问:
Orthonormal bases: Does a complex vector 3-space have twice as many bases as a real vector 3-space?
Apologies in advance if the title sounds weird. Anyway,

I'm brushing up on my math, reading "Modern Mathematical Analysis."  Early on it discussed vector spaces, mainly 3-space, and it wasn't difficult: the basis e$_1$ is the vector from 0 to 1 on the x-axis, etc.

Now it is extending to complex spaces, but it still talks in terms of e$_i$ without breaking it down into real and complex. So my guess is: e$_1$ is the vector from 0 to 1 on the real x-axis, e$_2$ is the vector from 0 to 1$\cdot$i on the imaginary x-axis, and so forth (so four more). Is that correct?

Also, regarding the title: is the complex analog of real 3-space still called 3-space? Or must it be called 6-space?

答:
No, the dimension of a vector space is just the number of basis vectors it has. So a 3 dimensional complex vector space has exactly 3 basis elements for any basis. In particular any basis of $\mathbb{C}^3$ has 3 elements. It is true that there is a sense in which $\mathbb{C}^3$ is related to $\mathbb{R}^6$ but that does not make it 6 dimensional as a vector space.  However, your intuition might tell you that its geometry is 6 dimensional. However, importantly that isn't a vector space quality.

For a concrete example, let's look at the 1 dimensional vector space of $\mathbb{C}$ over itself. This is indeed 1 dimensional, not 2 dimensional, so then {1, $i$} should not be a basis of this. and indeed it isn't, because 1 and i are not linearly independent, as $1=i\cdot(-i)$

However, it should be noted if its a complex vector space, then we are looking at it as a vector space over the field $\mathbb{C}$, we can also consider $\mathbb{C}$ or even $\mathbb{C}^3$ as a vector space over $\mathbb{R}$ in which case, the first one is 2 dimensional, and the second is 6 dimensional.


问:
what is the relation of smooth compact supported funtions and real analytic function?
What is the major difference between real analytic and test function (smooth compact supported functions). Can we find a real analytic function $f$ on $\mathbb R^n$ which is also smooth compact supported? If not then need a proof.

答:
A test function is a smooth and compactly supported function, while a real analytic function is smooth and given by a power series uniformly converging at every point.

Perhaps I will only give a hint as an answer to your second question: A test function must be equal to zero everywhere outside some compact set. If it were also analytic, then what would the coefficients of its power series have to be?

ContentLicense: CC BY-SA 4.0

ContentLicense: CC BY-SA 3.0


问:
$\overline{\mathrm{span} \{e_{2n} \ \mid \ n \}} + \overline{\mathrm{span} \{ e_{2n} + \frac{1}{n+1}e_{2n+1} \} }$ is dense but not the entire space

Regarding my previous question: $\overline{\mathrm{span} \{e_{2n} \ \mid \ n \}} \cap \overline{\mathrm{span} \{ e_{2n} + \frac{1}{n+1}e_{2n+1} \} } = 0$, but their sum is dense.

In the context of the above question: $\mathcal{H}$ is a separable Hilbert space, with a total orthonormal set $\{e_n \}_n$. I have able to prove that $$\overline{\mathrm{span} \{e_{2n} \ \mid \ n \}} \cap \overline{\mathrm{span} \{ e_{2n} + \frac{1}{n+1}e_{2n+1} \} } $$ is an internal direct sum and that $$\overline{\mathrm{span} \{e_{2n} \ \mid \ n \}} + \overline{\mathrm{span} \{ e_{2n} + \frac{1}{n+1}e_{2n+1} \} } $$ is dense in $\mathcal{H}$. However, I would now like to prove that $$\overline{\mathrm{span} \{e_{2n} \ \mid \ n \}} + \overline{\mathrm{span} \{ e_{2n} + \frac{1}{n+1}e_{2n+1} \} } \neq \mathcal{H}. $$

I have tried to do so by considering the sequence $$x_n = \frac{1}{n+1}e_{2n+1} = \left( e_{2n} + \frac{1}{n+1}e_{2n+1} \right) - e_{2n},$$ and at its sequence of partial sums $$s_n = \sum_{i=1}^n x_i. $$

The sequence $(s_n)_n$ does converge in $\mathcal{H}$ as, the sequence $(1/n)_n$ is an element of $\ell^2$. However, is its limit in the direct sum from above?

答:
You are on the right track: No, the limit is not contained in the direct sum. Which proves the question claim.

The proof depends on a fact in Banach space theory:

Let $B$ be a Banach space and consider closed subspaces $X,Z\subset B$. If $B$ is algebraically isomorphic to the direct sum $X\oplus Z$, then the isomorphism is also topological, i.e., an isomorphism of Banach spaces.

This is implied by the Inverse mapping theorem which states "The inverse of a continuous linear bijection between Banach spaces is continuous." (being a corollary to the Open mapping theorem).

Furthermore, the projectors onto the subspaces are then continuous operators on $B$.

Generalising your choice of $(s_n)$, let $\lambda\in\ell^2(\mathbb N)$ satisfy $\lambda_{2n+1}\neq 0\,$ for infinitely many $n$, but $\lambda_{2n}=0$ for all $n$.
This gives
$$v\:=\:\sum_{n=0}^\infty\lambda_{2n+1}e_{2n+1}\in\mathcal{H}$$
and it is shown, after appropriate choice of $\lambda$, that $v\notin F+G\,$ where $F,G$ denote the closed subspaces in question.

Look at the partial sums
$$\begin{align}v_N & \:=\:\sum_{n=0}^N\lambda_{2n+1}e_{2n+1}\:=\:
\sum_{n=0}^N\lambda_{2n+1}\big[(n+1)\big(e_{2n}+\tfrac1{n+1}e_{2n+1}\big)
 -(n+1)e_{2n}\big]\\[1ex]
 & \:=\:\sum_{n=0}^N\lambda_{2n+1}\sqrt{(n+1)^2+1}\; \frac{e_{2n}+\tfrac1{n+1}e_{2n+1}}{\sqrt{1+\frac1{(n+1)^2}}}
 \quad-\quad\sum_{n=0}^N\,(n+1)\lambda_{2n+1}\,e_{2n}\end{align}$$
which are finite linear combinations, thus certainly $v_N\in F+G$.

It is shown in another post that $F\cap G=\{0\}$, so that we really deal with a direct sum  decomposition $F\oplus G$ here. Then the splitting in the last formula line into $F$- and $G$-components is uniquely determined, and these components are the images under the two projections onto $F$ and $G$ of $v_N$.

If $F\oplus G$ were all of $\mathcal H$, then the projections are continuous by the above fact, and the convergence $v_N\rightarrow v$ implies convergence in $F$ and $G$ separately.

Now many possible choices for $\lambda\in\ell^2(\mathbb N)$ carry on sabotage of convergence in $F$ and $G$. Your choice made in the question  is one of them since the second summand then becomes $\sum_{n=0}^N e_{2n}\,$.

This proves $F\oplus G\not= \mathcal H.\qquad\blacksquare$


问:
Finding all the points on an elliptic curve over P(GF(16))
GF(16) is constructed using p(x) = x^4+x+1 and let β = x (mod p(x)). E is the elliptic curve over P(GF(16)) that satisfies y^2 + xy = x^3+(β^4)x^2+1. How do you figure out all the points on E?

Is it sufficient to solve y^2 + xy = x^3+(β^4)x^2+1 over the integers, in which case (-1,0), (-1,1), (0,1), (0,-1), (9,-44) and (9,35) are the points?

答:
$k = \Bbb{F}_2[\beta]/(\beta^4+\beta+1)$ is the field with 16 elements. 14 of its elements are not represented by integers.

Your affine elliptic curve is $$E = \{ (x,y)\in k^2, y^2+xy=x^3+\beta^4 x^2+1\}$$ and I guess you want to look at its projective closure $$C=\{ [x:y:z]\in \Bbb{P}^2(k), zy^2+zxy=x^3+\beta^4 zx^2+z^3\}$$
You can try all the $(16^3-1)/15$ points by hand, or use a CAS such as http://magma.maths.usyd.edu.au/calc/

```<code> k<b> := GaloisField(16);
 b^4+b+1 eq 0;
 C := EllipticCurve([k|1,-b^4,0,0,1]);
 Points(C);
</code>```

Body:
$k = \Bbb{F}_2[\beta]/(\beta^4+\beta+1)$ is the field with 16 elements. 14 of its elements are not represented by integers.

Your affine elliptic curve is $$E = \{ (x,y)\in k^2, y^2+xy=x^3+\beta^4 x^2+1\}$$ and I guess you want to look at its projective closure $$C=\{ [x:y:z]\in \Bbb{P}^2(k), zy^2+zxy=x^3+\beta^4 zx^2+z^3\}$$
You can try all the $(16^3-1)/15$ points by hand, or use a CAS such as http://magma.maths.usyd.edu.au/calc/

 k<b> := GaloisField(16);
 b^4+b+1 eq 0;
 C := EllipticCurve([k|1,-b^4,0,0,1]);
 Points(C);


问:
Show that $\lim\limits_{n\to\infty}\frac{2}{n(n+1)}\left(nz_1+(n-1)z_2+\cdots+2z_{n-1}+z_n\right)=z_0$

Let $(z_{n})$ be a sequence of complex numbers converging to $z_0$. Show that
$$\lim_{n\to\infty}\frac{2}{n(n+1)}\left(nz_1 + (n-1)z_2 + \cdots + 2z_{n-1} + z_n\right) = z_0$$

I try this using

$$a_{n} = \frac{2}{n(n+1)}\left(nz_1 + (n-1)z_2 +...+2z_{n-1}+ z_n\right) - z_{0}$$

and trying to demostrated that $\lim a_n = 0 $ but I'm stuck in a triangular inequality. How can I prove this?.

答:
You see
$$\frac{2}{n(n+1)}( nz_1+...+z_n) =\frac{2}{n(n+1)}(n+1)(z_1+z_2+...+z_n) -\frac{2}{n(n+1)}(z_1+2z_2+...+nz_n) $$
I think it's now clear for you to do the rest.

Update
Hm, I think I should put in a more clear representation.

So if you define $b_n=z_n-z_0$, you can see that $\lim |b_n| =0$ and
$$\| \frac{2}{n(n+1)}( nz_1+...+z_n) -z_0 \| \le 2\frac{ |b_1|+...+|b_n|}{n}+ \frac{ |b_1|+...+n|b_n|}{1+...+n}$$
From which you can apply Cesaro's theorem.


问:
Find all $M \in \mathbb{N}_{>0}$ s.t. $0 \equiv M^{n+1} + 12^{2n-1} \pmod{133}, \forall n \in \mathbb{N}_{>0}$
NB?: Holds for M = 11.

Not sure how to approach the above question?

答:
how to approach the above question?

Here's a way.  In order for $M$ to qualify, it would have to work for $n=1$, so we'd have $M^2+12\equiv0\bmod133$.  That happens iff $12+M^2\equiv0\bmod7$ and $12+M^2\equiv0\bmod19$; equivalently, $M^2\equiv2\bmod7$ and $M^2\equiv7\bmod19$; equivalently, $M\equiv\pm3\bmod7$ and $M\equiv\pm8\bmod19$; equivalently, $M\equiv 11, 46, 87, $ or $122\bmod133$.  We have already narrowed down the possibilities for $M$ considerably.  Now check if any of those $M$ values works for $n=2$, and, if one does, see if you can prove it works for all $n\in \mathbb N_{>0}$.

P.S.  After writing the above, I thought of this alternative approach.
In order for $M$ to qualify, it would have to work for $n=\phi(133)=108$.    Then we'd have $M^{109}+12^{215}\equiv0\bmod133$, which is equivalent to $M\equiv-144^{108}12^{-1}\equiv-(-11)=11.$  Now you just have to show that, if $M\equiv11\bmod133$, then it works for all $n\in \mathbb N_{>0}$, and it does, because $11^{n+1}+12^{2n-1}=121\cdot11^{n-1}+12\cdot144^{n-1}\equiv133\cdot 11^{n-1}\equiv0\bmod133$.


问:
Taylor vs. Laurent series.
So I'm trying to figure out methods to get a Laurent series instead of a Taylor series. Here is the initial problem:

$$\frac{1}{(z-1)}$$

So if I'm trying to find the Laurent series and I know they need negative exponents, what's the strategy? I know the answer is to factor out a $\frac{1}{z}$ but how would I get there?

$$\frac{1}{(z-1)} = \frac{1}{z} \cdot \frac{1}{1 - \frac{1}{z}} = \frac{1}{z}(1 + \frac{1}{z} + \frac{1}{z^2} + ... )$$

The above Laurent series (it has negative exponents) converges when $|z| > 1$ right?

And what's the corresponding Taylor series representation? Is the takeaway here that you can represent the same function with two different series?

答:
We discussed this before, with regards to the function $1/(1-z)$.

You're apparently wondering about uniqueness.  The Laurent (or Taylor) series around a given point can indeed be shown to be unique in a given region.

Title: Taylor vs. Laurent series.


问:
How to find Time when you know the Distance and Acceleration
The Distance is 0.1693 meters and the Acceleration is 10 m/s^2. What equation would I use to find time this certain question.</p>

答:
We assume the object starts at rest, so at $t=0$ we have $d(0)=0, v(0)=0, a(0)=10$.

so the initial displacement, and velocity is zero.

then we solve for $v(t)=\int_0^ta(x)dx=10t$ This is a function which gives us the velocity at a time $t$ in seconds. Then we can solve for $d(t)=\int_0^tv(x)dx=\int_0^t10xdx=\frac{10t^2}{2}$ which gives us the distance after $t$ seconds. Then we set $d(t)=0.1693$ and solve for $t$.

the reason we took both integrals from 0 is because of the initial conditions were assumed to be 0. If we had some other initial position or initial velocity, we would take the integrals from there.


答:
If by "test function" you mean a smooth function with compact support (as opposed to a Schwarz function), then the answer is no. Consider an analytic function $f \in C_c^\infty(\mathbb R^n)$ and a point $x$ outside of its support. Because $f$ is analytic, it must equal its own Taylor series developed at $x$, i.e. $0$, and since the radius of convergence is infinite, $f$ must be the zero function.

The upshot of this is that analytic functions are very rigid objects. Loosely speaking, perturbing an analytic function in a single point causes global changes to "ripple" out from the perturbation in order to preserve analyticity.

Title: what is the relation of smooth compact supported funtions and real analytic function?

Body: What is the major difference between real analytic and test function (smooth compact supported functions). Can we find a real analytic function $f$ on $\mathbb R^n$ which is also smooth compact supported? If not then need a proof.


问:
Computing the 0-dim cohomology group of a connected simplicial complex with coefficients in $\mathbb{Z}_{2}$
Let $L$ be a connected locally finite simplicial complex with coefficients in $\mathbb{Z}_{2}$. I want to prove that $H^{0}(L)=\mathbb{Z}_{2}$. We know that $H^{0}(L)$= $Ker\delta_{1} / Im\delta_{0}=Ker\delta_{1}$ and $\delta_{1}: Hom(C_{0},\mathbb{Z}_{2})\to Hom(C_{0},\mathbb{Z}_{2})$ with $\delta_{1}(f)=fd_{1}$ ($d_{1}$ the boundary). I also know that we can see the elements of $Hom(C_{m},\mathbb{Z}_{2})$ i.e. the $m$-dim cochains, as the subsets of the set of all m-simplices but I am not sure how can I use that to prove the statement. Any help? Why $Ker\delta_{1}\cong \mathbb{Z}_{2}$? Also if $L$ has infinitely many $0$-simplices why $H_{f}^{0}=0$? ($H_{f}^{0}$ the restriction in finite cocycles modulo coboundaries of finite cochains.) Thank you in advance.

答:
In the cochain complex $$ 0 \to \text{Hom}(C_0(L), \mathbb{Z}/2\mathbb{Z}) \to \text{Hom}(C_1(L), \mathbb{Z}/2\mathbb{Z}) \to \dots,$$ the group of 0-chains $C_0(L)$ is simply the free abelian group generated by all points of $L$, and so $\text{Hom}(C_0(L), \mathbb{Z}/2\mathbb{Z}) = \text{Fun}(X,\mathbb{Z}/2\mathbb{Z})$.
For $\phi \in \text{Hom}(C_0(L), \mathbb{Z}/2\mathbb{Z})$ and $\sigma: \Delta^1 \to L$ a 1-simplex, we have $$ \delta^1 \phi (\sigma) = \phi (\partial \sigma) = \phi (\sigma(v_1)) - \phi(\sigma(v_0)),$$ so in other words $\phi \in \ker \delta^1$ if and only if $$ \phi(\gamma(1)) = \phi(\gamma(0))$$ for all paths $\gamma: [0,1] \to L$.

In this way, $H^0(L;\mathbb{Z}/2\mathbb{Z})$ is the space of all functions $L \to \mathbb{Z}/2\mathbb{Z}$ which are constant on every path component.
In particular then, if $L$ is path connected then its zeroth cohomology group is $\mathbb{Z}/2\mathbb{Z}$.

Note that none of this required the simplicial complex structure.


答:
First note that $12^3 \equiv -1 \pmod{133}, 12^6 \equiv 1 \pmod{133}$, therefore $-12^{2n-1}$ cycles through $121, 1, 11, 121, 1, 11, \cdots, \pmod{133}$, and we must have $$M^2\equiv 121, M^3\equiv 1, M^4 \equiv 11 \pmod{133}$$
Then it follows immediately that $M\equiv M \cdot M^3 = M^4 \equiv 11 \pmod{133}$, and we can easily check that $M^{n+1}$ cycles through $121, 1, 11, 121, 1, 11, \cdots, \pmod{133}$.

Conclusion: $M=11+133(k-1), \forall k \in \mathbb N$.

Title: Find all $M \in \mathbb{N}_{>0}$ s.t. $0 \equiv M^{n+1} + 12^{2n-1} \pmod{133}, \forall n \in \mathbb{N}_{>0}$

Body:
NB?: Holds for M = 11.

Not sure how to approach the above question?


问:
limits proof by induction
Let $ \{a_n\}_{n \geq 0},\{b_n\}_{n \geq 0}  $ be two sequences. The sequences are given as $ b_n:= \dfrac{a_{n-1}+b_{n-1}}2 $ $a_n = \frac{c}{b_{n}} $ for $n \geq 1$ with $b_0 = b $ and $a_0 = \dfrac{c}{b}$.

I know that: $$ \frac{a_{0}+b_{0}}{2} > \sqrt{a_{0}b_{0}} $$

Now I want to prove that for all integers n: $$ a_n < \sqrt{c} < b_n  $$
I wanna prove this by induction over $n$.

For the base case: $n=1$

$$ a_1 = \frac{c}{b_{1}} = \frac{c}{0.5(a_{0}+b_{0})} < \frac{c}{\sqrt{a_{0}b_{0}}} = \frac{c}{\sqrt{c}} = \sqrt{c} = \sqrt{a_{1}b_{1}} < {0.5(a_{1}+b_{1})} = {0.5(\frac{c}{b_1}+b_{1})} = {0.5(\sqrt{c}+b_{1})}  $$

But for me it seems to go nowhere :/

Can somebody help me? :)

答:
There are several issues with your question. First, you should require $b \gt 0$ and $c \ge 0$ for it to work properly (e.g., if $c \lt 0$, then $\sqrt{c}$ is not even a real number). Second, your $\lt$ should be $\le$, e.g., if $b = c = 1$. Third, what you want to prove for "all integers $n$" should be for positive integers $n$ since your relation doesn't necessarily hold for $n = 0$ (e.g., $b = 1$ and $c = 2$).

This means you want to prove that for all integers $n \ge 1$ that

$$a_n \le \sqrt{c} \le b_n \tag{1}\label{eq1A}$$

Note you can prove this directly. However, to do it using induction, start with your base case of $n = 1$. As you stated, using the inequality of arithmetic and geometric means gives

$$b_1 = \frac{a_{0} + b_{0}}{2} \ge \sqrt{a_{0}b_{0}} = \sqrt{c} \tag{2}\label{eq2A}$$

Using $a_1 = \frac{c}{b_1} \implies a_1 b_1 = c$ and multiplying both sides above by $a_1$ (note $a_1 \ge 0$ so the inequality doesn't change) gives, for $c \gt 0$, that

$$\begin{equation}\begin{aligned}
a_1 b_1 & \ge a_1\sqrt{c} \\
c & \ge a_1\sqrt{c} \\
\sqrt{c} & \ge a_1
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$

Note if $c = 0$, then $a_n = 0$ for all $n \ge 0$ so \eqref{eq3A} (and \eqref{eq5A} below) still hold. This confirms the base case.

For the inductive step, assume \eqref{eq1A} is true for $n = k$ for some integer $k \ge 1$. Since $a_k b_k = c$, using the AM-GM inequality again gives, similar to \eqref{eq2A},

$$b_{k+1} = \frac{a_{k} + b_{k}}{2} \ge \sqrt{a_{k}b_{k}} = \sqrt{c} \tag{4}\label{eq4A}$$

Basically repeating the procedure used with \eqref{eq3A} gives

$$\begin{equation}\begin{aligned}
a_{k+1}b_{k+1} & \ge a_{k+1}\sqrt{c} \\
c & \ge a_{k+1}\sqrt{c} \\
\sqrt{c} & \ge a_{k+1}
\end{aligned}\end{equation}\tag{5}\label{eq5A}$$

This shows \eqref{eq1A} is also true for $n = k + 1$ so, by induction, it's true for all $n \ge 1$.

Title: limits proof by induction


问:
Proof about $T^2 = T\circ T$
I have the following statement to prove:

Proof if $T$ is a linear map defined as $T: V\to V$ with $Dim(V)=n$
and $T^2 = T \circ T=I_V$ with $I_V$ the identity linear map, then $Ker(T-I_V) \cap Ker(T+I_V) = \{0\}$

My attempt was:

If $\lambda_1, \lambda_2$ are different eigenvalues, then we have that $W_{\lambda_1} \cap W_{\lambda_2} = \{0\}$ and since $W_{\lambda_1}=Ker(T-\lambda_1I_n)$, $W_{\lambda_2}=Ker(T-\lambda_2I_n)$ , this will equivalent to prove that there exists eigenvalues equal to $\lambda_1=1, \lambda_2=-1$.

So i need to prove that there exists at least one $v_1,v_2 \in V$ such that $T(v_1)=v$ and $T(v_2)=-v_2$.

But from here i can't do more.

I tried to find from the information $T(v)\cdot T(v) = T(T(v))$ that $T(v)=v$ but i could not.

答: In general $\ker(A)\cap \ker(B)\subseteq \ker(B-A)$. Apply this to $A=T-I_V$ and $B=T+I_V$.

Title: Proof about $T^2 = T\circ T$


问:
Asymptotic Variance of MLE Exponential
Suppose we have a random sample $(X_{1},.....,X_{n})$, where $X_{i}$ follows an Exponential Distribution with parameter $\lambda$, hence:

$F(x) = 1 - exp(-\lambda x)$

$\mathbb{E}(X_{i}) = \dfrac{1}{\lambda}$

$ Var(X_{i}) =\dfrac{1}{\lambda^{2}}$

I know that the MLE estimator $\hat{\lambda} = \dfrac{n}{\sum_{i=1}^{n}X_{i}}$, asymptotically follows a normal distribution, but I'm interested in his variance.
So, since $\sqrt n (\hat{\lambda} - \lambda)\stackrel{D}{\rightarrow} \mathcal{N}(0, \sigma^{2}) $

I was thinking about dealing with:

\begin{align}
\sigma^{2} &= Var\left[\sqrt n (\hat{\lambda} - \lambda)\right]\\
&= n Var (\hat{\lambda})\\
&= n Var \left[ \dfrac{n}{\sum_{i=1}^{n}X_{i}} \right] \\
&= n^{3} Var \left[ \dfrac{1}{\sum_{i=1}^{n}X_{i}}\right]
\end{align}

But I don't know how to proceed from here. I think the sum of exponentials follows a gamma with parameters $(n,\lambda)$, and that $\dfrac{1}{\sum_{i=1}^{n}X_{i}}$ then follows an inverse gamma?

答:
Following the discussion with @tommik, I am still struggling with something:

Since $\bar{X}$ is the sample mean and given the properties of $(X_{1},....X_{n})$ in the problem set, the Central Limit Theorem tell us that:

$\sqrt{n}\left(\bar{X} - \dfrac{1}{\lambda}\right) \stackrel{D}{\rightarrow}  \mathcal{N}\left(0, \dfrac{1}{\lambda^{2}}\right)$

We can apply the Delta Method:

Consider the function $h(x) = x^{-1}$, where $h'(x)= -x^{-2}$, therefore:

$\sqrt{n}\left(h(\bar{X}) - h(\frac{1}{\lambda}) \right) \stackrel{D}{\rightarrow}  \mathcal{N}\left(0, \overbrace{\left[h'(\frac{1}{\lambda})\right]^{2} \dfrac{1}{\lambda^{2}}}^{\sigma^{2}}\right)$

$\sqrt{n}\left(\hat{\lambda} - \lambda\right) \stackrel{D}{\rightarrow}  \mathcal{N}\left(0, \left[h'(\frac{1}{\lambda})\right]^{2} \dfrac{1}{\lambda^{2}}\right)$

Where notice that the argument inside of the function $h'()$ is $\dfrac{1}{\lambda} = \lambda^{-1}$.

Let's find out what the variance $\sigma^{2}$ is:

\begin{align*}
\sigma^{2} &= \left[h'\left(\frac{1}{\lambda}\right)\right]^{2} \dfrac{1}{\lambda^{2}}\\
&= \left[ - \left(\dfrac{1}{\lambda}\right)^{-2} \right]^{2} \dfrac{1}{\lambda^{2}}\\
&= \dfrac{\lambda^{4}}{\lambda^{2}}\\
&= \lambda^{2}\\
\end{align*}

So, I don't know how to conciliate the comments and answers above.

Title: Asymptotic Variance of MLE Exponential


问:
Compact submanifolds of $\mathbb{R}^n$ without boundary
I'm having a little trouble seeing how to do Exercise 7.5 in Lee Smooth Manifolds:

Let $M$ be a smooth compact manifold. Show there is no submersion $F:M\rightarrow\mathbb{R}^k$ for any $k>0$.

If $F:M\rightarrow\mathbb{R}^k$ were a submersion, then $\dim(M)\geq k$. This rules out things like $\mathbb{S}^{k-1}\hookrightarrow\mathbb{R}^k$. Approaching it the other way, $\mathbb{B}^k\hookrightarrow\mathbb{R}^k$ is a submersion, but the open ball $\mathbb{B}^k$ is not compact. It seems like what's going on is that, since the image of $F$ would be a compact, hence closed, subset of $\mathbb{R}^k$, if $F(M)$ were "$k$-dimensional" it would require $M$ to have been a manifold with boundary, which isn't allowed. However, I'm not sure how to fill in the gaps here / make it rigorous.

On a possibly related note: Is it possible to have an immersed compact $k$-submanifold of $\mathbb{R}^k$?

答: Submersions are open maps; but the image of $M$ is compact in a Hausdorff space, and hence closed as well. So it's a clopen nonempty set. Since $\mathbf{R}^n$ is connected, it's the whole thing. But then $\mathbf{R}^n$ is the quotient of a compact space, so it's compact, which is not true.</p>


问:
For a (not necessarily affine) scheme $X = \bigcup_{i=1}^n X_{f_i}$, does $(f_1, \ldots, f_n) = (1)$ in $\mathcal{O}_X(X)$?
This is of course true in the affine case, so it seems like it should be true in general, because $\mathcal{O}_X(X)$ should be "smaller" for a non-affine scheme than for a similar affine scheme (e.g. all global sections over a projective scheme are constants).  Just to be clear on notation:

$X$ is a scheme, not necessarily affine.

$f_1, \ldots, f_n \in \mathcal{O}_X(X)$.

$X_f := \{ x \in X \; | \; f_x \not \in \mathfrak{m}_{X, x} \}$, where $\mathfrak{m}_{X, x}$ is the maximal ideal of the stalk $\mathcal{O}_{X, x}$

$X = X_{f_1} \cup \cdots \cup X_{f_n}$

Does it follow that $(f_1, \ldots, f_n) = (1)$ in $\mathcal{O}_X(X)$?

We have $(f_1|_U, \ldots, f_n|_U) = (1)$ in $\mathcal{O}_X(U)$ where $U \subset X$ is open affine, but I don't see how to extend this to all of $X$.

答:
This is a long comment, which might lead to a counterexample.

The question is equivalent to: Let $f : \mathcal{O}_X^n \to \mathcal{O}_X$ be a surjective homomorphism of quasi-coherent modules on $X$. Is it also surjective on global sections? If $K$ denotes the kernel of $f$, the long exact sequence of cohomology groups associated to (*) $0 \to K \to \mathcal{O}_X^n \to \mathcal{O}_X \to 0$ tells us that a sufficient condition is that $H^1(X,K)=0$. Note that $K$ is a vector bundle of rank $n-1$, since (*) splits locally.

This shows on the one hand that some schemes satisfy the property (affine schemes and $\mathbb{P}^1$, perhaps someone can add more examples), but on the other hand it shows how to construct counterexamples: Find $X$ which admits some vector bundle $K$ which fits into some sequence (*) such that $H^1(X,\mathcal{O}_X)=0$, but $H^1(X,K) \neq 0$.

Probably there are already examples when $X$ is a variety which is covered by two open affines.


问:
What is $Var[b]$ in multiple regression?
Assume a linear regression model $y=X\beta + \epsilon$ with $\epsilon \sim N(0,\sigma^2I)$ and $\hat y=Xb$ where $b=(X'X)^{-1}X'y$. Besides $H=X(X'X)^{-1}X'$ is the linear projection from the response space to the span of $X$, i.e., $\hat y=Hy$.

Now I want to calculate $Var[b]$ but what I get is an $k\times k$ matrix, not an $n \times n$ one. Here's my calculation:

\begin{align}
Var[b] =&\; Var[(X'X)^{-1}X'y]\\
=& \;(X'X)^{-1}X'\,\underbrace{Var[y]}_{= \sigma^2I}X(X'X)^{-1}\\\\
\text{Here you can }& \text{see already this thing will be k $\times$ k} \\\\
=&\; \sigma^2 \underbrace{(X'X)^{-1}X'X}_{I}(X'X)^{-1}\\
=& \sigma^2(X'X)^{-1}\, \in R^{k\times k}
\end{align}

What am I doing wrong?

Besides, are $E[b]=\beta$, $E[\hat y]=HX\beta$, $Var[\hat y]=\sigma^2H$, $E[y-\hat y]=(I-H)X\beta$, $Var[y-\hat y]=(I-H)\sigma^2$ correct (this is just on a side note, my main question is the one above)?

答:
The covariance matrix for $b$ (the estimator for $\beta$) should be $k\times k$. If the $X$ matrix is $n\times k$ then $\beta$ has to be $k\times 1$; otherwise the product $X\beta$ wouldn't be $n\times 1$.

So if $\beta$ is a constant vector of $k$ parameters, then its estimator $b$ is a random vector with $k$ elements. Therefore the covariance matrix for $b$ consists of covariances for all possible combinations of two members selected from the random vector, hence it must be a $k\times k$ matrix.

To answer your side notes, all your calculations are correct but some can be simplified further. Check that $HX=X$, so that $E[\hat y]=X\beta$, and $E[y-\hat y]=0$.


问:
Changing measures from counting measure to Lebesgue measure
Consider a Lebesgue measurable space $([0,1], F, \lambda)$, countable measure $\mu$ and a Riemann-integrable function $f:[0,1] \rightarrow \mathbb R (\in L^2([0,1]))$. Define $A_n = \{\frac{i}{2^n} | 0\leq i \leq 2^n\}$. Then,

$$
\int_0^1 fd\lambda = \lim_{n\rightarrow \infty} \sum_{k=1}^{2^n} f(k)\frac{1}{2^n} = \lim_{n\rightarrow\infty} \int_{A_n} f(k)\frac{1}{2^n} d\mu(A_n)
$$
where the first equality is from the definition of Riemann integration. My question is that can we find some a infinite set such that
$$
\int_0^1 f d\lambda = \int_{A_\infty} g(k) d\mu(A_\infty) = \sum_k^\infty g(k).
$$
My inutition says $g(k) = f'(k)$. So, $f'$ should decay fast at $1$.

I know my question is not rigorous and very vague.. Could you clarify what I am talking about?

答:
I think your expectation that $g$ might be $f'$ is misplaced, and you are really getting confused between $f$ and its integral. Remember that when you express the integral of $f$ as a Riemann sum, the terms in the sum don't in any way represent the derivative of $f$, rather by the fundamental theorem of calculus they represent the derivative of the integral of $f$, i.e. $f$ itself.

The integral of $f$ on $[0, 1]$ can be better and better approximated by the sum of $f$ on an "increasingly dense" finite set of points, weighted by $\frac1{2^n}$. With the above caveat on the confusion between $f$ and $f'$, you seem to be asking if there's a way to "take the limit", and express $\int f$ as a sum of $f$ over infinitely many, perhaps dense in $[0, 1]$, points, perhaps also weighted by some value or values.

If this were possible in a way that captured your intuition, presumably it should work on every interval. That is, we should have

$$\int_I f \text dx = \int_I g \text d\mu(A_\infty)$$

where $g$ is maybe $f$ times some weight function, or maybe something more exotic. But in any case this identity should hold for any $I$. Maybe $A_{\infty}$ is a dense set, maybe even the dyadic fractions, which are in some sense the limit of the sequence of point-sets used in your Riemann sums.

But this can't work. Take $I_n$, a nested sequence of intervals with lengths going to zero containing some point $x\in A_\infty$. Then the counting measure integrals should be converging to $g(x)$, but the standard integrals should be converging to $0$, since the intervals are going to zero in length.

The trouble in my view is that you're ignoring that $\frac1{2^n}$ in the Riemann sums. No reasonable "at infinity" version of that counting measure integral could be anything other than zero.


问:
Cayley tables for two non-isomorphic groups of order 4.
Not sure how to make tables, but:

For a binary operation $*$, and set $\{e,a,b,c\}$, in the Cayley table, $a*a$ can be filled with either the identity or an element different from both $e$ and $a$. If in the table the place for $a*a$ is filled with $e$, then the rest of the table can be filled out in two different ways.

What are the two different ways?

Thanks

答:
Two days ago I computed all Cayley tables for the Gorup $(G,*)$ with $|G|=4$ and came up with 4 distinct Cayley tables (!):

For the abelian group of the itegers modulo $4$: $(\mathbb{Z}/(4),+_4)$.

$1.$ $$\begin{array}{|c|c|c|c|c|}
\hline
+_4 & e & a & b & c\\
\hline
e & e & a & b & c \\
\hline
a & a & b & c & e \\
\hline
b & b & c & e & a \\
\hline
c & c & e & a & b \\
\hline
\end{array}$$
For the Group formed by the $\mathbb{C}$-roots of $z^4=1$: $(\xi_4,\cdot_\mathbb{C})$

$2.$ with $\xi_4=\{ 1,i,-i,-1 \}$

$$\begin{array}{|c|c|c|c|c|}
\hline
\cdot_\mathbb{C} & e & a & b & c\\
\hline
e & e & a & b & c \\
\hline
a & a & c & e & b \\
\hline
b & b & e & c & a \\
\hline
c & c & b & a & e \\
\hline
\end{array}$$
$3.$ with $\xi_4=\{1,-1,i,-i \}$
$$\begin{array}{|c|c|c|c|c|}
\hline
\cdot_\mathbb{C} & e & a & b & c\\
\hline
e & e & a & b & c \\
\hline
a & a & e & c & b \\
\hline
b & b & c & a & e \\
\hline
c & c & b & e & a \\
\hline
\end{array}$$
And for the Klein-$4$-Group: $(\mathbb{Z}_2\times\mathbb{Z}_2,+_{\mathbb{Z}_2\times\mathbb{Z}_2})$

$4.$ $$\begin{array}{|c|c|c|c|c|}
\hline
+_{\mathbb{Z}_2\times\mathbb{Z}_2} & e & a & b & c\\
\hline
e & e & a & b & c \\
\hline
a & a & e & c & b \\
\hline
b & b & c & e & a \\
\hline
c & c & b & a & e \\
\hline
\end{array}$$
The last two are the ones that you are looking for.


问:
Categorical Time
I always want to apply category theory to structures that involve "time" or "stepping"/"increment"(discrete "time").

I visualize it as a sequence of categories that are somehow connected(generally the objects are the same and only the morphisms will change, not sure if they are functorially related or not in all cases).

In my cases they are discrete systems in the sense that a morphism does not continuously flow, although, I suppose it should be valid for such systems too(analogous to the calculus of limits).

Is there anything in category theory that deals with this topic? I find that that category theory is more relevant to real systems when time seems to be included but I haven't been able to really apply it with time. In fact, one thing I sort of want to do is have Markovian like structures on categories. A set of categories with probabilistic transitions between them, this too is a sort of category with, I guess, the transitions being functors that are selected with a probability(but in a sequence of steps).

It seems to be category theory generally is thought of as a static structure but this is quite limiting to problems that are actually relatively simple statically but change over time. E.g., imagine a cell phone network where the objects are cell phones and the morphisms are connections. The connections may be off or on(a phone call) and also last for any length of time. Each instant in time there is a category/graph of the network. But over time the graph is dynamically changing. We can even do things like calculus over it.

So, in such ways it looks more like graph theory, BUT, of course, these happen to be categorical structures(identities, composition inclusion, etc)... so the goal is to extract some time-dependent structural variation. E.g., we can think of a typical "basic" category as static in time analogous to a constant function, it's derivative is 0. In this case the categorical structure is dynamic and its "derivative" is not constant but shows how the structure is evolving. Ultimately I want to be able to extract some meta-structure from the evolution of the changing categorical structure(and ultimately derive some type of meta^n-structural deductive chains which would, I hope, ultimately be categorical in nature).

答:
I don't know if the following are exactly what you're looking for, since they are about modelling dynamical systems within category theory, rather than having a dynamical system of categories, but they are likely to be useful for inspiration, if nothing else.

There's a classic way to look at discrete-time dynamical systems in category theory. One is just to model a dynamical system as a set $X$ with an endomap, that a function $f\colon X\to X$, which you think of as the update function. This can be seen as a functor $\mathcal{N}\to\mathsf{Set}$, where $\mathcal{N}$ is a monoid (i.e. a category with a single object) whose morphisms are the natural numbers and the composition of two morphisms is given by addition. Natural transformations between such functors give a natural notion of morphism between dynamical systems, and the resulting category (being a presheaf category and hence a topos) has a lot of very nice properties. This is explored in Lawvere and Schanuel's classic book 'Conceptual Mathematics', among other places.

For Markov processes, a nice place to start is Tobias Fritz' recent paper A synthetic approach to Markov kernels, conditional independence and theorems on sufficient statistics. This paper gives a modern framework for thinking about probability in category theory, based on what he calls "Markov categories", which generally behave like categories of sets and stochastic functions (i.e. Markov kernels). Similarly to the above construction discrete-time, time-homogeneous Markov process is a functor $\mathcal{N}\to \mathscr{M}$, where $\mathscr{M}$ is a Markov category. Fritz mentions Markov processes only briefly in his paper, but still I think it's a good place to start in thinking about them.

In both of these cases, if you want to consider non-time-homogeneous systems (i.e. where the update function can change on every time step) you can look at functors from $\mathbf{N}$ instead of $\mathcal{N}$, where $\mathbf{N}$ is the natural numbers seen as a preorder instead of a monoid. (I'm making this notation up - I'm not sure if there are standard symbols for these categories.)

This kind of approach can also be used for continuous-time systems, by replacing the monoid of natural numbers with a monoid of real numbers, although things tend to get significantly more complicated due to the need to consider topology. See this blog post by Jade Master for such a construction.

Since you seem to be looking for a discrete-time dynamical system where the state space consists of categories, it could be that you would be interested in functors $\mathcal{N}\to\mathsf{Cat}$ or $\mathbf{N}\to\mathsf{Cat}$. This seems like it should be related to the Grothendieck construction. I have no intuition for what it would be like, and I also don't immediately know how one would make it into a stochastic process.

Going in a somewhat different direction, there is a fair bit of current research in category theory in something called "open dynamical systems", which may be in discrete or continuous time. The idea here is to define dynamical systems with inputs and outputs (in various different senses), such that they can be composed with each other. If you search YouTube you can find some lectures by David Jaz Myers that give a good overview, and you can also find some papers on this topic by John Baez' group, among others.

Title: Categorical Time


问:
Notation and multiplication of a probability distribution
I am having some trouble with notation and multiplication of a probability distribution.

Basically, I have a model that estimates the probability that debris will travel a distance $x$ using the Weibull density function:

$$\begin{aligned}
P[X=x] = \frac{\kappa}{\lambda} \biggl(\frac{x}{\lambda}\biggl)^{\kappa-1}e^{-(x / \lambda)^\kappa} &  & x \in (0,\infty)\\
\end{aligned}$$

The probability of exceeding a distance $x$ is modeled as the Weibull distribution function:

$$\begin{aligned}
P[X>x] = e^{-(x / \lambda)^\kappa} &  & x \in (0,\infty)\\
\end{aligned}$$

There is an object located between a distance of $0$ and $s$, and the location of the object is independent of $x$. What is the probability that the object will be hit by the debris?

I think that this is the answer:

\begin{equation}
 P[HIT ] =
  \begin{cases}
    \frac{x}{s} e^{-(x / \lambda)^\kappa} & 0<x \leq s \\
    e^{-(x / \lambda)^\kappa} &  x > s
  \end{cases}
\end{equation}

However, I can't completely convince myself it is correct, because I am unsure of how to write the notation for the middle step.

How would I write a probability equation for of the object being a distance less than $x$? (assuming that the location is uniformly distributed between $0$ and $s$)

I have been overthinking this problem for a while, so any input would be appreciated.

答:
Condition on the location of the object: if $H$ is the event that the object is hit by the debris, then
$$P(H) = \int_0^s \frac{1}{s} P(H | t) dt $$
where $P(H|t)$ is the probability that there is a hit given that the object is at location $t$. Then
$$P(H|t) = P(X > t)$$
And you know $P(X > t)$ from above. Can you solve that integral now?


问:
Uniqueness of solution of an integral equation [Updated]
Problem. Be $f\in \mathcal{C}(\mathbf{R})$ and fix $c_0\in\mathbf{R}$ and $c_2<0$. Consider the integral equation
$$ f(y) = c_0 + c_2 y^2 + \int dz\, f(z) \exp(f(y-z)) $$
where the integral is over $(-\infty,\infty)$. Under some conditions on $c_0$ and $c_2$, it is possible to find $a_0$ and $a_2$ such that $f(y)=a_0+a_2 y^2$ is a solution. The problem is: Is this solution unique?

Let's be free to choose specific (nontrivial) values for $c_0$ and $c_2$ if that simplifies the analysis; in particular though, let's keep $c_2\neq 0$. Also, let's be free to assume that $f$ is smooth, e.g. $f\in \mathcal{C}^2(\mathbf{R})$.

Roughly speaking, the problem is how to prove that there is only one solution to an equation like $$ f * \exp f(y) = c_2 y^2 + f(y). $$

Attempt 1. From the form of the equation, it is natural to define the operator
$$ F_r(f)(y) \triangleq c_0 + c_2 y^2 + \int_{-r}^r dz\, f(z) \exp(f(y-z)) $$
and regard the solution as a fixed point of $F_r$ for any fixed $r>0$. So my first attempt would be to apply Banach fixed point theorem to $F_r$. I don't know if this works at all, even in principle, since $f\in \mathcal{C}(\mathbf{R})$, but for now this is what I have:
\begin{align} \| F_r(f)-F_r(g) \|_\infty
& \leq \sup_{y\in\mathcal{B}_r(0)} \int_{-r}^r dz  \left| f(z)\exp(f(y-z)) - g(z)\exp(g(y-z)) \right|
\end{align}
where $\mathcal{B}_r(0)$ is the ball of radius $r$ and center the origin. From here, I should try to end up with something like $C\|f-g\|_\infty$ for some $C>0$...

Attempt 2. A different approach is to show that, if $f^*$ and $g^*$ are two solutions, i.e., $F(f^*)=f^*$ and $F(g^*)=g^*$, then $f^*=g^*$. I tried the following: Since by assumption $f^*$ and $g^*$ are two solutions, then
$$ f^*(y) - g^*(y) = F(f^*)(y) - F(g^*)(y) = \int dz\, \big[f^*(y-z) \exp(f^*(z))-g^*(y-z) \exp(g^*(z))\big]. $$
I should show that the above quantity is identically equal to zero (for all $y$). I would proceed by contradiction. Suppose that $f^*\neq g^*$, that is, there exists $h=f^*-g^*$ that is not identically zero. Suppose that $h(y)\neq 0$ in a neighborhood of $y=y_0$. Then I try to expand $h$ in Taylor series around $y=y_0$. For any $y\neq y_0$ it results
\begin{multline} 0 \neq h(y_0) + h'(y_0)(y-y_0) + \cdots \\ = \int dz\, \bigg[ ( f^*(y_0-z) + (f^*)'(y_0-z)(y-y_0+z) + \cdots) \exp(f^*(z)) \\ - ( g^*(y_0-z) + (g^*)'(y_0-z)(y-y_0+z) + \cdots) \exp(g^*(z))  ) \bigg]. \end{multline}
In the end, I should lead to some contradiction, e.g. the integrand is zero...

Attempt 3. I don't know if it is just a coincidence or not, but the solution $f(y)=a_0 + a_2 y^2$ has the same form of the term $c_0+c_2 y^2$. In other words, if we define the operator

$$ G(f) \triangleq f * \exp f - f $$

then the problem is to find $f$ such that $$ G(f)(y) = c_0 + c_2 y^2 .$$

Therefore, an approach may be to find the solution to a “Green equation” like $$G(\varphi)(y)=\delta_y$$ where $\delta_y$ represents a Dirac mass at $y$. Such a $\varphi$ would (?) unlock the possibility of solving the “inhomogeneous” case.

答:
If $f(y) = c_0 + c_2 y^2 + \int f(z) e^{f(y-z)}\mathrm{d}z$
differentiating three times gives
$f'''(y)=\int f'''(z)e^{f(y-z)}\mathrm{d}z$. Thus,
$f'''=f'''*e^f= f'''*\underbrace{e^f \ast \cdots \ast e^f}_n$,
so that $f'''(y)=0$. (Maybe a bit more here.) Higher derivtives are also zero so
that $f$ is quadratic.  Let $f=a_0+a_2(y-\mu)^2$. If
$f(y) = c_0 + c_2 y^2 + \int f(z) e^{f(y-z)}\mathrm{d}z$, then
\begin{align*}
  -a_2\mu& = \frac12 f'(0) =  a_2\int(z-\mu) e^{a_0+a_2(z-\mu)^2}\mathrm{d}z
  %%
  = 0 \implies \mu=0\\[3mm]
  %%
a_0 & = f(0)   = c_0 +
  %%
 e^{a_0}\int (a_0+a_2z^2) e^{a_2z^2}\mathrm{d}z\\
  %%
 & = c_0 + a_0e^{a_0}\sqrt{-4\pi a_2}+
  %%
a_2e^{a_0}\sqrt{-4\pi a_2}/(-2a_2) \\
  %%
&  = c_0  + 2a_0e^{a_0}\sqrt{-\pi a_2}
  %%
-e^{a_0}\sqrt{-\pi a_2 }\\[3mm]
  %%
  a_2& = \frac12 f''(0) = c_2 + a_2\int e^{a_0+a_2z^2}\mathrm{d}z
  %%
  = c_2  + 2a_2e^{a_0}\sqrt{-\pi a_2}.
  %%
\end{align*}
Thus the only solution is $f(y)=a_0+a_2y^2$, where $a_0$ and $a_2$
give the unique (and here) solution to
\begin{align*}
  a_0& =  c_0  + 2a_0e^{a_0}\sqrt{-\pi a_2}
  %%
-e^{a_0}\sqrt{-\pi a_2 }\\
%%
a_2&=c_2  + 2a_2e^{a_0}\sqrt{-\pi a_2}.
%%
\end{align*}


问:
Factoring $x^{n}-1$ over the finite field $\mathbf F_q$.
If $\mathbf F_q$ is a finite field. Then how to factor $x^{n}-1$ in $\mathbf F_q[x]$?

I think we can suppose that $(q,n)＝1$,since if $n＝q^{e}n'$ then $x^{n}-1＝(x^{n'}-1)^{q^{e}}$ so that we just consider how to factor $x^{n'}-1$, which is the same.

答:
Since $q$ might not be prime, but only a power of a prime $p$ (e.g., $q = 8$), you should not be factoring out the biggest power of $q$ from $n$, but the biggest power of $p$ from $n$. Write $n = p^en'$ where $p \nmid n'$ and then in $\mathbf F_q[x]$, $x^n - 1 = (x^{n'}-1)^{p^e}$.

You can factor $x^{n'}-1$ further in $\mathbf Z[x]$ to cyclotomic polynomials indexed by the (positive) factors of $n'$. Then reduce the factorization mod $p$ to have a factorization in $\mathbf F_p[x] \subset \mathbf F_q[x]$. In order to factor each of those cyclotomic factors further in $\mathbf F_q[x]$, Galois theory for finite fields says the degree of a number over $\mathbf F_q$ is the number of its distinct $q$-power iterates.  So if $q$ is a power of the prime $p$ and $\alpha$ in a field of characteristic $p$ has multiplicative order $m$ (necessarily $p \nmid m$), then $[\mathbf F_q(\alpha):\mathbf F_q]$ is the number of distinct values among $\alpha, \alpha^q, \alpha^{q^2}, \ldots$, which is the order of $q \bmod m$.  So when $p \nmid m$, the $m$th cyclotomic polynomial (with coefficients reduced mod $p$, where $q$ is a power of $p$, and having degree $\varphi(m)$) will break up over $\mathbf F_q$ into irreducible factors all having degree equal to the order of $q \bmod m$.


问:
Why can the inner product be a criteria for the orthogonality of a matrix
I would like to verify if the matrix
$\left( \begin{array}{rr} a&b\\c&d \end{array}\right)$ is a Lorentz transformation.

Lorentz Transformations are orthogonal together with the inner product $(x,y) = x_0y_0 -x_iy_i$ . If $a =0$ and $d=0$ then the Matrix is invertible and therefor orthogonal. My script now states that because $(Lx,Lx) \leq 0$ for $x = \left( \begin{array}{r} 1\\0 \end{array}\right)$ L is not a Lorentz transformation for the case that $a=0$.

I would like to understand why the inner product of the resulting vector with itself has to be greater $0$ to be orthogonal?

答:
the Lorenz transformations in this dimension, having positive determinant and positive trace, are
$$
L =
\left(
\begin{array}{cc}
\cosh t & \sinh t \\
\sinh t & \cosh t
\end{array}
\right)
$$
The property required is
$$L^T H L = H,  $$
where
$$
H =
\left(
\begin{array}{cc}
1 & 0 \\
0 & -1
\end{array}
\right)
$$
You may also negate the right column of $L$ or the left column or both. The version I gave is symmetric.


问:
Definite integral of $\frac{1}{a\sin x+b}$ with upper limit $\pi$
I am really confused about the following, I hope it's not a stupid question with an obvious answer.

I want to solve the definite integral

$$\int\limits_c^\pi\frac{1}{a\sin(x)+b}dx$$

with $a>0$, $b>0$, $b>a$, $\pi/2\leq c <\pi$. Computer algebra systems (CAS) such as Mathematica and Matlab (symbolic math toolbox) can only solve the definite integral when the stated assumptions are provided, otherwise they don't yield a solution. (This is a general question about math, not the implementation details of CAS).

This seems weird to me, because the indefinite integral calculates to

$$\int\frac{1}{a\sin(x)+b}dx=\frac{2 \arctan\left(\frac{a+b \tan \left(\frac{x}{2}\right)}{\sqrt{b^2-a^2}}\right)}{\sqrt{b^2-a^2}},$$
and what is most important about this result, as far as I know, is that it holds regardless of any constraints on the constants. (Apart from the fact that the term inside the square root has to be $>0$ in order to avoid a complex solution). So what I'd do next is to insert my limits $c$ and $\pi$ into the antiderivative, but it's obvious that inserting $x=\pi$ into this antiderivative gives us issues, because $\tan(\pi/2)$ is undefined.

My questions are: why can the definite integral be solved under the stated constraints $a>0$, $b>0$, $b>a$, $\pi/2\leq c <\pi$, and how does one arrive at this solution, given that the antiderivative seems undefined at $x=\pi$?

A few thoughts:

Is there a different way to arrive at an equivalent formulation of the antiderivative which doesn't have this issue, either by performing the antiderivative calculation differently, or maybe by somehow simplifying $\arctan(d+e\tan(x/2))$ after it's calculated? After all, $\arctan$ is just the inverse function of $\tan$.

Maybe we can work with the antiderivative provided above, knowing that because of the fact that the lower integration limit $c$ is smaller than $\pi$ we approach the upper limit $\pi$ from below, so we know that $\tan(\pi/2)$ equals infinity and we're basically calculating $\arctan(\infty)$, which is $\pi/2$? That seems quite hand-waving to me, though. But at least that seems to yield the correct result for the upper limit, by the way.

(Why is the antiderivative undefined at all at some arguments? The original function $1/(a\sin(x)+b)$ with the stated assumptions is always $>0$ and continuous, so it's not obvious to me why it should not be possible to calculate the area under the curve within any interval.)

Thank you for your help, any insight is greatly appreciated!

答:
Notice that for $x\to k\pi/2$ that:
$$\arctan\left(d+e\tan(x/2)\right)\sim\arctan(\tan(x))=\pi/2$$


问:
Is every monoid isomorphic to a submonoid of a full transformation monoid?
We know that every group is isomorphic to a subgroup of a symmetric group. So, the question arises, is every monoid a submonoid of a full transformation monoid, where a full transformation monoid is the set of all functions from a set $X$ to $X$, the operation is composition of functions, and the identity element is the identity function.

答:
Elaborating on Eric's comment, the exact same argument works for monoids.

Specifically, given a monoid $\mathcal{M}=(M,*)$, we have a natural action of $\mathcal{M}$ on the set $M$ given by left multiplication: each $a\in M$ corresponds to the function $f_a: b\mapsto a*b$.

The map $act: a\mapsto f_a$ is easily checked$^1$ to be an injective monoid homomorphism from $\mathcal{M}$ to $Sym(M)$, so $\mathcal{M}$ is isomorphic to the submonoid of $Sym(M)$ with domain $im(act)$.

$^1$Note that the presence of an identity element is crucial here for showing injectivity (if $a\not=b$ then $f_a(e)=a\not=b=f_b(e)$). Consider the two-element semigroup  $\mathcal{S}$ with underlying set $\{0,1\}$ and multiplication given by the constant map $(x,y)\mapsto 1$. If we run the construction above we get a non-injective homomorphism from $\mathcal{S}$ to $Sym(\{0,1\})=S_2$. It's a good exercise to check whether $\mathcal{S}$ does in fact embed into any $Sym(X)$.

Associativity is also necessary, and we know it has to be: no non-associative magma can embed into any $Sym(X)$. So what goes wrong if we don't assume associativity, in the above argument?


问:
Getting consistent normals along a 3D (Bezier) curve
I'm trying to get consistent normals along a 3D Bezier curve $B(t)$, where for any point I compute the normal as:

$$
\begin{align}
\vec{a} &= B'(t) \\
\vec{b} &= B''(t) \\
\vec{c} &= \vec{a} + \vec{b} \\
\vec{r} &= \vec{c} × \vec{a} \\
\vec{n} &= \vec{r} × \vec{a} \\
\end{align}
$$

So, get the derivative at a point for time value $t$, and implicitly get the plane of curvature at the point by computing the cross product of the derivative vector at the point, and the "next" derivative vector we get from moving the derivative by the amount dictated by the second derivative. The cross product yields the axis of rotation, so to then form the normal at the point for time value $t$ I take the cross product of the axis of rotation, and the original derivative vector, since these three vectors are by definition perpendicular.

The problem is that normals computed this way are not consistent: they will "flip" around inflections, and I'm not sure what the right way is to go about making sure that does not happen.

As visual illustration, consider the following 3D cubic Bezier curve:

$$
B(t) =
\left[\begin{matrix}1&t&t^2&t^3\end{matrix}\right]
\left[\begin{matrix}1&0&0&0\\-3&3&0&0\\3&-6&3&0\\-1&3&-3&1\end{matrix}\right]
\left[\begin{matrix}
0 & 0 & 0\\
-0.38 & 2.68 & 0\\
-0.25 & 5.41 & 0\\
-0.15 & 8.21 & 0
\end{matrix}\right]
$$

Now, this happens to be a 3D curve that lies entirely on the x/y plane, but it illustrates the problem rather well. The above procedure yields the following normals:

However, this is rather different from the 2D normals we get when taking advantage of the 2D plane, where a normal can be constructed by simply rotating the (normalised) derivative vector a quarter turn clockwise, setting $(x,y)$ as $(-y,x)$:

I'd like to get something similar to the 2D case for the 3D case, but I don't know how to ensure that the cross products are unaffected by "which direction" the second derivative moves the derivative across its plane of curvature

(Effectively, how do I ensure that, when considering the triplet {normal,derivative,axis of rotation} that these always map to the local {x,y,z} axes, rather than sometimes mapping to {x,y,z} and somethings mapping to {y,x,z} axes)

Edit

While more "algorithmic" than I'd like, the only workable solution I've found so far is to compute the normals for two points $B(t)$ and $B(t+\varepsilon )$, then computing the angular difference in the plane for those two normals,

$$
\theta = \textit{acos} \left ( \frac{n_1 \cdot n_2 }{||n_1|| ||n_2||} \right )
$$

and then check whether that value is close to $\pi$ or not. Even in fast-changing curves, the angle between two "reasonable" normals is a relatively small value, so if the angle suddenly flips to "nearly $\pi$" then as of that time value the "desired normals" are negative actual normal.

While that works, it feels kind of hacky.

Without algorithmic flipping:

With algorithmic flipping:

Note this does not affect cuves with "reasonable twisting", e.g. when we set the $z$ values to $\{0,200,-200,600\}$ for the first, second, third and fourth control point respectively:

答:
(I came across this question trying to debug my implementation of essentially the same problem. The stated answer did not work for my case. It could be that I implemented it wrong, or maybe the stated answer works for less curvy curves than I need [see below]. I did manage to fix my own initial implementation, and it is conceptually simpler than the procedure in the accepted answer, so I thought it would be a good additional answer, regardless of whether the accepted answer is sometimes inferior.)

Given a curve approximated by a sequence of points and tangents, find a sequence of normals that follows the curve and follows some intuition of non-drifting. (Actually, we need a complete "frame", which consists of a sequence of three mutually orthogonal vectors, made from the tangents, the "consistent" normals, and their cross product.)

Let $t_i$ be the sequence of tangents.

We compute the three axes $x_0$, $y_0$, and $z_0|$ of the first frame using Method 1 described in this answer. It is simply a way to find a good set of orthogonal vectors in a plane, and we use the plane at the first point of our curve with the same normal as our first tangent.

We then compute each frame iteratively as follows:

$z_i = t_i$

$y_i = t_i \times x_{i - 1}$

$x_i = y_i \times z_i$

The idea is basically we start with some normal. Then the next normal is a vector that 1) is perpendicular to the tangent, and 2) lies in the same plane as the tangent and previous normal. The cross products are just calculating such a vector.

I use this frame to calculate a "thin tube" around a curve (represented by a sequence of points and tangents).

Below on the left, you can see the result of the implementation in the accepted answer (and remember my implementation may be wrong rather than the answer itself), and on the right is the implementation described here. Notice how the seem (the black line) drifts from the top to the bottom in the left mesh, but stays on the bottom in the right mesh.


问:
If $z_n$ are the zeros of the zeta function, what is the limit of $\Im{(z_n)}$ as $n$ goes to infinity?
Sorry if this question has already been asked, but it's a little difficult to look things up in Google if the statement of the problem is not very simple and involves symbols that Google doesn't recognize.

The question I have regards the zeta function. If $z_n$ is the sequence of non-trivial zeros of the zeta function with positive imaginary part and sorted by ascending imaginary part, what is the limit when $n$ goes to infinity of $\Im{(z_n)}$?

Does this explode out to infinity or is it finite?

Asking for a friend (paper here).
He has derived a new super simple equation whose solution is equivalent to the Riemann hypothesis.

答:
The Riemann-von Mangoldt formula asserts that the number of zeroes of the form $\frac{1}{2} + it$ where $t \in [0, T]$ is asymptotically

$$\frac{T}{2\pi} \log \frac{T}{2\pi} - \frac{T}{2\pi} + O(\log T)$$

from which it follows that $\text{Im}(z_n)$ grows something like $\frac{2 \pi n}{\log n} \left( 1 + \frac{\log \log n}{\log n} \right)$, but I haven't been too careful about that calculation.

Large tables of zeros are available to double-check this asymptotic against; for example, the millionth zero has imaginary part $\approx 600269$ whereas the asymptotic above gives $\approx 541230$, so it's a bit of an underestimate.

Working a bit more carefully, write $\text{Im}(z_n) = \frac{2 \pi n}{\log n} \left( 1 + e_n \right)$, where $e_n \to 0$ (slowly). Then to match the asymptotic above we need

$$\frac{n}{\log n} (1 + e_n) \log \left( \frac{n}{\log n} (1 + e_n) \right) - \frac{n}{\log n} (1 + e_n) = n + O(\log n).$$

Dividing by $\frac{n}{\log n}$, expanding out, and canceling the dominant term from both sides gives, after some simplification,

$$e_n \log n + (1 + e_n) \log (1 + e_n) - (1 + e_n) \log \log n - (1 + e_n) = O \left( \frac{(\log n)^2}{n} \right).$$

In order for the LHS to have limit $0$ as $n \to \infty$ we see that we need $e_n \approx \frac{\log \log n + 1}{\log n}$. This is already a noticeable improvement; it improves the estimate of the imaginary part of the millionth zero to $\approx 574149$. To do better than this we'll estimate

$$\log (1 + e_n) = e_n + O(e_n^2)$$

(keeping in mind that $O(e_n^2)$ is $O \left( \left( \frac{\log \log n}{\log n} \right)^2 \right)$ which is quite a bit slower than $O \left( \frac{(\log n)^2}{n} \right)$ so this is not best possible), which means the LHS becomes, after some simplification,

$$\left( e_n \log n - \log \log n - 1 \right) - e_n \log \log n + O(e_n^2)$$

so we can improve our estimate some more to $e_n \approx \frac{\log \log n + 1}{\log n - \log \log n}$. This is again a noticeable improvement; now the estimate for the imaginary part of the millionth zero is $\approx 602157$. We have two digits of accuracy now! Altogether, then,

$$\boxed{ \text{Im}(z_n) \approx \frac{2 \pi n}{\log n} \left( 1 + \frac{\log \log n + 1}{\log n - \log \log n} \right) }$$

and with a little more effort one could give a big-$O$ description of the error in this approximation but I'll stop here.


问:
Choose 1 item in a pair-- for 51 pairs. How do I get a certain number of "selected" responses for each of 102 items?
I'm new here, so please let me know if I'm not following proper etiquette.

I’m doing a project that requires gathering judgements for 100 items. These hundred items will be randomly paired without repeats, for a total of 50 randomly chosen pairs (order chosen doesn't matter). Each participant in the project will see these 50 pairs, and will select one item in each pair as their response. Each item in the pairs will have roughly equal probability of being chosen.

My question is about how many participants I would need in order to get 40 "selection" judgements per item. In other words, I want 40 "chosen" responses for each of my 100 items. How many participants does that require?

Thank you in advance!

答:
Unless I am misunderstanding you, it sounds like each participant chooses $51$ images. Also, if you want, on average, each of the $102$ images to be chosen $40$ times, then that means you are looking for $102 \cdot 40 = 4080$ choices. So, you need $\frac{4080}{51}=80$ participants.

Put differently: with each image being chosen with roughly the same likelihood, you need two participant choices to get each image selected once. So, for an image to be selected $40$ times, you need $40 \cdot 2 = 80$ participants.


问:
The basis for intersection of two subspaces
I know there are other posts on how to do this, but I was wondering if there is a slicker method.

$X = Span\{(1,1,0,-1),(1,2,3,0),(2,3,3,-1)\}$
$Y = Span\{(1,2,2,-2),(2,3,2,-3),(1,3,4,-3)\}$

I want to find a basis for $X \cap Y$

The method I have been trying to use is by letting $v \in X$ and $v \in Y$ and attempting to solve the equation $v - v = 0$. However it seems very long and tedious. I was thinking there may be a better method.

答:
Well, for this particular problem, there are some shortcuts that make life a bit easier. Let $X = \operatorname{span} \left\{v_1,v_2,v_3 \right\}$ and $Y = \operatorname{span} \left\{w_1,w_2,w_3 \right\}$, where $v_1=(1,1,0,-1)$, $v_2=(1,2,3,0)$ etc.

You could firstly note that
$$(1,1,0,-1)+(1,2,3,0)=(2,3,3,-1) \implies v_3 \in \operatorname{span}\left\{v_1, v_2\right\},$$
and
$$3(1,2,2,-2)-(2,3,2,-3)= (1,3,4,-3) \implies w_3 \in \operatorname{span} \left\{w_1, w_2 \right\},$$
and so your problem reduces to finding a basis for $X \cap Y$, where $X = \operatorname{span} \left\{v_1,v_2 \right\}$ and $Y = \operatorname{span} \left\{w_1,w_2 \right\}$.

Now, taking a similar approach with the resulting sets $X$ and $Y$, we could note that
$$(2,3,2,-3) - (1,2,2,-2)=(1,1,0,-1) \implies v_1 \in \operatorname{span} \left\{w_1, w_2 \right\},$$
and so on.


问:
Can countability coexist with infinity?
This question concerns the countability of the real numbers.
First I will show how I count the numbers between 0 and 1 on the real line.
It is done by reversing digits behind the coma, so
that e.g. 0,761 maps to 167. Obviously this is 1 to 1 mapping,
but there are infinite number of those unique mappings depending on the chosen reasonable radix.
In decimal number system I could count numbers like this:

$
0 , 0.1 , 0.2 , 0.3 ... 0.9 , \\
0.01 , 0.11 , 0.21 , 0.31 , ... , 0.91 , \\
0.02 , 0.12 , 0.22 , 0.32 , ... , 0.92 , \\
0.03 , 0.13 , 0.23 , 0.33 , \dots , 0.93 , \\
\vdots \\
0.09 , 0.19 , 0.29 , 0.39 , \dots , 0.99 , \\
0.001 , 0.101 , 0.201 , 0.301 , \dots , 0.901 , \\
\vdots \\
0.002 , \dots , 0.902, \\
\vdots \\
0.092, \dots , 0.992 , \\
0.003 , \dots , 0.903 , \\
\vdots \\
0.004, \dots , 0,904,\\
\vdots \\
\infty \\
$

Now, given that I can "succeed" to count to infinity, I would also count all irrational numbers.
There is no reason to haste. But then all irrational numbers are "somewhere" in the infinity.
So either counting to infinity allows me to write irrational numbers backwards, or infinity and countability can not coexist.
Which one is true ? How does your solution compare to rational numbers ?

答:
Let's denote the $n^\text{th}$ number in your list by $f(n)$.
To use your method to prove that $[0,1)$ is countable, you need to show that $f$ is a surjection, that is, for every real number $x \in [0,1)$ there is a natural number $n$ such that $f(n) = x$.  In particular, you have to show that there is a natural number $n$ such that $f(n) = 1/3$.
Note that $n$ here is just an ordinary, finite, natural number.
So you have to show that $1/3$ appears at some finite stage in your list (e.g. it is the tenth number listed, or the millionth number listed.)

This is not the case, and it is not correct to argue that it must appear sometime because $\mathbb{N}$ is so very large.  For no value $n \in \mathbb{N}$ do we have $f(n)=1/3$, any more than we have $f(n) = -1$ or $f(n) = 37$.  These numbers simply do not appear in the list.

Title: Can countability coexist with infinity?


问:
Integral $\int_{0}^{2\pi}\log|e^{i \theta}-1|d \theta$
Consider $$\int_{0}^{2\pi}\log|e^{i \theta}-1|d \theta$$

Is it equal to $0$ ? Why ?

Any hint ?

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{2\pi}
\ln\pars{\verts{\expo{\ic\theta} - 1}\dd\theta}} =
2\,\Re\int_{0}^{2\pi}
\ln\pars{\expo{\ic\theta} - 1}\dd\theta
\\[5mm] = &\
2\,\Re\oint_{\verts{z}\ =\ 1}\,\,
\ln\pars{z - 1}{\dd z \over \ic z} =
2\,\Im\oint_{\verts{z}\ =\ 1}\,\,
{\ln\pars{z - 1} \over z}\,\dd z
\\[5mm] \stackrel{z\ \mapsto\ z - 1}{=}\,\,\,\,\,&
2\,\Im\oint_{\verts{z + 1}\ =\ 1}\,\,
{\ln\pars{z} \over z + 1}\,\dd z
\end{align}
I'll choose the $\ds{\ln}$-branch cut
$$
\ln\pars{z} = \ln\pars{\verts{z}} +
\ic\arg\pars{z}\,,\quad
0 < \arg\pars{z} < 2\pi\,,\quad z \not= 0
$$
Then $\ds{\pars{~\mbox{with an}\ indent\ \mbox{around the origin}~}}$,
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{2\pi}
\ln\pars{\verts{\expo{\ic\theta} - 1}\dd\theta}}
\\ = &\ \
\overbrace{2\,\Im\braces{2\pi\ic\bracks{\ln\pars{\verts{-1}} + \ic\pi}}}^{\ds{=\ 0}}\ -\
\overbrace{2\,\Im\lim_{\epsilon \to 0^{+}}\,\,
\int_{2\pi}^{0}\bracks{\ln\pars{\epsilon} + \ic\theta}
\epsilon\expo{\ic\theta}\ic\,\dd\theta}^{\ds{=\ 0}}
\\[5mm] = & \ \bbx{0} \\ &
\end{align}


答: It depends on how sure you want to be that an image is picked $40$ times.  If you imagine that each participant throws a fair coin to choose the image, for $n$ participants you will have each image selected on average $\frac n2$ times.  The standard deviation on this is $\frac 12\sqrt n$.  If you want a $95\%$ chance that a given image has been selected $40$ times, you need $40$ to be the mean minus $1.96$ standard deviations or $40=\frac n2-1.96\cdot \frac 12\sqrt n$.  Alpha tells us you need $100$ people under this model.  You will still probably have a few images selected less than $40$ times, so you might want more people.

Title: Choose 1 item in a pair-- for 51 pairs. How do I get a certain number of "selected" responses for each of 102 items?

Body:
I'm new here, so please let me know if I'm not following proper etiquette.

I’m doing a project that requires gathering judgements for 100 items. These hundred items will be randomly paired without repeats, for a total of 50 randomly chosen pairs (order chosen doesn't matter). Each participant in the project will see these 50 pairs, and will select one item in each pair as their response. Each item in the pairs will have roughly equal probability of being chosen.

My question is about how many participants I would need in order to get 40 "selection" judgements per item. In other words, I want 40 "chosen" responses for each of my 100 items. How many participants does that require?

Thank you in advance!


问:
Finding eigenvalues from simple characteristic polynomial gone wrong - Solution explanation
Now I thought I had this understood, but apparently not. So I did some reduction and got a characteristic polynomial of the form:

$$-x^{3} + 2x^{2} + x - 2 = 0$$

Where I'm just using $x$ in place of $\lambda$ for typing purposes.

In my first attempt at solving it I went through this process:

$$-x(x^{2}-2x-1) = 2 \\ x = -2\ \text{or} \ (x^{2}-2x-1) = 2$$
From which I could rewrite (so I thought) as

$$x^{2}-2x-3 = 0 $$

From which we could factor to get

$$(x-3)(x+1)$$

Fortunately there is a solution I saw which gave the following factorization:

$$(x-1)(x-2)(x+1) = 0$$

Now I re worked this and the only way I could get this solution was by "guessing a root (using rational zeros)", then dividing my cubic polynomial, and finally factoring the quadratic that remains. Which for example would be of the form:

$$(x-1)(x^{2}-x-2) = 0$$

Now I know I should know the reason why but can't remember, but why didn't my first approach work? And what is a more efficient way of getting this factorization besides "guessing roots".

答:
What went wrong with your first attempt is that $a\times b=2$ does not imply that $a=2$ or $b=2$, the way $a\times b=0$ implies $a=0$ or $b=0$ in an integral domain (such as a field).  For further explication of that important point, see this answer.

Regarding how to find rational roots of a polynomial, see the rational root theorem, which says they will be factors of the constant term divided by factors of the leading coefficient; in your case the candidates would be $\pm1$ or $\pm2$.


问:
Proving that a infinite sum of random variables converges
Let $(X_n)_n$ be a sequence of independent random variables in a probability space $(\Omega,\mathcal{F},P)$ with $E(X_i) = \mu_i$ and $\text{var}(X_i) = \sigma^2_i$. Suppose that $\sum_n \mu_n <\infty$ and $\sum_n\sigma^2_n<\infty$. Then, I want to prove that
$$
\sum_{n\geq 1} X_i <\infty
$$
a.s.

What I am trying to do is to prove that $Y_n = \sum_{k=1}^n X_k$ is a supermartingale and applying Doob's theorem, but I can't since $\mu_i$ can either be positive or negative. Any hint? Thanks.

答:
We consider $Y_{m}:=\sum_{n=1}^{m}(X_{n}-\mu_{n})$, note that $\left\{Y_{m}\right\}_{m\geq 1}$ is a martingale respect to $\mathcal{F}_{m}:=\sigma(X_{1},\ldots,X_{m})$. Indeed, we have that
\begin{align}
\mathbb{E}[Y_{m+1}|F_{m}]&=\mathbb{E}[Y_{m}+(X_{m+1}-\mu_{m+1})|\mathcal{F}_{m}]\\
&=\mathbb{E}[Y_{m}|\mathcal{F}_{m}]+\mathbb{E}[X_{m+1}|\mathcal{F}_{m}]-\mathbb{E}[\mu_{m+1}|\mathcal{F}_{m}]\\
&=Y_{m}+\mathbb{E_{m+1}}-\mu_{m+1}\\
&=Y_{m}+\mu_{m+1}-\mu_{m+1}\\
&=Y_{m}.
\end{align}

We also have that $\left\{Y_{m}\right\}_{m\geq 1}$ is bounded in $L^{2}$, to prove this fact we use the following result:

Let $M=\left\{M_{n}\right\}_{n\geq 1}$ be a martingale, then $M$ is bounded in $L^{2}$ if and only if $\sum_{n=1}^{\infty}\mathbb{E}[(M_{n}-M_{n-1})^{2}]<\infty$.

In that sense, note that
\begin{align}
\sum_{m=1}^{\infty}\mathbb{E}[(Y_{m}-Y_{m-1})^{2}] &=\sum_{m=1}^{\infty}\mathbb{E}[(X_{m}-\mu_{m})^{2}] \\
&=\sum_{m=1}^{\infty}\sigma_{m}^{2} <\infty.
\end{align}
So, $\left\{Y_{m}\right\}_{m\geq 1}$ is bounded in $L^{2}$. But we know that bounded in $L^{2}$ implies uniformly integrable (UI), then  $\left\{Y_{m}\right\}_{m\geq 1}$ is uniformly integrable (UI). Therefore, $\lim_{m\rightarrow\infty}Y_{m}$ exists a.s, that is, $\sum_{n\geq 1} X_n <\infty$ a.s.


问:
$\frac{d}{dx}\int_{0}^{e^{x^{2}}} \frac{1}{\sqrt{t}}dt$
I'm having trouble understanding how to apply the $\frac{d}{dx}$when taking the anti-derivative.

$$\frac{d}{dx}\int_{0}^{e^{x^{2}}} \frac{1}{\sqrt{t}}dt$$
In class it was mentioned we'll end up taking the derivative of $e^{x^{2}}$ which is $2xe^{x^{2}}$

My guess of solving this is

$$\frac{d}{dx}\int_{0}^{e^{x^{2}}} \frac{1}{\sqrt{t}}dt = [2t^{\frac{1}{2}}]_0^{e^{x^2}}$$

Then
$$[2(e^{x^{2}})^{\frac{1}{2}}]-[2(0)^{\frac{1}{2}}]=2(e^{x^{2}})^{\frac{1}{2}}$$
Am i missing something?

答:
No you aren't. Just continue to get

$\Large2(e^{x^{2}})^{\frac{1}{2}}=2e^{\frac{x^{2}}{2}}$

and then differentiate with respect to $x$ to get

$\Large 2xe^{\frac{x^{2}}{2}}$


问:
Trying to get a bound on the tail of the series for $\zeta(2)$
$$\frac{\pi^2}{6} = \zeta(2) = \sum_{k=1}^\infty \frac{1}{k^2}$$ I hope we agree.

Now how do I get a grip on the tail end $\sum_{k \geq N} \frac{1}{k^2}$ which is the tail end which goes to zero?

I want to show that $\sqrt{x}\cdot \mathrm{tailend}$ is bounded as $x \to \infty$.

All this to show that $x\cdot \mathrm{tailend} = \mathcal O\sqrt{x}$

The purpose is to get the asymptotic formula for the distribution of square free integers.

p.269 Exercise 8 Stewart and Tall.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\sum_{k\ \geq N}\,{1 \over k^{2}}} =
\overbrace{\sum_{k = 1}^{\infty}{1 \over k^{2}}}
^{\ds{\zeta\pars{2}}}\ -\
\sum_{k = 1}^{N - 1}{1 \over k^{\color{red}{2}}}
\\[5mm] = &\
\zeta\pars{2} -
\bracks{\zeta\pars{\color{red}{2}} +
{\pars{N - 1}^{1 - \color{red}{2}} \over 1 - \color{red}{2}} +
\color{red}{2}\int_{N - 1}^{\infty}\,\,{\braces{x} \over x^{\color{red}{2} + 1}}\,\dd x}
\end{align}

The last term is an
identity related to the $\ds{\zeta}$-function.

Then,
\begin{align}
&\bbox[5px,#ffd]{\sum_{k\ \geq N}\,{1 \over k^{2}}} =
{1 \over N - 1} - 2\int_{N - 1}^{\infty}\,\,{\braces{x} \over x^{3}}\,\dd x
\end{align}
Note that $\ds{0 < 2\int_{N - 1}^{\infty}\,\,{\braces{x} \over x^{3}}\,\dd x < 2\int_{N - 1}^{\infty}\,\,{\dd x \over x^{3}} =
{1 \over \pars{N - 1}^{2}}}$ which yields
$$
\bbx{{1 \over N - 1} - {1 \over \pars{N - 1}^{2}} <
\bbox[5px,#ffd]{\sum_{k\ \geq N}\,{1 \over k^{2}}} <
{1 \over N - 1}} \\
$$


问:
A question about the proof of Borsuk's lemma in Munkres' topology

My problem is with the part where he uses the pasting lemma to conclude that $h$ is continuous.

The pasting lemma requires that both sets on which the corresponding functions are defined be either open or closed in the whole space $X$ which is $R^2$ in this case.

$A$ is a compact subspace of $R^2 - 0$ and hence is closed in it, but that doesn't guarantee it is closed in $R^2$.

$C$ and $D$ are both open in $R^2 - A$ because they are components and the space is locally path connected, but I haven't been able to prove anything about whether they are open or closed $R^2$.

It looks like a trivial detail among the main points in the proof so I must be missing something here.

答:
Compactness of a set is a topological property: it is independent of the space in which that set is embedded. $A$ is a compact subset of $\Bbb R^2\setminus\{\bf0\}$, so it is compact as a space and therefore compact as a subset of any space in which it is embedded. In particular, it is a compact subset of $\Bbb R^2$.

$C$ and $D$ are open subsets of $\Bbb R^2\setminus A$, which is open in $\Bbb R^2$, so $C$ and $D$ are open in $\Bbb R^2$. Specifically, for $C$ there is an open $U$ in $\Bbb R^2$ such that $C=U\cap(\Bbb R^2\setminus A)$, but that means that $C$ is the intersection of two open sets in $\Bbb R^2$ and hence is itself open in $\Bbb R^2$. The argument for $D$ is the same.


问:
Show that the set $U = \{(1,2), (1,3), (1,4)\}$ spans $\Bbb{R}^2$ but isn't a basis.
I really need help with that question, so I'd appreciate a solution with steps because this problem was given for practice and my exam is in a few days so I'd like to understand the steps to the solution. Thank you.</p>

答:
For instance, $(1,3)$ is not a multiple of $(1,2)$, so they are independent and span $\Bbb R^2$.

But any $3$ vectors in a two-dimensional space are dependent, and so can't form a basis.

I have taken as given that $\Bbb R^2$ is two-dimensional, say with basis $e_1=(1,0)$ and $e_2=(0,1)$.


问:
Find the maximum value of a function $f(x) = |x(x - 1)(x - 2)(x - 3)(x - 4)(x - 5)(x - 6)(x - 7)|$ where $3 \leq x \leq 4$
Find the maximum value of a function $f(x) = |x(x - 1)(x - 2)(x - 3)(x - 4)(x - 5)(x - 6)(x - 7)|$ where $3 \leq x \leq 4$.

This question is supposed to be solved without using derivatives.

答:
If the use of AM-GM inequality is allowed, then
$$
f(x)=x(7-x) \cdot (x-1)(6-x) \cdot (x-2) (5-x) \cdot (x-3)(4-x) \\
\le \left(\frac 72\right)^2 \cdot \left(\frac 52\right)^2 \cdot \left(\frac 32\right)^2 \cdot \left(\frac 12\right)^2 = \frac{11025}{256}.
$$
And all four equality hold simultaneously at $x=3.5$.


答:
As you have already shown, the eigenvalues of $P$ are $\lambda_1 = -1/\sqrt{2}$, $\lambda_2 = 0$, $\lambda_3 = 1/\sqrt{2}$ and $\lambda_4 =1$. Therefore, we can write $P$ as $P = SDS^T$ where $D = \text{diag}(\lambda_1, \lambda_2, \lambda_3, \lambda_4)$. We also know that $P^r = SD^r S^T$. Therefore, we have,
\begin{align*}
\lim_{n \to \infty} M^{(n)} & = \lim_{n \to \infty} \frac{1}{n + 1} \sum_{r = 0}^n P^r \\
& =  \lim_{n \to \infty} \frac{1}{n + 1} \sum_{r = 0}^n SD^r S^T \\
& = \lim_{n \to \infty} \frac{1}{n + 1} \sum_{r = 0}^n S \ \text{diag}(\lambda_1^r, \lambda_2^r, \lambda_3^r, \lambda_4^r) \  S^T \\
& =  \lim_{n \to \infty} \frac{1}{n + 1} S \ \text{diag} \left( \sum_{r = 0}^n \lambda_1^r, \sum_{r = 0}^n \lambda_2^r, \sum_{r = 0}^n \lambda_3^r, \sum_{r = 0}^n \lambda_4^r \right) \  S^T \\
& = \lim_{n \to \infty} \frac{1}{n + 1} S \ \text{diag} \left( \dfrac{1 - \lambda_1^n}{1 - \lambda_1}, 0, \dfrac{1 - \lambda_3^n}{1 - \lambda_3}, n+1 \right) \  S^T.
\end{align*}

Note that since $|\lambda_1|, |\lambda_3| < 1$, $\displaystyle \lim_{n \to \infty} \frac{1}{n + 1} \dfrac{1 - \lambda_i^n}{1 - \lambda_i}  = 0$ for $i = \{1, 3 \}$. Thus,
\begin{align*}
\lim_{n \to \infty} M^{(n)} & = S \ \text{diag}(0,0,0,1) \ S^T = \frac{1}{4} \mathbf{1}_{4 \times 4},
\end{align*}
where $\mathbf{1}_{n \times n}$ is the $n \times n$ matrix consisting of all ones.

Title: How to compute the limit of $M^{(n)}/(n+1)$ for the matrices P?

Body:
Let $M^{(n)}=\sum_{r=0}^{n}P^r$.
How to calculate $\lim_{n\to \infty}\frac{M^{(n)}}{n+1}$ for $P=\begin{bmatrix}
0.5 & 0.5 & 0 & 0 \\
0.5 & 0 & 0.5 & 0\\
0 & 0.5 & 0 & 0.5 \\
0 & 0 & 0.5 &0.5
\end{bmatrix}$ ?

First, I used the method of diagonalization to compute $P^n$. I used Matlab and I got reslut

```<code>A =[0.5,0.5,0,0;0.5,0,0.5,0;0,0.5,0,0.5;0,0,0.5,0.5]
[S,D]=eig(A)
A =

    0.5000    0.5000         0         0
    0.5000         0    0.5000         0
         0    0.5000         0    0.5000
         0         0    0.5000    0.5000

S =

    0.2706    0.5000    0.6533    0.5000
   -0.6533   -0.5000    0.2706    0.5000
    0.6533   -0.5000   -0.2706    0.5000
   -0.2706    0.5000   -0.6533    0.5000

D =

   -0.7071         0         0         0
         0   -0.0000         0         0
         0         0    0.7071         0
         0         0         0    1.0000

</code>```

But I have no idea what to do next.


问:
Trying to evaluate a complex integral?
I have a question where I must evaluate the following integral over a circle where $\lvert z \rvert = 2$.

$$I = \oint \frac{z^3e^{\frac{1}{z}}}{1+z}dz$$

I have tried the $z^3 = 8 \cdot \mathrm{e}^{3it}$ approach and ended up with:

$$I = i\int_0^{2\pi} \frac{16e^{4it}e^{\frac{1}{2e^{it}}}}{1+2e^{it}}dt$$

But I have no idea what to do from here! Or if this is even the correct approach. Thanks in advance to anyone who knows

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
I & \equiv \bbox[5px,#ffd]{\oint_{\verts{z}\ =\ 2}\, {z^{3}\expo{1/z} \over 1 + z}\,\dd z}
\,\,\,\stackrel{z\ \mapsto\ 1/z}{=}\,\,\,
\oint_{\verts{z}\ =\ 1/2}\,\,\,
{\expo{z} \over z^{4}\pars{1 + z}}\,\dd z
\\[5mm] = &
2\pi\ic\,{1 \over 3!}\,\lim_{z \to 0}\,\totald[3]{}{z}
\pars{\expo{z} \over 1 + z} =
\bbx{-\,{2\pi \over 3}\,\ic} \\ &
\end{align}

Title: Trying to evaluate a complex integral?


问:
Reducible and irreducible polynomials
Im stuck on something..
Why is $f(x)=2x$ irreducible in $\mathbb{Q}[x]$ but reducible in $\mathbb{Z}[x]$?

I know that $2$ is invertible in $\mathbb{Q}(x)$ but not in $\mathbb{Z}(x)$, the point of my question is:

Why are invertible elements not taken into consideration while decomposing something?

答:
Any invertible element divides every element; if invertible elements “counted”, then nothing would be irreducible, and then notion becomes worthless. It would be as if, in the integers, we said that $p$ is prime if and only if it cannot be written as $p=ab$. If we used that definition, then no element would be prime, because we would always be able to write any integer $p$ as $p=1\times p$.

Similarly, if $u$ is invertible in a domain, then there exists $v$ such that $uv=1$. Then any element $f$ could be written as $f=u(vf)$, so nothing would be irreducible.

Title: Reducible and irreducible polynomials


问:
What is the rule about the log of small o notation?
Say I have the MacLaurin Series expansion of a function and get
\begin{equation}
1+2f(0)c_is_i\theta+o(\theta)
\end{equation}
and I want to take the log of this function -i.e. $\log[1+2f(0)c_is_i\theta+o(\theta)]$. How would the last term be expressed?

答:
Write $a = 2 f(0) c_i s_i$ so your expression is $1 + a \theta + o(\theta)$. We want to compare the log of this expression to $\log (1 + a \theta)$. We have

$$\log (1 + a \theta + o (\theta)) - \log (1 + a \theta) = \log (1 + o \left( \frac{\theta}{1 + a \theta} \right) )$$

and the Taylor series of the logarithm gives $\log (1 + x) = x + O(x^2)$ which gives, altogether,

$$\log (1 + a \theta + o (\theta)) = \log (1 + a \theta) + o \left( \frac{\theta}{1 + a \theta} \right).$$

(This way of writing it is agnostic about whether you want $\theta \to 0$ or $\theta \to \infty$; it works either way. If $\theta \to 0$ it simplifies to $o(\theta)$ and if $\theta \to \infty$ it simplifies to $o(1)$.)


答:
Let $$f(x)=e^{x^2}$$ $$g(x)=\int_0^x \frac{dt}{\sqrt t}$$

Then $$h(x)=\int_0^{e^{x^2}}\frac{dt}{\sqrt t}=g\circ f(x)$$

Use the chain rule and FTC to find $h'(x)$, which is what you want.

Body:
I'm having trouble understanding how to apply the $\frac{d}{dx}$when taking the anti-derivative.

$$\frac{d}{dx}\int_{0}^{e^{x^{2}}} \frac{1}{\sqrt{t}}dt$$
In class it was mentioned we'll end up taking the derivative of $e^{x^{2}}$ which is $2xe^{x^{2}}$

My guess of solving this is

$$\frac{d}{dx}\int_{0}^{e^{x^{2}}} \frac{1}{\sqrt{t}}dt = [2t^{\frac{1}{2}}]_0^{e^{x^2}}$$

Then
$$[2(e^{x^{2}})^{\frac{1}{2}}]-[2(0)^{\frac{1}{2}}]=2(e^{x^{2}})^{\frac{1}{2}}$$
Am i missing something?


问:
What is the maximum number of eigenvalues of this $\mathbf M_{n, n} \to \mathbf M_{n, n}$ operator?
For two square matrices $A$ and $B$ of the same size $n$ let us define $A★B$ following matrix:

$ (A★B)_{ij} = \begin{cases} (AB)_{ij}, & \text{if $i$ is odd,} \\ b_{ij}, & \text{else.} \end{cases}$

For matrix $A$ let us define operator $\Phi_A : B \mapsto A \bigstar B$ on the $ n \times n $ matrices.

(a) Can matrix $A$ exist so this operator eigenvalue is $2$?

(b) What is the maximum number of eigenvalues this operator can have (for fixed n)?

I solved (a):

Yes, for $ A = 2 I $, where $I$ is identity matrix, and $ (B)_{ij} = \begin{cases} 1, \text{if i is odd}, \\ 0, \text{else}. \end{cases}$

I am unsure how to solve b and would appreciate a hint. Thanks very much!

答:
Hint. Let $(r,s)=(n,n-1)$ when $n$ is odd, or $(n-1,n)$ when $n$ is even. That is, $r$ and $s$ are respectively the last odd index and the last even index in $\{1,2,\ldots,n\}$.

Let $E_{ij}\in M_{n,n}$ be the matrix with a $1$ at the $(i,j)$-th position and zeroes elsewhere. Consider the ordered basis $\mathcal B$ of $M_{n,n}$ consisting of the $E_{ij}$s in the following order:
\begin{aligned}
&E_{11},E_{31},\ldots,E_{r1},E_{21},E_{41},\ldots,E_{s1},\\
&E_{12},E_{32},\ldots,E_{r2},E_{22},E_{42},\ldots,E_{s2},\\
&\ldots,\\
&E_{1n},E_{3n},\ldots,E_{rn},E_{2n},E_{4n},\ldots,E_{sn}.\\
\end{aligned}
The matrix representation of the linear operator $L(B)=A★B$ with respect to $\mathcal B$ is then
$$
M=\pmatrix{X&Y\\ 0&I_{s/2}\\ &&X&Y\\ &&0&I_{s/2}\\ &&&&\ddots\\ &&&&&X&Y\\ &&&&&0&I_{s/2}}
$$
where $\pmatrix{X&Y}$ is the submatrix consisting of the odd-indexed rows of $A$.


问:
Does anyone know how to calculate the following integral?
Consider the function (coming from a joint probability density):
$$
f(x,y) = \frac{1}{y}e^{-y-\frac{x}{y}}.
$$

I want to evaluate the definite integral (marginal):
$$
F(x) = \int_0^\infty f(x,y)\,dy.
$$

It is very hard to find a primitive in $y$, so what I did was the following workaround:
$$
\int f(x,y)\,dx = -e^{-y-\frac{x}{y}}.
$$

So if we let $g(x,y):= -e^{-y-\frac{x}{y}}$, we can write:
$$
f(x,y) = \frac{\partial g}{\partial x}(x,y).
$$

Therefore:
$$
F(x) = \int_0^\infty \frac{\partial g}{\partial x}(x,y)\,dy.
$$

Now the trick was: can we move the derivation out of the integral? Under what assumptions? If that were the case, we could write:
$$
F(x) = \frac{d}{dx} \int_0^\infty g(x,y)\,dy,
$$

which can be calculated in terms of Bessel functions:

$$
F(x) = \frac{d}{dx} \int_0^\infty -e^{-y-\frac{x}{y}}\,dy = -\frac{d}{dx} \big(2\sqrt{x}\,K_1(2\sqrt{x})\big).
$$

Deriving:
$$
F(x) = K_0(2\sqrt{x}) - \frac{1}{\sqrt{x}}\,K_1(2\sqrt{x})+K_2(2\sqrt{x}).
$$

(The last two passages according to Wolfram Alpha, for $x\ge 0$.)
The $K_n$ should be the "modified Bessel functions of the 2nd kind".

I would like to ask:

Is it "legal" to carry the derivative out of the integral sign?

Are the last two (Wolfram Alpha) passages correct?

Is there any other way of obtaining $F(x)$? What is the result?

Thanks.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}{1 \over y}
\exp\pars{-y - {x \over y}}\dd y}
\\[5mm] = &\
\int_{0}^{\infty}{1 \over y}
\exp\pars{-\root{x}\bracks{%
{y \over \root{x}}  + {\root{x} \over y}}}\,\dd y
\\[5mm] \stackrel{y\ =\ \root{x}\expo{\theta}}{=}\,\,\,&
\int_{-\infty}^{\infty}{1 \over \root{x}\expo{\theta}}
\expo{-2\root{x}\cosh\pars{\theta}}\,\,
\pars{\root{x}\expo{\theta}}\,\dd\theta
\\[5mm] = &\
2\int_{0}^{\infty}
\expo{-2\root{x}\cosh\pars{\theta}}\,\,\dd\theta =
\bbx{2\on{K}_{0}\pars{2\root{x}}} \\ &
\end{align}
$\ds{\on{K}_{\nu}}$ is a Modified Bessel Function.
The last result is given by the DLMF Bessel Library.


问:
Derivative of multilinear function
Theorem: Let $x_1, \dots, x_n$ be differentiable vector-valued functions. Then the multilinear function $M(x_1, ... ,x_n)$ is differentiable.

This is the proof: I know that $M$ is linear in $x_1, \dots, x_n$, so that if for all $x_i$ (for $i \neq j$) the function M is linear, then it is a linear function of $x_j$.
But I can't understand the given proof.

答:
By expanding along a basis, it is straightforward to show that $M$ is continuous.

If a function has continuous partials then it is differentiable. It is not hard to see
that
$M(x_1,...,x_k+h,...x_n) - M(x_1,...,x_k,...x_n) - M(x_1,...,h,...x_n) = 0$ so it follows
that
${\partial M(x_1,...,x_n) \over \partial x_k}h = M(x_1,...,h,...x_n)$ and so is continuous. Hence $M$ is differentiable and

$DM((x_1,...,x_n))(h_1,...,h_n) = \sum_k {\partial M(x_1,...,x_n) \over \partial x_k}h_k$.

Title: Derivative of multilinear function


问:
Need help with Fourier transform problem
I'm trying to calculate the Fourier transform of the unit step function,

$$\mathcal{F}[u(t)] \ = \int_{-\infty}^{\infty}u(t)e^{-i\omega t}dt \ = \int_{0}^{\infty}e^{-i\omega t} dt. \tag{1}$$

This simplifies to,

$$U(\omega) = (i\omega)^{-1},\ (\omega \not = 0). \tag{2}$$

However, my book claims that $(1)$ simplifies to $ \pi \delta(\omega) + (i\omega)^{-1}. \tag{3}$

Here, $\delta(\omega)$ is the unit impulse function. I don't have my book with me right now but I think they use the differentiation property to derive it by calculating the transform of the derivative of $u$ (which is $\delta$).

My question is, isn't the appearance of $\delta$ in the result they obtain irrelevant? Since, at $\omega  = 0$, $(3) = \infty$ and elsewhere $(3) = (2).$ So, why would they write $(3)$ instead of $(2)$? I should mention this is not in a mathematics textbook, but an engineering textbook.

答:
The reason why you get a delta function is because the step function is actually defined as

$$\theta(t) = \begin{cases} \\ 1 & t \gt 0\\1/2 & t=0\\0 & t \lt 0\end{cases}$$

That nonzero value at $t=0$ is a bit troublesome.  Better to consider the signum funciton $\text{sgn}(t) = 2 \theta(t)-1$.  The FT of $\text{sgn}(t)$ is

$$\int_0^{\infty} dt \, e^{-i \omega t} - \int_0^{\infty} dt \, e^{i \omega t} = \frac{2}{i \omega}$$

Note that the value at $t=0$, being zero, does not contribute to the FT.  The FT of $\theta(t)$ follows from this, because

$$\theta(t) =  \frac12 \text{sgn}(t) + \frac12$$

so that its FT is

$$\frac{1}{i \omega} + \frac12 \int_{-\infty}^{\infty} dt \, e^{-i \omega t} = \frac{1}{i \omega} + \pi \delta(\omega)$$


问:
Find the exact value of $\int_{-\infty}^{\infty} x^2[f(x)]^{10} \,dx$
Let $f(x)$ be the pdf of a normal random variable with mean $\mu$ and variable $\sigma^2$.
Find the exact value of $$\int_{-\infty}^{\infty} x^2[f(x)]^{10} \,dx$$

I understand that $f(x) = \frac{1}{\sigma \sqrt{2\pi}}\exp[-\frac{1}{2\sigma^2}(x-\mu)^2]$. I noticed that if $f(x)$ is a standard normal then that integral goes to $1$ but I am not sure how to find the exact value for a normal distribution. Any help would be greatly appreciated. Thank you.

答:
After expanding the power of $10$, it becomes
$$\int_{-\infty}^{\infty}x^2 \frac{1}{\sigma^{10}(2\pi)^5} e^{-\frac{5}{\sigma^2}(x-\mu)^2}dx$$

This is equal to $$\frac{1}{\sigma^9\sqrt{10}(2\pi)^5\frac{1}{\sqrt{2\pi}}}\int_{-\infty}^{\infty}x^2 \frac{1}{\sigma_1\sqrt{2\pi}} e^{-\frac{1}{2\sigma_1^2}(x-\mu)^2}dx$$

where $\sigma_1=\frac{\sigma}{\sqrt{10}}$. Can you finish from here?


问:
Application of Bayes Theorem in Dependent Events
Here is my attempt to solve this problem using Bayes's Theorem. I would appreciate it if anyone could identify a mistake (if any) in my approach.

Problem: There are two bins B1 and B2. In B1 there are 2 red balls and 8 blue balls. In B2 there are 7 red balls and 3 blue balls. Suppose Bob first picks a bin B1 or B2 with equal probability and then picks a ball from the bin he has chosen (without telling us the bin). Given that Bob has picked a red ball, what is the probability he chose B1?

My approach:

It is clear that B1 and r are not independent events. Let B1 be the event of picking a B1 bin and r be the event of picking a red ball.

Use Bayes theorem: $P(B1|r)=\frac{P(r|B1)*P(B1)}{P(r)} = \frac{\frac{1}{5}*\frac{1}{2}}{\frac{9}{20}} = \frac{2}{9}$

Everything here is intuitive to me apart from the fact that I had $P(r) = \frac{9}{20}$. Here I just combined balls in B1 and B2 and there are 9 red balls out of 20 total. I appreciate any insights into how my approach or application of Bayes Theorem is incorrect. Thank you.

答:
You can combine like that because there are ten balls in each bin and an equal choice between bins, therefore each ball has an equal chance of being selected.   Nine of the twenty balls are red.

If there were different amounts in each bin, there would be bias, and you would instead have to use the Law of Total Probability to get the correct result.   However, that is not the case, so you are okay.

Of course, you may still use it to obtain the same answer:

$$\begin{align}\mathsf P(r)&=\mathsf P(r\mid B_1)\mathsf P(B_1)+\mathsf P(r\mid B_2)\mathsf P(B_2)\\[1ex]&= \frac 2{10}\cdot\frac 12+\frac 7{10}\cdot\frac 12\\[1ex]&=\frac 9{20}\end{align}$$

So, yes, indeed.   The probability for having selected bin 1 when given that you obtained a red ball is $2/9$.   It is less likely that you picked from the bin with fewer red balls -- two among the nine red balls come from bin 1 and all balls are each equally likely to have been selected.

$$\begin{align}\mathsf P(B_1\mid r)&=\dfrac{\mathsf P(r\mid B_1)\mathsf P(B_1)}{\mathsf P(r\mid B_1)\mathsf P(B_1)+\mathsf P(r\mid B_2)\mathsf P(B_2)}\\[1ex]&=\frac 29\end{align}$$


问:
Prove that if $A$ is normal, then eigenvectors corresponding to distinct eigenvalues are necessarily orthogonal (alternative proof)
The problem statement is as follows:

Prove that for a normal matrix $A$, eigenvectors corresponding to different eigenvalues are necessarily orthogonal.

I can certainly prove that this is the case, using the spectral theorem.  The gist of my proof is presented below.

If possible, I would like to find a simpler proof.  I was hoping that there might be some sort of manipulation along these lines, noting that
$$
\langle Av_1,A v_2\rangle
= \langle v_1,A^*Av_2\rangle
= \langle v_1,AA^*v_2\rangle
= \langle A^* v_1,A^* v_2 \rangle
$$

Any ideas here would be appreciated.

My proof:

Let $\{v_{\lambda,i}\}$ be an orthonormal basis of eigenvectors (as guaranteed by the spectral theorem) such that
$$
A v_{\lambda,i} = \lambda v_{\lambda,i}
$$
Let $v_1,\lambda_1$ and $v_2,\lambda_2$ be eigenpairs with $\lambda_1 \neq \lambda_2$.  We may write
$
v_1 = \sum_{i,\lambda}a_{i,\lambda}v_{i,\lambda}
.$
We then have
$$
0 = Av_1 - \lambda_1 v_1 = \sum_{i,\lambda}(\lambda - \lambda_1)a_{i,\lambda}v_{i,\lambda}
$$
So that $a_{i,\lambda} = 0$ when $\lambda \neq \lambda_1$.  Similarly, we may write $v_2 = \sum_{i,\lambda}b_{i,\lambda}v_{i,\lambda}$, and note that $b_{i,\lambda} = 0$ when $\lambda \neq \lambda_2$.  From there, we have
$$
\langle v_1,v_2 \rangle = \sum_{i,\lambda}a_{i,\lambda}b_{i,\lambda}
$$
the above must be zero since for each pair $i,\lambda$, either $a_{i,\lambda}=0$ or $b_{i,\lambda} = 0$.

答:
I'll use the notation from Introduction to Linear Algebra by Strang. Let $A^H$ be the transpose conjugate of $A$, and suppose $A$ is normal, i.e. $AA^H = A^H A$.

Examine an eigenpair $Ax = \lambda x$. Multiply by $A^H$ and we have $A^H Ax = AA^H x = \lambda A^H x$. Assume only one $x$ with eigenvalue $\lambda$ (see note at end), so here we know that $A^H x$ is a multiple of $x$, i.e. $A^H x = c x$. Multiply by $x^H$ and we have $c = \bar \lambda$, i.e. the conjugate of $\lambda$.

Now consider another eigenpair $A y = \mu y$ with $\mu \neq \lambda$. From the result in the previous paragraph, we have $y^H A = \mu y^H$. Now multiply by $x$ and we obtain $\mu y^H x = \lambda y^H x$, which leads to $y^H x = 0$.

Note: if there are multiple $x$ with eigenvalue $\lambda$, then choose $x$ to be the linear combination that is also an eigenvector of $A^H$. Such a choice is always be possible given that there's at least one eigenvector with eigenvalue $\lambda$ (proof is left as an exercise for the reader).


问:
Divergence of Harmonic Series
I understand that the divergence of the harmonic series is a classic proof but what I don't understand is the way it seems to contradict standard methods of finding if a series converges or not.

For the series $\Sigma \frac{1}{n}$ as $n$ approaches infinity it would seem that $\frac{1}{n}$ would tend to $0$. We know this isn't the case though but why is this seemingly standard procedure wrong? Is there some extra proof that must be shown other than just inserting $n$ for infinity?

答:
If the terms of a series do not approach $0$, then the series diverges.

If the terms of a series do approach $0$, then the series may or may not converge.

The harmonic series is an example of a series that does not converge though the terms approach $0$.

See  this Wikipedia article.

Title: Divergence of Harmonic Series


答:
You just got a little "sloppy" with notation, and forgot that your aim is to take the derivative of the integral you found:

First, when integrating, after evaluating the indefinite integral, drop the integral sign:

$$\frac{d}{dx}\int_{0}^{e^{x^{2}}} \frac{1}{\sqrt{t}}dt = \frac d{dx}\left[ 2t^{\frac{1}{2}}\Big|_0^{e^{x^2}}\right]$$

Then
$$\dfrac d{dx}\left[[2(e^{x^{2}})^{\frac{1}{2}}]-[2(0)^{\frac{1}{2}}]\right]= \frac{d}{dx} \left[2(e^{x^{2}})^{\frac{1}{2}}\right]$$

Precisely...Now, use what you know about $\frac d{dx}\left(e^{x^2}\right) = \frac d{dx}\left(x^2\right)\cdot \left(e^{x^2}\right)$ and apply the same strategy (chain rule) to your result of evaluating the indefinite integral, as you did well.

$$\frac{d}{dx} \left[2(e^{x^{2}})^{\frac{1}{2}}\right] = \frac{d}{dx} \left[2\left(e^{\large\frac{x^{2}}{2}}\right)\right] = 2\frac{d}{dx}\left(\frac{x^2}{2}\right)\cdot e^{\large\frac{x^2}{2}}= 2\cdot \dfrac 12\cdot 2xe^{\large\frac{x^{2}}{2}} = 2xe^{\frac{x^{2}}{2}}$$

Title: $\frac{d}{dx}\int_{0}^{e^{x^{2}}} \frac{1}{\sqrt{t}}dt$


问:
(Use the Vitali covering lemma) to show that the union of closed intervals of real numbers is measurable
I have seen similar questions before but the replies given were less than satisfactory.

Let $\mathcal{F} = \{F_\alpha\}_{\alpha \in J}$ be an arbitrary family of closed intervals of real numbers. Let $E = \displaystyle \bigcup_{\alpha \in J} F_\alpha$, and let $E_n = E \cap [-n, n]$. Let $V$ be a Vitali covering of $E$; then $V$ is a Vitali covering of $E_n$. By the Vitali Lemma, we can find a finite, pairwise disjoint collection $\{V_i\}_{i = 1}^m$ of $V$ such that $m^\ast(E_n \smallsetminus \displaystyle \bigcup_{i = 1}^m V_i) < \varepsilon$. Since each $V_i$ is an interval of real numbers, and intervals are Lebesgue measurable, it follows that $\displaystyle \bigcup_{i = 1}^m V_i$ is measurable. Thus
$$\bigcup_{n = 1}^ \infty E_n$$
is a countable collection of measurable sets and is therefore measurable. Moreover, the Lebesgue measure of this union is equal to the Lebasgue measure of $E$, which is what we sought to prove.

Does this make any sense?

答: No, your proof is not correct. In your proof, you need to show that the union of Vi is in En, and your conclusion can hold. Thus, you cannot find an arbitrary Vitali cover of E. Did you realize that you did not even use the fact that E is the union of a family of closed interval? Now, you can select all closed intervals included in E. Then the collection of them is a Vitali cover of E (you can show it). And the union the the select sequence is in E.</p>