问:
Finding the coordinates of a point on a line closest to a given point
I have a line from A to B and point C. I have the coordinates of A, B, and C, and I need to find the location of D which is the closest point on the line to C:

Keep in mind that A, B, and C could be any point in space. I have tried many attempts and failed.

答:
What you need is the perpendicular projection of $\vec{C}$ on the line through $\vec{A}$ and $\vec{B}$. To learn more try this wiki: https://en.wikipedia.org/wiki/Vector_projection

We can use the dot product to compute this vector projection, keeping in mind that the vectors aren't normalized when we plug this vector into the formula $\frac{a\cdot b}{b\cdot b}b$ we get:

$$\vec{D}-\vec{A} = \frac{(\vec{B}-\vec{A})\cdot(\vec{C}-\vec{A})}{(\vec{B}-\vec{A})\cdot(\vec{B}-\vec{A})} (\vec{B}-\vec{A}).$$

If you bring $\vec{A}$ over to the other side of the equation, you get a formula for $\vec{D}$ in terms of $\vec{A}, \vec{B}$ and $\vec{C}$.

EDIT:

For example If we take $\vec{A} = (0, 0)$, $\vec{B} = (1, 1)$ and $\vec{C} = (1, 0)$.

In this case $\vec{B} - \vec{A} = (1, 1) - (0, 0) = (1, 1)$.

The squared norm of this vector is $(\vec{B}-\vec{A})\cdot (\vec{B}-\vec{A}) = (1, 1) \cdot (1, 1) = 1.1+1.1 = 2$

The other vector is $\vec{C}-\vec{A} = (1, 0) - (0, 0) = (1, 0)$.

The innerproduct of these vectors is $(\vec{B}-\vec{A})\cdot (\vec{C}-\vec{A}) = (1, 1)\cdot(1, 0) = 1.1 + 1.0 = 1$.

The quotient of these two innerproducts is $\frac{(\vec{B}-\vec{A})\cdot(\vec{C}-\vec{A})}{(\vec{B}-\vec{A})\cdot(\vec{B}-\vec{A})} = \frac{1}{2}.$

To get the vector $\vec{D}-\vec{A}$ we need to multiply this fraction by $\vec{B}-\vec{A}$. This means the following calculation $\frac{(\vec{B}-\vec{A})\cdot(\vec{C}-\vec{A})}{(\vec{B}-\vec{A})\cdot(\vec{B}-\vec{A})} (\vec{B}-\vec{A}) = \frac{1}{2} (\vec{B}-\vec{A}) = \frac{1}{2}(1, 1) = (.5, .5)$.

Finally we computed the vector $\vec{D}-\vec{A}$, to get the exact position of $\vec{D}$ we add $\vec{A}$. In other words $\vec{D} = (.5, .5) + \vec{A} = (.5, .5) + (0, 0) = (.5, .5)$.

As expected the point $\vec{D}$ lies at coordinates $(.5, .5)$.


问:
Probability that two tickets drawn out of $n$ tickets with and without replacement and are consecutive integers
I have a question given as follows:

A box contains $n$ tickets numbered $1$ to $n$. Two tickets are chosen at random. Find the probability that the numbers will be consecutive integers, if the tickets are chosen (i)
without replacement, and (ii) with replacement.

MY ATTEMPT

The ways in which consecutive integers can be drawn are as follows:

$$(1,2), (2,3), \dots , (n - 1, n)$$
$$(2,1), (3,2), \dots , (n, n - 1)$$

So number of favourable cases = $2(n - 1)$.

i) Total number of cases = $n(n - 1)$ (since without replacement).

So probability $$= \frac{2(n-1)}{n(n-1)} = \frac{2}{n}$$

ii) Total number of cases = $n^2$ (since with replacement).
So probability $$= \frac{2(n-1)}{n^2}$$

Is my attempt is correct? Any kind of help will be highly grateful. Thanking in advance!

答:
Yes, your method and answer are correct. As a sanity check, analyze the experiment for $n=2$.

In first case, one is sure to draw consecutive integers. Does your result give $1$?

In second case, the second card can be same as first or consecutive to it. These two are only possibilities. Does your result evaluate to $1/2$?

ContentLicense: CC BY-SA 4.0


问:
$R$ is an integral domain iff $R[x_1,...,x_n]$ is a integral domain
I would like to prove that $R$ is an integral domain iff $R[x_1,...,x_n]$ is a integral domain. The converse is trivial, since R can be viewed as a subring of $R[x_1,...,x_n]$. In order to prove the first implication, I'm trying to use the same argument we use when we prove the case $n=1$ looking to the highest degree coefficient, am I on the right way?

Thanks in advance

答: Did you prove the case $n=1$? If so, then you're done. Suppose $R$ in an integral domain, and inductively assume $R[x_1,\ldots,x_n]$ is an integral domain. Then $R[x_1,\ldots,x_n,x_{n+1}] = R[x_1,\ldots,x_n][x_{n+1}]$ is an integral domain since $R[x_1,\ldots,x_n]$ is.</p>

ContentLicense: CC BY-SA 3.0


问:
Theorem of class field theory in appendix of the book 'Cyclotomic fields' by Washington
I came across the following theorem of class field theory in the appendix of the book 'Cyclotomic fields' by Washington. I could not find it in Milne, Cassels Frohlich, Neukirch or Artin Tate.

Theorem 14. (a) If $K/k$ is abelian, then there is a closed subgroup $H$ with $D_k\subseteq H\subseteq C_k$, such that
$$C_k/H\simeq\mathrm{Gal}(K/k).$$
The prime $\mathfrak{p}$ is unramified $\Longleftrightarrow k^\times U_\mathfrak{p}/k^\times\subseteq H$.

(b) Given a closed subgroup $H$ with $D_k\subseteq H\subseteq C_k$ (equivalently, $C_k/H$ is totally disconnected), there is a unique abelian extension corresponding to $H$, as in (a).

Where can I find proof and discussion on this theorem?

答: this is called the existence theorem and I think it is mentioned in both Milne book and Cassels Frohlich, possibly with a different language. but It is discussed in length in the Bonn lecture of Neukirch https://www.springer.com/gp/book/9783642354366. the part about the ramification comes from local class field theory and compatibility of local and global class field theory.</p>


答: Your thought is correct, there is only a problem in your first formula for $n=1$. In that case, the probability should be $0$, not $2$.

Title: Probability that two tickets drawn out of $n$ tickets with and without replacement and are consecutive integers

Body:
I have a question given as follows:

A box contains $n$ tickets numbered $1$ to $n$. Two tickets are chosen at random. Find the probability that the numbers will be consecutive integers, if the tickets are chosen (i)
without replacement, and (ii) with replacement.

MY ATTEMPT

The ways in which consecutive integers can be drawn are as follows:

$$(1,2), (2,3), \dots , (n - 1, n)$$
$$(2,1), (3,2), \dots , (n, n - 1)$$

So number of favourable cases = $2(n - 1)$.

i) Total number of cases = $n(n - 1)$ (since without replacement).

So probability $$= \frac{2(n-1)}{n(n-1)} = \frac{2}{n}$$

ii) Total number of cases = $n^2$ (since with replacement).
So probability $$= \frac{2(n-1)}{n^2}$$

Is my attempt is correct? Any kind of help will be highly grateful. Thanking in advance!


问:
How to $2^n/n! \leq 4/n$ by induction
I'm trying to prove that $\frac{2^n}{n!} \leq \frac{4}{n}$ for all positive integers $n$.
It's obvious this holds for $n=1$.
My inductive hypothesis is that $\frac{2^n}{n!} \leq \frac{4}{n}$.
I want to show that $\frac{2^{n+1}}{(n+1)!} \leq \frac{4}{(n+1)}$.
I can see that $\frac{2^{n+1}}{(n+1)!} = \frac{2^n \cdot 2}{n! \cdot (n+1)} \leq
\frac{4 \cdot 2}{n(n+1)}$, where the inductive hypothesis is applied in the last step.
All I need to show is that $\frac{4 \cdot 2}{n(n+1)} \leq \frac{4}{n+1}$ and I'm done. But I can't think of a good justification for this.
Any hints or suggestions?

Title: How to $2^n/n! \leq 4/n$ by induction

Body:
$\frac{4\cdot 2}{n(n+1)} = \frac{2}{n}\frac{4}{n+1}\leq \frac{4}{n+1}$, since $\frac{2}{n}\leq 1$ for $n\geq 2$.

In the case $n=1$, $2/1 \leq 4/1$.


问:
Sturm-Liouville Problem
How could one prove that there are at most countably many eigenvalues of the Sturm-Liouville problem $-Lu = ju$, $j$ = eigenvalue, and $u$ is in $C^2[a,b]$? I have been attempting at this problem for a while and I am not sure how to proceed. Thank you very much. </p>

答:
Dear Carlotto, for each eigenvalue, you may assume that $u$ is a real function. If it were not, you could take the real part of $u$ to make it real and it would still be an eigenstate.

Take an eigenvalue $\lambda$, find the corresponding eigenstate $u_\lambda$, and - the critical point is coming - count the number $n$ of zeros of $u_\lambda$ in the interval $[a,b]$. You're finished if you can prove that for each number of zeros $n$, you can only find one eigenvalue (or two eigenvalues); the eigenvalues may be labeled by $n$ (and an extra bit if needed) and this map proves that the set of eigenvalues is countable.

Now, it's not hard to see that the statement is right. The function $u$ has to satisfy the S-L boundary conditions at $x=a$. It is evolved, while staying real, in the direction towards $b$ where some S-L boundary conditions have to hold, too. The evolution depends on $\lambda$. As you adjust $\lambda$, you're increasing the number of zeros and for each number of zeros, you will find just one value of $\lambda$ for which the boundary conditions are satisfied at $b$ again.

There are subtleties for a general S-L problem but I recommend you to start with the $-d^2/dx^2$ operator on an interval $[0,\pi]$ where the eigenstates are just standing waves $\sin(nx)$ and eigenvalues are $n^2$. The boundary conditions are $y=0$ at $x=0,\pi$. The general S-L problem is actually conceptually analogous although the detailed operators are shifted and the eigenvalues are changed. But you may continuously connect it to the standing waves case so the spectrum has to be countable - another way to see it.

There's a minimization method to find the functions. The lowest-eigenvalue eigenstate with the minimum number of zeros minimizes the expectation value of $L$ among all states; the next-to-lowest eigenstate minimizes the eigenvalue of $L$ between all states that are perpendicular to the ground state; the next one minimizes it among all states that are orthogonal to both the previous ones; and so on.

Title: Sturm-Liouville Problem


答:
$$R[x_1,x_2]=R[x_1][x_2]$$
So if you can handle the one-variable case, you're done by induction.</p>

Body:
I would like to prove that $R$ is an integral domain iff $R[x_1,...,x_n]$ is a integral domain. The converse is trivial, since R can be viewed as a subring of $R[x_1,...,x_n]$. In order to prove the first implication, I'm trying to use the same argument we use when we prove the case $n=1$ looking to the highest degree coefficient, am I on the right way?

Thanks in advance


问:
Find the locus of this point P

Find the locus of all points inside $\triangle ABC$ such that $PA^{2}+PB^{2}=PC^{2}$.

At first, i tried finding a right angled triangle and then tried to go on applying Pythagorean Theorem and finding other trivia but that didn't seem to work. I think the locus is most likely to be some line segment rather than an arc of some circle as that seems to be quite unrelated seen from an Euclidean perspective. Applying Apollonius' Theorem and Stewart's Theorem on some triangles might be the key to this problem although i could not find such triangles. I don't know whether this can be solved by trigonometry or not but since this is a problem from a chapter on euclidean geometry, i am sure that there is some clever way to look at this problem.

答:
We extend median $CM$ such that $CM = MD$. Then please note that $MP$ is median of $\triangle CPD$ and $\triangle APB$. Applying Apollonius's theorem,

$PA^2 + PB^2 = 2(MP^2 + AM^2)$, $PC^2 + PD^2 = 2(MP^2 + CM^2)$

Subtracting second from first,

$PA^2 + PB^2 - PC^2 = 0 = PD^2 + 2 (AM^2 - CM^2)$

i.e $PD^2 + 2(\frac{c^2}{4} - \frac{1}{2}(a^2 + b^2) + \frac{c^2}{4}) = 0$

That leads to $PD = \sqrt{a^2+b^2-c^2} \ $. So the locus of point $P$ satisfying $PA^2 + PB^2 = PC^2$ is a circular arc with point $D$ being the center and radius $\sqrt{a^2+b^2-c^2}$. We also note that this is possible only if $a^2 + b^2 \geq c^2$.

Title: Find the locus of this point P


问:
Number of squares of the form $a^2+x^2$ in a finite field
Suppose $a$ is a non-zero element in a finite field $GF(q)$ of odd characteristic.

How many $x \in GF(q)$ are such that $a^2+x^2$ is a square in $GF(q)$?

From some experiments with some small fields, it seems that the answer is $(q+\epsilon)/2$, where $\epsilon = q \mod 4$ , but I can't prove it.

Any help appreciated!

答:
Let $N$ be the number of solutions for $(x,y) \in \mathbb{F}_q$ of $a^2+x^2=y^2$. This equation is equivalent to $(y-x)(x+y)=a^2$. Since $q$ is odd, $(x,y) \in \mathbb{F}_q^2 \longmapsto (x+y,y-x) \in \mathbb{F}_q^2$ is a bijection, so $N$ is the number of solutions to $xy=a^2$. There are obviously $q-1$ of them.

Now, we can group the solutions of $x^2+a^2=y^2$ into two groups: when $y \neq 0$, and $y=0$. Let $N_1,N_2$ be the numbers of solutions in each group.

Then the number $Q$ of $x$ such that $x^2+a^2$ is a perfect square is $N_1/2+N_2$. Now, $N_2$ is zero or two, and two iff $-a^2$ is a square iff $-1$ is a square iff $4|q-1$.

So if $4|q-1$, $N_2=2,N_1=q-3$ so $Q=\frac{q+1}{2}$.
If not, $N_2=0,N_1=q-1$ so $Q=\frac{q-1}{2}$.


问:
How can $\frac{1}{\ln(x)}$ be equal to $0$?
High school calculus student checking in here – first time poster.

I got asked a question by one of my friends:
If $\cot x = \frac{1}{\tan(x)}$, how could $\cot(x) = 0$? For this to be possible, he reasoned, $\tan(x)$ would have to be equal to infinity – and division by infinity does not work.

Rewriting $\cot(x)$ as $\frac{\cos(x)}{\sin(x)}$ makes the problem make sense, but doesn't explain why it doesn't make sense in the previous form.

Furthermore, as I was looking for more examples of similar things, I found that (at least according to desmos) $\frac{1}{\ln(x)}$ is satisfied by the coordinate $(0, 0)$. I cannot wrap my head around how this is possible, particularly since $\ln(x)$ is not even defined at $0$.

I wasn't able to find a good answer to this online, and we would really appreciate an elegant (i.e. understandable for high school students) explanation of why this is.

答:
$$x \mapsto \frac{1}{\ln(x)}$$
is indeed not defined for $x=0$, but the point is that it can be extended to a continuous function $f : [0,1) \rightarrow \mathbb{R}$ satisfying $f(0)=0$. More precisely, define
$$f(x)=\left\{
    \begin{array}{ll}
        \frac{1}{\ln(x)} & \mbox{if } 0<x<1 \\
        0 & \mbox{if } x=0
    \end{array}
\right.
$$

You can prove that $f$, as defined, is a continuous function, and this is the only way to extend your original function to a continuous one at $x=0$.


问:
Tight upper bound on the sum of $\left( \frac{1}{k} \right)^k \exp \left(-\frac{\log(N)(\sqrt{N-k} - \sqrt{k})^2}{N}\right)$
Consider the following sum:

\begin{align}
      S = \sum_{k=\lceil\log(N)\rceil}^N\left(  \frac{1}{k} \right)^k \exp \left(-\frac{\log(N)(\sqrt{N-k} - \sqrt{k})^2}{N}\right).
\end{align}
Is it possible to show that this is sum is of order $1/N$ for  large $N$?

My attempt:
When $k$ is the same order as $N$, then $(1/k)^k$ is very small. It seems sufficient to consider $k \ll N$ and write
\begin{align*}
    S \leq N \frac{1}{(\log(N))^{\log(N)}} \exp( - \log(N)) \leq  \frac{1}{N},
\end{align*}
where I used the fact that $N \frac{1}{(\log(N))^{\log(N)}} \leq 1$ for $N$ large enough.

答:
You're absolutely right. Slightly more formally, we might do something like the following:

Expand out the exp term to get

$$
\sum k^{-k} \exp \left ( - \frac{\log N}{N} (N - 2 \sqrt{k(N-k)}) \right )
$$

Distributing and simplifying gives

$$
\sum k^{-k} \frac{1}{N} \exp \left ( \frac{2 \sqrt{k(N-k)} \log N}{N} \right )
$$

Then the stuff inside the $\exp$ is maximized when $k = \frac{N}{2}$. This is actually a terrible upper bound, but it's good enough for what we want. So our sum is at most

$$
\sum k^{-k} \frac{1}{N} e^{\log N}
$$

That is,

$$S_N \leq \sum_{k = \log N}^N k^{-k}$$

Now, as you've noticed, the first term will dominate this series. In fact, since we're already grossly overestimating from the last step, we might as well keep making our lives easy:

$$
\sum_{k = \log N}^N \left ( \frac{1}{k} \right )^k
\leq
\sum_{k = \log N}^\infty \left ( \frac{1}{\log N} \right )^k
$$

But now we're in the clear! Applying the formula for a geometric series gives

$$
S_N \leq
\left ( \frac{1}{\log N} \right )^{\log N} \frac{1}{1 - \frac{1}{\log N}}
$$

Of course, $\frac{1}{1 - (\log N)^{-1}} \leq 2$ for $N > 8$ or so, which means we can safely ignore it. So at long last

$$
S_N \leq O  \left ( \left ( \frac{1}{\log N} \right )^{\log N} \right ).
$$

This is actually better than $O \left ( \frac{1}{N} \right )$, even if it's not obvious at first. You can compute
$\lim_{n \to \infty} \frac{(\log N)^{-\log N}}{N^{-1}} = 0$ (or look at a graph) to convince yourself of this.

I hope this helps ^_^


答: If you already have proved this assertion for $n=1$ then you can use the induction since $R[x_1,...,x_n]=(R[x_1,...,x_{n-1}])[x_n]$.</p>

Title: $R$ is an integral domain iff $R[x_1,...,x_n]$ is a integral domain


答:
Desmos draws the graph of a function at the points at which it is defined. Indeed, $\frac1{\ln x}$ is undefined at $0$; but, since $\lim_{x\to0^+}\frac1{\ln x}=0$, you see nothing peculiar at $(0,0)$. If you ask Desmos to draw the graph of, say $\frac{\sin(x)}x$ or $\frac{\cos(x)-1}{x^2}$, the same thing will occur.

And the equality $\cot x=\frac1{\tan x}$ is valid only when $\tan(x)\ne0$.

Title: How can $\frac{1}{\ln(x)}$ be equal to $0$?

Body:
High school calculus student checking in here – first time poster.

I got asked a question by one of my friends:
If $\cot x = \frac{1}{\tan(x)}$, how could $\cot(x) = 0$? For this to be possible, he reasoned, $\tan(x)$ would have to be equal to infinity – and division by infinity does not work.

Rewriting $\cot(x)$ as $\frac{\cos(x)}{\sin(x)}$ makes the problem make sense, but doesn't explain why it doesn't make sense in the previous form.

Furthermore, as I was looking for more examples of similar things, I found that (at least according to desmos) $\frac{1}{\ln(x)}$ is satisfied by the coordinate $(0, 0)$. I cannot wrap my head around how this is possible, particularly since $\ln(x)$ is not even defined at $0$.

I wasn't able to find a good answer to this online, and we would really appreciate an elegant (i.e. understandable for high school students) explanation of why this is.


问:
Deriving probability density function
I'd like to derive the probability density of $\eta:=e^{-\log(\xi)}$ for $\xi$ being uniformly distributed on $[0,1]$.

I tried to figure it out by calculating according to

https://www.cl.cam.ac.uk/teaching/2003/Probability/prob11.pdf

which has led to $f_\eta(y)=\frac{1}{y^2}$ but I'm not sure if this is right and if so, which distribution it represents.

答:
As Kavi Rama Murthy commented,  $\eta = \frac1{\xi}$

$F_\eta(x) = \mathbb P(\eta \le x) = \mathbb P(\xi \ge \frac1x) = 1-\frac1x$,  so $f_\eta(x)=\frac1{x^2}$ as you found

Note that your density is for $x \ge 1$, since $0 \le \frac1x\le 1$

You have a Pareto distribution with shape parameter $\alpha=1$ and minimum $1$

Title: Deriving probability density function


答:
If you define

$$\cot(x):=\dfrac1{\tan(x)},$$ then the cotangent can never be $0$. But if you define it as

$$\cot(x):=\dfrac{\cos(x)}{\sin(x)},$$ then it is indeed zero for all $x$ such that $\cos(x)=0$. The second is the "best" definition, and

$$\cot(x)\equiv\dfrac1{\tan(x)}\equiv\dfrac{\cos(x)}{\sin(x)}$$ is not completely true.

Now if we look at

$$\frac1{\log(x)},$$

this expression is indeed undefined at $x=0$ because the logarithm is undefined. You can extend the definition in a way that makes the function continuous as follows:

$$\begin{cases}x=0\to\lim_{t\to0}\dfrac1{\log(t)},\\x>0\to\dfrac1{\log(x)}.\end{cases}$$

Note that a grapher cannot show you the difference between the definitions, because they differ in isolated points, which are "infinitely tiny".


答:
$\eta=\xi^{-1}$ so that for $x\geq1$ we find: $$P(\eta\leq x)=P(\xi\geq x^{-1})=1-x^{-1}$$
Differentiating this CDF we find a corresponding PDF prescribed by:$$x\mapsto x^{-2}\text{ if }x\geq1\text{ and }x\mapsto0\text{ otherwise}$$

Title: Deriving probability density function

Body:
I'd like to derive the probability density of $\eta:=e^{-\log(\xi)}$ for $\xi$ being uniformly distributed on $[0,1]$.

I tried to figure it out by calculating according to

https://www.cl.cam.ac.uk/teaching/2003/Probability/prob11.pdf

which has led to $f_\eta(y)=\frac{1}{y^2}$ but I'm not sure if this is right and if so, which distribution it represents.


答:
You have a removable singularity in your function definition using the division. While the function as written is not defined for some values, there is a single way how those gaps can be filled while making the function behave nicely around them.

Dealing with equations we tend to do such things all the time. When you simplify $\frac{a^2-b^2}{a-b}={a+b}$ wer tend to ignore the special case of $a+b$ where this is invalid. And we often can get away with it although it can come to bite us. In that sense we can say the two sides of the equation with the cot in it are the same, keeping in mind that in the case of two that's not literally true.


答:
As per the fact that $f_{\xi}=1$ to derive $f_{\eta}$ it is enough to calcualate the derivative

$$f_{\eta}=\Bigg|\frac{d \xi}{d \eta}\Bigg|=\frac{1}{\eta^2}$$

Obviously being $\xi \in (0;1]$ it is also $\eta \in [1;+\infty)$

Title: Deriving probability density function


问:
Law of Large Numbers, a confusion
According to Law of Large Numbers, if I throw a coin 1000 times approximately 500 will be head and 500 tail. Suppose that I throw the coin 700 times and I got 700 heads. Can I say that in the next 300 throws the probability of getting tails will be higher than probability of getting heads?

Edit: To make an analogy, imagine that you have a box that contains 500 black and 500 white balls. If 700 times (700 exaggerated) you choose black, than it is more probable that for 701 you get white. However if you say me that instead of one box you have 1000 similar box and every time you choose from one box that means the probability of choosing black or white will never change. With independent events you mean this?

Edit2: Imagine that there are billions of people that throwing coins 1000 times. For each person there is an empty box. When he throw the coin, if it is tail he puts a black ball in his box, when the coin is head he puts white ball. So at the end of the experiment there are billions of boxes that each box contains approximately 500 black 500 white ball. So they give me the opportunity to choose one box. The box that I choose represents the one possible coin throwing that I would make. I am asking what is the difference between throwing coins and choosing one box in billions of boxes? If there is no difference, than the first statement holds. For example I pick 400 black balls from my box than it is more probably to choose white ball from remaining 600 hundreds.

答:
The way the law of large numbers works isn't by cancelling variations you've already gotten.  Instead, it just keeps adding new values until the variations become insignificant.

So if you've gotten 700 heads, every flip still has a 50% chance of heads or tails, which means you'll on average have 700 more total heads than tails for the rest of your flips, no matter how many more you do.  The law of large numbers says this is true even if you do 1 million more flips, in which case you'll have about half heads and half tails, since at that point a difference of 700 isn't very much.

Title: Law of Large Numbers, a confusion


答:
Just to add to the other answers given (which I completely agree with), I wanted to mention that this is sometimes referred to as the gambler's fallacy. </p>

<p>And in response to your recent edits: say we have a box that contains 500 black and 500 white balls. Then (if we replace the ball we pick each time so that there are always 500 of each) there will always be an equal probability of picking black or white (assuming the picking process is truly random). Picking 700 black balls in a row doesn't change this probability ( although 700 in a row is indeed astonishingly unlikely to happen and we might then start to wonder if we are really picking randomly or if something, e.g. the way the balls are arranged in the box, is affecting our choice, etc). But if it is indeed random then it is not more probable to get white on pick 701. However, if we do not replace the balls then the more black balls we pick the less likely it becomes to pick another (although we could never pick 700 of them if there are only 500 to begin with). Hope that helps. </p>

Title: Law of Large Numbers, a confusion

Body:
According to Law of Large Numbers, if I throw a coin 1000 times approximately 500 will be head and 500 tail. Suppose that I throw the coin 700 times and I got 700 heads. Can I say that in the next 300 throws the probability of getting tails will be higher than probability of getting heads?

Edit: To make an analogy, imagine that you have a box that contains 500 black and 500 white balls. If 700 times (700 exaggerated) you choose black, than it is more probable that for 701 you get white. However if you say me that instead of one box you have 1000 similar box and every time you choose from one box that means the probability of choosing black or white will never change. With independent events you mean this?

Edit2: Imagine that there are billions of people that throwing coins 1000 times. For each person there is an empty box. When he throw the coin, if it is tail he puts a black ball in his box, when the coin is head he puts white ball. So at the end of the experiment there are billions of boxes that each box contains approximately 500 black 500 white ball. So they give me the opportunity to choose one box. The box that I choose represents the one possible coin throwing that I would make. I am asking what is the difference between throwing coins and choosing one box in billions of boxes? If there is no difference, than the first statement holds. For example I pick 400 black balls from my box than it is more probably to choose white ball from remaining 600 hundreds.

Body:
Just to add to the other answers given (which I completely agree with), I wanted to mention that this is sometimes referred to as the gambler's fallacy.

And in response to your recent edits: say we have a box that contains 500 black and 500 white balls. Then (if we replace the ball we pick each time so that there are always 500 of each) there will always be an equal probability of picking black or white (assuming the picking process is truly random). Picking 700 black balls in a row doesn't change this probability ( although 700 in a row is indeed astonishingly unlikely to happen and we might then start to wonder if we are really picking randomly or if something, e.g. the way the balls are arranged in the box, is affecting our choice, etc). But if it is indeed random then it is not more probable to get white on pick 701. However, if we do not replace the balls then the more black balls we pick the less likely it becomes to pick another (although we could never pick 700 of them if there are only 500 to begin with). Hope that helps.


问:
Atiyah-McDonald Exercise 4.11

If $\mathfrak{p}$ is a minimal prime ideal of a ring $A$, then $S_\mathfrak{p}(0)$ is the smallest $\mathfrak{p}$-primary ideal.

Let $\mathfrak{a}$ be the intersection of the ideals $S_{\mathfrak{p}}(0)$ as $\mathfrak{p}$ runs through the minimal prime ideals of $A$. If the zero ideal is decomposible, then $\mathfrak{a} = 0$ iff every prime ideal of $0$ is isolated.

Notation: $S_{\mathfrak{p}}(0)=\text{Ker}(A\to A_{\mathfrak{p}})$.

I know $r(S_{\mathfrak{p}}(0))=\mathfrak{p}$. I tried to show this is primary by showing if $xy\in S_{\mathfrak{p}}(0)$ and $y\notin S_{\mathfrak{p}}(0)$ then $x\in S_{\mathfrak{p}}(0)$ but I don't know how to prove.

Suppose every prime ideal of $0$ is isolated. Then every prime ideal is a minimal prime ideal of $A$ as every prime ideal contains $0$. Hence $\mathfrak{a} = \bigcap_{\mathfrak{p}}S_{\mathfrak{p}}(0) = 0$. Now suppose $\mathfrak{a}=0$ and there is a prime ideal $\mathfrak{p}$ of $0$ such that $\mathfrak{p}$ is embedded. Then there is some prime ideal $\mathfrak{p}'$ of $0$ such that $\mathfrak{p}'\subset\mathfrak{p}$. Hence, $S_{\mathfrak{p}'}(0)\supset S_{\mathfrak{p}}(0)$. From here, I don't know how to get further. Could you give any hint?

答:
for the first part you have to prove that $x^n\in S_p(0)$ not $x$. and then use $\sqrt{S_p(0)}=p$ which means that for every $x\in p$,$x^n\in S_p(0)$.

for the second part I think you are wrong that every prime ideal is minimal,every associated prime ideal is minimal, and from this you can conclude the first direction, for the other direction use that the set of radicals of the primes that apear in primary decomposition is equal to the set of associated prime and the fact you mentioned about radical of $S_p(0)$.(and of course the uniqueness of primary decomposition.) remember that you are only considering $S_p$ for minimal prime ideals

Title: Atiyah-McDonald Exercise 4.11


问:
Which is the signature of the matrix?
Let $A\in \mathbb{R}^{4\times 4}$ is a real symmetric matrix with characteristic polynomial $$\chi_A(x)=x^4-9x^3+cx^2+dx+37$$ If we know that all eigenvalues of $A$ have the same sign , which is the signature of the matrix $A$ ?

The signature is number of positive eigenvalues minus number of negative eigenvalues.

So is the signature either $4$ or $-4$, since the characteristic polynomial  is of degree $4$ ?

答:
Since the eigenvalues $\lambda_1,\lambda_2,\lambda_3,\lambda_4$ of $A$ are the roots of the characteristic polynomial, we have $$\chi_A(x) = x^4-9x^3+cx^2+dx+37 = (x-\lambda_1)(x-\lambda_2)(x-\lambda_3)(x-\lambda_4).$$ By comparing the $x^3$ coefficient of both sides, we get $\lambda_1+\lambda_2+\lambda_3+\lambda_4 = 9$. By comparing the constant term of both sides, we get $\lambda_1\lambda_2\lambda_3\lambda_4 = 37$.

We are given that all the eigenvalues have the same sign. Since $\lambda_1\lambda_2\lambda_3\lambda_4 = 37 > 0$, none of the eigenvalues can be zero. If all the eigenvalues were negative, we'd have $9 = \lambda_1+\lambda_2+\lambda_3+\lambda_4 < 0$, a contradiction. Hence, all the eigenvalues must be positive, i.e. the signature of $A$ is $4-0 = 4$.

Title: Which is the signature of the matrix?


问:
In $\Delta ABC$, prove $A'D=\frac{(c^2- b^2)}{2a}$ with $A'$ be the midpoint of $BC$, $AB>AC$, and other Information given.
Problem

In $\Delta ABC$, we have $AB>AC$.

Let $A'$ be the midpoint of $BC$

$D$ be the foot of altitude from  $A$ to $BC$

Internal and external Angle Bisectors of $\angle A$ meet $BC$ at $X$ and $X'$ respectively.

Prove that

$A'X = \frac{a(c-b)}{2(c+b)}$

$A'X' = \frac{a(c+b)}{2(c-b)}$

$A'D=\frac{(c^2- b^2)}{2a}$ where $a, b, c$ are the sides $BC, CA, AB$ as used commonly.

[Question Taken From Challenge and Thrill of Pre-College Mathematics]

My Approach

I could solve part 1. and 2. using the angle bisector theorem a few times, and with some pre-requisite knowledge, which were in-fact to be proven in questions in the exercise preceding this one:

$XC= \frac{ab}{(b+c)}$, consequently $XB= \frac{ac}{(b+c)}$

$CX'= \frac{ac}{(c-b)}$, consequently $BX'= \frac{ab}{(c-b)}$

$XX'= \frac{2abc}{(c^2- b^2)}$

While I could prove the given formulas separately, I have no idea how to solve part 3.. A hint to help me start off would be appreciated.

答:
If you are allowed to use law of cosines, then it is very straightforward.

$A'D = c \cos B - \frac{a}{2}$

and $\displaystyle \cos B = \frac{a^2 + c^2 - b^2}{2ac}$.

That is all it takes.


问:
How to compute $\int_0^\infty \frac{\sin t}{t^{s+1}} dt $?
How to compute $\displaystyle\int_0^\infty \frac{\sin t}{t^{s+1}}\;\text dt$ ?

Here, the real part of the complex number $s$ is negative and greater than $-1$.

答:
Using the definition of $\Gamma$-function from here, prove:
$$\int_0^\infty e^{-a t}\,t^{s-1}\,dt=\Gamma(s)\,a^{-s},$$
then represent
$$\sin t=\frac{e^{i t}-e^{-i t}}{2\,i}.$$</p>


问:
Let $F/K$ be a galois extension, $\alpha\in F$,and $H=\{\sigma\in G｜\sigma(\alpha)＝\alpha\}$ , to show $L^H＝K（α）$
Let $F/K$ be a finite galois extension.
Let $G＝\text{Gal}(F/K)$, and $\alpha\in F$ and $H＝\{\sigma\in G｜\sigma(\alpha)＝\alpha\}$.
I want to show $K(\alpha)＝F^H$, where $F^H＝\{x\in F｜\forall\sigma\in H,\ \sigma(x)＝x\}$.

$K(\alpha)\subseteq F^H$ is clear, because $\sigma$ fixes $K$ and $\alpha$.
Could you prove other direction, $F^H\subseteq K(\alpha)$?

Thank you for your help.

答:
Hint: Show that the map $$\sigma H\mapsto \sigma(\alpha)$$
defines a bijection between the set of left cosets of $H$ in $G$ and the set of conjugates of $\alpha$ over $K$. Conclude that $[K(\alpha):K]=[F^H:K]$ and hence $F^H=K(\alpha)$.


问:
Proving a quadratic equation identity
Continued question here

I encountered this question:
For quadratic polynomials such as $x^2\pm5x\pm6$ or $x^3\pm5x\mp6$, can be factorised over the integers. The main problem is to find a generator which can generate every polynomial that have this identity.

What have I tried:

For $x^2\pm mx+n=(x\pm p)(x\pm q)$ and $x^2\pm mx-n=(x\pm s)(x\mp t)$, solving a simultaneous equation for $q$ and $s$, getting
$$q=\frac {t(p+t)} {p-t}$$
$$s=\frac {p(p+t)} {p-t}$$

and applying into the general equation getting
$$x^2+(\frac {p^2+t^2} {p-t})x+\frac {pt(p+t)} {p-t}$$
but it doesn't suit for every integer $p$ and $t$, how can we get furthur than that?

答:
Let $m$ and $n$ be two positive integers such that the two polynomials
$$x^2+mx+n\qquad\text{ and }\qquad x^2+mx-n,$$
both factor over the integers. Note that then also the two polynomials
$$x^2-mx+n\qquad\text{ and }\qquad x^2-mx-n,$$
factor over the integers. It means there exist integers $p$, $q$, $s$ and $t$ such that
\begin{eqnarray*}
x^2+mx+n&=&(x+p)(x+q)=x^2+(p+q)x+pq,\\
x^2+mx-n&=&(x+s)(x+t)=x^2+(s+t)x+st,\\
\end{eqnarray*}
and so in particular $m^2-4n$ and $m^2+4n$ are both perfect squares, because
\begin{eqnarray*}
m^2-4n&=&(p+q)^2-4pq=(p-q)^2,\\
m^2+4n&=&(s+t)^2-4st=(s-t)^2,\\
\end{eqnarray*}
and hence $(p-q)^2$, $m^2$ and $(s-t)^2$ are three perfect squares in arithmetic progression, with common difference $4n$. Conversely, if $a^2$, $b^2$ and $c^2$ are three perfect squares in arithmetic progression with common difference $4d$ then
\begin{eqnarray*}
p&=&-\frac{b-a}{2},\qquad\text{ and }\qquad q&=&-\frac{b+a}{2},\\
s&=&-\frac{b+c}{2},\qquad\text{ and }\qquad t&=&-\frac{b-c}{2},
\end{eqnarray*}
are integers satisfying
$$(x-p)(x-q)=x^2+bx+d,$$
$$(x-s)(x-t)=x^2+bx-d.$$
So your question is equivalent to characterizing all arithmetic progressions of perfect squares of length $3$. This was somewhat famously done by Fermat, see wikipedia, for example. The parametrization given there  yields the parametrization
\begin{eqnarray*}
m&=&u^2+v^2,\\
n&=&uv(u^2-v^2),
\end{eqnarray*}
of the coefficients of such polynomials. Here $u$ and $v$ range over the integers.

Title: Proving a quadratic equation identity


问:
Compute the residue at a pole of a meromorphic function
I am working through a larger computation, and I have gotten stuck at a step where I need to compute the residue of $\frac{f(\zeta)}{(\zeta-z)\zeta^n}$ at the point $\zeta=0$.  We are given that $f$ is a holomorphic function and I know that the residue should be equal to
$$\frac{1}{(n-1)!}\lim_{\zeta\to0}\frac{d^{n-1}}{d\zeta^{n-1}}\frac{\zeta^nf(\zeta)}{(\zeta-z)\zeta^n}$$
But I have no idea how to get an expression for this.

答:
I will assume that $z\neq 0$ in the following. Then,
$ \frac{1}{(n-1)!} \frac{d^{n-1}}{d\zeta^{n-1}} \frac{f(\zeta)}{(\zeta - z)}$ is already a holomorphic function for $\zeta \neq z$ and we can obtain a result just by plugging in $\zeta = 0$.

To write it in terms of derivatives of $f$, you can use the product rule with binomial coefficients $\begin{pmatrix} n \\ k \end{pmatrix}$:
$$ \frac{d^n}{d\zeta^n} g(\zeta) f(\zeta) = \sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix} g^{(k)}(\zeta) f^{(n-k)}(\zeta).   $$
For derivatives of $g$, one obtains
$$  g(\zeta) = \frac{1}{(\zeta - z)} ~\Rightarrow~ g^{(k)}(\zeta) \Big|_{\zeta = 0} = \frac{(-1)^k k!}{(\zeta - z)^{k+1}} \Big|_{\zeta = 0} = -\frac{k!}{z^{k+1}}, $$
which leads to a fairly explicit result in terms of $f^{(k)}$:
\begin{align}
&\text{Res}_{\zeta = 0} \frac{f(\zeta)}{(\zeta - z)\zeta^n} = \frac{1}{(n-1)!} \frac{d^{n-1}}{d\zeta^{n-1}} \frac{f(\zeta)}{(\zeta - z)}\Big|_{\zeta = 0} \\
=& \frac{1}{(n-1)!} \sum_{k=0}^{n-1} \begin{pmatrix} n-1 \\ k \end{pmatrix} \frac{(-k!)}{z^{k+1}} f^{(n-1-k)}(0) = - \sum_{k=0}^{n-1} \frac{f^{(n-1-k)}(0)}{(n-1-k)!z^{k+1}} .
\end{align}
Let me know if there are any questions or I made any mistakes :)


问:
Books on shelf problem
Rashida keeps her collection of 162 books on 3 shelves.

There are 42 books on the first shelf, $x$ books on the second shelf, and the number of books on the third shelf is 2 times the number of books on the second shelf.

What is the value of $x$, the number of books on the second shelf?

答:
Total books $= 162$

books on 1st shelf $= 42$
books on 2nd shelf $= x$

then books on 3rd shelf $= 2x$

Thus $42 + x + 2x= 162$

thus $3x = 120$

implying $x = 40$

Total books on 2nd shelf $= 40$

Title: Books on shelf problem


答:
Easy, so we can know how much the number of books combined from the second and third shelf is $162 - 42 = 120$. The number of books on the third shelf is 2 times the number of books on the second shelf so there are 3 units of the second shelf, which basically means there are 3 times the number of books on the second shelf and the third shelf than is the number of books on the second shelf shelf. So basically $$x = \frac{120}{3} = 40$$

:)

Title: Books on shelf problem

Body:
Rashida keeps her collection of 162 books on 3 shelves.

There are 42 books on the first shelf, $x$ books on the second shelf, and the number of books on the third shelf is 2 times the number of books on the second shelf.

What is the value of $x$, the number of books on the second shelf?


问:
estimate population percentage within an interval, given a small sample
Given a small sample from a normally-distributed population, how  do I calculate the confidence that a specified percentage of the population is within some bounds [A,B]?

To make it concrete, if I get a sample of [50,51,52] from a normally-distributed set, I should be able to calculate a fairly high confidence that 50% of the population lies within the range of 0-100, even with such a small sample.

This is certainly related to the "tolerance interval", but differs in an important way. In all of the examples I can find for tolerance intervals, the required percentile and confidence is given, and the interval is found. In my problem, the interval and percentile are given, and I need to find the confidence.

The relevant equation is this one: (Guttman 1970)

$$1 - \gamma = P\left[P(X \geqq t_0) \geqq 1 - p\right] = P\left[T_{n-1}^*(\sqrt n z_p) \leqq \sqrt n K\right]$$

With definitions:

$1 - \gamma$ is the confidence

$n$ is the number of samples

$100p$ is the percentage of the population required to be within the interval, as estimated from the sample mean and sample variance.

$t_0 = x - K_{S, z_p}$ is the $(1 - p) 100$th percentile of the standard normal distribution

$T_v^*(\delta)$ is the noncentral Student’s t distribution with $v$ degrees of freedom and noncentrality parameter $\delta$.

This solves the one-sided problem, but I'm having trouble extending this to the two-sided problem. In confidence-interval land, I'd use the fact that $P(t_1 \leqq X \leqq t_2) = 1 - P(t_1 \gt X) - P(X \gt t_2)$, to break this into two one-sided problems, but in tolerance-interval land I need to relate these back to the confidence ($1-\gamma$), and I don't see how.

$$1 - \gamma = P\left[P(t_1 \geqq X \geqq t_2) \geqq 1 - p\right] = ??? $$

If I attempt to turn this into two one-sided problems:

$$1 - \gamma = P\left[1 - P(t_1 \lt X) - P(X \lt t_2) \geqq 1 - p\right] = ??? $$

And I'm utterly stuck there. I don't see how to relate this back to the one-sided tolerance interval solution.

I'm not certain this is useful for people to understand the question, but it might, so I'm putting it in this addenda.

In scipy, I'm able to pretty easily calculate $K$ given $p$ $\gamma$ and $n$ as:

```<code>def K(p, gamma, n):
    from scipy import stats
    return stats.nct.ppf(1-gamma, n-1, sqrt(n) * stats.norm.ppf(1-p)) / sqrt(n)
</code>```

I'm also able to find $\gamma$ given $K$ $p$ and $n$ as:

```<code>def gamma(p, n, K):
    from scipy import stats
    z_p = stats.norm.ppf(1-p)
    return 1 - stats.nct.cdf(sqrt(n) * K, n-1, sqrt(n) * z_p)
</code>```

Much less important, but is this a valid simplification of the Guttman's formula?

$$1 - \gamma = P\left[P(X \geqq t_0) \geqq 1 - p\right] = P\left[T_{n-1}^*(\sqrt n z_p) \leqq \sqrt n K\right]$$
$$\gamma = P\left[P(X \geqq t_0) \lt 1 - p\right] = P\left[T_{n-1}^*(\sqrt n z_p) \gt \sqrt n K\right]$$
$$\gamma = P\left[P(X \lt t_0) \lt p\right] = P\left[T_{n-1}^*(\sqrt n z_p) \gt \sqrt n K\right]$$

If so, this form seems way easier to understand, to me.

答:
The following R code simulates the posterior distribution of the Bayesian Gaussian model $y_i \sim_{\text{iid}} {\cal N}(\mu, \sigma^2)$ starting with the Jeffreys prior and returns the $100(1-\alpha)\%$ equi-tailed two-sided confidence interval about the probability $\Pr({\cal N}(\mu, \sigma^2) \in [a,b])$.

```<code># confidence inteval about  Pr(a < N(mu,sigma^2) < b)
bounds <- function(y, a, b, alpha=.05, nsims=10000){
        n <- length(y)
    ssq <- (n-1)*var(y)
    rho <- rchisq(nsims,n)/ssq
    mu <- rnorm(nsims, mean(y), 1/sqrt(n*rho))
    p <- pnorm(b,mu,1/sqrt(rho))-pnorm(a,mu,1/sqrt(rho))
return(quantile(p, c(alpha/2,1-alpha/2)))
}
</code>```

Example:

```<code>> n <- 100
> y <- rnorm(n,0,1)
> bounds(y,a=-1.96,b=1.96)
     2.5%     97.5%
0.9146264 0.9773575
</code>```

There's no theoretical guarantee that the confidence level is well controlled (that is, for example, the effective confidence level of the confidence interval with  $95\%$ nominal confidence level is not exactly $95\%$). You should use simulations to check, but I think the effective confidence level is rather close to the nominal one.

Body:
The following R code simulates the posterior distribution of the Bayesian Gaussian model $y_i \sim_{\text{iid}} {\cal N}(\mu, \sigma^2)$ starting with the Jeffreys prior and returns the $100(1-\alpha)\%$ equi-tailed two-sided confidence interval about the probability $\Pr({\cal N}(\mu, \sigma^2) \in [a,b])$.

# confidence inteval about  Pr(a < N(mu,sigma^2) < b)
bounds <- function(y, a, b, alpha=.05, nsims=10000){
        n <- length(y)
    ssq <- (n-1)*var(y)
    rho <- rchisq(nsims,n)/ssq
    mu <- rnorm(nsims, mean(y), 1/sqrt(n*rho))
    p <- pnorm(b,mu,1/sqrt(rho))-pnorm(a,mu,1/sqrt(rho))
return(quantile(p, c(alpha/2,1-alpha/2)))
}

Example:

> n <- 100
> y <- rnorm(n,0,1)
> bounds(y,a=-1.96,b=1.96)
     2.5%     97.5%
0.9146264 0.9773575

There's no theoretical guarantee that the confidence level is well controlled (that is, for example, the effective confidence level of the confidence interval with  $95\%$ nominal confidence level is not exactly $95\%$). You should use simulations to check, but I think the effective confidence level is rather close to the nominal one.


问:
Show that the equality $|z_2 -z_1|^2=|z_2-z_0|^2+|z_1-z_0|^2$ implies $z_2-z_0=iλ(z_1-z_0)$ where λ is real and vice versa
I tried this

$|z_2 -z_1|^2=|z_2-z_0|^2+|z_1-z_0|^2= with z_0, z_1, z_2 \in \mathbb{C}$

$=-z_2 \overline z_1-z_1\overline z_2 + z_1 \overline z_1 + z_2\overline z_2=-z_2 \overline z_0-z_0\overline z_2 + z_0 \overline z_0 + z_2\overline z_2 -z_1 \overline z_0-z_0\overline z_1 + z_1 \overline z_1 + z_0\overline z_0=$

$=-z_2 \overline z_1-z_1\overline z_2=-z_2 \overline z_0-z_0\overline z_2 + z_0 \overline z_0 +z_0 \overline z_0-z_0\overline z_1 - z_1 \overline z_0=$

$=-z_2 \overline z_1-z_1\overline z_2+ z_2 \overline z_0 + z_0\overline z_2 - z_0 \overline z_0 -z_0 \overline z_0+z_0\overline z_1 + z_1 \overline z_0=0=$

$=\overline z_1(z_0-z_2) + \overline z_0(z_2-z_0)+\overline z_0(z_1-z_0)+ \overline z_2(z_0-z_1)=0=$

$=(\overline z_0-\overline z_1)(z_2-z_0)+(\overline z_0-\overline z_2)(z_1-z_0)=$

$(z_2-z_0)=(z_1-z_0)\frac{\overline z_0- \overline z_2}{\overline z_0- \overline z_2}$

Can you help me from here? I don´t know how to continue

答:
$$|z_2 -z_1|^2=|z_2-z_0|^2+|z_1-z_0|^2~~~~(1)$$
Clearly $P(z_2),Q(z_1),R(z_0)$ are satisfying Pythagoaras, where $PQ=|z_2-z_1|$ is hypotnuse, $PR=|z_1-z_0|$ and $QR=|z_2-z_0|$ are lengths of normal sides, the angle $R$ is 90 deg. This means  PR and QR are perpendicular so their comlex slopes sum to 0.
$$\implies \frac{z_2-z_0}{\bar z_2-\bar z_0}+\frac{z_1-z_0}{\bar z_1- \bar z_0}=0 \implies \frac{z_2-z_0}{z_1-z_0}+\frac{\bar z_2-\bar z_0}{\bar z_1- \bar z_0}=0 \implies \frac{z_2-z_0}{z_1-z_0}=i \lambda, \lambda \in R~~(2).$$
So (2) follows from (1).
Next, we show the reverse by getting (1) from (2):
$$\frac{z_2-z_0}{z_1-z_0}-1=-1+i \lambda \implies |z_2-z_1|^2 =|z_1-z_0|^2+\lambda^2|z_1-z_0|^2=|z_1-z_0|^2+|z_2-z_0|^2.$$

It completes the proof.


问:
Trying to change the form of the multiplicative inverse of $a+b\sqrt2$ into the form $c+d\sqrt2$. [Serge Lang - Basic Mathematics]
I'm struggling with this problem from Serge Lang's Basic Mathematics

Let $a$, $b$ be rational numbers. Prove that the multiplicative inverse of $a+b\sqrt2$ can be expressed in the form $c+d\sqrt2$, where $c$, $d$ are rational numbers.

By multiplying the denominator and numerator of $\frac{1}{a+b\sqrt2}$ by $a-b\sqrt2$ I have been able to get $\frac{a-b\sqrt2}{a^2-2b^2}$, but after that I can't figure out how to express it in the form $c+d\sqrt2$. I tried to cancel out the $a-b\sqrt2$ but that just goes back to $\frac{1}{a+b\sqrt2}$, so I end up going in circles.

答: $c$ and $d$ are any rational numbers. You can just split your answer into two fractions. $c$ and $d$ will be rational because the set of rational numbers are closed under addition, subtraction, multiplication and division by non-zero numbers.


答:
$$\frac{a-b\sqrt2}{a^2-2b^2}=\frac{a}{a^2-2b^2}+\frac{b}{2b^2-a^2}\sqrt2.$$

Can you proceed ?

Observe that $a^2 \ne 2b^2$ (Why ?)

Title: Trying to change the form of the multiplicative inverse of $a+b\sqrt2$ into the form $c+d\sqrt2$. [Serge Lang - Basic Mathematics]

Body:
I'm struggling with this problem from Serge Lang's Basic Mathematics

Let $a$, $b$ be rational numbers. Prove that the multiplicative inverse of $a+b\sqrt2$ can be expressed in the form $c+d\sqrt2$, where $c$, $d$ are rational numbers.

By multiplying the denominator and numerator of $\frac{1}{a+b\sqrt2}$ by $a-b\sqrt2$ I have been able to get $\frac{a-b\sqrt2}{a^2-2b^2}$, but after that I can't figure out how to express it in the form $c+d\sqrt2$. I tried to cancel out the $a-b\sqrt2$ but that just goes back to $\frac{1}{a+b\sqrt2}$, so I end up going in circles.


问:
When does a surjective module map admit a section such that the diagram commutes?
Let $A$ be a polynomial ring over $\mathbb{R}$. Let $M, N$ be free $A$-modules.
Let $L$ be a finitely generated $A$-module.
Suppose we have $A$-module maps $f\colon M\to N$, $g\colon M\to L$ and $h\colon L\twoheadrightarrow N$ such that the following diagram commutes.

Do we have an $A$-linear section $s\colon N\to L$ such that the diagram commutes (i.e., $h\circ s = id$ and $s\circ f = g$)?

Body:
Take $M=A^2,N=A,L=A^2$.

For all $a_1,a_2\in A$, set  $f(a_1,a_2)=a_1$, $h(a_1,a_2)=a_2$ and $g(a_1,a_2)=(a_2,a_1)$.

Then for all $a_1,a_2\ni A$, $h(g(a_1,a_2))=h(a_2,a_1)=a_1=f(a_1,a_2)$, so $h\circ g=f$.

Since $s$ is $A$-linear, $s(a)=a(u_1,u_2)=(au_1,au_2)$ for some fixed $(u_1,u_2)\in A^2$ (in fact $(u_1,u_2)=s(1)$)

Now $h(s(a))=au_2$ for all $a\in A$. Since we want $h(s(a))=a$ for all $a\in A$, we have $u_2=1$. Now $s(f(a_1,a_2))=s(a_1)=(a_1u_1,a_1u_2)=(a_1u_1,a_1)$

Hence we want $(a_1u_1,a_1)=g(a_1,a_2)=(a_2,a_1)$ for all $a_1,a_2\in A$,that is $a_1u_1=a_2$ for all $a_1,a_2\in A$, which is not possible (take $a_1=0$).

Hence $s$ does not exist in this case.


答:
Perhaps the most likely explanation for a coin coming up heads 700 times in a row is that it is a two-headed coin. But let's ignore that. Suppose that the coin is fair and that heads and tails both have probability $\frac{1}{2}$ of occuring.

The Law of Large Numbers does not predict that the coin will "make up" for the 700 heads by being more likely to come up tails later.

The Law of Large Numbers does not say here that the gap
$$
|Number \ of \ Heads - Number \ of \ Tails| \to 0,
$$
as you do more and more coin tosses. This is what would happen if the coin could "remember" the past results and correct for them later. This does not happen. Each coin toss is independent and, no matter what happened in the past, the probability of getting tails in the next coin toss is still $\frac{1}{2}$.

Instead it says that
$$
\frac{Number \ of \ Heads}{Total\ Number \ of \ Coin \ Tosses} \to \frac{1}{2},
$$
as we do more and more coin tosses.

To see the difference, suppose we now flip the coin another $2,000,000$ times. Let's guess that there will be $1,000,000$ heads and $1,000,000$ tails (for a total of $1,000,700$ heads and $1,000,000$ tails.
Then we have
$$
\frac{Number \ of \ Heads}{Total \ Number \ of \ Coin \ Tosses}= \frac{1,000,700}{2,000,700} \approx \frac{1}{2}.
$$
The gap between heads and tails is still $700$; it did not make up for the initial 700 heads with extra tails later. But we see that final outcome is closer to the probability of $\frac{1}{2}$.

Title: Law of Large Numbers, a confusion


答:
Let $CD=x$. Apply Pythagorean Theorem on $\triangle ADC$ and $\triangle ADB$ to get $b^{2}-x^{2}=c^{2}-(a-x)^{2}$. Solve for $x$ here( the $x^{2}$-s will cancel out).

Now, $A'D=A'C-CD=\frac{a}{2}-x$

Title: In $\Delta ABC$, prove $A'D=\frac{(c^2- b^2)}{2a}$ with $A'$ be the midpoint of $BC$, $AB>AC$, and other Information given.

Body:
Problem

In $\Delta ABC$, we have $AB>AC$.

Let $A'$ be the midpoint of $BC$

$D$ be the foot of altitude from  $A$ to $BC$

Internal and external Angle Bisectors of $\angle A$ meet $BC$ at $X$ and $X'$ respectively.

Prove that

$A'X = \frac{a(c-b)}{2(c+b)}$

$A'X' = \frac{a(c+b)}{2(c-b)}$

$A'D=\frac{(c^2- b^2)}{2a}$ where $a, b, c$ are the sides $BC, CA, AB$ as used commonly.

[Question Taken From Challenge and Thrill of Pre-College Mathematics]

My Approach

I could solve part 1. and 2. using the angle bisector theorem a few times, and with some pre-requisite knowledge, which were in-fact to be proven in questions in the exercise preceding this one:

$XC= \frac{ab}{(b+c)}$, consequently $XB= \frac{ac}{(b+c)}$

$CX'= \frac{ac}{(c-b)}$, consequently $BX'= \frac{ab}{(c-b)}$

$XX'= \frac{2abc}{(c^2- b^2)}$

While I could prove the given formulas separately, I have no idea how to solve part 3.. A hint to help me start off would be appreciated.


问:
Let $F$ in $(N)$ be the set of all finite subsets of $N$. Prove that $F$ in $(N)$ is countably infinite.
Let $\mathcal{F}\subseteq\mathcal{P}(\mathbb{N})$ be the set of all finite subsets of $\mathbb{N}$. Prove that $\mathcal{F}$ in $\mathcal{P}(\mathbb{N})$ is countably infinite. Where $\mathbb{N}$ is the set of natural numbers and $\mathcal{P}(\cdot)$ denotes the power set.

答:
Hint: The set $F(\Bbb N_0)$ is the set of all finite words over the alphabet $\Sigma=\{0,1\}$, i.e., $\Sigma^*$.

Take the sequence of primes $(p_0,p_1,\ldots)$.

The mapping $f:\Sigma^* \rightarrow \Bbb N_0:w_0w_1\ldots w_k\mapsto p_0^{w_0}\cdots p_k^{w_k}$ is injective.

For instance: empty word is mapped to $0$, $1010$ is mapped to $p_0p_2$ and so on.


问:
Checking unbiasedness of $\dfrac{1}{2n^2}\sum_{i}\sum_{j}(X_i-X_j)^2$ for $\sigma^2$?
Let $X_1,X_2,...,X_n$, be i.i.d random variables having $N(\mu,\sigma^2)$ distribution, where $\mu ∈\mathbb{R}$ and
$ \sigma > 0$. Define

$$\dfrac{1}{2n^2}\sum_{i}\sum_{j}(X_i-X_j)^2$$

Then $W$, as an estimator of $\sigma^2$, is

$(A)$ biased and consistent

$(B)$ unbiased and consistent

$(C)$ biased and inconsistent

$(D)$ unbiased and inconsistent

I have figured out the solution by multiplying and diving by $\sqrt{2}\sigma$ making it chi-square etc. I was actually trying to figure this problem properly by opening squares. I am getting confused in summations.

$E\left(\dfrac{1}{2n^2}\sum_{i}\sum_{j}(X_i-X_j)^2 \right)=\dfrac{1}{2n^2}\sum_{i}\sum_{j}(E(X_i^2)+E(X_j^2)-2E(X_i)E(X_j))=\dfrac{1}{2n^2}\sum_{i}\sum_{j}(\mu^2+\sigma^2+\mu^2+\sigma^2-2\mu^2)$

When I solve it I get it unbiased but it's not. What am I doing wrong?

答:
It may be due to the fact that when $i=j$, $E(X_iX_i)\neq E(X_i)E(X_i)$, because $E(X_i)E(X_i)=\mu^2, E(X_iX_i)=\sigma^2+\mu^2$

$$E(\frac{1}{2n^2}\sum_i\sum_{j\neq i} (X_i-X_j)^2)=...\\
=\frac{1}{2n^2}\sum_i\sum_{j\neq i}(\mu^2+\sigma^2+\mu^2+\sigma^2-2\mu^2)\\
=\frac{n(n-1)}{n^2}\sigma^2 $$


问:
$\int_{-\frac{w}{2}}^\frac{w}{2} \cos\left(\sqrt{x^2 + (y + h)^2}\right)\,dh$
I'm working on a diffraction simulator. In order to take into account for slit-width, I need to take this integral:
$$\int_{-\frac{w}{2}}^\frac{w}{2} \cos\left(\sqrt{x^2 + (y + h)^2}\right)\,dh$$
Where $x$ and $y$ are the coordinates of the point for which the amplitude is computed and $w$ is the width of the slit. When developing and simplifying, I get:
$$\int_{-\frac{w}{2}}^\frac{w}{2} \cos\left(\sqrt{h^2 + Ah + B}\right)\,dh$$
Where $A = 2y, B = x^2 + y^2$.
I tried u-substitution, with $u^2 = h^2 + Ah + B$, yielding $dh = \frac{2udu}{2h + A}$ and thus being rather useless.

Note: I can't approximate the integral by small steps for performance reasons.

Also, if there is no general solution (which seems pretty likely), is there any computationally efficient way of approximating it?

Thanks in advance

答:
As you, I am more than skeptical about a closed form.

For an approximation, you  could try this $1,400$ years old approximation
$$\cos(t) \simeq\frac{\pi ^2-4t^2}{\pi ^2+t^2}\qquad \text{for} \qquad-\frac \pi 2 \leq t\leq\frac \pi 2$$

This would give
$$\int \cos \left(\sqrt{(h+y)^2+x^2}\right)\sim \frac{5 \pi ^2 }{\sqrt{x^2+\pi ^2}}\tan ^{-1}\left(\frac{h+y}{\sqrt{x^2+\pi
   ^2}}\right)-4 h$$

This would give for your integral
$$\frac{5 \pi ^2}{\sqrt{x^2+\pi ^2}}\tan ^{-1}\left(\frac{4 \omega \sqrt{x^2+\pi ^2} }{4 \left(x^2+y^2+\pi
   ^2\right)-\omega ^2}\right) -4 \omega$$

Edit

Concerning the approximation of $$\cos(x)\qquad \text{for} \qquad-\frac \pi 2 \leq t\leq\frac \pi 2$$ for easy integrations, we could use the $[2n,2]$ Padé approximants $(P_n)$. They write
$$\left(
\begin{array}{ccc}
 n & \text{numerator} & \text{denominator} \\
 1 & 12-5 x^2 & x^2+12 \\
 2 & 3 x^4-56 x^2+120 & 4 x^2+120 \\
 3 & -13 x^6+660 x^4-9720 x^2+20160 & 360 \left(x^2+56\right) \\
 4 & 17 x^8-1680 x^6+65520 x^4-887040 x^2+1814400 & 20160 \left(x^2+90\right)
\end{array}
\right)$$

To check their quality, considering the norm
$$\Phi_n=\int_{-\frac \pi 2}^{+\frac \pi 2} \Big[\cos(x)-P_n\Big]^2\, dx$$
$$\left(
\begin{array}{cc}
 n &  \Phi_n \\
 1 & 1.42 \times 10^{-4} \\
 2 & 9.09 \times 10^{-8} \\
 3 & 2.30 \times 10^{-11}\\
 4 & 5.47 \times 10^{-15}
\end{array}
\right)$$ while

$$\int_{-\frac \pi 2}^{+\frac \pi 2} \Bigg[\cos(x)-\frac{\pi ^2-4x^2}{\pi ^2+x^2}\Bigg]^2\, dx \sim  2.98 \times 10^{-6}$$


问:
Why the addition of the noise will almost certainly make the square matrix invertible?

Why the addition of the noise will almost certainly make the square matrix invertible?

The matrix is invertible if its determinant is not equal to $0$. The determinant of the matrix geometrically can be understood as a $n$-dimensional parallelepiped defined by its row/column
vectors.

Now let our non-invertible matrix $A$ be for simplicity $3 \times 3$ matrix. Obviously, $det(A) = 0$ i.e. the parallelepiped formed by the rows/columns of $A$ is flat. If we add some noise matrix $V$ where $v_{ij}$~$Norm(0,a)$ to each entry of $A$ i.e. $A+V$, we will change the the rows/columns of $A$. How can one show geometrically (or any way) that the new rows/columns would be independent?

答:
Denote by $V$ your random matrix of $v_{i,j}$'s. Note that $\textrm{det}(A+V)$ is a polynomial in the entries $V$.

Appealing to the implicit function theorem, the level sets of $B\mapsto \textrm{det}(A+B)$ are co-dimension $1$ manifolds (in $\mathbb{R}^{n^2}$) and hence, are Lesbegue null-sets. Since $V$ is Gaussian and hence, is absolutely continuous with respect to the Lesbegue measure, for any $x\in \mathbb{R}$, we have that $\mathbb{P}(\textrm{det}(A+V)=x)=0$. This holds, in particular, for $x=0$ and we conclude that $A+V$ is almost surely invertible.


问:
Prove the identity $k \choose 2$ + $k \choose k-2$ + $k^2$ = $2k \choose 2$, where $k\geq2$ using a combinatorial proof.
A combinatorial proof for an identity proceeds as follows: State a counting question. Then, answer the question in two ways: One answer must correspond to the left-hand side (LHS) of the identity. The other answer must correspond to the right-hand side (RHS). Conclude that the LHS is equal to the RHS.

Using this proof method, give a combinatorial proof of the identity:

$k \choose 2$ + $k \choose k-2$ + $k^2$ = $2k \choose 2$, where $k\geq2$

I understand you have to essentially split the $2k \choose 2$. I could treat it as $m +n \choose 2$ where m and n are equivalent and equal to k. And I get where the $k \choose 2$ and  $k \choose k-2$ come from but I have no idea how the $k^2$ comes into play. and why thats involved?

答:
Noting that $\binom{k}{k-2}=\binom{k}{2}$ (choosing $2$ out of $k$ is the same as choosing the $k-2$ in the complement within the $k$), we can see the right as the number of ways to pick $2$ persons out of $2k$ persons, $k$ of which are men and $k$ are women.

The left had side corresponds then to the mixed groups (1 man out of the $k$, 1 woman out of the $k$, so $k^2$ of those) and the all men groups $\binom{k}{2}$ and all women groups (also $\binom{k}{2} = \binom{k}{k-2}$). Together these combine also to all such choices of $2$ out of the $2k$).


问:
Ellipse formula
In the video https://youtu.be/eI4an8aSsgw?t=16354 the professor says that the following equation
$$\sqrt{(x+c)^2 + y^2} + \sqrt{(x-c)^2+y^2}=2a$$
simplifies to
$$
(a^2-c^2)x^2+a^2y^2=a^2(a^2-c^2)
$$
I don't get it. I need someone who can explain it step by step

Title: Ellipse formula

Body:
Let
$$\sqrt{(x+c)^2 + y^2} + \sqrt{(x-c)^2+y^2}=2a=\sqrt{P}+\sqrt{Q}~~~(1)$$
$$\implies P-Q=4cx \implies \sqrt{P}-\sqrt{Q}=2cx/a~~~(2)$$
Adding (1) and (2) get
$$\sqrt{P}=xc/a+a \implies (x+c)^2+y^2=(cx/a+a)^2 \implies (1-c^2/a^2)x^2+y^2=a^2-c^2$$
$$\implies \frac{x^2}{a^2}+\frac{y^2}{(1-c^2/a^2)}=1$$
Let $c=ae$, then we get ellips as
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$$


问:
Is there a way to show $\frac{1646-736\sqrt{5}}{2641-1181\sqrt{5}}=\frac{17+15\sqrt{5}}{7+15\sqrt{5}}$ without multiplying large numbers?
I need to show that the following equality holds $$\frac{1646-736\sqrt{5}} {2641-1181\sqrt{5}} =\frac{17+15\sqrt{5}} {7+15\sqrt{5}} $$ The only way I could prove it was via cross multiplication using a calculator.

Is there any simpler way to do this? Following prime factorization is given $$1646=2\times 823,736=2^5\times 23,2641=19\times 139$$ and $1181$ is prime. I strictly want to avoid multiplying large numbers. I don't know if that's possible here.

The above simplification is used to obtain a famous approximation to $\pi$ given by Ramanujan.

答:
There is another approach which is more in line with the answer linked in the question.

Let $\phi=(1+\sqrt {5})/2$ denote the golden ratio so that $\sqrt{5}=2\phi-1$ and $$\phi^2=\phi+1\tag{1}$$ The fraction in question can now be written as $$\frac{1191-736\phi}{1911-1181\phi}\tag{2} $$ Let us observe that $$ \frac{a-b\phi} {c-d\phi} =\frac{b\phi^2-a\phi}{d\phi^2-c\phi}=\frac{b-(a-b)\phi}{d-(c-d)\phi}\tag{3}$$ and apply this to $(2)$ repeatedly. We get the following sequence of fractions $$\frac{736-455\phi}{1181-730\phi}, \\ \frac{455-281\phi}{730-451\phi},\\ \frac{281-174\phi}{451-279\phi},\\ \frac{174-107\phi}{279-172\phi},\\ \frac{107-67\phi}{172-107\phi},\\ \frac{67-40\phi}{107-65\phi},\\ \frac{40-27\phi}{65-42\phi},\\ \frac{27-13\phi}{42-23\phi},\\ \frac{13-14\phi}{23-19\phi},\\ \frac{14-(-1)\phi}{19-4\phi},\\ \frac{-1-15\phi}{4-15\phi} $$ The last fraction equals $$\frac{15\phi+1}{15\phi-4}=\frac{30\phi+2}{30\phi-8}=\frac{17+15\sqrt{5}}{7+15\sqrt{5}}$$ The application of $(3)$ was stopped at a point when the numbers in the fraction were least in absolute value. Any further application would have led to larger numbers. This approach has advantage that we don't need to know beforehand the final simple answer.

If we have a fraction like $$\frac {a+b\phi} {c+d\phi} $$ then we can use the transformation $$\frac{a+b\phi} {c+d\phi} =\frac{a/\phi+b} {c/\phi+d} =\frac{a(\phi-1)+b}{c(\phi-1)+d}=\frac{b-a+a\phi}{d-c+c\phi}$$ and hope to get an expression with small numbers.


问:
Action of automorphisms on Nonforking Extensions
Let $T$ be a totally transcedental theory and $M$ a model of $T$ with $A \subset M$. Let $p(x) \in S(A)$ (a complete type with parameters in $A$) and let $q(x)$ be a nonforking extension of $p(x)$ with parameters in $M$. I am trying to understand how $q(x)$ behaves under an $A$-automorphism of $M$. </p>

<p>I know that given another non-forking extension $s(x)$ of $p(x)$ (to all of $M$), there exists an $A$-automorphism of $M$ , say $f$ such that $f(q(x)) = s(x)$ (This is result prop 2.33 in http://www.math.uiuc.edu/People/pillay/lecturenotes.stability.pdf). But what I am wondering is given an arbitrary $A$-automorphism $f$, is $f(p(x))$ going to be a nonforking extension of $p(x)$? That is, do the $A$-automorphisms permute the nonforking extensions of $p(x)$? </p>

答:
Yes, the $A$-automorphisms permute the nonforking extensions of $p(x)$.

The reason depends on which definition of nonforking you're using. Since you linked to Pillay's lecture notes, I'll assume that you make the definition $q$ doesn't fork over $A$ if and only if $q$ is definable almost over $A$.

Let $q$ be a nonforking extension of $p$, and let $f$ be an $A$-automorphism. Now the $\phi(x,y)$ part of $q$ is definable by $\psi(y,a)$, with $a\in \text{acl}^{eq}(A)$, so the $\phi(x,y)$ part of $f(q)$ is definable by $\psi(y,f(a))$. Any automorphism fixing $A$ fixes (setwise, not pointwise) $\text{acl}^{eq}(A)$, so $f(a)\in\text{acl}^{eq}(A)$, and $f(q)$ is definable almost over $A$. Finally, $q\restriction A = p$, since the formulas with parameters in $A$ are fixed by $f$. So $q$ is a nonforking extension of $p$.


问:
Is the set $A$ of anti-symmetric relations over $\mathbb{N}$ countable?
I think this set is uncountable, but I don't really know how to prove this. Maybe somehow find an injection from $\mathcal{P}( \mathbb{N} \times  \mathbb{N}) \rightarrow A$?

答: For any $S \subset \mathbb{N}$ take $\mathcal{R}_S = \bigcup_{s \in S} \{(s,s+1)\}$. It is clear that $\mathcal{R}$ is anti-symmetric, as it's a subset of anti-symmetric relation $\mathcal{R}=\{(x,y):x<y\}$, and that this transformation is injective, so there are at least $\# \mathcal{P}(\mathbb{N})$ anti-symmetric relations on $\mathbb{N}$, thus their number is uncountable.


答: For any $X\subseteq \mathbb N$, the relation $R_X = \{(x,x) | x \in X\}$ is anti-symmetric.

Title: Is the set $A$ of anti-symmetric relations over $\mathbb{N}$ countable?

Body: I think this set is uncountable, but I don't really know how to prove this. Maybe somehow find an injection from $\mathcal{P}( \mathbb{N} \times  \mathbb{N}) \rightarrow A$?


问:
Understanding weighted inner product and weighted norms
I am reading  this book where at page 27 following definitions about weighted inner product and weighted  norms are given.

Let $M$ and $N$ be Hermitian positive definite matrices of order $m$ and $n$ respectively. The weighted inner products in $\mathbb{C}^{m}$ and $\mathbb{C}^{n}$ are

$(x, y)_{M} = y^{*}Mx$ , $x, y \in \mathbb{C}^{m}$ and $(x, y)_{N} = y^{*}Nx$ , $x, y \in \mathbb{C}^{n}$ ....$(1)$

The definitions of weighted vector norms are

$\|x\|_{M} = (x, x)^\frac{1}{2}_{M} = (x^{*}Mx)^\frac{1}{2} = \|M^\frac{1}{2} x\|_{2}$, $x\in\mathbb{C}^{m}$ ....$(2)$

$\|x\|_{N} = (x, x)^\frac{1}{2}_{N} = (x^{*}Nx)^\frac{1}{2} = \|N^\frac{1}{2} x\|_{2}$, $x\in\mathbb{C}^{n}$ ....$(3)$

The definitions of weighted matrix norm are

$\|A\|_{MN} = \max_{\|x\|_{N} = 1}{\|Ax\|_{M}},\;  x \in\mathbb{C}^n and ~~A\in \mathbb{C}^{m\times n}$

$\|B\|_{NM} = \max_{\|x\|_{M} = 1}{\|Bx\|_{N}},\;  x \in\mathbb{C}^n and ~~B\in \mathbb{C}^{n\times m}$

Such a norm is sometimes called an operator norm subordinate to vector norm. It is easy to verify that

$\|A\|_{MN} =  \|M^\frac{1}{2} A  N^\frac{-1}{2} \|_{2}$ ....$(4)$

$\|B\|_{NM} =  \|N^\frac{1}{2} B  M^\frac{-1}{2} \|_{2}$ ....$(5)$

Could anybody explain me about the significance of weighted norms? Why we need weighted norm? In $(2)$ how we got $\|M^\frac{1}{2} x\|_{2}$? How could we find square root of matrix $M$?  How did we got equation $(4)$ and $(5)$.

I would be very much thankful for the help and suggestions.

答:
The above answers are perfectly nice. I just want to point out another example: energy norms.

I don't know how familiar you are with differential equations and/or calculus of variations, but I'll give it a try anyway.

Consider the following integral:

$$
E(v) = \frac{1}{2}\int_\Omega |\nabla v|^2 dx
$$
where $\Omega$ is a nice (say with a smooth boundary, no corners nor spikes) bounded domain in $\mathbb{R}^n$. This in many application represents the internal energy of a system in a configuration given by the function $v$. For instance, if $v$ is the displacement from a reference configuration, $E(v)$ represents the elastic energy of the system (assuming linear elasticity).

The above integral can be rewritten as

$$
E(v) = a(v,v)
$$
with

$$
a(u,v) = \frac{1}{2}\int_\Omega \nabla u\cdot\nabla v dx
$$

Now, suppose we have a finite dimensional representation of the function $v$ (if you know finite elements you know where I'm heading). This means

$$
v(x) = \sum_{i=1}^n v_i \varphi_i(x),
$$
where all the $\varphi_i(x)$ are fixed and known a priori.

If you plug this expression inside the definition of $E$ you get (being careful not to mess up the indices)

$$
E(v) = \frac{1}{2}\int_\Omega \nabla\left(\sum_{j=1}^n v_j \varphi_j(x)\right)\cdot\nabla\left(\sum_{i=1}^n v_i \varphi_i(x)\right)dx\\
= \cdots = \sum_{i=1}^n\sum_{j=1}^n v_iv_j\frac{1}{2}\int_\Omega \nabla \varphi_j \cdot \nabla \varphi_i dx
$$

Now let $\underline{v}$ be the vector of the coefficients $v_i$ and $A$ the matrix whose entries are

$$
a_{ij} = a(\varphi_j,\varphi_i) = \frac{1}{2}\int_\Omega \nabla \varphi_j\cdot\nabla\varphi_i dx
$$
Under certain assumptions on $v$ (for instance, $v=0$ on $\partial\Omega$, the boundary of $\Omega$), it can be shown that this is indeed a positive definite matrix.

Now, if the system is in a configuration described by $v$ and $v$ is expressed as above, then the energy of the system is given by

$$
E(v) = a(v,v) = \underline{v}^tA\underline{v}
$$
which is precisely a weighted norm of $\underline{v}$ (squared). Here the matrix is not exactly a weight, bur rather it encodes the physics of the phenomenon. It is possible to show that, if $v$ is expanded as before, and you pick your basis functions $\varphi_i$ in such a way that

$$
\int_\Omega \varphi_i\varphi_j dx =
\begin{cases}
0\mbox{ if }i\neq j\\
1\mbox{ if }i=j,
\end{cases}
$$
then the standard Eucledian norm of $\underline{v}$ is corresponds to the value of the integral

$$
I(v) = \int_\Omega v^2 dx
$$
which is an important norm of $v$, but it measures a different energy of the system.


问:
Evaluate the integral using a specified substitution
I'm asked to evaluate the following integral: $$\int \frac{x^2}{\sqrt{x^2+4}}dx$$ using

a) Trigonometric substitution

b) The substitution $x=t-t^{-1}$.

I did part a) just fine. The Integral is $\frac{1}{2}x\sqrt{x^2+4}-2\ln\left|\frac{\sqrt{x^2+4}+x}{2}\right|+C$ which is correct.

However, I can't figure out how to do part b). If I let $x=t-t^{-1}$, then I can make the following substitutions:

$x=t-t^{-1} \rightarrow x=t-\frac{1}{t} \rightarrow x=\frac{t^2-1}{t}$ or, $x^2=\frac{(t^2-1)^2}{t^2}$.

Differentiating, this gives $dx=(1+t^{-2})dt \rightarrow dx=\left( 1+\frac{1}{t^2} \right) dt \rightarrow dx = \left(\frac{t^2+1}{t^2}\right) dt$

So now, my integral becomes:

$$\int \frac{x^2}{\sqrt{x^2+4}}dx \rightarrow \int \frac{(t^2-1)^2}{t^2\sqrt{\frac{(t^2-1)^2}{t^2}+4}} \frac{t^2+1}{t^2} dt$$

However, I'm not really sure how to proceed from here... Can someone lend a hand and perhaps guide me through this? Thanks!

I should theoretically get the same answer as part a.

答: Note that\begin{align}\sqrt{\left(t-\frac1t\right)^2+4}&=\sqrt{t^2+2+\frac1{t^2}}\\&=t+\frac1t.\end{align}Can you take it from here?


Title: Evaluate the integral using a specified substitution

Body:
I'm asked to evaluate the following integral: $$\int \frac{x^2}{\sqrt{x^2+4}}dx$$ using

a) Trigonometric substitution

b) The substitution $x=t-t^{-1}$.

I did part a) just fine. The Integral is $\frac{1}{2}x\sqrt{x^2+4}-2\ln\left|\frac{\sqrt{x^2+4}+x}{2}\right|+C$ which is correct.

However, I can't figure out how to do part b). If I let $x=t-t^{-1}$, then I can make the following substitutions:

$x=t-t^{-1} \rightarrow x=t-\frac{1}{t} \rightarrow x=\frac{t^2-1}{t}$ or, $x^2=\frac{(t^2-1)^2}{t^2}$.

Differentiating, this gives $dx=(1+t^{-2})dt \rightarrow dx=\left( 1+\frac{1}{t^2} \right) dt \rightarrow dx = \left(\frac{t^2+1}{t^2}\right) dt$

So now, my integral becomes:

$$\int \frac{x^2}{\sqrt{x^2+4}}dx \rightarrow \int \frac{(t^2-1)^2}{t^2\sqrt{\frac{(t^2-1)^2}{t^2}+4}} \frac{t^2+1}{t^2} dt$$

However, I'm not really sure how to proceed from here... Can someone lend a hand and perhaps guide me through this? Thanks!

I should theoretically get the same answer as part a.

Body:
\begin{align}
\int \frac{x^2}{\sqrt{x^2+4}}dx
&= \int \frac{(t^2-1)^2}{t^2\sqrt{\frac{(t^2-1)^2}{t^2}+4}} \frac{t^2+1}{t^2} dt\\
&= \int \frac{(t^2-1)^2 }{t^2\sqrt{\frac{(t^2-1)^2+ 4t^2}{t^2}}} \frac{t^2+1}{t^2} dt\\
&= \int \frac{(t^2-1)^2 }{t^2\frac{1}{t}\sqrt{(t^2-1)^2+ 4t^2}} \frac{t^2+1}{t^2} dt\\
&= \int \frac{(t^2-1)^2 }{t^3\sqrt{(t^2-1)^2+ 4t^2}} (t^2+1) dt\\
&= \int \frac{(t^2-1)^2 }{t^3\sqrt{(t^4-2t^2 + 1)+ 4t^2}} (t^2+1) dt\\
&= \int \frac{(t^2-1)^2 }{t^3\sqrt{(t^4+2t^2 + 1)}} (t^2+1) dt\\
&= \int \frac{(t^2-1)^2 }{t^3\sqrt{(t^2 + 1)^2}} (t^2+1) dt\\
\end{align}
...and you can probably take it from here.


问:
Compact Hermitian Operator, closed range implies finite rank
Let $K$ be a compact Hermitian operator on a Hilbert space $H$. How to show that if $K$ has infinite rank then the range of $K$ is not closed in $H$? </p>

<p>My thought is that suppose range $K$ is closed in $H$ and then we need to prove that $K$ has finite rank. So far I think that spectral theorem can be applied. I want to find contradiction to compactness, but I still have no idea. Thank you for your help. </p>

答: Your idea is the right one, but I would not use the spectral theorem: If the range is closed, the open mapping theorem implies together with compactness that the image of the unit ball is a relatively compact neighborhood of zero, hence the range must be finite-dimensional. You don't need that you deal with a Hermitian operator.</p>


问:
Primality test based on initial conditions alone.
Let $m=252601$. Suppose we discover that
$$3^{126300} ≡ 67772 \pmod{252601}$$
$$3^{252600} ≡ 1\pmod{252601}$$
Is then $252601$ prime? composite? Or can we not decide for sure from the information given?</p>

Body:
Hint: $\,252601\mid 67772^2\!-1 = (67772-1)(67772+1)\ \Rightarrow\ \gcd(252601, 67772\pm1) > 1.$

In fact $\ \gcd(252601, 67772\!+\!1) = 41,\ \ \gcd(252601, 67772\!-\!1) = 6161,\ \ 41\cdot 6161 = 252601.$

Finding nontrivial square roots of $\,1\,$ is the key idea behind many integer factorization algorithms.


问:
When is the map $f$ diagonisable?
We have the linear map $f:\mathbb{R}_2[x]\rightarrow\mathbb{R}_2[x]$ with $$f(p(x))=10p(1)+5p(x)+3\lambda xp''(x)+4p'(x)$$ where $\mathbb{R}_2[x]=\{ax^2+bx+c:a,b,c\in \mathbb{R}\}$ is the vector space of polynomials of degree $\leq 2$ with real coefficients.

How can we choose $\lambda$ so that $f$ is diagonisable?

If we write this in matrix form we have $$ f(p(x))=\begin{pmatrix}10 & 0 & 0 & 0 \\ 0 & 5 & 0 & 0 \\ 0& 0 & 3x\lambda & 0\\ 0 & 0 & 0 & 4\end{pmatrix}\begin{pmatrix}p(1) \\ p(x) \\ p''(x) \\ p'(x)\end{pmatrix}$$ Is this the correct matrix form?

Now we have to check when the matrix is diagonisable, or not?

答:
The standard basis of $\mathbb R_2[x]$ is $\{1,x,x^2\}$ so that

$f(1)=15=15(1)+0(x)+0(x^2)$

$f(x)=14(1)+5(x)+0(x^2)$

$f(x^2)=10+(6\lambda+8) x+5x^2$

The matrix corresponding to $f$ is $T=\begin{pmatrix}15& 14& 10\\0& 5& 6\lambda+8\\ 0& 0& 5\end{pmatrix}$

For the repeated eigenvalue $t=5$, you need $AM=GM$ for which $Rank(T-5I)=1$ is required. This suggests $\lambda=-4/3$.

Title: When is the map $f$ diagonisable?


问:
Solve BVP $y''-3xy'-y=x$, $y(0)=y'(1)=0$ by power series, got differences to numerical solution
I was trying to solve the ODE $y''-3xy'-y=x$ by series. Well, we have regular points all over it so I thought it would be a nice shot, but I am having some problems finding the solutions.

By proposing a solution of the form $y(x)=\sum_{n=0}^{\infty}a_{n}x^{n}$, we will have $y'(x)=\sum_{n=1}^{\infty}a_{n}(n)x^{n-1}$, and $y''(x)=\sum_{n=2}^{\infty}a_{n}(n)(n-1)x^{n-2}$. Substituting in the ODE:
\begin{equation}
\sum_{n=2}^{\infty}a_{n}(n)(n-1)x^{n-2}-
3x\sum_{n=1}^{\infty}a_{n}(n)x^{n-1}-
\sum_{n=0}^{\infty}a_{n}x^{n}
=x
\end{equation}
\begin{equation}
\sum_{n=2}^{\infty}a_{n}(n)(n-1)x^{n-2}-
\sum_{n=1}^{\infty}3a_{n}(n)x^{n}-
\sum_{n=0}^{\infty}a_{n}x^{n}
=x
\end{equation}
\begin{equation}
\sum_{n=2}^{\infty}a_{n}(n)(n-1)x^{n-2}-
\sum_{n=3}^{\infty}3a_{n-2}(n-2)x^{n-2}-
\sum_{n=2}^{\infty}a_{n-2}x^{n-2}
=x
\end{equation}
After, I arrange the equation in a more direct form:
\begin{equation}
(2a_{2}-a_{0})+
[(3!)a_{3}-4a_{1}]x+
\sum_{n=4}^{\infty}[a_{n}(n)(n-1)x^{n-2}-3a_{n-2}(n-2)x^{n-2}-a_{n-2}]x^{n-2}
=x
\end{equation}
And I find the resulting system:
\begin{alignat*}{2}
   a_{2} & {}={} &  \frac{a_{0}}{2}       \\
   a_{3} & {}={} &  \frac{(1+4a_{1})}{3!}   \\
   a_{n} & {}={} &  \frac{3n-5}{n(n-1)}a_{n-2}
\end{alignat*}
Separating for odd, and even terms:

For even $k\geq2$
\begin{equation}
      a_{2k}=\frac{(6k-5)}{(2k)(2k-1)}a_{2k-2}=\frac{\Pi_{j=2}^{k}(6j-5)}{(2k)!}a_{0}
\end{equation}
For odd $k\geq2$:
\begin{equation}
      a_{2k+1}=\frac{(6k-2)}{(2k+1)(2k)}a_{2k-1}=\frac{\Pi_{j=2}^{k}(6j-2)}{(2k+1)!}[1+4a_{1}]
\end{equation}
And then comes the trick. I want to solve it for the conditions: $y(0)=0$ and $y'(1)=0$. The former says all my even terms are zero (Yey!), so I can abandon them, but the latter says I need to satisfy this relation $\sum_{k=0}^{\infty}a_{2k+1}(2k+1)=0$. Well, in this case I need to have $-0.25<a_{1}<0$, otherwise I couldn't make the sum zero.
\begin{equation}
     \sum_{k=0}^{\infty}a_{2k+1}(2k+1)=a_{1}+3a_{3}+\sum_{k=2}^{\infty}a_{2k+1}(2k+1)=0
\end{equation}
\begin{equation}
     a_{1}+\frac{(1+4a_{1})}{2!}+(1+4a_{1})\sum_{k=2}^{\infty}\frac{\Pi_{j=2}^{k}(6j-2)}{(2k)!}=0
\end{equation}
Making $S=\sum_{k=2}^{\infty}\frac{\Pi_{j=2}^{k}(6j-2)}{(2k)!}$, I get that $a_{1}=-\frac{\frac{1}{2}+S}{3+4S}$.

Apparently, I don't know if this is right because I am also solving this same problem using finite differences, and the results are really different one from the other as you can see on the image.

答:
Your formulas all check out. Let's try to replicate the numerical results. Simplify by shifting the inhomogeneity to the first coefficient, set $c=4a_1+1$, then $a_1=\frac{c-1}4$ and the series becomes
$$
y(x)=\frac{c-1}4x+c\sum_{k=1}^\infty\frac{\prod_{j=2}^{k}(6j-2)}{(2k+1)!}x^{2k+1}
\\
y'(x)=\frac{c-1}4+c\sum_{k=1}^\infty\frac{\prod_{j=2}^{k}(6j-2)}{(2k)!}x^{2k}
$$
First evaluate an approximation for $S=\sum_{k=0}^\infty\frac{\prod_{j=2}^{k}(6j-2)}{(2k)!}$ so that for $y'(1)=0$ one needs
$$
0=\frac{c-1}4+cS\iff c=\frac1{1+4S}
$$

```<code>k=1
term_k = 1/2
S_k = term_k
while S_k+term_k!=S_k :
    k+=1
    term_k *= (3*k-1)/(k*(2*k-1))
    S_k += term_k
c = 1/(1+4*S_k)
n=k
x = np.linspace(0,1.2,150)
term_k=x**3/6
y = term_k
for k in range(2,n):
    term_k *= x**2*(3*k-1)/(k*(2*k+1))
    y+=term_k
y = (c-1)/4*x+c*y
plt.plot(x,y, lw=3, label="series")
</code>```

For the numerical solution, use the second order approximations of the derivatives and use for expediency the general solver fsolve. Encode the equations and boundary conditions, the right via a ghost cell, and then plot everything

```<code>from scipy.optimize import fsolve

for N in [5,10]:
    x = np.arange(N+2)/N
    dx = 1/N
    def F(y):
        y = np.concatenate([[0],y]) # force BC at x=0
        diffeq = (y[2:]-2*y[1:-1]+y[:-2])-3*x[1:-1]*dx/2*(y[2:]-y[:-2]) - dx**2*(y[1:-1] + x[1:-1])
        bc1 = y[-1]-y[-3]
        return [bc1, *diffeq]

    y = fsolve(F,0*x[1:])
    plt.plot(x[1:],y,'-x', lw=0.5,label=f"N={N}")
plt.legend(); plt.grid(); plt.show()
</code>```

The resulting plot shows that the numerical solution is as close as can be expected to the series solution

As your series solution plot is the same as above, you must have some error in establishing the system for the numerical solution.

Body:
Your formulas all check out. Let's try to replicate the numerical results. Simplify by shifting the inhomogeneity to the first coefficient, set $c=4a_1+1$, then $a_1=\frac{c-1}4$ and the series becomes
$$
y(x)=\frac{c-1}4x+c\sum_{k=1}^\infty\frac{\prod_{j=2}^{k}(6j-2)}{(2k+1)!}x^{2k+1}
\\
y'(x)=\frac{c-1}4+c\sum_{k=1}^\infty\frac{\prod_{j=2}^{k}(6j-2)}{(2k)!}x^{2k}
$$
First evaluate an approximation for $S=\sum_{k=0}^\infty\frac{\prod_{j=2}^{k}(6j-2)}{(2k)!}$ so that for $y'(1)=0$ one needs
$$
0=\frac{c-1}4+cS\iff c=\frac1{1+4S}
$$

k=1
term_k = 1/2
S_k = term_k
while S_k+term_k!=S_k :
    k+=1
    term_k *= (3*k-1)/(k*(2*k-1))
    S_k += term_k
c = 1/(1+4*S_k)
n=k
x = np.linspace(0,1.2,150)
term_k=x**3/6
y = term_k
for k in range(2,n):
    term_k *= x**2*(3*k-1)/(k*(2*k+1))
    y+=term_k
y = (c-1)/4*x+c*y
plt.plot(x,y, lw=3, label="series")

For the numerical solution, use the second order approximations of the derivatives and use for expediency the general solver fsolve. Encode the equations and boundary conditions, the right via a ghost cell, and then plot everything

from scipy.optimize import fsolve

for N in [5,10]:
    x = np.arange(N+2)/N
    dx = 1/N
    def F(y):
        y = np.concatenate([[0],y]) # force BC at x=0
        diffeq = (y[2:]-2*y[1:-1]+y[:-2])-3*x[1:-1]*dx/2*(y[2:]-y[:-2]) - dx**2*(y[1:-1] + x[1:-1])
        bc1 = y[-1]-y[-3]
        return [bc1, *diffeq]

    y = fsolve(F,0*x[1:])
    plt.plot(x[1:],y,'-x', lw=0.5,label=f"N={N}")
plt.legend(); plt.grid(); plt.show()

The resulting plot shows that the numerical solution is as close as can be expected to the series solution

As your series solution plot is the same as above, you must have some error in establishing the system for the numerical solution.


问:
Finding the explicit formula for the given recurrence relation
This is the recurrence relation:$ a_n=4-a_{n-1}, a_2=-3$.I noticed that the result for even index of $a$ is always -3 and the result for odd index of $a$ is 7,But I don't know where to go with it.

答:
$$f(n):=\begin{cases}\text{even}(n)&\to-3,\\\text{odd}(n)&\to\ \ \ 7\end{cases}$$ is a valid answer.

If you really want a "formula",

$$2-5\cos(n\pi)$$
or
$$(10n)\bmod20-3.$$


问:
A set $E$ is closed if and only if $E = E^-$
Let $E$ be a subset of a metric space $(S,d)$. I want to show that the set $E$ is closed if and only if $E = E^-$ where $E^-$ is the closure of a set $E$.

First I assumed $E = E^-$. Then since I know $E^-$ is the intersection of all closed sets containing $E$, $E^-$ must be closed because the intersection of any closed sets is a closed set. So it follows that $E$ is closed.

Now if $E$ is closed, then $S \setminus E$ is open. That also means $E^- \setminus E$ is open since $E \subseteq E^-$ by definition. Suppose $E \subsetneq E^-$. Let $x$ be on the boundary of $E^-$. Then there does not exist an $r>0$ such that $\{s \in S \ | \ d(s, x) < r \} \subseteq E^- \setminus E$. That means $E^- \setminus E$ is not open which gives a contradiction. So it must be the case that $E^- = E$.

Could someone give me feedback on my proof? And is there a better/shorter proof?

答:
The second implication could be simpler I believe. It seems that your definition of $E^-$ is
$$
E^-=\bigcap_{C\supseteq E\ ;\ C\text{ closed}}C.
$$

If $E$ is closed, then $E$ is indeed a closed set containing $E$. Then $E$ is a member the above intersection, so $E^-\subseteq E$. Of course, $E\subseteq E^-$, so $E=E^-$.


答: The natural way is induction. You have $a_2=-3$. Now take $k\in\Bbb N$ and assume that$$a_k=\begin{cases}-3&\text{ if $k$ is even}\\7&\text{ if $k$ is odd.}\end{cases}$$Then,\begin{align}a_{k+1}&=4-a_k\\&=\begin{cases}7&\text{ if $k$ is even}\\-3&\text{ if $k$ is odd.}\end{cases}\\&=\begin{cases}-3&\text{ if $k+1$ is even}\\7&\text{ if $k+1$ is odd.}\end{cases}\end{align}

Title: Finding the explicit formula for the given recurrence relation

Body: This is the recurrence relation:$ a_n=4-a_{n-1}, a_2=-3$.I noticed that the result for even index of $a$ is always -3 and the result for odd index of $a$ is 7,But I don't know where to go with it.


答:
Use Iverson bracket.

$$a_n=-3[n \equiv 0\pmod 2]+7[n \equiv 1\pmod 2]$$


问:
$X \sim Beta(m,2)$ and $P \left(X\le\dfrac{1}{2}\right)=\dfrac{1}{2}$ find Variance

Let $X$ follow a beta distribution with parameters $m(<0)$ and $2$. If
$P \left(X\le\dfrac{1}{2}\right)=\dfrac{1}{2},$ then $Var(X)$ ?

$(A)\ \ \dfrac{1}{10}\space\ \ \ (B)\ \ \dfrac{1}{20}\ \ \ (C)\ \
 \dfrac{1}{25} \ \ \ (D)\ \ \dfrac{1}{40}$

$V(X)=\dfrac{mn}{(m+n)^2(m+n+1)}=\dfrac{2m}{(m+2)^2(m+3)} \ \ \ \ \ \ ---->(1)$

$\int_{0}^{\frac{1}{2}}x^{m-1}(1-x)dx=\left(\frac{x^{m}}{m}\right)_{0}^{\frac{1}{2}}-\left(\frac{x^{m+1}}{m+1}\right)_{0}^{\frac{1}{2}}=\left(\frac{1}{2^{m}m}\right)-\left(\frac{1}{2^{m+1}(m+1)}\right)=\frac{B(m,2)}{2}\ \ \ \ \ \ ---->(2)$

Now I plugged in values by hit and trial

For $m=2$

$\dfrac{2m}{(m+2)^2(m+3)}=\dfrac{1}{20}$

also satisfies $(2)$

$\dfrac{1}{2^2\cdot2}-\dfrac{1}{24}=\dfrac{1}{12}=\dfrac{B(2,2)}{2}$

For $m=3$

$\dfrac{2m}{(m+2)^2(m+3)}=\dfrac{1}{25}$

but

$\dfrac{1}{2^3\cdot3}-\dfrac{1}{2^4 \cdot 4}=\dfrac{1}{12}\ne\dfrac{B(3,2)}{2}$

This question came $2$ marks so there must be an easy way out which I am not able to figure out. Please suggest me alternate method.

答:
The simple approach is to say that if a $\mathrm{Beta}(a,b)$ distribution has $a>b$ then $\mathbb P(X \le \frac12) \lt \frac12$ and its median is above $\frac12$, while if $a<b$ then then $\mathbb P(X \le \frac12) \gt \frac12$ and its median is below $\frac12$.

This can be proved by comparing the densities at $x$ and $1-x$ and then integrating over the half intervals.  When $a=b$ you get $\mathbb P(X \le \frac12) = \frac12$ and a median of $\frac12$ by symmetry.

So here you have $m=2$ and $\mathrm{Beta}(2,2)$ is symmetric with a median of $\frac12$ and a variance of $\frac1{20}$.


问:
A path from $S^1$ to $\mathbb{C} \setminus \{0\}$ can be expressed in "polar" form if and only if, the winding number of the normalized path is zero
I'm trying to prove the following result:

Let $f: \mathbb{S}^1 \to \mathbb{C} \setminus \{0\}$ be a continuous map. Then there exists a continuous $g: \mathbb{S}^1 \to \mathbb{R}$ such that $f(z) = |f(z)|e^{ig(z)}$ for every $z \in \mathbb{S}^1$ if, and only if, the path $c: [0, 1] \to \mathbb{C} \setminus\{0\}$, defined by $c(s) = f(e^{2\pi i s})$, satisfies $n(c) = 0$.

Here $n(c)$ is the winding number of the path $[0, 1] \ni s \mapsto \alpha(s) =  \dfrac{f(e^{2\pi i s})}{|f(e^{2\pi i s} )|} \in \mathbb{S}^1$ (i.e, $2 \pi \cdot n(c) = \tilde{\alpha}(1) - \tilde{\alpha}(0)$, where $\tilde{\alpha}: [0, 1] \to \mathbb{R}$ is any lifting of $\alpha$).

Now, $\pi: \mathbb{R} \to \mathbb{S}^1, t \mapsto e^{it}$, is a covering map, and by the path lifting lemma there exists a lift $\tilde{\alpha}: [0, 1] \to \mathbb{R}$ such that $e^{i \tilde{\alpha}(s)} = \alpha(s)$ for every $s \in [0, 1]$, or equivalenty, such that $f(e^{2 \pi i s}) = |f(e^{2 \pi i s})|e^{i \tilde{\alpha}(s)}$ for every $s \in [0, 1]$. Every $z \in \mathbb{S}^1$ can be expressed as $z = e^{2 \pi i s}$ for some $s \in [0, 1]$, but I don't know how to get such a $g$ from $\tilde{\alpha}$ and how to relate this to the condition that the winding number is zero. The other direction (proving that if such a $g$ exists, then the winding number must be zero) is still unclear to me as well. Can anyone help me with this? I'd be grateful. Thanks in advance!

答:
Your result is equivalent to the following special case:

Let $\phi: \mathbb{S}^1 \to \mathbb{S}^1$ be a continuous map. Then there exists a continuous $g: \mathbb{S}^1 \to \mathbb{R}$ such that $\phi(z) = e^{ig(z)}$ for every $z \in \mathbb{S}^1$ if and only if the path $c : [0, 1] \to \mathbb{S}^1$, defined by $c(s) = \phi(e^{2\pi i s})$, satisfies $n(c) = 0$.

The general case follows from the special case by considering $\phi(z) = f(z)/\lvert f(z) \rvert$.

The map $q : [0,1] \to \mathbb{S}^1, q(s) = e^{2 \pi i s}$, is a closed  map, hence a quotient map. We have $c = \phi \circ q$.

Let $n(c) = 0$. Take any lift $\tilde c : [0,1] \to \mathbb R$. Then $\tilde c (1) = \tilde c (0)$. Since $q$ is a quotient map, we get a unique continuous $g : \mathbb{S}^1 \to \mathbb R$ such that $g \circ q = \tilde c $. Therefore
$$\pi \circ g \circ q = \pi \circ \tilde c = c = \phi \circ q .$$
Since $q$ is surjective, we conclude $\pi \circ g = \phi$ which means $e^{ig(z)} = \phi(z)$ for all $z \in \mathbb{S}^1 $.

Let $g : \mathbb{S}^1 \to \mathbb R$ be a map such that $\pi \circ g = \phi$. Then
$$c = \phi \circ q = \pi \circ g \circ q .$$
This means that $\tilde c = g \circ q$ is a lift of $c$. But clearly  $\tilde c (1) = \tilde c (0)$. Hence $n(c) = 0$.


问:
**Tensor field and divergence $\nabla.$

Let $A$ be continuously differentiable second-order tensor field defined as follows:
A=\begin{bmatrix}A_{11}&A_{12}&A_{13}\\A_{21}&A_{22}&A_{23}\\A_{31}&A_{32}&A_{33}\end{bmatrix}
the divergence in cartesian coordinate system is a first-order tensor field and can be defined as:\begin{bmatrix}{\dfrac {\partial A_{11}}{\partial x_{1}}}+{\dfrac {\partial A_{12}}{\partial x_{2}}}+{\dfrac {\partial A_{13}}{\partial x_{3}}}\\{\dfrac {\partial A_{21}}{\partial x_{1}}}+{\dfrac {\partial A_{22}}{\partial x_{2}}}+{\dfrac {\partial A_{23}}{\partial x_{3}}}\\{\dfrac {\partial A_{31}}{\partial x_{1}}}+{\dfrac {\partial A_{32}}{\partial x_{2}}}+{\dfrac {\partial A_{33}}{\partial x_{3}}}\end{bmatrix}

A)What is meant by 1st and 2nd order tensor?

B)And a second question is that I am little confused in general about $divU$, where $U$ is a vector or matrix, what shall be the nabla operator? i.e is it a column vector or a row vector? And how $divU$ is expressed in the two cases:
1)U is a matrix
2)U is a vector.

Thank you in advance!

答:
A) A first order tensor can be represented by a vector. A second order tensor can be represented as a matrix.

B) The divergence of a matrix $A$ is a vector, given by the formule you wrote which can be written

$$div(A)=\begin{bmatrix}{ \nabla \cdot A_{1} \\ \nabla \cdot A_{2} \\ \nabla \cdot A_{3} } \end{bmatrix} = \begin{bmatrix}{\dfrac {\partial A_{11}}{\partial x_{1}}}+{\dfrac {\partial A_{12}}{\partial x_{2}}}+{\dfrac {\partial A_{13}}{\partial x_{3}}}\\{\dfrac {\partial A_{21}}{\partial x_{1}}}+{\dfrac {\partial A_{22}}{\partial x_{2}}}+{\dfrac {\partial A_{23}}{\partial x_{3}}}\\{\dfrac {\partial A_{31}}{\partial x_{1}}}+{\dfrac {\partial A_{32}}{\partial x_{2}}}+{\dfrac {\partial A_{33}}{\partial x_{3}}}\end{bmatrix}$$
where $A_i$ stand for the line $i$ of $A$ and $\nabla =(\partial_1, \partial_2, \partial_3)^t$.

The divergence of a vector $U=( u_1,_2,u_3)^t$ is a real, given by

$$div(U) = \nabla \cdot U= \partial_1 U_1 + \partial_2 U_2 + \partial_3 U_3$$

Title: **Tensor field and divergence $\nabla.$


问:
Expansion of $\cos( a \log_e(n+1))$ where $a$ is a constant and $n\in \mathbb{N}$
Expansion of $$\cos( a  \log_e(n+1))$$ where a is a constant and $n\in \mathbb{N}$

My try-

$$\cos( a  \log_e(n+1))= \cos(a  \log_e n+a\log_e(1+\frac{1}{n}))
=\cos(a\log_e n) \cos(a\log_e(1+\frac{1}{n}))-\sin(a \log_e n)\sin(a\log_e(1+\frac{1}{n}))$$
Now, $$\log_e(1+\frac{1}{n})= \frac{1}{n}-\frac{1}{2n^2}+\mathcal{o}(\frac{1}{n^3})$$
How to proceed?

答:
If the expansion in powers of $n $ is meant you can use complex representation of $\cos (x) $ to obtain:

$$\cos (a\log (1+n))=\frac {(1+n)^{ia}+(1+n)^{-ia}}{2}
=\sum_{k\ge0}\frac12\left[\binom {ia}k+\binom{-ia}k  \right]n^k,
$$
where the binomial $\binom zk$ is understood as $\frac {(z)_k}{k!}$ with $(z)_k $ being the falling factorial $z (z-1)\cdots (z-k+1)$.


问:
$\dim _ x(X) = \dim _ y(Y) + d$ for a morphism of relative dimension $d$
Here is a lemma in tag01FE.

Let $f: X→Y$ be a morphism of locally Noetherian schemes which is flat
and locally of finite type such that all fibers are equidimensional of
dimension d. Then for every point x in X with image y in Y we have
$\dim_x(X)=\dim_y(Y)+d$.

Proof. After shrinking X and Y to open neighborhoods of x and y, we can assume that $\dim_x(X)=\dim X$ and $\dim_y(Y)=\dim Y$, by definition of the dimension of a scheme at a point. It remains to show that $\dim(X)=\dim(Y)+d$.

I think this process should go as follows: Choose an open neighborhood $V$ of $y$ such that $\dim_y(Y)=\dim V$ and choose an open neighborhood $U$ contained in $f^{-1}(V)$ such that $\dim_x(X)=\dim U$. Then we consider the morphism $U\rightarrow V$.

My question is that why the fibers of $U\rightarrow V$ are still equidimensional of dimension d? When we shrink $X$ and $Y$ into $U$ and $V$, I think the underlying topological space of fibers should become smaller.

Thanks.

答: The shrinking that goes on cannot change the dimension of the fiber in this case. Why? The fiber over $y\in Y$ is given by $\operatorname{Spec} \kappa(y)\times_Y X$, which means that the fiber is locally of finite type over a field. Now use that for an irreducible scheme locally of finite type over a field, any (nonempty) open subset has the same dimension as the whole scheme - applying this to each irreducible component given the reduced induced structure, we see that all of the fibers are still equidimensional of dimension $d$.


问:
Find $\int^\infty_0 \frac{\sin x}{x(1+\cos ^2 x)}dx$
find the value:
$$\int^\infty_0 \frac{\sin x}{x(1+\cos ^2 x)}dx$$
I try to integrate by parts:
$$-\int^\infty_0 \frac{1}{x}d\arctan(\cos x)$$
But it's not run,help me,thank you.

答: It can be computed as a Lobachevsky-type integral (see also [1] and [2] here on MSE): $$\int_0^\infty\frac{\sin x\,dx}{x(1+\cos^2 x)}=\int_0^{\pi/2}\frac{dx}{1+\cos^2 x}\underset{\tan x=y}{\phantom{\big[}\quad=\quad\phantom{\big]}}\int_0^\infty\frac{dy}{2+y^2}=\frac{\pi}{2\sqrt2}.$$


答:
Alternatively you can count such antisymmetric relations on a set having $n$ elements, which are

$2^n 3^{\frac{n^2-n}{2}}$

Set $n=\aleph_0$ for set of natural numbers, so that

$2^{\aleph_0} 3^{\frac{\aleph_0^2-\aleph_0}{2}}>2^{\aleph_0}=c(\text{cardinality of real numbers }\mathbb R)$


问:
Expressing Gamma Function as an euler constant
Q. Use the definition $$ \Gamma(x) = \lim_{n\to \infty} \frac{n!n^x}{x(x+1)\cdots(x+n)} $$ in this problem.

a) Show that $ \Gamma (x+1) = x\Gamma(x) $ for $ x>0 $

b) First, express $\mathbb{ln}\Gamma(x)$ as sum  of terms.
Then, differentiate to show that $$\Gamma'(1) = -\gamma$$ where $\gamma$ is the Euler constant.

For part a), I tried putting $(x+1)$ into the expression given and tried to simplify, but I didn't get much far. Same for multiplying x to the definition as well.
For part b), I am completely lost as I do not know how to express $\mathbb{ln}\Gamma(x)$ as a sum of terms.

答:
a) Separate out the required terms for the expression for x
$$
\frac{n!n^{x+1}}{(x+1)(x+1)\dots(x+1+n)}
=
\frac{n!n^x}{x(x+1)\dots(x+n)}\cdot\frac{nx}{x+n+1}
$$
Both factors converge independently for $n\to\infty$.

b) The logarithmic derivative of the term under the limit is
$$
\ln(n)-\frac1x-\frac1{x+1}-\dots-\frac1{x+n}
$$
Compare with the definition of the Euler-Mascheroni constant.

Now argue that you can exchange the limits of the derivative and the gamma function. The logarithm is known as continuous.


问:
Jacobson Radical and Finite Dimensional Algebra
In general, it is usually not the case that for a ring $R$, the Jacobson radical of $R$ has to be equal to the intersection of the maximal ideals of $R$.

However, what I do like to know is, if we are given a finite-dimensional algebra $A$ over the field $F$, is it true that the Jacobson radical of $A$ is precisely the intersection of the maximal ideals of $A$?

答:
You're probably familiar with the basic result that among Artinian rings, the semiprime rings (intersection of all prime ideals $P(R)$ is zero) and semiprimitive rings (intersection of all maximal right ideals $J(R)$ is zero) coincide. These are precisely semisimple rings. So, it makes sense that they should coincide for Artinian rings. By the same basic facts, the prime ideals of Artinian rings coincide with the maximal ideals, so that justifies us using $P(R)$ above as notation for the intersection of maximal ideals.

Here's an easy way to see they coincide for a finite dimensional algebra $R$:

Since $P(R)$ is a nil ideal and $R$ is Artinian, $P(R)$ is nilpotent, and hence is contained inside $J(R)$. Since $R/P(R)$ is semisimple, and $J(R)$ is minimal among ideals which make a semisimple quotient of $R$,  $J(R)\subseteq P(R)$. Thus, the two sets are equal.

Denoting the intersection of maximal ideals as $M(R)$, there's an easy way to see why $J(R)\subseteq M(R)$ for all rings. If $I$ is a maximal ideal of $R$, then $I+J(R)$ is also an ideal, but it cannot be $R$. Thus $I+J(R)=I$ implies $J(R)\subseteq I$. Taking the intersection over all $I$, $J(R)\subseteq M(R)$.


问:
Boundness in $W_0^{1, p}(\Omega)$ of the weak limit
Let $\Omega$ be an open bounded domain in $\mathbb{R}^N$ and $p\geq 2$. Let $(u_n)_n\subset W_0^{1, p}(\Omega)$ such that $u\in W_0^{1, p}(\Omega)$ exists so that
$$
u_n\rightharpoonup u \quad\mbox{ in } W_0^{1, p}(\Omega).
$$

I don't know if it is a stupid question or not, but in these hypotheses, we can say that $u$ is bounded in $W_0^{1, p}(\Omega)$? I mean, it is true that there exists $M>0$ such that $\| u\|_{W^{1, p}}\leq M$?

Since $u\in W_0^{1, p}(\Omega)$, thus $\| u\|_{W_0^{1, p}}<+\infty$, so I maybe my guess holds. Could anyone please confirm (or not) that?

Thank you in advance!

答:
One even have the following: If there exists $M \ge 0$ with $\|u_n\| \le M$, then $\|u\| \le M$.
This follows from the fact that
$$B_M := \{ v \in V | \|v\| \le M\}$$
is convex and closed, thus weakly closed.

(One even has
$$\|u\| \le \liminf_{n\to\infty} \|u_n\|.$$
)


问:
Why is the following operator independent of the choice of basis
Let $G$ be a lie group with lie algebra $\mathfrak{g}$ and let $M$ be a $G$-manifold. We denote the space of smooth differential forms on $M$ by $A(M)$. Let $E^i$ be a basis of $\mathfrak{g}$ and let $E_i$ $\in \mathfrak{g}^*$ be the dual basis. Let $S(\mathfrak{g}^*)$ be the symmetric algebra of $\mathfrak{g}^*$. Define the operator $\overline{d}$ on $S(\mathfrak{g}^*) \otimes A(M)$ by

$$\overline{d}(P \otimes \alpha)= \sum_i E_iP \otimes \iota (E^i_M) \alpha$$

for $P \in S(\mathfrak{g}^*)$ and $\alpha \in A(M)$ (note that $\iota (E^i_M)$ is the contraction by the vector field $E^i_M$ wich is defined by $$(E^i_M.f)(m)= \frac{d}{dt} \bigg|_{t=0} f(\exp(tE^i)m),$$ for $f \in C^\infty(M)$ and $m \in M$).

My question is why  the definition of $\overline{d}$ is independent of the choice of the basis $E^i$ ?

Actually, I get this question by reading section $2$ of the article of equivariant cohomology with generalized coefficients by Kumar and Vergne, and the authors have commented on this by saying the following

This expression is independent of the choice of the basis $E^i$, as the element $\sum_i E_i \otimes E^i \in \mathfrak{g}^* \otimes \mathfrak{g}$ is the canonical element $I \in \operatorname{End}(\mathfrak{g})$, where $I$ is the identity element of $\operatorname{End}(\mathfrak{g})$.

But I didn't understand their comment! Moreover, I tried to prove this independence in the case where $\mathfrak{g}$ is one dimensional without going any where.

Please help with this.

答:
When $V$ is a finite-dimensional vector space over a field $k$, there is a canonical isomorphism of vector spaces $\varphi : V \otimes V^* \to \operatorname{End}_k(V)$, such that $v \otimes f$ gets taken to the linear operator $\varphi(v \otimes f): u \mapsto v f(u)$ and extended linearly. This map is not hard to understand in terms of a basis: if $e_1, \ldots, e_n$ is a basis of $V$ and $e_1^*, \ldots, e_n^*$ the dual basis, and we identify $\operatorname{End}_k(V)$ with $n \times n$ matrices using this basis, then $\varphi(e_i \otimes e_j^*)$ is the matrix with a $1$ in the $(i, j)$ place and zeros elsewhere.

Inside the vector space $\operatorname{End}_k(V)$ we have the identity map $\operatorname{id}_V \colon V \to V$, and we can ask what it corresponds to in $V \otimes V^*$. Many authors call this the canonical element $\varphi^{-1}(\operatorname{id}_V) \in V \otimes V^*$. Once we fix a basis and its dual, it is easy to check that $\varphi^{-1}(\operatorname{id}_V) = \sum_i e_i \otimes e_i^*$. Moreover, since $\varphi^{-1}(\operatorname{id}_V)$ was defined independent of any basis, it must be true for every basis and its dual that $\varphi^{-1}(\operatorname{id}_V) = \sum_i e_i \otimes e_i^*$.