问:
Show that $f:[a,b]\to \Bbb R$ is a measurable function.
Let $f:[a,b]\to\Bbb R$. Suppose that for every $\epsilon>0$ there exists a measurable set $E\subseteq [a,b]$ such that $m(E)<\epsilon$ and $f$ is continuous on $[a,b]\setminus E$. We want to prove that $f$ is a (Lebesgue) measurable function.

Attempt. For any $n\in \Bbb Z_{>0}$ there exists a measurable set $E_n\subseteq [a,b]$ zuch that $m(E_n)<1/n$ and $f$ is continuous on $[a,b]\setminus E_n$. We also define $f_n:[a,b]\to\Bbb R$ by $f_n(x) = \left\{
  \begin{array}{lr}
    f(x) & , x \in [a,b]\setminus E_n\\
    0 & , x \in E_n
  \end{array}
\right.
$. We claim that the sequence $\{f_n\}_{n=1}^{\infty}$ converges to $f$ a.e pointwise on $[a,b]$ i.e. $m(C)=0$ where $C=\{x\in [a,b] : \lim_{n\to\infty} f_n(x)\neq f(x)\}$. Note that if $x\in C$ then there exist $\epsilon>0$ and $N_x\in \Bbb N$ such that $|f_n(x)-f(x)|>\epsilon >0$ for all $n\geq N_x$. So, $x\in E_n$ for all $n\geq N_x$ and $f(x)\neq 0$, so $x\in\cap_{n\geq N_x}E_n$. So we get $$C\subseteq \bigcup_{x\in C}\bigcap_{n\geq N_x}E_n.$$
But, I couldn't go further from here. Could anyone check my work and give any suggestions? Thanks!

-Update-

$$C\subseteq \bigcup_{x\in C}\bigcap_{n\geq N_x}E_n \subseteq \bigcup_{k=1}^\infty \bigcap_{n\geq k}E_n \implies 0\leq m(C)\leq m\left(\bigcup_{k=1}^\infty \bigcap_{n\geq k}E_n\right)=\lim_{k\to \infty}m(\cap_{n\geq k}E_n)\leq\lim_{k\to\infty} m(E_k)=\lim_{k\to\infty} 1/k=0.$$
So $m(C)=0$. So, $f$ is measurable since $f_n$ is measurable for all $n\in \Bbb Z_{>0}$.

答:
I am presuming that you wish to show that $f$ is Lebesgue measurable.

The trick is to define the approximations in a 'cumulative' way.

Suppose $f$ is continuous on $E^c$, then the function $f \cdot 1_{E^c}$ is measurable.

There is a sequence of $E_n$ such that $f$ is measurable on $E_n^c$ and $m E_n < {1 \over n}$. Let $E= \cap E_n$ and note that $m E = 0$.

Define $f_n = f \cdot 1_{E_1^c}+ f \cdot 1_{E_2^c \setminus E_1^c}+ \cdots + f \cdot 1_{E_n^c \setminus (E_1^c \cup \cdots \cup E_{n-1}^c)}$. We see that $f_n$
is measurable and $f_n(x) \to f(x)1_{E^c}(x)$. Hence $f \cdot 1_{E^c}$ is measurable.

Since $m$ is complete, we see that $f \cdot 1_E$ is measurable and hence
$f=f \cdot 1_{E^c}+ f \cdot 1_E$ is measurable.


问:
Unable to get to all permutations after $n-1$ transpositions
Problem: Give an example of a permutation of the first $n$ natural numbers from which it is impossible to get to the standard permutation $1,2,\ldots,n$ after less than $n-1$ transposition operations (i.e switching the place of two elements).

My attempt

Suppose we have a permutation $T$ and we perform one transposition on $T$ to get $T'$. That would mean $T(i) = j, T(i') = j'$ and $T'(i) = j' , T'(i') = j$ for some $i,j,i',j'$. It is easy to see that the permutation T contains 2 cycles (may be the same): $(i,j,...)$ and $(i',j',\ldots)$. The transposition operation would affect only these two cycles but keep all other cycles intact. Therefore, the number of cycles, if decreased, will not decrease more than $1$.

Now the permutation $[1,2,\ldots,n]$ has n cycles and the permutation $[2,3,\ldots,n,1]$ has 1 cycle only. So it is impossible to use less than $n-1$ operations to get $[2,3,\ldots,n,1]$ from $[1,2,\ldots,n]$. Keeping in mind that getting from permutation A to B is the same as getting from B to A, problem solved.

My question

Is my approach correct and are there any better solutions? Thank you.

答:
There is a quantifiable sense in which permutations with more motion require more transpositions to achieve.

For a sequence of transpositions of pairs in a set $S$, build an undirected graph, with vertices $S$ and an edge between every pair that is exchanged at some time in the sequence.

The vertex sets of the connected components of this transposition graph are permuted within themselves.  To reach all permutations the entire graph must be a single component.  This requires at least $n-1$ edges for $n$ vertices (at minimum, a tree).

This assumes that the transpositions can be chosen to fit the permutation one wants to achieve.   If there are $n-1$ transpositions given, then although any permutation can be reached using a chain of those transpositions, the length of the chain can be as high as $n(n-1)/2$, the number of exchanges of adjacent elements needed to reverse the order of an array of $n$ distinct objects.

Same argument shows that $\frac{n-1}{k-1}$ cycles of length $k$ are needed to sort arrays of length $n$.

ContentLicense: CC BY-SA 3.0


问:
Linear equations Systems and parameter
Consider
$$A=\begin{pmatrix}
1 & 2 & 3 \\
2 & 4 & 3\\
3 & 6 & a
\end{pmatrix}
$$

a) Wich is the range of $A$ in fuction of $a\in \mathbb{R}$?

b) What terms have to satisfy $b \in \mathbb{R}^{3}$ for the equation system $Ax=b$ has solution, with $x \in \mathbb{R}^{3}$ ? How it depends on $a\in \mathbb{R}$?

For a) I did elemental operations and we have that the equivalent matrix is
$$R=\begin{pmatrix}
1 & 2 & 0\\
0 & 0 & 1\\
0 & 0 & a-9
\end{pmatrix}$$
but it doesn't matter wich is the value of $a$ because $\begin{pmatrix} 0 & 0 & a-9 \end{pmatrix}= (a-9)\begin{pmatrix}  0 & 0 &1 \end{pmatrix}$, so the $range(A)=2$ always , or Am I wrong? Maybe I didn't uderstand the problem.

答:
Hint:  The range is a subspace of $\Bbb R^3$. So for each $a$, we get a two-dimensional subspace, $V_a=\{(x+3y,2x+3y,3x+ay):x,y\in\Bbb R\}$.  (The dimension is two because that is the rank.)

On $b)$, we have that $b$ must be in the range, which is the answer from $a)$.  Thus we need $b=(x+3y,2x+3y,3x+ay)$ for some $x,y\in\Bbb R$.  Say $b=(b_1,b_2,b_3)$.  Then we can get $x$ and $y$ in terms of $b_1$ and $b_2$.  We have $\begin{pmatrix}1&3\\2&3\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}b_1\\b_2\end{pmatrix}$.  You can solve for this because the matrix is invertible, and then get $b_3$ in terms of $b_1,b_2$ and $a$ (since $b_3=3x+ay$).

ContentLicense: CC BY-SA 4.0

Title: Linear equations Systems and parameter


问:
Why is Wolfram Alpha showing me this function has no roots; or does it actually have no roots?
I am currently trying to find the roots of this function below.

$$ f_s(c,s,\alpha,\rho, \gamma, us) =(\alpha s^{\rho-1} ((1-\alpha)c^\rho + \alpha s^\rho)^{((1-\gamma)/\rho) - 1}) - us $$

, where us is one whole variable.

I want to find the roots of this function, given fixed values of s,alpha,rho,gamma and us. The issue is, I have typed this function into wolfram trying different fixed values, but the result is always that it yields no roots.

s = 40, alpha = 0.5, rho = 0.5, gamma = 2, us = 3

s = 20, alpha = 0.2, rho = 0.5, gamma = 2, us = 2

Is there something I'm missing or does this equation truly have no roots? I've attached some more images below that might be of help, thanks a lot!

Wolfram Alpha Giving the Origin of that function above, and us is just a new variable I made up to subtract from it.

EDIT 1 : Thank you for all your suggestions; I have tried graphing it, and for sure there are no roots. This is supposed to work with either of the fixed values I chose, so I would conclude that this function really has no roots and I was down the wrong track. If anyone else has any insights, that would be great but if not, thanks for your help everyone and I'll close this thread.

答:
Let's rename $us$ to $u$, since it's more common to work with single-character variable names in math (and it makes it easier to read). The equation is
$$\alpha s^{\rho-1}\big[(1-\alpha)c^\rho+\alpha s^\rho\big]^{\frac{1-\gamma}\rho-1}=u.$$
Now, for every quintuple $(\alpha,s,c,\rho,\gamma)$, there is (exactly) one value of $u$ for which the equation is satisfied. This means that, for at least some values of $(\alpha,s,\rho,\gamma,u)$, there is some value $c$ that satisfies the equation: pick some real $c_0$ and some parameters $(\alpha,s,\rho,\gamma)$, set
$$u_0=\alpha s^{\rho-1}\big[(1-\alpha)c_0^\rho+\alpha s^\rho\big]^{\frac{1-\gamma}\rho-1},$$
and then $c=c_0$ will be a root of the equation with parameters $(\alpha,s,\rho,\gamma,u_0)$.

This may not be that satisfying, since it doesn't tell you much about when a solution might exist. Let's try to solve the system. We can first divide by $(\alpha s^{\rho-1})$ to get
$$\big[(1-\alpha)c^\rho+\alpha s^\rho\big]^{\frac{1-\gamma}\rho-1}=\frac{u}{\alpha s^{\rho-1}}.$$
Now, we can take each side to the power of $\frac{1-\gamma-\rho}{\rho}$ to get
$$(1-\alpha)c^\rho+\alpha s^\rho = \left(\frac{u}{\alpha s^{\rho-1}}\right)^{\frac{1-\gamma-\rho}\rho}.$$
This gives
$$c^\rho=\frac1{1-\alpha}\left(\frac{u}{\alpha s^{\rho-1}}\right)^{\frac{1-\gamma-\rho}\rho}-s^\rho\frac{\alpha}{1-\alpha}.$$
So, depending on your value of $\rho$, this might be solvable (if $\rho$ is an odd integer, there is a root somewhere, while otherwise you probably want the right side to be positive). In general, though, if
$$\left(\frac{u}{\alpha s^{\rho-1}}\right)^{\frac{1-\gamma-\rho}\rho}\geq \alpha s^\rho,$$
then there is a (positive real) solution for $c$.

Edit (to explain the remark on how the solvability depends on $\rho$): Say $\rho$ is some arbitrary real number. Where is $c^\rho$ defined? A priori, it's hard to say how to define exponentation exactly -- it's easy when $\rho$ is an integer and we can just multiply $c$ together a bunch of times, but harder for arbitrary $\rho$. It turns out (and Desmos, for example, will give you a graph of this) that you can define $c^\rho$ for any real $\rho$, as long as $c$ is positive (for an example of why this constraint is reasonable, think about $(-1)^{1/2}$; it turns out we can define exponentiation $x^y$ for all reals, but we lose some nice properties, like the fact that $x^y$ is a real number). These values are always positive, so there is only a solution to $c^\rho=x$ in $c$ for arbitrary $\rho$ when $x$ is positive.

If we're in the special case where $\rho$ is an integer, then we can define $c^\rho$ for all real $c$, and then try to solve for $c$. If $\rho$ is even, $c^\rho\geq 0$ for all $c$, and so we can't solve $c^\rho=x$ for negative $x$. However, if $\rho$ is odd, we can!


问:
How to prove $f$ can be uniformly approximated on $[1,\infty )$ by a function $g$, using Weierstrass theorem.

Let $f:[1,\infty )\to \Bbb R$ be continuous, and suppose that $\lim \limits_{x \to \infty}f(x)$ exists. Use the Weierstrass approximation theorem to prove that $f$ can be uniformly approximated on $[1,\infty )$ by a function $g$ of the form $g(x)=p(\frac{1}{x})$, where $p$ is a polynomial.

By the assumption that $\lim \limits_{x \to \infty}f(x)=L<\infty$, given $\epsilon>0$ there exists $R>0$ such that $\forall x>R$, we have $|f(x)-L|<\varepsilon$. If we restricts $f$ to $[R,a]$ for any $a>R$, then by the Weierstrass theorem, there exists a sequence of polynomials $\{P_n\}$ converges uniformly to $f$ on $[R,a]$. We find that $P_n(x)=L$ on $[R,a]$, when $n$ is large enough. So, I wonder if we restrict $f$ to $[1,R]$, can we find a sequence of polynomials $\{p_n\}$ such that $\lim \limits_{n\to \infty} p_n(x)=f(x)$ uniformly on [1,R] and $p_n(x)=L$ on $[R,\infty)$ such that $p_n\to f$ uniformly on $[1,\infty)$?

I have not figured out how to find the function $g$ mentioned in the statement, I 'm not sure how to prove this statement. Appreciate any suggestion.

答:
Define $\tilde{f}(x) = \begin{cases} f({1 \over x}),& x \in (0,1]\\
\lim_t f(t), & x=0 \end{cases}$. Note that $\tilde{f}$ is continuous on the compact interval $[0,1]$ hence for any $\epsilon>0$ there is some polynomial $p$ such that
$\|\tilde{f}-p\|_\infty < \epsilon$.

For $t \ge 1$ we have $|\tilde{f}({ 1 \over t}) - p({ 1 \over t})|< \epsilon$ and
since $\tilde{f}({1 \over t}) = f(t)$, we get the desired result.


问:
Integrate $\int_{-\infty}^{\infty} \frac{e^{-x^2}}{(x + jp)^M (x - jp)^M } dx$
I want to solve:

$$ \int_{-\infty}^{\infty} \frac{e^{-x^2}}{(x + jp)^M (x - jp)^M } dx$$

where $M \in \mathbb{Z}_{++}$ is even. To start, I create a square contour $\Gamma$ in the upper-half plane: $[-R, R], [R, R+ja], [R+ja, -R+ja], [-R+ja, -R]$, where $a > p$ and $p \in \mathbb{R}_{++}$. Thus I have:

$$ \int_{\Gamma} f(z) dz = \int_{\Gamma} \frac{e^{-z^2}}{(z + jp)^M (z - jp)^M } dz =   \left[ \int_{-R}^{R} + \int_{R}^{R + ja} + \int_{R+ja}^{-R + ja} + \int_{-R + ja}^{-R}   \right] \frac{e^{-z^2}}{(z + jp)^M (z - jp)^M } dz$$

where $ \int_{\Gamma} f(z) dz = \text{Res}_{f(z)}\left( z = jp \right)$ and I will let $R \rightarrow \infty$.

It is easy to show that $ \left[ \int_{R}^{R + ja} + \int_{-R + ja}^{-R}  \right] \left( \cdot \right) = 0 $. We just let $z = \pm r + jb$, respectively, for $b \in [0, a]$, and we get a $e^{-R^2}$ term in front of the integrals, which goes to zero as R goes to infinity.

I am having problems with the integral $\int_{R+ja}^{-R + ja} \frac{e^{-z^2}}{(z + jp)^M (z - jp)^M } dz$ however. Letting $z = r+ j\alpha$ for $r \in [-R, R]$ doesn't seem to do it.

Questions:

How can I compute $\int_{R+ja}^{-R + ja} \frac{e^{-z^2}}{(z + jp)^M (z - jp)^M } dz$?

Am I correct that I can use the residue theorem here, as I have done it?

Edit: I am a complete n00b so please be patient and explain like I am five.

Edit2: For the entire integral, mathematica gives me $\sqrt{\pi} \cdot \text{HypergeometricU}\left[M, M + \frac{1}{2}, p^2 \right]$

答:
There's nothing to expect for the residue theorem to yield here (especially since a CAS reports special functions in the answer). But there is another (perhaps easier) approach: $$f_n(a):=\int_{-\infty}^\infty(a+x^2)^{-n-1}e^{-x^2}\,dx=\frac{(-1)^n}{n!}\frac{d^n}{da^n}f_0(a)$$ for $a>0$, and $f_0(a)=e^a g(a,1)$, where for $t\geqslant 0$

\begin{align*}
g(a,t)&=\int_{-\infty}^\infty\frac{e^{-t(a+x^2)}}{a+x^2}\,dx=\int_{-\infty}^\infty\int_t^\infty e^{-y(a+x^2)}\,dy\,dx
\\&=\int_t^\infty e^{-ay}\int_{-\infty}^\infty e^{-yx^2}\,dx\,dy=\sqrt\pi\int_t^\infty e^{-ay}\frac{dy}{\sqrt y}
\\&=2\sqrt{\frac\pi a}\int_{\sqrt{at}}^\infty e^{-x^2}\,dx=\frac{\pi}{\sqrt a}\operatorname{erfc}\sqrt{at}.
\end{align*}

The given integral is $f_{M-1}(p^2)$.


问:
identity of polylogarithm
let be the function defined by a series

$$ f(x)= \sum_{n=0}^{\infty}g(n)x^{n} $$

assume also that $ g(n)= \sum_{k=0}^{\infty}a(k)n^{k} $

then we have the double series

$$ f(x)= \sum_{n=0}^{\infty}\sum_{k=0}^{\infty}a(k)n^{k}x^{n} $$

if we interchange the order of summation and make the sum over 'n' plus zeta regularization with the polylogarithm $ Li _{s}(x) = \sum_{n=0}^{\infty} \frac{x^{n}}{n^{s}} $

we find $$ f(x)= \sum_{k=0}^{\infty}a(k)Li_{-k}(x) $$

if the number of 'k' is limited so the infinite series is finite we can evaluate always $ f(x) $ by using a finite sum of terms $ Li_{-k}(x) $ for k = 0,1,2,3, N $

are my calculations correct ??

答:
They are correct. It should be noted, however, that polylogarithms of negative orders are elementary functions. In fact, one doesn't need to know anything about polylogarithms, it suffices to differentiate geometric series an appropriate number of times:
$$\sum_{n=0}^{\infty}n^kx^n=\left(x\frac{d}{dx}\right)^k\sum_{n=0}^{\infty}x^n=
\left(x\frac{d}{dx}\right)^k\frac{1}{1-x}.$$
So, for example,
$$\mathrm{Li}_{-2}(x)=\left(x\frac{d}{dx}\right)^2\frac{1}{1-x}=\frac{x(1+x)}{(1-x)^3}.$$</p>

Title: identity of polylogarithm


问:
The Euclidean projection of $x_0$ on a hyperplane $C = \{x | a^Tx = b\}$

Find the Euclidean projection of $x_0$ on a hyperplane $C = \{x | a^Tx = b\}$.

My thoughts:
We can view the point as a vector from the origin. The projection of a vector to the line is a multiple of the line $tx$. From the right triangle formed by the line and the vector we can say that $x^t(x_0-tx), t = \frac{x^tx_0}{|x|_2^2}$. So, the projection is $tx = \frac{x^tx_0}{|x|_2^2}x$.

However, the solution of the problem is $x_0+\frac{(b_0-a^Tx_0)}{|a|_2^2}a$.

A similar question that I don't understand is the projection of a point $x_0$ to the halfspace $C=\{x|a^Tx \le b\}$.

答:
I think I got it. Let $x \in C$. Let's find the projection of a vector $x_0-x$. By the same logic as in my question the projection is a sum multiple of the line $ta$. Then from the right triangle we get $a^t(x_0-x-ta) = \frac{a^t(x_0-x)}{|a|_2^2}$. So, the projection itself is $ta = \frac{a^t(x_0-x)}{|a|_2^2}a$. However, we does not want the projection of a vector $(x_0-x)$ we seek the projection of a point $x_0$. From the right triangle it is known that $x_0-x = (P_C(x_0)-x_0) + \frac{a^t(x-x_0)}{|a|_2^2}a$. Where $P_C(x_0)$ is the projection of a point $x_0$ to the set $C$. Therefore, $P_C(x_0) = x + \frac{a^t(x-x_0)}{|a|_2^2}a = x + \frac{b-a^tx_0}{|a|_2^2}a$.

However, as you can see the answers do not coincide fully. Why? Did I make mistake?


问:
For the given function $f(x)=\frac{1}{\sqrt {1+x}} +\frac{1}{\sqrt {1+a}} + \sqrt{\frac{ax}{ax+8}}$, prove that $1<f(x)<2$ for positive a and $x\ge 0$

For the given function $f(x)=\frac{1}{\sqrt {1+x}} +\frac{1}{\sqrt {1+a}} + \sqrt{\frac{ax}{ax+8}}$, prove that $1<f(x)<2$ for +ve $a$ and $x\ge 0$

I tried the most rudimentary method ie. differentiating wrt x and hoped to get at least 2 solutions for $f’(x)$, unfortunately I wasn’t able to find a solution for implying function is monotonous . Now whether it’s sis increasing or decreasing still remains a mystery, given that we don’t know what $a$ is.

What method can be used to solve this?

答:
Partial answer

Knowing if the map is increasing or decreasing if it is known to be monotonous is easy. Just take two values $u \lt v$ and look at the sign of $f(v)-f(u)$.

We have $$f(0)=1 +\frac{1}{\sqrt {1+a}} $$ and

$$\lim\limits_{x \to \infty} f(x)=\frac{1}{\sqrt {1+a}} + 1 $$

Knowing that the two values are equal... implies that $f$ is not monotonous and that you made an error computing the derivative.


问:
How to get the necessary acceleration for a block to remain in place when pushed by a wedge?
The problem is as follows:

The figure from below represents a wedge which has a mass $M$ which is
being pushed by a force $F$. On the top of the wedge there is a small
block whose mass is $m$. Assume the wedge $M$ and the block $m$ are
frictionless. Given this information. Find the value of the force $F$
so that the block $m$ does not slide over the wedge.

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{45 N}\\
2.&\textrm{60 N}\\
3.&\textrm{75 N}\\
4.&\textrm{80 N}\\
\end{array}$

So far, below are the forces in the FBD which I made. I'm not sure if these are the only ones acting, thus I need help if the vectors are put properly.

As it can be seen my source of concern is. The problem does not indicate anything about the floor. It doesn't say that it is rugged or whatsoever. So what should I suppose?.

This is the reason of my doubt?. Because as it stands it seems that the only way to get $F$ might be using terms of equilibrium.

$R_M\cos 37^\circ=mg$

$F=R_m\sin 37^\circ$

But are the following equations correct?. Do they make sense?.

Then:

By Newton's third law: $R_m=R_M$

$F=\frac{mg}{\cos 37^\circ}\cdot\sin 37^\circ$

$F=mg\tan 37^\circ$

Hence by introducing the given values it would be:

$F=2\cdot 10 \cdot \frac{3}{4}=15\,N$

But this doesn't appear in the alternatives. What's exactly the right interpretation here?.

Can someone help me here?. I'm stuck exactly if is correct my interpretation of the vectors. Does it exist a way to know if should I include the component of the friction on the floor?.

The misleading part is, how to account for that?. I mean it is not given the value of the coefficient of friction for the surface. In my question I used the word acceleration, but is this necessary in this scenario?.

答:
The $M+m$ system is accelerating to the right with an acceleration $$a=\frac{F}{M+m}$$

Then the force equations for $m$ would be

\begin{align}
R_m \sin 37^\circ &= m \cdot \frac{F}{M+m}\\
R_m \cos 37^\circ &= mg
\end{align}

Dividing and substituting for values, we get
$$F=75 \, N$$

We have assumed the floor is frictionless (since there is no mention of $\mu$), $F_{\text{fric}}=0$ so that the net horizontal force acting on $M+m$ system is only $F$ (and not $F-F_{\text{fric}}$).


问:
scaling of a convex function
I have the following result. Let $X\subset \mathbb{R}^d$ containing the origin.

Let $f: X\rightarrow [0,\infty)$ be a differentiable concave function. Then for any $x\in\mathbb{R}^d$ and $c\geq 1$ such that $cx\in X$, we have $f(cx)\leq cf(x)$.

To prove the statement, first note that

$0\leq f(0) \leq f(x) - \langle\nabla f(x),x\rangle \Rightarrow \langle\nabla f(x),x\rangle \leq f(x) $

Now

$f(cx)\leq f(x) + \langle \nabla f(x), (c-1)x\rangle. \leq f(x) + (c-1)f(x) = cf(x)$

Now suppose $f: \mathbb{R}^d\rightarrow [0,\infty)$ be a differentiable convex function. Does there exists a similar relation between $f(cx) $ and $f(x)$, where $c\in\mathbb{R}$?

Thanks

答:
If you were to have a general relation between $f(cx)$ and $cf(x)$, it would have to be
$$
f(cx)\ge cf(x)
$$
since this is true about $f(x)=\|x\|^2$:
$$
f(cx)=\|cx\|^2=c^2\|x\|^2=c^2f(x)\ge cf(x),
$$
since $c\ge 1$.

Now consider the shifted version (in two dimensions, for simplicity)
$$
g(x_1,x_2)=\|(x_1-1,x_2-1)\|^2
$$
Take the point $x=(1-\sqrt{2}/2, 1-\sqrt{2}/2)$ and $c=(2+\sqrt{2})/(2-\sqrt{2})>1$. Then you can easily check that
$$
g(x)=1, g(cx)=1
$$
and, therefore, $g(cx)<cg(x)$.

Conclusion: there is no such general property.

Title: scaling of a convex function


问:
Proof sketch verification: Given $f : [0,1] \to \Bbb R$ with a special property, show that $\mathrm{supp}(f)$ is countable.
$
\newcommand{\R}{\mathbb{R}}
\newcommand{\N}{\mathbb{N}}
\newcommand{\supp}{\mathrm{supp}}
\newcommand{\set}[1]{\left\{ #1 \right\}}
\newcommand{\setb}[2]{\left\{ #1 \; \middle| \; #2 \right\}}
\newcommand{\ve}{\varepsilon}
\newcommand{\abs}[1]{\left| #1 \right|}
\newcommand{\FF}{\mathcal{F}}
$
The Problem: Let $f : [0,1] \to \R$ be such that $\exists \ve > 0$ such, for any choice of finitely many distinct $x_1, \cdots, x_n$ in $[0,1]$,

$$\left| \sum_{i=1}^n f(x_i) \right| \le \ve$$

Show that $\supp(f) := \setb{ x \in [0,1] }{ f(x) \ne 0 }$ is countable.

Brief Overview & My Goal: I have played with this problem a little bit, and have found it quite easy to show that $\supp(f)$ can be finite for some $f$, and sometimes $\supp(f)$ can be countably infinite. So the question remains: can $\supp(f)$ be uncountably infinite? Or, rather, what contradiction does that make?

So my goal in this approach was trying to do so via contradiction: suppose $f$ is such that $\supp(f)$ is uncountable, and see what contradiction arises. I think I found an approach where the boundedness of a finite sum is violated, but I wanted to make sure I have the right idea.

I will admit this proof does have some details that need fleshed out for this to be a "proper" proof (some statements are made without proof), but hopefully at least I have the overall "big picture" of what needs to be done.

Of course, if there are any logical flaws/inconsistencies, I would appreciate them being pointed out!

My Approach So Far:

By way of contradiction, suppose $\supp(f)$ as defined for an applicable $f$ is uncountable. Then $\FF := \set{ |f(x)| \mid x \in [0,1] }$ is nonempty trivially. Moreover, $\FF$ must be bounded above - otherwise, there will always exist some $x$ such that $|f(x)| > \ve$ for any $\ve > 0$. Since $\varnothing \ne \FF \subseteq \R$ is bounded above, then $M := \sup \FF$ exists, is positive, and is finite, due to the least-upper-bound property of $\R$. It thus holds that $M \in \FF'$ (the set of limit points of $\FF$), and thus $\exists \set{p_n}_{n \in \N}$ such that $|f(p_n)| \to M$.

Moreover, $\exists N \in \N$ such that the subsequence $\set{|f(p_n)|}_{n=N}^\infty \subseteq [M-\delta,M]$ for any given $\delta$. $(\ast)$

With this in mind, we will take a finite number of points $\set{x_i}_{i=1}^n$, where $x_i = p_{N+i}$. Note that the maximum element that $|f|$ outputs of this sequence, due to the monotonicity, is $|f(x_n)|=|f(p_{N+n})|.$ Then it holds
\begin{align*}
\abs{ \sum_{i=1}^n f(x_i) }
&\le \sum_{i=1}^n \abs{ f(x_i) } \tag{triangle inequality} \\
&\le \sum_{i=1}^n \abs{ f(x_n) } \tag{def. of maximum} \\
&= n \abs{ f(x_n) } \tag{simplify}
\end{align*}
Observe that, for $n$ sufficiently large, $nf(x_n) > \ve$, contradicting the notion that $f$ can satisfy the property for an uncountable $\supp(f)$.

Potential Problems:

This feels overly complicated, for one ... I feel there is a more elementary solution, but I can't figure out what it is.

I especially have doubts about the claim at $(\ast)$. I've tried a few ways to phrase what I'm trying to intuitively grasp while also being useful for breaking the bounds eventually: essentially, I want the sequence to eventually be within some certain distance of the supremum. (For instance, another I used at one point is that the sequence would be "eventually monotonic" after some $n=N$, but I don't think that's necessarily true. For instance, the sequence $1,0,1/2,0,1/3,0,1/4,\cdots$ converges to $0$ and is bounded below, but isn't monotonic.)

Of course I doubt these are the only problems ... any thoughts?

答:
Working with $|f(x)|$ is  a bad idea since inequalities go  the wrong way. You will have to consider positive and negative values of $f$ separately.

Let $A_n=\{x: f(x) \geq \frac 1  n\}$. Let us show that this is set is necessarily a finite set. If $x_1,x_2,...,x_N$ are distinct members of this set then $\epsilon \geq \sum\limits_{k=1}^{N} f(x_k) \geq \sum\limits_{k=1}^{N}  \frac  1n=\frac N n$. It follows that $N \leq n\epsilon$. Hence $A_N$ has at most $[n\epsilon]+1$ elements in it.

Let $A=\bigcup_N A_N$. Then $A$ is at most countable. If $x \notin A$ then  $f(x) <\frac 1 N$ for every $N$. Hence $x\notin A \implies f(x) \leq 0$. Similarly, we get another at most countable set $B$ such that $x\notin B \implies f(x) \geq 0$. It follows that $x\notin A\cup B \implies f(x)= 0$. This finishes the proof.


问:
Anonymous graphs and graph embeddedness
What are anonymous graphs, what is graph embeddedness, and how do they relate to each other? Very confused - I could not find short answer. Thanks.</p>

答:
Anonymous graphs are graphs that do not contain any labels and vertices cannot be identified.
According to wikipedia, embeddedness describes how close the neighbors of a vertex are to being a clique and this can be measured by calculating the local clustering coefficient.</p>

OwnerDisplayName: Aubergine

Title: Anonymous graphs and graph embeddedness


问:
How many solutions does $2 (\sin^{-1}x)^2 -5\sin^{-1}x+2=0$ have?

Number of solutions of the equation

$2 (\sin^{-1}x)^2 -5\sin^{-1}x+2=0 $

Let
$t=\sin^{-1}x $

$2 t^2 -5t+2=0$

$(t-2)(2t-1)=0 $

$t=2$    or $t=1/2$

Then

$\sin^{-1}x=2\quad$  or$\quad   \sin^{-1}x=1/2$

$x=\sin 2\quad$  or$\quad  x=\sin(1/2)$

Did I do something wrong here? I found a solution here =>  https://www.toppr.com/ask/question/number-of-solutions-of-the-equation-2sin1x25sin1x20/

The solution given on this site says only one solution exists. But I can't understand why $x=\sin 2$   is rejecting. isn't $t$ is an angle and $x$ is a value? Please help.

答: The (real-valued) arcsine returns a number in $[-\pi/2,\pi/2]$. In particular $\arcsin\sin2$ is not $2$, but $\pi-2$, so it doesn't work.


答:
You have to reject $x=\sin 2$ as $\arcsin x$ is defined on $[-1,1]$ and takes values in $[-\frac{\pi}{2}, \frac{\pi}{2}]$ by definition. As $2$ doesn’t belong to the last interval, this solution has to be rejected.

See inverse trigonometric functions.

Title: How many solutions does $2 (\sin^{-1}x)^2 -5\sin^{-1}x+2=0$ have?

Body:
Number of solutions of the equation

$2 (\sin^{-1}x)^2 -5\sin^{-1}x+2=0 $

Let
$t=\sin^{-1}x $

$2 t^2 -5t+2=0$

$(t-2)(2t-1)=0 $

$t=2$    or $t=1/2$

Then

$\sin^{-1}x=2\quad$  or$\quad   \sin^{-1}x=1/2$

$x=\sin 2\quad$  or$\quad  x=\sin(1/2)$

Did I do something wrong here? I found a solution here =>  https://www.toppr.com/ask/question/number-of-solutions-of-the-equation-2sin1x25sin1x20/

The solution given on this site says only one solution exists. But I can't understand why $x=\sin 2$   is rejecting. isn't $t$ is an angle and $x$ is a value? Please help.


问:
Expectation of Product of Normal Random Variable Exponential and Some Function?
$$X \sim \mathcal{N}(\mu, \sigma^{2})$$
$$X = \mu + \sigma Z$$

Show that:
$$\text{E}\bigg(\text{exp}(\theta X) \cdot f(X)\bigg) = \text{exp}(\theta \mu+ \frac{1}{2} \theta^{2}\sigma^{2}) \cdot \text{E}\bigg(f(X + \theta \sigma^{2})\bigg)$$

My approach is:

Assume the two functions multiplied are independent, hence can conveniently decompose expectation.
$$\color{blue}{\text{E}\bigg(\text{exp}(\theta X) \cdot f(X)\bigg) = \text{E}\bigg(\text{exp}(\theta X)\bigg) \cdot \text{E}\bigg(f(X)\bigg)}$$

I basically have an expectation of lognormal random variable in the first factor; which is a standard result.
$$\color{blue}{= \text{exp}(\theta \mu + \frac{1}{2} \theta\sigma^{2}) \cdot \text{E}\bigg(f(X)\bigg)}$$

This does not look like the expected answer ☹.

答:
First write
\begin{align*}
    \mathbb{E} \left [ e^{\theta X} f(X) \right ] &= \frac{1}{\sqrt{2 \pi \sigma^2}} \int_{-\infty}^\infty e^{\theta t} e^{-\frac{(t - \mu)^2}{2\sigma^2}} f(t) dt
\end{align*}

then do a change of variable $t = y + \theta \sigma^2$ to get

\begin{align*}
\mathbb{E} \left [ e^{\theta X} f(X) \right ]  &= \frac{1}{\sqrt{2 \pi \sigma^2}} \int_{-\infty}^\infty e^{\theta (y + \theta \sigma^2 )} e^{-\frac{(y + \theta \sigma^2 - \mu)^2}{2 \sigma^2}} f(y + \theta \sigma^2 ) dy
\end{align*}

Simplifying this gets you

\begin{align*}
    e^{\theta \mu  + \frac{1}{2} \theta^2 \sigma^2} \frac{1}{\sqrt{2 \pi \sigma^2 }}\int_{-\infty}^\infty f(y + \theta \sigma^2) e^{- \frac{(y - \mu)^2}{2 \sigma^2}} dy  =  e^{\theta \mu  + \frac{1}{2} \theta^2 \sigma^2} \mathbb{E} [f(X + \theta \sigma^2)]
\end{align*}
which is the desired answer.


问:
Help me factorise
Please help me factorize $9x^2-4a^2+4ay +y^2$. I took out "-" as a common factor from $(- 4a^2+4ay +y^2)$ but I cannot factorize it further using difference of 2 squares.

答:
Well, here goes nothing....

Let $(ux + vy + wa)(\alpha x+ \beta y + \gamma a) = 9x^2 -4a^2 +4ay + y^2$ so

So $u\alpha x^2 + (v\alpha + u\beta)xy + (u\gamma + w\alpha)xa +$

$v\beta y^2 + (w\beta + v\gamma)ay + w\gamma a^2=9x^2 -4a^2 +4ay + y^2$

So $u\alpha =9; v\alpha + u\beta =u\gamma + w\alpha = 0; v\beta =1; w\beta +v\gamma =4; w\gamma = -4$.

So $\beta = \frac 1v$ and

$u\alpha =9; v\alpha + \frac uv =u\gamma + w\alpha = 0;  \frac wv +v\gamma =4; w\gamma = -4$

Let $\gamma = -\frac 4w$ and

$u\alpha =9; v\alpha + \frac uv = w\alpha-\frac {4u}w = 0;  \frac wv -\frac {4v}w =4;$

Let $\alpha = \frac 9u$ and

$\color{red}{\frac {9v}u + \frac uv =0}; \frac {9w}u-\frac {4u}w = 0;  \frac wv -\frac {4v}w =4;$

And so $9w^2=4u^2; \color{red}{9v^2=-u^2}; w^2-4v^2-4wv=(w-2v)^2 =0$

But we have unresolvable error.  $\frac {9v}u$ and $\frac uv$ must be the same sign and can't add up to $0$ unless they are both $0$ which they can't be.

We could continue to solve this using complex numbers (That would require $3v = \pm ui$ and $w=2v$ etc....) but it can not be factored over the reals.

Title: Help me factorise


答:
I think that the derivative gives a good indication.

$$f(x)=\frac{1}{\sqrt {1+x}} +\frac{1}{\sqrt {1+a}} + \sqrt{\frac{ax}{ax+8}}$$
$$f'(x)=4 \sqrt{\frac{a}{x (a x+8)^3}}-\frac{1}{2 (x+1)^{3/2}}$$ Without any constraint, the derivative cancels at four points
$$x_{1,2}=\pm \frac{2 \sqrt{2}}{\sqrt{a}}\qquad x_{3,4}=\frac{2 \left(16-6a\pm \sqrt{2(a-8)^2 (2-a)}\right)}{a^2}$$ $x_{3,4}$ do not exist in the real domain if $a >2$ but, in any case, remains $x_{1}= \frac{2 \sqrt{2}}{\sqrt{a}}$.

On the other side $f''(x_1)$ is positive if $0 <2$ and negative if $a>2$.

Add all of that to @mathcounterexamples's answer.

Title: For the given function $f(x)=\frac{1}{\sqrt {1+x}} +\frac{1}{\sqrt {1+a}} + \sqrt{\frac{ax}{ax+8}}$, prove that $1<f(x)<2$ for positive a and $x\ge 0$

Body:
For the given function $f(x)=\frac{1}{\sqrt {1+x}} +\frac{1}{\sqrt {1+a}} + \sqrt{\frac{ax}{ax+8}}$, prove that $1<f(x)<2$ for +ve $a$ and $x\ge 0$

I tried the most rudimentary method ie. differentiating wrt x and hoped to get at least 2 solutions for $f’(x)$, unfortunately I wasn’t able to find a solution for implying function is monotonous . Now whether it’s sis increasing or decreasing still remains a mystery, given that we don’t know what $a$ is.

What method can be used to solve this?

Body:
I think that the derivative gives a good indication.

$$f(x)=\frac{1}{\sqrt {1+x}} +\frac{1}{\sqrt {1+a}} + \sqrt{\frac{ax}{ax+8}}$$
$$f'(x)=4 \sqrt{\frac{a}{x (a x+8)^3}}-\frac{1}{2 (x+1)^{3/2}}$$ Without any constraint, the derivative cancels at four points
$$x_{1,2}=\pm \frac{2 \sqrt{2}}{\sqrt{a}}\qquad x_{3,4}=\frac{2 \left(16-6a\pm \sqrt{2(a-8)^2 (2-a)}\right)}{a^2}$$ $x_{3,4}$ do not exist in the real domain if $a >2$ but, in any case, remains $x_{1}= \frac{2 \sqrt{2}}{\sqrt{a}}$.

On the other side $f''(x_1)$ is positive if $0 <a <2$ and negative if $a>2$.

Add all of that to @mathcounterexamples's answer.


问:
A upper bound to the distribution of the sum of random variables
Let be $X,Y$ random variables and $\epsilon>0$. Then

$$
\mathbb{P}[X+Y\leq x]\leq \mathbb{P}[X\leq x] +\mathbb{P}[|Y|>\epsilon]+\epsilon.
$$

My idea to prove the inequality is using the next:
\begin{align*}
 \{X+Y\leq x\}&= \{X+Y\leq x, X\leq x\}\cup  \{X+Y\leq x ,X\geq x\}\\
&=\{X+Y\leq x, X\leq x\}\cup  \{Y\leq 0 ,X\geq x\}\\
&\subset \{ X\leq x\}\cup  \{Y\leq 0 \}\\
&\subset \{ X\leq x\}\cup  \{|Y|\leq  \epsilon \}\cup\{|Y|>  \epsilon \}.
\end{align*}
Integrating and using subadditivity, we get
$$
\mathbb{P}[X+Y\leq x]\leq \mathbb{P}[X\leq x] +\mathbb{P}[|Y|>\epsilon]+\mathbb{P}[|Y|\leq \epsilon].
$$
But I think  we need to use something like $\mathbb{E}[|Y|1_{A\cap \{|Y|\leq \epsilon\}}]\leq  \epsilon$ (for any event $A$) to get the desired inequality. Any suggestion woul be appreciated.

答: The stated inequality is false. For a counter-example take $X=\frac  1 4, Y=-\frac 1 4. x=\frac1  5$ and $\epsilon =\frac1  2 $.


问:
Integrating this expression over two spheres
I'd like to analytically perform the following integral:
$$ I=\int\limits_V d^3\textbf x\,d^3\textbf y \frac{1}{(\textbf x-\textbf y)^2}, $$
where the volume $V$ of the integration is such that: $|\textbf x|\in [0,A]$ and $|\textbf y|\in [0,B]$, i.e. spherical for $x$ and $y$ separately.

I tried to get a solution from Mathematica, but the output confuses me a bit:
$$ I = \begin{cases} \frac{\pi^2}{3}\left[ 10A^3B+18AB^3+3\text{i} B^4\pi+6(A^4+2A^2B^2-3B^4)\text{ arctanh}(\frac{B}{A}) \right] & A>B \\ 4A^4\pi^2 & A=B \\ 2\pi^2\left[ AB(A^2+B^2)-(A^2-B^2)^2\text{ arctanh}(\frac{A}{B}) \right] & B>A \end{cases} $$
I'd appreciate any help. Thanks in advance!

I think I've managed to get a better expression out of Mathematica:
$$ I = 2\pi² \left[ AB(A^2+B^2) -(A^2 -B^2)^2 \text{ arctanh}\left(\frac{<}{>}\right)  \right] $$
...where $>$ and $<$ are placeholders for $A$ and $B$, depending on which is larger or smaller.

Still, I'd love to know how to do this analytically!

答:
Let's start with $I=\int\limits_{V_{x}}\int\limits_{V_y} d^3\textbf x\,d^3\textbf y \frac{1}{(\textbf x-\textbf y)^2}$$=\int\limits_{V_{y}}(\int\limits_{V_x} d^3\textbf x\, \frac{1}{(\textbf x-\textbf y)^2})d^3\textbf y$$=\int\limits_{0}^A\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi}r^2dr\sin\theta_1d\theta_1d\phi_1\left(\int\limits_{0}^B\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi}R^2dR\sin\theta_2d\theta_2d\phi_2\frac{1}{R^2+r^2-2rR\cos\theta_2}\right)$
$$I=\int\limits_{0}^A\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi}r^2dr\sin\theta_1d\theta_1d\phi_1F(B,r)$$

For the convenience we define $B>A$ and orient the $z$ axis of the polar system of $X$ coordinates along the vector $\textbf y$. We can do this, because, as soon as we intergate over $R^3$ it is our choice of how to orient the coordinates system in the space. Integrating over $\theta_2$ and $\phi_2$ we get:

$F(B,r)=2\pi\int\limits_{0}^B\int\limits_{0}^{\pi}R^2dR\sin\theta_2d\theta_2\frac{1}{R^2+r^2-2rR\cos\theta_2}$$=2\pi\int\limits_{0}^B\frac{R^2}{2Rr}\log\frac{(R+r)^2}{(R-r)^2}dR$

Evaluating the last integral we have to be careful with the singular point $R=r$ (in fact we evaluate the integral principal value). To show our actions step by step let's regularize the integral by means of small $\epsilon$ near $R=r$, and in the end we will set $\epsilon$ to zero.

$F(B,r)=\pi\int\limits_{0}^B\frac{R}{r}\log\frac{(R+r)^2}{(R-r)^2}dR$$=\frac{2\pi}{r}\int\limits_{0}^BR\log(R+r)dR-\frac{2\pi}{r}\int\limits_{0}^{r-\epsilon}R\log(r-R)dR-\frac{2\pi}{r}\int\limits_{r+\epsilon}^BR\log(R-r)dR$.

Integrating and taking  limit at $\epsilon\to0$ we get $$F(b,r)=\frac{\pi}{r}(B^2-r^2)\log\frac{B+r}{B-r}+2\pi{B}$$ And, integrating over $\theta_1$ and $\phi_1$we finally get the desired integral in the form $$I=4{\pi}^2\int_0^A(B^2-r^2)\log(\frac{B+r}{B-r})rdr+8{\pi}B\int_0^Ar^2dr$$
Integrating by part (to get rid of logarithm) we get$$I=4{\pi}^2(\frac{B^2r^2}{2}-\frac{r^4}{4})\log(\frac{B+r}{B-r})|_0^A-4{\pi}^2\int_0^A(\frac{B^2r^2}{2}-\frac{r^4}{4})(\frac{1}{B+r}+\frac{1}{B-r})rdr+\frac{8{\pi}^2BA^3}{3}=$$ $$={\pi}^2(2B^2A^2-A^4)\log(\frac{B+A}{B-A})-4{\pi}^2B\int_0^A\frac{\frac{B^2}{2}-\frac{r^2}{2}+\frac{B^2}{2}}{B^2-r^2}r^2dr+\frac{8{\pi}^2BA^3}{3}$$ Finally, $$I={\pi}^2(2B^2A^2-A^4)\log(\frac{B+A}{B-A})-{\pi}^2\frac{2BA^3}{3}-{\pi}^2B^3\left(-2A+B\log(\frac{B+A}{B-A})\right)+\frac{8{\pi}^2BA^3}{3}$$ $$I=2{\pi}^2BA(B^2+A^2)-{\pi}^2(B^2-A^2)^2\log\frac{B+A}{B-A};   B>A$$


问:
Show that the symmetric positive semidefinite cone $K = S_+^n$ is a proper cone.
I am reading on Convex Optimization by Stephen Boyd.

I want to show that the symmetric positive semidefinite cone $K = S_+^n$, where $S_+^n$ is a set of symmetric $ n \times n$ matrices, is a proper cone

i am using the definition that a cone $K \subseteq \mathbb{R}^{n}$ is called a proper cone if
K is convex, closed, pointed (contains no line) and nonempty.

Looking through, $S_+^n$ is indeed a convex cone, but I can't seem to prove to myself that it is also indeed a proper cone.

This is what I know so far:

To prove $S_+^n$ is convex, we can prove it by using:

$X \in S_+^n \iff Z^TXZ \geq 0, \forall Z$

$X \geq 0$, $Y \geq 0$

$Z^T(\theta_1 X + (\theta_2 Y)Z$

$ = \theta_1Z^TXZ + \theta_2Z^TYZ$

Hence, since $X$ and $Y$ are affine, they are convex and hence, $K$ is convex.

答:
Pointed (contains no line.)

We prove this by contradiction. If $K=S_+^n$ contains a line. Then there exists $A,B\in S_+^n$ and $A \ne B$ such that for any $\theta\in \mathbb{R}$, $\theta A+(1-\theta)B\in S_+^n$. Now if there exists $x\in \mathbb{R}^n$ s.t. $x^TAx\ne x^TBx$, without loss of generality. we assume $x^TAx > x^TBx$ then we can find $\theta$ s.t. $x^T(\theta A+(1-\theta)B)x<0$.

\begin{align}
x^T(\theta A+(1-\theta)B)x = \theta x^T Ax+(1-\theta) x^TBx< 0
\end{align}
if and only if
$$\theta(x^TAx-x^TBx)<-x^TBx\Leftrightarrow \theta <-\frac{x^TBx}{x^TAx-x^TBx} $$
which implies $\theta A+(1-\theta)B\not \in S_+^n$. So there doesn't exists $x$ s.t. $x^TAx=x^TBx$ which $A=B$ is the case. Contradiction!

$S_+^n$ is closed.

Here is a link from Math stackexchange, proving $S_+^n$ is closed

$S_+^n$ has non-empty interior.

We can use example $I_{n\times n}\in S_+^n$ and $I_{n\times n}\succ 0$


答:
I disagree with the other responses as well as the official answer, because I consider the normal range of the ArcSin function as irrelevant.

Consider the equation $t^2 = 2.$  The fact that the square root function has a normal range of non-negative values only is irrelevant.  The equation $t^2 = 2$ still has two solutions, not one.

Just because the ArcSin function normally has a restricted range does not render the equation ArcSin$(x) = 2$ as gibberish.  Here, $x$ can be reasonably interpreted as representing the $\sin(2)$.

Ultimately, it comes down to interpretation.  Do you interpret the equation ArcSin$(x) = 2$ to mean that $2$ is the result of the ArcSin function at some value $x$ (which would imply that this equation has no solution), or do you interpret the equation as find $x$ such that $\sin(2) = x.$

Note that the $\sin^{(-1)}$ syntax is being used instead of ArcSin.  It is not that unusual, when faced with a non-injective function $f(x)=y$ to interpret $f^{(-1)}(y)$ to represent all of the values of $x$ such that $f(x) = y.$

Very similarly, do you interpret the equation $t^2 = 2$ as find $t$ such that $t^2 = 2$, or do you interpret it as find $t$ such that $t = \sqrt{2}.$


问:
Assumption check on proving a theorem
Let $\mathcal{B}=\{[a,b):a\in \Bbb R, b\in \Bbb Q,a<b\}$

$\mathcal{B}$ is the basis for a topology $\tau_{1}$ on $\Bbb R$.

The space $(\Bbb R,\tau_1)$ is called the
Sorgenfrey line

Prove
If $\tau$ is the Euclidean topology on $\Bbb R$ then $\tau_1 \supsetneq \tau$.

Strategy

If I can show the union of open sets of the
Sorgenfrey line is inside  $\tau$

If I can show the intersections of the
Sorgenfrey line is inside $\tau$

Then I think I have proved it

The empty set and $X$ are obvious.

Are my assumptions correct

If not what do l have to do?

Thanks

答:
$[0,1) \in \mathcal{B}$ so $[0,1) \in \tau_1$, but $[0,1) \notin \tau$ as $0$ is not an interior point of $[0,1)$ in the Euccidean topology (any neighbourhood of $0$ contains some open interval $(-\varepsilon, \varepsilon)$, but such sets cannot be contained in $[0,1)$ as they always contain negative reals.)

Every basic open set $(a,b)$ ($a < b$) of $\tau_2$ is in $\tau_1$, as $$(a,b) = \bigcup \{[c,d)\mid c \in (a,b), d \in \Bbb Q, c < d < b\}$$
which is a union of sets from $\mathcal{B}$ so lies in $\tau_2$.

As all sets in $\tau$ are unions of these intervals, all Euclidean open sets are in $\tau_1$ and you'rre done showing the inclusion.

Title: Assumption check on proving a theorem


问:
Show that $\mathcal{S}=\{\cup_{n\in K}(n,n+1]: K \subset \mathbb{Z}\}$ is a $\sigma$-algebra on $\mathbb{R}$
Clearly $\emptyset \in \mathcal{S}$ by definition. Now, I am struggling to prove that if $E \in \mathcal{S}$ then $\mathbb{R}\setminus E \in \mathcal{S}$. My idea was the following:

$E\in \mathcal{S} \Rightarrow E=\bigcup_{n\in K}(n,n+1]$ for some $K\subset \mathbb{Z}$. This implies that $\mathbb{R}\setminus E=\bigcap_{n\in K}(\mathbb{R}\setminus (n,n+1])$. Not exactly sure where to go from here.

答: It's just in bijection with the powerset of $\Bbb Z$, sending $K$ to $f(K):=\bigcup_{n \in K} [n,n+1) \in \mathcal{S}$. As $f$ preserves unions and intersections and complements, this is an isomorphism of algebras. It's obvious $\mathscr{P}(\Bbb Z)$ is a $\sigma$-algebra, hence so is $\mathcal{S}$. $\mathcal{S}$ is just the powerset of $\Bbb Z$ with all points of $\Bbb Z$ "swollen" to an interval in a disjoint way, you could say.


问:
Defining an unusual subspace of $c_0$
This is going to be a long post, so I'm giving a description first:

I recently came across the following exercise: Let $(X,\mathcal{A},\mu)$ be a measure space. If $(f_n)\subset L^p(\mu)$ and $f\in L^p(\mu)$, $p\geq 1$, such that $\|f_n-f\|_p\leq n^{-c}$ with $c>1/p$, prove that $f_n\to f$ a.e. on $X$.  The solution is not a big deal, but leads to an interesting question: which rates are good enough for convergence in norm to imply convergence a.e.? Recall that with no further assumptions, convergence in norm only implies the existence of a subsequence that converges a.e.

Anyway, the following definition is necessary: Define $\text{gr}^p(\mu)$ as the set $$gr^p(\mu):=\{(a_n)\in c_0|\text{ for all } (f_n)\subset L^p(\mu): \big{(}\forall n\in\mathbb{N}: \|f_n\|_p\leq |a_n|\big{)}\implies f_n\to0\text{ a.e.}\}$$

What I would like is to describe this set. My progress is the following:

1) For any measure space and any $p\geq 1$ it is $(0)\in\text{gr}^p(\mu)$, therefore this set is never empty.

2) If $(a_n)\in\text{gr}^p(\mu)$ and $\lambda\in\mathbb{C}$ then $\lambda\cdot(a_n)\in\text{gr}^p(\mu)$.

Indeed, if $\lambda=0$ it is obvious; otherwise if $(f_n)\subset L^p(\mu)$ with $\|f_n\|_p\leq|\lambda a_n|$ for all $n$ we have that $\displaystyle{\|\frac{1}{\lambda}f_n\|_p\leq|a_n|}$ for all $n$ therefore $\frac{1}{\lambda}f_n\to 0$ a.e. which is true iff $f_n\to 0$ a.e.

3) For any measure space, $\ell^p\subset\text{gr}^p(\mu)$.

Let $(a_n)\in\ell^p$ and $(f_n)\subset L^p(\mu)$ with $\|f_n\|_p\leq|a_n|$ for all $n$. We have $\displaystyle{\int_X|f_n|^pd\mu\leq|a_n|^p}$ for all $n$ and by summing and using the Monotone convergence theorem we have that $\displaystyle{\int_X\sum_{n}|f_n|^pd\mu\leq\|(a_n)\|_{\ell^p}<\infty}$, therefore the series $\sum_n|f_n|^p$ converges a.e. hence $|f_n|^p\to0$ a.e. which implies $f_n\to 0$ a.e.

4) In any measure space, if $(a_n)\in\text{gr}^p(\mu)$ and $(a_{n_k})\subset(a_n)$, we have $(a_{n_k})\in\text{gr}^p(\mu)$.

Let $(a_{n_k})\subset(a_n)\in\text{gr}^p(\mu)$ and $(f_k)\subset L^p(\mu)$ s.t. for all $k$ it is $\|f_k\|_p\leq |a_{n_k}|$; Define $g_n$ as $0$ if $n\not\in\{n_k: k\in\mathbb{N}\}$ and $g_{n_k}=f_k$ for all $k$. Then $\|g_n\|_p\leq |a_n|$ for all $n$, hence $g_n\to 0$ a.e. which of course implies $f_k\to0$ a.e.

5) $\text{gr}^p(\mu)$ is a linear subspace of $c_0$.

We need only to prove that it is closed under addition. Let $(a_n),(b_n)\in\text{gr}^p(\mu)$ and $(f_n)\subset L^p(\mu)$ with $\|f_n\|_p\leq|a_n+b_n|$ for all $n$. We partition $\mathbb{N}$ in $S=\{n: a_n=0\}$ and its complement $\mathbb{N}-S$.

Case 1: $S$ is an infinite set. By 4), $(b_n)_{n\in S}\in\text{gr}^p(\mu)$ and $\|f_n\|_p\leq|b_n|$ for all $n$ in $S$; therefore the subsequence $(f_n)_{n\in S}$ converges a.e. to $0$. Now for $n\not\in S$ we can find $\lambda_n\in\mathbb{C}$ such that $b_n=\lambda_n\cdot a_n$. We have to deal with two sub-cases:

Sub-case 1: There exists $M>0$ s.t. for all $n\in\mathbb{N}-S$ it is $|\lambda_n|\leq M$.

In this sub-case, for $n\not\in S$ we have $\|f_n\|_p\leq |a_n|+|b_n|\leq (1+M)|b_n|$. But $(b_n)_{n\in\mathbb{N}-S}\in\text{gr}^p(\mu)$ by 4), and by 2) we have $((1+M)b_n)_{n\in\mathbb{N}-S}\in\text{gr}^p(\mu)$. Hence $(f_n)_{n\in\mathbb{N}-S}$ converges to $0$ a.e.

Sub-case 2: $|\lambda_n|\to\infty$ as $n\to\infty$ through $\mathbb{N}-S$ (note that if $\mathbb{N}-S$ is finite we are automatically in sub-case 1).

We can find $n_0\in\mathbb{N}$ such that for all $n\geq n_0$ and $n\not\in S$ it is $|\lambda_n|>1$. For those $n$ it is $a_n=\frac{1}{\lambda_n}b_n$ therefore $\|f_n\|_p\leq|1+1/\lambda_n|\cdot|b_n|\leq2|b_n|$; now since $(b_n)_{n\geq n_0, n\in\mathbb{N}-S}\in\text{gr}^p(\mu)$ it is $(f_n)_{n\geq n_0, n\in\mathbb{N}-S}\to0$ a.e. and we are done.

Case 2: S is finite; we can do exactly what we did in the two sub-cases above for $\mathbb{N}-S$ and we are done.

Anyway, my questions to the community are these:

Are these spaces any interesting in your opinion?

What would be a good norm for these spaces? I can't think of anything that is of interest.

In this post I prove that for a series of Dirac point-mass measures the space $\text{gr}^p(\mu)$ is the entire $c_0$ for all $p$ and that for the measure space $(\mathbb{R}^d, \mathcal{L}^d, \lambda_d)$ the space $\text{gr}^p(\mu)$ is only $\ell^p$.

EDIT: An easy argument that shows that $\text{gr}^p(\mu)$ is not closed under the supremum norm, unless $\text{gr}^p(\mu)=c_0$: obviously $c_{00}\subset\text{gr}^p(\mu)$, where $c_{00}$ denotes the subspace of sequences that are eventually $0$. Thus $\text{gr}^p(\mu)$ is dense in $c_0$ with the supremum norm.

答:
Per your profile, this is so you can sleep at night.
Your space is exactly $\ell^p$. Here is a proof:

Suppose that $\|a\|_p<\infty$ and $\|f_n\|_p\le |a_n|$ for each $n$. Then $\|f_n^p\|_1\le |a_n|^p$ for each $n$, so that $\|\sum |f_n|^p\|_1<\infty$. In particular, $\sum_n |f_n|^p$ is finite almost everywhere, so that $|f_n|^p$ converges to 0 almost everywhere and hence $f_n$ converges to 0 almost everywhere.

For the converse, suppose that $\|a\|_p=\infty$. Then we construct a sequence of functions $f_n$ on $[0,1]$ with $\|f_n\|_p\le a_n$ so that $f_n(x)$ does not converge for every $x\in [0,1]$. Let $b_n=\min(|a_n|^p,\frac 12)$ and notice that if $f_n$ is the indictator function of an interval of length $b_n$, then $\|f_n\|_p=b_n^{1/p}\le |a_n|$. By assumption, $\sum b_n=\infty$.

Now we just use the standard construction of a sequence of functions that converges to zero in $L^1$, but not pointwise. Let $t_0=0$ and $t_{n}=(t_{n-1}+b_n)\bmod 1$ for each $n$. Then let $f_n=\mathbf 1_{[t_{n-1},t_n]}$ (where if $t_n<t_{n-1}$, this means $\mathbf 1_{[t_{n-1},1]} + \mathbf 1_{[0,t_n]}$. Now it is known that $\limsup f_n(x)=1$ for all $x$ and $\liminf f_n(x)=0$ for all $x$. (This has been described to me as the "Goodyear blimp": the support of the function continually moves to the right, each one starting where the previous one finished. The blimp passes overhead infinitely often because the $b_n$ are not summable).

Title: Defining an unusual subspace of $c_0$


问:
Asif wants to build tower of cubes
Statement:

Asif has cubes of three colors. He builds a tower from them, placing
each next cube on the previous one. It is forbidden to use more than
$4$ cubes of each color. Asif finishes building the tower as soon as
it contains $4$ cubes of some two colors. How many different towers
can Asif build?

My thoughts:

For me, let's think the color are red(r), green(g), blue(b) and so I need to find the permutations of $rrrrggggbbbb$ which is $ \frac{12!}{(4!)^3}$ but later I understood it can be with 1 cube, 2 cubes, 3 cubes ....12 cubes tower. I can't find out how to encounter it.

Thank you so much for your help in advance.

答:
Please note that you can make a tower of minimum $8$ cubes and maximum $11$ cubes. Lower bound is obvious. Now take the upper bound. This is because the moment you have $4$ cubes of two colors, you stop. So as an example, if I have an arrangement of $(4R, 3B, 3G, B)$, I stop.

So I first choose which two colors, I will have $4$ cubes of. I can have $0, 1, 2, 3$ cubes of the last color.

So number of ways to choose colors $ = 3$

Interesting thing to note is that if I take $11$ cubes with $4$ each of two colors and $3$ of one, it should give me all permissible cubes. Say we have $4 R$, $4 B$, $3 G$. Now if one $G$ is on top and rest two are below one of the $R, B$, those are valid arrangements of $10$ cubes. If all $3$ greens are on top, those are valid arrangements of $8$ cubes.

So the answer is $ = 3 \times \frac{11!}{4!4!3!} = 34650$

If this is not convincing, here is another approach you can take -

Fix $2$ colors $R \ B$ with $4$ cubes each.

Number of towers with $8$ cubes $ = \frac{8!}{4!4!} = 70$

Now for towers with $9$ cubes, the green cube can be in any of the $8$ positions out of $9$ (except the last)

So number of towers with $9$ cubes $ = 8 \times \frac{8!}{4!4!} = 560$

Now for towers with $10$ cubes, the last cube has to be either blue or red.

So number of towers with $10$ cubes $ = 2 \times \frac{9!}{4!3!2!} = 2520$

Similarly, number of towers with $11$ cubes $ = 2 \times \frac{10!}{4!3!3!} = 8400$

Adding all of them, we get $11550$. Now multiply this by $3$ as the color with less than $4$ cubes can be any of the three. That is your answer. Same as earlier.

Title: Asif wants to build tower of cubes


问:
probabilities distribution-find the right answer
I have a company that produces 1000 pounds of plastic. It follows the normal distribution with mean 297.9 and standard deviation is 3.5. One method  has been used 22 times and the average is 299.9. It must be pass the 0.05 significance test. Choose the right answer:

a) we can't decline the $H_0$ with $z=2.6802$

b) we decline the $H_0$ with $z=2.6802$

c) we decline the $H_0$ with $z= 1.3401$

d) we can't decline the $H_0$ with $z= 1.3401$

e) something else

I wasn't able to find those results it asks, and I DON'T think it is something else.

What I have tried:

$s=s^2 /\sqrt n =22/ \sqrt{3.5}$ and I find $s$ then I make a replacement in the normal distribution below, but I can't write it in the proper form.

$m=299.9$, $x=297.9$, and

```<code>x-tn-1:1-a/2 Sx <= m <= x+tn-1:1-a/2 Sx  = 1-a
</code>```

答:
The $22$ independent samples individually follow the law ${\mathcal N}(297.9, 3.5^2)$, hence their average $X$ follows ${\mathcal N}(297.9, \frac{3.5^2}{22})$, hence $Z = \frac{X-297.9}{3.5/\sqrt{22}}$ follows ${\mathcal N}(0,1)$. In our case
\begin{equation}
z=\frac{299.9-297.9}{3.5/\sqrt{22}}=2.6802
\end{equation}
For the normal law, the $0.05$ significance test declines $H_0$ if $|z|>1.96$. It is the case here, so the correct answer is b)

Here is how to compute the value of $z$ in Python

```<code>>>> from math import sqrt
>>> (299.9 - 297.9)/(3.5/sqrt(22))
2.68023757704196
</code>```

Body:
The $22$ independent samples individually follow the law ${\mathcal N}(297.9, 3.5^2)$, hence their average $X$ follows ${\mathcal N}(297.9, \frac{3.5^2}{22})$, hence $Z = \frac{X-297.9}{3.5/\sqrt{22}}$ follows ${\mathcal N}(0,1)$. In our case
\begin{equation}
z=\frac{299.9-297.9}{3.5/\sqrt{22}}=2.6802
\end{equation}
For the normal law, the $0.05$ significance test declines $H_0$ if $|z|>1.96$. It is the case here, so the correct answer is b)

Here is how to compute the value of $z$ in Python

>>> from math import sqrt
>>> (299.9 - 297.9)/(3.5/sqrt(22))
2.68023757704196


问:
How to use "results from partial fractions"?
Let ${a_n}$ be a sequence whose corresponding power series $A(x)=\sum_{i\geq 0}a_ix^i$ satisfies

$$A(x)=\frac{6-x+5x^2}{1-3x^2-2x^3}$$

The denominator can be factored into $(1-2x)(1+x)^2$. Using results from partial fractions, it can be shown that there exists constants $C_1,C_2,C_3$ such that

$$A(x)=\frac{C_1}{1-2x}+\frac{C_2}{1+x}+\frac{C_3}{(1+x)^2}$$

Determine these constants and find $a_5$ using this new expression.

What exactly are "results from partial fractions" and how should I use them?

答:
Hints:

Solve the resulting polynomial equation

$$A(x)=\frac{6-x+5x^2}{(1-2x)(1+x)^2}=\frac{C_1}{1-2x}+\frac{C_2}{1+x}+\frac{C_3}{(1+x)^2}\implies$$

$$5x^2-x+6=C_1(1+x)^2+C_2(1-2x)(1+x)+C_3(1-2x)$$

For example, if you substitute $\,x=-1\,$ in both sides , you get

$$12=3C_3\implies C_3=4$$

Now $\,x=\frac12\,$ :

$$\frac{27}4=\frac94C_1\implies C_1=3\;,\ldots\text{etc.}$$


问:
ring operation and commutivity
Is the following set $R$ with the indicated binary operations are commutative ring with unity?

$R$ is the set of integers with the usual addition and multiplication defined $m \star n=m^{3n}$

I am studying RING THEORY right now and countered with this question. I know there are two operations in ring - usual addition and multiplication. But if such a specific binary operation is mentioned, can you please tell me how it works ?? and please correct me if I have any wrong concept.

答:
This is not a ring because in a ring multiplication is distributive with respect to addition, meaning that:
$(b + c) * a = (b * a) + (c * a)  ,\forall a,b,c \in R$

For example, take $a=3,b=4,c=6$

We have $(4+6)*3=(4+6)^9=10^9\neq4*3+6*3=4^9+6^9=10339840$

Title: ring operation and commutivity


问:
Writing a function as sum of even and odd functions requires continuity?
Is continuity a requirement for a function to be decomposed into the sum of an even and odd component? I thought you can do that with any function.</p>

答:
You can always do it. If $f:\Bbb{R}\to\Bbb{R}$ is any function (doesn't have to be continuous), then
\begin{align}
f_E(x):=\frac{f(x)+f(-x)}{2} \quad \text{and}\quad f_O(x):=\frac{f(x)-f(-x)}{2}
\end{align}
are even and odd respectively, and $f=f_E+f_O$. Furthermore, this decomposition is unique.


问:
Functions with property $1<2f(x)-f(x-1)-f(x+1)<2$
Let $x>0$ and let $f(x)$ be an increasing function on the domain $x>0$. Are there any $f(x)$ with the property  $$1<2f(x)-f(x-1)-f(x+1)<2?$$
Since $f(x-1)+f(x+1)<2f(x)$, $f(x)$ is a concave function. But I don't know how to construct $f(x)$ satisfying above inequality. Do you know how to find such $f(x)$ or are there some references on this kind of problems? Thanks!!

答:
Write the condition in the form
$$
1 < \left[ {f(x) - f(x - 1)} \right] - \left[ {f(x + 1) - f(x)} \right] < 2.
$$
Summing this over the integers from $2$ to $n$ gives
$$
n-1 < f(2) - f(1) + f(n) - f(n + 1) < 2n-2.
$$
This implies that for sufficiently large $n$ (more precisely, $n > f(2) - f(1)+1$),
$$
0 < f(n) - f(n + 1),
$$
i.e., $f$ cannot be an increasing function.


答:
Intuitively, if we set $2f(x)-f(x-1)-f(x+1)=c$ for some constant $c$, our function will look something like $a+bx-c\frac{x^2}2$, and won't be able to be increasing (this is a linear recurrence relation, for which there are many references). Your question is slightly different, but can be solved by essentially the same techniques.

Consider
$$\sum_{i=1}^n (2f(x+i)-f(x+i-1)-f(x+i+1)).$$
On one hand, this at least $n$ and on the other hand it's
$$2\sum_{i=1}^n f(x+i)-\sum_{i=1}^n f(x+i-1)-\sum_{i=1}^n f(x+i+1)=f(x+1)+f(x+n)-f(x)-f(x+n+1)\leq f(x+1)-f(x),$$
since $f$ should be increasing. However, $f(x+1)-f(x)$ can't be arbitrarily large, a contradiction.

(We never used the upper bound of $2$ for the given constraint.)

Title: Functions with property $1<2f(x)-f(x-1)-f(x+1)<2$

Body:
Let $x>0$ and let $f(x)$ be an increasing function on the domain $x>0$. Are there any $f(x)$ with the property  $$1<2f(x)-f(x-1)-f(x+1)<2?$$
Since $f(x-1)+f(x+1)<2f(x)$, $f(x)$ is a concave function. But I don't know how to construct $f(x)$ satisfying above inequality. Do you know how to find such $f(x)$ or are there some references on this kind of problems? Thanks!!


问:
Show that there is only one real root of the equation $1 + a^2 + ax - x^3$
I have to show that for every $a>0$ the equation $1 + a^2 + ax - x^3 = 0$, has exactly one solution. I have made the graph of the function $f(x) = 1 + a^2 + ax - x^3$ on desmos and I can clearly see that. But, how can I prove it explicity?

Thanks.

答: Write $f(x)=x^3-ax-a^2-1.$ Then we have that $f'(x)=3x^2-a.$ Hence the function has its local extrema at $$x=\pm\sqrt {\frac a3}.$$ It follows that there is exactly one real root if the relative extrema have equal sign. Hence we compute $$f\left(\sqrt {\frac a3}\right)f\left(-\sqrt {\frac a3}\right)$$ and confirm that it is positive. This gives $$\left[a^2+1 +a\sqrt {\frac a3}-\left(\sqrt {\frac a3}\right)^3\right]\left[a^2+1-\left(a\sqrt {\frac a3}-\left(\sqrt {\frac a3}\right)^3\right)\right]=(a^2+1)^2-\left[a\sqrt {\frac a3}-\left(\sqrt {\frac a3}\right)^3\right]^2=(a^2+1)^2-\frac a3\left(a-\frac a3\right)^2=a^4-\frac{4}{27}a^3+2a^2+1=\left(a^2+\frac{2a}{27}\right)^2+\left(2-\frac{4}{27^2}\right)a^2+1,$$ which is clearly positive for all $a.$


答:
If $1$ is the unit element, then $m\star 1 = 1\star m$. But $m\star 1 = m^{3\cdot 1}$ and $1\star m= 1^{3\cdot m}.$

It's not clear what e.g. $m^{3\cdot 1}$ means. The 3-fold of $n$ is $3\cdot n =n+n+n$. But $1$ could be the unit matrix in a matrix ring. Then $m^{3\cdot 1}$ makes no sense.

Title: ring operation and commutivity

Body:
Is the following set $R$ with the indicated binary operations are commutative ring with unity?

$R$ is the set of integers with the usual addition and multiplication defined $m \star n=m^{3n}$

I am studying RING THEORY right now and countered with this question. I know there are two operations in ring - usual addition and multiplication. But if such a specific binary operation is mentioned, can you please tell me how it works ?? and please correct me if I have any wrong concept.


问:
Closed form for a series of combinations
Let $M$, $N$, and $n$ be three integer numbers such that $n$ is smaller than both $M$ and $N$ ($n < M$ and $n < N$). I would like to know if $\sum_{i = 0}^{n - 1}{N-1 \choose i} {M \choose n - 1 - i}$ has a closed-form solution.

Here is why I am interested in this sum: Assume we have $M$ white and $N$ black balls in an urn. If only one of the black balls has a dot on it and we randomly select $n$ balls from the urn, I need to calculate the probability that the dotted black ball is among the selected ones.

$\Pr(\text{dotted black ball is selected}) = \frac{\sum_{i = 0}^{n - 1} \text{(dotted selected)} \times \text{($i$ of the non-dotted blacks selected)} \times \text{($n-1-i$ of the white selected)}}{{\text{selecting $n$ out of $M+N$}}} =\frac{\sum_{i = 0}^{n - 1} 1 \times {N-1 \choose i} \times {M \choose n - 1 - i}}{{M + N\choose n}}$

Actually, I need to study the sensitivity of this probability to $M$ when both $M$ and $N$ are large. For example, when $N = 100,000$, I want to know how the probability changes by $M$ if it varies between $20,000$ and $50,000$, given $n = 10,000$. That is why I need to be able to have a closed-form for the numerator to be able to calculate the numerical value of the probability.

答:
This is just a special case of Vandermonde's identity:

$$\sum_{i = 0}^{n - 1}{N-1 \choose i} {M \choose n - 1 - i} = {M+N-1 \choose n-1}$$

and the result follows immediately if you think that once chosen the spotted black ball you remain with ${M+N-1 \choose n-1}$ choices for the other balls.

And note that the probability is equal to:

$$\frac{n}{M+N}$$

because drawing one ball at a time, the probability of finding the spotted black one within attempt $k$ is:

$$P_k=P_{k-1}+(1-P_{k-1})\frac{1}{M+N-k+1}=\frac{k}{M+N}$$

and:

$$P_1=\frac{1}{M+N}$$.


答:
We describe what we do once we have the coefficients $C_1,C_2,C_3$. It all comes from the expansion
$$\frac{1}{1-t}=1+t+t^2+t^3+\cdots.\tag{$1$}$$
Putting $t=2x$, we find that $\frac{C_1}{1-2x}$ has expansion
$$C_1+2C_1 x+4C_1x^2+8C_1x^3+\cdots.$$
Putting $t=-x$, we can in a similar way get the power series expansion of $\frac{C_2}{1+x}$.

For $\frac{C_3}{(1+x)^2}$ we need an additional idea. Look at Equation $(1)$, and differentiate both sides. We get
$$\frac{1}{(1-t)^2}=1+2t+3t^2+4t^3+\cdots.$$
Put $t=-x$, and multiply by $C_3$.

Now that we have the three power series expansion, we can read off the coefficient of $x^n$ in their sum.

Details: Almost all the work in finding the $C_i$ has been done by DonAntonio. Bringing $\frac{C_1}{1-2x}=\frac{C_2}{1+x}+\frac{C_3}{(1+x)^2}$ to the common denominator $(1-2x)(1+x)^2$, we find that the numerator is $C_1(1+x)^2+C_2(1-2x)(1+x)+C_3(1-2x)$. Thus, identically, we must have
$$C_1(1+x)^2+C_2(1-2x)(1+x)+C_3(1-2x)=6-x+5x^2.$$
 There are various ways to find the constants. Put $x=-1$. We get $C_3(3)=12$, so $C_3=4$. Put $x=1/2$. We get $C_1(9/4)=27/4$, so $C_1=3$. Finally, the constant term on the left is $C_1+C_2+C_3$, while on the right it is $6$. Since $C_1+C_3=7$, we get $C_2=-1$. So our original function is equal to
$$\frac{3}{1-2x}-\frac{1}{1+x}+\frac{4}{(1+x)^2}.\tag{$2$}$$

Finally, we compute the coefficient $a_5$ of $x^5$. By the discussion in the main answer, the coefficient of $x^5$ in the expansion of $\frac{1}{1-2x}$ is $2^5$.

The coefficient of $x^5$ in the expansion of $\frac{1}{1+x}$ is $(-1)^5$.

The expansion of $\frac{1}{(1-t)^2}$ was obtained by differentiating $1+t+t^2+\cdots$. so the $t^5$ term is $6t^5$. Putting $t=-x$ we get that the coefficient of $x$ is $6(-1)^5$. Putting things together, and remembering our $C_i$, we get
$$a_5=(3)(2^5)+(-1)(-1)^5+(4)(6)(-1)^5.$$

Title: How to use "results from partial fractions"?

Body:
Let ${a_n}$ be a sequence whose corresponding power series $A(x)=\sum_{i\geq 0}a_ix^i$ satisfies

$$A(x)=\frac{6-x+5x^2}{1-3x^2-2x^3}$$

The denominator can be factored into $(1-2x)(1+x)^2$. Using results from partial fractions, it can be shown that there exists constants $C_1,C_2,C_3$ such that

$$A(x)=\frac{C_1}{1-2x}+\frac{C_2}{1+x}+\frac{C_3}{(1+x)^2}$$

Determine these constants and find $a_5$ using this new expression.

What exactly are "results from partial fractions" and how should I use them?


问:
Help Understanding the paragraph in page 16 of Humphreys Lie Algebras
The following paragraph is in page 16 of "Introduction to Lie Algebras and Representation Theory -  Humphreys"

$K$ is a subalgebra of $\mathfrak gl$($V$). Denote $W$={$w\in V$ : $x.w=\lambda(x)w$, for all $x \in K$}, $i.e.$ the set of common eigenvectors of $K$.
Let $L=K+Fz$ (for any $z\in L\setminus K$) and [use the fact that $F$ is algebraically closed to find and eigenvector $v_0\in W$ of $z$.] Then $v_0$ is obviously a common eigenvector for $L$.

I can't understand the part [ ]. I mean, I don't know how to use the algebraically closedenss of $F$. I guess it's quite easy part but I'm stucked. Thanks for any help in advance.

答: If $F$ is an algebraically closed field and $V$ is a finite-dimensional vector spae, then every linear map $L\colon V\longrightarrow V$ has some eigenvalue, since the eigenvalues are the roots of the characteristic polynomial of $L$. If $F$ is not algebriacally cloed, then $L$ may not have eigenvalues.


问:
Without calculus or integrals, how can you prove that the area under 1/x stays the same, if decomposed into increasingly large segments?
Can someone please elaborate the embolded sentence  below? How can you prove that "the area under the curve stayed the same" without integrals?

I'm going to go back in time to the development of the logarithm. The log was a way of performing multiplication (in some base) by simply using addition. Meaning, every time you add one number, your output is proportional to the last output.

At some point in time shortly afterwards, mathematicians were working on integration techniques. They noticed that the basic methods of finding anti derivatives worked for everything except for 1/x. However it was realized, that if you break down the graph of 1/x, into increasingly large segments (1-2, 2-4, 4-8, 8-16...) That the area under the curve stayed the same. this is actually very easily provable! But now look what they found..... They found that the area under the curve increases by simple addition as you double the interval. (Meaning adding the area interval (1-2) + (2-4) + (4-8), is simply the same area added together 3 times).

This should stick out to you, because this is exactly how we defined a logarithm to work. Addition carrying out multiplication with some base number.

Because this logarithm seemed to be so "natural" (it just came about all by itself), it was called the "natural logarithm". So the relation was known that ln(x) = integral of 1/x, before it was known what that base number was. Euler was actually able to calculate the number, and the proof for that is very interesting as well! (See my paper for the proof).

答:
We can try to use the definition of the integral, not the fundamental theorem of algebra (which tells us that we can calculate integrals by using antiderivatives). Let $A= \{x=x_0 < x_1 < \ldots < x_n=y\}$ be any subdivision of $[x, y]$. Then $2A = \{2x_0 < \ldots < 2x_n \}$ is a subdivision of $[2x, 2y]$. The corresponding lower sums are
$$ S_A = \sum_{k=0}^{n-1} \frac{x_{k+1} - x_k}{x_{k+1}} $$
and
$$ S_{2A} = \sum_{k=0}^{n-1} \frac{2x_{k+1} - 2x_k}{2x_{k+1}} =
\sum_{k=0}^{n-1} \frac{x_{k+1} - x_k}{x_{k+1}} = S_A $$
The same holds for the upper sums. Taking the limit for finer subdivisions, we get that the area under $x \mapsto x^{-1}$ over the intervals $[x,y]$ and $[2x,2y]$ is the same.


答:
The definition of a ring does not require specific symbols to be used but the conventions of familiar examples, e.g. the integers, are usually adopted.  So, addition is almost always denoted by $+$.  Multiplication varies a bit, as it does for integers, $\times$ is sometimes used but also $.$ or just juxtaposition (nothing).

A common case for using an alternative symbol is when you want to use a familiar set but with an unfamiliar operation, especially if the definition of the non-standard operation depends on the standard one.  This is your case.  Your example has the usual addition but not the usual multiplication.  So, a different symbol has been used partly as a reminder that it is not the familiar operation but also to allow the definition of the non-standard operation to refer to the familiar one.

As Eminem says, in this case, it fails to be a ring.  I guess that this is the point of the exercise: for you to check.

You may know some group theory.  The operation is most often written in a multiplication style but sometimes other symbols are used.  When the group is abelian (commutative), an additive style is common.  If you forget the multiplication of a ring then you have an abelian group with its operation written as addition.

Title: ring operation and commutivity


问:
Minimal value of the product $(x+y)(y+z)$

Find the min. value of the product $\left(x + y\right)\left(y + z\right)$ given that $xyz\left(x + y + z\right) = 1$ and $x,y,z > 0$.

I want to solve this problem using geometry of a triangle:

Let $a,b,c$ be the sides of the triangle and we can substitute
$a = x + y\,,\,\, b = y + z\,,\,\, c = z + x$.

$A^{2} = xyz\left(x + y + z\right)$ is just a squared area of a triangle and $\left(x + y\right)\left(y + z\right)$ is a part of another formula of triangle  $$
\mbox{i.e.}\quad A = {1 \over 2}\,ab\sin\left(\theta\right)\quad \mbox{or}\quad
A = {1 \over 2}\left(x + y\right)\left(y + z\right)\sin\left(\theta\right)
$$
I' ve tried the same with other formulas but I don't know how to "extract" inequality.

Before the solution I would appreciate that you give a hint, and then a solution. Any help appreciated.

答:
Let $a=x+y, \, b=y+z ,\, c = z+x, \, p = \frac{a+b+c}{2},$ then
$$(x+y)(y+z) = ab,$$
and
$$1 = \sqrt{xyz(x+y+z)} = \sqrt{p(p-a)(p-b)(p-c)} = S = \frac{1}{2} ab \sin C.$$
But $\sin C \leqslant 1,$ so
$$2 = ab \sin C \leqslant ab.$$
Therefore $ab \geqslant 2.$ Equality hold for $x = 1, \, y = \sqrt 2  -1, \, z = 1.$

P/s. Using the AM-GM inequality, we have
$$(x+y)(y+z) = xz+y(x+y+z) \geqslant 2\sqrt{xyz(x+y+z)} = 2.$$


答:
"Results from partial fractions" presumably refers to the following theorem about partial fraction decompositions.

Theorem: Let $\frac{P(x)}{Q(x)}$ be a rational function such that $\deg P < \deg Q$ and such that $Q(x)$ factors as $\prod (1 - r_i x)^{m_i}$ where the $r_i$ are distinct. Then there exist unique constants $c_{i, j}$ such that

$$\frac{P(x)}{Q(x)} = \sum_{i, j : 1 \le j \le m_i} \frac{c_{i, j}}{(1 - r_i x)^j}.$$

The special case of this result when $Q$ has distinct roots is probably familiar but the case of repeated roots is a little more subtle.

In any case, you don't need to know this; the result has already been used for you.


问:
Find all the angles between $-2\pi$ and $-\pi$ which satisfy the equation $\sec \frac{7}{5}x = -2$.

Find all the angles between $-2\pi$ and $-\pi$ which satisfy the equation $\sec \frac{7}{5}x = -2$.

Can anybody help me with this question? I have never attempted a question where the angle range is in the negatives. I'm aware of the ASTC quadrants.

答:
Given:
$$\sec\Bigl(\frac{7x}{5}\Bigr)= -2$$
$$\cos\Bigl(\frac{7x}{5}\Bigr)= \frac{-1}{2}$$
in the range $[-2\pi,-\pi]$

So, $$\frac{7x}{5}=\cos^{-1}\Bigl(\frac{-1}{2}\Bigr)$$
in the interval $[-2\pi,-\pi]$
$$\cos(-\pi+\frac{-\pi}{3})=\cos\Bigl(\frac{-4\pi}{3}\Bigr)=\cos\Bigl(\frac{4\pi}{3}\Bigr)=\frac{-1}{2}$$
$$\cos^{-1}\Bigl(\frac{-1}{2}\Bigr)=\frac{-4\pi}{3}=\frac{7x}{5}$$
Therefore, $$x=\frac{-20\pi}{21}=-2.99$$

Here, x doesn't have two solutions and you will find that to be obvious by having a look at the graph.


问:
Finding a singular point for the given differential equation
I'm trying to show that $x = 0$ is a regular point for this ODE:

$xy'' + (1-x)y' + \lambda y = 0$

What I have tried:

writing the equation above in these two general forms:

$y'' = p(x)y' + q(x)y = 0$

$x^2y'' xP(x)y' + Q(x)y = 0$

we then get:

$p(x) = 1-\frac{1}{x}$ or $\frac{(1-x)}{x}$

$q(x) = \frac{\lambda}{x} $

Then

$P(x)= x(1-\frac{1}{x})=x-1$

$Q(x) = \frac{\lambda}{x}\cdot x^2 = \lambda x $

How should I understand the process of a regular singular point? do I replace every $x$ value as $0$ and see if the equation is equal to $0~$? or assume as $x \to 0$ they remain finite hence $x = 0$ is there an inductive process behind understanding this?

答:
Definition: Consider the general homogeneous second order linear differential equation $$u''+P(x)u'+Q(x)u=0\tag1$$
where $x \in D \subseteq \mathbb{C}$.

The point $x_0 \in D$ is said to be an ordinary point of the above given differential equation if $P(x)$ and $Q(x)$ are analytic at $x_0$.

If either $P(x)$ or $Q(x)$ fails to be analytic at $x_0$, the point $x_0$ is called a singular point of the given differential equation.

A singular point $x_0$ of the given differential equation is said to be regular singular point if the function $(x-x_0)P(x)$ and $(x-x_0)^2 Q(x)$ are analytic at $x_0$ and irregular otherwise.

${}$

The given differential equation $~xy'' + (1-x)y' + \lambda y = 0~$ can be written as $$y'' + \left(\dfrac 1x-1\right)y' + \lambda \dfrac yx = 0$$which is of the form equation $(1)$ where $~P(x)=1/x-1~$ and $~Q(x)=\lambda/x~.$

Clearly both the $~P(x)~$ and $~Q(x)~$ are not analytic at $~x_0=0~.$ Therefore $~x_0=0~$ is the singular point of the given differential equation.

Now since $$(x-x_0)P(x)=x \left(\dfrac 1x-1\right)=1-x$$ and

$$(x-x_0)^2 Q(x)=x^2 \frac{\lambda}{x}=\lambda x$$

so both are analytic at $x_0=0$, and hence $x_0=0$ is regular singular point.


答:
Your formula is more complicated than it needs to be.  If you have $\ B\ $ balls of which exactly one is marked with a dot, and you choose $\ n\ $ of them at random, then the probability that the dotted ball will be among the $\ n\ $ chosen is $\ \displaystyle\frac{n}{B}\ $. In your case $\ B=M+N\ $, so the probability that the dotted black ball is chosen is $\ \displaystyle\frac{n}{M+N}\ $ for any $\ n=1,2,\dots,\ $$M+N\ $.

This tells you that you must have
\begin{align}
\sum_{i=0}^{n}{N-1\choose i}{M\choose n-i}&=
\frac{n+1}{N+M}{N+M\choose n+1}\\
&={M+N-1\choose n}\ ,
\end{align}
for $\ n=0,1,\dots,\min(M-1,N-1)\ $, and, indeed, this identity is not difficult to prove using generating  functions.

Title: Closed form for a series of combinations

Body:
Let $M$, $N$, and $n$ be three integer numbers such that $n$ is smaller than both $M$ and $N$ ($n < M$ and $n < N$). I would like to know if $\sum_{i = 0}^{n - 1}{N-1 \choose i} {M \choose n - 1 - i}$ has a closed-form solution.

Here is why I am interested in this sum: Assume we have $M$ white and $N$ black balls in an urn. If only one of the black balls has a dot on it and we randomly select $n$ balls from the urn, I need to calculate the probability that the dotted black ball is among the selected ones.

$\Pr(\text{dotted black ball is selected}) = \frac{\sum_{i = 0}^{n - 1} \text{(dotted selected)} \times \text{($i$ of the non-dotted blacks selected)} \times \text{($n-1-i$ of the white selected)}}{{\text{selecting $n$ out of $M+N$}}} =\frac{\sum_{i = 0}^{n - 1} 1 \times {N-1 \choose i} \times {M \choose n - 1 - i}}{{M + N\choose n}}$

Actually, I need to study the sensitivity of this probability to $M$ when both $M$ and $N$ are large. For example, when $N = 100,000$, I want to know how the probability changes by $M$ if it varies between $20,000$ and $50,000$, given $n = 10,000$. That is why I need to be able to have a closed-form for the numerator to be able to calculate the numerical value of the probability.


问:
Definitions of uniform space
Let $X$ be a topological space. The collection $\mathcal{A}=\{U_\lambda(x)\}_{\lambda\in \Lambda, x\in X}$, with $U_\lambda(x)\subseteq X$, is called a uniform structure of $X$ if

If $x\in X$ and $\lambda\in \Lambda$, then $x\in U_\lambda(x)$.

If $\alpha, \beta\in\Lambda$, then there is some $\gamma\in \Lambda$ such that whenever $U_\gamma(x)\subseteq U_\alpha(x)\bigcap U_\beta(x)$ for all $x\in X$.

3)For every $\epsilon\in \Lambda$ there is $\lambda\in\Lambda$ such that whenever $x,y, z\in X$ with $x,y \in U_\lambda(z)$ there follows $x\in U_\epsilon(y)$

Also Let $(X, \mathcal{V})$ be as uniform space $\mathcal{V}=\{V_\alpha\}$ with entourage $\Delta_X\subseteq V_\alpha$.

What is difference between $(X, \mathcal{V})$ and $(X, \mathcal{A})$ in above?

Let uniform space $(X, \mathcal{V})$ with $\mathcal{V}=\{V_\alpha\}_{\alpha\in \Lambda}$ be given. Define $U_\alpha(x)= V_\alpha[x]=\{y: (x, y)\in V_\alpha\}$. I think that $\mathcal{A}= \{U_\alpha(x)\}_{\alpha\in \Lambda, x\in X}$ is satisfies items 1, 2, 3 in above.
What can say about reverse inclusion?
Is it true that if  $\mathcal{A}=\{U_\lambda(x)\}_{\lambda\in \Lambda, x\in X}$ satisfies in items 1, 2, 3 and take $U_\alpha=\bigcup_{x\in X}U_\alpha(x)\times U_\alpha(x)$, then $\mathcal{U}=\{U_\alpha\}_{\alpha\in \Lambda}$ is an an uniformity for $X$?

Would you please help me to know it or introduce a reference to study difference between them.

答:
Having a "uniformity" $(X,\mathcal{A})$ as described, gives rise to a
base for a uniformity $(X,\mathcal{U})$ in "entourage style" (see the Wikipedia page). Just let $\mathcal{B}(A) = \bigcup\{U_\lambda(x) \times U_\lambda(x): x \in X\}$ where $A = A_\lambda= \{U_\lambda(x): x \in X\}$ is in $\mathcal{A}$ (so $\lambda \in \Lambda$), and check that this forms a base for an entourage uniformity.

And this $\mathcal{A}$ is itself a base for a uniformity in "cover style" : just take the collection of all covers (any kind) that are refined by a member of $\mathcal{A}$. And conversely too if $\mathcal{U}$ is a (base for an) entourage uniformity then  all covers $\{U[x]\mid x \in X\}$, where $U$ ranges over $\mathcal{U}$, form a uniformity as described in your first paragraph.

Title: Definitions of uniform space


问:
How can can you write a prime number as a product of prime numbers?
According to the fundamental theorem of arithmetic (unique factorization theorem), you can write every number as the product of some prime numbers, for example $33 = 11 \cdot 3$.

However, how can you do this when you're dealing with a prime number? If you write $29 = 29 \cdot 1$ you use 1 and that isn't a prime number. Should you just write $29 = 29^1$?

答:
A single number, like $31$ or $7$, is in fact a product as far as mathematics is concerned.  It is a product of $1$ integer.

Indeed, you can even have a product of $0$ integers.  This is defined to be $1$, because $1$ is the identity element of multiplication.  (See Qiaochu's comment.)

When we say that an integer has a unique prime factorization, we mean it can be written as a product of some nonnegative integer number of primes.  Thus, "$2 \cdot 2 \cdot 23$", "$31$", "$7$", and "$\quad$" are all valid prime factorizations.


问:
Conditional expectation $\mathbb E(Y \mid X \in B)$ instead of $\mathbb E(Y \,|\, X = x)$ (generalization of Shorack's PFS Notation 7.4.1)
In Shorack's Probability for Statisticians Notation 7.4.1, he notes that the conditional expectation (defined in the measure theoretic way) $\mathbb E(Y\mid X)$ is $g(X)$ for some measurable $g :(\mathbb R, \mathcal B_\mathbb R) \to (\mathbb R, \mathcal B_\mathbb R)$. He then defines $\mathbb E(Y \mid X=x)$ as simply $g(x)$. For conditional probabilities, I'm pretty sure this means that $P(A \mid X=x)$ will be defined as $\mathbb E(1_A \mid X=x)$. I'm not entirely confident that this measure-theoretic definition of conditional probability conditioned on $X=x$ matches with the classical notion of conditional probability, so if someone could shed some light on that too that'd be great.

Here's a picture of the relevant section from the book.

My question is: is there a generalization of this definition to $\mathbb E(Y\mid X\in B)$ for some Borel set $B \in \mathcal B_\mathbb R$?

Looking at this question/answer When do the measure-theoretic and elementary definitions of conditional probability/expectation coincide?, it seems like the generalization would require dividing $P(X\in B)$, which may not be possible since it could be $0$. In that case, why is it that we can have a general definition of $\mathbb E(Y\mid X=x)$ but not of $\mathbb E(Y\mid X\in B)$?

答:
The measure theoretic approach to conditional probability / expectation provides a unifying framework that avoids separate argumentation for "discrete" and "continuous" random variables. One of the first examples/exercises that you will encounter is a verification that the measure theoretic $P(Y|X)$ and $E(Y|X)$ are indeed generalizations of the classical concepts. For example:

Claim: Let $X$ be a discrete random variable taking finitely many values $x_1,\ldots,x_k$, each with positive probability. Then $$E(Y\mid X)= \sum_{i=1}^k \frac{E(Y; X=x_i)}{P(X=x_i)} I(X=x_i).\tag1$$

Formula (1), which follows from the definition of conditional expectation, is an explicit representation of the measure-theoretic $E(Y\mid X)$ as a linear combination of indicators of the events $\{X=x_1\}$, $\ldots$, $\{X=x_k\}$.  Another way to write (1) is:
$E(Y\mid X) = g(X)$ where the function $g$ is defined by
$$g(x)=\begin{cases}\displaystyle\frac{E(Y;X=x_i)}{P(X=x_i)}&\text{if $x=x_i$}\\0&\text{otherwise}\end{cases}.$$ We next remark that if we adopt the notation
$$E(Y\mid X=x):=g(x),$$ then this notation coincides with the elementary concept of conditional expectation given $X=x$ for discrete $X$. You can prove that this correspondence holds in the continuous case too; this is the motivation for Notation 4.1 in the passage that you excerpted.

As another exercise, you can prove a measure-theoretic analog of (1) for $P(A\mid X)$, namely $$P(A\mid X)=\sum_{i=1}^k \displaystyle\frac{P(A; X=x_i)}{P(X=x_i)}I(X=x_i).$$  So yes, the notation $P(A\mid X=x_i)$ matches the classical notion.

In the measure-theoretic development the quantity $E(X\mid A)$, for $A$ an event, retains the same meaning as in the elementary approach: it's still $E(X;A)/P(A)$. There's no need to redefine in a "measure-theoretic way", since the numerator and denominator are well defined and their ratio makes sense as long as $P(A)$ is not zero. This is not a limitation, since typically the only events we condition on that have probability zero are those of the form $\{X=x\}$; and we've already given a rigorous interpretation to the notation $E(Y\mid X=x)$. Either that, or our arguments involve conditional expectations in their original form $E(Y\mid X)$ and we avoid conditioning on events at all.


问:
Uniform boundedness principle (Do we need the Banach space assumption ? )
Im reading the proof for Banach–Steinhaus theorem. I'll explain my question:

The theorem is as following:

Let $ V $ be a Banach space and $ W $ a normed vector space. Assume that for $ \mathcal{F}\subseteq B\left(V,W\right) $, a set of bounded linear transformations $ T: V \to W $, there exists a set $ \varOmega $ of second Baire category, such that:

$ \forall v\in\varOmega:\sup_{T\in\mathcal{F}}||Tv||<\infty $

Then, $ \underset{T\in\mathcal{F}}{\sup}||T||<\infty $.

Where $ ||T|| $ denotes $ \sup_{\underset{||v||\leq1}{v\in V}}||Tv|| $.

Now in the proof we define a sequence of sets $ A_n $ as following:

$ A_{n}=\left\{ v\in V:\sup_{T\in\mathcal{F}}||Tv||\leq n\right\}  $

And we notice that

$ A_{n}=\bigcap_{T\in\mathcal{F}}\left(T^{-1}\left(B_{c}\left(\underline{0},n\right)\right)\right) $

Where $ B_c $ denotes a closed ball.

Since $ T $ is bounded for any $ T\in \mathcal{F} $ we also know that $ T $ is continuous and thus $ A_n $ is an intersection of closed sets, and specifically $ A_n $ is a closed set for any $ n $.

Now, here's what I dont get: According to any source that I found, we know claim that

$ \varOmega\subseteq\bigcup_{n=1}^{\infty}A_{n} $ (and others may take $\varOmega=V $) And using Baire Category Theorem we know that there exists $n_0 $ such that $ \text{Int}\left(A_{n}\right)\neq\emptyset $.

My question is why do we need exactly Baire Category Theorem for? (Also since I think we dont need it I also dont get why would we want $ V $ to be a Banach space).

If we assume that $ \varOmega $ is a set of second Baire category, meaning it is not a set of the first Baire category, and if we agree that a set is a set of first Baire category by definition, if it is a subset of a union of closed sets with empty interior, then assuming for contradiction that $ \text{Int}\left(A_{n}\right)=\emptyset $ for any $ n\in \mathbb{N} $, gives us that, by definition, $ \varOmega$ is a set of first Baire category.

So I'll be happy if someone can clarify why those assumption are needed. Thanks in advance

答:
Because you already assume you have such a $\Omega$ of second category, you don't need BCT in the proof anymore.

But in practice, to apply the theorem, you can maybe show $V=\Omega$ but then you need $V$ to be a Banach space to conclude $\Omega$ is of second category.


问:
$\lim\limits_{x \to \infty} {x}(a^{1/x}-1)$, such that $a > 1$
$\lim\limits_{x \to \infty} {x}(a^{1/x}-1)$, such that $a > 1$

My Approach:

Since $x$ approaches infinity I presume we have to 'get rid' of the $x$ on the outside because the root $-1$ will be more than 0 but I have no idea how to reduce the $x$ so please help me out!

Body:
$$L=\lim_{x \to \infty} x(a^{1/x}-1)$$
let $x=1/t$, then
$$L=\lim_{t\to 0} \frac{a^t-1}{t}=\lim_{t \to 0} \frac{1+t\ln a+(t\ln a)^2/2+...-1}{t} =\ln a.$$


问:
How to prove that a $3 \times 3$ Magic Square must have $5$ in its middle cell?
A Magic Square of order $n$ is an arrangement of $n^2$ numbers, usually distinct integers, in a square, such that the $n$ numbers in all rows, all columns, and both diagonals sum to the same constant.

How to prove that a normal $3\times 3$ magic square must have $5$ in its middle cell?

I have tried taking $a,b,c,d,e,f,g,h,i$ and solving equations to calculate $e$ but there are so many equations that I could not manage to solve them.

答:
Once 15 is established as the magic number the easiest way to see this is to reason as follows. If a row/column/diagonal is 8,6,1, we say that 8 is opposite 1.

If a number smaller than five is in the centre consider the number opposite the remaining smallest number. It will have to be greater than 10.

Similarly a number greater than five will not do.


问:
coefficient $[q^n]\sum_{m\ge1}\frac{q^{\alpha m}}{1-q^{\beta m}}$ where $0<\alpha<\beta$
Problem:
I am looking for a finite-sum expression for the coefficient $c_n=c_n(\alpha,\beta)$, where
$$C(\alpha,\beta;q)=\sum_{m\ge1}\frac{q^{\alpha m}}{1-q^{\beta m}}=\sum_{n\ge1}c_n(\alpha,\beta)q^n,$$
with $\alpha,\beta\in\Bbb Z$, $0<\alpha<\beta$, and $|q|<1$.

Context: I know such $c_n$ exist, because $|q|<1$ ensures that $\frac{q^{\alpha m}}{1-q^{\beta m}}$ is analytic about $q=0$ for $m\ge1$.

I was motivated to attempt this because of the well-known
$$\begin{align}
\vartheta_3^2(q)&=1+4\sum_{m\ge1}\frac{q^m}{1+q^{2m}}\\
&=1+4\left(C(3,4;q)-C(1,4;q)\right),
\end{align}$$
which implies that $r_2(n)=4c_n(3,4)-4c_n(1,4)$ for $n\ge1$. I know that finite-sum expressions for $c_n(3,4)$ and $c_n(1,4)$, along with $c_n(\alpha,\beta)$ exist because the following (more general) result is true:
$$A(a,b;q)=\sum_{m\ge1}\frac{(aq)^m}{1-bq^m}=\sum_{n\ge1}q^n\sum_{d|n}a^db^{n/d-1}.\tag1$$

My Attempts:
Let $i(n,m)=1$ when $n|m$ and $i(n,m)=0$ otherwise, and note that for positive integer $v$,
$$\frac{x^v}{1-x^v}=\sum_{r\ge v}x^ri(v,r).$$
Then let $u=\beta-\alpha$ so that
$$\begin{align}
C(q)&=\sum_{m\ge1}\frac{q^{\alpha m}}{1-q^{\beta m}}\\
&=\sum_{m\ge1}q^{-um}\frac{q^{\beta m}}{1-q^{\beta m}}\\
&=\sum_{m\ge1}q^{-um}\sum_{r\ge\beta}q^{mr}i(\beta,r)\\
&=\sum_{r\ge\beta}i(\beta,r)\sum_{m\ge1}q^{(r-u)m}\\
&=\sum_{t\ge\alpha}i(\beta,t+u)\sum_{m\ge1}q^{tm}\\
&=\sum_{t\ge\alpha}i(\beta,t+u)\sum_{s\ge t}q^{s}i(t,s)\\
&=\sum_{t\ge\alpha}\sum_{s\ge t}q^{s}i(\beta,t+u)i(t,s)\\
&=\sum_{t\ge\alpha}\sum_{j\ge 0}q^{j+t}i(\beta,t+u)i(t,j+t)\\
&=\sum_{j\ge0}\sum_{t\ge \alpha}q^{j+t}i(\beta,t+u)i(t,j+t)\\
&=\sum_{j\ge0}\sum_{l\ge 0}q^{j+l+\alpha}i(\beta,l+\beta)i(l+\alpha,j+l+\alpha).
\end{align}$$
Then write $n=j+l+\alpha$. Since $C(0)=0$ we know that $n\ge1$. Thus
$$\begin{align}
C(q)&=\sum_{n\ge1}\sum_{\,\,\,j,l\ge0\\ j+l=n-\alpha}q^{j+l+\alpha}i(\beta,l+\beta)i(l+\alpha,j+l+\alpha)\\
&=\sum_{n\ge1}\sum_{\,\,\,j,l\ge0\\ j+l=n-\alpha}q^{n}i(\beta,l+\beta)i(l+\alpha,n)\\
&=\sum_{n\ge1}q^n\sum_{\,\,\,j,l\ge0\\ j+l=n-\alpha}i(\beta,l+\beta)i(l+\alpha,n).
\end{align}$$
Since $j,l\ge0$ and $j+l=n-\alpha$, we have $0\le l\le n-\alpha$. Then upon writing $d=l+\alpha$, we have
$$\begin{align}
C(q)&=\sum_{n\ge1}q^n\sum_{l=0}^{n-\alpha}i(\beta,l+\beta)i(l+\alpha,n)\\
&=\sum_{n\ge1}q^n\sum_{d=\alpha}^{n}i(\beta,d+u)i(d,n)\\
&=\sum_{n\ge1}q^n\sum_{d|n,\,d\ge\alpha}i(\beta,d+u).\tag{*}
\end{align}$$
This would imply that $$c_n(\alpha,\beta)=\sum_{d|n,\,d\ge\alpha}i(\beta,d+u).$$

Question:
Is $(*)$ correct? If not, what is the expression for $c_n$ that I'm looking for?

Thanks :)

答:
Yes, that's fine. I think your arrangement is a bit overcomplicated and less intuitive than it needs to be. Here's a more direct approach that fills in a little gap.

For $|q| < 1$, we have

$$\left|\frac{q^{\alpha m}}{1-q^{\beta m}}\right| \leq \frac{1}{1-|q|^{\beta m}} \leq \frac{1}{(1-|q|)^{bm}}.$$

By comparison with a geometric series, it follows from the Weierstrass M-test that indeed $\sum_{m \geq 1} \frac{q^{\alpha m}}{1-q^{\beta m}}$ converges uniformly on compact subsets of $|q|<1$, and your limit is indeed analytic on $|q|<1$. By the Cauchy integral formula and the geometric series,
\begin{align*}
c_n(\alpha, \beta)
&:= [q^n] \sum_{m \geq 1} \frac{q^{\alpha m}}{1-q^{\beta m}} \\
&= \sum_{m \geq 1} [q^n] \frac{q^{\alpha m}}{1-q^{\beta m}} \\
&= \sum_{m \geq 1} [q^n] (q^{\alpha m} + q^{\alpha m + \beta m} + q^{\alpha m + 2\beta m} + \cdots) \\
&= \#\{m \geq 1 : \exists k \geq 0 \text{ s.t. }n=\alpha m + \beta m k\}\\
&= \#\{m \geq 1 : m \mid n \text{ and }\beta \mid n/m - \alpha \geq 0\}\\
&= \#\{d \mid n : \beta \mid d - \alpha \geq 0\}.\\
\end{align*}


问:
$\lim\limits_{n\to\infty}n\big(\sum_{k=1}^n\frac{k^2}{n^3+kn}-\frac{1}{3}\big)$?
calculate $$\lim\limits_{n\to\infty}n\left(\sum\limits_{k=1}^n\dfrac{k^2}{n^3+kn}-\dfrac{1}{3}\right).$$
I got it $$\lim\limits_{n\to\infty}\sum\limits_{k=1}^n\dfrac{k^2}{n^3+kn}=\lim\limits_{n\to\infty}\dfrac{1}{n}\sum\limits_{k=1}^n\dfrac{(\frac{k}{n})^2}{1+\frac{k}{n^2}}.$$
Use Squeeze theorem we have
$$\frac{1}{n+1}\sum\limits_{k=1}^n(\frac{k}{n})^2<\dfrac{1}{n}\sum\limits_{k=1}^n\dfrac{(\frac{k}{n})^2}{1+\frac{k}{n^2}}<\dfrac{1}{n}\sum\limits_{k=1}^n(\frac{k}{n})^2$$
So $$\lim\limits_{n\to\infty}\sum\limits_{k=1}^n\dfrac{k^2}{n^3+kn}=\int_0^1x^2\mathrm{d}x=\frac{1}{3}.$$
Use $$\lim\limits_{n\to\infty}n\left(\int_0^1f(x)\mathrm{d}x-\frac{1}{n}\sum\limits_{k=1}^{n}f\left(\frac{k}{n}\right)\right)=\frac{f(0)-f(1)}{2}.$$
Hence $$\lim\limits_{n\to\infty}n\left(\sum\limits_{k=1}^n\dfrac{k^2}{n^3+kn}-\dfrac{1}{3}\right)=\frac{1}{2}.$$
If our method is correct, is there any other way to solve this problem? Thank you

答:
There is another using generalized haromonic numbers since
$$\sum\limits_{k=1}^n\dfrac{k^2}{n^3+kn}=n^3 \left(H_{n^2+n}-H_{n^2}\right)+\frac{1}{2} \left(-2 n^2+n+1\right)$$
$$S_n=-\dfrac{1}{3}+\sum\limits_{k=1}^n\dfrac{k^2}{n^3+kn}=\frac{1}{6} \left(-6 n^2+3n+1\right)+n^3 \left(H_{n^2+n}-H_{n^2}\right)$$ Using asymptotics
$$S_n=\frac{1}{4 n}-\frac{2}{15 n^2}+\frac{1}{12 n^3}+O\left(\frac{1}{n^4}\right)$$
$$nS_n=\frac{1}{4 }-\frac{2}{15 n}+\frac{1}{12 n^2}+O\left(\frac{1}{n^3}\right)$$ and we do not agree !

Try with $n=10$; the exact value is
$$10 S_{10}=10 \times \frac{140325051799081}{5909102214621606}\sim 0.237473$$ while the approximation gives $\frac{19}{80} \sim 0.237500$.


问:
Finding the covariance matrix of two random gaussian vectors and their characteristic function
Let $(W,Z)$ be the gaussian random variable vector that we want to find its covariance matrice and characteristic function.

We have $(X,Y)$ a guassian random variable vector with a mean of $m=(1,2)$ and a covariance matrix: $$
    \begin{bmatrix}
    4 & 1  \\
    1 & 4\\
    \end{bmatrix}
$$

We have that $Z=2X+Y-2$ and $W=\alpha$X+Y.

How do I calculate the covariance matrix of Z and W?

I calculated the following :

The expected value of W is $\alpha +2$.

The expected value of Z is $2$

and both of their characteristics functions.

How to calculate the characteristic function of $(W,Z)$ given both of their characteristic function and we don't know anything about their independency?

答:
$E(ZW)=E(2X+Y-2) (\alpha X+Y)=2\alpha E(X^{2}+2E(XY)+\alpha (XY)+E(Y^{2})-2\alpha EX-2EY$ You know $EX$ and$EY$. Now $E(XY)=4
$, $EX^{2}=5$ and $EY^{2}=8$. Independence is not required in this calculation. Finally, the covariance of $Z$ and $W$ is $E(ZW)-(EZ)(EW)$.


问:
Find the Cramer-Rao bound for an unbiased estimator of $b^2$
$X$ is a RV with pdf $f(x,b) = \frac{x}{b^2} \exp \{-\frac{x^2}{2b^2} \}$

I've got two different estimates: $\hat{b^2} =\frac{2}{\pi} (\frac{1}{n} \sum_{i=1}^n X_i)^2 $ using MME, and $\hat{b^2} = \frac{1}{n} \sum_{i=1}^n X_i^2$ using MLE. MLE is generally considered unbiased, right? So I'll go with that.

The Cramer-Rao bound is defined as $\mathbb{E}[T(X)-g(b)]^2 \ge \frac{(g'(b))^2}{J(b)}$ where $J(b)$ is the Fisher information and $g(b) = \mathbb{E}T(X) =\int_0^{\infty} T(x) \frac{x}{b^2} \exp \{-\frac{x^2}{2b^2} \} dx$.

Of course, $T(X)$ is the estimate : $T(X)=\hat{b^2} = \frac{1}{n} \sum_{i=1}^n X_i^2$.

I've got $J(b) = \frac{4}{b^2}$ for the Fisher info (God, the calculations are a pain!) and now I'm a little unsure whether I'm proceeding correctly to compute the $g'(b)$.

\begin{align}
g'(b) &=\frac{\partial}{\partial b}\int_0^{\infty} \frac{1}{n} \sum_{i=1}^n x_i^2 \frac{x_i}{b^2} \exp \{-\frac{x_i^2}{2b^2} \} dx\\
&=\frac{1}{n} \sum_{i=1}^n \int_0^{\infty}\Bigg(\frac{-2x^3}{b^3}+\frac{x^5}{b^5} \Bigg) e^{-\frac{x_i^2}{2b^2} } dx\\
&= 4b\\
\end{align}

is the setup correct?

This way my ultimate answer for the CRB would be $\mathbb{E}[T(X)-g(b)]^2 \ge 4b^4$.

Thanks for any help. Also : unless you really love calculus, I suggest you use Wolframalpha ;-)

答:
One can check one's work on such things with a computer algebra system. For this problem, you have a random variable $X$ ~ Rayleigh($b$) with pdf $f(x)$:

(source: tri.org.au)

Then, for random samples of size $n$ drawn on $X$, the CRLB for all unbiased estimators of $b^2$ is:

$$\text{CRLB}=\frac{1}{n *\text{FisherInformation}\left[b ^2,f\right]}$$

where FisherInformation is a mathStatica funkeh monkeh, and $f$ is the pdf. Here we go:

(source: tri.org.au)

Alternatively, since you have correctly derived yourself that the Fisher Information on $b$ (rather than $b^2$) is:
$$J(b) = \frac{4}{b^2}$$

and if $g(b)$ is some differentiable function of $b$ that we are interested in estimating, then, subject to some regularity conditions, the CRLB on $g(b)$ is:

$$\left(\frac{\partial g(b) }{\partial b }\right)^2 \frac{1}{n J(b)} $$

In your case, $g(b)=b^2$ ... so after differentiating, you get the same result $\frac{b^4}{n}$ .

P.S. Your title should be:  CRLB for an unbiased estimator of $b^2$  ...not estimate


问:
Change of Variable finding simple bounds
I have a very simple problem that I've solved before (and it gave me a lot of trouble then) and I somehow can not come up with the right bound again.

In Michael Corral's book he uses a change of variable transformation to make $x = \frac{1}{2}(u+v)$ and $y = \frac{1}{2}(v-u)$.

The original substitutions are:
$x - y = u$

$x + y = v$

The bounds on the $x,y$ plane are $(1,0)$, $(0,1)$ and $x = 1-y$
Setting x to $0$ I can come up with the first bound of $u = -v$
I did so by plugging in $0$ for $x$ and $\frac{1}{2}(v-u)$ for $y$ and set it equal to $u$
i.e. $u = -\frac{1}{2}(v-u)$ It also works simply pluggin in the numbers.

But the second bound for when $x = 1-y$ should be $u = v$
When plugging in I always get $0$ (i.e. the u's go away)
or I get $v = 1$ which is a correct line but I have no bound for u on this?
I'm doing something wrong and I'd be so grateful if you could help me out.
I have a total black out here.
I weirdly am super comfortable with polar and spherical substitution bounds but they don't apply here.

Thank you very much!

Edit: also if someone could recommend a ressource that has lot's of practice for finding these bounds that'd be great!

答:
Your region on $xy-$plane is bounded by a right angled triangle with boundaries $x=0, y=0 $ and $x+y=1$. You just need to know the boundaries of the transformed region in $uv-$plane using the given transformations:

$x-y=u$ and $x+y=v$

$x=0$ maps to $u=-y, v=y$ or equivalently $v=-u$

$y=0$ maps to $u=x, v=x$ or equivalently $v=u$

$x+y=1$ maps to $v=1$

So the new transformed region is a triangle in $uv-$plane bounded by the edges $v=-u, v=u$ and $v=1$.


问:
Minimum of $\left|1-\left(ab+bc+ca\right)\right|+\left|1-abc\right|$

If $a,b,c\in\mathbb{R}$ and $a+b+c=1$, then what is the minimum value
of $\left|1-\left(ab+bc+ca\right)\right|+\left|1-abc\right|$.

I used Wolfram Alpha and it says the minimum value is $\dfrac{44}{27}$ for $\left(a,b,c\right)=\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)$. Obviously $uvw$ method doesn't help. and since we have absolute value, I think The Buffalo doesn't help.

I wrote $ab+bc+ca$ this way: $abc\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$.

We know that
$$\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\le\frac{a+b+c}{3},$$
so
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge9\Rightarrow ab+bc+ca\ge9abc$$

I don't know if it helps (or is even true).

I think finding the minimum of $\left(1-\left(ab+bc+ca\right)\right)^2+\left(1-abc\right)^2$ has the same procedure. If anyone knows how to find it, it helps a lot (I think).

Body:
partial proof when  $a,b,c\ge -1$

WLOG $c=\min(a,b,c)\implies 3c-13<0$

$$|1-abc|+|1-ab-bc-ca|\ge |2-abc-ab-bc-ca|\ge 2-abc-ab-bc-ca$$ It remains to prove $$2-abc-ab-bc-ca\ge 44/27$$
$$\iff 2-ab(1+c)-c(a+b)-44/27\ge 0$$ $$2-\frac{{(a+b)}^2}{4}(1+c)-c(1-c)-44/27\ge 0$$
$$\iff 2-\frac{{(1-c)}^2}{4}(1+c)-c(1-c)-44/27\ge 0$$ $$\iff \frac{-1}{108}(3c-13){(3c-1)}^2\ge 0$$ which is obvious.

Update: River Li has a complete proof


问:
Convexity over the specified domain
I have the following problem:

My attempts:

for i: Hessian matrix I found for 1 is given below.

$$
    \begin{bmatrix}
    \frac {x_2^2}{4(x_1x_2)^\frac 32}+4& \frac {8(x1x2)^{\frac 32}-x1x2}{4( {x1x2)^{\frac 32}}}& 0 \\
    \frac {8(x1x2)^{\frac 32}-x1x2}{4( {x1x2)^{\frac 32}}} & \frac {x_1^2}{4(x_1x_2)^\frac 32}+4 & -2 \\
    0 & -2 & 6 \\
    \end{bmatrix}
$$
for iii: I could not figure out how to approach to this as I know ln function is convex.

for v: Hessian matrix I found for second part of 5 is given below.
$$
    \begin{bmatrix}
    12x_1^2+12x_1+4x_2^2+4x_2+10 & 8x_1x_2+4x_1+4x_2+2 \\
    8x_1x_2+4x_1+4x_2+2 & 12x_2^2+12x_2+4x_1^2+4x_1+10 \\
    \end{bmatrix}
$$

for vi: Hessian matrix I found for 6 is given a

$$
    \begin{bmatrix}
    12x_1^2+\frac {13}3x_2^2 & \frac {26}3x_1x_2 \\
    \frac {26}3x_1x_2 & 12x_2^2+\frac {13}3x_1^2 \\
    \end{bmatrix}
$$

答: I found it. I showed each Hessian is positive semidefinite on the mentioned domain, no matter which values $x_1, x_2$ and $x_3 $ take. I did that by finding determinants of sub square matrices and the Hessians itself. That way, I was able to prove that.

Title: Convexity over the specified domain