问:
Evaluating $(1^n + 2^n + 3^n + 4^n) \mod 5.$
What will be the reminder of (1n +2n +3n +4n)/5? Simply (1n +2n +3n +4n) mod 5=? Where n<=10100000.

My idea: The reminder may be 0 or 1 or 2 or 3 or 4 when a number is divided by 5.

We can easily calculate it when n is too small but when n  is too big like  n<=10100000 how can we calculate it?

答:
Hint:

If $n$ is odd, $$1^n+4^n$$ is divisible by  $1+4$  using Proof of $a^n+b^n$ divisible by $a+b$ when $n$ is odd

If $4\mid n, r^4\equiv1\pmod5$  for  $r=1,2,3,4$ generally for  $(r,5)=1$

If $n\equiv2\pmod4,$

$$1^n+2^n+3^n+4^n\equiv1^2+2^2+3^2+4^2\pmod5\equiv?\pmod5$$


答:
We can reduce $n$ mod $4$ using Fermat's little theorem,

which indicates that $a^4\equiv1\bmod5$ when $\gcd(a,5)=1$.

ContentLicense: CC BY-SA 4.0

Title: Evaluating $(1^n + 2^n + 3^n + 4^n) \mod 5.$

Body:
What will be the reminder of (1n +2n +3n +4n)/5? Simply (1n +2n +3n +4n) mod 5=? Where n<=10100000.

My idea: The reminder may be 0 or 1 or 2 or 3 or 4 when a number is divided by 5.

We can easily calculate it when n is too small but when n  is too big like  n<=10100000 how can we calculate it?


问:
Inducting on the sum of products of terms of a sequence
Heres a nice problem that I've been working on for a while and it's statement is as follows:

Suppose $n$ is a positive integer. We let $\mathcal{F}$ to be the set of positive integer sequences of the form $a_1 = 1$ and $a_{i+1} \leq a_i + 1$ for $1 \leq i \leq n-1$. Find $\sum_\mathcal{F} (a_1 a_2 \cdots a_n)$.

Here's what I have so far; I've divided it into four paragraphs for it to look nice stylistically, so I had to use 4 "Reveal spoilers", apologies on that part.

 For $n=j$, denote the desired sum as $f(j)$. I claim that the answer is $f(n)=(2n-1)!!$. We will prove this by induction with the base case $n=3$, with the lower cases manually checked; suppose now that this result holds for $n=k$; we now prove that it holds for $n=k+1$.

 There are three cases to examine: the first case holds when the sequence begins with $1,1,\ldots$, the second one holds if the sequence begins with $1,2,\ldots$ with the third digit less than $3$, whereas the third case holds if the sequence begins with $1,2,3,\ldots$.

 1. For this case, it suffices to take all the sequences for the case $n=k$ and insert the digit $1$ between the pre-existing first and second digits. (for instance, when inducting from $n=3$ to $n=4$, we would take the set of sequences $\{111,112,121,122,123\}$ and change them into $\{1111,1112,1121,1122,1123\}$) Its not hard to check that this generates all such sequences that satisfy the first case.

2. Similarly to the first case, we take all sequences for the case $n=k$ and insert the digit $2$ between the pre-existing first and second digits. (for instance, when inducting from $n=3$ to $n=4$, we would take the set of sequences $\{111,112,121,122,123\}$ and change them into $\{1211,1212,1221,1222,1223\}$) Its not hard to check that this generates all such sequences that satisfy the second case.

Anybody have an idea on how to complete the proof (ie. complete case 3 and wrap the entire induction by calculating that from the hypothesis $n=k$, it follows that for $n=k+1$, $f(n)=(2k+1)!!$? What I believe we should do is take all sequences of the form 123... (for $n\ge 3$) and append digits to the end, but I'm not quite sure now how to complete this argument. A solution for case 3 would be appreciated, and have a good day.

答:
The induction that you'd want to do is on the last digit, since that gives you full control over what the digit you can append is, thereby avoiding the issue that you're facing.

Specifically, let $ T(n,k)$ be the sum of the product of sequences of length $n$ that end with the digit $k$. It satisfies the recurrence
$$ T(n+1, k) = k \times \sum_{i=k-1}^n T(n,i). \quad \quad (1) $$

This is equivalent to

$$ \frac{ T(n+1, k) } { k } - \frac{ T (n+1, k-1)} { k-1} = T(n,k-2). \quad \quad (2) $$

Here are some initial values. Each row corresponds to a value of $n$, in increasing $k$.

1

1, 2

3, 6, 6

15, 30, 36, 24

105, 210, 270, 240, 120

(Full disclosure: I used OEIS to find the expression for the individual terms. Otherwise, you can stare at it and guess the form, since each term looks like some multiple of the term above it. E.g. We can see that $T(n,1) = (2n-3)!!$ )

We can verify that $ T(n,k) = \frac{ k (2n-k-1)!}{2^{n-k} (n-k)!}$ satisfies the recurrence and initial starting values.

Note: It's easier to work with recurrence relation (2), since it avoids the summation on terms that don't seem to be easily factorable. IE

$$ \frac{ T(n+1, k) } { k } = \frac{ T (n+1, k-1)} { k-1} + T(n,k-2) = \ldots $$

From there, we can show that $ f(n) = \sum_{k=1}^n T(n,k) = T(n+1,1) = \frac{ (2n)!}{ 2^n n!} = (2n-1)!!$


问:
Taylor expansion of exponential around 0 for $x = O(1)$
Let $x$ be some real value that is $O(1)$, find the taylor expansion of $e^tx$ at $0$.

The text I am following writes:

$$e^tx = 1 + tx + O(t^2x^2e^{O(t)})$$

I'm confused why there is an additional factor of $e^{O(t)}$.

See the proof of lemma 1: https://terrytao.wordpress.com/2010/01/03/254a-notes-1-concentration-of-measure/

答:
You see, by Taylor, there is a number $c$ between $0$ and $tx$ (thus $c=O(t)$ )such that:
$$ e^{tx}=1+tx+\dfrac{1}{2} (tx)^2 e^{c}$$

Hence, the conclusion.


问:
Integrate over the region bounded by two regions. Using Polar coordinates.
Using Polar Coordinates integrate over the region bounded by the two circles:

$$x^2+y^2=4$$

$$x^2+y^2=1$$

Evaluate the integral of $\int\int3x+8y^2 dx$

So what I did was said that as $x^2+y^2=4$ and $x^2+y^2=1$

That $1 \le r \le 2$. And as there is a symmetry in the four quadrants

$0 \le \theta \le \frac{\pi}{2}$

which gave me $\int_0^\frac{\pi}{2}\int_1^2 3r^2\cos(\theta) +8r^3\sin^2(\theta) ~dr d\theta$

The answer it gives in the book is $30\pi$.

I'm getting
$28 +30\pi$

答:
There is less symmetry than you think because of the $x$. Well, there is symmetry there too, but it is cancellation symmetry: the contribution of $x$ to the integral is $0$. You can either note that, or integrate from $0$ to $2\pi$, or integrate from $0$ to $\pi$ and double.

Note that we need to replace $dx\,dy$ by $r\,dr\,d\theta$.

ContentLicense: CC BY-SA 3.0


问:
Units are homogeneous in $\mathbb Z$-graded domains
I am confused by an exercise from Tom Marley which is:

Let $R$ be an arbitrary $\mathbb Z$-graded domain:

$1)$ Prove that all units in $R$ are homogeneous.

$2)$ By using $1$, if $R$ is a field, then $R_0=R$ and $R_n=0$ for all $n\neq 0$.

If we remove the domain condition, I know $1$ is not true, but I cannot find the counter example.

Any comments and guidance would be highly appreciated.

答: Let $x,y\in R$ with $xy=1$ and $x=\sum x_i, y=\sum y_j$. We know that $1\in R_0$. So, $x_iy_j=0$ for all $i,j$ with $i+j\neq 0$. Let $x$ be not homogeneous, and has two non zero components $x_{i_a}, x_{i_b}$, where $i_a\neq i_b$. It is clear that $j+i_a\neq 0$ or, $j+i_b\neq 0$ f0r all $j$. Hence, $y_j=0$ for all $j$ and so $y=0$, which is a contradiction.


问:
Is there a clever solution to Arnold's "merchant problem"?
There is a problem that appears in an interview$^\color{red}{\star}$ with Vladimir Arnol'd.

You take a spoon of wine from a barrel of wine, and you put it into your cup of tea. Then you return a spoon of the (nonuniform!) mixture of tea from your cup to the barrel. Now you have some foreign substance (wine) in the cup and some foreign substance (tea) in the barrel. Which is larger: the quantity of wine in the cup or the quantity of tea in the barrel at the end of your manipulations?

This problem is also quoted here.

Here's my solution:

The key is to consider the proportions of wine and tea in the second spoonful (that is, the spoonful of the nonuniform mixture that is transported from the cup to the barrel). Let $s$ be the volume of a spoonful and $c$ be the volume of a cup. The quantity of wine in this second spoonful is $\frac{s}{s+c}\cdot s$ and the quantity of tea in this spoonful is $\frac{c}{s+c}\cdot s$. Then the quantity of wine left in the cup is $$s-\frac{s^2}{s+c}=\frac{sc}{s+c}$$ and the quantity of tea in the barrel now is also $\frac{cs}{s+c}.$ So the quantities that we are asked to compare are the same.

However, Arnol'd also says

Children five to six years old like them very much and are able to solve them, but they
may be too difficult for university graduates, who are spoiled by formal mathematical training.

Given the simple nature of the solution, I'm going to guess that there is a trick to it. How would a six year old solve this problem? My university education is interfering with my thinking.

$\color{red}{\star}\quad$ S. H. Lui, An interview with Vladimir Arnol′d, Notices of the AMS, April 1997.

答:
First We have a $B_{wine}$ and a $C_{tea}$ and a $S$poon

Now we have $B_{wine}-S_{wine}$ and $C_{tea}+S_{wine}$

Then we have $B_{wine}-S_{wine}+(\frac{k}{100}S_{wine}+\frac{100-k}{100}S_{tea})$ and $ C_{tea}+S_{wine}-(\frac{k}{100}S_{wine}+\frac{100-k}{100}S_{tea})$

Which shows that in cup of tea we have $\frac{100-k}{100}S_{wine}$ and in wine barrel we have $\frac{100-k}{100}S_{tea}$. Of course $S_{tea}=S_{wine}$. (Both are one spoon)!


问:
When does $\kappa^{<\lambda}=\kappa^\lambda$?
Can cardinals $\kappa$ and $\lambda$ with $$\bigcup_{\delta<\lambda}\kappa^\delta=\kappa^\lambda$$
be characterized? Or at least a sufficient condition for this equality be given.
I thing that regularity of both of them is not sufficient.

答:
One necessary condition is that the values $\kappa^\delta$ for $\delta<\lambda$ must stabilize.  That is, there must exist some $\delta<\lambda$ such that for all $\delta'$ with $\delta\leq\delta'<\lambda$, $\kappa^{\delta'}=\kappa^\delta$.  Indeed, suppose no such $\delta$ exists (note that this in particular implies $\lambda$ is a limit cardinal).  Then we can pick a cofinal sequence $(\delta_\alpha)$ of the cardinals below $\lambda$ such that $\kappa^{\delta_{\alpha}}<\kappa^{\delta_{\alpha+1}}$ for each $\alpha$.  We then have $$\kappa^{<\lambda}=\sum_\alpha \kappa^{\delta_{\alpha}}$$ and $$\kappa^\lambda=\kappa^{\sum_{\alpha}\delta_{\alpha+1}}=\prod_\alpha \kappa^{\delta_{\alpha+1}}$$ and so $\kappa^{<\lambda}<\kappa^\lambda$ by König's theorem.

Now assuming that $\lambda$ is a limit cardinal and the values of $\kappa^{\delta}$ for $\delta<\lambda$ do stabilize (say, with stable value $\theta$), we have $$\kappa^{<\lambda}=\theta\cdot\operatorname{cf}(\lambda)$$ and $$\kappa^{\lambda}=\theta^{\operatorname{cf}(\lambda)}.$$  Note though that clearly $\kappa^{<\lambda}\geq\lambda\geq\operatorname{cf}(\lambda)$ as long as $\kappa>1$ (and the cases $\kappa\leq 1$ are trivial), so $\kappa^{<\lambda}=\theta$.  If $\lambda$ is singular, then $\operatorname{cf}(\lambda)<\lambda$ so $\theta=\kappa^\delta$ for some $\delta\geq \operatorname{cf}(\lambda)$ and so $\kappa^\lambda=\theta$.  So, when $\lambda$ is singular, $\kappa^{<\lambda}=\kappa^{\lambda}$ iff the values of $\kappa^\delta$ for $\delta<\lambda$ stabilize.


问:
Find limits for functions of natural numbers
So I am dealing with some problems about random graphs where we find limits of functions of natural numbers. A simple example can be the limit of $\ln(n)$ as $n \rightarrow \infty$, where $n$ is the number of vertices of a graph. My concern is how do we intepret this type of limit? Can we view it as a special type of limits in the calculus setting, where we have the variable $x \in \mathbb{R}$? In particular, can we still apply tools from calculus (say the l'Hospital's rule) to find limits of $f(n)$, where $n \in \mathbb{N}$?</p>

答:
Well, if we know that $f(x)$ converges as $x \to \infty$, then it doesn't matter how we get there (we can take $ x_n = n$). But if we aren't sure there can be problems. For example $\lim_{n \to \infty} \sin{n\pi x} = 0$, so if we know convergence "in $\mathbb{N}$" it doesn't necessarily buy us anything in the real sense. You aren't sampling all the possible sequences, in a sense. </p>

<p>So yes, if we know a limit exists, we can apply calculus tools. But if we aren't sure, we might get gobbledygook.  </p>

Body:
Well, if we know that $f(x)$ converges as $x \to \infty$, then it doesn't matter how we get there (we can take $ x_n = n$). But if we aren't sure there can be problems. For example $\lim_{n \to \infty} \sin{n\pi x} = 0$, so if we know convergence "in $\mathbb{N}$" it doesn't necessarily buy us anything in the real sense. You aren't sampling all the possible sequences, in a sense.

So yes, if we know a limit exists, we can apply calculus tools. But if we aren't sure, we might get gobbledygook.


问:
What am I missing in this proof with uniform boundedness principle
I have the following exercise:  Suppose $y=\left(y_{j}\right)_{j=1}^{\infty}$ is a sequence of complex numbers. Prove that if the series $\sum_{j=1}^{\infty} y_{j} \bar{x}_{j}$ is convergent for any sequence $x=\left(x_{j}\right)_{j=1}^{\infty} \in c_{0}$ then $y \in \ell_{1}.$

My attempt: Consider the following:
\begin{align*}
        T_n(x) : & c_0  \longrightarrow \mathbb{R}  \\
         & x\longmapsto \sum_{j=1}^{n} \bar{x}_j y_j
         \end{align*}
Note that, due to the hypotesis, $\sup_{n\in \mathbb{N}}\left|\left| T_n(x) \right|\right|= \left|\left| \sum_{j=1}^{\infty} \bar{x}_j y_j \right|\right| < \infty$. Now we can apply the uniform boundedness principle, that tells us that $\sup \nolimits _{{n\in \mathbb{N},\|x\|=1}}\|T(x)\|= \sup \nolimits _{n \in \mathbb{N}}\|T_n\|<\infty$.

In this part is where I am confused, because $\sup \nolimits _{{n\in \mathbb{N},\|x\|=1}}\|T(x)\| =\left|\left| \sum_{j=1}^{\infty} \bar{x}_j y_j\right|\right|\leq \sum_{j=1}^{\infty} |\bar{x}_j y_j|\leq \sum_{j=1}^{\infty} |y_j|$. This last sum is what we want to be finite in order to see that $y\in \ell_1$, but how can I finish this argument?

答:
Besides observing that $T_n$ is continuous for obvious reasons,  there is no need to compute its norm.   Once the
hypothesis guarantees  that
$$
  T(x):= \lim_{n\mapsto \infty } T_n(x)
  $$
exists for every $x$, a well known Corollary of the Uniform Boundedness Principle, known as the Banach-Steinhauss
Theorem (https://en.wikipedia.org/wiki/Uniform_boundedness_principle#Corollaries), implies that $T$ is continuous.

Since we know that $c_0' =\ell^1$, it follows that there exists some $z = (z_j)_{j=1}^\infty $ in $\ell^1$, such that
$$
  T(x) = \sum_{j=1}^\infty x_jz_j,
  $$
for all $x$.
By computing both sides on  the standard basis vectors of $c_0$,  one easily deduces that $y_j=z_j$,  from where it follows
that $y$ lies in $\ell^1$.


答: At the end the tea cup is as full as at the start. This implies that the added  wine is exactly outweighed by the tea that has disappeared.</p>

Title: Is there a clever solution to Arnold's "merchant problem"?

Body:
There is a problem that appears in an interview$^\color{red}{\star}$ with Vladimir Arnol'd.

You take a spoon of wine from a barrel of wine, and you put it into your cup of tea. Then you return a spoon of the (nonuniform!) mixture of tea from your cup to the barrel. Now you have some foreign substance (wine) in the cup and some foreign substance (tea) in the barrel. Which is larger: the quantity of wine in the cup or the quantity of tea in the barrel at the end of your manipulations?

This problem is also quoted here.

Here's my solution:

The key is to consider the proportions of wine and tea in the second spoonful (that is, the spoonful of the nonuniform mixture that is transported from the cup to the barrel). Let $s$ be the volume of a spoonful and $c$ be the volume of a cup. The quantity of wine in this second spoonful is $\frac{s}{s+c}\cdot s$ and the quantity of tea in this spoonful is $\frac{c}{s+c}\cdot s$. Then the quantity of wine left in the cup is $$s-\frac{s^2}{s+c}=\frac{sc}{s+c}$$ and the quantity of tea in the barrel now is also $\frac{cs}{s+c}.$ So the quantities that we are asked to compare are the same.

However, Arnol'd also says

Children five to six years old like them very much and are able to solve them, but they
may be too difficult for university graduates, who are spoiled by formal mathematical training.

Given the simple nature of the solution, I'm going to guess that there is a trick to it. How would a six year old solve this problem? My university education is interfering with my thinking.

$\color{red}{\star}\quad$ S. H. Lui, An interview with Vladimir Arnol′d, Notices of the AMS, April 1997.


问:
Does $c^n(n!+c^n)\lt (n+c^2)^n$ hold for all positive integers $n$ and $c\gt 0$?
I am  not sure whether the following inequality is true? Some small $n$ indicates it is true. </p>

<p>Let $n$ be a positive integer and $c\gt0$, then
$$c^n(n!+c^n)\lt(n+c^2)^n.$$ </p>

答:
I believe this is false.

Take $n=25, c = 5$.

I came up with this example by setting $c = \sqrt{n}$ and using Stirling's approximation formula for $n!$.

Wolfram alpha link showing the computation.

In fact for $c = \sqrt{n}$, we have that

$$\frac{c^n(n!+c^n)}{(n+c^2)^n}  \sim \sqrt{2 \pi n}\  \left(\frac{n}{\sqrt{2e}}\right)^{n/2}$$

which goes to $\infty$ as $n \to \infty$.


问:
Cauchy formula problem: undefined function
I have this problem where it is supposed to use the Cauchy integral formula: $$\oint_{C} \frac{f(z)}{z(2z+1)^2}dz.$$ The poles clearly are at $z=0 ,z=\frac{-1}{2}. $ The contour is the unitary circle centered at the origin. I tryed to factorize de denominators, but for that i need $f(z)$ to find the coeficientes. How can one manage to evaluate this problem? Can i evaluate the formula over one pole first then the other or should i work with the contour?

答:
Hint: Separate the expression using partial fractions. That is, use the fact that
$$
\frac 1{z(2z+1)^2} = \frac 1z - \frac 2{2z + 1} - \frac 2{(2z + 1)^2}.
$$
It should now be a bit clearer how Cauchy's integral formula applies to this integral.


问:
Help with arguing against symmetry proofs.
I am asking for the help of mathematicians here to show my colleague that her arguments are wrong. I tried my best to explain that inequalities and equations can be solved like that, but maybe my arguments were not good enough.

Some of the examples of her solutions:

Example 1. Over the real numbers, if $x+y=2$, what is the maximum value of $xy$?

Solution: By symmetry, there must be a maximum or minimum when $x=y$ so $x=y=1$ and $xy=1$. This is clearly a maximum, since if say $x=0$ and $y=2$, you get a lower value.

Example 2. Let $x, y$ and $z$ be positive real numbers such that $x + y + z = 1$. Prove that $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2} \geq 27$.

Solution: By symmetry, there is a minimum at x=y=z hance all = 1/3 and the expression equals 27.

Example 3. For positive numbers $a,b,c$ prove that $\sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{\frac{c}{c+a}} \leq \frac{3\sqrt{2}}{2}$?

Solution: Another of those symmetry arguments.  The LHS is a maximum for $a=b=c$ and then equals the RHS.

It is very clear, that these are not even close to be called solutions. When I (or my other collegaues) try to explain that it not that simple to show that extreme points will be at $x=y=z$ we get answers like

It's a symmetry thing.  If you imagine that say $a$ tends to 0 or infinity you see that the LHS is going to decrease.

have repeatedly said that symmetry arguments do not always work.  However, it must be obvious to any bright 18 year old that the argument does work in this particular case.

Much to my amazement, many people in this group do not understand how to solve problems using symmetry. In England, this is taught in schools.

I am not sure what will help us to explain that it does not work like that. If we will get many reponces here, I believe it will help to convince my colleagues. Also, if you are from UK or have degree from prestigous college it will be helpfull to mention that.

答:
Example 1 does not seem so bad at first sight. It is true that by symmetry (and smoothness of the functions) $x=y$ is a stationary point, hence it probably forms a local maximum or minimum.

Anyway, exhibiting a lower point is not enough to prove that we have a global maximum.

If we absorb the constraint with $x=1-t,y=1+t$, we only have to maximize $1-t^2$ and it turns out that the unique stationary point is a maximum, which favors your colleague.

Now, on the opposite, the function $t^2-t^4$, corresponding to the maximization of

$$(4xy+1)(x^2+y^2)-x^4-y^4-6x^2y^2-2xy$$ under $x+y=2$, induces an asymmetric solution.


问:
How many 5-digit numbers such that the sum of their digits is a multiple of 4?
My textbook has the same problem, except for 5-digit numbers the sum of whose digits is a multiple of 5. The general approach is:

Select the first 4 digits. Number of possible ways of doing this = $9\times 10 \times 10 \times 10$ since the ten thousands-place digit cannot be zero. The sum of these 4 digits is $5k$, $5k+1$, $5k+2$, $5k+3$, or $5k+4$. If it's $5k$, the last digit may only be $0$ or $5$. Sum is $5k+1 \implies$ last digit is $4$ or $9$, and so on. In each case, there are only $2$ possible last digits.

Therefore the total number of such 5-digit numbers is $9\times 10 \times 10 \times 10 \times 2 = 18000$.

When we apply this approach to the "sum of digits is a multiple of 4" problem, we get,

Sum of first 4 digits is $4k \implies$ last digit is $0$, $4$, or $8$.

Sum of first 4 digits is $4k+1 \implies$ last digit is $3$ or $7$.

Sum of first 4 digits is $4k+2 \implies$ last digit is $2$ or $6$.

Sum of first 4 digits is $4k+3 \implies$ last digit is $1$, $5$ or $9$.

In two of these cases, we get three possible last digits and in the other two, we get two possible last digits. Our general approach clearly breaks down here. One way to go around this would be to ask ourselves, "how many ways can we select four digits such their sum is $4k$ or $4k+3$", and multiply that number by 3; repeat for $4k+2$ and $4k+1$ but multiply by 2. Add both these numbers to get the answer.

This seems too long. Is there a simpler way to do this?

答:
Let $S$ be the set of numbers where at least one digit is in the range $\{1,\dots,8\}$. I claim that exactly one fourth of the numbers in $S$ have a digit sum which is a multiple of $4$. To do this, we will partition $S$ into groups of four, where the numbers in each group have different remainders modulo $4$.

Let $S_1$ be the set of numbers in $S$ whose first digit is in $\{1,\dots,8\}$, and let $x$ be some number in $S_1$. If the first digit of  $x$ is in $\{1,2,3,4\}$, the group containing $x$ is made by changing that digit to the other numbers in $\{1,2,3,4\}$. For example, the group containing $34682$ would be $\{14682, 24682, 34682, 44682\}$. You do the corresponding thing if the first digit is in $\{5,6,7,8\}$.

But what do we do if the first digit is $0$ or $9$? You then consider $S_2$, the set of numbers whose first digit is $0$ or $9$, and whose second digit is in $\{1,\dots,8\}$, and do the same procedure in the first bullet to the second digit.

Then, you let $S_3$ be the set of numbers each of whose first two digits are $0$ or $9$, and whose third digit is in $\{1,\dots,8\}$, and do the same thing. Same with $S_4$.

Since $S$ is the non-overlapping union of $S_1,S_2,S_3$ and $S_4$, we have partitioned $S$ into groups of four as desired. Therefore, we need to count the number of numbers in $S$ and divide by $4$ to find the number of desired numbers in $S$. To count $S$, use complementary counting.

What about the numbers outside $S$? There are only $16$ of these, and you can count the number of these whose digit sum is a multiple of four directly.


问:
Elementary embeddings
Im not quite sure I understand elementary embeddings right. Say M elementary embeds into N and $M \models\varphi$, does it follow that $N \models \varphi$ or is there a problem with existential formulas?

答:
Short answer: yes that follows.

This is really direct from the definition of elementary embedding.

Definition. A function $f: M \to N$ is an elementary embedding if for every formula $\phi(x_1, \ldots, x_n)$ and all $a_1, \ldots, a_n \in M$ we have
$$ M \models \phi(a_1, \ldots, a_n) \quad \Longleftrightarrow \quad N \models \phi(f(a_1), \ldots, f(a_n)).$$

So if $\phi$ is a sentence, i.e. contains no free variables, then this simplifies to
$$
M \models \phi \quad \Longleftrightarrow \quad N \models \phi.
$$
In particular this means that $M$ and $N$ are elementarily equivalent, which means that they satisfy exactly the same sentences.

You might be confused with embeddings. Let me recall that definition (well, one of the equivalent definitions).

Definition. A function $f: M \to N$ is an embedding if for every quantifier-free formula $\phi(x_1, \ldots, x_n)$ and all $a_1, \ldots, a_n \in M$ we have
$$ M \models \phi(a_1, \ldots, a_n) \quad \Longleftrightarrow \quad N \models \phi(f(a_1), \ldots, f(a_n)).$$

Note that in this definition we only care about quantifier-free formulas. The following fact is where existential formulas and universal formulas come into play.

Fact. Let $f: M \to N$ be an embedding and let $\phi(x_1, \ldots, x_n)$ be a formula and let $a_1, \ldots, a_n \in M$.

If $\phi$ is an existential formula then $M \models \phi(a_1, \ldots, a_n) \implies N \models \phi(f(a_1), \ldots, f(a_n))$.

If $\phi$ is a universal formula then $M \models \phi(a_1, \ldots, a_n) \Longleftarrow N \models \phi(f(a_1), \ldots, f(a_n))$.

Proof. Exercise.

Title: Elementary embeddings


问:
Triangle Area Problem with intersecting lines
I recently saw this geometry problem and it's been killing me. Given $\overline{AB}=\overline{AC}=20$, $\overline{AD}=\overline{AE}=12$, and $[ADFE]=24$ with $[\cdot]$ denoting area, find the area of $\triangle BCF$ (see diagram)

I started by considering line $\overline{DE}$. It's pretty clear to see that $\triangle DEF \sim \triangle CBF$. We also have $\triangle {ADE}\sim\triangle{ABC}$, with $9[ABC]=25[ADE]$. But once i've figured out these parts, I get stuck. What would be the next steps in this situation? I assume I'd gave to take advantage of the area ratios I've gotten, but I can't workout the details of that argument.

答:
$\triangle ADF = \frac{1}{2} \times 24 = 12$

$\triangle ADF$ shares the same altitude with $\triangle CDF$ (from point $F$ to $AC$). So their area is in the ratio of their bases $12:8$.

$\triangle CDF = 8 \implies \triangle ACE = (8 + 24) = \frac{1}{2} \times AC \times EH$

As $EH = \frac{12}{20} BG$

$\frac{1}{2} \times AC \times BG = \frac{160}{3} = \triangle ABC$

So $\triangle BCF = \frac{160}{3} - 24 - 8 -8 = \frac{40}{3}$


问:
Easy question about $H_0^1$ space
I have some trouble with proper understanding of $H_0^1(0,1)$ space. Consider the following space $$H_D = \{u\in H^1(0,1): u(0) = u(1) = 0\}.$$ What can we say about the connection between $H_D$ and $H^1_0(0,1)$. Is $H_D$ in $H^1_0(0,1)$? In literature stays, that functions in $H_0^1(0,l)$ are interpreted as $$''u = 0\; \text{on}\; \partial\Omega''.$$ </p>

答:
Do you know anything about the trace operator? There is a theorem that says that if $U$ is a bounded domain and $\partial U$ is $C^1$ and $u \in W^{1,p}$, then
$$
u \in W_0^{1,p}(U) \iff Tu=0 \text{ on } \partial U,
$$
where $T$ is the trace operator. You want $p = 2$ here.</p>

Title: Easy question about $H_0^1$ space


问:
Unpicking proof that det($AB$) = det$(A)$det$(B)$
I'm trying to understand this proof that $\text{det}(AB) =\text{det}(A)\text{det}(B)$, but I'm a little stuck on the third line. Please could someone explain to me what the mapping $\kappa$ does?

Let $\mathfrak{T}_n = {\rm Maps} (\{1, \ldots , n\}, \{1, \ldots , n\})$
\begin{eqnarray*}
{\sf det}(AB) & = & \sum_{\sigma \in \mathfrak{S}_n} {\sf sgn}(\sigma) \prod_{i=1}^n (AB)_{i\sigma(i)} \\ & = & \sum_{\sigma \in \mathfrak{S}_n} {\sf sgn}(\sigma) \prod_{i=1}^n \sum_{j=1}^n a_{ij} b_{j\sigma(i)} \\
& = & \sum_{\sigma\in \mathfrak{S}_n, \kappa \in \mathfrak{T}_n} {\sf sgn}(\sigma) a_{1\kappa(1)}b_{\kappa(1) \sigma(1)} \ldots a_{n \kappa(n)} b_{\kappa (n) \sigma (n)} \\
& = &\sum_{\kappa\in \mathfrak{T}_n} a_{1\kappa(1)} \ldots a_{n \kappa(n)} \sum_{\sigma \in \mathfrak{S}_n} {\sf sgn}(\sigma) b_{\kappa(1) \sigma(1)}\ldots b_{\kappa (n) \sigma (n)} \\
& = & \sum_{\kappa\in \mathfrak{T}_n}  a_{1\kappa(1)} \ldots a_{n \kappa(n)} {\sf det} (B_{\kappa})
\end{eqnarray*} where $B_{\kappa}$ denotes the matrix which is obtained by using the rows of $B$ labelled $\kappa(1), \ldots , \kappa (n)$ for its rows $1$ to $n$.
. Therefore: $$ \sum_{\kappa\in \mathfrak{T}_n}  a_{1\kappa(1)} \ldots a_{n \kappa(n)} {\sf det} (B_{\kappa})
 = \sum_{\kappa\in \mathfrak{S}_n}  {\sf sgn}(\kappa) a_{1\kappa(1)} \ldots a_{n \kappa(n)} {\sf det} (B) = \det (A) \det (B).$$

Thanks

答:
Sure, notice that
$$\sum_{\sigma \in \mathfrak{S}_n} {\sf sgn}(\sigma) \prod_{i=1}^n \sum_{j=1}^n a_{ij} b_{j\sigma(i)}=\sum_{\sigma \in \mathfrak{S}_n} {\sf sgn}(\sigma) \prod_{i=1}^n \left (a_{i1} b_{1\sigma(i)}+a_{i2} b_{2\sigma(i)}+\cdots +a_{in} b_{n\sigma(i)}\right ),$$
so when you unfold the product, for each one of the factors, you are choosing one of the summands to put together. So for the $i-$th factor, let's say that you pick t the $j-$th summand (i.e., $a_{ij} b_{j\sigma(i)}$). This creates a pairing $\{(i,j):1\leq i,j \leq n\},$ where $i$ in the first coordinate just appears once! So the pairing is functional that is the function $\kappa :\{1,\cdots ,n\}\longrightarrow \{1,\cdots ,n\}$ that you are looking at.

For general purposes:
In general if you have a sequence $(x_{i,j})_{i\in X,j\in Y}$ (apriori with $X,Y$ finite) then $$\prod _{i\in X}\sum _{j\in Y}x_{i,j}=\sum _{f\in Y^X}\prod _{i\in X}x_{i,f(i)},$$ where $X^Y$ denotes the functions from $Y$ to $X.$


问:
Using the converse if Cayley Hamilton theorem
My textbook says

: Let $M$ be a $3 \times 3$ Hermitian matrix which satisfies the matrix equation
$$
M^{2}-5 M+6 I=0
$$
Where $I$ refers to the identity matrix. Which of the following are possible eigenvalues of the matrix $M$
(a) (1,2,3)
(b) (2,2,3)
(c) (2,3,5)
(d) (5,5,6)

Then it proceeds as:

According to Cayley-Hamilton theorem, we can write $\lambda^{2}-5 \lambda+6=0 \Rightarrow \lambda=2,3$ Correct option is (b)

It's clear that the author has used the Cayley Hamilton theorem but in reverse but how can we use the converse Cayley Hamilton theorem? I've read that the converse of Cayley Hamilton theorem doesn't hold in general so what's the author doing here?

I'd be glad if someone pointed out my mistake. Much thanks.

答: What the author is doing is using that if a matrix $M$ satisfies a polynomial $p(t)$, the minimal polynomial of $M$ divides $p(t)$. As all the eigenvalues of $M$ appear as roots of the minimal polynomial, you get that the eigenvalues of $M$ are contained in the set $\{2,3\}$.


问:
Simple question about a closed interval in $\Bbb{R^2}$
Let $I=[a,b]\times[c,d]\subset$$\Bbb{R^2}.$ How does one infer, that $$I=(a,c) + [0,b-a]\times[0,d-c]?$$

I find this odd, since these intervals do not seem to have the same area, since we are adding the open interval $(a,c)$ here. With some concrete numbers, I did find a few examples that do work, but I don't see where the generalization comes from.

答:
$(a,c)$ doesn't mean an open interval there; it's referring to a point in $\mathbb R^2$.

And it's using the notation for adding an element to a set, where $x + A = \{x+a | a \in A \}$. This can be thought of, and is sometimes called, "shifting $A$ by $x$".

Basically,

$$(a,c) + ([0,b-a]\times[0,d-c])$$

can be read as "build a rectangle of the correct width, $(b-a)$, and height, $(d-c)$, at the origin, and then shift it so the lower left hand corner moves from $(0,0)$ to $(a,c)$". And this is the same as building it in place, i.e. $[a,b] \times [c,d]$.


问:
Integration via projection
Let $A=\{(x,y,z) \space\space s.t \space\space x^2+y^2<1 \space\space and \space\space 0<z<1\}$, and $f(x,y,z)=z$. I want to determine $\int_A f$ with integration by projection. More precisely, I have the projection of $A$ on to the $xy$-plane, the set $$A(x)=\{(x,y)\in \Bbb{R^2}\vert(x,y,z)\in A\}=B(0,1).$$

The projection of $A$ onto the $z$-axis is $$P_1 = \{z\in \Bbb{R}\vert\exists(x,y) \space\space s.t \space\space (x,y,z)\in A\}.$$

Then I the integral by projection should be $$\int_A f=\int_{P_1}\Biggr(\int_{A(x)}f(x,y)dy\Biggr)dx.$$

I dont know how to develop the integral from here. The projections are used to establish bounds for integration. The theorem I am using requires that $f(x,y,z)\in \cal{R(A)}$, and $f(x,y) \in \cal{R(A(x))}.$ I don't see why we need to establish this kind of function. Do we need a different function with respect to each projection?

答:
I have never seen the phrase "integration by projection" before.  The region of integration is, of course, that part of the cylinder, with axis along the z-axis and radius 1, between z= 0 and z= 1.   In this particular case, the function to be integrated is simply f(x,y,z)= z so we could write this as $\int_{x=-1}^1\int_{y= -\sqrt{1- x^2}}^{\sqrt{1- x^2}} \int_{z= 0}^1 z dzdydx$.  You could also do the integration in x and y in polar coordinates as $\int_{r= 0}^{1}\int_{\theta= 0}^{2\pi}\int_{z= 0}^1 z dzd\theta dr$.  Or you could just write the region in the xy-plane as "R" and call it $\int_R \int_0^1 z dzdA$ where dA is the differential in the xy-plane.

Of course that is the same as $\left(\int_R dA\right)\left(\int_0^1 z dz\right)$.   Of course, $\int_R dA$ is just the area of the cirle with radius 1, $2\pi$ and $\int_0^1 z dz= \left[\frac{1}{2}z^2\right]_0^1= \frac{1}{2}$.

The value of your integral is $(2\pi)\left(\frac{1}{2}\right)= \pi$.

Title: Integration via projection


问:
How to express the sum of a set?
Suppose I have a set of numbers. How can I express in set-theory terms the sum of the elements in that set?</p>

答:
There is no set-theoretic notation for that since a sum of numbers (presumably real or complex numbers, but more generally could be elements in a ring) is not a set-theoretic notion.

If the numbers you have are indexed by some set, that is if you have a function $f:I\to \mathbb R$ (just taking the reals here to simplify things), then (if $I$ is finite), the sum of the multi(!)set $\{f(i)\mid i\in I\}$ is commonly denoted by $\sum_{i\in I}f(i)$. If $I$ is infinite then some limiting notion is required to make sense of the sum.

Title: How to express the sum of a set?


问:
Example of a topological space $X$ such that $C_0 (X)$ is not a $C^*$-sub-algebra of $C^b (X)$
Let $X$ be an arbitrary topological space. If $X$ is locally compact and Hausdorff, then $C_0 (X)$ (space of continuous functions vanishing at infinity) is a $C^*$-sub-algebra of $C^b (X)$ (space of all bounded continuous functions).

Now, my question is how necessary these conditions "locally compact" and "Hausdorff" are for the above statement to be true.

If we take $X= \mathscr{H}$, an infinite dimensional Hilbert space, and consider the closed unit ball $B(0,1)\subset \mathscr{H}$, then the following function $$f(x)= \frac{1}{||x||},\, x\in B(0,1)^c\\
=1,\,x\in B(0,1)$$
is in $C^b (X)$, but not in $C_0 (X)$.

But it merely shows the proper containment of $C_0 (X)$ in $C^b (X)$. I think it is a right direction, but really I am confused here. I actually need an example of a topological space $X$ such that $C_0 (X)$ is not a $C^*$-sub-algebra of $C^b (X)$.

Thank you.

答:
For any topological space $X$,  whatsoever,  one has that $C_0(X)\subseteq C^b(X)$.  This is because if  $f$ is in $C_0(X)$ then
$f$ is:

continuous by definition,

bounded because,  given any $\varepsilon >0$,  say $\varepsilon =1$,  there exists a compact set $K\subseteq X$ such that $|f(x)|<\varepsilon $ for
all $x\in X\setminus K$. In particular $f$ is bounded on $X\setminus K$,  and it is also clearly bounded on $K$ because $K$ is compact.

Even though what follows has no bearing on the question itself, it should be remarked that $C_0(X)$ might be very very small for certain choices of $X$.  This is because if $f$ lies in $C_0(X)$,  then
$$
  \{x∈ X: |f(x)|>ε\}
  $$
is a relatively compact open set, but some topological spaces, such as
infinite dimensonal normed spaces,  have no nonempty sets with these properties.  In this case one would therefore have that $C_0(X) = \{0\}$.


问:
How do I solve these problems using the intersection of the union and other event properties?
A product X might show 2 types of manufacturing defects, A and B, if P(A)=0,1, P(B)=0,2 and P(A intersection B) = 0.05. formalize and compute the probability that a unit has:

at least one of the defects

defect A but not defect B

none of the defects

just one of the defects A and B.

答:
You are told $\Pr(A), \Pr(B)$ and $\Pr(A\cap B)$.  This is enough information to learn everything else requested about how $A$ and $B$ interact.

The following facts will be useful:

Inclusion-Exclusion (for two sets): $\Pr(X\cup Y) = \Pr(X)+\Pr(Y)-\Pr(X\cap Y)$

Total probability: $\Pr(X) = \Pr(X\cap Y)+\Pr(X\cap Y^c)$

De Morgan's Laws: $(X\cap Y)^c = X^c\cup Y^c$  and  $(X\cup Y)^c = X^c\cap Y^c$

and then just basic definitions and the ability to rewrite each of the statements in terms of sets (For example "Exactly one of the defects" corresponds to $(A^c\cap B)\cup (A\cap B^c)$)

 1) At least one defect: $\Pr(A\cup B) = \Pr(A)+\Pr(B)-\Pr(A\cap B)$.  2) Defect A but not defect B: $\Pr(A\cap B^c)=\Pr(A)-\Pr(A\cap B)$.  3) None of the defects: $\Pr(A^c\cap B^c)=1-\Pr(A\cup B) = 1 - \Pr(A)-\Pr(B)+\Pr(A\cap B)$.  4) Exactly one of the defects: $\Pr(A^c\cap B)+\Pr(A\cap B^c) = (\Pr(B)-\Pr(A\cap B))+(\Pr(A)-\Pr(A\cap B))$ or equivalently $\Pr(A\cup B) - \Pr(A\cap B)$.


问:
Upper bound on number of eigenvalues of an endomorphism
I am reading lecture notes on abstract algebra and came across a proof that doesn’t appear to be valid. The main theorem states that SpecT$\leq$DimV. It then says that this can be proven by showing that eigenvectors with distinct eigenvalues are linearly independent. This would be acceptable, however it requires that the number of eigenvalues of a given transformation(in this case an endomorphism) be finite. Is that the case or is there something I’m misunderstanding?

答: Suppose $V$ is a finite-dimensional vector space and let $n=\dim V$. The characteristic polynomial of $T$ is $p_T(X)=\ker(XI_n - T)$. We can show (by induction for example) that $p_T$ is indeed a polynomial of degree $n$. We can also show that $\lambda \in \text{Sp}(T) \iff p_T(\lambda) = 0$. Since the number of roots of a polynomial is inferior or equal to its degree, we have that the number of eigenvalues of $T$ is inferior or equal to $n$.


问:
On the proof of a result of Bayer and Stillman
I'm reading through the paper A criterion for detecting m-regularity of Bayer and Stillmann and came across a proof, where I don't understand an implication.

The following things may need to be mentioned:

$S = k[x_1,\ldots,x_n]$, $I \subset S$ is a homogeneous ideal and $M$ a graded $S$-module

$\mathfrak{m}$ = $(x_1,\ldots,x_n)$ denotes the irrelevant maximal ideal of $S$

$H_{\mathfrak{m}}^{i}(M)_d$ will denote the degree $d$ part of the $i$-th local cohomology group of $M$

$I$ is $m$-regular if $H_{\mathfrak{m}}^{i}(I)_d = 0 $ for all $i$ and $d \geq m-i+1$.

$I$ is $m$-saturated if $I^{\textit{sat}}_d = I_d$ for all $d \geq m$.

$h \in S$ is called generic for $I$ if $h$ is not a zero-divisor on $S/I^{\textit{sat}}$

It concerns the following Lemma:

Lemma 1.8: Let $I \subset S$ be an ideal, and suppose $h\in S_1$ is generic for $I$. TFAE:

(a) $I$ is $m$-regular

(b) $I$ is $m$-saturated, and $(I,h)$ is $m$-regular

The proof of the direction $(a)\implies(b)$ looks like this:

Proof.
Since $I$ is $m$-regular, it follows that $I$ is $m$-saturated by a previous remark (1.3 in paper). (Thus, it remains to be shown that $(I,h)$ is $m$-regular).

Let $Q = (I:h)/I$ to get an exact sequence
\begin{equation}
0 \rightarrow I \rightarrow (I:h) \rightarrow Q \rightarrow 0.
\end{equation}
Since $I$ is $m$-saturated and $h$ generic for $I$, by a previous Lemma (1.6 in the paper) it follows that $I_d = (I:h)_d$ for all $d \geq m$ and therefore $\dim(Q) = 0.$ Thus, $H_{\mathfrak{m}}^{i}(Q) = 0$ for $i \neq 0$, and $H_{\mathfrak{m}}^{0}(Q) = Q$.

Thus, by the long exact sequence for local cohomology we obtain
\begin{equation}
\tag{$\star$} H_{\mathfrak{m}}^{i}(I)_d \cong H_{\mathfrak{m}}^{i}((I:h))_d \text{ for } d \geq m -i+1 \text{ and all } i.
\end{equation}
Considering the exact sequence
\begin{equation}
0 \rightarrow I \cap (h) \rightarrow I \oplus (h) \rightarrow (I,h) \rightarrow 0.
\end{equation}
and the fact that $I\cap(h) = (I:h)h$, we get
\begin{equation}
0 \rightarrow (I:h)(-1) \rightarrow I\oplus(h) \rightarrow (I,h) \rightarrow 0,
\end{equation}
which leads to
\begin{equation}
 \tag{$\star \star$}H_{\mathfrak{m}}^{i}(I\oplus(h))_d \rightarrow H_{\mathfrak{m}}^{i}((I,h))_d\rightarrow H_{\mathfrak{m}}^{i+1}((I:h))_{d-1}.
\end{equation}

So far I think I can follow, but in the paper the proof gets concluded in the following way:

From $(\star\star)$ and the isomorphism $(\star)$ it follows that $(I,h)$ is $m$-regular.

Question:
If I have that $H_{\mathfrak{m}}^{i}((I,h))_d = 0 $ for all $i$ and $d \geq m-i+1$, then the conclusion follows by definition. However, I don't see how $(\star\star)$ and $(\star)$ imply this.

Since $I$ is $m$-regular $H_{\mathfrak{m}}^{i}(I)_d = H_{\mathfrak{m}}^{i}((I:h))_d = 0 $ for all $i$ and $d \geq m-i+1$. But I think that is not enough to conclude the result.

What am I missing resp. why is the above enough to conclude the proof?

I'm rather unfamiliar with the concept of local cohomology, so the implication may be obvious, but I can't see it.

答:
Expanding on Youngsu's comment:

We have the isomorphism $$H_{\mathfrak{m}}^{i}(I)_d \cong H_{\mathfrak{m}}^{i}((I:h))_d \text{ for } d \geq m -i+1 \text{ and all } i\tag{$\star$}$$
and the exact sequence
$$H_{\mathfrak{m}}^{i}(I\oplus(h))_d \rightarrow H_{\mathfrak{m}}^{i}((I,h))_d\rightarrow H_{\mathfrak{m}}^{i+1}((I:h))_{d-1}.$$
We need to show that
$H_{\mathfrak{m}}^{i}((I,h))_d = 0 $ for $d \geq m-i+1$ and all $i$.

Notice that $H_{\mathfrak{m}}^{i}(I\oplus(h))\cong H_{\mathfrak{m}}^{i}(I)\oplus H_{\mathfrak{m}}^{i}((h))$ (see for instance [1]). Next, we have $(h)\cong S(-1)$ as $S$-modules.
So,
\begin{align}
H_{\mathfrak{m}}^{i}(I\oplus(h))_d
&=H_{\mathfrak{m}}^{i}(I)_d\oplus H_{\mathfrak{m}}^{i}(S(-1))_d\\
&=H_{\mathfrak{m}}^{i}(I)_d\oplus H_{\mathfrak{m}}^{i}(S)_{d-1}.\\
\end{align}
Recall that the local cohomologies of the polynomial ring $S=k[x_1,\ldots,x_n]$ is given by
$$H_{\mathfrak{m}}^{i}(S)_d=\begin{cases}\operatorname{Hom}_k(S_{-n-d},k),&i=n\\0,&\text{otherwise}\end{cases}$$
(see [Eisenbud, Cor.10.9]).
Thus $H_{\mathfrak{m}}^{i}(I\oplus(h))_d=0$ for  all $d\geq m-i+1$ and $i\neq n$. Suppose that $i=n$ and $d\geq m-n+1$. Then we have $$H_{\mathfrak{m}}^{n}(I\oplus(h))_d=H_{\mathfrak{m}}^{n}(I)_d\oplus H_{\mathfrak{m}}^{n}(S)_{d-1}=0\oplus \operatorname{Hom}_k(S_{-n-d+1},k).$$
Since $d\geq m-n+1$, we have $0\geq -m\geq -n-d+1$, so $S_{-n-d+1}=0$. Therefore, $H_{\mathfrak{m}}^{n}(I\oplus(h))_d=0$ for all $d\geq m-n+1$. Thus the left-hand term of the exact sequence vanishes for all $d\geq m-n+1$.

Finally, consider the right-hand term. For $d\geq m-i+1$, we have $d-1\geq m-(i+1)+1$, so  $H_{\mathfrak{m}}^{i+1}((I:h))_{d-1}$ is isomorphic to $H_{\mathfrak{m}}^{i+1}(I)_{d-1}$ by $(\star)$. As $I$ is $m$-regular, it follows that $H_{\mathfrak{m}}^{i+1}((I:h))_{d-1}=0$ for all $d\geq m-i+1$. Hence the middle term $H_{\mathfrak{m}}^{i}((I,h))_d$ also vanishes for $d \geq m-i+1$ and all $i$.


问:
Convergence of Neumann-series
I need to proof the following:

If $\\$  $\sum_{k=0}^\infty$ $D^k$ with $D=diag(d_1, \ldots, d_n)$ and $|d_1|, \ldots, |d_n| < 1$ converges, the Neumann series $\sum_{k=0}^\infty$ $B^k$ to any similar matrix B also converges.

I know I can write $B=S^{-1}DS$, but how does this help me?

答:
By definition of Neumann series you stated, $B^k=(S^{-1}DS)^k=S^{-1}DS\ \cdot S^{-1}DS\ ...\ S^{-1}DS=S^{-1}D^kS$, Therefore $\sum_{k=0}^n B^k=\sum_{k=0}^n S^{-1}D^kS=S^{-1}(\sum_{k=0}^nD^k)S$.

Since you know that $\sum_{k=0}^nD^k$ converges already (actually the limit is $(I-D)^{-1}$), we know $\sum_{k=0}^n B^k=S^{-1}(\sum_{k=0}^nD^k)S$ also converges, and the limit is $S^{-1}(I-D)^{-1}S=(S^{-1}(I-D)S)^{-1}=(I-B)^{-1}$.

Let me know if this solves your question.

Title: Convergence of Neumann-series


问:
Does complex analysis generalize to analysis of $2\times2$ invertible matrices?
We can encode the complex numbers as matrices by letting $1$ be the identity matrix and $i = \left( \begin{array}{clcr} 0 & -1\\1 & 0 \end{array} \right)$. From this fact, this means that complex analysis is equivalent to the analysis of matrices of the form $ \left( \begin{array}{clcr} a & -b\\b & a \end{array} \right)$.

If we removed the restriction on the form of the matrices, and instead studied all invertible $2\times2$ matrices, how much of the theorems from complex analysis would generalize to this new system? Is there any place where I could read more about this?

答:
The set of real matrices of the form $\left(\begin{smallmatrix}a & -b \\ b & a\end{smallmatrix}\right)$ together with matrix addition and matrix multiplication form a field, that is, the field of complex numbers.

The set of all real $2\times 2$ matrices with the same addition and multiplication forms a unitary ring that is not commutative, does not admit inverses for all non-zero elements and hence misses a lot of properties that a field has.

Now the analysis for complex functions and such gadgets heavily depends on $\mathbb C$ being a field, so most of it will fail for functions over the ring of all real $2\times 2$ matrices.


答:
A lot of times, these kinds of formal manipulations become clearer if we look at an example. Let's look at $n = 2$, so that $\mathfrak T_n$ only has four elements to be considered. We have
$$
\begin{align}
\sum_{\sigma \in \mathfrak S_2} \mathsf{sgn}(\sigma) &\prod_{i=1}^2\sum_{j=1}^2 a_{ij} b_{j\sigma(i)}  =
\sum_{\sigma \in \mathfrak S_2} \mathsf{sgn}(\sigma) \prod_{i=1}^2
(a_{i1}b_{1\sigma(i)} + a_{i2}b_{2\sigma(i)})
\\ & = \sum_{\sigma \in \mathfrak S_2}\mathsf{sgn}(\sigma)(a_{11} b_{1\sigma(1)} + a_{12}b_{\sigma(2)})(a_{21}b_{2\sigma(1)} + a_{22}b_{2\sigma(2)})
\\ & = \sum_{\sigma \in \mathfrak S_2}\mathsf{sgn}(\sigma)
[a_{\color{red}{1}\color{green}1} b_{\color{green}1 \sigma(\color{red}1)}
a_{\color{red}{2}\color{green}1} b_{\color{green}1 \sigma(\color{red}2)}
+ a_{\color{red}{1}\color{green}1} b_{\color{green}1 \sigma(\color{red}1)}
a_{\color{red}{2}\color{green}2} b_{\color{green}2 \sigma(\color{red}2)}
%
\\ & \qquad + a_{\color{red}{1}\color{green}2} b_{\color{green}2 \sigma(\color{red}1)}
a_{\color{red}{2}\color{green}1} b_{\color{green}1 \sigma(\color{red}2)}
+ a_{\color{red}{1}\color{green}2} b_{\color{green}2 \sigma(\color{red}1)}
a_{\color{red}{2}\color{green}2} b_{\color{green}2 \sigma(\color{red}2)}].
\end{align}
$$
In the last line, we see that for each $\kappa : \{1,2\} \to \{1,2\}$, there is a term of the form $a_{1\kappa(1)}b_{\kappa(1)\sigma(1)}a_{2\kappa(2)}b_{\kappa(2)\sigma(2)}$, with $1,2$ in red and $\kappa(1),\kappa(2)$ in green.

Title: Unpicking proof that det($AB$) = det$(A)$det$(B)$

Body:
I'm trying to understand this proof that $\text{det}(AB) =\text{det}(A)\text{det}(B)$, but I'm a little stuck on the third line. Please could someone explain to me what the mapping $\kappa$ does?

Let $\mathfrak{T}_n = {\rm Maps} (\{1, \ldots , n\}, \{1, \ldots , n\})$
\begin{eqnarray*}
{\sf det}(AB) & = & \sum_{\sigma \in \mathfrak{S}_n} {\sf sgn}(\sigma) \prod_{i=1}^n (AB)_{i\sigma(i)} \\ & = & \sum_{\sigma \in \mathfrak{S}_n} {\sf sgn}(\sigma) \prod_{i=1}^n \sum_{j=1}^n a_{ij} b_{j\sigma(i)} \\
& = & \sum_{\sigma\in \mathfrak{S}_n, \kappa \in \mathfrak{T}_n} {\sf sgn}(\sigma) a_{1\kappa(1)}b_{\kappa(1) \sigma(1)} \ldots a_{n \kappa(n)} b_{\kappa (n) \sigma (n)} \\
& = &\sum_{\kappa\in \mathfrak{T}_n} a_{1\kappa(1)} \ldots a_{n \kappa(n)} \sum_{\sigma \in \mathfrak{S}_n} {\sf sgn}(\sigma) b_{\kappa(1) \sigma(1)}\ldots b_{\kappa (n) \sigma (n)} \\
& = & \sum_{\kappa\in \mathfrak{T}_n}  a_{1\kappa(1)} \ldots a_{n \kappa(n)} {\sf det} (B_{\kappa})
\end{eqnarray*} where $B_{\kappa}$ denotes the matrix which is obtained by using the rows of $B$ labelled $\kappa(1), \ldots , \kappa (n)$ for its rows $1$ to $n$.
. Therefore: $$ \sum_{\kappa\in \mathfrak{T}_n}  a_{1\kappa(1)} \ldots a_{n \kappa(n)} {\sf det} (B_{\kappa})
 = \sum_{\kappa\in \mathfrak{S}_n}  {\sf sgn}(\kappa) a_{1\kappa(1)} \ldots a_{n \kappa(n)} {\sf det} (B) = \det (A) \det (B).$$

Thanks


问:
Show that there is no linear transformation $T: R^2 \rightarrow R^2$ such that $T (1, π) = (0, 1)$, , $T (e, −π) = (1, 1)$ and $T (1, 0) = (2, 3)$.
I can't seem to find a way to prove this, I started last week with Lineal Transformations

Can someone explain me?

答:
The three vectors $(0,1), (e,-\pi)$ and $(1,0)$ are linearly dependent. Knowing the image of any two of them under $T$ determines the third, using the linearity property that

$$T(\alpha v + \beta w) = \alpha T(v) + \beta T(w).$$

Is this enough of a hint?


问:
what's the distribution of CDF(X) when X belong to normal distribution
Say I sample X from a standard normal distribution.
Then send those X to a standard normal CDF and get Y, then I plot the histogram of Y

it looks like they belong to a uniform distribution, but how do I express this in formula?

code:

```<code>import numpy as np
import scipy
import matplotlib.pyplot as plt
import seaborn as sns

x = np.random.normal(0,1,100000)
norm_cdf = scipy.stats.norm.cdf(x)

# plot the cdf
plt.subplot(211)
sns.distplot(x)
plt.subplot(212)
sns.distplot(norm_cdf)
plt.show()
</code>```

答: Pick $p\in [0, 1]$. What is the probability that $CDF(X)\leq p$? It is the same as that of $X\leq CDF^{-1}(p)$, which is $p$ by definition of the CDF. Thus $CDF(X)$ follows the uniform distribution on $[0, 1]$. This holds for any RV with a continuous pdf.


问:
If sequences comprised of every nth element converge, does the original sequence converge?
For instance, if the sequence made up of all the second elements converge to some N, and all the third elements, and all the fourth elements to some N, etc, does the original sequence converge to N?

I considered using the fact that if subsequences converge then the original sequence converges but I don't know if I can apply that here because this is a nonstandard definition of subsequence.

答: Define the sequence $x_n$ that is $1$ for prime indices and $0$ otherwise.  Then $x_{2n}$, $x_{3n}$, $x_{4n}$, etc all converge to $0$. However, the sequence does not converge.


问:
sufficient conditions for a scheme to be integral
Suppose $X$ is a scheme and we have an open affine cover {$\operatorname{Spec} A_i$} of $X$, such that each $A_i$ is integral domain. Also we have that any intersection of $\operatorname{Spec} A_i$ with $\operatorname{Spec} A_j$ is non empty. Can we say that $X$ is integral scheme?

If not, any counter example? What other conditions do I need to make sure $X$ is integral?

答: Yes, that's sufficient. Being reduced is a local property; you can check it on the stalk at each point, or on any affine open covering. (This is a good exercise in sheaf theory; this answer gives a hint on how to prove it.) That your hypotheses are sufficient for irreducibility is proved in Stacks Project, Lemma 28.3.3 (tag 01OM). Since a scheme is integral iff it's reduced and irreducible, this proves what you want.</p>


问:
Number of real roots of $\sum_{k=0}^{n}\frac{x^{k}}{k!}=0$
Prove the following, without induction. Is it possible?

The equation $\sum_{k=0}^{n}\frac{x^{k}}{k!}=0$ has no real root if $n$ is even.

  And if $n$ is odd, it has only one real root.

I also tried searching the proofs many times with search key words "number of real root" or "exponential function" or "$\sum_{k=0}^{n}\frac{x^{k}}{k!}$".
But failed. What is the way to search the related topics?

答:
Consider the function $f(x)=\sum_{k=0}^{2n}\frac{x^{k}}{k!}$

Note that $f'(x)+\frac{x^{2n}}{(2n)!}=f(x) \large \star$

Suppose there exist a minimum value for the function(at $x=c$) then $f'(c)=0$.

From $\star$ as $2n$ is even $f(c)=\frac{c^{2n}}{(2n)!} \ge 0 \quad\forall c \in \mathbb{R}$ (at $c=0$ $f(0)=1$ and the value of function is positive )

So $ \large {\sum_{k=0}^{2n}\frac{x^{k}}{k!}}>0 \quad \forall x\in \mathbb{R}$

Now consider the case of $f(x)=\sum_{k=0}^{2n+1}\frac{x^{k}}{k!}$

Noting that $f'(x)=\sum_{k=0}^{2n}\frac{x^{k}}{k!} >0$ the function is monotonous over real numbers and as the degree is odd the function has exactly one root.


答:
Such transformation is invertible since $T(1,\pi)=(0,1)$ and $T(e,-\pi)=(1,1)$ generates $\mathbb{R}^2$.

We have $T(1,\pi)+2T(e,-\pi)=T(1,0)$ implies that $(1,\pi)+2(e,-\pi)-(1,0)$ is in $KerT=\{0\}$ impossible.

Title: Show that there is no linear transformation $T: R^2 \rightarrow R^2$ such that $T (1, π) = (0, 1)$, , $T (e, −π) = (1, 1)$ and $T (1, 0) = (2, 3)$.

Body:
I can't seem to find a way to prove this, I started last week with Lineal Transformations

Can someone explain me?


问:
$C=\{\sum_{i=1}^m\lambda_iv_i$ | $\lambda_i\geq 0\text{ and }\sum_{i=1}^n\lambda_i=1\}$ a compact subset of $\mathbb{R}^n$.
Let $v_i\in\mathbb{R}^n$. Why is $C=\{\sum_{i=1}^m\lambda_iv_i\mid\lambda_i\geq 0\text{ and }\sum_{i=1}^n\lambda_i=1\}$ a compact subset of $\mathbb{R}^n$.

So this is like the convex hull of $\mathbb{R}^n$, which of course is bounded. However I don't see why it's necessarily closed?

I actually think I've read it somewhere, that this set should be compact, however I don't remember where. Does it have another name?

答:
Let $w=\max_i \| v_i \|$. We have
$$
\forall \lambda_i, \,\left\| \sum_{i=1}^m\lambda_iv_i\right\| \le mw.
$$
So $C$ is bounded.

Let $(\lambda_i^{(n)})_n$ be a convergent sequence of real numbers such that
$$
\forall n,\, \sum_{i=0}^m \lambda_i^{(n)}=1 \text{ and } \forall n, \forall i, \, \lambda_i^{(n)}\ge 0.
$$
and $\lambda_i^{(n)} \to \lambda_i$.
Taking the limits, we have
$$
\sum_{i=0}^m \lambda_i=1 \text{ and } \forall i, \, \lambda_i\ge 0,
$$
so $\sum_{i=0}^m \lambda_i v_i\in C$. This shows that $C$ is closed, so $C$ is closed and bounded, so $C$ is a compact subset of $\Bbb R^n$.


问:
Lineal Transformation $T : V \rightarrow V$ injective implies that $T$ is surjective
Show that if $V$ is a vector space over a finite-dimensional field $K$ and $T: V → V $ is a linear transformation, then if $T$ is injective it implies that $T$ is surjective.

Hi, I have been stuck with this problem can someone help me please?

答:
This might look very complicated but its really not, I have just written everything in quite a lot of detail. So lets look at what the words we are going to be using mean:

$\textbf{Surjective:}$ "Everything in V can be reached from T". i.e for all v in V there is a x in V such that T(x) = v

$\textbf{Injective:} $ "If $v_1 \not = v_2 $ then $T(v_1) \not = T(v_2) $

For this proof we will the $\textbf{Rank-Nullity Theorem}$. This states that: Dimension(V) = Rank(T) + Nullity(T). Where Rank(T) and Nullity(T) are respectively the dimension of the Image and Kernal of T. Let us again define this two things.

$\textbf{Image:}$ This is the set of points that T maps to. i.e Image(T) = $\{ x \in V :  x = T(v)$ for some $v \in V \}$

$\textbf{Kernel:}$ This is the set of points that get mapped to $0_V$ i.e Kernal(T) ] $\{x \in V : T(x) = 0_V \} $

Clearly a map T is surjective if (and only if) the dimension of the image = dimension of the place it maps to. Firstly T maps V to itself so $T(V) \subset V $

Secondly Dim(T(V)) = Dim(V) $\implies $ T(V) = V

So all we have to do is prove that Dim(Image) (= Rank(T)) = Dim(V) and we are done. From the Rank Nullity theorem we could alternatively show: Nullity(T) = $0$ as Dimension(V) = Rank(T) + Nullity(T) from the RNT

We are given that T is injective and want to show that $T(v) = 0 \implies v = 0_V$ i.e the kernel = {0} and thus the Dimension(Kernel) = Nullity(T) = $0 $

Let us prove this via a contradiction. Assume that there is another element to the kernel, call it $k$ and this $k \not = 0_V$ And thus $T(k) = 0$ This violates that T is injective as $k \not = 0_V $ Yet $T(k) = T(0_V) = 0 $

Hence the $\textbf{Kernel} $ of T is {$0$} and thus the $\textbf{Nullity}$ of T, the Dimension of the $\textbf{Kernel}$, is also $0$ and thus from the $\textbf{Rank-Nullity Theorem}$ the Rank of T, the Dimension of the $\textbf{Image} $ = Dimension(V) - Nullity(T) = Dimension(V) and so $T(V) = V$ And thus it is surjective.


问:
A paradox in the formula for propagation of uncertainty?
My question is the following: what is the uncertainty propagated to area of a square?

The usual answer is as follows. Let $(l\pm\sigma_l)$ denote an experimental measure of the sides of the square, with $l$ being the size of the sides of the square and $\sigma_l$ its associated uncertainty. The area of the square is given by the function $A_{\textrm{square}}(l)=l^2$, with the corresponding uncertainty given by the formula for the propagation of uncertainty in indirect measures:

$\sigma_{A_{\textrm{square}}} = \sqrt{\left(\frac{dA}{dl}\right)^2 \sigma_l^2} = 2 l \sigma_l.$

Now let us think on the uncertaint propagated to the area of a rectangle with sides $(l_1\pm\sigma_{l_1})$ and $(l_2\pm\sigma_{l_2})$. The area is given by the function $A_{\textrm{rectangle}}(l_1,l_2)=l_1 l_2$ and the associated uncertainty propagated to the area of the rectangle is:

$\sigma_{A_{\textrm{rectangle}}} = \sqrt{\left(\frac{\partial A}{\partial l_1}\right)^2 \sigma_{l_1}^2 + \left(\frac{\partial A}{\partial l_2}\right)^2 \sigma_{l_2}^2} = \sqrt{l_2^2 \sigma_{l_1}^2 + l_1^2 \sigma_{l_2}^2}.$

These are the usual answers. However, I can think on a square as a limiting case of a rectangle where $l_1 = l_2 \equiv l$. Correspondingly, it follows that in this limiting case, $A_{\textrm{rectangle}}(l,l) = l^2 = A_{\textrm{square}}(l)$. Then, I would expect that the uncertainty of the area of a rectangle would reduce to that of the area of a square, when I take $l_1 = l_2 \equiv l$, but that is not what happens:

$\sigma_{A_{\textrm{rectangle}}}\Bigr\rvert_{l_1=l_2\equiv l} = \sqrt{l^2 \sigma_{l}^2 + l^2 \sigma_{l}^2} = \sqrt{2} l \sigma_l,$

and this is different than the first formula obtained above for the uncertainty propagated to the area of a square! What is the source of this paradox???

================

EDIT: the user Henry has solved the problem! I will only provide below the details on the solution.

The point is that the formula for the propagation of uncertainty used above assumes that the two sides of the rectangle are uncorrelated. This is no longer the case when we take both sides to be equal, in which case they are obviously correlated (that was my mistake!). In this case, the formula for the propagation of uncertainty to the area of the rectangle with both sides equal is:

$\sigma_{A_{\textrm{rectangle}}} = \sqrt{\left(\frac{\partial A}{\partial l_1}\right)^2 \sigma_{l_1}^2 + \left(\frac{\partial A}{\partial l_2}\right)^2 \sigma_{l_2}^2 + 2 \frac{\partial A}{\partial l_1}\frac{\partial A}{\partial l_2} \sigma_{l_1 l_2}} = \sqrt{l_2^2 \sigma_{l_1}^2 + l_1^2 \sigma_{l_2}^2 + 2 l_1 l_2 \sigma_{l_1 l_2}},$

where $\sigma_{l_1 l_2}$ is the sample covariance between $l_1$ and $l_2$,

$\sigma_{l_1 l_2} = \sum_{i=1}^N \frac{(l_{1_i}-\langle l_1\rangle) (l_{2_i}-\langle l_2\rangle)}{N-1}$.

When $l_1 = l_2 \equiv l$, the above formula reduces to the sample variance, $\sigma_{l_1 l_2} \to \sigma_l^2$, therefore,

$\sigma_{A_{\textrm{rectangle}}}\Bigr\rvert_{l_1=l_2\equiv l} = \sqrt{l^2 \sigma_{l}^2 + l^2 \sigma_{l}^2 + 2 l l \sigma_l^2} = 2 l \sigma_l = \sigma_{A_{\textrm{square}}},$

as expected!

答:
It seems that in the case of the rectangle you have assumed that errors on one side are independent of errors on the other: this leads you to add the variances before you take the square root.

In the case of the square you have assumed that errors on one side are equal to errors on the other: it is this dependency (implicitly the associated covariance) that leads to the higher propagation of error into the area.


答:
converting comment to answer

Since $T$ is linear we should have had that $T(u+v)=T(u)+T(v)$ and also that $T(\alpha u)=\alpha T(u)$.

Then what does $T(1+e,0)$ equal taking both of these into account and that $(1,\pi)+(e,-\pi)=(1+e,0)$ as well as that $(1+e,0)=(1+e)(1,0)$

 $$(1,2)=(0,1)+(1,1)=T(1,\pi)+T(e,-\pi) =^\dagger T(1+e,0)=^\star (1+e)T(1,0)\\=(1+e)(2,3)=(2+2e,3+3e)$$ where at the equality marked with a $\dagger$ we used $T(u)+T(v)=T(u+v)$ and at the equality marked with a $\star$ we used $T(\alpha u)=\alpha T(u)$.  But of course $(1,2)\neq (2+2e,3+3e)$, so we have a contradiction.


问:
Definition of a modular form and a survey of the main examples
I am trying to solve the following exercise:

Let  k $\in \mathbb {Z} $ and $f: \mathbb{H}\longrightarrow \mathbb {C} $ be a holomorphic function with
$f \mid_k\gamma = f$
for all
$\gamma \in\Gamma_1$
and
$$f(x+iy) = O(1)    (y \rightarrow+\infty).$$
show: for all $x\in \mathbb {Q} $ the following holds
$$f(x+iy)= O(y^{(-k)})   (y\downarrow 0)$$
hint: show
$f(\frac{a}{c}+iy)=({\frac{i}{c}})^k y^{-k} f(-\frac{d}{c}+\frac{i}{c^2y})$.

can you pls help me, i have approaches but I don't get continue

答:
Let $f$ be a weight $k$ modular form for $SL(2,\mathbb Z)$. Let $\mathbb Q\ni x=\frac ac$ with $a,c\in\mathbb Z$, and
$$\gamma=\begin{pmatrix}d&0\\ -c&a\end{pmatrix}\in SL(2,\mathbb Z).$$
In fact, $\gamma$ is just the inverse of $\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)$ with $b=0$. The relation $f|_k\gamma=f$ for $\tau=x+iy$ then reads
$$f(\tfrac ac+i y)=\frac{1}{(-i c y)^k}f\left(-\tfrac dc+\tfrac{  ad}{c^2 y}i\right).$$
The limit $y\downarrow0$ of the second factor is a constant,
$$\lim_{y\downarrow0}f\left(-\tfrac dc+\tfrac{i  ad}{c^2 y}\right)=\lim_{y\uparrow\infty}f\left(-\tfrac dc+\tfrac{  ad y}{c^2 }i\right)=\mathcal O(1).$$
Therefore, the rhs diverges as
$$\lim_{y\downarrow0}f(\tfrac ac+i y)=\mathcal O(y^{-k}).$$


问:
Study if $f$ is differentiable in its domain
Study if $f$ is differentiable in its domain, where $f(x) = \sqrt{x} -x$

I tried to have the derivative by the definition but I'm going nowhere with this.

$$\lim_{h \rightarrow 0} \frac{f(x+h) -f(x)}{h} = \lim_{h \rightarrow 0} \frac{\sqrt{x+h} -(x+h) - (\sqrt{x}-x)}{h} = \lim_{h \rightarrow 0} \frac{\sqrt{x+h} -h - \sqrt{x}}{h}$$

And now I don't know what to do. I tried to multiply by the conjugate but didn't work.

答:
With your method, you write $\sqrt{x+h}=\sqrt{x}\sqrt{1+\frac{h}{x}}$ and use the expansion of $(1+a)^{1/2}=1+\frac{a}{2}+\text{o}(a)$ which is valid when $a \to 0$.
Another approach can be the following: $\sqrt{x}$ is differentiable everywhere in its domain except for $x=0$ and $-x$ is differentiable everywhere, so the difference $\sqrt{x}-x$ is differentiable in $(0,\infty)$.


答:
We have
$$
\frac{\sqrt{x+h} -h - \sqrt{x}}{h}
=
\frac{\sqrt{x+h} - \sqrt{x}}{h} -1 = \frac{h}{h(\sqrt{x+h} + \sqrt{x})}-1
=
\frac{1}{\sqrt{x+h} + \sqrt{x}}-1.
$$
Now, if $x\neq 0$, you can take the limit and you get $\frac{1}{2\sqrt{x}}-1$. But if $x=0$, we get
$$
\frac{1}{\sqrt{h}}-1 \to_{h \to 0} + \infty
$$
so the limit isn't finite.

Body:
Study if $f$ is differentiable in its domain, where $f(x) = \sqrt{x} -x$

I tried to have the derivative by the definition but I'm going nowhere with this.

$$\lim_{h \rightarrow 0} \frac{f(x+h) -f(x)}{h} = \lim_{h \rightarrow 0} \frac{\sqrt{x+h} -(x+h) - (\sqrt{x}-x)}{h} = \lim_{h \rightarrow 0} \frac{\sqrt{x+h} -h - \sqrt{x}}{h}$$

And now I don't know what to do. I tried to multiply by the conjugate but didn't work.


问:
how to prove transitive relation for this one?
$R=\{(x,y) \mid 3 |(x-y^2) \}$ while R is relation on set A=$\mathbb{Z}$

I need to check if it's reflective , symmetric , transitive .

I figured already that it's not reflective or symmetric with (2,1) or (10,2),(2,10)

and I don't how to check if it's transitive or not.

I think we need to prove that, we have $(x,y), (y,c)$ thus $(x,c) $ then $3|(x-y^2)  = 0$ and

3|$(y-c^2) = 0$ but $3|(x-c^2) \neq  0$

答:
If $3|x-y^2$ and $3|y-z^2$ does it follow that $3|x-z^2$?

$3|x-y^2\implies z\equiv y^2 \pmod 3$ and $3|y-z^2\implies y\equiv z^2 \pmod 3$.

So if $x\equiv y^2$ and $y\equiv z^2$ then $x \equiv z^4\equiv \pmod 3$.

Must it follow that $z^4 \equiv z^2 \pmod 3$?

If $z\equiv 0$ then $z^2 \equiv z^4 \pmod 3$ and, yes,  $x\equiv y^2\pmod 3$ and $y\equiv 0^2\pmod 3\implies x\equiv 0\equiv z^2 \pmod 3$.

If $z \equiv \pm 1$ then $1\equiv z^2\equiv z^4 \pmod 3$ and yes $x\equiv y^2 \pmod 3$ and $y\equiv (\pm 1)^2 \equiv 1\pmod 3 \implies y^2 \equiv 1 \pmod 3$ and so $x\equiv y^2 \equiv 1\equiv z^2 \pmod 3$.

So yes, it is transitive.

......

Alternatively.  If $3|x-y^2$ then we have $x-y^2 = 3k$ for some $k$.  And $3|y-z^2$ then we have $y-z^2 = 3j$ for some $j$.  So $y=3j-z^2$.

So $x-y^2 = x-(9j^2 - 6jz^2 + z^4)= 3k$

So $x-z^2 = 3k + 9j^2 - (6j+1)z^2 + z^4=$

$3(k + 3j^2 -2jz^2) +z^4-z^2$.

so $3|x-z^2$ if and only if $z^4-z^2 = z^2(z^2 - 1)=z^2(z+1)(z-1)$ is divisible by $3$.

But either $z, z+1$ or $z-1$ must be divisible by $3$ as they are three consectutive integers and one of them must be divisible by $3$.

So $3|z^2(z+1)(z-1)$.


问:
Laplace function and how to integrate it in order to find the cumulative distribution function.
I have a probability density function: $f(x) = \frac{\beta}{2} e^{-\beta |x|}$ with $\beta \gt 0$ and I've been asked to find the cumulative distribution function F of X.
So I know in order to get from a pdf to cdf that I need to integrate.

But when I integrate I'm not 100% sure what to do, does it just become:
$F(x) = -\frac{\beta e^{-\beta |x|}}{2\beta}$

I wasn't sure whether I needed to consider it for x being positive and x being negative. But I didn't think it would affect the outcome, so I just integrated normally.

Finally I need to sketch the CDF F of X. & I have no idea how to do this. I think I know what the PDF would look like but I'm unsure on what the CDF would look like.

答:
For the cumulative distribution function, you need to integrate the density from $-\infty$ (or the lowest point in the support) to the point of interest.  So you want $$F(x)=\int\limits_{y=-\infty}^{x} \frac \beta 2 e^{-\beta |y|} \,dy$$

For $x \le 0$ you can say $$F(x)=\int\limits_{y=-\infty}^{x} \frac \beta 2 e^{\beta y}\, dy=\frac12 e^{\beta x} =\frac12 e^{-\beta |x|}$$ but for $x > 0$ you need to deal with the change in sign of $|y|$ to get $$F(x)=\int\limits_{y=-\infty}^{0} \frac \beta 2 e^{\beta y}\, dy + \int\limits_{y=0}^{x} \frac \beta 2 e^{-\beta y}\, dy =\frac12 +\left(-\frac12 e^{-\beta x} +\frac12\right)= 1 - \frac12 e^{-\beta |x|} $$


问:
Probability that subsets intersect
Given a set $N$ I would like to calculate the probability that two arbitrarily chosen and equally likely subsets $K\subseteq N$ and $J\subseteq N$ both of fixed size intersect. Let's say $n=\#N$, $k=\#K$, and $j=\#J$ denote the number of elements of the respective sets. The probability should be called $\rho(p,j,k)$ and so far I think I have two formulas which do the job.

The first is recursively defined as:

$$ \rho(p,j,k)=\begin{cases}\frac{k}{n}+\frac{n-k}{n}\cdot\rho(j-1,k,n) & \text{if $j$>0},\\ 0 & \text{otherwise}.\end{cases} $$

The idea is that taking the first element of $J$ there is a chance of $\frac{k}{n}$ that it will be an element of $K$. If not, i.e. with a probability of $1-\frac{k}{n}=\frac{n-k}{n}$ there might be the probability that the next element will be in $K$ and so on.

I have an explicit formula as well:

$$ \rho(p,j,k)=\frac{k}{n}+\frac{k(k-1)}{n^2}+\cdots+\frac{k(k-1)\cdots(k-j+1)}{n^j}$$

The idea here is that either one element of $J$ is in $K$ (Probability: $\frac{k}{n}$) or two elements of $J$ are in $K$ (Probability: $\frac{k(k-1)}{n^2}$) and so on...

Now I have two questions:

First, I think I need to assume that $j\le k$ and I would like to not put any restrictions on the sizes of the sets.

Second, the problem seems very general so I wonder whether there is some known obvious result to it which I am overlooking.

答:
I suppose that $k=\#K$ and $j=\#J$ are given. The number of all $J$'s is $\binom{n}{j}$, the number of those $J$'s that don't intersect $K$ is $\binom{n-k}{j}$, so the probability that $K$ and $J$ don't intersect is $\binom{n-k}{j}/\binom{n}{j}$.

edit: oops, you actually know the answer - but I'll leave it here anyway, just in case someone finds it useful

Title: Probability that subsets intersect


问:
Show that the ‖‖$u$+‖‖$v$ vector is perpendicular to the ‖‖$u$−‖‖$v$ vector.
Show that the ‖‖$u$+‖‖$v$ vector is perpendicular to the ‖‖$u$−‖‖$v$ vector.

What should I do here? If it's perpendicular, the dot product of the two vectors should be 0. But I could not use this.

答:
The definition of perpendicular of two vectors $v,u$ is defined to mean $\langle v,u\rangle=0$. So we need to show $\langle ‖‖+‖‖,‖‖−‖‖\rangle=0$, then by definition those two are perpendicular.

Actually,
$$
\begin{eqnarray} \big\langle‖‖+‖‖,‖‖−‖‖\big\rangle&=&‖‖^2\langle ,u\rangle+‖‖‖‖ \langle v,\rangle−‖‖‖v‖\langle u,v\rangle-‖‖^2\langle ,v\rangle\\
&=&0\end{eqnarray}$$ since $‖‖^2=\langle u,u\rangle$ and $‖v‖^2=\langle v,v\rangle$.

let me know if this solves your questions.


问:
Find the values of $x$ which will make each of the following expression a perfect square: $x^4+ 6x^3+13x^2+13x-1$.
$$x^4+ 6x^3+13x^2+13x-1=\textrm{perfect square}$$
I don't know how to approach this. I have tried to factor this but was unable to, and equating it to squares of numbers is a tedious process, and I am not sure how to solve this problem.

By the way, this question is from Elementary Algebra by Hall and Knight, Exercise XXX b question 15.

答:
Hint:

Can you show that $(x^2+3x+2)^2<x^4+6x^3+13x^2+13x-1<(x^2+3x+3)^2$ for $x>5$?

(A suggestion in this direction was provided by player3236 in a comment.)


问:
If $A$ is invertible show that $\det(A) \neq 0$
If $A \in M_{n\times n}(F)$ is not invertible then the rank of $A$ is less than $n$,

thus $\det(A) =0$.

I proved that way, but looks like too simple so I think maybe there is a trick that I missed.

Or I proved it right?

Modification:

($\Rightarrow$) If $A$ is invertible, $AB=I$, $\det(AB)=\det(A)\det(B)=1$ so $\det(A) \neq 0$

($\Leftarrow$) If $\det(A) \neq 0$ , by Cramer's rule, $Ax=b$ has a unique solution and it means $A$ is invertible ($x=A^{-1}b$)

Now I think I complete the proof.

答:
Perhaps the simplest way to prove this, and in two lines, is the following... but it depends on what definitions you have (all the time square matrices of the same order):

A matrix $\,A\,$ is invertible iff there exists a matrix $\,B\,$ s.t. $\;AB= BA=I\;$ iff when reduced (by rows or columns) no row (column) is all zeros iff $\,\det A\neq 0\,$ .

The last double implication follows from the fact that the determinant of an upper (lower) triangular matrix is the product of the entries in its main diagonal (and, of course, on the fact that interchanging rows only multiplies the matrix by $\,\pm 1\,$)

If your definitions cover the above there you have an easy proof of what you want.


问:
How to prove $\int_{0}^{\infty} \frac{\sin x}{x(x^2+1)}dx=\frac{\pi}{2}(1-\frac{1}{e})$ by residue calculus?
How to prove $$I=\int_{0}^{\infty} \frac{\sin x}{x(x^2+1)}dx=\frac{\pi}{2}(1-\frac{1}{e})\, ?$$

Let $$f(z)=\frac{\sin z}{z(z^2+1)}=\frac{e^{iz}-e^{-iz}}{2i\,z(z^2+1)}.$$

Then I have no idea how to deal with $[0,\infty)$.

Can anyone please give me a hint?  Thanks in advance.

答:
You can notice since your $f$ is even it is
$$\int_0^{\infty} f(x) \text{d}x=\frac{1}{2} \int_{-\infty}^{\infty} f(x) \text{d}x$$


问:
Good source to study complex analysis
I am taking a course on Analytic Number Theory. For that, as a pre-requisite, complex analysis is needed. Can anyone give me a video lecture series to start complex analysis? I need to know about holomorphic functions and their singularities, and contour integration(Mentioned by my course instructor).

If anyone can refer a good book, then also it will be helpful.

答:
You may take a look at Complex Analysis written by Bak and Newman. The short proof of the prime number theorem that Newman discovered in 1980 is replicated in this book.

You may also try Ash and Novinger's Complex Variables (this link is legitimate; the book is freely available from Prof. Ash's website; a very cheap printed version was published by Dover Publications). It contains Newman's proof of the prime number theorem as well as Dixon's short proof of Cauchy's theorem, the key theorem in every undergraduate course on complex analysis.

Title: Good source to study complex analysis


问:
The Mayer-Vietoris sequence
If $X$ is a space with a pair of subspaces $A, B \subset X$ such that $X$ is the union of the interiors of $A$ and $B$, then there is a long exact sequence of homology groups

$\displaystyle \ldots\to H_n(A\cap B) \quad \xrightarrow{\Phi}\quad  H_n(A) \oplus H_n(B) \quad \overset{\Psi}{\to} \quad H_n(X) \quad \overset{\partial}{\to} \quad H_{n-1} (A \cap B)\to \ldots$

The explicit action of the boundary map $\partial : H_n(X)\to H_{n-1}(A \cap B)$ escapes me. Allen Hatcher, in his book, writes

A class $\alpha \in H_n(X)$ is represented by a cycle $z$, and by barycentric subdivision or some other method we can choose $z$ to be a sum $x+y$ of chains in $A$ and $B$, respectively.

I think I can understand how $\partial \alpha$ can then be identified with $H_{n-1}(A \cap B)$, but can someone make clear to me how we choose such a sum? If you can go through all the steps taking an element in $H_n(X)$ to $H_{n-1}(A\cap B)$ I will be even more grateful.

答:
Let $\mathcal U=\{\operatorname{int}A, \operatorname{int}B\}$, which is an open covering of $X$. According the Proposition 2.21 in Hatcher, the "inclusion" map $\iota:C^{\mathcal U}_\bullet(X)\to C_\bullet(X)$, where $C^{\mathcal U}_\bullet(X)$ is the subcomplex of $C_\bullet(X)$ spanned by the singular simplices whose image is completely contained in one of the elements of $\mathcal U$, induces an isomorphism in homology.

If $\alpha$ is a homology class, represented by a cycle $z\in C_p(X)$, then the previous paragraph implies that there is a $z'\in C^{\mathcal U}_p(X)$ such that $z$ and $\iota(z')$ are homologous. Now $z'$ is a sum of singular simplices which are either in $A$ or in $B$ (or both): let $z'_1$ be the subsum of those in $A$, and let $z'_2$ the subsum of those not in $A$. Then $z'=z'_1+z'_2$ is the decomposition you want.

Title: The Mayer-Vietoris sequence


问:
Average number of guesses for simple game
The guess-a-number game starts with one player (the chooser) picking a number
between 1 and 100 (inclusive) and secretly writing it down. The other player
(the guesser) attempts to guess the number. After each guess, the chooser responds
with “correct” (the guesser guessed the number and the game is over),
“higher” (the actual number is higher than the guess), or “lower” (the actual
number is lower than the guess).

What is the average number of guesses the guesser will need to find the number given that he uses the optimal, most efficient strategy?

I found that it will be 6 guesses $29/32$ times and 7 guesses $3/32$ times using the most clumsy method ever which I won't post. It can be no more than 7 since $2^7=124>100$.

It was like

$(100-1)/2 = 49.5, \\(49-1)/2 = 24,  \\(50-1)/2 = 24.5 \\-> 24, 25 $

with 24 $3/4$ of the time and 25 $1/4$ of the time, etc etc etc.

What is a better way (since the method had so many opportunities for error I'm probably wrong anyway)?

答:
The average number of guesses would be ~5.80905. As the analytical approach to solving this would be rather extensive, the answer could be estimated with a simple simulation, in which the program follows the optimal strategy of cutting the set of values in half by selecting the mid point, and then we average the number of trials after a large number of iterations.

```<code>n <- 100
count<-c()

for (i in 1:100000){

  s <- round(runif(1,0.5,n+0.5),0)
  iter <- 1
  guess <- trunc(n /2)
  iter_par <-trunc(guess/2)

  while (guess!= s){

    ifelse(guess<s,guess<-guess+iter_par,guess<-guess-iter_par)
    iter_par<-round(iter_par/2,0)
    ifelse(iter_par<1,iter_par<-1,iter_par<-iter_par)
    iter<-iter+1
  }

  count<-append(count,iter)

}

mean(count)
</code>```

Body:
The average number of guesses would be ~5.80905. As the analytical approach to solving this would be rather extensive, the answer could be estimated with a simple simulation, in which the program follows the optimal strategy of cutting the set of values in half by selecting the mid point, and then we average the number of trials after a large number of iterations.

n <- 100
count<-c()

for (i in 1:100000){

  s <- round(runif(1,0.5,n+0.5),0)
  iter <- 1
  guess <- trunc(n /2)
  iter_par <-trunc(guess/2)

  while (guess!= s){

    ifelse(guess<s,guess<-guess+iter_par,guess<-guess-iter_par)
    iter_par<-round(iter_par/2,0)
    ifelse(iter_par<1,iter_par<-1,iter_par<-iter_par)
    iter<-iter+1
  }

  count<-append(count,iter)

}

mean(count)


问:
Boundary Map of Bar Resolution vs. Face Map of the Nerve of a Group
For a discrete group $G$ I have the following two definitions, which I think are correct:

The nerve of $G$ is $NG$, a simplicial set whose $n$-simplices are $G^n$ ($G^0$ being the trivial group $\{1\}$) and whose face maps are $d_0(g_1,...,g_n)=(g_2,...,g_n)$, $d_n(g_1,...,g_n)=(g_1,...,g_{n-1})$, and $d_i(g_1,...,g_n)=(g_1,...,g_i g_{i+1},...,g_n$ for all $0<i<n$. The degeneracy maps are $s_0(g_1,...,g_n)=(1,g_1,...,g_n)$ and $s_i(g_1,...,g_n)=(g_1,...,g_i,1,g_{i+1},...,g_n)$ for $i>0$. The geometric realization of $NG$ is a space which may be viewed as a CW complex with $n$-cells that can be identified with the nondegenerate simplices of $NG$, that is, those that don't have $1$ as any of the $g_i$ in $(g_1,...,g_n)$. Generally this space is considered to be the (a?) classifying space for $G$.

The bar resolution of $G$ is the sequence $0\leftarrow \mathbb{Z}\leftarrow G^0 \leftarrow G^1 \leftarrow...$ with boundary maps $\partial_n(g_1,...,g_n)$$=g_1*(g_2,...,g_n)$$+\overset{n-1}{\underset{i=1}{\sum}}(-1)^i(g_1,...,g_i g_{i+1},...,g_n)$$+(-1)^n(g_1,...,g_{n-1})$.

I'm looking for a way to understand the cohomology groups of $G$, as defined using the bar resolution, in terms of the cellular cohomology groups of $BG$. It seems like there should be a very direct relationship between the face maps $d_i$ which can be thought of as determining the cellular structure of $BG$ and the bar resolution's boundary map $\partial_n$; in fact $\partial_n$ looks almost exactly like $\overset{n}{\underset{i=0}{\sum}}(-1)^i d_i(g_1,...,g_n)$, except that the first term is $g_1*(g_2,...,g_n)$ instead of just $(g_2,...,g_n)$. Essentially, my question is where this "extra multiplication by $g_1$" comes from. I also wonder if my definitions are missing something or my understanding of them has a gap. I'd like to understand this in a way that doesn't appeal to the "other" usual construction of $BG$ as a quotient of $EG$ if at all possible.

答:
I believe I have settled this issue for myself, and the key ingredient is indeed the Dold-Kan correspondence. However, I find a synthesis of these two expositions to be much more comprehensible than anything nlab has to offer: (1),  (2) They also both refer to the same book reference, namely this one, which is also helpful in understanding the result.

The basic idea is that any chain complex corresponds to a simplicial abelian group, and any simplicial abelian group $A$ has a corresponding chain complex $A_*$. The image of the degenerate simplices under this correspondence forms a subcomplex $DA_*$, and there is another subcomplex $NA_*$ which turns out to have the property that $NA_*\oplus DA_*\cong A_*$. In other words, $A_*/DA_*\cong NA_*$. The Dold-Kan correspondence theorem itself says that $A\to NA_*$ gives an equivalence of categories between chain complexes and simplicial abelian groups, which then gives a way to "normalize" certain chain complexes, including the chain complex defining the simplicial homology of a topological space.

In order to apply this to my situation, which involves cohomology rather than homology, it is necessary to "dualize" this result. The key insight, I think, is that actually any simplicial set  $X$ can be "upgraded" into an abelian simplicial group $\mathbb Z (X)$, whose $n$-simplices $\mathbb Z(X)[n]$ are $\mathbb Z (X[n])$, which consists of formal sums $\sum_{x\in X}k_x x$ such that $k_x\in \mathbb Z$ and only finitely many of them are nonzero. Then any cochain complex $C^n(X,A)$ defined as $\lbrace f:X[n]\to A \rbrace$ can instead be thought of as $Hom_\mathbb{Z}(\mathbb Z(X)[n],A)$, and if $C_0^n(X,A)$ is the functions which are zero on any nondegenerate $n$-simplex, then  $C^n(X,A)/C^0_n(X,A)\cong Hom_\mathbb{Z}(N\mathbb Z(X)[n],A)$ and the Dold-Kan correspondence gives a chain equivalence between $Hom_\mathbb{Z}(\mathbb Z(X)[n],A)$ and $ Hom_\mathbb{Z}(N\mathbb Z(X)[n],A)$, which makes the natural projection $C^n(X,A)\to C^n(X,A)/C^0_n(X,A)$ into a chain equivalence also.

After this, I found that it does appear necessary, or at least convenient, to go through $\mathcal E G$, the simplicial set with $\mathcal E G[n]=G^{n+1}$ and $d_i$, $s_i$ given by deletion and repetition of the $i$th element respectively, and whose geometric realization is $EG$. Even though this (or the bar resolution) may be viewed as a simplicial group, $G$ may not be abelian, so that isn't particularly helpful. It's better to think of it as a simplicial set and use a slightly modified version of the result above (essentially, the same formulation can be done for subcomplexes of $C^n(X,A)$ to "normalize" them). Then (in broad strokes) if $C^n(G,A)=\lbrace f:G^{n+1}\to A\text{ such that }gf=fg\text{ for all }g\in G \rbrace$, there is an injective map $C^n(G,A)/C^n_0(G,A)\to C^n_\text{cell}(EG,A)$, and an injective map $C^n_\text{cell}(BG,A)\to C^n_\text{cell}(EG,A)$; if $g$ acts trivially on $A$ then both of which have the same subcomplex as their image. The isomorphism follows from that, and if $G$ doesn't act trivially on $A$ then that first injective map can still be used to get an isomorphism to the cohomology with local coefficients instead.


问:
Real and complex analysis Rudin. Question 6
On page 33 of Rudin's Real And Complex Analysis, question 6 asks us " let X be an uncountable set, let M be the collection of all sets E in X such that either E or E's complement is countable. Define u(E)=0 (u is a measure on M) in the first case and u(E)=1 in the second. ...
Describe the corresponding measurable functions and their integrals."

I have no idea how the integrals would be defined. I'm  pretty sure the measurable functions send uncountable sets to uncountable sets and countable sets to countable sets. So the integrals would zero? Because the preimage of any point on the real line is a countable set, and the measure of a countable set is 0??? Any help would be appreciated.

答:
So the set-up is this: Let $X$ be an uncountable set and let
$$\mathcal{M} = \{E\subseteq X:E\text{ is countable or }X\setminus E\text{ is countable}\}.$$
Let $\mu$ be a measure on $\mathcal{M}$ such that $\mu(E)=0$ if $E$ is countable and $\mu(E)=1$ is $X\setminus E$ is uncountable.

We first want to find all the measurable functions on $X$. By definition, a function $f:X\to\mathbb{R}$ is measurable if, for all $a\in\mathbb{R}$, the set $\{x\in X:f(x)>a\}$ is measurable in $X$. We'll use $\{f>a\}$ as a shorthand for that set from here on.

Claim 1: $f$ is measurable if and only if $f$ is constant except on a countable set. That is, $f$ is measurable if and only if there exists some $E\subseteq X$ such that $E$ is countable and $f$ is constant on $X\setminus E$.

Proof. The $\impliedby$ direction is more or less trivial, you should just take it as an exercise. Now, we will show the $\implies$ direction. First notice that, for all measurable $f$ and $a\in\mathbb{R}$, we either have $\{f>a\}$ is countable or $\{f\leq a\}$ is countable.

Let $f:X\to\mathbb{R}$ be a measurable function. Consider the set
$$A= \{a\in\mathbb{R}:\{f>a\}\text{ is countable}\}.$$
First, notice that $A\ne\emptyset$ because, if for all $a\in\mathbb{R}$, the set $\{f>a\}$ is uncountable, then that would imply $\{f\leq a\}$ is countable for all $a\in\mathbb{R}$ and we would have a contradiction (i.e. we would have $\{f\leq n\}$ to be countable for all $n\in\mathbb{N}$ and thus $f^{-1}(\mathbb{R})$ would be countable, which contradicts the uncountability of $X$). On the other had, also note that $A$ is bounded below by a similar argument.

Let $c=\inf A$. We claim that $f^{-1}(\mathbb{R}\setminus\{c\})$ is countable and that completes our proof. To see this, first notice that
$$\{f>c\} = \bigcup_k\left\{f>c+\frac1k\right\}$$
and hence $\{f>c\}$ is countable. On the other hand, consider
$$\{f<c\} = \bigcup_k\left\{f\leq c-\frac1k\right\}.$$
Note that each of $c-\frac1k\notin A$, hence $\left\{f>c-\frac1k\right\}$ is uncountable, which implies that $\left\{f\leq c-\frac1k\right\}$ is countable. Thus, $\{f<c\}$ is countable as well. The above two results imply that $f^{-1}(\mathbb{R}\setminus\{c\})$ is countable. QED

Now that we have characterized all the measurable functions, the integration should be relatively straightforward. Let $f$ be a measurable function and $c\in\mathbb{R}$ be such that $f^{-1}(\mathbb{R}\setminus \{c\})$ is countable (the existence of such $c$ is shown above). Then we have
$$\int_X f = \int_{f^{-1}(\{c\})}f + \int_{f^{-1}(\mathbb{R}\setminus \{c\})}f = c$$
because the set $f^{-1}(\mathbb{R}\setminus \{c\})$ has measure zero.


问:
Calculating Nil(Z_n) faster
As far as I know, in order to calculate Nil(Z_n) I need to calculate each number raised to different values and see if I get 0. That is a very slow and inefficient way, in my opinion.

Is there any other faster way to do this?

For example, U(Z_n) is equal with the numbers with the property (number, n) = 1.

Is there any similar thing for Nil(Z_n)?

Thanks.

答:
For any integer $n$,

the nilpotent elements of the finite ring   $ \mathbb {Z} /n\mathbb {Z}$ are all of the multiples of the radical of $n$,

which is  the product of the distinct prime numbers dividing $ n$.

For example, rad$(12)=$rad$(2^2\cdot3)=2\cdot3=6$,

so the nilpotent elements of $\mathbb Z/12\mathbb Z$ are multiples of $6$.

Title: Calculating Nil(Z_n) faster


问:
Is $[0, 1) \times (0, 1] \to D \setminus \{ 0 \}$, $(\varphi, r) \mapsto r e^{2 \pi i \varphi}$ not a homeomorphism?
In our lecture, it was claimed that
$$
\Psi \colon [0, 1) \times (0, 1] \to D \setminus \{ 0 \}, \
(\varphi, r) \mapsto r e^{2 \pi i \varphi},
$$
where $D$ denotes the unit disk in $\mathbb R^2 \cong \mathbb C$, is a homeomorphism.
On Wikipedia it states that "The fundamental group is a homotopy invariant—topological spaces that are homotopy equivalent (or the stronger case of homeomorphic) have isomorphic fundamental groups.", which I understand to mean: "If $f \colon X \to Y$ is a homeomorphism, then $\pi_1(X) \cong \pi_1(Y)$", where $X$ and $Y$ are connected topological spaces, so the fundamental group $\pi_1$ is independent of its basepoint.

It is well known, that $\pi_1(D \setminus \{ 0 \}) \cong \mathbb Z$ and we should have $\pi_1([0, 1) \times (0, 1]) \cong \{ e \}$, where $e$ is the identity element, since $[0, 1) \times (0, 1]$ is a convex set. Is my reasoning correct?

Nevertheless, the map $\Psi$ seems to be continuous as composition of continuous maps, and I think the inverse is given by
$$
\Psi^{-1}(t)
= \left( \frac{1}{2 \pi i} \log\left( \frac{t}{| t |}\right), | t |\right)
$$
which looks continuous also.

答: Yes, your reasoning is correct. If you want a direct proof of the fact that $\Psi$ is not a homeomorphism, consider the sequence $\left(1-\frac1n,\frac12\right)_{n\in\Bbb N}$ of points of $[0,1)\times(0,1]$. It diverges in that space, but it is mapped by $\Psi$ into the sequence $\left(\frac12e^{2\pi i(1-1/n)}\right)_{n\in\Bbb N}$, which converges to $\frac12$.


问:
Conditions for the identity to be true
I'm writing an article about some useful techniques of integration. In the article I make use of the following ( well known ) identities.Here $p$ is a polynomial:

$$\int_{a}^{b} p(x)\cot x \, \mathrm{d}x = 2\sum_{n=1}^{\infty}\int_{a}^{b} p(x)\sin 2nx \, \mathrm{d} x \tag{1}$$

$$\int_{a}^{b} p(x) \csc x \, \mathrm{d}x= 2 \sum_{n=0}^{\infty} \int_{a}^{b} p(x) \sin (2n+1)x \, \mathrm{d}x  \tag{2}$$

$$\int_{a}^{b} p(x) \tan x \, \mathrm{d}x \overset{(?)}{=} -2 \sum_{n=1}^{\infty} (-1)^{n} \int_{a}^{b} p(x) \sin 2nx \, \mathrm{d}x \tag{3}$$

*I'm not sure about the 3rd one if the index of summation starts at 1 or 0. *

The proof of $(2)$ for instance comes from the fact that

$$\sum_{n=0}^{N} \sin (2n+1)x  = \frac{\csc x}{2} - \frac{\cos (2N+1) x }{2 \sin x}$$

in combination with the Riemann-Lebesgue lemma.

Now, the question is what are the conditions for the above $3$ identities to hold? Can we generalize for an arbitrary $f$? If so, what must $f$ satisfy for the identities to hold? twice differentiable , bounded , only integrable , etc?

答:
I think I found the answer.

Let $p:[a, b] \rightarrow \mathbb{R}$ be a polynomial such that $p(a)=0$ and $\lim \limits_{x \rightarrow b^-} p(x) \csc x = \ell \in \mathbb{R}$.  It holds that
$$\int_{a}^{b} p(x) \csc x \, \mathrm{d}x= 2 \sum_{n=0}^{\infty} \int_{a}^{b} p(x) \sin (2n+1)x \, \mathrm{d}x$$

Proof:  We start from the well known identity

$$\sum_{n=0}^{N} \sin (2n+1)x = \frac{\csc x}{2} - \frac{\cos (2N+1)x}{2 \sin x}$$

Integrating we have

$$2 \sum_{n=0}^{N} \int_{a}^{b} p(x) \sin (2n+1) x \, \mathrm{d}x = \int_{a}^{b} p(x) \csc(x) \, \mathrm{d}x -\int_{a}^{b} p(x) \cdot \frac{\cos (2N+1)x}{\sin x} \, \mathrm{d}x$$

Now let $N \rightarrow +\infty$. Using the Riemann - Lebesgue Lemma we have that

$$\lim_{N \rightarrow +\infty} \int_{a}^{b} p(x) \cdot \frac{\cos (2N+1)x}{\sin x} \, \mathrm{d}x = 0 $$

Hence,

$$\int_{a}^{b} p(x) \csc x \, \mathrm{d}x= 2 \sum_{n=0}^{\infty} \int_{a}^{b} p(x) \sin (2n+1)x \, \mathrm{d}x$$

Other similar identities are:

$\displaystyle \int_{a}^{b} p(x)\cot x \, \mathrm{d}x = 2\sum_{n=1}^{\infty}\int_{a}^{b} p(x)\sin 2nx \, \mathrm{d} x$

$\displaystyle \int_{a}^{b} p(x) \tan x \, \mathrm{d}x = -2
   \sum_{n=1}^{\infty} (-1)^{n} \int_{a}^{b} p(x) \sin 2nx \,
   \mathrm{d}x$

Title: Conditions for the identity to be true


答:
First note that the integrand is an even function so that

$$\int_0^\infty \frac{\sin(x)}{x(x^2+1)}\,dx=\frac12 \int_{-\infty }^\infty \frac{\sin(x)}{x(x^2+1)}\,dx\tag1$$

Next, we can use Euler's Formula $e^{ix}=\cos(x)+i\sin(x)$ in $(1)$ to write

$$\frac12 \int_{-\infty }^\infty \frac{\sin(x)}{x(x^2+1)}\,dx=\frac12 \text{Im}\left(\int_{-\infty }^\infty \frac{e^{ix}}{x(x^2+1)}\,dx\right)\tag2$$

where the integral on the right-hand side of $(2)$ is interpreted as a Cauchy Principal Value.

Then, we analyze the contour integral $I$ where

$$I=\oint_C \frac12  \frac{e^{iz}}{z(z^2+1)}\,dz$$

where the closed contour $C$ is comprised of $(i)$ the straight line segments from $-R$ to $-\varepsilon$ and from $\varepsilon$ to $R$ (where $R>1$ and $R>\varepsilon>0$), $(ii)$ the circular arc in the upper-half plane, centered at the origin, with radius $\varepsilon$ from $-\varepsilon$ to $\varepsilon$, and $(iii)$ the circular arc in the upper-half plane, centered at the origin, with radius $R$ from $R$ to $-R$.

The value of this integral is equal to $2\pi i $ times the residue of the integrand at $z=i$.

As $R\to \infty $ show that the contribution to the value of $I$ from integration over the semicircular arc of radius $R$ approaches $0$.

As $\varepsilon\to 0^+$, show that contribution to the value of $I$ from integration over the semicircular arc of radius $\varepsilon$ approaches $i\pi$ times the residue of the integrand at $z=0$.

Finally, put everything together to arrive at the coveted result.

Can you proceed now?

Title: How to prove $\int_{0}^{\infty} \frac{\sin x}{x(x^2+1)}dx=\frac{\pi}{2}(1-\frac{1}{e})$ by residue calculus?

Body:
How to prove $$I=\int_{0}^{\infty} \frac{\sin x}{x(x^2+1)}dx=\frac{\pi}{2}(1-\frac{1}{e})\, ?$$

Let $$f(z)=\frac{\sin z}{z(z^2+1)}=\frac{e^{iz}-e^{-iz}}{2i\,z(z^2+1)}.$$

Then I have no idea how to deal with $[0,\infty)$.

Can anyone please give me a hint?  Thanks in advance.


问:
Integral expression for $\log (\zeta (s))$
I read here (and in many other places) that

$$\log (\zeta (s))=s \int_0^{\infty } J(x) x^{-s-1} \, dx$$

where $J(x)=\sum _{n=1}^{\infty } \frac{\pi \left(x^{1/n}\right)}{n}$ is Riemann's prime counting function (link), and $\pi (x)$ is the usual prime counting function. Clearly, the RHS has poles wherever $\zeta (s)=0$.

My question is: what kind of integral is this?

Does it in any way relate to the area under the curve $f(x,s)=J(x) x^{-s-1}$? (I doubt it; it seems more likely It's a measure some path in the complex plane.)

In what sense can it be said to converge? From rough plotting in Mathematica, it looks to me as though $f(x,s)=J(x) x^{-s-1}$ diverges for values of $s$ less than roughly $-\frac{1}{4}$.

Riemann was creating an analytic continuation of the zeta function: so, how does this integral representation cope with complex values of ${x,s}$?

I'm sure these are naive questions, but I am sincerely trying to understand the mechanics of this equation. I'd of course be very grateful for any explanation. In addition, if someone could point me to a source that would allow me to learn and understand this kind of integral, I'd really appreciate it.

Many thanks.

答:
For $\Re(s) \ge 1,s\ne 1$ and not for any other $s$

$$\frac{\log \zeta(s)}{s}=\int_0^\infty J(e^u) e^{-su}du$$ is the Laplace-Fourier transform of $J(e^u) = \sum_{p^k\le e^u} \frac1k$ and $\log \zeta(s)$ is the Laplace-Fourier transform of its distributional derivative $\sum_{p^k} \frac1k \delta(u-k\log p)$.


问:
Uniqueness of "Punctured" Tubular Neighborhoods (?)
Here is a question that has been haunting me for a while:  Let $\mathbb{R}^{n-1} \times [0, \infty)$ be the upper half space of $\mathbb{R}^n$ and suppose we have a smooth homeomorphism (not a diffeo)

$f : \mathbb{R}^{n-1} \times [0, \infty) \rightarrow \mathbb{R}^{n-1} \times [0, \infty)$

such that $f$ is the identity on $\mathbb{R}^{n-1} \times \{0\}$ and $f$ restricts to a diffeomorphism from the punctured (at the origin) half-space to itself. In other words, $f$ is the identity on the boundary and it would be a diffeomorphism from the half-space to itself except that the derivative is singular at the origin.

Question : Is the restriction of $f$ to the punctured half space smoothly isotopic to the identity?

A little background: If $Df$ were nonsingular at 0 this would just be an application of the uniqueness of collars, which is a special case of the uniqueness of tubular neighborhoods.  That proof uses the morse lemma to write the coordinates of the imbedding as a linear combination of coordinate functions. The coefficients are given by smooth functions whose values at zero are the entries of the Jacobian of the imbedding at the origin.  The isotopy one constructs takes the imbedding to the linear map $Df_0$.  The issue here is of course that this linear map is degenerate.  But perhaps we can construct a different isotopy if we remove the bad point?

This is mainly motivated by my (borderline obsessive and unhealthy) thinking about this question: Different definitions of handle attachment

Thanks for reading my question and I hope It's not too cranky.

答:
I don't really see the connection to your handle attachment question -- it looks to me like you're approaching it as a more general problem than it actually is.

But this question has a reasonable answer.  Think of your half-space as sitting inside the Euclidean space $\mathbb R^n$.  Then you can intersect the half-space with spheres of radius $r$ centred around the origin.  This converts your punctured half-space (via a diffeomorphism) to:

$$ D^{n-1} \times \mathbb R $$

So you are studying the group of diffeomorphisms of this manifold where the diffeomorphism restricts to the identity on the boundary.  I think in full generality this space is not connected.  Aside from dimension $4$ I think this space is known to have the homotopy-type of $\Omega Diff(D^n)$, where $Diff(D^n)$ is the group of diffeomorphisms of the $n$-disc $D^n$ which restrict to the identity on the boundary.

The basic idea for why you'd expect something like this is that a diffeomorphism of $D^{n-1} \times \mathbb R$ restricts to an embedding of $D^{n-1} \times \{0\}$ into $D^{n-1} \times \mathbb R$.  That map is a fibre bundle.  There's a standard sequence of relations between this embedding space and $Diff(D^n)$, it's outlined at the end of Hatcher's paper on the Smale conjecture.   Needless to say, the spaces $Diff(D^n)$ for $n$ large tend not to be contractible, and the fundamental groups are also known to be non-trivial in a large number of cases.  Partial computations of the fundamental groups were done by Milnor and Kervaire.


问:
What is a "symmetric Boolean function"?
Below are two different definitions of a "symmetric boolean function".

Question: Are these equivalent and how?

Definition 1:

The $k$th elementary symmetric function of $n$ variables is the sum of all possible products of $k$ distinct variables = $\Sigma x_{i_1} \cdot \cdot \cdot x_{i_k}$ where the subscripts are taken from the first $n$ integers. (Cunkle, p. 834)

Definition 2:

A symmetric Boolean function is a Boolean function whose value does not depend on the permutation of its input bits, i.e., it depends only on the number of ones in the input. (wikipedia)

References:

Cunkle, C. (1963). Symmetric Boolean Functions. The American Mathematical Monthly, 70(8), 833-836. doi:10.2307/2312663

答:
They are related, but not equivalent. Definition 1 gives the "building blocks" for more complicated functions, hence elementary symmetric functions. Also, as Jair Taylor noted above, this definition is not explicitly about Boolean functions.

Given inputs $x_1$ and $x_2$, consider that these meet Definition 1:
$$f(x) = x_1+x_2\\
g(x) = x_1x_2$$
while this next one, being the sum of the above, doesn't:
$$h(x) = x_1+x_2+x_1x_2$$

All of the above are symmetric functions, however, in that the inputs could be permuted without affecting the result.

Title: What is a "symmetric Boolean function"?


问:
In graph G finding a cycle with no duplicate vertices of length at least k is NP complete.
I know that this problem is similar to the hamming cycle problem, But how can I show that the hamming cycle problem reduces to this problem in a polynomial number of steps?</p>

答:
You mean the Hamilton cycle problem.

To reduce the Hamilton cycle problem to this problem, simply set $k$ to the number of vertices in the graph.