinality of this set is at most $\frac{n(n-1)}{2}$ This question is from Apostol introduction to analytic number theory 144 page. Question 12
Let $G$ be a group and let $S$ be a subset of $n$ distinct elements of $G$ with the property that $a \in S$ implies $a^{-1} \notin S$. Consider the $n^2$ products of form $ab$, where $a$ and $b$ both belong to $S$. Then prove that at most $n(n-1) /2$ of these products belong to $S$.
Attempt : I assumed that let a, b, c belongs to S and a=bc then a$b^{-1}$ = c, but $b^{-1}$ doesn't belongs to S.There will also be a case when a doesn't belongs to S.
Then how to use it to prove the result in question? Unfortunately, I am not able to move ahead.
It is my humble request to help me
| Presumably Apostol means there are at most $(n^2-n)/2$ ordered pairs $(a,b) \in S^2$ such that $ab\in S.$
Let $T=\{(a,c)\in S^2: a\ne c\}.$ Let $U=\{(a,c)\in T:a^{-1}c\in S\} .$
Now $T$ has $n^2-n$ members. And for any $(a,c)\in T,$ at most one of $(a,c),(c,a)$ belongs to $U.$ Because $a^{-1}c$ and $(a^{-1}c)^{-1}=c^{-1}a$ cannot both belong to $ S.$ So $U$ has at most half of the members of $T.$
So $U$ has at most $(n^2-n)/2$ members.
Let $V=\{(a,b)\in S^2: ab\in S\}.$ The function $F(\,(a,b)\,)=(a,ab)$ is a bijection, and $$\forall (a,b)\in S^2\,(\,(a,b)\in V\iff F(\,(a,b)\,)\in U\,).$$ So $V$ has exactly the same number of members that $U$ has.
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Evaluation of a telescoping sum I have come to a problem in a book on elementary mathematics that I don't understand the solution to. The problem has two parts :
a.) Factorize the expression $x^{4} + x^{2} + 1$
b.) Compute the value of the sum $\sum_{k=1}^{n} \frac{k}{k^{4} + k^{2} + 1}$ in terms
of $n \in \mathbb{N}$.
I was able to perform part (a.) to get :
\begin{equation}
x^{4} + x^{2} + 1 = (x^{2} + x + 1)(x^{2} - x + 1)
\end{equation}
I wasn't able to do part (b.), but in the answer key the first part of the solution is :
\begin{align}
\frac{k}{k^{4} + k^{2} + 1} & = \frac{k}{(k^{2} + k + 1)(k^{2} - k + 1)} \\
& = \frac{1}{2} \left( \frac{1}{k^{2} -k + 1} - \frac{1}{k^{2}+k+1} \right)
\end{align}
I can see that the first transformation above comes from the answer to part (a.). I do not understand how they did the second transformation. Could someone show me how to go from the second expression to the third above ?
| We can start from
$$ \frac{A}{k^{2} -k + 1} + \frac{B}{k^{2}+k+1}=\frac{A(k^{2}+k+1)+B(k^{2} -k + 1)}{(k^{2} -k + 1)(k^{2}+k+1)} = \frac{k}{(k^{2} + k + 1)(k^{2} - k + 1)} $$
from which we obtain $A=-B=\frac12$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3873819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
An involute of a shortest path in a strictly convex surface in $\mathbb{R}^3$. Consider a smooth closed surface $\Sigma$ in $\mathbb{R}^3$ of positive Gaussian curvature, homeomorphic to a sphere. Here $U$ is the interior of the convex hull of $\Sigma$. When $c:[0,l]\rightarrow \Sigma$ is a minimizing geodesic of unit speed in it, then we defines $$a :[0,l]\rightarrow \mathbb{R}^3,\ a(t)=c(t)-t c'(t)$$ and $$f(t)=d(x,a(t))$$ for any fixed $x\in \Sigma$. Here $d$ is the intrinsic distance on $ \mathbb{R}^3- U$.
Problem : Then prove that $f(t)$ is increasing.
Thank you in advance.
| Edit: the current answer only shows that the distance function on $\mathbb{R}^3$ is increasing. The OP wants to show that the distance on $\mathbb{R}^3 \setminus U$ is increasing.
Let $g(t)$ be the squared distance $\langle x - a(t), x-a(t) \rangle$. Its derivative is
$$
\begin{align*}
g'(t) &= 2 \langle x - c(t) + tc'(t), - c'(t) + c'(t) + tc''(t) \rangle \\
&= 2 t\langle x - c(t) + tc'(t), c''(t) \rangle \\
&= 2 t \langle x - c(t), c''(t) \rangle.
\end{align*}
$$
In the last step we used that $\langle c'(t), c''(t) \rangle = 0$ since $c$ has unit speed.
Since $c(t)$ is a geodesic, the acceleration vector $c''(t)$ is parallel with the surface normal of $\Sigma$ at $c(t)$. So the tangent plane at $c(t)$ is given by $\langle p - c(t), c''(t)\rangle = 0$. Also note that $c''(t)$ is an inward pointing normal vector at $c(t)$.
Finally, by the convexity of $\Sigma$, all points of $\Sigma$ lie on one part the tangent plane. Since $c''(t)$ is inward pointing, $\langle x - c(t), c''(t)\rangle\geq 0 $ and consequently
$$
g'(t) = 2 t \langle x - c(t), c''(t) \rangle \geq 0.
$$
This proves that the Euclidean distance function is increasing.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3873974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
How to prove $F_{n+1}^2 - F_{n-1}^2 = F_{2n} $ by combinatorial-proofs? I'm working on a problem that asks us to prove for any n $\geq$ 1, $F_n$ the Fibonacci number satisfy $F_{n+1}^2 - F_{n-1}^2 = F_{2n}$.
I learned Fibonacci identity $F_{2n} = F_{1}+F_{3}+F_{5}+...+F_{2n-1}$ by combinatorial-proofs but I don't know how to deal with the LHS.
Any help would be appreciated.
| Let
$$\mathcal{F} =
\begin{pmatrix}
1 & 1\\
1 & 0
\end{pmatrix}
$$
and
$$\mathcal{F}_n =
\begin{pmatrix}
F_{n+1}\\
F_n
\end{pmatrix}
$$
we have
$$\mathcal{F}_{n+1} = \mathcal{F}\mathcal{F}_n$$
so
$$\mathcal{F}^n =
\begin{pmatrix}
F_{n+1} & F_n\\
F_n & F_{n-1}
\end{pmatrix}
$$
now write $\mathcal{F}^{2n} = \mathcal{F}^n \mathcal{F}^n$
we deduce among other identities
\begin{align}
F_{2n} &= F_nF_{n+1} + F_nF_{n_1}\\
&= F_n\left(F_{n+1} + F_{n-1}\right)\\
&= \left(F_{n+1} - F_{n-1}\right)\left(F_{n+1} + F_{n-1}\right)\\
&= F_{n+1}^2 - F_{n-1}^2
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3874156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Good upper bound on $f(n)$ This is an elementary question. I'm trying to understand the function (a discrete),$$f(n) = \left(\displaystyle \dfrac{\sum_{k=0}^n \frac{a^k}{k!}}{\sum_{k=0}^{n-1}\frac{a^k}{k!}}\right)^n\,$$ where $a>0$ is a constant.
I used WolframAlpha to check the behavior of a graph of a function similar to this and it turns out it's a bell curve (like the Binomial Distribution), which means that this's neither monotonically increasing nor decreasing.So, there must be a local maximum but I don't think there's a proper way to figure out the exact maximum value. Anyways, I'm interested in a good upper bound on $f(n)$.
Also, for a given $a>0$, I noticed that $f(n)\to 1$ as $n\to \infty.$ It can be proved that $f(n)^{1/n} \to 1$ as $n\to \infty$ but not sure how to show that for $f(n)$.
Any help would be apreciated.
| Let $g(n):=\sum_{k=0}^n\frac{a^k}{k!}$ so that $g(n)=g(n-1)+\frac{a^n}{n!}$ and $\lim_{n\to\infty}g(n)=e^a$. Then
$$f(n)=\left(\frac{g(n)}{g(n-1)}\right)^n
=\left(\frac{g(n-1)+\tfrac{a^n}{n!}}{g(n-1)}\right)^n
=\left(1+\frac{a^n}{n!g(n-1)}\right)^n.$$
This shows that for $h(n)=\frac{a^n}{(n-1)!g(n-1)}$ we have
$$\lim_{n\to\infty}f(n)
=\lim_{n\to\infty}\left(1+\frac{h(n)}{n}\right)^n
=\lim_{n\to\infty}e^{h(n)}=1.$$
because $\lim_{n\to\infty}h(n)=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3874307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to calculate the arc length of a difficult radical function I have been struggling with an arc length question, and I want to make sure I get this right. I have the function of:
\begin{align}
f(x) = \sqrt{7.2 (x-\frac {1}{7}}) - 2.023, [0.213, 0.127].
\end{align}
I have found the derivative of the function and set up my integral this way:
\begin{align}
I &= \int_{0.127}^{0.213} \sqrt{1 + \frac{12.96}{7.2x-\frac{7.2}{7}}}~dx
\end{align}
Letting A = 12.96 and simplifying:
\begin{align}
I &= \int_{0.127}^{0.213} \sqrt{\frac{7.2x-\frac{7.2}{7}+A}{7.2x-\frac{7.2}{7}}}~dx
\end{align}
$u=7.2x-\frac{7.2}{7}, du= 7dx, dx=\frac{du}{7}$:
\begin{align}
I &= \int_a^b \sqrt{\frac{{u}+A}{u}}~\frac{du}{7}
\end{align}
\begin{align}
I &= \frac{1}{7}\int_a^b \sqrt{\frac{{u}+A}{u}}~du
\end{align}
$u = C\tan^2v\\ du = 2C \tan v \sec^2 v ~ dv$
\begin{align}
I
&= \frac{1}{7}\int_{x=a}^{x=b} \sqrt{\frac{u + A}{u}}~du\\
&= \frac{1}{7}\int_{x=a}^{x=b} \sqrt{\frac{A(\tan^2 v + 1)}{A \tan^2 v}}~2A\tan v \sec^2 v ~ dv\\
&= \frac{2A}{7}\int_{x=a}^{x=b} \sqrt{\frac{\sec^2 v}{\tan^2 v}}\tan v \sec^2 v ~ dv; & \\
&= \frac{2A}{7}\int_{x=a}^{x=b} \sqrt{\frac{\frac{1}{\cos^2 v}}{\frac{\sin^2 v}{\cos^2 v}}}\frac{\sin v}{\cos^3 v}~ dv; \\
&= \frac{2A}{7}\int_{x=a}^{x=b} \sqrt{\frac{1}{\sin^2 v}}\frac{\sin v}{\cos^3 v}~ dv \\
&= \frac{2A}{7}\int_{x=a}^{x=b} \frac{1}{\sin v}\frac{\sin v}{\cos^3 v}~ dv \\
&= \frac{2A}{7}\int_{x=a}^{x=b} \frac{1}{\cos^3 v}~ dv \\
&= \frac{2A}{7}\int_{x=a}^{x=b} \frac{\cos v}{\cos^4 v}~ dv \\
&= \frac{2A}{7}\int_{x=a}^{x=b} \frac{\cos v}{(1-\sin^2(v))^2} dv~
\end{align}
This is where I am stuck. Could I make a substitution such as:
$t = \sin v\\dt=\cos v\ dt\\\frac{dt}{cos\ v}=dv$
and then:
\begin{align}
&= \frac{2A}{7}\int_{x=a}^{x=b} \frac{\cos v}{(1-t^2)^2} \frac{dt}{cos\ v}~\\
&= \frac{2A}{7}\int_{x=a}^{x=b} \frac{1}{(1-t^2)^2} dt~
\end{align}
which gives me an ordinary partial fractions integral.
Could I make this substitution or is it not possible because I would have 2 different variables in my integral, and if it's not possible, how else could I solve this integral?
| I actually found the answer to my question.
\begin{align}
& I=\frac{2A}{7}\int_{x=a}^{x=b} \frac{1}{\cos^3 v}~ dv \\
& =\frac{2A}{7}\int_{x=a}^{x=b} sec^3v~ dv \\
& =\frac{A}{7}\int_{x=a}^{x=b} \sec v \tan v+ln \lvert\sec v + \tan v\rvert~ dv \\
\end{align}
The new bounds would be:
$arctan\frac{\sqrt{7.2*1.27-\frac{7.2}{7}}}{\sqrt{A}}$ and
$arctan\frac{\sqrt{7.2*0.213-\frac{7.2}{7}}}{\sqrt{A}}$
It was much simpler than I thought.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3874485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
A certain small town, whose population consists of 100 families, has 30 families .... A certain small town, whose population consists of 100 families, has 30 families with 1 child, 50 families with 2 children, and 20 families with 3 children. The birth rank of one these children is 1 if the child is the firstborn, 2 if the child is the secondborn, and 3 if the child is the thirdborn.
a) A random family is chosen (with equal probabilities), and then a random child within that family is chosen (with equal probabilities). Find the PMF, mean, and the variance of the child’s birth rank.
b) A random child is chosen in the town (with equal probabilities). Find the PMF, mean, and variance of the child’s birth rank.
*
*For part A I got $P(X = 1) = 37/60, P(X = 2) = 19/60, P(X = 3) = 4/60$, $E(X) = 1(37/60) + 2(19/60) + 3(4/60) = 1.45$, and $\operatorname{Var}(x) = 149/60 - (1.45)^2= 457/1200$
*For part B I got $E(x) = 1(100/190) + 2(70/190) + 3(20/190) = 1.579\dots$ and $\operatorname{Var}(x) = 2.947 -(1.579)^2= 0.454\dots$
*However, I feel like part B is off but I am not sure where, it just seems incorrect. Any help would be greatly appreciated!
| No, your calculations are all correct. That's just how the numbers fall here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3874639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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For any prime p>5 proving the existence of consecutive quadratic residues and consecutive quadratic non residues This question was asked in an assignment which I am solving and I couldn't solve it.
Prove that for any prime $p>5$ there exist integers $1\leq a,b \leq p-1$ for which $\binom{a}{p}=\binom{a+1}{p}=1$ and $\binom{b}{p}=\binom{b+1}{p} =-1$.
Quadratic residues and non-residues are equal in number , if I prove that any 2 residues are consecutive then non-residues will be automatically proved consecutive.
But I am unable to prove and 2 residues to be consecutive. For theory, I am studying David M Burton's book.
Can you please help?
| Here is a rather interesting solution involving a bit of combinatorical tricks:
We wil start with some basic theory:
Lemma $(1)$:
For any odd prime $p$, there are exactly $\frac{p-1}{2}$ quadratic residues (greater than $0$) and exactly $\frac{p-1}{2}$ non-quadratic residues.
Lemma $(2)$:
For any integer $n$, $n^2$ is a quadratic residue$\pmod{p}$. in other words, $\big(\frac{n^2}{p}\big)=1$ (I used Legendre's symbol here)
Lets now start the proof!
Consider the residues$\pmod{p}$: $(1,2,...,p-1)$. Assign $1$ to a residue $k$ if it is a quadratic residue, and $0$ if it is not a quadratic residue. We will now get a new sequence of $1$s and $0$s out of the originl sequence.
As an example, for $p=7$ we would get $(1,2,3,4,5,6)\mapsto(1,1,0,1,0,0)$.
Observe that to prove your question, we must show that for any prime $p\geq 5$, in this sequence, there are $2$ consecutive $1$s and $2$ consecutive $0$s (at least).
Because we have a sequence of $p-1$ $1$s and $0$s and there are exactly $\frac{p-1}{2}$ $1$s and $\frac{p-1}{2}$ $0$s $($this is an implication of lemma $(1)$, because there are $\frac{p-1}{2}$ quadratic residues and $\frac{p-1}{2}$ non-quadratic residues$)$ we can easily see that if there are 2 consecutive $0$s, then there must be $2$ consecutive $1$s too.
Suppose for this prime $p$ there are no $2$ consecutive $1$s or $0$s. Then, using lemma $(2)$, because $1$ is a quadratic residue, our sequence should look like this: $(1,0,1,0,......,0)$, in other words all odd numbers from $1$ to $p-1$ are quadratic residues and all even numbers from $1$ to $p-1$ are non-quadratid residues. But using lemma $(2)$ again, we know $4$ is a quadratic residue, thus leding in a contradiction for $p\geq 5$ $(p$ must be $\geq 5$ so $4\leq p-1$).
So for $p\geq 5$ we will always find $2$ consecutive $1$s and $2$ consecutive $0$s.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3875005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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The proof of $\frac{f(a)+f(b)}{2}\le\frac{1}{b-a}\int^b_af(x)dx$ provided that $f''(x)<0,x \in [a,b]$ and $f\in \mathcal C^2[a,b]$ This question is geometrically obvious, but I got a little trouble when proving it.
Let's consider the Taylor expansion of $F(x)=\int^x_af(x)\mathrm dx$ at $a$ and $b$ and subsitute $b,a$ into it respectively.
$$\int_a^bf(x)\mathrm dx=F_a(b)=f(a)(b-a)+f'(a)(b-a)^2/2+f''(\xi_1)(b-a)^3/6\\0=F_b(a)=\int^b_af(x)\mathrm dx+f(b)(a-b)+f'(b)(a-b)^2/2+f''(\xi_2)(a-b)^3/6$$
And subtract the second from the first
$$\int_a^bf(x)\mathrm dx=F_a(b)-F_b(a)=-\int_a^bf(x)\mathrm dx+(b-a)(f(a)-f(b))+(f'(a)-f'(b))(b-a)^2/2+(f''(\xi_1)+f''(\xi_2))(b-a)^3/6$$
And divide both side by 2(b-a) and subsitute $f'(a)-f'(b)=-f(\xi_3)(b-a)$ into it
$$\frac{1}{b-a}\int_a^bf(x)\mathrm dx=\frac{f(a)-f(b)}2-f''(\xi_3)(b-a)^2/4+(f''(\xi_1)+f''(\xi_2))(b-a)^2/12$$
since $f''(\xi_3)<0$ we have
$$\frac{1}{b-a}\int_a^bf(x)\mathrm dx\ge\frac{f(a)-f(b)}2+(f''(\xi_1)+f''(\xi_2))(b-a)^2/12$$
I guess my approximation is too rough so I get an additional term $(f''(\xi_1)+f''(\xi_2))(b-a)^2/12$.
Maybe the glitch is that I use $f''(x)<0$ just at one point i.e. $\xi_3$ but actually it holds at every point in $[a,b]$ which is a decisive condition.
| Since $f''(x)<0$, the function $f$ is concave, i.e. for all $t\in [0,1]$, \begin{align*}
f(a)+t\big(f(b)-f(a)\big)&=tf(b)+(1-t)f(a)\\
&\leq f\big(tb+(1-t)a\big).
\end{align*}
Integrating both sides yields
$$\frac{f(a)+f(b)}{2}\leq \int_0^1 f\big(tb+(1-t)a\big)\,\mathrm d t=\frac{1}{b-a}\int_a^b f(x)\,\mathrm d x,$$
as wished.
Remark that the result hold for all concave functions, no need to be $\mathcal C^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3875161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Solve equation $log_{\frac{1}{2}}\left | x \right |=\frac{1}{4}\left ( \left | x-2 \right |+\left | x+2 \right | \right )$ Solve equation $log_{\frac{1}{2}}\left | x \right |=\frac{1}{4}\left ( \left | x-2 \right |+\left | x+2 \right | \right )$
I tried solving 4 separate cases
1.) $x< -2$
2.) $x\in[ \,-2,0\rangle$
3.) $x\in[ \,0,2\rangle$
4.) $x\geqslant2$
But I dont know how to solve this
in first case I get
$2^{\frac{1}{2}x}=-x$
And I dont know how to solve this.
| Since $x < -2$, we see that $2^{\frac12 x} < 2^0 = 1$ while $-x > 2$, so the equation is never satisfied. The case for $x \ge 2$ is very similar.
As to case 2, that is, $-2 \le x < 0$, we see that
$$\frac 14(|x-2|+|x+2|) = \frac14(2-x+x+2) = 1$$
and $\log_{\frac12}|x| = 1$ is easy to solve. Same goes for case 3: $0 \le x < 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3875360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to find the limit of $x_n$ if $x_1=1; x_{n+1}=2+\frac {3}{x_n}+\frac{1}{n}$. How to find the limit of $x_n$ if $x_1=1; x_{n+1}=2+\frac {3}{x_n}+\frac{1}{n}$.
My idea is to limit $x_n$ from both sides with something, and prove that these sequences' limits are the same.
So, I can say that:
$y_{n+1} = 2 + \frac {3}{x_n} \leq 2+\frac {3}{x_n}+\frac{1}{n} = x_{n+1}$. Аnd its limit is 3.
But I cannot find any sequence $z_n$ whose values satisfy the inequality $x_n \leq z_n$ and $\lim { z_n } = 3$.
Can someone help me with this task, please. Maybe exist any other solution of this task. I have no more ideas so that's why I'm here.
| Hint: Set up a double sided inequality to bound $x_{n+1}$ for $n$ large enough. This can be proved via induction.
$ 3 < x_n < 3 + \frac{1}{n-1}$.
Hence conclude that $\lim x_n = 3 $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3875558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Proving there exists an $n \times m$ matrix $S$ such that $ST=I_n$, where $I_n$ is the $n×n$ identity matrix I'm learning linear algebra and came across this problem which I am having a difficult time trying to solve: The problem is:
Suppose $T$ is a $m\times n$ matrix with linearly independent columns. Prove there exists an $n \times m$ matrix $S$ such that $ST=I_n$, where $I_n$ is the $n×n$ identity matrix.
I was thinking of starting the proof with something like supposing that $T{\bf x}={\bf z}$ and if $S$ were such that $ST=I_n$, then finding $S{\bf z}$ but I'm not sure if that helps to prove the question. Do I need to find the number of solutions to $T{\bf x}={\bf z}$?
Any help would be appreciated!
| Hint:
Interpret $T$ as the matrix of a linear map $f:K^n\longrightarrow K^m$ ($K$ is the base field). The hypothesis implies $f$ is injective. Prove an injective linear map has a linear retraction, i.e. a linear map $r:K^m\longrightarrow K^n$ such that $r\circ f=\operatorname{id}_{K^n}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3875669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Closed formula to $N:=\sum_{j=0}^{k/2}\left(\begin{array}{c} n \\ k-j \end{array}\right)\left(\begin{array}{c} k-j \\ j \end{array}\right) $ I gave mysel the following
Problem: For $k$ even and $n\geq k$, can one find a closed formula for the sum
$$
N:=\sum_{j=0}^{k/2}\left(\begin{array}{c}
n \\
k-j
\end{array}\right)\left(\begin{array}{c}
k-j \\
j
\end{array}\right)?
$$
At first, I thought that some sort of Vandermonde's identity could solve this. Now I believe this is far from trivial, if possible at all. I want to make sure I am not mistaken.
| $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\on}[1]{\operatorname{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
Note that $\ds{{k - j \choose j} = 0}$ when
$\ds{j > {k \over 2}}$: The sum can be performed
$\ds{\mbox{" up to $\ds{+\infty"}$}}$:
\begin{align}
{\cal N}\pars{k,n} & \equiv \bbox[5px,#ffd]{%
\sum_{j = 0}^{k/2}{n \choose k - j}{k - j \choose j}}
\\[5mm] & =
\sum_{j = -\infty}^{0}{n \choose k + j}
{k + j \choose -j}
\\[5mm] & =
\sum_{j = -\infty}^{k}{n \choose j}
{j \choose k - j} =
\sum_{j = 0}^{\infty}{n \choose j}{j \choose k - j}
\\[5mm] & =
\sum_{j = 0}^{\infty}{n \choose j}\bracks{z^{k - j}}
\pars{1 + z}^{j}
\\[5mm] & =
\bracks{z^{k}}\sum_{j = 0}^{\infty}{n \choose j}
\bracks{z\pars{1 + z}}^{j}
\\[5mm] & =
\bracks{z^{k}}\pars{1 + z + z^{2}}^{n}
\\[5mm] & =
\bracks{z^{k}}{1 \over
\bracks{1 -2\pars{\color{red}{-1/2}}z + z^{2}}
^{\color{red}{-n}}}
\\[5mm] & =
\sum_{j = 0}^{\infty}\on{C}_{j}^{\color{red}{-n}}\,
\pars{\color{red}{-{1 \over 2}}}z^{j}
\end{align}
$\ds{C_{j}^{\pars{\alpha}}}$ is a
Gegenbauer Polynomial.
$$
\implies \bbx{{\cal N}\pars{k,n} =
C_{k}^{\pars{\color{red}{-n}}}\,
\pars{\color{red}{-{1 \over 2}}}} \\
$$
*
*$\ds{{\cal N}\pars{k,n} = 0}$ when $\ds{k \geq 2n + 1}$.
*$\ds{{\cal N}\pars{k,n} =
{\cal N}\pars{k,n - k}}$.
Plot of $\ds{{\cal N}\pars{k,10}}$:
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3875978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Suppose $M⊆\mathbb{R}^n$ is connected and $M⊆N⊆\overline{M}$. Prove that N is connected. I have gotten to a point where I need to show there exists an element in $M\cap V$$\neq$$\emptyset$ and $M\cap U$$\neq$$\emptyset$, where V and U are relatively open sets in $R^n$. A friend helped me earlier and said $\overline{M}=\partial{M}\cup M$. This fact hasn't been defined in the book as of yet though, so I need help going in another direction.
| If $\overline{M}$ is disconnected, say $\overline{M} \subseteq U\cup V$ where $U, V$ are disjoint open sets with $\overline{M}\cap U, \overline{M}\cap V$ non empty, then $M\subseteq U\cup V$. Because $M$ is connected, it must be contained in $U$ or $V$, say $M\subseteq U$.
Then, $V\cap \overline{M} = \emptyset$ because every point in $V$ has a neighbourhood, namely $V$, that is disjoint from $M$, and this is a contradiction.
Hence $\overline{M}$, and similarly any $M\subseteq N\subseteq\overline{M}$, is connected.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3876179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Which is larger $(101!)^{100}$ or $(100!)^{101}$ I am supposed to tell which one of $(101!)^{100}$ and $(100!)^{101}$ is larger. I am trying to use the behavior of the function $f(x)=x^{1/x}$ as is a standard technique to dealing with questions of this sort. Here is what I have so far.
$$\begin{aligned}(101!)^{100!}&\lt (100!)^{101!}\\ (101!)^{100} &\lt (100!)^{101\times 100}\end{aligned}$$
Any ideas on how to proceed. Thanks.
| Well
$$(101!)^{100}\cdot (101!)=\color{green}{(101!)^{101}}$$
while:
$$(100!)^{101}\cdot (101)^{101}=\color{green}{(101!)^{101}}$$
the one you have to multiply by the larger number is smaller
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3876294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How many times does p appears in this sequence I'm trying to understand the proof for this theorem, page 233. One of the steps is to prove that every term in this expression:
$$
\frac{p^\alpha}{j} \binom{p^\alpha}{j-1} k^{j-1}p^{(j-1)\beta}, \;\;\; (j>1)
$$
is divisible by $p^\alpha$. As $p^\alpha$ appears in the numerator and $\binom{p^\alpha}{j-1}$ and $ k^{j-1}p^{(j-1)\beta}$ are integers, $j$ is the only factor which can "take" some part of $p^\alpha$. Now we have to prove that $p^{(j-1)\beta}$ divides $j$. To do that, it says that the number of times the factor $p$ can appear in $j\in\{2, \ldots, p^\alpha\}$, is:
$$
\frac{j}{p}+\frac{j}{p^2}+\frac{j}{p^3}+\cdots=\frac{j}{p-1}
$$
But the factor $p$ appears at least this many times in $p^{(j-1)\beta}$, since $\beta\geq1$.
Those are the two statements I cannot fully understand. How has he counted that? I do not know where that formula comes from. Any help?
| This is a bound of the Legendre's formula i.e.,
$$\nu _p(n!)=\sum _{k=1}^{\infty}\left\lfloor \frac{n}{p^k}\right \rfloor =\frac{n-s_p(n)}{p-1},$$
where $s_p(n)$ is the sum of the digits of $n$ in base $p.$
The bound is by dropping the ceiling function i.e., $\lfloor x\rfloor \leq x$ and geometric series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3876462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why is $\sum_{t=0}^{n-1} \exp \{ -t \Delta^2 \} \geq \frac{1-e^{-1}}{\Delta^2}$ for $n \geq \frac{1}{\Delta^2}$? This is a follow-up question on this question. I was reading a paper about lower bounds for bandit problems (https://arxiv.org/abs/1302.1611). In Theorem 5, they prove a lower bound with an example problem with two arms. In the comments/answers of the previous question on stackexchange, it turned out that the theorem in the paper contains a few errors: the lower bound in Thrm. 5 should be $\frac{1-e^{-1}}{4\Delta}$, is should only hold for $n \geq 1/\Delta^2$, and the sum should start at 0.
I still don't understand the following step in the corrected proof:
$\sum_{t=0}^{n-1} \exp \{ -t \Delta^2 \} \geq \frac{1-e^{-1}}{\Delta^2}$ for $n \geq 1 / \Delta^2$.
I've tried to use:
*
*Jensen's inequality,
*Taylor expansion,
*infinite sums,
all leading to an upper bound in stead of a lower bound.
| Use$$\sum_{t=0}^{n-1}\exp(-t\Delta^2)=\frac{1-e^{-n\Delta^2}}{1-e^{-\Delta^2}}\ge\frac{1-e^{-1}}{1-e^{-\Delta^2}}\ge\frac{1-e^{-1}}{\Delta^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3876610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to come up with an example of a triangulation with two adjacent vertices of odd degree? We define a triangulation as a planar graph with all faces(including the outer one) as triangles. What is an example of a triangulation with exactly two vertices of odd degree that are adjacent to each other? The number of vertices of even degree is unimportant.
| Vertices: A B C D E F G H I J K L
Edges: AB AC AD AE AL BE BF BG BL CD CH CL DE DH EF EH EI EJ EK FG FK GK GL HI HL IJ IL JK JL KL
The vertices A B H K are of odd degree, but A and B are the only vertices of odd degree that are adjacent.
If you allow only two vertices of odd degree in the whole graph, there is no such graph where they are adjacent.
Proof:
Let's assume the vertices A B C D are part of the smallest possible graph where A and B are the only vertices of odd degree. We remove vertex A and the edges AB AC, then we triangulate the graph by replacing the edge AD by the remaining even number of edges. So A disappeares, B turns even, C and D turn odd and all other verices' grades remain the same. So we have again exactly two adjacent vertices C and D in a smaller graph, which contradicts our assumption that our original graph were the smallest one.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3876734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Significance of Rayleigh quotient of a non-hermitian matrix? I've got a non-hermitian, but diagonalizable matrix $A$ (with respect to a matrix $K$); all its eigenvalues are real, but they need not be simple.
At the same time, the quadratic form associated with $A$ is positive-valued and it dominates the quadratic form of another, Hermitian matrix $B$. Can one conclude that the eigenvalues of $A$ are greater/equal than the eigenvalues of $B$? (Actually, I am only interested in the lowest eigenvalue.) Can I enforce this behavior by a suitable assumption on the diagonalizing matrix $K$?
| By saying the quadratic form associated with $A$ dominates the quadratic form of $B$, I presume you mean $v^* A v \ge v^* B v$ for all vectors $v$.
If $v$ is an eigenvector for $A$ with eigenvalue $\lambda$, normalized so $\|v\|=1$, then
$v^* B v \le v^* A v = \lambda$. That implies the lowest eigenvalue of $B$ is at most $\lambda$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3876891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Struggling with basic limit problem As part of some other proof, I'm trying to show the following:
Let $A \subset \mathbb{R}^n$ and $a \in \overline{A}-A$ (ie. in the closure of A but not in A) so that the function $f : A \to \mathbb{R}$ defined by $f(x) = \frac{1}{|x-a|}$ is well-defined on $A$. I want to show that the function is continuous everywhere on $A$. Now I went through the usual rigmarole which I'll outline below, but essentially I
am struggling to find a way to bound the $\frac{1}{|y-a|}$ that you'll see in the attempted proof below. I seem to have forgotten how to deal with cases like this from single-variable calculus.
Attempt:
Fixing any $x \in A$ and given any $\epsilon$, we want to be able to bound, for any other $y \in A$ (subject to that $y$ being in the ball defined by the $\delta$ we're trying to determine),
$$|f(y)-f(x)|=\left|\frac{1}{|y-a|}-\frac{1}{|x-a|}\right|=$$
$$\frac{1}{|x-a||y-a|}\left||y-a|-|x-a|\right|\leq\frac{1}{|x-a||y-a|}|y-x|$$
so that we have it in the nice form where if we require $\delta$ to be the $min(\epsilon|x-a|, B)$, where $B$ is that something I'm missing to put a bound on the $\frac{1}{|y-a|}$, then we are done. Can someone help point out what I need?
Roughly, it seems clear to me that I need to show $|y-a|>C$ for some $C$ that depends only on $x, \epsilon, a$.
| For any $\epsilon\gt0$, and $x\in A$, let $\delta=\min\left(\frac{|x-a|}2,\frac{|x-a|^2}2\epsilon\right)$. If $|y-x|\le\delta$, then
$$
\begin{align}
|y-a|
&\ge|x-a|-|y-x|\tag1\\[3pt]
&\ge|x-a|-\frac{|x-a|}2\tag2\\
&=\frac{|x-a|}2\tag3
\end{align}
$$
Explanation:
$(1)$: triangle inequality
$(2)$: $|y-x|\le\delta\le\frac{|x-a|}2$
$(3)$: arithmetic
Therefore,
$$
\begin{align}
\left|\frac1{|x-a|}-\frac1{|y-a|}\right|
&=\frac{|\,|y-a|-|x-a|\,|}{|x-a||y-a|}\tag4\\
&\le\frac{|y-x|}{|x-a|\frac{|x-a|}2}\tag5\\
&\le\epsilon\tag6
\end{align}
$$
Explanation:
$(4)$: arithmetic
$(5)$: triangle inequality and $(3)$
$(6)$: $|y-x|\le\delta\le\frac{|x-a|^2}2\epsilon$
This says that $\frac1{|x-a|}$ is continuous at any $x\in A$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3877055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Construct an isomorphism between $V$ and $\Bbb F^2$ and justify Consider the subspace of $M_2(\Bbb F):V = \biggl\{\begin{pmatrix}a+b&a\\0&b\end{pmatrix}\Biggm\vert a,b\in \Bbb F\biggr\}$
Construct an isomorphism between $V$ and $\Bbb F^2$ and justify
| I assume $F$ is a general field and $M_2(F)$ is the set og $2 \times 2$ matrices under that field. An isomorphism $T: F^2 \rightarrow V$ could simply be the function that maps a point $(a,b) \in F^2$ to the associated matrix in $V$, i.e.
$$T(a,b) = \begin{bmatrix}
a+b & a \\
0 & b \end{bmatrix}.$$
It is clear that $T$ is a linear tranformation: Let $(a,b), (\alpha, \beta) \in F^2$ and $x, y \in F$. We have
$$T(x(a,b)+y(\alpha, \beta)) = \begin{bmatrix}
xa + y\alpha + xb + y\beta & xa+ y\alpha \\
0 & xb + y\beta \end{bmatrix} = xT(a,b)+yT(\alpha, \beta)$$.
We can also show that $T$ is injective by assuming $T(a,b) = T(\alpha, \beta)$. We get
$$\begin{bmatrix}
a+b & a \\
0 & b \end{bmatrix} = \begin{bmatrix}
\alpha+\beta & \alpha \\
0 & \beta \end{bmatrix} \implies a=\alpha \wedge b=\beta$$
It is also clear that $T$ is surjective by construction. Thus, $T$ is an isomorphism between $V$ and $F^2$.
I may have misinterpreted the question. I hope this could be of some help anyway
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3877213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Combinatorics Urn Problem I have 208 unique marbles:
*
*From this 208, I select 15 marbles without replacement and place them into an urn.
*I do this again with a new urn, 3 more times, but using the original
208 marble set, so I apply replacement after each selection of 15.
*I now have 4 urns of 15 marbles each, with no duplicate marbles within
a particular urn, but possibly duplicates across urns.
What is the probability that there are exactly 14 marbles that appear in more
than 1 urn $?$.
| Following the advice from Quester I simulated the problem with Monte Carlo
import numpy as np
from tqdm import tqdm
import matplotlib.pyplot as plt
count = np.zeros(208)
rng = np.arange(208).astype(int)
urns=np.zeros([4,15]).astype(int)
exactly14 = 0
morethan14 = 0
lessthan14 = 0
history = []
for i in tqdm(range(1000000)):
for j in range(4):
urns[j, :] = np.random.choice(rng, 15, replace=False)
n_unique=len(np.unique(np.ravel(urns)))
n_multi=len(np.ravel(urns))-n_unique
history.append(n_multi)
if n_multi == 14:
exactly14 += 1
if n_multi > 14:
morethan14 += 1
else:
lessthan14+=1
print(exactly14/1000000)
0.000489
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3877305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Determine whether $\sum_{n=1}^\infty (-1)^n \Big(1-\frac{2}{n}\Big)^n$ converges or diverges $$\sum_{n=1}^\infty (-1)^n \Big(1-\frac{2}{n}\Big)^n$$
If these terms were all positive I would have no problem concluding that the series diverges since the limit would be $e^{-2}>0$. However, I don't understand how the divergence test works with $(-1)^n$ added in. Is this still divergent? Both the Ratio and Root tests are inconclusive, and Wolfram Alpha is scratching its head. When I put partial sums above 1000, it spits out imaginary numbers somehow. Yet it seems to stay around 1.
| If it's convergent then $\left | a_{n} \right |\rightarrow 0$
$(1-\frac{2}{n})^{n}=((1+\frac{1}{-2^{-1}n})^{-2^{-1}n})^{2}\rightarrow e^{2}\neq 0$
and therefore isn't convergent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3877478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
If ${\lvert{E(G)}\rvert}>{\lvert{V(G)}\rvert}+3$, graph $G$ has at least two disjoint cycles. Assume that $G$ is a graph which ${\lvert{E(G)}\rvert}>{\lvert{V(G)}\rvert}+3$. Prove $G$ has two disjoint cycles.
Answer:
I tried to prove it when $G$ is connected. consider the smallest cycle of length $k$. It exists because ${\lvert{E(G)}\rvert}>{\lvert{V(G)}\rvert}-1$.
Now, I divided vertices into $2$ groups. Group $A$ consists of $k$ vertices in the smallest cycle and group $B$ consists of the other $n-k$ vertices.
Now, between $n-k$ vertices are exactly $n-k-1$ edges. Because, If it is more then we have another cycle (the proof is done) and if it is less then $G$ is disconnected (contradiction). So there are at least $5$ edges between group $A$ and $B$.
Now, what should I do? should I write every possibility to prove it? Are there any easy ways to continue the proof?
What about if $G$ is disconnected? I think we should do it in each component. Am I right?
Thank you so much for your answers.
| Suppose the result is false; take a minimal counterexample $G$ with $n$ vertices and exactly $n+4$ edges (if there were more edges, we could delete some).
Then $G$ cannot contain a $3$-cycle or $4$-cycle. Otherwise, removing the edges of such a cycle, we have a graph with $n$ vertices and at least $n$ edges, which must still contain another cycle.
Also, $G$ has minimum degree $3$. If $G$ had a vertex $v$ of degree $1$, we could delete it and the edge out of it, getting a smaller counterexample (since the leaf was not part of any cycle). If $G$ had a vertex $v$ of degree $2$, with neighbors $w_1, w_2$, we could replace $v$ and its two edges with the single edge $w_1w_2$. This also gives a smaller counterexample, since any cycle containing $v$ would have had to use both edges, and now it can use the edge $w_1w_2$ instead. Also, because $G$ had no $3$-cycle, this still gives a simple graph.
Since $G$ has minimum degree $3$ and $n$ vertices, it must have at least $\frac 32n$ edges, but we know it has exactly $n+4$. Therefore $n+4 \ge \frac32n$, or $n \le 8$.
However, if we take a vertex $v$, we know it must have at least $3$ neighbors $w_1, w_2, w_3$. These have no edges between them (because $G$ has no $3$-cycles) and each has $2$ more neighbors (by minimum degree $3$) which are all distinct (because $G$ has no $4$-cycles), giving us at least $10$ vertices already. This is a contradiction, so the result must be true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3877654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
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} |
prove AM-GM inequality for $n=k-1$ The question is assume the AM-GM inequality holds for $n$, prove for $n=k-1$.
After a bit of moving, I'm stuck at the following.
$\frac{x_1+...+x_{k-1}}{k-1}\geq \frac{k}{k-1}x_k^{1/k} ((x_1...x_{k-1})^{1/k}-x_k^\frac{k-1}{k})$
What do I do next? Also please elaborate if you are giving hint.
| Hint: let
\begin{eqnarray*}
X=x_k= \frac{x_1+\cdots+x_{k-1}}{k-1}.
\end{eqnarray*}
We have
\begin{eqnarray*}
\frac{x_1+\cdots+x_{k}}{k}= \frac{x_1+\cdots+x_{k-1}+ \frac{x_1+\cdots+x_{k-1}}{k-1}}{k} =\frac{x_1+\cdots+x_{k-1}}{k-1} =X.
\end{eqnarray*}
So
\begin{eqnarray*}
X \geq \sqrt[k] {x_1 \cdots x_{k-1} X}
\end{eqnarray*}
raise this to the power of $k$, divivde through by $X$ & then take the $(k-1)^{th}$ root.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3877817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the general term for sequence ($a_n$) which equates to the recursive equation $a_{n+3}=5a_{n+2}-7a_{n+1}+3a_n+16+24n^2+36*3^n$ Find the general term for sequence ($a_n$) which equates to the recursive equation $a_{n+3}=5a_{n+2}-7a_{n+1}+3a_n+16+24n^2+36*3^n$ with $a_0=3$, $a_1=5$ and $a_2=27$
I tried doing this question by working out how much $a_k$ is for some $a_k$.
$a_3=5*27-7*5+3*3+16+24*9+36*3^3=1383$
As soon as I saw this huge result, I realized that I was going down the wrong path. I then thought that maybe it is a function like $f(x)=Ax^2+Bx+C$ and I tried substituting some values (I know that this is not correct mathematical thought, but I was hoping for some inspiration on what to do, inspiration which unfortunately did not come). All of these routes I attempted did not work out for me. This is the first time I am seeing a question of this type, could you please explain to me how to solve it, how you intuitively thought of each step and also what general thought pattern I should follow in the future when confronted with a question like this?
| You can put it in WA e.g.
the recurrence
See the Recurrence equation solution section there.
Then just use the values for $a_0, a_1, a_2$ to find the constants $c_i$.
Probably there's good amount of theory behind that and that's what WA implemented.
And here is the solution with the constants found.
complete solution
All in all, I don't think this is a problem well-suited for humans, it's hard to guess the solution or find any pattern by just observing. Maybe there's some trick (or theoretical apparatus) to simplify things and solve it in the general case... but unless you know it, you have no chance.
At the end of the day the formula is:
$a(n) = -n (n ((n - 2) n + 9) - 3^{n + 1} + 1) + 3^n + 2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3877962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Has this energy a maximum in $B$? Suppose $\mathcal{H}$ is an Hilbert space and $\langle ., . \rangle$ his scalar product.
Suppose $T:\mathcal{H} \to \mathcal{H}$ is linear, continuos, compact and self-adjoint.
If we define $J:\mathcal{H} \to \mathbb{R}$ as $J(x)=\langle Tx, x \rangle$ for all $x \in \mathcal{H}$, can we conclude $J$ has a maximum in $B=\{x \in \mathcal{H}: \|x\| \leq 1\}$?
I know that if $\mathcal{H}$ is separable I can use the fact that $B$ is sequentially compact and I can work with a maximizing sequence.
But If I do not know the separability of $\mathcal{H}$?
Does the same statement hold?
I tried proving that $J_{|B}:B \to \mathbb{R}$ is weakly superior semi-continuos. In fact if $u_n \rightharpoonup u$ then $J(u_n) \to j(u)$. Is it enough to conclude?
| By the spectral Theorem there exists an orthonormal set $\{x_n\}_{n\in \mathbb N}$, and a sequence
$\{\lambda_n\}_{n\in \mathbb N}$ of real numbers converging to zero, such that
$$
T(x) = \sum_{n\in \mathbb N} \lambda _n\langle x,x_n\rangle x_n, \quad \forall x\in \mathcal H
$$
It follows that
$$
J(x) = \langle T(x),x\rangle = \sum_{n\in \mathbb N} \lambda _n|\langle x,x_n\rangle |^2,
$$
so the supremum of $J$ on $B$ is clearly equal to the maximum of the $\lambda _n$, as long as there exists at least one
positive $\lambda _n$, and zero otherwise.
In the first case $J$ attains its maximum at $x_n$, where $n$ is such that $\lambda _n = \max_k\lambda _k$, and in the second case,
$J$ attains its maximum at zero.
Alternatively you may choose a maximizing sequence $\{x_n\}_n$ and set $K$ to be the Hilbert subspace generated by that sequence. Notice that $K$ is separable, even if the whole space isn't. So you may now employ your argument based on sequencial compactness of the unit ball of $K$!
| {
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"url": "https://math.stackexchange.com/questions/3878088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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cyclic rational inequalities $\frac{1}{a^2+3}+\frac{1}{b^2+3}+\frac{1}{c^2+3}\leq\frac{27}{28}$ when $a+b+c=1$ I've been practicing for high school olympiads and I see a lot of problems set up like this:
let $a,b,c>0$ and $a+b+c=1$. Show that
$$\frac{1}{a^2+3}+\frac{1}{b^2+3}+\frac{1}{c^2+3}\leq\frac{27}{28}$$
Any problem that involves cyclic inequalities like these always stump me. I know I'm supposed to use Cauchy-Schwarz or AM-GM at some point, but I can never get to a place where this might be useful. My first instinct is to get common denominators and hope stuff simplifies, but I can never get farther than that. For example, in this problem I did the following:
$$\frac{1}{a^2+3}+\frac{1}{b^2+3}+\frac{1}{c^2+3}$$
$$=\frac{(a^2+3)(b^2+3)+(b^2+3)(c^2+3)+(c^2+3)(a^2+3)}{(a^2+3)(b^2+3)(c^2+3)}$$
$$=\frac{a^2b^2+b^2c^2+c^2a^2+6(a^2+b^2+c^2)+27}{(a^2+3)(b^2+3)(c^2+3)}$$
but this is where I get stuck. I've tried using Cauchy-Schwarz on parts of this fraction to simplify it, but I can never get anything to work. How could you prove this inequality, and what are the important things to look out for in problems of this nature
| Tangent Line method helps.
Indeed, let $a=\frac{x}{3},$ $b=\frac{y}{3}$ and $c=\frac{z}{3}.$
Thus, $x+y+z=3$ and
$$\frac{27}{28}-\sum_{cyc}\frac{1}{a^2+3}=\sum_{cyc}\left(\frac{9}{28}-\frac{9}{x^2+27}\right)=\frac{9}{28}\sum_{cyc}\frac{x^2-1}{x^2+27}=$$
$$=\frac{9}{28}\sum_{cyc}\left(\frac{x^2-1}{x^2+27}-\frac{1}{14}(x-1)\right)=\frac{9}{392}\sum_{cyc}\frac{(x-1)^2(13-x)}{x^2+27}\geq0.$$
| {
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Negating a statement with an arbitrary variable Consider the statement $P(a)$, where $a$ is an arbitrary element from $D$.
Which of these is its negation?
*
*$\neg P(a)$, where $a$ is an arbitrary element from $D$
*$\neg P(a)$, where $a$ is some element from $D$
|
Consider the statement $P(a)$, where $a$ is an arbitrary element from $D$.
Although we can apply Universal Generalisation to $P(a),$ $P(a)$ itself remains an open formula (i.e., propositional function, i.e., predicate), not a statement.
Which of these is its negation?
*
*$\neg P(a)$, where $a$ is an arbitrary element from $D$
*$\neg P(a)$, where $a$ is some element from $D$
The negation of $P(a)$ is simply $\,¬P(a)\,;$ the denotation of $a$ as an arbitrary element of $D$ is unaffected.
Based on previous answers given, I think I have come to a conclusion:
If I negate the proposition like this "For an arbitrary integer $a$, $\neg P(a)$",
No. Here, you've negated $P(a)$ and applied Universal Generalisation after.
If I negate the proposition like this "$\neg$ (For an arbitrary integer $a$, $P(a)$)",
No. Here, you've derived $\forall a\,P(a)$ and then negated the result.
Now, remember, negation is a logical operation, not an inferential rule. So, there is no reason to expect these four formula from above to be equivalent to one another. Indeed, no pair is equivalent to each other:
*
*$P(a)$
*$\lnot P(a)$
*$\forall\lnot P(a)$
*$\lnot\forall P(a)$
| {
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$\int_R xy^2ds$ where $R$ is the upper half of the circle $x^2+y^2=25$. Is my procedure correct for the following problem?
Calculate
$$\int_R xy^2ds$$
where $R$ is the upper half of the circle $x^2+y^2=25$.
What I did was parametrize the circle which gives $g(t) = (5\cos(t), 5\sin(t))$ and $0 \leq t \leq \pi $. This ends in the integral
$$\int_0^{\pi} (125 \sin^2(t)\cos(t))(5)dt=0$$
Is it correct that it's $0$ or did I make an incorrect parametrization? Thanks.
| It's unclear whether you want the surface integral over the region enclosed or the line integral over the boundary of the semicircle, but you can use a symmetry argument in any case.
Here is how you would do it for the surface integral. Let us split your region into two disjoint parts $A$ and $B$, where $A = \{(x, y) ~|~ x = 0\}$ and $B = \{(x, y) ~|~ x \neq 0\}$. Note that for every point $(x, y)$ with $x \neq 0$ in region $B$, the point $(-x, y)$ also lies in the region. Since the integrand is odd in $x$ (i.e. $xy^2 = -((-x)y^2$), and hence $\int_B xy^2 ds = 0$. In addition, the integrand is zero on every point in $A$, so it follows that $\int_A xy^2 ds = 0$. Hence,
$$\int_R xy^2 ds = \int_A xy^2 ds + \int_B xy^2 ds = 0$$
The symmetry argument for the line integral is similar.
| {
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Find $\sum_{k=1}^n P(z_k)$ where P is a polynomial and $z_k$ the $k$th root of $z^{13}=1$ Question: Given a complex number z such that $z^{13}=1$, find the sum of all possible values of $z+z^3+z^4+z^9+z^{10}+z^{12}$.
I know we have to use roots of unity and try to manipulate the polynomial we want to evaluate, but can't find a pattern.
| As stated in comments, for a fixed $z=e^{i\theta}$ the given expression is equal to $2(\cos\theta+\cos3\theta+\cos4\theta)$. As $z$ runs through its $13$ possible values, so does each term ($1,3,4$ are all coprime to $13$) and the sum over all possibilities is
$$6\sum_{k=0}^{12}\cos\frac{2\pi k}{13}$$
but the summation is well-known to evaluate to zero, so the final answer is $0$.
| {
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Probability of finding the fly inside a range A fly is traversing the non-negative x-axis. It starts at $x_0=k$. At the $i^{th}$ step (starting from the zeroth step), it uniformly randomly jumps to a point in the range $[0,x_i]$.
Probability that the fly is in the range $[a,b]$ after $n$ jumps is denoted by $P_n(a,b)$. Find
$$\lim_{y \to 0}\frac{P_n(1,1+y)}{y}$$
Any help will be appreciated.
EDIT:
Sorry for the late response. Here is my brief try. If we consider the condition that after $n$ jumps, the fly is at a position $> a$, but exclude the condition that it should be $< b$. Let $p=a/k$
\begin{align}
\begin{split}
P_n(a)&=\int\limits_{x1=a}^k ~\int\limits_{x_2=a}^{x_!}....\int\limits_{x_n=a}^{x_{n-1}}\frac{1}{k}\frac{1}{x_1}...\frac{1}{x_{n-1}}\ dx_1 \ dx_2 ... \ dx_n\\
&=1-p+p\sum_{i=1}^{n-1}\frac{(\log p)^i}{i!}
\end{split}
\end{align}
| EDIT-2: Let $p=a/k$ and $q=b/k$
\begin{align}
\begin{split}
P_n(a,b)&=\int\limits_{x1=a}^k ~\int\limits_{x_2=a}^{x_!}....\int\limits_{x_{n-1}=a}^{x_{n-2}}\int\limits_{x_n=a}^{\min (x_{n-1},b)}\frac{1}{k}\frac{1}{x_1}...\frac{1}{x_{n-1}}\ dx_1 \ dx_2 ... \ dx_n \\
&=q-p+\sum_{i=1}^{n-1}\frac{(-1)^i}{i!}\left[q(\log q)^i-p(\log p)^i\right]
\end{split}
\end{align}
Substituting $a=1$ and $b=1+y$, and evaluating the limit,
\begin{align}
\begin{split}
\lim_{y \to 0}\frac{P_n(1,1+y)}{y}=\frac{(-1)^{n+1}}{(n-1)!k}\left(\log \frac{1}{k}\right)^{n-1}
\end{split}
\end{align}
| {
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Factoring $X^5 - X^4 - X^3 + X^2 + X$ into irreducible factors: Is my solution correct? I'm trying to reduce $f(x) = x^5 - x^4 - x^3 + x^2 + x$ in $F_3[X]$ in irreducible factors.
What I came up with was $(x^2 + x + 2)(x^2 - 2x + 2)(x)$
How I did it was I first found all the irreducible polynomials of degree $2$ in $F_3[X]$. Then I did long divisions on $f(x)$ with one of the polynomials I found (brute force), until found a factor. Which was $x^3 - 2x^2 + 2x$. Dividing it by $x$ gives $x^2 - 2x + 2$. And that's how I found the factors.
Is this a correct solution to the question? I'm especially having doubts about the last part of my solution where I just divide it by X. Hopefully this is okay to ask here.
| Your answer is right.
I like the following way: $$x^5-x^4-x^3+x^2+x=x(x^4-x^3-x^2+x+1)=$$
$$=x(x^4+2x^3-x^2-2x+1)=x(x^2+x-1)^2.$$
| {
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2 questions related to generating function of partition function in number theory I am self studying chapter partitions (chapter number-14) from Apostol Introduction to analytic number theory.
I had studied that chapter earlier also and had questions but as I don't have anyone to guide so I couldn't ask anyone about it.
For |x|<1 ,as we have partition function $\prod_{m=1}^{\infty} \frac{1}{1-x^m} =\sum_{n=0}^{\infty} p(n) x^n$ , where p(0)=1.
But then Apostol in table on page 310 writes that generating function for number of partitions of n into parts which are odd is $\prod_{m=1}^{\infty}\frac{1} {1-x^{2m-1}}$ . He doesn't give an explanation and I don't know how to deduce it. Only intutively , I can think of the reasoning that due to odd parts requirements , in the product author is using 2m-1 .
But that can't be said rigorious by any means. Can you please tell how to rigoriously prove it?
Also , in the same table autor writes in number of partitions of n into parts which are unequal the generating function is $\prod_{m=1}^{\infty}(1+x^m)$ . Unfortunately for this part I don't have any intution.
So, Its my humble request can you please provide reasoning behind these 2 cases so that I can understand them . As of now I have no idea on how it works.
| A partition into odd parts
\begin{eqnarray*}
N= \underbrace{1+\cdots+1}_{k_1 ones} + \underbrace{3+\cdots+3}_{k_3 threes}+ \cdots.
\end{eqnarray*}
The generating function
\begin{eqnarray*}
\prod_{n=1}^{\infty} \frac{1}{1-x^{2n-1}}= \left(1+x+\cdots +x^{k_1}+\cdots \right) \left(1+x^3+\cdots +x^{3k_3}+\cdots \right)\cdots
\end{eqnarray*}
For distinct parts, a part is either not used $1$, or the part $p$ is used $x^p$ ... but the part can only be used once .. so $1+x^p$
The generating function
\begin{eqnarray*}
\prod_{p=1}^{\infty} (1+x^p).
\end{eqnarray*}
| {
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Prove that $|\operatorname{Gal}(F(\lambda)/F)|=\left|\frac{\langle a \rangle}{\langle a \rangle \cap H}\right|.$ Question: Let $F$ be a field contains primitive nth root of unity $\zeta$. Define the multiplicative group with order $n$ to be $u_n=\{z\in F : z^n=1\}$. Let $a\in F\setminus\{0\}$ and $\lambda \in \bar F$ such that $\lambda$ is a root of $x^n-a=0$.
Suppose $G=\operatorname{Gal}(F(\lambda)/F)$.
I want to prove
*
*$\phi:G\to u_n$ with $\sigma\mapsto \sigma(\lambda)\lambda^{-1}$ is
an injective homomorphism.
*Suppose $H=\{y^n: y^n\in F^*\}$. Prove that $$|\operatorname{Im}(\phi)|=\left|\frac{\langle a \rangle}{\langle a \rangle \cap H}\right|.$$
Here is my work on 1. : Let $\sigma,\tau \in G.$ $$\phi(\sigma\circ\tau)=\sigma(\tau(\lambda))\lambda^{-1}$$ $$\phi(\sigma)\phi(\tau)=\sigma(\lambda)\lambda^{-1}\cdot\tau(\lambda)\lambda^{-1}.$$
They seems to be not equal... Is the multiplication $\sigma(\lambda)\tau(\lambda)$ just the notation of permutation decomposition?
For checking the injectivity, if $\sigma \in ker(\phi)$, then $\sigma(\lambda)\lambda^{-1}=1$. So $\sigma(\lambda)=\lambda$. Can I say $\sigma$ is just the identity permutation?
For 2.,
By the last part and fundamental homomorphism theorem 1, $\operatorname{Im}(\phi) \cong G$.
Then I divide the problem into 2 cases
*
*$a \neq 1$
*$a=1$.
The case $a=1$ is trivial because it makes $F(\lambda)=F$, then $|G|=[F(\lambda):F]=1$ and also $\left|\frac{\langle a \rangle}{\langle a \rangle \cap H}\right|=1$. But when $a\neq 1, |\operatorname{Im}(\phi)|=[F(\lambda)/F]=n$ but $\left|\frac{\langle a \rangle}{\langle a \rangle \cap H}\right|=\frac{|\langle a \rangle|}{\operatorname{lcm}(|\langle a \rangle|,|H|)}.$ They look very different...
| $G$ is cyclic generated by $g(\lambda)=\zeta^m \lambda$. Take $\sigma=g^a,\tau=g^b$.
For a primitive $n$-th root of unity to exist $F(\lambda)/F$ is separable (thus Galois)
$\lambda$'s $F$ minimal polynomial is $\prod_{d=1}^{|G|} (x-\zeta^{md} \lambda)=x^{|G|} -\lambda^{|G|}$ where $|G|=n/\gcd(n,m)$ must be the least integer such that $\lambda^r\in F$.
| {
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Values of the expression $ax \pmod b$ as a function of $x$ I'm trying to better understand behavior of the expression $(ax\pmod b)$ as a function of $x \in \Bbb N$. Both constant numbers $a \in \Bbb N$ and $b \in \Bbb N$, and also we can assume that the number $a$ is prime. For example, the graph below shows the function $(31 x \mod 12)$ for $x \in [0,11]$.
We can see three monotonically increasing subsequences here, which fit on three straight lines:
*
*$(0,0),(2,2),(4,4),(6,6),(8,8),(10,10)$ - for even $x \in [0,10]$
*$(1,7),(3,9),(5,11)$ - for odd $x \in [1,5]$
*$(7,1),(9,3),(11,5)$ - for odd $x \in [7,11]$
Of course, there are many other straight lines here, but I need to consider a minimal number of such lines. It's clear that this picture depends on values of $a$ and $b$. For example, the expression $(37x \mod 12)$ will give us a single straight line:
Is it any theory, which can predict configuration of these lines in general case, for any $a$ and $b$?
| The graph of $y \equiv 31x \pmod{12} : x \in \mathbb{Z^+},
y \in \{0,1,2, \cdots, 11\}$ can be interpreted as follows.
For $y \in \mathbb{Z^+},$ let $r(y)$ denote the unique value $k$ in
$\{0,1,2, \cdots, 11\}$
such that $y \equiv k\pmod{12}.$
First, since $31 \equiv 7\pmod{12},$
start with the graph of $y = 7x.$
Then, replace each point $(x,y)$ with $(x,r(y))$.
Addendum
Per OP's request.
Consider the general case of $y = ax \pmod{b}.$
The specification of $a =$ a prime number
is covered under the more general specification
that $a$ and $b$ are relatively prime.
Let the set $U \equiv \{0,1,2, \cdots, (b-1)\}.$
For $y \in \mathbb{Z^+},$ let $r(y)$ denote the unique value $k$ in $U$ such that $y \equiv k\pmod{b}.$
Let the set $T \equiv \{r(0), r(1), r(2), \cdots, r(b-1)\}.$
It is well settled, that since
$a$ and $b$ are relatively prime,
the set $T$ is the same as the set $U$, except for the
ordering of the elements.
Further, for $m \in \mathbb{Z^+},~$
let the set
$$T_m \equiv
\{r([b\times m] + 0), r([b\times m] + 1), r([b\times m] + 2), \cdots, r([b\times m] + b-1)\}.$$
Then, the set $T_m$ will be the same as the set $T$, with
the elements appearing in the same order.
To graph the function:
start with the graph of $y = ax,$
then, replace each point $(x,y)$ with $(x,r(y))$.
| {
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Quadrilateral with given angles
We are looking for angles x and y.
I have found the values of the following angles:
BEA = 74,
BDA = 64,
ACD = 68,
ECD = 112,
plus the relationship $x+y = 68$.
All other angles equations, from triangles or the sum of angles in the quadrilateral (360) end up in the same equation!
I have found through Geogebra that $x=18$ and $y=50$ but I can't figure out a second relationship to determine them geometrically!
Does anyone have any ideas?
Thank you!
| Sinx / Sin(x + 64) = Sin38 / Sin84 *Sin22 / Sin48
Multiply both denom and nom by 4Sin82
Sinx /Sin(x + 64) = 4Sin22Sin38Sin82 / (Cos6Sin48Sin82)
(4Sin22Sin38Sin82 = Sin66 REPLACEMENT)
Sinx / Sin(x + 64) = Cos24 / (4Sin48Cos6Sin82)
Sinx / Sin(x + 64) = 1 / (8Sin24Cos6Sin82)
(2Sin24*Cos6 = Sin54 REPLACRMENT)
Sinx / Sin(x + 64) = 1 / (4Sin54*Sin82)
(1/4Sin54 = Sin18 REPLACEMENT)
Sinx / Sin(x + 64) = Sin18 / Sin82
X = 18
Y = 50
| {
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"timestamp": "2023-03-29T00:00:00",
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Complex trig function example. If $\cos z=5$, why does $e^{2iz}-10e^{iz}+1=0$? The example in my book says this:
equation 1 is:
$$\cos{z} = \frac{1}{2} ( e^{iz} + e^{-iz} )$$
Where is $e^{2iz} - 10e^{iz} + 1 = 0 $ coming from? Can someone show me how they are doing this solution?
| You have:
$$
\cos z = \dfrac{1}{2} \left( e^{iz} + e^{-iz} \right) \tag{1}
$$
Multiplying both sides by $ e^{iz} $ as said (and replacing $ \cos z $ with $5$):
$$
5 e^{iz} = \dfrac{e^{iz}}{2} \left( e^{iz} + e^{-iz} \right) \\
\implies 10 e^{iz} = e^{2iz} + 1
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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stationary distribution equation I have a question to solve a stationary distribution equation for a DMC. My equations look like this
$$\pi_0 = (1-p)\pi_1$$
$$\pi_1 = \pi_0 + (1-p)\pi_2$$
$$\pi_i = (1-p)\pi_{i+1} + p\pi_{i-1}, 2\le i\le n-1$$
$$\pi_n = \pi_n+p\pi_{n-1}$$
for $0<p<1$. I know as well $\sum \pi_i = 1$. But how can I solve such a system? I've tried to write down some cases but can't see a particular form for a generalization.
| EDIT
There has been some confusion in the comments regarding the boundary conditions at $n$, so I want to clarify what I think is going on here:
The relations should be viewed as $\pi_i = F_i(\pi_{i-1},\pi_{i-2}) $ that is, each $\pi_i$ is determined as a function of the two lower $\pi$'s. When the equations are written in that way, we see that the value of $\pi_0$ determines the entire solution. Writing relationships as $\pi_i = G_{i}(\pi_{i+1},\pi_{i-1}) $ is confusing and leads to various contradictions when we try to write both an equation for $\pi_{n-1}$ and $\pi_{n}$.
Bottom line: having both special conditions for $i=0$ and $i=n$ and normalization causes the equations to become over determined ( to make an analogy to a second order ode with 2 boundary conditions and a normalization condition). The $\pi_n$ equation is spurious in my opinion.
END OF EDIT
Start from the recursion for $2\le i \le n-1$:
$$(1-p)\pi_{i+1}-\pi_i +p\pi_{i-1}=0$$ and consider solutions of type $a\lambda^{i}$, solve the quadratic equation to get $\lambda_{+},\lambda_{-}$.
Consider the general solution
$$ \pi_i = a\lambda_{+}^{i}+b\lambda_{-}^{i}$$
and set $a,b$ by using the normalization and the boundary conditions.
| {
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Can't figure this out: $dy/dx = \cos(y)\sin(x)$ Problem: Find the solution and domain of validity for the following differential equation:
$\dfrac{dy}{dx} = \cos(y) \cdot \sin(x)$
What I tried: I noticed that this is a separable equation, so I wrote it in the form: $\dfrac{1}{\cos(y)} \cdot dy = \sin(x) \cdot dx$
Next, I found out that if $\cos(y) = 0$ then $\dfrac{1}{\cos(y)}$ is undefined. So I separate into two following cases;
Case1: $\cos(y) \neq 0$;
Then $\int\dfrac{dy}{\cos(y)} = \int\sin(x)dx$. By solving these integrals I get that $\tan(y) \cdot \sec(y)+ c_0 = -\cos(x)$ which implies that
$\arccos(-\tan(y) \cdot \sec(y)+ c) = x$, for $c=-c_0$ which is an arbitrary constant.
I wonder if this is a general solution.
Case2: $\cos(y)=0$;
This implies that $y\in \{ \mathbb{R} -\dfrac{(2k+1)\pi}2:k\in \mathbb{Z}\}$
Pick an arbitrary $y_0\in \{ \mathbb{R} -\dfrac{(2k+1)\pi}2:k\in \mathbb{Z}\}$, substituting this into the differential equation gives that;
$\dfrac {dy_0}{dx} = \cos(y_0)\cdot \sin(x)$.
I am pretty sure that $\cos(y_0) = 0$. But I do not know how can I show that $\dfrac {dy_0}{dx} = 0$ to find out if there is a singular/lost solution or not.
Moreover, I have been taught that $dx$ means derivative, for instance $x^2\cdot dx = 2x$. Then why can't we solve a separable differential equation just by taking derivative of both sides for instance in $\dfrac{1}{\cos(y)} \cdot dy = \sin(x) \cdot dx$
| $$\int \frac{dy}{\cos y}=\int \sin x \,dx=-\cos x +C $$
Then for the integral on the LHS:
$$I=\int \frac{dy}{\cos y}=\int \frac{d \sin y}{\cos^2 y}$$
$$I=\int \frac{d \sin y}{1-\sin^2 y}=\int \frac{d u}{1-u^2}$$
$$I=\dfrac 12 \left (\int \frac{du }{u+1}-\int \frac{du}{u-1}\right )$$
$$ I=\dfrac 12 \ln \left |\dfrac {u+1}{u-1}\right |+C= \dfrac 12\ln \left |\dfrac {(u+1)^2}{u^2-1}\right |+C$$
Where $u =\sin y$.
$$ I= \ln \left |\dfrac {1+\sin y}{\cos y}\right |+C$$
You can also use this result:
$$ I=\dfrac 12 \ln \left |\dfrac {u+1}{u-1}\right |+C=\tanh^{-1} u+C$$
$$ \implies I=\tanh^{-1} ( \sin y)+C$$
So that we have:
$$\tanh^{-1} ( \sin y)=(C-\cos x)$$
$$ y(x)=\arcsin (\tanh(C-\cos x))$$
| {
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Is a closed interval an open set? I read somewhere that a closed interval is not an open set but I don't see why not?
Some definitions in metric space: $(X,d)$
Open Ball: Let $p \in X$ and $r>0$ then $B(p,r) = \{ x \in X : d(x,p) < r\}$
Open set: A $S \subseteq X$ is open set if $\forall p \in S, \exists r>0$ such that $B(p,r) \subseteq S$
Now let's take $X= [a,b]$ and $d(x,y)= |x-y|$ then clearly an open ball can be made for $a<x<b$, for $x= a$, we see $B(a,r) = [a,r) \subseteq X$ if $0<r<b$ and similarly it works for $x=b$, so the closed interval is open set, yeah?
| It depends on what you mean. If $X=[a,b]$ then the open sets (in the subspace topology) are of the form
$[a,b] \cap U$ where $U$ is open (in the usual sense) in $\mathbb{R}$. For example, $(a,b]$ is open.
If $X=\mathbb{R}$, then $[a,b]$ is not open because any open set that contains $b$ also contains points in $(b,\epsilon)$ for any $\epsilon>0$.
| {
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References for linear advection system with constant coefficients I'm searching for references where I can found the study of finite difference schemes for linear advection system with constant coefficients.
More specifically, I would like to read about:
The Symbol, Phase and Amplitude Errors, dissipation and dispersion, and some MATLAB examples
Thanks you a lot for your help
| You might be interested by Chap. 3 of the book [1]:
3 Finite Difference Methods for Hyperbolic Equations
3.1 Introduction
3.2 Some basic difference schemes
3.3 Dissipation and dispersion errors
3.4 Extensions to conservation laws
3.5 The second-order hyperbolic PDEs
3.6 Numerical examples with MATLAB codes
3.7 Bibliographical remarks
3.8 Exercises
Besides the absence of Matlab examples, you might also enjoy Chap. 5 of the monograph [2] which includes many exercises & detailed examples.
[1] J. Li, Y.-T. Chen: Computational Partial Differential Equations Using MATLAB®, 2nd ed., Chapman & Hall/CRC, 2020.
[2] J.C. Strikwerda: Finite Difference Schemes and Partial Differential Equations, 2nd ed., SIAM, 2004.
| {
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How to know if the solution for this angle ϕ is in units of radians or degrees? I am studying out of interest a modelling approximation for a plectrum plucking a string (eg. guitar, piano, harpsichord) as given by these articles:
*
*http://recherche.ircam.fr/pub/dafx11/Papers/24_e.pdf (relevant part on page 3 of pdf)
*https://stacks.stanford.edu/file/druid:wp454hs7976/Jack%20Perng%20Thesis-augmented.pdf (relevant part on page 75 of pdf)
The model works roughly like this:
The string is represented here as the yellow circle in cross-section. The red part on the left is the component the plectrum is attached to that moves (moving upwards along the y-axis here). The black bendy part is obviously the plectrum.
$F_p$ (the force from the plectrum) can be broken down in terms of $x$ and $y$ vector components as per:
Where E is the Young's Modulus, I is moment of inertia, and L' is how far along the length of the plectrum the string is contacting.
The only trouble I'm having in understanding this model is concerning the units of $\phi$. Is this angle solution going to be in radians or degrees? In C++, sin and cos take radians by default and you have to convert degrees to radians first if working from degrees. So I need to know which one I'm getting from this equation for $\phi$.
Is $\phi$ here in radians or degrees?
Thanks for any guidance.
| Oddly enough, the angle should usually be given in radians. However, looking at the first referenced paper, I am 99% sure that they are using angles in degrees. The authors speak of ( after (6) )
For small-angle approximations ($\phi<<1$), this reduces [...].
If this would be in radians, this would mean roughly 57° which is not really a "small-angle approximation".
Also, they say that
when $\theta=90°$, there is no bending [...].
Which also indicates the use of angles in degrees. I would suggest contacting the authors. Usually, authors are very happy if someone is interested in their work and are willing to help. You could also try to replicate the given results since the authors seem to have provided all necessary data.
| {
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Existence of independent events Let $(\Omega, \mathcal {F},\mathbb {P})$ be a probability space such that $\Omega$ is countable, and $\mathcal {F}=2^{\Omega}$. I want to show that it is impossible to exist a countable collection of events $A_{1},A_{2},\cdots\in \mathcal {F}$ which are independent, such that $\mathbb {P}(A_{i})=\frac {1}{2}$ for each $i$. I think showing $\mathbb {P}(\omega)\leq \frac {1}{2^n}$ for $\omega\in\Omega$ and $n\in \mathbb {N}$ might help?
| Hints:
Let $A_\infty$ be the set of $\omega$ which are contained in infinitely many
$A_i$.
First, show that each $\omega\in A_\infty$
satisfies $\Bbb P(\{\omega\})=0$.
Then, use a version of the Borel-Cantelli lemma (one where independence or pairwise independence plays an important role), to show that $\Bbb P(A)=1$ holds.
Since $\Omega$ is countable, this will result in a contradiction.
| {
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How to find $E(\bar X_n(1-\bar X_n))$ (Bernoulli random variables) I'm trying to solve this exercise:
Let $X_1,\ldots, X_n$ be i.i.d. Bernoulli random variables, with
unknown parameter $p\in (0, 1)$. The aim of this exercise is to
estimate the common variance of the Xi’s.
*
*Show that $var(X_i) = p(1 − p)$
*Let $\bar X_n$ be the sample average of the $X_i$’s. Prove that $\bar X_n(1 − \bar X_n)$ is a consistent estimator of $p(1 − p)$.
*Compute the bias of this estimator.
*Using the previous question, find an unbiased estimator of $p(1 − p)$.
I solved in this way:
*
*We have $E(X_i)=p$ and $E[X_i^2]=1^2\cdot p+ 0^2\cdot(1-p)=p$ by LOTUS. Then $var(X_i)=p - p^2=p(1-p)$.
*By WLLN we have $\bar X_n\xrightarrow{P} p$ and by the mapping theorem we have $\bar X_n(1-\bar X_n)\xrightarrow{P}p(1-p)$
*I don't know how to find $E(\bar X_n(1-\bar X_n))$. I Tried to expand $\bar X_n$ but it gives me ugly calculations.
I'm stuck in the question 3, I need help how to proceed.
| $$E[\bar{X}_n(1-\bar{X}_n)]=E[\bar{X}_n]-E[\bar{X}^2_n]=E[\bar{X}_n]-\left[Var(\bar{X_n})+E[\bar{X}_n]^2\right]$$
| {
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What is a simple way to locate the foci of an ellipse? I have DIY project going, and ran into the same issue as posted in the question here: stove jack for a tent
The answer was helpful, but somewhere along the line I got lost in the math... (Perhaps because the example was over simplified, using a 45 triangle with units of 1, and the distance between foci equaled the minor axis?)
Because my roof pitch isn't 45 degrees, I wanted to work the problem on a 30-60-90 triangle to make sure I understood the concept before I unfold my tent and take measurements of the actual slope. I plan on using a 4" diameter stove pipe, and created the diagram below. (apologies for the image quality)
I realize that I could make a template using a French curve like I did for my diagram above and be "close enough", but I would like to try the string method as described in the answer on the other question I linked to.
Is there a relatively easy way to calculate distance X given the angles, and/or dimensions for someone like me who is not a math whiz?
ADDENDUM:
I am accepting Parcly Taxel’s answer because it was concise, first, correct, and it pointed me in the right direction even if it didn’t state the obvious.
I understand that equations are the language of mathematics, and I didn’t ask for a plain English answer, but as I alluded to in a comment – for many of us knowledge of anything beyond the basic arithmetic we use regularly may have been hard won, and could have atrophied in the intervening decades. Even equations that are very simple for the left brained intelligentsia frequenting this venue may create anxiety for us simpletons cutting holes in our tents who need to review whether to use COS or TAN to calculate the opposite side!
I don’t want this to morph into a critique about how I was taught math, but now that I “get it” I wanted to offer an example below of how I might answer the question for an admitted non-math person. FWIW, and I welcome any feedback in the comments…
| Say the pipe has radius $r$ and the roof has angle $\theta$ to the horizontal. The ellipse's semi-axes will be $b=r$ and $a=r\sec\theta$. Now, checking Wikipedia we find
Linear eccentricity
This is the distance from the center to a focus: $c=\sqrt{a^2-b^2}$.
So the distance between foci in our case will be $2\sqrt{r^2(\sec^2\theta-1)}=2r\tan\theta$.
| {
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How to evaluate this tricky combinatorial sum? According to Mathematica,
$$\sum _{i=s}^p (-1)^i \binom{p}{i} \binom{i}{i-s}\frac{1}{2 i+1}
=(-1)^s \frac{p!\,\Gamma \left(s+\frac{1}{2}\right)}{2 s! \,\Gamma \left(p+\frac{3}{2}\right)}.$$
How can we prove this? I'd especially like a solution method that can be generalized to other sums of this type.
The assumptions are that $s,p\in\mathbb N$ with $0≤s≤p.$
| Recall that $\binom{a}{b}\binom{b}{c}=\binom{a}{c}\binom{a-c}{b-c},$ so
$$\sum _{i=s}^p(-1)^i\binom{p}{i}\binom{i}{i-s}\frac{1}{2i+1}=\binom{p}{s}\sum _{i=s}^p(-1)^i\binom{p-s}{i-s}\frac{1}{2i+1}$$
$$=(-1)^s\binom{p}{s}\sum _{i=0}^{p-s}(-1)^i\binom{p-s}{i}\frac{1}{2(i+s)+1}=\frac{(-1)^s}{2s+1}\binom{p}{s}\sum _{i=0}^{p-s}(-1)^i\binom{p-s}{i}\frac{1}{\frac{2i}{2s+1}+1}$$
Using geometric sum and assuming $p<2s+1/2$(Thanks to the OP who pointed this out)
$$\frac{(-1)^s}{2s+1}\binom{p}{s}\sum _{k=0}^{\infty}(\frac{-2}{2s+1})^k\sum _{i=0}^{p-s}(-1)^i\binom{p-s}{i}i^k=\frac{(-1)^s}{2s+1}\binom{p}{s}\sum _{k=0}^{\infty}(\frac{-2}{2s+1})^k(p-s)!{k\brace p-s}$$
where the last step is one of the forms of Stirling numbers of second kind
$$\frac{(-1)^s(p-s)!}{2s+1}\binom{p}{s}\sum _{k=0}^{\infty}(\frac{-2}{2s+1})^k{k\brace p-s}$$
Now, recall that $\displaystyle \sum _{n=0}^{\infty}{n\brace k}x^n=\frac{1}{(k+1)!x\binom{1/x}{k+1}}$ is the generating function for Stirling numbers, so
$$=\frac{(-1)^s(p-s)!}{2s+1}\binom{p}{s}\frac{1}{(p-s+1)!(-2/(2s+1))\binom{\frac{-1}{2}(2s+1)}{p-s+1}},$$
which if I am not mistaken, should be that expression using that $\binom{n}{k}=(-1)^k\binom{-n+k-1}{k}$ for $n<0.$
| {
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calculate size of sample to reach wanted probability
You are recording neural activity in a cortical brain region. This
brain region is known to contain excitatory and inhibitory neurons
randomly distributed in space. In the cortex, the number of excitatory
neurons is 4 times than the inhibitory neurons. You record blindly the
neural activity, that is, you don’t know the type of the recorded
neurons. Note that each time you stick your electrode you might
record the same neuron
*
*Your recorded 100 neurons. How many neurons you expect to see from each type?
*How many neurons must you record so that you will have at least one neuron from each type with probability of >0.95? (use matlab to
estimate the number, it is not easy to solve the equation
analytically)
*
*I think that the answer for this is 80/20, am I right? (seems to easy to be true)
*I'm not sure how to start with this. am I suppose to calculate the CDF? I was thinking that if the probability of hitting inhibitory neuron is the smallest(0.2?) than I should just calculate the value of X for CDF(X) =0.95. is this the correct way to solve this?
| I think you're right for the first part.
For the second part, say we record n neurons.
What's the probability that we don't have at least one of each type?
That's P(no Es) + P(no Is) (we can't have none of either so these cases are disjoint)
P(no Es) = P(all Is)
P(no Is) = P(all Es)
So our probability is $ (\frac15)^n + (\frac45)^n $
So the probability there is at least one of each type, P(n):
$$ P(n) = 1 - (\frac15)^n - (\frac45)^n $$
We want the smallest integer n so $P(n) > 0.95 $
I'm not sure how we could solve this with algebra so I just used a table of values, to find that the smallest is $n = 14$
| {
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Algebra - Piecewise Function Let $p(x)$ be defined on $2 \le x \le 10$ such that$$p(x) = \begin{cases} x + 1 &\quad \lfloor x \rfloor\text{ is prime} \\ p(y) + (x + 1 - \lfloor x \rfloor) &\quad \text{otherwise} \end{cases}$$where $y$ is the greatest prime factor of $\lfloor x\rfloor.$ Express the range of $p$ in interval notation.
I do not not where to start.
I know at $x \in [2, 3) $, the range is $[3, 4)$. Should I continue with this? Like continuing with $x\in [3,4)$, then $x \in [4, 5)$ and so on?
| You are on the right path.. Just a small observation away from the solution. In case you don't get it, I am providing the solution below.
Answer:
The range is $\cup_{p \in S} [p+1, p+2)$ where S is the set of all prime numbers.
To see this, take two cases-
*
*[x] is prime: The largest prime factor would be the number itself. So the range is [p+1,p+2).
*[x] is not prime: [x] can be decomposed into ${p_1}^{a_1}{p_2}^{a_2}...{p_n}^{a_n}$ for this, the range is [$max(p_i)$+1,$max(p_i)$+2).
By 1. we get all [p+1,p+2), where p is prime, in our range. By 2., we remove the possibility of having any [n+1,n+2), where n is not prime, in our range.
Edit: I did not see the constraints put on x. Upon adding the constraints, the answer would include just the prime numbers less than 10. This makes the question kinda trivial though :/
| {
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Determining all distinct equivalence classes Let
A = {−6, −5, −4, −3, −2, −1, 0, 1, 2, 3, 4}
and define a relation R on A as follows:
For all
m, n is in Z,
$m R n ⇔ 3|(m^2 − n^2)$
It is a fact that R is an equivalence relation on A. Use set-roster notation to list the distinct equivalence classes of R.
I believe I can go through each element and find all the pairs that satisfy $3|m^2-n^2$, however, this seems like it will take a while. I know I just have to find $3|(m-n)$ or $3|(m+n)$, but this still doesn't seem like the best method. What is the most efficient way of doing this?
| $m,n$ belong to the same equivalence class iff $m^2\equiv_3n^2$ i.e. $m^2,n^2$ belong to the same equivalence class with respect to modulo $3$ relation. So simply square all the elements in $A$ and form classes of the squares according to modulo $3$ relation. Then simply replace each square number by the original element in each class.
Edit: In effect, replace each element of $A$ by the non-negative remainder its square leaves when divided by $3$.
$$\begin{align*}A&=\{-6,-5,-4,-3,-2,-1,0,1,2,3,4\}\\
B&=\{~~~0,~~~1,~~~1,~~~0,~~~1,~~~~1,0,1,1,0,1\}\end{align*}$$
Now two elements of $A$ belong to the same equivalence class iff they have the same $B$ values i.e. just group the elements with $B$ value $1$ and $0$ in two equivalence classes.$$E_1=\{-6,-3,0,3\},E_2=A-E_1$$
| {
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Bump function and Gevrey class So firstly we defined Gevrey Class as following:
"A function $f$ is in the Gevrey class of order $θ$ if for every $r > 0,$ there are some constants $M,\; a$ such that
$$|f(m)^{(t)}| ≤ Ma^m(m!)^θ,\;\, |t| < a"$$
Problem:
Show that
$$ϕ(t) := \left\{\begin{matrix}
e^{−1/t^2}, & t>0\\ 0,
& t\leq 0
\end{matrix}\right.,$$
is in the Gevrey class of order $θ =\frac{3}{2}.$
My Approach:
So I tried using Cauchy Integral
$$ \phi^{(m)}(t)=\frac{m!}{2\pi i}\oint_{\gamma}^{}\frac{e^{-z^{-2}}}{(z-t)^{m+1}}dz $$
with
$$ \gamma=\left\{ t+\frac{t}{2}e^{i\omega} \Bigg|0\leq\omega \leq2\pi \right\}. $$
Firstly I tried to calculate the minimum of $\Re(z^{-2}):$
I substituted $\omega=z-2$, which means I needed to minimize $\Re(\omega + 2)^{-2}$ on the unit circle, $x$ is a minimum for it, one must have $xf′(x)<0$, where $f(\omega)=(\omega+2)^{−2}$ so $\frac{x}{(x+2)^3}>0$ or reversing $x^2+6x+12+8\bar{x}=c>0$ since $|x|=1$ and taking imaginary parts we get $\Im x^2−2\Im x=0$ which easily gives $sinθ=0$ or $cosθ=1$, where $x=e^{iθ}$ so $θ=0,π$ and only $θ=0$ gives the expected $\frac{1}{9}$ minimum.
My Problem:
So this is the part, where I got stuck and couldn't go further because I am not sure wheater my calculation of the minimum was the good one. I would appreciate any kind of help!
| Noting as above that the minimum depends on $t$, so $1/9$ is for $t=2$ but for general $t$ it is $4/(9t^2)$, the problem reduces to maximize $|t^{-m}|e^{-4/(9t^2)}$ or equivalently $x^me^{-4x^2/9}$ for $x>0$
By calculus it is clear that the maximum happens at $x^2=9m/8$ so we need to estimate $m^{m/2}e^{-m/2}$ as the $(9/8)^{m/2}$ gets comingled into the constant $a^m$ with the extra $2^m$ from the integral ${2\pi i}\oint_{\gamma}^{}\frac{1}{(z-t)^{m+1}}dz=(2/t)^m$ once we estimate the numerator of the original integral away
But by Stirling $m!\ge \sqrt {2\pi m}(m/e)^m$ so $m^{m/2}e^{-m/2} \le (m!)^{1/2}$ and we finally get our required approximatiom:
$|\phi^{(m)}(t)| \le (2\sqrt {9/8})^m (m!)^{3/2}$
| {
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Distinct fixed points of Möbius transformation I am attempting the following problem.
Let $f:\mathbb{C}\cup\{\infty\}\to \mathbb{C}\cup\{\infty\}$ be a Möbius transformation. Prove that if $f(\alpha)=\alpha,f(\beta)=\beta$ with $\alpha\neq\beta\in\mathbb{C}\cup\{\infty\}$, then $f'(\alpha)f'(\beta)=1$.
I know $f$ must have the form
$$
f(z)=\frac{az+b}{cz+d}
$$
and $f'$ has the form
$$
f'(z)=\frac{ad-bc}{(cz+d)^2},
$$
but I do not know how to proceed.
I wonder if I can use $f$ to define a composition $g$ of conformal maps from $\mathbb{D}\to\mathbb{D}$, such that $g$ has two distinct fixed points, one at $0$, then apply Schwarz's Lemma somehow. But I do not know how to do so. Thank you for any help!
| Fixed point of the Möbius transformation are given by $f(z)=z,$ and hence by the quadratic equation $$cz^2+(d-a)z−b= 0.$$ Assume $c\neq0,$ then $$\alpha+\beta=\dfrac{a-d}{c},\qquad\qquad \alpha\beta=-\dfrac{b}{c}.$$ Consider $$f'(\alpha)f'(\beta)=\dfrac{(ad-bc)^2}{(c\alpha+d)^2(c\beta+d)^2}=\dfrac{(ad-bc)^2}{(c^2\alpha\beta+cd(\alpha+\beta)+d^2)^2}.$$ From here I will leave the rest of the simplification for you.
Finally consider the case $c=0$ case separately, in which case $\infty$ is a fixed point.
However I believe that there must bee a simple geometrical reasoning for this nice formula.
| {
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$\lim_{x\rightarrow\infty} \sqrt{x+1}-\sqrt{x}$ by L'Hospital rule Consider evaluation of the limit $\lim_{x\rightarrow\infty} \sqrt{x+1}-\sqrt{x}$. By a direct way, we have
$$
\lim_{x\rightarrow\infty} \sqrt{x+1}-\sqrt{x}=\lim_{x\rightarrow\infty} \frac{1}{\sqrt{x+1}+\sqrt{x}}=0.
$$
On the other hand, if $f(x)=\sqrt{x+1}$ and $g(x)=\sqrt{x}$, then $\lim_{x\rightarrow\infty} f(x)=\lim_{x\rightarrow\infty} g(x)=\infty$, so the above limit is in indeterminate form $\infty-\infty$. If we want to apply L'Hospital's rule, we try to convert it into $\infty\cdot 0$ form by
$$f(x)-g(x)=f(x)g(x)\Big{(} \frac{1}{g(x)}-\frac{1}{f(x)}\Big{)}=\sqrt{x+1}\sqrt{x} \Big{(} \frac{1}{\sqrt{x}}-\frac{1}{\sqrt{x+1}}\Big{)}.$$
But, after this, I was unable to compute the limit of this by L'Hospital's rule. Any hint for it?
| No problem!
$$\lim_{x\rightarrow+\infty}(\sqrt{x+1}-\sqrt{x})=\lim_{x\rightarrow+\infty}\frac{x}{\sqrt{x^3+x^2}+\sqrt{x^3}}=\lim_{x\rightarrow+\infty}\frac{1}{\frac{3x^2+2x}{2\sqrt{x^3+x^2}}+\frac{3}{2}\sqrt{x}}=0.$$
| {
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Assume that $ab \mid (a+b)^2.$ Show that $ab \mid (a-b)^2$.
Assume that $ab \mid (a+b)^2.$ Show that $ab \mid (a-b)^2$.
If $ab \mid (a+b)^²$, then $ab\mid a^2+2ab+b^2 \Longrightarrow ab\mid a^2, ab\mid 2ab$ and $ab\mid b^2$ right?
So since $(a-b)^2 = a^2-2ab+b^2$ from the assumption we have that $ab \mid a^2$ and $ab \mid b^2$. Now only remains to show that $ab \mid -2ab$ which is clearly true.
Is this valid? I'm not sure about the implication that $ab$ would divide all the terms in $a^2+2ab+b^2$.
| We have $2 \mid 6=3+3$ but $2 \not \mid 3$, so the fact that if $x \mid y +z$ then $x \mid y$ and $x \mid z$ isn't valid (at least in general).
What you just have to use here is the fact that if
$$
x \mid y \text{ and } x \mid z
$$
therefore
$$
x \mid \lambda y + \mu z.
$$
Here, if $ab \mid (a+b)^2$ and since it's obvious that $ab \mid ab$, we have
$$
ab \mid (a+b)^2 -4ab=(a-b)^2.
$$
By the way, if $ab \mid (a+b)^2$, you can show that $a=b$ (see here).
| {
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Clarification about one of Stanley's problems on Catalan numbers I have a question about the statement of problem (aa) in Stanley's list of problems on Catalan numbers (see here), in which he lists 66 sets whose elements are counted by the $n$th Catalan number $C_n$.
The statement seems to be imprecise, or incomplete. I am copying it here for ease of reference:
[We consider] equivalence classes $B$ of words in the alphabet [$n-1$] such that any three consecutive letters of any word in $B$ are distinct, under the equivalence relation $uijv \sim ujiv$ for any words, $u, v$ and any $i, j \in$ [$n-1$] satisfying $|i-j|\geq 2$. For $n=3$, equivalence classes are {$\varnothing$}, {1}, {2}, {12}, {21}. For $n=4$ a representative of each class is given by $\varnothing$, 1, 2, 3, 12, 21, 13, 23, 32, 123, 132, 213, 321, 2132.
Now, while this is not stated, we are clearly interested in the smallest equivalence relation containing those ordered pairs. Furthermore, it seems that we are only considering words of length at most $n$. Even taking into account this, it is still not clear to me why for $n=4$ we only have one equivalence class for words of length $4$. For instance why, in addition to $[2132]$, do we not also have the four pairwise distinct equivalence classes $[1231], [1321], [3123], [3213]$?
For example, let's consider $[1231]$. Then $1231$ is not equivalent to $1321$, since we are only considering permutations of pairs $ij$ with $|i-j|\geq 2$. In particular, it seems that $1231$ is not equivalent to any other word such that any three consecutive letters are all distinct.
Please note that I am not asking for a solution to the counting problem, but simply trying to understand the statement. Since these problems are quite well-known and used in many combinatorics classes, I am a bit surprised at the fact that the statement appears to be so imprecise.
| I think that if any word in the equivalence class is an invalid word, then all words in the equivalence class are invalid. So 1231 is invalid since it's equivalent to 1213, which is invalid. For $n=4$ you cannot have a word of length $5$ or greater. This is not an additional assumption, but follows from the constraint. To see this, note that the middle three letters in a five letter word cannot all be 2s. Any 3 in the interior of the word must either be preceded or succeed by a 1. Likewise any 1 must be preceded or succeeded by a 3. In either casethe surrounding letters are forced to both be 2s, giving either 2132 or 2312 (which are equivalent). Any letter preceding or succeeding this word must be a 1 or 3, but both are impossible.
| {
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The function $f(x)=\frac{2x^3+7x}{x^2+4}$ is invertible. Calculate the derivative $(f^{-1})(\frac{-9}{5})$. I was supposed to find the derivative of the inverse of the function f, and then its derivative and evaluate it at $\frac{-9}{5}$.
I initially tried by hand and it became really messy but I got a solution that seemed reasonable, however it was not correct. So I turned to WolframAlpha and got the same solution: $\frac{32761}{65047}$. I used the the following theorem: $(Df^{-1})(y)=\frac{1}{f'(x)}$.
Any suggestion on how to solve it, and/or perhaps if the theorem I'm using is useless in this case?
We have an online module that checks if your answer is correct so I don't know the actual solution.
Thanks :)
| If $f^{-1}$ exists, then
$y=f(x), y_0=f(x_0) \implies x_0=f(y_9)$ Here for $y_0=-9/5$ $x_0=-1 \implies f^{-1}(-9/5)=-1.$
$$f(x)=\frac{2x^3+7x}{x^2+4} \implies f'(x)=2-\frac{8}{x^2+4}^2+\frac{1}{x^2+4}$$
$$\frac{d f^{-1}(y)}{dy}|_{y=y_0}=\frac{1}{f'(x_0)} \implies \frac{d f^{-1}(y)}{dy}|_{y=-9/5}=\frac{25}{47}.$$
| {
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What is the best way to express the set of sets that has no element in common with another set? I am trying to express the following: I have a set $A$ and the powerset (set of all subsets of $A$) $P(A)$. I have another set $S \in P(A)$, and I want to get the sets in $P(A)$ with no element in common with $S$.
The way I express this is $$X \in P(A): X \cap S = \{\}$$
Is there a more concise way or a special notation from set theory?
| You could use union of disjoint subsets to achieve the same thing: $X, Y \in P(A), X \sqcup Y \equiv A$
| {
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Consider 8 points on a circle of radius 1. Show that at least two points have a distance less than $\frac{9}{10}$ from each other. Consider 8 points on a circle of radius 1. Show that at least two points have a
distance less than $\frac{9}{10}$ from each other.
What I know so far:
I think I start off with 8 points on the circumference which creates an octagon with side lengths $\frac{1}{2}$. Hence all points have at most a maximum distance of $\frac{1}{2}$.
However I have no idea what to do next. Any help would be great!
| We use the following claim:
Claim 1: Let $d_0,\ldots, d_m$ be $m$ real numbers. Then if $\sum_{i=0}^m d_i=K$ then there exists an $i$ such that $d_i \le \frac{K}{m+1}$
So now we use Claim 1. Write the points $x_0,\ldots, x_7$, in the order around the circle. Let us write as $d'(x_i,x_j)$ the length of the shortest walk between $x_i$ and $x_j$ on the circle.
The circumference of the circle is $2\pi$, so $\sum_{i=0}^7 d'(x_i,x_{i+1}) = 2\pi$ the circumference of the circle. By Claim 1, there exists an $i$ such that $d'(x_i,x'_{i+1}) \le \frac{2\pi}{8} <.8 < \frac{9}{10}$.
As the euclidian distance between $x_i$ and $x_{i+1}$ is shorter than the length of the shortest segement on the circle between $x_i$ and $x_{i+1}$, if the inequality $d'(x_i,x'_{i+1}) \le \frac{2\pi}{8} < \frac{9}{10}$ is satisfied then the euclidena distance between $x_i$ and $x_{i+1}$, which is less than $d'(x_i,x_{i+1})$, must also be less than 9/10.
| {
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Finding stable, unstable and centre manifold
Find the stable, unstable, and center manifolds of the following system
\begin{align}
\dot{x}& = -x,\\
\dot{y}& = -2y + 2z^2(1-z^2), \\
\dot{z} &= -z^3
\end{align}
at the origin. Prove that origin is the local attractor.
Solving the above system I got: $$x = x_0 \ e^{-t}, \ y(t) = \frac{1}{2}\bigg[\frac{1}{t+z_0} + \bigg(\frac{2y_0 z_0 -1}{z_0}\bigg)e^{-2t} \bigg], \ z(t) = \sqrt{\frac{1}{2(t+z_0)}}$$
Stable manifold: $$W^{s}(0,0,0) := \{(x,y,z):x(t), y(t) \text{ and } z(t) \to 0 \text{ as } t \to +\infty\} =\{(x,0,z)\}$$
Unstable manifold: $$W^{u}(0,0,0) := \{(x,y,z):x(t), y(t) \text{ and } z(t) \to 0 \text{ as } t \to -\infty\} = \{(x,y,z):
x=0, yz=\frac{1}{2} \} $$
I do not whether my calculations are right or not. I do not know how to prove the second part.
| First note that the third equation has the solution
$$
z(t)^{-2}-z_0^{-2}=2t\implies z(t)=\frac{z_0}{\sqrt{1+2z_0^2t}}
$$
where now indeed $z(0)=z_0$.
Next, you can be a little creative in solving the second equation by inserting the third to reduce the 4th degree term,
$$
\dot y+2y=2z^2-2z^4=2z^2+2z\dot z=\frac{d}{dt}(z^2)+2z^2.
$$
This means that "luckily" the terms turn out to be that the same differential operator $(D+2)$ is applied to both sides. The solution then is
$$
y(t)-z(t)^2=(y_0-z_0^2)e^{-2t},\\
y(t)=\frac{z_0^2}{1+2z_0^2t}+(y_0-z_0^2)e^{-2t}.
$$
I do not see how there can be any unstable manifold. From points outside the set with $x_0=0$ and $y_0=z_0^2$ the solutions will converge exponentially fast towards points within that set, and then slowly along that set towards the origin. In my mind this was the characteristic of a center manifold?
| {
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How can we define $\mathbb{R}^0$? I'm reading a paper which asks to consider some box B $\subset \mathbb{R}^{v-1}$, for all $v\geq 1$. So for example, if we let $v=2, $ we would have that B $\subset \mathbb{R}$, so B would just be some interval $I$. I'm wondering what the case would be for $v=1$. Then we would be considering some box $B \subset \mathbb{R}^0$. I'm not sure what to make of this. Would it be empty...or just a singular point?
I can link the paper if its helpful. Thanks in advance.
| $\mathbb R^{v-1}$ is the set of all $v-1$-tuples of real numbers, i.e. all functions from
$\{1,2,\ldots, v-1\}$ to $\mathbb R$. In the case $v=1$, that's all functions from the empty set to $\mathbb R$. There is one function from the empty set to anything, namely a function that doesn't do anything.
I'd have to look closer at the paper you're referring to, to see what is done with $B$, but my guess is that the author didn't really bother much about the case $v=1$. Either didn't think about it at all, or is leaving this up to the reader.
| {
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What does the golden spiral converge to? I am looking at pictures of spirals associated with the Fibonacci sequence and the golden ratio and I am seeing several different spiral diagrams.
For example one image I saw shows the spiral with the following numbers $8, 5, 3, 2, 1, 1$.
Then another has $34, 21, 13, 8, 5, 3, 2, 1$.
So, since the Fibonacci sequence continues forever is it possible to figure out where the golden spiral converges to in an image with Fibonacci sequence of $F_n, F_{n-1}, F_{n-2}, \dots$ ?
| The Fibonacci sequence usually begins $0, 1, 1, 2, 3, 5, \ldots$ with $F_0 = 0$. The sequences above are this in reverse. The second one excludes $F_1$. It seems that $F_n$ where $n < 0$ are being excluded and you are asking for the coordinates of the the start of the spiral at $F_1$. That depends on the diagram, and the coordinate system. In general, the coordinates can be found.
| {
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sequence/series limits I am struggling a bit with the following two questions I got from a sister who is doing BSC in Applied Mathematics. I always try my best to help her but was stuck on these questions and so any answer and explanation is much appreciated:
Here's the question:
Decide which of the following sequences have limits when $n\rightarrow \infty$ , and find the relevant limits if they exist. In the case of $C(n)$ it is enough to show that the limit is bounded from above and below.
Sequence $B(n)$ such that
$B(0) = 0$ and $B(n) = B(n-1) + 1/2^n$ , for any positive integer $n$.
Sequence $C(n)$ such that $C(1) = 1$ and $C(n) = C(n-1) + 1/n^2$ , for any integer $n>1$
(Hint: Use the fact $\frac{1}{n-1} -\frac{1}{n} =\frac{1}{(n(n-1))}$ with the sandwich reasoning that if $C'(n)<C(n)<C''(n)$ for all $n$, then the limit of $C(n)$ is between the limits of $C'(n)$ and $C''(n)$, if such limits exists.)
On the first question , I tried to plug some values of n and got the limit of B(n) as 1, although I am not 100% sure. But on the second one, I couldn't do much and the hint didn't help me a lot.
| $B(0)=0$, $B(n)=B(n-1)+\frac 1 {2^{n}}$,
Thus $$B(1)= \frac 1 2,B(2)= \frac 1 2+\frac 1 4,B(3)=\frac 1 2+ \frac 1 4+\frac 1 8,\cdots B(n)= \frac1 2+\cdots \frac 1 {2^{n}} $$
Now see $$\frac{1}{2} B(n)= \frac 1 4+\frac1 8+\cdots \frac 1 {2^{n+1}}\\\implies \left(1-\frac 1 2\right)B(n)=\frac 1 2-\frac 1 {2^{n+1}}\\\implies \lim_{n\to +\infty}B(n)=1$$
$C(1)=1,C(n)=C(n-1)+\frac 1 {n^2}$
Similarly we get, $C(n)=1+\frac 1 {2^2}+\frac 1 {3^2}+\cdots +\frac 1 {n^2}$
Observe $$\frac 1 {2^2}+\frac 1 {3^2}<\frac 2 {2^2}=\frac 1 {2}\\\frac 1 {4^2}+\frac 1 {5^2}+\frac 1 {6^2}+\frac 1 {7^2}<\frac 4 {4^2}=\frac 1 {4}\\\cdots\\$$
Similarly we can say that $C_{2^{n+1}-1}<1+\frac 1 {2}+\frac 1 {4}+\cdots +\frac 1 {2^n}=B(n)+1$
Therefore $\lim C(n) <\lim(B(n))+1$, Using the fact that $C(n)$ is increasing you can conclude that $\lim C(n)$ exists.
| {
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How to show that ${\omega \in \Omega : X(\omega) \neq Y(\omega)}$ is a measurable set, $X,Y$ being $\overline{\mathbb{R}}$-valued. How to show that $\{\omega \in \Omega : X(\omega) \neq Y(\omega)\}$ is a measurable set, $X,Y$ being $\overline{\mathbb{R}}$-valued. I mean, how to fix the case when $X$ and $Y$ = $\infty$? Here i can't pass to the $\{\omega \in \Omega : X(\omega) - Y(\omega) \neq 0\}$ because $\infty - \infty$ isn't defined. Thank you!
| $\{\omega: X(\omega)=\infty, Y(\omega)<\infty\} =\bigcup_n \{\omega: X(\omega)=\infty\}\bigcap \{\omega: Y(\omega)<n\}$ which is measurable. Simialrly, $\{\omega: Y(\omega)=\infty, X(\omega)<\infty\}$ is measurable. Can you finish?
| {
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Why is $\sec^2 (\tan^{-1} (\frac{x}{2})) = 1 + \tan^2 (\tan^{-1} (\frac{x}{2}))$ and not $(1 + \tan^2 )(\tan^{-1} (\frac{x}{2}))$?
Why is $\sec^2 (\tan^{-1} (\frac{x}{2})) = 1 + \tan^2 (\tan^{-1} (\frac{x}{2}))$ and not $(1 + \tan^2 )(\tan^{-1} (\frac{x}{2}))$?
If you replace $\sec^2$ with $\tan^2 + 1$, it should be $(1 + \tan^2 )(\tan^{-1} (\frac{x}{2}))$, right?
However, it seems that it is not multiplication here : $\sec^2 (\tan^{-1} (\frac{x}{2}))$, when the $1$ becomes independent of the $\tan^2$ somehow.
Could someone explain why?
| The identity
$$\sec^2\theta\equiv1+\tan^2\theta$$
holds for all values of $\theta$, not just some values- that's why it's an identity, as opposed to an equation. Replacing $\theta$ by $\tan^{-1}(\frac{x}{2})$ yields the correct identity:
$$\sec^2(\tan^{-1}(\frac{x}{2}))\equiv1+\tan^2(\tan^{-1}(\frac{x}{2}))\equiv1+\left(\tan(\tan^{-1}(\frac{x}{2}))\right)^2\equiv1+\frac{x^2}{4}$$
Edit
Thinking that $\sec^2\theta$ means $\sec^2\times\theta$ is meaningless- $\sec^2$ or indeed any other trigonometrical function has no meaning unless it has an argument, ie the value the that is an input into the trigonometical function (in your case $\tan^{-1}\frac{x}{2}$).
| {
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Proving a f is continuous at 1/3 Show that $f(x) = \frac{1}{5x}$ is continuous at $x = \frac{1}{3}$. I have to use an $\varepsilon - \delta$ proof? I am having trouble choosing my delta because I am confused as to how $|\frac{1}{x} - \frac13|$ can get to $|x - \frac{1}{3}|$?
| A function is continuous at a given point $x_0$ if for every $ϵ>0$ you can find some $δ>0$ such that $|x−x_0|<δ⟹|f(x)−f(x_0)|<ϵ.$
What is useful with these problems is to "solve them backwards". Start with
$$\left|\frac{1}{5x} - \frac{1}{15}\right| < \epsilon$$
What happens when $0<x<3$?
$$\frac{1}{5x} - \frac{1}{15} < \epsilon$$
$$\frac{1}{x} - \frac{1}{3} < 5 \epsilon$$
$$\frac{1}{x} < \frac{1}{3} + 5 \epsilon = \frac{1+15 \epsilon}{3}$$
$$x > \frac{3}{1+15\epsilon}$$
What happens when $x>3?$
$$-\frac{1}{5x} + \frac{1}{15} < \epsilon$$
$$-\frac{1}{x} + \frac{1}{3} < 5\epsilon$$
$$ \frac{1}{3} - 5\epsilon <\frac{1}{x}$$
Now we're gonna want to guarantee that the left side of this inequality is positive, so choose $\epsilon <\frac{1}{15}.$
$$ \frac{1-15\epsilon}{3} <\frac{1}{x}$$
$$ \frac{3}{1-15\epsilon} >x$$.
In conclusion, we see that when $\epsilon <\frac{1}{15}$ and $x>0,$ then $x$ lies in the range
$$\frac{3}{1+15\epsilon} < x < \frac{3}{1-15\epsilon}.$$
So
$$\frac{3}{1+15\epsilon} -3 < x -3 < \frac{3}{1-15\epsilon} -3.$$
Now think about how you can choose $\delta$ so that $|x-3| < \delta$ implies the above condition. (Hint: just make sure $\delta$ is the minimum of the two). Do you need to worry about $x<0$? (Hint: you could just choose $\delta$ to guarantee this doesn't happen).
| {
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On the zeros of Bessel functions of parameter $\nu=-1/3$ Working in a problem I found with the task of finding the zeros of a sum of the Airy functions ( See for example) $\sqrt{3}\text{Ai}(x)+\text{Bi}(x)$, it result that these are exactly the zeros for the modified Bessel function of the First Kind of parameter $\nu=-1/3$ this is $$I_{-1/3}(x)=\sum_{n=0}^{\infty} \frac{(-1)^{n}}{n !} \frac{(x / 2)^{2 n-1/3}}{\Gamma(n-\frac{1}{3}+1)}$$.
Do you know if there is an expression for the zeros of this function with this specific value of parameter?
Added: With computational calculation it looks like that if $\lambda_n$ are the zeros then $|\lambda_n-\lambda_{n+1}|\to \pi.$ Looking for I already found that the zeros have the following expresion $z_{\nu, k}=\left\{k+\frac{1}{2}\left(\nu-\frac{1}{2}\right)\right\} \pi-\frac{4 \nu^{2}-1}{8\left\{k+\frac{1}{2}\left(\nu-\frac{1}{2}\right)\right\} \pi}+O\left(\frac{1}{k^{3}}\right)$.
| If you consider Wolfram language functions a closed form, then there is Bessel Y Zero $\text y_{v,x}$:
$$\def\Ai{\text{Ai}} \def\Bi{\text{Bi}} \sqrt3\Ai(x)+\Bi(x)=0\implies -\sqrt3=\frac{\Bi(x)}{\Ai(x)}$$
Now use Bessel J and Bessel Y:
$$\frac{\text J_\frac13(x)}{\text Y_\frac13(x)}=\frac4{\frac{\Ai\left(-\left(\frac32\right)^\frac23x^\frac23\right)}{\Bi\left(-\left(\frac32\right)^\frac23x^\frac23\right) }+\sqrt3}-\sqrt3\implies \frac{\Bi(x)}{\Ai(x)}=\frac{\frac4{1-\sqrt3 \frac{\text J_\frac13\left(\frac23(-x)^\frac32\right)}{\text Y_\frac13\left(\frac23(-x)^\frac32\right)}}-1}{\sqrt3}=\sqrt3$$
Here is the solution
$$\begin{align}\frac{\text J_\frac13\left(\frac23(-x)^\frac32\right)}{\text Y_\frac13\left(\frac23(-x)^\frac32\right)} =\sqrt3\iff \sqrt3\Ai(x)+\Bi(x)=0\implies x=-\left(\frac32\text y_{\frac13,\Bbb N+\frac16}\right)^\frac23\end{align}$$
which is true with the $n$th natural number giving the $n$th solution.
| {
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How to write a polynomial function that has the roots $-2$ and $\sqrt7$? I need to write a polynomial function with integer coefficients that has the roots $-2$ and $\sqrt7$. I'm able to do this correctly when I'm given roots like $-3+i$ & $-3-i$, in which I set the roots equal to zero and then multiply them by one another. However, when I try this with $-2$ and $\sqrt7$ and multiply $x+2$ by $x-\sqrt7$, I get $x^2+2x-\sqrt7 x-2\sqrt7$. I don't know where to go from here, and I don't think that this is the correct next step. What do I do next?
| Try multiplying by $x+\sqrt7,$ the conjugate of $x-\sqrt7$, as well.
$(x+\sqrt7)(x-\sqrt7)=x^2-7$ has integer coefficients, and its product with $x+2$ will too.
| {
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Derivative of a function $f({\bf X})$ w.r.t. matrix ${\bf X}$ when ${\bf X}^\prime {\bf X}$ is a diagonal matrix I wish to compute the derivative and the Hessian of the function ${\bf f}(X)$ where
$$ {\bf f}({\bf X}) = {\bf X} \, {\bf a}, $$
${\bf X}$ is an $(m \times n)$ matrix and ${\bf a}$ is a vector of constants of size $n$. From "Matrix Differential Calculus" by Magnus and Neudecker, it is relatively straightforward to obtain the derivative and the Hessian of ${\bf f}({\bf X})$ w.r.t. ${\bf X}$. For example, the first differential is
\begin{align}
\partial {\bf f}({\bf X}) &= (\partial {\bf X}) {\bf a} = {\rm vec} (\partial {\bf X}) {\bf a} = ({\bf a}^\prime \otimes {\bf I}_m) \,\partial {\rm vec} {\bf X} \\
\end{align}
In my case however the matrix ${\bf X}$ has a special structure. Specifically, I know that that ${\bf X}^\prime {\bf X}$ is a diagonal matrix, say, ${\bf X}^\prime {\bf X} = {\rm diag}(d_1, \ldots, d_n)$, which is not necessarily equal to the identity matrix. In other words, I know that the columns of ${\bf X}$ are mutually orthogonal, but each column vector can be of any length. How do I compute the derivative and the Hessian of $f({\bf X})$ w.r.t. ${\bf X}$ to take this structure into account?
Any references on how to proceed would be greatly appreciated.
I could not find a similar question posted before. The closest I found was Hessian of $f(X)$ when $X$ is a symmetric matrix, but I was not able to apply it to my problem.
| $\def\p{\partial}$
Use an orthogonal matrix $Q$ and a vector $y$ to construct $X$
$$\eqalign{
Y &= {\rm Diag}(y) &\;=\; \sqrt{X^TX} \\
X &= QY &\implies X^TX = YQ^TQY = Y^2 \\
Q &= X(X^TX)^{-1/2} \\
}$$
Also define the diagonal matrix $\;A = {\rm Diag}(a)$.
Write the function in terms of these new variables
and calculate its gradient.
$$\eqalign{
f &= Xa = QYa = QAy\\
\p f &= QA\,\p y \\
\frac{\p f}{\p y} &= QA \\
}$$
This can be interpreted as a gradient wrt $X$ by using
$$\eqalign{
\p{\rm vec}(X) &= (I\otimes Q)\,{\rm vec}\Big({\rm Diag}(\p y)\Big) \\
}$$
| {
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Pigeon Hole Principle - 14 days, 10 physical exercises. How will you do PE on consecutive days at least once in the next fortnight.? I was given this question. Maths is a weakness of mine and I'm really struggling to understand what this question is asking, and how to solve it. My notes indicate it's a pigeon hole principle but I'm unsure.
"You prepare a schedule of your physical exercises for the next fortnight (14 days). You don’t do physical exercise more than once a day. If you have 10 PE sessions planned, explain using the counting principles covered in class how this means you will do PE on consecutive days at least once in the next fortnight."
Thanks in advance :)
| Let $\ n\ $ be the number of PE lessons scheduled for days immediately followed by a day free of any PE lesson. These lessons plus the following lesson-free day must occupy a total of $\ 2n\ $ days, at most one of which (if it is the day following the last lesson) can lie outside the allowed $14$-day period. This leaves at most $\ 15-2n\ $ days available to be occupied by the remaining $\ 10-n\ $ lessons. Thus, $\ 10-n\le15-2n\ $, giving $\ n\le5\ $. Thus, there must be at least $\ 5\ $ lessons on days that are immediately followed by a day with another lesson.
| {
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Finding the range of $f(x)=2\csc(2x)+\sec x+\csc x$ Hi this is the question:
Find the range of $$f(x)=2\csc(2x)+\sec x+\csc x$$
What I've tried:
I know that the range of $\csc(x)$ which is $R\setminus (-1,1) $, the range of $\sec(x)$ is $R\setminus (-1,1)$ too. And I've managed to simplify the expression to have in terms of $\sec x$ and $\csc x$ as such:
$$f(x)=\sec(x)\cdot \csc(x)+\sec(x)+\csc(x).$$
But then, what do I do when I have 3 terms and not just $\sec$ or $\csc$?
| Starting with $$f(x)=\frac{2}{\sin x+ \cos x-1} \implies y=\frac{1+t^2}{t-t^2} \implies (1+y)t^2-yt-1=0,$$
As $t=\tan(x/2)$ lies in $(-\infty, \infty)$, the range will be all values of $y$ when the above quadratic has real roots: $B^2 \ge 4AC$
$$\implies y^2-4y-4 \ge 0 \implies y\ge 2+\sqrt{2} ~or~ y\le2-\sqrt{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The set of all subsequential limits of a sequence is closed.
Let $X$ be a metric space and $\{x_n\}$ be a sequence in $X$. Let $E$ be the set of all subsequential limits of $\{x_n\}$. Then $E$ is closed.
I have a idea of this proof.
Assume $E$ is a infinite set.
Let $p\in X-E$. Then $\exists r>0$ such that $B(p,r)\cap E=\emptyset$. The reason is if $p\in \bar E$, then there is some subsequential limit $y$ close to $p$, and there exists a subsequence $x_{n_k}$ converges to $y$. So, $d(x_{n_k},p)<d(x_{n_k},y)+d(y,p)<\varepsilon$ which is a contradiction.
So, $B(p,r)\subset X-E$ and $E$ is closed.
| A more direct but conceptual way is the following. Suppose $(x_n)\in X^{\mathbb{N}}$ is a sequence in the metric space $X$. Denote $L$ the set of all limits of the subsequences of $(x_n)$. Then
\begin{align}
L = \bigcap_{k\geqslant 0}\overline{\{x_n ~|~ n \geqslant k\}}
\end{align}
It is an intersection of closed subset, thus it is a closed subset.
This is because a point $l$ is in $L$ if and only if for every $\varepsilon >0$ and every $n \geqslant 0$, there exists $k \geqslant n$ such that $d(l,x_k) \leqslant \varepsilon$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Probability distribution with $p(x) = p(1/x)$ I am looking for a probability distribution defined on $x\in(0,\infty)$ that satisfies, for all $x$:
$$p(x) = p(1/x)$$
Naturally, we must have:
$$\int_0^\infty p(x) \, dx= 1$$
and thereore:
$$
\int_0^1 p(x) \, dx + \int_1^\infty p(x) \, dx = 1
$$
Does such a probability distribution exist?
| It is not true that $\int^1_0 p(x)\, dx$ and $\int^\infty_1 p(x) dx$ necessarily both need to be equal to $0.5$, just that their sum needs to be equal to $1$.
As easy example could be $$p(x) = \begin{cases} \exp(-\alpha/x)&\text{for $0<x<1$},\\
\exp(-\alpha x) & \text{for $x \geq 1$},\end{cases}$$
where $\alpha$ is a parameter such that
$$\left[\int^1_0\exp(-\alpha/x)\, dx\right] + \left[\int^\infty_1 \exp(-\alpha x) \, dx\right] = 1.$$
A quick numerical check yields $\alpha \approx 0.68$.
| {
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Product of CDFs at different points Suppose that $X$ is a continuous random variable with finite moments, and $\mathbb a=(a_i) \in \mathbb R^n$. In my calculations, I have reached an expression as
$$ f(\mathbb a )=\prod_i \Pr \{ X\le a_i\} $$
but I am not sure what to do with it to make it a simple expression (i.e. a CDF of a univariate random variable or a limiting theorem that make the whole thing tractable at least asymptotically, etc.). Let's suppose that $n$ is large enough.
One idea was to take another random variable $ Y_i = X - a_i$ , then $$ f(\mathbb a )= \prod_i \Pr \{ Y_i\le 0\} = \Pr \{ \max_i (Y_i)\le 0\} $$ and try to use results from Extreme Value Theory but I only know this theory for i.i.d. case (here we have different first moments).
| I can only help you with a special case: Let's X is Gumbel-distributed
$$\mathrm{Pr}\{X\leq a_{i}\}=e^{-e^{-\frac{x-a_{i}}{\beta}}}$$
with the same variance $\beta$. Then
$$f(a)=\prod_{i=1}^{n}\mathrm{Pr}\{X\leq a_{i}\}=\prod_{i=1}^{n}e^{-e^{-\frac{x-a_{i}}{\beta}}}=e^{-e^{-\frac{x}{\beta}}\left(\sum_{i=1}^{n}e^{\frac{a_{i}}{\beta}}\right)}=e^{-e^{-\frac{x-\beta\log\left(\sum_{i=1}^{n}e^{\frac{a_{i}}{\beta}}\right)}{\beta}}}=\mathrm{Pr}\left\{ X\leq \bar{a}\right\} $$
where the new mean
$$\bar{a}=\beta\log\left(\sum_{i=1}^{n}e^{\frac{a_{i}}{\beta}}\right)$$
is often called the mellowmax.
| {
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Symbolic definition for $i$? Please help!
I am trying to find a definition for $i$ that doesn't work for $-i$.
let $j$ be either $i$ or $-i$.
*
*saying $j^2=-1$ doesn't help since $(i)^2=-1$ and $(-i)^2 = -1$
*saying $j=\sqrt{-1}$ doesn't halp since $(-1)^{\frac{1}{2}}$ is multivalued $i$ and $-i$
*saying $j=e^{i\pi/2}$ doesnt help since $e^{i\pi/2}=i$ but $e^{(-i)\pi/2}=-i$ too (yeah I know I said j=$e^{i\pi/2}$ and not $j=e^{j\pi/2}$ I did that to hopefully make it more intuitive)
*saying $j=ln(i)/(\pi/2)$ doesn't work since $i=ln(i)/(\pi/2)$ and $-i=ln(-i)/(pi/2)$ (check with a phase plotter, ln is multivalued but includes i and -i in their respective functions shown above http://davidbau.com/conformal/#log(z)%2F(pi%2F2)-z)
*saying $Im(j) > 0$ doesn't work because $Im(z)=Re(z/i)$ and $Re(i/i)>0$ but $Re((-i)/(-i))>0$
Is this parity some sort of law? And yet $i\neq-i$ since $i=(-1)\cdot(-i)$. Unless...
Post Scriptum: Irionically, it is very easy to symbolically represent $1$ vs $-1$ even though $(1)^2=1$ and $(-1)^2=1$, we can just say $x$ such that $x=x^2$
EDIT: can quaternions help? Is $i$ a set of 2 numbers? Are they always equal?
| It sounds like you have picked up on the fact that conjugation $$f: a+bi \mapsto a-bi$$ is a ring isomorphism where $i$ and $-i$ correspond to each other.
*
*Addition preserving: $f((a+bi)+(c+di))=f((a+c)+(b+d)i)=(a+c)-(b+d)i=(a-bi)+(c-di)=f(a+bi)+f(c+di)$
*Multiplication preserving: $f((a+bi)(c+di))=f((ac-bd)+(ad+bc)i)=(ac-bd)-(ad+bc)i=(a-bi)(c-di)=f(a+bi)f(c+di)$
*$f(1)=1$
Conjugation is clearly bijective too.
| {
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Propositional calculus logical proof help I am asked to prove:
(Y => Z) V X ≡ (Y V X) =>(Z V X)
This is what I have so far,
(Y V X) =>(Z V X)
(Definition of => X,Y := Y V X, Z V X)
(Y V X) V (Z V X)≡ (Z V X)
(V SYMMETRY twice )
(X V Y) V (X V Z)≡ (Z V X)
(V/V X,Y,Z := X,Y,Z)
X V (Y V Z) = (Z V X)
How do I get rid of the (Z V X) so that I have the full proof Picture of the proofs I am using
| $$\begin{align} (Y \to Z) \lor X &\equiv (Y\lor X) \to (Z\lor X)\\ \\
\iff (\lnot Y \lor Z \lor X) &\equiv \lnot (Y \lor X) \lor (Z \lor X)\\ \\
\iff (\lnot Y \lor Z \lor X) &\equiv (\lnot Y \land \lnot X) \lor (Z \lor X)\\ \\
\iff (\lnot Y \lor Z \lor X) &\equiv (\lnot Y \lor Z \lor X) \land \underbrace{(\lnot X \lor Z \lor X)}_{true}\\ \\\iff (\lnot Y \lor Z \lor X) &\equiv (\lnot Y \lor Z \lor X)\end{align}$$
I leave it to you to justify the steps. If you are unsure of any of them, feel free to comment below the answer. I've made use of the equivalence between $(P\to Q)\equiv (\lnot P \lor Q)$ (and more than once). I've also used DeMorgan's law, and the fact that $\lnot X \lor X$ is always true ($\top$). I used the fact that $\top \lor Z$ is always true, and the fact that $\top \land P \equiv P$.
| {
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Showing a summation of powers of a prime power $q$ I calculated that the number of rank $k$ matrices in $\mathbb{F}_q^{n \times m}$ (where we may assume $n \leq m$) for a prime power $q$ is equal to
$$W_{n,m}(k) := \prod_{i = 1}^k \frac{(q^m - q^{i-1})(q^n - q^{i-1})}{(q^k - q^{i-1})}$$
And so summing all $W_{n,m}(k)$ from rank $k = 0$ to full rank $k = n$ should gives us all matrices of $\mathbb{F}_q^{n \times m}$
$$\sum_{k = 0}^n W_{n,m}(k) = q^{nm}$$
Can we also show this equality using pure arithmetic and combinatorial tricks, i.e. using just the formulas for $W_{n,m}$?
| I managed to solve this, but it took a number of pages. The full answer can be found here.
In short: one can show that the following $q$-analogue of Pascal's rule holds:
$$\prod_{i=1}^k \frac{q^{n+1} - q^{i-1}}{q^k - q^{i-1}} = q^k \cdot \prod_{i=1}^k \frac{q^{n} - q^{i-1}}{q^k - q^{i-1}} + \prod_{i=1}^{k-1} \frac{q^{n} - q^{i-1}}{q^{k-1} - q^{i-1}}$$
Then, this allows you to show that
$$
W_{n+1,m}(k) = q^k\cdot W_{n,m}(k) + (q^m - q^{k-1})\cdot W_{n,m}(k-1)
$$
by multiplying both sides with $\prod (q^m - q^{i-1})$.
Lastly, use induction on $n$ to get the final equality.
(Fun challenge, try to find the linear algebra interpretation of each statement)
| {
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Orthogonal complement and shortest distance on $L^{2}[-1,1]$ Consider the subset $A\subset L^{2}[-1,1]$ defined by
$$
A=\left\{f\in L^{2}[-1,1]:\int_{-1}^{1}f(t)dt=0\right\}
$$
I want to find the set $A^{\perp}=\{g\in L^{2}[-1,1]:(f,g)=0\}$ for all $f\in A$ and $(,)$ is the inner product on $L^{2}$. Also I want to know what is the shortest distance between $t^{2}$ and $A$.
My attempt to find $A^{\perp}$: Let $f\in A$ fixed and let $g\in L^{2}[-1,1]$, then, using Cauchy-Schwarz inequality I get
$$
0\leq (f,g)\leq \|f\|\|g\|
$$
So I conclude that $A^{\perp}=\left\{g\in L^{2}[-1,1]:\int_{-1}^{1}|g(t)|^{2}dt=0\right\}$, am I correct?
I have troubles finding the shortest distance between $t^{2}$ and $A$, how can I calculate it?
| First notice that any constant function is in $A^\perp$.
Take any $g \in A^\perp$. Then $(f,g) = 0$ for all $f \in A$. Set $\overline g = \frac 12 \int^1_{-1} g(t) dt$, and take $$f(x) = g(x) - \overline g, \,\,\,\,x\in [-1,1].$$ Then $f \in A$ (you should check this), and so $$0 = (f,g) = (f,g) - (f,\overline g) = (f,g-\overline g) = (g-\overline g, g-\overline g) = \| g - \overline g\|_2^2.$$ Thus $g = \overline g$, and so $g$ is constant. Thus $A^\perp$ is precisely the constant functions.
To find the closest constant $c$ to $h(t) = t^2$ simply minimize $$\phi(c) = \int^1_{-1} (t^2 - c)^2 dt = \frac{2}{5} - \frac{4c}{3} + 2c^2$$ using a calculus argument: $$\phi'(c) = 4c - 4/3 \,\,\,\,\, \implies \,\,\,\,\, c = 1/3.$$ This isn't a coincidence, the best constant approximation to $h(t) = t^2$ is the average value $\overline h = \frac 1 2\int^1_{-1} h(t) dt = 1/3$. The distance is then $$\sqrt{\phi(1/3)} = \sqrt{\frac 8 {45}}.$$
| {
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Basis of Field F over $\mathbb{Q}$ I am self studying Field theory and got struck on this problem.
If $\;F=\mathbb{Q} \left(\sqrt{2}, \sqrt{3}\right),\;$ find
$\;\left[F:\mathbb{Q}\right]\;$ and a basis of $F$ over $\mathbb{Q}$.
I have proved that $\left[F:\mathbb{Q}\right] =4$, but there is a problem in basis elements.
My basis set is $\left\{a, b\sqrt{2}, c\sqrt{3}\right\}$ such that $a, b, c$ belongs to $\mathbb{Q}$. But what should be fourth element and why ?
I am not able to see .
Kindly help.
| A basis of $F$ over $\mathbb{Q}$ is $\;\left\{1,\sqrt{2},\sqrt{3},\sqrt{6}\right\},$ indeed for any element $\;x\in F\;$ there exist $\;r_1,\;r_2,\;r_3,\;r_4,\;r_5,\;r_6,\;r_7,\;r_8\in\mathbb{Q}\;$ such that
\begin{align}
x&=\dfrac{r_1+r_2\sqrt{2}+r_3\sqrt{3}+r_4\sqrt{6}}{r_5+r_6\sqrt{2}+r_7\sqrt{3}+r_8\sqrt{6}}\\
&=\dfrac{r_1+r_2\sqrt{2}+r_3\sqrt{3}+r_4\sqrt{6}}{r_5+r_6\sqrt{2}+r_7\sqrt{3}+r_8\sqrt{6}}\cdot\dfrac{r_5+r_6\sqrt{2}-r_7\sqrt{3}-r_8\sqrt{6}}{r_5+r_6\sqrt{2}-r_7\sqrt{3}-r_8\sqrt{6}}\\
&=\dfrac{\left(r_1+r_2\sqrt{2}+r_3\sqrt{3}+r_4\sqrt{6}\right)\left(r_5+r_6\sqrt{2}-r_7\sqrt{3}-r_8\sqrt{6}\right)}{\left(r_5+r_6\sqrt{2}\right)^2-\left(r_7\sqrt{3}+r_8\sqrt{6}\right)^2}\\
&=\dfrac{s_1+s_2\sqrt{2}+s_3\sqrt{3}+s_4\sqrt{6}}{s_5+s_6\sqrt{2}}\\
&=\dfrac{s_1+s_2\sqrt{2}+s_3\sqrt{3}+s_4\sqrt{6}}{s_5+s_6\sqrt{2}}\cdot\dfrac{s_5-s_6\sqrt{2}}{s_5-s_6\sqrt{2}}\\
&=\dfrac{\left(s_1+s_2\sqrt{2}+s_3\sqrt{3}+s_4\sqrt{6}\right)\left(s_5-s_6\sqrt{2}\right)}{\left(s_5+s_6\sqrt{2}\right)\left(s_5-s_6\sqrt{2}\right)}\\
&=q_11+q_2\sqrt{2}+q_3\sqrt{3}+q_4\sqrt{6}
\end{align}
where $\;q_1,\;q_2,\;q_3,\;q_4\in\mathbb{Q}\;$.
| {
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Notation $\ln|x|$ vs. $\ln(x)$ Years ago, when I was taught about the function $\ln$ (logarithm in base $e$), all of my teachers and our book, too, insisted that we should write input of this function inside the absolute value notation and I am doing this since now. But, now when I am reading some university books or some answers on this site, I see in most answers, people write $\ln(x)$ using parentheses. It's been a question for a long time to me why people just use parentheses instead and how it is not wrong conventionally? I am sure if I used $\ln(x)$ in high school, it would've always been possible to get a minus point! I am a university student now. Can I write $\ln(x)$ safely and is it conventionally acceptable in mathematics? (I mean, in general, for $\ln(f(x))$ not only $\ln(x)$)
| Your book is an outlier. $\ln(x)$ is different from $\ln|x|$; the latter is practically only encountered as the indefinite integral of $\frac1x$ from $0$ and is the result of precomposing the absolute value function to $\ln(x)$.
An increasing number of sources I read, and my own answers on this site, are lazy enough to go all the way and omit brackets: $\ln x$.
| {
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Trouble understanding equation of an ellipse I'm self-studying algebra by going through Sheldon Axler's algebra book. I'm a bit stuck on equation of an ellipse topic. Here's how it's explained in the book:
Example: Find an equation describing the ellipse in the xy-plane produced by stretching a circle of radius 1 centered at the origin horizontally by a factor of 5 and vertically by a factor of 3.
To find an equation describing ellipse, consider a typical point (u,v) on the circle of radius 1 centered at the origin. Thus $u^2 + v^2 = 1$. Stretching horizontally by a factor of 5 and stretching vertically by a factor of 3 transforms the point (u, v) to the point (5u, 3v). Rewrite the equation $u^2 + v^2 = 1$ in terms of this new point, getting $$\frac{(5u)^2}{25} + \frac{(3v)^2}{9} = 1$$
Shouldn't the resulting equation be $(5u)^2 + (3v)^2 = 1$? Where did 25 and 9 denominators come from?
The rest of the solution in the book:
Write the transformed point $(5u, 3v)$ as $(x,y)$, thus setting $x = 5u$ and $y = 3v$, getting $$\frac{x^2}{25} + \frac{y^2}{9} = 1,$$ which is the equation of the ellipse
| What you quote is not the full solution, is it?
The point $(u,v)$ on the circle satisfies
$$u^2+v^2=1$$
which we just rewrite in equivalent form:
$$\frac{(5u)^2}{25}+\frac{(3v)^2}{9}=1$$
and from this we see that the point $(x,y)=(5u,3v)$, which belongs to the ellipse, satisfies
$$\frac{x^2}{25}+\frac{y^2}{9}=1$$
and so this is the equation of the ellipse.
It would be much more natural to think about it the other way around. Take the point $(x,y)$ on the ellipse, note that $(x/5, y/3)$ (squeezing inverse to your stretching) belongs to the circle, and therefore
$$\left(\frac{x}{5}\right)^2+\left(\frac{y}{3}\right)^2=1$$
which gives the same answer.
| {
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connected but not path-connected sets Let $S\in[0,1]^2$ be defined as follows:
$$(x,y)\in S \Leftrightarrow \begin{cases}y\in[0, \frac{1}{q}], &x=\frac{p}{q}\in\mathbb{Q} \text{ with } q \text{ even}\\
y\in[1-\frac{1}{q}, 1], &x=\frac{p}{q}\in\mathbb{Q} \text{ with } q \text{ odd}\\
y\in[0,1], &\text{otherwise}\end{cases}$$
, where for every $x\in\mathbb{Q}$, the fractional representation $\frac{p}{q}$ is lowest in terms of $q>0$. Prove that $A$ is connected, but not pathwise connected.
I found that proving the connectedness of $S$ directly from definition is hard. So I'm trying to prove it using the property of connectedness that if $A\subset S\subset \bar{A}$, where $A$ is connected, then $S$ is connected. However, I'm having trouble finding this connected subset $A$. Any suggestions?
| Suppose we had two non-empty clopen sets $U, V$ partitioning $S$. Each set of the form $\{x\} \times [0, 1] \subseteq S$, where $x \in [0, 1] \setminus \Bbb{Q}$, must be entirely contained in $U$ or entirely contained in $V$, since each such subset is connected.
Let $A = \{x \in [0, 1] \setminus \Bbb{Q} : \{x\} \times [0, 1] \subseteq U\}$ and similarly define $B$ in relation to $V$. I claim that $A$ and $B$ are clopen in $[0, 1] \setminus \Bbb{Q}$. To see $A$ is open, suppose $x \in A$. Then $(x, 0) \in U$, which is open, hence a ball $B_S((x, y); r)$ exists in $U$. It's straight forward to verify that $B_A(x; r) \subseteq A$. Similarly, $B$ is open, and given both partition $[0, 1] \setminus \Bbb{Q}$, $A$ and $B$ are both clopen.
Due to the density of the irrationals, and the fact that $U, V$ are open and non-empty, we can also verify that $A$ and $B$ are both non-empty.
Consider now the closures of $A$ and $B$ in $[0, 1]$. The density of the irrationals imply that the closures of these sets cover $[0, 1]$. By the connectedness of $[0, 1]$, there must exist some point in their intersection. That is, there exists some $x$ that is simultaneously the limit (in $[0, 1]$) of a sequence $a_n \in A$ and a sequence $b_n \in B$. As both $A$ and $B$ are closed and disjoint, we must have $x \in \Bbb{Q}$.
Now, if $x$ has even lowest denominator, then consider the sequences $(a_n, 0)$ and $(b_n, 0)$. If $x$ has odd lowest denominator, then consider the sequences $(a_n, 1)$ and $(b_n, 1)$. In either case, each sequence respectively belongs to $A$ and $B$, converging to a point in $S$. As $U$ and $V$ are closed, this would produce a point in $U \cap V$, contradicting $U$ and $V$ being disjoint, and proving connectedness.
| {
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Why does the Big O notation of summation give log I’m not understanding the following line in a proof
I believe it should be $O(h^2\psi(2h))$ since $\sum_{j=1}^{h}\frac{1}{j}\leqslant h\cdot1=h$
| You found the correct statement
$$\sum_{j=1}^h \frac1j \in O(h).$$
The authors of the paper used the correct statement
$$\sum_{j=1}^h \frac1j \in O(\log h),$$
as explained in the answer by Parcly Taxel. Since $\log h \in o(h)$, the paper uses the sharper, "better" bound. Yours isn't wrong, it's just not as good as could be.
You can find details about why $\sum_{j=1}^h \frac1j \in O(\log h)$ in the wikipedia entry for the Harmonic numbers.
| {
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Numerical Approximation of the derivative In high school students are typical presented with the one sided definition of the derivative. Let $\epsilon > 0 $ and $f(x)$ be a mapping between two vector spaces then by definition
$$f'(x) = \lim_{\epsilon \to 0} \frac{f(x + \epsilon) - f(x)}{\epsilon}$$
Another definition is given as
$$f'(x) = \lim_{\epsilon \to 0} \frac{f(x + \epsilon) - f(x - \epsilon)}{2\epsilon}$$
From a theoretical point of view these two are the same since $\epsilon$ truly approaches zero. However, from a practical point of view in the world of numerical computation we can only approximate the limit. It turns out that the second definition yiels an error of $O(\epsilon^2)$ whereas the first one yields an error of $O(\epsilon)$. Why is this the case? Is there a known proof of this?
| Recall Taylor expansion: for $f \in C^3$, you have:
*
*$f(x+h) = f(x) + f'(x)h + \dfrac{1}{2} f''(x) h^2 + O(h^3)$
*$f(x-h) = f(x) - f'(x)h + \dfrac{1}{2} f''(x) h^2 + O(h^3)$
Then with appropriate differences, you have:
*
*$\dfrac{f(x+h) - f(x)}{h} - f'(x) = \dfrac{1}{2} f''(x) h + O(h^2) = O(h)$
*$\dfrac{f(x+h) - f(x-h)}{2h} - f'(x) = O(h^2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3888155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Relationship between the length of the tangent line through a point on sphere and great-circle distance As an aviator I'm familiar with the concept of great-circle navigation because when we fly a route between 2 points on the globe we know the shortest distance between these two points is the great circle distance.
I'm developing a navigation app in Google Earth and I need to calculate the shortest distance from the surface of the "spherical" Earth to any point on the tangent line through A (origin) when flying the great circle path.
Also, I'm using a mean earth radius of 6,371.009 km for WGS84 ellipsoid.
Just to be clear, I'd like to refer to the diagram in the following link:
http://www.alaricstephen.com/main-featured/2017/5/22/the-haversine-formula
I use the Haversine formula to calculate the distance, d, between the points A and D (see diagram). What I'd like to calculate is the distance D to E as a function of d.
In the diagram this is referred to as the external secant (exsec) which is the portion DE of the secant exterior to the circle.
| HaverSine Formula is used routinely to compute long distances in navigation along shortest path great circles of the Earth between two points of given latitude and longitude.
First find $d$ on the Earth. Next air distance along a tangent if the flight point $B$ is above the Earth: $ t= r \tan \dfrac{d}{r}.$
$-------------------------$
After clarification the above can be ignored.
For perfect sphere model of earth it is simple trig. calculation.
Distance $ AD= r \theta = $ the arc distance you calculated using Haversine formula along a great circle arc of earth radius $=r$ as shown. Calculate $ \theta $ in radians in the plane of kite shape $OAEB$ if we imagine $B$ on another tangent point below. We have $ \theta= \dfrac{\text{arc} AD}{r}$
Central dimension is length $OE$. From this subtract earth radius.
$$ DE = r \sec \theta - r\; = r (\sec \theta -1 )$$
This is the red height above target/destination/landing place which should vanish on landing at $D$. It is indicated by exsec in the supplied link for unit earth radius.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3888279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Mean & variance of normal distribution Let $Y = \ln(X)$, where $Y$ follows a normal distribution with mean $μ$ and variance $σ^2$. What is the mean and variance of $X$?
I know that if $Z$ follows a normal distribution with mean $μ$ and variance $σ^2$
, then $M_z(t) = e^{μt+ \tfrac{1}{2}σ^2 t^2}$
| If you know that a normal distribution $Y \sim \mathcal{N}(\mu,\sigma^2)$ has a moment generating function of $M(t) = \mathbb{E}[e^{tY}] = e^{\mu t + \sigma^2t^2/2}$, then you can use that to calculate the mean and variance of $X = e^Y$ as follows:
$$\mathbb{E}[X] = \mathbb{E}[e^Y] = M(?) \cdots$$
$$\text{Var}[X] = \mathbb{E}[X^2] - \mathbb{E}[X]^2 = \mathbb{E}[e^{2Y}] - \mathbb{E}[e^Y]^2 = M(?) - M(?)^2 = \cdots$$
I'll let you fill in the $?$'s and continue the calculations from there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3888580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate: $\lim_{n\to \infty}\left(\frac{a+n}{b+n}\right)^n$ I tried this way : $$\lim_{n\to \infty}\left(\frac{a+n}{b+n}\right)^n=\lim_{n\to \infty}\left(1+\frac{a-b}{b+n}\right)^n$$
We know $\lim_{n\to \infty}\left(1+\frac1u\right)^u=e$. therefor in this case we have: $$\lim_{n\to \infty}\left(1+\frac{a-b}{b+n}\right)^{\tfrac{b+n}{a-b}}=e$$
Here $n$ goes to infinity so we can ignore the number ($b$)added to it in numerator of the exponent. So $\lim_{n\to \infty}\left(1+\frac{a-b}{b+n}\right)^n=e^{a-b}$
Is my answer right? and is there any other approach to evaluate the limit?
| I think it is easier to write the expression as $$\frac{(1+\frac{a}{n})^{n}}{(1+\frac{b}{n})^{n}}$$ and take it from there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3888712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
A limit with an arctan I need to calculate another limit, $$ \underset{x\rightarrow\infty}{\lim}\left(\frac{2}{\pi}\arctan x\right)^{\sin\left(2/x\right)}$$
I have no idea how to start or what to do because I've never dealt with an arctan in a limit before.
What should I do first?
| First of all: don't panic! :-)
There is a nice identity for the arctangent:
$$
\arctan x+\arctan\frac{1}{x}=
\begin{cases} \pi/2 & x>0 \\[4px] -\pi/2 & x<0 \end{cases}
$$
Since the limit is for $x\to\infty$, it's not restrictive to assume $x>0$, so
$$
\arctan x=\frac{\pi}{2}-\arctan\frac{1}{x}
$$
and, setting $y=1/x$, the limit becomes
$$
\lim_{y\to0^+}\Bigl(1-\frac{2}{\pi}\arctan y\Bigr)^{\sin2y}
$$
which is not an indeterminate form because it is $1^0=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3888883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
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