问:
Proof by induction on two variables
Suppose we have relation

$T(n,k)=\begin{cases}
  T(n,k)=T(n-1,k-1)+T(n-1,k)+1, & \\
  1,\text{$k=0  \text{ or } n=k$}
 \end{cases}
$.

Now we relate above relation to $C(n,k)$.

Let $C(n,k)=T(n,k)+1$ then
$C(n,k)=C(n-1,k-1)+C(n-1,k)$ this is exactly relation binomial coefficients.

Now i want prove that $C(k,n)=2\binom{n}{k}$
, But when i use iduction on $k,n$ get stuck. How i can use induction on this problem?

My attempt:(i read this post)

First i define $P(t)$ such that $t=k+n$, and $0\leq k\leq n$ , and $C(n,k)=2\binom{n}{k}$, now main problem is how i can relate $P(t+1)$ to $P(t)$? Any hint be appreciated.

答:
As your question states,

$$C(n,k) = T(n,k) + 1, \; C(n,k) = C(n-1,k-1) + C(n-1,k) \tag{1}\label{eq1A}$$

You've also defined $P(t)$ for $t \ge 0$ as

$$k + n = t, \; 0 \le k \le n, \; \implies \; C(n,k) = 2\binom{n}{k} \tag{2}\label{eq2A}$$

Due to the use of $(n - 1) + (k - 1) = (n + k) - 2 = t - 2$ and $(n - 1) + k = (n + k) - 1 = t - 1$ on the right side of \eqref{eq1A}, this requires using strong induction, with $2$ base cases.

First, $P(0)$ requires $k = n = 0$. This gives $C(0,0) = T(0,0) + 1 = 1 + 1 = 2 = 2\binom{0}{0}$. Next, $P(1)$ has $k = 0$ and $n = 1$, with $C(1, 0) = T(1, 0) + 1 = 1 + 1 = 2 = 2\binom{1}{0}$. Thus, both of the base cases, i.e., $P(0)$ and $P(1)$, are satisfied.

Next, the inductive step is to assume $P(t)$ is satisfied for all $t \le m$ for some integer $m \ge 1$. For $P(m + 1)$, which has

$$k + n = m + 1, \; 0 \le k \le n \tag{3}\label{eq3A}$$

there are $3$ cases to consider. First, if $k = 0$, then $n = m + 1$, with $C(m + 1, 0) = T(m + 1, 0) + 1 = 1 + 1 = 2 = 2\binom{m + 1}{0}$, so $P(m + 1)$ applies here. The second case is with $k = n$, giving $C(n, n) = T(n, n) + 1 = 1 + 1 = 2 = 2\binom{n}{n}$, so $P(m + 1)$ is also satisfied. The third case is with $1 \le k \le n - 1$. Using \eqref{eq3A}, we have

$$(n - 1) + (k - 1) = n + k - 2 = m - 1, \; 0 \le k - 1 \lt n - 1 \tag{4}\label{eq4A}$$

$$(n - 1) + k = n + k - 1 = m, \; 0 \lt k \le n - 1 \tag{5}\label{eq5A}$$

Thus, by the inductive step, $P(m - 1)$ and $P(m)$ applies to the terms on the right side in \eqref{eq1A}. This therefore gives, while also using Pascal's rule, that

$$\begin{equation}\begin{aligned}
C(n,k) & = C(n-1,k-1) + C(n-1,k) \\
& = 2\binom{n-1}{k-1} + 2\binom{n-1}{k} \\
& = 2\left(\binom{n-1}{k-1} + \binom{n-1}{k}\right) \\
& = 2\binom{n}{k}
\end{aligned}\end{equation}\tag{6}\label{eq6A}$$

which shows $P(m+1)$ applies in this third case as well.  This therefore proves using strong induction that $P(t)$ in \eqref{eq2A} is true for all $t \ge 0$.

Title: Proof by induction on two variables


问:
How to evaluate $ \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...$?
I was given the series:
$$ \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...$$
Making some observations I realized that the $ a_{n} $ term would be the following:
$$ a_{n} = \frac{1}{ 3^{n} } \left(\frac{3}{2}+ \frac{(-1)^{n}}{2} \right)$$
What I wanted to do is to find the result of the series, so the answer would be:
$$\sum_{n=1}^{ \infty } \frac{1}{ 3^{n} } \left(\frac{3}{2}+ \frac{(-1)^{n}}{2} \right) = \frac{3}{2} \sum_{n=1}^{ \infty } \left[ \frac{1}{3^{n} } \right]+ \sum_{n=1}^{ \infty } \left[\frac{1}{3^{n} } \frac{(-1)^{n}}{2}\right]$$
I can tell that the first term is convergent because it is a geometric series, in fact, the result is $\frac{3}{4}$. However, I have no clue in how to solve the second term series. I should say that the series given in the beginning is convert and its result is 5/8. How to arrive to it is a mystery to me.

ContentLicense: CC BY-SA 4.0

Body:
Your sum can be put as

$$(1+\frac 23)\Bigl(\frac 13+\frac{1}{3^3}+\frac{1}{3^5}...\Bigr)=$$
$$\frac 53\frac 13\Bigl(1+\frac 19+\frac{1}{9^2}+...\Bigr)=$$
$$\frac 53\frac 13 \frac{1}{1-\frac 19}=\frac 58$$


问:
Sub Gaussian concentration of correlated random variables
Let $X_1, X_2, \dots, X_n$ be $n$ zero mean random variables. It is given that $X_i$ is sub-Gaussian with parameter $\sigma_i^2$ (assumed known) for $i = 1,2, \dots, n$. Consider the random variable $Y = \sum_{i = 1}^n X_i$. It is known that $Y$ is a sub-Gaussian random variable. We denote the corresponding sub-Gaussian parameter as $\sigma_Y^2$.

If $X_i$'s are given to be independent, then it is well-known that $\sigma_Y^2 = \sum_{i = 1}^n \sigma_i^2$. Similarly, a well known upper bound for the general case is $\sigma_Y^2 = \left( \sum_{i = 1}^n \sigma_{i}\right)^2$. However, when the random variables $X_i$'s are correlated such that the covariance matrix is known, can this bound on $\sigma_Y$ be improved using this information? In particular, if we know that the random variables are negatively correlated for all $i$, I would expect the bound on $\sigma_Y$ to be closer to that obtained in the independent case than the one obtained in the general case.

I was wondering if there was any way to prove the same or any standard results that account for correlation between random variables. I was also interested in the case where the random variables are sub-Exponential. Any leads or references will be appreciated. Thanks!

答:
Here is a special case for Gaussians. I'm sceptical whether such a nice result can hold very generally.

If $X\sim \mathcal{N}(0,\sigma^2)$, its sub-gaussian norm is $||X||_{\psi_2}=\sigma$.
If $(X_1,\dots,X_n)$ is jointly Gaussian-distributed, then $\sum_{i=1}^n X_i$ is also Gaussian.
So, you can determine its sub-gaussian norm by analyzing the variance of $\sum_{i=1}^n X_i$.
Here, the covariance matrix will indeed show up.

For example, if $(X,Y)$ are jointly Gaussian distributed with mean zero, then:
$$
||X+Y||_{\psi_2}=\sqrt{Var[X+Y]}=\sqrt{E[X^2]+E[Y^2]+2E[XY]}=\sqrt{\sigma^2_X+\sigma^2_Y+\rho_{XY}\sigma_X\sigma_Y}
$$
Note that for $\rho_{XY}=0$, we have $||X+Y||_{\psi_2}^2=||X||_{\psi_2}^2+||Y||_{\psi_2}^2$ and for $\rho_{XY}=1$, $||X+Y||_{\psi_2}^2=(||X||_{\psi_2}+||Y||_{\psi_2})^2$.
So, for the Gaussian case, we can indeed nicely interpolate between the two extreme cases you describe. As stated in the beginning, I doubt that the general case can be this terrific.


问:
Combinatorial identity on decreasing dice throws
Suppose I repeatedly throw fair $n$-sided dice until I throw a $1$, at which point I stop.

I want to know the probability $p(n)$ that my sequence of throws will be decreasing, such as $5-4-2-1$ or just $1$ immediately, but not $2-3-1$ which would count as a failure.

The answer clearly depends on whether I am considering strictly decreasing sequences in which case $4-2-2-1$ would count as a failure, or weakly decreasing sequences in which case it would be a success. Allowing weakly decreasing sequences would increase the probability and would mean there is no bound on the potential length of successful sequences.

For strictly decreasing sequences I have $$p_s(n)=\sum\limits_{k=1}^n {n-1 \choose k-1}\frac{1}{n^k} = \frac{(n+1)^{n-1}}{n^n}$$

For weakly decreasing sequences and $n>1$ I have $$p_w(n)=\sum\limits_{k=1}^\infty {n+k-3 \choose k-1}\frac{1}{n^k} = \frac{n^{n-2}}{(n-1)^{n-1}}$$

For example with $n=6$, I get $p_s(6)=\frac{7^5}{6^6}\approx 0.36023$ and $p_w(6)=\frac{6^4}{5^5}\approx 0.41472$. In general $$p_s(n)=p_w(n+1).$$  Is there a combinatorial argument that leads to this equality between the finite sum and the infinite sum?

答:
We can get more specific. For each subset $K\subseteq[n]$, what is the probability that the sequence will be strictly decreasing, and the set of values appearing will be $K$? Call this $p_s(n,K)$ and $p_w(n,K)$, depending on the interpretation of "decreasing."

$p_s(n,K)=\frac1{n^{|K|}}$, obviously.

$p_w(n,K)=\frac1n \cdot \frac1{(n-1)^{|K|-1}}$. There is a $1/n$ chance the first number rolled is the highest value of $K$. You then keep rolling until you get a different number, and the next new number has a $1/(n-1)$ chance of being the second highest value of $K$. Similarly, each subsequent new number has a $1/(n-1)$ chance of being the next value of $K$.

Using these equations, you can show that for any $K$ with $[1]\subseteq K\subseteq [n]$,
$$
p_s(n,K)=p_w(n+1, K)+p_w(n+1,K\cup\{n+1\}).
$$
Summing over all valid $K$, you get $p_s(n)=p_w(n+1)$.

Intuitively, when allow weakly decreasing sequences, the rerolled values can be ignored, reducing the number of sides of the die by one for all rolls except the first one. For the first roll, we have the equality $1/n=1/(n+1)+1/((n+1)\cdot n)$ which allows everything to work out.


问:
How to prove two subspaces are complementary
To give some context, I'm continuing my question here.

Let $U$ be a vector space over a field $F$ and $p, q: U \rightarrow U$ linear maps. Assume $p+q = \text{id}_U$ and $pq=0$. Let $K=\ker(p)$ and $L=\ker(q)$. From the previous question, it is proven that $p^2=p$, $qp=0$ and $K=\text{im}(q)$.

Now I want to prove that $U=K \oplus L$.

By definition, I know that if they are complementary, then $K \cap L = \left\{0 \right\}$ and $K + L =U$. Also, from my lecture notes it mentions a certain proposition that when applied to this question, I can say that subspaces $K$ and $L$ of a vector space $U$ are complementary if and only if each vector $u \in U$ can be written uniquely as $u = k + l$, where $k \in K$ and $l \in L$.

How do I go about proving this? It's hard for me to know what's going on. I would appreciate it if someone could help me draw a diagram representing the linear transformations of $p$ and $q$ as it is easier for me to see. Thank you for your time.

答:
$\Rightarrow$

Let $u\in U$. Since $U=K+L$ so $u=k+l$ where $k\in K$ and $l\in L$ and for uniqueness suppose that $u=k'+l'$ then
$$k-k'=l'-l\in K\cap L=\{0\}$$
hence $k=k'$ and $l=l'$.

$\Leftarrow$

Let $u\in U$ then there's $k\in K$ and $l\in L$ s.t. $u=k+l$ so $U=K+L$ and if $x\in K\cap L$ then
$$x=x+0=0+x$$
and by uniqueness $x=0$ so
$$K\cap L=\{0\}$$

ContentLicense: CC BY-SA 3.0


问:
Pdf and Mgf of random variable
Suppose that $X$ and $Y$ are independent random variables that follow an exponential distribution with mean $1$ and let $W = Y - X$.

(i) Using the cumulative distribution function (cdf), find the probability density function (pdf) of $W$, $f_W(w)$ for $w > 0$. (No need to compute $w \leq 0$).

(ii) Determine the moment generating function (mgf) of $W$ by finding the mgfs of $Y$ and $-X$. Justify where the mgf is defined.

I'm not quite sure how to do this problem but here is what I did:

(i)

Suppose $X, Y$ have means $\lambda_1$ and $\lambda_2$. If $W = Y - X$ then (I think)

$$P(W \leq w) = \int_0^\infty \int_{0}^{x+z} \lambda_1 e^{-\lambda_1 x} \lambda_2 e^{-\lambda_2 y}\,dy \ \ dx$$

If we substitute $\lambda_1 = \lambda_2 = 1$ then

$$P(W \leq w) = \int_0^\infty \int_{0}^{x+z} e^{-x - y}\,dy \ \ dx$$

(ii)

The moment generating function for an exponential distribution is $M(t) = \frac{1}{1- \theta t}$ where $\theta > 0$ and $t < \frac{1}{\theta}$.

If the mean $\theta = 1$ then we know that for $Y$

$M_Y(t) = \frac{1}{1-t}$

As for $-X$, I'm not sure.

I'm not sure about the formula in (i) or how to derive it, and I'm confused on finding the mgf of $-X$ in (ii). But in any case I am stuck. Any assistance will be much appreciated.

答:
For the formula in (i),
$$ \mathbb{P}(W \leq w) = \mathbb{P}(f(X,Y) \in ]-\infty,w]) = \mathbb{P}((X,Y) \in f^{-1}(]-\infty,w]))$$
where $f(x,y) = y-x$. Since $X \sim \mathcal{E}(\lambda_1)$ and $Y \sim \mathcal{E}(\lambda_2)$ are independent and admit both density, denoted $f_X$ and $f_Y$, the couple $(X,Y)$ admites the density
$f_{(X,Y)}(x,y) = f_X(x)f_Y(y)$ and thus
\begin{align}
\mathbb{P}((X,Y) \in f^{-1}(]-\infty,w])) &= \int_{f^{-1}(]-\infty,w])} \lambda_1e^{-\lambda_1x}\lambda_2e^{-\lambda_2y}1_{x >0}1_{y>0}dxdy\\
&= \int_{\mathbb{R}^2} 1_{y-x \leq w} \lambda_1e^{-\lambda_1x}\lambda_2e^{-\lambda_2y}1_{x >0}1_{y>0}dxdy\\
&= \int_0^{\infty} \int_0^{\infty} 1_{y \leq x+w}\lambda_1e^{-\lambda_1x}\lambda_2e^{-\lambda_2y}dxdy \\
&= \int_0^{\infty} \int_0^{\infty} 1_{y \leq x+w}\lambda_1e^{-\lambda_1x}\lambda_2e^{-\lambda_2y}dydx \\
&= \int_0^{\infty} \int_0^{\infty} 1_{y \leq x+w}1_{x+w \geq 0}\lambda_1e^{-\lambda_1x}\lambda_2e^{-\lambda_2y}dydx \\
\end{align}
and now you can adapt the rest of the computations depending on the sign of $w$.

For (ii), for all $t \in ]-1,1[$,
$$ M_{-X}(t) = \mathbb{E}[e^{-tX}] = M_X(-t) = \frac{1}{1+t}.$$

Title: Pdf and Mgf of random variable


答:
There is a much simpler way to solve this.
You know how to compute the geometric series
$$\frac{1}{3} + \frac{1}{3^2}+ \frac{1}{3^3}+ \frac{1}{3^4}+ \cdots = \frac{1}{2}$$
you also know how to compute the geometric series
$$\frac{1}{9} + \frac{1}{9^2}+ \frac{1}{9^3}+ \frac{1}{9^4}+ \cdots= \frac{1}{8}$$
But this can be rewritten in terms of $3^2$, so that you get
$$\frac{1}{3^2} + \frac{1}{3^4}+ \frac{1}{3^6}+ \frac{1}{3^8}+ \cdots$$
Now the answer to your question is simply $\frac{1}{2} +\frac{1}{8}$.

Title: How to evaluate $ \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...$?

Body:
I was given the series:
$$ \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...$$
Making some observations I realized that the $ a_{n} $ term would be the following:
$$ a_{n} = \frac{1}{ 3^{n} } \left(\frac{3}{2}+ \frac{(-1)^{n}}{2} \right)$$
What I wanted to do is to find the result of the series, so the answer would be:
$$\sum_{n=1}^{ \infty } \frac{1}{ 3^{n} } \left(\frac{3}{2}+ \frac{(-1)^{n}}{2} \right) = \frac{3}{2} \sum_{n=1}^{ \infty } \left[ \frac{1}{3^{n} } \right]+ \sum_{n=1}^{ \infty } \left[\frac{1}{3^{n} } \frac{(-1)^{n}}{2}\right]$$
I can tell that the first term is convergent because it is a geometric series, in fact, the result is $\frac{3}{4}$. However, I have no clue in how to solve the second term series. I should say that the series given in the beginning is convert and its result is 5/8. How to arrive to it is a mystery to me.


答:
The series can also be written as
$$
\begin{align}
\overbrace{\ \sum_{k=1}^\infty\frac1{3^k}\ }^{\substack{\text{each power}\\\text{of $3$}}}+\overbrace{\ \sum_{k=1}^\infty\frac1{9^k}\ }^{\substack{\text{even powers}\\\text{of $3$}}}
&=\frac{\frac13}{1-\frac13}+\frac{\frac19}{1-\frac19}\\
&=\frac12+\frac18\\[6pt]
&=\frac58
\end{align}
$$


问:
Finding the change of basis matricies
Not quite sure what to do for this question.

Let $B_1$= {$\vec{u}_1,\vec{u}_2,\vec{u}_3$} and $C_1$= {$\vec{w}_1,\vec{w}_2,\vec{w}_3$} be two bases for $\mathbb{R}^3$ such that
$\vec{w}_1=\vec{u}_1-\vec{u}_2,$  $\vec{w}_2=\vec{u}_2-\vec{u}_3, $ $\vec{w}_3=\vec{u}_2+\vec{u}_1 $.
Find the change of coordinates matrices $P_{{C_1}\to {B_1}}$ and $P_{{B_1}\to {C_1}}$

Any hints would be greatly appreciated.

答:
(I'm not sprinking all the little \vec s through this answer.)

Interpreted in the $C_1$ basis, the vector $\begin{pmatrix}1\\0\\0\end{pmatrix}$ is $w_1$.  To indicate with respect to which basis a vector is written, we subscript it with that basis.  So we could write
$$  \begin{pmatrix}1\\0\\0\end{pmatrix}_{C_1} = w_1  \text{.}  $$

Notice that
$$  P_{C_1 \rightarrow B_1} \cdot \begin{pmatrix}1\\0\\0\end{pmatrix}_{C_1}  $$
gives $1$ times the first column of $P_{C_1 \rightarrow B_1}$ plus zero times the other columns.  We are given that (the same quantity since the $P$ matrix is being multiplied by an alias for the same vector)
$$  P_{C_1 \rightarrow B_1} \cdot w_1 = u_1 - u_2 = \begin{pmatrix}1\\-1\\0\end{pmatrix}_{B_1}  \text{.}  $$

So the first column of $P_{C_1 \rightarrow C_2}$ must be $\begin{matrix}1\\-1\\0\end{matrix}$.  Similarly, the second column must be $\begin{matrix}0\\1\\-1\end{matrix}$.  You should be able to find the last column.

To go in the other direction, solve for $u_1$, $u_2$, and $u_3$ in terms of the $w_j$s and then repeat the above analysis using these three solutions to write down the columns of $P_{B_1 \rightarrow C_1}$.

Title: Finding the change of basis matricies


问:
Tracial states on $C_\text{r}^\ast(\mathbf{D}_\infty)$, $\mathbf{D}_\infty$ being the infinite dihedral group
Let $\mathbf{D}_\infty$ be the infinite dihedral group, i.e. the group generated by two elements $s$ and $t$ with $s^2=t^2 =e$ which are free with respect to each other. Consider the reduced group $C^\ast$-algebra $C_\text{r}^\ast (\mathbf{D}_\infty)$. There exist two canonical tracial states on this $C^\ast$-algebra, namely the state $\tau$ with $\tau(1)=1$ and $\tau(\lambda_s)=\tau(\lambda_t)=0$ and (since $\mathbf{D}_\infty$ is amenable) the character coming from the trivial representation of the group $\mathbf{D}_\infty$ (i.e. the corresponding state maps $1\mapsto1$, $\lambda_s\mapsto1$, $\lambda_t \mapsto 1$). Of course convex combinations of these two states are also tracial, but are these all the tracial states? If no, is there a complete description of all tracial states on $C_\text{r}^\ast(\mathbf{D}_\infty)$? How do they look like?

答:
G. K. Pedersen  proved that  $C^*(D_\infty )$ is isomorphic to the subalgebra of $C([-1, 1], M_2)$ formed by all
continuous functions
$$
  f:[-1,1]\to M_2,
  $$
such that $f(-1)$ and $f(1)$ are diagonal, via an isomorphism taking the two generators $s$ and $t$ to the functions
$f_s$ and $f_t$ defined by
$$
  f_s(x) = \pmatrix{ 1 & 0 \cr 0 & -1},
  \quad \text {and} \quad
  f_t(x) = \pmatrix {x & \sqrt{1-x^2} \cr \sqrt{1-x^2} & -x}.
  $$

If $\mu $ is any probability measure on $[-1, 1]$ then the functional
$$
  \tau (f) = \frac 12\int_{-1}^1 \text{tr}(f(x))\,  d\mu (x)
  $$
is a trace.  There are four other traces on this algebra, actually characters, not covered by the  above class of
examples, given by
$$
  \tau ^-_{1}(f) = f(-1)_{1, 1},
  $$
$$
  \tau ^-_{2}(f) = f(-1)_{2, 2},
  $$
$$
  \tau ^+_{1}(f) = f(1)_{1, 1},
  $$
$$
  \tau ^+_{2}(f) = f(1)_{2, 2},
  $$
so there are many more traces  than the ones described by the OP.  On the other hand,  I believe the above span all
traces of    $C^*(D_\infty )$.


问:
Splitting a circle into 3 equal parts using 2 lines
The picture probably explains my question best.

I need to find a way to divide a circle into 3 parts of equal area with only 2 lines that intersect each other on the outline of the circle.

Also I need to check, if whatever diameter is between those lines, also splits circles with a different diameter into equal parts.

And lastly, and probably the most difficult question: How do I have to calculate the angle between x lines that all intersect in one point, so that the circle is split into x+1 parts with area = 1/(x+1) of the circle?

I tried my best, but couldn't even find a single answer or the right strategy to tackle the question...

1

答:
To find the angle $x$ between the red lines you need to solve the equation $$\sin(x)+x=\frac\pi3.\tag1$$
This can be done only numerically.

Yes it divides also the green circle in three equal parts.

Observe the essential difference between the cases with even and odd $n$. Whereas in former case the diameter drawn from the common intersection point is one of the "red" lines, in the latter case it is the bisector of the central angle. Let $x_k$ be the angle of a red line with the diameter. Then the following are the equations to find $x_k$:
$$
x_k+\frac12\sin2x_k=\frac\pi{n}k, \quad\text{with}\quad
k=-\frac n2+1\dots \frac n2-1.\tag2
$$
Observe that the numbers $k$ are integer for even $n$ and half-integer for odd $n$. For $n=3$ the equation $(2)$  reduces to $(1)$ with $x=2x_\frac12$.


答: Connect the ends of the chord (red line) with the center. You get isosceles triangle. Let $x$ be the angle opposite to the base (i.e. chord) in that triangle. You want: $$\frac{r^2x}{2}-\frac{r^2 \sin x}{2}=\frac{r^2\pi}{3}$$ or $$x-\sin x=\frac{2\pi}{3}$$ I do not think it can be solved analytically but it can be solved numerically with the answer $x \approx 2.60533$ radian. The same logic can be applied to more than 3 pieces. The angle you are looking for is $\pi-x$

Title: Splitting a circle into 3 equal parts using 2 lines

Body:
The picture probably explains my question best.

I need to find a way to divide a circle into 3 parts of equal area with only 2 lines that intersect each other on the outline of the circle.

Also I need to check, if whatever diameter is between those lines, also splits circles with a different diameter into equal parts.

And lastly, and probably the most difficult question: How do I have to calculate the angle between x lines that all intersect in one point, so that the circle is split into x+1 parts with area = 1/(x+1) of the circle?

I tried my best, but couldn't even find a single answer or the right strategy to tackle the question...

1


问:
Assuming that $P(x)$ is a polynomial and it is true that $P(3x)=26 \cdot P(x+1)$, then of what degree is the polynomial?
Assuming that $P(x)$ is a polynomial and it is true that $P(3x)=26 \cdot P(x+1)$, then of what degree is the polynomial?

This is the first time that I've seen such a question and hence I don't know how to attack it. What I attempted to do was the following:

If it is of degree 1, then $P(x)=ax+b$ hence $P(3x)=3ax+b$, but $P(3x)=26*P(x+1)$, hence
\begin{align*}
3ax+b &= 26 \cdot 3a(x+1)+b \\
3ax &= 26 \cdot 3a(x+1) \\
x & = 26 \cdot (x+1) \\
x &= 26 \cdot x+26,
\end{align*}

Hence $x$ needs to be a certain value, something which can't be true.

I then tried the same thing for degree $2, 3, 4$ in the hope of finding some sort of pattern, but I couldn't find one. Could you please explain to me how to solve this question as well as how to solve other similar style questions?

答: Hint: $P(3x)=26 P(x+1)$ implies $3^n a_n = 26 a_n$ and so $a_n=0$ because $26$ is not an integer power of $3$.


问:
Implicit Runge-Kutta-methods alternative formulation
Implicit Runge-Kutta-methods with $s$ stepts and with constant lenght $h$ for
$$
x'(t)=f(x(t),t), \quad x(t_0)=x_0
$$
have the form
$$
X_i=f(x_{n-1}+h \sum_{j=1}^s a_{ij}X_j,t_{n-1}+hc_i), \quad i=1,...,s \\ x_n=x_{n-1}+h\sum_{i=1}^s b_iX_i
$$
Now my lecture notes state that this system is equivalent to the following one
$$
Z_i=x_{n-1}+h \sum_{j=1}^s a_{ij}f(Z_j,t_{n-1}+hc_i), \quad i=1,...,s \\ x_n=Z_s
$$
with $c_s=1$ and $a_{si}=b_i$ if the Runge-Kutta-Method has order $d\geq1$ and $A=(a_{ij})_{i,j=1}^s$ is regular.

答:
This is just setting
$$
Z_i=x_{n-1}+h\sum a_{ij}X_j,
$$
where as before
$$
X_i=f(Z_i,t_{n-1}+c_ih).
$$
If you put the two identities together to eliminate $Z_i$, then you get the first form, if you eliminate $X_i$, then you get the second form.

This is valid for all Runge-Kutta methods, also the explicit ones.


问:
Proof of Existence of Algebraic Closure: Too simple to be true?
Having read the classical proof of the existence of an Algebraic Closure (originally due to Artin), I wondered what is wrong with the following simplification (it must be wrong, otherwise why would we bother with the extra complications in Artin's proof?):

Theorem: Let $K$ be a field, then $K$ has an algebraic closure $\bar{K}$ (i.e an algebraic extension that is algebraically closed).

"Proof": Define $A=\{ F \supset K | F \text{ is an algebraic extension of } K\}$ and inherit this with the usual partial order of inclusion. One can check that Zorn's lemma applies (union of a nested chain of algebraic extensions is itself algebraic). Thus take $\overline{K}$ to be a maximal element. It must be algebraically closed for otherwise there is an irreducible polynomial with root in some strictly bigger field.  $\blacksquare$

Now here is what I suspect is false about this proof: The definition of $A$ smells like your usual set theory paradoxes like Russell's paradox. In fact one could just as well use the same technique to prove that there exists a "largest set" which of course there does not. I am however under the impression that "most" working mathematicians ignore set theory foundations and just "do" mathematics, so is there a safe way of doing this (i.e: do "concrete everyday mathematics" by avoiding set theory) without getting burnt?

答:
This is a comment on a common error made in proves of existence of algebraic closures sin Artin's method (I don't have Artin's reference at hand, but I am sure he had a flawless argument)

The construction as suggested by @Dustan Levenstein is on the right track however, the implication from the arguments laid down there, specifically that the fields $K_n$ constructed iteratively satisfy $K=K_0\subset K_1\subset\ldots K_n\subset K_{n+1}\subset\ldots$ is not correct! Each $K_n$ is embedded isomorphically into a quotient field $K_{n+1}$,  but $K_n$ is not a subset of $K_{n+1}$.

To fix that little nuance, it seems one needs to take the colimit also known as (direct limit) of the $K_n$'s.

This a learnt from a note by Kevin Buzzard. The construction outlined  by @Dustan Levenstein produces a sequence of fields $\{K_n:n\in\mathbb{Z}_+\}$ with $K_0=K$ (the original field) and (field) homomorphisms $F_n\stackrel{\pi_n}{\longrightarrow} F_{n+1}$ such that $\pi_{n}(F_n)$ and $F_n$ are isomorphic (as fields). Let $L:=\bigsqcup_n K_n$ the disjoint union of the $F_n's$. In $L$ considers the equivalence relation $x Ry$ iff there is $z\in \overline{K}$ such that
\begin{aligned}
z&=\pi_{n,m}(x)=\pi_{m-1}\circ\ldots\circ\pi_n(x)\\
&=\pi_{\ell,m}(y)=\pi_{m-1}\circ\ldots\circ\pi_\ell(y)
\end{aligned}
where $\ell,n\leq m$ ($x\in K_n$ and $y\in K_\ell$ for some $n,\ell\in\mathbb{Z}_+$. The quotient space $\overline{K}:=L/R$ will be the desired object with the algebraic operations transported from the spaces $K_n$ to $L$: For $[x],\,[y]\in L$ suppose $x\in K_m$ and $y\in K_n$ with $m\leq m$. Define

$[x]+[y]=[\pi_{m,n}(x)+y]$

$[x]\cdot[y]=[\pi_{m,n}(x)\cdot y]$

where $ \pi_{m,n}=\pi_{m-1}\circ\ldots\circ\pi_n$. This can be show to define a field structure in the colimit $\overline{K}$ and that $K$ is embedded in $\overline{K}$ (say as $K\simeq K'\subset\overline{K}$), and $\overline{K}$ is algebraically closed (every non constant polynomial in $\overline{K}[x]$ has a root in $\overline{K}$). Now, consider the set $A$ of all algebraic elements in $\overline{K}$ over the $K'$. This is the algebraic closure of $K$ (or rather $K'$), that is $A$ is algebraically closed and every element of $A$ is algebraic over $K$.


问:
How do morphisms between affine varieties induce ring homomorphisms?
.An algebraic set defined by $S \subseteq k[x_1,...,x_n]$ is given by

$$X = \{a \in \mathbb{A}^n(k) | f(a) = 0, \forall f\in S\}$$

Suppose $X \subseteq \mathbb{A^n}(k)$ and $Y \subseteq \mathbb{A^m}(k)$ are algebraic sets. The morphisms between $X$ and $Y$ are the polynomial mappings:

$$\phi: X \rightarrow Y$$

given by:

$$p \mapsto (f_1(p),..,f_m(p))$$ for some $f_1,...,f_m \subset k[x_1,..,x_n]$.

Now what I don't understand is how do morphism induce a ring homomorphism:

$$k[y_1,..,y_m]/I(Y) \rightarrow k[x_1,..,x_n]/I(X)$$ given by

$$f \mapsto f \circ \phi$$.

I don't even get the intuition behind this, let alone why this would be the case in first place.
I'm really struggling to see why the map above induces a map between the coordinate rings of the two varieties. Can someone try to explain this in a simple language?

答:
One way to think about the elements of the coordinate ring $k[x_1,...,x_n]/I(X)$ is as functions from $X$ to $k$. Indeed, note that every coset $f+I(X)$ defines a function $\hat{f}:X\to k$ by $(a_1,...,a_n)\to f(a_1,...,a_n)$. The definition of $I(X)$ tells us that $\hat{f}$ is indeed well defined. Moreover, this map $\hat{f}$ fully determines the coset $f+I(X)$ in the sense that we have $f+I(X)=g+I(X)$ if and only if $\hat{f}=\hat{g}$ . So usually we don't really denote the map by $\hat{f}$, but rather just think of the elements of $k[x_1,...,x_n]/I(X)$ themselves as of functions from $X$ to $k$. Also, many times we just write $f$ instead of $f+I(X)$.

Once you understand this, the homomorphism from $k[y_1,...,y_m]/I(Y)\to k[x_1,...,x_n]/I(X)$ should become clear. If $f\in k[y_1,...,y_m]/I(Y)$ then $f$ is map $Y\to k$. Thus the composition $f\circ\phi:X\to k$ is defined. Moreover, $f\circ\phi$ is exactly the function which corresponds to the element $f\circ (f_1,...,f_m)+I(X)\in k[x_1,...,x_n]/I(X)$. So if, as before, we identify the elements of the coordinate ring with the corresponding functions, we get that $f\circ\phi$ is indeed an element of $k[x_1,...,x_n]/I(X)$.

Note: By $f\circ (f_1,...,f_m)$ I mean the polynomial $f(f_1(x_1,...,x_n),...,f_m(x_1,...,x_n))\in k[x_1,...,x_n]$.


问:
Double covering $SL(2, \mathbb{C}) \to SO^{+}(1,3)$ preserves adjoints
Consider the usual double covering $\kappa : SL(2, \mathbb{C}) \to SO^{+}(1,3)$ defined by $\kappa(A)x_{*} = A x_{*} A^{\dagger}$ where $x \in \mathbb{R}^4$ is identified with the $2\times2$ Hermitian matrix $x_{*} = x^0 I + x^1 \sigma_1 + x^2 \sigma_2 + x^3 \sigma_3$ where $\sigma_i$ are the Pauli matrices. In his Quantum Field Theory book, Folland claims (p. 71) that $\kappa(A^\dagger) = \kappa(A)^\dagger$ where $A^\dagger$ is the conjugate transpose of $A$ and $\kappa(A)^\dagger$ is the transpose of $\kappa(A)$. Is there a simple way to see this or need one work out a formula for $\kappa(A)$ in terms of the entries of $A$?

答: I think I figured it out. Given that the $(I, \sigma)$ basis for the space of Hermitian $2 \times 2$ matrices is orthogonal under the trace inner product, it follows that (up to a factor of $2$) the $(j, k)$ entry of $\kappa(A)$ is $\text{tr}(A \sigma_j A^{\dagger} \sigma_k)$ (here I'm using $\sigma_0 = I$). Similarly the $(k, j)$ entry of $\kappa(A^{\dagger})$ is (up to the same factor of $2$) given by $\text{tr}(A^{\dagger} \sigma_k A \sigma_j)$. These are the same due to the ciclicity of the trace, so that $\kappa(A^\dagger)$ is the transpose of $\kappa(A)$.


问:
Define a relation ∼ on Z as x ∼ y if and only if |x− y| ≤ 1. Say whether ∼ is reflexive. Is it symmetric? Transitive?
Just checking to see if I'm doing this right.

$|x-x| = 0 \le 1$ and similarly for y, so this is reflexive.

If $x,y \in Z$ then $|x - y | \le 1$ so $x\le 1+y$ and $y \ge x-1$

$|x- (x-1)| \le 1 \Rightarrow |-1|\le 1$. And for y: $|(1+y)-y|\le 1 \Rightarrow |1|\le1.$ So this is symmetric.

I'm a little stuck on proving transative. This is what I think the start looks like.

If $x,y,z \in Z$ for $|x-y|\le1$ and $|y-z|\le1$ then $y\ge x-1$ and $y\le 1+z$.

答:
You are right that it is symmetric but I would simplify the proof even further by noting that $\lvert x-y\rvert = \lvert -(y-x) \rvert = \lvert y-x\rvert$.

As for transitivity, your relation is actually not transitive. Consider a simple counterexample of $x=2,y=3,z=4$. Clearly $\lvert 2-3\rvert \leq 1$ and $\lvert 3-4\rvert \leq 1$, but $\lvert 2-4\rvert = 2>1$, so it's not transitive.


问:
The largest negative root of the trigonometric equation
$$
f(x) = 2\cos^2 x + 9\sin x - 6
$$

Find (in degrees) the largest negative root of the equation $f(x) = 0$.

Find (in degrees) the smallest positive integer value of $x$ for which  inequality $f(x) > 0$.

Body:
hint
\begin{align*}
f(x) &= 2\bigl(1-\sin^2(x)\bigr)+9\sin(x)-6 \\
&= -2\sin^2(x)+8\sin(x)+\sin(x)-4 \\
&= -2\sin(x)\bigl(\sin(x)-4\bigr)+\bigl(\sin(x)-4\bigr) \\
&= \bigl(\sin(x)-4\bigr)\bigl(1-2\sin(x)\bigr)
\end{align*}

So
$$f(x)=0\iff \sin(x)=\frac 12$$
and
$$f(x)>0\iff \sin(x)>\frac 12$$


问:
Find closest point, subject to linear inequality constraints
Given a point $p\in \mathcal{R}^2$, I want to compute the closest point $x \in \mathcal{R}^2$, subject to linear inequality constraints $Ax \leq b$. That is,

$$\begin{array}{ll} \text{minimize} & \|x-p\|_2\\ \text{subject to} & A x \leq b\end{array}$$

I believe that this can be done with an off-the-shelf quadratic programming solver, but I'm wondering if there is a more efficient algorithm specialized to two variables ($x \in \mathcal{R}^2$) and Euclidean distance ($\min \|x - p\|$).

I think that this question is a bit different than this one because I don't have a direct representation of the feasible region as a list of vertices.

答:
This problem can be set up as a minimization problem using the calculus of variations and then solved as a linear equation.  Basically, you minimize the equation

$$J = \frac{1}{2}\sum_{i=1}^{N} (x_i - p_{i})\cdot(x_i - p_{i}) + \vec{\lambda}^T (A\vec{x} - \vec{b})$$

where $\vec{x}$ is the original vector of length $N$ and $\vec{\lambda}$ is a vector of unknown constants with a length equal to the number of equations in the set $A\vec{x} = \vec{b}$ (we will call this length $M$).

Once you have written this equation, take the partial derivatives of J with respect to the vectors $\vec{x}$ and $\vec{\lambda}$, set them equal to $0$

$$ \begin{bmatrix}
\frac{\partial J}{\partial \vec{x}} \\
\frac{\partial J}{\partial \vec{\lambda}}
\end{bmatrix} = 0$$

and then solve for $\vec{x}$ and $\vec{\lambda}$.  This should give you a linear equation of the form $C\vec{y}=\vec{d}$ where the values for $C$, $\vec{y}$, and $\vec{d}$ are respectively:

$$ \begin{bmatrix}
{I}_{NxN} & A^T \\
A & {0}_{MxM}
\end{bmatrix} \begin{bmatrix}
\vec{x} \\
\vec{\lambda}
\end{bmatrix} = \begin{bmatrix}
\vec{p} \\
\vec{b}
\end{bmatrix}$$

$N$ = number of variables in $\vec{x}$

$M$ = number of constraint equations in $A\vec{x}=\vec{b}$

$\vec{y} = C^{-1} \vec{d}$ , and the first $N$ values of $\vec{y}$ will be your answer for $x_i$ (the remaining $M$ values will be the values of $\vec{\lambda}$, and are not actually necessary for your solution).


问:
Properties of an alternating bilinear form its coordinate matrix
I found that I lack many basic knowledge about linear algebra, so read the wiki article about Bilinear Forms. Especially this Paragraph. I tried to proof of "Every alternating form is skew-symmetric." was quite easy. And I found a counter-example for the inverse if $char(F) = 2$.

However, I am currently trying to find a proof for this:

A bilinear form is alternating if and only if its coordinate matrix is skew-symmetric and the diagonal entries are all zero

I looked at the following equations, to understand that.

$\begin{pmatrix}v_1 & v_2\end{pmatrix} \cdot \begin{pmatrix}a & b \\ c & d\end{pmatrix} \cdot \begin{pmatrix}v_1 \\ v_2\end{pmatrix} = 0$

$\begin{pmatrix}v_1 & v_2\end{pmatrix} \cdot \begin{pmatrix}a & b \\ c & d\end{pmatrix} \cdot \begin{pmatrix}w_1 \\ w_2\end{pmatrix} = - \begin{pmatrix}w_1 & w_2\end{pmatrix} \cdot \begin{pmatrix}a & b \\ c & d\end{pmatrix} \cdot \begin{pmatrix}v_1 \\ v_2\end{pmatrix}$

And they are both solvable for $a=0, d=0, c=-b$, thus the matrix $\begin{pmatrix}0 & b \\ -b & 0\end{pmatrix}$.

However, I am not any nearer to a proof. Can someone please point me into the right direction?

答:
If $B(x,y)$ is alternating, clearly the coordinate matrix of $B$ has a zero diagonal. Also, from the identity
$B(x+y,x+y) = B(x,x) + B(y,y) + B(x,y) + B(y,x)$, we see that $B$ is skew-symmetric. So is its coordinate matrix (why?).

Conversely, if the coordinate matrix $(b_{ij})$ of $B$ is skew-symmetric, then $b_{ij}x_ix_j+b_{ji}x_jx_i=0$ and hence $B(x,x)=\sum_i b_{ii}x_i^2$. So, if $(b_{ij})$ also has a zero diagonal, then $B(x,x)=0$ for all $x$, regardless of $\operatorname{char}(\mathbb{F})$.


问:
Understanding the bounds of integration.
This is our homework question:

5. Let $X$ and $Y$ be random variables with joint pdf
$$ f(x,y) = \begin{cases} 1/4 &-1\le x,y \le 1 \\
0 &\text{otherwise} \end{cases} $$
Find $P(2X - Y \gt 0)$.

The solution given to us is the following:

$\displaystyle P(2X - Y \gt 0) = \int_{-1}^1 \int_{\frac{y}{2}}^1 \frac{1}{4} dxdy = \frac{1}{2} $.

I don't quite get the bounds of integration used. I get that we'd do $P(2X>Y)$ leading to $P(X>Y/2)$. But why is upper bounded by 1? Don't the constraints mean it should go to infinity? Why is y bound between -1 and 1?

答:
The usual short answer to these is "draw the picture".

Here, $X \in [-1,1]$ and $Y \in [-1,1]$ and we want the subset where $2X-Y > 0$.  This region is shown in blue below.

Partitioning along the $Y$-axis (into horizontal strips), $Y$ ranges from $-1$ to $1$, the left bound is $X = Y/2$, and the right bound is $X = 1$.


问:
Why is CurlF a vector field
I am having a really hard time understanding the concept of CurlF. Could someone help explain what it is and how we know that it's a vector field?</p>

答: For 3 dimensional space the definition is $$ \text{curl}\overrightarrow{F}=\overrightarrow{\nabla}\times\overrightarrow{F}=\left(\frac{\partial F_{z}}{\partial y}-\frac{\partial F_{y}}{\partial z}\right)\hat{x}+\left(\frac{\partial F_{x}}{\partial z}-\frac{\partial F_{z}}{\partial x}\right)\hat{y}+\left(\frac{\partial F_{y}}{\partial x}-\frac{\partial F_{x}}{\partial y}\right)\hat{z} $$ And by definition it is a vector-field

Title: Why is CurlF a vector field


问:
Let A a Invertible n.n matrix and Let X and Y be vectors In C^n
Let A a Invertible n.n matrix and Let X and Y be vectors In C^n Why A^-1.X.Y^* is a matrix of rank 1 and why the eigen values of this matrix different from 0 is Y^*.A^-1^.X</p>

答:
It is of rank one because it sends everything orthogonal to $Y$ to $0,$ and it sends $Y$ to $A^{-1} X \|Y\|^2.$

For eigenvalues, see above.


问:
Distributions of differences and sums of normal random variables
1) Bean bags are automatically completed by a machine, with an average weight per bag of $60$ kg, standard deviation of $1.5$ kg and normal distribution. In the storage and transport process, the average loss per bag is $1.2$ kg and standard deviation of $0.4$ kg, also with normal distribution.

a) Calculate the probability that in a consignment of $140$ bags of beans, the total weight will not exceed $8230$ kg.

b) Calculate the probability that in a shipment of $140$ bags of beans, the total weight is between $8000$ and $8400$ kg

Can anyone give me some tips on the above issues? I don't know anything like that

答:
I begin with some basic facts. You should look for
the corresponding formulas in your text or class notes. (I predict this is not the last use you will have for these formulas.)

The adjusted weight $A$ of a single bag after shipping
has $E(A) = 60 - 1.2 = 58.8,$ $Var(A) = 1.5^2 + .4^2 = 2.41,$ and $SD(A) = \sqrt{2.41} = 1.5524.$

Also, the weight $C$ of a consignment of 140 bags
has $E(C) = 140(58.8) = 8232$ and $SD(C) = \sqrt{140}(1.5524) = 18.3682.$

And the weight $C$ of the
consignment has $C \sim \mathsf{Norm}(\mu = 8232,
\sigma = 18.3682).$

You seek $P(C \le 8230) = 0.4566$ and
$P(8000 < C < 8400) \approx 1.$ These probabilities can be found directly in R as shown below, or by standardizing
and using printed normal CDF tables (possibly a little less precisely, where rounding is necessary).

```<code>pnorm(8230, 8232, 18.3682)
[1] 0.4566473
diff(pnorm(c(8000,8400), 8232, 18.3682))
[1] 1
</code>```

I'm pretty sure you're expected to
get the probabilities using normal tables. I will leave that to you. (What are z-scores corresponding to consignment values 8230, 8000, and 8400?)

Mainly for visualization, here is a simulation of a million consignments with summary statistics and a histogram. (Well... also it's a way to check my
arithmetic.)

Mean, standard deviation, and proportions from a million consignments match population mean, population SD, and probabilities (to about two decimal places).

```<code>set.seed(2021)
c = replicate(10^6, sum(rnorm(140,60,1.5)-rnorm(140,1.2,.4)))
summary(c);  sd(c)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
   8147    8220    8232    8232    8244    8321
[1] 18.36845  # SD of simulated values

mean(c <= 8230)
[1] 0.456224   # aprx P(C < 8230)
mean(c >=8000 & c <= 8400)
[1] 1          # aprx P(8000 < C < 8400)
</code>```

```<code>hdr = "Simuulated Consignment Weights with Normal Density"
hist(c, prob=T, col="skyblue2", main=hdr)
 curve(dnorm(x, 8232, 18.37), add=T, lwd=2)
 abline(v = 8230, col="orange", lwd=3, lty="dotted")
</code>```

Body:
I begin with some basic facts. You should look for
the corresponding formulas in your text or class notes. (I predict this is not the last use you will have for these formulas.)

The adjusted weight $A$ of a single bag after shipping
has $E(A) = 60 - 1.2 = 58.8,$ $Var(A) = 1.5^2 + .4^2 = 2.41,$ and $SD(A) = \sqrt{2.41} = 1.5524.$

Also, the weight $C$ of a consignment of 140 bags
has $E(C) = 140(58.8) = 8232$ and $SD(C) = \sqrt{140}(1.5524) = 18.3682.$

And the weight $C$ of the
consignment has $C \sim \mathsf{Norm}(\mu = 8232,
\sigma = 18.3682).$

You seek $P(C \le 8230) = 0.4566$ and
$P(8000 < C < 8400) \approx 1.$ These probabilities can be found directly in R as shown below, or by standardizing
and using printed normal CDF tables (possibly a little less precisely, where rounding is necessary).

pnorm(8230, 8232, 18.3682)
[1] 0.4566473
diff(pnorm(c(8000,8400), 8232, 18.3682))
[1] 1

I'm pretty sure you're expected to
get the probabilities using normal tables. I will leave that to you. (What are z-scores corresponding to consignment values 8230, 8000, and 8400?)

Mainly for visualization, here is a simulation of a million consignments with summary statistics and a histogram. (Well... also it's a way to check my
arithmetic.)

Mean, standard deviation, and proportions from a million consignments match population mean, population SD, and probabilities (to about two decimal places).

set.seed(2021)
c = replicate(10^6, sum(rnorm(140,60,1.5)-rnorm(140,1.2,.4)))
summary(c);  sd(c)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
   8147    8220    8232    8232    8244    8321
[1] 18.36845  # SD of simulated values

mean(c <= 8230)
[1] 0.456224   # aprx P(C < 8230)
mean(c >=8000 & c <= 8400)
[1] 1          # aprx P(8000 < C < 8400)

hdr = "Simuulated Consignment Weights with Normal Density"
hist(c, prob=T, col="skyblue2", main=hdr)
 curve(dnorm(x, 8232, 18.37), add=T, lwd=2)
 abline(v = 8230, col="orange", lwd=3, lty="dotted")


问:
Isomorphism of representation spaces
The 4-dimensional spin group $Spin(4)=SU_{+}(2) \times SU_{-}(2)$, denote a typical element as $(A_+,A_-)$. We have for the 4-dimensional Euclidean space $V=\mathbb{R}^4 \simeq \mathbb{H}  $ we can have the following representation
$$f: Spin(4) \mapsto GL(V)$$ given by $$f(A_+,A_-)(Q)=A_-Q(A_+)^{-1}$$
Where $Q$ is a typical element of $V$ identified with $\mathbb{H}$ in the matrix form
, i.e. via the following identification
$$t1+xi+yj+zk \leftrightarrow
\begin{bmatrix}
    t+xi       & y+zi \\
    -y+zi      & t-xi \\
\end{bmatrix}
$$
By calculation we can see that $f$ actually gives $f: Spin(4) \mapsto SO(V)=SO(4)$

We can also regard $Spin(4)$ as subspace of $4\times4$ matrices $\begin{bmatrix}
    A_+       & 0 \\
    0      & A_- \\
\end{bmatrix} $

With this, consider two copies of $\mathbb{C}^2$, i.e. $W_+$ and $W_-$ with standard hermitian metric, we have $Spin(4)$ acts on them by
$$f_+(\begin{bmatrix}
    A_+       & 0 \\
    0      & A_- \\
\end{bmatrix})(w_+)=A_+w_+$$
$$f_+(\begin{bmatrix}
    A_+       & 0 \\
    0      & A_- \\
\end{bmatrix})(w_-)=A_-w_-$$

Then we can have an isomorphism of representation spaces
$$V\otimes \mathbb{C} \simeq Hom(W_+,W_-)$$

Why is this true? Psychologically I understand this is simply the $\mathbb{R}^4 \otimes \mathbb{C} \simeq \mathbb{C}^4 \simeq \mathbb{C}^2 \otimes (\mathbb{C}^2)^*$, but how to establish the isomorphism in this concrete case and make it compactible with the representation given above?

答:
By definition, $\hom(W_+,W_-)$ consists of linear maps $W_+\to W_-$ with the action

$$ (A_+,A_-)T:=A_-\circ T\circ A_+^{-1} $$

The carrier spaces (underlying sets) of $W_+$ and $W_-$ are both simply $\mathbb{C}^2$, so the linear maps $T$ may be represented by (arbitrary) $2\times 2$ complex matrices, and the $A_\pm$ are certain $2\times2$ matrices too, and composition becomes matrix multiplication. That is, $(A_+,A_-)T=A_-TA_+^{-1}$ describes the action of $\mathrm{Spin}(4):=\mathrm{SU}(2)\times\mathrm{SU}(2)$ on $\hom(W_+,W_-):= M_2(\mathbb{C})$.

$V$ is the image of the algebra embedding $\mathbb{H}\to M_2(\mathbb{C})$, with $\mathrm{SU}(2)$ the image of $S^3=\mathrm{SU}(2)$. (For comparison, $\mathrm{SO}(2)$ is the image of $\mathrm{U}(1)$ under the algebra embedding $\mathbb{C}\to M_2(\mathbb{R})$.) The action is still conjugation coming from $\mathrm{Spin}(4)$, of course, but $V$ is a real subspace of $M_2(\mathbb{C})$. However, notice that $M_2(\mathbb{C})=V\oplus iV$ is an internal direct sum of real subspaces, so we may identify $V\otimes\mathbb{C}$ with all of $M_2(\mathbb{C})$, and again the action remains conjugation!

Title: Isomorphism of representation spaces


问:
Variance of sum of deviations
Suppose I have an i.i.d. sample $\{X_i\}_{i=1}^M$ for some positive integer $M$, and suppose that $X_i \sim X$ for some random variable $X$ with finite variance. Then, denote by
$$
E_M = \frac1M\sum_{i=1}^MX_i
$$
the empirical mean. The variance of $E_M$ (with respect to the sample) is by independence
$$
\mathrm{Var}(E_M)=\mathrm{Var}\left(\frac1M\sum_{i=1}^MX_i\right) = \frac1{M^2}\sum_{i=1}^M\mathrm{Var}\left(X_i\right) = \frac1M \mathrm{Var}(X).
$$
All clear and standard. Now consider one point (e.g. $X_1$) and the quantity $E_M - X_1$:
$$
E_M - X_1 = \frac1M\sum_{i=1}^MX_i - X_1 = \frac1M\sum_{i=1}^M (X_i - X_1).
$$
What about its variance? The increments $X_i - X_1$ are now not independent. Can I still show that
$$
\mathrm{Var}(E_M-X_1)\leq \frac{C}M \mathrm{Var}(X),
$$
for a constant $C$ independent of $M$?

答:
\begin{align}
V(E_M-X_1) &= V\left(\frac{1}{M}\sum_{i=1}^nX_i - X_1 \right) \\
&= V\left(-\frac{M-1}{M}X_1+\frac{1}{M}\sum_{i=2}^nX_i  \right) \\
&= \left(\frac{M-1}{M} \right)^2V(X_1)+ \frac{1}{M^2}\sum_{i=2}^nV\left(X  \right) \\
&= \left(\left(\frac{M-1}{M} \right)^2 + \frac{M-1}{M^2} \right)V(X) \\
&= \frac{M-1}{M} V(X) \\
\end{align}

You can then deduce easily $C = M-1$ but this $C$ depends on $M$. If $M$ is bounded as $M<b$, for example, then
$$C = b-1$$

Title: Variance of sum of deviations


问:
$-1$ is a quadratic residue modulo $p$ if and only if $p\equiv 1\pmod{4}$
I came across this problem and I believe Lagrange's theorem is the key to its solution.  The question is:

Let $p$ be an odd prime. Prove that there is some integer $x$ such that $x^2 \equiv −1 \pmod p$ if and only if $p \equiv 1 \pmod 4$.

I appreciate any help.  Thanks.

答:
If we are allowed to use the theory of Frobenius of prime ideals, then we could also show as follows:

A prime ideal $(p)$ splits in $\mathbb Q(i)$ if and only if $x^2+1\pmod p$ has a solution in integers. Now this is equivalent with the triviality of the Frobenius of $(p)$ in $\mathbb Q(i)$. Namely, it is equivalent with the condition $\zeta^p=\zeta$ where $\zeta=i$. This tells us that, when $p\not=2$, $(\dfrac{-1}{p})=1$ if and only $p\equiv 1\pmod 4$.

P.S. The reason to require $p\not=2$ is to guarantee that $(p)$ does not ramify in $\mathbb Q(i)$, so that we can apply the theorem of Dedekind to conclude as above.

Any error is welcomed to be pointed out.

Thanks for the attention.


问:
Matrix derivative of $x^\top A$ with respect to $A$
I'm interested in the following notation-problem. Assume we have a transpose vector equals $x^\top A$, where $x^\top \in \mathbb{R}^n$ and $A \in \mathbb{R}^{n \times m}$. I'm interested in $\dfrac{d (x^\top A)}{d A}$. I understand that it will be 3rd order tensor. But does there any chance (or operation) to represent it with a close form?

答:
First, I don't like multiplying by vectors on the left, so I will talk about $y=A x$ (transpose at your own peril). Second, we just need to compute:

$$\frac{\partial y_i}{\partial{a_{kl}}}$$
Notice that $y_i = \sum x_j a_{ij},$ so the partial in question is equal to $x_l$ if $i=k,$ and $0$ otherwise.


答:
Your implications aren't strong enough.  $|x-y| \le 1$ does mean $y \ge x-1$ but it also means $x \ge y -1$ (you can't have $y=10,000$ and $x - 2$ even though $10,000 \ge 2-1$).

But you are well on your way....

If $|x-y | \le 1$ then $-1 \le x-y \le 1$ and if $|y-z| \le 1$ then  $-1 \le y -z \le 1$.  If we add those together we ge $-2 \le x - z \le 2$ and we can't seem to get any stronger.  And that gives us a germ of an idea for a counter example.

Suppose we had a case where $x-y < 1$. and $y-z < 1$ so $x-y + y-z = x-z < 2$ but what if we have $1< x-z < 2$.  Is that possible?

Then answer is .... sure, why not?

Let $x - y = \frac 34$ and $y- z = \frac 34$ so $x-z = 1\frac 12$.  We can easily to that, can't we?

$y -z = \frac 34$ so $y = z + \frac 34$ and $x-y =\frac 34$ so $x -(z+\frac 34) = \frac 34$ so $x = z + 1\frac 12$.  Just let $z = 0, y = \frac 34, z =1\frac 12$.

Counter example. Done.

.....

Oh,... I didn't see that the relation was on $\mathbb Z$ and not $\mathbb R$.  Well, no matter the idea is the same.  I only to $x-y = \frac 34$ to show there was some "leeway".  We could use $z=0;y=1; x=2$ just as easily.

......

But if $x,y \in \mathbb Z$ then $x-y\in \mathbb Z$ so $|x-y| \le 1$ means $x-y = -1,0,1$ which means in turn $x = \begin{cases}y-1\\y\\y+1\end{cases}$.

Viewing it that way we can note that

It is reflexive as $x =\begin{cases}\color{red}{x-1}\\\color{blue}x\\\color{red}{x+1}\end{cases}$.

And it is symmetric an $x=\begin{cases}\color{green}{y-1}\\\color{blue}{y}\\\color{orange}{y+1}\end{cases}$ would imply

$\begin{cases}\color{green}{y-1}\\\color{blue}{y}\\\color{orange}{y+1}\end{cases} = x$ which implies

$\begin{cases}\color{green}{y}\\\color{blue}{y}\\\color{orange}{y}\end{cases} =\begin{cases}x+1\\x\\x-1\end{cases} $ or

$y = \begin{cases}\color{green}{x+1}\\\color{blue}{x}\\\color{orange}{x-1}\end{cases}$

But it's not transitive as

$x = \begin{cases}y-1\\y \\y+1\end{cases}$ and $y = \begin{cases}z-1\\z\\z+1\end{cases}$ would imply

$x =\begin{cases}\begin{cases}z-2\\z-1\\z\end{cases}\\\begin{cases}z-1\\z\\z+1\end{cases}\\\begin{cases}z\\z+1\\z+2\end{cases}\end{cases}=\begin{cases}z-2\\z-1\\z\\z+1\\z+2\end{cases}$

Title: Define a relation ∼ on Z as x ∼ y if and only if |x− y| ≤ 1. Say whether ∼ is reflexive. Is it symmetric? Transitive?

Body:
Just checking to see if I'm doing this right.

$|x-x| = 0 \le 1$ and similarly for y, so this is reflexive.

If $x,y \in Z$ then $|x - y | \le 1$ so $x\le 1+y$ and $y \ge x-1$

$|x- (x-1)| \le 1 \Rightarrow |-1|\le 1$. And for y: $|(1+y)-y|\le 1 \Rightarrow |1|\le1.$ So this is symmetric.

I'm a little stuck on proving transative. This is what I think the start looks like.

If $x,y,z \in Z$ for $|x-y|\le1$ and $|y-z|\le1$ then $y\ge x-1$ and $y\le 1+z$.


答:
Some treatments of this subject distinguish between polar vectors, which are conventional vectors defining a quantity and a direction, and axial vectors (sometimes called pseudovectors) which describe a rotational quantity.

An axial vector is like this.

Imagine a wheel spinning round an axle. This can be described by an axial vector. How fast the wheel is spinning is identified by the magnitude of the vector. The direction of the axle is then identified with the direction of the vector. The direction is defined by the right hand rule: if you imagine your hand curled around the axle with your thumb sticking out, and your fingers pointing in the direction the wheel is spinning, the positive direction is the direction in which your thumb is pointing.

Now, you can loosely describe the curl of a field as an axial vector defining the spinning motion of a small element of the field. If this small element is turning, then there is an axial vector defining that turn. And the curl of a vector field is the amount of turning at each point.

The above is possibly inaccurate in order to give you a general feel for what is going on.

Title: Why is CurlF a vector field

Body: I am having a really hard time understanding the concept of CurlF. Could someone help explain what it is and how we know that it's a vector field?</p>


问:
Galois group of $x^4+2x^2-2$ and intermediate fields.
Let $\alpha_i,\; 1 \leq i \leq 4$ be the four roots of the polynomial. I know that the Galois group is isomorphic to $D_4$. To find the intermediate fields of the splitting field of the polynomial, I find the fixed fields of the corresponding subgroups. However I was not able to find the intermediate field corresponding to $H = \langle r_{180} \rangle$ (the rotation by $180^\circ$). Any hints or solutions would be appreciated.

Body: Hint: $x^4 + 2 x^2 -2 = (x^2+1)^2 - 3.$


问:
Is there a Laplace transform for an unknown squared function?
I was solving the differential equation
$$y'-y = e^x y^2$$
I know that this is a simple differential equation, and I would solve it with a simple change of the variable, however, I was wondering if I could use Laplace transform to make it easier and more direct.
I asked my professor, and she said she didn't know.

So is there a way to do $\mathscr{L}([y(x)]^2)$?

答:
Unless if you are able to make a clever substitution you cannot solve this ODE with the Laplace Transform.

The Laplace Transform is only used for solving linear equations, as it's a linear transform.

Trust me, i have tried solving nonlinear equations, i have failed.

What a Laplace Transform does to an ODE is that it basically finds the characteristic polynomial for a linear ODE.

$$ay''(t)+by'(t)+cy(t)=f(t)$$

$$\downarrow\mathscr{L}$$

$$a[s^2Y(s)-sy(0)-y'(0)]+b[sY(s)-y(0)]+cY(s)=F(s)$$

where $y(0), y'(0)$

are the intial conditions, as we use the Laplace Transform to see how a system dampens, so we always need those.

As you can see we get an equation of the form:

$$Y(s)=\xi(s)$$

where $\xi$ is a polynomial made of $s$-s.

For $y(0)=0, y'(0)=0$

The above equation will result in:

$$Y(s)=\dfrac{F(s)}{as^2+bs+c}$$

$$\downarrow\mathscr{L}^{-1}$$

Now after transforming back we'll get the solution to $y(t)$ and of course we get the same solution as if we use the classical method to solve the ODE.

Edit: For your case even with a substitution you'll still have to plug in initial conditions which sadly make this method kind of bad for general solutions (or you could assume y(0)=y'(0)=0, but that will give you a solution to an IVP)


答:
$$y'-y = e^x y^2$$
Substitute $u=-\dfrac 1y$ then solve with Laplace transform:
$$u'+u=e^x$$
Otherwise you can't use Laplace transform with non linear DE.

Title: Is there a Laplace transform for an unknown squared function?

Body:
I was solving the differential equation
$$y'-y = e^x y^2$$
I know that this is a simple differential equation, and I would solve it with a simple change of the variable, however, I was wondering if I could use Laplace transform to make it easier and more direct.
I asked my professor, and she said she didn't know.

So is there a way to do $\mathscr{L}([y(x)]^2)$?


答:
The components of $x^T A$ are $(x^T A)_j = x_i A_{ij}$ using Einstein summation convention on the index $i$. Using the fact that $ \frac{\partial A_{ij}}{\partial A_{lm}} = \delta_{i,l} \delta_{j,m}$, you can see that
\begin{equation}
\frac{\partial (x_i A_{ij})}{\partial A_{lm}} = x_i \delta_{i,l} \delta_{j,m}= x_l \delta_{j,m}
\end{equation}which, as you predicted, is a tensor with three indices.

Title: Matrix derivative of $x^\top A$ with respect to $A$

Body: I'm interested in the following notation-problem. Assume we have a transpose vector equals $x^\top A$, where $x^\top \in \mathbb{R}^n$ and $A \in \mathbb{R}^{n \times m}$. I'm interested in $\dfrac{d (x^\top A)}{d A}$. I understand that it will be 3rd order tensor. But does there any chance (or operation) to represent it with a close form?


问:
Better way to solve $\cos\left(\frac{\gamma'}{2}\right) = g_0$ and $e^{i\beta'} \sin\left(\frac{\gamma'}{2}\right) = g_1$ for $\gamma',\beta'$?
I'm trying to solve the two equations to solve for:
$\gamma',\beta'$:
\begin{align}
\cos\left(\frac{\gamma'}{2}\right) = g_0\qquad e^{i\beta'}\sin\left(\frac{\gamma'}{2}\right) = g_1
\end{align}
Where $g_0$ is real, and  $g_1$ is some complex number. I tried
\begin{align*}
\gamma'=2\cos^{-1}(g_0),\qquad\cos(\beta')\sin\left(\frac{\gamma'}{2}\right)=Re(g_1),\ \text{(or)}\ \sin(\beta')\sin\left(\frac{\gamma'}{2}\right)=Im(g_1)
\end{align*}
Then for each $\gamma'$, there should be two corresponding $\beta'$s. However, it turned out that my solution doesn't quite work. I'm wondering is there another way I can solve this?

Thanks!

答: Solve one for $\sin{\frac{\gamma '}{2}}$, solve the other for $\cos{\frac{\gamma '}{2}}$, and then use $\sin^2 x + \cos^2 x = 1$.


问:
Walks on graphs and tensor products
In Qiaochu's article https://qchu.wordpress.com/2010/03/07/walks-on-graphs-and-tensor-products/#more-4807, he gives the following setup:

Construct a directed graph $\Gamma(V)$ as follows: its vertices are the irreducible representations of $G$, and the number of arrows from $A$ to $B$ is the multiplicity of $B$ in $A \otimes V$. Then the multiplicity of the trivial representation in $V^{\otimes n}$ is, essentially by definition, the number of walks on $\Gamma(V)$ of length $n$ from the trivial representation to itself!

If in addition $V \simeq V^{*}$, then $A$ $\otimes V \simeq A \otimes V^{*} = \textbf{Hom}(V, A)$ (the internal hom), so a form of Frobenius reciprocity then implies that $\Gamma(V)$ is an undirected graph.

When $G$ is finite, $\Gamma(V)$ has finitely many vertices and its adjacency matrix has coefficients

$\displaystyle \frac{1}{|G|} \sum_{g \in G} \chi_A(g) \chi_V(g) \overline{\chi_B}(g)$.

Three thing are unclear to me here:

How does the fact that the multiplicity of an irreducible representation in $V^{\otimes n}$ equals the number of walks on $\Gamma(V)$ of length $n$ from it to itself come from the definition of multiplicity above? I encountered the claim that every irreducible representation of a group $G$ is contained in some tensor power of a representation of $G$, but the proof didn't seem trivial.

How does the undirectness follow from Frobenius reciprocity? I guess it creates a symmetry of multiplicities in the opposite directions but how can one see it rigorously?

Correct me if I'm wrong but as far as I know, "matrix coefficients" refers to "matrix entries". Adjacency matrices have entries $\{0,1\}$, how come we get entries of different values here?

答:
(2) Technically if you're constructing a directed graph, then by fiat you get a directed graph, not an undirected one. But if you get the same number of edges in both directions between every pair of vertices, then you might as well have constructed an undirected graph since having directions may be unnecessary information in that case. (Or, you can tweak the definition of an undirected graph to simply be a directed one with the appropriate symmetry.) That's all that's being said here.

(3) For a simple graph, sure, you get $0$s and $1$s as the matrix entries. But these are not (in general) simple graphs, since they (in general) have multiple edges between vertices. As you ought to be able to guess from context, that means the matrix entry designates not merely if there is an edge from one vertex to another but how many there are from one to another.

The multiplicity of an irrep $U$ in an irrep $V$ is $\dim\hom_G(U,V)=\tfrac{1}{|G|}\sum\overline{\chi_U(g)}\chi_V(g)$, which ought to be in any rep thry text, and which we can apply here to the multiplicity of an irrep $B$ in $V\otimes A$. (And the character of a tensor product is a product of characters.)

(1). The number of walks of length $n$ from one vertex to another is the corresponding entry of the $n$th power of the adjacency matrix. This is a textbook graph theory fact, and makes a good exercise to see this.

Any rep is a direct sum of irreps, so we can specify a rep by the irreps' multiplicity in a vector. Applying the adjacency matrix to this vector produces the same result as tensoring with $V$ and then expressing as a vector!

Title: Walks on graphs and tensor products


问:
Is there an method similar to FFT for Fractionally Iterated Fourier Transforms?
FFT is one of the 20th Century's greatest inventions, running as $O(n \log(n))$ rather than as $O(n^2)$ as a simple implementation of a discrete Fourier transform would.  But what about half-order Fourier transforms, or arbitrary real order iterated Fourier transforms?

Literature on this topic is quite sparse.

答: I don't think the Mehler kernel has a FFT equivalent. An easy way to create a fractional FFT is to project $x$ onto the eigenspaces through $$P_k(x) = \frac14\sum_{l=0}^3 i^{-lk} F^l(x),\qquad P_k^2=P_k,P_mP_k=0$$ where $F^l$ is the FFT iterated $l$ times, then $S(x) = \sum_{k=0}^3 e^{2i\pi k/8} P_k(x)$ satisfies $$S^2(x)=\sum_{k=0}^3 i^k P_k(x)=F(x)$$


Title: Better way to solve $\cos\left(\frac{\gamma'}{2}\right) = g_0$ and $e^{i\beta'} \sin\left(\frac{\gamma'}{2}\right) = g_1$ for $\gamma',\beta'$?

Body:
I'm trying to solve the two equations to solve for:
$\gamma',\beta'$:
\begin{align}
\cos\left(\frac{\gamma'}{2}\right) = g_0\qquad e^{i\beta'}\sin\left(\frac{\gamma'}{2}\right) = g_1
\end{align}
Where $g_0$ is real, and  $g_1$ is some complex number. I tried
\begin{align*}
\gamma'=2\cos^{-1}(g_0),\qquad\cos(\beta')\sin\left(\frac{\gamma'}{2}\right)=Re(g_1),\ \text{(or)}\ \sin(\beta')\sin\left(\frac{\gamma'}{2}\right)=Im(g_1)
\end{align*}
Then for each $\gamma'$, there should be two corresponding $\beta'$s. However, it turned out that my solution doesn't quite work. I'm wondering is there another way I can solve this?

Thanks!

Body: $$e^{i\beta}=\pm\frac{g_1}{\sqrt{1-g_0^2}}$$ and $$\beta=-i\log\frac{g_1}{\sqrt{1-g_0^2}}+k\pi.$$


问:
Sufficient condition for a Markov chain with tridiagonal transition matrix to be null recurrent
Consider the random walk induced by the Markov matrix
\begin{equation}
\begin{Vmatrix}P_{ij}\end{Vmatrix}=
\begin{Vmatrix}
r_0 & p_0 & 0 & 0 & \cdots \\
q_1 & r_1 & p_1 & 0 & \cdots \\
0 & q_2 & r_2 & p_2 & \cdots \\
\end{Vmatrix}
\end{equation}
Let
\begin{equation}
\pi_0=1,\pi_n=\frac{p_0p_1\cdots p_{n-1}}{q_1q_2\cdots q_n}.
\end{equation}
Assume that the Markov chain is recurrent. Show that
\begin{equation}
\sum^\infty_{i=0}\pi_i=\infty\Longrightarrow\mathrm{null\;reccurent}
\end{equation}
\begin{equation}
\sum^\infty_{i=0}\pi_i<\infty\Longrightarrow\mathrm{positive\;reccurent}
\end{equation}

I have completely no idea. Please give me a hint.

答:
If $\lim_{n\rightarrow\infty}P^n_{ii}=b_i>0$ for one $i$ in an aperiodic recurrent class, then $b_j>0$ for all $j$ in the class of $i$. In this case, we call the class positive recurrent or strongly ergodic. If each $b_i=0$ and the class is recurrent we speak of the class as null recurrent or weakly ergodic.

Theorem In a positive recurrent aperiodic class with states $j=0,1,2,\cdots$,
\begin{equation}
\lim_{n\rightarrow\infty}P^n_{jj}=b_j=\sum^\infty_{i=0}b_iP_{ij},\;\sum^\infty_{i=0}b_i=1
\end{equation}
and the $b$'s are uniquely determined by the set of equations
\begin{equation}
b_i\geq0,\;\sum^\infty_{i=0}b_i=1,\;\mathrm{and}\;b_j=\sum^\infty_{i=0}b_iP_{ij}
\end{equation}

We investigate the existence of a stationary probability distribution, i.e., we wish to determine the positive solutions of
\begin{equation}
(p_i+q_i)x_i=\sum^\infty_{j=0}x_jP_{ji}=p_{i-1}x_{i-1}+q_{i+1}x_{i+1},\;i=0,1,\cdots,\;(1)
\end{equation}
under the normalization
\begin{equation}
\sum^\infty_{i=0}x_i=1,
\end{equation}
where $p_{-1}=0$, and thus $x_0=q_1x_1+r_0x_0$. Using Equation (1) for $i=1$, we could determine $x_2$ in terms of $x_0$. Equation (1) for $i=2$ determines $x_3$ in terms of $x_0$, etc. It is immediately verified that
\begin{equation}
x_i=\frac{p_{i-1}p_{i-2}\cdots p_0}{q_iq_{i-1}\cdots q_1}x_0=x_0\pi_i,\;i\geq1,
\end{equation}
is a solution of (1), with $x_0$ still to be determined. Now since
\begin{equation}
1=x_0+\sum^\infty_{i=1}x_0\pi_i=\sum^\infty_{i=0}x_0\pi_i,
\end{equation}
we have
\begin{equation}
x_0=\frac{1}{\sum^\infty_{i=0}x_0\pi_i}
\end{equation}
and so
\begin{equation}
x_0>0\;\mathrm{if\;and\;only\;if}\;\sum^\infty_{i=0}x_0\pi_i<\infty.
\end{equation}

Combining this result, the definition, and the theorem, we get the desired sufficient condition.


答: Clearly $\gamma^\prime\pm2\arccos g_0(\mod4\pi)$, which is consistent with $\sin\tfrac{\gamma^\prime}{2}=\pm|g_1|$ iff $g_0^2+|g_1|^2=1$, in which case $e^{i\beta^\prime}=\operatorname{sgn}g_1$ (except for the case $g_0=\pm1,\,g_1=0$, where $\beta^\prime$ is arbitrary). This constrains $\beta^\prime$ to either the even or odd multiples of $\pi$.


答:
Suppose $y$ is analytic at $0$, having power series
$$  y(x) = \sum_{k=0}^\infty a_k x^k  \text{.}  $$
Then
$$  y^2(x) = \sum_{n=0}^\infty \left( \sum_{i=0}^n a_i a_{n-i} \right) x^n  $$
and
$$  \mathcal{L}\{y^2\}(s) = \sum_{n=0}^\infty \left( \sum_{i=0}^n a_i a_{n-i} \right) \frac{n!}{s^{n+1}}  \text{,}  $$
wherever that series converges (if it converges anywhere).

I don't imagine this would ever be useful.  (Just how badly does applying the Frobenius method have to go to try taking a Laplace transform in the middle of it?)  But, a calculation can be done and there is a (formal) result.

One can certainly expand somewhere other that $0$, but this doesn't improve the situation.


问:
Maclaurin expansion of $f(x)=e^{ax^{1/2}}$ with undefined derivative at $0$
I am asked to find the Maclaurin expansion of the following function:

\begin{equation}
 f(x)=e^{ax^{1/2}}
\end{equation}

I have trouble understanding how to do that given that the first derivative of the function (for example) is undefined at $0$.
Any help would be greatly appreciated.

Thank you in advance!

答:
The function doesn't have a Maclaurin series for the reason that you have said (the first derivative is undefined at $0$), but we still know that for any $x$ we have
$$e^x=\sum_{r=0}^{\infty}\frac{x^r}{r!}$$
Hence,
$$f(x)=e^{ax^{1/2}}=\sum_{r=0}^{\infty}\frac{(a\sqrt x)^r}{r!}$$
I hope that is somewhat helpful.


问:
Orthogonal transformations and matrices having similar eigenvalues
If there is an orthogonal similarity between symmetric matrices $A$ and $B$ by having $B=OAO'$ for an orthogonal matrix $O$ ($'$ is transpose) we infer $A$ and $B$ are having identical eigenvalues (since $A$ is symmetric there is orthogonal $O_1$ satisfying $O_1\Lambda O_1'=A$ where $\Lambda_1$ is diagonal and $B=OO_1\Lambda O_1'O'\iff O_1'O'BOO_1=\Lambda$ diagonalizes $B$).

If $A$ and $B$ are symmetric and are having identical eigenvalues is there always an orthogonal $O$ satisfying $B=OAO'$?

答:
Yes there is, if $A$ and $B$ are symmetric and have the same eigenvalues then there exists two ortogonal matrices $O_{A}$ and $O_{B}$ and a diagonal matrix $\Lambda$ such that the following equations holds:

$$A=O_{A}\Lambda O_{A}^{T}$$
and $$B=O_{B}\Lambda O_{B}^{T}$$
this implies
$$\Lambda=O_{A}^{T}A O_{A}$$
and $$\Lambda=O_{B}^{T}B O_{B}$$
so we have
$$O_{A}^{T}A O_{A}=O_{B}^{T}B O_{B}$$
$$A=O_{A}O_{B}^{T}B O_{B}O_{A}^{T}$$,
but $O_{A}O_{B}^{T}$ is an ortogonal matrix becuase is product of two ortogonal matrices.


问:
Every closed ideal in $C(X)$ is the set of functions that vanish on some closed subset
This is an exercise from Banach Algebra Techniques in Operator Theory by Douglas

My attempt:

I've seen a similar argument that shows that every maximal ideal is of the form $I_x = \{ f \in C_0(X) : f(x)= 0\}$. I tried the same approach, by considering the evaluation map $\phi_x : C_0(X) \rightarrow \mathbb{C}$, where $\phi_x(f) = f(x)$ for $x \in K$.

Then it seemed like that the same argument wouldn't work, because we are dealing with closed ideals.

So I'm not sure how to proceed.

Any help will be appreciated!

Thank you!

答:
Typically the key to proving statements of this sort is the Stone-Weierstrass theorem.  Suppose $I$ is a closed ideal in $C(X)$.  First I claim that $I$ is automatically a $*$-ideal.  The idea is simple: if $f\in I$, then we can multiply it by the function $\overline{f}/f$ to conclude that $\overline{f}\in I$ as well.  Unfortunately, this doesn't quite work since it may not be possible to extend $\overline{f}/f$ continuously to the points where $f$ vanishes.  But you can take a function $h$ of norm $1$ which agrees with $\overline{f}/f$ except where $f$ is small, and then $hf$ will be close to $\overline{f}$.  Since $I$ is closed, you can then take a limit of such approximations to conclude that $\overline{f}\in I$.

Now let $K$ be the set of points of $X$ where every element of $I$ vanishes; we want to show that $I$ actually contains all functions that vanish on $K$.  If $K=\emptyset$, then $I$ is not contained in any maximal ideal of $C(X)$ so it must be all of $C(X)$.  So we may assume $K$ is nonempty.

Now the trick is, consider the quotient space $Y$ of $X$ that identifies all the points of $K$ together.  We can naturally consider $C(Y)$ as a subalgebra of $C(X)$ (the functions that are constant on $K$), and it contains $I$ since every element of $I$ is $0$ on $K$.  Now Stone-Weierstrass implies that $\mathbb{C}\cdot 1+ I$ must actually be all of $C(Y)$ (exercise: use the fact that $I$ is an ideal to show that this algebra separates points of $Y$).  That is, every function on $X$ that is constant on $K$ is a constant plus an element of $I$.  In particular, this implies that every function that vanishes on $K$ is in $I$, since the constant would have to be $0$.

Alternatively, here is a quicker proof if you know a bit more theory of $C^*$-algebras.  The quotient $C(X)/I$ is a commutative unital $C^*$-algebra, so it is $C(K)$ for some compact Hausdorff space $K$.  Moreover, the quotient homomorphism $\pi:C(X)\to C(X)/I\cong C(K)$ is induced by some continuous map $i:K\to X$.  It follows that $I=\ker(\pi)$ is just the set of functions that vanish on the image of $i$.


问:
Real part of a primitive root of unity is not an algebraic integer for n>4.
I'm trying to prove the following.

Suppose that $\alpha$ is an $n$th root of unity whose real part is an algebraic integer. Then $\alpha^4 = 1$.

I've thought about this for a while, and have the following ideas/questions.

If I can show that the algebraic norm $\frac{1}{2}(\alpha + \alpha^{-1})$ is not an integer for $n> 4$, then I will be done. Since $x^2-(\alpha +\alpha^{-1})x +1$ is the minimal polynomial of $\alpha$ over $\mathbb{Q}(\alpha+\alpha^{-1})$, we have that
$$
N\left(\frac{1}{2}(\alpha + \alpha^{-1})\right) = \frac{1}{2^{\phi(n)/2}}N(\alpha+\alpha^{-1})
$$
where $\phi(n)$ is the Euler $\phi$ function. So we just need to consider $N(\alpha+\alpha^{-1})$. Is there a way that we can bound $N(\alpha + \alpha^{-1})$? If we can show that this norm is less than $2^{\phi(n)/2}$ then we're done.

Another thought is that $\frac{1}{2}(\alpha+\alpha^{-1})$ is a root of $T_n(x)-1$ where $T_n$ is the $n$th Chebyshev polynomial. This however is not generally irreducible, so the fact that $T_n$ is not monic does not tell us that $\frac{1}{2}(\alpha+\alpha^{-1})$ is not an algebraic integer.

Thanks for any thoughts or hints.

答:
Here is a more direct argument that shows that the norm of $(\alpha + \alpha^{-1})/2$ is less than one unless $\alpha^2 = 1$. The norm is the product of the conjugates. Note that the conjugates of this are all of the form $(\beta + \beta)/2$ for some other root of unity $\beta = \alpha^k$. But note by the triangle inequality that

$$\left| \frac{\beta + \beta^{-1}}{2} \right| \le \frac{1}{2} + \frac{1}{2} = 1.$$

In particular, either the norm has absolute value strictly less than one, or equality holds for all conjugates and particular for $\beta = \alpha$. But the triangle inequality $|x + y| \le |x| + |y|$ for complex numbers is strict unless $\arg(x) = \arg(y)$. Since $\alpha$ and $\alpha^{-1}$ both have absolute value one, that means the triangle inequality is strict unless $\alpha = \alpha^{-1}$, which implies that $\alpha^2 = 1$.

To finish the original problem, if $(\alpha + \alpha^{-1})/2$ is an algebraic integer then its norm is an algebraic integer. Since the norm is less than $1$ unless $\alpha^2 = 1$, the only remaining possibility is that the norm is $0$, in which case $\alpha + \alpha^{-1} = 0$ and $\alpha^2 = -1$, or $\alpha^4 = 1$.


问:
How to check if $x^TCx\geq0$?
I have the next 3x3 block matrix C, where each block is a square matrix.

$$
C = \begin{bmatrix}
0 & A & B \\
0 & A+K_1 & B \\
0 & A & B+K_2
\end{bmatrix},
$$
where $K_i$ is also a squared diagonal matrix (I have freedom choosing their elements). I want to be sure that $x^TCx \geq 0$ choosing the appropriated $K_i$.

EDIT

The answer is no

答:
$$
C = \begin{bmatrix}
0 & A & B \\
0 & A+K_1 & B \\
0 & A & B+K_2
\end{bmatrix},
$$

Let $A$ be a $n\times n$ matrix.

Subtract $1$st row from $(n+1)$ th row then from By a matrix multiplying the matrix $P_1$ from left of $C$(Let $C_1=PC$).

Now we will do the same thing but now we will subtact $2$ nd row from $n+2$ th row by multiplying with the matrix $P_2$.In this way we will go on i.e we will subtract the $i$ th row from $(n+i)$th row using the matrix $P_i$.So we have $P_nP_{n-1}\dots P_{1}C=C_n=\begin{bmatrix}
0 & A & B \\
0 & K_1 &0 \\
0 & A & B+K_2
\end{bmatrix}$

Now again,

Subtract $1$st row from $(2n+1)$ th row then from By a matrix multiplying the matrix $Q_1$ from left of $C$(Let $C_{n+1}=PC_1$).

Now we will do the same thing but now we will subtact $2$ nd row from $2n+2$ th row by multiplying with the matrix $Q_2$.In this way we will go on i.e we will subtract the $i$ th row from $(2n+i)$th row using the matrix $Q_i$.So we have $Q_nQ_{n-1}\dots Q_{1}P_nP_{n-1}\dots P_{1}C=C_{2n}=\begin{bmatrix}
0 & A & B \\
0 & K_1 &0 \\
0 & 0 & K_2
\end{bmatrix}$

Now Let $D=Q_nQ_{n-1}\dots Q_{1}P_nP_{n-1}\dots P_{1}$

Here $C_{2n}$ is triangular ,so $C_{2n}D^t=\begin{bmatrix}
0 & A & B \\
0 & K_1 &0\\
0 & 0 & K_2
\end{bmatrix}$ is also triangular(Use the fact that multiplying $C_{i-1}$ by $P_i^t$ from right which will subtract the ist column from the $n+i$ column of $C_{i-1}$.And so for $Q_i$)

So we have $DCD^t=\begin{bmatrix}
0 & A & B \\
0 & K_1 &0\\
0 & 0 & K_2
\end{bmatrix}$

Title: How to check if $x^TCx\geq0$?


问:
Extensions of $\mathbb{Z}$ by $\mathbb{Z}$
Given two groups $G$ and $H$, an extension of $G$ by $H$ is a group $K$ with a normal subgroup $N$ isomorphic to $H$ such that the quotient group $K/N$ is isomorphic to $G$ (equivalently, $1 \to H \to K \to G \to 1$ is an exact sequence).

So, up to isomorphism, how many (not necessarily abelian) extensions of $\mathbb{Z}$ by $\mathbb{Z}$ are there?

One obvious extension is the direct product $\mathbb{Z} \times \mathbb{Z}$. Another extension is the group with the same underlying set as $\mathbb{Z} \times \mathbb{Z}$ and $(a,b) \ast (c,d)$ still defined to be $(a+c,b+d)$ if $b$ is even, but it is defined to be $(a-c,b+d)$ if $b$ is odd (or simply, $(a+(-1)^bc,b+d)$). This nontrival extension is the semidirect product $\mathbb{Z} \rtimes \mathbb{Z}$ corresponding to the homomorphism $\mathbb{Z} \to \operatorname{Aut}(\mathbb{Z})$ where even integers act as the identity and odd integers act as negation.

The answer would be $2$ if the direct product and the even/odd semidirect product were the only extensions up to isomorphism. Is this in fact the case?

答:
Conveniently, there is a standard line of attack for such questions which works when the subgroup in the extension is abelian: classify actions $\rho$ of the quotient $G$ on the subgroup $A$, and for each such $G$-module structure on $A$, compute $H^2(G, A_\rho)$ (group cohomology).

How many group actions are there of $\mathbb{Z}$ on $\mathbb{Z}$ (compatible with $+$ on the right): answer: 2, since there are two maps from $\mathbb{Z}$ to $\text{Aut}(\mathbb{Z}) = \pm 1$.

For each of these $\mathbb{Z}$-modules, compute the group cohomology $H^2(\mathbb{Z}, M)$. Since $\mathbb{Z}$ has homological dimension $1$, these are both trivial, and so the only two extensions are the ones you wrote down.


答: Yes they are the only two. Since the infinite cyclic group is free, any such extension must split, and $|{\rm Aut}({\mathbb Z})| = 2$, so there are only two possible actions in the semidirect product: the trivial action and inversion.

Title: Extensions of $\mathbb{Z}$ by $\mathbb{Z}$

Body:
Given two groups $G$ and $H$, an extension of $G$ by $H$ is a group $K$ with a normal subgroup $N$ isomorphic to $H$ such that the quotient group $K/N$ is isomorphic to $G$ (equivalently, $1 \to H \to K \to G \to 1$ is an exact sequence).

So, up to isomorphism, how many (not necessarily abelian) extensions of $\mathbb{Z}$ by $\mathbb{Z}$ are there?

One obvious extension is the direct product $\mathbb{Z} \times \mathbb{Z}$. Another extension is the group with the same underlying set as $\mathbb{Z} \times \mathbb{Z}$ and $(a,b) \ast (c,d)$ still defined to be $(a+c,b+d)$ if $b$ is even, but it is defined to be $(a-c,b+d)$ if $b$ is odd (or simply, $(a+(-1)^bc,b+d)$). This nontrival extension is the semidirect product $\mathbb{Z} \rtimes \mathbb{Z}$ corresponding to the homomorphism $\mathbb{Z} \to \operatorname{Aut}(\mathbb{Z})$ where even integers act as the identity and odd integers act as negation.

The answer would be $2$ if the direct product and the even/odd semidirect product were the only extensions up to isomorphism. Is this in fact the case?


问:
Connections and Projection Maps
Let $M$ be a manifold and let $\pi: TM\rightarrow M$ be the projection map.  Taking the pushforward of $\pi$ we obtain a bundle map
$$
\pi_\ast: T(TM)\rightarrow TM.
$$
Let
$$
V=\ker \pi_\ast
$$
be the vertical subbundle of $T(TM)$.

Let $\nabla$ be a covariant derivative on $M$.  How does $\nabla$ define a smooth bundle map $K: T(TM)\rightarrow T(TM)$ such that $K(T(TM))=V$ and $K^2=K$? Also, is it true that there is a one to one correspondence between covariant derivatives on $M$ and smooth bundle maps $K:T(TM)\rightarrow T(TM)$ satisfying $K^2=K$ and $K(T(TM))=V$?

Of course, the bundle map $K$ with the above properties defines a decomposition
$$
T(TM)=H\oplus V
$$
where $H:=\ker K$ is the so-called horizontal distribution.

I have seen similar questions on Stack Exchange but the constructions given in the comments or answers either do not seem well defined or they do not appear to work.

答:
Let $M$ be a manifold and $\nabla$ a connection on $M$.  Let $D/dt$ denote the induced covariant derivative for vector fields on $M$ defined along curves.  Let $\sigma: (a,b)\rightarrow M$ be a (smooth) curve and let $X:(a,b)\rightarrow TM$ be a vector field along $c$.  In other words, if $\pi: TM\rightarrow M$ is the projection, then
$$
c(t)=\pi(X(t))
$$
or $X(t)\in T_{c(t)}M$. Let $t_0\in (a,b)$ and choose a coordinate neihborhood $(U,x_i)$ of $p=c(t_0)$.  Let $\varepsilon>0$ be such that $(t_0-\varepsilon, t_0+\varepsilon)\subset (a,b)$ and
$$
c(t_0-\varepsilon, t_0+\varepsilon)\subset U.
$$
Write $c(t)$ and $X(t)$ for $t\in (t_0-\varepsilon, t_0+\varepsilon)$ in local coordinates:
$$
c(t)=(c^i(t)),\hspace{0.1in} X(t)=X^i(t)\frac{\partial}{\partial x^i}\big|_{c(t)}
$$
where Einstein notation is employed in the expression for $X(t)$. The covariant derivative of $X(t)$ is then
$$
\frac{DX}{dt}(t)=\left(\frac{dX^k}{dt}+\frac{dc^i}{dt}X^j\Gamma^k_{ij}\right)\frac{\partial}{\partial x^k}\big|_{c(t)}\in T_{c(t)}M
$$
where $\Gamma^k_{ij}$ are the Chrisoffel symbols of $\nabla$.

The idea now is to use $D/dt$ to define a map $K: T(TM)\rightarrow T(TM)$ with the desired properties.

For convenience, we will denote the tangent vector $v\in T_p M$ as $(p,v)$.  Note that
$$
V_{(p,v)} = T_{(p,v)}(T_pM) \simeq T_p M
$$
and the isomorphism between $V_{(p,v)}$ and $T_pM$ is canonical because $T_pM$ is a vector space.  Specifically, given a tangent vector $Y\in T_pM$, we associate it to the vector which in $Y_v\in V_{(p,v)}$ which is specified by the curve
$$
\alpha(t)=(p,v+tY)\in T_pM\subset TM.
$$
In other words,
$$
\alpha(0)=(p,v),\hspace{0.1in} \dot{\alpha}(0)=Y_v\in V_{(p,v)}.
$$
We can express $Y_v$ in terms of local coordinates as follows.  Writing
$$
Y=Y^i\frac{\partial}{\partial x^i}\big|_p,
$$
let $(TU,x^i,y^i)$ be the coordinate system induced by $(U,x^i)$.  Then one sees that
$$
Y_v = Y^i\frac{\partial}{\partial y^i}\big|_{(p,v)}
$$
The map $K: TTM\rightarrow TTM$ is then defined as follows.  For $\xi\in T_{(p,v)}TM$, let $X:(-\varepsilon,\varepsilon)$ be any curve such that
$$
X(0)=(p,v),\hspace{0.1in}\dot{X}(0)=\xi.
$$
We then define
$$
K\xi = \left(\frac{DX}{dt}(0)\right)_v\in V_{(p,v)}.
$$
This definition is independent of the choice of curve $X(t)$.  This is clear by expressing $K\xi$ in local coordinates.  Write
$$
\xi = \xi^i_{(1)} \frac{\partial}{\partial x^i}\big|_{(p,v)}+\xi^i_{(2)} \frac{\partial}{\partial y^i}\big|_{(p,v)}
$$
Let $c_X(t)=\pi(X(t))$ and write
$$
X(t) = X^i(t)\frac{\partial}{\partial x^i}\big|_{c(t)},\hspace{0.2in} v=v^i\frac{\partial}{\partial x^i}\big|_{p}
$$
Since $X(0)=(p,v)$ and $\dot{X}(0)=\xi$, we have
$$
X^i(0)=v^i,\hspace{0.1in} \frac{dc_X^i}{dt}(0)=\xi^i_{(1)}, \hspace{0.1in}\frac{dX^i}{dt}(0)=\xi^i_{(2)}
$$
So
$$
\begin{align}
\frac{DX}{dt}(0)&=\left(\frac{dX^k}{dt}(0)+\frac{dc^i}{dt}(0)X^j(0)\Gamma^k_{ij}(p)\right)\frac{\partial}{\partial x^k}\big|_{p}\\
&=\left(\xi^k_{(2)}+\xi^i_{(1)}v^j\Gamma^k_{ij}(p)\right)\frac{\partial}{\partial x^k}\big|_{p}
\end{align}
$$
$K\xi$ is then
$$
K\xi=\left(\frac{DX}{dt}(0)\right)_v=\left(\xi^k_{(2)}+\xi^i_{(1)}v^j\Gamma^k_{ij}(p)\right)\frac{\partial}{\partial y^k}\big|_{(p,v)}
$$
The definition of $K\xi$ in local coordinates makes it clear that $K$ is linear on the fibers of $TTM$ and that its smooth.  Also, note that if $\xi\in V_{(p,v)}$, then the components $\xi^i_{(1)}=0$ and
$$
K\xi=\xi^k_{(2)}\frac{\partial}{\partial y^k}\big|_{(p,v)}=\xi.
$$
By construction, the image of $K$ is all of $V$.  Also, since $K$ is the identity on $V$, we also have $K^2=K$.  $K$ is then the desired smooth bundle map.

Title: Connections and Projection Maps


问:
When is $ u v^* \succeq 0$, for two complex vectors $u$ and $v$?
It seems like maybe $u$ has to equal $v$, or maybe $ u = cv $ for some $ c > 0 $. Is this true?

If $ u v^* $ is positive semidefinite, then $ x^* u v^* x \ge 0$. I observe then that $ x^* u v^* x = (x^* u) (v^* x)$, the product of two inner products. If this is greater than equal to zero, then the two inner products must always have the same sign, for all $x$. Geometrically, this means $u$ and $v$ are always "pointing in the same direction" as $x$, for all $x$ (or both pointing "away" from $x$). I think this implies necessarily that $ u = cv $ for some $ c > 0 $. If not, we could choose an $x$ pointing in the same direction as $u$ but not $v$, and then $ (x^* u) (v^* x) < 0 $, violating the positive semidefinite condition.

答: Notice that the matrix annihilates the orthogonal complement of $v,$ and sends $v$ itself to a positive multiple of $u.$ It's adjoint annihilates everything orthogonal to $u$ and sends $u$ itself to a positive multiple of $v.$ So, the matrix is only self-adjoint when $u$ is a real multiple of $v,$ and only PSD if the multiple is positive. In which case it is just a positive multiple of the projection onto the subspace generated by $u$ (or $v$ if you prefer).


问:
Group topology and free topological groups
The following result is in Topological Groups and Related Structures, Arhangel'skii, Tkachenko;

Notation: $F_a(X)$ denotes the free group on $X$ and $F(X)$ the free topological group on $X$.

Claim: Let $f:X\to Y$ be a contiuous mapping of Tychonoff spaces. Then $f$ admits an extension to continuous homomorphism $F(f):F(X)\to F(Y)$. In addition, if $f$ is a quotient, then $F(f)$ is open.

We recall we already proved that $F(X)$ and $F_a(X)$ are the same of algebraic point of view.

Proof: Since $Y$ is identified with the corresponding subspace of $F(Y)$, we can consider $f$ as a continuous mapping of $X$ to the free topological group $F(Y)$. Therefore, by the definition of $F(x)$, $f$ can be extended to a continuous homomorphism $F(f):F(X)\to F(Y)$ wich we shall denote by $\overline{f}$.

Now suposse that $f$ is a quotient. Denote $\tau _q $ the family of all $\overline{f}(U)$ where $U$ is open in $F(X)$. Then $\tau _q $ is a group topology on the abstract group $F_a(Y)$.

Here are my questions:

1.- Why is $\tau _q$ a topology? I don't see why is closed under finite intersections.

2.- Why is $(F_a(Y),\tau _q)$ a topological group? I don't know how to prove that the map $(x,y)\mapsto xy$ is continuous on $F_a(Y)\times F_a(Y)$.

I recall that $f$ is a quotient map if it is surjective, continuous and $f^{-1}(V)$ is open implies $V$ is open.

Thanks.

答: It seems that $(F_a(Y),\tau_q)$ is a quotient group (see section 1.5 of “Topological Groups and Related Structures” about this construction) of the topological group $F(X)$, and $\bar f$ is the quotient map.</p>


问:
How to show that ${m \choose n} + {m \choose n-1} = {m+1 \choose n}$?
This is from Serge Lang's book Basic Mathematics. I've stuck at this for over six hours now. Read multiple posts, forums, and even checked the answer key.

The main thing I'm confused about is how can we get this common denominator: $n!(m - n + 1)!$

Everything else is easy.

答:
You have $m + 1$ biscuits and you want to pick $n$ of them.

Put butter on one of the biscuits. Now to pick $n$, you can either pick the biscuit with the butter, and then you still have to pick $n-1$ more (a total of $m \choose n-1$ ways to do this), or you can avoid the biscuit with the butter, in which case you have to pick all $n$ from the remaining $m$ (a total of $m \choose n$ ways to do this).


答:
Every root of unity is an algebraic integer. The "real part" (not really so well defined) of a root of unity is (however we interpret it) (EDIT)! half the sum of it and one of its conjugates (also an algebraic integer...), so is (EDIT) at worst half an algebraic integer.

Or... are you asking the question you mean to?

(Thanks to comments for the correction about "half"...)

Title: Real part of a primitive root of unity is not an algebraic integer for n>4.

Body:
I'm trying to prove the following.

Suppose that $\alpha$ is an $n$th root of unity whose real part is an algebraic integer. Then $\alpha^4 = 1$.

I've thought about this for a while, and have the following ideas/questions.

If I can show that the algebraic norm $\frac{1}{2}(\alpha + \alpha^{-1})$ is not an integer for $n> 4$, then I will be done. Since $x^2-(\alpha +\alpha^{-1})x +1$ is the minimal polynomial of $\alpha$ over $\mathbb{Q}(\alpha+\alpha^{-1})$, we have that
$$
N\left(\frac{1}{2}(\alpha + \alpha^{-1})\right) = \frac{1}{2^{\phi(n)/2}}N(\alpha+\alpha^{-1})
$$
where $\phi(n)$ is the Euler $\phi$ function. So we just need to consider $N(\alpha+\alpha^{-1})$. Is there a way that we can bound $N(\alpha + \alpha^{-1})$? If we can show that this norm is less than $2^{\phi(n)/2}$ then we're done.

Another thought is that $\frac{1}{2}(\alpha+\alpha^{-1})$ is a root of $T_n(x)-1$ where $T_n$ is the $n$th Chebyshev polynomial. This however is not generally irreducible, so the fact that $T_n$ is not monic does not tell us that $\frac{1}{2}(\alpha+\alpha^{-1})$ is not an algebraic integer.

Thanks for any thoughts or hints.


Title: How to show that ${m \choose n} + {m \choose n-1} = {m+1 \choose n}$?

Body:
This is from Serge Lang's book Basic Mathematics. I've stuck at this for over six hours now. Read multiple posts, forums, and even checked the answer key.

The main thing I'm confused about is how can we get this common denominator: $n!(m - n + 1)!$

Everything else is easy.

Body:
Quite simple:
\begin{align}
\binom mn+\binom m{n-1}&=\frac{m!}{n!(m-n)!}+\frac{m!}{(n-1)!(m-n+1)!}\\
& =\frac{m!}{\color{red}n\,(n-1)!(m-n)!}+\frac{m!}{(n-1)!(m-n)!\,(\color{blue}{m-n+1})} \\
&=\frac{[(\color{blue}{m-n+1})+\color{red}n]m!}{n!\,(m-n+1)!}=\frac{(m+1)!}{n!\,(m-n+1)!}.
\end{align}


问:
understanding the uniqueness of a mobius transformation.
When given the mobius transformation and three points $f(z)=\frac{az+b}{cz+d}$ for $z,a,b,c,d\in \mathbb{C},ad-bc\neq0, f(z_1)=f1,f(z_2)=f_2,f(z_3)=f_3$,

what is the approach to determine the unique transformation? Because during the substitutions I always end up with a quadratic equation involving one of the variables. How do I then definitively disprove the other solution?

答:
A good idea is to use $(0,\infty,1)$ as a "way station" (midpoint) between two triples.

To go from $(z_1,z_2,z_3)$ to $(0,\infty,1)$, we may put $z-z_1$ in a numerator to ensure $f(z_1)=0$, we may put $z-z_2$ in a denominator to ensure $f(z_2)=\infty$, and then normalize to ensure $f(z_3)=1$:

$$ f(z)=\frac{z-z_1}{z-z_2}\cdot\frac{z_3-z_2}{z_3-z_1}, $$

the "cross-ratio." The corresponding matrix for this transformation is the product

$$ \begin{bmatrix} z_3-z_2 & 0 \\ 0 & z_3-z_1 \end{bmatrix}
 \begin{bmatrix} 1 & -z_1 \\ 1 & -z_2 \end{bmatrix}. $$

To go from $(z_1,z_2,z_3)$ to $(f_1,f_2,f_3)$, then, we compose with the inverse of the transformation that goes from $(f_1,f_2,f_3)$ to $(0,\infty,1)$, which (using $(AB)^{-1}=B^{-1}A^{-1}$) would yield

$$ \begin{bmatrix} 1 & -f_1 \\ 1 & -f_2 \end{bmatrix}^{-1}
 \begin{bmatrix} f_3-f_2 & 0 \\ 0 & f_3-f_1 \end{bmatrix}^{-1}
\begin{bmatrix} z_3-z_2 & 0 \\ 0 & z_3-z_1 \end{bmatrix}
 \begin{bmatrix} 1 & -z_1 \\ 1 & -z_2 \end{bmatrix}. $$

The title suggests you also want to understand why there is a unique transformation between triples. Suppose there were two transformations between a pair of triples. We may compose both with a map that turns the second triple into $(0,\infty,1)$, so we end up with two transformations that turn a triple into $(0,\infty,1)$. But then we may precompose with a map that turns $(0,\infty,1)$ into the first triple, and we end up with two transformations from $(0,\infty,1)$ to itself. But it is an easier exercise to show there is only one such transformation.


问:
Finding the change of basis matricies
Not quite sure what to do for this question.

Let $B_1$= {$\vec{u}_1,\vec{u}_2,\vec{u}_3$} and $C_1$= {$\vec{w}_1,\vec{w}_2,\vec{w}_3$} be two bases for $\mathbb{R}^3$ such that
$\vec{w}_1=\vec{u}_1-\vec{u}_2,$  $\vec{w}_2=\vec{u}_2-\vec{u}_3, $ $\vec{w}_3=\vec{u}_2+\vec{u}_1 $.
Find the change of coordinates matrices $P_{{C_1}\to {B_1}}$ and $P_{{B_1}\to {C_1}}$

Any hints would be greatly appreciated.

答:
Given $\vec{w_1} = \vec{u_1}-\vec{u_2},\:\:\:\vec{w_2} = \vec{u_2}-\vec{u_3},\:\:\: \vec{w_3} = \vec{u_2}+\vec{u_1}$

Every vector in $\mathbb{R}^3$ can be written as a linear combination of the vectors in $C_1$, and also the vectors in $B_1$, and given a linear combination of the vectors in $C_1$, what is the corresponding vector as a linear combination of the vectors in $B_1$? There is a translation of coefficients:

$a_1\vec{w_1}+a_2\vec{w_2}+a_3\vec{w_3}= (a_1+a_3)\vec{u_1}+(-a_1+a_2+a_3)\vec{u_2}+(-a_2)\vec{u_3}$, which implies

$\begin{bmatrix} \vec{w_1}\; \big|\; \vec{w_2}\;\big|\;  \vec{w_3}\: \end{bmatrix} \begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix} = \begin{bmatrix} \vec{u_1}\; \big|\; \vec{u_2}\;\big|\;  \vec{u_3}\: \end{bmatrix}\begin{bmatrix} a_1+a_3 \\ -a_1+a_2+a_3 \\ -a_2 \end{bmatrix} = \begin{bmatrix} \vec{u_1}\; \big|\; \vec{u_2}\;\big|\;  \vec{u_3}\: \end{bmatrix} \begin{bmatrix} 1 & 0 & 1 \\ -1 & 1 & 1 \\ 0 & -1 & 0 \end{bmatrix}\begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix}$

thus $P_{C_1\rightarrow B_1} = \begin{bmatrix} 1 & 0 & 1 \\ -1 & 1 & 1 \\ 0 & -1 & 0 \end{bmatrix}$, and similarly $P_{B_1\rightarrow C_1} = (P_{C_1\rightarrow B_1})^{-1}$.

If you don't believe me:

solving for $\vec{u_1}=\frac{1}{2}(\vec{w_1}+\vec{w_3}),\vec{u_2}=\frac{1}{2}(\vec{w_3}-\vec{w_1}),\vec{u_3}=\frac{1}{2}(\vec{w_3}-\vec{w_1}-2\vec{w_2})$ then

$c_1\vec{u_1}+c_2\vec{u_2}+c_3\vec{u_3} = \left(\frac{c_1 - c_2 - c_3}{2}\right)\vec{w_1} +(-c_3)\vec{w_2}+\left(\frac{c_1+c_2+c_3}{2}\right)\vec{w_3}$ and

$\begin{bmatrix} \vec{u_1}\; \big|\; \vec{u_2}\;\big|\;  \vec{u_3}\: \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} = \begin{bmatrix} \vec{w_1}\; \big|\; \vec{w_2}\;\big|\;  \vec{w_3}\: \end{bmatrix}\begin{bmatrix} \frac{c_1 - c_2 - c_3}{2} \\ -c_3 \\ \frac{c_1+c_2+c_3}{2} \end{bmatrix} = \begin{bmatrix} \vec{w_1}\; \big|\; \vec{w_2}\;\big|\;  \vec{w_3}\: \end{bmatrix} \begin{bmatrix} \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} \\ 0 & 0 & -1 \\ \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \end{bmatrix}\begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix}$

where $$\begin{bmatrix} 1 & 0 & 1 \\ -1 & 1 & 1 \\ 0 & -1 & 0 \end{bmatrix}\begin{bmatrix} \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} \\ 0 & 0 & -1 \\ \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \end{bmatrix}= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$

P.S. I may or may not of mixed up the meanings of $P_{B_1\rightarrow C_1}$ and
$P_{C_1\rightarrow B_1}$

Title: Finding the change of basis matricies


问:
Immersed Submanifold Requirements
Wald's General Relativity textbook reads

We can use the above definition of integration on manifolds to define the integral of $p-$forms on $M$ over well behaved, orientable $p-$dimensional surfaces in $M.$ First, we must define more precisely the notion of a "well behaved surface." Let $S$ be a manifold of dimension $p<n$. If $\phi: S \to M$ is $C^{\infty}$, is locally one to one - i.e., each $q \in S$ has an open neighborhood $O$ such that $\phi$ restricted to $O$ is one-to-one - and $\phi^{-1}: \phi[O] \to S$ is $C^{\infty}$, then $\phi[S]$ is said to be an immersed submanifold of $M.$ If, in addition, $\phi$ is globally one-to-one (i.e., $\phi[S]$ does not "intersect itself"), then $\phi[S]$ is said to be an embedded submanifold of $M.$

I am having trouble understanding the necessity of these requirements. What would be an example of a submanifold which is not "immersed"? That is, what would a pathological manifold not satisfying the "$\phi: S\to M$ being locally one-to-one" requirement look like?

答:
Sadly, the terminology "an immersed submanifold" is suboptimal which leads to questions such as yours. The thing to realize is that an "immersed submanifold" in general is not a submanifold!

My preferred terminology is one of an immersion, which is a triple
$$
\phi: S\to M
$$
where $\phi$ is a smooth map such that the differential $d\phi_x$ is 1-1 for every $x\in S$. Equivalently, for each $x\in S$ there exists a neighborhood $U$ of $x$ in $S$ such that:

a. $\phi(U)$ is a smooth submanifold $N$ in $M$.

and

b. $\phi: U\to  N$ is a diffeomorphism, i.e. a map which has a smooth inverse.

Given this, one can say that a subset $A\subset M$ is an "immersed submanifold" if there exists an immersion $\phi: S\to M$ whose image $\phi(S)$ equals $A$.

To make things worse, Wald's definition of an embedding (an embedded submanifold) is plain wrong! The correct definition is that $\phi$ is an immersion which is a homeomorphism to its image. (This includes, the 1-1 requirement, but  is stronger.)

Examples of immersed submanifolds were discussed many times at MSE, for instance here.

Title: Immersed Submanifold Requirements


问:
Why is choosing a connected component of $\Lambda^{n}(V^{*})\setminus\{0\}$ equiv. to choosing an orientation $[(v_{1},...,v_{n})]$ of V, Lee, p.380
I am trying to understand the following from Lee's book "Introduction to smooth manifolds":

Question:

What exactly does choosing a connecetd component mean? Does it mean choosing an element of one component?

Is the above excerpt supposed to mean something like:  Having an orientation $[(v_{1},...,v_{n})]$   $\iff$ choosing a connected component of $\Lambda^{n}(V)^{*}\setminus\{0\}$? The "$\Leftarrow$-direction" makes sense I think. I can not make sense of the other direction. Do I have to come up with an n-form that gives rise to the same equivalence class? However in Tu's book there is the following, which I think is only the "$\Leftarrow$-direction". It made me dought that the "$\Rightarrow$-direction" makes any sense.

Many thanks in advance!

答:
The space of $n$ forms on a vector space $V$ of dimension $n$ is 1-dimensional. If $v_i$ is a basis of $V^*$ then any $n$-form can be written
\begin{equation}
c\ v_1 \wedge \cdots\wedge v_n
\end{equation}
with $c\in \mathbb{R}$. Hence $\Lambda ^n(V^*)=\mathbb{R}$. If you exclude 0-forms, so $c\neq 0$ you get $\mathbb{R}\setminus 0$ which has two connected components, positive and negative multiples of some arbitrarily selected non-zero $n$-form. The orientation does not see (define an equivalence relation if you want to say it in a fancier way) the value of $c$ but only its sign.


问:
Poisson process condition
A certain toy store sells toys. During a particular hour, customers arrive at the store according to a Poisson Process with mean $10$ per hour. Let $X$ be the number of customers who arrive during this hour. For each customer, there is a $\frac{1}{5}$ chance that the customer will buy exactly one toy and a $\frac{4}{5}$ chance that the customer will not buy any toy, independent of other customers. Let $Y$ be the number of toys sold in the one hour period. By conditioning $Y$ on $X$, use the law of iterated expectation to determine $\mathbb{E}(Y)$ and then use the law of total variance to determine $\mathrm{Var}(Y)$.

I have been trying to solve this problem for hours but without success. Any assistance you can provide is much appreciated.

答:
Since customer arrival is a Poisson process with an intensity of $10$, the Number of customers in an hour will have distribution Poisson$(10)$, and each customer buys a toy with probability $1/5$ therefore expected number of toys sold given $x$ many customers arrived ($\mathbb{E}(Y|X=x)$) is $x/5$. Thus

$$
\mathbb{E}(Y) = \sum_{x=0}^{\infty}\mathbb{E}(Y|X=x)\mathbb{P}(X=x) = \sum_{x=0}^{\infty}\frac{x}{5}\frac{e^{-10}10^{x}}{x!} = \frac{1}{5}\sum_{x=0}^{\infty}\frac{x}{5}\frac{e^{-10}10^{x}}{x!} =2,
$$
since $\sum_{x=0}^{\infty}{x}\frac{e^{-10}10^{x}}{x!}$ is the expectation of Poisson$(10)$, which is $10$.

You can write
$$
Var(Y) = Var(\mathbb{E}(Y|X)) +\mathbb{E}(Var(Y|X)).
$$
you can use this formula and compute the two terms separately. Try it!

Hint 1: $Y|X \sim$ Binomial$(X,\frac{1}{5})$.

Hint 2: $X=$ the Number of customers in an hour. It has distribution Poisson$(10)$

Title: Poisson process condition


问:
Small (elementary) confusion regarding affine algebraic sets.
If we have a non-constant polynomial $f_\in k[T_1]$ and we look at the affine variety $D(f)\subseteq \mathbb{A}^1_k$, then it's affine coordinate ring is $\Gamma(D(f))=k[T_1]_f\cong k[T_1,T_2]/(1-fT_2)$ as $k$-algebras, hence we have that by the equivalence of reduced finitely generated $k$-algebras and affine algebraic sets, that $D(f)\cong V(1-fT_2)\subseteq \mathbb{A}^2_k$ as affine algebraic sets. However, the "obvious" map $x\mapsto (x,f(x)^{-1})$ is in general not given by polynomials, so it can't serve as the isomorphism. My question is if everything that I wrote so far is correct - or rather where the mistakes are - and maybe to get some help to resolve my confusion, because for my eyes it seems as though the $k$-algebra isomorphism is canonical, so I would expect also a canonical isomorphism on the level of affine algebraic sets. I'm quite sure that I have a substantial misunderstanding here, so that's why I'm asking. Thanks!

答: Your issue is that $D(f)\subset \Bbb A^1$ is not an affine algebraic set, because it's not the zero locus of some collection of polynomials in $k[T_1]$. On the other hand, everything you say is true if you enlarge your category a bit by looking at quasi-affine algebraic sets (open subsets of algebraic sets) where the morphisms are given by regular functions (each component is locally the quotient of two polynomials where the denominator does not vanish). Once you make this upgrade, it's clear that the map you write down is in fact a morphism in this category, and it's actually an isomorphism.