问:
Convergence of infinite products flaw in this derivation of Euler product formula?
I am trying to write a derivation of the Euler product formula in a way that is accessible to younger students and those not mathematically trained at university.

Because of this I am starting from the simplest starting point that I can, and use the simplest possible steps in the derivation.

I arrive at the following which I know is wrong as it is missing the exponent $s>1$ to ensure convergence.

$$ \prod_{p}\frac{1}{(1-\frac{1}{p})}=\sum_{n}\frac{1}{n} $$

My question is to ask where I am introducing the error in the following simple derivation.

Step 1

Start with a familiar series, valid for $|x| < 1$.

$$ \frac{1}{(1-x)}=1+x+x^{2}+x^{3}+\ldots $$

We're interested in primes, and might be tempted to set $x$ to a prime $p$, but we can't because $|x|$ needs to be $<1$. Let's try $\frac{1}{p}$ which is always $<1$.

$$ \frac{1}{(1-\frac{1}{p})}=1+\frac{1}{p}+\frac{1}{p^{2}}+\frac{1}{p^{3}}+\ldots $$

This series converges to a finite value.

Step 2

Let's now pick two different primes $p_1$ and $p_2$, and multiply the analogous series for each.

$$ \frac{1}{(1-\frac{1}{p_{1}})}\cdot\frac{1}{(1-\frac{1}{p_{2}})}=\left(1+\frac{1}{p_{1}}+\frac{1}{p_{1}^{2}}+\ldots\right)\cdot\left(1+\frac{1}{p_{2}}+\frac{1}{p_{2}^{2}}+\ldots\right) $$

I don't think there is a problem here.

We are multiplying two infinite series, each one known to converge to a finite value. Am I wrong?

Step 3

Now we extend from two primes, $p_1$ and $p_2$, to all primes $p_i$.

$$ \prod_{p_{i}}\frac{1}{(1-\frac{1}{p_{i}})}=\prod_{p_{i}}\left(1+\frac{1}{p_{i}}+\frac{1}{p_{i}^{2}}+\ldots\right) $$

I think this is the flaw in my derivation. We have an infinite product of finite values.

Each factor $\frac{1}{(1-\frac{1}{p_{i}})}$ is $>1$ so the infinite product looks like it might not converge.

However each factor tends to $\rightarrow 1$ as $i \rightarrow \infty$. This suggests the infinite product is ok. Am I wrong?

Step 4

If we multiply out those brackets, each of the form $\left(1+\frac{1}{p_{i}}+\frac{1}{p_{i}^{2}}+\ldots\right)$, we will obtain terms which are all combinations of primes, in all the combinations of powers for each prime.

$$ \prod_{p_{i}}\left(1+\frac{1}{p_{i}}+\frac{1}{p_{i}^{2}}+\ldots\right) = 1 + \frac{1}{p_1} + \frac{1}{p_2} + \ldots + \frac{1}{p_1 p_2} + \frac{1}{p_1 p_3} + \ldots + \frac{1}{p_1^2} + \frac{1}{p_2^2} + \ldots  $$

Importantly, each combination occurs only once.

Step 5

The fundamental theorem of arithmetic tells us that each integer can be written as a unique product of primes. This means that each integer is represented in the terms we arrived at in the previous step.

Because each combination of primes occurs only once, each integer is represented only once.

That leads us to the Euler product formula.

$$ \boxed{\prod_{p}\frac{1}{(1-\frac{1}{p})}=\sum_{n}\frac{1}{n}} $$

Discussion

If my suspicion is correct that the error is in the infinite product because it doesn't converge, how does changing $p$ for $p^s$ for $s>1$ help?

答:
I'll give another treatment of the problem in step 4 - plenty of other answers cover step 3 but there's a relatively accessible way of explaining why step 4 is invalid that may be useful for the kind of teaching you're doing.

It comes down to what we mean by an infinite sum. We have to remember that a sum of infinite terms is defined as the limit of the sequence of all its partial sums.

Let's look at that infinite product you defined in step 4:

$$ \prod_{p_{i}}\left(1+\frac{1}{p_{i}}+\frac{1}{p_{i}^{2}}+\ldots\right) = 1 + \frac{1}{p_1} + \frac{1}{p_2} + \ldots + \frac{1}{p_1 p_2} + \frac{1}{p_1 p_3} + \ldots + \frac{1}{p_1^2} + \frac{1}{p_2^2} + \ldots  $$

The right-hand side is expressed as an infinite sum, but as I've mentioned, what that actually means is that it's the limit of the sequence:

$$ 1, 1 + \frac{1}{p_1}, 1 + \frac{1}{p_1} + \frac{1}{p_2}, \ldots  $$

The problem is that for any natural number i, the ith term of this sequence is only affected by the terms up to $$ \frac{1}{p_i} $$ - the later terms in the sum expressed at the top never appear. This means that when we take the limit, only those initial terms appear, and it's actually just giving us:

$$ \sum \frac{1}{p_i} $$

In general, reordering infinite sums like this often introduces problems: you can't put any infinite set of terms first, as that makes any terms that would appear 'after' them no longer actually appear in the sequence whose limit we find to calculate the sum. That's not to say this is the only way such reorderings can cause problems, either, but examples of such are further from the scope of this question.


问:
What is the remainder when $1^{2016} + 2^{2016} + ⋯ + 2016^{2016}$ is divided by $2016$?
What is the remainder when $1^{2016} + 2^{2016} + ⋯ + 2016^{2016}$ is divided by $2016$?

My Approach:

started breaking 2016 into product of pairwise coprime positive integers to get $2016 = 32 * 7 * 9$
This suggests that we want to find the sum modulo 32,7, 9 separately so that we can apply the chinese remainder theorem.

but I couldn't solve it any further than this so please help me out.

ContentLicense: CC BY-SA 4.0

Body:
$$S_7=\sum_{r=0}^{287}\left((7r+1)^{2016}+(7r+2)^{2016}+(7r+3)^{2016}+(7r+4^{2016}+(7r+5)^{2016}+(7r+6)^{2016}\right)$$

As $2016\equiv0\pmod6,a^{2016}\equiv1\pmod7$ if $(a,7)=1$

$$S_7\equiv288\equiv\sum_{r=0}^{287}(1+1+1+1+1+1+0)\equiv1(-1)\text{ as }288\equiv1\ \ \ \  (1)$$

Again using Carmichael Function $\lambda(32)=2^{32/4}$,

$$S_{32}=\sum_{r=1}^{2016/2}(2r)^{2016}+\sum_{r=1}^{2016/2}(2r-1)^{2016}\equiv\sum_{r=1}^{2016/2}0+\sum_{r=1}^{2016/2}1\pmod{32}\equiv1008\equiv?\ \ \ \  (2)$$

Finally,

$$S_9=\sum_{r=1}^{2016/3}(3r)^{2016}+\sum_{r=1}^{2016/3}((3r-2)^{2016}+(3r-1)^{2016})$$
$$\equiv\sum_{r=1}^{2016/3}0++\sum_{r=1}^{2016/3}(1+1)\equiv?\ \ \ \  (3)$$

Now apply  Chinese remainder theorem on $(1),(2),(3)$


问:
Show that the function $f:[a,\infty] \to \mathbb R$ defined by $x \mapsto \frac{1}{x}$ for $a$ a positive number is uniformly continuous.
Show that the function $f:[a,\infty] \to \mathbb R$ defined by $x \mapsto \frac{1}{x}$ for $a$ a positive number is uniformly continuous.

The function is continuous if and only if $$\forall \epsilon>0 (\exists \delta >0(\forall x,y \in [a,\infty] (\left|x-y\right|<\delta \implies \left|f(x)-f(y)\right|< \epsilon)))$$

$$\left|f(x)-f(y)\right|=\left|\frac{1}{x}-\frac{1}{y}\right|=\frac{\left|x-y\right|}{xy}\le\frac{\left|x-y\right|}{a^{2}}$$

Taking $\delta= a^2\epsilon$ follows uniform continuity of $f$ over the given interval.

I also used the sequential criterion :

The function is continuous if and only if
$$\forall (x_n)_n,(y_n)_n \subset [a,\infty]:\lim_{n\to \infty} (x_n-y_n)=0 \implies \lim_{n\to \infty} (f(x_n)-f(y_n))=0$$

Let $(x_n)_n,(y_n)_n \subset [a,\infty]$ be arbitrary sequences with the property that $\lim_{n\to \infty} (x_n-y_n)=0 $ then $$\lim_{n\to \infty} (f(x_n)-f(y_n)) =\lim_{n\to \infty} \frac{y_n-x_n}{x_ny_n}\le -\frac{1}{a^2}\lim_{n\to \infty} (x_n -y_n)=0$$

But I'm not able to use squeeze theorem to conclude that $\lim_{n\to \infty} (f(x_n)-f(y_n))=0$.

答:
Note that $\lim a_n=0\Leftrightarrow \lim |a_n|=0$.

The last expression should be
$$\lim_{n\to \infty} |f(x_n)-f(y_n)|=\lim_{n\to \infty} \frac{|y_n-x_n|}{x_ny_n}\le \frac{1}{a^2}\lim_{n\to \infty} |x_n -y_n|=0.$$

Now, the squeeze theorem allows us to conclude that $\lim_{n\to \infty} (f(x_n)-f(y_n))=0$.


问:
Prove that $\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n}>\frac{2}{3}$
Prove that $\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n}>\frac{2}{3}$ for any $n \in N$

I used AM $\geq$ HM and got $$\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n}\geq\frac{(n+1)^2}{n+n+1+...+2n}=\frac{(n+1)^2}{\frac{3}{2}n(n+1)}>\frac{2}{3}$$
My question is whether this solution is correct and also why $n+n+1+...+2n = \frac{3}{2}n(n+1)$, since I used wolfram to compute that and cannot see it myself.

Any tips on how to compute sums like this or other possible solutions for this problem would be appreciated as well.

答:
That is correct, and $$n+(n+1) +(n+2) +\dots +(n+n) \\ =(n+1)\times n +(1+2+3+\dots +n) \\ = n(n+1) + \frac{n(n+1)}{2} \\ =\frac 32 n(n+1) $$

Alternatively, you can use the formula for the sum of an arithmetic  progression $$S=\frac {\text{number of terms}}{2} (\text{first term + last term}) $$


答:
Yes, it's correct.

Your sum is an arithmetic sequence.

To sum them up, you can use

$$\sum_{i=1}^m a_i = \frac{m}2 (a_1 + a_m) $$

Here we have $m+1$ terms, the first term is $n$ and the last term if $2n$.

$$\sum_{i=n}^{2n} i  = \frac{(n+1)}2 (n + 2n)=\frac{3n}{2}(n+1)$$

We have

$$\frac23 \cdot \frac{n+1}{n} = \frac23 + \frac2{3n} > \frac23$$

Body:
Prove that $\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n}>\frac{2}{3}$ for any $n \in N$

I used AM $\geq$ HM and got $$\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n}\geq\frac{(n+1)^2}{n+n+1+...+2n}=\frac{(n+1)^2}{\frac{3}{2}n(n+1)}>\frac{2}{3}$$
My question is whether this solution is correct and also why $n+n+1+...+2n = \frac{3}{2}n(n+1)$, since I used wolfram to compute that and cannot see it myself.

Any tips on how to compute sums like this or other possible solutions for this problem would be appreciated as well.


问:
Proof for $\lim\limits_{x \to a} f\left(x\right)$, as $f= \dfrac{1}{n}$, if $x \in A_{n}$, and $f=0$ if $x \notin A_{n}$
I just wanted to ask, if this is the correct way to prove this limit. My confusion comes from the fact, that I found one solution, and it was much more complex than mine. So I am not sure if I misunderstood the exercise, or just did something wrong.

The exercise is:
Suppose that $A_{n}$ is, for each natural number $n$, some $finite$ set of numbers in $\left[0,1\right]$, and that $A_{n}$ and $A_{m}$ have no members in common if $m \ne n$. Define $f$ as follows:

$f\left(x\right)=\dfrac{1}{n}$, $x$ in $A_{n}$.

$f\left(x\right)=0$, $x$ not in $A_{n}$, for any $A_{n}$.

Prove that $\lim\limits_{x \to a} f\left(x\right)=0$ for all $a$ in $\left[0,1\right]$.

As I understood this problem we can just appeal that any set $A_{n}$ cannot have any elements, so the proof follows as:

Let us assume that there is some $a_{\alpha}$, and this $a_{\alpha} \in A_{\alpha}$, and let say that it is the only element in this set. Then $f\left(a_{\alpha}\right)=\dfrac{1}{\alpha}$, and for all other $x$, as $x \ne a_{\alpha}$, $f\left(x\right)=0$.

Then there too is set $A_{\beta}$, this set has element $a_{\beta} \in A_{\beta}$ and, again, it is the only element in this set. So $f\left(a_{\beta}\right)=\dfrac{1}{\beta}$, and for all other $x$, as $x \ne a_{\beta}$, $f\left(x\right)=0$.

But, by a set $A_{\alpha}$, $f\left(a_{\beta}\right)=0$, and by $A_{\beta}$, $f\left(a_{\alpha}\right)=0$. This cannot happen, as then one $x$ would have 2 different values, which cannot be an function. By this, for any set $A_{n}$, $a \notin A_{n}$, for all $a \in \left[ 0,1\right]$.

By this for all $f \left(x\right)$, as $x \in \left[0,1\right]$, $f \left(x\right)=0$.

So, $\left| f \left(x\right)-l \right| <\epsilon$, then $\left|l \right| <\epsilon$, and as $ 0< \epsilon$, then $l=0$.

In conclusion, $\lim\limits_{x \to a} f\left(x\right)=0$ for all $a$ in $\left[0,1\right]$.

答:
You may have misunderstood the exercise: for each $x \in [0,1]$, $f(x)$ has a well-defined value.

Either $x$ belongs to one of the $A_n$. Then $f(x) = 1/n$. Note that there is no ambiguity here, since $x$ cannot belong to two different sets $A_n$ and $A_m$.

Or $x$ does not belong to any of the $A_n$, and then $f(x) = 0$.

Think for instance that $A_n$ is just the one-element set $A_n = \{1/n\}$. Then $f(x) = 0$ for any $x$, except $f(1) = 1$, $f(1/2) = 1/2$, $f(1/3)  = 1/3$, etc. In this case, can you prove the given statement? In particular, show that
$$
\lim_{x \to 0} f(x) = 0.
$$
You probability need to go back to the $\epsilon / \delta$ definition of limit for this. Then, try to generalize.


问:
How easily to check: $f$ is a closed map?
Assume that $f:\mathbb R\to\mathbb R$ is a function where $\mathbb R$ is the real number and the usual topology is defined on $\mathbb R$.

I have two questions:
   1. Let $C$ be closed. Then what is the proper form of $C$? For example, without the loss of generality, open set is replaced by the open interval.
   2. To check that $f$ is a closed map, we need some information: $f(C)$ is closed. How to easily check this?

答:
To check that $f(C)$ is closed you can either prove that all the limit points of $f(C)$ lie in $f(C)$ or $\mathbb{R}\setminus f(C)$ is open. The second one is easy most of the times but it really depends on what is $f$.
Also open sets are unions of open intervals not only open intervals.</p>

ContentLicense: CC BY-SA 3.0


问:
Solving Binomial probability question- what's wrong with my solution?

A company establishes a fund of $120$ from which it wants to pay an amount, $C$, to any of its $20$ employees who achieve a high-performance level during the coming year. each employee has a $2\%$ chance of achieving a high-performance level during the coming year, independent of any other employees. Determine the maximum value of $C$ for which the probability is less than $1\%$ that the fund will be inadequate to cover all payments for high performance.

Multiple-Choice Options: $24, 30, 40, 60, 120$.

My attempt:

Let $X$ be the RV denoting the number of employees that need to be paid the amount $C$. $X \sim Bin(20, 0.02)$. We need to find $P(X>120/C)<0.01$.

Note that $P(X= 0) = 0.667$, $P(X= 1) = 0.272$, $P(X= 2) = 0.052$, $P(X= 3) = 0.006$. Since $P(X= 3)<0.01$, we obtain $$3>\frac{120}{C} \implies C>40$$

I chose the option for $40$, however, the correct answer is $60$. Now, since my answer said $C>40$, I should have chosen $60$ or $120$, but I suspect that the solution has major flaws. Can someone please review my solution and let me know where I went wrong? Thanks.

答:
The fund will be insufficient if $X$ is "too large."  We want to limit the probability of this.  This suggests finding some integer $c$ such that $$\Pr[X > c] < 0.01. \tag{1}$$ Since $X \sim \operatorname{Binomial}(n = 20, p = 0.02)$, we calculate a table:
$$\begin{array}{c|c|c|c}
x & \Pr[X = x] & \Pr[X \le x] & \Pr[X > x] \\
\hline
0 & 0.667608 & 0.667608 & 0.332392 \\
1 & 0.272493 & 0.940101 & 0.059899 \\
2 & 0.0528303 & 0.992931 & 0.00706869 \\
3 & 0.00646901 & 0.9994 & 0.000599679
\end{array}$$
The third column is computed by successively adding up entries in the second column, and the fourth column is $1$ minus the third column.  We observe that the first value of $x$ for which the desired inequality $(1)$ is satisfied is $x = 2$; therefore, this is our value of $c$.  Hence the probability of having more than $2$ high-performing employees is less than $0.01$, and if the award is $C$, the company can make $C = 60$ without having the probability of insufficient funds exceed $0.01$, since the probability of three or more high-performing employees is less than $0.01$.

Your error is in concluding $c = 3$ rather than $c = 2$.  But there is a more fundamental error, which is that you did not compute cumulative probabilities.  Instead, you computed individual probabilities and stopped when $\Pr[X = c] < 0.01$.  This is not correct, as the inequality $(1)$ above explains.


问:
Finding the kernel of homomorphism between two cyclic groups
I have two cyclic group $\langle G,*\rangle $ with order $7$ and a generator $a\in{G}$,  $\langle H=\{\bar2,\bar4,\bar6,\bar8\},\cdot_{10}\rangle $.

I also have a homomorphism $\phi:G\rightarrow H , $where $ \phi(a)=\bar8$.

I have tried to see where each power of $a$ maps to and saw that :

$$\begin{align}
\phi(a)&=\bar8,\\
\phi(a^2)&=\bar4,\\
\phi(a^3)&=\bar2,\\
\phi(a^4)&=\bar6=e_H,\\
\phi(a^5)&=\bar8,\\
\phi(a^6)&=\bar4,\\
\phi(a^7)&=\bar2
\end{align}$$

I am a bit confused becauase I know that if $\phi$ is homomorphism then the identity element $a^7=e_G$ must map to the identity element of $H$. But I have $\phi(a^7)=\bar2$.

答:
Note that if $\phi : G \to H$ is a homomorphism and for $g \in G$, $o(g) = n$, then $o(\phi(g))\mid n$.

So, if we suppose $\phi$ is a homomorphism in you example, then since $a$ is a generator of a cyclic group of order $7$, we have $o(a) = 7$. So, we must have $o(\phi(a)) \mid 7 \implies o(\phi(a)) = 1$ or $o(\phi(a)) = 7$. Latter case is not possible since $|H| = 4$. And for the former case, we must send $a$ to $\bar{6}$ (identity of $H$), which is not the case. So, our assumption of $\phi$ being a homomorphism is contradictory. From this, we can also conclude that in your example, if $\phi: G \to H$ is a homomorphism, then it must be trivial.


问:
A category $I$ is filtered iff any colimits of $I$ in the category of sets commutes with any finite limits
A category is said to be filtered if:

it is not empty

for any objects $i, j$, there exists an object $k$ and
maps $i \to k, j \to k $.

for any objects $i, j$ and for any maps $u, v : i \to j $, there exists $j \to k$ such that the two maps $i \to k$ coincides.

The “only if” part is fundamental.
I want to show the “if” part.
The condition 2 is easy: it follows from the commutativity of filtered colimits with direct products.

But I can’t show 1 and 3.
I think that 3 follws from the commutativity with equalizers.

And 1. seems to be wrong:
The empty category satisfies the condition of the title, but does not satisfies the condition 1.

答:
Here's a general result, which covers the case of filtered colimits and finite limits.

Proposition. Let $\mathcal{I}$ and $\mathcal{J}$ be small categories. If colimits of shape $\mathcal{I}$ commute with limits of shape $\mathcal{J}$ in $\textbf{Set}$, then  every diagram $\mathcal{J}^\textrm{op} \to \mathcal{I}$ admits a cocone in $\mathcal{I}$.

Proof. Let $F : \mathcal{J}^\textrm{op} \to \mathcal{I}$ be a diagram and consider the diagram $\mathcal{I} (F -, -) : \mathcal{J} \times \mathcal{I} \to \textbf{Set}$.
We have a comparison map:
$$\textstyle \varinjlim_\mathcal{I} \varprojlim_\mathcal{J} \mathcal{I} (F -, -) \longrightarrow \varprojlim_\mathcal{J} \varinjlim_\mathcal{I} \mathcal{I} (F -, -)$$
For each object $i$ in $\mathcal{J}$, $\varinjlim_\mathcal{I} \mathcal{I} (i, -) \cong 1$.
(This is basically a consequence of the Yoneda lemma.)
Hence, the RHS is $\varprojlim_\mathcal{J} 1$, which is just $1$.
On the other hand, the LHS is a quotient of the disjoint union of $\varprojlim_\mathcal{J} \mathcal{I} (F -, j)$ as $j$ varies, so if the comparison map is a bijection  then there exists some object $j$ in $\mathcal{J}$ for which $\varprojlim_\mathcal{J} \mathcal{I} (F -, j)$ is not empty.
But $\varprojlim_\mathcal{J} \mathcal{I} (F -, j)$ is the set of cocones from $F$ to $j$, so we are done. ■

The above argument actually only uses the hypothesis that the comparison map is a surjection. If we use the hypothesis that the comparison map is a bijection, we get a stronger conclusion:

Proposition. Let $\mathcal{I}$ and $\mathcal{J}$ be small categories. If colimits of shape $\mathcal{I}$ commute with limits of shape $\mathcal{J}$ in $\textbf{Set}$, then for every diagram $F : \mathcal{J}^\textrm{op} \to \mathcal{I}$, the category ${}^{F /} \mathcal{I}$ of cocones under $F$ is (inhabited and) connected.

Finally, a word of warning: these are necessary conditions, but not sufficient.


问:
Struggling with a recursive expression
So I was looking into recursive expressions and recently dealt with this expression:

$$ f(i) = q \times (f(i-1)+1) + \bar{q} \times f(i-1) $$

where $q$ is a probability between $0$ and $1$ and $\bar{q} = 1-q$.

A non-recursive formula is quite simple ($ = q \times i$)

But when I change the recursive expression to:

$$ f(i) = q \times (f(i-1)+1) + \frac{\bar{q} \times f(i-1)}{2} $$

the solution I found becomes:

$$ \frac{q-q\times(q+\frac{\bar{q}}{2})}{1-(q+\frac{\bar{q}}{2})} $$

There might be other solutions though.
I do find it strange that a relative small change in the recursive expression leads to a significantly more complex non-recursive formula.

I now got another recursive expression:

$$ f(i) = q \times i + \bar{q} \times f(i-1) $$

The solution to this one (again, there are other ones) is:

$$ i-\frac{\bar{q}\times(1-\bar{q}^i)}{q} $$

But now when I do the same 'thing' as with the other expression:

$$ f(i) = q \times i + \frac{\bar{q} \times f(i-1)}{2} $$

I get completely stuck in finding a solution (i.e. a non-recursive formula).

Does anyone have an idea to solve this recursion or any hints?

答:
Since I am unfamiliar with generating functions, I am forced to ignore the comment of ancientmathematician, and proceed very inelegantly.

$$ f(i) = q \times i + \frac{\bar{q} \times f(i-1)}{2} $$

I get completely stuck in finding a solution (i.e. a non-recursive formula).

Assume that $f(0) = 0$ and let $r = \frac{1-q}{2}.$

Then:

$f(1) = q.$

$f(2) = 2q + rq.$

$f(3) = 3q + r(2q + rq) = 3q + 2rq + r^2q.$

Inductively assume that

$$f(n) = \sum_{k=1}^n qkr^{(n-k)}.$$

Then

$$f(n+1) ~=~ (n+1)q + \sum_{k=1}^n qkr^{([n+1] - k)}
~=~ \sum_{k=1}^{n+1} qkr^{([n+1]-k)}.$$

Title: Struggling with a recursive expression


问:
Results for elements in Dihedral Groups: $x \notin \langle r\rangle  \Rightarrow rx = xr^{-1}$
As a follow-up to my previous question, the next two exercises state:

Use the given generators and relations to show that if $x$ is any element of $D_{2n}$ which is not a power of $r$, then:

a) $rx = xr^{-1}$, and

b) $x$ has order 2, and hence $D_{2n}$ is generated by $s$ and $sr$, each with order 2.

Here, $D_{2n}$ has the "usual" presentation
$$D_{2n} = \left\langle r, s \mid r^n = s^2 = 1, rs = sr^{-1} \right\rangle.$$

For part a), since $x$ is not a power of $r$, I have written $ x = sr^k$ for some $0 \leq k \leq n - 1$.

So then $rx = r(sr^k) = (rs)r^k = (sr^{-1})r^k = sr^{k - 1}$, but this doesn't seem to be moving me in the right direction.

Should I be using induction on the power of $r$ in the element $x$ ?

Then for part b), I simply wrote $(sr^k)(sr^k) = s(r^ks)r^k = s(sr^{-k})r^k = s^2 = 1$, but I doubt that I have sufficiently justified my work.

From the comments to the previous question, I get the feeling that this is a very simple verification, but I'm having trouble.

Thanks in advance for your help!

答:
For (a), I reckon induction is the way to go. You have the case $k=0$ in the group presentation.

For (b), your argument looks sound. The only part you might want to think about is justifying $r^ks=sr^{-k}$, but that's an almost immediate consequence of (a).


问:
Integral independent of the parameter $y$
Consider the integral
$$\int_a^b \frac{K(x)}{1+yx^2}dx$$
where $a$, $b$ are fixed constants and $y$ is a parameter.

Question:

Is there a smooth function $K(x)\neq 0$,
s.t.
$$\int_a^b \frac{K(x)}{1+yx^2}dx=\int_a^b \frac{xK(x)}{1+yx^2}dx=0,$$
for arbitrary $y$?

For example, when $K(x)=x$ and $[a,b]=[-1,1]$
$$\int_{-1}^1 \frac{x}{1+yx^2}dx=\ln (1+yx^2)/(2y)|_{-1}^1=0$$
which is independent of $y$.

I don't know if there exists a function $K(x)$ satisfying the conditions in my question above.

答:
So you could try taking a derivative with respect to $y$ and commuting with the integral to derive a condition for independence. Set

$$P(y)=\int\limits_a^b\frac{K(x)}{1+yx^2}dx$$

Then if $\frac{dP}{dy}=0$ we have functional independence. Well assuming $a$ and $b$ are indepenent of $y$,

$$\frac{dP}{dy}=\int\limits_a^bK(x)\frac{\partial}{\partial y}\left(\frac{1}{1+yx^2}\right)dx=-\int\limits_a^b\frac{K(x)x^2}{(1+yx^2)^2}dx$$

So the condition becomes

$$\int\limits_a^b\frac{K(x)x^2}{(1+yx^2)^2}dx=0 \tag{1}$$

In particular if K(x) is chosen to be an odd function and the limits chosen such that $b=-a$ as you did in your example we can see $P(y)=0$ always regardless of the value of y, this is why your example works.

To answer this question in general however it might be a good idea to try an solve $(1)$. One way of doing this is to solve for the integrand being zero, however this results in $K(x)=0$. I do not suspect there is a solution to $(1)$ outside of the special cases I have mentioned for nonzero $K$ independent of $y$.

If you allow your limits to depend on $y$ then there will be much more freedom in the resulting condition for independence and thus more likely for such a K(x) to exist.


问:
Determining the local extreme values of a function
I am currently studying analysis and I am facing a problem with understanding how to determine the local extreme values of a function. I just started studying this, but our prof is already giving us mandatory homeworks with extremely hard examples and I am really struggling, if anyone could please help me it would be very helpful.

The function is:

I would really appreciate the help, I really need it.

答:
$\frac{\partial f}{\partial x}=3x^2-12=0$ $\Rightarrow x=2 $ or $x=-2$

$\frac{\partial f}{\partial y}=3y^2-3=0 \Rightarrow  y=1$ or $y=-1$

$(2,1), (2,-1), (-2,1) ,(-2,-1)$ are extreme points

$\frac{\partial^2 f }{\partial x^2}\frac{\partial^2 f }{\partial y^2}-(\frac{\partial^2 f }{\partial x \partial y})^2=36xy$

at point $(2,1) $ $\frac{\partial^2 f }{\partial x^2} >0$ and $\frac{\partial^2 f }{\partial x^2}\frac{\partial^2 f }{\partial y^2}-(\frac{\partial^2 f }{\partial x \partial y})^2=36xy>0$ which means in this point function reaches a local minimum

at point $(-2,-1)$ $\frac{\partial^2 f }{\partial x^2}< 0 \frac{\partial^2 f }{\partial x^2}\frac{\partial^2 f }{\partial y^2}-(\frac{\partial^2 f }{\partial x \partial y})^2=36xy>0 $ that means in this point function reaches a local maximum

and the others are saddle points because $\frac{\partial^2 f }{\partial x^2}\frac{\partial^2 f }{\partial y^2}-(\frac{\partial^2 f }{\partial x \partial y})^2=36xy<0$


问:
If $x$ and $y$ are nonzero vectors in $R^n$ whose lengths are equal to each other ,then $x-y$ is orthogonal to $x+y$.Is it true ,if true ,show it
If $x$ and $y$ are nonzero vectors in $R^n$ whose lengths are equal to each other, then $x-y$ is orthogonal to $x+y$.Is it true, if true, show it.

I am stuck in showing this question , it seems like true but I cannot prove it.

Can you help me?

答:
The inner product of $x-y$ and $x+y$ is $(x-y)^T(x+y)=x^Tx + x^Ty -y^Tx -y^Ty=|x|^2-|y|^2=0$

so, $(x-y)$ and $(x+y)$ are orthogonal


问:
support for composition of continuous function
Let $f:X\to \Bbb{R}$ be some continuous function from topological space $X$.Define $\varphi:X'\to X$ be some continuous function also.

If we further assume $\varphi$ is homeomorphism,can we claim $\text{supp}(f\circ \varphi) = \varphi^{-1}(\text{supp}(f)).$

I prove it as follows:Denote $Z_f= \{x\mid f(x) = 0\}$ ,then we can show $f\circ \varphi(\varphi^{-1}(Z_f)) = f(Z_f）,f\circ \varphi(\varphi^{-1}(Z_f^c)) = f(Z_f^c）$

Hence $Z^c_{f\circ \varphi} = \varphi^{-1}(Z_f^c)$ hence taking closure we have the desired result(since $\varphi $ is homeomorphism )

If we do not assume $\varphi$ is homeomorphism ,do we have relation between $\text{supp}(f\circ \varphi) $ and $ (\text{supp}(f)).$ ?

The converse direction is also not so hard,since any $x\notin \varphi^{-1}(\text{supp}(f))$ then exist some neiborhood $U_{\varphi(x)}$ that also not contains in $\text{supp}(f)$ , hence exist neiborhood of $x$ which is $\varphi^{-1}(U_{\varphi(x)})$ does not contains in the support of $f\circ\varphi$.Which means :

$$\text{supp}(f\circ \varphi) \subset \varphi^{-1}(\text{supp}(f))$$

This inclusion is strict in general taking $f(x) = x$ ,$\varphi(x') = 0$
then $\text{supp}(f) = \Bbb{R}$, then we have $\emptyset=\text{supp}(f\circ \varphi) \subset \varphi^{-1}(\text{supp}(f)) = \Bbb{R}$

Is my interpretation correct?

答:
Consider $X=\mathbb{R}$ (with the usual topology) and $f(x)=x$. Then $\operatorname{supp}(f)=\mathbb{R}$.

On the other hand, if $X'$ is the set of real numbers with the discrete topology and $\varphi$ is the identity map, then $\operatorname{supp}(f\circ\varphi)=\mathbb{R}\setminus\{0\}$.


答:
For the original problem
$$S_n=\sum_{k=0}^n \frac 1{n+k}$$ you also could using harmonic numbers since
$$S_n=\sum_{k=0}^n \frac 1{n+k}=\sum_{k=1}^{2n} \frac 1{k}-\sum_{k=1}^{n-1}\frac 1{k}= H_{2 n}-H_{n-1}$$ Now, the asymptotics of
$$H_p=\log (p)+\gamma +\frac{1}{2 p}-\frac{1}{12p^2}+O\left(\frac{1}{p^4}\right)$$ Applying twice and using long division to make it simpler
$$S_n=\log (2)+\frac{3}{4 n}+\frac{1}{16 n^2}+O\left(\frac{1}{n^4}\right)$$ and tis very few terms are always larger than $\frac 23$

Title: Prove that $\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n}>\frac{2}{3}$


问:
Exercise 1.5.7 (b) in Abbott’s Understanding Analysis
In the second edition, this exercise is as follows:

Find a 1-1 mapping from $S$, the open unit square $\{(x,y) : 0<x,y<1\}$, to the open interval $(0,1)$. Use the fact that each real number has a decimal expansion.

The answer I’ve been seeing the most, and is also the one I thought of, was to consider the decimal expansions of $x$ and $y$, $0.x_1x_2x_3...$ and $0.y_1y_2y_3...$, respectively. Map $(x,y)$ to the number in $(0,1)$ of the form $0.x_1y_1x_2y_2...$

Since numbers of the form $0.a_1a_2...a_n$ are equal to $0.a_1a_2...a_{n-1}(a_n-1)99999...$, I noticed that this creates the following issue:

Suppose I have $(.2,.7)$ which is equal to the pair $(.1999...,.6999...)$. Upon applying the previous mapping to both, they map to $.27$ and $.16999...$, respectively. Since $.16999...$ is the same as $.17$, this means the ordered pairs $(.1,.7)$ and $(.2,.7)$ map to the same element of $(0,1)$. So it isn’t 1-1? Is this problem avoided by just referring to $x$ and $y$‘s finite decimal expansion if it exists?

答:
Maybe "1-1 mapping" just means injective.

In this case you omit all "finite" decimals, and represent ${1\over2}$ as $0.49999\ldots\ $. Doing your construction you will then obtain a unique decimal for each pair $(x,y)\in S$, but you will not obtain all numbers in $(0,1)$. For example, the number $0.3510401050902060\ldots$ will not appear in the image.


答:
Hint:

Simplify the expression $\langle x-y, x+y\rangle$ using the following rules:

$\langle z, u+v\rangle = \langle z,u\rangle + \langle z, v\rangle$

$\langle u, v\rangle = \langle v, u\rangle$

$\langle \alpha u, v \rangle = \alpha \langle u, v\rangle$.

also, use the fact that $\langle x,x\rangle = \langle y,y\rangle$.

Title: If $x$ and $y$ are nonzero vectors in $R^n$ whose lengths are equal to each other ,then $x-y$ is orthogonal to $x+y$.Is it true ,if true ,show it

Body:
If $x$ and $y$ are nonzero vectors in $R^n$ whose lengths are equal to each other, then $x-y$ is orthogonal to $x+y$.Is it true, if true, show it.

I am stuck in showing this question , it seems like true but I cannot prove it.

Can you help me?


问:
If $Y_t=\int_0^t X_sdW_s$ where $\int_0^t X_s^2ds<\infty $ a.s., do we have that $\mathbb EY_t=0$?
Let $Y_t=\int_0^t X_sdW_s$. I know that if $\mathbb E\int_0^tX_s^2ds<\infty $, then $(Y_t)$ is a martingale, and $\mathbb EY_t=0$ for all $t$. But what if we only have $\int_0^t X_s^2ds<\infty $ ?

I know that in this case, $(Y_t)$ is a local martingale only, but do we have that $\mathbb E Y_t=0$ or not ? What I know, is thata there is a sequence of stopping time $\tau_n\nearrow \infty $ s.t. $(Y_{\tau_n\wedge t})$ is a martingale. Therefore, if $\mathbb E Y_{t\wedge \tau_n}=0$ for all $t$ and all $n$. But I think that we can't switch $\mathbb E$ and limit, so I think that that $\mathbb EY_{t\wedge \tau_n}$ won't hold, but I can't find a concrete example. Any idea ? And if it doesn't work, is there a condition on $Y=(Y_t)$ to have that $\mathbb E Y_t=0$ ?

答:
Obviously, $\int_0^t e^{2B_s^4}ds<\infty $ a.s. According to Itô formula,
$$\int_0^te^{B_s^4}\,\mathrm d B_s=\int_0^t e^{B_s^4}\,\mathrm d s-\int_0^t 2B_s^3e^{B_s^4}\,\mathrm d s.$$

Since the RHS is not integrable, $\mathbb E\left[\int_0^t e^{B_s^4}\,\mathrm d s\right]$ doesn't exist (see Wolfram Alpha).


问:
True or False: if $(a_n)^2$ converges and $a_n\le 0$ for every n, then $a_n$ converges.
My attempt: if $(a_n)^2$ converges, then it converges to a finite limit, then by the definition of the limit we get that : $|(a_n)^2 - L| < \epsilon $
 then I opened up the absolute value and got: $L-\epsilon < a_n^2 < L + \epsilon $. (I really wanted to take a square root and end it here but if any sides of the inequation has a negative value then my proof is false, I tried to come up with a counter example and I couldn't find one too, but I think here I can choose the $\epsilon $ as I like since the limit exists so if I took $\epsilon = \frac {L} 2$, I would have all positive and be able to take square root on all the inequality sides.
 I wanna make sure if my proof is correct, and if it is, then why choosing one specific $\epsilon $, holds for a proof.
 if it's not any explanation would be pretty much appreciated.
 Thanks in advance.

答: Suppose $a_n^2 \to L$. Note that $L \geq 0$ since $a_n^2 \geq 0$ for all $n$. Then by continuity of $\sqrt{\cdot}$, we obtain $-a_n= \sqrt{a_n^2} \to \sqrt{L}$ and thus $a_n \to -\sqrt{L}$.


问:
Parabolic projectile equation demonstration question
I was looking at a book of physics and, it will sound dumb, but while I know that the maximum height equation of a projectile is max=(v·senα)/2g, I can't understand how do you get there from 0=v·senα-gt. I need someone to explain it to me step by step.</p>

答:
Suppose you shoot an arrow with initial speed $v$ at the angle $\alpha$ to horizon. We also suppose that there's no interaction with air. We study the second Newton law applied to our arrow: $m\frac{d^2}{dt^2}{\vec x}=m\vec g$ or $ \frac{d^2}{dt^2}{\vec x}= \vec g$.

$\vec x = (h,l)$ - horizontal and vertical coordinate.
$\vec g$ has only vertical component , hence we study only the vertical part of our equation:

$\begin{cases}\frac{d^2}{dt^2}h(t) = -g\\ h'(0)=v\sin\alpha\\h(0)=0\end{cases}$

Apparently, $h(t)=tv\sin\alpha-gt^2/2 $. Now it's evident, that the maximum height is when $h'(t)=0$ and $h''(t)<0$ (i.e. the height was increasing before and will decrease after, and right now it doesn't change), which is true for $t_\ast = \frac{v\sin\alpha}{g}$. The height is $h(t_\ast) = \frac{(v\sin\alpha)^2}{2g}$.


问:
For what values of $x_0$, $x_n:=\frac{x_{n-1}}{2}+\frac{3}{x_{n-1}}$ where $n \geq 1$ with $x_0>0.$ converges?
$x_n:=\frac{x_{n-1}}{2}+\frac{3}{x_{n-1}}$ where $n \geq 1$ with $x_0>0.$

For what values of $x_0$ does this sequence converge?

If $(x_n) \to L$, then assuming $L>0$, I have
$L=\lim_\limits{n \to \infty} x_n=
\lim_\limits{n \to \infty}\left[\frac{x_{n-1}}{2}+\frac{3}{x_{n-1}}\right]=
\lim_\limits{n \to \infty}\left[\frac{x_{n-1}^2+6}{2x_{n-1}}\right]=
\frac{\lim_\limits{n \to \infty}x_{n-1}^2+6}{\lim_\limits{n \to \infty}2x_{n-1}}=
\frac{L^2+6}{2L} \implies 2L^2=L^2+6 \implies L=\sqrt{6}.$

How to determine what values of $x_0$ will give the convergence of $(x_n)$?

答:
$x_n > 0 $ for all $n$

$x_{n} \ge \sqrt{6}$ for all $n \ge 1$ by the AM-GM inequality

$x_n=\frac{x_{n-1}}{2}+\frac{3}{x_{n-1}}=x_{n-1}$ when $x_{n-1} = \sqrt{6}$

$x_n=\frac{x_{n-1}}{2}+\frac{3}{x_{n-1}}<x_{n-1}$ when $x_{n-1} > \sqrt{6}$

Basically its going decrease till $x_n = \sqrt{6}$

Therefore, it converges to $\sqrt{6}$ irrespective of the value of $x_0$


问:
conjugate of the sum of a real number and a complex number in polar form
I have an expression which I have a hard time understanding how they found the conjugate of $1 + e^{it}$ in

$\frac{1}{1 + e^{it}}$

In order to have a fraction without a complex denominator.

答: Since$$(e^{it})^\ast=(\cos t+i\sin t)^\ast=\cos t-i\sin t=e^{-it},$$the conjugate of $1+e^{it}$ is $1+e^{-it}$, so your fraction is$$\frac{1+e^{-it}}{(1+e^{it})(1+e^{-it})}=\frac{1+e^{-it}}{2+e^{it}+e^{i-t}}=\frac{1+e^{-it}}{2(1+\cos t)},$$which you're free to rewrite with more trigonometric identities. In fact$$1+e^{it}=2e^{it/2}\cos\tfrac{t}{2}\implies\tfrac{1}{1+e^{it}}=\tfrac12e^{-it/2}\sec\tfrac{t}{2}=\tfrac{1-i\tan(t/2)}{2}.$$


问:
How many elements of order p (p,q are primes) in finite abelian group $\mathbb{Z}_p \times \mathbb{Z}_p \times \mathbb{Z}_{q^3}$
How many elements of order p (p,q are primes) in finite abelian group $\mathbb{Z}_p \times \mathbb{Z}_p \times \mathbb{Z}_{q^3}$?

My answer is:

Edit:

As p is a prime we have p-1 elements of order p in a cyclic group

So we have $(p-1)^2 -1 $

Is this correct?

答:
If $p\not = q$:

The subgroup $H_1=\mathbb{Z}_p \times 1 \times 1$ contains already $p-1$ elements of order $p$. The same for $H_2=1 \times \mathbb{Z}_p \times 1$, as those groups are isomorphic to $\mathbb{Z}_p$.

This is due to the fact, that a cyclic group of order $n$ contains $\varphi(n)$ generators of order $n$, where $\varphi$ is the euler-totient-function.

Now any non-trivial element of $H_1H_2$ has order p, which are $p^2-1$ elements.


答:
The key fact here is that multiplying a complex number by its conjugate gives a real number, so it makes sense to set the conjugate of $1+e^{it}$ to be $\frac{a}{1+e^{it}}$ for some (yet to be determined) real number $a$.

To do this, use the fact that their moduli have to be equal, i.e. $$|1+e^{it}| =\left| \frac{a}{1+e^{it}} \right|\implies a= |1+e^{it} | ^2 \\ = (1+\cos t)^2 +\sin^2 t = 2(1+\cos t) = 4\cos^2 \frac t2 $$ So, $$(1+e^{it})^*= \frac{4\cos^2\frac t2}{1+e^{it}} $$

Title: conjugate of the sum of a real number and a complex number in polar form

Body:
I have an expression which I have a hard time understanding how they found the conjugate of $1 + e^{it}$ in

$\frac{1}{1 + e^{it}}$

In order to have a fraction without a complex denominator.


答:
You can treat the case $L=0$ separately using $\epsilon'=\sqrt{\epsilon}$, so let assume $L>0$.

WLOG you can set $b_n=-a_n$ to work with all positive quantities and get rid of annoyances with sign and inequalities.

Let set $L=\ell^2$ with $\ell>0$, we get

$$|b_n-\ell|=\dfrac{\overbrace{|{b_n}^2-\ell^2|}^{\to 0}}{\underbrace{|b_n+\ell|}_\text{bounded}}\le \frac 1\ell|{b_n}^2-L|\to 0$$

The proof essentially relies on showing that the denominator is bounded but this is trivial since $|b_n+\ell|=b_n+\ell\ge \ell>0$ using the positivity of $b_n$.

You can try writing the proof directly with $a_n$ but it requires a bit of gymnastic to deal with inequalities properly.

Title: True or False: if $(a_n)^2$ converges and $a_n\le 0$ for every n, then $a_n$ converges.

Body:
My attempt: if $(a_n)^2$ converges, then it converges to a finite limit, then by the definition of the limit we get that : $|(a_n)^2 - L| < \epsilon $
 then I opened up the absolute value and got: $L-\epsilon < a_n^2 < L + \epsilon $. (I really wanted to take a square root and end it here but if any sides of the inequation has a negative value then my proof is false, I tried to come up with a counter example and I couldn't find one too, but I think here I can choose the $\epsilon $ as I like since the limit exists so if I took $\epsilon = \frac {L} 2$, I would have all positive and be able to take square root on all the inequality sides.
 I wanna make sure if my proof is correct, and if it is, then why choosing one specific $\epsilon $, holds for a proof.
 if it's not any explanation would be pretty much appreciated.
 Thanks in advance.


问:
Are two graphs isomorphic?
Are the two graphs isomorphic?

$$G_1=\begin{bmatrix}
a & b & c & d & e & f \\
b & a & a & c & d & a \\
c & c & b & e & f & d \\
f &   & d & f &   & e
\end{bmatrix}\
\quad\quad
G_2=\begin{bmatrix}
u & v & w & x & y & z \\
v & u & v & u & x & u \\
x & w & x & w & z & w \\
z &   & z & y &   & y
\end{bmatrix}$$

$a,b,c$ creates a triangle in $G_1$, but no triangle is created in $G_2$.

Is that enough to claim that they are not isomorphic?

答:
"Isomorphic" does not only mean that the (sorted) degree-sequences must coincide.

For every set of vertices in one graph there must be a corresponding set of vertices in the other graph with the property that two vertices in one set are connected if and only if the corresponding vertices in the other set are connected.

In short, every subgraph must occur also in the other graph, if they are isomorphic.

So, if one graph contains a triangle, quadrilateral, pentagonal and so on, but not the other, they cannot be isomorphic.

Title: Are two graphs isomorphic?


问:
is there a continuous $f\colon [1,\infty) \to [0,\infty)$ where $\lim f$ is $0$ at $\infty$ but integral of $f(x) / x$ is divergent?

I've been struggling with finding a counter example to a question regarding to the limit comparison test.

The false statement is that for every $f(x)$ that is defined and continuous $[1, \infty) \to [0, \infty)$, if : $$\lim_{x \to \infty} f(x) = 0$$ then the following integral exists (does not diverge):
$$\int_1^\infty\frac{f(x)} xdx$$

I've been trying to use weird functions like $\frac{\ln(x)}{x^{0.1}}$ and to choose a function that will make $\frac{f(x)}x$ have an asymptote at $x=1$ but couldn't find anything robust to counter this false statement.

Thanks in regards!

答:
Here is one: $f\colon[1,\infty)\to [0,\infty)$ defined by
$$
f(x) = \frac{1}{\ln(1+x)}
$$
Note that
$$\begin{align*}
\int_1^\infty \frac{f(x)}{x}dx &= \int_1^\infty \frac{1}{x\ln(1+x)}dx\\
&\geq \int_1^\infty \frac{1}{(1+x)\ln (x+1) }dx \\
&= \int_2^\infty \frac{1}{x\ln x }dx = \infty
\end{align*}$$
since a primitive of $\frac{1}{x\ln x}$ is $\ln\ln x$.


问:
Is $X:=\{a\}$ Hausdorff space?
I wonder if $X:=\{a\}$ is a Hausdorff space.

Definition;

$A$ is a Hausdorff space $\iff
\forall x,y \in A(x\neq y);
\exists \text{ open set } U,V s.t. x\in U,y\in V, U\cap V=\phi.$

But we cannot pick up two elements from $X=\{a\}.$

How should I judge whether $X$ is a Hausdorff space or not?

答:
It becomes easier when you understand what $\forall x,y\in A(x\neq y)$ really means here. This is an example where the shorthand notation may cause confusion (whereas it usually comes to clarify over-formality).

When we write $(\forall x\in A)\varphi(x)$, what we really mean to say is $\forall x(x\in A\to\varphi(x))$, so that if $x\in A$, then $\varphi(x)$, but if $x\notin A$, then the statement is vacuously true. This extends to the $(x\neq y)$ part in this case. Since we're quantifying over elements of $A$ anyway let me omit it from the notation at the moment, and just write: $$\forall x,y(x\neq y\to\exists U,V(U,V\text{ are open}\land x\in U\land y\in V\land U\cap V=\varnothing)).$$

In other words, if $x\neq y$, then there are disjoint open sets, etc. Now it's easy to see what happens with a singleton: the implication is vacuously true, there are never any $x\neq y$.

Title: Is $X:=\{a\}$ Hausdorff space?


问:
How to find the $\delta$ for some $\varepsilon$ such that: $|x-0| < \delta \Rightarrow |g(x)-g(0)|<\varepsilon$
$g(x)$ is defined as:

$$g(x)=\frac{\sqrt{x+1}-1}{x};\  x\in [-1,0)\ \cup\ (0,\infty)$$

$$g(x)=\frac{1}{2};\ x=0$$

How can I find the $\delta$ for some $\varepsilon$ such that: $|x-0| < \delta \Rightarrow |g(x)-g(0)|<\varepsilon$.

I tried a lot of things, but I seem to be completely stuck. Rationalizing the numerator doesn't help, because x cancels out.

答:
Hint:
$$|g(x) - g(0)| = \left|\frac{\sqrt{x+1}-1}{x} - \frac{1}{2}\right| = \left|\frac{1}{\sqrt{x+1}+1} - \frac{1}{2} \right| < \varepsilon$$
Solving the inequality for $x$ you get
$$\varepsilon - \frac{1}{2} < \frac{1}{\sqrt{x+1}+1} < \varepsilon + \frac 12 \\
\frac{1}{\varepsilon + 1/2} < \sqrt{x+1} + 1 < \frac{1}{\varepsilon - 1/2} \\
\left(\frac{1/2 - \varepsilon}{\varepsilon + 1/2}\right)^2 -1 < x < \left(\frac{3/2-\varepsilon}{\varepsilon - 1/2} \right)^2 - 1$$


答:
As Professor Vector noted in the comments, you can apply AM-GM inequality for all $n\geq 1$:
$$x_n = \frac{x_{n-1}}{2}+ \frac 3{x_{n-1}} \geq 2 \sqrt{\frac{x_{n-1}}{2}\cdot \frac 3{x_{n-1}}} = \sqrt 6.$$
(To apply AM-GM, you need an observation that $x_n>0$, for all $n$, since $x_0>0$.)

Therefore, we can disregard $x_0$ from the sequence and look at $(x_n)_{n\geq 1}$ instead - convergence doesn't change, but we now have $x_n\geq \sqrt 6,\ n\geq 1$.

Now, notice that $x_n\geq \sqrt 6$ implies
$$x_{n+1} = \frac{x_n}{2}+\frac{3}{x_n} \leq \frac{x_n}{2} + \frac 3{\sqrt 6}\leq x_n.$$ (The last inequality is equivalent to $x_n\geq \sqrt 6.$)

Thus, the sequence $(x_n)_{n\geq 1}$ is decreasing and bounded from below, hence convergent. Therefore, the original sequence is convergent as well, for all values $x_0 > 0$.

Title: For what values of $x_0$, $x_n:=\frac{x_{n-1}}{2}+\frac{3}{x_{n-1}}$ where $n \geq 1$ with $x_0>0.$ converges?

Body:
$x_n:=\frac{x_{n-1}}{2}+\frac{3}{x_{n-1}}$ where $n \geq 1$ with $x_0>0.$

For what values of $x_0$ does this sequence converge?

If $(x_n) \to L$, then assuming $L>0$, I have
$L=\lim_\limits{n \to \infty} x_n=
\lim_\limits{n \to \infty}\left[\frac{x_{n-1}}{2}+\frac{3}{x_{n-1}}\right]=
\lim_\limits{n \to \infty}\left[\frac{x_{n-1}^2+6}{2x_{n-1}}\right]=
\frac{\lim_\limits{n \to \infty}x_{n-1}^2+6}{\lim_\limits{n \to \infty}2x_{n-1}}=
\frac{L^2+6}{2L} \implies 2L^2=L^2+6 \implies L=\sqrt{6}.$

How to determine what values of $x_0$ will give the convergence of $(x_n)$?


答:
Alternative (very informal) approach.  The (complex) conjugate of the complex number $z$ is the number $\overline{z}$ that represents the mirror image of $z$ with respect to the $x$-axis.

Further when you have any number $w \in \mathbb{C}$ and any number $r \in \mathbb{R}$ then $(w + r)$ is simply the horizontal translation of $w$, by $r$ units (i.e. if $r < 0,$ then the translation is to the left).

Also, you have that for any two complex numbers $w_1, w_2$ then $\overline{(w_1 + w_2)} ~=~ \overline{w_1} + \overline{w_2}.$

Finally, you have that for any real number $r,$ the complex conjugate $\overline{r} = r.$

Putting this all together, it is immediate that

$$\overline{1 + e^{it}} = 1 + e^{-it}.$$


问:
Characterization of groups of order 110 (understanding semidirect product)
I try to characterize all groups $G$ of order $110=2 \cdot 5 \cdot 11$.

Sylow implies the existence of a subgroup $N \trianglelefteq G$ of order $55$ and a subgroup $H$ of order $2$. It follows, that $G$ is a semidirect product $N \rtimes H$.

As there exists only two groups of order $55$, namely $C_{55}$ and $C_{11}\rtimes C_5$, it follows that $G$ is of the form $G_1=C_{55} \rtimes C_2$ or $G_2=(C_{11}\rtimes C_5) \rtimes C_2$.

Via GAP I found that there exist $4$ groups of the form $G_1$. I was wondering how this is possible. To my understanding, the semidirect product ist defined by an homomorphism $\phi : C_2 \mapsto Aut(C_{55})=(C_{55})^\times=C_{18}$. Of course there is the trivial homomorphism leading to the group $C_{110}$. Am I right, that there exists only one other homomorphism, sending $1$ to $9$, which leads to $D_{110}$?

How is it possible, that there are two more semidirect products of this kind, abstractly $C_5 \times D_{22}$ and  $C_{11} \times D_{10}$, both with normal $C_{55}$-Subgroup. To which mapping do belong?

Furthermore I would like to know how it can be seen, that there exist only two groups of type $G_2$, namely $(C_{11}\rtimes C_5) \times C_2$ and $(C_{11}\rtimes C_5) \rtimes C_2 \cong C_{11} \rtimes C_{10}$. Do I need to calculate the autmorphism group of $C_{11}\rtimes C_5$?

Thank you for your help!

答:
In "Die Gruppen der Ordnungen p^3,pq^2,pqr,p^4", O.Hoelder, Math. Annalen, XLIII, 1894

you can find all necessary informations.

Shortly, the are six groups (GAP lists all these groups). They are

Z110

D110

Z5 x D22

Z11 x D10

Z2 x  (Z11 semidirect Z5)

Z11 semidirect Z10

D. Kaesbauer

dieter.kaesbauer@googlemail.com


问:
An exact and an approximate confidence interval for a Poisson distribution
$X_{1}, ..., X_{10}$ ~ $Pois(\theta)$

Observations:
$x_{1} = x_{3} = x_{6} = x_{8} = x_{9} = 0$; $x_{2} = x_{5} = x_{10} = 1$; $x_{4} = 2$; $x_{7} = 3$.

I want to determine an exact (numerical) and an approximate (numerical) confidence interval for $\theta$ of confidence level 0.9:

exact confidence interval: I don't know how to do this. On the internet I can only find things about an approximate confidence interval.

approximate confidence interval:

$X = (X_{1}, ..., X_{10})$

MLE:  $\hat{\theta} = \bar{X} = 0.8$

And: $Var(X) = \theta$. So an estimate for the variance is also $\bar{X} = 0.8$.

Then according to the answer: An approximate confidence interval is $\bar{X}$ $\pm$ $\sqrt{\bar{X}/n}$ $\xi_{1 - \alpha/2}$.

But this means that: T = $\sqrt{n}$ $\frac{\theta - E[X]}{\sqrt{Var(X)}}$ ~ N(0,1).
But this does not hold for X ~ Pois($\theta$), right?
Because a Poisson distribution does not have to be symmetric.

答:
Actually, according to the central limit theorem, it does (in terms of limit distribution). You can check that the poisson distribution checks every requirement :
$X_1,..X_n $ are independant, iid, $\bar{X}, \sigma$ are finite and $\sigma\neq0 $


问:
Proving $\,2\sqrt 2 + 1\,$ is irrational by contradiction
I am working on some review questions for my Discrete Structures final and I needed some assistance for the problem "Prove by contradiction that $\,2\sqrt 2 +1$ is irrational".

Now I know how prove that $\sqrt 2$ is irrational on its own, by showing that $\sqrt 2 = a/b$ and showing they are not in lowest terms. My issue is that other stuff is messing up my proof and I am not sure how to solve it.

Here is what I have so far. Assume that $\,2\sqrt 2 +1$ is rational, therefore $\,2\sqrt 2 +1 =$ $\frac {a}{b}$. Then:
$$2 \sqrt 2 +1 = \frac {a}{b}$$
$$ \sqrt 2 = \frac {a-b}{2b}$$
$$ a = 2(b \sqrt 2 + \frac{b}{2})$$
$$ a= 2k$$
$$\sqrt 2 = \frac{2k-b}{2b}$$

And since they are not in lowest terms therefore $\sqrt 2$ is irrational.

Any help would greatly appreciated. Thanks.

答: Let $\mathbb{Q}$ denote the set of rational numbers. If $2\sqrt2+1 = x$ (say) is in $\mathbb{Q}$, then $(x-1)/2$ is in $\mathbb{Q}$, and so $\sqrt2$ is in $\mathbb{Q}$, which is a contradiction.


问:
Show that $\sqrt{\frac{1+2\sin\alpha\cos\alpha}{1-2\sin\alpha\cos\alpha}}=\frac{1+\tan\alpha}{\tan\alpha-1}$
Show that $\sqrt{\dfrac{1+2\sin\alpha\cos\alpha}{1-2\sin\alpha\cos\alpha}}=\dfrac{1+\tan\alpha}{\tan\alpha-1}$ if $\alpha\in\left(45^\circ;90^\circ\right)$.

We have $\sqrt{\dfrac{1+2\sin\alpha\cos\alpha}{1-2\sin\alpha\cos\alpha}}=\sqrt{\dfrac{\sin^2\alpha+2\sin\alpha\cos\alpha+\cos^2\alpha}{\sin^2\alpha-2\sin\alpha\cos\alpha+\cos^2\alpha}}=\sqrt{\dfrac{(\sin\alpha+\cos\alpha)^2}{(\sin\alpha-\cos\alpha)^2}}=\sqrt{\left(\dfrac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}\right)^2}.$

Using the fact that $\sqrt{a^2}=|a|$ the given expression is equal to $\left|\dfrac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}\right|.$ I think that in the inverval $\left(45^\circ;90^\circ\right) \sin\alpha>\cos\alpha$ but how can I prove that? What to do next?

答:
Apparently your second question hasn't been answered, about what to do next. Here's what to do:
$$\dfrac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}\equiv\frac{\frac{1}{\cos \alpha}}{\frac{1}{\cos \alpha}}\times\dfrac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}\equiv\dots$$

I hope that's helpful.

If you need any more help please don't hesitate to ask.


问:
bayesian inference -- multinomial likelihood with unknown number of categories.
It is simple to do the Bayesian inference using a Multinomial likelihood with an Dirichlet conjugate prior. However, what if the $k$ categories in the multinomial likelihood is unknown, and the number of $k$ is changing (increasing in my case..) depends on new oberservations.

I guess there must be some papers doing this kind of inference, any suggestions? Thanks.

答:
I found this paper answers the question well:

https://www.sciencedirect.com/science/article/pii/S0888613X04001069


问:
Why does the Minus-1 trick to get particular and general solutions work?
Minus-1 trick: A practical trick where the reduced row-echelon form of a system of equations is augmented with -1 rows. It is used to get particular and general solutions.

While it is a handy trick, I was wondering why does it work.

The following is an example of the minus-1 trick from the book Mathematics for Machine Learning Book

答:
In the example, set the non-pivot variables to $x_2=-\lambda_1$, $x_5=-\lambda_2$ and solve the system for other variables. You will get the exact same result.

The minus-1 trick is just a schematic way to achieve the same with less writing.

Another way to look at this: The columns $1$, $3$ and $4$ of $A$ are the standard basis of the column space. Using these you can uniquely express the remaining two columns:
\begin{align*}
\color{blue}{\begin{pmatrix} 3\\0\\0\end{pmatrix}} &= 3 \begin{pmatrix}1\\0\\0\end{pmatrix} + 0 \begin{pmatrix}0\\1\\0\end{pmatrix} + 0 \begin{pmatrix}0\\0\\1\end{pmatrix},
\\
\color{blue}{\begin{pmatrix} 3\\9\\-4\end{pmatrix}} &= 3 \begin{pmatrix}1\\0\\0\end{pmatrix} + 9 \begin{pmatrix}0\\1\\0\end{pmatrix} -4 \begin{pmatrix}0\\0\\1\end{pmatrix}.
\end{align*}
We can rewrite these two equations as
\begin{align*}
3 \begin{pmatrix}1\\0\\0\end{pmatrix} \color{blue}{{}- 1 \begin{pmatrix} 3\\0\\0\end{pmatrix}} + 0 \begin{pmatrix}0\\1\\0\end{pmatrix} + 0 \begin{pmatrix}0\\0\\1\end{pmatrix} + \color{red}{0 \begin{pmatrix} 3\\9\\-4\end{pmatrix}} &= \begin{pmatrix}0\\0\\0\end{pmatrix},
\\
3 \begin{pmatrix}1\\0\\0\end{pmatrix} + \color{red}{0 \begin{pmatrix} 3\\0\\0\end{pmatrix}} + 9 \begin{pmatrix}0\\1\\0\end{pmatrix} - 4 \begin{pmatrix}0\\0\\1\end{pmatrix} \color{blue}{{}- 1 \begin{pmatrix} 3\\9\\-4\end{pmatrix}} &= \begin{pmatrix}0\\0\\0\end{pmatrix},
\end{align*}
Hence, we have two (linearly independent) solutions to $Ax=0$ given by
$$
\begin{pmatrix} 3\\\color{blue}{-1}\\0\\0\\\color{red}{0} \end{pmatrix},
\begin{pmatrix} 3\\\color{red}{0}\\9\\-4\\\color{blue}{-1} \end{pmatrix}.
$$
Again, the minus-1 trick is just a schematic way to achieve the same result mechanically.


问:
In a topology space, a set is open iff the complement is closed?
I am learning topology spaces and in the notes it states a theorem:

For a topology space $(X,\tau)$, and $A\subseteq X$, $A$ is closed iff $X\setminus A$ is open.

In the notes, the definition of topology is that it should follow the following three rules:

The empty set and the whole set should be in the topology.

Finite, countable, and uncountable union of elements in the topology is still in the topology.

Finite intersection of elements in the topology is still in the topology

And in the notes it says that the proof of this theorem is similar for that of metric space, but we don't have distance in metric space, so we cannot order all set of open sets containing $x$ to get a limit and have a contradiction. I have the proof for cases if $\tau$ is finite and countable, how to prove the uncountable case?

Also, in the notes, the definition of closed set is that any limit of any convergent sequence is still in the set, so I need to prove this definition is equivalent to what the theorem states.

答: You cannot prove it, since that's not a theorem. It's a definition instead. That is how we define a closed set in a topological space.


答: You've got it wrong; $\mathbb Z_{55}^×$ has order $\varphi(55)=40$ and is isomorphic to $\mathbb Z_4×\mathbb Z_{10}$. The three non-direct semidirect products are derived from the three elements of order $2$ in $\mathbb Z_{55}^×$: $34$, $21$ and $-1$.

Title: Characterization of groups of order 110 (understanding semidirect product)

Body:
I try to characterize all groups $G$ of order $110=2 \cdot 5 \cdot 11$.

Sylow implies the existence of a subgroup $N \trianglelefteq G$ of order $55$ and a subgroup $H$ of order $2$. It follows, that $G$ is a semidirect product $N \rtimes H$.

As there exists only two groups of order $55$, namely $C_{55}$ and $C_{11}\rtimes C_5$, it follows that $G$ is of the form $G_1=C_{55} \rtimes C_2$ or $G_2=(C_{11}\rtimes C_5) \rtimes C_2$.

Via GAP I found that there exist $4$ groups of the form $G_1$. I was wondering how this is possible. To my understanding, the semidirect product ist defined by an homomorphism $\phi : C_2 \mapsto Aut(C_{55})=(C_{55})^\times=C_{18}$. Of course there is the trivial homomorphism leading to the group $C_{110}$. Am I right, that there exists only one other homomorphism, sending $1$ to $9$, which leads to $D_{110}$?

How is it possible, that there are two more semidirect products of this kind, abstractly $C_5 \times D_{22}$ and  $C_{11} \times D_{10}$, both with normal $C_{55}$-Subgroup. To which mapping do belong?

Furthermore I would like to know how it can be seen, that there exist only two groups of type $G_2$, namely $(C_{11}\rtimes C_5) \times C_2$ and $(C_{11}\rtimes C_5) \rtimes C_2 \cong C_{11} \rtimes C_{10}$. Do I need to calculate the autmorphism group of $C_{11}\rtimes C_5$?

Thank you for your help!


问:
Expected measure of a ball in a probability space with a metric
Assume we are given a probability space $(\mathbb{X}, \mathcal{X}, \mathbb Q)$ and a measurable distance function defined on it $d:\mathbb{X}\times \mathbb{X}\to \mathbb{R}^+\cup\{0\}$ that conforms to the usual definition of distance on metric spaces. I am trying to understand (in lay terms) the "expected value of the measure of a ball with fixed radius $r> 0$ centered at a randomly chosen point in $\mathbb{X}$ with measure $\mathbb{Q}$". That is more formally
\begin{equation}
\int_{\mathbb{X} }q(x)\mathbb{Q}(\mathrm{d}x)
\end{equation}
where $q(x) = \mathbb{Q}(B(x,r))$ and $B(x,r) = \{y\in \mathbb{X}: d(x,y)\leq r\} $. The first thing to ask here is if $q(x)$ is measurable but for now I assume $\mathcal{X}$ is large enough that it is.

My problem originates from this question: "find a lower bound for the probability that a randomly chosen iid set of points in a space are close-by". I am wondering what assumptions are needed from $\mathbb{Q}$ and $d$ such that this can be lower bounded. Of course I would not expect a bound valid for all $\mathbb{Q}$ and $d$ but any tip about how to treat this integral (or you know if anybody thought about this before and gave it a name so I can search) would be helpful. The trivial bound $\inf_{x\in\mathbb X} \mathbb{Q}(B(x,r))$ is not useful for my purposes.

答:
How is it defined the probability density function $\mathbb{Q}$? Is it discrete or continuous?

If it's continuous then $\lim_{r\rightarrow0}\int_{B_r}q(x)dx=0$

And if it's discrete then $\lim_{r\rightarrow0}\int_{B_r}q(x)dx=q(y)$


问:
If $A= \begin{pmatrix} -4 & 1\\ 3 & 2 \end{pmatrix}$ find $f(A)$ if $f(x) =x^2 - 2x+3$
If $A= \begin{pmatrix} -4 & 1\\ 3 & 2 \end{pmatrix}$ find $f(A)$ if $f(x) =x^2 - 2x+3$.

Actually while solving for $f(A)$, my teacher said to change $3$ into $3I$ where $I$ is an identity matrix, so that matrix can add up but I am still confused about how can an integer change into the matrix and what is its significance?

答:
I agree ... arbitrarily changing a number into a matrix doesn't make much sense, and you're right to question it.

In my opinion, the problem is just poorly written, and you shouldn't worry about that. The function $f$ is pointless, and doesn't need to be mentioned at all. What the problem is really asking you to do is calculate $A^2 - 2A + 3I$, and I expect you know how to do that.


问:
How to evaluate the following (rather complex) integral between $x=0$ and $x=\infty$
I've found a function that describes the concentration function of a polymer around a sphere. I want to evaluate the following integral that contains that function but I'm not sure how to go about it. Can someone send me in the right direction/help me out. Thanks in advance.

$$\displaystyle\int_0^\infty4\pi(x+R)^2\left(1-\left(\frac{\frac{x}{R}+\tanh{\frac{x+p}{D}}}{\frac{x}{R}+1}\right)^2\right)dx$$

答:
Note that
$$\frac{\frac{x}{R}+\tanh{\frac{x+p}{D}}}{\frac{x}{R}+1}\equiv\frac{x+R\tanh\frac{x+p}{D}}{x+R}$$
$$\implies1-\left(\frac{\frac{x}{R}+\tanh{\frac{x+p}{D}}}{\frac{x}{R}+1}\right)^2\equiv1-\frac{x^2+2Rx\tanh\frac{x+p}{D}+R^2\tanh^2\frac{x+p}{D}}{(x+R)^2}$$
$$\equiv\frac{2Rx+R^2-2Rx\tanh{\frac{x+p}{D}}-R^2 \tanh^2{\frac{x+p}{D}}}{(x+R^2)}$$
$$\implies\pi(x+R)^2\left(1-\left(\frac{\frac{x}{R}+\tanh{\frac{x+p}{D}}}{\frac{x}{R}+1}\right)^2\right)\equiv\pi(x+R)^2\frac{2Rx+R^2-2Rx\tanh{\frac{x+p}{D}}-R^2 \tanh^2{\frac{x+p}{D}}}{(x+R^2)}$$
which simplifies nicely, and then the integral is easy to evaluate.

If you need anymore help please ask :)


问:
Results for elements in Dihedral Groups: $x \notin \langle r\rangle  \Rightarrow rx = xr^{-1}$
As a follow-up to my previous question, the next two exercises state:

Use the given generators and relations to show that if $x$ is any element of $D_{2n}$ which is not a power of $r$, then:

a) $rx = xr^{-1}$, and

b) $x$ has order 2, and hence $D_{2n}$ is generated by $s$ and $sr$, each with order 2.

Here, $D_{2n}$ has the "usual" presentation
$$D_{2n} = \left\langle r, s \mid r^n = s^2 = 1, rs = sr^{-1} \right\rangle.$$

For part a), since $x$ is not a power of $r$, I have written $ x = sr^k$ for some $0 \leq k \leq n - 1$.

So then $rx = r(sr^k) = (rs)r^k = (sr^{-1})r^k = sr^{k - 1}$, but this doesn't seem to be moving me in the right direction.

Should I be using induction on the power of $r$ in the element $x$ ?

Then for part b), I simply wrote $(sr^k)(sr^k) = s(r^ks)r^k = s(sr^{-k})r^k = s^2 = 1$, but I doubt that I have sufficiently justified my work.

From the comments to the previous question, I get the feeling that this is a very simple verification, but I'm having trouble.

Thanks in advance for your help!

答:
Why do you think you are not going in the right direction in part (a)? You've shown that $rx = sr^{k-1}$. Now, how much is $xr^{-1}$? Since $x=sr^k$, then $xr^{-1} = (sr^k)r^{-1} =\ldots$

Remember what you are trying to prove!

For part (b), you have to show that
$$r^ks = sr^{-k}$$
which can be done by induction on $k$. Alternatively, why not use part (a)? You have $x=sr^k$, so then
$$(sr^k)^2 = (sr^{k-1})r(sr^{k-1})r = sr^{k-1}sr^{k-1}r^{-1}r = (sr^{k-1})^2.$$
Repeating this until $k=0$ we get $x^2=1$.

Once you do that, you've correctly shown that $x$ has order dividing 2; now simply note that it cannot have order $1$ (it's not the identity, because the identity does lie in $\langle r\rangle$), so it has order exactly $2$.

But you aren't done, because you haven't addressed the second part. Can you show that $s$ and $sr$ generate $D_{2n}$? And that they each have order $2$? (The latter part is what you just finished; the former is easy: just show that you can obtain all elements from a generating set by using $s$ and $sr$).


问:
Why do you use the inverse matrix to find the image of a curve in the plane?
Given a $2\times2$ linear transformation matrix and the equation of a curve in the plane (e.g. $x^2+y^2=1$), why does one use the inverse matrix to find the equation of the image of the equation after the transformation?

I recently watched a video that explained how $2\times2$ matrices represent linear transformations in the plane.  For example, if I wanted to find the image of the point $(1,0)$ after the transformation given by the matrix

$$
\left[\begin{matrix}2&1\\-1&0\\\end{matrix}\right],
$$

I would simply multiply:

$$
\left[\begin{matrix}2&1\\-1&0\\\end{matrix}\right]\left[\begin{matrix}1\\0\\\end{matrix}\right]=\left[\begin{matrix}2\\-1\\\end{matrix}\right]\rightarrow(2,-1).
$$

To extend this idea, I wanted to see the image of the circle $x^2+y^2=1$ after the transformation given by the same matrix.  I figured that since

$$
\left[\begin{matrix}2&1\\-1&0\\\end{matrix}\right]\left[\begin{matrix}x\\y\\\end{matrix}\right]=\left[\begin{matrix}2x+y\\-x\\\end{matrix}\right],
$$

the equation of the transformed circle would be $(2x+y)^2+(-x)^2=1$.  But after testing a few points on the circle, I quickly realized that it was not the correct equation.  Instead, the correct equation turned out to be

$$
(-y)^2+(x+2y)^2=1,
$$

which is obtained by multiplying $x$ and $y$ by the inverse matrix:

$$
\left[\begin{matrix}2&1\\-1&0\\\end{matrix}\right]^{-1}\left[\begin{matrix}x\\y\\\end{matrix}\right]=\left[\begin{matrix}0&-1\\1&2\\\end{matrix}\right]\left[\begin{matrix}x\\y\\\end{matrix}\right]=\left[\begin{matrix}-y\\x+2y\\\end{matrix}\right]
$$

So why do we use the original matrix to find the coordinates of the image of individual points (i.e. $(1,0)\rightarrow(2,-1)$), but the inverse matrix to find the equation of the image of the equation of a curve?  Is there a geometric explanation for this (or how do I explain this to my students)?

答:
A nice way to think about it is to consider the transformed curve as a geometric locus. Fist of all let the starting curve be $S$ and the transformed curve be $C$ and $\alpha$ the transformation itself. What we need to obtain is a function $f(X,Y)=0$ defined as the set $C$ of points $(X,Y) \implies (x,y) \in S$, namely:  $(X,Y) \in (C) \iff (x,y) \in (S)$.
Generally, when we calculate a geometric locus, we choose an equation that implies that a point is on the curve, and then we write the variables in that equation in terms of $x$ and $y$.
For example: the condition for a point to be on the unit circle is $d=1$, where $d$ is the distance from the origin. At this point we start writing the distance from the origin in terms of $(x,y)$.
For the transformed curve it is exactly the same. The condition for a point to be on the transformed curve is: $(x,y) \in S$, which is the equation of the original curve. Now we just need to write $(x,y)$ in terms of $(X,Y)$, and the job is done :)


问:
Decomposing a polynomial ring as a graded ring
Let's say I have a ring $A$. Then if we want to decompose $A[x]$ as a graded ring, then correct me if I'm wrong, but the construction goes as follows - we have an isomorphism
$$A[x] \cong \bigoplus_{d \geq 0}A_d$$
where $A_0 = A$ (technically this is identifying the image of the injection $A \hookrightarrow A[X]$ with $A$) and $A_d = \left(x^d\right)$.

But if this were the case then there would be elements of the direct sum which I could express non-uniquely. For example take $x^2 \in A[x]$. Then I could express $x^2$ as either
$$x^2 = 0 + x\cdot x + 0 + \dots \in \bigoplus_{d \geq 0}A_d$$
or as $$x^2 = 0 + 0 + 1\cdot x^2 + \dots \in \bigoplus_{d \geq 0}A_d$$
which contradicts the fact that elements of a direct sum of rings are always expressed uniquely. Which part of my understanding is incorrect and what is the correct way to decompose $A[x]$ as a graded ring?

答: $A_d$ is an $A_0$-module, not an $A[x]$-module. Hence $x\cdot x$ can't be an element of $A_1$.


问:
Infinite sum of inverse of products of successive primes
If we iteratively remove the multiples of the succesive prime numbers from the natural numbers, starting from 2, i.e.,

$1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12,  ...  \rightarrow$

$1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21,  ...  \rightarrow$

$1, 5, 7, 11, 13, 17, 19, 23, 25,  ...  \rightarrow$

we are consecutively removing fractions

$$\frac{1}{2}, \; \frac{1}{2·3}, \; \frac{1}{2·3·5}, \; \frac{1}{2·3·5·7},\;  ... $$

from the total amount of natural numbers. This suggests that the infinite sum

$$\frac{1}{2}+ \frac{1}{2·3}+ \frac{1}{2·3·5}+ \frac{1}{2·3·5·7}+ ...$$

is equal to 1, but I couldn't figure out a proof for this result. Some references about this particular sum would be appreciated too.
Thank you in advance

答: The denominators are called primorials, the series have been studied and proved irrational in this article.</p>


问:
If $f:U\to \mathbb{R}$ is continuous and $(x^2+y^4)f(x,y) + (f(x,y))^3=1$, then $f$ is $C^\infty$

Let $f:U\to \mathbb{R}$ be continuous in $U \subset\mathbb{R}^2$, such that
  $$(x^2+y^4)f(x,y) + (f(x,y))^3=1$$
  for all $(x,y) \in U$. Prove that $f\in C^{\infty}$.

I'm learning the implicit function theorem, I read it, and its proof, and I thought I had understood it. But when I face a problem like this I just don't know how to use it yet. If someone could help me I would appreciate it.

答: Hint: Consider $F\colon \mathbb R^3\to\mathbb R$, $F(x,y,z)=(x^2+y^4)z+z^3$. On the level set $F(x,y,z)=1$, when can you solve for $z$ locally as a (smooth) function of $(x,y)$?</p>


问:
A question based on proving a result Related to Klein J function
I was unable to solve some questions asked in my mid term of number theory exam and so I am asking it here.

If $\tau \in H$ and $x= e^{2\pi i \tau}$, prove that $[{504 \sum_{n=0}^{\infty}\sigma_{5}(n) x^n}]^2 = [j(\tau) -{12}^3 ] \sum_{n=1}^{\infty} \tau(n) x^n$.

Here $j(\tau)={12}^3 J(\tau)$ . I used $j(\tau) = 1/x + 744 + \sum_{n=1}^{\infty} c(n) e^{2\pi in \tau}$ but it doesn't have $\sigma$ term in it . The problem I am facing is that I don't think that we have been thaught a result that has both $\sigma$ and $J(\tau)$ in it.

Please help me in proving  this question.

Ty

答:
This is simply a reformulation of the identity
$$j=1728 \frac{E_4^3}{E_4^3-E_6^2},$$ where
$$E_{k}(\tau)=1-\frac{2k}{B_k}\sum_{n=1}^\infty \sigma_{k-1}(n)\,q^n$$ and $\Delta(\tau)=E_4(\tau)^3-E_6(\tau)^2=\sum_{n=1}^\infty \tau(n)q^n$.


问:
Spectrum of $*$- endomorphisms of finite-dimensional $C^*$-algebras
Let $\mathfrak{A}$ be a unital finite dimensional $C^*$- algebra. Then it is isomorphic to a direct sum of matrix algebras i.e. there exist positive integers $m_1,m_2, \dots,m_r$ such that
\begin{equation*}
\mathfrak{A} \simeq M_{m_1}(\mathbb{C})\, \oplus \, M_{m_2}(\mathbb{C})\, \oplus \dots\oplus M_{m_r}(\mathbb{C}).
\end{equation*}

Let $\phi : \mathfrak{A} \rightarrow \mathfrak{A}$ be a unital $*$- endomorphism. In general, can one say something about its spectrum? Of course it will be $\sigma(\Phi) \subset \overline{\mathbb{D}}$, also, $0 \in \sigma(\Phi)$ if the map is not injective.
But for example, is it possible to have $\lambda \neq 0$ with $\lambda \in \sigma(\Phi)$ and $|\lambda|<1$?

答:
the answer to your question is no:
in fact, the spectrum of any unital -endomorphism of a finite-dimensional
C-algebra is always peripheral, and 0 is the only eigenvalue with absolute value less than 1 that can appear. The proof of this is not too hard, yet it isn't to be found in the literature as far as I know. In a paper in preparation I'm writing with F. Fidaleo and F. Ottomano we include a detailed proof; as soon as the work is ready, I can let you have the draft if you're still interested.


问:
Proof of a corollary for the Strong Law of Large Number
Here is Theorem 1.13 from Mathematical Statistics Jun Shao.

$X_i$ are i.i.d. A necessary and sufficient condition for the existence of constant $c$ for which
\begin{align}
\frac{1}{n}\sum_{i=1}^n X_i \longrightarrow c\quad a.s.
\end{align}
is that $E|X_1|<\infty$ in which case $c=EX_1$. And
\begin{align}
\frac{1}{n}\sum_{i=1}^nc_i(X_i-EX_1)\longrightarrow 0 \quad a.s.
\end{align}
for every bounded real $\{c_i\}$

The first part is just the SLLN. The proof of the second part is marked as an exercise. However, I am having trouble to see this. My attempt is the following: To prove the result, it is sufficient to prove $P(|c_i(X_i-EX_1)|\geq \epsilon \ i.o.)=0 $ for every $\epsilon$. And this is equivalent to prove $P(|X_i-EX_1|\geq \epsilon \ i.o.) = 0$ since $c_i$ is bounded. At this step, I try to use Borel-Cantelli, but it seems does not work. Does anyone have any idea for this?

Thanks in advance!

答:
Expanding a bit on my comment: replace $X_i$ by $X_i-\mathbb{E}X_i$. Let $X$ be distributed as the $(X_i)_{i\geq 1}$ and $Y_i=c_iX_i\mathbf{1}_{\lvert c_iX_i\rvert\leq i}$. By dominating convergence, $\lim\mathbb{E}Y_i=0$. Denoting $M=\sup_{i\geq 1}\lvert c_i\rvert$,
$$\sum_{i=1}^\infty\mathbb{P}(c_iX_i\neq Y_i)=\sum_{i=1}^\infty\mathbb{P}(\lvert c_iX\rvert>i)\leq\sum_{i=1}^n\mathbb{P}(M\lvert X\rvert>i)\leq\int_0^\infty\mathbb{P}(M\lvert  X\rvert>t)dt=\mathbb{E}M\lvert X\rvert<\infty$$
So the Borel-Cantelli lemma applies and a.s., there exists $J\geq 1$ such that for all $i\geq J$ the equality $c_iX_i=Y_i$ holds. Finally,
$$\sum_{i=1}^\infty\frac{\operatorname{Var}(Y_i)}{i^2}\leq\sum_{i=1}^n\frac{\mathbb{E}Y_i^2}{i^2}=\mathbb{E}\left[\lvert X\rvert^2\sum_{i=1}^\infty\frac{\mathbf{1}_{\lvert c_iX\rvert>i}}{i^2}\right]\leq c\mathbb{E}\lvert X\rvert<\infty$$
for some constant $c\geq 0$. We can now apply the following lemma with $Z_n=Y_n-\mathbb{E}Y_n$.

Let $(Z_n)_{n\geq 1}$ be a sequence of independent r.v. with $\mathbb{E}Z_n=0$ and $\sum_{n\geq 1}\operatorname{Var}(Z_n)/n^2<\infty$. Denoting $S_n=Z_1+...+Z_n$, the sequence $(S_n/n)_{n\geq 1}$ converges a.s. to $0$.

Since $(\mathbb{E}Y_i)_{i\geq 1}$ converges to $0$
$$\frac{Y_1+...+Y_n}{n}=\frac{Z_1+...+Z_n}{n}+\frac{\mathbb{E}Y_1+...+\mathbb{E}Y_n}{n}$$
converges to $0$ as well a.s. Finally, a.s., for all $n\geq J$,
$$\frac{c_1X_1+...+c_nX_n}{n}=\frac{c_1X_1+...+c_{J-1}X_{J-1}}{n}+\frac{Y_J+...+Y_n}{n}\longrightarrow 0$$


问:
A work problem on complex variable
Let be  $f:\mathbb{C}\to \mathbb{C}$ a continuous function satisfying

a) $\lim_{z\to \infty} |f(z)|=\infty$;

b) $f(\mathbb{C})$ is open.

Prove that $f(\mathbb{C})=\mathbb{C}$.

Can someone give some hint?

答:
Hint

Using the fact that $\lim_{z\to \infty }|f(z)|=\infty $, you can prove that $f(\mathbb C)$ is sequentially closed, and thus closed.

Since $f(\mathbb C)$ is open as well, using the fact that $\mathbb C$ is connected gives you the wished result.

Title: A work problem on complex variable


问:
Difference of homomorphism and diffeomorphism between two smooth manifolds
I'm a beginner of a smooth manifold.
I can not understand the difference of homomorphism and diffeomorphism between two smooth manifolds.

Let $M$ and $N$ be smooth manifolds. If $M$ and $N$ are homomorphic, then there exists a homomorphism $F:M\to N$.
I think $F$ is also a diffeomorphism between $M$ and $N$ because, for a chart $(U,\phi)$ of $M$, a pair $(F(U), \phi\circ F^{-1})$ becomes a chart of $N$ and a composite function $(\phi\circ F^{-1})\circ F \circ \phi^{-1}=\mathrm{Id}$ is smooth over $\phi(U)$.

Is there anything wrong with the above discussion?

答:
Your confusion is between homeomorphisms and diffeomorphisms between smooth manifolds. Smooth manifolds are topological spaces with extra structure, which allows us to talk about smooth functions (not just continuous) on our spaces. Two topological spaces are 'the same' (in the sense that they possess the same topological properties) if they are homeomorphic, while two smooth manifolds are so if they are diffeomorphic.

Since smooth manifolds are very special topological space, we should expect 'equivalent' smooth manifolds, that is, diffeomorphic ones, to also be topologically equivalent. This is so: this is proposition 2.15c in John M. Lee's 'Introduction to Smooth Manifolds.'

But homeomorphic smooth manifolds are not neccesarily diffeomorphic:
there exist different smooth structures on $S^7$, which make it into truly different, non-diffeomorphic smooth manifolds, although these different smooth manfiolds are homeomorphic to each other. See the wiki page on exotic spheres for more information.

Finally, to address the flaw in your argument, the chart which you define on $N$ doesn't have to be smoothly compatible with the atlas on $N$.


问:
How long will it take Marie to saw another board into 3 pieces?
So this is supposed to be really simple, and it's taken from the following picture:

Text-only:

It took Marie $10$ minutes to saw a board into $2$ pieces. If she works just as fast, how long will it take for her to saw another board into
  $3$ pieces?

I don't understand what's wrong with this question.  I think the student answered the question wrong, yet my friend insists the student got the question right.

I feel like I'm missing something critical here. What am I getting wrong here?

答: The topologists among us may perhaps enjoy the following defense of the teacher's answer: if the board is in the shape of a ring, it will take two cuts to get two pieces, and three cuts to get 3 pieces.</p>


问:
Confused about combination of linear transformations
I'm trying to solve the following exercise:

Consider the vectorspace $V$ of polynomials of the form $ax + b$, $a, b \in \mathbb{R}$. Find two linear transformations $f_1, f_2 : V \rightarrow V$ for which $f_1 \circ f_2 = \operatorname{id}$ and $f_2 \circ f_1 \neq \operatorname{id}$.

Surely this must be impossible, right? Let $B = \{1, x\}$ be a basis of $V$ and thus:

$$
M_B^B(f_1 \circ f_2) = M_B^B(f_1) M_B^B(f_2) = I
$$

such that:

$$
M_B^B(f_2) = M_B^B(f_1)^{-1}
$$

and thus:

$$
M_B^B(f_2 \circ f_1) = M_B^B(f_1)^{-1} M_B^B(f_1) = I \implies f_2 \circ f_1 = \operatorname{id}
$$

Where am I wrong here?

答:
I think your argument is correct.

But it involves a bit of a leap ... you have two matrices, which I'll call $P$ and $Q$, for short, and you show that $PQ=I$. From this, you immediately conclude that $Q = P^{-1}$. That is true, in fact, but it's not at all obvious. See here.

In fact, the given problem is essentially equivalent to the problem of finding two $2 \times 2$ matrices $P$ and $Q$ with $PQ=I$ and $QP \ne I$, and that's not possible.


问:
mean value theorem of harmonic function
The mean value theorem of harmonic function tells us: If $U \subset \Bbb R^n$ open, $f: U \to \Bbb R$ harmonic then for all $y \in U$ and $r>0$ s.t. $\overline{B}(y,r) \subset U$ the following is true:
$$ f(y) = \frac{1}{vol_{n-1}( \partial B(yr))}\int_{\partial B(y,r)}f d vol_{n-1}$$

Now the proof starts with the following observation: $vol_{n-1}(\partial B(y,r))= n \cdot w_n \cdot r^{n-1}$, where $w_n:=\mu(B^n(0,1))$ (the jordan measure of the unit disc). First how we can see this?

The next point is a strange integral transformation:

$$\frac{1}{S^{n-1}}\int_{\partial B(y,r)} f d vol_{n-1}=\int_{S^{n-1}}f(y+rw)d vol_{n-1}(w)$$
How can we conclude this?

Many thanks for some help!

答:
Hint

For your first question, use polar coordinates.

For your second question, first of all, I guess that $\frac{1}{S^{n-1}}$ is wrong. It should be $\frac{1}{|\partial B(y,r)|}$, and on the RHS, a $\frac{1}{|S^{n-1}|}$ is missing (or something like that). Then, $$\partial B(y,r)=\{y+r\sigma \mid \sigma \in S^{n-1}\},$$
and $$\mathrm d z=d(y+r\sigma )=r^{n-1}\,\mathrm d \sigma,$$
where $\mathrm d z$ is an area element of $\partial B(y,r)$ and $\mathrm d \sigma $ an area element of $S^{n-1}$.

Title: mean value theorem of harmonic function


答:
For the antiderivative, if we expand the integrand (as @A-Level Student did), it is
$$4 \pi  R^2+8 \pi  R x-8 \pi  R x \tanh
   \left(\frac{p+x}{D}\right)-4 \pi  R^2 \tanh ^2\left(\frac{p+x}{D}\right)$$

For the difficult antiderivatives, let $x=Dt-p$ to face
$$I_1=\int x \tanh
   \left(\frac{p+x}{D}\right)\,dx=D^2\int  t \tanh (t)\,dt-D p \int\tanh (t)\,dt$$
$$\int  t \tanh (t)\,dt=\frac{1}{2} \left(t \left(t+2 \log \left(e^{-2
   t}+1\right)\right)-\text{Li}_2\left(-e^{-2 t}\right)\right)$$
$$\int   \tanh (t)\,dt=\log (\cosh (t))$$

$$I_2=\int\tanh ^2\left(\frac{p+x}{D}\right)\,dt=D \int\tanh ^2(t)\,dt=D (t-\tanh (t))$$

Now, we can easily go back to $x$ but the problem is to compute the limits.

Edit

Aking a CAS to compute the limits, the result for
$$I= \displaystyle\int_0^\infty4\pi(x+R)^2\left(1-\left(\frac{\frac{x}{R}+\tanh{\frac{x+p}{D}}}{\frac{x}{R}+1}\right)^2\right)dx$$ is
$$\frac{3 }{2 \pi  R}I=
D \left(\pi ^2 D+6 R\right)+12 p^2-3 D^2 \log ^2\left(1+e^{\frac{2 p}{D}}\right)-$$ $$6 D \left(D\,
   \text{Li}_2\left(\frac{1}{2} \left(\tanh
   \left(\frac{p}{D}\right)+1\right)\right)+R \tanh
   \left(\frac{p}{D}\right)\right)$$

Title: How to evaluate the following (rather complex) integral between $x=0$ and $x=\infty$

Body:
I've found a function that describes the concentration function of a polymer around a sphere. I want to evaluate the following integral that contains that function but I'm not sure how to go about it. Can someone send me in the right direction/help me out. Thanks in advance.

$$\displaystyle\int_0^\infty4\pi(x+R)^2\left(1-\left(\frac{\frac{x}{R}+\tanh{\frac{x+p}{D}}}{\frac{x}{R}+1}\right)^2\right)dx$$


问:
An exact and an approximate confidence interval for a Poisson distribution
$X_{1}, ..., X_{10}$ ~ $Pois(\theta)$

Observations:
$x_{1} = x_{3} = x_{6} = x_{8} = x_{9} = 0$; $x_{2} = x_{5} = x_{10} = 1$; $x_{4} = 2$; $x_{7} = 3$.

I want to determine an exact (numerical) and an approximate (numerical) confidence interval for $\theta$ of confidence level 0.9:

exact confidence interval: I don't know how to do this. On the internet I can only find things about an approximate confidence interval.

approximate confidence interval:

$X = (X_{1}, ..., X_{10})$

MLE:  $\hat{\theta} = \bar{X} = 0.8$

And: $Var(X) = \theta$. So an estimate for the variance is also $\bar{X} = 0.8$.

Then according to the answer: An approximate confidence interval is $\bar{X}$ $\pm$ $\sqrt{\bar{X}/n}$ $\xi_{1 - \alpha/2}$.

But this means that: T = $\sqrt{n}$ $\frac{\theta - E[X]}{\sqrt{Var(X)}}$ ~ N(0,1).
But this does not hold for X ~ Pois($\theta$), right?
Because a Poisson distribution does not have to be symmetric.

答:
Let's start with the approximate CI:

For $n$ greater enough ($n=10$ is borderline but actually enough to get your CI with a Gaussian distribution) you can apply CLT in the following way

$$\frac{\overline{X}_{10}-\theta}{\sqrt{\theta}}\sqrt{10}\sim \Phi$$

In fact this is a pivotal quantity with Standard Gaussian distribution so you have two ways to calculate an appropriate approximate CI

(suggested procedure) Estimate the standard deviation of $\overline{X}_{10}$ with $\sqrt{\frac{\overline{X}_{10}}{10}}$ finding the following CI

$$\Bigg(\overline{X}_{10}-1.64\sqrt{\frac{\overline{X}_{10}}{10}};\overline{X}_{10}+1.64\sqrt{\frac{\overline{X}_{10}}{10}}\Bigg)$$

That is

$$\Bigg(0.8-1.64\sqrt{0.8/10};0.8+1.64\sqrt{0.8/10}\Bigg)$$

$$\Bigg(0.3348;1.2652\Bigg)$$

(less common procedure) Soving the following double inequality

$$-1.64<\frac{0.8-\theta}{\sqrt{\theta}}\sqrt{10}<1.64$$

Leading to the following CI

$$0.8+\frac{1.64^2}{20}\pm \sqrt{\Bigg(\frac{1.64^2}{20}\Bigg)^2+0.8\frac{1.64^2}{10}}$$

Exact Confidence interval

It is easy to use the following estimator (that is Complete and sufficient as the sample mean)

$$T=\Sigma_i X_i\sim Po(10\theta)$$

now let's set the two probability

$$0.05=\sum_{t=0}^{8}\frac{e^{-10\theta}(10\theta)^t}{t!}$$

$$0.05=\sum_{t=8}^{\infty}\frac{e^{-10\theta}(10\theta)^t}{t!}$$

In order to solve these two equation w.r.t $\theta$ with some attempts (you can start with the approximate bounds found before) you (quite) easy will find that an exact CI for $\theta$ at 90% is the following

$$\Big(0.3980;1.4435\Big)$$

Further explanation for the example "in the lecture": finding CI for bernulli with the Statistical Method. They suppose $n=20$ and $\Sigma_i X_i =4$

Graphically:

Left tail: $5.1\%$

Right tail: $1.6\%$

Confidence interval %: $100-1.6-5.1=93.3\%$


问:
Is automorphism group of lie group a lie group?
Let $G$ be a lie group. Is its automorphism group $\text{Aut}(G)$ of smooth automorphisms then a lie group?

According to G. Hochschild is $\text{Aut}(G)$ when equipped with the compact-open topology a lie group if the component group $G/G_0$ is finitely generated. Here is link https://www.jstor.org/stable/1990752?seq=1#metadata_info_tab_contents

What if $G/G_0$ is not finitely generated? I have tried to find an example of lie group with non-finitely generated component group whose automorphism group is not a lie group when equipped with the compact-open topology, but haven't found one. I suspect a discrete countable group whose automorphism group is uncountable can be such an example.

答:
The compact-open topology is the coarsest topology on $\text{Aut}(G)$ so the action $\text{Aut}(G)\times G\longrightarrow G$ is continuous. With that topology the answer is NO.

Take $G$ to be any countable discrete group with uncountable automorphism group $\text{Aut}(G)$, for example $G=\mathbb{Q}/\mathbb{Z}$. If $X$ and $Y$ are topological spaces and $Y$ discrete, then $\mathcal{C}(X,Y)$ is totally disconnected wrt. the compact-open topology. To see this, let $f\in\mathcal{C}(X,Y)$ and $A\subseteq \mathcal{C}(X,Y)$ be a connected subset so $f\in A$. Then $\forall x \in X \quad B_x\equiv \{h\in A \mid h(x)=f(x)\}$ is non-empty and clopen in $A$. So $A=B_x \quad\forall x\in X$. $A=\bigcap_{x\in X}B_x=\{f\}$. That means $\mathcal{C}(X,Y)$ is totally disconnected.

Since $G$ is discrete, $\text{Aut}(G)$ is totally disconnected wrt. the compact-open topology. Since $\text{Aut}(G)$ is also uncountable, it is not a manifold.


问:
What is the remainder when $1^{2016} + 2^{2016} + ⋯ + 2016^{2016}$ is divided by $2016$?
What is the remainder when $1^{2016} + 2^{2016} + ⋯ + 2016^{2016}$ is divided by $2016$?

My Approach:

started breaking 2016 into product of pairwise coprime positive integers to get $2016 = 32 * 7 * 9$
This suggests that we want to find the sum modulo 32,7, 9 separately so that we can apply the chinese remainder theorem.

but I couldn't solve it any further than this so please help me out.

答:
Define $$S = \sum_{i=1}^{2016} i^{2016}$$

Notice that $2016 = 32\cdot 9\cdot7$

and $\phi(32) | 2016,\; \phi(9) |2016,\; \phi(7) | 2016$, so we have

$$i^{2016} \equiv \begin{cases}
 1 \pmod n  \text{ if } \gcd(i,n)=1\\ 0 \pmod n \text{ if } \gcd(i,n)>1 \end{cases}$$
for $n=32,9,7$.

Then each group of $n$ has $\phi(n)$ values coprime to $n$ and there are $2016/n$ such groups in each case so:
$$S \equiv \frac{2016}{32}\cdot \phi(32) \equiv 7\cdot 9 \cdot \phi(32) \pmod{32} \tag 1$$
$$S \equiv \frac{2016}{9}\cdot \phi(9) \equiv 32\cdot 7 \cdot \phi(9) \pmod{9} \tag 2$$
$$S \equiv \frac{2016}{7}\cdot \phi(7) \equiv 9 \cdot 32 \cdot \phi(7) \pmod{7} \tag 3$$

Can you end it now?

Update:

There's a shortcut to CRT. Simply add RHS of $(1), (2), (3)$,

$$x = 7 \cdot 9 \cdot \phi(32) + 32 \cdot 7 \cdot \phi(9) + 9 \cdot 32 \cdot \phi(7) = 4080$$

Then $S \equiv x \pmod{32,9,7} \implies S\equiv x = 4080 \equiv 48 \pmod{2016}$


问:
How to find the $\delta$ for some $\varepsilon$ such that: $|x-0| < \delta \Rightarrow |g(x)-g(0)|<\varepsilon$
$g(x)$ is defined as:

$$g(x)=\frac{\sqrt{x+1}-1}{x};\  x\in [-1,0)\ \cup\ (0,\infty)$$

$$g(x)=\frac{1}{2};\ x=0$$

How can I find the $\delta$ for some $\varepsilon$ such that: $|x-0| < \delta \Rightarrow |g(x)-g(0)|<\varepsilon$.

I tried a lot of things, but I seem to be completely stuck. Rationalizing the numerator doesn't help, because x cancels out.

答:
Observe that
\begin{align*} |g(x) - g(0)| &= \left| \dfrac{\sqrt{x+1} - 1}{x} -\dfrac{1}{2} \right| \\
&= \left| \dfrac{1}{\sqrt{x+1} + 1} - \dfrac{1}{2} \right| \\
&= \left| \dfrac{1 - \sqrt{x+1}}{2(\sqrt{x+1} + 1)} \right| \\
&= \left| \dfrac{-x}{2(\sqrt{x+1} + 1)^2} \right| \\
&= \dfrac{|x|}{2(\sqrt{x+1} + 1)^2} \\
&< \dfrac{\delta}{2}
\end{align*}

So we can choose $\delta < 2 \epsilon$.

However, the domain of $g$ is $[-1,\infty)$. If $\delta > 1$, any $c$ in $(-\delta,-1)$ such as $c = -\dfrac{\delta + 1}{2}$ satisfies $|c| < \delta$ but $g(c)$ does not exist, thus $|g(c) - g(0)| < \epsilon$ is false.

So one of the answer is $\delta < \min\{ 1, 2 \epsilon \}$.


问:
How to prove $ \prod_{n=1}^{\infty} \left(1+\frac{2}{n}\right)^{(-1)^{n+1}n} \,= \frac{\pi}{2e}$
How can I prove that

$$ \prod_{n=1}^{\infty} \left(1+\frac{2}{n}\right)^{\large{(-1)^{n+1}n}} \,= \frac{\pi}{2e}$$

The result is given here (result 48).

The source: Prudnikov et al. 1986, p. 757 is given, however I have been unable to find the book online.

Some of my attempts include:

Multiplying known infinite products for $\frac{\pi}{2}$  and
$\frac{1}{e}$

$$ \frac{\pi}{2} = \frac{2\cdot2\cdot4\cdot4\cdot6\cdot6\cdots}{1\cdot3\cdot3\cdot5\cdot5\cdot7\cdots} $$
$$ \frac{1}{e} =  \frac{1}{2} \left(\frac{3}{4}\right)^{\large{\frac{1}{4}}}\left(\frac{5\cdot7}{6\cdot8}\right)^{\large{\frac{1}{4}}} \cdots$$

as well as others given in https://arxiv.org/pdf/1005.2712.pdf

Trying to find the partial products from ${n=1} $ to $k $, Mathematica Gives:

$$\scriptsize{ \frac{\exp\left(2\left(-\zeta(-1,k+\frac{1}{2})^{(1,0)}+\zeta(-1,k+\frac{3}{2})^{(1,0)}+\zeta(-1,k+1)^{(1,0)}-\zeta(-1,k+2)^{(1,0)}\right)\right)\pi\, \Gamma(k+2)^2}{2\,\Gamma(k+\frac{3}{2})^2}} $$

However, I have been unsuccessful producing partial products on my own.

Taking the $\ln$ of the product to try to find the partial sums;

Trying to transform the following integral into infinite product.

$$\int_{0}^{\infty} \frac{\cos(x)}{1+x^2} = \frac{\pi}{2e} $$

From what I have seen in some papers, the partial products are found, then the Stirling's Approximation is used to find the limit.

Question: How can I prove the value of the given Infinite Product? Proofs or hints are both welcome.

Thank you kindly for your help and time.

答:
This answer is spliced from the paper linked in comments

Melzak (1961) shows that the largest-volume cylinder (Cartesian product of a hypersphere and a line) in an $n$-dimensional unit sphere occupies this proportion of the sphere:
$$\rho_n=\frac{2(n\pi)^{-1/2}(1-1/n)^{(n-1)/2}\Gamma(n/2+1)}{\Gamma((n+1)/2)}$$
We have $\rho_2=\frac2\pi$ and $\lim_{n\to\infty}\rho_n=\sqrt{\frac2{\pi e}}$. Now define
$$\sigma_n=\frac{\rho_{n+2}}{\rho_n}=\sqrt{\left(\frac n{n+2}\right)^n\left(\frac{n+1}{n-1}\right)^{n-1}}$$
Telescoping on $\sigma_n$ for $n=2,4,6\dots$ then gives
$$\sqrt{\frac\pi{2e}}=\prod_{n=1}^\infty\left(\frac n{n+1}\right)^n\left(\frac{2n+1}{2n-1}\right)^{(2n-1)/2}$$
Squaring gives
$$\frac\pi{2e}=\prod_{n=1}^\infty\left(\frac {2n}{2n+2}\right)^{2n}\left(\frac{2n+1}{2n-1}\right)^{2n-1}=\prod_{n=1}^\infty(1+2/n)^{(-1)^{n+1}n}$$


问:
What is the idea behind sheaves of rings on distinguished open sets
In the book on algebraic geometry of Mumford (which can be found here), he said :

We want to enlarge the ring $R$ into a whole sheaves of rings on $SpecR$, written $\mathcal{O}_{SpecR}=\mathcal{O}_{X}$.

So he need to define $\mathcal{O}_{X}(U)$ for $U$ is an open set in $X$, in particular he need to define $\mathcal{O}_{X}(X_f)$ for some distinguished open subset $X_f$. And he said

We want to define : $\mathcal{O}_{X}(X_f)=R_f$=localization of the ring $R$ w.r.t multiplicative system $\lbrace 1,f,f^2,..\rbrace$

My question is : Why do we want to define the sheaf $\mathcal{O}_{X}(X_f)$ to be $R_f$ ?

Back to case of affine variety $V$, $V\setminus V(f)$ is a quasi-affine variety with the ring of regular function $k[V]_f$. So can we think about the sections of $\mathcal{O}_{X}(X_f)$ as the regular function on some spaces ?

答:
To add just a little bit to Martin's already excellent answer.

Philosophically, we like the following idea: given a space $X$, we can learn a lot from $X$ by a ring of functions on this space. However, the whole ring loses a ton of information. So instead, lets remember what happens locally: so we get a sheaf of functions on a space.

Now given a ring $R$, we do the math-y thing and reverse the above construction. We want a space $X$ on which $R$ is the ring of functions. We want it to be as natural a construction as possible. Grothendieck's insight was that the "set of prime ideals" forms a good "set of points of a space" to do geometry. Again though, if we only remember the global information, we lose a huge amount of information that is intrinsically.

So working with $Spec(R)$ and the ring of functions $R$ gives you natural motivation to not only study a ring, but all its quotients (functions on closed sets) and all its localizations (functions on open sets) at once.

Why localization? It really comes down to what localization does to the behaviour under localization of the lattice of prime ideals. Think first about the process of passing from $R$ to $R/I$ for some ideal $I$. What are the prime ideals in $R/I$ - they are those prime ideals of $R$ containing $I$. Containing a prime ideal is what we take to be "vanish at that point". This is the key idea: the property of containing a prime ideal is the right replacement for vanishing in the classical setting.

It is a good idea to take a good look at the Nullstellensatz, and extract the above statement in the classical setting.

Now why localization? In a similar manner to the above discussion, the prime ideals that survive localization are the ones that dont intersect the set that you are inverting. If you want to think about a single function $f$, then localizing at $f$ to get $R_f$ this tells us that the prime ideals of the localization are (more precisely, correspond to) those in $R$ which do not contain $f$: i.e. the set of points where $f$ doesnt vanish.

One should write down and chase the appropriate diagrams to make sure that the "correspondence between prime ideals" for the two cases I mentioned above are functorial, that they behave well with morphisms, but thats the intuition.


问:
Is this regular matrix?
Lets suppose, that for k:

$A^k=I$ , k is natural number:

Does it mean that matrix A is regular?

I think yes. Every regular matrix is invertible:

$AA^{-1}=I$

$A^k=I$

then
$A^k=AA^{-1}$

Is it right or am I missing something?

答:
It is right since, for example:

$$1=\det I=\det A^k=\left(\det A\right)^k\implies \det A \neq0\iff A\;\;\text{is invertible}$$

Title: Is this regular matrix?


问:
When is $f(f^{-1}(B)) \neq B$?
$f: X \longrightarrow Y$ and $B \subseteq Y$

I am currentry trying to give an example where $f(f^{-1}(B)) \neq B$.

I know, that for a function to have an inverse, it must be bijective.

My thought process is, that the answer can only be equal to $B$
if the function $f$ is bijective, because only then can every element in $f^{-1}(B)$ be directly traced back to B.

Here's where I struggle: How can $f(f^{-1}(B)) \neq B$ if the function needs to be bijective in the first place? Wouldn't this be a contradiction? Or is that exactly what I need to disprove here?

答:
The inverse of a mapping may not necessarily be a mapping. $f^{-1} [B]$ (the notation here is superior) is merely "the set of all elements of $X$ whose image under $f$ is in $B$".

$f^{-1} [B] = \{x \in X: f(x) \in B\}$

So $f [f^{-1} [B]]$ is the intersection of $B$ with the image of $f$.

Hence if you have elements of $B$ which are not mapped to by $f$, then you have $f [f^{-1} [B]] \ne B$.

Title: When is $f(f^{-1}(B)) \neq B$?


问:
Given $x,y,z\geq0$ and $x^2+y^2+z^2+x+2y+3z=13/4$. Find the minimum of $x+y+z$.

Given $x,y,z\geq0$ and $x^2+y^2+z^2+x+2y+3z=13/4$. Find the minimum of $x+y+z$.

I tried many method, such as AM-GM, but all of them failed.

Thank you.

Body:
use this
$$(x+y+z)^2+3(x+y+z)\ge x^2+y^2+z^2+x+2y+3z$$
and let $x+y+z=t$,
then $$t^2+3t\ge \dfrac{13}{4}$$


问:
A Combinatorial/Algebraic Solution to Combinatorics Identities Exercise
The exercise:

$$\sum_{k=r}^n {n \choose k} {k \choose r} 2^k = {n \choose r} 2^r 3^{n-r}$$

I have tried everything. Combinatorial proof, differentiating a function to reach both sides, finding patterns of combinatorial identities, binomial coefficient identity and mixtures of all the above. Nothing. One idea which I thought was promising is summing both sides with $\sum_{r=0}^n$ so the right-hand side becomes $5^{n}$ by binomial identity but (a) I'm not sure that is even valid and (b) I couldn't make anything of the left-hand side.

Thanks.

答: Hint: Show that $\binom{a}{b}\binom{b}{c}=\binom{a}{c}\binom{a-c}{b-c}.$ Then multiply and divide by $2^r.$