问:
Integral of Binomial $f(z,r) = \int_0^\infty {\binom{ r}{t} z^{t} dt} $
The binomial theorem
$$
(1 + z)^r = \sum_{k \geq 0}\binom{r}{k}z^{k} , \quad |z|<1,
$$
deals with the sum of the binomial coefficient times $z^{k}$ .

What function would instead result from taking the integral
$$
f (z,r) =
\int_{0}^{\infty}\binom{r}{t}z^{t}\,\mathrm{d}t\ ?
$$
(intending the binomial expressed through the Gamma function:
$\displaystyle\binom{r}{t} = \frac{\Gamma(r+1)}{\Gamma(t+1)\Gamma(r-t+1)}$.)

I am just at the beginning of analyzing this curiosity of mine, and before attempting that
I would ask your advise, thoughts and references if any already exist.
-- update  --

Following the interesting @metamorphy's answer, I found another way to demonstrate that,
for the bilateral integral
$$
I_B (w,z) = \int\limits_{ - \infty }^\infty
 {\left( \matrix{  z \cr   s \cr}  \right)w^{\,\,s} ds}
$$
since
$$
I_B (1,0) = \int\limits_{ - \infty }^\infty  {\left( \matrix{  0 \cr   s \cr}  \right)ds}
  = \int\limits_{ - \infty }^\infty  {{{\sin \left( {\pi s} \right)} \over {\,\pi s}}ds}  = 2^{\,0}  = 1
$$
then, by the Vandermonde convolution we get
$$
\eqalign{
  & I_B (1,z) = \int\limits_{ - \infty }^\infty
 {\left( \matrix{  z \cr   s \cr}  \right)w^{\,\,s} ds} \quad \left| {\;0 \le } \right.
{\mathop{\rm Re}\nolimits} \left( z \right)\quad  =   \cr
  &  = \int\limits_{ - \infty }^\infty  {\sum\limits_k
 {\left( \matrix{  z \cr   k \cr}  \right)\;\left( \matrix{  0 \cr  s - k \cr}  \right)} \;ds}
  = \sum\limits_k {\left( \matrix{  z \cr   k \cr}  \right)\int\limits_{ - \infty }^\infty
  {\left( \matrix{  0 \cr   s - k \cr}  \right)\;ds} \;}  =   \cr
  &  = \sum\limits_k {\left( \matrix{  z \cr   k \cr}  \right)\;}  = 2^{\,z}  \cr}
$$
where for the convolution to be valid it is required that $0 \le \Re (z)$, while the approach
used by @metamorphy requires that $-1< \Re (z)$.

In any case it seems that the bilateral integral be more interesting to analyze, but only formally, because it converges only for $|z|=1$.

答:
Not an answer, but too long for a comment: I wouldn't expect a closed form.

If $r$ is a nonnegative integer, then (as I show in this answer) $$\binom{r}{t}=\frac{\sin\pi t}{\pi}\sum_{k=0}^r\binom{r}{k}\frac{(-1)^k}{t-k};$$ thus, even at $z=1$, $\int_0^\infty\binom{r}{t}\,dt$ is a combination of values of the sine integral.

Let's replace $z$ with $e^{-z}$ (now we assume $\Re z>0$). So, if $r$ is a nonnegative integer, $\int_0^\infty\binom{r}{t}e^{-zt}\,dt$ expresses in terms of $\int_0^x\frac{\sin\pi t}{\pi t}e^{zt}\,dt$ at integer values of $x\leqslant r$ (and some elementary extra things). Clearly, the situation when $r$ is not an integer is even more complicated. Anyway, alternative integral representations come from the ones for $\binom{r}{t}$, that is, for the "reciprocal beta".

Say, if $\Re(a+b)>-1$ and $0<c<1$, we have $$\int_{c-i\infty}^{c+i\infty}w^{-a-1}(1-w)^{-b-1}\,dw=\frac{2\pi i\,\Gamma(a+b+1)}{\Gamma(a+1)\Gamma(b+1)},\tag{I}\label{intrep}$$ which gives a representation for $\binom{r}{t}$, $\Re r>-1$ if we put $a=t$ and $b=r-t$; this results in $$\int_0^\infty\binom{r}{t}e^{-zt}\,dt=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}w^{-1}(1-w)^{-r-1}\big(z+\log w-\log(1-w)\big)^{-1}dw$$ for $1/2<c<1$ (at least). If we deform the path of integration to encircle any of the branch cuts of the integrand, we obtain real alternative integral representations. The same can be done for $\Re r\leqslant-1$.

Update. The integral representation \eqref{intrep} can be used to show that
$$\int_{-\infty}^\infty\binom{s}{x}\,dx=2^s\qquad(\Re s>-1)$$ (stated in this answer). We take $c=1/2$ and substitute $w=\displaystyle\frac12\left(1-i\tan\frac\phi2\right)$: $$\binom{s}{x}=\frac{2^s}{2\pi}\int_{-\pi}^\pi\left(1+i\tan\frac\phi2\right)^{-s}e^{ix\phi}\,d\phi,$$ which, integrated over $x\in(-y,y)$, after substituting $\phi=t/y$, gives $$\int_{-y}^y\binom{s}{x}\,dx=\frac{2^s}{\pi}\int_{-y\pi}^{y\pi}\left(1+i\tan\frac{t}{2y}\right)^{-s}\frac{\sin t}{t}\,dt.$$ Now take $y\to\infty$ (DCT is clearly applicable here).


问:
How to evaluate this integral $I= \int_a^t {(t-s)^{\alpha-2}f (s)\, ds} $ with $0 <\alpha <1$?
I have a positive real valued function say $f$. Suppose this function is also differentiable. Then is the following integral defined for any $t>a$ and $0 <\alpha <1$?
$$I= \int_a^t {(t-s)^{\alpha-2}f (s)\, ds} $$
The problem is if i apply integration by parts by taking $dv=(t-s)^{\alpha-2}ds$, $u=f (s)$ then i could not apply the upper limit. Any idea on how to evaluate this integral would be of great help.

答: Since $f$ is differentiable, it is also continuous, and therefore bounded on $[a,t]$, say $|f|\le M$ for some $M\gt 0$. Therefore $|\int_a^t (t-s)^{\alpha-2}f(s)ds|\le M|\int_a^t (t-s)^{\alpha-2}ds|=\dfrac{M(t-a)^{\alpha-1}}{\alpha-1}$.

ContentLicense: CC BY-SA 4.0


问:
Probability question - finding the minimum number of people at an event of 100
At an event, 100 people are invited. 50% are coming alone and the rest will bring either 1 or 2 plus ones. What is the minimum number of 6 people tables that should be at booked at the event?

So, I'm confused on what it should be.

If I assume that 50 people are single, and the other 50 only bring 1 extras, then the total will be 150, giving us 25 tables. And if I assume, that there is a probability of 50/50 for plus 1 or 2, then there will be 75 extra, which would then be a total of 175, so ~30 tables. Or take the max number of people, 200 and so the minimum number of tables would be 34 tables.

Which should be correct?

Thank you for the help!

答: Your conditions allow for the case of 50 alone and 50 with 2 guests, or $50 + 50 \cdot 3 = 200$.  So you need $\lceil \frac{200}{6} \rceil = 34$.


问:
Does there exist a matrix $P$ such that $P^n=M$ for a special matrix $M$?
Consider the matrix
$$
M=\left(\begin{matrix}
0&0&0&1\\
0&0&0&0\\
0&0&0&0\\
0&0&0&0\\
\end{matrix}\right).
$$
Is there a matrix $P\in{\Bbb C}^{4\times 4}$ such that $P^n=M$ for some $n>1$?

One obvious fact is that if such $P$ exists, then $P$ must be nilpotent. However, I have no idea how to deal with this problem. Furthermore, what if $M$ is an arbitrary nilpotent matrix with index $k$?

答:
Purely by trial and error, let
$$P=\left(\begin{matrix}
0&1&1&1\\
0&0&0&1\\
0&0&0&0\\
0&0&0&0\\
\end{matrix}\right)$$ which squares to $M$.

The matrix
$$P=\left(\begin{matrix}
0&1&0&0\\
0&0&0&1\\
0&0&0&0\\
0&0&0&0\\
\end{matrix}\right)$$
and its anti-transpose are more simple and also work.

ContentLicense: CC BY-SA 3.0


问:
Get Point on the Perimeter of a Rectangle from Topleft Corner
I stumbled upon the need to get a random point on the perimeter of a rectangle while making a game, the solution programmatically is easy enough, tho i was curious of a mathematical solution too.

Say we have a random Distance we want to travel from the TopLeft corner of the rectangle along all its sides.
We known the Width and Height of the rectangle, including the Distance or Length that we'll travel around it.

Our point will only ever be on the edge of the rectangle and never inside it.

Is there a simple equation to get the X,Y coordinates?

Here's an illustration in mspaint

Note that our random Distance can be the full perimeter of the rectangle, having us travel all around it completely.

答:
Assuming positive $x$ is right and positive $y$ is down, and the rectangle has width $w$ and height $h$, and the distance travelled is $0\le d\le 2(w+h)$ then

$$(x,y) = \begin{cases}(d,0) &\text{for } 0\le d\le w\\
(w,d-w) &\text{for } w\le d\le w+h\\
(2w+h-d,h) &\text{for } w+h\le d\le 2w + h\\
(0,d-2w-h) &\text{for } 2w+h\le d\le 2(w+h)\\\end{cases}$$

If $d$ exceeds $2(w+h)$ replace $d$ by its remainder when divided by $2(w+h)$.


问:
Periodic sequence problem
Given sequence $a_n$ defined such that $a_1=3$, $a_{n+1}=\begin{cases}\frac{a_n}{2},\quad 2\mid a_n\\ \frac{a_n+1983}{2},\quad 2\nmid a_n\end{cases}$. Then prove that the sequence $a_n$ is periodic and find the period.

It's easy to prove that $0<a_n<1983$ by induction. By pigeonhole principle, there exist $i,j$ such that $a_i=a_j\implies a_{i+1}=a_{j+1}$. By induction, we can prove $a_{i+k}=a_{j+k},\forall k\in\mathbb{N}$. Otherwise, $a_n\begin{cases}2a_{n+1}, \quad a_{n+1}\le 991\\ 2a_{n+1}-1983, \quad a_{n+1}\ge 992\end{cases}$. So we can prove also $a_{i-k}=a_{j-k} $ for $min(i,j)>k, \forall k\in\mathbb{N}$. So it's periodic. But I can't find the period. In my opinion, the period is $660$. Because $3\mid a_n$ and $0<a_n<1983$. But I can't prove $\forall k, \exists i$ such that $a_i=3k$, Can anyone help me?

Title: Periodic sequence problem

Body:
Put $p=661=1983/3$ and for each natural $i$ put $b_i\equiv a_i/3 \pmod p$. Then $b_1\equiv 1\pmod p $ and $b_{i-1}=2 b_i\pmod p$ for each $i>1$. Counting $\{b_i\}$ backwards from sufficiently large $i$, we see that its period $N$ is the smallest integer $n$ such that $2^n\equiv 1\pmod p$. Since $p$ is prime, by the Fermat little theorem, $2^{p-1}\equiv 1\pmod p$, so $N|p-1=2^2\cdot 3\cdot 5\cdot 11$. So to show that $N=p-1$ it suffices to check that $2^n\not\equiv 1\pmod p$ for each $n\in \{(p-1)/2, (p-1)/3, (p-1)/5, (p-1)/11\}$. Calculating modulo $p$, we see that

$2^{11}\equiv 2048\equiv 65$, $65^3\equiv 310$, $65^5\equiv 309$.

$2^{(p-1)/2}-1\equiv 2^{330}-1\equiv 65^{30}-1\equiv (65^{15}+1) (65^{15}-1)$,

$65^{15}+1\equiv (65^5+1)(65^5(65^5-1)+1) \equiv 310\cdot (309\cdot 308+1)\not\equiv 0$,

$65^{15}-1\equiv (65^5-1)(65^5(65^5+1)+1) \equiv 308\cdot (309\cdot 310+1)\not\equiv 0$.

$2^{(p-1)/3}-1\equiv 2^{220}-1\equiv 65^{20}-1\equiv (65^{10}+1) (65^5+1) (65^5-1),$

$65^{10}+1\equiv 309^2+1\not\equiv 0.$

$2^{(p-1)/5}-1\equiv 2^{132}-1\equiv 65^{12}-1\equiv (65^6+1) (65^3+1) (65^3-1),$

$65^6+1\equiv 310^2+1\not\equiv 0.$

$2^{(p-1)/11}-1\equiv 2^{60}-1\equiv (2^{30}+1)(2^{15}+1) (2^{15}-1),$

$2^{15}\equiv 2^{11}\cdot 2^4 \equiv 65\cdot 16\equiv 379\not\equiv \pm 1,$

$2^{30}+1\equiv (2^{15})^2+1\equiv 379^2+1\not\equiv 0.$


问:
Show that $x^{2} \equiv 1 \pmod{15}$ has four incongruent solutions mod 15.
Show that $x^{2} \equiv 1 \pmod{15}$ has four incongruent solutions mod 15.

My instinct was to find the primitive root and then use a theorem to directly show the number of incongruent solutions, which follows from knowing the primitive root. But, apparently there are no primitive solutions to $mod 15.$ So what's another option?

答:
Let $\phi$ denote Euler's totient function.

We have

$\quad \phi(3) = 2$

$\quad \phi(5) = 4$

$\quad \phi(15) = \phi(3) \cdot \phi(5) = 2 \cdot 4 = 8$

So the multiplicative group of integers modulo $15$,  ${\displaystyle (\mathbb {Z} /15\mathbb {Z} )^{\times }}$, contains $8$ elements. Since the solutions to

$\tag 1 x^2 \equiv 1\pmod{15}$

forms a subgroup, the number of solutions has to divide $8$.

Since $2^2 \equiv 4 \pmod{15}$, the order of $[2] \in {\displaystyle (\mathbb {Z} /15\mathbb {Z} )^{\times }}$ is greater than $2$. It follows that the number of solutions for $\text{(1)}$ must divide $4$.

Calculating further we determine that the order of $2$ is equal to $4$.

Observing that $-1 \equiv 14 \pmod{15}$, we now have $3$ solutions:

$\quad 1^2 \equiv 1 \pmod{15}$

$\quad 4^2 \equiv 1 \pmod{15}$

$\quad 14^2 \equiv 1 \pmod{15}$

We conclude that there are exactly $4$ solutions for $\text{(1)}$.


问:
Find the limit $\lim\limits_{n \to \infty} \left ( \left (1 + \frac {1} {2} + \cdots + \frac {1} {n} \right ) - \ln n \right ).$

Prove that the following limit exists and find the limit $:$ $$\lim\limits_{n \to \infty} \left ( \left (1 + \dfrac {1} {2} + \cdots +  \dfrac {1} {n} \right ) - \ln n \right ).$$

I know that the above sequence is strictly decreasing and bounded below by $0$ and hence by monotone convergence theorem it has to converge to it's infimum which is known as Euler-Mascheroni's constant. But how do I evaluate the limit?

Any help or suggestion in this regard will be highly appreciated. Thanks in advance.

Source $:$ ISI (Indian Statistical Institute) PhD entrance test in Mathematics, TEST CODE : MTA (FORENOON SESSION) (Question No. $1$) held in $20$th September this year.

答:
It is known that
$$
1 + \frac{1}{2} +  \cdots  + \frac{1}{n} - \log n = \gamma  + \frac{1}{{2n}} - \sum\limits_{k = 1}^{N - 1} {\frac{{B_{2k} }}{{2k}}\frac{1}{{n^{2k} }}}  - \theta _N (n)\frac{{B_{2N} }}{{2N}}\frac{1}{{n^{2N} }}
$$
where $0<\theta_N(n)<1$ is a suitable number and the $B_m$ denotes the Bernoulli numbers. You can re-arrange this formula to obtain accurate approximations for $\gamma$. If you take $N = \left\lfloor {\pi n} \right\rfloor$,
$$
\left| {\frac{{B_{2N} }}{{2N}}\frac{1}{{n^{2N} }}} \right| = \mathcal{O}\!\left(\frac{e^{ - 2\pi n}}{\sqrt n} \right).
$$
Hence, the error will be exponentially small in $n$.


问:
Distinguishing between a parabola and (one branch of) a hyperbola by looking at the graph

If I were to show you this curve that is quite zoomed in is it possible to tell if it is a hyperbola of a parabola

This is a picture my friend sent me and I cannot tell which conic section it is.

答:
This looks like a hyperbola.

For $x > 4$, the gradients of the curve are nearly constant for both positive and negative values of $y$. This suggests it is approaching an asymptote, something which does not happen for a a parabola.


问:
Proving that $Aut_{\mathbb{Q} } { \mathbb{R}} $
The following question is from my Field Theory assignments.

Prove that  $Aut_{\mathbb{Q} } { \mathbb{R}} $  is identity group.

I took $f:\mathbb{R} \to \mathbb{R}$ be an automorphism.
Now  which result I should use to prove that all the automorphism which are $\mathbb{Q}$ -module homomorphisms are identity.

Let $f'$ be such an automorphism and q belongs to $\mathbb{Q}$. So, $f'(q) = qf'(1) =q$.but how to prove that $f'(q) =1$ .

Kindly guide.

答:
First off, note that it would be enough to show that any such automorphism must be continuous. (and then since any element of $\Bbb R$ is a limit of elements in $\Bbb Q$ you are done)

To do this, note that for $a\in\Bbb R$, we have $f(a)=f(\sqrt a ^2)=f(\sqrt a)^2\geq0$, so $f$ must be increasing, and its continuity follows because if $f(a^+)>f(a^-)$ then the elements in the open interval $(f(a^-),f(a^+))$ don't appear in the image of $f$ (since it is strictly increasing).

We thus get that $f(a^+)=f(a^-)$ for any $a\in\Bbb R$ and thus $f$ is continuous, proving the claim using $f(q_n)=q_n\to r$ for $q_n$ a sequence of rationals converging to $r$.


答:
From the definition of the Riemann's integral, it is easy to see that:
$$
\int_1^{n+1}\frac1xdx\lt\sum_{k=1}^n\frac 1k\lt1+\int_1^{n}\frac1xdx
$$
We arrive at
$$
\ln(n+1)\lt\sum_{k=1}^n\frac 1k\lt1+\ln n$$
As a rough approximation we can take the average of the 2 bounds:
$$
\sum_{k=1}^n\frac 1k\approx\frac12+\frac12\ln [n(n+1)]
$$
and the appoximation for $\gamma$ is:
$$
\gamma\approx\frac12+\frac12\ln [n(n+1)]-\ln n=\frac12+\frac12\ln(1+\frac1n)
$$
As $n$ increases we get $\gamma\approx0.5$. The actual value is $\gamma=0.577...$ so it is not highly accurate, but considering the effort (not much) I think it should be satisfying...

Title: Find the limit $\lim\limits_{n \to \infty} \left ( \left (1 + \frac {1} {2} + \cdots + \frac {1} {n} \right ) - \ln n \right ).$

Body:
Prove that the following limit exists and find the limit $:$ $$\lim\limits_{n \to \infty} \left ( \left (1 + \dfrac {1} {2} + \cdots +  \dfrac {1} {n} \right ) - \ln n \right ).$$

I know that the above sequence is strictly decreasing and bounded below by $0$ and hence by monotone convergence theorem it has to converge to it's infimum which is known as Euler-Mascheroni's constant. But how do I evaluate the limit?

Any help or suggestion in this regard will be highly appreciated. Thanks in advance.

Source $:$ ISI (Indian Statistical Institute) PhD entrance test in Mathematics, TEST CODE : MTA (FORENOON SESSION) (Question No. $1$) held in $20$th September this year.


问:
Critical points of the moment map
Let $(M,\omega)$ be a symplectic compact connexe manifold endowed with a hamiltonian action of a torus $T$. Let $\mu : M \longrightarrow {Lie(T)}^*,$  be a moment map associated to this action.

We say that a point $m \in M $ is critical point of $\mu$, if $T_m \mu : T_m M \longrightarrow {Lie(T)}^*$,  the tangent map of $\mu$ is not surjective.

How can one prove that the above definition of critical points of the moment map $\mu$ is equivalent to the following statements:

(1).  $m \in M $ is critical point of $\mu$ if and only if the stabilizer of m, stab(m), contains a sub-torus of T of dimension 1.

(2). The tangent map $T_m \mu$ is not surjective if there exists an element $X \in Lie(T)-\lbrace 0 \rbrace$, such that $\langle T_m \mu, X \rangle = 0$.

Any feedback would be appreciated.

答: More information in this book of Jean-Louis Koszul, especially for the case of non-null cohomology based on Jean-Marie Souriau works: https://link.springer.com/book/10.1007%2F978-981-13-3987-5</p>

Title: Critical points of the moment map


问:
$\sum_{n=0}^{\infty}(-1)^n a_n = \pi$, $a_n\in \mathbb Q$ and not $a_n$ not monotonic
How can I construct a sequence $\{a_n\}$ of positive rational numbers, which is not monotonic such that $$\sum_{n=0}^{\infty}(-1)^n a_n = \pi$$
I thought a lot on this question but I always come up with a monotonic sequence. Any help will be appreciated.</p>

答:
There are a few methods possible for achieving this. One is to make use of Newton's method as it applies to $\sin(x)$ (which will repeatedly overestimate the step by a small amount). That is,

$$
\Delta x = -\frac{\sin(x)}{\cos(x)} = -\tan(x)
$$
Choose a starting value slightly above $\pi$, say $a_0=3.5$. Then let $a_n=-\tan(\sum_{i=0}^{n-1} a_i)$. Each term will then be of opposing sign relative to the one before it, and it will converge rapidly to $\pi$.

EDIT:

Oh, just noticed the "$a_n\in\mathbb{Q}$" condition. This can be enforced by simply using a rational approximation to $\tan$ for each step, simply requiring that it be, for instance, $\frac1m$ for the integer $m$ that makes it nearest to the value of $\tan$ that is a longer step than $\tan$ itself. For instance, the first iteration has $a_1=-\tan(3.5)\approx -0.374$. The nearest value of $m$ is thus $3$, but that makes it a smaller step, so choose $2$, and you have $a_1=-\frac12$.

Then, $-\tan(3)\approx 0.142$, which is nearest to $\frac17$, but that's too short a step, so we take $\frac16$. Now $-\tan(3+1/6)\approx -0.0251$, so the nearest is $\frac1{40}$, but again, too big a step, so use $m=39$. And so on. The sequence of values of $m$ is thus (in absolute value, for each step):
$$
2, 6, 39, 1763...
$$


问:
"Negative" versus "Minus"
As a math educator, do you think it is appropriate to insist that students say "negative $0.8$" and
not "minus $0.8$" to denote $-0.8$?

The so called "textbook answer" regarding this question reads:

A number and its opposite are called additive inverses of each other because their sum is zero, the identity element for addition. Thus, the numeral $-5$ can be read "negative five," "the opposite of five," or "the additive inverse of five."

This question involves two separate, but related issues; the first is discussed at an elementary level here. While the second, and more advanced, issue is discussed here. I also found this concerning use in elementary education.

I recently found an excellent historical/cultural perspective on What's so baffling about negative numbers? written by a Fields medalist.

答:
I had chanced upon this page with nary a doubt that “negative $3$” is the superior choice (and “minus $3$” mildly puerile) to express the negative difference between $5$ and $2$. However, while scrolling through the responses, I became increasingly swayed in the minus direction.

The strongest arguments were

the chorus of international observations that “negative $3$” is
linguistically nonstandard (which supports that “minus $3$” is the
more logical/fundamental cognitive construct while “negative $3$” is
perhaps cultural);

that $\pm{3}$ is always read as “plus minus $3$” anyway;

and that the $-$ symbol when used as a prefix is generally more
instructive (and less inconsistent) when construed as a (unary)
arithmetic operator than as an adjective: $$7=-(-7)$$ “negative
negative $7$”(?!) versus “minus minus $7$”.

Interestingly though, it appears that Mandarin Chinese runs counter to the European languages in reading $$2-5=(-3)$$ as “$2$ minus $5$ equals negative 3” (“2 减 5 得 负 3”; the character 负 is opposed to the character 正 meaning positive). (However, “minus $3$” is colloquially used in some Mandarin-speaking regions.)

All things considered, I shall henceforth be saying “minus $3$” instead of “negative $3$”.

Title: "Negative" versus "Minus"


问:
Integral of a discontinuous function

Define
$$
 f(x)=\begin{cases}
      0 & x=\frac{n}{n+1}, n\in \mathbb{N} \\
      1 & \text{elsewhere}
   \end{cases}
$$
Find $$\int_0^1f(x)dx$$

One could argue geometrically that $f(x)=1$ occurs more frequently than $f(x)=0$, so the area must tend to $1\times1=1$

How do I solve this mathematically(rigorously)?

I tried to define the area as a Riemann sum, but failed miserably.

答:
Let $N$ be a positive integer. The interval $B_N = [0, 1-\frac{1}{N}]$ contains $N$ points of the form $\frac{n}{n+1} = 1 - \frac{1}{n+1}$. Let us partition $[0, 1]$ by partitioning $B_N$ in $N^2$ intervals of equal lengths and the last interval of the partition is $[1 - \frac{1}{N}, 1]$. Let $L_N$ and $U_N$ be the lower and upper Darboux sums of $f$ for this partition of $[0,1]$. Then
\begin{equation}
1 -\frac{2N}{N^2}-\frac{1}{N}\le L_N \le U_N\le 1
\end{equation}
It follows that $\int_0^1 f(x) d x = 1$ by letting $N\to\infty$.

Title: Integral of a discontinuous function


问:
Clarification on probability problem involving groups and combinations
Here's a question that I've thought of myself.

On an online Zoom meeting, the host divides $15$ participants into five groups of three people each. The next day, the host does the same. (The participants' names, for simplicity, are $A, B, C, D, E, F, G, H, I, J, K, L, M, N, \text{ and } O$.) What's the probability that if $ABC$ were in the same group the first time, they are in the same group the second time?

Here's my approach:

There are $\binom{15}{3} \cdot \binom{12}{3} \cdot \binom{9}{3} \cdot \binom{6}{3} \cdot \binom{3}{3}$ total groups. If we define one group to have the same members as yesterday, we have $12$ members left to form four different groups. We have $\binom{12}{3} \cdot \binom{9}{3} \cdot \binom{6}{3} \cdot \binom{3}{3}$ ways for the numerator. However, we need to use PIE since we overcount, so we calculate the total number of ways two groups have the same people.

This is just $\binom{9}{3} \cdot \binom{6}{3} \cdot \binom{3}{3}$ if we define two teams to be the same. And so on, for PIE.

I'm just not sure whether to multiply by $\binom{5}{2}$ in that last expression. Normally in PIE you would do so, but since I am defining a specific group to be chosen again, do I multiply? Maybe it would be $4$, since we already have one defined team but the second team can be any of the four others?

Thanks in advance.

答:
The question is only asking about ABC, they aren't bothered about the rest.
Imagine 15 slots divided into 5 groups,
$\boxed{.}\boxed{.}\boxed{.}$  $\boxed{.}\boxed{.}\boxed{.}$ $\boxed{.}\boxed{.}\boxed{.}$  $\boxed{.}\boxed{.}\boxed{.}$  $\boxed{.}\boxed{.}\boxed{.}$

$If$ $the$ $groups$ $are$ $labelled$

A has 3 places available out of 15, B has 2 remaining places out of 14, etc.

so Pr = $\frac 3{15} \cdot\frac 2{14} \cdot\frac 1{13}$

$If$ $the$ $groups$ $are$ $unlabelled$

A can be anywhere, so Pr = $1 \cdot\frac2{14} \cdot\frac 1{13}$


问:
proof for that exactly one normal can be drawn to a parabola from an exterior point.
It is well known that a maximum of three normals can be drawn from the interior of a parabola. But I am interested in the proof for the fact that exactly one normal can be drawn to a parabola from an exterior point wrt to the parabola.

So far I have done this.

WLOG we consider $y^2=4ax$.

Let the exterior point be $(h,k)$.

We must have $$k^2-4ah> 0\tag1$$

Also the equation of a normal to a parabola is $$y=mx-2am-am^3$$
so $(h,k)$ satisfies $$am^3+2am-mh+k=0$$

All I have to do is show that there is only one real root possible for this cubic using $(1)$.
Can anyone complete the proof?

Alternative methods are welcome!

NoteBy a normal i mean that a segment that is external to parabola

Thank you!

答:
The question claims that exactly one normal can be drawn to a parabola from a point outside the parabola. This is incorrect.

Presumably the OP is imposing the restriction that the segment of the normal joining the point to the parabola must itself be outside the parabola. In this case, here is a proof not involving cubics:

WLOG, assume $a > 0$. Like in the question, let $(h, k)$ be a point outside the parabola.

Let $(x_p, y_p)$ be a point on the parabola. Consider the normal to the parabola at $(x_p, y_p)$; let $(x_n, y_n)$ be a point on the normal. We can divide the normal into two “rays”: one has $x_n > x_p$ and is (at least partly) inside the parabola; the other has $x_n < x_p$ and is outside the parabola. We are only interested in the second. From now on, we refer to this “ray” as the normal.

We also want the possibility that $(h, k)$ is on the normal, so we now require $x_p > h$.

Let $k'$ be the value of $y_n$ when $x_n = h$. That is, let $(h, k')$ be a point on the normal. Also let $m = -\frac{1}{2a}y_p$ be the gradient of the normal. Then

$$k' = y_p + m(h - x_p) = y_p(1 + \frac{1}{2a}(x_p - h))$$

so

$$|k'| = |y_p|(1 + \frac{1}{2a}(x_p - h))$$

is a continuous, strictly increasing, unbounded function of $|y_p|$ that takes on the value $|k|$ for exactly one value of $|y_p|$.

Since $k'$ has the same sign as $y_p$, $k'$ is a one-to-one function of $y_p$ that takes on the value $k$ for exactly one value of $y_p$.

This shows that exactly one normal contains the point $(h, k)$, as required.


问:
derivative of logarithmic arcsine yields unexpected result
So, I begin with knowing that:
$$ \sin{x} = \frac{e^{ix} - e^{-ix}}{2i} = L,~\text{s.t}~i=\sqrt{-1}$$
And so continuing to solve for $x$, yielding the standard result:
$$ \arcsin{L} = -i \ln{[Li + \sqrt{1 - L^2}]},~\text{assm. principal branch}$$
Then I attempt to find $\frac{d}{dL} \arcsin{L}$ from the leading definition, and I come up with:
$$\frac{d\arcsin{L}}{dL} = -i\left[(Li + \sqrt{1 - L^2})' * (\sqrt{1 - L^2} - Li)\right]
\\ = -i\left[i + \frac{L}{(1 - L^2)^{3/2}}\right]\left[\sqrt{1 - L^2} - Li\right]
\\ = -i \left[i\sqrt{1 - L^2} + L + \frac{L}{1 - L^2} - \frac{Li}{(1 - L^2)^{3/2}}\right]
\\ = \sqrt{1 - L^2} - Li - \frac{Li}{1 - L^2} - \frac{L}{(1 - L^2)^{3/2}}$$

Which, seems completely different to the regular result I find on the internet of:
$$ \frac{d\arcsin{x}}{dx} = \frac{1}{\sqrt{1 - x^2}}$$
I've tried to look for a similar derivation, but to no avail, so I wonder whether differentiating the complex arcsine function is intended to give a different answer to its (presumably) real counterpart? I'm sorry if it's obvious, but I've spent a few hours on this and I seem to be making mistakes in and out.

答:
There is an error in the second step of differentiation.

$$\begin{align*}\frac{d\arcsin{L}}{dL}&= -i\left[(Li + \sqrt{1 - L^2})' * (\sqrt{1 - L^2} - Li)\right]
\\&= -i\left[i \color{red}-\frac{L}{(1 - L^2)^{\color{red}{1/2}}}\right]\left[\sqrt{1 - L^2} - Li\right]\end{align*}$$which yields the correct result.


问:
Show that $V = G(\lambda_1,T)\oplus ... \oplus G(\lambda_m,T)$ where $G(\lambda,T)$ is the generalized eigenspace for $\lambda$

Suppose $V$ is a finite-dimensional complex vector space and $T:V\to V$ is a linear operator on $V$. Let $\lambda_1,...\lambda_m$ be the distinct eigenvalues of $T$. Then, show that $V = G(\lambda_1,T)\oplus ... \oplus G(\lambda_m,T)$, where $G(\lambda,T)$ is the generalized eigenspace corresponding to $\lambda$.

We know that $G(\lambda, T) = \ker (T-\lambda I)^{\dim V}$ (let $\dim V$ = n for simplicity). Also, we know that generalized eigenvectors corresponding to the eigenvalues $\lambda_1,...,\lambda_m$ are linearly independent. Using this, I was able to show that $G(\lambda_i,T) \bigcap G(\lambda_j,T) = \{0\}$ for $i\neq j$.

Now, it would suffice to show that $\sum_{k=1}^m \dim G(\lambda_k,T) = n$, right? Alternatively, we could also just show that if $0 = u_1 + u_2 + ... + u_m$ for $u_k \in G(\lambda_k,T)$, then $u_k = 0$ for all $k=1,2,...,m$. I'm not sure how to proceed, and would appreciate any hints.

In addition, I was able to show that each $G(\lambda_k,T)$ is $T$-invariant, and $(T-\lambda_k I)\vert_{G(\lambda_k,T)}$ is a nilpotent operator. I don't know if these will help.

P.S.
I came across an inductive proof, that does induction on the dimension of $V$ - but I'm really looking for something direct and non-inductive as above, if possible. Thanks!

答:
Let me denote the generalized eigenspace for $\lambda_k$ simply as $G_k$. You've shown that distinct generalized eigenspaces have trivial intersection, so to show that $V$ decomposes as a direct sum, it suffices to show that the sum of the generalized eignespaces span $V$, i.e., that every $\mathbf{v} \in V$ can be written as
$$\mathbf{v} = \sum_{k=1}^m\mathbf{v}_k,\tag{1}$$
where $\mathbf{v}_k \in G_k$. One way of doing this is to use Bezout's identity.

Let $p(x)$ be the characteristic polynomial, factorized as
$$p(x) = \prod_{k=1}^m (x-\lambda_k)^{a_k},$$
where $a_k$ denotes the algebraic multiplicities of each eigenvalue. For each $1 \le k \le m$, let $p_k$ denote the polynomial obtained from $p$ by dividing out the factor associated with $\lambda_k$, i.e.,
$$p_k(x) = \frac{p(x)}{(x-\lambda_k)^{a_k}}.$$
Let $F_k = \ker\left[(T-\lambda_k)^{a_k}\right]$. Then $F_k \subseteq G_k$. It will turn out that $F_k = G_k$, but we haven't shown this yet. We will obtain this fact automatically by showing that the $\mathbf{v}_k$ in equation $(1)$ can be taken to be elements of $F_k$.

Now, note that the collection of polynomials $\{p_k\}_{k=1}^m$ are collectively coprime. Therefore by Bezout's identity, there exists polynomials $\{f_k\}_{k=1}^m$ such that
$$\sum_{k=1}^mf_kp_k = 1.$$
Evaluating at $T$, we get the operator equation
$$\sum_{k=1}^mf_k(T)p_k(T) = I.$$
Now, let $\mathbf{v} \in V$ be arbitrary. Acting with the equation above on $\mathbf{v}$, we get
$$\sum_{k=1}^m f_k(T)p_K(T)\mathbf{v} = \mathbf{v}.$$
Let $\mathbf{v}_k = f_k(T)p_K(T)\mathbf{v}$. I claim that $\mathbf{v}_k \in F_k$. Indeed, we have
$$(T-\lambda_k)^{a_k}\mathbf{v}_k = f_k(T)p_k(T)(T-\lambda_k)^{a_k}\mathbf{v} = f_k(T)p(T)\mathbf{v} = \mathbf{0},$$
where the last equality follows from the Cayley-Hamilton theorem since $p(T) = 0$. Since $\mathbf{v}$ was arbitrary, it follows that
$$F_1 + \cdots + F_m = V.$$
Moreover, since you've already shown that $G_k$ are independent, it follows that the $F_k$ are also independent, and we can upgrade the sum above to a direct sum
$$F_1 \oplus \cdots \oplus F_m = V.$$
Finally, this also allows us to conclude that we must have $F_k = G_k$.


问:
If the columns of AB are linearly independent, how can I prove the columns of B must be linearly independent?
So a matrix, A, is linearly independent if Ax = 0 has only the trivial solution (x=0)

so let A = AB, then if the columns of AB are linearly independent, the equation (AB)x = 0 has only the trivial solution

by the associative law of multiplication, (AB)C = A(BC), so (AB)x = A(Bx) = 0

I'm not sure where to go from here, or even if this is a proper way to go about proving that the columns of B are linearly independent

答: If $Bx=0$ has a nonzero solution $x$, then it also solves $ABx=0$.


问:
Union of open balls around rationals - topologically
Assuming $(q_n)_{n \in \mathbb N  } $ is an enumeration of all the rational numbers, let $F$ be a subset of $ \mathbb R $
$$ F = \bigcup_{n \in \mathbb N} (q_n - \frac{1} {2^n},q_n + \frac{1} {2^n}) $$
Prove of $F \subsetneq \mathbb R $

My idea:
$F $ is a dense subset of $\mathbb R $, since it contains $\mathbb Q $.
The n-th open interval "becomes very small very fast",
and, if all these intervals were to be disjoint (which is not true, of course), then the sum of the measures of each interval would be finite (using the convergence of the geometric series $ \sum_{n=0}^{\infty} \frac{1} {2 ^n} $).
Though, I suspect the exercise requires an explicit (or constructive) proof. I found it in a general topology book, and I don't know even if I can use arguments from measure theory.
I saw other answers to a similar problem but using Lebesgue measure. Is it possible not to use it?

Anyway, what should I be considering? Thanks

答: Without any result from  measure theory we can show that if $[a,b]$ is covered by  a finite number of open intervals then $b-a$ cannot exceed the sum of then lengths of the intervals. [Use induction on the number of intervals].  If $\mathbb R =\bigcup (q_n-\frac 1  {2^{n}},q_n+\frac 1  {2^{n}})$ then, by compactness $[-2,2]$ is covered by  a finite number of the intervals $(q_n-\frac 1  {2^{n}},q_n+\frac 1  {2^{n}})$ which leads to the contradiction $4 \leq 2$ since $\sum \frac 2 {2^{n}}=2$.


问:
How to calculate first and second order partial derivatives of $g(x,y)=\phi(x^2+y^2)$
How to calculate first and second order partial derivatives of $g$ where:

$g(x,y)=\phi(x^2+y^2)$

Here's what I did :
$g(x,y)=\phi(x^2+y^2)=(\phi\circ h)(x,y)$ where :  $h(x,y)=x^2+y^2$

$\frac{\partial g(x,y)}{\partial x}$=$\frac{\partial \phi(h(x,y))}{\partial x}$$\times$$\frac{\partial h(x,y)}{\partial x}$ =$\frac{\partial \phi(x^2+y^2)}{\partial x}\times 2x$

$\frac{\partial g(x,y)}{\partial y}$=$\frac{\partial \phi(h(x,y))}{\partial y}$$\times$$\frac{\partial h(x,y)}{\partial y}$ =$\frac{\partial \phi(x^2+y^2)}{\partial y}\times 2y$

and :

$\frac{\partial^2 g(x,y)}{\partial x^2}$=$\frac{\partial^2 \phi(h(x,y))}{\partial x^2} \times 2x +2\times \frac{\partial \phi(h(x,y))}{\partial x}$

$\frac{\partial^2 g(x,y)}{\partial y^2}$=$\frac{\partial^2 \phi(h(x,y))}{\partial y^2} \times 2y +2\times \frac{\partial \phi(h(x,y))}{\partial y}$

are these steps correct ?

答:
Your second derivatives are wrong because you made a small mistake in taking the first partial derivative.

$g(x,y)=\phi(h(x,y))\implies\frac{\partial g}{\partial x}=\frac{\partial \phi(h)}{\partial \color{red}h}\times \frac{\partial h}{\partial x}=2x\phi'(h)$

So the second derivative is$$\begin{align*}\frac{\partial g^2}{\partial^2x}&=2\phi'(h)+2x\times\frac{\partial \phi'(h)}{\partial x}=2\phi'(h)+2x\times\frac{\partial \phi'(h)}{\partial h}\times\frac{\partial h}{\partial x}\\&=2\phi'(h)+4x^2\phi''(h)\end{align*}$$

By symmetry you can obtain $g_y,g_{yy}$ by substituting $y$ instead of $x$ in the above expressions. The mixed partial derivative will be$$\frac{\partial g^2}{\partial y\partial x}=\frac{\partial}{\partial y}[2x\phi'(h)]=4xy\phi''(h)$$


问:
Prove/Disprove linear independence
Consider following question:

Prove/Disprove:

1.If vectors {a,b,c} and {d,e} are  linearly independent then {a,b,c,d,e} are linearly independent.

2.If vectors {a,b,c,d,e} are  linearly independent then vectors {d,e} are linearly independent.

3.If vectors {a,b,c,d,e} are  linearly independent then vectors {d,e,f} are linearly independent.

For Q2:  I think that this is true beacuse if you take Ax=0 then you can simplify it into uppertriangular form and then {d,e} must be linearly independent.Then vectors {d,e} are linearly independent.

For Q1,Q3:I am not sure how to deal with it.Any help would be appreciated. Thanks!

答:
Q1)

The vectors d or e can be equal to any of the vector in {a,b,c}. That is counter example.

Q2)

You are right!

Q3)

It is the same as Q1. The vector f can depend on any of the vectors  d and e.

Title: Prove/Disprove linear independence


问:
Calculating the Galois group of an (irreducible) quintic
On my homework I have been asked to compute the Galois group of a quintic.  I have no idea how to do this, except

(a) I calculated that it was irreducible (brute-force)

(b) Since it is irreducible, its splitting field must have degree divisible by $5$

(c) The Galois group must be a subgroup of $S_5$.

There is also a fact that looks helpful, about $S_n$ only having one normal subgroup for $n \geq 5$.  Does this mean there's only one possible Galois extension, or is that only if $S_n$ was already some sort of Galois group?

Aside from that, I have no idea how to do the problem, so can you help me?

Thanks!

P.S. The quintic in question is $x^5 + x - 1 \in \mathbb{Q}[x]$

答: Here's a general theorem that may help: If $f$ is an irreducible rational polynomial of prime degree $p$ with exactly two non-real roots, then the Galois group of $f$ is the full symmetric group $S_p$.</p>


问:
A question with infinity
I am a sophomore in high school and my math teacher did a very short lesson on infinity, here's how it went:  (Try to solve each part yourself the first two are easy)

Part 1

You have an inf. number of boxes each labeled like so

```<code>[1][2][3][4].....
</code>```

and that there is a person in each one of these boxes.  But you have another set of five boxes like so

```<code>[1][2][3][4][5]
</code>```

You want to only have one set of boxes what do you do?  You must be able to tell each person what box they should goto next.

Spoiler

Each person is told to move down five boxes.  Or n+5 is you new box, and the five people from the other boxes are able to move into the first five boxes or just n+0.

Part 2

Well you actually have two of these inf. box sets, but you really only want one.  What now? (same rules apply)

Each person from set one is told to go to box 2n and each person from set two is told to box 2n-1

Part 3

Now you have an inf. number of these inf. box sets arranged like so

```<code>1: [1][2][3][4][5]...
2: [1][2][3][4][5]...
3: [1][2][3][4][5]...
4: [1][2][3][4][5]...
5: [1][2][3][4][5]...
</code>```

and you want to fit them all into one inf. box set, how do you do it.  She refused to tell us the answer and no one in my class figured it out, but I think I just did (correct me if I'm wrong, I'm also looking for alternate solutions).

How I went about solving this problem

Step 1

I assumed that there had to be some sort of pattern so I tried a few

Step 2

I picked the following patter.

1: [1][3][6][10][15]...

2: [2][5][9][14][]...

3: [4][8][13][][]...

4: [7][12][][][]...

5: [11][][][][]...

The numbers in the box represent the new box of each person

Step 3

I started to create a formula for the people to find there new room based of there current room number and box set number.  I found the following sequence in for people in the first row, 1 3 6 10 15.  Or +2 +3 +4 +5.  Which told me that it was quadratic so the equation had to be ax^2 + bx + c = new room number

Step 4

I found three equation by plugging in room numbers and their correct new room numbers to find these three equations:

(1^2)a + 1b + c = 1 or a + b + c = 1

(2^2)a + 2b + c = 3 or 4a + 2b + c = 3

(3^2)a + 3b + c = 6 or 9a + 3b + c = 6

Step 5

After solving this system I got the following equation 1/2x^2 + 1/2x + 0 = new room number where x = current room number.  This of course only tell the people in the boxes on the bottom where to go. Now I needed to find how to place the rest of the people.  The first thing I noticed was that people on the same diagonal as seen bellow were very much related to each other when you look at the rooms they will end up in.

[x-4][][][][]

[][x-3][][][]

[][][x-2][][]

[][][][x-1][]

[][][][][x]

The solution

Working this knowledge into my equation was fairly easy (I was able to do it while going for a run so I won't explain how I got my equation).  The equation is:

 (1/2(r+c-1)^2 + 1/2(r+c-1) + 0)-r+1 = new room number | r = row | c = column

Now my question to all of you is

Is my equation correct?

Are there any alternate solutions to this problem?

How could you fit a 3D array of boxes (inf boxes of people in all direction) into a 1D line of boxes?

Can you find a way to fit a n dimensional array of boxes into a 1D line of boxes?

For any of you into JAVA here's a a short snippet of code I made to prove my answer

```<code>    package main;

public class Main {

    /**
     * @param args
     */
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        int f = 2;
        int r = 1;
        int c = 1;
        while(true){
            double x = (.5*(r+c-1)*(r+c-1) + .5*(r+c-1) + 0)-r+1;
            System.out.print(x +", ");
            c++;
            r--;
            if(r < 1){
                r = f;
                f++;
                c = 1;
            }
        }
    }
}
</code>```

答: Let me comment on your solution and some of those proposed in the other answers.  First of all, both of Michael Hardy's pictures will lead to formulas more complicated than yours, because he goes back and forth rather than always going in the same direction as you did.  In particular, his first picture is just the back-and-forth version of yours except that you tabulated numbers instead of drawing a picture.  (For related non-mathematical amusement,you might look up "boustrophedron".)  Hagen von Eitzen, on the other hand, has a really simple formula, and it actually admits a rather nice description in terms of moving people from infinitely many rows of boxes to a single row; I'll call that single row the "target" row.  I haven't checked whether the following gives exactly Hagen's formula, but it'll be close.  Tell the people in the first row of boxes to occupy every second box in the single target row, just as you did when you stated with two rows of boxes.  They occupy target boxes $0,2,4,\dots$, leaving infinitely many empty boxes in the target row, namely $1,3,5,\dots$.  Now tell the people in the second of your original rows to occupy every second one of those still-empty boxes, namely $1,5,9,\dots$, again leaving infinitely many boxes empty, namely $3,7,11,\dots$.  Continue in the same way with each successive row, putting its occupants into every second one of whatever target boxes remain empty; that leaves infinitely many target boxes empty, so you can continue to the next row.  </p>

Title: A question with infinity


问:
How to get the center of this circle
I now realize that I have over-simplified the problem in my last post because it is hard to explain in words.

This is the original problem that I am fighting with for more than a week:

drawing

There are two lines originating from the origin. The angle of each line in relation to the diagonal is the same and is known. A circle is centered on the diagonal and only a 90 degrees arc of the circle is drawn between the lines. On the arc of the circle is a point P of which the coordinates are known. The dimension and location of the circle is not known, the only thing that is known is that point P lies on the arc and the arc is a 90 degrees segment of a circle with its center on the diagonal.

How can I find the coordinates of the circle?

答:
A Straightedge and compass construction

After algebra, let's come back to geometry. Here is a Straightedge and compass construction.

The construction comes from following basic idea:

It is easy to construct an arc $A_0$ having all the required properties... except that it isn't passing through $P$.

From such an arc, the desired arc is the one obtained as the image of previous arc under the homothetic transformation $T$.

The homothety $T$ is the one having for center the origin $O$ and that transforms $P_0$ into $P$.

Where $P_0$ is the intersection between the arc $A_0$ and the line $(O, P)$.

Base on that, the center $C$ of the desired arc $A$ is the intersection of the line parallel to $(C_0, P_0)$ passing through $P$ with the main diagonal.

See picture below.

Geogebra file

Title: How to get the center of this circle


问:
Radius of convergence for matrix exponential
In the context of systems of linear ODE with constant coefficients, my lecture notes on ODE mention that the matrix exponential $e^{tA}$ has an infinite radius of convergence. This shows up in a proof that $e^{tA}$ is a fundamental matrix of $y'(t)=Ay(t)$, where the derivative of a matrix-valued function can be defined component-wise.

The matrix exponential function is defined by the series

$exp(A):=e^A:=\sum \limits_{n=0}^{\infty} \frac{A^n}{n!}$.

Now I know that in the case of the power series in $\mathbb{C}$ we define the radius of convergence of a power series $\sum \limits_{n=0}^{\infty} c_n z^n$ as

$R:=\sup \{r \geq0:(c_n r^n)_{n \in \mathbb{N}} \text{ is a bounded sequence}\}$

where bounded means that $|c_n r^n| \leq M$ for some $M>0$.

How can we generalize this to matrix power series like the matrix exponential? In general, it does not make sense to take the supremum over a set of matrices. But if we have a power series where $z$ is still a complex number, and only the coefficients can be matrices, then we could use the same definition as before

$R:=\sup \{r \geq0:(C_n r^n)_{n \in \mathbb{N}} \text{ is a bounded sequence}\}$

where $C_n$ are matrices and bounded then means that $\|C_n r^n\|| \leq M$ for some $C$.

It follows that each component of the power series has the same radius of convergence as the matrix series.

So we can differentiate term-wise in each component to obtain the derivative of the matrix exponential.

Is this a common definition? I could not find any source that gives a general definition.

Thanks for any help and suggestions!

答:
You are correct to observe that we need some suitable function from matrices to real numbers to define convergence. This is called a norm. Unfortunately for a vector space (and the set of matrices here can be seen as a vector space) there is more than one norm. Fortunately, there is a theorem that says on a finite dimensional vector space all norms are equivalent (see also the wikipedia article). Here this means that all norms will give us the same definition of convergence.

One possible norm for matrices is the sum of the absolute values of the eigenvalues of the matrix. Using this norm (maybe there is a more convenient choice of norm) one can show that the radius of convergence of the exponential series is indeed infinite.


答:
It is useful to adopt a different notation for derivatives when a function is of one or several variables.

$\phi$ is mapping $\Bbb R\to\Bbb R$, so we can just denote its derivatives as $\phi',\;\phi''$ and so on.

$h:\Bbb R^2\to\Bbb R$, so here it is the case to distinguish which variable we are differentiating with respect to.

Hence $g:=\phi\circ h:\Bbb R^2\to\Bbb R$ and the chain rule says
\begin{align*}
\frac{\partial g}{\partial x}(x,y)
&=\phi'(h(x,y))\cdot \frac{\partial h}{\partial x}(x,y)\\
&=\phi'(x^2+y^2)\cdot 2x
\end{align*}
hence
\begin{align*}
\frac{\partial^2 g}{\partial x^2}(x,y)
&=\phi''(h(x,y))\cdot \frac{\partial h}{\partial x}(x,y)\cdot2x+\phi'(x^2+y^2)\cdot2\\
&=4x^2\cdot\phi''(x^2+y^2)+2\cdot \phi'(x^2+y^2)\;.
\end{align*}
Similarly for $y$-derivatives.

Title: How to calculate first and second order partial derivatives of $g(x,y)=\phi(x^2+y^2)$

Body:
How to calculate first and second order partial derivatives of $g$ where:

$g(x,y)=\phi(x^2+y^2)$

Here's what I did :
$g(x,y)=\phi(x^2+y^2)=(\phi\circ h)(x,y)$ where :  $h(x,y)=x^2+y^2$

$\frac{\partial g(x,y)}{\partial x}$=$\frac{\partial \phi(h(x,y))}{\partial x}$$\times$$\frac{\partial h(x,y)}{\partial x}$ =$\frac{\partial \phi(x^2+y^2)}{\partial x}\times 2x$

$\frac{\partial g(x,y)}{\partial y}$=$\frac{\partial \phi(h(x,y))}{\partial y}$$\times$$\frac{\partial h(x,y)}{\partial y}$ =$\frac{\partial \phi(x^2+y^2)}{\partial y}\times 2y$

and :

$\frac{\partial^2 g(x,y)}{\partial x^2}$=$\frac{\partial^2 \phi(h(x,y))}{\partial x^2} \times 2x +2\times \frac{\partial \phi(h(x,y))}{\partial x}$

$\frac{\partial^2 g(x,y)}{\partial y^2}$=$\frac{\partial^2 \phi(h(x,y))}{\partial y^2} \times 2y +2\times \frac{\partial \phi(h(x,y))}{\partial y}$

are these steps correct ?


问:
prove thatt $\sum_{cyc} \frac{1}{{(x+y)}^2}\ge 9/4$

prove that $$\sum_{cyc} \frac{1}{{(x+y)}^2}\ge 9/4$$ where $x,y,z$ are positives such that $xy+yz+xz=1$

By Holder;$$\left(\sum_{cyc} \frac{1}{{(x+y)}^2} \right){\left(\sum yz+zx \right)}^2\ge {\sum \left(z^{2/3} \right)}^{3}$$.

Hence it suffices to prove $$\sum {\left(z^{2/3} \right)}^{3}\ge 3^{2/3}$$ which is falsse.

I was able to get a weaker bound.since $xy+yz+zx=1$ we have $x^2+y^2\le 2$. By C-S and $x^2+y^2<2$
$$\sum_{cyc} \frac{1}{{(x+y)}^2}\ge \frac{1}{2}\sum \frac{1}{x^2+y^2}>3/4$$

I am interested on a solution using standard inequalities (C-S AM-GM,chebyshov etc) rather than fully expanding and using uvw/pqr/schur.

答:
This is the solution by Vo Quoc Ba Can.

Need to prove:
$$(xy+yz+zx)\left(\frac{1}{(x+y)^2}+\frac{1}{(y+z)^2}+\frac{1}{(z+x)^2}\right)\geqq \frac{9}{4}$$
Wlog, assume $x\geq y \geq z,$ we have the following lemma:
$$\sum\limits_{cyc} \frac{1}{(x+y)^2}\geqq \frac{1}{4xy}+\frac{2}{(x+z)(y+z)}\quad \quad(1)$$
$$\Leftrightarrow  \left(\frac{1}{x+z}-\frac{1}{y+z}\right)^2\geqq \frac{(x-y)^2}{4xy(x+y)^2}$$
$$\Leftrightarrow 4xy(x+y)^2 \geqq (x+z)^2(y+z)^2$$
Since $x\geqq y \geqq z \therefore 4xy\geqq 4y^2\geqq (y+z)^2;(x+y)^2 \geqq (x+z)^2.$

So $(1)$ is true, need to prove:
$$(xy+yz+zx)\left(\frac{1}{4xy}+\frac{2}{(x+z)(y+z)}\right)\geqq\frac{9}{4}$$
But it's $$(x+y)(y+z)(z+x)\geq 8xyz,$$
which is true by AM-GM.


问:
Trying to find a recursive solution to an infinite series
I have an infinite series that I need to write a function for (in Python) to calculate. To do so, I need to find the recursive definition of the formula, which I am having trouble with.

$$\frac{(-1)^nz^{2n+1}}{n!(2n+1)}$$

So far I have figured out the numerator to be

$$ t_{n+1} = t_n \times -z^2 $$

and the $n!$ turns into $(n+1)$, but I have no idea what the rest of the denominator turns into.

Any help would be much appreciated!

答:
Looks awfully familiar... start with:
$$
exp(- z^2) = \sum_{n \ge 0} \frac{(-1)^n z^{2 n}}{n!}
$$
Integrate with respect to $z$, and you have your terms. But that integral is essentially the error function, which has good implementations in the math library (I believe it is even mandated by POSIX).</p>


答:
Write $$A\le aU$$ where $a$ is the largest element of $A$ and $U$ is a $d\times d$ matrix of all ones. The comparison is made element-wise, ignoring the signs.

Then

$$A^n\le(aU)^n= d^{n-1}a^nU$$ and

$$\sum_{n=0}^\infty\frac{A^n}{n!}\le\frac{e^{da}}dU=:MU.$$

Title: Radius of convergence for matrix exponential

Body:
In the context of systems of linear ODE with constant coefficients, my lecture notes on ODE mention that the matrix exponential $e^{tA}$ has an infinite radius of convergence. This shows up in a proof that $e^{tA}$ is a fundamental matrix of $y'(t)=Ay(t)$, where the derivative of a matrix-valued function can be defined component-wise.

The matrix exponential function is defined by the series

$exp(A):=e^A:=\sum \limits_{n=0}^{\infty} \frac{A^n}{n!}$.

Now I know that in the case of the power series in $\mathbb{C}$ we define the radius of convergence of a power series $\sum \limits_{n=0}^{\infty} c_n z^n$ as

$R:=\sup \{r \geq0:(c_n r^n)_{n \in \mathbb{N}} \text{ is a bounded sequence}\}$

where bounded means that $|c_n r^n| \leq M$ for some $M>0$.

How can we generalize this to matrix power series like the matrix exponential? In general, it does not make sense to take the supremum over a set of matrices. But if we have a power series where $z$ is still a complex number, and only the coefficients can be matrices, then we could use the same definition as before

$R:=\sup \{r \geq0:(C_n r^n)_{n \in \mathbb{N}} \text{ is a bounded sequence}\}$

where $C_n$ are matrices and bounded then means that $\|C_n r^n\|| \leq M$ for some $C$.

It follows that each component of the power series has the same radius of convergence as the matrix series.

So we can differentiate term-wise in each component to obtain the derivative of the matrix exponential.

Is this a common definition? I could not find any source that gives a general definition.

Thanks for any help and suggestions!


问:
Classification of Continuous Group Homs from Circle to $SL_2(\mathbb{R})$
Problem. Classify all continuous group homomorphisms $f:\mathbb{S}^1\to SL_2(\mathbb{R})$.

Attempt 1. My initial thought was to look at the induced Lie algebra map $df:\mathbb{R}\to \mathfrak{sl}_2$. Since we have a basis $E,F,H$ (denoted $e,f,h$ in the link), I thought we can classify $f$ according to what $df(1)=:v$ is. However, since $\mathbb{S}^1$ is not simply connected, Lie's Second Theorem does not hold, so we do not know if there exists an $f$ given $v$.

Attempt 2. My other idea was to look at the rational points $x$ on $\mathbb{S}^1$ which have order $q$ ($f$ is determined by these points by continuity). Since $f$ is a homomorphism, $f(x)$ must have order dividing $q$. But then I do not know which elements in $SL_2(\mathbb{R})$ has finite order. I thought of using Iwasawa decomposition, but I could not make it work.

I guess other things one could try is to try looking at the universal cover $\widetilde{SL_2(\mathbb{R})}$, but that restricts us to a map $f$ that lifts.

Question 1. How does one approach the above problem? Can we salvage either of my approaches?

Question 2. More generally, how does one approach these classification problems when Lie's second theorem is not available to us?

答: Looking at the induced Lie algebra map will work fine. $df(1)$ is some $X \in \mathfrak{sl}_2(\mathbb{R})$ which exponentiates to a one-parameter subgroup $\varphi : \mathbb{R} \to SL_2(\mathbb{R})$ and next you want to figure out which of these one-parameter subgroups are periodic with period $2\pi$ (or $1$ depending on your preferred conventions). Thinking about the eigenvalues of $X$ it's not hard to see that this happens iff the eigenvalues of $X$ are purely imaginary, so they are necessarily two conjugate pairs $is, -is$ (because $\text{tr}(X) = 0$), and moreover $s$ is an integer (or an integer multiple of $2\pi$, again depending on your preferred conventions). Can you finish it from here?


问:
Let $f : {\bf R}^2 \to {\bf R}^2$ and $f(x,y) = (x^2+2xy+y^2, 2x + 2y).$ Determine if $f$ is locally injective.

Let $f : {\bf R}^2  \to {\bf R}^2$ and $f(x,y) = (x^2+2xy+y^2, 2x + 2y).$ Determine if $f$ is locally injective.

Noting first that $f$ is $C^1$ and computing the Jacobi one has that $$J_f(x,y) = \begin{bmatrix} 2x+2y && 2y+ 2x \\ 2 && 2\end{bmatrix}.$$

Thus the determinant $\det(J_f(x)) = 0$. However, this only means that the Jacobian matrix isn't invertible. The function might still be injective.

Any hints on how should I go about this and determine if the function is actually injective or not?

答:
Write
$$x^2+ 2xy +y^2 =(x+y)^2, \ \ \ 2x+2y =2(x+y),$$
then your function depends only on $x+y$. So for example, changing $(x,y)$ to $ (x-\epsilon, y+\epsilon)$ do not change the function value. Thus it is not even locally injective.


问:
To find the x and y-intercepts of the line $ax+by+c=0$
Please check if I've solved the problem in the correct way:

The problem is as follows:

Find the points at which the line $ax+by+c=0$ crosses the x and y-axes. (Assume that $a \neq 0$ and $b \neq 0$.

My solution:

We have to find the x and y-intercepts of the line. At the 'x-intercept' the ordinate must be equal to $0$ and at the 'y-intercept' the abscissa must be equal to $0$.

Now we solve the equation $ax+by+c=0$ for $y$:

$ax + by + c=0$

$ax + by = -c$

$by = -ax -c$

$y = \frac {-ax-c}{b}$

$\because x = 0$ at y-intercept,

$\therefore y = \frac {-a(0)}{b} -\frac{c}{b}$

$y = -\frac cb$.

The point at which the line crosses the y-axis is $(0,-\frac cb)$

Now we solve the equation $ax+by+c=0$ for $x$:

$ax+by+c=0$

$ax+by=-c$

$ax = -by-c$

$x = \frac {-by-c}{a}$

$\because y = 0$ at x-intercept

$\therefore x = \frac {-b(0)}{a} -\frac{c}{a}$

$x = -\frac ca$

The point at which the line crosses the x-axis is $(-\frac ca,0)$

答: Being lazy, I'd simplify the equation first and do algebraic manipulations later.  When looking for the $x$-intercept, I'd know that I'll need to plug in $0$ for $y$, so I'd do that before any complicated algebraic work.  Plugging it into $ax+by+c0$, I'd get $ax+c=0$.  Then I'd solve, getting $x=-c/a$.  Similarly, to get the $y$-intercept, I'd plug in $x=0$ first, getting $by+c=0$, and then solve, getting $y=-c/b$.


问:
Is inline summation index has different functionality, such as $\sum_{x,y} f(X,Y)$?
I find different papers use summation with a different style than other summation in the same paper. This thing repeats in more than one paper. Therefore I believe this is summation do different work than normal one.

 <- from the paper 1

2<-from the paper 2

What does this summation call?

What does it do?

Paper use this kind of summation:

Jeffrey–divergence eq, https://www.researchgate.net/publication/3703241_Non-parametric_similarity_measures_for_unsupervised_texturesegmentation_and_image_retrieval

9+ 10 eq, http://www.sciencedirect.com/science/article/pii/S0924271616304324

答:
Your first example seems to be the one that’s worrying you. The “$ij$” is written to the lower right of the $\sum$ symbol. I think this is what you mean by “in-line” positioning.

The positioning of the $ij$ doesn’t mean anything. It doesn’t matter whether the $ij$ is to the right of the $\sum$ symbol or below it. The meaning of the summation is the same in both cases.


问:
What prevents the Expectation conditional on a sub sigma-algebra to be the the variable being conditioned?
I have come to a very simple issue which I have not figured out yet.
In fact there are two questions.
Could you please help me out on this?

Let $(\Omega,\mathcal{F},P)$ be a probability space, $X:\Omega\rightarrow R$ be a random variable, and $\mathcal{A}$ be a sub $\sigma$-algebra in $\mathcal{F}$. The conditional expectation of $X$ given $\mathcal{A}$ is the a.s. unique random variable, denoted as $E[X/\mathcal{A}]:\Omega\rightarrow R$ that satisifies: (i) $E[X/\mathcal{A}]$ is $\mathcal{A}/\mathcal{B}$ measurable and (ii) $\int_BE[X/\mathcal{A}]dP=\int_B XdP \hspace{5mm } \forall B\in\mathcal{A}$.

What prevents $X$ to be this random variable regardless of $\mathcal{A}$?

$X$ seems to satisfy both conditions, (i) $\mathcal{A}/\mathcal{B}$ measurability and (ii) $\int_BE[X/\mathcal{A}]dP=\int_B XdP$.

The second question is:

In the same setting as above, it is defined $P[A/\mathcal{A}]=E[I_A/\mathcal{A}] \hspace{5mm}\forall A\in \mathcal{F}$, and if $\mathcal{A}=\sigma(Y)$ for a random element $Y:\Omega\rightarrow\Lambda$.

How does $P[A/Y]=P[A/\mathcal{A}]$ relate with $P[A/Y=y]$ for $A\in\mathcal{F}$ and $y\in\Lambda$.

Best regards,

Juan Manuel

答: I'm not sure what is meant by $\mathcal A/\mathcal B$-measurability.  The definition of conditional expectation that I learned requires it to be $\mathcal A$-measurable, which $X$ usually isn't.  </p>


Title: Let $x_1,x_2,x_3,x_4$ denote the four roots of the equation $x^4 + kx^2 + 90x - 2009 = 0$. If $x_1x_2 = 49$, then find the value of $k$.

Body:
Let $x_1,x_2,x_3,x_4$ denote the four roots of the equation $$x^4 + kx^2 + 90x - 2009 = 0.$$ If $$x_1x_2 = 49,$$ then find the value of $k$.

What I tried :- From Vieta's Formula for quartic equations I get that
$$ x_1 + x_2 + x_3 + x_4 = 0$$

$$x_1x_2 + x_1x_3 + x_2x_3 + x_2x_4 + x_3x_4 + x_4x_1 = k$$

$$x_1x_2x_3 + x_2x_3x_4 + x_3x_4x_1 + x_4x_1x_2 = (-90)$$

$$ x_1x_2x_3x_4 = -2009$$

Since $x_1x_2 = 49$, we have $x_3x_4 = (-41)$.
Hence in equation $3$ I get

$$x_1x_2(x_3 + x_4) + x_3x_4(x_1 + x_2) = (-90)$$

$$\Rightarrow 49(x_3 + x_4) - 41(x_1 + x_2) = (-90)$$

As $(x_1 + x_2 + x_3 + x_4) = 0$, $(x_1 + x_2) = -(x_3 + x_4)$, so $$49(-x_1 - x_2) - 41(x_1 + x_2) = (-90)$$

$$\Rightarrow (-90)(x_1 + x_2) = (-90)$$

$$\Rightarrow (x_1 + x_2) = 1,$$ so $(x_3 + x_4) = (-1)$.

From here I don't know how to proceed, can anyone help?

Body:
We have :- $x_1 + x_2 + x_3 + x_4 = 0$

$\rightarrow  x_3 + x_4 = -(x_1 + x_2)$

$(x_1 * x_2 * x_3 * x_4 = -2009)$ and $(x_1 * x_2 = 49)$

$\rightarrow x_3*x_4 = -41$

Now, $$k = (x_1+x_2)(x_3+x_4) + (x_1*x_2) + x_3*x_4 = 8 + (x_1+x_2)(x_3+x_4)$$

$$-90 = x_1x_2(x_3+x_4) + x_3x_4(x_1+x_2)$$

$$\rightarrow -90 = 90(x_3+x_4)$$

$\rightarrow x_3 + x_4 = -1$ and $x_1 + x_2 = 1$.

So :- $k = (8 - 1) = 7$.


问:
A is a symmetric matrix of order n. Are all bases for the eigenspaces of A orthogonal?
Suppose A is a symmetric matrix of order n. Which of the following statements are true?

$\quad $ (1) Let k be an eigenvalue of A. Any basis $ S = \{u_1, ...u_k\} $ for the eigenspace $ E_k, $ must be orthogonal

$\quad $ (2) There is always a basis of $ \mathbb{R}^n $ consisting of eigenvectors of A.

$\quad $ (3) For every distinct eigenvalues $ k_1, k_2 $ of A, $ E_{k_{1}} $ is orthogonal to $ E_{k_{2}} $

$\quad $ (4) A can be orthogonally diagonalized, after applying Gram-Schmidt process to the chosen basis for each of the eigenspaces of A.

I can only prove that (3) and (4) is true. (4) is true because a symmetric matrix is a diagonalizable matrix, so each of the bases will be transformed to an orthonormal basis. (3) is true because the dot product of 2 eigenvectors of 2 distinct eigenvalues is zero.

How do I go about proving that (1),(2) are also true? Please correct me if my reasoning is incorrect. Thank you.

答:
Hint:

For $1$, take $A=I_n$ and $k=1$ is its unique eigenvalue. We see that $S=(e_1+e_2,e_2, \ldots, e_n)$ is a basis for the eigenspace $E_1$ but isn't orthogonal.

For $2$, if $A \in \Bbb M_n(\Bbb R)$ and $A$ is symmetric, we know that $A$ is diagonalizable, so this is true (it is for every diagonalizable matrix on $\Bbb R$).

For $3$, by computing the dot product we see that it's true.

For $4$, since $A$ is diagonalizable and its eigenspaces are orthogonal, it's true.


问:
Linear dependency in $\mathbb{R}^4$
Show that the three vectors are linearly dependent in $\mathbb{R}^4$.

$u=(0,3,1,-1)\ v=(6,0,5,1)\ w=(4,-7,1,3)$

They are linearly independent if $c_{1}u+c_{2}v+c_{3}w=0\Rightarrow c_{1}=c_{2}=c_{3}=0$
$$
c_{1}u+c_{2}v+c_{3}w\Rightarrow \begin{pmatrix}
0 & 6 &4 \\
 3&0  &-7 \\
1 &5  &1 \\
-1 &1  &3
\end{pmatrix}
\Rightarrow
\begin{pmatrix}
1 &0  &-\frac{7}{3} \\
0 &1  &\frac{2}{3} \\
0 &0  &0 \\0
 &0  &0
\end{pmatrix}
$$

How does that show that the vectors are linearly dependent?

Write each vector as a linear combination of the other two.

EDIT.
$$\begin{pmatrix}
0\\3
\\1
\\-1
\end{pmatrix}=a\begin{pmatrix}
6\\0
\\5
\\1
\end{pmatrix}\cdot b\begin{pmatrix}
4\\-7
\\1
\\3
\end{pmatrix}\Rightarrow a=\frac{2}{7},\ b=-\frac{3}{7}$$

Have i understood 2. correct?

答: We have found, by row reduced echelon form (RREF), that the system $Ac=0$ has a not trivial solution (i.e. a solution with $c\neq 0$) that is precisely the definition for linear dependency for the columns of $A$.

Title: Linear dependency in $\mathbb{R}^4$


问:
Proof that a statistic is complete
I need some help with this exercise:

The density function of the random variable $X_i$ is:

\begin{equation*}
f_\theta(x)=\exp(\theta-x)1_{[\theta,\infty)}(x)
\end{equation*}

The maximum likelihood function is:

\begin{equation*}
L(x,\theta)=\prod_{i=1}^n\exp(\theta-x_i)1_{[\theta,\infty)}(x_i)=exp\bigg(n\theta-\sum_{i=1}^n x_i\bigg)1_{[0,\min\{X_1,...,X_n\}]}(\theta)
\end{equation*}

And its (unique) maximum is attended for $\hat\theta_{ML}=\min\{X_1,...,X_n\}$.

I find that the statistic $Y=\min\{X_1,...,X_n\}$ is sufficient using the factorization theorem and I now have to prove that it is also complete.

First of all I've calculated the density of $Y$ which is:

\begin{equation*}
g_\theta(x)= n\exp(n\theta-nx)
\end{equation*}

Then for the definition of complete statistic I have to prove that :

\begin{equation*}
E_\theta(h(Y))=0 \Rightarrow h=0
\end{equation*}

and this implies that:

\begin{equation*}
E_\theta(h(Y))=\int_\theta^\infty h(y)n\exp(n\theta-ny)dy=0
\end{equation*}

How can I conclude from the last expression that $h(y)$ has to be zero $\forall y$

答:
$f(\theta)=\int_\theta^\infty h(y)e^{-ny}dy$ is identically zero $\forall\theta>0$.

$\implies f'(\theta)=-h(\theta)e^{-n\theta}=0$ which gives $h(\theta)=0$ over $\Bbb R^+$ so $P_\theta[h(x)=0]=1$.

Title: Proof that a statistic is complete


问:
If $f$ is continuous and bounded, and $\text{span}\{x \mapsto f(x+k) \mid k \in \Bbb Z\}$ is finite-dimensional, what can we say about $f$?

If $f : \Bbb R \to \Bbb R$ is continuous and bounded, and $\text{span}\{x \mapsto f(x+k) \mid k \in \Bbb Z\}$ is finite-dimensional, what can we say about $f$?

I found this problem difficult (I found examples of functions for which the hypothesis is true like $n$-periodic functions, where $n \in \Bbb N^*$, but I am struggling for the reverse implication). Can someone help me solve it and does someone have a reference from where it comes?

答:
Claim. If is $f$ is continuous and bounded, then $\,\mathrm{span}\,\{f(x+k): k\in\mathbb Z\}\,$ is a finite dimensional space if and only if $f$ is of the form
$$
f(x)= \exp\big(i\varphi_1 x\big)e_1(x)+\cdots+
\exp\big(i\varphi_k x\big)e_k(x), \tag{$\star$}
$$
where $e_1,\ldots,e_k$ are continuous $1-$periodic functions and $\varphi_1,\ldots,\varphi_k\in [0,2\pi)$, distinct.

The inverse is straight-forward since if we assume that $f$ is of the form $(\star)$, then $f(x),f(x+1),\ldots,f(x+k)$ are linearly dependent.

First, let's study the discrete problem:

If a vector $(a_j)_{j\in\mathbb Z}$ satisfies the recursive relation
$$
a_{j+n}+c_{n-1}a_{j+n-1}+c_{n-2}a_{j+n-2}+\cdots+c_{0}a_{j}=0, \quad j\in\mathbb Z, \tag{1}
$$
where $c_0,\ldots,c_{n-1}$ are given constants (complex in general), then $(a_j)_{j\in\mathbb Z}$ is uniquely expressed in the form
$$
a_j=p_1(j)w_1^j+\cdots+p_k(j)w_k^j,\quad j\in\mathbb Z,
$$
where $w_1,\ldots,w_k$ are the distinct roots of the polynomial
$$
P(x)=x^n+c_{n-1}x^{n-1}+\cdots+c_1x+c_0,
$$
and $p_1,\ldots,p_k$ are polynomials with $\,\deg p_\ell<$ multiplicity of the root $w_\ell$ in the polynomial $P(x)$. If $|w_\ell|>1$ or $0<|w_\ell|<1$, for
some $\ell=1,\ldots,k$, then $(a_j)_{j\in\mathbb Z}$ would not be bounded. Also, if $|w_\ell|=1$ and $\,\deg p_\ell\ge 1$, then again $(a_j)_{j\in\mathbb Z}$
would not be bounded.

Hence, a vector $(a_j)_{j\in\mathbb Z}$ satisfying $(1)$
is bounded if and only if it is of the form
$$
a_j=d_1w_1^j+\cdots+d_k w_k^j,\quad j\in\mathbb Z,
$$
where $d_1,\ldots,d_k$ constants and $|w_1|=\cdots=|w_k|=1$ and of course, the $w_\ell$'s are still distinct roots of $P(x)$.

If  $a_j$'s and $c_k$'s are real, then
$(a_j)_{j\in\mathbb Z}$ should be of the form
$$
a_j=\sum_{\ell=1}^k d_\ell\cos (w_\ell j)+e_\ell\sin (w_\ell j), \quad j\in\mathbb Z.
$$

Back to the continuous problem, assume that $f(x)$ is continuous and bounded and satisfies
$$
f(x+n)+c_{n-1}f(x+n-1)+c_{n-2}f(x+n-2)+\cdots+c_{0}f(x)=0, \quad x\in\mathbb R.
$$
Fixing $x\in \mathbb R$, and setting $a_j(x)=f(x+j)$, we obtain, based on our previous discussion, that
$$
a_j(x)=d_1(x)w_1^j+\cdots+d_k(x) w_k^j=f(x+j),\quad j\in\mathbb Z, \tag{2}
$$
where, the $w_\ell$'s are the distinct roots of the polynomial $P(x)$
and $|w_1|=\cdots=|w_k|=1$.
In $(2)$, if we set $j=1,\ldots,k$, we obtain a $k\times k$ linear system with unknowns $d_1(x), \ldots, d_k(x)$. The matrix of the system is $(w_k^j)_{j,k}$, and it is invertible since it is a Vandermonde matrix. Hence the system possesses a unique solution vector $\big(d_1(x),\ldots,d_k(x)\big)$, with is expressed uniquely as a linear combination of the values of $f(x+1),\ldots,f(x+k)$. Hence the $d_\ell$'s are all continuous functions.

So far we have shown that, is $f$ satisfies the conditions of the question, and $a_j(x)=f(x+j)$, then $a_j(x)$ is expressed as in $(2)$, uniquely (once we know $f$ and the unit roots of $P$), with $d_\ell$'s continuous functions and $w_\ell$ discrete unit complex numbers.

Also
$$
d_1(x+1)w_1^{j}+\cdots+d_k(x+1) w_k^{j}=
a_{j}(x+1)=f(x+j+1) \\ =a_{j+1}(x)=d_1(x)w_1^{j+1}+\cdots+d_k(x) w_k^{j+1} \\=
w_1d_1(x)w_1^{j}+\cdots+w_kd_k(x) w_k^{j}
$$
and hence, due to the uniqueness
$$
d_\ell(x+1)=w_\ell d_\ell(x), \quad x\in\mathbb R,\,\,\,\ell=1,\ldots,k.
$$
Clearly, there exist numbers $\varphi_\ell\in [0,2\pi)$, such that $w_\ell=\exp\big(i\varphi_\ell\big)$. Then the functions
$$
e_\ell(x)=\exp\big(-i\varphi_\ell x\big)d_\ell(x), \quad \ell=1,\ldots,k,
$$
are clearly $1-$periodic.

So, we have shown that $f$ is of the form
$$
f(x)=a_0(x)=d_1(x)+\cdots+d_k(x) \\ = \exp\big(i\varphi_1 x\big)e_1(x)+\cdots+
\exp\big(i\varphi_k x\big)e_k(x),
$$
where $e_\ell(x)$ are continuous $1-$periodic functions and $\varphi_\ell$'s distinct numbers in $[0,2\pi)$.

Note. If we need to restrict ourselves in $\mathbb R$, then $(\star)$ should be of the form
$$
f(x)=\sum_{\ell=1}^k \big(\cos(\varphi_\ell x)C_\ell(x)+
\sin(\varphi_\ell x)S_\ell(x)\big)
$$
where the $C_\ell$'s and $S_\ell$'s are $1-$periodic functions.


问:
"$X$ is an ordered set in the order topology." What is the meaning of this sentence? (James R. Munkres "Topology 2nd Edition")
I am reading "Topology 2nd Edition" by James R. Munkres.

There is the following sentence on p.90 in this book:

Now let $X$ be an ordered set in the order topology, and let $Y$ be a subset of $X$.

I am very poor at English.

What is the meaning of the following sentence?:

$X$ is an ordered set in the order topology.

Does the above sentence mean the following?:

$X$ is an ordered set and the topology on $X$ is the order topology.

答:
Yes, it means exactly that.

Munkres considers an ordered set $X$ and puts the order topology on it.

If you take a subset $Y \subseteq X$, there are two possible topologies on $Y$:

(1) $Y$ is itself an ordered set, so it obtains an order topology as well.

(2) $Y$ is a subset of $X$, so it obtains a subspace topology.

A natural question is whether these topologies are the same. This turns out to be false, as Munkres demonstrates.


问:
Generalize this notation for an ODE?
I have

$$
\frac{dN(t)}{dt}=a(t)N(t) \tag 1
$$
where $N(0)=N_0 $ (constant).

Question 1:
From the information in $(1)$ I assume I have the functions $a, N:\mathbb R\rightarrow \mathbb R$? And $N_0\in \mathbb R$? Is that correct?

Suppose I now want to generalize the notation in $(1)$ by replacing the right hand side with a function $f$.

Question 2:
Does this mean I should write
$$
\frac{d N(t)}{dt}=f(t) \tag 2
$$
where $N, f:\mathbb R\rightarrow \mathbb R$ ?

Or maybe

$$
\frac{dN(t)}{dt}=f(a(t),N(t)) \tag 3
$$
where $f:\mathbb R \times \mathbb R \rightarrow \mathbb R$?

Or maybe

$$
\frac{dN(t)}{dt}=f(t,a(t),N(t)) \tag 4
$$
where $f:\mathbb R \times \mathbb R \times \mathbb R \rightarrow \mathbb R$ and $t \in \mathbb R$?

$\times$ is the Cartesian product.

Which one is correct? Thanks!

答:
An explicit first order ODE has the following form (see wiki)
$$
\frac{dN(t)}{dt} = f(t, N(t)).
$$

The form (2) is too narrow, since does not allow dependence on $N(t)$ in the right hand side. Form (3) does not allow dependence on raw $t$, and the form (4) has extra $a(t)$ which may be easily built in the definition of $f(t, N)$ itself.

Few examples:

$\frac{dN(t)}{dt} = g(t)$ is a special case of $f(t, N) = g(t)$

$\frac{dN(t)}{dt} = a(t) N(t)$ is a special case of $f(t, N) = a(t) N$

$\frac{dN(t)}{dt} = \frac{a(t)}{N(t) + t}$ is a special case of $f(t, N) = \frac{a(t)}{N + t}$

Title: Generalize this notation for an ODE?


问:
Why is it not correct to add up the random variables to find this probability?
I'm working on this problem on exponential distribution:

Consider a post office with two clerks. Three people, A, B, and C, enter simultaneously. A and B go directly to the clerks, and C waits until either A or B leaves before he begins service. What is the probability that A is still in the post office after the other two have left when the service times are exponential with mean 1/μ?

I want to use this property
on probability of inequality of exponential random variables

I want to sub in $T_A$, time that person A spends in the post office, for $X_2$ in the formula, and $T_B + T_C$ for $X_1$. And then I would simplify $T_B + T_C$ as $2T_B$, which by the property of exponential random variable would have a rate of μ/2. So I would get μ/2 in the numerator and μ/2 + μ in the denominator, which simplifies to 1/3. However, the solution I've found uses a different approach and gives a different result, for example #4 on here: https://www2.isye.gatech.edu/~sman/courses/6761/hw6f12solns.pdf. Can someone let me know why my method is incorrect?

答: The first step of your calculation is incorrect, the formula you have given only holds if $X_1$ and $X_2$ are exponentially distributed. However, $X_2:=T_B+T_C$ is not exponentially, but $\Gamma(2,1/\mu)-$distributed (you cannot just simplify $T_B+T_C$ as $2T_B$, the distribution of the sum of two independent random variables is given by its convolution). If you use the PDF of this distribution instead of the PDF of the exponential distribution in equation 1 of your source, the calculation should work.


问:
Power series representation of $(1+x^2)\arctan x$
so i've been looking for a cleaner way of solving this question, finding a Power series representation of the following:
$$
\left(1+x^2\right)\arctan\left(x\right)
$$
knowing the power series of arctan i got:
$$
\sum_{n=0}^{\infty \:}\frac{\left(-1\right)^n\cdot x^{2n+1}}{2n+1}\:+\:\sum_{n=0}^{\infty}\frac{\left(-1\right)^n\cdot \:x^{2n+3}}{2n+1}
$$
which after opening up the sum expressions amounts to :
$$
x+\:\:\sum_{n=1}^{\infty}\frac{\left(-1\right)^n\cdot \:2\cdot x^{2n+1}}{\left(2n\right)^2-1}
$$

I am sure there is a better way of doing this, maybe differentiation or integration? I've tried but i can't figure it out.
Any help would be greatly appreciated :]

答:
Yes, there is indeed a solution by differentiation + integration:

Differentiate your expression:

$$2x \arctan(x) +1$$

replace $\arctan(x)$ by its expansion, then integrate it term by term... taking into account the integration constant in such a way that there is an agreement in $x=0$ of the initial expression and your expansion.


问:
Distinguishing between a parabola and (one branch of) a hyperbola by looking at the graph

If I were to show you this curve that is quite zoomed in is it possible to tell if it is a hyperbola of a parabola

This is a picture my friend sent me and I cannot tell which conic section it is.

答: Draw two or more parallel chords (not perpendicular to the axis). If their midpoints are aligned parallel to the axis, then it's a parabola. Otherwise it's a hyperbola and the intersection between the line of the midpoints and the axis is its center.


问:
Proving that $Aut_{\mathbb{Q} } { \mathbb{R}} $
The following question is from my Field Theory assignments.

Prove that  $Aut_{\mathbb{Q} } { \mathbb{R}} $  is identity group.

I took $f:\mathbb{R} \to \mathbb{R}$ be an automorphism.
Now  which result I should use to prove that all the automorphism which are $\mathbb{Q}$ -module homomorphisms are identity.

Let $f'$ be such an automorphism and q belongs to $\mathbb{Q}$. So, $f'(q) = qf'(1) =q$.but how to prove that $f'(q) =1$ .

Kindly guide.

答:
First of all, note that the identity is not the map that maps alls real numbers to $1$; that is not even a homomorphism. Instead, it is the map that maps every real number to itself.

Hint to prove this:

$f$ is increasing. Hint: consider $b > a$ and look at $f(\left(\sqrt{b-a}\right)^2$).

$f$ is continuous. Hint: use (only) that $f$ is an increasing bijection.

$f = \mathop{id}_{\mathbb R}$. Hint: $f|_{\mathbb Q} = \mathop{id}_{\mathbb Q}$ and ${\mathbb Q}$ is dense in ${\mathbb R}$.


问:
Prove the convergence of a sequence with the Mean Value Theorem
I am given the following problem:

Let f: R$\rightarrow$R be differentiable such that,  for some $\alpha$ $\in$ R, $\left|f'(x)\right|$$\leq$ $\alpha$ $<1$ for all $x$$\in$ R.
 Let  $a_{1}$$\in$ R and $a_{n+1}$ $=$ $f$($a_{n}$), for $n$ $\in$ R.
 Show that the sequence $a_{n+1}$ converges.

Since it is an exercise of the chapter of the book about the Mean Value Theorem, I think I have to apply it, but I do not know where and what I can obtain in order to prove the convergence of the given sequence.

答:
Hint

Let $x,y\in\mathbb R$, then, there is $c_{x,y}\in [x,y]$ s.t. $$|f(x)-f(y)|\leq |f'(c_{x,y})||x-y|\leq \alpha |x-y|.$$
Therefore, $f$ is a contraction. Your job now :

1) Prove that $$|a_{n+1}-a_n|\leq \alpha ^n|a_1-a_0|,$$
for all $n$.

2) Deduce that $(a_n)$ is a Cauchy sequence.


问:
Prove that $\mathcal U(\mu, \nu)$ is compact for the weak topology of measures
In the book Computational Optimal Transport, by Peyré and Cuturi, the following assertion is made.

Let $X$ and $Y$ be two compact probability spaces, with $c$ continuous and
$$
\mathcal U(\mu, \nu):= \left\{
\gamma \in \mathcal P(X \times Y) \quad :  \quad
(\text{Proj}_X)_\# \gamma = \mu,
(\text{Proj}_Y)_\# \gamma = \nu
\right\}
$$

The author then claims that $\mathcal U(\mu, \nu)$ is compact for the weak topology of measures.

How does one proves this assertion?

答:
Are X and Y considered to be Polish?

If yes: every probability measure on a Polish space is tight. Using this you can show that $\mathcal{U}(\mu,\nu)$ is also tight. By Prokhorov's Theorem its closure is compact. Using how $\mathcal{U}(\mu,\nu)$ is defined you can show that it is closed w.r.t. the weak topology and therefore it is compact.


问:
Declarative statements
Is there some branch of mathematics that works with truth bearing statements instead of variables, and defines operations between them?

Basically I am looking for some well known system that defines true or false statements like variables and has some short notation used to denote the truth/false hood of these statements as well as the logical conjunction of several statements.

答: Your description is rather vague, but propositional logic (sometimes called sentential logic or propositional calculus) seems to fit.  If it's not what you have in mind, then please edit the question to make it more specific.</p>

Title: Declarative statements


问:
Prove or disprove: $(A-B)-C=(A-C)- B$ for sets $A$, $B$, $C$

Prove or disprove: For sets $A$, $B$, $C$,
$$(A-B)-C=(A-C)- B$$

Now my attempt is

$$(x\in A \;\text{and}\; x\notin B) \;\text{and}\; x\notin C \tag1$$

We can write it as

$$(x\in A  \;\text{and}\; x\notin C)  \;\text{and}\; x\notin B \tag2$$

which proves the statement.

Is it right?

答:
A valid Venn diagram-based proof:

This can equivalently be expanded to a proof of $(A\land\lnot B)\land\lnot C=(A\land\lnot C)\land\lnot B$ using a truth table.


问:
Is It Hopeless for Me to Try to Understand the Derivations of All Formulas I Use?
Recently, I've been doing some relatively more advanced math (multivariable and vector calculus) and the derivations of formulas are becoming increasingly complicated. I don't like using a formula if I don't understand why it works and where it comes from, so I have understood the derivation of all formulas I've used so far. My question is: is it hopeless for me to keep learning derivations? Will I eventually reach a point where understanding why the formulas I use work becomes too time-consuming and unreasonable? To clarify, I prefer to understand proofs in order use them, I don't memorize them. Any help is appreciated. Thank you.

答:
First of all, you need very much of focus , without which all that you grasp may be lost eventually. Focus your mind when attending lectures, lessons or seminars. Keep a sharp ear.

Next, practice - deriving something once needn't always help you fix stuff in mind, so practising it as many times as you can fixes whatever you are learning in your mind.

Third - a strong heart. Never feel disheartened by the derivations you have to learn, be it complex or way above your level. Just give it a go, try again if you don't get it. Also, you can discuss it in here (I mean, the difficulties you find in trying the derivations and such stuff). As the lines in Bastille's 'Laura Palmer' goes - "Walking out into the dark, cutting out a different path, led by a beating heart..." - keep it beating, keep it brave, keep it right and you are ready to face the walls ahead of you. Never lose hope or faith in yourself.

Fourth - basics. I know you have the basics, but still it is inevitable that I say that anyone must have at least a small bit of the basics to get started.


问:
how to show that $a_{n+1}=\sqrt{6+a_{n}}$ is bounded?
I have to show that the sequrnce $\{a_n\}$ with $a_{n+1}=\sqrt{6+a_{n}}$ is bounded, i know : $a_{n-1}<a_{n}$ because i demonstrated this. I tied to use this relation: $ab\leq \frac{a^2}{2}+\frac{b^2}{2}$. I don't kow how to increase $\sqrt{6+a_{n}}$

答:
$a_n\leq 3$ can be proved by induction:

Induction base: $a_1=0\leq 3$.

Assume that $a_n\leq 3$, then we have $a_{n+1}=\sqrt{6+a_n}\leq \sqrt{6+3}=3$


问:
If $a$ is sufficiently large as compared with $b,$ and $\sqrt \frac{a}{a-b}+\sqrt \frac{a}{a+b}=2+k(\frac{b}{a})^2$,then what is the value of $k?$

If  $a$ is sufficiently large as compared with $b,$ and $\sqrt
 \frac{a}{a-b}+\sqrt \frac{a}{a+b}=2+k(\frac{b}{a})^2$,then what is the
value of $k?$

Query

Can we find it using particular value?,like putting $a=4,b=2,$ then we get

$k=4\sqrt2+4\sqrt \frac{2}{3}-8$

This problem is given in  Multiple Choice Question exercise,the possible answers given are

A.$\frac{2}{3}$

B.$\frac{3}{4}$

C.$\frac{4}{5}$

D.$\frac{5}{6}$

None of the answer mathches with mine....

Please give some hint to determine value of $k?$

答:
Use the Negative Binomial Theorem.

$(1-a/b)^{-1/2}+(1+a/b)^{-1/2}=2+2\left(\frac{a}b\right)^2\left[\binom{-1/2}2+\binom{-1/2}4\left(\frac{a}b\right)^2+\binom{-1/2}6\left(\frac{a}b\right)^4+...\right]$

Since we want $k$ (constant) to be independent of $a/b$, we reject the higher order terms in the bracket on the ground that $a>>b$, giving $k=2\binom{-1/2}2=(-1/2)\times(-1/2-1)=3/4$.

Note that with the obtained $k$, $(1-a/b)^{-1/2}+(1+a/b)^{-1/2}\approx2+k(a/b)^2$ i.e. $(1-a/b)^{-1/2}+(1+a/b)^{-1/2}$ is not actually equal to $2+k(a/b)^2$ in general. As you have seen, for equality $k$ will have to be taken as a function of $a,b$ and not constant. Thus plugging in specific values of $a,b$ to find $k$ (which gives the exact value of $k$) will not yield the desired answer.

If you wanted to take a risk, you could check which option for $k$ yields the closest answer to the value of $(1-a/b)^{-1/2}+(1+a/b)^{-1/2}$ for the selected $a,b$. But remember that you still might not get $k=3/4$ since it is not necessary that $3/4$ gives the closest approximation of $(1-a/b)^{-1/2}+(1+a/b)^{-1/2}$ for all $a,b$.


问:
Solve recursion $a_{n}=ba_{n-1}+cd^{n-1}$
Let $b,c,d\in\mathbb{R}$ be constants with $b\neq d$. Let
$$\begin{eqnarray}
a_{n} &=& ba_{n-1}+cd^{n-1}
\end{eqnarray}$$
be a sequence for $n \geq 1$ with $a_{0}=0$. I want to find a closed formula for this recursion. (I only know the german term geschlossene Formel and translated it that way I felt it could be right. So if I got that wrong, please correct me)

First I wrote down some of the chains and I got
$$\begin{eqnarray}
a_{n} &=& ba_{n-1}+cd^{n-1}\\
&=& b\left(ba_{n-2}+cd^{n-2}\right)+cd^{n-1}\\
&=& b\left(b\left(ba_{n-3}+cd^{n-3}\right)+cd^{n-2}\right)+cd^{n-1}\\
&=& b\left(b\left(b\left(ba_{n-4}+cd^{n-4}\right)+cd^{n-3}\right)+cd^{n-2}\right)+cd^{n-1}\\
&=& \dots\\
&=& \sum_{k=0}^{n}b^{k}cd^{n-k-1}\\
&=& \sum_{k=0}^{n}b^{k}cd^{n-\left(k+1\right)}
\end{eqnarray}$$

So I catched the structure in a serie. Now I am asking myself how to proceed. I took the liberty to have a little peek at what WolframAlpha wood say to this serie. I hoped for inspiration and I got

$$\sum_{k=0}^{n-1}b^{k} c d^{n-(k+1)} = (c (b^n-d^n))/(b-d)$$

How did this came to be? And more important: Is my approach useful?
Thank you in advance for any advice!

Edit: My final Solution (recalculated)

$$\begin{eqnarray}
a_{n} &=& ba_{n-1}+cd^{n-1}\\
&=& b\left(ba_{n-2}+cd^{n-2}\right)+cd^{n-1}\\
&=& b\left(b\left(ba_{n-3}+cd^{n-3}\right)+cd^{n-2}\right)+cd^{n-1}\\
&=& b\left(b\left(b\left(ba_{n-4}+cd^{n-4}\right)+cd^{n-3}\right)+cd^{n-2}\right)+cd^{n-1}\\
&=& b^{4}a_{n-4}+b^{3}cd^{n-4}+b^{2}cd^{n-3}+bcd^{n-2}+cd^{n-1}\\
&=& b^{5}a_{n-5}+b^{4}cd^{n-5}+b^{3}cd^{n-4}+b^{2}cd^{n-3}+bcd^{n-2}+cd^{n-1}\\
&=& b^{n}a_{0}+b^{n-1}c+\dots+b^{4}cd^{n-5}+b^{3}cd^{n-4}+b^{2}cd^{n-3}+cbd^{n-2}+cd^{n-1}\\
&=& \dots\\
&=& 0+b^{n-1}c+\dots+b^{4}cd^{n-5}+b^{3}cd^{n-4}+b^{2}cd^{n-3}+cbd^{n-2}+cd^{n-1}\\
&=& \sum_{k=0}^{n-1}b^{k}cd^{n-1-k}\\
&=& cd^{n-1}\sum_{k=0}^{n-1}b^{k}d^{-k}\\
&=& cd^{n-1}\sum_{k=0}^{n-1}\left(\frac{b}{d}\right)^{k}\\
&=& cd^{n-1}\frac{1-\left(\frac{b}{d}\right)^{n}}{1-\left(\frac{b}{d}\right)}\\
&=& cd^{n-1}\frac{1-\frac{b^{n}}{d^{n}}}{1-\frac{b}{d}}\\
&=& cd^{n-1}\frac{\frac{d^{n}-b^{n}}{d^{n}}}{\frac{d-b}{d}}\\
&=& cd^{n-1}\frac{d^{n}-b^{n}}{d^{n}}\cdot\frac{d}{d-b}\\
&=& \frac{c\left(d^{n}-b^{n}\right)}{d-b}
\end{eqnarray}$$

答:
Argh... use Wilf's techniques from "generatingfunctionology". Start defining:
$$
A(z) = \sum_{n \ge 0} a_n z^n
$$
Also write:
$$
a_{n + 1} = b a_n + c d^n
$$
Multiply by $z^n$, add over $n \ge 0$:
$$
\frac{A(z) - a_0}{z} = b A(z) + c \frac{1}{1 - d z}
$$
Solve for $A(z)$, express as partial fractions. The resulting terms are of the forms:
$$
(1 - \alpha z)^{-m}
  = \sum_{n \ge 0} \binom{-m}{n} (- \alpha)^n z^n
  = \sum_{n \ge 0} \binom{n + m - 1}{m - 1} \alpha^n z^n
$$
Note that:
$$
\binom{n + m - 1}{m - 1}
  = \frac{(n + m - 1) (n + m - 2) \ldots (n + 1)}{(m - 1)!}
$$
This is a polynomial in $n$ of degree $m - 1$.
$$
$$</p>


问:
Trouble undestanding notation $\left\{\sup_n f_n > a \right\} = \cup_n\left\{f_n>a\right\}$
I am reading a paper on measure and integration where the following notation is used:
$$\left\{\sup_n f_n > a \right\} = \cup_n\left\{f_n>a\right\}$$
$(f_n:(X, \mathcal{A})\longrightarrow \mathbb{R})$

I don't think I understand this notation. The RHS is just a function and the LHS is a union of functions. I don't see why this gives the $\sup_n$. Similarly, the infimum is defined
$$\left\{\inf_n f_n > a \right\} = \cap_n\left\{f_n>a\right\}$$
Can someone help me undestand this?

答:
You have to show appropriate set inclusions, let $x\in X$ such that $\sup_n f_n(x) >a$, show that there exists $n$ such that $x\in  \{ z\in X: f_n(z)>a \}$.

Likewise show that if $x\in X$ where $f_n(x)>a$, then $x\in \{ z\in X: \sup_n f_n(z)>a \}$.


答:
Yes, that is the core of a valid proof.  All that is left is to tidy up the presentation.

The set of $(A\smallsetminus B)\smallsetminus C$ is the set of elements that are in $A$ but not in $B$ and not in $C$, which is one and the same as the set $(A\smallsetminus C)\smallsetminus B$.

$$\begin{align}(A\smallsetminus B)\smallsetminus C&=\{x:x\in A\smallsetminus B~\land~x\notin C\}\\&=\{x:(x\in A~\land~x\notin B)~\land~ x\notin C\}\\&=\{x:(x\in A~\land~x\notin C)~\land~ x\notin B\}\\&=\{x:x\in A\smallsetminus  C~\land~x\notin B\}\\&=(A\smallsetminus C)\smallsetminus B \end{align}$$

Or... whatever format you prefer.

Title: Prove or disprove: $(A-B)-C=(A-C)- B$ for sets $A$, $B$, $C$

Body:
Prove or disprove: For sets $A$, $B$, $C$,
$$(A-B)-C=(A-C)- B$$

Now my attempt is

$$(x\in A \;\text{and}\; x\notin B) \;\text{and}\; x\notin C \tag1$$

We can write it as

$$(x\in A  \;\text{and}\; x\notin C)  \;\text{and}\; x\notin B \tag2$$

which proves the statement.

Is it right?


问:
Show that $\mathrm{min}(\frac{a}{b},\frac{a'}{b'}) \leq \frac{a+a'}{b+b'} \leq \mathrm{max}(\frac{a}{b},\frac{a'}{b'})$
The general inequality is : \begin{equation}\mathrm{min}(\frac{a_1}{b_1},\dots,\frac{a_n}{b_n}) \leq \frac{a_1+\dots+a_n}{b_1+\dots+b_n} \leq \mathrm{max}(\frac{a_1}{b_1},\dots,\frac{a_n}{b_n})\end{equation}where $a_1,\dots,a_n,b_1,\dots,b_n$ are real non-zero positive numbers. I thought of starting by the easier case $n=2$ but I still have no idea how to compare $\frac{a}{b}$ and $\frac{a'}{b'}$.

答:
Idea: Go case by case

Case 1:

$\mathrm{min}(\frac{a}{b},\frac{a'}{b'}) = \frac{a}{b}$ and $\mathrm{max}(\frac{a}{b},\frac{a'}{b'}) = \frac{a}{b}$

Case 2:

$\mathrm{min}(\frac{a}{b},\frac{a'}{b'}) = \frac{a}{b}$ and $\mathrm{max}(\frac{a}{b},\frac{a'}{b'}) = \frac{a'}{b'}$

Case 3:

$\mathrm{min}(\frac{a}{b},\frac{a'}{b'}) = \frac{a'}{b'}$ and $\mathrm{max}(\frac{a}{b},\frac{a'}{b'}) = \frac{a}{b}$

Case 4:

$\mathrm{min}(\frac{a}{b},\frac{a'}{b'}) = \frac{a'}{b'}$ and $\mathrm{max}(\frac{a}{b},\frac{a'}{b'}) = \frac{a'}{b'}$


问:
Study the convergence of the series $\sum_{n \ge 1} \sin \frac{1}{n^{5/4}}$.
I have to show whether the series

$$\sum_{n \ge 1}\sin \frac{1}{n^{5/4}}$$

is convergent or not.

This is what I tried, but I am not sure if it's correct:

We know:

$$\sin x \le x, \hspace{1cm} \forall x \ge 0$$

So then

$$\hspace{6cm} \sin \frac{1}{n^{5/4}} \le \frac 1{n^{5/4}}, \hspace{.5cm} \forall n \in \mathbb{N} \hspace{3cm}(1)$$

By the generalized harmonic series, we also know

$$\hspace{7cm} \sum_{n \ge 1} \frac{1}{n^{5/4}} \hspace{.25cm} \text{convergent} \hspace{3cm} (2)$$

Now, using $(1)$ and $(2)$, we can conclude by the First Comparison Test that the series

$$\sum_{n \ge 1} \sin \frac{1}{n^{5/4}}$$

is convergent.

Is this correct?

答:
When you wrote that $\displaystyle\sin\left(\frac1{n^{5/4}}\right)\leqslant n^{5/4}$, what you should have written was that $\displaystyle\sin\left(\frac1{n^{5/4}}\right)\leqslant\frac1{n^{5/4}}$.

Besides, the comparison test is for series of non-negative numbers. So, you should add to your proof that$$(\forall n\in\Bbb N):\frac1{n^{5/4}}\in\left(0,\frac\pi2\right)\implies(\forall n\in\Bbb N):\sin\left(\frac1{n^{5/4}}\right)>0.$$


答:
Hint: Let $f(x)=\sqrt{6+x}$ for $x \ge -6$. Prove that

$f(x) \ge x$ when $ -6 \le x \le 3$

$f(x) \le x$ when $ 3 \le x$

Apply this result to $a_{n+1}=f(a_n)$ and conclude that the sequence is monotone.

Title: how to show that $a_{n+1}=\sqrt{6+a_{n}}$ is bounded?

Body: I have to show that the sequrnce $\{a_n\}$ with $a_{n+1}=\sqrt{6+a_{n}}$ is bounded, i know : $a_{n-1}<a_{n}$ because i demonstrated this. I tied to use this relation: $ab\leq \frac{a^2}{2}+\frac{b^2}{2}$. I don't kow how to increase $\sqrt{6+a_{n}}$


问:
prove thatt $\sum_{cyc} \frac{1}{{(x+y)}^2}\ge 9/4$

prove that $$\sum_{cyc} \frac{1}{{(x+y)}^2}\ge 9/4$$ where $x,y,z$ are positives such that $xy+yz+xz=1$

By Holder;$$\left(\sum_{cyc} \frac{1}{{(x+y)}^2} \right){\left(\sum yz+zx \right)}^2\ge {\sum \left(z^{2/3} \right)}^{3}$$.

Hence it suffices to prove $$\sum {\left(z^{2/3} \right)}^{3}\ge 3^{2/3}$$ which is falsse.

I was able to get a weaker bound.since $xy+yz+zx=1$ we have $x^2+y^2\le 2$. By C-S and $x^2+y^2<2$
$$\sum_{cyc} \frac{1}{{(x+y)}^2}\ge \frac{1}{2}\sum \frac{1}{x^2+y^2}>3/4$$

I am interested on a solution using standard inequalities (C-S AM-GM,chebyshov etc) rather than fully expanding and using uvw/pqr/schur.

Body:
We need to prove that:
$$\sum_{cyc}\frac{1}{(x+y)^2}\geq\frac{9}{4(xy+xz+yz)}$$ or
$$\sum_{cyc}\left(\frac{1}{(x+y)^2}-\frac{3}{4(xy+xz+yz)}\right)\geq0$$ or
$$\sum_{cyc}\frac{4xz+4yz-2xy-3x^2-3y^2}{(x+y)^2}\geq0$$ or
$$\sum_{cyc}\frac{(z-x)(3x+y)-(y-z)(3y+x)}{(x+y)^2}\geq0$$ or
$$\sum_{cyc}(x-y)\left(\frac{3y+z}{(y+z)^2}-\frac{3x+z}{(x+z)^2}\right)\geq0$$ or
$$\sum_{cyc}(x-y)^2(3xy+xz+yz-z^2)(x+y)^2\geq0.$$
Now, let $x\geq y\geq z$.

Thus, $$\sum_{cyc}(x-y)^2(3xy+xz+yz-z^2)(x+y)^2\geq$$
$$\geq\sum_{cyc}(x-y)^2(xz+yz-z^2)(x+y)^2\geq$$
$$\geq(x-z)^2y(x+z-y)(x+z)^2+(y-z)^2x(y+z-x)(y+z)^2\geq$$
$$\geq(y-z)^2y(x-y)(x+z)^2+(y-z)^2x(y-x)(y+z)^2=$$
$$=(x-y)^2(y-z)^2(xy-z^2)\geq0.$$


问:
Boundary orientation for a cylinder

Please help me.I am think that I can use stokes theorem but ı could not apply.This question is very benefical for me to learn the subject please help me :(

答: Unless your book/class is using some strange conventions, the orientations of the boundary components should match the little almost-circle drawn in Figure 21.8.  This means that on $C_0$ the orientation goes from left to right across the front of the picture and (therefore) from right to left across the back. That's because the little circle in the picture is oriented from left to right where it goes near $C_0$ on the front of the cylinder.  So $C_0$ would be oriented counterclockwise as seen from above.  $C_1$ would have the opposite orientation, clockwise as seen from the top.  (In this answer, I've assumed that you're working with reasonable definitions of boundary orientations and that the little circle in hte picture indicates the 2-dimensional orientation in a reasonable way.)</p>

Title: Boundary orientation for a cylinder


答:
Suppose that for small $\epsilon$:

$$(1+\epsilon)^{-1/2}=1+\binom{-1/2}{1}\epsilon+\binom{-1/2}{2}{\epsilon^2}+O(\epsilon^3)=1-\frac12\epsilon+\frac{3}{8}\epsilon^2+O(\epsilon^3)$$

If you put $-\epsilon$ instead of $\epsilon$ into the last expression you get:

$$(1-\epsilon)^{-1/2}=1+\frac12\epsilon+\frac{3}{8}e^2+O(\epsilon^3)$$

It means that:

$$(1+\epsilon)^{-1/2}+(1-\epsilon)^{-1/2}=2+\frac34\epsilon^2+O(\epsilon^3)$$

If you put $\epsilon=b/a$, you are done. Obviously $k=\frac34$

Body:
If  $a$ is sufficiently large as compared with $b,$ and $\sqrt
 \frac{a}{a-b}+\sqrt \frac{a}{a+b}=2+k(\frac{b}{a})^2$,then what is the
value of $k?$

Query

Can we find it using particular value?,like putting $a=4,b=2,$ then we get

$k=4\sqrt2+4\sqrt \frac{2}{3}-8$

This problem is given in  Multiple Choice Question exercise,the possible answers given are

A.$\frac{2}{3}$

B.$\frac{3}{4}$

C.$\frac{4}{5}$

D.$\frac{5}{6}$

None of the answer mathches with mine....

Please give some hint to determine value of $k?$


问:
Sum with two indices, how to handle the condition $i\neq j$?
I'm given
$$
\sum_{i,j=1, i \neq j}^{N} ij \tag 1
$$
I assume this is a short hand for

$$
\sum_{i=1}^{N}
\Bigg (
\sum_{j=1}^{N} ij
\Bigg )
\qquad? \tag 2
$$
But where does $i\neq j$ belong, to the inner or outer sum?

Example with $N=2$ and $i\neq j$ in the inner sum:
\begin{align}
\sum_{i,j=1 \\i\neq j}^{2} ij
&=
\sum_{i=1}^{2}
\Bigg (
\sum_{j=1, i \neq j}^{2} ij
\Bigg )
\tag 3
\\
&=\sum_{i=1}^{2}
\Bigg (
i1+i2
\Bigg )
\tag 4
\\
&=
(11+12 + 21 + 22)
\tag 5
\\
&\text{\{Now discard 11 och 22\}}\\
&=23
\end{align}

I guess the end result is correct, but I don't think my approach is correct ("Now discard"). How should I handle the condition $i\neq j$?

答:
If you're familiar with the Kronecker delta You can write that sum as the following:

$$\sum_{i=1}^{N}\sum_{j=1}^N (1 -\delta_{ij}) ij$$

If $i = j$ then $\delta_{ij}= 1$ and that will cancel that turm of the sum


问:
How to determine the partial derivative of a function, given its gradient evaluated at a point, and its composition.
Assume that the function $f$ is $C^1$ and that $\nabla f(1, -4) = (3, -2)$. Consider the function: $h(x,y) = f(3y^2 - x, 2xy)$. What is $\frac{\partial h}{\partial y}(2, -1)$?

I've seen another question asking how to find $f$ given the symbolic gradient. But I haven't been able to find a question similar to mine (where the gradient evaluated at a point is given). My first guess is that if we were given the symbolic gradient we could integrate it w.r.t one variable (as seen in the previously mentioned question). Then we could apply the chain rule to find $\frac{\partial h}{\partial y}$. But I'm not sure how to approach this using only the gradient evaluated at a point. So I was hoping somebody could tell me how to approach this :)

EDIT:
So it seems I wasn't explicit enough in laying out my understanding of the question. So I'm aware that in order to evaluate $\frac{\partial h}{\partial y}(2, -1)$, we need to use the multivariable chain-rule. Let $u = 3y^2 - x$ and $v = 2xy$. Then we have:
$$
\frac{\partial h}{\partial y} = \frac{\partial f}{\partial u}\frac{\partial u}{\partial y} + \frac{\partial f}{\partial v} \frac{\partial v}{\partial y}
$$
The partial derivatives are then:
$$
\frac{\partial u}{\partial y} = 6y \quad \text{and} \quad  \frac{\partial v}{\partial y} = 2x
$$
Inserting into $\frac{\partial h}{\partial y}$ we get:

$$
\frac{\partial h}{\partial y} = \frac{\partial f}{\partial u}6y + \frac{\partial f}{\partial v} 2x
$$
But at this point I'm still not fully sure how to use the fact that $\nabla f(1, -4) = (3, -2)$ to determine $\frac{\partial h}{\partial y}$?

答:
We can write:

$$ F(x,y)= f \big(u(x,y) , v(x,y) \big) $$

Differentiating both sides with $y$ by the multivariable chain rule

$$ \frac{ \partial F}{\partial y}  = \frac{ \partial f}{ \partial u} \frac{d u}{ dy} + \frac{ \partial f}{ \partial v} \frac{ d v}{dy}$$

Observe that $$ ( \frac{ \partial f}{\partial u} , \frac{ \partial f}{\partial v})  = \nabla f|_{(u,v)}$$

Hence:

$$ \frac{ \partial F}{\partial y} = \nabla f|_{(u,v) }  \cdot ( \frac{du}{dy} , \frac{dv}{dy})$$

Can you finish?


答:
Let us call $u=3y^2-x$, $v=2xy$ and $z=h(x,y)=f(u,v)$. Then, the chain rule is neatly:
$$
\frac{\partial h}{\partial y}=\frac{\partial z}{\partial y}=\frac{\partial z}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial y}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial y}
$$

Now, let us particularize everything at the given point $(x,y)=(2,-1)$. First of all, you have $(u,v)=(1,-4)$, so you can write the chain rule as:
$$
\left.\frac{\partial h}{\partial y}\right|_{(2,-1)}=\left.\frac{\partial f}{\partial u}\right|_{(1,-4)} \left.\frac{\partial u}{\partial y}\right|_{(2,-1)}+ \left.\frac{\partial f}{\partial v}\right|_{(1,-4)} \left.\frac{\partial v}{\partial y}\right|_{(2,-1)}
$$

And you also have the information that $\left.\frac{\partial f}{\partial u}\right|_{(1,-4)}=3$, and $\left.\frac{\partial f}{\partial v}\right|_{(1,-4)}=-2$.

All that remains is to compute the partial derivatives of $u,v$ and substitute $(x,y)=(2,-1)$. Can you follow from here?

Title: How to determine the partial derivative of a function, given its gradient evaluated at a point, and its composition.

Body:
Assume that the function $f$ is $C^1$ and that $\nabla f(1, -4) = (3, -2)$. Consider the function: $h(x,y) = f(3y^2 - x, 2xy)$. What is $\frac{\partial h}{\partial y}(2, -1)$?

I've seen another question asking how to find $f$ given the symbolic gradient. But I haven't been able to find a question similar to mine (where the gradient evaluated at a point is given). My first guess is that if we were given the symbolic gradient we could integrate it w.r.t one variable (as seen in the previously mentioned question). Then we could apply the chain rule to find $\frac{\partial h}{\partial y}$. But I'm not sure how to approach this using only the gradient evaluated at a point. So I was hoping somebody could tell me how to approach this :)

EDIT:
So it seems I wasn't explicit enough in laying out my understanding of the question. So I'm aware that in order to evaluate $\frac{\partial h}{\partial y}(2, -1)$, we need to use the multivariable chain-rule. Let $u = 3y^2 - x$ and $v = 2xy$. Then we have:
$$
\frac{\partial h}{\partial y} = \frac{\partial f}{\partial u}\frac{\partial u}{\partial y} + \frac{\partial f}{\partial v} \frac{\partial v}{\partial y}
$$
The partial derivatives are then:
$$
\frac{\partial u}{\partial y} = 6y \quad \text{and} \quad  \frac{\partial v}{\partial y} = 2x
$$
Inserting into $\frac{\partial h}{\partial y}$ we get:

$$
\frac{\partial h}{\partial y} = \frac{\partial f}{\partial u}6y + \frac{\partial f}{\partial v} 2x
$$
But at this point I'm still not fully sure how to use the fact that $\nabla f(1, -4) = (3, -2)$ to determine $\frac{\partial h}{\partial y}$?