问:
Combining Cayley transform and Fourier series
If one has a function $F(x)$ defined on the real line ($x \in \mathbb{R}$) then one can study it by means of its Fourier transform. Because $\mathbb{R}$ is not compact one has a Fourier integral rather than a Fourier series (assuming $F$ is sufficiently nice that it can be expressed as such). However $\mathbb{R}$ can be mapped by the Cayley transform to the unit circle $C: x \mapsto \frac{i-x}{i+x}$ and so composing $F$ with the Cayley transform one can define $F$ on the unit circle by $F \circ C^{-1}$. One then can compute Fourier coefficients
$$a_n = \frac{1}{2\pi} \int_{\mathbb{T}} F \circ C^{-1}(e^{i\theta}) e^{-in\theta} \, d\theta = \frac{1}{\pi} \int_{\mathbb{R}} F(x) \left( \frac{i+x}{i-x} \right)^n \frac{1}{1+x^2} \, dx $$
$F \circ C^{-1}$ is of course not defined at $-1$ but $\{ -1 \}$ is a set of measure zero and so we could let $F \circ C^{-1}$ take some arbitrary value at this point.

If $F$ is continuously differentiable, and if $F(\infty) = F(-\infty)$ and $F^\prime(\infty) = F^\prime(-\infty)$ then it would seem the Fourier series of $F \circ C^{-1}$ converges pointwise uniformly on the circle and so the series
$$\sum_{n \in \mathbb{Z}} a_n \left( \frac{i-x}{i+x} \right)^n$$
should converge pointwise uniformly to $F(x)$ on $\mathbb{R}$. Is this analysis correct and has this approach ever been studied before?

答:
Since $F$ is defined on $\mathbb R$ and its Cayley transform $C$ is defined on the circle $\mathbb T,$ I think it might be good to define $\tilde F = F\circ C^{-1}.$

Note that with your conditions on $F,$ $\tilde F$ extends to a function continuous on $\mathbb T$ from the $F(\infty)=F(-\infty)$ condition. Secondly, since $F'(\infty)=F'(-\infty),$ then $F'(\infty)=0.$ If not, then $|F(x)|$ must grow to $\infty$ as $x\to \pm \infty.$ That violates $F(\infty)=F (-\infty).$ Thus, intuitively, $F$ is close to being constant on $\{|x|>R\}$ if $R$ is large.

On the convergence question: I tried but did not answer this fully. One thing we can say: The Fourier series of  $\tilde F$ converges at each point of $\mathbb T\setminus \{-1\}.$ That's because i) $\tilde F= F\circ C^{-1}$ is bounded on $\mathbb T,$ hence has a well defined Fourier series, and ii) $\tilde F$ is differentiable at every point of $\mathbb T\setminus \{-1\}.$ Here I'm using the fact that if $g$ is an $L^1$ function on the circle, and $g$ is differentiable at some $e^{it},$ then the Fourier series of $g$ converges to $g(e^{it}).$

We can say more: The Fourier series of $\tilde F$ converges uniformly to $\tilde F$ on the arc $\{e^{it}: t\in [-\pi+\delta,\pi -\delta]\},$ for any $\delta, 0<\delta<\pi.$ That's because $\tilde F$ is $C^1$ on such arcs. (Actually the result will hold if $\tilde F$ is merely Lipschitz on such arcs.)

Now, if we knew $\tilde F$ had a bounded derivative on $\mathbb T\setminus \{-1\},$ then with the other conditions we have on $F,$ we could view $\tilde F$ as a function in $C^1(\mathbb T).$ This would give the desired uniform convergence.

I looked at two examples: $F(x)=x/(1+x^2)$ and $F(x)=1/(1+x^2).$ If I've done the calculations correctly, the first example gives $\tilde F'$ blowing up at $-1.$ The second one gives a bounded $\tilde F'.$


问:
What is the asymptotic expansion of $x_n$ where $x_{n+1} = x_n+1/x_n$?
Let
$x_{n+1} = x_n+1/x_n,
x_0 = a \gt 0$
and
$y_n = x_n^2$.

What is the asymptotic expansion
of $x_n$ ($y_n$ will do)?

I can show that
$y_n
=2n+\dfrac12 \ln(n) + O(1)
$.

Is there an explicit form
for the constant
implied by the $O(1)$?

What is the asymptotic form
of the terms following that constant
(e.g.,
$O(\frac{\ln(n)}{n}),
O(\frac1{n}), ...
$)?

答:
Not a rigorous answer, and it is rather a continuation of Mr Cohen's work.

My idea

My idea is standard, I just try to put all the approximate terms aside to see what will be left.

Main work

We start by giving the following equation.

$y_{n+1}-2(n+1)-\frac{1}{2}\ln(n+1) = -\frac{\ln(n)}{4ny_n}+\frac{\frac{1}{2}\ln(n) +2n-y_n}{2ny_{n}}-\frac{1}{2}\left[ \ln(1+\frac{1}{n}) -\frac{1}{n}\right]+( y_{n}-2n-\frac{1}{2}\ln(n))$

So if the approximation given by Mr Cohen is true, we have:
$$|e_{n+1}-e_n|=\left| -\frac{\ln(n)}{4ny_n}+\frac{\frac{1}{2}\ln(n) +2n-y_n}{2ny_{n}}-\frac{1}{2}\left[ \ln(1+\frac{1}{n}) -\frac{1}{n}\right]\right| \le c_1\frac{ln(n)}{n^2} $$
for some constant $c_1>0$, where $e_n:= y_n-2n-\frac{1}{2}\ln(n)$ as in the post of Mr Qiaochu.

Hence, $e_{n}$ converges to some limit, noted by $C$ also as in the post of Mr Qiaochu

Let  $u_n:=e_n-C$
then, the very fist equation is equivalent to
$$u_{n+1}-u_n= \frac{-u_n}{2ny_n}-\frac{\ln(n)}{4ny_n}-\frac{C}{2ny_n}-\frac{1}{2}\left[ \ln(1+\frac{1}{n})-\frac{1}{n} \right]$$
Thus,
$$-u_N = \sum_{n \ge N}\left\{ \frac{-u_n}{2ny_n}-\frac{\ln(n)}{4ny_n}-\frac{C}{2ny_n}-\frac{1}{2}\left[ \ln(1+\frac{1}{n})-\frac{1}{n} \right] \right\}$$

Clearly,
$$ \lim_{N \rightarrow +\infty} \frac{N}{\ln(N)}\sum_{n \ge N}\frac{-u_n}{2ny_n}=0$$
$$\lim_{N \rightarrow +\infty} \frac{N}{\ln(N)}\sum_{n \ge N}\frac{C}{2ny_n}=0$$
$$\lim_{N \rightarrow +\infty} \frac{N}{\ln(N)}\sum_{n \ge N}\left[ \ln(1+\frac{1}{n})-\frac{1}{n}\right]=0$$

And though it is not really straightforward, it is also not hard to see that:
$$\lim_{N \rightarrow +\infty} \frac{N}{\ln(N)}\sum_{n \ge N}\frac{\ln(n)}{4ny_n} = \dfrac{1}{8}$$

So, in conclusion
$$y_n= 2n+\frac{1}{2}\ln(n)+C+\frac{1}{8}\frac{\ln(n)}{n}+o(\frac{\ln(n)}{n}) $$

Comment: It looks like the next terms are $\frac{1}{n}$ something and  $\frac{\ln(n)^2}{n^2}$ something. So, it would be a surprise to me if one can find an analytic form for this series.

Updated comment
So,
$$x_n=\sqrt{2n}\sqrt{ 1+\frac{y_n-2n}{2n}}= \sqrt{2n}\left[1+\frac{\ln(n)}{8n}+\frac{C}{4n}+\frac{\ln(n)}{32n^2}-\frac{1}{8}\left( \frac{\ln n}{4n}+\frac{C}{2n} \right)^2 +o(\frac{\ln(n)}{n^2}) \right]$$

ContentLicense: CC BY-SA 4.0


问:
How to interpret summation signs
I'm taking a course in statistics, and I really need to brush up my math to be able to follow the book at times.

I'm looking at formulas for sum of squares, and I am slightly confused about the capital sigma letter and how to interpret an equation with several signs like this:

$$SS_a = \sum^a_{i=1} \sum^b_{j=1} \sum^n_{k=1} (\bar{X}_i - \bar{X})^2 $$

I mean, it's only i which is used  the summation signs, but when I try it out (ignoring b and n) it seems like I get it wrong, and that it should be something like this:

$ b * n * \sum^a_{i=1}(\bar{X}_i -\bar{X})^2 $

Is this correct, and what is the rule behind it?

PS. I wasn't sure what tags really fits into this question, might it be basic algebra? DS

答: Yes, as $(\bar X_i-\bar X)^2$ does not depend on $k$, the innermost sum is simply $n\cdot(\bar X_i-\bar X)^2$. AS this expresion does not depend on $j$, we get $b\cdot n\cdot(\bar X_i-\bar X)^2$ next and then can use distributivity to obtain $b\cdot n\cdot \sum_{i=1}^a(\bar X_i-\bar X)^2$.</p>

ContentLicense: CC BY-SA 3.0

Title: How to interpret summation signs


问:
Notation for applying a function recursively.
Consider a function $f: \mathbb{R} \to \mathbb{R}$ that is monotonically increasing in $x$ such that $x_t = f(x_{t-1})$ where $t$ represents time.

Let $x_t \leq x'_t$, so
$$ f(x_t) \leq f(f(x'_{t-1})) \leq f(f(f(x'_{t-2}))) \leq \cdots$$

Is there a more elegant way to write the expression with inequalities?

Note: I am unsure what tags to add.

答:
$$f^n(x_{t-n+1}) \le f^{n+1}(x_{t-n}),$$
where the superscript indicates $n$-fold composition.


问:
Direct formula to calculate total amount for compound interest with a compounding interest rate
```<code>Principal amount: $100
Compound interest rate: 24%
Compound interest on the original interest: 2%

Total amount after first year: 100 * 1.24 = <span class="math-container">$124
Total amount after second year: 100 * 1.24 * (1 + (0.24*1.02)) = $</span>154.3552 (Because interest rate also grows)
Total amount after third year: 100 * 1.24 * (1+(0.24*1.02)) * (1+(0.24*1.02*1.02)) = $192.89707
</code>```

Is it possible to derive a direct formula to calculate the total accumulated amount after $n$ number of years? A formula without iterative steps (summation etc), one where I can directly feed in $n$ and get the answer using a basic calculator.

The formula that I tried was $P \times(1.24^n) \times (1.02^{n(n-1) / 2})$, this is wrong for obvious reasons, one can not take out 1.02 factor directly outside, it adds extra interest for original principal rather than just to the interest rate. Is it possible to achieve a direct formula for this?

Note: I am new to the math stack exchange, kindly adjust with the math conventions that I might not have followed

答:
I believe the 'formula' is..

$$ P * \prod_{n=1}^{n} (1 + (0.24){(1 + 0.02)}^{n-1} )$$

As Dave Says, I do not think much can be done to simplify it to create a formula without using the product notation. The 1 +  parts in the rate, make it hard to simplify it, unlike when principal compounds by number and also that number compounds by a number.

Your formula in the question

$$ P \times(1.24^n) \times (1.02^{n(n-1) / 2}) $$

will work when

```<code>P is principal amount
1.24 is the base factor by which the amount compounds every month
1.02 is the factor by which the base factor compounds every month
</code>```

Edit 1:

Assuming $0.24$ as $r$, and $1.02$ as $q$ I tried the product series in wolfram to see if it offers a simplification. It offered a new notation called $q-Pochhammer$, which I am not familiar with. But looking around about it, I do not think even that offers a simplification to a formula that looks like the one in the question.

Body:
I believe the 'formula' is..

$$ P * \prod_{n=1}^{n} (1 + (0.24){(1 + 0.02)}^{n-1} )$$

As Dave Says, I do not think much can be done to simplify it to create a formula without using the product notation. The 1 +  parts in the rate, make it hard to simplify it, unlike when principal compounds by number and also that number compounds by a number.

Your formula in the question

$$ P \times(1.24^n) \times (1.02^{n(n-1) / 2}) $$

will work when

P is principal amount
1.24 is the base factor by which the amount compounds every month
1.02 is the factor by which the base factor compounds every month

Edit 1:

Assuming $0.24$ as $r$, and $1.02$ as $q$ I tried the product series in wolfram to see if it offers a simplification. It offered a new notation called $q-Pochhammer$, which I am not familiar with. But looking around about it, I do not think even that offers a simplification to a formula that looks like the one in the question.


问:
Prove that an irrotational vector field is conservative by showing the $\nabla$ of the scalar potential of $v$ written as an integral is equal to $v$
I am trying to prove that if $\vec v$ is a continuously differentiable vector field define on $\mathbb R^n$ and satisfies

$$\frac {\partial v_i} {\partial x_j} = \frac {\partial v_j} {\partial x_i}$$

then $\nabla f(x) = \vec v(x)$ where $f(x) = \int_0^1 x \cdot \vec v(tx) \, \mathrm d t$.

This is my attempt: It suffices to prove $v_i(x) = \frac {\partial f(x)} {\partial x_i}$.

I shall use implicit summation over $j$ here when I apply the chain rule. I first use the fundamental theorem of calculus: $$\begin{align} v_i(x) & = \int _0^1  \frac {\partial v_i(tx)} {\partial t}\, \mathrm d t + v_i(0) \\ & = \int_0^1 \frac {\partial v_i(tx)} {\partial x_j} \frac {\partial x_j} {\partial t}\, \mathrm d t + v_i(0) \\ & = \int_0^1 \frac {\partial v_j(tx)} {\partial x_i}  \frac {\partial x_j} {\partial t}\, \mathrm d t + v_i(0) \end{align}$$

I shall also use implicit summation over $j$ here when I expand out the dot product: $$\begin{align} \frac {\partial f(x)} {\partial x_i} & = \int_0^1 \frac {\partial } {\partial x_i} \left( x_j v_j(tx)\right) \, \mathrm d t \\ & = \int_0^1 v_i(tx) + x_j \frac {\partial v_j(tx)} {\partial x_i}\, \mathrm d t\end{align}$$

I am rather stuck here though and not sure how to proceed. Any hints would be appreciated. Have I made any mistakes so far?

答:
You drop a t taking a partial derivative. The derivation goes like this:

$\begin{align}
\frac {\partial f(x)} {\partial x_i} & = \int_0^1 \frac {\partial } {\partial x_i} \left( x_j v_j(tx)\right) \, \mathrm d t \\
& = \int_0^1 v_i(tx) + x_j \frac {\partial} {\partial x_i}[v_j(tx)]\, \mathrm d t =
\int_0^1 v_i(tx) + x_j \frac {\partial v_j} {\partial x_i}(tx)*t \,\, \mathrm d t  \\
& = \int_0^1 v_i(tx) + t\,x_j \frac {\partial v_i} {\partial x_j}(tx)\, \mathrm d t  \\
& = \int_0^1 v_i(tx) + t \ \frac{d}{dt} [v_i(tx)]\, \mathrm d t =
\int_0^1  \frac{d}{dt} [t \ v_i(tx)]\, \mathrm d t
\\
& = \ \ t \ v_i(tx)\Big{|}_0^1  = v_i(x)\\
\end{align}$


答: You are correct.  The summation signs are nested.  First you sum over $k$ from $1$ to $n$.  As you say, the quantity does not depend on $k$, so you can just multiply by $n$.  Then you sum that result over $j$ from $1$ to $b$, and it does not depend on $j$, so you can multiply by $b$, and you have your result.</p>

Title: How to interpret summation signs

Body:
I'm taking a course in statistics, and I really need to brush up my math to be able to follow the book at times.

I'm looking at formulas for sum of squares, and I am slightly confused about the capital sigma letter and how to interpret an equation with several signs like this:

$$SS_a = \sum^a_{i=1} \sum^b_{j=1} \sum^n_{k=1} (\bar{X}_i - \bar{X})^2 $$

I mean, it's only i which is used  the summation signs, but when I try it out (ignoring b and n) it seems like I get it wrong, and that it should be something like this:

$ b * n * \sum^a_{i=1}(\bar{X}_i -\bar{X})^2 $

Is this correct, and what is the rule behind it?

PS. I wasn't sure what tags really fits into this question, might it be basic algebra? DS


问:
Sporadic simple groups in GAP
I sincerely apologize in advance for this simple question.

I recently started learning GAP and I am having difficulties finding Sporadic simple groups in GAP. For example, if I want to find the order of the symmetric group $S_{10}$, I know that Order(SymmetricGroup(10)) works. I also know how to call certain sporadic simple groups, to be precise, Mathieu groups. eg:-Order(MathieuGroup(12))

Could anyone please tell me how to find the remaining Sporadic groups.

Thank you!

答:
There is a command SimpleGroup that takes a name. e.g.
SimpleGroup(“HS”);
Note that Monster and Baby Monster are not available yet, and some of the larger groups require the atlasrep package.

Title: Sporadic simple groups in GAP


问:
Showing or refuting that two monic irreducible polynomials are coprimes.
Let $F$ field and $p(x),q(x)\in F[x]$. If  $p(x)\neq q(x)$ are monic and irreducible polynomial. Why $p(x),q(x)$ are coprimes?

I have this:  Assume that $(p(x),q(x))\neq 1$. Because $(p(x),q(x)):=d(x)$ divides $p(x)$ and $q(x)$ then $p(x)=d(x)s(x)$ and $q(x)=d(x)r(x)$ some $s(x), r(x)$.

$p(x)$ irreducible implies that $d(x)$ unit or $s(x)$ unit

$q(x)$ irreducible implies that $d(x)$ unit or $r(x)$ unit

If $d(x)$ unit then $d(x)=d$ some $d\in F^{*}$

therefore $p(x)=ds(x)$ and $q(x)=dr(x)$

but I don't know what else I can do with this

analogously, if $s(x)$ unit and ($d(x)$ unit or $r(x)$ unit)

$s(x)=s$ some $s\in F^{*}$ invertible therefore
$p(x)=sd(x)$ and $q(x)=dr(x)$ or $q(x)=d(x)r$ some $d,r\in F^{*}$

but I don't know what else I can do with this too.

答: If $d$ is not a unit, then it divides $p$, hence $d = p$. Similarly, it divides $q$, hence $d = q$ which is a contradiction. Therefore, $d$ is a unit and applying the division algorithm there are polynomials $r, s$ such that $rp+sq = d$. Divide by $d$ to get $1$ as a linear combination of $p, q$.


问:
If $\theta :G\rightarrow H$ is a surjective homomorphism, then $\theta(\Phi(G))\leq\Phi(H)$.
This is a claim when I try to solve another problem related to the Fratinni group of a p-group, and I saw an answer Frattini subgroup of a finite elementary abelian $p$-group is trivial. I am stuck at the second step where the author claims that If $\theta :G\rightarrow H$ is a surjective homomorphism, then $\theta(\Phi(G))\leq\Phi(H)$, where $\Phi(G)$ and $\Phi(H)$ are the Frattini subgroup of $G$ and $H$. Any help is appreciated.

答:
Welcome to MSE!

This follows from the correspondence theorem.

Let $h \not \in \Phi(H)$. Then for some maximal subgroup $M \leq H$, we must have $h \not \in M$ (since $\Phi$ is the intersection of the maximal subgroups). But then any preimage $\theta^{-1} h \not \in \theta^{-1} M$, which is a maximal subgroup by correspondence. So $\theta^{-1}h \not \in \Phi(G)$.

Taking contrapositives gives the desired claim: $\theta [ \Phi(G) ] \leq \Phi(H)$.

I hope this helps ^_^


问:
Is $\{0\}$is borel set in $\mathcal{B}(\mathbb{R})?$ yes/no
Is   $\{0\}$is  borel set in   $\mathcal{B}(\mathbb{R})?$

My attempt : since  $\mathcal{B}(\mathbb{R})$ is generated by $(-\infty, a)$ and $(b, \infty )$

so my answer is  No

答:
Hint:

The correct answer is Yes.

To see why, remember $\mathcal{B}$ allows countable intersections of generating sets. Can you write $\{ 0 \}$ as such a countable  intersection? You might want to consider the sequence $\frac{1}{n}$.

I hope this helps ^_^


问:
Square of stationary variable
I'm given a stationary process $x_t$:
$$\Bbb E(x_t) = \mu $$
$$Cov(x_t,x_{t+k}) = \xi_k \text{ some function of } k \text{ and not of } t$$
I need to find $ \Bbb E(x_t^2)$ using $\mu$ and $\xi_k$.

答:
We can use the formula for variance:

$$ \Bbb Var(x_t) = \Bbb E(x_t^2) - \Bbb E^2(x_t)$$
From this follows that
$$ \Bbb E(x_t^2) = \Bbb Var(x_t) + \Bbb E^2(x_t)$$

Knowing that the process is stationary, we have
$$\Bbb E^2(x_t) = \mu^2$$
$$\Bbb Var(x_t) = \Bbb Cov(x_t, x_t) = \Bbb Cov(x_t, x_{t+0}) = \xi_0$$

Therefore,
$$ \Bbb E(x_t^2) = \xi_0 + \mu^2$$

Title: Square of stationary variable


问:
If every non-trivial subset of a set C has non-empty boundary in C, then C is connected
I came across this proposition in Functional analysis book by Joseph Muscat. I am having problem in applying this to following example. Let's say C = [-1,1] \ {0}, then C is not connected. But I am not able to find a subset with non-empty boundary. Subset A = [-1,0) has non-empty boundary {-1} in C. Please help me understand what I am missing. Thanks.</p>

答: The boundary of your set $A$ is empty. $-1$ is an interior point of $A$ in your space $C$ since $[-1,0)$ is an open subset of $C$ containing $-1$ which is contained in $A$. .


问:
Deduce that there exists a prime $p$ where $p$ divides $x^2 +2$ and $p≡3$ (mod 4)
I am revising for a number theory exam and have a question that I am struggling with, any help would be greatly appreciated.

First I am asked to show that for an odd number $x$, $x^2+2 ≡3$(mod 4).

I can do this part of the question, but next I am asked to deduce that there exists a prime $p$ where $p$ divides $x^2 +2$ and $p≡3$ (mod 4)

I am struggling to see how to attempt the second part and how the first part relates.

My thoughts so far are that I want to show $x^2≡-2$(mod p) ?
 And perhaps Fermat's Little Theorem could be of use here somehow?

Not sure if I'm barking up the wrong tree though.

Thanks in advance.

答:
The question, as asked, has been answered.  Nevertheless, I'll show how quadratic residues can help in problems like these, and show that moreover $x^2+2$ ($x$ odd) has a prime factor congruent to $3 \!\!\!\mod 8$:

Suppose that $x$ is odd.  If $p$ divides $x^2+2$, then $p$ is odd and $x^2 \equiv -2 \!\!\mod p$.  In particular, $-2$ is a square mod $p$, hence
$$1=\left(\frac{-2}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{2}{p}\right)=(-1)^{(p^2-1)/2} (-1)^{(p-1)/2}=(-1)^{(p^2+p-2)/2}.$$
This implies $p \equiv 1,3 \!\!\mod \!8$.  If each prime $p$ dividing $x^2+2$ were congruent to $1 \!\!\mod 8$, then $x^2+2$ would be congruent to $1\!\! \mod 8$.  This is a contradiction, hence there exists some $p \mid x^2+2$ congruent to $3 \!\!\mod 8$.  (In fact, there must be an odd number of such primes.)

Of course, any one of the primes is congruent to $3 \!\!\mod 4$, which gives your result.

We can ignore quadratic residues in your particular problem because there are only two odd residues mod $4$.  In more difficult questions, looking at quadratic residues can give more information.


问:
Simplification of gradient computation form $O(n^2)$ to $O(n)$
Let y and x be n−dimensional vectors related by $y=f(x)$ , $L$ be a differentiable loss function. According to the chain rule of calculus,$∇_xL= (∂y∂x)>∇_yL$, which takes up $O(n^2)$ computational time in general (as it requires a matrix-vector multiplication).

Show that if $f(x) = σ(x)$ or $f(x) =Softmax(x)$, the above matrix-vector multiplication canbe simplified to a $O(n)$ operation. Note that here, we used the sigmoid function for a vector input $x= (x_1,...,x_n) \in R^n →σ(x) = (σ(x1),...,σ(xn))$

答: Well, if $i \neq j$ then $\frac{ \partial \sigma(x_i)}{\partial x_j} = 0 $ so $\nabla f$ is sparse, only the n entries along the diagonal are non-zero, etc.


问:
Why are Machin-like formulas with more terms "better"
As in the title, when looking at the Wikipedia page, it seems that the more terms the Machin-like formula the better. Could someone explain this to me?

I think it is for two reasons:

It is easier to "split up" the computation. I can compute one term on one computer, another term on another etc. I have tenuous grasp on computational efficiency so I do not know if this makes any difference for computers.

The smaller the numbers are inside arctan the smaller the number of terms needed to evaluate arctan to a given degree of accuracy.

Are my reasons correct? IS there a better reason as to why the formulas used to approximate pi have more and more terms?

答: It is mostly the second reason. To be able to use smaller arguments, more terms are in general needed. If the arguments are mostly dyadic powers, then the computation of the series is greatly simplified on a binary computer. So one might accept a few additional terms to get that.</p>


问:
Ball-counting problem (Combinatorics)
I would like some help on this problem, I just can't figure it out.

In a box there are 5 identical white balls, 7 identical green balls and 10 red balls (the red balls are numbered from 1 to 10). A man picks 12 balls from the box. How many are the possibilities, in which:

a) exactly 5 red balls are drawn --
b) a red ball isn't drawn --
c) there is a white ball, a green ball and at least 6 red balls

Thanks in advance.

答:
Hints: (a) How many ways can we choose $5$ numbers from $1,2,...,9,10$? (This will tell you how many different collections of $5$ red balls he may draw.) How many distinguishable collections of $7$ balls can he draw so that each of the seven is either green or white? Note that the answers to those two questions do not depend on each other, so we'll multiply them together to get the solution to part (a).

(b) Don't overthink it. How many ways can this happen?

(c) You can split this into $5$ cases (depending on the number of red balls drawn) and proceed in a similar way to what we did in part (a) for each case (bearing in mind that we've already drawn one green ball and one white ball). Then, add up the numbers of ways each case can happen.

Title: Ball-counting problem (Combinatorics)


问:
Proving that the least upper bound of  is less than or equal to the greatest lower bound of B
I'm failing to understand the basics concepts of upper/lower bounds and I was hoping I could get some guidance through a simple question.

Suppose that $A$ and $B$ are two non-empty bounded sets of real numbers, such that $x \leq y$ for every $x \in A$ and $y \in B$. Show $lubA \leq glbB$

I have no idea how to even start this proof. I obviously need to somehow relate $x \leq y$ to the boundaries of their sets, that is $lubA \leq glbB$. Although I'm not even sure if my idea is correct. Any help?

答:
The least upper bound of a set is called its supremum ($\text{sup}$) and the greatest lower bound of a set is called its infimum ($\text{inf}$).

Firstly, we have,

$\forall x\in A\text{ and } y\in B, x\leq y
\implies y\text{ is an upper bound for every }x
\implies\text{sup}A\leq y,\text{ }\forall y\in B$ as $\text{sup}A$ is the least upper bound of the set $A$.

Now we have,

$\forall y\in B, \text{ sup}A\leq y\implies\text{sup}A\leq\text{inf}B$ as $\text{inf} B$ is the greatest lower bound of the set $B$.

Thus, proved.


问:
Find all the singularities in the finite plane and the corresponding residues. Show the details.
I am a bit stuck on this problem:

Find all the singularities in the finite plane and the corresponding residues. Show the details.

$$\frac{8}{1+z^2}$$

So I know the residual is going to be the coefficient associated with the first negative exponent in the laurent series.

So the singularities are at $z = i $ because $i^2 = -i$ and that's when the denominator will equal 0.

But how do I find the residuals?

So I remember that:
$$\frac{1}{1+z} = 1 - z + z^2 - ...$$

So I now multiply each element by 8:

$$\frac{8}{1+z} = 8(1 - z + z^2 - ...)t$$

But none of these terms have an exponent of -1 so I cannot find the coefficient?

Sigh, what am. I doing wrong?

Next problem, same as above but different function: $$\frac{1}{1-e^z}$$

Body:
If f has a simple zero at a, then, near a, $f(z) = (z-a)f'(a) + (z-a)^2 \frac{f''(a)}{2}+... = (z-a)f'(a) (1 + O(z-a)) $ ,

so:

$\frac{1}{f(z)} = \frac{1}{(z-a)f'(a)(1 + O(z-a))} = \frac{1}{(z-a)f'(a)} \frac{1}{(1 + O(z-a))} = \frac{1}{(z-a)f'(a)} (1 + O(z-a)) =  \frac{1}{(z-a)f'(a)}  + O(1) $

from where you can read the residue of 1/f at a.


问:
I am not getting this in binomial?
No ways to arrange ABC by using C(3,2) is AB BC AC . Not BA CB OR CA including.

When we solve $(a+b)^2$

a^2 + b^2 + 2ab

Is it that no ways to arrange for the 3rd term since it would be C(2,1) . it is no ways to arrange them would be ab or ba.
How do we write this here then?

答:
Okay,  I see what your book is getting at.

${3\choose 2}$ is the coefficient of $ab^2$.  That is one $a$ and two $b$s.  So you need the number of ways to arrange $a,b,b$.  They are not three different things.

There are three ways to do it.   they are 1) $a,b,b$ and 2) $b,a,b$ and 3) $b,b,a$.

There are two ways to think of ${3\choose 2}$.  It can be thought of as the number of ways to pick two items out of three different items. (That's how I think of it and that is why the operation is call "$3$ CHOOSE $2$).  But you book is thinking of it as: how many ways are there to arrange two different types of items if you have $2$ of one type, and the remaining out of $3$ items are all the second type.

Those two ways of thinking are equivalent because if you think of it the second way, all you are really doing is choosing $2$ out of three positions to place the first type of item.

=========

${3 \choose 2}$ is not the number of ways to arrange $A,B,C$.

${3\choose 2}$ is the number of ways to choose two elements from $A,B,C$.  And the three ways are: 1) The two elements you choose are $A, B$. 2) the two things you choose are $A,C$ or 3) the two things you choose are $B,C$.  Notice that choosing $A$ and $B$ are the exact same two things as choosing $B$ and $A$....

The number of ways to arrange $A,B,C$ are $3!=6$.  The six ways to arrange are i)$ABC$, ii) $ACB$, iii) $BAC$, iv) $BCA$, v) $CAB$, vi) $CBA$.  The number of ways to choose two items is a completely different question.

${2\choose 1}=2$ is the number of ways  to choose ONE item from two.  So the number of ways to choose one item from $A,B$ are two ways,  1) You can choose $A$, or 2) You can choose $B$.  Those are the only two ways to choose one item from $A,B$.

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

The binomial Thereom states if you expand out $(a + b)^k$ you will get a sum with $k+1$ terms and the coefficient the $j$th term will be ${k\choose j-1}$.

So the three terms when you expand $(a+b)^2$ will be $a^2 + 2ab + b^2$ (note the order of the coefficients).  The first coefficient of the $a^2$ term will be ${2\choose 0} = 1$.  There is one way to choose $0$ items from $2$ items that one way is to choose no items.

The second coefficient of the $ab$ term will be ${2\choose 1} = 2$.  There are $2$ ways  to choose one item from $2$ items. (You can chose one... or the other).

And the third coefficient of the $b^2$ term will be ${2\choose 2} = 1$.  There is $1$ way to choose two items from $2$ items.  That one and only way is to choose both the items.

So if we expand $(a+b)^2$ we get $(a+b)^2 = {2\choose 0}a^2 + {2\choose 1}ab + {2\choose 2}b^2 = 1\cdot a^2 + 2\cdot ab + 1\cdot b^2 = a^2 + 2ab +b^2$.

And if we expand $(a+b)^3$ we get $(a+b)^3 = {3\choose 0}a^3 + {3\choose 1}a^2b + {3\choose 2}ab^2 + {3\choose 3}b^3 = 1a^3 + 3a^2b + 3ab^2 + 1b^3 = a^3 + 3a^2b + 3ab^2 + b^3$.

Do you see why ${3\choose 0} = 1$?  The only $1$ way to choose zero items is not to choose any.

Do you see why ${3\choose 1} =3$.  The  $3$ ways to choose one item, is to choose one item, and there are three to choose from.

Do you see why ${3\choose 2} =3$?  There are three pairs we can choose.

Do you see why ${3\choose 3} =1$?  The $1$ way to choose three items is to choose all of them.

Title: I am not getting this in binomial?


问:
Help for solving a PDE: $\partial_1 u(x) + u(x)b(x) = 0$
Let $b : \mathbb R^N \rightarrow \mathbb R$ be a $C^1(\mathbb R^N)$ known function. We are looking for a solution $ u : \mathbb R^N \rightarrow \mathbb R$ of the following PDE :

$$\partial_1 u(x) + u(x)b(x) = 0 \ \ \ (1)$$
where $\partial_1 $ is the partial derivative of u with respect to $x_1$.

And so, more generally, let now $ c : \mathbb R^N \rightarrow \mathbb R^N$ be a $C^1(\mathbb R^N)$ vector field, we want to solve :
$$\nabla u(x) + u(x)c(x) = 0 \ \ \ (2) $$

But I guess (2) follows from (1).

Thank you in advance for your answers.

答:
Observe that the components $(x_2,x_3,\dots x_n)$ can be considered parameters, since your equation does not involve differentiation with respect to them. You can call $f(t)=u(t,x_2,x_3,\dots x_n)$ and $g(t)=b(t,x_2,x_3,\dots x_n)$. Your equation reads
$$
f'(t)+f(t)g(t)=0
$$
which is linear and has solution
$$
f(t)=Ce^{-\int_0^tg(s)ds}
$$
Therefore, your original equation has solution
$$
u(x_1,x_2,\dots x_n)=Ce^{-\int_0^{x_1}b(s,x_2,x_3,\dots x_n)ds}
$$


问:
Can there be a bump function with bounds on some partial derivatives?
I am looking for a smooth function $h:\mathbb{R}^2\to\mathbb{R}$ such that,

$Supp(h)\subset B(0,1)$

$|h(0)| \ge 1$

$|\partial_x h| < \epsilon$ everywhere, where $\epsilon > 0$ is a given constant

I have seen this and this, where the answer is negative for bounded derivatives. But note that I don't require any control over the partial derivative $\partial_y h$.

So can such a function $h$ exist? Or is the situation still the same and we cannot expect any kind of derivative bound at all?

If it helps, geometrically I am trying to think of climbing a hill of height $1$ and base $B(0,1)$, by using a road with many hairpin bends : as I am traversing the road the inclination (i.e $\partial_x h$) doesn't grow too much; but as I am crossing a bend, the $\partial_y h$ value gets arbitrarily large. I apologize if this makes no sense at all!

Any help is appreciated. Cheers!

答:
For any ${\bf a}, {\bf b} \in \mathbb R^2$, and any polygonal path $\Gamma$ from $\bf a$ to $\bf b$,
$$h({\bf b}) - h({\bf a}) = \int_\Gamma (\nabla h)({\bf r}) \cdot d{\bf r}$$

In particular, take ${\bf a} = (-1,0)$ and ${\bf b} = (0,0)$, and $\Gamma$ the straight
line segment from $\bf a$ to $\bf b$.  We get

$$1 \le h({\bf b}) - h({\bf a}) = \int_{-1}^0 \partial_x h(t,0)\; dt
\le \max_{0 \le t \le 1} \partial_x h(t,0)$$

i.e. there must be some $t \in [0,1]$ with $\partial_x h(t,0) \ge 1$.

The point is that the first equation is true for every polygonal path,
not just for some.  You don't get to choose the road!


答:
Yes, the third term, $2ab$ stems from the fact that the # of ways of secting one element from a choice of 2 is $\binom{2}{1} = \frac{2!}{1!\times(2-1)!} = 2.$

Write down $(a + b)$.

Directly below it, write down $(a + b)$ again.

When you manually multiply it out, you see that the $2ab$ term comes from :

First you select one of the $a$'s from either the first or second line.

Once that $a$ is chosen, considering that you are focusing on the $2ab$ term, you will then choose the $b$ from the other line.

For a clearer understanding of why combinations rather than permutations are involved, perform the same manual computation of $(a + b)^3$ by writing $(a + b)$ on three separate lines, and work through the multiplication.

For example, the $3a^2b$ term will arise because there are $\binom{3}{2}$ ways of choosing which 2 of the 3 lines, you will select the $a$ factor from.  Then, each selection of 2 $a$'s and 1 $b$ results in a $a^2 b$ product.

In fact, this type of analysis can be construed to be a demonstration of the binomial theorem, re for $n \in \mathbb{Z^+},$

$$(a + b)^n = \sum_{k=0}^n \binom{n}{k}a^k b^{(n-k)}.$$

The binomial theorem can alternatively be demonstrated through straight algebra, based on the idea that

$$\binom{n}{k} + \binom{n}{k+1} = \binom{n + 1}{k + 1}.$$

Addendum

Responding to the OP's request to re-check his query.

The image shown matches the binomial theorem.

Is it that no ways to arrange for the 3rd term since it would be C(2,1) . it is no ways to arrange them would be ab or ba.

If this is your question, then no.  This is the wrong way to consider it.  The easiest way to demonstrate how you calculate the coefficient of each term in the binomial expansion is to consider $(a + b)^4$, which can be represented as

$(a + b) ~\times$

$(a + b) ~\times$

$(a + b) ~\times$

$(a + b).$

Consider the terms:

$(a^4)$

There are $\binom{4}{4}$ ways of choosing 4 $a$'s from the 4 factors above.

$(a^3b)$

There are $\binom{4}{3}$ ways of choosing 3 $a$'s from the factors above.

$(a^2b^2)$

There are $\binom{4}{2}$ ways of choosing 2 $a$'s from the factors above.

$(ab^3)$

There are $\binom{4}{1}$ ways of choosing 1 $a$ from the factors above.

$(b^4)$

There are $\binom{4}{0}$ ways of choosing 0 $a$'s from the factors above.

Addendum-1

Responding to the OP's last comment.

If there 2 a’s , then there are also 2b’s.That means no of ways to arrange them is $\frac{4!}{2!}{2!}.$

Okay, now, I see what you are saying.  Instead of focusing on the fact that in the Addendum multiplication, you were choosing 2 of the 4 $a$'s, you are focusing on the fact that if you have 2$a$'s and 2$b$'s and 4 slots to drop them in, there are $\binom{4}{2}$ ways of choosing which slots to assign to the $a$'s.

There is a bijection between arranging the 2$a$'s and 2$b$'s into 4 slots on the one hand, and choosing which of the 4 factors will contribute the 2$a$'s on the other hand.

Therefore, if I have understood you correctly, yes, you can explain the coefficient of any term $a^kb^{(n-k)}$ by asking how many ways can you select $k$ slots out of $n$ to position the k$a$'s.

Title: I am not getting this in binomial?

Body:
No ways to arrange ABC by using C(3,2) is AB BC AC . Not BA CB OR CA including.

When we solve $(a+b)^2$

a^2 + b^2 + 2ab

Is it that no ways to arrange for the 3rd term since it would be C(2,1) . it is no ways to arrange them would be ab or ba.
How do we write this here then?


答: Consider the prime factorization of $x^2+2$. Since $x$ is odd, $x^2+2$ is odd implying $2$ will not show up in the factorization. Now consider the primes that DO show up in the prime factorization. If they are all $1$ in modulo 4, then their product will also be one in modulo $4$. This is not true though, since you know that $x^2+2$ is $3$ modulo $4$. Therefore, there must be a prime that in the prime factorization of $x^2+2$ s.t. it is not $1$ modulo 4. Since primes other than 2 that are not $1$ modulo 4 are $3$ modulo $4$, this completes the reasoning. </p>

Title: Deduce that there exists a prime $p$ where $p$ divides $x^2 +2$ and $p≡3$ (mod 4)

Body:
I am revising for a number theory exam and have a question that I am struggling with, any help would be greatly appreciated.

First I am asked to show that for an odd number $x$, $x^2+2 ≡3$(mod 4).

I can do this part of the question, but next I am asked to deduce that there exists a prime $p$ where $p$ divides $x^2 +2$ and $p≡3$ (mod 4)

I am struggling to see how to attempt the second part and how the first part relates.

My thoughts so far are that I want to show $x^2≡-2$(mod p) ?
 And perhaps Fermat's Little Theorem could be of use here somehow?

Not sure if I'm barking up the wrong tree though.

Thanks in advance.


问:
Don't understand proof that interior of a set is open
Say we want to show that the interior of a set $A$ is open.

If $x \in Int(A)$, then there exists an open ball $B_r(x) \subseteq A$.

Since $B_r(x)$ is open, $y \in B_r(x)$ also has an open ball $B_s(y) \subseteq B_r(x) \subseteq A$, so $y \in Int(A)$.

Now, somehow we have to show that the ball $B_r(x) \subseteq Int(A)$, and that would complete the proof.

All of the proofs I read say this is obvious, but I don't see how $B_r(x) \subseteq Int(A)$ immediately follows here.

答:
If $ x \in int(E) $ then $ \exists r >0 $ , $ B_r(x) \subset E $

Now $ \forall y \in B_r(x) \subset E $, $ \exists r_y = r - d(x,y) $, $ \ni B_{r_y} \subset B_r (x) \subset E $ Hence $ y \in int(E) $

Hence $ B_r(x) \subset int(E) $ or x is interior point of int(E)


问:
Unsure of steps of how to solve infinite summation
I am trying to find the convergent value for the following summation:

$$\sum_{a=0}^∞ a \left( \frac{x-1}{x}\right)^{\!a}$$

This sum seems to be convergent by ratio test as $$\frac{x-1}{x} < 1$$ but I am unsure of how to deal with the auxiliary $a$ term being multiplied in the summation. I am aware of the geometric series formula for infinite sums in the general case, but since we have a varying constant inside the summation, the series summation does not seem to apply. Could someone elaborate the steps to take to solve this summation?

Thanks in advance -- all help is greatly appreciated!

答:
Let $y= \frac{x-1}{x}$
$$\sum_{a=0}^\infty a \left( \frac{x-1}{x}\right)^{a}=\sum_{a=0}^\infty a y^{a}=y\sum_{a=0}^\infty a y^{a-1}=y\frac d {dy}\left(\sum_{a=0}^\infty y^{a} \right)$$


问:
Difference equation with the annihilator method $y(n+2)-4y(n)=2^{n+3}-1$, $n\geq 0$
I want to find the general solution of the difference equation $y(n+2)-4y(n)=2^{n+3}-1$, $n\geq 0$ with the annihilator method. I can rewrite the left-hand side as $L(y)(n)$ where $L=\tau^2-4$ and $\tau$ is the backshift operator. But I do not know how to rewrite the right-hand side as a solution of some difference equation. Could someone help me out? Thanks in advance!</p>

答: The right-hand side (RHS) is the linear superposition of $c_1 2^n$ and $c_2$ for constants $c_1$ and $c_2$. The annihilator for the RHS is $(\tau -2)\tau$, where $\tau$ is the timeshift operator. Like you said, the annihilator for the LHS (if the difference equation was homogenous) is $(\tau^2 - 4)$. Knowing that, we can make the original equation homogenous by multiplying the difference equation by the annihilator of the RHS: $(\tau-2)^2(\tau+2)\tau = 0$. The complete solution is $y(n) = a2^n + bn2^n + c(-2)^n + d$. The annihilator for the LHS corresponds to the homogenous solution $y_h(n) = a2^n + c(-2)^n$, leaving the particular solution $y_p(n) = bn2^n + d$. The coefficients $b$ and $d$ can be solved for, while $a$ and $c$ depend on initial conditions.


问:
Find Res $(f,0)$ if $f(z) = \frac{\sin z}{z^8}$.
I also have a question that I ask for help to check it.

Find Res $(f,0)$ if $f(z) = \frac{\sin z}{z^8}$.

Given that $f(z) = \frac{\sin z}{z^8}$, I found out that the isolated singularity $z = 0$ is a pole of order 8 and so what I did is

$Res (f,0) = \frac{1}{(8-1)!} \cdot \frac{d^7}{dz^{7}} \left(z^8 f(z) \right)_{z=0} = \frac{1}{7!} \cdot \frac{d^7}{dz^{7}} \left(z^8 \frac{\sin z}{z^8} \right)_{z=0} = \frac{1}{7!} \cdot \frac{d^7}{dz^{7}} \left(\sin z \right)_{z=0}$.

If $g(z) = \sin z$. Then $g^{(7)} = -\cos z$. So,

$Res (f,0) = \frac{1}{7!} (-\cos z)_{z=0} = - 1/7!$.

Is this correct?  Or can somebody point out my mistake?  I would truly appreciate your kindness.

答:
It usually could be much easier to use the Taylor series of $sin(z)$. Write
$$\frac{sin(z)}{z^8}=\frac{1}{z^8}\sum_{n=0}^{\infty}\frac{(-1)^nz^{2n+1}}{(2n+1)!}=\sum_{n=0}^{\infty}\frac{(-1)^nz^{2n-7}}{(2n+1)!}$$
You got a Laurent series of a meromorphic function. The residue is the coefficient of $z^{-1}$, which in our case, corresponds to $n=3$. So we get:
$$Res(f,0)=\frac{(-1)^3}{(2\cdot 3+1)!}=-\frac{1}{7!}$$
And as you see, this is identical to your result.


问:
Non-uniqueness of the characterization of $H^{-1}$ (Brezis Prop. 8.14)
Let $I$ be an interval and $W^{-1,p'}(I)$ be the dual space of $W^{1,p}_0(I)$ ($1\leq p<\infty$). Brezis says

My question is on

Remark 19. The functions $f_0$ and $f_1$ are not uniquely determined by $F$.

Is the following argument correct?

Here is my counterexample: let $I=(0,1)$. Let $g_{0}=f_{0}$ and $g_{1}=f_{1}+c$ for some constant $c\neq0$.

From $f_{1}\neq g_{1}$, it suffices to show $\int_{I}g_{1}v'=\int_{I}f_{1}v'$ for $H^1_0(I)$.
But from $c,v\in H^{1}(I)$ we have
$$\int_{I}cv'=cv(1)-cv(0)-\int_{I}0\times v,$$
but from $v\in H_{0}^{1}(I)$ we have $v(1)=0=v(0)$. Hence $$\int_{I}g_{1}v'=\int_{I}f_{1}v'+0.$$
Thus, we conclude $\langle F,v\rangle = \int_{I}g_{0}v+\int_{I}g_{1}v'$ with $g_1\neq f_1$.

One concern is that the above calculation implies we have $\int_{I}v'=0$ for $v\in H^1_0(I)$, which I never thought about (but now I think of e.g. the hat function, the integral of its derivative Heaviside function is indeed $0$).

Another concern is that an answer in characterization of the dual space of the Sobolev space $H_0^1$
contains a "quote" from Brezis Prop. 9.20, saying that "$(1)$: $ \, v^0,v^1,
\dots , v^n$ is NOT determined uniquely by $v\in H^{-1}$, that is, $v_0$ can be varied". Note that this (that $v_0$ can be varied) is not claimed in Brezis book.

Anyway, in my example above, $v_0$ is the same but it is $v_1$ that is different.

答:
Your argument is correct.

Here is a another possibility to find those $g_i$'s. Let $c\ge0$. Define $u_c$ as the weak solution of
$$
-\Delta u + cu = F
$$
in $H^1_0(\Omega)$. Then $g_0=cu_c$ and $g_1=\nabla u_c$.

Note that it is possible to vary $g_0$. But then $g_1$ has to be adapted. It is not possible to only vary $g_0$ (and let $g_1$ be fixed).


问:
Is my proof for $T_1=T_2$ correct?
So the question says:

Suppose that $V$ and $W$ are vector spaces and $T_1\colon V\longrightarrow W$ and $T_2\colon V\longrightarrow W$ are linear. Show that if $u_1,\ldots,u_n\in V$ span $V$ and $T_1u_i=T_2u_i$ for all $1\le i\le n$, then $T_1=T_2$, i.e., $T_1x=T_2x$ for all $x\in V$.

For my proof I just wrote:

Suppose $u_1,\ldots,u_n\in V$ span $V$ and $T_1u_i=T_2u_i$ for all $1\le i\le n$. Then $aT_1u_i=aT_2u_i$ for all $a \in\Bbb R$ and $1\le i\le n$. Then $T_1x=T_2x$ for all $x\in V$. So $T_1=T_2$.

I'm unsure if my proof is correct since I'm still learning how to write proofs. I feel like I should add more details, but I'm not sure what.

答:
No, that is not enough. Your proof didnt consider $x$ that cant be described as $au_i$ for any $i$ or $a$.

Instead, try to use the definition of the spanning ability to find a description for a general $x$ and then use the linearity of the operators to show the desired result.

Title: Is my proof for $T_1=T_2$ correct?


答:
No, it is not correct. When you write “Then $T_1x = T_2x$ for all $x\in V$”, you provide no justification for that assertion.

Take $v\in V$. You can write $v$ as $\alpha_1u_1+\alpha_2u_2+\cdots+\alpha_nu_n$. But then\begin{align}T_1(v)&=T_1(\alpha_1u_1+\alpha_2u_2+\cdots+\alpha_nu_n)\\&=\alpha_1T_1(u_1)+\alpha_2T_1(u_2)+\cdots+\alpha_nT_1(u_n)\\&=\alpha_1T_2(u_1)+\alpha_2T_2(u_2)+\cdots+\alpha_nT_2(u_n)\\&=T_2(\alpha_1u_1+\alpha_2u_2+\cdots+\alpha_nu_n)\\&=T_2(v).\end{align}Since this occurs for every $v\in V$, $T_1=T_2$.

Title: Is my proof for $T_1=T_2$ correct?

Body:
So the question says:

Suppose that $V$ and $W$ are vector spaces and $T_1\colon V\longrightarrow W$ and $T_2\colon V\longrightarrow W$ are linear. Show that if $u_1,\ldots,u_n\in V$ span $V$ and $T_1u_i=T_2u_i$ for all $1\le i\le n$, then $T_1=T_2$, i.e., $T_1x=T_2x$ for all $x\in V$.

For my proof I just wrote:

Suppose $u_1,\ldots,u_n\in V$ span $V$ and $T_1u_i=T_2u_i$ for all $1\le i\le n$. Then $aT_1u_i=aT_2u_i$ for all $a \in\Bbb R$ and $1\le i\le n$. Then $T_1x=T_2x$ for all $x\in V$. So $T_1=T_2$.

I'm unsure if my proof is correct since I'm still learning how to write proofs. I feel like I should add more details, but I'm not sure what.


问:
Behaviour of the Gamma function near zero
The Gamma function on the positive real half-line is defined via the reknown formula
$$
\Gamma(z)=\int_0^\infty x^{z-1}e^{-x}dx, \quad z>0.
$$
A classical result is Stirling's formula, describing the behaviour of $\Gamma(z)$ as $z$ diverges to infinity,
$$
\Gamma(z)\sim \sqrt{\frac{2\pi}{z}} \left( \frac{z}{e}\right)^z, \quad z \to \infty.
$$
Is there any such approximation formula for $z \downarrow 0$, describing the speed at which the Gamma function diverges near the origin?

答:
Around $z=0$
$$\Gamma(z)=\frac{1}{z}-\gamma +\sum_{n=1} a_n z^n$$
$$a_1=\frac{1}{12} \left(6 \gamma ^2+\pi ^2\right)\qquad a_2=-\frac{1}{6} \left(2 \zeta (3)+\gamma ^3+\frac{\gamma  \pi ^2}{2}\right)$$
$$a_3=\frac{1}{24} \left(8 \gamma  \zeta (3)+\gamma ^4+\gamma ^2 \pi ^2+\frac{3 \pi
   ^4}{20}\right)$$

Edit

A quite good approximation is given by
$$\Gamma(z)\sim \frac 1 z \frac{1+\frac{\left(\pi ^2-6 \gamma ^2\right) }{12 \gamma }z } {1+\frac{\left(\pi ^2+6 \gamma ^2\right) }{12 \gamma }z  }$$


问:
How to find details about function in $\mathbb{R}^2$ if $\frac{df} {dx}= \frac{df}{dy}$ for all $(x,y)\in \mathbb{R}^2$
I am trying assignment questions of real analysis and I was unable to solve this question. In fact, I don't have any idea on how to start.

So, I am looking for help here.

Let $f$  be a function on $\mathbb{R}^2$ such that $\frac{df} {dx}= \frac{df}{dy}$ for all $(x,y)\in  \mathbb{R}^2$. Now which of the following options are true:

1 $f$ is constant on all lines parallel to the line $x=-y$.

2 $f(x,y)=0 \quad\forall (x,y) \in \mathbb{R}^2$.

3 $f(x,y)=f(-y,x) \quad\forall (x,y) \in \mathbb{R}^2$.

4 $f(x,y)-f(y,x)=(x-y) \frac {df}{dx}(x*,y*) + (y-x) \frac {df}{dy} (x*,y*) \in \mathbb{R}^2$.

I am unable to get any idea on how to use the hypothesis $\frac{df} {dx}= \frac{df}{dy}$ to deduce or contradict  any of the options.

I studied multivariable calculus from Tom M Apostol in case it is of any help.

Please help me. Kindly just give outlines not complete solution.

答:
Here is a somewhat more geometric explanation of @Yiorgos or @Ted-Shrifin answers.

$f$ satisfies a first order partial differential equation. This equation says something about a directional derivative, namely that the derivative in the direction $ (-1,1)^t $ is 0. To make things clear, we can change variables so the direction $(-1,1)^t$ is a coordinate direction. There are many ways of doing this, for example:
$ (x,y) = u*(-1,1) + v*(1,0) $

i.e.: $x = -u + v$ and $y = u$.

The inverse transformation is $u = y$, $v=x+y$. Think of $f$ as a function of points, that can be written in terms of different coordinate systems. To avoid confusion, let $f^*$ be $f$ in the new coordinates. Explicitly:

$ f^*(u,v) = f(-u + v, u)$

And so, $f(x,y) = f^*(y, x+y)$

Then:

$ \frac{\partial f^*}{\partial u}(u,v) = -\frac{\partial f}{\partial x}(-u+v,v) + \frac{\partial f}{\partial y}(-u+v,v) = 0$

an equation which is easy to solve: $f^*$ is a function of v only (constant in $u$):
$f^*(u,v) = h(v)$ for some function $h$ (for example $h(v) = f^*(0,v)$ ), and going back to the original coordinates: $f(x,y) = f^*(y,x+y) = h(x+y)$.

So, 1) is true, 2) and 3) are false (they may hold for some solution of the PDE, but not for all solutions), and (assuming (x*,y*) means an intermediate point between (x,y) and (y,x)) 4) is also true (the expressions at both sides of the equation are both 0).


问:
What is the integral of $\int e^x\,\sin x\,\,dx$?
I'm trying to solve the integral of $\left(\int e^x\,\sin x\,\,dx\right)$ (My solution):

$\int e^x\sin\left(x\right)\,\,dx=$

$\int \sin\left(x\right) \,e^x\,\,dx=$

$\left(\sin(x)\,\int e^x\right)-\left(\int\sin^{'}(x)\,\left(\int e^x\right)\right)$

$\left(\sin(x)\,e^x\right)-\left(\int\cos(x)\,e^x\right)$

$\left(\sin(x)\,e^x\right)-\left(\cos(x)\,e^x-\left(\int-\sin\left(x\right)\,e^x\right)\right)$

$\left(\sin(x)\,e^x\right)-\left(\cos(x)\,e^x-\left(-\sin\left(x\right)\,e^x-\int-\cos\left(x\right)\,e^x\right)\right)$

I don't know how to complete because the solution gonna be very complicated.

答:
Another way via leibniz theorem,

$$ \frac{d^4 (e^x \sin x)}{dx^4} =e^x \left[\binom{4}{0} \sin x+ \binom{4}{1} \cos x - \binom{4}{2} \sin x - \binom{4}{3} \cos x+ \binom{4}{4}  \sin x\right]$$

Or,

$$ \frac{1}{ \sum_{k=0}^2 \binom{4}{2k} } \frac{d^4 (e^x \sin x)}{dx^4} = e^x \sin x$$

Integrate both sides and apply leibniz for third derivative again,

$$ \int e^x \sin x dx =  \frac{1}{ \sum_{k=0}^2 (-1)^k \binom{4}{2k} } \int  \frac{d^4 (e^x \sin x)}{dx^4} dx=e^x \frac{\sum_{j=0}^3 \binom{3}{j} \sin( x + j \frac{\pi}{2})}{ \sum_{k=0}^2 (-1)^k \binom{4}{2k} }  $$

Since,

$$ \frac{d^i}{dx^i} \sin x = \sin( x +i \frac{\pi}{2} )$$


问:
Help show that $E[h'(X)] = E[h(X)(\frac{(1+ln(X))}{X})]$
$$f(x) = \frac{1}{x\sqrt{2\pi}}exp[(\frac{-((ln(x))^2}{2})]$$ is the PDF of the random variable of X

Let h(x) be a known function and suppose that h'(x) exists. Assuming all expectations exist, show that
$$E[h'(X)] = E[h(X)(\frac{(1+ln(X))}{X})]$$
(If needed, you may assume that $\lim_{x\to \infty}$h(x)f(x) = $\lim_{x\to 0}$h(x)f(x) = 0 )

I have started the proof and got to the part that where E[h'(x)] cannot equal 0 by using the definition of expectation but I am not sure how that helps or where to go from E[h'(X)] to E[h(X)$(\frac{(1+ln(X))}{X})$]

Any help would be great, thank you.

答:
Take $Y:= \ln(X)$. Clearly, $Y$ has a standard normal distribution.

Take $g(y):= h(e^y)e^{-y}$, clearly, $g(Y)=\frac{h(X)}{X}$ and $g'(y)=h'(e^y)-g(y)$

Thus, your equation is equivalent to
$$ \mathbb{E}(g'(Y)+g(Y)) = \mathbb{E}( g(Y) (1+Y))$$
or
$$\mathbb{E}(g'(Y)) = \mathbb{E}( g(Y) Y)$$
Which is the following Stein's lemma
https://en.wikipedia.org/wiki/Stein%27s_lemma

If you want to reprove that lemma, the integration by parts is enough.


问:
Can an equilateral triangle be dissected into 5 congruent convex pieces?
There is a rather surprising dissection of an equilateral triangle into 5 congruent pieces:

However, these pieces aren't very "nice", consisting of 2 or 6 connected components depending on how one counts single-point overlaps.

I expect that the analogous question for arbitrary connected pieces is quite difficult, but I wonder if the case of convex pieces is sufficiently restricted to allow for a proof of impossibility (which I strongly suspect is the case).

One aspect of this problem that makes it much more tractable is that the pieces have to be polygons, because any border between two pieces must be a straight line to preserve convexity of both pieces.

In fact, this analysis can be extended a little further to show that the pieces have at most 5 sides. The number of edges on a piece is at most the number of edges of the triangle it touches, plus the number of other pieces it touches. $K_5$ is non-planar, so there is a piece touching at most 3 others. If this piece touches all three sides of the triangle, then its complement has at least two disconnected regions (since it can't take up an entire side), so the pieces in those regions are at most pentagons by a similar analysis. If it does not touch all three sides of the triangle, then it is itself at most a pentagon.

If the polygons are triangles, then I can go through some casework on the edges to show that the triangles must be $30-60-90$ triangles, and from there derive a contradiction.

So the only remaining cases are those of quadrilaterals and pentagons. Can such dissections be shown impossible?

Interestingly, the answer is known to be yes for some squarefree multiples of $5$ like $180$; see this MathOverflow thread for a picture.

答:
I have made some partial progress on the problem: The pentagon case is impossible! In fact, we can show the stronger statement that a triangle is not the union of any five strictly convex pentagons.

Suppose we had such an arrangement. Then we can create a graph on $8$ vertices as follows: we have a vertex for each pentagon, where we connect two pentagons with an edge if they share a positive-measure part of their borders. We also have three vertices positioned above each face of the triangle, connected to one another and to every pentagon whose border occupies a positive-measure part of the corresponding side of the triangle. Here is an image to illustrate the idea, though (by necessity) not all of the regions are pentagons.

Clearly, such a graph is planar. However, a planar graph on $n>2$ vertices has at most $3(n-2)$ edges, which can be seen by triangulating and using $V-E+F=2$. So this graph has at most $18$ edges, which means it has at most $36$ half-edges (which we can think of as ordered pairs of adjacent vertices).

Each vertex on the inside of the triangle has degree at least 5 (one for each side of the pentagon), so they contribute at least $25$ half-edges. However, a planar graph on $5$ vertices has at most $9$ edges, so at most $18$ of these half-edges are between two red vertices (to use the coloring in the diagram); at least $7$ must be between a red and a blue vertex.

Each of these red-blue edges (of which we now know there are at least $7$) contributes another half-edge (since we have only counted the red "side" of them), bringing our total to $32$. But then the three blue-to-blue edges (and thus six half-edges) takes us to $38$, yielding a contradiction.

This sort of analysis will not work on the quadrilateral case, however, as a triangle can be decomposed into five strictly convex quadrilaterals; the congruence assumption must be utilized somehow.


问:
Show that $f$ has a fixed point at $\overline{B_1(0)}$.
I really don't know how to start with this.  Can you please help me with this?

Suppose $\displaystyle{f: \overline{B_1(0)} \rightarrow \overline{B_1(0)}}$ is continuous and $f$ is analytic in $\displaystyle{B_1(0)}$.  Prove that $f(z)$ has a fixed point in $\displaystyle{\overline{B_1(0)}}$.

How to do this?  Please. I would appreciate any help.  Thanks a lot.

答: This is true by Brouwer's fixed-point theorem, simply because $f$ is continuous. The fact that it is analytic on $B_1(0)$ does not matter.


问:
Can the set of positive rational numbers be decomposed into two non-empty, disjoint parts such that closed by the addition for?
I can't get started, and I'm pushing for a deadline. I can't start, Thank you!

Can the set of positive rational numbers be decomposed into two non-empty, disjoint parts such that closed by the addition for?

答: I assume you are asking whether $\mathbb{Q}^+$ can be partitioned into two nonempty disjoint sets, i.e., $\mathbb{Q}^+=A\cup B$ and $A\cap B=\varnothing$ and $A,B$ are closed under addition. If so, then the answer is no, you cannot do that. Suppose you can, then $1$ is either in $A$ or $B$. WLOG, assume $1\in A$, then $1+1=2\in A$ and keep doing this you can see all positive integers should be in $A$. For any rational number $q$, it can be written as $q=m/n$ for positive coprime integers $m,n$, so if $q\in B$, then $q+q=2q\in B$, and if you keep doing this, $nq\in B$. However, $nq=m$, so $m\in A\cap B$, which is a contradiction to $A\cap B=\varnothing$. This shows any rationals should be in $A$, which implies $B=\varnothing$, again contradiction...


问:
Maximal consistent set. Decomposition lemma

Let $\Gamma$ be a maximal consistent set. Prove: $\varphi \lor \psi
 \in \Gamma \iff \varphi \in \Gamma $ or $ \psi \in \Gamma$.

Now define $V_{\Gamma}: Q \to \{ 0, 1 \}$ as follows:

$V_{\Gamma}(p):= \cases{{1  \mbox{ if } p \in \Gamma} \\ 0 \mbox{ if }
 p \notin \Gamma}$.

Show that for every formula $\varphi$ we have: $\varphi \in \Gamma
 \iff V_{\Gamma} \models \varphi$.

I'm at a loss here. Could anyone please help me out?

Thanks in advance.

答: Here's an answer for the more difficult direction of the first part.  Suppose, toward a contradiction, that $\Gamma$ contains $\phi\lor\psi$ but contains neither $\phi$ nor $\psi$.  Since it doesn't contain $\phi$ but is maximal consistent, $\Gamma\cup\{\phi\}$ must be inconsistent.  So $\neg\phi$ is a logical consequence of $\Gamma$.  Similarly, so is $\neg\psi$.  But then so is $\neg(\phi\lor\psi)$ because that's a logical consequence of $\{\neg\phi,\neg\psi\}$.  But since $\phi\lor\psi$ is in $\Gamma$, this contradicts the assumption that $\Gamma$ is consistent.</p>


问:
Euler Lagrange problem - find wave equation
I started learning calculus of variations and found this question on my textbook:

The Lagrangian for a π-meson is given by $$L(x,t)=\frac{1}{2}(\phi'^2-|\triangledown\phi|^2-\mu^2\phi^2),$$ where x is a vector and $\mu$ is the menson mass and $\phi(x,t)$ is its wavefunction. Assuming Hamilton’s principle find the wave equation satisfied by $\phi$.

I think there are n Euler-Lagrange equations one for each component of the form:

$$\frac{\partial L}{\partial q_i}=\frac{d}{dt}(\frac{\partial L}{\partial q_i'}),\,\ i=1,2,..,n$$

How can I apply this to the problem?

答:
The Euler-Lagrange equation for fields is given by
\begin{align}
\frac{\partial \mathcal{L}}{\partial \phi} = \sum^3_{\mu = 0}\partial_\mu\frac{\partial \mathcal{L}}{\partial (\partial_\mu\phi)}.
\end{align}
Hence, in this case, we see that
\begin{align}
\frac{\partial \mathcal{L}}{\partial \phi}=-\mu^2\phi.
\end{align}
Likewise, we have that
\begin{align}
\frac{\partial \mathcal{L}}{\partial(\partial_0 \phi)} = \dot\phi \ \ \text{ and } \ \ \frac{\partial \mathcal{L}}{\partial(\partial_i \phi)}= -\partial_i\phi
\end{align}
which means
\begin{align}
\partial_0\frac{\partial \mathcal{L}}{\partial(\partial_0 \phi)} = \ddot \phi \ \ \text{ and } \ \ \partial_i\frac{\partial \mathcal{L}}{\partial(\partial_i \phi)} = -\partial^2_i\phi.
\end{align}
Hence it follows
\begin{align}
-\mu^2\phi = \ddot\phi -\Delta\phi \ \ \implies \ \ \ddot\phi -\Delta\phi +\mu^2\phi = 0
\end{align}
which is the Klein-Gordon equation.


问:
$V$ is finite dimensional vector space. $G$ is a spanning set of $V$, $I \subseteq G$ is linearly independent
I'm totally new to the subjects like spanning set, bases and dimension. Here is a theorem that I could not understand.

Let $V$ be a finite dimensional vector space. If $G$ is a  finite spanning set of $V$ and if $I \subseteq G$ is linearly independent, then there is a basis $B$ of $V$ such that $I \subseteq B  \subseteq G$.

If $\text{Span}(I)= V$ then $B = I$ and we have nothing left to prove.

I know that if  $\text{Span}(I)= V$ and since  $\text{Span}(B)= V$ follows  $\text{Span}(I)= \text{Span}(B)$ but how do we know $I=B$ and why we have nothing left to prove?

Suppose  $\text{Span}(I) \ne V$ then $I \ne G$ and $G \setminus \text{Span}(I) \ne \emptyset$ (if $G \subseteq \text{Span}(I)$ we would have $V=\text{Span}(G) \subseteq \text{Span}(I)$).Thus $\exists g_1 \in G$ such that $g_1 \notin \text{Span}(I)$. Then $I'= I \cup \left\{g_1 \right\} \subseteq G$ is linearly independent.

Why $\text{Span}(I) \ne V$ does imply that $I \ne G$ and also $I'= I  \subseteq V$?

If $\text{Span}(I')=V$ we are done. If not,$\exists g_2 \in G$ such that $g_2 \notin \text{Span}(I')$ and we get a linearly independent set $I''= I' \cup \left\{g_1 ,g_2\right\} $. If $\text{Span}(I'')=V$ we are done. If not, we continue. Since $G$ is a finite set, this process will come to an end.Finally we will get a linearly independent set  $I'= I \cup \left\{g_1 ,g_2,\ldots,g_r\right\} \subseteq G$  which spans $V$ . This is the basis $B$ and obviously $I \subseteq B  \subseteq G$.

I know that $G$ is a spanning set of $V$ but why a subset of $G$ like $I'= I \cup \left\{g_1 ,g_2,\ldots,g_r\right\} $ is also a spanning subset of $G$?

答:
We don't know that $B=I$; we define $B$ as being equal to $I$.

Since $\operatorname{Span}(G)=V$, if $\operatorname{Span}(I)\ne V$, then $I\ne G$. So, $I\varsubsetneq G$. And $I'\subseteq G$ because $I'=I\cup\{g_1\}$, where $I$ is a subset of $G$ and $g_1\in G$.

Because otherwise $I'\varsubsetneq G$ and then we could take an element $g$ of $G$ and add it to $I'$ so that $I'\cup\{g\}$ would still be linearly independent. But we cannot keep on doing that forever, since $G$ is finite. So, sooner or later, that process must come to an end.


问:
Solution to a definite integral as $\int_0^{1}(x^{m}\left(1+x)^{n}dx\right)$
While solving a question based on Binomial series, I encountered a definite integral as : $$\int_0^{1}(x^{m}\left(1+x)^{n}dx\right)$$ where $m,n$ are natural numbers.
My approach : wrote the above expression in two ways (1)- $$\int_0^1\sum_{r=0}^n \binom{n}{r} \left( {x}\right)^{m+r}dx $$ and also as (2)- $$\int_0^1\sum_{r=0}^m \binom{m}{r} \left(-1)^{r}( {1+x}\right)^{m+n-r}dx$$ and tried to solve using both the above mentioned expressions. Did some other simplification of expression (1) , But was unable to proceed further. Can anyone guide me to a "general answer based on any m,n" ? Any help would be highly appreciated. Thanks in advance.!

Body:
I think about another IDEA to approach (may be useful)

take $$I(m,n)=\int_0^{1}x^{m}(1+x)^{n}dx$$ so $$I(m,n+1)=\int_0^{1}x^{m}(1+x)^{n+1}dx=\\
\int_0^{1}x^{m}(1+x)^{n}(1+x)dx=\\
\int_0^{1}x^{m}(1+x)^{n}dx+\int_0^{1}x^{m+1}(1+x)^{n}dx=\\
I(m,n)+I(m+1,n)$$


问:
How to reverse a rotated rectangle?
Say there's a rectangle at (20, 20)(fig. 1), then it's rotated 45 degree (fig. 2), if I just have the rotated rectangle and know it's size and rotated degree, is there a way to reverse the rotation and get the original starting point (20, 20)?

fig. 1
fig. 2

答:
Hint

Let point $A$ denote (14,105) and let point $B$ denote (120,70)

You know the angle that $\overline{AB}$ makes with the horizontal.  Therefore, you should be able to add $45^{\circ}$ to that to get the new angle that $\overline{AB}$ makes with the horizontal.

Now you have a ray, emanating from point $B$.  Also, you know the distance of $\overline{AB}$.  Therefore, you know how far to traverse along the ray to identify the original point $A$.

Title: How to reverse a rotated rectangle?


问:
I've a small problem solving this inequality
$\sqrt{9 - x^2} \geq \sqrt{3}*x$

I used the standard method (meaning, it involves two systems of inequalities).

At the end I've got (on the number line):
$x>-3$, $x<3$, $-1.5<x<1.5$, $x>0$, and $x < 0$.

I don't understand the reason why in the solution set I have to include "$-3<x<-3/2$" because this set isn't satisfied by all the other set of solutions. (because of union, it's not the case of intersection).

答:
For real $x,\sqrt{9-x^2}\ge0$ and $x^2\le9\iff-3\le x\le3$

So, if $-3\le x<0, \sqrt{9-x^2}>\sqrt3x$

If $x\ge0,$  $$ \sqrt{9-x^2}\ge\sqrt3x\iff9-x^2\ge(\sqrt3x)^2\iff x^2\le\dfrac94$$

Now if for  $a\ge0,x^2\le a^2,-a\le x\le a$

and if $x\ge0, 0\le x\le a$


答:
Define $$f(x)=\sqrt{9-x^2},\quad g(x)=\sqrt3x,\quad h(x)=f(x)-g(x).$$  We want to find where $f(x)\geq g(x)\implies h(x)\geq0$.  We can first find where $h(x)=0$, then determine whether $h(x)$ is positive or negative on the intervals that have been created by the above solutions.

\begin{align}
h(x)&=0\\
\sqrt{9-x^2}-\sqrt3x&=0\\
\sqrt{9-x^2}&=\sqrt3x\\
9-x^2&=3x^2\\
0&=4x^2-9\\
x&=\pm\frac32
\end{align}

$h(x)$ is only defined on $[-3,3]$, so from that we can create the following table of values:
\begin{array}{|c|c|c|c|c|}
\hline\text{Interval}&h(x)\\
\hline x=-3 & 3\sqrt3\\
\hline-3<x<-\frac32 & +\\
\hline x=-\frac32 & 0\\
\hline -\frac32<x<\frac32 & +\\
\hline x=\frac32 & 0\\
\hline \frac32<x<3 & -\\
\hline x=3 & -3\sqrt3\\
\hline
\end{array}

Therefore, the solution is $\boxed{x\in\left[-3,\frac32\right]}$.

Title: I've a small problem solving this inequality

Body:
$\sqrt{9 - x^2} \geq \sqrt{3}*x$

I used the standard method (meaning, it involves two systems of inequalities).

At the end I've got (on the number line):
$x>-3$, $x<3$, $-1.5<x<1.5$, $x>0$, and $x < 0$.

I don't understand the reason why in the solution set I have to include "$-3<x<-3/2$" because this set isn't satisfied by all the other set of solutions. (because of union, it's not the case of intersection).


问:
Is the dual of a finitely generated module finitely generated?
I recently thought of this and have no idea whether over a general commutative unitary ring the dual of a finitely generated module is finitely generated. This must be known.</p>

答:
If $I$ is an ideal of $R$, the dual of $R/I$ is isomorphic to $\mathrm{Ann}(I) = \{r \in R : rI = 0\}$, and this doesn't have to be finitely generated. Take for instance $R = k[y,x_1,x_2,\dotsc]/(y x_i : i \geq 1)$ and $I=(y)$.

A more natural question would be: How can we characterize commutative rings with the property that duals of f.g. modules over that ring are f.g.? Even more natural: How can we characterize commutative rings with the property that hom modules between f.g. modules are f.g.? For example, noetherian commutative rings satisfy this property (see the comment by Keenan Kidwell). But I think that there are more examples (perhaps coherent rings?).


问:
Keras lstm equations, matrices and diagram
I was reading this post, mainly the LSTM equation section for understand the structure diagram and relate it with the matrices and I can deducted this cell diagram:

I made this homemade diagram because in the Colah's article it's not the exactly same cell that in keras.

I need know if it ok, can someone confirm me that? This are the equations

W are weights matrices for input data and U are weights matrices for previous hidden state

b are biases

In the pink circles there are operations. The point it's multiply element by element and X is matrix multiplication

答:
It's ok. The picture and all the formulas in the question describe a standard LSTM network. The part that confuses you may be the content of Keras code: such as dropout and recurrent dropout.

As long as you don't set these parameters when calling LSTM cells, you won't be confused about these advanced structures.


Title: Solution to a definite integral as $\int_0^{1}(x^{m}\left(1+x)^{n}dx\right)$

Body:
While solving a question based on Binomial series, I encountered a definite integral as : $$\int_0^{1}(x^{m}\left(1+x)^{n}dx\right)$$ where $m,n$ are natural numbers.
My approach : wrote the above expression in two ways (1)- $$\int_0^1\sum_{r=0}^n \binom{n}{r} \left( {x}\right)^{m+r}dx $$ and also as (2)- $$\int_0^1\sum_{r=0}^m \binom{m}{r} \left(-1)^{r}( {1+x}\right)^{m+n-r}dx$$ and tried to solve using both the above mentioned expressions. Did some other simplification of expression (1) , But was unable to proceed further. Can anyone guide me to a "general answer based on any m,n" ? Any help would be highly appreciated. Thanks in advance.!

Body:
$$I=\int_0^{1}x^{m}\,(1+x)^{n}\,dx$$ Let $x=-y$ to face the beta function
$$I=(-1)^m \int_0^{-1}y^{m}\,(1-y)^{n}\,dy=(-1)^m\, B_{-1}(m+1,n+1)$$


问:
How is this read correctly in maths , and meaning (about: sets)?

$J \subseteq \left\{1,2,..,n\right\}$ is an index set ....   Fix $J
\subseteq \left\{1,2,..,n\right\}$ and let $S=\left\{x_j : j \in
J\right\}$

The problem I'm having here is "Fix" , what does that mean ? I think "fix" refers to some constant value, or in this case SOME set we are chosing?

My other question is how is the set $S$ read correctly? I would read it like that: The set $S$ contains elements $x_j$ where each index $j$ is from index set $J$"

I hope someone can help :)

答:
I would not use the word "fix", since logically speaking what is going on here is that we want to consider any arbitrary given $J ⊆ \{1..n\}$. Much better would be something like "Take any $J ⊆ \{1..n\}$, and let $S = \cdots$. Then . . .". But why do people use this word? It is because in their mind they are thinking that they want to temporarily focus on one specific $J$ and reason about that set together with another set $S$ associated with that particular $J$. So this becomes "let's fix $J$ for now, and define $S$ in terms of this $J$, and ...". Of course, in the end we are not talking about only one constant fixed $J$. This notion can be made clear in Fitch-style natural deduction:

[We want to prove $∀n{∈}ℕ\ ∀J{⊆}\{1..n\}\ ( \ Q(n,J) \ )$.]

Given $n{∈}ℕ$ and $J{⊆}\{1..n\}$:

  [It suffices to prove $Q(n,J)$ in this subcontext (i.e. with $n{∈}ℕ$ and $J{⊆}\{1..n\}$ given to you).]

  [Note that in this subcontext you have no control over $n,J$; they are fixed here.]

  $\vdots$

  $Q(n,J)$.

$∀n{∈}ℕ\ ∀J{⊆}\{1..n\}\ ( \ Q(n,J) \ )$.   [In the conclusion $n,J$ are of course not fixed.]


问:
Proving limit of an analytic function $f(z)$ on $z_0$, $\lim_{z \to z_0}\frac{f(z)}{z-z_0} = f'(z_0)$
I'm trying to prove the following

If $f$ analytic on $z_0$ and $f(z_0) = 0$, prove that
$\lim_{z \to z_0}\frac{f(z)}{z-z_0} = f'(z_0)$

So far I've been reading about poles, residue and zeros, I'm thinking to prove this, I need to argue about the zeroes of the function, is this correct?

Any insight would really help, thanks beforehand

edit: it seems like I'm taking this too far, should've tried using simple means first, that is the definition of $f'(z_0)$

答:
I think this follows directly from the definition of the derivative:

$$f'(z) = \lim_{z \rightarrow z_0} \frac{f(z)-f(z_0)}{z-z_0} = \lim_{z \rightarrow z_0} \frac{f(z)}{z-z_0}$$


问:
Differences between “for each” and “for all” according to Pugh's book
On page 6 of C. C. Pugh's Real Mathematical Analysis (2017)：

Avoid reading $\forall$ as "for all," which in English has a more inclusive connotation.

Instead, one should read $\forall$ always as for each, as the writer suggests.

Could anyone please point out for me what the more inclusive connotation Pugh is referring to?

答:
I'm not really sure what Pugh's getting at here. My best guess is that he's worried that one might have trouble distinguishing

"For all $x$ there is some $y$ such that [stuff]"

from

"There is some $y$ such that for all $x$ [stuff]"

on the grounds that "for all" sort of sounds like it's describing a single condition which must be met all at once.

OK this doesn't quite match up with Pugh's "more inclusive connotation," but it really is the best guess I have.

But to me this is a huge reach: quantifier alternation is a genuinely difficult topic for many people, language be darned, and I don't expect "For each" to do much better than "For all" in this respect. Certainly nobody will look at you askance if you read "$\forall$" as "for all."

Conversely, nobody will look at you askance if you read it as "for each" either. If that choice of language makes quantifier structure easier to parse, you should absolutely do it. But there's nothing wrong with "for all" here.


问:
A hard geometry problem involving harmonic divisions

Let acute triangle $ABC$. Let $A_1$ and $A_2$ the intersections of the circle of diameter $(BC)$ and the altitude from $A$ to $BC$ ($A_1$ is closer to $A$ than $A_2$). Similarily define points $B_1$, $B_2$, $C_1$, $C_2$. Let $A'$ the intersection of $B_1C_2$ and $B_2C_1$. Similarily define points $B'$ and $C'$. Prove that $AA'$, $BB'$ and $CC'$ concur.

My idea: Maybe it helps to see that $(A,H;A_1,A_2)$ (where $H$ is the orthocenter of triangle $ABC$) is a harmonic division. Also $H$ is the radical center of the 3 drawn circles, so $B_1C_1B_2C_2$ is cylcic (by power of a point). It is obvious by the figure that $A'$ must lie on $BC$. But I don't know how to prove it. Can you please help me? Thanks in advance!

If it helps, the problem comes from a Romanian book about harmonic divisions, but the solution is ommited.

答:
We solve the problem using trilinear coordinates. The altitude $AD$ is the set of points whose coordinates $x:y:z$ satisfy
$$y\cos B=z\cos C$$
The circle with diameter $BC$ is defined analogously, the points satisfying
$$yz=x(x\cos A-y\cos B-z\cos C)$$
(See exercise (174) here for the reference.) Arbitrarily setting $x=1$ (since trilinear coordinates are ratios) and then solving for $y,z$ gives us the coordinates of $A_1$ and $A_2$:
$$A_{1,2}=1:
-\cos C\pm\sqrt{\frac{\cos C}{\cos B}(\cos A+\cos B\cos C)}:
-\cos B\pm\sqrt{\frac{\cos B}{\cos C}(\cos A+\cos B\cos C)}$$
The plus sign gives $A_1$ and the minus sign gives $A_2$; $B_1,B_2,C_1,C_2$ can be obtained by cyclically permuting $A,B,C$ in the above equation.

Now associate the vector $(u,v,w)^T$ with both the point at coordinates $u:v:w$ and the line $ux+vy+wz=0$. It is well-known that the line through points $P_1$ and $P_2$ is $(\mathbf P_1×\mathbf P_2)\cdot(x,y,z)^T=0$ and that the intersection of lines $l_1$ and $l_2$ is $\mathbf l_1×\mathbf l_2$. Based on this, the intersection of the lines $B_1C_2$ and $C_1B_2$ is
$$A'=(\mathbf B_1×\mathbf C_2)×(\mathbf C_1×\mathbf B_2)$$
$$=0:(\cos A\cos C+\cos B)\sqrt{\cos C(\cos A\cos B+\cos C)}:(\cos A\cos B+\cos C)\sqrt{\cos B(\cos A\cos C+\cos B)}$$
Thus $A'$ lies on $BC$ as you suspected. The line $AA'$ then has normal vector $\mathbf l_A=\mathbf A'×(1,0,0)^T$, and similarly for $\mathbf l_B=BB'$ and $\mathbf l_C=CC'$ by cyclically permuting $A,B,C$; the determinant of the matrix formed by these three vectors is
$$\begin{vmatrix}\mathbf l_A&\mathbf l_B&\mathbf l_C\end{vmatrix}=0$$
Hence the lines $AA',BB',CC'$ concur, as was required to be shown, at the point with trilinear coordinates
$$X=\sqrt{1+\frac{\cos B\cos C}{\cos A}}:\sqrt{1+\frac{\cos C\cos A}{\cos B}}:\sqrt{1+\frac{\cos A\cos B}{\cos C}}$$
$$=\frac1{a\sqrt{b^2+c^2-a^2}}:\frac1{b\sqrt{c^2+a^2-b^2}}:\frac1{c\sqrt{a^2+b^2-c^2}}$$
$$=\frac1{\sqrt{a\cos A}}:\frac1{\sqrt{b\cos B}}:\frac1{\sqrt{c\cos C}}$$

Here is the SymPy code I used to derive all the expressions above:

```<code>#!/usr/bin/env python3
from sympy import *
cA, cB, cC = symbols('cA cB cC', positive=True) # cos A, cos B, cos C
x, y, z = symbols('x y z', real=True)

def cycB(p): # ABC -> BCA
    q = p.subs({cA: cB, cB: cC, cC: cA}, simultaneous=True)
    return Matrix([q[2], q[0], q[1]])
def cycC(p): # ABC -> CAB
    q = p.subs({cA: cC, cB: cA, cC: cB}, simultaneous=True)
    return Matrix([q[1], q[2], q[0]])

f1 = y*cB - z*cC
f2 = cA - y*cB - z*cC - y*z
sols = solve([f1, f2], [y, z])
A1 = Matrix([1, sols[1][0].expand(), sols[1][1].expand()])
A2 = Matrix([1, sols[0][0].expand(), sols[0][1].expand()])
print("A1 =", A1)
print("A2 =", A2)
B1 = cycB(A1)
B2 = cycB(A2)
C1 = cycC(A1)
C2 = cycC(A2)
Ap = simplify(  B1.cross(C2).cross(B2.cross(C1))  ) # A'
Ap *= sqrt(cA*cB*cC)/2
print("A' =", Ap)
lA = Ap.cross(Matrix([1, 0, 0]))
lB = cycB(lA)
lC = cycC(lA)
D = Matrix([lA.T, lB.T, lC.T])
pprint(D)
print("det(D) =", D.det()) # 0

X = D.nullspace()[0] * sqrt(cA*cB + cC) / sqrt(cC)
a, b, c = symbols('a b c', positive=True)
X = X.subs(cA, (b**2+c**2-a**2)/(2*b*c))
X = X.subs(cB, (c**2+a**2-b**2)/(2*c*a))
X = X.subs(cC, (a**2+b**2-c**2)/(2*a*b))
Delta = sqrt(-(a - b - c)*(a - b + c)*(a + b - c))*sqrt(a + b + c)/sqrt(2) # area of triangle
X = factor(X, deep=True) / Delta
print("X =", X.simplify())
</code>```

Body:
We solve the problem using trilinear coordinates. The altitude $AD$ is the set of points whose coordinates $x:y:z$ satisfy
$$y\cos B=z\cos C$$
The circle with diameter $BC$ is defined analogously, the points satisfying
$$yz=x(x\cos A-y\cos B-z\cos C)$$
(See exercise (174) here for the reference.) Arbitrarily setting $x=1$ (since trilinear coordinates are ratios) and then solving for $y,z$ gives us the coordinates of $A_1$ and $A_2$:
$$A_{1,2}=1:
-\cos C\pm\sqrt{\frac{\cos C}{\cos B}(\cos A+\cos B\cos C)}:
-\cos B\pm\sqrt{\frac{\cos B}{\cos C}(\cos A+\cos B\cos C)}$$
The plus sign gives $A_1$ and the minus sign gives $A_2$; $B_1,B_2,C_1,C_2$ can be obtained by cyclically permuting $A,B,C$ in the above equation.

Now associate the vector $(u,v,w)^T$ with both the point at coordinates $u:v:w$ and the line $ux+vy+wz=0$. It is well-known that the line through points $P_1$ and $P_2$ is $(\mathbf P_1×\mathbf P_2)\cdot(x,y,z)^T=0$ and that the intersection of lines $l_1$ and $l_2$ is $\mathbf l_1×\mathbf l_2$. Based on this, the intersection of the lines $B_1C_2$ and $C_1B_2$ is
$$A'=(\mathbf B_1×\mathbf C_2)×(\mathbf C_1×\mathbf B_2)$$
$$=0:(\cos A\cos C+\cos B)\sqrt{\cos C(\cos A\cos B+\cos C)}:(\cos A\cos B+\cos C)\sqrt{\cos B(\cos A\cos C+\cos B)}$$
Thus $A'$ lies on $BC$ as you suspected. The line $AA'$ then has normal vector $\mathbf l_A=\mathbf A'×(1,0,0)^T$, and similarly for $\mathbf l_B=BB'$ and $\mathbf l_C=CC'$ by cyclically permuting $A,B,C$; the determinant of the matrix formed by these three vectors is
$$\begin{vmatrix}\mathbf l_A&\mathbf l_B&\mathbf l_C\end{vmatrix}=0$$
Hence the lines $AA',BB',CC'$ concur, as was required to be shown, at the point with trilinear coordinates
$$X=\sqrt{1+\frac{\cos B\cos C}{\cos A}}:\sqrt{1+\frac{\cos C\cos A}{\cos B}}:\sqrt{1+\frac{\cos A\cos B}{\cos C}}$$
$$=\frac1{a\sqrt{b^2+c^2-a^2}}:\frac1{b\sqrt{c^2+a^2-b^2}}:\frac1{c\sqrt{a^2+b^2-c^2}}$$
$$=\frac1{\sqrt{a\cos A}}:\frac1{\sqrt{b\cos B}}:\frac1{\sqrt{c\cos C}}$$

Here is the SymPy code I used to derive all the expressions above:

#!/usr/bin/env python3
from sympy import *
cA, cB, cC = symbols('cA cB cC', positive=True) # cos A, cos B, cos C
x, y, z = symbols('x y z', real=True)

def cycB(p): # ABC -> BCA
    q = p.subs({cA: cB, cB: cC, cC: cA}, simultaneous=True)
    return Matrix([q[2], q[0], q[1]])
def cycC(p): # ABC -> CAB
    q = p.subs({cA: cC, cB: cA, cC: cB}, simultaneous=True)
    return Matrix([q[1], q[2], q[0]])

f1 = y*cB - z*cC
f2 = cA - y*cB - z*cC - y*z
sols = solve([f1, f2], [y, z])
A1 = Matrix([1, sols[1][0].expand(), sols[1][1].expand()])
A2 = Matrix([1, sols[0][0].expand(), sols[0][1].expand()])
print("A1 =", A1)
print("A2 =", A2)
B1 = cycB(A1)
B2 = cycB(A2)
C1 = cycC(A1)
C2 = cycC(A2)
Ap = simplify(  B1.cross(C2).cross(B2.cross(C1))  ) # A'
Ap *= sqrt(cA*cB*cC)/2
print("A' =", Ap)
lA = Ap.cross(Matrix([1, 0, 0]))
lB = cycB(lA)
lC = cycC(lA)
D = Matrix([lA.T, lB.T, lC.T])
pprint(D)
print("det(D) =", D.det()) # 0

X = D.nullspace()[0] * sqrt(cA*cB + cC) / sqrt(cC)
a, b, c = symbols('a b c', positive=True)
X = X.subs(cA, (b**2+c**2-a**2)/(2*b*c))
X = X.subs(cB, (c**2+a**2-b**2)/(2*c*a))
X = X.subs(cC, (a**2+b**2-c**2)/(2*a*b))
Delta = sqrt(-(a - b - c)*(a - b + c)*(a + b - c))*sqrt(a + b + c)/sqrt(2) # area of triangle
X = factor(X, deep=True) / Delta
print("X =", X.simplify())


问:
I don't think W/O replacement events are dependent.

Let say there're cards numbered 1 to 10. P(A) is the probability you pick a card whose number is the multiple of 2. P(B) is the probability you pick a card whose number is the multiple of 5. So now you draw two cards without replacement. You don't place the first card you just drew back in the original deck. Then, you draw again from that 9 remaining cards.

In the picture, P(A) is shown as the ordered pairs colored red, and P(B) is shown as the ordered pairs shaded with light blue.

Then I counted them all.

P(A) = (9+9+9+9+9) / (100-10) = 45/90 = 1/2

P(B) = (9+9) / (100-10) = 18/90 = 1/5

P(A∩B) = 9 / 90 = 1/10

So here, I found out that P(A∩B) is actually the multiple of P(A) and P(B), which shouldn't be if they are dependent, meaning they are independent. I've also found out, in a different model, that the probabilities could be dependent each other in the case of W/O replacement such as this one.

I know events with replacements are independent. But it feels like events without replacements doesn't mean dependency. It's more like the case of contraposition of the first one. Events with replacements mean independent, which means, by the contraposition rule, dependent events are the ones done without replacements.

But all the other internet pages say like "if they are done without replacement, then they are dependent". This claim is so prevalent. But one of us, they or I, must be wrong. What am I missing here?

答:
first, "with\without replacement" always appears with "sampling" 。I think the event A and event B you described is not sampling itself.

Normally, if P(A=i) is the probability you pick a card whose number is i,P(B=j) is the probability you pick a card whose number is j. apperantly,  P(B|A) != P(B) without replacement.


问:
Poisson distribution question, tips needed!
A car dealership opens every day with a fresh stock of $A$ cars. Let $N$ be the r.v. corresponding to the number of purchases per day. Suppose $N$ is distributed according to the Poisson distribution with parameter $\lambda=2$. What should $A$ be for the dealership to run out of cars to sell at most once every 10 days?

I have no clue how to begin to answer this question. How should I approach this problem? I'm not looking for a ready solution, but rather an explanation/tips on how to deal with this sort of problem. All I know is how the Poisson distribution looks like, and what the parameter means, but beyond that, I'm stuck.

答:
We need to interpret "at most once every $10$ days." With bad luck one could run out every day for a week, no matter how big $A$ is.

So we interpret "at most once every $10$ days" as meaning "with daily probability $\frac{1}{10}$."

So our stock should be such that $\Pr(N\le A) \ge 0.9$. Presumably we want the smallest such $A$.


问:
Evaluating $\lim_{x\to 0}\dfrac{(1+x)^{\tfrac1x}-e}{x}$
Since we have $\frac00$,I Applied L'Hopital rule :

$$\lim_{x\to 0} (1+x)^{\tfrac1x}\times\left(\cfrac{-\ln(1+x)}{x^2}+\cfrac{1}{x(x+1)}\right)$$$$=\lim_{x\to 0}\cfrac{x^2(x+1)(1+x)^{\tfrac1x}-(x+1)\ln(1+x)+x}{x^2(x+1)}$$

But as you can see it is getting very ugly.

Body:
\begin{aligned}
&\lim_{x\to 0}\frac{(1+x)^{\frac{1}{x}}-e}{x}\\
&=\lim_{x\to 0}\frac{e^{{\frac{1}{x}}\ln(1+x)}-e}{x}\\
&=e\lim_{x\to 0}\frac{e^{{\frac{1}{x}}\ln(1+x)-1}-1}{x}\\
&=e\lim_{x\to 0}\frac{{\frac{1}{x}}\ln(1+x)-1}{x}\\
&=e\lim_{x\to 0}\frac{\ln(1+x)-x}{x^2}\\
&=e\lim_{x\to 0}\frac{\frac{1}{1+x}-1}{2x}\\
&=e\lim_{x\to 0}\frac{-1}{2(1+x)}\\
&=\frac{-e}{2}
\end{aligned}


问:
the minimal Hausdorff space
Let

$$X=\{a_{ij},b_{ij},c_i,a,b\mid i=1,2,\ldots;\,j=1,2,\ldots\}$$

where all these elements are assumed to be distinct. Define the following neighborhood systems on $X$ :
Each $a_{ij}$ is isolated and each $b_{ij}$ is isolated:

$$\mathscr{B}(c_i)=\left\{V^n(c_i)=\bigcup_{j=n}^\infty\{a_{ij},b_{ij},c_i\}\mid n=1,2,\ldots\right\}$$

$$\mathscr{B}(a)=\left\{V^n(a)=\bigcup_{j=1}^\infty\bigcup_{i=n}^\infty\{a_{ij},a\}\mid n=1,2,\ldots\right\}$$

$$\mathscr{B}(b)=\left\{V^n(b)=\bigcup_{j=1}^\infty\bigcup_{i=n}^\infty\{b_{ij},b\}\mid n=1,2,\ldots\right\}\,.$$

Let us denote this topology by $\tau$.

Why is $(X,\tau)$ is minimal Hausdorff but not compact?

答:
$X$ is clearly not compact: $\{V^1(a),V^1(b)\}\cup\{V^1(c_i):i\in\Bbb Z^+\}$ is an open cover with no finite subcover.

A space is minimal if and only if it is semiregular and H-closed, as is noted in the remarks at the end of the answer to this question. It is straightforward to verify that $X$ is H-closed. If $\mathscr{U}$ is an open cover of $X$, there are $U_a,U_b\in\mathscr{U}$ such that $a\in U_a$ and $b\in U_b$. From the definition of $\tau$ it’s clear that there is a $k\in\Bbb Z^+$ such that

$$a_{ij},b_{ij},c_i\in\operatorname{cl}U_a\cup\operatorname{cl}U_b$$

for all $i>k$ and all $j\in\Bbb Z^+$. For $i=1,\ldots k$ there are a $U_i\in\mathscr{U}$ and an $n_i\in\Bbb Z^+$ such that $c_i,a_{ij},b_{ij}\in U_i=\operatorname{cl}U_i$ for all $j\ge n_i$. Thus,

$$X\setminus\left(\operatorname{cl}U_a\cup\operatorname{cl}U_b\cup\bigcup_{i=1}^k\operatorname{cl}U_i\right)$$

is at most a finite set of isolated points and is covered by the closures of some finite $\mathscr{F}\subseteq\mathscr{U}$, so that

$$X=\operatorname{cl}U_a\cup\operatorname{cl}U_b\cup\bigcup_{i=1}^k\operatorname{cl}U_i\cup\bigcup\{\operatorname{cl}U:U\in\mathscr{F}\}\,,$$

and $X$ is therefore H-closed.

Finally, all of the basic open nbhds $\{a_{ij}\}$, $\{b_{ij}\}$, and $V^n(c_i)$ are clopen, and the basic open nbhds $V^n(a)$ and $V^n(b)$ are regular open, so $X$ is semiregular.

Title: the minimal Hausdorff space


问:
An interesting application of the Hahn-Banach Theorem.
Let $X$ be a normed space and $x_1,x_2\in X$ nonzero elements. Show that there are functionals $F_1,F_2\in X'$ such that $F_1(x_1)F_2(x_2)=\lVert x_1\rVert \lVert x_2\rVert$ and $\lVert F_1\rVert \lVert x_1\rVert =\lVert F_2\rVert \lVert x_2\rVert$.

My attempt:

My idea was to define a functional $f_1:\langle \{x_1\}\rangle\to \mathbb{R}$, by $f_1(\alpha x_1)=\alpha \lVert x_2\rVert.$ Then, by using Hahn-Banach, I extend $f_1$ to a functional $F_1:X\to \mathbb{R}$ such that $F_1(x_1)=\lVert x_2\rVert$ and $\lVert F_1\rVert=\lVert f_1\rVert=\frac{\lVert x_2\rVert}{\lVert x_1\rVert}$. After that, I tried to define a functional $F_2:X\to \mathbb{R}$ such that $F_2(x_2)=\lVert x_1\rVert$ and $\lVert F_2\rVert=1$, but I coundn't do that.

答:
What you're doing is great!  Observe the symmetry between the indices $1$ and $2$ in the question.  I think that $$\lVert F_1\rVert=\lVert f_1\rVert=\frac{\lVert x_2\rVert}{\lVert x_1\rVert}\tag{*}\label{*}$$ breaks the symmetry.
Your goal is $\lVert F_1\rVert \lVert x_1\rVert =\lVert F_2\rVert \lVert x_2\rVert$, so the denominator of \eqref{*} should be $1$ instead.  To acheive this, replace $f_1$ with $f_1(ax_1) = a$ in your proof.  Then $\| F_1 \| = 1/\|x_1\|$.  Similarly, $\|F_2\| = 1/\|x_2\|$.

But $F_1(x_1)F_2(x_2) = 1$.  To fix this, it's simple.  Just multiply the everything by $\sqrt{\|x_1\|\|x_2\|}$.

To wrap up, define $f_1(ax_1) = a \sqrt{\|x_1\|\|x_2\|}$ and $f_2(ax_2) = a \sqrt{\|x_1\|\|x_2\|}$.  Then $\|F_1\| = (1/\|x_1\|) \; \sqrt{\|x_1\|\|x_2\|}$ and $\|F_2\| = (1/\|x_2\|)\sqrt{\|x_1\|\|x_2\|}$, so that $\|F_1\|\|x_1\| = \|F_2\|\|x_2\| = \sqrt{\|x_1\|\|x_2\|}$ and $F_1(x_1)F_2(x_2) = \sqrt{\|x_1\|\|x_2\|} \; \sqrt{\|x_1\|\|x_2\|} = \|x_1\|\|x_2\|$.


问:
A question about a contractible manifold
A couple of questions. $X$ is a compact manifold, $Z\subset Y$ is a closed submanifold, then

(i) If $Y$ is contractible (i.e. $id_Y$ is homotopic to a constant map), then any map $f: X\rightarrow Y$ is homotopic to the constant map. May I know why this is true?

$X\xrightarrow{f}Y\xrightarrow{id_Y}Y$ and $id_Y\sim \{y\}$, $\{y\}$ being a constant map.

(ii) No compact manifold is contractible. The reason given is let $Z$ be a single point and let $f:X\rightarrow X$ be an identity map. Then since $X$ is contractible, $f$ is homotopic to a constant map and hence $I_2(f,Z) = 0$. But $\#(f(X)\cap Z) = \#Z = 1$ and hence a contradiction.

I know that if $g:X\rightarrow Y$ is homotopic to a constant map, then $I_2(g,Z) = 0$. So, how does $f\sim \{y\}\implies g\sim \{y\}$?. Thanks!

答:
For (i), it follows from the result Perturbative mentions and noting that being homotopic is a transitive property. Perturbative is also right in saying that there are certainly compact manifolds that are contractible, [0,1] is clearly an example as given, but if we consider only manifolds without boundaries, then (ii) is very much true (in fact, it's exercise 2.4.6 in Guillemin & Pollack).

As the chapter would have you work out, you would start by proving 2.4.4:

If $f:X \longrightarrow Y$ is homotopic to a constant map, show that $I_{2}(f,Z) = 0$
for all complementary dimensional closed $Z$ in $Y$, except perhaps if $\dim X = 0$.

The idea is that we can always push a point on the manifold into the surrounding ambient space with an arbitrarily small deformation, which will allow its preimage to yield the empty set (zero intersection number). Since this arbitrarily small deformation can be made via a straight line homotopy (manifolds are locally path connected, at least in G&P), the result follows since $I_{2}(f,Z) = I_{2}(g,Z)$ for $f \sim g$.

Use this to prove 2.4.5:

If $Y$ is contractible and $\dim Y > 0$, then $I_{2}(f,Z) = 0$ for every
$f:X \longrightarrow Y$, $X$ compact and $Z$ closed, $\dim X + \dim Z = \dim Y$.

(Since all maps into a contractible space are homotopic, in particular, any map is homotopic to the constant map.)

Finally, we approach 2.4.6 by assuming, for a contradiction, the existence of a compact, contractible manifold $Y$ (without boundary) that isn't already the single point space. We then take $Z = \{y_{0}\}$ with $y_{0} \in Y$, and by 2.4.5, we have that $I_{2}(Id_{Y},Z) = 0$, but this is a contradiction since $|Id_{Y}^{-1}(Z)| = |Z| = 1$ by nature of the identity, as you've noted.


问:
$L^{p_1}(X,\mu) \cap L^{p_3}(X,\mu) \subset L^{p_2}(X,\mu)$ for $1 \le p_1 \lt p_2 \lt p_3 \le \infty$
Let $(X,\sum,\mu)$ be a measure space and let $1 \le p_1 \lt p_2 \lt p_3 \le \infty$. Show that $L^{p_1}(X,\mu) \cap L^{p_3}(X,\mu) \subset L^{p_2}(X,\mu)$.

From Holder's Inequality we know that for $1 \le p \lt q \le \infty$ that $L^q(X,\mu) \subset L^p(X,\mu)$.

Thus, $L^{p_3}(X,\mu) \subset L^{p_1}(X,\mu) \implies L^{p_3}(X,\mu) \cap L^{p_1}(X,\mu) \subset L^{p_1}(X,\mu)$. But from the statement above, since $ 1 \le p_1 \lt p_2 \lt \infty$ we should have that $L^{p_2}(X,\mu) \subset L^{p_1}(X,\mu) \implies  L^{p_2}(X,\mu) \subset L^{p_3}(X,\mu) \cap L^{p_1}(X,\mu)$. I'm not quite sure what I'm doing wrong here.

答:
$\textbf{Proposition } $(Interpolation inequality):  If $ (X,\Sigma,\mu) $ is a measurable space, $ 1 \leq p \leq q \leq \infty $ and $ f \in L_p(X) \cap L_q(X) $, then $ f \in L_r(X)  $ for all $ p \leq r \leq q $.

$\textbf{Proof}:$
First assume that $ q = \infty $. If  $ t=\frac{p}{r} $, then   $ |f|^r \leq ||f||_{\infty}^{r-p} |f|^p $. Hence,
$$ ||f||_r  \leq ||f||_{\infty}^{1-\frac{p}{r}} ||f||_p^{\frac{p}{r}} < \infty .$$
Now assume that $ q< \infty $. Let $t \in [0,1]$ such that
$$1= \frac{tr}{p}+\frac{(1-t)r}{q} .$$ Consider the conjugate exponents
$ p/(tr) $ and $ q/((1-t)r) $. By Holder's inequality, it follows that
$$ ||f||_r^r = \int |f|^r = \int |f| ^{tr} |f|^{(1-t)r} \leq ||f||_p^{tr} ||f||_q^{(1-t)r} < \infty.$$


问:
What is the reason for the name *left* coset?
Let $G$ be a group and let $H \leq G$ be a subgroup. It seems that it is now standard to call the cosets
$$gH=\{gh \ | h \in H \}$$ the left cosets of $H$ in $G$. I have to admit to being slightly annoyed with this convention: these are the orbits for the right action of $H$ on $G$. Therefore I am perpetually tempted to refer to them as right cosets. Is there a second good reason (I happily admit convention is a very good reason) for calling these left cosets? (The fact that the "g" is written on the left does not count, as far as I am concerned).

答: Don't think of them as the orbits of the right action of $H$ on $G$. The set of left cosets itself admits a natural left $G$-action, and every transitive left $G$-action arises in this way. </p>


答:
I think he wants to make sure the meaning is not read as "for all at once" instead of the intended "for each separately (whichever one you care to choose)".

For those who have used the notation for years the meaning is intuitive. Perhaps there is a point of confusion at the early learning stage. Most would not be dogmatic about it - "For all ..." is after all a colloquial rendering of a precise mathematical statement (which is why we have the notation in the first place).

Title: Differences between “for each” and “for all” according to Pugh's book

Body:
On page 6 of C. C. Pugh's Real Mathematical Analysis (2017)：

Avoid reading $\forall$ as "for all," which in English has a more inclusive connotation.

Instead, one should read $\forall$ always as for each, as the writer suggests.

Could anyone please point out for me what the more inclusive connotation Pugh is referring to?


Title: Evaluating $\lim_{x\to 0}\dfrac{(1+x)^{\tfrac1x}-e}{x}$

Body:
Since we have $\frac00$,I Applied L'Hopital rule :

$$\lim_{x\to 0} (1+x)^{\tfrac1x}\times\left(\cfrac{-\ln(1+x)}{x^2}+\cfrac{1}{x(x+1)}\right)$$$$=\lim_{x\to 0}\cfrac{x^2(x+1)(1+x)^{\tfrac1x}-(x+1)\ln(1+x)+x}{x^2(x+1)}$$

But as you can see it is getting very ugly.

Body:
Hint

Taylor series of $${(1+x)}^{1/x}=e(1-\frac{1}{2}x+\frac{11}{24}x^2+...)$$

If used you get $-e/2$


问:
Exercise in Chapter 1 of Switzer's Algebraic Topology book
I'm trying to answer the following question: (I've added details to make the post self-contained.)

Fix two pointed spaces $(K,k_0)$ and $(L,l_0)$ and define two functors $F_K$, $F_L$ from the category of pointed spaces + continuous functions preserving base point $\mathscr{PT}$ to the category of pointed sets + functions preserving base point $\mathscr{PS}$ as follows:

$(X,x_0)  \longmapsto F_K(X,x_0) = [ K,k_0; X,x_0]$, here $[ \cdot, \cdot ]$ represents homotopy classes of maps.

$f:(X,x_0) \rightarrow (Y,y_0) \longmapsto F_K(f)$,
where $F_K(f)[g] = [f \circ g] \in [K,k_0;Y,y_0]$ for $[g]\in [K,k_0;X,x_0]$

The functor $F_L$ is similarly defined (replacing $(K,k_0)$ with $(L,l_0)$).

Next, define a natural transformation as follows. Suppose
$\phi: (K,k_0) \rightarrow (L,l_0)$ is a morphism in $\mathscr{PT}$. Then
define $T_\phi:F_L \rightarrow F_K$ by

$T_\phi (X,x_0)[g] = [g\circ \phi] \in [K,k_0;X,x_0]$,

for every $(X,x_0)$ and $[g] \in F_L(X,x_0)$. (Switzer shows that this satisfies the properties of a natural transformation.)

Finally, the actual question: Show that $\mathbf{ \phi:(K,k_0) \rightarrow (L,l_0)}$ is a homotopy equivalence if and only if $T_\phi$ is a natural equivalence ($\mathbf{T_\phi(X,x_0)}$ is a bijection for every $\mathbf{ (X,x_0)}$).

My attempt at a solution for the $(\Rightarrow)$ direction: (for the sake of reducing typing, I am going to ignore writing the base points)

Let $\phi: K \rightarrow L$ be a homotopy equivalence. Then there exists $\psi: L \rightarrow K$ such that $\psi \circ \phi \simeq id_K$ and $\phi \circ \psi \simeq id_L$. I will now define a bijection $\Phi: [L;X] \cong [K;X]$. For $[h] \in [L;X]$ define $\Phi[h] := [h\circ \phi]\in [K;X]$. To see that this is 1-1, let $[h_1]$ and $[h_2]$ be in $[L;X]$ and assume that $\Phi[h_1] = \Phi[h_2]$ then $[h_1 \circ \phi] = [h_2 \circ \phi] \Rightarrow [h_1 \circ \phi \circ \psi ] = [h_2 \circ \phi \circ \psi ] \Rightarrow [h_1] = [h_2]$. Next, for onto, let $[g] \in [K;X]$, and note $[g] = [g \circ \psi \circ \phi] = \Phi[g \circ \psi]$.

So, if $\phi : K \rightarrow L$ is a homotopy equivalence then $[K;X] \cong [L;X]$ for all $X$.

Question: Does anyone see anything wrong with my solution to this direction?

For the $(\Leftarrow)$ direction I am completely stuck at: Let $T_\phi: F_L \rightarrow F_K$ be an equivalence for all $X$; that is $[K;X] \cong [L;X]$ for all $X$. :)

Question: Can anyone give me a nudge for this direction?

答:
Your proof is correct, but I suggest to use the notation introduced in your question. That is, use $T_\phi(X)$ instead of $\Phi$ (base point suppressed). Proceed as follows:

Switzer's construction works for any $\theta : A \to B$ to give $T_\theta(X) : [B,X] \to [A,X]$.

Clearly $T_\theta(X)$ only depends on the pointed homotopy class $[\theta]$ of $\theta$.

Obviously  $T_{id}(X) = id$ and $T_{\chi \circ \theta}(X) = T_\theta(X) \circ T_\chi(X)$.

If $\phi$ is a pointed homotopy equivalence and $\psi$ is a homotopy inverse, then  $T_\phi(X) \circ T_\psi(X) = T_{\psi \circ \phi}(X) = T_{id}(X) = id$, similarly $T_\psi(X) \circ T_\phi(X) = id$. This means that $T_\phi(X)$ and $T_\psi(X)$ are bijections which are inverse to each other.

For the converse, take $X = K$ and let $\psi : L \to K$ represent the unique homotopy class in $[L;K]$ such that $T_\phi(K)([\psi]) =  [id] \in [K;K]$ (recall that $T_\phi(K) : [L;X] \to [K;X]$ is a bijection for all $X$). Then $[\psi] \circ [\phi] = [\psi \circ \phi] = T_\phi([\psi]) = [id]$. By 2. we get $T_\phi(L) \circ T_\psi(L) = T_{\psi \circ \phi}(L) = id$, thus $T_\psi(L) = T_\phi(L)^{-1}$, i.e. $T_\psi(L) : [K;L] \to [L;L]$ is a bijection. Let $\phi' : K \to L$ represent the unique homotopy class in $[K;L]$ such that $T_\psi(L)([\phi']) =  [id] \in [L;L]$. This means $[\phi'] \circ [\psi] = [id]$. Thus $[\psi]$ has a rigt and a left inverse. It is a general theorem of category theory that if a morphism $v : B \to A$ has a right inverse $u : A \to B$ (i.e. $v \circ u = id_A$) and a left inverse $u' : A \to B$ (i.e. $u' \circ v = id_B$), then $v$ is an isomorphism and $u = u' = v^{-1}$.

To see this, note that $u = id_B \circ u = (u' \circ v) \circ u = u' \circ (v \circ u) = u' \circ id_A = u'$.


问:
Evaluating $\oint_C (\arctan{y} - 5y)dx + \left(\frac{x}{1+y^2} + 2x\right)dy$
Evaluate integral of

$$\oint_C (\arctan{y} - 5y)dx + \left(\frac{x}{1+y^2} + 2x\right)dy$$

where $C$ is the ellipse $x^2/16 + y^2/25 = 1$.

So I used the Green's Theorem and got double integral of $D$ of $7 \; dy\; dx$. What do I do now? How do I get the $x$ and $y$ bounds? Please show steps to how to solve this using Green's Theorem. Please don't use some crazy methods to solve it. I just want to keep it simple. Thank you

答:
If you do not want to argue geometrically, using the area of ellipse, use Green's formula twice (for double credit?)

$$\iint_D 7\,dx\,dy  = \oint_C \frac{7}{2}(-y\,dx+x\,dy) $$
and use the parametric equation $x=4\cos t$, $y=5\sin t$  to finish it off:
$$ \int_0^{2\pi} \frac{7}{2}(-5\sin t \cdot (-4\sin t) +4\cos t\cdot 5\cos t) \,dt
$$
where the integrand simplifies a lot.


Body:
Without series, only L'Hospital
$$
\lim_{x\to 0}\dfrac{(1+x)^{\tfrac1x}-e}{x}
$$
we get
$$
\lim_{x\to 0}\left[\left(-\frac{\ln(1+x)}{x^2}+\frac{1}{x}(1+x)^{-1}\right)(1+x)^{\frac{1}{x}}\right] \\
=\lim_{x\to 0}\left(-\frac{\ln(1+x)}{x^2}+\frac{\frac{1}{1+x}}{x}\right)\lim_{x\to 0}(1+x)^{\frac{1}{x}} \\
=\lim_{x\to 0}\left(\frac{\frac{x}{1+x}-\ln(1+x)}{x^2}\right)e
$$
another L'Hospital
$$
= \lim_{x\to 0}\left(\frac{\frac{1}{1+x}+\frac{-x}{(1+x)^2}-\frac{1}{1+x}}{2x}\right)e
= \lim_{x\to 0}\frac{-1}{2(1+x)^2}e = -\frac{e}{2}\\
$$


问:
Problem Finding the limit with 2 parameters where $I(n,m) = \lim\limits_{x \to \pi} \frac{\sin (nx)}{\sin (mx)}$
Here i have a problem since x approaches pi

So here i have to calculate the Limit and then add it

I (2019,2020) + I (2018,2020) + I (2019,2021) rounded upto 4 decimal places

I got a answer of 2.9975 which is wrong i need  to know how to approach this problem

答:
Let $x-\pi=y$

$$\sin(mx)=\sin m(\pi+y)=\sin(my)\cos(m\pi)$$

Now for integer $m,\cos(m\pi)=(-1)^m$

For integer $m,n;$
$$\lim_{x\to\pi}\dfrac{\sin mx}{\sin nx}=(-1)^{m-n}\dfrac mn\cdot\dfrac{\lim_{y\to0}\dfrac{\sin my}{my}}{\lim_{y\to0}\dfrac{\sin ny}{ny}}$$