问:
Does the series $\sum_{n=1}^{\infty}\frac{\sin{(n\alpha)}}{\ln(10)^n}$ converge?
$$\sum_{n=1}^{\infty}\frac{\sin{(n\alpha)}}{\ln(10)^n}$$
For what $\alpha\in \text{Real numbers}$ does the series converge ?

What I did:

Using the nth root test we get:

$$\lim_{n\rightarrow\infty}\frac{(\sin{(n\alpha))^{1/n}}}{\ln(10)} = \frac{1}{\ln(10)}<1 \implies \text{convergence for all alpha} $$

Is it correct ?

答:
Since proving that $$\lim_{n\to\infty} (\sin(n\alpha))^{1/n} = 1$$
may be difficult, and for some $\alpha$ it's just not true, I'd first use the comparison test.
We have
$$ 0 \le \left|\frac{\sin(n\alpha)}{(\ln (10))^n}\right| \le \frac{1}{(\ln (10))^n} $$
Since $$ \sum_{n=1}^\infty \frac{1}{(\ln (10))^n} $$
is convergent (by the root test), then it means that
$$ \sum_{n=1}^\infty \left|\frac{\sin(n\alpha)}{(\ln (10))^n}\right| $$
is convergent, and this means that
$$ \sum_{n=1}^\infty \frac{\sin(n\alpha)}{(\ln (10))^n} $$
is convergent.


问:
Ring of integers of $f(X)=X^3+X^2-2X+8$ is principal
Let $K=\Bbb Q[x]$ be a cubic number field with $x$ be a root of $f(X)=X^3+X^2-2X+8$. I want to show that $A$, the ring of integers of $K$ is principal.

What I have shown is that $A=\Bbb Z+\Bbb Zx+\Bbb Zy$ where $y=4/x$(edited) and $(2)=(2,x,y)(2,x+1,y)(2,x,y+1)$ is the product of distinct ideals of norm 2 (by brutal method), and Minkowski bound is more than $6$, so it suffices to show that every prime factor $(2),(3),(5)$ whose norm $\le 6$ is principal.

But I don't know how to proceed. The hint says that $(x-2)/(x+2), (x-1)/(x+3)$ are the generators of two of the three prime ideals of norm 2, and $(3)$ is prime, and $(5)$ is a product of prime ideal of norm 5 and norm 25, and compare $(5)$, $(x+1)$. How can I proceed?

答:
Let $\alpha$ be a root of $f(x)=x^3+x^2-2x+8$. Are you sure that $1$, $\alpha$, $\frac{1}{4}\alpha$ is an integral basis? You might want to take a look at the minimal polynomial of $\frac{1}{4}\alpha$.

An integral basis is actually given by $1$, $\alpha$, $\beta=\frac{1}{2}(\alpha+\alpha^2)$. You then factor the ideals $(2)$, $(3)$, and $(5)$.

Let $\mathfrak{p}_2=(2,\beta)$, and $\mathfrak{q}_2=(2,\beta+1)$. Then $(2)=\mathfrak{p}_2\mathfrak{q}_2^2$.

$(3)$ is inert.

Let $\mathfrak{p}_5=(5,\alpha+1)$, and $\mathfrak{p}_{25}=(5,\alpha^2+3)$. Then $(5)=\mathfrak{p}_5\mathfrak{p}_{25}$.

The hint then seems to have done most of the hard work for you. You've already said that it's given you $\frac{\alpha-2}{\alpha+2}$ as a generator for one of the prime ideals of norm $2$, and $\frac{\alpha-1}{\alpha+3}$ as a generator of the other.

You also say to look at $(\alpha+1)$, and $\text{Nm}(\alpha+1)=10=2\cdot 5$, so you instantly find two elements of norm $\pm 5$, namely
$$
(\alpha+1)\frac{\alpha+2}{\alpha-2}\text{ and }(\alpha+1)\frac{\alpha+3}{\alpha-1}
$$
Can you show that one of these generates $\mathfrak{p}_5$?

ContentLicense: CC BY-SA 3.0


问:
Disprove that for all integers $a,b$ that if $d=gcd(a,b)$ then $d=gcd(a^2,b)$ or $d=gcd(a,b^2)$
Question:
Disprove that for all integers $a,b$ that if $d=gcd(a,b)$ then $d=gcd(a^2,b)$ or $d=gcd(a,b^2)$

Attempts.

I thought to first try $a=b$, then $d=a$ and $gcd(a^2,b)=b=d$

Ok, how about $a=0, b \neq 0$, then $d=b$ and $gcd(a^2,b)=b=d$

Ok, if $a,b$ are co-prime, then the $d=1$ and $gcd(a^2,b)=1$

Ok, maybe $a=b^2$, then $d=b$ and $gcd(a^2,b)=b$

Not sure what else to try. Maybe the problem is wrong? I copied it down carefully. Hints appreciated.

答:
Take two different primes $p,q$ and then set $a=p^2q$ and $b=pq^2$. This will give you:

$d=(a,b)=pq$

$(a^2, b)=(p^4q^2, pq^2)=pq^2$,

$(a, b^2)=(p^2q, p^2q^4)=p^2q$.

For example, for $p=2, q=3$ you will have $a=12, b=18$.

ContentLicense: CC BY-SA 4.0


问:
Does a completion exist in general?
Let $\mathcal H_0$ be a space of functions equipped with an inner product. Is it possible in general to define a “completion” $\mathcal H$ such that $\mathcal H_0 \subset \mathcal H$ and cauchy sequences converge in this space? I thought that it was in fact possible, but I am working on a HW problem that asks me to show (for a specific definition of $\mathcal H_0$ and the inner product) that the completion exists and that cauchy sequences do actually converge.

答:
Basically, yes, but there are some details to pay attention to. Importantly, the particular construction of the completion is irrelevant (any one would do, all leading to isometric models). Further, the entire inner product space structure extends uniquely to the completion. This is true by a general principle: any uniformly continuous function on a space extends uniquely to all of the completion. And, when that is done, the completed space's metric agrees with the one induced by the extended norm (obtained from the inner product).

In short: any inner product space (resp. normed space) has an essentially unique completion and that completion is uniquely an inner product space (resp. normed space) in a way that extends the original structure and agreeing with the metric. This is a standard fact of Banach space and Hilbert space theory showing that any inner product space embeds essentially uniquely in a Hilbert space and similarly for normed spaces in Banach spaces. It should be noted though that very often having a particular model of the completed space is very important. Working directly with any of the standard methods that produce a completion of a metric space (or, more generally, a uniform space) is not a good idea. In particular, the completion of the space of continuous functions with respect to the $L_p$ norm is the $L_p$ space --- that space simply exists as a completion but its model in terms of measurable functions is very important.

Title: Does a completion exist in general?


问:
Probability of extinction in generation 3
The branching process has offspring distribution $μ(0) = a, μ(1) = b, μ(2) = c$ and $μ(n) = 0$, what is the probability that the population is extinct in the 3rd generation, given that it is not extinct in the 2nd generation?

How would i solve this? How is it different from finding the probability of extinction in exactly generation 3?

答:
If $X_n$ is the event of extinction in exactly generation $n$

then the probability of extinction in generation $3$ is $\mathbb P(X_3)$

but the probability of extinction in generation $3$ given not extinct by generation $2$ (i.e. no extinction in generation $1$ or in generation $2$) is $$\mathbb P(X_3 \mid X_1^c, X_2^c) = \dfrac{\mathbb P(X_3 ,X_1^c, X_2^c)}{\mathbb P(X_1^c, X_2^c)}\text{ which is }\dfrac{\mathbb P(X_3 )}{1-\big(\mathbb P(X_1)+\mathbb P(X_2)\big)}$$ since $X_3 \subset X_1^c \cap X_2^c$ and $X_1 \cap X_2 = \emptyset$


问:
Finding the common modulus given many congruences
Consider the following relationships:

$a_1 \equiv b_1 \mod{g}$

$a_2 \equiv b_2 \mod{g}$

$\cdots $

$a_n \equiv b_n \mod{g}$

Where:

All values $a_1, a_2, ... , a_n$ are known

All values $b_1, b_2, ... , b_n$ are unknown

$g$ is an unknown modulus

$n$ is an unknown limit

It is known that such a relationship exist for all values $a_1, a_2, ... , a_n$

All the numbers are generated randomly (AKA: $g$ is not intentionally prime and there is no intentional coprime relationships), so an RSA common modulus attack is not intentional)

Question is: what would be the fastest way to find $g$ and/or all of $b_1, b_2, ... , b_n$ values?

I do realize that a brute-force attack would definitely reveal it, especially if the bit space is not high, but I'm looking for an easier and less noisy way. Does such a way exist?

答:
"I do realize that a brute-force attack would definitely reveal it"

No, you don't, and no it wouldn't.

"but I'm looking for an easier and less noisy way. Does such a way exist?"

Not only does no easier way exist, no hard way exist.

Note: each state gives you absolutely no information at all.

Let $a_k$ be a known constant.  Well for every possible $b_k$ we can always have $N$ be any factor of $|b_k - a_k$ and we will have $a_k \equiv b_k \pmod N$.

And with any possible $g$ we can have $W = a + mg$ for any integer $m$ and have $a_k \equiv W \pmod g$.

So knowing that $a_k \equiv b_k \pmod g$ for some $g$ and for some $b_k$ tells us absolutely nothing.

Maybe if we knew some of the $b_k$ we could figure out the $g$ (as $g$ would have to be a factor of $|b_k -a_k|$).  But knowing none of the $b_k$ and just that $g$ we can never calculate the $b_k$ (as there are infinitely many $a_k + mg$).

[On the other hand we would also never care what integer $b_k$ is; $\mod g$ is an equivalence relation so if $b_k = a_k + mg$ then $b_k \equiv a_k \pmod g$ and for all intents and purposes $a_k$ and $b_k$ are completely equivalent and interchangeable and can be considered to be the exact same thing.]

But if you don't know either you know utterly nothing.

====

Howeve if the question is that the $a_k$ and $b_k$ are known, and the question is how to find $g$ the answer is:

Find the $\gcd(a_k- b_k)$.  $g$ must be a factor of $\gcd(a_k-b_k)$.

(which is probably why they used the letter $g$....)


问:
How to Evaluate $\int_{0}^{1} \frac{x^2 \ln(1-x^4)}{1+x^4}dx$?
How to evaluate

$$ \int_{0}^{1} \frac{x^2 \ln(1-x^4)}{1+x^4} \,dx \approx -0.162858 \tag{1}$$

The integral arises in the computation of

$$\left( \sum_{n=1}^{\infty} \frac{(-1)^n}{n}\right)\left(\sum_{n=1}^{\infty} \frac{(-1)^n}{4n-1}\right)$$
as

$$ \scriptsize{\frac{\pi \ln(2)}{4\sqrt{2}} + \frac{\ln(2) \ln(3-2\sqrt{2})}{4\sqrt{2}} -3\ln(2) + \frac{\pi}{2}= \int_{0}^{1} \frac{x^2 \ln(1-x^4)}{1+x^4} \,dx + \int_{0}^{1} \frac{x^{1/4}}{2(1+x)}\left(\tan^{-1}(x^{1/4}) - \tanh^{-1}(x^{1/4}) \right)}dx $$

A similar Integral

$$ \int_{0}^{1}\left( \frac{x^2 \ln(2)}{x^4-1} - \frac{x^2 \ln(1+x^4)}{x^4-1}\right)dx = C-\frac{\pi^2}{16}+\frac{\ln^2(\sqrt{2}-1)}{4}+\frac{\pi \ln (\sqrt{2}-1)}{4} \tag{2} $$

Where $ C $ = Catalan Constant

Unfortunately the same techniques I used to evaluate $(2)$ have not worked for $(1)$.
I know only for integration - By parts, U-Sub, and  using Taylor Series as well as Mathematica.

Q = Is there a closed form for Integral $(1)$?

EDIT

$$ (1) = \frac{-\pi^2}{8\sqrt{2}} + \frac{\pi \ln(8)}{4\sqrt{2}} +\frac{\ln(8) \ln(3-2\sqrt{2})}{4\sqrt{2}} -\frac{\pi \ln(3-2\sqrt{2})}{8\sqrt{2}} + 4\sum_{k=1}^{\infty} \frac{(-1)^k}{4k-1}\sum_{n=1}^{k} \frac{1}{4n-1} $$

$$\sum_{k=1}^{\infty} \frac{(-1)^k}{4k-1}\sum_{n=1}^{k} \frac{1}{4n-1} = \frac{1}{64}\left(\psi^{(1)}\left(\frac{7}{8}\right)-\psi^{(1)}\left(\frac{3}{8}\right)\right) + W $$

Where $W$ is some value.

答:
This may be counterproductive but we can obtain another form from @yarchik's solution.
$$
\begin{aligned}
S%
&=-\sum _{n=0}^{\infty } \frac{(-1)^n }{4n+3}H_{n+\frac{3}{4}},\\
&=-\frac{1}{4}\sum _{n=0}^{\infty } \frac{(-1)^n }{n+3/4}H_{n+\frac{3}{4}},\\
&=-\frac{1}{4}\int_0^1x^{-1/4}\sum _{n=0}^{\infty } (-x)^nH_{n+\frac{3}{4}}\,\mathrm dx,\\
\end{aligned}
$$
The remaining series can be solved exactly yielding
$$
\begin{aligned}
S%
&=\frac{1}{7}\int_0^1\frac{x^{3/4}}{1+x}F\left({1,7/4\atop 11/4};-x\right)\,\mathrm dx-\frac{H_{3/4}}{4}\int_0^1\frac{x^{-1/4}}{1+x}\,\mathrm dx,\\
&=\frac{1}{7}\int_0^1\frac{x^{3/4}}{1+x}F\left({1,7/4\atop 11/4};-x\right)\,\mathrm dx-\frac{H_{3/4}(-1)^{1/4}}{2}(\operatorname{arctan}((-1)^{1/4})- \operatorname{arctanh}((-1)^{1/4})),
\end{aligned}
$$
The hypergeometric function in the integral is zero-balanced and may reduce further.


问:
Solve 1 equation and 3 unknowns
So I know that I should have infinite solutions when this is the case, but I don't know how to derive the solution without just using some logic and it seems like there should be a way to solve it mathematically (maybe some simple linear algebra?)

$$
s^2 + 2s = b(s^2+3s+2) + a_1(s+2) + a_2
$$
Clearly the only term with an $s^2$ on it it is $b$ and since we only want one $s^2$ we see that $b=1$. Now we need to cancel out one of the $3s$ we got because we only want $2s$ so $a_1$ must be $-1$. Conveniently this canceled out the $2$ for us so $a_2 = 0$.

While this makes sense, I want to use math to solve it.

Alright although the question is already solved I'll provide a little bit more background for the question if future people are curious.

I'm given

$$
G(s) = \begin{bmatrix} \frac{1}{s+1} \\ \frac{1}{(s+1)(s+2)}\end{bmatrix}
$$

$G(s) = C(sI-A)^{-1}B + D$ but in the problme I'm given that $D$ is a vector of zeros, and C is the identity matrix.

My goal was to find A and B that satisfied.

$$
\begin{bmatrix}
\dot x_1 \\
\dot x_2
\end{bmatrix}
=
\begin{bmatrix}
a_1 & a_2 \\
a_3 & a_4
\end{bmatrix}
\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}
+
\begin{bmatrix}
b_1 \\
b_2
\end{bmatrix}
$$
Given $G(s)$.

答:
Precisely.
All you need to do is expand the RHS and collect like terms: $$b(s^2+3s+2)+a_1(s+2)+a_2=(b)s^2+(3b+a_1)s+(2b+2a_1+a_2)$$ and match co-efficients with the LHS, i.e.: $$\begin{cases} b=1 \\ 3b+a_1=2 \\ 2b+2a_1+a_2=0\end{cases}$$
Which is easily solved for $(b, a_1,a_2)=(1,-1,0)$

Title: Solve 1 equation and 3 unknowns


答:
It is not very clear from the question what is known and what is unknown, but in this answer I presume you want to find $b, a_1, a_2$ such that:

$$s^2 + 2s = b(s^2+3s+2) + a_1(s+2) + a_2$$

as a polynomial in $s$, i.e. that all the coefficients (multiplying the powers of $s$) in the LHS are the same as all the coefficients in the RHS.

For that, re-origanise RHS as a polynomial in $s$:

$$s^2 + 2s +0= \underbrace{b}_1 s^2+\underbrace{(3b+ a_1)}_2 s +\underbrace{(2b+2a_1+ a_2)}_0$$

which yields a system of three equations in three unknowns:

$$\begin{array}{rrrcl}b&&&=&1\\3b&+a_1&&=&2\\2b&+2a_1&+a_2&=&0\end{array}$$

which has a unique solution: $b=1, a_1=-1, a_2=0$.

Title: Solve 1 equation and 3 unknowns

Body:
So I know that I should have infinite solutions when this is the case, but I don't know how to derive the solution without just using some logic and it seems like there should be a way to solve it mathematically (maybe some simple linear algebra?)

$$
s^2 + 2s = b(s^2+3s+2) + a_1(s+2) + a_2
$$
Clearly the only term with an $s^2$ on it it is $b$ and since we only want one $s^2$ we see that $b=1$. Now we need to cancel out one of the $3s$ we got because we only want $2s$ so $a_1$ must be $-1$. Conveniently this canceled out the $2$ for us so $a_2 = 0$.

While this makes sense, I want to use math to solve it.

Alright although the question is already solved I'll provide a little bit more background for the question if future people are curious.

I'm given

$$
G(s) = \begin{bmatrix} \frac{1}{s+1} \\ \frac{1}{(s+1)(s+2)}\end{bmatrix}
$$

$G(s) = C(sI-A)^{-1}B + D$ but in the problme I'm given that $D$ is a vector of zeros, and C is the identity matrix.

My goal was to find A and B that satisfied.

$$
\begin{bmatrix}
\dot x_1 \\
\dot x_2
\end{bmatrix}
=
\begin{bmatrix}
a_1 & a_2 \\
a_3 & a_4
\end{bmatrix}
\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}
+
\begin{bmatrix}
b_1 \\
b_2
\end{bmatrix}
$$
Given $G(s)$.


问:
When to use these fractions properly?
I'm self-taught in mathematics, however, I have missed many basic ideas of arithmetic, where I'll usually use symbolab to help me get through some simplification of equations. although, I'm starting to get annoyed by this, as its impeding my learning. I would really like your support in helping me understand the distinction in fractions likeso:

$$[a] \space \cfrac{x}{1-x\over x}$$ when this could be simplified as either:

$$[1] \space\frac{x}{(1-x)x} $$
or $$[2] \space \frac{x^2}{1-x}$$

Though, I've also used symbolab and it's given me fractions likeso:

$$[b] \space \cfrac{x \over1+x}{1-x\over x}$$

Where it will be split like the first fraction into either (1) or (2). Are there rules that help me decide if its either (1) or (2) given equation (a), and when is it acceptable to use these for equation (b).

Or does it not matter for (b), and that the result should ensue likeso:

$$[b] \space \cfrac{x \over1+x}{1-x\over x} = \frac{x}{1+x}\frac{1-x}{x}$$

Thank you for the support! this should help me from relying on symbolab all the time for this.

答:
$$\frac{x}{\frac{y}{z}}=\frac{(x)}{\left(\frac{y}{z}\right)}=x\left/\frac{y}{z}\right.=\frac{xz}{y}$$

$$\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{\left(\frac{a}{b}\right)}{\left(\frac{c}{d}\right)}=\frac{a}{b}\left/\frac{c}{d}\right.=\frac{ad}{bc}$$

In both cases, the middle fraction line is the "longest", so there are implicit brackets around the numerators and the denominators.

Title: When to use these fractions properly?


答:
A good rule of thumb: to divide by a fraction $\frac{a}{b}$ is the same as to multiply by the reciprocal fraction $\frac{b}{a}$.

Thus:

$$\frac{x}{\frac{1-x}{x}}\left(\text{ which is }x\div\frac{1-x}{x}\right)=x\cdot \frac{x}{1-x}=\frac{x^2}{1-x}$$

as the reciprocal of $\frac{1-x}{x}$ is $\frac{x}{1-x}$.

Make sure, however, that you always write clearly which division is done first, and which division is done next! A very similarly looking expression:

$$\frac{\frac{x}{1-x}}{x}=\frac{x}{1-x}\div x$$

is actually completely different, the order of division is swapped, and the result is different: $\frac{x}{(1-x)x}$.

Hope this helps.

Title: When to use these fractions properly?

Body:
I'm self-taught in mathematics, however, I have missed many basic ideas of arithmetic, where I'll usually use symbolab to help me get through some simplification of equations. although, I'm starting to get annoyed by this, as its impeding my learning. I would really like your support in helping me understand the distinction in fractions likeso:

$$[a] \space \cfrac{x}{1-x\over x}$$ when this could be simplified as either:

$$[1] \space\frac{x}{(1-x)x} $$
or $$[2] \space \frac{x^2}{1-x}$$

Though, I've also used symbolab and it's given me fractions likeso:

$$[b] \space \cfrac{x \over1+x}{1-x\over x}$$

Where it will be split like the first fraction into either (1) or (2). Are there rules that help me decide if its either (1) or (2) given equation (a), and when is it acceptable to use these for equation (b).

Or does it not matter for (b), and that the result should ensue likeso:

$$[b] \space \cfrac{x \over1+x}{1-x\over x} = \frac{x}{1+x}\frac{1-x}{x}$$

Thank you for the support! this should help me from relying on symbolab all the time for this.


答:
In general, $$\frac{a}{\bigg(\frac bc\bigg)}=\frac{ac}{b}$$
$$\frac{\bigg(\frac ab\bigg)}{c} =\frac{a}{bc}$$
$$\frac{\bigg(\frac ab\bigg)}{\bigg(\frac cd\bigg)}=\frac{ad}{bc}$$

I recommend using simple examples to show you, and recall that dividing by a fraction is the same as multiplying by its reciprocal. I don't know how many times I chucked $\frac{\frac 12}{\frac 35}$ into a calculator to make sure I was right about how to evaluate those fractions

Title: When to use these fractions properly?


问:
A variant of the Hungarian algorithm / assignment problem
Is there an algorithm for the following variant of the assignment problem:

Suppose we have $n$ workers $A_i$ and $m$ tasks and each worker $A_i$ has to do exactly $t_i$ tasks. (The numbers are chosen such that $m= \sum_{i=1}^n t_i$.) For any $i$ let $C_i$ be the sum of the costs of worker $A_i$. I am searching for an assignment such that $\max_i C_i$ is minimal.

I have found this variant, but this deals only with the maximal cost of one task, not with the sum of the tasks for one worker.

答:
If all $t_i$s are $1$, then (as said in the linked post), the problem becomes a perfect matching problem. Which is easy to solve.

But here, because you take a $\max$ of a sum of costs, this would translate into a hypergraph matching (where the hyperedges are, for a worker, the groups of tasks whose sum is below the threshold), which (in general) is much harder. You can't directly prove that the hardness of general hypergraph matching translates to this problem, because you can't encode general hypergraphs as arising from this problem ; but it looks like bad news...

If the $t_i$ aren't fixed (but instead we only insist that each task is assigned exactly once), this is the Santa Claus problem (but instead of giving tasks, you give presents ! Way nicer), which is difficult but has an approximation algorithm.


问:
About asymptotic in series
I solved a lot of convergence problems, but now I realized I wasn't thinking  too much applying tests and rules.

Question

Suppose we have a series:
$$
\sum_{n=2}^{\infty} \frac{\cos n}{n^{2/3}+(-1)^n}
$$
It looks like for big $n$
$$
\frac{\cos n}{n^{2/3}+(-1)^n} \approx \frac{\cos n}{n^{2/3}}
$$
And then I think we can reasonably conclude that both the series of LHS and of RHS converges or diverges simultaneously.

But how we can show that more rigorously? I mean, what Test or Rule should I apply? For example, we can't use Limit Comparison Test now, since both sequences may be negative.

答:
You can do a Taylor development (at $\infty$) and see how that goes. Namely, write
$$
\frac{1}{n^{2/3}+(-1)^n}
= \frac{1}{n^{2/3}} - \frac{(-1)^n}{n^{4/3}} + o\!\left(\frac{1}{n^{4/3}}\right) \tag{1}
$$
using that $\frac{1}{1+x} = 1-x + o(x)$ when $x\to 0$. This means that
$$
\frac{\cos n}{n^{2/3}+(-1)^n}
= \underbrace{\frac{\cos n}{n^{2/3}}}_{a_n} - \underbrace{\frac{(-1)^n\cos n}{n^{4/3}} + o\!\left(\frac{1}{n^{4/3}}\right)}_{b_n} \tag{2}
$$
the $o()$ since $|\cos n|\leq 1$. You can then write
$$
\sum_{n=1}^N \frac{\cos n}{n^{2/3}+(-1)^n}
= \sum_{n=1}^N a_n +\sum_{n=1}^N b_n \tag{3}
$$

Note that $ \sum_{n=1}^N b_n$ is absolutely convergent now, so the series on the RHS converges if, and only if, $\sum_{n=1}^N a_n$ converges.

In general, you can/should do the Taylor series development to as many terms you need to obtain as the last part an absolutely convergent series. It (very) often works.

Title: About asymptotic in series


问:
How to prove that $\neg (p \land \neg q) \lor (\neg p \land q) \equiv \neg p \lor q$?
This is what i got so far:

(~pvq) v (q∧~p)  ≡

I like using association property, but it's not possible in this case, because of (q∧~p), if I use Morgan's law on this part, this is what will happen

(~pvq) v ~(~qvp) ≡

But I do not know if we can do Distributive in this case since theres a negation mark.

How do i do this? I have to finish this ASAP. Help me please.

答:
Simply distribute $(\neg p \lor q)$ into $q \land \neg p$:

\begin{align}(\neg p \lor q) \lor (\neg p \land q) &\equiv (\neg p \lor q \lor \neg p) \land (\neg p \lor q \lor q) \\&\equiv (\neg p \lor q)\land (\neg p \lor q) \\&\equiv (\neg p \lor q)\end{align}


问:
Basic property of Hausdorff distance
Let $(M, d)$ be a compact metric space, where $d$ is a distance metric. If I define the Hausdorff distance, $d_H$ as:

\begin{equation}\nonumber
    d_H(X,Y) = \inf\{\epsilon \geq 0: X\subseteq (Y)_\epsilon\text{ and }Y\subseteq (X)_\epsilon\}
\end{equation}
where $(Z)_\epsilon$ represents the $\epsilon$-fattening of $Z$ defined as $(Z)_\epsilon=\{m \in M:\exists z \in Z \text{ such that } d(z,x)\leq \epsilon\}$. Is it true that for arbitrary compact sets $X,Y$ and $Z$ that:

\begin{equation}\nonumber
    d_H(X\cup Z,Y) \leq d_H(X,Y)+ d_H(Z,Y)
\end{equation}

I believe this is true and has something to do with the metric subadditivity property but am unsure how to proceed proving it.

答:
First, note that the "$\inf$" in the definition of $d_{H}$ is actually a "$\min$" since we are dealing with compact subsets. Thus, for all compact subsets $X$ and $Y$ of $M$, we have $X \subset (Y)_{d}$ and $Y \subset (X)_{d}$, where $d = d_{H}(X, Y)$.

Now, let $X$, $Y$ and $Z$ be compact subsets of $M$. Set $$d = \max\left\lbrace d_{H}(X, Y), d_{H}(Z, Y) \right\rbrace \, \text{.}$$ On the one hand, we have $$Y \subset (X)_{d_{H}(X, Y)} \subset (X)_{d} \subset (X \cup Z)_{d} \, \text{.}$$ On the other hand, we have $$X \subset (Y)_{d_{H}(X, Y)} \subset (Y)_{d} \quad \text{and} \quad Z \subset (Y)_{d_{H}(Z, Y)} \subset (Y)_{d} \, \text{,}$$ and hence $$X \cup Z \subset (Y)_{d} \, \text{.}$$ Therefore, we have $$d_{H}(X \cup Z, Y) \leq \max\left\lbrace d_{H}(X, Y), d_{H}(Z, Y) \right\rbrace \, \text{.}$$

Title: Basic property of Hausdorff distance


问:
Is the completion of a commutative Noetherian local ring Noetherian?
Maybe for some straightforward, but not for me:

Let $(R,\mathfrak{m})$ be  a commutative Noetherian local ring with maximal ideal $\mathfrak{m}$. Why is the completion $\widehat{R}$ of $R$ with respect to the maximal ideal $\mathfrak{m}$ again a Noetherian ring?

Thanks.

答: Yes; if $A$ is a Noetherian ring, and $\mathfrak a$ is an ideal of $A$, then the $\mathfrak a$-adic completion of $A$ is Noetherian.  This is Theorem 10.26 in Atiyah-MacDonald. (I would write the proof here for you, but it is rather involved; hopefully, this standard reference will suffice!)</p>


答:
Let $r>\max\{d_H(X,Y),d_H(Z,Y)\}$.

It is clear that $(X\cup Z)_r=(X)_r\cup (Z)_r$. By assumption,
$$
(X)_r\subseteq Y \wedge (Y)_r\subseteq X \wedge (Z)_r\subseteq X\wedge (X)_r\subseteq Z
$$
Therefore,
$$
(X\cup Z)_r\subseteq Y \wedge  (Y)_r\subseteq X\cup Z
$$
By definition, $d_H(X\cup Z,Y)\le r$. Since $r>\max\{d_H(X,Y),d_H(Z,Y)\}$ was arbitrary, it follows that
$$
d_H(X\cup Z,Y)\le \max\{d_H(X,Y),d_H(Z,Y)\}
$$

Title: Basic property of Hausdorff distance

Body:
Let $(M, d)$ be a compact metric space, where $d$ is a distance metric. If I define the Hausdorff distance, $d_H$ as:

\begin{equation}\nonumber
    d_H(X,Y) = \inf\{\epsilon \geq 0: X\subseteq (Y)_\epsilon\text{ and }Y\subseteq (X)_\epsilon\}
\end{equation}
where $(Z)_\epsilon$ represents the $\epsilon$-fattening of $Z$ defined as $(Z)_\epsilon=\{m \in M:\exists z \in Z \text{ such that } d(z,x)\leq \epsilon\}$. Is it true that for arbitrary compact sets $X,Y$ and $Z$ that:

\begin{equation}\nonumber
    d_H(X\cup Z,Y) \leq d_H(X,Y)+ d_H(Z,Y)
\end{equation}

I believe this is true and has something to do with the metric subadditivity property but am unsure how to proceed proving it.


问:
How to explain these features of the squares from 01-99?
The last two digits of the square of numbers from 1-99 have some interesting  features as observed below. Do we have simple explanations for them? Or are there any other patterns we find?

Let  the square be $(10X+Y)^2$  (X from 0-9 and Y from 1-9)

[Case X=0 and 5]The last two digits of the squares of $({Y})^2$ and  $({50+Y})^2$ are the same (example: $01^2= 01,51^2=2601$):     List 1
\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline  \
\ 01 & 04&09&16&25&36&49&64&81
 \\\hline\hline
\end{array}

[Case X=4 and 9]The last two digits of the squares of $({40+Y})^2$ and  $({90+Y})^2$  are the same((example: $41^2= 1681,91^2=8281$): List 1'
\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline  \
\ 81 & 64&49&36&25&16&09&04&01
 \\\hline\hline
\end{array}
Note List 1' is List1 in reverse order.

[Case X=1 and 6]The last two digits of the squares of $({10+Y})^2$ and  $({60+Y})^2$ are the same(example: $11^2= 121,61^2=3721$): List 2
\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline  \
\ 21& 44&69&96&25&56&89&24&61
 \\\hline\hline
\end{array}

[Case X=3 and X=8]The last two digits of the squares of $({30+Y})^2$ and  $({80+Y})^2$ are the same(example: $31^2= 961,81^2=6561$):
List 2'
\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline  \
\ 61& 24&89&56&25&96&69&44&21
 \\\hline\hline
\end{array}
Note List 2' is List 2 in reverse order.

[Case X=2 and 7]The last two digits of the squares of $({20+Y})^2$ and  $({70+Y})^2$ are the same(example: $21^2= 441,71^2=5041$): List 3
\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline  \
\ 41 & 84&29&76&25&76&29&84&41
 \\\hline\hline
\end{array}

Note List 3 is symmetric about the center value of 25 (example: $21^2= 441,29^2=841$).

Summary:

When the tenth digits differ  by 5, last two digits of squares of  two digit numbers are the same.

When the tenth digits sum to 4,9, or 14, the lists of last two digits of squares of  two digit numbers are in reverse order.

The list of last two digits of squares for $({20+Y})^2$ and  $({70+Y})^2$  is symmetric about its enter of 25.

答:
First, to make your observations a little more rigorous, let's define the last two digits of any one-digit number $D$ as $0$ and $D$. For example, the last two digits of the number $4$ are $04.$ This justifies writing $01$ for the last two digits of $(10\times 0 + 1)^2.$

Now observe that under this definition, if you have a non-negative integer $Q$ and you add any multiple of $100$ to $Q$ (or subtract any multiple of $100$), as long as the result is non-negative it will have the same last two digits as $Q$.
(I stipulate "non-negative" in order to avoid questions such as what the last two digits of $33 - 100 = -67$ are.)

Notice that for any integer $N,$
$$
(50 + N)^2 = 2500 + 100N + N^2 = (25+N)\times100 + N^2.
$$
The number $(25+N)\times100 + N^2$ is definitely not negative, because it's the square of an integer and the square of an integer cannot be negative.
But $(25+N)\times100$ is a multiple of $100$, so
$(25+N)\times100 + N^2$ has the same last two digits as $N^2.$

Now observe that $$10(X+5) + Y = 50 + (10X + Y).$$
Consider what happens if we use $10X + Y$ as $N$ in the previous paragraph.
Whatever the last two digits are when we square $10X + Y,$
increasing $X$ by $5$ will result in the same last two digits.
That's why $X=5$ gives the same results as $X=0,$
$X=6$ gives the same results as $X=1,$ etc.

Next, notice that for any integer $M,$
$$
(50 - M)^2 = 2500 - 100M + M^2 = (25-M)\times100 + M^2.
$$
Again, we have a non-negative number (because it is the square of an integer);
and since $(25-M)\times100$ is a multiple of $100$,
it follows that $(25-M)\times100 + M^2$ has the same last two digits as $M^2.$
This is true even if $(25-M)\times100$ is negative; as long as we begin and end with a non-negative integer, we can subtract a multiple of $100$ without changing the last two digits of the result.

Now observe that
$$
10(4-X) + (10 - Y) = 40 - 10X + 10 - Y = 50 - (10X + Y).
$$
This tells us that $(10(4-X) + (10 - Y))^2$ has the same last two digits as
$(10X+Y)^2.$

So let's take any digit $X$ from $0$ to $4$ and look at the sequence we get by taking the last two digits of $(10X+Y)^2$ for $Y = 1, 2, 3, \ldots, 9.$
We can replace $X$ by $4-X$ and replace $Y$ by $10-Y$ and get the exact same sequence of two-digit combinations; but $10 - Y$ counts down from $9$ when $Y$ counts up from $1$; to make $10 - Y$ count from $1$ up to $9$ we need $Y$ to count downward from $9$, and so we will go through the sequence "backwards."

That explains why the the list of last two digits for $X=4$ is the reverse of the list for $X=0$ and the list for $X=3$ is the reverse of the list for $X=1$.
It also says the list for $X=2$ is the reverse of itself, because $4 - 2 = 2$;
therefore the list of $X=2$ must be symmetric around its center entry.

And of course it is still true that adding $5$ to $X$ doesn't change the results of the last two digits, so the list for $X=9$ is the same as for $X=4$ and is the reverse of the list for $X=0$, etc.

The part of this answer above this line is the elementary explanation using the idea of multiples of an integer but not referring to modular arithmetic.
If you understand arithmetic modulo $b$ where $b$ is an integer greater than $1$,
we can shorten the proofs:

\begin{align}
(50 \pm N)^2 = 2500 \pm 100N + N^2 &\equiv N^2 \pmod{100},\\
(10(X+5)+Y)^2
 = (50+(10X+Y))^2
& \equiv (10X+Y)^2 \pmod{100}, \\
(10(4-X)+(10-Y))^2
 = (50-(10X+Y))^2
& \equiv (10X+Y)^2 \pmod{100}.
\end{align}


问:
$\int \sin( \pi x) \cos( \pi x)\ dx$
I would like some help computing this integral:

$$\int \sin(\pi t)\cos(\pi t)dt$$

Apparently, the answer should be $\frac{\sin^2(\pi t)}{2\pi}+C$ but I get $\frac{-\cos(2 \pi t)}{4 \pi}$+C instead. Here's what I did:

I rewrote the integrand as $\frac{1}{2}\sin(2\pi t)$.

I took out the constant $\frac{1}{2}$ out of the integral, so my expression became $\frac{1}{2} \int \sin(2\pi t)dt$.

I evaluated the antiderivative of $\sin(2\pi t)$, which I believe is $-\frac{\cos(2\pi t)}{2 \pi}+C$.

Lastly, I multiplied the antiderivative with $\frac{1}{2}$, yielding my final answer.

Would it be wrong to assume that both answers are equivalent, since the integral can either be rewritten using the half-angle formula or solved by U-substitution? If not, could you please explain why?

答:
$$I=\int\sin(\pi t)\cos(\pi t)\,dt$$
now use your double angle formula which says:
$$\sin(2y)=2\sin(y)\cos(y)\Rightarrow \sin(y)\cos(y)=\frac{\sin(2y)}{2}$$
and in your case you have $y=\pi t$ so:
$$I=\int\frac{\sin(2\pi t)}{2}dt$$
now if you want you could make a substitution so its clearer: $u=2\pi t\Rightarrow dt=\frac{du}{2\pi}$ so:
$$I=\frac1{4\pi}\int\sin(u)du=-\frac{\cos(u)}{4\pi}+C=-\frac{\cos(2\pi t)}{4\pi}+C$$
which is what you got. Now, remember that the constant of integration can "absorb" other numbers and so we could make manipulations like:
$$\cos(2y)=1-2\sin^2(y)$$
which would give us:
$$\frac{-\cos(2\pi t)}{4\pi}+C=-\frac{1-2\sin^2(\pi t)}{4\pi}+C=\frac{\sin^2(\pi t)}{4\pi}-\frac{1}{4\pi}+C$$
now you can just define then constants at the end as a new constant, and you see that the too solutions are actually equivalent

Title: $\int \sin( \pi x) \cos( \pi x)\ dx$


问:
What does the comma notation mean?
I am confused by the notation comma.

I know that the comma means 'AND' in Set theory as gate ($a \land b=a$ AND $b$),

But we write solution of equation as $x=1$, $2$ (the equation: $x^2-3x+2=0$)

The question is whether $x=1,2$ is WRONG?

$x=1,2$ $\iff$ $x=1$ AND $x=2$, so rewritten as $x=1$ or $x=2$.

But in so many books, it is written as $x=1$, $2$.

I am confused.....

(PS. I think the comma of $x=1$, $2$ is only notation of classification...? Is it ok?)

答:
As far as I know,

$$ x = 1, 2 $$

is rather unconventional, but most people will still understand what you mean. So it is not necessarily wrong, but it is admittedly somewhat confusing, since it appears to claim that $ x $ equals two distinct values simultaneously, which of course is impossible.

Either of these are clearer:

$$ x \in \{1, 2\} $$
$$ x = 1.5 \pm 0.5 $$

Title: What does the comma notation mean?


问:
Why does an exponential function eventually get bigger than a quadratic
I have seen the answer to this question and this one.

My $7$th grade son has this question on his homework:

How do you know an exponential expression will eventually be larger than any quadratic expression?

I can explain to him for any particular example such as $3^x$ vs. $10 x^2$ that he can just try different integer values of $x$ until he finds one, e.g. $x=6$. But, how can a $7$th grader understand that it will always be true, even $1.0001^x$ will eventually by greater than $1000 x^2$? They obviously do not know the Binomial Theorem, derivatives, Taylor series, L'Hopital's rule, Limits, etc,

Note: that is the way the problem is stated, it does not say that the base of the exponential expression has to be greater than $1$. Although for base between $0$ and $1$, it is still true that there exists some $x$ where the exponential is larger than the quadratic, the phrase "eventually" makes it sound like there is some $M$ where it is larger for all $x>M$. So, I don't like the way the question is written.

答: For one example, look at how $2^x$ grows in comparison with $x^2$. If $f(x)=2^x$, then $f(2x)=2^{2x}=(2^x)^2$. So doubling the input means that the output is squared (!). By contrast, if $g(x)=x^2$, then $g(2x)=(2x)^2=4x^2$, and so the output is quadrupled. Ultimately, squaring is much more powerful than quadrupling, and so the exponential function grows faster.


问:
Unitization of a $C^{*}$-algebra - completeness of the constructed norm
I have a question about the unitization of $C^{*}$-algebras. More precisely, a question about the proof of the following statement:

If $A$ is a (possibly unital) $C^{*}$-algebra, then there is a unique $C^{*}$-norm on the unitization $A_{1}:=A\oplus\mathbb{C}$.

Here $A_{1}$ is endowed with the component wise linear operations. The multiplication is given by $(a\oplus\lambda)(b\oplus\mu):=ab+\lambda b+\mu a\oplus\lambda\mu$ and the involution by $(a\oplus\lambda)^{*}:=a^{*}\oplus\bar{\lambda}$.

Let us restrict to the common proof of the non-unital case. Consider the bounded linear operator $L_{a}\colon A\to A$ defined by $L_{a}(b):=ab$. Using the fact that $A$ is non-unital, one can prove that $$A_{1}\to B(A),\qquad a\oplus\lambda\mapsto L_{a}+\lambda I$$ is injective. So $\|a\oplus\lambda\|:=\|L_{a}+\lambda I\|$ is a well-defined norm on $A_{1}$. With some effort one can show that it is submultiplicative and that it satisfies the $C^{*}$-identity.

In order to conclude existence (and automatically uniqueness), one also needs that this norm is complete. The proofs I found ignore completeness of this norm.

Is completeness of the above norm meant to be obvious? I really don't see it. Any help would be greatly appreciated!

答:
If you have a Cauchy sequence $(a_n \oplus \lambda_n) \subset A_1$, then you have two Cauchy sequences $(a_n)$ and $(\lambda_n)$ in $A$ and $\mathbb{C}$ respectively.

Indeed,
$$
||a_n - a_m|| = ||a_n \oplus 0 - a_m \oplus 0|| = ||P_A (a_n \oplus \lambda_n - a_m \oplus \lambda_m)|| \leq ||P_A||||a_n \oplus \lambda_n - a_m \oplus \lambda_m||
$$
where $P_A$ is a projector on the first summand. $P_A$ is bounded because it is continuous at zero: if $L_{b_n} + I\beta_n \to 0$, then $L_{b_n} \to 0$ since $A$ is strictly nonunital.

Then by triangle inequality $$ |\lambda_n - \lambda_m| \leq ||a_n - a_m|| + ||a_n \oplus \lambda_n - a_m \oplus \lambda_m||.$$

$A$ and $\mathbb{C}$ are Banach, so you can say $a_n \to a$ and $\lambda_n \to \lambda$. Then $a \oplus \lambda \in A_1$ is the limit of the original sequence:
$$
||a_n \oplus \lambda_n - a \oplus \lambda|| = ||(a_n - a) \oplus (\lambda_n - \lambda)|| \leq ||a_n - a|| + ||\lambda_n - \lambda|| \to 0.
$$

The triangle inequality follows from $||a\oplus\lambda|| = ||L_a + I\lambda||$. Also identities $||L_a|| = ||a||$ and $||I\lambda|| = |\lambda|$ were used.


答:
Is the $s$ an unknown variable to solve?

If so there really isn't much to say you have $1$ equation and $4$ unknowns and although you can solve for $s$ with that quadratic equation and have the solution in terms of $a_1, a_2, b$ so long as they are so that the radicand is not negative, there isn't much point as $a_1, a_2, b$ are for all practical purposes unlimited.

(That is to say $s^2(1-b) + (2-3b -a_1)s -a_2 = 0$ so

$s = \frac {3b +a_1-2 \pm \sqrt{(2-3b-a_1)^2 + 4a_2(1-b)}}{2-2b}$

so long as $(2-3b-a_1)^2 + 4a_2(1-b) \ge 0$ and $b\ne 1$ we can have $a_1, a_2, b$ be anything at all)

.....

Or is this supposed to be true for all $s$ no matter what $s$ is?

Then you don't have $1$ equation and $3$ unknowns; you have infinite equations and three unknowns.

So $s=0$ and $a_2 = 0$.

Let $s = 1$ and you have $(1-b) + (2-3b - a_1) = -4b-a_1 + 3 = 0$ so $a_1 = 3-4b$.

Let $s = -1$ and you have $(1-b)-(2-3b -a_1) = 2b +a_1 -1 = 0$ so $a_1 = 1-2b$

So $3-4b = 1-2b$ so $2b = 2$ so $b =1$ and $a_1=3-4 =1-2 = -1$.

....

Saying it true for all $s$ is equivalent so say we have an equation between polynomials.

$s^2 + 2s = b(s^2+3s+2) + a_1(s+2) + a_2$

On the LHS we have the polynomial $P(s) = s^2 + 2s$ and on the RHS we have the polynomial $R(x) = bs^2 + (3b + a_1) s + (2a_1 + 2b + a_2)$

Two say two polynomials are equal is to say 1) the always have the same value for all $s$ OR it is to say 2) the coefficients are equal.

ANd if the coefficients are equal then ... well, then its three equations (one for each coefficient) and $3$ unknowns.

The three equations are

1)$1=b$
2)$2=3b+a_1$
3)$0 = 2a_1 + 2b+a_2$

So $b= 1$ and $2=3+a_1$ so $a_1 = -1$ and $2(-1)+2 + a_2 = 0$ so $a_2 = 0$.

Title: Solve 1 equation and 3 unknowns


问:
Derangement problem but with functions instead of permutations, using inclusion-exclusion.
How many functions $f : [n] \to [n]$ are there with no fixed points, i.e. such that $f(i) \ne i$ for all $i$? Use inclusion-exclusion to solve it.

答:
For $k\in[n]$ let $A_k$ be the set of functions from $[n]$ to $[n]$ that have $k$ as a fixed point. Then the principle of inclusion-exclusion says that

$$\begin{align*}
\left|\bigcup_{k\in[n]}A_k\right|&=\sum_{\varnothing\ne I\subseteq[n]}(-1)^{|I|+1}\left|\bigcap_{k\in I}A_k\right|\\
&=\sum_{k=1}^n(-1)^{k+1}\binom{n}kn^{n-k}\,:
\end{align*}$$

there are $\binom{n}k$ subsets $I\subseteq[n]$ such that $|I|=k$, and there are $n^{n-k}$ functions from $[n]\setminus I$ to $[n]$ whose set of fixed points includes all of $I$. There are $n^n$ functions from $[n]$ to $[n]$, so there are

$$\begin{align*}
n^n-\sum_{k=1}^n(-1)^{k+1}\binom{n}kn^{n-k}&=n^n+\sum_{k=1}^n(-1)^k\binom{n}kn^{n-k}\\
&=\sum_{k=0}^n(-1)^k\binom{n}kn^{n-k}
\end{align*}$$

functions from $[n]$ to $[n]$ with no fixed points.

As a sanity check let $n=3$. If we list $f(1),f(2)$ and $f(3)$ in that order, the $8$ functions from $[3]$ to $[3]$ with no fixed points can be described by the triples $231$, $312$, $211$, $311$, $212$, $232$, $331$, and $332$, and indeed

$$\sum_{k=0}^3(-1)^k\binom3k3^{3-k}=27-27+9-1=8\,.$$


答:
Via Boolean algebra,

$$\overline{ p \bar{q} } + \bar{p} q = \bar{p} + q + \bar{p} q = \bar{p} + \underbrace{(1 + \bar{p})}_{=1} q = \color{blue}{\bar{p} + q}$$

Title: How to prove that $\neg (p \land \neg q) \lor (\neg p \land q) \equiv \neg p \lor q$?

Body:
This is what i got so far:

(~pvq) v (q∧~p)  ≡

I like using association property, but it's not possible in this case, because of (q∧~p), if I use Morgan's law on this part, this is what will happen

(~pvq) v ~(~qvp) ≡

But I do not know if we can do Distributive in this case since theres a negation mark.

How do i do this? I have to finish this ASAP. Help me please.


答: I assume you are talking about the completion of a local Noetherian ring $A$ with respect to the topology induced by its maximal ideal $m$. Then $\hat{A}$ is again Noetherian local ring with maximal ideal $m \hat{A}$. Reference: Matsumura's Commutative Ring Theory p. 63.</p>

Title: Is the completion of a commutative Noetherian local ring Noetherian?

Body:
Maybe for some straightforward, but not for me:

Let $(R,\mathfrak{m})$ be  a commutative Noetherian local ring with maximal ideal $\mathfrak{m}$. Why is the completion $\widehat{R}$ of $R$ with respect to the maximal ideal $\mathfrak{m}$ again a Noetherian ring?

Thanks.


问:
The Basel problem
As I have heard people did not trust Euler when he first discovered the formula (solution of the Basel problem)
$$\zeta(2)=\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$$
However, Euler was Euler and he gave other proofs.

I believe many of you know some nice proofs of this, can you please share it with us?

答:
The following proof is by Khalaf Ruhemi ( he is not a MSE member)

By partial fraction decomposition, we have
$$\frac{y}{(1+y^2)(y^2+x^2)}=\frac{1}{x^2-1}\left(\frac{y}{1+y^2}-\frac{y}{y^2+x^2}\right).$$
Integrate both sides from $y=0$ to $y=\infty$,
\begin{gather*}
\int_0^\infty\frac{y}{(1+y^2)(y^2+x^2)}\mathrm{d}y=\frac{1}{x^2-1}\int_0^\infty\left[\frac{y}{1+y^2}-\frac{y}{y^2+x^2}\right]\mathrm{d}y\\
=\frac{1}{x^2-1}\left[\frac12\ln(1+y^2)-\frac12\ln(y^2+x^2)\right]_0^\infty=\frac{1}{2(x^2-1)}\left[\ln\left(\frac{1+y^2}{y^2+x^2}\right)\right]_0^\infty\\
=\frac{1}{2(x^2-1)}\left[\ln(1)-\ln\left(\frac{1}{x^2}\right)\right]=\frac{1}{2(x^2-1)}\left[2\ln(x)\right]=\frac{\ln(x)}{x^2-1}.
\end{gather*}
Next, integrate both sides from $x=0$ to $x=\infty$
\begin{gather*}
\int_0^\infty\frac{\ln(x)}{x^2-1}\mathrm{d}x=\int_0^\infty\int_0^\infty\frac{y}{(1+y^2)(y^2+x^2)}\mathrm{d}y\,\mathrm{d}x\\
\{\text{change the order of integration}\}\\
=\int_0^\infty\frac{1}{1+y^2}\left[\int_0^\infty\frac{y\,\mathrm{d}x}{y^2+x^2}\right]\mathrm{d}y\\
=\int_0^\infty\frac{1}{1+y^2}\left[\arctan\left(\frac{x}{y}\right)\right]_0^\infty dy=\int_0^\infty\frac{1}{1+y^2}\left[\frac{\pi}{2}-0\right] \mathrm{d}y\\
=\frac{\pi}{2}\int_0^\infty\frac{1}{1+y^2} dy=\frac{\pi}{2}\arctan(y)\bigg|_0^\infty=\frac{\pi}{2}\cdot\frac{\pi}{2}=\frac{\pi^2}{4}.
\end{gather*}
Thus,
\begin{gather*}
\frac{\pi^2}{4}=\int_0^\infty\frac{\ln(x)}{x^2-1}\mathrm{d}x=\left(\int_0^1+\int_1^\infty\right)\frac{\ln(x)}{x^2-1}\mathrm{d}x\\
=\int_0^1\frac{\ln(x)}{x^2-1}\mathrm{d}x+\underbrace{\int_1^\infty\frac{\ln(x)}{x^2-1}\mathrm{d}x}_{x\to1/x}\\
=2\int_0^1\frac{\ln(x)}{x^2-1}\mathrm{d}x=-\int_0^1\frac{\ln(x)}{1-x}\mathrm{d}x-\int_0^1\frac{\ln(x)}{1+x}\mathrm{d}x\\
\left\{\text{use $\frac{1}{1+x}=\frac{1}{1-x}-\frac{2x}{1-x^2}$ in the second integral}\right\}\\
=-2\int_0^1\frac{\ln(x)}{1-x}\mathrm{d}x+2\underbrace{\int_0^1\frac{x\ln(x)}{1-x^2}\mathrm{d}x}_{x^2\to x}\\
=-2\int_0^1\frac{\ln(x)}{1-x}\mathrm{d}x+\frac12\int_0^1\frac{\ln(x)}{1-x}\mathrm{d}x\\
=-\frac32\int_0^1\frac{\ln(x)}{1-x}\mathrm{d}x\overset{1-x\to y}{=}-\frac32\int_0^1\frac{\ln(1-y)}{y}\mathrm{d}y\\
\{\text{expand $\ln(1-y)$ in series}\}\\
=\frac32\sum_{n=1}^\infty \frac{1}{n}\int_0^1 y^{n-1}\mathrm{d}y=\frac32\sum_{n=1}^\infty\frac{1}{n^2}=\frac{3}{2}\zeta(2).
\end{gather*}
So we have
$$\frac{\pi^2}{4}=\frac{3}{2}\zeta(2)\Longrightarrow \zeta(2)=\frac{\pi^2}{6}.$$

Title: The Basel problem


问:
Lipschitz constant of rescaled function
Let $f:\mathbb{R}^n \to \mathbb{R}$ be an $L$-lipschitz function. Now, for all $x \in \mathbb{R}^{n}$, define $$\tilde{f}(x) = \left\{ \begin{array}{cc}
\|x\|_2 f\left(\frac{x}{\|x\|_2}\right) & x \neq 0\\
0 & x = 0
\end{array}\right.
$$ where $\|\cdot\|_2$ is the Euclidean norm.

Two questions:

(1) I have seen it cited that the lipschitz constant of $\tilde{f}$ is at most $3L$, but I have not been able to prove this. Is there some simple trick I am missing?

(2) If we require that $f$ also satisfies $af(x) = f(a x)$ for constants $a > 0$, it seems obvious that $\tilde{f}$ is $L$-lipschitz, so what is the disconnect?

Any insight would be helpful here!

答:
The statement is not generally true. Take for example $f(x) = \cos(x) + 10$ with $L=1$, then:
$$
\tilde{f}(x) = \lvert x\rvert(\cos(x/\lvert x\rvert) + 10) = \lvert x\rvert(\cos(1) + 10) ,
$$
which has smallest Lipschitz constant $\tilde{L} = \cos(1) + 10 > 3$.

However, let $x,y\in\mathbb{R}^n\setminus\{0\}, x\neq y,$ such that $\lVert x\rVert = \lVert y\rVert = a$, then
$$
\lvert\tilde{f}(x) - \tilde{f}(y)\rvert=a\lvert f(x/a) - f(y/a)\rvert\leq aL\lVert x/a - y/a\rVert = L\lVert x-y\rVert.
$$

Now, let $x,y\in\mathbb{R}^n\setminus\{0\}, x\neq y,$ be arbitrary and assume w.l.o.g. that $a = \lVert x\rVert > \lVert y \rVert$, and let $z = ay/\lVert y \rVert$, $t=\lVert y\rVert/a, 0 < t < 1$, then
$$
\lvert\tilde{f}(z) - \tilde{f}(y)\rvert = \lvert af(z/a) - taf(z/a)\rvert = a\lvert f(z/a)\rvert\lvert t-1\rvert \leq a(1-t)(\lvert f(z/a) - f(0)\rvert + \lvert f(0)\rvert)\leq a(1-t)(L + \lvert f(0)\rvert).
$$

Therefore
$$
\lvert\tilde{f}(x) - \tilde{f}(y)\rvert \leq \lvert\tilde{f}(x) - \tilde{f}(z)\rvert + \lvert\tilde{f}(z) - \tilde{f}(y)\rvert \leq L\lVert x-z\rVert + a(1-t)(L + \lvert f(0)\rvert)\leq\\ L\lVert x-y\rVert + L\lVert y-z\rVert + a(1-t)(L + \lvert f(0)\rvert) = L\lVert x-y\rVert + taL\lvert 1/t - 1 \rvert  + a(1-t)(L + \lvert f(0)\rvert)=\\
L\lVert x-y\rVert + aL(1-t) + a(1-t)(L + \lvert f(0)\rvert) = L\lVert x-y\rVert + a(1-t)(2L + \lvert f(0)\rvert).
$$

Now,
$$
\lVert x \rVert = a \leq \lVert x-y\rVert + at \implies \frac{1}{ \lVert x-y\rVert}\leq \frac{1}{a(1-t)},
$$
thus
$$
\frac{\lvert\tilde{f}(x) - \tilde{f}(y)\rvert}{\lVert x-y\rVert}\leq L + \frac{a(1-t)(2L + \lvert f(0)\rvert)}{\lVert x-y\rVert} \leq 3L + \lvert f(0)\rvert.
$$

Because the choice of $x$ and $y$ is arbitrary, we conclude that the Lipschitz constant of $\tilde{f}$ is no larger than $3L + \lvert f(0)\rvert$.

Title: Lipschitz constant of rescaled function


问:
Can I apply Bayes' Theorem here?
I'm quite inexperienced in mathematics so I apologize in advance for using incorrect terms or fundamentally misunderstanding the application of Bayes' Theorem.

My question is about whether you can apply Bayes' Theorem in the given scenario:

A student is attempting to answer a question.

The underlying probability that a student will get a question in this category wrong is $50\%$. This is based on past data of how all questions have been answered so far.

A teacher thinks that the probability a student will get one specific question wrong is $70\%$. The teacher is accurate $90\%$ of the time.

What is the overall probability the student will get this specific question wrong?

My current understanding is that Bayes' Theorem gives this relationship:

$$
\text{Q wrong} = \text{"Student will get this specific question wrong"} \\
\text{T's guess} = \text{"Teacher thinks they will get this question wrong"}  \\
$$

$$
P(\text{Q wrong}|\text{T's guess}) = \frac{P(\text{T's guess}|\text{Q is wrong})P(\text{Q is wrong})}{P(\text{T's guess})}
$$

The interpretation being:

$P(\text{Q wrong})$ = Prior belief or "base rate"

$P(\text{Q wrong}|\text{T's guess})$ = Posterior, updated belief based on teacher's estimate

However, this model doesn't seem right. It doesn't fit the normal examples of Bayes' Theorem I've seen, like a medical test with a false positive rate since the teacher isn't just stating that they will get the question wrong, instead giving an estimate of the probability.

Very un-mathematically blindly plugging the numbers in doesn't work unfortunately:

$$
\frac{0.7 \times 0.5}{0.9} \approx 0.389
$$

Which isn't correct because the probability should be above the baseline estimate.

Is it possible to answer this question using Bayes' Theorem or using some other method? Am I missing something crucial that makes this question impossible to ask without other information?

Thanks in advance.

答:
You actually need the law of total probability here. The sample space is partitioned into when the teacher is right and when the teacher is wrong. If the teacher is right, the student should get this specific question wrong with chance 70%. If the teacher is wrong, with probability 10%, we use the baseline probability that the student gets the question wrong (for there is no other estimate). We write

$$\begin{split}P(question\_wrong)&=P(question\_wrong|teacher\_right)P(teacher\_right)+P(question\_wrong|teacher\_wrong)P(teacher\_wrong)\\
&=.7(.9)+.5(.1)\\
&=.68\end{split}$$

This is between the baseline estimate and the teacher's shenanigans.

Title: Can I apply Bayes' Theorem here?


问:
How to factor $a^2+3b^2+4ab+2ac+6bc-4b+4c-4$?
Can someone please explain in the simplest language they can manage strategies (best if one that applies to most polynomial factorization) to complete the formula above? And how in general to improve factorizing? I'm only fourteen with limited knowledge (duh) so I might not understand what you're saying but say it anyways cause then at least I'll know where to look/ get the gist of it. Thanks in advance!</p>

Body:
To factor it, I will use the "Crossover Method", I don't know if it exists, but it's the way I call it.
Well, let's go!

\begin{align}
a^2+3b^2+4ab+2ac+6bc-4b+4c-4\\
= a^2+4ab+2ac+3b^2+6bc-4b+4c-4\\
= a^2+2(2b+c)a+(3b^2+6bc-4b+4c-4)\\
= a^2+2(2b+c)a+[3b^2+(6c-4)b+4(c-1)]\\
= a^2+2(2b+c)a+[3b^2+(6c-4)b+2.2(c-1)]\\
= a^2+2(2b+c)a+[b+2(c-1)] (3b+2)\\
= a^2+2(2b+c)a+(b+2c-2)(3b+2)\\
= [a+(b+2c-2)][a+(3b+2)]\\
= (a+b+2c-2)(a+3b+2)
\end{align}

So, that's it, I hope I have helped. Thank you!


问:
How to Evaluate $\int_{0}^{1} \frac{x^2 \ln(1-x^4)}{1+x^4}dx$?
How to evaluate

$$ \int_{0}^{1} \frac{x^2 \ln(1-x^4)}{1+x^4} \,dx \approx -0.162858 \tag{1}$$

The integral arises in the computation of

$$\left( \sum_{n=1}^{\infty} \frac{(-1)^n}{n}\right)\left(\sum_{n=1}^{\infty} \frac{(-1)^n}{4n-1}\right)$$
as

$$ \scriptsize{\frac{\pi \ln(2)}{4\sqrt{2}} + \frac{\ln(2) \ln(3-2\sqrt{2})}{4\sqrt{2}} -3\ln(2) + \frac{\pi}{2}= \int_{0}^{1} \frac{x^2 \ln(1-x^4)}{1+x^4} \,dx + \int_{0}^{1} \frac{x^{1/4}}{2(1+x)}\left(\tan^{-1}(x^{1/4}) - \tanh^{-1}(x^{1/4}) \right)}dx $$

A similar Integral

$$ \int_{0}^{1}\left( \frac{x^2 \ln(2)}{x^4-1} - \frac{x^2 \ln(1+x^4)}{x^4-1}\right)dx = C-\frac{\pi^2}{16}+\frac{\ln^2(\sqrt{2}-1)}{4}+\frac{\pi \ln (\sqrt{2}-1)}{4} \tag{2} $$

Where $ C $ = Catalan Constant

Unfortunately the same techniques I used to evaluate $(2)$ have not worked for $(1)$.
I know only for integration - By parts, U-Sub, and  using Taylor Series as well as Mathematica.

Q = Is there a closed form for Integral $(1)$?

EDIT

$$ (1) = \frac{-\pi^2}{8\sqrt{2}} + \frac{\pi \ln(8)}{4\sqrt{2}} +\frac{\ln(8) \ln(3-2\sqrt{2})}{4\sqrt{2}} -\frac{\pi \ln(3-2\sqrt{2})}{8\sqrt{2}} + 4\sum_{k=1}^{\infty} \frac{(-1)^k}{4k-1}\sum_{n=1}^{k} \frac{1}{4n-1} $$

$$\sum_{k=1}^{\infty} \frac{(-1)^k}{4k-1}\sum_{n=1}^{k} \frac{1}{4n-1} = \frac{1}{64}\left(\psi^{(1)}\left(\frac{7}{8}\right)-\psi^{(1)}\left(\frac{3}{8}\right)\right) + W $$

Where $W$ is some value.

答:
Proof:

Let
$$
I=\int_0^1\frac{x^2 \log(1-x^4)}{1+x^4}\,\mathrm dx.
$$
Expanding the factor of $(1+x^4)^{-1}$ and integrating termwise yields
$$
I=-\frac{1}{4}\sum_{k=0}^\infty \frac{H_{k+3/4}}{k+3/4}(-1)^k.
$$
Working with the properties of the Pochhammer symbol we may write
$$
I=\frac{1}{3}\partial_c F\left({1,3/4 \atop c};-1\right)\bigg|_{c=7/4}-\frac{H_{3/4}}{3} F\left({1,3/4 \atop 7/4};-1\right).
$$
We need to find an explicit expression for the remaining derivative. Using this relationship we have
$$
F\left({1,3/4 \atop c};-1\right)=2^{-3/4}(c-1)\left(-\frac{1}{2}\right)^{1-c}\operatorname{B}_{-1}(c-1,7/4-c),
$$
where $\operatorname{B}_z(a,b)$ is the incomplete beta function. It follows that
$$
\partial_c F\left({1,3/4 \atop c};-1\right)\bigg|_{c=7/4}%
=\frac{1}{4}(-1)^{3/4} (3\pi+\mathrm i(4+\log 8))\operatorname{B}_{-1}(3/4,0)
-\frac{3}{4}(-1)^{1/4}\partial_c\operatorname{B}_{-1}(c-1,7/4-c)|_{c=7/4},
$$
where
$$
\partial_c\operatorname{B}_{-1}(c-1,7/4-c)|_{c=7/4}=\partial_a\operatorname{B}_{-1}(a,b)|_{a=3/4, b=0}-\partial_b\operatorname{B}_{-1}(a,b)|_{a=3/4, b=0}.
$$
We can evaluate the remaining derivatives using the derivatives of $\operatorname{B}_z(a,b)$ w.r.t. $a$ and $b$ here. We have for the first term
$$
\partial_a\operatorname{B}_{-1}(a,b)|_{a=3/4, b=0}=
\frac{1}{4} (-1)^{3/4} \left(\psi ^{(1)}\left(\frac{7}{8}\right)-\psi
   ^{(1)}\left(\frac{3}{8}\right)\right)+i \pi  B_{-1}\left(\frac{3}{4},0\right).
$$
The second term is more complicated as it has poles of the form $b^{-2}$ as $b\to 0$. The derivatives gives
$$
\partial_b\operatorname{B}_{-1}(a,b)|_{a=3/4, b=0}=\lim_{b\to 0}\left(
\frac{2^b}{b^2}{_3F_2}\left({\frac{1}{4},b,b\atop b+1,b+1};2\right)+\left(\psi ^{(0)}(b)-\psi^{(0)}\left(b+\frac{3}{4}\right)\right) \operatorname B\left(\frac{3}{4},b\right)-\log (2)\operatorname{B}_2\left(b,\frac{3}{4}\right)\right).
$$
We make the following observations as $b\to 0$:
$$
\left(\psi ^{(0)}(b)-\psi ^{(0)}\left(b+\frac{3}{4}\right)\right)
   B\left(\frac{3}{4},b\right)\sim -\frac{1}{b^2}+4 C-\frac{7 \pi ^2}{24}+\frac{\log ^2(8)}{2}-\frac{1}{2} \pi  \log (8)+\mathcal O(b),
$$
and
$$
\frac{2^b}{b^2}{_3F_2}\left({1/4,b,b\atop b+1,b+1};2\right)\sim \frac{1}{b^2}+\frac{\log 2}{b}+\frac{1}{2}\log^2 2+\frac{1}{2}{_4F_3}\left({1,1,1,5/4\atop 2,2,2};2\right)+\mathcal O(b),
$$
and
$$
\log (2)\operatorname{B}_2\left(b,\frac{3}{4}\right)\sim\frac{\log 2}{b}+\log^22+\frac{\log 2}{2}{_3F_2}\left({1,1,5/4\atop 2,2};2\right)+\mathcal O(b).
$$
These three results then give us
$$
\partial_b\operatorname{B}_{-1}(a,b)|_{a=3/4, b=0}=
4\log^22-\frac{3}{2}\pi\log 2+4 C-\frac{7 \pi ^2}{24}
+\frac{1}{2}{_4F_3}\left({1,1,1,5/4\atop 2,2,2};2\right)
-\frac{\log 2}{2}{_3F_2}\left({1,1,5/4\atop 2,2};2\right).
$$
Bringing all results together gives a final expression for $I$.

Edit:

Much of the special functions in this solution reduce to elementary functions by way of the differentiation formula
\begin{align}\left(z\frac{d}{dz}+\beta_k-1\right){}_pF_q\biggl[
\begin{array}{c}\alpha_1,\ldots,\alpha_p \\
\beta_1,\ldots,\beta_k,\ldots,\beta_q\end{array};z\biggr]&=\\
=\left(\beta_k-1\right)
{}_pF_q\biggl[
\begin{array}{c}\alpha_1,\ldots,\alpha_p \\
\beta_1,\ldots,\beta_k-1,\ldots,\beta_q\end{array};z\biggr]&.
\tag{$\spadesuit$}\end{align}

For example, this formula gives
$$
(z\partial_z+1){_3F_2}\left({1,1,5/4\atop 2,2};z\right)={F}\left({1,5/4\atop 2};z\right)=\frac{4}{z}((1-z)^{-1/4}-1).
$$
Letting $y(z)={_3F_2}(1,1,5/4; 2,2;z)$ the above formula is written as
$$
z y^\prime+y=\frac{4}{z}((1-z)^{-1/4}-1),\quad y(0)=1.
$$
Solving the differential equation we find
$$
y(z)=\frac{2}{z}\left((-2-2 i) \log \left(-\sqrt[4]{1-z}+i\right)-(2-2 i) \log \left(\sqrt[4]{1-z}+i\right)-4 \log
   \left(\sqrt[4]{1-z}+1\right)-(1-2 i) \pi +\log (64)\right).
$$
Substituting $z=2$ and simplifying then gives the result
$$
{_3F_2}\left({1,1,5/4\atop 2,2};2\right)=-\mathrm i\pi+4\log 2+(1-i) \log (3-2 \sqrt{2}).
$$
I suspect my solution for $I$ here can be reduced down to to elementary functions without the need to beta and hypergeometric functions.


问:
Classifying subgroups of $\mathbb{Z}^n$ up to isomorphism
I believe that the only subgroups of $\mathbb{Z}^n$ up to isormorphism are $\{0\}$ and $\mathbb{Z}^m$, with $m\leq n$.

This because if $z\neq 0\hookrightarrow H<\mathbb{Z}^n$ so $\langle z\rangle\sim \mathbb{Z}\hookrightarrow H$.

However, I'd like to get this formally (for instance, by a result on a lemma or a more formal proof). Could you help me?

Thank you so much

答: Since $\Bbb Z^n$ is a finitely-generated abelian group, the result follows from the Fundamental Theorem of Finitely-generated Abelian Groups, which is a classification result described in this Wikipedia article.


问:
Homeomorphisms between a circle and a square
If $A$ is $S^1$, $B$ is a unit square (only the boundary) and $f:A\rightarrow B$ is any homeomorphism then does $x$ and $y$ being antipodal points on the circle imply their images $f(a)$ and $f(b)$ will be ‘antipodal’ on the square in the sense that the length of the path connecting $f(a)$ and $f(b)$ is equal to $2$ ?

答:
No. You can easily construct homeomorphisms of $S^1$ to $S^1$ that do not preserve antipodal points, and then compose one of those with a map that radially projects points on $S^1$ to the square to get a map which also doesn't preserve antipodal points.

For the former, for example, map the upper half of the circle to only the first quarter, and then map the lower half ("$3\times$ faster") to the other three quarters.


问:
Show that the equation $\log(1+e^x)=\cos(x)$ has infinitely many negative solutions.
Show that the equation $\log(1+e^x)=\cos(x)$ has infinitely many negative solutions. Find out if there is a positive solution and if it is unique.

From the graph I can see that it has an infinitely many negative solutions and that it has one positive solution. I don't know how to prove it rigorously. Could you help me? Thank you for your time.

答: The fact is more general: given any function $f:(-\infty,0]\to[0,1)$, the equation $f(x)=\cos x$ has infinitely many solutions. Hint for a proof: consider the intermediate value theorem applied to the function $f(x)-\cos x$, in particular e.g. at the values where $\cos $ attains its max or min.</p>


问:
$0 = \sqrt{2}\cos(x) - 2\sin(x)\cos(x)$, why do I have to factor out $\cos(x)$, not to divide by $\cos(x)$?

Solve for $x$.
$$0 = \sqrt{2}\cos(x) - \sin(2x)$$
$$0 = \sqrt{2}\cos(x) - 2\sin(x)\cos(x)$$

Here, I thought I could divide all terms by $\cos(x)$ to get

$$0 = \sqrt{2} - 2\sin(x)$$

But the solution says I need to factor out $\cos(x)$ to get

$$0 = \cos(x)(\sqrt{2} - 2\sin(x))$$

not to divide by it.

Could you tell me why?

答:
The $\cos x$ term might be also a solution to this equation. If you divide it you'll loose it.

You have $$\cos(x)\cdot (\sqrt 2-2\sin x) = 0$$
Now we have three possibilities:

$\sqrt 2-2\sin x = 0, \, \cos x \neq 0$

$\cos x = 0,\,\sqrt 2-2\sin x \neq 0$

$\sqrt 2-2\sin x=  0 ,\,\cos x = 0$

If you divide by $\cos x$ you will only find zeros for the first case.
You'll loose solutions like $\pi/2+k\pi, k\in\mathbb{Z}.$ If both expressions were zero, you would lose them too since you divided by $0$.


答:
Suppose that you were trying to solve the quadratic
$$
x^2+x=0
$$
by dividing both sides by $x$. Then we get
$$
x+1=0
$$
and so $x=-1$. But quadratic equations are meant to have two solutions! One of them was lost along the way because when you divided by $x$, you did not consider what might happen if $x$ was equal to $0$. The same principle applies here. Dividing both sides by $\cos x$ is only valid when $\cos x \neq 0$, since division by zero is undefined. When you blindly divide both sides by $\cos x$, you are ignoring this fact. If you insist upon dividing by $\cos x$, then you can split the problem into two cases: (a) if $\cos x = 0$, and (b) if $\cos x \neq 0$. When dealing with case (b), we can then safely divide by $\cos x$. But this method is more cumbersome than factoring.

Title: $0 = \sqrt{2}\cos(x) - 2\sin(x)\cos(x)$, why do I have to factor out $\cos(x)$, not to divide by $\cos(x)$?

Body:
Solve for $x$.
$$0 = \sqrt{2}\cos(x) - \sin(2x)$$
$$0 = \sqrt{2}\cos(x) - 2\sin(x)\cos(x)$$

Here, I thought I could divide all terms by $\cos(x)$ to get

$$0 = \sqrt{2} - 2\sin(x)$$

But the solution says I need to factor out $\cos(x)$ to get

$$0 = \cos(x)(\sqrt{2} - 2\sin(x))$$

not to divide by it.

Could you tell me why?


问:
Give a regular grammar for L
Give a regular grammar for $L = \{a^n b^n : n \leqslant 100\}$

I would do something like this :

$S \to A \ |\ \text{empty string}$

$A \to aB \ |\  \text{empty string}$

$B \to Ab$

but How do we keep count of the number in the grammar? meaning How does it know when there are more that $100$ $a$'s. Also I'm not even sure if my way makes sense.

Any help would be appreciated.

答: Hint: Use non-terminals as memory.  In particular (and this envisions a right regular grammar) for each $n$ let $A_n$ be a nonterminal which means that (so far) $n$ $\mathtt{a}$'s have been written to the string, and let $B_n$ be a nonterminal which means that $n$ more $\mathtt{b}$'s need to be written to the string.

Title: Give a regular grammar for L


问:
Doubt in my proof of Hartshorne I (4.7).
Let $X,Y$ be varieties, and choose points $P$ and $Q$ so that there exists an isomorphism of k-algebras $f^*\colon \mathcal{O}_{P,X} \to \mathcal{O}_{Q,Y}$. Then show that there exists open sets $P \in U \hookrightarrow X$ and $Q \in V \hookrightarrow Y$, so that there is an isomorphism $f\colon U \to V$.

This was my approach:
Let $f\colon X \to Y$. Suppose there exists a k-algebra isomorphism $f^*\colon \mathcal{O}_{P,X} \to \mathcal{O}_{Q,Y}$. Then using corollary 1.4.5, and using that fact that $\mathcal{O}_{P,X} \hookrightarrow K(X)$, we then have that $K(X) \cong K(Y)$. This is precisely the definition of birational equivalence between $X$ and $Y$, by corollary 1.4.5. Thus also noting that $\mathcal O_{P,X}$ is an invariant for any open set $U \hookrightarrow X$, we obtain a birational equivalence between two open sets $U$ and $V$. Namely, there exists a point $Q \in Y$, so that a point $P \in X$ gets mapped through some morphism $f$ to $Q$. //

Is this proof correct? If not, could I have some intuition or motivation (not an answer, I want to do it myself) for a correct proof?

答:
There are couple of issues I notice in your proof:

First, a smaller issue of giving proper justifications: To conclude that $K(X) \cong K(Y)$, you're using more than just that $\mathcal{O}_{P, X}$ is a subset of $K(X)$; you're using that $K(X)$ is the field of fractions of $\mathcal{O}_{P, X}$ (and likewise for $\mathcal{O}_{Q, Y}$ and $K(Y)$).

Second, a bigger issue with the proof: At this point you've proved that $X$ and $Y$ are birationally equivalent, and hence there are some dense open subsets $U \subseteq X$ and $V \subseteq Y$ such that $U \cong V$. But how do you know you can find such subsets such that $P \in U$ and $Q \in V$? You've essentially swapped some quantifiers here—the points $P$ and $Q$ are fixed from the outset, and cannot depend on the open subsets you construct.

In fact, it's not enough to know that $X$ and $Y$ are birationally equivalent. For example, if $X$ is a singular curve with smooth projective model $Y$, and $P \in X$ is a singular point, then $\mathcal{O}_{P, X}$ is not isomorphic to the local ring $\mathcal{O}_{Q, Y}$ for any point $Q \in Y$, because the local ring detects whether a point is singular.

Here's a hint for a different approach to the problem: You'll likely need to use the definition of the local ring of a variety at a point, and not just the fact that it's a subring of the field of fractions. How is this local ring constructed, and what can you deduce from an isomorphism between such local rings?


问:
If $\sum {a_n}$ converges, does $\sum {o(a_n)}$ converges as well?
Suppose we have an arbitrary sequence $a_n$. And assume that it's series

$$
\sum a_n
$$
converges conditionally. Does that mean that the series
$$
\sum b_n
$$
also converges if $\lim_\limits{n \to \infty} b_n/a_n=0$ ?

答: This is false in general, take $a_n=\frac{(-1)^n}{\log n}$ and $b_n=\frac{1}{n}$, then $\frac{b_n}{a_n}=(-1)^n\frac{\log n}{n}\rightarrow 0$, $\sum a_n$ converges but $\sum b_n$ diverges.


问:
If you roll a fair 6-sided die and then flip a fair coin that number of times, what is the probability that you will get at least two heads?
My idea is to use disjoint events and calculating the probability of getting at least two heads for each number rolled. For example, if I roll a 3, I would calculate the probability with the expression $(\frac{1}{6}) (\frac{1}{2})^3 \binom{3}{2} + (\frac{1}{6}) (\frac{1}{2})^3\binom{3}{3})= \frac{1}{12}$ and then add up the probabilities of getting at least two for each rolls, since the events are disjoint, summing to $\frac{67}{128}$. Is this a valid solution? Is there a better approach to solving this problem?

答:
You can compute this much simpler. The probability that you get not more than one head out of $n$ flips is $\frac{n+1}{2^n}$. Therefore the probability in question is:
$$
\frac16\sum_{n=1}^6\left(1-\frac{n+1}{2^n}\right).
$$


答:
You might let $x$ get to infinity through only a thin sequence.  If you want to compare $1.01^x$ with $100x^2$, suppose $x = 1.01^n$ and let $n$ go to infinity.  Now you're comparing

$$100\cdot 1.01^{2n} \mbox{ with }  (1.01)^{1.01^n}.$$

Now it's a matter of which exponent $2n$ or $1.01^n$  get bigger;  does an exponential grow faster than a linear function?  If you have to argue yes, you can repeat the above, and reduce to whether an exponential grows faster than a constant.

Title: Why does an exponential function eventually get bigger than a quadratic

Body:
I have seen the answer to this question and this one.

My $7$th grade son has this question on his homework:

How do you know an exponential expression will eventually be larger than any quadratic expression?

I can explain to him for any particular example such as $3^x$ vs. $10 x^2$ that he can just try different integer values of $x$ until he finds one, e.g. $x=6$. But, how can a $7$th grader understand that it will always be true, even $1.0001^x$ will eventually by greater than $1000 x^2$? They obviously do not know the Binomial Theorem, derivatives, Taylor series, L'Hopital's rule, Limits, etc,

Note: that is the way the problem is stated, it does not say that the base of the exponential expression has to be greater than $1$. Although for base between $0$ and $1$, it is still true that there exists some $x$ where the exponential is larger than the quadratic, the phrase "eventually" makes it sound like there is some $M$ where it is larger for all $x>M$. So, I don't like the way the question is written.


问:
What is $f^{-1}(1)$ where $f(x,y,z)=(\sqrt{x^2+y^2}-2)^2+z^2$

I was doing a problem that requires finding the inverse image of $1$ by the function $f:\mathbb{R}^3\to \mathbb{R}$ where
$$f(x,y,z)=(\sqrt{x^2+y^2}-2)^2+z^2$$

$(x,y,z)\in f^{-1}(1) \iff (\sqrt{x^2+y^2}-2)^2+z^2=1$

$(\sqrt{x^2+y^2}-2)^2+z^2=1$

$\implies (\sqrt{x^2+y^2}-2)^2=1-z^2$

I am not sure how to complete. Any help would be appreciated. Thanks

答:
Since $f:\mathbb{R}^3\to\mathbb{R}$, then $f^{-1}:\mathbb{R}\to\mathbb{R}^3$, then the solution of $f^{-1}(1)$ is the 3-tuple $(x,y,z)$ such that:
$$
(\sqrt{x^2+y^2}-2)^2=1-z^2
$$
That seems pretty much the answer. If you wish you can expand the above expression and recover some form of conic section, if there exists (I doubt it does, due to the sqrt term).


问:
How to prove that $\neg (p \land \neg q) \lor (\neg p \land q) \equiv \neg p \lor q$?
This is what i got so far:

(~pvq) v (q∧~p)  ≡

I like using association property, but it's not possible in this case, because of (q∧~p), if I use Morgan's law on this part, this is what will happen

(~pvq) v ~(~qvp) ≡

But I do not know if we can do Distributive in this case since theres a negation mark.

How do i do this? I have to finish this ASAP. Help me please.

答:
A short version of the solution looks for example like this:

\begin{align}\neg \ (p \ \land \ \neg \ q) \lor (\neg \ p \ \land \ q) &\equiv \ (\neg \ p \ \lor q) \lor (\neg \ p \ \land \ q) \\ &\equiv \ ((\neg \ p \ \lor q) \lor \neg \ p) \ \land ((\neg \ p \ \lor q) \lor \ q) \\ &\equiv \ (\neg \ p \ \lor q) \ \land (\neg \ p \ \lor  \ q) \\ &\equiv \ \neg \ p \ \lor q \  \end{align}

A lengthy version for the above mentioned short version of the solution looks for example like this:

First some theoretical background on boolean algebra. In boolean algebra there are two distributive laws and they read as follows for 3 boolean variables $x, y$ and $z$:

$\color{blue}{x \ \land} \ ( y \ \lor \ z) \ \equiv \ (\color{blue}{x \ \land} \ y) \lor  (\color{blue}{x \ \land} \ z) $

$\color{red}{x \ \lor} \ ( y \ \land \ z) \ \equiv \ (\color{red}{x \ \lor} \ y) \land(\color{red}{x \ \lor} \ z) $

Note: In each distributive law, the boolean variable $x$ is distributed with whatever operation next to it. These laws can be proven for example with a truth table.

There are also two so called De Morgan's laws, which read as follows:

$ \neg \ ( y \ \lor \ z) \ \equiv \ (\neg \ y) \ \land \ (\neg \ z)  $

$ \neg \ ( y \ \land \ z) \ \equiv \ (\neg \ y) \ \lor\ (\neg \ z)  $

Note: The Operations exchange and the negation operates on the individual variables. This law is also valid for more than two variables. These laws can be proven for example with a truth table.

Also, the double negation of a variable $\neg \ \neg \ p $ is the variable $ p $ itself (double negation rule).

There are also associative laws for both operations $\land$ and $\lor$ and some other laws like idempotent laws, commutative laws etc. which you should be able to check, for example here: https://www.mi.mun.ca/users/cchaulk/misc/boolean.htm

A proof for the statement:

$$\neg \ (p \ \land \ \neg \ q) \lor (\neg \ p \ \land \ q) \ \equiv \ \neg \ p \ \lor \ q$$

could look like this: The first expression $\neg \ (p \ \land \neg \ q)$ can be simplified with the help of the 2nd de Morgan law as mentioned above and the double negation rule to be $\neg \ p \ \lor \ q \ $, so $$\color{red}{\neg \ (p \ \land \neg \ q)} \ \equiv \ (\neg \ p) \ \lor (\neg \ \neg \ q)\ \ \equiv \ \neg \ p \ \lor q\ . $$ If we substitute $ \color{red}{x} \ \equiv \ \neg \ p \ \lor \ q \ $, then the left side of the statement $$ \neg \ (p \ \land \ \neg \ q) \lor (\neg \ p \ \land \ q) \ $$ becomes $$ x \lor (\neg \ p \ \land \ q) \ . $$

Using the 2nd distributive law as mentioned above: $$ x \lor (\neg \ p \ \land \ q) \ \equiv \  (x \ \lor \ (\neg \ p)) \land (x \ \lor \ q) \ . $$

Plugging back $ \ \neg \ p \ \lor \ q \ \ \equiv \ x$ in this last expression, we get

$$  ((\neg \ p \ \lor \ q ) \ \lor \ (\neg \ p)) \land ((\neg \ p \ \lor \ q ) \ \lor \ q) \  $$

According to the associative law of the $\lor$-operation we can drop the brackets and get:

$$  (\neg \ p \ \lor \ q  \ \lor \ \neg \ p) \land (\neg \ p \ \lor \ q  \ \lor \ q) \  $$

Changing the positions of $ \ \neg \ p$ and $ q $ inside the left bracket (Commutative law of the $\lor$-operation) of the above expression, we get:

$$  (\neg \ p  \ \lor \ \neg \ p \ \lor \ q ) \land (\neg \ p \ \lor \ q  \ \lor \ q) \  $$

According to the idempotent law of the $\lor$-operation, we get $\neg \ p  \ \lor \ \neg \ p \ \equiv \neg \ p $ and $ \ q  \ \lor \ q \ \equiv \ q$, the above expression becomes

$$  (\neg \ p  \ \lor \ q ) \land (\neg \ p \ \lor \ q  ) \  $$

We can substitute $y \ \equiv \ \neg \ p  \ \lor \ q $ and the above expression becomes

$$  y \land y \  $$

According to the idempotent law of the $\land$-operation the above expression becomes

$$  y $$

Substituting back $  \neg \ p  \ \lor \ q \ \equiv \  y $ the left side of the statement becomes

$$  \neg \ p  \ \lor \ q   $$

the right side of the statement.

Another short version of the solution looks for example like this:

\begin{align}\neg \ (p \ \land \ \neg \ q) \lor (\neg \ p \ \land \ q) &\equiv \ (\neg \ p \ \lor q) \lor (\neg \ p \ \land \ q) &&(l_1)
\\ &\equiv \neg \ p \ \lor q \lor (\neg \ p \ \land \ q) &&(l_2)
\\ &\equiv \neg \ p \ \lor \ (q \ \land \ 1) \ \lor (q \ \land \ \neg \ p) &&(l_3)
\\ &\equiv \neg \ p \ \lor \ q \ \land \ (1 \ \lor \neg \ p) &&(l_4)
\\ &\equiv \neg \ p \ \lor \ q \ \land \ 1 &&(l_5)
\\ &\equiv \neg \ p \ \lor \ q \    &&(l_6) \end{align}

Tip: Read old texts on boolean algebra, which are free to find on the internet for example here: archive.org. In some old texts the operation "$\lor$" is called "$+$", the operation "$\land$" is called "$\cdot$" and the negation $\neg \ q$ is called $\overline{q}$. So, the same problem could also read like this: $ \overline{p \cdot \overline{q}} + \overline{p} \cdot q \ \equiv \ \overline{p}+q $ and the solution like this:

$$\begin{align}(\overline{p \cdot \overline{q}}) + (\overline{p} \cdot q) & \equiv (\overline{p}+q)+(\overline{p}\cdot q) &&(l'_1)
\\ & \equiv  \overline{p}+q+(\overline{p}\cdot q) &&(l'_2)
\\ & \equiv  \overline{p}+(q\cdot 1)+(q \cdot \overline{p}) &&(l'_3)
\\ & \equiv  \overline{p}+q\cdot (1 + \overline{p}) &&(l'_4)
\\ & \equiv  \overline{p}+q\cdot 1 &&(l'_5)
\\ & \equiv  \overline{p}  + q &&(l'_6)\end{align}  $$

As in arithmetic multiplication of variables, sometimes the "$\cdot$" is ommited, e. g. $p \cdot q \equiv pq$.

Some hints to the above solution:

In the first line $(l_1)$ or $(l'_1)$, which is the same, only De Morgan's law for the OR-operation is used on the expressions $\neg \ (p \ \land \ \neg \ q)$ or $\overline{p \cdot \overline{q}}$, which are the same thing, to get $\neg \ p \ \lor q$ or $\overline{p}+q$, which again are equal things, respectively.

In the second line we use the associative law for the OR-operation or in other words, we neglect the brackets. Here we could also neglect the brackets around the variables joined by the AND-operation, because in boolean algebra: the AND-operation has precedence over the OR-operation, like for example multiplication has precedence over addition in arithmetics (precedence rules)

In the third line we substitute for $q$ the expression $q \land 1$. It's just the same thing by the definition of the AND-operation. If $q = 0$, then $q \land 1 \equiv 0 \land 1 \equiv 0 \equiv q$ and if $q \equiv 1$, then $q \land 1 \equiv 1 \land 1 \equiv 1 \equiv q$. Now $q$ is a boolean variable and can only be either $0$ or $1$. So whatever $q$ might be, it is always $q \land 1 \equiv q$.

In the fourth line the distributive law of the AND-operation is used in the reverse direction.

To get from the fourth line to the fifth line, we use the fact that $1 \lor x \equiv 1$ regardless of what $x$ might be. This can be verified by plugging $0$ and $1$ for $x$ and checking the results as in point 3.

In general there are 4 rules for the operation of boolean variables, e. g. $x$, with boolean constants, these are the constants $0$ and $1$. The rules are: $x \land 1 \equiv x$, $x \land 0 \equiv 0$, $x \lor 1 \equiv 1$, $x \lor 0 \equiv x$.

You might would like to try finding a way of forming the right hand side $\neg \ p \ \lor \ q$ into the left handside $ \neg \ (p \ \land \ \neg \ q) \lor (\neg \ p \ \land \ q) \ $ to get a third possible solution, maybe like this:

\begin{align}
\neg \ p \ \lor q &\equiv \ ((\neg \ p) \ \lor q) \ \land ((\neg \ q) \ \lor  \ q)
\\ &\equiv  \ (((\neg \ p) \ \lor q) \ \land (\neg \ q)) \ \lor  \ (((\neg \ p) \ \lor q) \ \land ( \ q))
\\ &\equiv  \ ((\neg \ p)\land (\neg \ q)) \ \lor (q \land (\neg \ q)) \  \ \lor  \ ((\neg \ p) \land  \ q) \ \lor \ (q \ \land  \ q)
\\ &\equiv  \ ((\neg \ p)\land (\neg \ q)) \  \lor  \ 0 \ \lor \ ((\neg \ p) \land  \ q) \ \lor \ q
\\ &\equiv  \ ((\neg \ p)\land (\neg \ q)) \  \lor  \ ((\neg \ p) \land  \ q) \ \lor \ q
\\ &\equiv  q \ \lor \ ((\neg \ p)\land (\neg \ q)) \  \lor  \ ((\neg \ p) \land  \ q)
\\ &\equiv  (q \ \lor \ (\neg \ p)) \land (q \ \lor \ (\neg \ q)) \  \lor  \ ((\neg \ p) \land  \ q)
\\ &\equiv  (q \ \lor \ (\neg \ p)) \land 1 \  \lor  \ ((\neg \ p) \land  \ q)
\\ &\equiv  (q \ \lor \ (\neg \ p)) \  \lor  \ ((\neg \ p) \land  \ q)
\\ &\equiv  \neg \ ((\neg q) \ \land \ p) \  \lor  \ ((\neg \ p) \land  \ q)
\\ &\equiv  \neg \ (p \ \land \ (\neg q)) \  \lor  \ ((\neg \ p) \land  \ q)
\\ &\equiv  \neg \ (p \ \land \ \neg \ q) \  \lor  \ (\neg \ p \land  \ q)
\end{align}

using many different rules of boolean algebra and the fact that $(\neg \ q) \lor \ q \equiv 1$ regardless of the value of $q$ and also the fact that $x \land 1 \equiv x \lor 0 \equiv x$ regardless of the value of $x$.


问:
Show that $\{P_n\}_{n \geq 1}$ does not converge in norm topology.

Let $\mathcal H = \ell^2 (\Bbb N)$ and $P_n$ be the projection onto $\text {span}\{e_0,e_1, e_2, \cdots, e_n \}.$ Show that the sequence $\{P_n\}_{n \geq 1}$ does not converge in norm topology.

Attempt $:$ Here for any $x \in \ell^2$ we have $$P_n (x) = \sum\limits_{i=0}^{n} \left \langle e_i, x \right \rangle e_i.$$

So if $P_n \xrightarrow {\text {NORM}} P$ then $P_n \xrightarrow {\text {SOT}} P,$ where SOT denotes the strong operator topology. But then $$P(x) = \sum\limits_{n=0}^{\infty} \left \langle e_n, x \right \rangle e_n$$ The series on the right is the Fourier-Bessel series for $x$ which converges and since $\{e_n\ |\ n \geq 0 \}$ is an orthonormal basis for $\ell^2$ it follows that the Fourier-Bessel series for $x$ converges to $x.$ Hence $Px = x,$ for all $x \in \mathcal H.$ But then $P_n \xrightarrow {\text {NORM}} I,$ where $I$ is the identity operator. But this in turn implies $$\sup\limits_{\|x\| = 1} \sum\limits_{i \gt n} \left \lvert \left \langle e_i, x \right \rangle \right \rvert^2 \longrightarrow 0\ \text {as}\ n \to \infty.$$ Can it happen in any circumstances?

Any help or suggestion in this regard will be highly appreciated. Thanks for your time.

答:
For fixed $n$, choose some $j>n$, and choose $x=e_j$. Then $\|x\|=\|e_j\|=1$, so $\sum_{i>n}|\langle e_i,x\rangle|\geq |\langle e_j,x\rangle|=|\langle e_j,e_j\rangle|=1$.
Hence $\sup_{\|x\|=1}\sum_{i>n}|\langle e_i,x\rangle|\geq 1$ independently of $n$. So indeed, we cannot have $\sup_{\|x\|=1}\sum_{i>n}|\langle e_i,x\rangle|\to 0$ if $n\to\infty$.


问:
Show that a matrix has eigenvalues at zero
Let $A\in\mathbb{R}^{n\times n}$, $B\in\mathbb{R}^{n\times m}$, $C\in\mathbb{R}^{m\times n}$, and define
\begin{equation}
X:=\begin{bmatrix}A&BC\\I_n&0\end{bmatrix}\in\mathbb{R}^{2n\times 2n}.
\end{equation}
Show that if $m<n$, then $X$ has $n-m$ eigenvalues at zero.

答:
Exploit the fact that there is an identity matrix in the bottom left and the zero matrix in the bottom right. By adding multiples of the bottom n rows to the top n rows, you can make A zero without affecting the rest of the matrix. Then permute columns to get the matrix into the form

$$\begin{bmatrix} BC & 0 \\ 0 & I_n \end{bmatrix} $$

From this block diagonal form you can see the rank is n plus the rank of BC.


问:
Difference between a measurable set in $\mathbb{R}$ and $\mathbb{R}^n$ (Folland Theorem 2.40)

In the proof of Theorem 2.40 b), it says the proof follows from Theorem 1.19. But in Theorem 1.19, we have $E \subset \mathbb{R}$, not $\mathbb{R}^n$. So why does Theorem 2.40 still hold with $E \subset \mathbb{R}^n$?

Thanks!

答:
Attention: In the proof of Theorem 2.40 b), it does not say the proof follows from Theorem 1.19.

The exact sentence in Folland is : "This proves the first equation in part (a); the second one, and part (b), then follows as in the proofs of Theorem 1.18 and 1.19."

Note that the expression used is "...follows as in the proofs...", and it  means the second equation in part (a), and part (b), can be proved in a similar way to the proofs of  Theorem 1.18 and 1.19. But it was never implied that Theorem 2.40 b) is a consequence of Theorem 1.19.


问:
Invers functions on manifolds
I know that For a function $f:\mathbb{R} \to \mathbb{R} $, any function $g:\mathbb{R} \to \mathbb{R}$ that satisfies the condition

$(f \circ g) (x)=x$

$(g \circ f) (x)=x$

$g$ would be the inverse function of $f$. I was wondering weather that work for manifolds or not.

I have this function

$$F:\mathbb{R}P^2 \to \mathbb{R}P^2$$

$$F([x,y,z])=[-x,z,y]$$

I notice that

$(F\circ F) ([x,y,z])=F([-x,z,y])=[x,y,z]$

Then I conclude that $F$ is its own inverse. But then I had some questions:

Can a function be its own inverse (on $\mathbb{R}$ or any other manifolds)?

Is the definition of inverse function on $\mathbb{R}$ holds for manifolds?

I feel like that $(F\circ F) ([x,y,z])=F([-x,z,y])=[x,y,z]$ does not guarantee that $F$ is its own inverse because of the equivalence classes.

Any help would be appreciated. Thank you all.

答:
A function can be its own inverse, yes. On the reals, consider $f(x) = -x$.

On $\Bbb R^2$, reflection about the line $x = y$ is another such map, defined by
$$g(x, y) = (y, x).$$

The same is true for manifolds --- there can be functions that are their own inverses ---  and you've constructed an example, one that's mostly made up as a combination of the functions $f$ and $g$ I mentioned above.

There's no problem with equivalence classes at all.

Title: Invers functions on manifolds


问:
Show the intersection of 2 subspace topologies is a subspace

Background

The following problem is out of my head. You can refer to Topology Without Tears by Sidney A. Morris for definitions:

Theorem

Let (X $\tau$) Y,Z $\subseteq X $ be a subspace of X with thier
respective topologies inherited from X.
Show $Y\cap Z $ is a subspace  .

My Understanding

${\tau}_Y$={Y$\cap$ O:O$\in\tau$} and ${\tau}_Z$={Z$\cap$ O:O$\in\tau$}

$\exists$ U open in Y s.t U=Y$\cap O$

$\exists$ V open in Z s.t V=Z$\cap O$

Then

U$\cap V$ =(Y$\cap O )\cap( Z\cap O$)=(Y$\cap$ Z)$\cap$ O

Since Y ,Z are sets, they are subspaces of X
Hence Y$\cap Z $ is a subspace too

答:
You made up this problem yourself I believe (I couldn't find it in the mentioned book), so the problem is unclear. An interpretation:

$Y \cap Z $ can inherit the subspace topology from $Y$, from $Z$ and from $X$ itself. The result is the same in all cases:

$O \subseteq Y \cap Z$ open when seen from $X$, means that we can write $O = O' \cap (X \cap Y)$ where $O'$ is open in $X$. But then $(O'\cap Y) \in \mathcal{T}_Y$ and $(O' \cap Y) \cap Z = O$ and so $O$ is also open seen as a subspace of $Y$, and similarly from $\mathcal{T}_Z$.

It also follows from exercise 4 in paragraph 4.1 where subspaces are first introduced.


问:
Help proving that $\lim(s_n+t_n)=+\infty$

If $\lim s_n=+\infty$ and if $(t_n)$ is a bounded sequence, then $\lim(s_n+t_n)=+\infty$.

My proof:

Suppose $\lim s_n=+\infty$ and $(t_n)$ is bounded.

Since $\lim s_n=+\infty$, by definition we have that for all $M\in\mathbb{R}$, there exists $N\in\mathbb{N}$ such that $n\geq N$ implies that $s_n>M$.

Since $(t_n)$ is bounded, we have that for all $n\in\mathbb{N}$, $k\leq t_n\leq k'$ for some $k,k'\in\mathbb{R}$.

This is where I get stuck. I'm not sure how to put these two definitions together coherently. It's obvious to me that $t_n$ being bounded means that $s_n + t_n >M$. However, I'm having trouble arriving at this result from the definitions that I've stated.

Maybe since $\lim s_n=+\infty$, I could say that $s_n>M-t_n$ which implies that $s_n+t_n>M$. But I'm not sure how the definition of $t_n$ being bounded plays into this.

Am I on the right track? Any hints/advice would be helpful. Thank you!

答:
You want to prove that $\lim (s_n + t_n) = \infty$.

Given some $A \in \mathbb{R}$.

We know that there exists $K$ such that $|t_n| < K$ for all $n$.

Since $\lim s_n = \infty$, we can pick $N$ such that $n\ge N\implies s_n \ge A+K$.

Thus, $n\ge N \implies s_n+t_n\ge A \ \ \blacksquare$.

Comment: I used to struggle with these proofs as well. Make sure to think about what exactly you want to prove. I always start these proofs with "given some $A$ / given some $\epsilon$ (as appropriate)". Consider the definition of the statement as a "challenge": if an adversary gives you some $A$, how do you come up with $N$ that beats the challenge?

Further comment on your attempt:

The way you did it almost worked (you got $s_n>M-t_n$). You just have to see that $|t_n|<K$.


答:
Consider the function $g:\mathbb{C} \to \mathbb{R}$ given by:

The domain of $g$ is $\{z : 1 \leq |z| \leq 3\}.$

$g(z) = |z| - 2.$

Then, similar to the idea that $\sin^2(\theta) + \cos^2(\theta) = 1$, you imagine the points on the unit circle in $\mathbb{R^2}$, where the $x$ coordinate is $g(x + iy)$, and the $y$ coordinate is the value $z$ from the original function $f$.

Then, these points on the unit circle will represent all possible ordered triplets $(x,y,z)$ from the original $\mathbb{R^3}$ such that $f(x,y,z) = 1.$

Title: What is $f^{-1}(1)$ where $f(x,y,z)=(\sqrt{x^2+y^2}-2)^2+z^2$

Body:
I was doing a problem that requires finding the inverse image of $1$ by the function $f:\mathbb{R}^3\to \mathbb{R}$ where
$$f(x,y,z)=(\sqrt{x^2+y^2}-2)^2+z^2$$

$(x,y,z)\in f^{-1}(1) \iff (\sqrt{x^2+y^2}-2)^2+z^2=1$

$(\sqrt{x^2+y^2}-2)^2+z^2=1$

$\implies (\sqrt{x^2+y^2}-2)^2=1-z^2$

I am not sure how to complete. Any help would be appreciated. Thanks


问:
Identities with Adjoints
The classical adjoint $\operatorname{adj}(A)$ of a square matrix $A$ has its $(i,j)$-th entry equal to the $(j,i)$-th cofactor (signed minor) of $A$. If $\det(A)\neq0$ we can define the inverse $A^*$ of $A$ as $\operatorname{adj}(A)/\det(A)$ and use $\operatorname{adj}(A)=\det(A)A^*$. It is known that, for arbitrary $n\times n$ matrices $A$ and $B$, $\operatorname{adj}(AB)=\operatorname{adj}(B)\operatorname{adj}(A)$. The property is either not stated in popular Linear Algebra books or, if stated, it is only proven for invertible matrices by using the properties:
\begin{align*}
\operatorname{adj}(AB)
&= \det(AB)(AB)^*\\
&= (\det A)(\det B)B^*A^*\\
&= (\det(B)B^*)(\det(A)A^*)\\
&= \operatorname{adj}(B)\operatorname{adj}(A).
\end{align*}
MY POINT is that no new proof is required when $A$ and $B$ are not necessarily invertible. Just look at the equality as at a collection of n.n polynomial type identities, each in terms of n.n entries of and $A$ and n.n entries of $B$. It is enough to consider generic matrices $A$ and $B$, having independent variables as entries. In such a case $A$ and $B$ have nonzero determinants and so they are invertible. Thus, the proof mentioned above applies and the identity must be true for each pair of $n\times n$ matrices $A$ and $B$.
My method would also apply to a few other adjoint identities, which are true in general but only proven for invertible matrices.
A few of my colleagues (teachers) could not easily agree with my approach although one finally did. What is your opinion ?</p>

答:
Let $R$ be a (commutative) domain and consider its quotient field $F$; build the field $F(\mathcal{X})$ where $\mathcal{X}$ is a set of the $2n^2$ indeterminates $X_{11},\dots,X_{1n},\dots,X_{n1},\dots,X_{nn}$ and $Y_{11},\dots,Y_{1n},\dots,Y_{n1},\dots,Y_{nn}$. Then the matrices $\mathbb{A}=[X_{ij}]$ and $\mathbb{B}=[Y_{ij}]$ are obviously invertible in $F(\mathcal{X})$ because their columns are certainly linearly independent.

Notice now that $\def\adj{\mathrm{adj}}\adj(\mathbb{A})$, $\adj(\mathbb{B})$ and $\adj(\mathbb{AB})$ have their coefficients in $R[\mathcal{X}]$ (the polynomial ring). Then

$$\adj(\mathbb{AB})=\adj(\mathbb{B})\adj(\mathbb{A})$$

because the matrices are invertible.

Now, if you have two matrices $A=[a_{ij}]$ and $B=[b_{ij}]$ with coefficients in $R$, just specialize the above relation with the evaluation $X_{ij}\mapsto a_{ij}$ and $Y_{ij}\mapsto b_{ij}$.

Title: Identities with Adjoints


问:
Condition for value function to be strictly concave
Consider the optimization problem
$$
c(p) = \min_x \sum_{i=1}^n x_ip_i
$$
subject to $f(x)\geq 1$ where $f:\mathbb{R}^n_+\mapsto \mathbb{R}$ is increasing and concave. I know that $c$ is concave (see here for a proof). My question is: are there conditions on $f$ such that $c$ is strictly concave? Many thanks for any help!

答: Notice that since $f$ is increasing, $p_i \geq 0$ for all $i$; otherwise we do not have the solution for problem. Then at the optimal solution $x(p)$, we must have $f(x(p))=1$. Using this remark, one sufficient condition for $c(p)$ is strictly concave is that $f$ is strictly concave.


问:
Express $\operatorname{sech}^{-1}(x)$ in terms of logarithms
I'm trying to express the following  $\operatorname{sech}^{-1}(x)$ in terms of logarithms, and would warmly appreciate feedback towards my approach. The solution should be :

$$\ln\left(\dfrac{1+\sqrt{(1-x^2)}}{x}\right)$$ when $x > 0.$ However, I cannot seem to get this. My working out:

$y = \operatorname{sech}^{-1}(x)\implies \operatorname{sech}(y) = x$

$$-\operatorname{sech}(y)\operatorname{tan}(y) = x; $$

Once I got to here, I decided to transform the $\operatorname{sech}$ and $\tan$ into their hyperbolic identities.

$$-\dfrac{\operatorname{sinh}(y)}{\operatorname{cosh}^2(y)}=\dfrac{e^y-e^{-y}}{2}(\dfrac{e^y+e^{-y}}{2})^2 = x$$

Though I've tried reworking this into a quadratic form $\dfrac{b \space \pm \space \sqrt{b^2-4ac}}{2a}$, however, I couldn't manage to get the right form. I would greatly appreciate some help on the next steps towards this.

答:
Note

\begin{align}
\text{sech}\left(\ln\frac{1+\sqrt{1-x^2}}x \right)
=& \frac2{ \exp\left(\ln\frac{1+\sqrt{1-x^2}}x \right)+\exp\left(-\ln \frac{1+\sqrt{1-x^2}}x\right)}\\
 = &\frac2{ \frac{1+\sqrt{1-x^2}}x + \frac x{1+\sqrt{1-x^2}}}
=\frac2{ \frac{1+\sqrt{1-x^2}}x + \frac{1-\sqrt{1-x^2}}x }
=x
\end{align}

Thus

$$\text{sech}^{-1}x =\ln\frac{1+\sqrt{1-x^2}}x $$


问:
Integral comparison test for $\frac{1}{x \ln(x)^2} $ or $\frac{1}{x \log(x)^2}$ on $(0, \frac{1}{2})$
I am trying to show $f(x) = \frac{1}{x \ln(x)^2}$ on the interval $(0,\frac{1}{2})$ is in $L^p$ only for $p=1$ and not for $p>1$. I can show $||f||_{L^1( (0,\frac{1}{2}) )} < \infty $ but for $p>1$, I have a hard time figuring out how to evaluate $\int_0^{\frac{1}{2}} \frac{1}{x^p \ln(x)^{2p}} dx$. Change of variable with $u= \ln(x)$ gives me $\int_{-\infty}^{\ln(\frac{1}{2})} \frac{1}{(e^u)^{p-1} u^{2p}} du$ which looks like I should try an integral test instead.

答:
HINT:

For any $\alpha>0$, note that $\frac{x^\alpha -1}{\alpha x^\alpha}\le\log(x)$.  For $0<x<1$, this gives the inequalities

$$\frac1{\log^2(x)}\ge \frac{\alpha^2 x^{2\alpha}}{(x^\alpha -1)^2}$$


问:
Geodesic Uniqueness in the Hyperbolic Plane
I am studying Hyperbolic Geometry. At this part, I have proved that semicircles and straight lines orthogonals to the real axis are geodesics in the hyperbolic plane. But how I proof that this geodesics are uniques? That it does not exist others geometric places between two different points that minimizes the hyperbolic length? </p>

<p>I have seen a topic related to this, but I do not know nothing about Riemannian Geometry. If someone could indicate to me some references to this proof(or the idea behind) I will be very grateful! </p>

答:
There is a general fact in Riemannian geometry which follows from "standard" existence and uniqueness theorems in the study of ODEs.  Mainly, given any point $p$ in a Riemannian manifold $M$ and given any $v\in T_pM$, there is a unique geodesic defined for some open neighborhood of time around 0 starting at $p$ in the direction of $v$.

Assuming this, it's not too hard to show that you've found all the geodesics.  You just need to verify that for any point $p$ in the hyperbolic plane and tangent vector,  either the vector points up/down (so a vertical line geodesic has $v$ as tangent vector), or some semicircle through $p$ hitting the $x$-axis perpendicularly does.

Finally, to show there is a unique geodesic between any two points, one argues as follows:

Given $p$ and $q$, if $q$ and $p$ have the same $x$ coordinate, a vertical line passes through them.  Further, no semicircle hitting the $x$-axis perpendicularly can go through both $p$ and $q$ since the semicircle is the graph of a function, so passes the vertical line test.

Next, given $p$ and $q$ with different $x$ coordinates, it's clear that a vertical line can't connect them, so we need only find all semicircles passing through both perpendicular to the $x$-axis.  Algebraically, proceed by noting the equation of such a circle is $(x-a)^2 + y^2 = r^2$ for some $a$ and $r$.  Plugging in $(x,y) = (p_1,p_2)$ and $(x,y) = (q_1,q_2)$ respectively and setting the two equal to each other (since they both equal $r^2$), one gets $$(p_1-a)^2 + p_2^2 = (q_1-a)^2 + q_2^2.$$

By simplifying this, one sees the $a^2$ cancels out and so one gets an equation linear in $a$, so $a$ is uniquely determined.  (The assumption that $p_1\neq q_1$ means that the equation has the linear piece).  Knowing $a$ easily implies that $r$ is uniquely determined (keeping in mind that $r \geq 0$).  This shows that there is a unique semicircle perpendicular to the $x$-axis going through $p$ and $q$.


答:
The upshot is that you have an fourth-degree surface in $3$-space. Probably too hard for a purely algebraic analysis. But I think that the approach of @user2661923 is the right one. The whole thing looks as if it might be a torus sitting in space.

Here, however is how that purely algebraic approach would work, going one more step than @TheVal has: we start with his displayed formula,
\begin{align}
\left(\sqrt{x^2+y^2}-2\right)^2&=1-z^2\\
x^2+y^2-4\sqrt{x^2+y^2}+4&=1-z^2\\
x^2+y^2+z^2+3&=4\sqrt{x^2+y^2}\\
x^4+y^4+z^4+\cdots+9&=16x^2+16y^2\,,
\end{align}
where I did expand the left-hand side of the last two lines, but the result is probably not helpful to the understanding. The steps are, first, to expand the parenthesis of the first line; second, to isolate the radical on the right; third, to square both sides. We’re doing algebraic geometry at this stage, folks, so you don’t worry about extraneous solutions. Gather all terms to the left and get a quartic polynomial in three variables. I would not presume to proclaim authoritatively that this is irreducible, but I would at least bet a nickel on the claim.


问:
Is it possible to eliminate a variable from these two differential equations?
I have two second order differential equations in three variables that I need to use to establish a relation between two of the variables.

The equations are

$\alpha$ $ x $ + $\beta$$y$ = $A$$(cx + x'')$$ $ + $\gamma$r

$\alpha$ $ x $ + $\beta$$y$ = $B$$(cx + x'')$$ $ + $\delta$r

where $x$, $y$, $r$ are functions of $z$, and $\alpha$, $\beta$, $\gamma$, $\delta$, $A$, $B$ are constants, and $x''$ = $\frac{\partial^2x}{\partial^2z}$.

I need to get a relation between $y$ and $r$ by eliminating $x$ from the equations, but cannot manage to do this. Is it actually possible to eliminate $x$ from these equations?

Any help would be greatly appreciated.

答:
[If $c$ isn't constant this doesn't work.]

Rearranging your first Eqn gives
\begin{align}
x''+(c-\frac{\alpha}{A})x=\frac{\beta y-\gamma r}{A}\equiv f(z)
\end{align}
Letting
\begin{align}
u&=\kappa x+x'\\
u'&=\kappa x'+x'',
\end{align}
and substituting into your ODE gives that
\begin{align}
u'+\frac{Ac-\alpha}{A\kappa}u-\left[\frac{Ac-\alpha}{A\kappa}+\kappa\right]x'=f(z),
\end{align}
define $\kappa$ such that your $x'$ term goes to zero. . .
\begin{align}
u'-\kappa u=f(z),
\end{align}
using integrating factor arrive at the solution for $u$
\begin{align}
u(z)=e^{\kappa z}\left(c_1+\int e^{-\kappa z}f(z)\mathrm dz\right)\equiv f_2(z).
\end{align}
Then solve for $x$ using the definition of $u$
\begin{align}
x(z)=e^{-\kappa z}\left(c_2+\int e^{\kappa z}f_2(z)\mathrm dz\right),
\end{align}
or written out completely,
\begin{align}
x(z)=e^{-\kappa z}\left(c_2+\int\left[ e^{2\kappa z}\left(c_1+\int e^{-\kappa z}\left[\frac{\beta y-\gamma r}{A}\right]\mathrm dz\right)\right]\mathrm dz\right).
\end{align}
I hope this helps.

Pro tip: You can use this method to solve linear, non-homogeneous, constant-coefficient ODE's of any order!


答:
Here is another way: decompose $X$ into:
$$
X =\begin{bmatrix}A & I\\ I & 0\end{bmatrix}\begin{bmatrix}I & 0\\0 & BC\end{bmatrix} = \tilde{A}\tilde{B},
$$
and observe that $\tilde{A}$ is invertible. Then you have the generalized eigenvalue problem
$$
\tilde{B} x = \lambda\tilde{A}^{-1}x.
$$
You already concluded that $BC$ has at least $n-m$ zero eigenvalues, which implies that $\tilde{B}$ also has at least $n-m$ zero eigenvalues. Let $x$ be a nonzero eigenvector corresponding to such an eigenvalue, then we have:
$$
\lambda \tilde{A}^{-1}x = 0,
$$
but this is only possible whenever $\lambda = 0$, which is an eigenvalue of $X$.

Title: Show that a matrix has eigenvalues at zero

Body:
Let $A\in\mathbb{R}^{n\times n}$, $B\in\mathbb{R}^{n\times m}$, $C\in\mathbb{R}^{m\times n}$, and define
\begin{equation}
X:=\begin{bmatrix}A&BC\\I_n&0\end{bmatrix}\in\mathbb{R}^{2n\times 2n}.
\end{equation}
Show that if $m<n$, then $X$ has $n-m$ eigenvalues at zero.


答: For $p=1,$ the integral can be evaluated in closed form. For $p>1,$ your integral is greater than the integral of $1/x$ which diverges at $0$ (since $\log(x)^2$ grows slower than any power of $x$ at $0$).

Title: Integral comparison test for $\frac{1}{x \ln(x)^2} $ or $\frac{1}{x \log(x)^2}$ on $(0, \frac{1}{2})$

Body: I am trying to show $f(x) = \frac{1}{x \ln(x)^2}$ on the interval $(0,\frac{1}{2})$ is in $L^p$ only for $p=1$ and not for $p>1$. I can show $||f||_{L^1( (0,\frac{1}{2}) )} < \infty $ but for $p>1$, I have a hard time figuring out how to evaluate $\int_0^{\frac{1}{2}} \frac{1}{x^p \ln(x)^{2p}} dx$. Change of variable with $u= \ln(x)$ gives me $\int_{-\infty}^{\ln(\frac{1}{2})} \frac{1}{(e^u)^{p-1} u^{2p}} du$ which looks like I should try an integral test instead.


问:
Losing solutions when dividing by an expression
There's something that's been  aggravating me when solving implicit differentiation. For instance, take the implicitly defined curve

$$\cos x+\cos y=\frac{1}{2}$$

(Differentiate both sides w.r.t x)

$$-\sin x-\sin y \frac{dy}{dx}=0$$

$$-\frac{dy}{dx}\sin y= \sin x$$

$$\frac{dy}{dx}=-\frac{\sin x}{\sin y}$$

The last step is the one which really confuses me, which when we divide by $(-\sin y)$: aren't we saying by doing this step that $(-\sin y)$ won't be equal to zero; aren't we kind of erasing or canceling certain values on the graph / function?

Like when we do in questions such as:

$3\cos x-12\sin x\cos^2x=0$,

$0<x<\pi$

We can make the equation look neater by dividing by $\cos x$ but we will lose solutions, as $\cos x$ won't be equal to zero and $x=\pi / 2$ won't be among our answers.

答:
You should note that "dividing won't always work, but factorising always works"

$$(\sin(y))\cdot \frac{dy}{dx} = -\sin x$$ is always true, but

$$\frac{dy}{dx}=-\frac{\sin x}{\sin y}$$ is only true where $\sin y \neq 0$.

Likewise, if you factorise your trig equation as $\cos x (3-12\sin x \cos x) = 0$, you won't miss any solutions, because it is possible that $\cos x = 0$.

Sidenote: I encourage you to plot your graph on Desmos (or other software). It's actually kind of pretty.