问:
$xy'' + y' = 0$, Why is this in Euler-Cauchy form?
I am having difficulty figuring out why $xy''(x) + y'(x) = 0$ is in Euler-Cauchy form. Can anyone explain this to me?  Thank you!

答:
The differential equation $$a_nx^ny^{(n)} + a_{n−1}x^{n−1}y^{(n−1)} +\cdots + a_0y = 0\tag1$$
is called the Cauchy-Euler differential equation of order $n$. The symbols $~a_i,~~ i = 0, \cdots , n~$ are constants and $~a_n\ne 0~.$

Now let us come to your differential equation,$$xy''(x) + y'(x) = 0\tag2$$Multiplying both side by $~x~,$ we have $$x^2y''(x) + xy'(x) = 0\tag3$$ Clearly here $~n=2~$ and $~a_2=1\left(\ne0\right),~a_1=1,~a_0=0~$ are constants.
Hence the differential equation $(3)$ is same as that of equation $(1)$ and it comes from of equation $(2)$.
Therefore the given differential equation $(2)$, is Cauchy-Euler differential equation of order $2$.


问:
How do I solve $\lim_{n \to \infty} \frac{n!}{2^{n+1}}$
I am trying to solve the limit $\lim_{n \to \infty} \frac{n!}{2^{n + 1}}$.

Since both $n!$ and $2^{n + 1}$ approach $\infty$ as $n \to \infty$, my thinking was that I had to take the derivative of both the numerator and the denominator. However, from my research, the derivative of $n!$ is very complicated, involving some gamma function that we haven't covered in the course yet.

Is there an easier, more intuitive approach to solving this problem without using L'Hopital's Rule?

答: Hint: use the Stirling's approximation, which says that $n! \approx \sqrt{2\pi n}\left(\frac{n}{e}\right)^n$

ContentLicense: CC BY-SA 4.0


Title: How do I solve $\lim_{n \to \infty} \frac{n!}{2^{n+1}}$

Body:
I am trying to solve the limit $\lim_{n \to \infty} \frac{n!}{2^{n + 1}}$.

Since both $n!$ and $2^{n + 1}$ approach $\infty$ as $n \to \infty$, my thinking was that I had to take the derivative of both the numerator and the denominator. However, from my research, the derivative of $n!$ is very complicated, involving some gamma function that we haven't covered in the course yet.

Is there an easier, more intuitive approach to solving this problem without using L'Hopital's Rule?

Body: $n! >(1)(2)(3^{n-2})$. Can you finish?


问:
Multiplying Binomial Terms
$(x+2)(x-1$)

Totally forgot how to do it.
Isn't this the rule:

$(a+b)(c+d)$

$a*c +a*d +b*c +b*d$

The answer I'm getting is $x^2-x+2x-2$ which is wrong.
Please help me out, thanks!

答:
It's not wrong. You just haven't noticed that $-x+2x = x$, so that, if you simplify, you get
$$(x+2)(x-1) = x^2 -x + 2x - 2 = x^2 + x - 2.$$

ContentLicense: CC BY-SA 3.0

Title: Multiplying Binomial Terms


问:
Cardinality of Center of p Groups $ |Z(G)| \neq p^{n-1}$
Question

Let G be a p group, such that $ |G|=p^n$. Prove that $ |Z(G)|\neq p^{n-1}$.

Beginning of Proof

So, I began by writing out $ p^n=|G|=\sum_{i\in I}[G:Stab_G(x_i)]+|Z(G)|=\sum_{i\in I}[G:Stab_G(x_i)]+p^{n-1}$, when $ \left\{x_i\right\}_{i\in I}$ are representatives in the following way $ \cup_{i\in I}x_iGx_i^{-1}\cup Z(G)=\cup_{x\in G}xGx^{-1}$, when $ \forall i\in I, |x_iGx_i^{-1}|\geq p$ (because otherwise $x_i \in Z(G)$). Therefore, $ \sum_{i\in I}[G:Stab_G(x_i)]=p^n(p-1)$.

Then, we can say that $ \forall i\in I, \exists j_i\in [1,n-1]: [G:Stab_G(x_i)]=p^{j_i}$, according to Lagrange's theorem. Since p-1 is even (if $ p\neq 2$), then in order for this sum of odd numbers to be even, then |I| must be even. Therefore, we can write that $|I|=2k$, when $ 2k \in [2,\frac{p^n(p-1)}{2}]$. $2k\neq p^n(p-1)$, because then, we'll have $p^n(p-1)$ numbers in the sum, and therefore every number in the sum must be 1, such that $\forall i\in I, |Stab_G(x_i)|=|G| \Rightarrow x_i\in Z(G)$. Therefore, $k\in [1,\frac{p^n(p-1)}{4}]$, and therefore, $p-1\equiv 0 (mod 4) \Rightarrow p \equiv 1 (mod 4)$. Therefore, $ |Z(G)|\neq p^{n-1}$, for $p: p \equiv 3(mod 4)$.

I haven't been able to prove this for all other primes p. Thanks a lot for the help!!!

答: You need know some basic facts. If $G$ is abelian so $G=Z(G)$. But $|G|\neq|Z(G)|$. So assume that $G$ is non-abelian. Since $Z(G)$ is a normal subgroup so $G/Z(G)$ is defined and $[G:Z(G)]=p$. In this case using another answer hint you get $G$ is abelian which is a contradiction.</p>


问:
Inradius given radius of three circles tangent to incircle

The incircle $T$ of the scalene $\triangle ABC$ touches $BC$ at $D$, $CA$ at $E$ and $AB$ at $F$. lf $R_1$ be the radius of the circle inside $\triangle ABC$ which is tangent to $T$ and the sides $AB$ and $AC$. Define $R_2$ and $R_3$ similarly. If $R_1=16$, $R_2=25$, $R_3=36$, determine radius of $T$.

My Effort
I tried drawing diagram but felt completely clue-less.

答:
It's quite easy to show that if $R$ is the radius of $T$,
$$\sin\frac A2=\frac{R-16}{R+16}$$
$$\sin\frac B2=\frac{R-25}{R+25}$$
$$\sin\frac C2=\frac{R-36}{R+36}$$
We may now use the following identity since $A+B+C=\pi$:
$$\sin^2\frac A2+\sin^2\frac B2+\sin^2\frac C2+2\sin\frac A2\sin\frac B2\sin\frac C2=1$$
Substituting the expressions for $\sin\frac X2$ in terms of $R$ and solving the resulting equation yields the only positive solution as $R=74$.

More generally, if $R_1=a^2$, $R_2=b^2$ and $R_3=c^2$ then $R=ab+bc+ca$.


答:
This sequence diverges and there's no need for any fancy techniques.

I will prove a more general statement: Let $(x_n)_{n\in\mathbb{N}}$ be a sequence with $x_n=\frac{a^{n+1}}{n!}$ with $a>0,\ a\in\mathbb{R}$.

Now let us take the natural number $N\geq2|a|$. For any $n\geq N$ the following is true:
\begin{equation*}
\frac{a^{n+2}}{(n+1)!}=\frac{a^{n+1}}{n!}\cdot \frac{a}{n+1}\leq \frac{1}{2}\frac{a^{n+1}}{n!}
\end{equation*}
Throug induction we can prove that $\frac{a^{n+1}}{n!}\leq 2^{-(n-N)}\frac{a^{N+1}}{N!}$. The right side converges to $0$. Therefore $x_n\to 0$. This yields that your sequence diverges.


问:
How to find the inverse function of $f(x)=\sin^2\left(\frac{2x+1}{3}\right)$ and find its domain.
$f(x)=\sin^2\left(\frac{2x+1}{3}\right)$ which is restricted on $-\frac{3\pi+1}{2}\le x< -\frac{3\pi+2}{4}$

I know I have to switch the $f(x)$ and the $y$:

$x=\sin^2\left(\frac{2f^{-1}(x)+1}{3}\right) \to \arcsin^2(x)=\frac{2f^{-1}(x)+1}{3} \to f^{-1}=\frac{3\arcsin^2(x)-1}{2}$

To find the domain:

$$-\frac{3\pi+1}{2}\le \frac{2x+1}{3}< -\frac{3\pi+2}{4}$$

$$-\frac{9\pi+3}{2}\le 2x+1< -\frac{9\pi+6}{4}$$

$$-\frac{9\pi+5}{2}\le 2x< -\frac{9\pi+10}{4}$$

$$-\frac{9\pi+5}{4}\le x< -\frac{9\pi+10}{8}$$

The inverse function is $f^{-1}=\frac{3\arcsin^2(x)-1}{2}$ and its domain is $-\frac{9\pi+5}{4}\le x< -\frac{9\pi+10}{8}$

But I am slightly confused since Desmos will not map it correctly. I'm asking if this is the correct solution or I went wrong somewhere.

Thanks

答:
As IV_ said, your inverse function should be $y=\pm\frac{3}{2}\arcsin(\sqrt{x})-\frac{1}{2}$.

Your original function $ f(x)=\sin^2\left(\frac{2x+1}{3}\right)$ is equal to a sine function squared. If you graph f(x), you can see that its biggest value is $1$, and it achieves its smallest value at $x=(-3\pi+1)/2$ because the domain is restricted. Calculating the smallest $y$-value, we get $f((-3\pi+1)/2)$, which is approximately $0.382$ according to the graph. Thus, the range of $f(x)$ is $[0.382,1]$.

So, your function maps $x$-values in its domain to the range $[0.382,1]$. An inverse function reverses this process, so it maps $[0.382,1]$ back to the domain of the original function. That means that the domain of your inverse function is $[0.382,1]$.

Notice how we didn't have to find the domain of the original function to find the domain of the inverse?


问:
What is the proper way of notating all elements in an equation are real?
Assuming c = a + ib and c* = a - ib where a and b are real

I want to show that c + c* is real

To do this I showed

(c + c*) = a + ib + a - ij = 2a

but I don't know how to state 2a is real. I'm thinking there is a way to state: "each element in the equation 2a belong to the set of real numbers, therefore c+c* is real" is what I have to say, I just don't know how to put that into proper math notation.

答:
You're splitting each number into real and imaginary parts:

$c=a+ib$

By definition, $a$ and $b$ are real. Hence, the first term on the right side is real, and the second term is imaginary.

So when you get

$c + c^* = a + ib + a - ib = 2a,$

you're done. The value $2a$ is real (because the reals are closed under multiplication).


问:
A variant of diagonal intersection
I am doing an exercise about a variant of diagonal intersection：$\bigtriangleup_{s\in {\left[\kappa\right]}^{<\omega}}X_s=\{\alpha\in\kappa:\alpha\in\bigcap_{s\in {\left[\alpha\right]}^{<\omega}}X_s\}$.

Here $\kappa$ is a cardinal, $X_s\in U$, where $U$ is a normal filter (We can assume $U$ is $\kappa$-complete or ultra if needed).

The problem is showing $\bigtriangleup_{s\in {\left[\kappa\right]}^{<\omega}}X_s\in U$. I tried to construct a useful bijection from ${\left[\kappa\right]}^{<\omega}$ to $\kappa$ but failed, can someone give me a hint? Thanks in advance.

答:
Observe that your diagonal intersection is equal to
$$\bigcap_{n<\omega}\left\{ \alpha<\kappa\> \middle| \> \alpha\in \bigcap_{s\in [\alpha]^n}X_s\right\}.$$

Now I claim that $U$ is closed under the diagonal intersection over $[\kappa]^n$, that is, if $\{Y_s\mid s\in[\kappa]^n\}\subseteq U$ is a family of sets, then $\triangle_{s\in[\kappa]^n}Y_s = \{\alpha<\kappa\mid \alpha\in\bigcap_{s\in[\alpha]^n}Y_s\}\in U$.

The case $n=1$ is obvious (it is just a usual diagonal intersection.) Assume that our claim holds for $n$. Let $\{Y_s\mid s\in[\kappa]^{n+1}\}\subseteq U$ be a given family of sets.
We can see that
$$\triangle_{s\in [\kappa]^{n+1}}Y_s= \{\alpha<\kappa\mid \alpha\in \bigcap_{t\in[\alpha]^n}\bigcap_{\max t<\beta<\alpha} Y_{t\cup\{\beta\}}\},$$

which is equal to $\triangle_{t\in[\kappa]^n}\{\alpha<\kappa\mid \alpha\in \bigcap_{\max t<\beta<\alpha} Y_{t\cup\{\beta\}}\}$. You can see that each
$$\left\{\alpha<\kappa\>\middle|\> \alpha\in \bigcap_{\max t<\beta<\alpha} Y_{t\cup\{\beta\}}\right\}$$
is a member of $U$, since it is a diagonal intersection of sets of $U$.

Title: A variant of diagonal intersection


问:
Can all naturals can be expressed as sums of prime powers of $\{3,5,7,13\}$?
Define the set $$S=\left\{(p,k) \mid p\in\{3,5,7,13\} \land k \in\mathbb Z^{0+}\right\}.$$

The question is whether every natural number $n$ can be expressed as the sum of some subset $s\subseteq S$ of distinct pairs $(p,k)$ such that $$n=\sum_{(p,k)\in s}{p^k}.$$ To be clear, any pair in $S$ may be used at most once in the sum, though $k=0$ is allowed as an exponent. As an example, when using only the primes $\{3,5,7\}$, $24$ has two valid representations:

$$ 3^1+3^2+5^1+7^1 = 3^0+3^2+5^0+5^1+7^0+7^1 = 24. $$

Another way to describe the problem is to ask whether every natural can be described as the sum of four numbers, one each in bases 3, 5, 7, and 13, using only $0$ and $1$ digits. So equivalently, in that formulation, $$110_3+10_5+10_7=101_3+11_5+11_7=24.$$

Using $\{3,5,7,13\}$ gives 20 representations for $24$, which seemed a little spammy to include.

This post is a direct follow-up to this one, which posed the same question for $p\in\{3,5,7\}$. A nice proof was given that that set isn't sufficient to cover the naturals; the bottom line is that this arises because $$\frac{1}{3-1}+\frac{1}{5-1}+\frac{1}{7-1}=\frac{11}{12}<1.$$

This leaves open the interesting edge case of whether it would be possible for $\{3,5,7,13\}$, as $$\frac{1}{3-1}+\frac{1}{5-1}+\frac{1}{7-1}+\frac{1}{13-1}=1.$$

Note: To explain comments below, I originally thought I had a counterexample, but that was a rounding error. I'm leaving this question up as it seems like it still has merit. Clever arguments as to why $\{3,5,7,13\}$ should or should not work are also welcome.

Update

It actually seems like you may be able to limit this to proper prime powers after all, i.e. disallowing $p^0$ terms, for all $n\geq 113$. The only numbers I can find at first glance that require them are $\{1, 2, 4, 6, 11, 26, 31, 107, 112\}$ (though this is not proven!) So if you confine yourself to $n \geq 113$, each term $p^k$ in the sum must be distinct, no caveats required.

This seems like a cleaner formulation of the problem, but it's unclear to me which version would is more likely to be tractable or of greater interest. Input on this welcome.

答:
Let $n_1, n_2, n_3, \dots$ be a nondecreasing sequence of integers $\geq 1$.
Let $s_0 = 0, s_k = s_{k-1} + n_k$ for $k \geq 1$.

Then every integer is the sum of a finite subsequence of the $n_k$ if and only if $n_k \geq s_{k-1} + 1$. Proof: if the condition is met then by induction over $k$ all $n \leq s_k$ are the sum of a subsequence of the first $k$ numbers. Otherwise if $k$ is the smallest $k$ such that the condition is not met, then $n_k - 1$ is not the sum of any subsequence.

If we sort the prime powers in non-descending order then $\frac{1}{p_1-1} + \dots + \frac{1}{p_4-1} \geq 1$ guarantees the condition above. If  the sum is less than $1$ then the condition above will fail if there are $4$ prime powers very closely together. A counter example can then be found by sorting the prime powers and calculating the sequence $n_k$ and $s_k$.

Proof: Assume we calculated up to $n_k$ and $s_k$, and the next $4$ powers are $a, b, c$ and $d$. Then $s_k = \frac{a-1}{2} + \frac{b-1}{4} + \frac{c-1}{6} + \frac{d-1}{12}$. Let $x$ be the smallest of $a, b, c, d$ then $s_k \geq \frac{x-1}{2} + \frac{x-1}{4} + \frac{x-1}{6} + \frac{x-1}{12} = x-1$, and $n_{k+1} = x \geq s_k +1$ as required.

There is no need for any or all of the numbers to be primes at all, as long as they are integers.


问:
Applications of Double/Triple Integrals
This is the question that I need to solve using mathematica:

The concentration of an air pollutant at a point $(x,y,z)$ is given by: $$p(x,y,z) = x^2y^4z^3 \text{ particles}/m^3$$ We're interested in studying the air quality in a region in 3-space which satisfies $x\ge0$ and is bounded by the $xy$-plane, the surface $z=\sqrt{4-x^2-y^2}$, the plane $y=-2x$, and the $xz$-plane.  Find the total amount of pollutant in this region.

I don't want a full answer to this question, obviously (I need to work it out myself), but I'm just having trouble getting started.  How do I use the bounds here to set up an integral for the problem?  What exactly do the bounds mean/represent in the first place?

Title: Applications of Double/Triple Integrals

Body: $\int_{y=-4/\sqrt 5}^{y=0} dy  \int_{x=-y/2}^{x=\sqrt{4-y^2}} dx \int_{z=0}^{z=\sqrt{4-x^2-y^2}}x^2 y^4 z^3 dz$


问:
Prove that $f(x)=0$ for all $x∈\Bbb R$ if all rationals are mapped to $0$ using only topology
Prove that if all rationals are mapped to $0$ and f is continuous then f(x) = 0 for all x in R.

I have done this proof before using epsilon delta definition. I was wondering if this proof can be done using only topology and open and closed sets.

答:
Note that the topology on $\Bbb R$ is first countable.

Recall the limit point definition of continuity: $f$ is continuous at $a$ iff $\lim_{x\to a}f(x)=f(a)$.

Let $x\in\Bbb R$.  By density of $\Bbb Q$, there is a sequence $(x_n)$ of rationals with $x_n\to x$.

Then by continuity, $f(x)=\lim_{n\to\infty}f(x_n)=\lim_{n\to\infty}0=0$.


问:
Prove function is continuous using only topology
Let(X,τ) be the subspace of R given by X= [0,1]∪[2,4].
Define Define f: (X,τ)→R by f(x) =1,if x∈[0,1] and 2,ifx∈[2,4]

So I am not allowed to use the epsilon delta definition. I can show that all preimage of open sets are open or preimage of all closed sets are closed which seems easier. So for any closed set in R not containing 1 or 2, the preimage is the empty set which is closed. If any closed set contains 1 or 2, then the preimage are the closed intervals given above which are closed. Same if we take the preimage of the entire real line which is also closed, thus the mapping is continuous.

答:
The argument is correct: for (closed) subsets $A$ of $\Bbb R$ there are $4$ options:

$1 \in A$, $2 \notin A$, then $f^{-1}[A] = [0,1]$.

$2 \in A$, $1 \notin A$, then $f^{-1}[A] = [2,4]$.

$1 \in A$, $2 \in A$, then $f^{-1}[A] = X$.

$1 \notin A$, $2 \notin A$, then $f^{-1}[A] = \emptyset$.

All of those optiosn give open and closed pre-images ($[0,1]$ is open in $X$ as it $X\setminus [0,1] = [2,4]$ is closed, e.g. or because it equals $(-1,3) \cap Y$ etc.)

So in this case going via open or closed pre-images comes down to the same thing.


问:
How to resolve this algebra equation?
$$f = X^3 - 12X + 8$$ $a $-  complex number, $a$ is a root for $f$

$b = a^2/2 - 4 $.

Show that $f(b) = 0$

This is one of my theme exercises ... Some explanations will be appreciated ! Thank you all for your time .

Title: How to resolve this algebra equation?

Body:
\begin{align*}
f(b) &= f(\frac{a^2}{2} - 4) \\
&= (\frac{a^2}{2} - 4)^3 - 12(\frac{a^2}{2} - 4) + 8  \\
&= \frac{a^6}{8} - 3a^4 + 24 a^2 - 64 - 6a^2 + 48 + 8 \\
&= \frac{a^6}{8} -3a^4 + 18a^2 -8
\end{align*}

On the other hand,
$$
f(a) = a^3 - 12a + 8 = 0
$$
and therefore
\begin{align*}
0 = 0^2 &= (a^3 - 12a + 8)^2 \\
&= a^6 - 24a^4 + 16a^3 + 144 a^2 - 192a + 64 \\
&= (a^6 - 24a^4 + 144 a^2 - 64) + (16a^3 - 192a + 128) \\
&= 8(\frac{a^6}{8} -3a^4 + 18a^2 -8) + 16(a^3 - 12a + 8)\\
&= 8f(b) + 16f(a) \\
&= 8f(b)
\end{align*}
Hence,
$$
f(b)=0.
$$


问:
Can I modify the vector mutiplied by a matrix to have specific resultant vector?
Given a matrix $W$ and a vector $d$ where $W*d = g$.  the dimension of $W$ is $n$x$n$ and of $d$ and $g$ is $n$x$1$.

Assume the $W$ is known and half of the vector $d$ is known too,  my question is,  can I set the other half of the vector $d$ to have the resultant of multiplication $g$ known ?

That can be summarized as below :

what I known is $k_{n/2 + 1 } ... k_n$ , The question that I set that part of the vector in order to have the resultant vector $g$ known ?

EDIT

Following the solution in the answer below, I have built the matlab code and used Gaussien elimination to solve $k$, but when I multiply the matrix again $W*[d;k]$ I can't get the same value of $g$. here is the code

```<code>close all
clc
W = hadamard(8);  %Taking the matrix of hadamard as matrix W
W1 = W(:,1:4);
W2 = W(:,5:8);
g = ones(8,1);
d = [0.7000 + 0.7000i; 0.7000 - 0.7000i; -0.7000 - 0.7000i; -0.7000 - 0.7000i];

A = W2;
b = g - W1*d;

n= size(A,2);
k= zeros(n,1);   %initialize x
%code for forward elimination to correct the matrix A to upper triangular
%form
for i =1: n-1
    m=A(i+1:n,i)/A(i,i);
    A(i+1:n,:)=A(i+1:n,:)-m*A(i,:);
    b(i+1:n,:) = b(i+1:n,:)-m*b(i,:);
end

%code for back substitution to find unknowns
k(n,:)= b(n,:)/A(n,n);
for i= n-1:-1:1
    k(i,:) =(b(i,:)- A(i,i+1:n)*k(i+1:n,:))/A(i,i);
end
D = [d;k];

W*D       %% that should be equal to g !!
</code>```

答:
$$\begin{bmatrix}W_1 & W_2 \end{bmatrix}\begin{bmatrix}d \\ k\end{bmatrix}=g$$

$$W_1 d + W_2k = g$$

$$W_1 d = g-W_2k$$

Now you can solve a linear system for $d$ if it is consistent.

Edit:

For your code, if you check the RREF of $[A, b]$, you will notice that it is not a consistent system.

In particular, hadamard matrix is invertible, hence if we solve $WD=g$ $D$ is unique. If you choose the first few component of $D$ arbirarily, it need not be consistent.


问:
Gram-Schmidt-like "orthogonalization" on a normed space
Let $X$ be a normed vector space and $w_1,\ldots,w_n\in V$ be linearly independent. Assume $k\in\{1,\ldots,n\}$, $u_1,\ldots,u_k\in X$ with $\operatorname{span}\{u_1,\ldots,u_k\}=\operatorname{span}\{w_1,\ldots,w_k\}$ and $u_1^\ast,\ldots,u_k^\ast\in X'$ with $u_i^\ast(u_j)=\delta_{i,j}$ and $\left\|u_i\right\|_X=\left\|u_i^\ast\right\|=1$.

Can we construct $u_{k+1}\in X,u_{k+1}^\ast\in X'$ so that the former properties remain true?

We can clearly define $v_{k+1}:=w_{k+1}-\sum_{i=1}^ku_i^\ast(w_{k+1})u_i$ and $u_{k+1}:=\frac{v_{k+1}}{\left\|v_{k+1}\right\|}$. Then $U:=\operatorname{span}\{u_1,\ldots,u_{k+1}\}=\operatorname{span}\{w_1,\ldots,w_{k+1}\}$ and $u_i^\ast(u_{k+1})=0$ for all $i\le k$.

Now, we can clearly find $u_{k+1}^\ast\in U^\ast$ with $u_{k+1}^\ast(u_i)=\delta_{i,k+1}$ for all $i\le k$. Then by the Hahn-Banch theorem, $u_{k+1}$ has an extension to $X$ with the same operator norm.

The only missing property is $\left\|u_{k+1}^\ast\right\|=1$. Can we fix that?

答:
Since Hahn-Banach does not increase the norm of the functional, it is enough to compute the norm of $u_i^*$ on the subspace $span(u_1\dots u_i)$:
$$
\|u_i^*\| = \sup_{\|u\|\le1} u_i^*(u)
=\sup_{\| \sum_{k=1}^i a_ku_k \|\le1} u_i^*(\sum_{k=1}^i a_ku_k)
=\sup_{\| \sum_{k=1}^i a_ku_k \|\le1} a_i
\ge \sup_{\sum_{k=1}^i |a_k|\le1} a_i =1.
$$
The last equality is due to $\| \sum_{k=1}^i a_ku_k \| \le \sum_{k=1}^1i |a_k|$, hence last supremum is taken over a smaller set than the second-to-last.

The reverse inequality is a consequence of the dual basis property:
$$
1\le u_i^*(u_i) \le \|u_i^*\|\cdot \|u_i\| = \|u_i^*\|.
$$


问:
Theorem 46.8 Munkres Topology
In the proof of Theorem 46.8 Munkres Topology following is claimed (not exact copy but exact interpretation)

If A is any compact subset of a metric space Y and V is any  open subset of Y containing A then there is a $\epsilon>0 $ such that the $\epsilon-$neighborhood of A is contained in V.

The mentioned claim is true provided $d(A, X-V) >0$ is positive. 1- But how to prove that? And 2- how any open set containing a compact set is a proper superset?

答:
A direct proof which does not relyy on the continuity of the distance function (a handy fact though that is..):

If $A \subseteq V$ for each $a \in A$ find $r_a>0$ such that $B(a,2r_a) \subseteq V$ (by openness of $V$ and the fact that $a \in A$ implies $a \in V$).

By compactness of $A$ there are finitely many $a_1, \ldots a_n \in A$ so that

$$A \subseteq \bigcup_{i=1}^n B(a_i, r_{a_i})$$

and define $r=\min_{i=1}^n r_{a_i} >0$.

If now $x$ in $U(A,r)$ there is some $a \in A$ such that $d(x,a) < r$. This $a$ lies in some $B(a_i, r_{a_i})$ so that $d(x,r_{a_i}) \le d(x,a) + d(a,a_i) < r + r_{a_i} < 2r_{a_i}$ and so $x \in B(a_i, 2r_{a_i}) \subseteq V$. This shows that $U(a,r) \subseteq V$ as required.

Title: Theorem 46.8 Munkres Topology


问:
How to prove that $\lim_{n \to \infty} \frac{1}{2^{2n}} \binom{2n}{n} \sim \frac{1}{\sqrt{\pi n}}$
I am presented with the following problem:

Prove that $\lim_{n \to \infty} \frac{1}{2^{2n}} \binom{2n}{n} \sim \frac{1}{\sqrt{\pi n}}$

After simplifying the left hand side expression:

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

And using Stirling's approximation, I get:

$\frac{\sqrt{2\pi n}(\frac{n}{e})^n}{2^{2n + 1}}$

However, I'm not sure how to get from:

$\lim_{n \to \infty} \frac{\sqrt{2\pi n}(\frac{n}{e})^n}{2^{2n + 1}}$

to

$\frac{1}{\sqrt{\pi n}}$

答:
It seems you might have simplified the expression wrong.

$$\binom{2n}n=\frac{(2n)!}{(n!)^2}$$

After applying Stirling's approximation, we have

$$\frac{(2n)!}{(n!)^2}\approx\frac{\sqrt{4\pi n}\left(\frac{2n}e\right)^{2n}}{\left(\sqrt{2\pi n}\left(\frac ne\right)^n\right)^2}=\frac{2\sqrt{\pi n}\left(\frac{n}e\right)^{2n}2^{2n}}{2\pi n\left(\frac ne\right)^{2n}}$$

Note that $\frac{\sqrt{\pi n}}{\pi n}=\frac{1}{\sqrt{\pi n}}$, while the $\left(\frac ne\right)^{2n}$ and $2$ on both sides of the fraction cancel out, giving us

$$\binom{2n}n\approx\frac{2^{2n}}{\sqrt{\pi n}}$$

Can you continue from here?


问:
Does $ \sum_1^\infty {(-1)^n \over n \ \log_x(p_n)}$ converge for all $x$?
Does
$$ \sum_1^\infty {(-1)^n \over n \ \log_x(p_n)}$$
converge for all x, where $p_n$ is the nth prime?

答:
$$\sum_{n=1}^\infty {(-1)^n \over n \,\log_x(p_n)}=\log(x)\sum_{n=1}^\infty {(-1)^n \over n \, \log(p_n)}$$

Use the bounds of the prime numbers and the alternating series test ad the squeeze theorem.


答:
You seem to be asking what is the notation for saying "$M$ is a real number".

The notation is:  $M\in \mathbb R$.

What that means literally is: $\mathbb R$ represents the set of all real numbers.  $\in$ is the symbol indicate "is an element of a set" or "is a member of a set" so $x\in A$ means that $x$ is an element of the set $A$.

So $2a \in \mathbb R$ means that $2a$ is an element of the set of real numbers.  Or..... in plain English, $2a$ is a real number.

But speaking of plain English, it's not clear to me why you think you can't use plain english in your proof.

Your proof is fine as: "$c =a + bi$ and $c^* = a-bi$ where $a,b$ are real numbers and $i$ is the imaginary unit.  So $c+c^* = (a+bi) + (a-bi) = 2a$ which is a real number."

It's good to know the notation $2a\in \mathbb R$ but as you seem to have gone this far without it you could go another day or two without it.

Title: What is the proper way of notating all elements in an equation are real?

Body:
Assuming c = a + ib and c* = a - ib where a and b are real

I want to show that c + c* is real

To do this I showed

(c + c*) = a + ib + a - ij = 2a

but I don't know how to state 2a is real. I'm thinking there is a way to state: "each element in the equation 2a belong to the set of real numbers, therefore c+c* is real" is what I have to say, I just don't know how to put that into proper math notation.


问:
$f(Y|X=x) \sim B(n,\frac{1}{2})$ is function of $x$ its pmf is ${{n} \choose {x}}(\frac{1}{2})^{x}(\frac{1}{2})^{n-x} $?

Let $X$ be the number of tosses of a fair coin required to get the
first head. If $Y | X = n$ is distributed as Binomial$(n, \frac{1}{2}
)$, then what is $P(Y = 1)$?

Before attempting this I have a question in my mind?

$f(Y|X=x)$ is function of $x$, so I can write its pmf as ${{n} \choose {x}}(\frac{1}{2})^{x}(\frac{1}{2})^{n-x} $ or is it ${{n} \choose {y}}(\frac{1}{2})^{y}(\frac{1}{2})^{n-y} ?$

Clear this doubt for me please.

Now from question $X$ follows geometric distributions.

$P(X=x)=(\frac{1}{2})^{x-1}(\frac{1}{2})$

$f(Y|X)=\dfrac{f(x,y)}{f_X(x)} \implies f(Y|X)\cdot {f_X(x)}=f(x,y)$

Then I have to find marginal of $Y$ and $P(Y=1)$ but I am stuck initially

答:
From the question, that $X=x,\:Y|X=x\sim B(x,0.5)$. For a binomial distribution $B(n,p)$, the pmf is given by $$p(k)=\binom{n}{k}p^k(1-p)^{n-k}$$
In this case we have $n=x,\:p=0.5$. So the pmf is $$\mathbb P(Y=y|X=x)=\binom{x}{y}(0.5)^y(1-0.5)^{x-y}=\binom{x}{y}(0.5)^x$$
Using the law of total expectation,
$$\mathbb P(Y=1)=\mathbb E(\mathbb P(Y=1|X))=\mathbb E\left(\binom{X}{1}(0.5)^X\right)\\ =\sum_{k=1}^\infty\binom{k}{1}(0.5)^k\cdot (0.5)^{k-1}(0.5)=\sum_{k=1}^\infty k(0.5)^{2k}=\frac{4}{9}$$


答:
This problem is conducive to using spherical coordinates.  If you draw a picture, you will see that the effect of the plane $y=-2 x$ is to extend the range of the azimuthal angle to $[0,\pi/2+\arctan{2}]$.  (The azimuthal angle is measured with respect to the positive $x$ axis.)  Therefore, the integral you seek is

$$\int_0^2 dr \, r^2 \int_0^{\pi/2} d\theta \, \sin{\theta} \, \int_0^{\pi/2 + \arctan{2}} d\phi \, p(r \sin{\theta} \cos{\phi}, r \sin{\theta} \sin{\phi},r \cos{\theta}) \\ = \int_0^2 dr \, r^{11} \int_0^{\pi/2} d\theta \, \sin^7{\theta} \cos^3{\theta} \, \int_0^{\pi/2 + \arctan{2}} d\phi \, \cos^2{\phi} \sin^4{\phi}$$

Note that these integrals may be evaluated separately and then multiplied together for the final result.

Title: Applications of Double/Triple Integrals

Body:
This is the question that I need to solve using mathematica:

The concentration of an air pollutant at a point $(x,y,z)$ is given by: $$p(x,y,z) = x^2y^4z^3 \text{ particles}/m^3$$ We're interested in studying the air quality in a region in 3-space which satisfies $x\ge0$ and is bounded by the $xy$-plane, the surface $z=\sqrt{4-x^2-y^2}$, the plane $y=-2x$, and the $xz$-plane.  Find the total amount of pollutant in this region.

I don't want a full answer to this question, obviously (I need to work it out myself), but I'm just having trouble getting started.  How do I use the bounds here to set up an integral for the problem?  What exactly do the bounds mean/represent in the first place?


问:
show that $-(\vec{r} \cdot \nabla)^2\psi = -r^2\frac{\partial^2\psi}{\partial r^2} -2r\frac{\partial\psi}{\partial r}$
So, I'm trying to proof the relation $(\vec{r}\times\nabla)\cdot (\vec{r}\times\nabla)\psi = r^2\nabla^2\psi -r^2\frac{\partial^2\psi}{\partial r^2} -2r\frac{\partial\psi}{\partial r}$, where $\psi$ is a scalar (wave) function. That's is the problem 1.9.9 from "George B Arfken, Hans J Weber - Mathematical Methods For Physicists- Sixth edition". I know that it would be easier to proof using spherical polar coordinates, but until that chapter of the book he hasn't mention anything about spherical polar coordinates, so because of that, I would like to try that using only cartesian coordinates.

So, I was thinking about use the relation $(\vec{A}\times\vec{B})\cdot (\vec{A}\times\vec{B}) = (AB)^2 - (\vec{A}\cdot\vec{B})^2$, so the first term come easily out:

$\begin{equation}(\vec{r}\times\nabla)\cdot (\vec{r}\times\nabla)\psi = r^2\nabla^2\psi - (\vec{r}\cdot\nabla)^2\psi \end{equation}$

But now I'm having some trouble tho show that:

$\begin{equation}-(\vec{r} \cdot \nabla)^2\psi = -r^2\frac{\partial^2\psi}{\partial r^2} -2r\frac{\partial\psi}{\partial r}\end{equation}$

I think that it should be possible to show using only cartesian coordinates, but I'm not seeing this throw... Could someone please help me?

答:
Let $r_1,r_2,r_3$ denote the Cartesian components of $\vec{r}$. All that is needed about spherical coordinates is

$$r = |\vec{r}|= \sqrt{r_1^2+ r_2^2+r_3^2}, \quad \frac{\partial r}{\partial r_j}= \frac{2r_j}{2 \sqrt{r_1^2+ r_2^2+r_3^2}}= \frac{r_j}{r}, \quad \frac{\partial f}{\partial r_j} = \frac{\partial f }{\partial r} \frac{\partial r}{\partial r_j} = \frac{\partial f }{\partial r}\frac{r_j}{r}$$

Using the Levi-Civita symbol, $\varepsilon_{ijk}$ and its property $\varepsilon_{ijk}\varepsilon_{imn} = \delta_{jm}\delta_{kn} - \delta_{jn}\delta_{km} $ we have with the Einstein summation convention

$$(\vec{r}\times\nabla)\cdot (\vec{r}\times\nabla)\psi = \varepsilon_{ijk}r_j\frac{\partial}{\partial r_k}\varepsilon_{imn}r_m\frac{\partial \psi}{\partial r_n} = (\delta_{jm}\delta_{kn} - \delta_{jn}\delta_{km} )r_j\frac{\partial}{\partial r_k}\left(r_m\frac{\partial \psi}{\partial r_n}  \right) \\ = (\delta_{jm}\delta_{kn} - \delta_{jn}\delta_{km} )r_jr_m\frac{\partial}{\partial r_k}\left(\frac{\partial \psi}{\partial r_n}  \right) + (\delta_{jm}\delta_{kn} - \delta_{jn}\delta_{km} )r_j\frac{\partial r_m}{\partial r_k}\frac{\partial \psi}{\partial r_n}\\ = \underbrace{r_j^2\frac{\partial\psi}{\partial r_k^2}}_A- \underbrace{r_jr_k\frac{\partial}{\partial r_k}\left(\frac{\partial \psi}{\partial r_j}  \right)}_B+ \underbrace{r_j\frac{\partial r_j}{\partial r_k}\frac{\partial \psi}{\partial r_k}}_C- \underbrace{r_j\frac{\partial r_k}{\partial r_k}\frac{\partial \psi}{\partial r_j}}_D $$

As a reminder, the Einstein summation convention gives $r_j^2 = r_1^2 + r_2^2 + r_3^2 = r^2$. Simplifying each of the four terms on the RHS, we get

$$A= r_j^2\frac{\partial\psi}{\partial r_k^2} = r^2 \nabla^2\psi$$

$$B = r_jr_k\frac{\partial}{\partial r_k}\left(\frac{\partial \psi}{\partial r_j}  \right) = r_jr_k\frac{\partial}{\partial r_k}\left(\frac{\partial \psi}{\partial r}\frac{r_j}{r}  \right)\\= r_jr_k\frac{\partial}{\partial r_k}\left(\frac{\partial \psi}{\partial r} \right)\frac{r_j}{r} + r_jr_k\frac{\partial \psi}{\partial r}\frac{\partial r_j}{\partial r_k}\frac{1}{r}+  r_j^2r_k\frac{\partial \psi}{\partial r}\frac{-1}{r^2}\frac{\partial r}{\partial r_k}\\ = \frac{r_j^2r_k^2}{r^2}\frac{\partial^2 \psi}{\partial r^2} + \frac{r_j^2}{r}\frac{\partial \psi}{\partial r}- \frac{r_j^2r_k^2}{r^3} \frac{\partial \psi}{\partial r}= r^2 \frac{\partial^2 \psi}{\partial r^2}$$

$$C= r_j\frac{\partial r_j}{\partial r_k}\frac{\partial \psi}{\partial r_k}= r_j\frac{\partial \psi}{\partial r}\frac{r_j}{r} = r \frac{\partial \psi}{\partial r}$$

$$D = r_j\frac{\partial r_k}{\partial r_k}\frac{\partial \psi}{\partial r_j}=3 r_j \frac{\partial \psi}{\partial r} \frac{r_j}{r}= 3r \frac{\partial \psi}{\partial r}$$

Thus,

$$(\vec{r}\times\nabla)\cdot (\vec{r}\times\nabla)\psi = A- B+ C- D = r^2 \nabla^2\psi -  r^2 \frac{\partial^2 \psi}{\partial r^2}+ r \frac{\partial \psi}{\partial r}- 3r \frac{\partial \psi}{\partial r} \\ =r^2 \nabla^2\psi -  r^2 \frac{\partial^2 \psi}{\partial r^2}- 2r \frac{\partial \psi}{\partial r}$$


问:
Algorithm to create "Spot It"/"Dobble" cards, but with two common images from any two cards
In the ordinary "Spot It"/"Dobble" game, there is exactly one common image between any two cards. At the moment I am trying to modify the algorithms already available on the internet so that any two cards have two, three, etc. elements in common.

Unfortunately, I have failed and am now looking for ideas on how such an algorithm could look in concrete terms. Do you have any ideas for such an algorithm?

答:
One can generate such a deck with a symmetric block design, which consists of

a set $\mathcal S$ (of symbols)

a collection $\mathcal C$ of subsets of $\mathcal S$ called blocks---in our case, cards---such that there are as many cards as there are symbols, and each $2$-subset of symbols occurs on equally many, say, $\lambda$, cards.

This concept is a special case of the concept of combinatorial design.

If we denote by

$v$ the number of symbols, equivalently the number of cards, and

$k$ the number of cards in which any given symbol appears---which for symmetric designs is the same as the number of symbols on each card---

then $$\lambda (v - 1) = k (k - 1) ,$$

and we call a design with given parameters a $(v, k, \lambda)$-design. The order of a symmetric design is $k - \lambda$.

It turns out that every two distinct cards have $\lambda$ symbols in common, so $(v, k, 1)$-designs (with $k > 2$) correspond to projective planes and therefore to the Spot-It like games outlined in MJD's blog post mentioned in the comments, and the question is asking about the construction of $(v, k, \lambda)$-designs with $\lambda > 1$.

There are lots of constructions for producing these:

Example 1 (faces of $n$-simplices) One easy construction is taking a set of $n$ symbols and making $n$ cards, each of which omits one of the symbols, so that each pair of cards intersects in $n - 2$ symbols, giving a $(n, n - 1, n - 2)$ design, which thus has order $1$. (We can think of such a design as an $n$-simplex, the cards as the codimension-$1$ faces of the simplex, and the sets of symbols common to two given cards as the codimension-$2$ face at which the corresponding faces meet.)

Example 2 (complementary designs) Given any $(v, k, \lambda)$-design, we can produce a $(v, v - k, \lambda + v - 2 k)$-design by replacing the set of symbols on each card with the complement of that set. The new design is called the complement of the old one and the two designs have the same orders.

Example 3 (Hadamard designs) For any Hadamard matrix of size $(4 m) \times (4 m)$, we can construct a $(4 m - 1, 2 m - 1, m - 1)$-design (called a Hadamard $2$-design), which thus has order $m$: By rearranging rows and columns of the Hadamard matrix, we can arranging for the first row and first column to have all entries $+1$. Delete that row and column, that is, consider the lower-right $(4 m - 1) \times (4 m - 1)$ minor $\Lambda$. Then, for each $(i, j)$ the card $C_j$ includes the $i$th symbol if and only if $\Lambda_{ij} = -1$.

Example 4 (Grassmannians of hyperplanes over finite fields) For any finite field $\Bbb F_q$ and any dimension $d \geq 2$, let the symbols and cards be the lines and hyperplanes, respectively, in $\Bbb F_q^{d + 1}$, and declare that a symbol $\ell$ is on a card $H$ if and only if $\ell \subset H$. This defines a $\left(\frac{q^{d + 1} - 1}{q - 1}, \frac{q^d - 1}{q - 1}, \frac{q^{d - 1} - 1}{q - 1}\right)$-design, which thus has order $q^{d - 1}$. The special case $d = 2$ yields projective planes.

A design with $\lambda = 2$---that is, a deck so that any two distinct cards have exactly two symbols is common---is called a biplane. A biplane of order $m$ corresponds to a deck with $\frac{1}{2} (m^2 + 3 m + 4)$ symbols and as many cards.

The only biplane of order $0$ corresponds to a trivial deck of $2$ cards, each of which have the same $2$ symbols:
$$01, \qquad 01$$

The only biplane of order $1$ is given by taking $n = 4$ in Example 1:
$$012, \qquad 013, \qquad 023, \qquad 123$$

The only biplane of order $2$ is the complement (Example 2) of the Fano plane, the smallest of the projective planes, i.e., the projective plane $\Bbb P(\Bbb F_2^3)$ corresponding to the $2$-element field $\Bbb F_2$:
$$0136, \qquad 0145, \qquad 0235, \qquad 0246, \qquad 1234, \qquad 1256, \qquad 3456 .$$

The only biplane of order $3$ arises from the unique Hadamard design (Example 3) with $m = 3$. Since $11$ is prime, we can apply the Paley construction to produce a $12 \times 12$ Hadamard matrix and hence an $11$-card deck:
\begin{multline}
01247, \quad
0136\textrm{A}, \quad
0259\textrm{A}, \quad
03789, \quad
04568, \quad
12358, \\
1489\textrm{A}, \quad
15679, \quad
23469, \quad
2678\textrm{A}, \quad
3457\textrm{A} .
\end{multline}
This configuration is called the Paley biplane, and it corresponds to the exceptional inclusion $PSL(2, 5) \hookrightarrow PSL(2, 11)$.

There are several distinct biplanes of order $4$, including one arising from the classic Kummer configuration, which corresponds to a deck of $16$ cards and as many symbols.

The Bruck-Ryser-Chowla Theorem implies that there are no biplanes of order $5$ or $6$.

I haven't seen the term before, but one might as call a design with $\lambda = 3$ a triplane; a triplane of order $m$ corresponds to a deck with $\frac{1}{3} (m^2 + 5 m + 9)$ symbols and as many cards, so a triplane cannot have order equivalent to $2 \pmod 3$.

There is a triplane of order $0$ analogous to the trivial biplane.

Taking $n = 5$ in Example $1$ gives a triplane of order $1$.

The dual of the Paley biplane is a triplane of order $3$.

Taking $q = 2, d = 3$ in Example 4 gives a triplane of order $4$, and it corresponds to a deck of $15$ cards and as many symbols.


问:
Constructing a continuous simple approximation of the identity map for a totally disconnected compact metric space
In Petersen's Ergodic Theory, when talking about totally disonnected spaces (Section 4.4), he says that if $D$ is a compact, totally disconnected subspace of $\mathbb{R}$, then for every $\delta > 0$, there exists a continuous simple function $\phi : D \to D$ taking values $\{ d_1, \ldots, d_n \} \subseteq D$ such that $\phi(d_j) = d_j$ and $\operatorname{diam} \phi^{-1} \{d_j\} < \delta$ for all $j$. How does he conclude this? If I'm not mistaken, this is equivalent to saying that there's a clopen neighborhood basis for each point in $D$, but I don't know how to prove this. I understand that this latter property is called being zero-dimensional, and that it's in general a stronger property than being totally disconnected, but I'm not sure how to show that in this case, the weaker claim implies the stronger.

答:
According to https://en.wikipedia.org/wiki/Zero-dimensional_space

[See the heading "Properties of spaces with small inductive dimension zero"] a locally compact space is totally disconnected iff it is zero dimensional. A reference for the proof is also given there.


问:
Is there a way to combine a product or sum of two cosines into one?
Is there a way to combine a product or sum of two cosines into one? More specifically:

$$
A \cos \alpha \cos \beta = B \cos \gamma
$$

or:

$$
A \cos \alpha + B \cos \beta = C \cos \gamma
$$

I've been doing some research and I found "linear combination of cosine and sine with equal periods". This "would" work for the latter case if one of the cosines was transformed into a sine, but still (even after the transformation):

$$
\alpha \neq \beta
$$

and:

$$
\frac{\pi}{2} - \alpha \neq \beta
$$

Thus, it will not work. I also know about the product-to-sum and sum-to-product identities, but that does not really help here. Is there a way of doing it? What if:

$$
\alpha \propto \beta
$$

where the ratio is a rational number, but not necessarily an integer? Any suggestions and guidance would greatly be appreciated.

答:
We have
$$
 \cos \alpha \cos \beta =  \cos \gamma
$$
where
$$
\gamma=\arccos(\cos \alpha \cos \beta)
$$
The same for sum.


问:
Linear matrix equation involving $\sum_i A_i X B_i$
I am dealing with a non-standard linear matrix equation:
\begin{equation}
X \quad+\quad \sum_{i=1}^r \Big( (A_i X B_i) + (A_i X B_i)^T + (B_i X A_i) + (B_i X A_i)^T \Big) \quad=\quad C
\end{equation}
The symbols in capital letters, i.e. $X, A_i, B_i, C$, denote $n \times n$ real matrices. $X$ is the unknown that we wish to solve for. Additional considerations:

The number of terms $r$ is large.

$C$ is symmetric

$A_i$'s are all rank one matrices (not sure if this information is helpful for expressing $X$ though).

Is there an elegant way to express the solution form $X$?

Thanks for your help!

Golabi

答:
Each summand in the summation sign is symmetrised and $C$ is symmetric. Hence $X$ must be symmetric and the equation is equivalent to
$$
X+\sum_i\left(A_iXB_i+B_i^TXA_i^T+B_iXA_i+A_i^TXB_i^T\right)=C,
$$
which can be rewritten as (see Wikipedia)
$$
\left[I+\sum_i\left(B_i^T\otimes A_i+A_i\otimes B_i^T+A_i^T\otimes B_i+B_i\otimes A_i^T\right)\right]\operatorname{vec}(X)=\operatorname{vec}(C).
$$
Call the matrix inside the pair of square brackets $M$. The equation $M\operatorname{vec}(X)=\operatorname{vec}(C)$ is solvable if and only if $MM^+\operatorname{vec}(C)=\operatorname{vec}(C)$, where $M^+$ denotes the Moore-Penrose pseudoinverse of $M$. In case it is solvable, $\operatorname{vec}(X)=M^+\operatorname{vec}(C)$ is always a solution.


问:
Generalization of Doob Dynkin for Stochastic processes
Let $\{X_t\}_{t\geq 0}$ be continuous time stochastic process and $\{\mathcal{F}_t^X\}_{t \geq 0}$ be the filtration generated by it. If the process $Y$ is $\{\mathcal{F}_t^X\}_{t \geq 0}$ adapted, is there any simple representation of the process $Y$ in terms of $X$. This will be a generalization of Doob Dynkin Lemma</p>

答: I don't think there is a non-trivial result in the general setting, but for some special cases of $X_t$ and $Y_t$ it is possible to represent $Y_t$ as an Ito integral w.r.t. $X_t$, see for example here.</p>


问:
Solving a PDE by method of characteristics
I solved the following:
\begin{align}
\frac{dy}{y}&=\frac{dy'}{2y'+wy}\\
\ln y + \ln C &= \frac{1}{2} \ln \left(2y'+wy\right)\\
2\ln y + \ln C &= \ln \left(2y'+wy\right)\\
C &= \frac{2y'}{y^2} + \frac{w}{y}\\
\end{align}

I isolated the constant since the purpose of solving is to find the differential invariant. However the answer given is:
$$
\frac{y'}{y^2}+\frac{w}{y}
$$

where am I going awry?

答:
Using the rule $\frac ab=\frac cd\implies \frac ab=\frac cd=\frac{a+c}{b+d}$, add the $ω$ extended first term to the second to find
$$
\frac{dy}{y}=\frac{dy'}{2y'+ωy}=\frac{dy'+ωdy}{2y'+2ωy}
$$
Now taking the first and last term this can be directly integrated to
$$
\ln|y|=\frac12\ln|y'+ωy|+c\implies Cy^2=y'+ωy
$$


问:
Evaluating $ \lim_{x\to1^-}\frac1{x^2-1}\,\exp(\frac1{x^2-1})$. Isn't this $\infty\times\infty$?
I need to find the following limit,
$$
\lim_{x\to1^-}\frac1{x^2-1}\,\exp\left(\frac1{x^2-1}\right)
$$

My attempt: I think the form is $\infty\times\infty$ form and hence limit goes to $\infty$, but Wolfram says I am wrong. Please help.

答:
$$
L=\lim_{x\to1^-}\frac1{x^2-1}\,\exp\left(\frac1{x^2-1}\right)
$$
$$\implies L=\lim_{h \to 0} \frac{1}{(1-h)^2-1}\exp\left( \frac{1}{(1-h)^2-1}\right)=\lim_{h \to 0} \frac{-1}{2h}e^{-1/(2h)}$$
We have ignored $h^2$ in cparision ro $h$. It amounts to using binomial approcimation that $(1-z)^k \approx 1-kz, if |z|<<1.$
$$L=\lim_{t \to \infty} -te^{-t}= -\lim_{t\to \infty}\frac{t}{e^t}=0$$
By L'hospital.


问:
What does it really mean when the probability of multiple independent events sum to 1? More than 1?
For example, say I roll a fair six-sided die six times, and ask what the likelihood is that I will see a result of "6" among the six outcomes. Each roll, ostensibly, has a 1/6 probability of giving me a "6", and future rolls are not dependent on past rolls. Summing the probability of the events, 1/6+1/6+...+1/6, or, in this case, (1/6)*6, I calculate a probability of 1. But in real life, I know that it is perfectly possible to roll a fair six-sided die six times and not see a "6" among the outcomes. Going a step further, I can roll a fair six-sided die nine times, and ask what the likelihood of seeing a "6" among the outcomes is, for a probability of (1/6)*9 or 1.5, but I can ALSO roll a fair six-sided die nine times in real life and never see a "6". So am I missing something? Does it have to do with the word "fair"? Am I using the wrong formula? Is probability just an inherently flawed description of reality that calls for some subjectivity? Something else?</p>

答:
Here's another angle:

Since the die rolls are independent (i.e., the outcome of one doesn't
affect the probabilities of the outcomes of any other), the probability that the first '4' is obtained on the $\left(n+1\right)^\text{th}$ roll is
$\left(\frac56\right)^n\left(\frac16\right).$

(So the chance of getting a '4' only on the $7^\text{th}$ roll is
$6\%.$)

Thus, there will always be a nonzero chance of still not getting a
'4' however many die rolls.

Addressing the question directly:

The event of obtaining a '4' on the $p^\text{th}$ roll is not
mutually exclusive of the event of obtaining a '4' on the
$q^\text{th}$ roll, because they can both†
occur. So, the probability of obtaining at least one '4' on the
$p^\text{th}$ and $q^\text{th}$ rolls is smaller than
$\frac16+\frac16.$

   † regardless of whether the rolls
are happening concurrently or in succession

As the number of rolls $n$ increases, the number of such common
outcomes among the rolls increases at an increasing rate, i.e.,
the probability of getting at least one '4' increases at a
decreasing rate. The desired probability is always smaller than $\frac n6$ and $1.$


问:
Solving a linear ODE using Frobenius' method
The given ODE is

$2y''$ - $y'/(x-1)$ + $y/(x-1)^2$ = 0.
Solving we get,

$\sum_{n=0}$$[2(n+r)(n+r-1)a_n - (n+r)a_n](x-1)^{(n+r-1)}$.

Now I have few basic doubts. Please bear with me.
Can't we directly equate the coefficient of $(x-1)^{(n+r-1)}$ to zero?

Because the solution given here shows that we need to first expand to get:

$[2r(r-1)-r+1]a_0(x-1)^{r-2} + [2r(r+1)-(r+1)+1]a_1(x-1)+...$

Then substituting the coefficient of $(x-1)^{r-2}$ to zero to get the indicial equation as:

$2r(r-1)-r+1=0$.

My second doubt is that, after solving, we get $r=1$ or $r=0.5$

Then what should we do after this?
The solution says substituting r, only coefficient of $a_0$ goes to zero, that means $a_1,a_2....= 0$

Therefore, the general solution should be $a_0(x-1)+a_2(x-1)^{0.5}$

Can someone elaborate on this please

答:
Yes, that looks correct. The equation is of Euler-Cauchy type,
$$
2(x-1)^2y''-(x-1)y'+y=0
$$
this implies that basis solutions $(x-1)^r$ exist and $r$ solves the characteristic equation
$$
0=2r(r-1)-r+1=(2r-1)(r-1)
$$
that is identical to the indicial equation. So indeed this gives a complete basis $(x-1)^{1/2}, (x-1)$.

To the first point, in a more general series expansion the first terms would have coefficients with negative indices. One can avoid that by treating these terms separately or by expanding the coefficient sequence with $a_{k}=0$ for $k<0$.


问:
how to derive the steps of this intergral for the product of two exponential terms?
For homework, I was looking for closed form solutions of exponential functions from this wiki site.
https://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions

But I was not sure how to derive the steps for this particular integral,

$\int_{-\infty}^{\infty}e^{-ax^{2}}e^{-\frac{b}{x^{2}}}dx={\sqrt {\frac {\pi }{a}}}e^{-2{\sqrt {ab}}}\quad (a,b>0)$

I am fairly new to normal distribution but I assume this has to do with Taylor series..

Any pointers would be greatly appreciated!

答:
Start completing the square
$$ax^2+\frac b {x^2}=a\left(x+\sqrt{\frac{b}{a}}\frac{1}{x}\right)^2-2\sqrt{ab}$$
$$x+\sqrt{\frac{b}{a}}\frac{1}{x}=t \implies x=\frac{1}{2} \left(t\pm\sqrt{t^2+4 \sqrt{\frac{b}{a}}}\right)$$

Making the calculation for the $\color{red}{-}$ branch.

This makes now the integrand to be
$$\frac{1}{2} e^{2 a \sqrt{\frac{b}{a}}} \left(e^{-a t^2}-\frac{t e^{-a t^2}}{\sqrt{t^2-4 \sqrt{\frac{b}{a}}}}\right)$$ Now, we face two "simple" integrals
$$\int e^{-a t^2}\,dt=\frac{\sqrt{\pi } \text{erf}\left(\sqrt{a} t\right)}{2 \sqrt{a}}$$
$$\int \frac{t e^{-a t^2}}{\sqrt{t^2-4 \sqrt{\frac{b}{a}}}}\,dt=\frac{\sqrt{\pi } e^{-4 a \sqrt{\frac{b}{a}}} \text{erf}\left(\sqrt{a} \sqrt{t^2-4
   \sqrt{\frac{b}{a}}}\right)}{2 \sqrt{a}}$$

Back to $x$
$$\frac{\sqrt{\pi } e^{-2 \sqrt{a} \sqrt{b}}
   \left(\text{erfc}\left(\frac{\sqrt{b}}{x}-\sqrt{a} x\right)-e^{4 \sqrt{a}
   \sqrt{b}} \text{erfc}\left(\sqrt{a} x+\frac{\sqrt{b}}{x}\right)\right)}{4
   \sqrt{a}}$$ Integrateing between $-p$ and $+p$
$$\frac{\sqrt{\pi } e^{-2 \sqrt{a} \sqrt{b}} \left(e^{4 \sqrt{a} \sqrt{b}}
   \text{erf}\left(\sqrt{a}
   p+\frac{\sqrt{b}}{p}\right)-\text{erf}\left(\frac{\sqrt{b}}{p}-\sqrt{a}
   p\right)\right)}{2 \sqrt{a}}$$ and the limit is
$$\frac{\sqrt{\pi } e^{-2 \sqrt{a b}}}{\sqrt{a}}$$


问:
Is the complement of a closed and nowhere dense set a dense set?
1)Suppose A is a closed and nowhere dense subset of X, is $A^c$ dense?

2)Suppose B is open and dense in X, is $B^c$ nowhere dense?

I tried proving both but it was long and messy so I am not sure it is right ... I wonder if there is a quick way to show this...

Also, is it necessary that A is closed or that B is open?

Here is what I did for the moment :

A closed --> cl(A) = A , and A is nowhere dense --> $int(cl(A))=int(A)=\emptyset$
so A contains no open set.
Take x in X, we want to show x is in $cl(A^c)$:
Take U open containing x, thus X is not entirely in A so $U\cap A^c$ is not the empty set so x is in $cl(A^c)$ so X=$cl(A)^c$

Thanks

答:
It’s very easy if you know that $D\subseteq X$ is dense in $X$ if and only if $D\cap U\ne\varnothing$ for every non-empty open set $U$ in $X$; if you’ve not yet seen this characterization of denseness, you should definitely try proving it.

Suppose that $A$ is closed and nowhere dense in $X$, and let $U=X\setminus A$. If $V$ is any non-empty open set in $X$, then $V\nsubseteq A$, so $V\cap U\ne\varnothing$. Thus, $U$ intersects every non-empty open set in $X$ and is therefore dense in $X$.

Conversely, if $U$ is a dense open subset of $X$, then $U$ intersects every non-empty open set in $X$, so $\operatorname{int}(X\setminus U)$ must be empty, and $X\setminus U$ is therefore closed and nowhere dense.


问:
(Understanding the proof of) Riesz representation theorem in $L^p$
I'm studying $L^p$ spaces for the first time. I am following, among others, the book by Stein & Shakarchi Functional Analysis. Introduction to Further Topics in Analysis. Before the proof of the famous theorem they introduce the following lemma (p.14).

Lemma 4.2 Suppose $1\leq p,q\leq\infty$ are conjugate exponents.

(i) If $g\in L^q$, then $\displaystyle\|g\|_{L^q}=\sup_{\|f\|_{L^p}\leq1}\left|\int fg\right|$.

[...]

In the proof of the above lemma for the case $p=1$ and $q=\infty$ they say:

Let $\epsilon>0$, and $E$ a set of finite positive measure, where $|g(x)|\geq\|g\|_{L^\infty}-\epsilon$. (Such a set exists by the definition of $\|g\|_{L^\infty}$ and the fact that the measure $\mu$ is $\sigma$-finite.)

I have a couple of doubts with what is said in the previous paragraph.

Doubt 1. When they say $|g(x)|\geq\|g\|_{L^\infty}-\epsilon$, should I understand
$$\forall x\in E:\ |g(x)|\geq\|g\|_{L^\infty}-\epsilon?$$
Doubt 2. I'm failing to see why such set $E$ must exists. Suppose we are talking about $L^1(X,\mu)$ space. I know that if the measure $\mu$ is $\sigma$-finite there exists a sequence of subsets $(X_n)$ such that
$$X=\bigcup_{n=1}^\infty X_n,\quad\mu(X_n)<\infty,\ \forall n\in\mathbb{N}.$$
I also know that if $g\in L^\infty(X,\mu)$ then
$$\|g\|_{L^\infty}=\mathrm{ess}\,\sup g<\infty\Longrightarrow\mu\{g>\mathrm{ess}\,\sup g\}=0.$$
I think I should choose the set $E$ with the help of said sequence $(X_n)$, but I don't see how to do it.

答: Let $\epsilon >0$ and $E=\{x: |g(x)| >\|g\|_{\infty} -\epsilon\}$. It is easy to that this set is measurable. If this set has measure $0$ then $|g| \leq \|g\|_{\infty} -\epsilon$ almost everywhere. But this contradicts the definition of $\|g\|_{\infty}$ as the  smallest number $M$ such that $|g(x)| \leq M$ almost everywhere so we must have $\mu (E) >0$. Of course, $|g(x)| >\|g\|_{\infty} -\epsilon$ for every $x \in E$.


问:
Prove or disprove that $f$ is convex if and only if $\frac1{b-a}\int_a^b f(x)\,dx \le \frac{f(a)+f(b)}2 .$

Let $f:\mathbb{R} \to \mathbb{R}$ be a continuous function.
Prove or disprove that $f$ is convex if and only if
$$\frac1{b-a}\int_a^b f(x)\,dx \le \frac{f(a)+f(b)}2 $$
for any $a<b \in \mathbb{R}$.

It is easy to prove that $f$ is convex implies the inequality
$$\frac1{b-a}\int_a^b f(x)\,dx \le \frac{f(a)+f(b)}2 $$
for any $a<b \in \mathbb{R}$.
Just use the substitution $x=(1-t)a+tb, t\in[0,1]$,
\begin{align*}
\frac1{b-a}\int_a^b f(x)\,dx
  &=\int_0^1 f((1-t)a+tb)\,dt\\
  &\le\int_0^1(1-t)f(a)+tf(b)\,dt\\
  &=\frac{f(a)+f(b)}2.
\end{align*}

What I want to know is that: does the inequality
$$\frac1{b-a}\int_a^b f(x)\,dx \le \frac{f(a)+f(b)}2,\forall a<b \in \mathbb{R} $$
imply that $f$ is convex.

The following is a characterization for convex functions: Characterization convex function.

Any help and hint will welcome!

答:
Assume $f$ is not convex. Then there are $x>y$ and $\lambda\in (0,1)$ such that
$$
f(\lambda x + (1-\lambda)y ) > \lambda f(x) + (1-\lambda)f(y).
$$
This inequality will be satisfied if we perturb $\lambda$ a little bit. The idea is now to take $[a,b]$ to be the maximal interval of these points, where this inequality is true (or convexity fails). Then evaluate the integral, to get a contradiction.

Define $g$ by
$$
g(t):=f(t x + (1-t)y ) -( t f(x) + (1-t)f(y) ).
$$
Due to continuity,  $g(t)>0$ in a neighborhood of $\lambda$. In addition, $g(0)=g(1)=0$.
Define
$$
A:=\{s\le\lambda: \ g(t)>0 \quad \forall t\in [s,\lambda]\}
$$
and
$$
B:=\{s\ge \lambda: \ g(t)>0\quad \forall t\in [\lambda,s]\}.
$$
Set $s_a:=\inf A$, $s_b:=\sup B$. Then $0\le s_a<\lambda<s_b\le 1$ and $g(s_a)=0=g(s_b)$.
Set $a:=s_a x+ (1-s_a)y$, $b:=s_b x+ (1-s_b)y$, which implies $b-a = (s_b-s_a)(x-y)>0$.
Then
$$\begin{split}
\int_a^b f(x)dx &= (x-y) \int_{s_a}^{s_b} f(sx+(1-s)y)ds \\
&> (x-y) \int_{s_a}^{s_b} sf(x)+(1-s)f(y)ds\\
&= \frac12(x-y)[ (s_b^2-s_a^2)f(x) + (2s_b-2s_a - (s_b^2-s_a^2)) f(y)]\\
&= \frac12(x-y)(s_b-s_a)( (s_b+s_a)f(x) + (2-s_b-s_a)f(y))\\
&= \frac{b-a}2 (f(a) + f(b)),
\end{split}
$$
which contradicts the assumption.


问:
$\{X_t\}$ Brownian motion implies that $X_{t_1},X_{t_2},\dotsc, X_{t_n}$ are independent when $t_k<t_{k+1}$?
If $\{X_t\}_{t\in [0,1]}$ is a one-dimensional Brownian motion with $X_0 = 0$ a.s. and $0\leq t_1<t_2<\cdots <t_n \leq 1$. Does it then hold that $X_{t_1},\dotsc, X_{t_n}$ are independent, and if so what is the intuition behind it?

I would say that they are independent based on the following argument but maybe there is something wrong with it and would be happy to be corrected in that case.

Let $A = \begin{pmatrix}1 & 0 & \cdots & 0\\ 1 & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & \cdots & 1\end{pmatrix}$, $Y = (X_{t_1}, X_{t_2}-X_{t_1},\dotsc, X_{t_n}-X_{t_{n-1}})$ and $X = (X_{t_1},X_{t_2},\dotsc,X_{t_n})$. Then $AY = X$ and hence it suffices to show that the distribution of $AY$: $\mathbb{P}_{AY}$ is a product measure on $\mathbb{R}^n$ to conclude that the components of $X$ are independent.

This however follows from the following computation: Let $B\in \mathscr{B}(\mathbb{R}^n)$ (Borel set) then
$$\mathbb{P}(AY\in B) = \int_{\mathbb{R}^n}1_{B}(x)\mathrm{d}\mathbb{P}_{Y}A^{-1} = \int_{\mathbb{R}^n}1_{B}(Ax)\mathrm{d}\mathbb{P}_{Y} = \int_{\mathbb{R}^n}1_{B}(x)\underbrace{|\det(A^{-1})|}_{=1}\mathrm{d}\mathbb{P}_{Y} = \mathbb{P}(Y\in B).$$
Since $\mathbb{P}_Y$ is a product measure this concludes the argument.

答: $X_t$ and $X_s$ are not independent for $0<t<s$: $Cov (X_t,X_s)=EX_tX_s=\min \{s,t\} \neq 0$.


问:
If $\lim\sup a_n = -\infty$, then $a_n\to -\infty$, and if $\lim\sup a_n = \infty$ there exists a subsequence of $a_n$ that $\to\infty$

Show that if $\lim\sup a_n = -\infty$, then $a_n\to -\infty$, and if $\lim\sup a_n = \infty$ there exists a subsequence of $a_n$ that $\to\infty$. What if $\lim\inf a_n = \pm\infty$? We're working with the extended real number system, $\bar{\mathbb{R}} = \mathbb{R} \cup\{\infty,-\infty\}$.

My work:
$$\lim\sup a_n = \inf_{n\ge1}\sup_{m\ge n} a_m  = -\infty$$
is known. This means that there exists some $n'\in\mathbb{N}$ such that $\sup_{m\ge n'}a_m = -\infty$. So, for all $m \ge n'$, we have $a_m \le -\infty$. This can happen iff $a_m  = -\infty$ for all $m\ge n'$. Hence, $a_n\to -\infty$.

Is this proof alright? I'm concerned about steps where I've written $a_m=-\infty$ and such, even though we know that we are working with the extended real number system. Is this okay to do? Just some uneasiness since I haven't used $=$ with infinities before.

Now assume that $$\lim\sup a_n = \inf_{n\ge1}\sup_{m\ge n} a_m  = \infty \implies \sup_{m\ge n} a_m = \infty \text{ for all }n\in\mathbb{N}$$
I can intuitively see that the least upper bound of all tail sequences of $a_n$ is $\infty$, so maybe $a_n\to\infty$ itself? I'm not sure because the problem explicitly asks to produce a subsequence of $a_n$ (or show that one exists) that goes to $-\infty$. How do we do this?

Similarly for the case when $\lim\inf a_n = \pm \infty$, what can we say? I'd guess that $\lim\inf a_n = \infty$, then $a_n\to\infty$ and if $\lim\inf a_n = -\infty$ then there is a subsequence of $a_n$ that goes to $\infty$.

Here's my definition for diverging to $+\infty$:
$$x_n\to +\infty:= \forall c\in\mathbb{R}\ \exists m\in\mathbb{N}\ \forall n > m\ (x_n > c)$$
$$x_n\to -\infty:= \forall c\in\mathbb{R}\ \exists m\in\mathbb{N}\ \forall n > m\ (x_n < c)$$

Second Attempt:

$$\lim\sup_{n\to\infty} a_n = -\infty \implies \lim_{n\to\infty} \sup\{a_k:k\ge n\} = -\infty$$ $$\implies \forall c\in\mathbb{R}\ \exists m\in\mathbb{N}\ \forall n>m\ (\sup\{a_k:k\ge n\} < c)$$ $$\implies \forall c\in\mathbb{R}\ \exists m\in\mathbb{N}\ \forall n>m\ (a_k < c, k\ge n)$$
$$\implies \forall c\in\mathbb{R}\ \exists m\in\mathbb{N}\ \forall n>m\ (a_n < c) \implies \lim_{n\to\infty} a_n = -\infty$$
Similarly, $$\liminf_{n\to\infty} a_n = \infty \implies a_n\to\infty$$

答:
The statement
$\sup_{m\geq n'}\alpha_m =-\infty,$
for some $n'$ is false in general. For example if $\alpha_m = -m$ then for every $n'$ we have $\sup_{m\geq n'}=\alpha_{n'}=-n'>-\infty$ and $\alpha_n\to -\infty$ as $n\to \infty$. To prove that $\alpha_n\to -\infty$ when $\limsup_{n}\alpha_n = -\infty$ fix some $c\in\mathbb{R}$. Since $\limsup_n \alpha_n = - \infty$ there is $n'$ such that
$$\sup_{m\geq n'}\alpha_m < c.$$
This means that for every $m\geq n'$ we have $\alpha_m<c$ which is equivalent to $\alpha_n\to -\infty$. Now, if $\limsup_{n}\alpha_n =\infty$ it does not necessarily implies that $\alpha_n \to \infty$. For example consider the sequence
$$
\alpha_n=\begin{cases}
n,\ n \text{ is even }\\
-1,\ n \text{ is odd}
\end{cases}
$$
Then, the subsequence $\alpha_{2n-1}$ is constant and equal to $-1$, hence $\alpha_n$ cannot converge to $\infty$. Although, it is true that there exists a subsequence $\alpha_{k_n}$ such that $\alpha_{k_n}\to \infty$. Since $\limsup_n \alpha_n =\infty$ for every $n$ you have that $\sup_{m\geq n}\alpha_m =\infty$. In particular,
$$\sup_{m\geq 1 }\alpha_m >1.$$
This means, that there exists some $k_1\geq 1$ such that $\alpha_{k_1}>1$. Now, using again the fact that $$\sup_{m\geq k_1 +1}\alpha_m>2$$
we may find $k_2 >k_1$ such that $\alpha_{k_2}>2$. Continuing this way, recursively, we may find indices
$$k_1<k_2<...<k_n<k_{n+1}<...$$
in ascending order such that $\alpha_{k_n}>n$ for every $n$. This means that the subsequence $\alpha_{k_n}$ converges to $\infty$. Similarly, if $\liminf_n \alpha_n=\infty$ then $\alpha_n\to \infty$ and if $\liminf_n \alpha_n =-\infty$ then there exists a subsequence $\alpha_{k_n}\to -\infty$.


答:
No, your proof is not correct. There exists in general no $n'\in\mathbb{N}$ such that $\sup_{m\ge n'}a_m = -\infty$, but there exists a subsequence $(n_k)$ so that $a_{n_k}\to-\infty(k\to\infty)$ and $\sup_{m\ge n}a_m \to -\infty (n\to\infty)$. The first comes from the fact that $\limsup a_n=c$ means your biggest cluster point is $c$ and cluster points are exactly the limits of converging subsequences. So $\limsup$ is the biggest extended real number a subsequence converges. The second can either be implied by the first or you can use $\limsup_{n\to\infty} a_n=\lim_{n\to\infty}\sup_{m\geq n} a_m$, which comes from the fact that $(\sup_{m\geq n} a_m)_{n\in\mathbb N}$ is a decreasing sequence so you can exchange $\inf$ and $\lim$ in
$$\limsup_{n\to\infty} a_n=\inf_{n\in\mathbb N}\sup_{m\geq n} a_m=\lim_{n\to\infty}\sup_{m\geq n}a_m.
$$

What you have to show (as boink already mentioned in the comments) is that for every $M\in\mathbb R$ (especially for big negative $M$) exists a $N\in\mathbb N$ so that for all $n\geq N$ holds $a_n<M$. So choose any $M$ and now let's find such a $N$. Your first step to consider the sequence $(\sup_{m\ge n}a_m)_{n\in\mathbb N}$ was in the right direction. As already mentioned above we have $\sup_{m\ge n}a_m \to -\infty (n\to\infty)$. This means we find a $N'$ so that for every $n\geq N'$ holds $\sup_{m\ge n}a_m <M$. This implies that for every $n\geq N'$ holds $a_n<M$. This number is the $N=N'$ we were looking for and we have finished the proof.

Now consider
$$\lim\sup a_n = \inf_{n\ge1}\sup_{m\ge n} a_m  = \infty \implies \sup_{m\ge n} a_m = \infty \text{ for all }n\in\mathbb{N}.$$
The $\lim\sup$ means the bigest cluster point of your sequence is $\infty$, which directly implies the statement, since you find for an arbitrary big $n$ a $m>n$ so that $a_m$ is arbitrary big. If you don't wanna handle with cluster points here, you can use the fact from above $$\infty=\limsup_{n\to\infty} a_n=\lim_{n\to\infty}\sup_{m\geq n}a_m$$
and then apply that the decreasing sequence $(\sup_{m\ge n}a_m)_{n\in\mathbb N}$ converges to $\infty$ which means it already has to be $\infty$.

Your last two guess regarding $\lim\inf a_n = \pm \infty$ are correct.

What you wrote under "Second Attempt" is comletely correct.

To give you an alternative proof to the first part, you could also use the property
$$
\lim\inf a_n\leq \lim\sup a_n,\tag{1}\label{tag}
$$
where equality holds if and only if the limit $\lim a_n$ exists and $$\lim\inf a_n= \lim\sup a_n= \lim a_n.$$ So if $\lim\sup a_n=-\infty$ already holds, it follows form \eqref{tag} that $\lim\inf a_n=-\infty$ and the statement follows from the 'iff'-part.

Title: If $\lim\sup a_n = -\infty$, then $a_n\to -\infty$, and if $\lim\sup a_n = \infty$ there exists a subsequence of $a_n$ that $\to\infty$

Body:
Show that if $\lim\sup a_n = -\infty$, then $a_n\to -\infty$, and if $\lim\sup a_n = \infty$ there exists a subsequence of $a_n$ that $\to\infty$. What if $\lim\inf a_n = \pm\infty$? We're working with the extended real number system, $\bar{\mathbb{R}} = \mathbb{R} \cup\{\infty,-\infty\}$.

My work:
$$\lim\sup a_n = \inf_{n\ge1}\sup_{m\ge n} a_m  = -\infty$$
is known. This means that there exists some $n'\in\mathbb{N}$ such that $\sup_{m\ge n'}a_m = -\infty$. So, for all $m \ge n'$, we have $a_m \le -\infty$. This can happen iff $a_m  = -\infty$ for all $m\ge n'$. Hence, $a_n\to -\infty$.

Is this proof alright? I'm concerned about steps where I've written $a_m=-\infty$ and such, even though we know that we are working with the extended real number system. Is this okay to do? Just some uneasiness since I haven't used $=$ with infinities before.

Now assume that $$\lim\sup a_n = \inf_{n\ge1}\sup_{m\ge n} a_m  = \infty \implies \sup_{m\ge n} a_m = \infty \text{ for all }n\in\mathbb{N}$$
I can intuitively see that the least upper bound of all tail sequences of $a_n$ is $\infty$, so maybe $a_n\to\infty$ itself? I'm not sure because the problem explicitly asks to produce a subsequence of $a_n$ (or show that one exists) that goes to $-\infty$. How do we do this?

Similarly for the case when $\lim\inf a_n = \pm \infty$, what can we say? I'd guess that $\lim\inf a_n = \infty$, then $a_n\to\infty$ and if $\lim\inf a_n = -\infty$ then there is a subsequence of $a_n$ that goes to $\infty$.

Here's my definition for diverging to $+\infty$:
$$x_n\to +\infty:= \forall c\in\mathbb{R}\ \exists m\in\mathbb{N}\ \forall n > m\ (x_n > c)$$
$$x_n\to -\infty:= \forall c\in\mathbb{R}\ \exists m\in\mathbb{N}\ \forall n > m\ (x_n < c)$$

Second Attempt:

$$\lim\sup_{n\to\infty} a_n = -\infty \implies \lim_{n\to\infty} \sup\{a_k:k\ge n\} = -\infty$$ $$\implies \forall c\in\mathbb{R}\ \exists m\in\mathbb{N}\ \forall n>m\ (\sup\{a_k:k\ge n\} < c)$$ $$\implies \forall c\in\mathbb{R}\ \exists m\in\mathbb{N}\ \forall n>m\ (a_k < c, k\ge n)$$
$$\implies \forall c\in\mathbb{R}\ \exists m\in\mathbb{N}\ \forall n>m\ (a_n < c) \implies \lim_{n\to\infty} a_n = -\infty$$
Similarly, $$\liminf_{n\to\infty} a_n = \infty \implies a_n\to\infty$$


问:
Whether two sets are equal when the mapping function is same
From this link
I understood when we can say that two functions are equal. My doubt is can I interpret the same as follows:
Let $X$ , $Y$ and $Z$ be three sets, Let $f : Z \rightarrow X$ and $g : Z \rightarrow Y$, be two functions, then if f = g, we can conclude that $X=Y$.

答:
No. If $f: [0,1]\to [0,1]$ and $g: [0,1]\to \mathbb R$ are defined by $f(x)=x$ and $g(x)=x$ for all $x$ then $f=g$.

PS What is involed here is a matter of conventions. Some authors may not consider my functions as equal.


问:
Show the inclusion $\mathcal{B}(\mathbb R)^{I}\supseteq Q$
Let $I$ be an uncountable set and consider the product sigma algebra on $\mathcal{B}(\mathbb R)^{I}$

I want to show that $\mathcal{B}(\mathbb R)^{I}=\{ \{f \in \mathbb R^{I}\lvert (f(t_{1}),f(t_{2}),...)\in A\}\lvert (t_{n})_{n\in \mathbb N}\subseteq I,\; A \in \mathcal{B}(\mathbb R)^{(t_{n})_{n \in \mathbb N}}\}=: Q$

I have already shown the inclusion $\mathcal{B}(\mathbb R)^{I}\subseteq Q$ by using the fact that $Q$ is a sigma-algebra and the preimages of projections of measurable sets indeed lie in $Q$

Apparently the direction $\mathcal{B}(\mathbb R)^{I}\supseteq Q$ is the easier one but I am lost on how to go about it.

Attempt: Assuming that I have  $M:=\{f \in \mathbb R^{I}\lvert (f(t_{1}),f(t_{2}),...)\in A\}$ such that $(t_{n})_{n\in \mathbb N}\subseteq I,\; A \in \mathcal{B}(\mathbb R)^{(t_{n})_{n \in \mathbb N}}$. Then if $A$ is of the form $A=\prod\limits_{n=1}^{\infty}A_{t_{i}}$ for $A_{t_{i}}\in \mathcal{B}(\mathbb R)$ clearly:

$M:=\mathbb R^{I\setminus (t_{n})_{n \in \mathbb N}}\times A= \bigcap\limits_{n=1}^{\infty}p_{t_{j}}^{-1}(A_{t_{j}})\in \mathcal{B}(\mathbb R)^{I}$. But my issue is that not every $A \in \mathcal{B}(\mathbb R)^{(t_{n})_{n \in \mathbb N}}$ is of the form $A=\prod\limits_{n=1}^{\infty}A_{t_{i}}$. But I know that the set

$$\mathcal{C}:=\{ \prod\limits_{n=1}^{\infty}A_{t_{i}}\lvert A_{t_{i}}\in \mathcal{B}(\mathbb R)\; \forall i \in \mathbb N\} $$ is a generator of $\mathcal{B}(\mathbb R)^{(t_{n})_{n \in \mathbb N}}$. Is there any way I could use this generator to extend the result, like one usually does with the monotone class theorem?

答: If my understanding of your notation is correct then $M$ is a Borel subset of $\Bbb R^I$, being a preimage of a Borel set $A$ by the (continuous) projection map from $\Bbb R^I$ to $\Bbb R^{\{t_n\}}$.


问:
How to find $B=50-\overline{(n-1)n}$ when $n=\operatorname{Im}(z+w)$ and it requires the simplification of each complex?
The problem is as follows:

First simplify the following complex numbers:

$z=3\left(\frac{1+i}{1-i}\right)+\left(\frac{1+i}{1-i}\right)^2+3\left(\frac{1+i}{1-i}\right)^3+\left(\frac{1+i}{1-i}\right)^4+...+3\left(\frac{1+i}{1-i}\right)^{4n+1}+\left(\frac{1+i}{1-i}\right)^{4n+2}$

$w=\left(\frac{1-2i}{2+i}\right)+\left(\frac{2-3i}{3+2i}\right)^2+\left(\frac{3-4i}{4+3i}\right)^3+\left(\frac{4-5i}{5+4i}\right)^4+...+\left(\frac{(4n+2)-(4n+3)i}{(4n+3)+(4n+2)i}\right)^{4n+2}$

After doing this find the value of $B$ from $n$:

$n=\operatorname{Im}(z+w)$

$B=50-\overline{(n-1)n}$

The alternatives given on my book are as follows:

$\begin{array}{ll}
1.&\textrm{5}\\
2.&\textrm{27}\\
3.&\textrm{38}\\
4.&\textrm{16}\\
\end{array}$

I'm very confused with how to approach this problem?. My best effort was to spot this:

$\left(\frac{1+i}{1-i}\right)=i$

This makes the earlier equation simplified a little bit, hence becoming the expression as follows:

$z=3\left(\frac{1+i}{1-i}\right)+\left(\frac{1+i}{1-i}\right)^2+3\left(\frac{1+i}{1-i}\right)^3+\left(\frac{1+i}{1-i}\right)^4+...+3\left(\frac{1+i}{1-i}\right)^{4n+1}+\left(\frac{1+i}{1-i}\right)^{4n+2}$

$z=3\left(i\right)+\left(i\right)^2+3\left(i\right)^3+\left(i\right)^4+...+3\left(i\right)^{4n+1}+\left(i\right)^{4n+2}$

But the thing is what to do from here?.

The rest, I mean for the second complex number seems that is more convoluted:

$w=\left(\frac{1-2i}{2+i}\right)+\left(\frac{2-3i}{3+2i}\right)^2+\left(\frac{3-4i}{4+3i}\right)^3+\left(\frac{4-5i}{5+4i}\right)^4+...+\left(\frac{(4n+2)-(4n+3)i}{(4n+3)+(4n+2)i}\right)^{4n+2}$

This second complex number doesn't seem to be easy to simplify.

I'm not really sure what sort of algebraic manipulation should be done here?. Can someone help me here?.

Since I'm not very savvy with this I really appreciate that someone could help me here with the most details as possible.

答:
The complicated expressions are only there to confuse you. You've noticed that $\frac{1+i}{1-i}=i$, so
$$z=[3(i)+(-1)+3(-i)+(1)]_n+3(i)+(-1)$$
where the part in square brackets, which is zero, repeats $n$ times so it may be ignored. Similarly, every fraction in brackets in $w$ is of the form $\frac k{ik}=-i$ for some $k$, so
$$w=[(-i)+(-1)+(i)+(1)]_n+(-i)+(-1)$$
and again the part in square brackets is repeated and is zero.

Thus $z=3i-1$, $w=-i-1$, $z+w=2i-2$, $n=2$ and $B=50-1\cdot2=48$.


问:
Exercise 2.7.1 of J. Norris, "Markov Chains"
I am working though the book of J. Norris, "Markov Chains" as self-study and have difficulty with ex. 2.7.1, part a.

The exercise can be read through Google books. My understanding is that the probability is given by (0,i) matrix element of exp(t*Q). Setting up forward evolution equation leads to differential difference equation which I was hinted admits no closed form solution (see my question on mathoverflow).

I obviously need an nudge in a right direction, and the exercise must admit a simple solution. Any directions are much appreciated.

答:
It seems you misread the question. One is not interested in the probability of the event $[X_t=i]$ for a given time $t$ but in the probability of the event that there exists a time $t$ such that $X_t=i$.

Here is a nudge...

One approach is to consider the sequence $(Y_n)_n$ of the successive different states visited by $(X_t)_t$. You could try to show that $(Y_n)_n$ is a discrete Markov chain, to compute its transition probabilities (these are very simple and, hint, they do not depend on the numbers $q_i$) and, finally, to compute the probability that there exists an integer time $n$ such that $Y_n=i$.


问:
proving that symplectic lie algebra is a subalgebra of GL
Suppose $S$ is n by n matrix over a field F. Define

$gl_S(n,F)=\{A \in gl(n,F): SA+A^TS= 0\}$

Show that this is a subalgebra of $gl(n,F)$

I get as far as:

$A \in gl_S(n,F)$ and $B \in gl_S(n,F)$

$S[AB] +[AB]^T S =SAB -SBA + (AB)^TS-(BA)^TS $
$=SAB + B^TA^TS - (SBA + A^TB^TS)$

I guess it's supposed to equal to zero somehow.

答:
Since $SA=-A^TS$ and $SB=-B^TS$,
$$
S[A, B]=SAB-SBA=-A^T SB+B^T SA=A^T B^T S-B^TA^TS=[A^T, B^T] S=-[A, B]^TS.
$$</p>


答:
By differentiation under the integral sign, the integral $I(a, b)$ satisfies the PDE
\begin{equation}
\frac{\partial}{\partial a}\frac{\partial}{\partial b} I(a,b) = I(a,b)
\end{equation}

The substitution $t = x \sqrt{a}$ shows that $I(a,b)$ has the form $a^{-1/2}\varphi(a^{1/2} b^{1/2})$ for some function $\varphi$.

Substituting this form into the PDE yields the ODE
\begin{equation}
\varphi''(u) = 4 \varphi(u)
\end{equation}
Hence $\varphi(u) = A e^{2 u}+ B e^{-2u}$ for some constants $A$ and $B$.

As $I(a,b)\to 0$ when $b\to \infty$, we have $A=0$

As $I(a, 0)= \sqrt{\pi/a}$, we have $B =\sqrt{\pi}$, hence $\varphi(u) = \sqrt{\pi}e^{-2u}$
It follows that
\begin{equation}
I(a, b) = \frac{1}{\sqrt{a}}\varphi(\sqrt{a}\sqrt{b}) = \frac{1}{\sqrt{a}}\sqrt{\pi}e^{-2\sqrt{a b}}
\end{equation}

Title: how to derive the steps of this intergral for the product of two exponential terms?

Body:
For homework, I was looking for closed form solutions of exponential functions from this wiki site.
https://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions

But I was not sure how to derive the steps for this particular integral,

$\int_{-\infty}^{\infty}e^{-ax^{2}}e^{-\frac{b}{x^{2}}}dx={\sqrt {\frac {\pi }{a}}}e^{-2{\sqrt {ab}}}\quad (a,b>0)$

I am fairly new to normal distribution but I assume this has to do with Taylor series..

Any pointers would be greatly appreciated!


问:
For what values of $l$ and $j$ (simultaneously) is the following congruence solvable?
Here is the congruence I want to know for what values of $l$ and $j$ (simultaneously) is it solvable?

$$ l (xp^s)^2 \equiv xp^{s}j p^{a-b} + yp^t\pmod{p^b} $$ where $x,y$ are not divisible by $p.$ and $s,a, b$ are any integers and $a \geq b.$

We have delt here what values of $j$ makes this congruence solvable? with the case when $l=0,$ but I need now to know for what values of $l$ and $j$ (simultaneously) is it solvable?

Could anyone help me in studying this?

答:
A slicker way of tackling this problem is to apply the following general result about the solution of equations of this type. This equation is symmetrical which allows a much easier case-by-case analysis.

Theorem 1

Let $M=\min\{A,B,C,D\}$. Then the equation
$$Xp^A+Yp^B+Zp^C \equiv 0\pmod {p^D}$$

can be solved with $XYZ \not\equiv 0\pmod p$ if and only if either

$\bullet \hspace{1cm}M=D$

$\bullet \hspace{1cm}M$ is equal to precisely two of $\{A,B,C\}$

$\bullet \hspace{1cm}M$ is equal to all three of $\{A,B,C\}$ and $p\ne 2$.

Proof

CASE 1. If $M=D$

Then each of $Xp^A,Yp^B$ and $Zp^C$ is divisible by ${p^D}$ whatever the values of $X,Y$ or $Z$.

CASE 2. If $M=A=B,M<C$

The equation is solved by, for example, $Y=Z=1,X=-1-p^{C-M}$. By symmetry, we have similar solutions when $M=A=C$ and $M=B=C.$

CASE 3. If $M=A=B=C,M<D$

If $p\ne 2$ then the equation is solved by, for example, $Y=Z=1,X=p^{D-M}-2$. However, if $p=2$ then $X+Y+Z$ has to be even and so at least one of $X,Y,Z$ is even. This contradicts the fact that $p=2$ cannot divide $XYZ$.

CASE 4. If $B,C$ and $D$ are all greater than $M=A$

Dividing the equation through by $p^M$ gives
$$X+Yp^{B-M}+Zp^{C-M} \equiv 0\pmod {p^{D-M}}$$ and then we have the contradiction that $p$ divides $X$.

By symmetry, we have similar contradictions when $M=B$ and $M=C.$

The transformation

Let $l=Lp^U$ and $j=Jp^V$ where $LJ \not\equiv 0\pmod p$. Then let $$A=a-b+s+V,B=t,C=2s+U,D=b.$$

Theorem 2

For non-negative integers $a\ge b,s$ and $t$, the equation
$$l (xp^s)^2 \equiv xp^{s}j p^{a-b} + yp^t\pmod{p^b}$$
can be solved with $xy \not\equiv 0\pmod p$ if and only if the equation
$$Xp^A+Yp^B+Zp^C \equiv 0\pmod {p^D}$$
can be solved with $XYZ \not\equiv 0\pmod p$.

Proof

If the $x,y$ equation can be solved then simply take $$X=xJ,Y=y,Z=-Lx^2.$$

Conversely, suppose we have a solution of the $X,Y,Z$ equation. Since $X$ and $L$ are coprime to $p$ we can solve
$$x \equiv -\frac {JZ}{LX}\pmod {p^D},y \equiv -\frac {J^2YZ}{LX^2}\pmod {p^D}.$$
Then $$  xp^{s}j p^{a-b} + yp^t-l(xp^s)^2=xJp^A+yp^B-Lx^2p^C$$
and this is
$$ -\frac {J^2Z}{LX^2}(Xp^A+Yp^B+Zp^C )\equiv 0\pmod {p^D}.$$

Conclusion

We can now see that the original equation can be solved if and only if the quantities $$A=a-b+s+V,B=t,C=2s+U,D=b$$ satisfy the conditions of Theorem 1. Let $v_p(n)$ represent the highest power of $p$ which divides the number $n$. You will notice that the precise values of $j$ and $l$ do not affect the existence of solutions; all that matters is $v_p(j)$ and $v_p(l)$ .

CASE 1. $t\ge b$,  $v_p(l)\ge b-2s$ and  $v_p(j)\ge 2b-a-s$.

CASE 2. There are three possibilities.

$v_p(l)> t-2s$ and  $v_p(j)=b-a-s+t$.

$v_p(l)< t-2s$ and  $v_p(j)=b-a+s+v_p(l)$.

$v_p(l)= t-2s$ and  $v_p(j)>b-a+t-s$.

CASE 3. If $p\ne 2$ then $t< b$,  $v_p(l)= t-2s$ and  $v_p(j)= b-a+t-s$.


问:
Application of the Portmanteau Theorem for weak convergence
I've got the following sequence of functions
\begin{align*}
F^N=(X^N_t-X^N_s)\prod_{i=1}^{n}f_i(X^N_{s_i}).
\end{align*}
for $n\in\mathbb{N}$, $(X_t^N)_t$ a stochastic process converging weakly to $X^*$ and $f_i$'s being continuous, bounded functions on $\mathbb{R}$ (i.e. $f_i\in C_b(\mathbb{R})$. By the continuous mapping theorem $F^N$ converges weakly.

Now that I have weak convergence, shouldn't there immediatly follow
\begin{align*}
\lim_{N\rightarrow\infty}\mathbb{E}[F^N]=\mathbb{E}[F^*],
\end{align*}
from the Portmanteau Theorem? The paper I got this from states that we need to make use of the boundedness and continuity of the $f_i's$ and I don't understand why this is necessary (although I see that it works).

答: Weak convergence does not give convergence of expectations (which may not even exist). On $(0,1)$ with Lebesgue measure let $f_n(x)=nI_{(0,\frac 1 n)}$. Then $f_n \to 0$ a,.s, hence weakly. But $Ef_n=1$ for all $n$. Portmanteau's Theorem is not useful in proving that expectations converge.


问:
How do I expand this product
How do I expand the following product:

\begin{equation}
\prod_{j=i+1}^n (1-\frac{1}{j})
\end{equation}

Wolfram Alpha tells me the answer is $i/n$, which is the answer I was looking for, but I have no idea how to derive this by hand.

答:
The product expands as
$$\frac i{\color{red}{i+1}}\cdot\frac{\color{red}{i+1}}{i+2}\cdots\frac{n-2}{\color{blue}{n-1}}\frac{\color{blue}{n-1}}n$$
which telescopes down to $\frac in$ (numerator and denominator of adjacent fractions cancel).

Title: How do I expand this product


问:
A Question about Doctoral Theses in Mathematics
This is most definitely a soft question, which I'm sure may get some negative attention, and perhaps even be voted closed. However, I genuinely would like to generate answers on this matter as it concerns anyone that decides to get a doctorate in mathematics. (Since I believe this is the best forum to do so, here goes...)

Question: Does a doctoral thesis in mathematics have to contain more than an abstract, a proposition and proof of a new and exciting result?

I strongly believe that brevity is beauty to mathematicians and see no problem with a thesis that contains a lengthy abstract (~350 words) to generate enthusiasm and then jumps right into the heart of a novel and far-reaching result. The proof provided is extremely condensed in the spirit of Zagier, leaving almost all details that can be found elsewhere to be found elsewhere, in the works cited. If the result has merit, I see no reason why it can't be submitted or published in such a manner. To support my stance, I offer the following condensed theses:

David Rector, "An Unstable Adams Spectral Sequence", MIT (1966), 9 pgs.

Burt Totaro, "Milnor K-Theory is the simplest part of algebraic K-theory", Berkeley (1989), 12 pgs.

Herman Buvik, A New Proof of Torelli's Theorem, NYU (1962), 12 pgs.

Eva Kallin "A non-local function algebra" Berkeley (1963), 13 pgs.

Edmund Landau "Neuer Beweis der Gleichung $\sum_{k = 1}^{\infty} \frac{\mu(k)}{k} = 0$", Berlin University (1899), 14 pgs.

Barry Mazur, "On Embeddings of Spheres", Princeton (1959), 26 pgs.

John F. Nash "Non-Cooperative Games", Princeton (1950), 27 pgs.

Kevin Walker, "An Extension of Casson's Invariant to Rational Homology Spheres", Berkeley (1989). 29 pgs.

Although I'd appreciate to hear from anyone worth giving his two cents, I'm specifically eager to hear from those senior members of the community.

Thanks!

答:
I'm not 100% sure that you yourself are the author of the thesis rather than the supervisor or some other interested party, but since you are offering to send it out by email I am probably safe in assuming the former.

In either case, I think you should consider carefully whether this is the best place for you to air a dispute with another faculty member. It seems likely that involved parties could recognize themselves or you and that this could be seen as unprofessional behavior.

In any case, what do you hope to achieve by asking this question? You are clearly hoping that M.SE users judge your own case, of which they know only your side, on its merits, rather than simply start a theoretical discussion about thesis length. Even if many or all of the answers agree with you, will this really get you any further in your dispute? The responses of strangers on the Internet are not universally considered to be valuable. If they are helpful for mathematics or programming questions, that is partly because they involve questions of information and verifiable fact.

You should also be aware that comparing your work to that done by a list of great mathematicians might not help your case either. This is frequently done by cranks. As you know, those theses might have been short because they were great, but they were not great because they were short.

Maybe a better approach would be to find some faculty member whom you know reasonably well and trust and try to get them to see your side of the story, and advise you on how to go forward. If you can't find anyone to take your side, you need to ask yourself is there is a reason for that.

If your result is great it will be recognized as such, regardless of how it is written up. It would be a shame if the matter became a distraction from your hard work, or if it led to the perception that you were arrogant or disrespectful to your senior colleagues.


问:
Term for repeated exponentiation?
Is there a term (or operation) for repeated exponentiation?

I.e. Repeated addition is multiplication, repeated multiplication is exponentiation, repeated exponentiation is X?

Is there even such a term for such a "higher order" operation, seeing as we could potentially deal with this by increasing the exponent?

答:
The word you are looking for is probably tetration

at least if you are looking for something like $$\underbrace{a^{(a^{(\cdot^{\cdot^{a)\cdots)}}}}}_n$$

Title: Term for repeated exponentiation?


问:
If $x^2\equiv 1 \pmod{n}$ and $x \not\equiv \pm 1 \pmod{n}$, then either $\gcd(x-1,n)$ or $\gcd(x+1,n)$ is a nontrivial factor of n
I'm reading elementary number theory and trying to understand the following problem: If $x^2\equiv 1 \pmod{n}$, $n=pq$, $p$ and $q$ are odd primes and $x \not\equiv \pm 1 \pmod{n}$, then either $\gcd(x-1,n)$ or $\gcd(x+1,n)$ is a nontrivial factor of $n$.

EDIT: Andreas Caranti wrote an updated, corrected version of my problem, so I wrote the definition again.

答:
I assume your primes are odd and distinct. Also, your statement is not correct as written, you have to add $x \not\equiv \pm 1 \pmod{n}$. The latter assumption implies that $\gcd(x-1, n), \gcd(x+1, n) < n$.

You have $n \mid x^{2} - 1 = (x-1)(x+1)$. Now if $\gcd(x - 1, n) = 1$, then $n$ divides $x + 1$, a contradiction, as we have taken $x \not\equiv -1 \pmod{n}$. So $\gcd(x - 1, n) > 1$. Similarly, $\gcd(x + 1, n) > 1$.

It follows that $\gcd(x-1,n)$ and $\gcd(x+1, n)$ are nontrivial, proper divisors of $n$, hence $p$ or $q$.


问:
Finding unique number and value of p.
Function = f(x) = $\frac{x^2+px+1}{x^2+x+1}.$

Q 1For any value of p , a unique number will be there in the range . Find it.

Q 2 If nine integers are in the range of f(x) , then solve for p.

Q3If  range of function is non negative , then find maximum no of integers in the range can be .

For the 1st one , i thought unique number as  neither a prime number nor a composite number. It has only one factor, i.e, the number itself. Is it correct ?

For the 2nd , I am not getting

For 3rd , no of integers will change as value of p will change . Then how to find the amswer

答:
Question 1

let $x=0$. No matter what the value of $p$ is, this value will be in common with all possible functions.

Question 2

Question 2 is asking the following:

If we have the function $f(x)=\frac{x^2+px+1}{x^2+x+1}$ and we look at range of $f(x)$, for some value of $p$ the range of $f(x)$ includes exactly $9$ integers. Find this value of $p$.

For question 2, note that
$$\frac{x^2+px+1}{x^2+x+1}=\frac{x^2+x+1-x+px}{x^2+x+1}=1+\frac{x(p-1)}{x^2+x+1}$$
I will assume that you want an integer value for $p$ (unless you want an inequality; if you do please specify), and that you mean exactly $9$ integers are in the range of $f(x)$.

If you actually look at this graph for different values of $p$ it is clear that the function always has a clear maximum and minimum point. We can find the coordinates of these stationary points by differentiating with respect to $x$:
$$\frac{d}{dx}\left(1+\frac{x(p-1)}{x^2+x+1}\right)=\frac{(p-1)(1-x^2)}{(x^2+x+1)^2}$$
after some rearranging and using the quotient rule. Now let this derivative be equal to $0$ for the stationary points.
$$\frac{(p-1)(1-x^2)}{(x^2+x+1)^2}=0\implies 1-x^2=0\implies x+\pm1$$
Hence the stationary points have coordinates $(1,f(1))$ and $(-1,f(-1))$ which comes to the coordinates $(-1,1-(p-1))$ and $(1,1+\frac{p-1}{3})$ which simplifies to
$$(-1,~2-p)~~\text{and}~~(1,\frac{p+2}{3})$$
If the range only includes exactly $9$ integers, then let the difference between the maximum point and minimum point of the function be $8$, as this is the smallest possible amount that could include $9$ integers. Hence
$$\frac{p+2}{3}-(2-p)=8\implies p+2-6+3p=24\implies 4p=28$$
so $p=7$. Check this indeed includes $9$ integers by finding the $y$-values of the coordinates of the stationary points for this value of $p$, and we are done.

Question 3

Let $f(x)>0$:
$$\frac{x^2+px+1}{x^2+x+1}>0$$
Note that $x^2+x+1>0$ for all real  $x$, so we can multiply through by $x^2+x+1$:
$$x^2+px+1>0$$
But $x^2+px+1=(x+\frac{p}{2})^2+1-\frac{p^2}{4}$. So if $f(x)>0$ for all values of $x$ then $\frac{p^2}{4}<1$
$$\implies (p+2)(p-2)<0$$
Hence $-2<p<2$.

Now plug in $p=2$ and $p=-2$ respectively into your general coordinates for the stationary points and see how many integers it includes.

Title: Finding unique number and value of p.


问:
Is the complement of a closed and nowhere dense set a dense set?
1)Suppose A is a closed and nowhere dense subset of X, is $A^c$ dense?

2)Suppose B is open and dense in X, is $B^c$ nowhere dense?

I tried proving both but it was long and messy so I am not sure it is right ... I wonder if there is a quick way to show this...

Also, is it necessary that A is closed or that B is open?

Here is what I did for the moment :

A closed --> cl(A) = A , and A is nowhere dense --> $int(cl(A))=int(A)=\emptyset$
so A contains no open set.
Take x in X, we want to show x is in $cl(A^c)$:
Take U open containing x, thus X is not entirely in A so $U\cap A^c$ is not the empty set so x is in $cl(A^c)$ so X=$cl(A)^c$

Thanks

答:
If $A$ is nowhere dense then $A^{c}$ is dense whether or not $A$ is closed. Proof: Suppose $A^{c}$ is not dense. Then there is an open set $U\neq \emptyset$ such that $U\cap A^{c}=\emptyset$ or $U \subseteq A\subseteq \overline A$. This is a contradiction to the fact that $A$ is nowhere dense.

If $B$ is open and dense then $B^{c}$ is closed and contains no interior points. (Why?)
Hence $B^{c}$ is  nowhere dense. But this implicatin fails if $B$ is not open: Take $B=\mathbb Q$.


问:
Full strength of Hausdorff axiom
Hi i am reading Topology by Munkres and there in one section under Hausdorff Spaces it is written that

One reason for our lack of interest in the $T_1$ axiom is the fact
that many of the interesting theorems of topology require not just
that axiom, but the full strength of the Hausdorff axiom.

My doubt is that what does he mean by the full strength of the Hausdorff axiom?. For reference i am writing the $T_1$ axiom definition as they've given:The condition that finite point sets be closed has been given a name called $T_1$ axiom.I am also attaching the screenshot where i have highlighted the part(Page 99)What are we missing here that he has to say "full strength of the Hausdorff axiom"?

答:
The technical meaning of "full strength" here is that a theorem requires the full strength of the assumption that a space is Hausdorff if it is false for every space that is not Hausdorff. So it does not suffice to assume anything that is "weaker" than the Hausdorff assumption.

A generalization of Theorem 17.10, which comes immediately after the bit you quote, is an example of such a theorem: a space $X$ is Hausdorff if and only if every net converges to at most one point in $X$. The "if and only if" means exactly that this statement fails in every non-Hausdorff space.

Here is another example: using the fact that compact Hausdorff spaces are normal, together with Urysohn's lemma, you can show that a compact space $X$ is Hausdorff if and only if continuous functions $f : X \to \mathbb{R}$ separate points (meaning that if $x \neq y$ then there exists a continuous function $f : X \to \mathbb{R}$ such that $f(x) \neq f(y)$). So this result requires the full strength of the Hausdorff assumption among compact spaces, and is false for every compact non-Hausdorff space.

I am actually not sure if Munkres intends the technical meaning of "full strength," though. A weaker version of what he means is that it does not suffice to assume a weaker separation axiom like $T_1$.

We can also give a precise technical meaning to the usage of "strength" and "weaker" here, which is often not made explicit. It refers to the following:

an assumption $A$ is stronger than an assumption $B$ (and $B$ is weaker than $A$) if $A$ implies $B$, meaning that in every case that $A$ is true, $B$ is also true; and

$A$ is strictly stronger than $B$ (and $B$ is strictly weaker than $A$) if $A$ strictly implies $B$, meaning that in every case that $A$ is true, $B$ is also true, and there is also at least one case in which $A$ is false and $B$ is true.

So to say that Hausdorff is a stronger assumption than $T_1$ is to say that it implies $T_1$, and to say that it is strictly stronger means that there exists a $T_1$ space which is not Hausdorff. "Strength" refers to the fact that stronger assumptions let you prove more facts, because you are proving them about a more restricted set of cases.

For another example of this usage of "strength," some statements are commonly proven using the axiom of choice but "do not require the full strength of this assumption," which formally means that there exists a model of ZF set theory in which the axiom of choice is false but these statements are still true. An example of such a statement is "a product of compact Hausdorff spaces is compact Hausdorff," which turns out to only require (and I think be equivalent to?) the ultrafilter lemma, which is known to be "strictly weaker" than the axiom of choice (which is equivalent to Tychonoff's theorem, that a product of compact spaces is compact).

Title: Full strength of Hausdorff axiom


问:
$\text{diag}(M^t,\ldots,M^t)=[\text{diag}(M,\ldots,M)]^t$ with $t$ is real and $M$ is a positive operator
Let $M$ be positive operator on a complex Hilbert space $E$.

Since the square root of a positive operator is unique, then

$$\begin{pmatrix} M^{1/2} &  \\
 & \ddots & \\
 & & M^{1/2}
 \end{pmatrix}
=\begin{pmatrix} M &  \\
 & \ddots & \\
 & & M
 \end{pmatrix}^{1/2}
.$$

Now if $t\in (0,\infty)$ be real number. Is
$$\begin{pmatrix} M^{t} &  \\
 & \ddots & \\
 & & M^{t}
 \end{pmatrix}
=\begin{pmatrix}M &  \\
 & \ddots & \\
 & & M
 \end{pmatrix}^{t}
?$$

答:
If $A$, $B$ are unital $C^\ast$-algebras and $\pi\colon A\to B$ is a unital $\ast$-homomorphism, then $\pi(f(x))=f(\pi(x))$ for every normal $x\in A$ and continuous function $f$ on the spectrum of $x$. If $f$ is a polynomial in $z$ and $\bar z$, this follows easily from the properties of a unital $\ast$-homomorphism, and for general continuous $f$ one can approximate by polynomials in $z$ and $\bar z$.

Now you can apply this result to the case when $B=M_n(A)$, $\pi(x)=\mathrm{diag}(x,\dots,x)$ and $f(\lambda)=\lambda^t$.


问:
Prove that $\Bbb{R^3}\setminus\Bbb{Q^3}$ is path connected in the euclidian metric space $\Bbb{E^3}$.
I understand the concept of one coordinate moving while the rest don't change, however I can't make up the  exact mapping that would prove this. Can anyone give me the concrete mapping?</p>

答:
Let $(a,b,c)$, $(d,e,f)$ be two points in $\mathbb{R}^3\setminus\mathbb{Q}^3$. Then at least one of $a,b,c$ is not rational. By symmetry, we may assume WLOG that it is $a$.

If $d$ is not rational, we can follow the following path which is entirely contained in  $\mathbb{R}^3\setminus\mathbb{Q}^3$:

$$(a,b,c) \to (a,\sqrt2,c)\to(d,\sqrt2,c)\to(d,e,f)$$
where $P\to Q$ signifies a straight line path from $P$ to $Q$.

If $e$ is not rational, we follow:
$$(a,b,c) \to (a,e,c) \to (d,e,f)$$

If $f$ is not rational, we follow:
$$(a,b,c) \to (a,b,f) \to (d,e,f)$$


答: Take $(a_1,a_2,a_3),(b_1,b_2,b_3)\in\Bbb R^3\setminus\Bbb Q^3$. Assume, for instance, that $a_1,b_3\notin\Bbb Q$. Then consider$$\begin{array}{rccc}\gamma\colon&[0,2]&\longrightarrow&\Bbb R^3\\&t&\mapsto&\begin{cases}\bigl(a_1,tb_2+(1-t)a_2,tb_3+(1-t)a_3\bigr)&\text{ if }t\in[0,1]\\\bigl((t-1)b_1+(2-t)a_1,b_2,b_3\bigr)&\text{ if }t\in[1,2].\end{cases}\end{array}$$Since $a_1,b_3\notin\Bbb Q$, $(\forall t\in[0,2]):\gamma(t)\in\Bbb R^3\setminus\Bbb Q^3$.

Title: Prove that $\Bbb{R^3}\setminus\Bbb{Q^3}$ is path connected in the euclidian metric space $\Bbb{E^3}$.

Body: I understand the concept of one coordinate moving while the rest don't change, however I can't make up the  exact mapping that would prove this. Can anyone give me the concrete mapping?</p>


问:
Solving $f(x+1)=xf(x)f\left(\frac1x\right)$
Recently, while scribbling around, I came up with the functional equation $f(x+1)=xf(x)f\left(\frac1x\right)$. It is somewhat similar to the functional equation for which the Gamma function is a solution, since $\Gamma(x+1)=x\Gamma(x)$, but it has the extra factor of $f\left(\frac1x\right)$.

If we introduce the restriction that $x\geq 1$, then I noticed that there is at least one solution (with this restriction put in place). That solution turns out to be, amazingly, $f(x)=\{x\}$, or the fractional part of $x$.

Edit: In fact, if $x\geq 1$, then $f(x)=\{x+a\}$ is a solution for all $a\in\mathbb{R}$ (I'm using the convention that $\{x\}=x-\lfloor x\rfloor$, where $\lfloor x\rfloor$ is defined as the greatest integer that is lower than $x$, therefore $a$ can be negative as well).

I have no idea how to approach this problem at all. The only thing that I was able to deduce is that for $n\in\mathbb{N}$, the following holds:

$$f(n)=f(1)\cdot (n-1)!\prod_{k=1}^{n-1}f\left(\frac1k\right)$$

I don't know if this helps in any way, but it might be a good start.

Knowing that when $x$ is restricted to be greater or equal to $1$, $\{x\}$ is a solution, I wonder if there can even exist a solution that is continuous. However, I don't know how to prove or disprove this, nor do I know how to find any solutions for the equation.

Is there a way to at least gain insight on how a possible solution for the equation must behave?

答:
I'll assume the restriction that $x \geq 1$ in the functional equation and consider functions with domain $\mathbb{R}^+$.

By applying the functional equation repeatedly we see that for any $x \geq 1$ and positive integer $n$, $f(x+n)=f(x)\prod_{k=0}^{n-1}(x+k)f(\frac{1}{x+k})$. Therefore if $f$ is defined on $(0, 2)$ this will extend uniquely to all of $\mathbb{R}^+$.

For arbitrary $f$ on $(0, 2)$, its extension will satisfy the functional equation. Note that $f(x)=\{x\}$ is a "nice" solution because the $(x+k)$ and $f(\frac{1}{x+k})$ terms cancel out above and it is periodic with degree $1$. In fact, if we define any $f$ with
$$f(x)=\begin{cases} x & 0 < x < 1 \\ g(x) & 1 \leq x < 2 \end{cases}$$
for arbitrary $g$, then $f$ extends to $g$ periodized with period $1$ on $[1, \infty)$. This will be continuous iff $g(1)=g(2)=1$. For example, the function $f(x) = \min(x, 1)$ is a continuous solution to the functional equation.

By the functional equation, $f(2) = f(1+1) = f(1)^2$. Any function $f$ that is continuous on $(0, 2)$ and such that $\lim_{x \to 2} f(x) = f(1)^2$ will extend continuously to $\mathbb{R}^+$. Of course $f(x)=0$ is a trivial continuous solution. Another solution from extending the constant function $f(x)=1$ on $(0, 2)$ to $\mathbb{R}^+$ is

$$f(x) = \begin{cases} 1 & x < 1 \\ \frac{\Gamma(x)}{\Gamma(1+\{x\})} & x \geq 1 \end{cases}$$

You can construct any continuous solution with this method.

The Bohr-Mollerup theorem suggests that you might find a more interesting family of solutions by considering log-convexity and the boundary condition $f(1)=1$, but I didn't see a nice way to use this.

Also, note that if $f$ and $g$ satisfy the functional equation, then $h = f/g$ satisfies $h(x+1) = h(x)h(1/x)$. This functional equation will again extend from $(0, 2)$ to $\mathbb{R}^+$ by $h(x+n) = h(x)\prod_{k=0}^{n-1}h(\frac{1}{x+k})$ for $x \geq 1$ and positive integer $n$, and will have the same continuity condition as above. Maybe with some stricter conditions this functional equation or $H = \log h$ satisfying $H(x+1) = H(x) + H(1/x)$ might have some interesting solutions.