问:
Proving if limit exists then function is continuous
The question I'm trying to solve:

Let $(X, d)$ and $(Y, e)$ be metric spaces, where $(Y, e)$ is complete.

Let $A \subseteq X$ be dense in $X$. Suppose $f: A \to Y$ is uniformly continuous.

Prove there exists a (unique) continuous map $F: X \to Y$ such that $F(x) = f(x)$ for all $x \in A$.

One of the parts defines $F: X \to Y$ by
$$
F(x) = \lim_{n \to \infty} f(a_n)
$$
for any sequence $(a_n) \in A$ which converges to $x$ in $(X, d)$.

The part that I'm struggling is:

Prove that $F: X \to Y$, as defined above, is continuous.

It clear that $F$ is continuous for all $x \in A$ as $F(x) = f(x)$ for $x \in A$ and it preserves uniform continuity on $A$ from $f$.

I'm trying to show continuity using definition in terms of limits of sequences which is as follows:
$$
\forall (a_n) \in A \lim_{x \to \infty} a_n = x \implies \lim_{x \to \infty} f(a_n) = f(x)
$$
I have shown before that sequence $(f(a_n))$ is Cauchy and therefore convergent in $(Y, e)$ as it's complete.

Therefore I know that the $\lim_{x \to \infty} f(a_n)$ exists and I denote it $L$.

All that is left to show is that $L = f(x)$ where $x \in A'$ ($A'$ denotes limit points of $A$).

I suspect this is true due to uniform continuity of $f$ which I suppose show us that $d(a_n, x) \implies e(f(a_n), f(x))$.

Yet, I'm not sure we could do this since f is defined on A and cannot take values of A'.

Thus I don't really understand how to show this.

答:
Let's prove this in stages. First, a simple fact (you stated this, but we'll prove it anyway):

Lemma. If the sequence $\{a_n\} \subset A$ converges to $x \in X$, then $\{f(a_n)\}$ is Cauchy in $(Y,e)$.

Proof. Indeed, since $f$ is uniformly continuous, for every $\varepsilon > 0$ we have $e(f(a),f(b)) < \varepsilon$ whenever $d(a,b) < \delta$. And since $\{a_n\}$ is Cauchy, there exists $N \geq 0$ such that $d(a_m, a_n) < \delta$ for all $m,n \geq N$. Thus, $e(f(a_m), f(a_n)) < \varepsilon$ for all $m,n \geq N$; this means that $\{f(a_n)\}$ is Cauchy. $\square$

Therefore, $\lim f(a_n)$ is a nice, well-defined object for sequences $a_n \to x$, since $(Y, e)$ is complete.

Next, let's show that $F$ is well-defined. That is, for $x \in X$ and sequences $\{a_n\}, \{b_n\} \subset A$  with $a_n \to x$ and $b_n \to x$, we show that

$$\lim f(a_n) = \lim f(b_n).$$

The important observation is that the sequence $\{c_n\} = \{a_1, b_1, a_2, b_2, \dots\}$, or formally by
$$
c_n =
\begin{cases}
a_{(n+1)/2} & \text{if $n$ is even,} \\
b_{n/2} & \text{if $n$ is odd,}
\end{cases}
$$
also converges to $x$. The proof is to choose $N_1, N_2 \geq 0$ such that $d(x, a_n) < \varepsilon / 2$ when $n \geq N_1$ and $d(x, b_n) < \varepsilon / 2$ when $n \geq N_2$. Taking $N = \max(N_1,N_2)$ therefore suffices. It follows that $\{f(c_n)\}$ is Cauchy in $Y$ by our lemma, so $\lim f(c_n)$ exists, and it equals $\lim f(a_n)$ and $\lim f(b_n)$ by taking subsequences.

Now we turn to continuity of $F$. For $x_n \to x$ in $X$, we fix the sequences $\{a^n_k\} \subset A$ such that $a^n_k \to x_n$. Construct a new sequence $b_n$ as follows. Let $k_n \geq 0$ be the smallest integer such that $e(f(a^n_k), F(x_n)) < 1/n$ for all $k \geq k_n$. Set $b_n = a^n_{k_n}$ and observe that $\{b_n\} \subset A$. Then it follows from the triangle inequality that
$$
e(F(x_n), F(x)) \leq e(F(x_n), f(b_n)) + e(f(b_n), F(x)) < \tfrac{1}{n} + e(f(b_n), F(x))
$$
for each $n=1,2,\dots$. On the other hand, $b_n \to x$, so $e(f(b_n), F(x)) \to 0$.
Therefore $F(x_n) \to F(x)$ as $x_n \to x$.

The last part to show is uniqueness of $F$. Suppose $G$ is also a continuous extension of $f$ from $A$ to $X$. Let $x \in X$, and fix $\{a_n\} \subset A$ such that $a_n \to x$. Then
$$
F(x) = \lim F(a_n) = \lim f(a_n) = \lim G(a_n) = G(x),
$$
so in fact $F=G$.


问:
Are there new possibilities in higher dimensions?
The question is about spatial dimensions.

It seems to me that every dimension adds something fundamentally new to space:

If you have one dimension, you have different positions and the ability to move an object (like a point).
By adding the second dimension, the objects have the possibility to go around each other - they mustn't go through each other to change positions.
The third dimension allows to tie a knot.
Is there anything new like this if you add the fourth, fifth, ... dimension?

答:
In four dimensions, you can not knot a line.

On the other hand, it is possible to build a city with roads, railways and footpaths, on the level with none of these crossing each other.

Seasons are still possible in four dimensions, but the day is 12 hours everywhere, and the world would be perfectly round.  What does happen though, is that one gets 'season zones' like 'time zones'.  One can think of 3d world as a line where the S hemisphere is one season, and the N hemisphere is six months later.  In four dimensions, there is always somewhere that is in early spring, or late autumn, or mid-winter, or whatever.

You can weave 2d sheets in 4d, in general, to cut and weave one needs N-2 dimensional sheets and blades.

In four dimensions, it is possible for topologically different figures to have the same topological surface.

In five dimensions, it is possible to knot, but not weave, 2d sheets.

ContentLicense: CC BY-SA 3.0


问:
Given $p \in \mathbb R^N$ and $X$ a symmetric matrix, find $\varphi \in C^2$ such that $D\varphi = p, D^2\varphi = X$ at a given point
In their paper User's guide to viscosity solutions of second order differential partial equations, Crandall, Ishii and Lions define the superjet of an upper semicontinuous function $u$ at $\hat x \in \mathcal O$, $\mathcal O$ a locally compact subset of $\mathbb R^N$, to be the set of $(p, X) \in \mathbb R^N \times \mathcal S_N$ ($\mathcal S_N$ the set of symmetric matrices) such that
\begin{equation}
u(x) \leq u(\hat x) + \langle p, x - \hat x \rangle + \frac 12 \langle X(x - \hat x), x - \hat x \rangle + o(|x - \hat x|^2) \quad \quad (*)
\end{equation}
holds. The superjet is denoted $J_{\mathcal O}^{2, +}u(\hat x)$. It is left as an exercise that
$$
J_{\mathcal O}^{2, +}u(\hat x) = \{(D\varphi(\hat x), D^2 \varphi(\hat x)) \ : \ \varphi \in C^2 \text{ and } u - \varphi \text{ has a local maximum at } \hat x\}.
$$

My question is: given $(p, X) \in \mathbb R^N \times \mathcal S_N$ such that $(*)$ holds, how to find such a $\varphi$?

Thanks in advance.

答:
This is proved in Proposition 12.11 in the Calder's Viscosity Solutions notes: http://www-users.math.umn.edu/~jwcalder/viscosity_solutions.pdf

The proof is similar to the one given by your professor, copied in another answer. If there is some step in the proof that is not clear, I can elaborate on it in this answer to explain it more clearly.

EDIT: My guess is your confusion is near the end of the proof, where one verifies that the constructed test function has the correct derivatives. I'll add some details here, and I'll follow the proof in Calder's
\ notes above.

At the end of the proof we've shown that
$$u(x) \leq p \cdot x + \frac{1}{2}x^T X x + \sigma(3|x|)=:\varphi(x)$$
in a neighborhood of the origin (we've taken $\hat{x}=0$). Earlier in the proof we defined
$$\sigma(r) = \int_0^r \int_0^s \rho(t)\, dt\, ds,$$
where $\rho$ is continuous, nondecreasing, and $\rho(0)=0$. From this, we immediately have
$$(*) \ \ \ \sigma(r) \leq \frac{1}{2}r^2\rho(r),$$
$$(**) \ \ \ \sigma'(r) = \int_0^r \rho(t) \, dt \leq r\rho(r),$$
and
$$(***) \ \ \ \sigma''(r) \leq \rho(r).$$
In particular, $\sigma$ is $C^2$ on $[0,\infty)$ and
$$\sigma(0)=\sigma'(0)=\sigma''(0)=0.$$

I assume what is not clear is to show that $D \varphi(0)= p$ and $D^2\varphi(0)=X$, as is claimed (without details) in the notes. To prove this, we just need to consider the perturbative term
$$\psi(x) := \sigma(3|x|),$$
and show that its gradient and Hessian vanish at $x=0$.
By $(*)$ we have
$$|\psi(x)| \leq \frac{9}{2}\rho(3|x|)|x|^2.$$
It follows by writing the difference quotient definition of the derivative that $D\psi(0)=0$. We also compute
$$D\psi(x) = 3\sigma'(3|x|)\frac{x}{|x|} \ \ \text{for }x\neq 0.$$
By $(**)$ we see that $D\psi(x)$ is continuous at $x=0$. Similarly, by $(**)$ we see that
$$|D\psi(x)| \leq 9\rho(3|x|)|x|.$$
Again using the difference quotient definitions of the derivative we have $D^2\psi(0)=0$, since $\rho$ is continuous and vanishes at the origin. We also compute
$$D^2\psi(x) = 3\sigma'(3|x|)\frac{I}{|x|} + 3\left(3\sigma''(3|x|) - \frac{\sigma'(3|x|)}{|x|}\right) \frac{xx^T}{|x|^2}$$
for $x\neq 0$.
As before, by $(**)$ and $(***)$ we find that $D^2\psi(x)\to 0$ as $x\to 0$, so we have continuity at zero. This verifies that $\psi$ is indeed $C^2$ and has vanishing gradient and Hessian.

I hope this helps clear things up. Let me know if you have other questions.

ContentLicense: CC BY-SA 4.0


问:
Limit of a simple Integral
Need to show that $$\lim_{t\to 1^-}(1-t)\int_0^t\frac{g(s)}{(1-s)^2}ds=g(1)$$ for any continuous function $g(s)$.

I tried a variable change $u=1/(1-s)$ which gives $du=ds/(1-s)^2$ with $$(1-t)\int_1^{\frac{1}{1-t}}g(1-\frac{1}{u})du$$

I cannot seem to figure out how to get to the limit, given that thus far it's correct.

答:
According to Theorem 5.13 in Rudin's "Principles of mathematical analysis" (slightly restated)

If $f$ and $h$ are real and differentiable in $(0,1)$ and if $h'(x)\neq 0$ in a neighborhood of $1$, and $h(x)\to +\infty$ as $x\to 1^-$, and in addition $f'(x)/h'(x)\to A$ as $x\to 1^-$, then $f(x)/h(x)\to A$ as $x\to 1^-$.

Applying this to
$$
f(x)=\int_0^x\frac{g(s)}{(1-s)^2}\,ds\quad\text{and}\quad h(x)=\frac{1}{1-x},
$$
we get (here we use the fundamental theorem of calculus and the property that $g$ is continuous)
$$
\frac{f'(x)}{h'(x)}=\frac{g(x)/(1-x)^2}{1/(1-x)^2}=g(x)\to g(1)\quad (x\to 1^-).
$$
Hence, the limit you look for (check the conditions) is also $g(1)$.

Title: Limit of a simple Integral


问:
Number of closed walks of length $r$ through a vertex on a graph with this condition.
Let $X$ be a regular graph of degree $d$ and let $v_{0}$ be a vertex such that the subgraph induced by the vertices $\{ v' \in X : dis(v_{0}, v') \leq r/2 \}$ is acyclic. Furthermore let $\theta(r)$ be the number of closed walks starting at $v_{0}$ of length $r$ in $X$

I was reading an old paper (Lemma 2.1) where the following is proved:

$$\theta(r) = \begin{cases} 0 &\text{$r$ is odd} \\ \sum^{s}_{k=1} {{2s-k}\choose{s}} \frac{k}{2s-k} d^{k}(d-1)^{s-k} &{r=2s}
 \end{cases}$$

When $r$ is even, we write $r=2s$

Unfortunately, many of the counting details are omitted in the proof I'm reading and I was wondering if someone could either provide a detailed proof or a helpful reference or even an example or strategy to gain more insight into this.

答:
The graph is connected, regular, undirected and acyclic.

These conditions imply that the grap is a tree.

You can characterize a tree by its property that for every two nodes, there's exactly one path connecting them.

It further follows that there are no shortcuts, and every edge to take either brings you closer or farther away from the root.

From this follows that there are no  closed walks for odd distances.

Now let $r=2s$.

Given that every edge you take either increases or decreases the distance to $v_0$ by exactly 1, every closed path from $v_0$ has to take exactly $s$ steps increasing the distance, and $s$ steps decreasing the distance.

Furthermore, with the exception of when you are at $v_0$, there's always exactly one edge that decreases the distance to $v_0$, and all other $d-1$ increase it.

And this now gives you the recipe:

Partition the set of all closed walks into how many times they arrive back at $v_0$.

Let's say we pick one partition $M$ in the set that arrives $k$ times back at $v_0$.  Let $P_i$ be the set of paths of length $i$.

Then every path in $M$ can be cut into $k$ segments by cutting whenever we arrive back at $v_0$. By cutting this way, we obtain a strong even partition $x_1,...,x_k$ of $2s$ in $k$ parts, which gives us the length of the cut walks.

Inversely, for every strong  partition $x_1,...,x_k$ of $s$ in $k$ parts we can count how many closed walks of length $2x_1$, then length $2x_2$,..., then length $2x_k$ there are, so that each of these sub-paths returns only to $v_0$ at the last node.

The number of closed walks of length $2x_i$ that return only to $v_0$ at the last node is:
$$
\frac{1}{x_i+1}\binom{2x_i}{x_i}d\cdot (d-1)^{x_i-1}
$$

This product comes as follows:

Every Dyck-path from 0 to 2s tells us for each step, whether we increase or decrease the distance to $v_0$. Whenever we increase the distance, we have $d-1$ options (besides at the starting point where we have $d$ options), and whenever we decrease the distance, we have $1$ option.

The number of closed walks of length $2x_1$, then length $2x_2$,..., then length $2x_k$ , where $x_1,...,x_k$ is a strong partition of $s$ is then:
$$
\prod_{i=1}^k \frac{1}{x_i+1}\binom{2x_i}{x_i}d\cdot (d-1)^{x_i-1}
$$

For the set $M$ of the partition, we therefore have
$$
|M| = \sum_{x_1+...+x_k=s\\x_1,...,x_k\ge 1}
\prod_{i=1}^k \frac{1}{x_i+1}\binom{2x_i}{x_i}d\cdot (d-1)^{x_i-1}
$$

And finally, for the set $S$ of all partitions, or equivalently, the set of all closed walks from $v_0$ of length $2s$, we have:
$$
|S|=  \sum_{k=1}^s \sum_{x_1+...+x_k=s\\x_1,...,x_k\ge 1}
\prod_{i=1}^k \frac{1}{x_i+1}\binom{2x_i}{x_i}d\cdot (d-1)^{x_i-1}
$$

Now all that's left to do is simplify:
$$
\sum_{k=1}^s \sum_{x_1+...+x_k=s\\x_1,...,x_k\ge 1} \frac{1}
\prod_{i=1}^k \frac{1}{x_i+1}\binom{2x_i}{x_i} d\cdot (d-1)^{x_i-1}
\\=
\sum_{k=1}^s \sum_{x_1+...+x_k=s\\x_1,...,x_k\ge 1}
 d^k\cdot (d-1)^{s-k}
\prod_{i=1}^k \frac{1}{x_i+1}\binom{2x_i}{x_i}
\\=
\sum_{k=1}^s d^k\cdot (d-1)^{s-k} \sum_{x_1+...+x_k=s\\x_1,...,x_k\ge 1} \prod_{i=1}^k \frac{1}{x_i+1}\binom{2x_i}{x_i}
$$
Here,  $\sum_{x_1+...+x_k=s\\x_1,...,x_k\ge 1} \prod_{i=1}^k \frac{1}{x_i+1}\binom{2x_i}{x_i} $  counts the number of Dyck-paths of length $2s$ that hit 0 exactly $k+1$ times ($k$ internal touches).

Therefore this is equal to:
$$
\sum_{k=1}^s d^k\cdot (d-1)^{s-k}\frac{k}{2s-k}\binom{2s-k}{s}
$$


问:
How many 5-digit palindromes are divisible by 9?
What I already have,

Palindrome in form XYZYX, where X can’t be 0.

Divisibility rule of 9: sum of digits is divisible by 9. So, we have 2(X+Y)+Z = 9M.

The first part is divisible by 9 if and only if X+Y is divisible by 9. So, we have 10 pairs out of 90. And each such pair the total sum is divisible by 9 when Z is also divisible by 9. There are 2 such Zs: 0, 9. So, there are 20 divisible palindromes.

If (X+Y) mod 9 = 1, then 2(X+Y) mod 9 = 2; and in order for the total sum to be divisible by 8, Z must have the remainder of 1 when divided by 9. There is 1 such Z: 1. And again, we have 10 xy pairs with the given remainder. So, this case yields 10*1 = 30 more palindromes.

Same logic as on previous step applies to the case when 2(X+Y) mod 9 = 2.

So, total number of divisible palindromes = 80?

When using this method, I only get 80 numbers of 5-digit palindromes that are divisible by 9(?) i dont think im doing this method correctly, can someone show me whats going on here

答:
The answer is $100$, my argument is as follows.

$X$ is choosen as a digit from $1$ to $9$, and $Y$ from $0$ to $9$,
now you want that $9|(2(X+Y)+Z)$.

Two cases:

(1) if $9|2(X+Y)$, then $2(X+Y)=9k$ for some natural number $k>0$, and since $Z$ is an integer from $0$ to $9$ you would have that both $Z=0,9$ work.

(2) if $9\not|2(X+Y)$, then you would have that there exists $k$ natural number such that $9(k-1)<2(X+Y)<9k$ and still becouse $0\le Z\le 9$ you obtain this time that there exists only one value of $Z$ that satisfies the problem, and it is $9k-2(X+Y)$.

So we want to find when $9|2(X+Y) \Rightarrow 9|(X+Y)$: so if $1\le X\le8$, then $Y=9-X$; if $X=9$, then $Y=0,9$ work. So $10$ cases where (1) happens

To conclude: there are $9\cdot 10=90$ possibilities between $X,Y$ of which $10$ cases are (1) and $80$ are (2), so the final number is $10\cdot 2 + 80\cdot 1=100$


问:
Showing that $\zeta(2)=\frac{\pi^2}{6}$ using the squeeze theorem and a few inequalities
I need to show that $\zeta(2)=\frac{\pi^2}{6}$ using the following inequalities:
$$ \frac{\pi^2}{(2n+1)^2} \frac{n(2n-1)}{3} \leq \sum_{k=1}^{n}\frac{1}{k^2} \leq \frac{\pi^2}{(2n+1)^2}\left(n+\frac{n(2n-1)}{3}\right) $$

$$ \zeta(2):=\sum_{k=1}^{\infty} \frac{\pi^2}{6} $$

When I compute the limit of the left side I end up either finding $+\infty * 0$ or $\frac{+\infty}{+\infty}$. How can I proceed?

答:
If you expand out the left-hand side, you get

$$\frac{2\pi^2 n^2 - \pi^2 n}{12n^2 + 12n + 3}$$

Since this is a ratio of polynomials of the same degree, the limit as $n \to \infty$ is given by the ratio of the leading coefficients. Thus,

$$\lim_{n \to \infty} \frac{2\pi^2 n^2 - \pi^2 n}{12n^2 + 12n + 3} = \frac{2 \pi^2}{12} = \frac{\pi^2}{6}$$

If the fact that the ratio of polynomials induces this limit isn't obvious, consider the general case:

$$\lim_{x \to \infty} \frac{a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0}{b_n x^n + b_{n-1} x^{n-1} + \cdots + b_1 x + b_0}$$

Then multiply the top and bottom by $1/x^n$. You get

$$\lim_{x \to \infty} \frac{a_n + a_{n-1}/x + a_{n-2}/x^2 + \cdots + a_1/x^{n-1} + a_0/x^n}{b_n + b_{n-1}/x + b_{n-2}/x^2 + \cdots + b_1/x^{n-1} + b_0/x^n}$$

Clearly this limit is given by $a_n/b_n$.


问:
Show that the metric d(x, y) = |x−y| / 1+|x−y| is not induced by any norm on $\mathbb{R}$.
Is it because the enlargement property is not satisfied?
Also, is there a way to prove this by contradiction?</p>

答: Your question essentially contains the answer. Is the distance from $0$ to $2$ twice the distance from $0$ to $1$?


问:
When are two vectors parallel if the vectors are $5e_1-3e_2+\alpha e_3$ and $\beta e_1 + 2e_2 + 3e_3$
When are two vectors parallel if the vectors are  $$5e_1-3e_2+\alpha e_3$$ and $$\beta e_1 + 2e_2 + 3e_3$$
The alpha and beta are parameters.</p>

答:
Two vector $\vec {n_1}$ and $\vec{n_2}$ are parallel when we can write them in this way

$\vec {n_1}=k\cdot \vec{n_2}\,$ where k  is any scaler quantity.

so given vectors are parallel when
$$
\dfrac5{\beta}=\dfrac{-3}{2}=\dfrac{\alpha}{3}
$$
In one of the other answer shown:  $\vec {n_1}\times \vec {n_2}=\vec 0\implies \vec{n_1}||\vec{n_2}$

$\vec {n_1}\times \vec {n_2}$ can be calculated as:

$\vec {n_1}\times \vec {n_2}=\begin{vmatrix} {e_1} &{e_2} & {e_3}\\
5 & -3 & \alpha \\
\beta & 2 & 3
\end{vmatrix}$
$$\vec {n_1}\times \vec {n_2}=e_1((-3\times 3)-(2\times \alpha))-e_2((5\times 3)-(\alpha\times\beta))+e_3(5\times 2-(-3\times \beta))$$
$$\vec {n_1}\times \vec {n_2}=e_1(-9-2\alpha)-e_2(15-\alpha\beta)+e_3(10+3\beta)$$


问:
A Problem on circular permutation.
We need to find the number of arrangements of 6 people at three identical round tables if each table must be occupied.

I tried to break it down into cases, like (1,1,4) , (1,2,3) , (2,2,2) as they are identical, the order won't matter in terms of number of people on a table. Also i am aware of the fact that number of ways of arranging $n$ people in a circle is $(n-1)!$. Also, i know if we need to arrange x objects in a line, where $ x_1 $ are of one kind, $x_2 $are of another kind and so , number of ways would be $= \frac{x!}{x_1!x_2!...x_k!}$ but i am not sure how to string these facts together for this problem.

Like for example of case (1,2,3) , first $\binom{6}{3}$ to select three people for first table , then $\binom{3}{2}$ to select two people for second table which automatically selects 1 person for first table. Then $\frac{\binom{6}{3} \cdot \binom{3}{2} \cdot 2}{z}$ but i am not able to figure out what that $z$ is going to be.

I would appreciate it if along with the answer you could also tell me method to handle these problems.

答:
Looking first at the case $(1,2,3)$, you could call the table with $3$ people $A$, the table with $2$ people $B$, and the table with $1$ person $C$. The number of assignments of people to tables equals the number of arrangements of the word $AAABBC$ and is $\frac{6!}{3!\,2!\,1!}$ or $\binom{6}{3}\binom{3}{2}\binom{1}{1}$. I see both sorts of expressions in your attempted solution, except that your expression involving binomial coefficients has a factor $\frac{2}{z}$ that I don't understand.

At any rate, the next step is to arrange the people at each table, which, as you mention can be done in $2!\,1!\,0!$ ways. So the result is
$$
\frac{6!}{3!\,2!\,1!}\cdot2!\,1!\,0!=\frac{6!}{3\cdot2\cdot1}.
$$

For a case like $(1,1,4)$ you have to account for the interchangeability of the two tables with $1$ person, so you get
$$
\frac{6!}{2!\cdot4!\,1!\,1!}\cdot3!\,0!\,0!=\frac{6!}{2!\cdot4\cdot1\cdot1}.
$$

You can do the third case by the same principle. You should know that there is a systematic analysis of this problem: your problem is equivalent to finding the number of permutations of $6$ objects that have $3$ cycles in their cycle decomposition. This is the Stirling number of the first kind $\left[6 \atop 3\right]$.

Title: A Problem on circular permutation.


问:
How do I find the Inverse of $f(x)$?
Given
$f(x+1)+f(x-1)=x^2$

I have subtituted $(a=x+1)$ and $(a=x-1)$ and got
$$f(x)+f(x-2)=(x-1)^2 \text{ and } f(x+2)+f(x)=(x+1)^2$$
Combining those equations, I got

$$f(x+2)-f(x-2)=4x$$

I could not even find $f(x)$

Please help me

答:
HINT:

WLOG let $$f(x)=g(x)+\sum_{r=0}^na_rx_r$$

$$x^2=g(x-1)+g(x+1)+2a_0+2a_1x+a_2((x+1)^2+(x-1)^2)+\cdots$$

Set $a_r=0$ for $r\ge3$

$$x^2=g(x-1)+g(x+1)+2a_0+2a_1x+2a_2(x^2+1)$$

Compare the coefficients of $x$ to find $a_1=0$

that of $x^2,1=2a_2$

those of constants,  $0=a_2+a_0\iff a_2=-a_0$ so that

$$f(x)=g(x)+\dfrac{x^2-1}2\text{ and } g(x-1)+g(x+1)=0\iff g(x+1)=-g(x-1)=\cdots=g(x+3)$$

So, $g(x)$ could be any periodic function with period $=4$

Title: How do I find the Inverse of $f(x)$?


问:
Construct a matrix with the specified characteristics
I am taking a signals processing class online and we are covering some basic linear algebra. Here is a question that I'm struggling with from the class.

Find a matrix $A$ which satisfies all of the following characteristics:

$\text{dim }\text{col}(A) = 2$

$\begin{bmatrix} 1 & 1 & 2 & -1\end{bmatrix}^T \in \text{null}(A)$

$\begin{bmatrix} 3 & 3 & 6 \end{bmatrix}^T \in \text{col}(A)$

$\begin{bmatrix} 3 & 0 & 0 & 3\end{bmatrix}^T \in \text{col}(A^T)$

Attempt

So far, I've started by making the first two columns of my matrix $A$ independent.

$$\begin{bmatrix}
1 & 0 & \alpha_1 & \beta_1 \\
1 & 0 & \alpha_2 & \beta_2 \\
0 & 1 & \alpha_3 & \beta_3
\end{bmatrix}$$

From this I can clearly fulfill the requirement that $\begin{bmatrix} 3 & 3 & 6 \end{bmatrix}^T \in \text{col}(A)$. Then I start filling in the $\alpha_i$ and $\beta_i$ values to fulfill the other characteristics.

The problem is, whenever I manage to satisfy a couple of the characteristics, I break the third one. I know how to individually satisfy these characteristics, but not how to satisfy them together. How can I go about constructing a matrix that satisfies all these characteristics?

答:
Here's an idea. Construct a suitable matrix of the form
$$
\pmatrix{3&0&0&3\\3&0&0&3\\6&?&?&6}.
$$
The first row and first column ensure that conditions 2 and 4 are satisfied. The fact that the middle row is the same as the first ensures that the rank is at most 2. In order to make sure that the rank is exactly $2$, it suffices to choose non-zero entries for the unknown entries. Note, however, that these must be selected in such a way that condition 2 holds.


问:
Trig substitution in vector calculus
I have  gotten the $$|T'(t)| = \sqrt{0^2+\left(\frac{-48}{12}\sin(4t)\right)^2+\left(\frac{-48}{12}\cos(4t)\right)^2}$$

Can I replace (using trig identity) $${\left(\frac{-48}{12}\sin(4t)\right)^2+\left(\frac{-48}{12}\cos(4t)\right)^2}$$ for $$\left(-\frac {48}{12}\right)^2\cdot 1$$

Is this valid?

My final $$|T'(t)| = \frac{48}{12}$$
Can I use this answer to calculate $K$ which is $\frac {|T'(t)|}{r'(t)}$

答: Yes, but be careful, your $12$ turned into a $13$, so you should have $$|T'(t)| = \sqrt{\left(\frac{48}{12}\right)^{2}} = 4$$

Title: Trig substitution in vector calculus


问:
4 balls with different colors, there are 3 identical boxes, how many ways?
4 balls with different colors, there are 3 identical boxes. the box can be empty. of course, all balls must be used. how many ways to distribute the balls?

The striling number of the second kind S(n,k) counts the number of ways to partition a set of n objects into k non-empty subsets. so my answer is,
S(4,1)+S(4,2)+S(4,3)
but I don't know how to get numeric answer from here like how do I find what S(4,1) is?

答:
There are closed formulas for those Stirling numbers, or you can just consult a table, e.g. just go here: https://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind

But, this problem isn't so hard that you couldn't just brute this yourself:

Label the balls $1$ through $4$, and put the first ball in a group: $1$

Now we can add the second ball to that one:  $12$, or not: $1,2$

To each of those possibilities, we find the different ways to add the third ball:

That is, $12$ leads to $123$, or $12,3$,  and $1,2$ leads to $13,2$, or $1,23$, or $1,2,3$. So note that this also tells us that if you had $3$ different balls, then there are $5$ ways to put them in $3$ boxes, possibly empty.

Finally, we add the fourth ball. Note that n the case where we still have only one group, there are $2$ possibilities, whereas when you have two or three groups, you have $3$ possibilities each. This gives us $14$ possibilities.  Here they are explicitly:

$1234$

$123,4$

$124,3$

$12,34$

$12,3,4$

$134,2$

$13,24$

$13,2,4$

$14,23$

$1,234$

$1,23,4$

$14,2,3$

$1,24,3$

$1,2,34$


问:
How to know whether to use "a" or "b" substitution when solving for ellipse's center
If I have the following problem:
Vertices: (1,5),(1,-3) and c=3
With the given vertices I know that h=1
I first try to solve for k to finish solving for my center. To this I can use the following format:
(h,k+-a) or (h,k+-b)

If I do not know whether the given vertices are from a major/minor axis, so how do I know whether I should substitute for a or b when solving for k?

答:
In any ellipse, by definition, the minor and major axis have the center of the ellipse as their midpoint. Recalling that the midpoint between $2$ points $(x_1,y_1)$ and $(x_2,y_2)$ is given by
$$
\left(\frac{x_1 +x_2}{2},\frac{y_1 +y_2}{2}\right)
$$
we see that the center of the ellipse has to be at
$$
C = \left(\frac{1+1}{2}, \frac{5 -3}{2}\right) = (1,1)
$$
So, the equation of our ellipse has the form
$$
\frac{(x-1)^2}{a^2} + \frac{(y-1)^2}{b^2} = 1 \tag{1}
$$
We'll also note that the distance between the 2 vertices given is
$$
d\left[(1,-3),(1,5)\right] = \sqrt{(1-1)^2 + (5-(-3))^2} = 8
$$
Now, to find out if these vertices correspond to the major or minor axis, we recall that, since the vertices need to be on the ellipse, they need to satisfy equation $(1)$ is we plug them in. Choosing one of the vertices, let's say $(1,5)$, by substituting in $(1)$ we get
\begin{align}
&\frac{(1-1)^2}{a^2} + \frac{(5-1)^2}{b^2} = 1\\
\implies&\frac{4^2}{b^2} = 1 \\
\implies &b = 4 \tag{2}
\end{align}
And recalling the relation
$$
c^2 = a^2 - b^2
$$
we can find the value of $a$ to be
\begin{align}
a &= \sqrt{c^2 + b^2}\\
&\overset{\color{blue}{(2)}}{=}\sqrt{c^2 + 16}\\
&= \sqrt{9 + 16}\\
& = 5
\end{align}
From here we get that the major axis has to have length $2a = 2(5) = 10$ and the minor axis has lenght $2b = 2(4) = 8$. So since the distance between our vertices is $8$, it has to be the case that they correpond to vertices on the minor axis.

Another way of looking at it is as follows. If we just guess that the vertices correspond to the major axis, recalling that the distance between them is $8$ we see that from this assumption we get $a = \frac{8}{2} = 4$. But using the relation $c^2 = a^2 -b^2$ we get that $b$ has length
$$
b = \sqrt{a^2 -c^2} = \sqrt{4^2 - 3^2} = \sqrt{7}
$$
And since we now know both $a$ and $b$, we see by equation $(1)$ that the equation of this assumed ellipse is given by
$$
\frac{(x-1)^2}{4^2} + \frac{(y-1)^2}{\sqrt{7}^2} = 1
$$
However, if we plug in one of the vertices (let's say $(1,5)$) into the equation above we get that
\begin{align}
&\frac{(1-1)^2}{4^2} + \frac{(5-1)^2}{\sqrt{7}^2} = 1 \\
\implies & \frac{4^2}{7} = 1 \\
\implies & 16 = 7
\end{align}
since this is clearly a contradiction, the conclusion is that the point $(1,5)$ is not on the ellipse! But this is a problem, since by it being a vertex it needs to be on the ellipse. Hence the solution where the points correspond to the major axis is impossible.

In conclusion, the condition that $c=3$ forces the vertices to be of the minor axis, and if we were to assume that they correspond to a major axis, we would get an impossible solution. We can visually verify this by plotting the 2 equations we get from assuming each case

where the green ellipse is given by the equation
$$
\frac{\left(x-1\right)^{2}}{5^{2}}+\frac{\left(y-1\right)^{2}}{4^{2}}=1
$$
and the red one is given by
$$
\frac{\left(x-1\right)^{2}}{4^{2}}+\frac{\left(y-1\right)^{2}}{\sqrt{7}^2}=1
$$
both of them satisfy that $c=3$, that their center is at $(1,1)$ and that one of their axis has length $8$, but only one of the 2 solutions passes through the vertices we're given.


答: For $c\in\{-1,\,2\}$,$$\lim_{n\to\infty}\frac{n(2n+c)}{(2n+1)^2}=\lim_{n\to\infty}\frac{2n^2(1+\tfrac{c}{2n})}{4n^2(1+\tfrac{1}{2n})^2}=\frac12\lim_{n\to\infty}\frac{1+\tfrac{c}{2n}}{(1+\tfrac{1}{2n})^2}.$$Now two famous results complete the proof.

Title: Showing that $\zeta(2)=\frac{\pi^2}{6}$ using the squeeze theorem and a few inequalities

Body:
I need to show that $\zeta(2)=\frac{\pi^2}{6}$ using the following inequalities:
$$ \frac{\pi^2}{(2n+1)^2} \frac{n(2n-1)}{3} \leq \sum_{k=1}^{n}\frac{1}{k^2} \leq \frac{\pi^2}{(2n+1)^2}\left(n+\frac{n(2n-1)}{3}\right) $$

$$ \zeta(2):=\sum_{k=1}^{\infty} \frac{\pi^2}{6} $$

When I compute the limit of the left side I end up either finding $+\infty * 0$ or $\frac{+\infty}{+\infty}$. How can I proceed?


问:
Proving Arnold's cat map has only rational periodic points.
I was trying to prove that all periodic points of the map are rational points. I started with $\begin{bmatrix}
           x_1 \\
           x_2 \\
         \end{bmatrix}= {\begin{bmatrix}
           1 & 1 \\
           1 & 2 \\
         \end{bmatrix}}^n \begin{bmatrix}
           x_1 \\
           x_2 \\
         \end{bmatrix}$ and manipulated it using diagnolization and arrived at $\begin{bmatrix}
           x_1 \\
           x_2 \\
         \end{bmatrix}=
         \begin{bmatrix}
           a & b \\
           b & c \\
         \end{bmatrix} \begin{bmatrix}
           x_1 \\
           x_2 \\
         \end{bmatrix}$ where a,b,c are fibonacci numbers. This seems to me to imply the only periodic point is the zero vector. But I know that isn't correct. I think it might have to do with me no including mod 1, but how do I account for that throughout the problem?

答:
Denote
$$
x = (x_1,x_2)^\top, \quad A = \pmatrix{1&1\\1&2}.
$$
Suppose that $x$ is periodic. That is, there exists an integer $n$ such that $A^nx - x \in \Bbb Z^2$. In other words, $(A^n - I)x \in \Bbb Z^2$. Let $y = (A^n - I)x$, and let $M = A^n - I$. We see that $M$ is a matrix with integer entries and, importantly, $M$ is invertible (since neither of the eigenvalues of $A$ satisfy $\lambda^n = 1$). It follows that $M^{-1}$ is a matrix with integer entries, and
$$
Mx = y \implies x = M^{-1}y.
$$
Because $M^{-1}$ and $y$ have rational entries, we conclude that $x$ must have rational entries.

The converse of this statement (that every rational point is periodic) is a bit more interesting. Suppose that $x$ has rational entries; we want to show that there exists a positive integer $n$ for which $(A^n - I)x$ has integer entries. We can write $x$ in the form $x = \frac 1d z$ for some common denominator $d \in \Bbb Z$ and vector $z \in \Bbb Z^2$. With that in mind, it suffices to show that there exists an $n$ for which the entries of $A^n - I$ are a multiple of $d$.

In other words, it suffices to show that there necessarily exists an $n$ for which
$$
A^n = I \pmod d.
$$
We can prove this as follows: by the pigeonhole principle, there exist positive integers $n_1<n_2$ for which
$$
A^{n_1} \equiv A^{n_2} \pmod d.
$$
However, $A$ has an inverse with integer entries. So, we can deduce that
$$
[A^{-1}]^{n_1} A^{n_1} \equiv [A^{-1}]^{n_1} A^{n_2} \pmod d \implies\\
I \equiv A^{n_2 - n_1} \pmod d.
$$
Thus, the desired conclusion holds with $n = n_2 - n_1$.


问:
Expectation of $\sqrt{r_t}$ in $\textrm{d}r_t=\kappa(\theta-r_t)\,\textrm{d}t+\sigma\sqrt{r_t}\,\textrm{d}B_t$
I'm currently looking at the CIR model and trying to calculate the expectation of $r_t$. I know that by Fubini's theorem, $\mathbb{E}\left(\int_0^t\kappa(\theta-r_s)\,\textrm{d}s\right)=\kappa\theta t-\int_0^t\mathbb{E}(r_s)\,\textrm{d}s$ and we can create an ODE from it, but as for the second term I know I can't use Fubini's theorem, so what can I do? I know one way is to multiply the SDE by $\textrm{e}^{\kappa t}$ but I haven't seen how it helps with the square root term, and Shreve claims it is an Ito integral but I don't see how $\sqrt{r_t}\in\mathcal{L}^2_{\mathcal{F}}$. All help is appreciated, thanks!

答:
We need the following Lemma.

Lemma.
Consider the SDE
$$dX_t = f(X_t)dt + g(X_t)dB_t, \quad X_0 = x_0 \in \mathbb R^{+}. \tag{1}$$

with solution $\{x_t; 0 \leq t \leq T\}.$ Assume that the linear growth condition holds, i.e.

There exists a constant $K \in \mathbb R$ such that for all $x \in \mathbb R$
$$|f(x)|^2 \bigvee |g(x)|^2 \leq K(1+x^2).$$
Then there exists a constant $C$ such that
$$E[\sup_{0 \leq t \leq T} X_t^2] \leq C.$$
(For a proof of this lemma see for example lemma 3.2 - page 51 in Xuerong Mao's book)

Now going to your problem: your SDE is of the SDE (1) type with $f(x)= \kappa(\theta -x)$ and $g(x) = \sigma \sqrt{x}.$ We can see that clearly the linear growth condition holds, so the lemma applies and there exists $C>0$ such that $E[\sup_{0 \leq t \leq T} r_t^2] \leq C.$ This implies that $E\int_0^T (\sigma \sqrt{r_t})^2dt < \infty.$ Thus the process $I_t:= \int_0^t (\sigma \sqrt{r_t})dB_t, 0 \leq t \leq T$ is a martingale and has constant expectation. Therefore, $E[I_T]=E[I_0]=0.$


问:
SVD of the $3 \times 2$ matrix of ones

Find the SVD form of the matrix $$A = \begin{pmatrix} 1 & 1  \\ 1& 1 \\ 1& 1 \end{pmatrix}$$

First, I calculated

$$A^TA = \begin{pmatrix} 3 & 3  \\ 3 & 3\end{pmatrix}$$

whose eigenvalues are $\lambda_1=6$, $\lambda_2=0$, so we get the singular values $\mu_1=\sqrt 6$ and $\mu_2=0$.
Then the orthonormal eignvectors are:
$u_1=1/\sqrt2 (1,1)$
And $u_2= 1/\sqrt2 (1,-1)$.
So, I found the two matrix  $\sum$ and $U^T$ where the SVD for A is of the form:
$A=V \sum U^T$.
I'm not sure how to find the orthogonal matrix V since I have a singular value which equals 0!
Moreover i cannot multiply matrix because $\sum$ is not invertible.
How can I finish it and get the SVD form for matrix $A$?

I am using information about SVD from the wikipedia page.

答:
We can notice that we can write the matrix $A$ as follows:

$$A = \begin{bmatrix} 1 & 1 \\ 1 & 1\\ 1 & 1\end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix}^\top = \sqrt6\left(\frac{1}{\sqrt{3}}\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \right) \left( \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \end{bmatrix} \right)^\top$$

So, we succeeded to write $A$ as $A=\sigma u_1 v_1$. In addition, $\sigma_2=0$ since $\mbox{rank}(A)=1$. Thus, we get

$$\Sigma= \begin{bmatrix} \sqrt{6} & 0 & 0\\ 0 & 0 & 0\end{bmatrix}$$

Then in order to find the matrix $U$ and $V^T$ as in SVD A can be written as $A=U \Sigma V^T$. We can choose any vector $v_2 \in \Bbb R^2$ which is orthonormal and orthogonal to $v_1$ and choose vectors $u_2, u_3 \in \Bbb R^3$ which are also orthonormal and orthogonal to $u_1$.

Title: SVD of the $3 \times 2$ matrix of ones


问:
Can "for any" be rephrased as "infinetly often"
If it is stated that an event $A(t)$, i.e. depening on $t$, happens for all $t > 0$ where $t \in \mathrm{IR}$.

Can this be rephrased as $A(t)$ happens infitely often?

答: No, it cannot. For example, if $A(t)$ happens if and only if $t$ is an integer, then it would happen infinitely often, but it does not happen for all positive real numbers.


问:
Wire basket shape for maximum Moment of Inertia
Attempted to find the meridian shape for optimum moment  of inertia with a minimum mass of wire constraint and like to share in the iso-perimetric problem in the plane.

Assuming constant density per unit length

$$ \int y^2 ds - \lambda^2 \int ds= \int( y^2-\lambda^2) ds ,\; {ds}/{dx}=\sqrt{1+y^{'2}}\tag1$$

Lagrangian is

$$ ( y^2-\lambda^2) \sqrt{1+y^{'2}}$$

Using Euler-Lagrange equations on the Lagrangian we obtain

$$( y^2-\lambda^2) \sqrt{1+y^{'2}} -y'\cdot  ( y^2-\lambda^2)\dfrac{y'}{\sqrt{1+y^{'2}}} = c^2 $$

$$\dfrac{y^2-\lambda^2}{\sqrt{1+y^{'2}}} = c^2 \tag2 $$

To eliminate $c^2$ raising to second order with one differentiation with respect to arc

$$\dfrac{1}{c^2}
 =\dfrac{\sec \phi}{y^2- \lambda^2} = \dfrac{ \sec \phi \tan \phi\;k(s)}{ 2 y \sin \phi } $$

$$ \text{curvature} \;k(s)= \dfrac{d \phi}{ds} , \quad  k(s)= \dfrac{ 2 y \cos ^2\phi}{c^2} \tag3 $$

changing  $y=r$ as symbol for radius in cylindrical coordinates and since $ \sin \phi = \dfrac{dr}{ds},$

$$\sec^2 \phi \sin \phi\; d\phi=\sec \phi \tan \phi\; d\phi=\dfrac{2 r dr}{c^2} \tag4 $$

Integrating

$$ \sec\phi =\dfrac{r^2}{c^2}+\text{const.} \tag 5 $$

and  initial condition  $ \phi=0, r = r_1$

$$ \sec \phi -1 = \dfrac{r^2-r_1^2}{c^2} \tag 6 $$

which is the ODE for meridians of the basket. For the given graph initial conditions are at vertex $ \phi_{i}=0, r_i=0.8 $. Maximum curvature  at vertex is $ k_{max}=\dfrac{2r_1}{c^2}$.

Please comment  about correctness and or whether the curve is already known.

答:
You formulate
$$
I[y]=\int L(y,y')ds=\int (y^2-λ^2)\sqrt{1+y'^2}\,dx
$$
with the start position fixed at the rotation axis and the other end free.

The Euler-Lagrange equations are $\newcommand{\pd}[2]{\frac{\partial#1}{\partial#2}}$
\begin{align}
0=\frac{d}{dt}\pd{L}{y'}-\pd{L}{y}
&=\frac{d}{dt}\left((y^2-λ^2)\frac{y'}{\sqrt{1+y'^2}}\right)-2y\sqrt{1+y'^2}
\\
&=-\frac{2y}{\sqrt{1+y'^2}}+(y^2-λ^2)\frac{y''}{\sqrt{1+y'^2}^3}
\end{align}
Separating the derivatives gives
$$
0=\frac{y'y''}{1+y'^2}-\frac{2yy'}{y^2-λ^2}
\\
(y^2-λ^2)=C\sqrt{1+y'^2}
$$
As for $y=0$ at the rotation axis the left side is negative, so must be the constant $C=-c^2$
$$
λ^2=c^2\sqrt{1+y'^2}+y^2
$$


问:
Question about curvature
that is my trouble:

Let $\alpha : I \rightarrow R^3 $ a regular curve parameterized by its
arc lenght, that is contained in the unit sphere.

If $k(s)>0$ and $\tau(s)\neq 0$ , $\forall s \in I$ then

$$ \frac{1}{k^2(s)} + \frac{(k'(s))^2}{k^4(s) \tau^2(s)} = 1 $$

my efforts were useless

答:
We may easily generalize to the case in which $\alpha(s)$ lies in the sphere of radius $R$ centered at some point $C \in \Bbb R^3$, that is, to

$(\alpha(s) - C) \cdot (\alpha(s) - C) = R^2. \tag 1$

The key idea here is to express $\alpha(s) - C$ itself in terms of $T(s)$, $N(s)$, and $B(s)$, the Frenet frame of the curve $\alpha(s)$.  We obtain, upon differentiating (1) with respect to the arc-length $s$,

$(\alpha(s) - C)' \cdot (\alpha(s) - C) = 0, \tag 2$

or

$\dot \alpha(s) \cdot (\alpha(s) - C) = 0; \tag 3$

since

$\dot \alpha(s) = T(s), \tag 4$

(3) becomes

$T(s) \cdot (\alpha(s) - C) = 0, \tag 5$

which shows that the $T(s)$ component of $\alpha(s) - C$
is zero.  We next differentiate (5):

$\dot T(s) \cdot (\alpha(s) - C) + T(s) \cdot T(s) = 0, \tag 6$

and using the first Frenet-Serret equation

$\dot T(s) = \kappa(s) N(s) \tag 7$

and the fact that $T(s)$ is a unit vector,

$T(s) \cdot T(s) = 1, \tag 8$

we transform (6) into

$\kappa(s) N(s) \cdot (\alpha(s) - C) + 1 = 0, \tag 9$

and readily express the $N(s)$ component of $\alpha(s) - C$ as

$N(s) \cdot (\alpha(s) - C) = -\dfrac{1}{\kappa(s)}; \tag{10}$

taking the $s$ derivative of this equation yields

$\dot N(s) \cdot (\alpha(s) - C) + N(s) \cdot T(s) = \dfrac{\dot \kappa(s)}{\kappa^2(s)}; \tag{11}$

since

$N(s) \cdot T(s) = 0, \tag{12}$

and the second Frenet-Serret equation affirms that

$\dot N(s) = -\kappa(s) T(s) + \tau(s) B(s), \tag{13}$

we may now write (11) in the form

$(-\kappa(s) T(s) + \tau(s) B(s)) \cdot (\alpha(s) - C) = \dfrac{\dot \kappa(s)}{\kappa^2(s)}, \tag{14}$

and so by virtue of (5),

$ \tau(s) B(s) \cdot (\alpha(s) - C) = \dfrac{\dot \kappa(s)}{\kappa^2(s)}, \tag{15}$

immediately leading to

$B(s) \cdot (\alpha(s) - C) = \dfrac{\dot \kappa(s)}{\kappa^2(s) \tau(s)}, \tag{16}$

the $B(s)$ component of $\alpha(s) - C$; we combine (5), (10) and (16) and arrive at

$\alpha(s) - C = -\dfrac{1}{\kappa(s)} N(s) +  \dfrac{\dot \kappa(s)}{\kappa^2(s) \tau(s)} B(s), \tag{17}$

now using

$N(s) \cdot B(s) = 0, \tag{18}$

it follows from (1) and (17) that

$R^2 = (\alpha(s) - C) \cdot (\alpha(s) - C) = \dfrac{1}{\kappa^2(s)} +  \dfrac{(\dot \kappa(s))^2}{\kappa^4(s) \tau^2(s)}, \tag{19}$

the requested formula; we observe that when $R = 1$ we obtain the exact relationship given in the text of the question itself. $OE\Delta$.

See also this question.

Title: Question about curvature


问:
Prove that $f(\overline C ) \subset \overline{ f(C)}$ .
Prove that $f: A \to B$ is continuous iff its graph is compact where $A$ is compact and $A,B$ are metric spaces.

My attempt: I have already proved it. But somehow i am not satisfied with my proof. Implies part is Ok. But converse part i want to prove f is continuous by using $f(\overline C) \subset \overline{ f(C)}$.

For this i took $C\subset E$ arbitrary. Let $T=\{(x, f(x)):\, x  \in C\}$, $F=\{(x, f(x)): x \in \overline{C}\}$. Now $\overline T$ is closed in graph(f),  hece it is compact. I cant proceed further. Please give me hint.

答:
I'll try proving what the title states when $f$ is continuous.

If $x\in f(\overline{A})$ then there exists $a\in \overline{A}$ such that $f(a)=x$, now since it is in the closure, we have
$$\forall U \mbox{ open set containing } a \mbox{, } U\cap A\not=\emptyset$$
Now, taken an open set $V$ containing $x$, since $f$ is continuous, $f(V)^{-1}$ is a an open set containing $a$, so as we saw $A\cap f(V)^{-1} \not=\emptyset$.

Now, taking it's immage, $f(\emptyset)\not=f(A\cap f(V)^{-1}) \subseteq f(A)\cap V$, so every $V$ intersects $f(A)$: by definition $x$ is in the closure of $f(A)$

I have show that $x\in f(\overline{A}) \Rightarrow x\in \overline{f(A)}$; this means $f(\overline{A})\subseteq \overline{f(A)}$.


问:
Calculate the integral of the pyramid.
Calculate the integral of the pyramid.

$S_{n} = \left \{ \sum_{k=1}^{n}x_{k} \leq  a, x_{k} \geq  0, k = 1, ... , n\right \}$

$\int_{S_{n}}  \left (\sum_{k=1}^{n}x_{k} \right )^p dx , p\geq  0 $

can enter new variables $y_1=x_1, y_2=x_1+x_2,..., y_n=x_1+x_2+...+x_n.$ The area will be defined by the constraints $0≤y_1≤y_2≤....≤y_n≤a$. The Jacobian of this substitution is 1.

The integrand has the form $y_n$.

$\int_{0}^{a}dy_{n}\int_{0}^{y_{n}} dy_{n-1} ... \int_{0}^{y_2} y_ndy_{1}$

Is the integral written correctly?

Please tell me how to calculate it.

答:
Well, the $y_n$ near the end should be $y_n^p$, but apart from that it's right.

Before we solve the problem, let's lay down three sanity checks. By dimensional analysis, your integral must be some multiple of $a^{n+p}$. The case $n=1$ needs to get $\int_0^ay_n^pdy_n=\frac{a^{p+1}}{p+1}$. Finally, the case $p=0$ is famous; it's the hypervolume, $\frac{a^n}{n!}$. With any luck, we'll get a result honouring all these constraints.

Now evaluate the integrals innermost first (or as you've written it, from right to left). Since $\int_0^{y_2}y_n^pdy_1=y_2y_n^p$,$$\int_0^{y_3}dy_2\int_0^{y_2}y_n^pdy_1=\int_0^{y_3}dy_2y_2y_n^p=\frac12y_3^2y_n^p,$$and$$\int_0^{y_4}dy_3\int_0^{y_3}dy_2\int_0^{y_2}y_n^pdy_1=\int_0^{y_4}\frac12y_3^2y_n^pdy_3=\frac{1}{3!}y_4^3y_n^p.$$As we continue in this pattern, we eventually get$$\int_0^ady_n\int_0^{y_n}\frac{1}{(n-2)!}y_{n-1}^{n-2}y_n^pdy_{n-1}=\int_0^ady_n\frac{1}{(n-1)!}y_n^{n+p-1}=\frac{a^{n+p}}{(n-1)!(n+p)}.$$Oh, good; that passes all of our checks.

Title: Calculate the integral of the pyramid.


问:
Intersection of two different analytic curves.
I hardly know about Differential Geometry. Recently, I came across the following:

An analytic curve in the plane $\mathbb{C}=\mathbb{C}^1$ of the complex variable $z=x_1+ix_2$ can be represented as a complex-analytic function of a real parameter $z=f(t)$, $\alpha\leq t \leq \beta$, $f'(t) \neq 0$ on $[\alpha,\beta]$. If the analytic curve is located in a domain $D\subset \mathbb{C}$, then a conformal mapping of $D$ into any domain will also yield an analytic curve. If the set of intersection points of two analytic curves is infinite, these analytic curves coincide.

I cannot prove the last line "If the set of intersection points of two analytic curves is infinite, these analytic curves coincide"

Can we prove the same, without using Differential Geometric terms, or, at least "a sort of" simple proof.

Thanks in advance.

答:
I will give a sketch of the proof of the following result: let $f,g:[0,1] \to \mathbb C$ regular analytic curves with geometric images $C_1, C_2$. Then if $C_1 \cap C_2$ is infinite, it is either one or two common arcs with the endpoints, being among the endpoints of the two curves.

Taking $C_1$ the upper half of a circle extended a little at both ends and $C_2$ the lower half extended, shows that we can have two arcs, while of course if $C_1 \subset C_2$ say we have one arc. In particular, if the curves are closed, they indeed coincide if they have infinite points in common.

So by hypothesis $f,g$ extend to analytic functions in an open neighborhood $U$ of $[0,1]$ and $f'(z), g'(z)$ do not vanish on $U$ for small enough $U$ by regularity ($f'(t) \ne 0, t \in [0,1]$ means $|f'(t)| \ge \delta >0$, pick neighborhoods of each $t$ where $|f'(z)| \ge \delta/2$ and cover $[0,1]$ by finitely many such etc).

Assume $M=C_1 \cap C_2$ infinite; it is compact hence closed so $M_1=f^{-1}(M)$ is a closed hence compact subset of $[0,1]$ and there is an accumulation point $w$ of $M_1$. As $f'(w) \ne 0$ there is a small open neighborhood $G$ of $w$ for which $f$ is invertible with image some open neighborhood of $f(w)=P \in M$.

Since $f$ is $1-1$ on $G$ there are points $t_n \to w, t_n \in G\cap [0,1], f(t_n)=P_n \in M$ which are distinct so by passing to a subsequence if needed we can assume that there are $s_n \to s \in [0,1], g(s_n)=P_n$ hence $g(s)=P$

Let $K_1, K_2$ the arcs of $C_1, C_2$ determined by $P$ containing the $t_n, s_n$ and belonging to $f(G)$, so $P_n \in K_1 \cap K_2$. Clearly $f^{-1}(K_1)$ is a segment of $[0,1]$ (and if $w \ne 0,1$, $w$ is inside it) while $f^{-1}(K_2)$ is an arc that intersects the real axis in infinitely many points $t_n= f^{-1}(P_n) \to t$. But this means that $f^{-1} \circ g$ is an analytic function in a neighborhood of $s$ that is real at infinitely many points $s_n \in [0,1]$ with an accumulation point $s$ there, hence by the identity theorem, it is real on a segment containing $s$, since $\Im f^{-1} \circ g$ restricted to $[s-\epsilon, s+\epsilon]$ is real analytic so if it has infinitely many zeroes with an acummulation point, is zero (here of course we can have $s=0,1$ when we adjust the interval correspondingly)

Hence we have shown that the two curves coincide on an arc containing $P$ and the proof shows that the arc contains $P$ inside unless $P$ is an endpoint of either curve. But now a connectivity argument immediately shows that we can extend the common arc as long as we do not hit endpoints of either curve at both ends of it, and since those are at most $4$ we can have at most $2$ such arcs, while the example above shows that indeed that situation can occur.


答: Of course you can just account for mod 1 at the end and have Our equation tells us that $(a-1)x_1+bx_2=0$ mod 1, that is to say that for some integer k, $(a-1)x_1+bx_2=k$. In fact if we let the first equation equal a $k_1$ and the second, $k_2$, in general we see $x_1=\frac{k_1}{a-1}-\frac{b(k_2+\frac{bk_1}{1-a})}{\frac{b}{1-a}+c-1}$ and $x_2=\frac{k_2+\frac{bk_1}{1-a}}{\frac{b}{1-a}+c-1}$ both of which must be rational numbers.

Title: Proving Arnold's cat map has only rational periodic points.

Body:
I was trying to prove that all periodic points of the map are rational points. I started with $\begin{bmatrix}
           x_1 \\
           x_2 \\
         \end{bmatrix}= {\begin{bmatrix}
           1 & 1 \\
           1 & 2 \\
         \end{bmatrix}}^n \begin{bmatrix}
           x_1 \\
           x_2 \\
         \end{bmatrix}$ and manipulated it using diagnolization and arrived at $\begin{bmatrix}
           x_1 \\
           x_2 \\
         \end{bmatrix}=
         \begin{bmatrix}
           a & b \\
           b & c \\
         \end{bmatrix} \begin{bmatrix}
           x_1 \\
           x_2 \\
         \end{bmatrix}$ where a,b,c are fibonacci numbers. This seems to me to imply the only periodic point is the zero vector. But I know that isn't correct. I think it might have to do with me no including mod 1, but how do I account for that throughout the problem?


答:
Two vectors $\vec u$ and $\vec v$ are parallel if
$$
\vec u = t \vec v
$$
for some $t \in \mathbb{R}$.

You have
$$
\vec u = 5 \vec {e_1} - 3 \vec {e_2} + \alpha \vec {e_3} = (5,-3,\alpha)
$$
and
$$
\vec v = \beta \vec {e_1} + 2\vec {e_2} + 3 \vec {e_3} = (\beta,2,3)
$$
Your condition is then
$$
(5,-3,\alpha) = t (\beta,2,3) = (\beta t, 2t, 3t)
$$
and therefore
\begin{align*}
5 &= \beta t \\
-3 &= 2t \\
\alpha &= 3t
\end{align*}
You have 3 equations and 3 unknowns $t,\alpha,\beta$...

Title: When are two vectors parallel if the vectors are $5e_1-3e_2+\alpha e_3$ and $\beta e_1 + 2e_2 + 3e_3$

Body:
When are two vectors parallel if the vectors are  $$5e_1-3e_2+\alpha e_3$$ and $$\beta e_1 + 2e_2 + 3e_3$$
The alpha and beta are parameters.</p>


问:
Why $n^{\underline k} \le n^k$?
As in the title,

I cannot prove why $n^{\underline k} \le n^k$

$n^{\underline k}$ is denoted as falling factorial. That is $n^{\underline k} = n \times (n-1) \times \cdots \times (n-k+1)$

Can anyone help me out this, please?

答:
There are $k$ terms in the product $n \times (n-1) \times \dots \times (n - k +1)$. Each one of them is positive and inferior to $n$. So the product is inferior to $n^k$.

One way of seeing there are $k$ terms is remarking that $n - (n-k+1) + 1 = k$.

Title: Why $n^{\underline k} \le n^k$?


问:
Derivative of arcsin, question on provided proof
The following is proof of the derivative of $\arcsin$.  Does anyone know why and with what axiom the $y'$ is introduced into the equation in the following step?

$\sin y = x$

$y'\cos y = 1$

答:
$$\sin y=x\implies y'\cos y=1$$ is just the differentiation of the two members of an identity. Then from

$$\cos^2y+\sin^2y=1$$ the above equation can be written

$$y'\sqrt{1-\sin^2y}=\sqrt{1-x^2}=1.$$


问:
Calculate a point from the end of a line, on a particular angle and distance
Considering a cartesian plane; There is a line from point A to B, at any angle from the horizontal axis. We need to draw another line from point B to C. The distance d of the new line is given as well as the angle a it will have from the first line. We need to find the x and y coordinates of the point C.

Explanation:
Lets suppose I have point A on the cartesian plane as such { Ax : 10 , Ay : 10 }. I have another point B such that { Bx : 20 , By : 20}. Now I need to find a point C in terms of Cx and Cy. While I have the length of the line BC or the magnitude of the vector BC; that is d = 20. And also the Angle AC that is a = 135. The Image attached also clarifies it further.

I found a solution to draw a line at 90 degrees from the end of a line, but it doesnt gives a solution for other angles. I also found a solution to find a point from a particular point at an angle relative to the X-axis, but it doesnt gives a solution for the angle being relative to a line instead.
Image Attached

答:
Position vector of point $A \,(x_1, y_1) \,$ and position vector of point $B \,(x_2, y_2) \,$ are known.

We assume position vector of point $C$ as $(x_3, y_3)$.

We now take two directional vectors from point $B$,

$\vec{a} = (x_1 - x_2) \hat{i} + (y_1-y_2)\vec{j}$

$\vec{c} = (x_3 - x_2) \hat{i} + (y_3 - y_2)\vec{j}$

We know $\vec{a}\,,$ the angle $\, \theta \,$ between $\vec{a}$ and $\vec{c} \,,$  and we also know $|\vec{c}|$.

So, $\vec{a} \cdot \vec{c} = |\vec{a}| \, |\vec{c}| \, \cos \theta$

This gives one equation in $x_3, y_3$.

We also know,

$|\vec{c}| = \sqrt{(x_3-x_2)^2 + (y_3-y_2)^2} \,$ and that is your second equation in $x_3, y_3$.

As you have two equations in two variables, you should be able to find $(x_3, y_3)$.


问:
Is it possible to simplify this version of a geometric series? $f(x) = \sum\limits_{k=1}^n \frac{x^k}{k}$
Is it possible to simplify this series?

$$
f(x) = \sum\limits_{k=1}^n \frac{x^k}{k}
$$

If I derive it, I get the geometric series

$$
f'(x) = \sum\limits_{k=1}^n x^{k-1} = \sum\limits_{k=0}^{n-1} x^{k} = \frac{1-x^n}{1-x}
$$

But this seems to lead to nothing as

$$
f(x) = \int \frac{1-x^n}{1-x} dx
$$

can't expressed simpler as a comment in my previous question Antiderivative of $\frac{1-x^n}{1-x}$ mentioned.

I know, that $\lim\limits_{n \to \infty} \int \frac{1-x^n}{1-x} dx = -\ln{(|x-1|)}, \quad -1 < x < 1$

But what about without these restrictions of $n$ and $x$?

答:
Mathematica tells me it is
$$ f(x) = \sum_{k=1}^n \frac{x^k}{k} = \int\frac{1-x^n}{1-x}\ \mathrm{d}x =  -\mathrm{B}(x,1+n,0) - \ln(1 - x),$$
where $\mathrm{B}(x,a,b)$ is the incomplete Beta function. Not sure if that satisfies you. Also not sure what you mean by simplifying.


问:
Answer check: $\int_{\lvert z\rvert=1}\lvert z-1\rvert^2 dz$ - stuck, I (HOPE) I've missed a trick
$\int_{\lvert z\rvert=1}\lvert z-1\rvert^2 dz$ where z is a complex number (so the integral over a circle effectively)

I'm still working it out (I'm doing lots of show thats in the margin, I've got $j\cos(\theta)^3-\sin(\theta)^3$ and I'm showing if that's $e^{3j(-\theta+\frac{\pi}{2})}$ so I am checking myself!)

I am interested in what method you use because I'd love to see a nice short-cut!

Working

The gist is: $dz=dx+jdy$, let $x=\cos(\theta)$ and $y=\sin(\theta)$ for $\theta\in[0,2\pi]$ and tidy up. It's not hard I'm just out of practice (~2 years without a complex number)

Next

I'm going to assume (where $s=\sin(\theta)$ and $c=\cos(\theta)$) you can get to this without me doing every step:

I=$\int^{2\pi}_0(jc^3-s^3+jc-s+2sc-sc^2-2jc^2+js^2c)d\theta$

This "tidies" up into:

$I=\int^{2\pi}_0(e^{3j(-\theta+\frac{\pi}{2})}+e^{j(-\theta+\frac{\pi}{2})}+\sin(2\theta)-2j\cos^2(\theta)+j\sin(\theta)^2\cos(\theta)-\sin(\theta)\cos(\theta)^2)d\theta$

That last bit looks like it ought to tidy up.

Actually

I'm kind of stuck at that point. I could do it but it'd be long winded, is there a short cut I have missed?

Because I really don't want to evaluate that:

I'm exploring $z=e^{j\theta}$ and seeing where that gets me.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
Note that
$\ds{\verts{z - 1}^{2} = \pars{z - 1}\pars{\overline{z} - 1} =
\pars{z - 1}\pars{{1 \over z} - 1} = {2 - z - {1 \over z}}}$.

\begin{align}
&\bbox[5px,#ffd]{\int_{\verts{z}\ =\ 1}\,\,
\verts{z - 1}^{2}\,\,\dd z} =
\int_{\verts{z}\ =\ 1}\,\,
\pars{2 - z - {1 \over z}}\,\,\dd z = \bbx{-2\pi\ic} \\ &
\end{align}


问:
Prove a set has cardinality $\aleph$
Let A be the set of all numbers between $0$ and $1$ so that each number's infinite decimal representation is comprised of only odd numbers after the decimal point (For example: $0.1351379...$)

Prove that $|A| = \aleph$

答:
For clarity, in my answer, I am not working in binary, I'm working in usual decimals base $10$.

It is well-known that the set $B$ of all numbers between $0$ and $1$ whose decimal expansion contains only $0$’s and $1$’s is an uncountable set, by Cantor’s diagonalisation argument. For example, this is proven in Rudin's PMA Theorem 2.14.

Your question is not so different...

For example map every $0$ digit after the decimal point to $3$ and every $1$ digit after the decimal point to $7$. e.g. $f: 0.00110100101001... \to 0.33773733737337...$.

This is a bijection.

Therefore the set of numbers between $0$ and $1$ whose decimal expansions only contain $3$ and $7$ is uncountable.

$A$ is a superset of this set, so A must also be uncountable.

[I learned in the comments that "uncountable" does not mean the same thing as "cardinality at least $|\mathbb{R}|$", because this would assume the c.h. which we do not implicitly assume to be true (if we assume the c.h. is true (or false) then we must state so). But for this question we don't need to address the c.h. at all.]

It can also be shown that there exists an injective function $f: \mathbb{R} \to B$, therefore $ \mathbb{R}$ and $B $ do have the same cardinality.

So by the "sandwich theorem" $|B| = |A| = |\mathbb{R}|$.

Title: Prove a set has cardinality $\aleph$


问:
Integral extensions and number of generators
here's a doubt which arised from a previous question:

Suppose $R$ is a ring and $S \subseteq R$ is a subring. Moreover suppose $R$ is integral over $S$ and $R$ is finitely generated as an $S$-module. In a previous post Matt E showed that we always have Specmax(S) is bounded above by Specmax(R). My question is: is it true that for any maximal ideal of S there are at most t maximal ideals of R?

Thanks again

答:
Let $\mathfrak{m} $ be a maximal ideal of $R$; we want to show that the fiber of the map $\mathrm{Spec} S \to \mathrm{Spec} R $ has cardinality at most $t$, where $S$ is generated by $t$ elements as an $R$-module. By making the  base-change $R \to R/\mathfrak{m}$, $S \to S/\mathfrak{m}S$, we are reduced to the case of $R$ a field. In this case, $S$ is a $t$-dimensional vector space.
We need to bound the size of $\mathrm{Spec} S$ by $t$.

Now $S$ is an artinian ring, and it factors as a product of local artinian rings (uniquely). Each of these local factors is an $R$-vector space of dimension at least one, so there are at most $t$ factors. By definition, each factor has one prime ideal. Since the spectrum of a direct product is the disjoint union, this shows that $\mathrm{Spec} S$ has at most $t$ elements.


答:
If there are $n$ objects and we are to pick $k$ objects to assign to a list of length $k$,

if we pick with replacement, there are $n^k$ different lists of length $k$;

if we pick without replacement, there are $n^{\underline k}$ different lists of length $k$.

And ways to pick with replacement include every way to pick without replacement, so

$$n^{\underline k} \le n^k$$

Title: Why $n^{\underline k} \le n^k$?

Body:
As in the title,

I cannot prove why $n^{\underline k} \le n^k$

$n^{\underline k}$ is denoted as falling factorial. That is $n^{\underline k} = n \times (n-1) \times \cdots \times (n-k+1)$

Can anyone help me out this, please?


问:
Quadratic variation of a process
I am given a filtered probability space and $(B_t)_{t \geq 0}$ as a standard $\mathcal{F}_t$ Brownian Motion. In addition, I have the following process: $X = (B_t^2 - t)_{t \geq 0}$. I am asked to show that the quadratic variation (aka sharp brackets) of X is given by:
$$\langle X \rangle _ t = \int_0^t4 \cdot B_s^2 \ ds$$
Is there any theorem that can help me working through this? If not, would someone give me a good hint on how to proceed?

答:
For a suitable integrand $Y$, the quadratic variation of the process $I_t := \int_0^t Y_u dB_u, 0 \leq t \leq T$ is the process $\langle I \rangle_t = \int_0^t Y^2_u du,  0 \leq t \leq T.$ (*)

Now, applying Ito's formula with the function $f(x)=x^2 -t,$ we obtain $dX_t = 2B_tdB_t.$ Applying this together with (*) we get the result.

Title: Quadratic variation of a process


问:
Relation between weak topology on space of signed measures on a metric space $E$ and the weak* topology on $C_b(E)'$
In general, if $E_i$ is a $\mathbb R$-vector space and $\langle\;\cdot\;,\;\cdot\;\rangle$ is a duality pairing between $E_1$ and $E_2$ and $$p_{x_2}(x_1):=q_{x_1}(x_2):=\left|\langle x_1,x_2\rangle\right|\;\;\;\text{for }(x_1,x_2)\in E_1\times E_2,$$ then $p_{x_2}$ is a seminorm on $E_1$ for all $x_2\in E_2$ and $q_{x_1}$ is a seminorm on $E_2$ for all $x_1\in E_1$.

Hence, $\left\{p_{x_2}:x_2\in E_2\right\}$ is inducing a locally convex topology $\sigma(E_1,E_2)$ on $E_1$ and $\left\{q_{x_1}:x_1\in E_1\right\}$ is inducing a locally convex topology $\sigma(E_2,E_1)$ on $E_2$.

If $\varphi\in E_1^\ast$, we can show that $\varphi$ is $\sigma(E_1,E_2)$-continuous iff $\varphi=\langle\;\cdot\;x_2\rangle$.

By the aforementioned result, we are able to identify $$E_1':=\left\{\varphi\in E_1^\ast:\varphi\text{ is }\sigma(E_1,E_2)\text{-continuous}\right\}$$ with $E_2$. Does that mean $E_1'$ is again a locally convex topological vector space when endowed with the topology $\sigma(E_2,E_1)$?

I'm mainly interested in the following instance: Let $E_1:=C_b(E)$ be the set of real-valued bounded continuous functions on a metric space $E$, $E_2:=\mathcal M(E)$ denote the set of finite signed measures on $\mathcal B(E)$ and $$\langle f,\mu\rangle:=\int f\:{\rm d}\mu\;\;\;\text{for }(f,\mu)\in C_b(E)\times\mathcal M(E).$$

If $E_1=C_b(E)$ is equipped with the supremum norm, we may identify $\mathcal M(E)$ with a subset of $C_b(E)'$. How is $\sigma(E_2,E_1)$ related the the subspace topology on $\mathcal M(E)$ induced from the weak* topology on $C_b(E)'$?

答:
Q1: Yes, $E_1'$ is a locally convex space when endowed with the $\sigma(E_1',E_1)$-topology. (It is just $E_2$ with the $\sigma(E_2,E_1)$-topology.)

Q2: The weak-$*$ topology respects subspaces. In general, if $\langle E, G \rangle$ is a dual pair and if $F \subseteq G$ is a subspace, then the relative $\sigma(G,E)$-topology on $F$ coincides with the $\sigma(F,E/F^\perp)$-topology.¹ If $F$ separates points on $E$, so that $\langle E,F\rangle$ is again a dual pair, then one has $F^\perp = \{0\}$, so now the $\sigma(F,E/F^\perp)$-topology is simply the $\sigma(F,E)$-topology.

¹: For a textbook reference, see [Sch99, §IV.4.1, Corollary 1 (p. 135)] or [Köt83, Proposition 22.2.(1) (p. 276)], among others. This is basically just a special case of the transitive law of initial topologies (see halfway through this very long answer), since weak topologies and subspace topologies are examples of initial topologies.

In your example, we would have $E = C_b(\Omega)$, $F = \mathcal M(\Omega)$ and $G = C_b(\Omega)'$. It is easy to see that $\mathcal M(\Omega)$ separates points on $C_b(\Omega)$, so $F^\perp = \{0\}$. Furthermore, we know from this other question of yours that $C_b(\Omega)$ separates points on $\mathcal M(\Omega)$, so $\mathcal M(\Omega)$ can be viewed as a subspace of $C_b(\Omega)'$. Hence we are in the setting described above.

References

[Köt83] Gottfried Köthe, Topological Vector Spaces I, Second revised printing, Grundlehren der mathematischen Wissenschaften 159, Springer, 1983.

[Sch99] H.H. Schaefer, with M.P. Wolff, Topological Vector Spaces, Second Edition, Graduate Texts in Mathematics 3, Springer, 1999.


问:
Why does non-full rankness imply $M(x_1, \dots, x_n)g(x_1, \dots, x_n)=0$?
Based on the answer to this question, if we have a matrix
$$M(x_1, \dots, x_n)= \begin{bmatrix} f_1(x_1) & \cdots & f_n(x_1) \\ \vdots & \ddots & \vdots \\f_1(x_n) & \cdots & f_n(x_n) \end{bmatrix} $$ and this matrix is not full rank, then there exists a function $g:\mathbb{R}^n \rightarrow \mathbb{R}^n\setminus \{0\}$ such that
$$ M(x_1, \dots, x_n)g(x_1, \dots, x_n)=0\,\,.$$
why is this the case -- why is $g$ denoted to depend on $x_1,\dots, x_n$? I thought that a square matrix not being full rank implies that its columns are linearly dependent and would this not imply that
$$ M(x_1, \dots, x_n)\mathbf{v}=0$$
for some non-zero constant vector $\mathbf{v}$? Is it just that the vector depends on the values the functions operate on, or is there something else going on?

答:
Consider two separate sets of points, $x_{1,1},\dots,x_{n,1}$ and $x_{1,2},\dots,x_{n,2}$. Let $M_j = M(x_{1,j},\dots,x_{n,j})$ for $j = 1,2$.

Because $M_1$ does not have full rank, there is necessarily a vector $\mathbf v_1$ for which $M_1 \mathbf v_1 = 0$. Similarly, because $M_2$ does not have full rank, there is a vector $\mathbf v_2$ for which $M_2 \mathbf v_2 = 0$. However, the fact that $M_1$ and $M_2$ fail to have full rank is not enough to deduce that there is a single vector $\mathbf v$ for which $M_1 \mathbf v = 0$ and $M_2 \mathbf v = 0$.

Likewise, the fact that $M(x_1,\dots,x_n)$ fails to have full rank for all sets tells us that for each set $x_1,\dots,x_n$, there is a vector $\mathbf v$ for which $M(x_1,\dots,x_n)\mathbf v = 0$ (i.e. there is a possible value for $g(x_1,\dots,x_n) = \mathbf v$). However, we have no reason to believe (without proof) that there exists a single vector $\mathbf v$ such that for all $x_1,\dots,x_n,$ $M(x_1,\dots,x_n)\mathbf v = 0$.


问:
solution verification :$\int_{\gamma}Re(z)+Im(z)dz$ on the parabola $y=x^2$ between $x\in[-1,1]$
$\int_{\gamma}\operatorname{Re}(z)+\operatorname{Im}(z)\,dz$ on the parabola $y=x^2$ between $x\in[-1,1]$
my try:
$z=x(t)+iy(t)\\x=4(t-0.5)^{2}\\y=2x-1$

whereas gamma is from the range [0,1]

$\int_{0}^{1}x(t)x'(t)\,dt+\int_{0}^{1}y(t)y'(t)\,dt$

答:
It is incorrect to write $\int_\gamma (x+y)\,dz=\int_0^1 x(t)x'(t)\,dt +\int_0^1 x(t)x'(t)\,dt$.

Rather, we have $x(t)=t$, $y(t)=t^2$, and $dz=(x'(t)+iy'(t))\,dt$, where $t$ begins at $-1$ and terminates at $1$.  Hence,

$$\begin{align}
\int_\gamma (x+y)\,dz&=\int_{-1}^1(t+t^2)\,(1+i2t)\,dt
\end{align}$$

And the interested reader can finish the exercise now.


问:
When are two vectors parallel if the vectors are $5e_1-3e_2+\alpha e_3$ and $\beta e_1 + 2e_2 + 3e_3$
When are two vectors parallel if the vectors are  $$5e_1-3e_2+\alpha e_3$$ and $$\beta e_1 + 2e_2 + 3e_3$$
The alpha and beta are parameters.</p>

答:
Hint:

Two vectors are parallel iff their cross product is zero:

$$(5e_1-3e_2+\alpha e_3)\times(\beta e_1+2e_2+3e_3)=(-9-2\alpha)e_1+(\alpha\beta-15)e_2+(10+3\beta)e_3\implies$$

$$\implies \alpha=-\frac92\;\;\ldots\; etc.$$

Or:

Two vectors are parallel iff their linearly dependent iff one is a scalar multiple of the other one. You'll get, of course, the same as above.


问:
Building a non linear ascending function that passes through 0,0 and an arbitrary point that grows fast at the beginning and very slow at the end.
I need to build a function for one of my games to be applied to my player experience. The domain of the experience is $[0, +\infty)$ but I need to constrain the function to pass through an exact point $(x,y)$.

The player is rewarded a semi random quantity of experience depending on the number of objectives he has completed. As you see, there's not a limit on the total quantity of experience he can get. But, I want to reward the player with high quantities at the beginning and low quantities at the end.

Another constraint exists and it is that I need to be able to make this function pass through a point $(x,y)$ that I will change over time as my game grows.

As an example, at this time I need my function pass through $(100, 10000)$ because I want to make sure that when the player completes $100$ objectives it must have $10000$ points of experience. Later I will tweak this as more objectives are added.

I would be happy with something like a logarithmic behavior because of its strong growing nature at the beginning and slow one at the end but wasn't able to construct anything usable around it.

How would you build something like this?

答: How about $y = a{\sqrt x} + b x$? Then suppose you want $x=100$ objectives to yield $y=10,000 $ points, you need to choose $a, b$ such that $10,000 = a\sqrt(100)+b100$. For example, $b=0, a=1000$ will give $y=1000\sqrt x$. You can play with desmos to fine tune. https://www.desmos.com/calculator/dyaear5la1


问:
One Theorem and example in Recursive Formula
I read a book as follows:

and following example:

$T(n)=T(\frac{n}{4})+O(\log^2n)=O(\log^3n)$

How we can fit this theorem in this example? Or this is cannot be handle in this way what is the most good method for solving this reuccrence?

Update: I need more quick version for finding solution of this recurrence.

答:
Assuming feasible $a,b,c$, a simple way to solve this recurrence:

$$
T(n) =a T\left(\frac nb\right)+n^c
$$

$$
T\left(b^{\log_b n}\right)= aT\left(b^{\log_b\frac nb}\right)+n^c
$$

now making

$$
\mathcal{T}(\cdot)=T\left(b^{(\cdot)}\right),\ \ \ z=\log_b n
$$

$$
\mathcal{T}(z)= a \mathcal{T}(z-1)+b^{zc}
$$

which is a linear recurrence with solution

$$
\mathcal{T}(z) = \mathcal{T}(z)_h + \mathcal{T}(z)_p,\ \ \cases{\mathcal{T}(z)_h=a \mathcal{T}(z-1)_h\\
\mathcal{T}(z)_p=a \mathcal{T}(z-1)_p+b^{zc}}
$$

The homogeneous has as solution

$$
\mathcal{T}(z-1)_h=a^{z-1}C_0
$$

for the particular, making $\mathcal{T}(z-1)_p=a^{z-1}C_0(z)$ and substituting we have

$$
C_0(z)-C_0(z-1) = \frac{b^{z c}}{a^{z-1}}
$$

and then

$$
C_0(z) = \frac{a^{1-z} \left(b^{c (z+1)}-a^z b^c\right)}{b^c-a}
$$

so

$$
\mathcal{T}(z)_p = \frac{b^{c (z+1)}-a^z b^c}{b^c-a}
$$

hence

$$
\mathcal{T}(z) = C_0 a^{z-1}+\frac{b^{c (z+1)}-a^z b^c}{b^c-a}
$$

now making the backwards substitution  $z\to\log_b n$ we can recover $T(n)$

$$
T(n) = C_0 a^{\log_b n-1}+\frac{b^c \left(n^c-a^{\log_b n}\right)}{b^c-a}
$$


问:
Why can I estimate the variance if I already estimated the mean in this way?
In my lectures it's discussed the estimation of $Var(X)$ in the case I have already estimated $E(X)$, but there is something I don't know if it's a problem or not.

Suppose $Var(X) = f(E(X))$ , for example in the case of a bernoulli r.v. we have $Var(X) = E(X)(1-E(X))$ , besides I have estimated $ E(X) $ to be $ \hat{\theta} $ then I could use as a estimate of $Var(X)$ the function $f(\hat{\theta})$ .

However, in this reasoning I think there is something not so clear about the error in the estimate because if the estimate of $E(X)$ is not exact (which is always the case) then the estimate of $Var(X)$ could be much more inaccurate depending on the actual function $f$.

Is this method used because it's assumed that the estimate $\hat{\theta}$ is equal to the actual value of $E(X)$? I don't have much knowledge in statistical inference so I don't know if this is a real problem. Is this a true problem?

答:
What @Ian said, is true, though not complete: for special distributions (like Bernoulli), there may be a sufficient statistic which (loosely speaking) contains all the information in the sample about the paramer(s). For a more formal discussion, see here: https://en.wikipedia.org/wiki/Sufficient_statistic#Bernoulli_distribution

So while in general, you'd estimate a variance by the sample variance, you'd better use a function of the sufficient statistic in this special case. Now $\hat{\theta}(1-\hat{\theta})$ would be biased, because an easy calculation shows $\displaystyle\mathbb{E}\hat{\theta}(1-\hat{\theta})=\frac{n-1}n\,p(1-p)$, but the remedy is obvious: use $\displaystyle\frac{n}{n-1}\,\hat{\theta}(1-\hat{\theta})$ as an estimator, then.


问:
Is an group determined by its torsion part and torsion-free part?
Let $G$ be an abelian group.  Let $G_T$ be the torsion part of $G$, i.e. the set of all elements of $G$ of finite order.  And let $G_F$ be the torsion-free part of $G$, i.e. the set containing $0$ along with all elements of $G$ of infinite order.  Then my question is, is the isomorphism type of $G$ uniquely determined by the isomorphism types of $G_T$ and $G_F$?

I think this is definitely true in the case when $G$ is finitely generated, but I'm asking about the general case.

答:
(Still not quite a complete answer.)

An abelian group $A$ does not have a canonically defined torsion-free subgroup in general (the elements of infinite order usually aren't a subgroup). What is canonically defined is a short exact sequence

$$0 \to A_T \to A \to A/A_T \to 0$$

where $A/A_T$, which we'll write $A_F$, is the universal torsion-free abelian group admitting a map from $A$; you might call this the "torsion-free part" but keep in mind that it's a quotient, not a subgroup.

$A$ is not determined by the isomorphism type of $A_T$ and $A_F$, and the reason is that the short exact sequence above does not split in general, and so in general defines a nontrivial class in $\text{Ext}^1(A_F, A_T)$. It does split if $A_F$ is free (equivalently, projective as a $\mathbb{Z}$-module), which in particular happens whenever $A$ is finitely generated; in this case we have

$$A \cong A_T \oplus A_F$$

so $A$ is in fact determined up to isomorphism by $A_T$ and $A/A_T$. So to exhibit a counterexample we need to find a torsion abelian group $A_T$ and a torsion-free abelian group $A_F$ such that $\text{Ext}^1(A_F, A_T) \neq 0$; in particular $A_F$ must be torsion-free (equivalently, flat as a $\mathbb{Z}$-module) but not free, and hence infinitely generated.

Take $A_F = \mathbb{Z} \left[ \frac{1}{p} \right]$ where $p$ is a prime. If we write $A_F$ as a filtered colimit of the diagram $\mathbb{Z} \xrightarrow{p} \mathbb{Z} \xrightarrow{p} \dots $ and try to compute $\text{Ext}^1(A_F, A_T)$ where $A_T$ is unspecified then we get a $\lim^1$ exact sequence

$$0 \to \lim^1_n A_T \to \text{Ext}^1 \left( \mathbb{Z} \left[ \frac{1}{p} \right], A_T \right) \to 0$$

so we want to find a torsion abelian group such that $\lim^1$ of the diagram $\dots \xrightarrow{p} A_T \xrightarrow{p} A_T$ is nonzero. I believe that we can take $A_T = \bigoplus_i \mathbb{Z}/p^i$ (which showed up in this previous math.SE question, also about extensions of torsion-free abelian groups by torsion abelian groups), but it seems annoying to verify that this works; I believe it does not suffice to take either $A_T = \mathbb{Z}/p$ or $A_T = \mathbb{Z} \left[ \frac{1}{p} \right]/\mathbb{Z}$ (the Prufer $p$-group), which I was hoping would work.

Edit: Apparently it's simpler to just construct $A$ directly. Supposedly the abelian group $A = \prod_i \mathbb{Z}/p^i$ has the property that its torsion sequence $0 \to A_T \to A \to A/A_T \to 0$ doesn't split, but I'm not sure how to prove it off the top of my head.


问:
maximum and minimum of the function $ y=2 \sin3x+\sin 6x$
I have to find maximum and minimum of the function
$$ y=2 \sin3x+\sin 6x$$

$y'= (\cos3x+\cos6x)= 6 \cos 3x (1+2 \sin 3x)=0 \Rightarrow \cos3x=0  \lor \sin3x=- \frac{1}{2}  \Rightarrow 3x= \frac{\pi}{2} \lor 3x= \frac{3\pi}{2} \lor 3x= \frac{7\pi}{6}\lor 3x= \frac{11\pi}{6}$.
Analyzing the sign of the derivative we can say that in $[0,2\pi]$:

$y'>0 \Rightarrow 0< 3x< \frac{\pi}{2} \lor \frac{7}{6} \pi<3x < \frac{3}{2}\pi \lor \frac{11}{6} \pi<3x <2 \pi $.

For $3x= \frac{\pi}{2}$ and $3x= \frac{3}{2} \pi$ there are two maxima.

For $3x= \frac{7}{6}\pi$ and $3x=  \frac{11}{6}\pi$ there are two maxima.

The first two positions can be summarized in $3x= \frac{\pi}{2}+k\pi \Rightarrow x= \frac{\pi}{6}+\frac{k \pi}{3}= \frac{\pi+2k \pi}{6}= \frac {1+2k}{6} \pi$.

The other two disequalities can be summarized in $3x= \frac{\pi}{2}+ \frac{2 \pi}{3} k \Rightarrow x=  \frac{\pi}{6}+ \frac{2 \pi}{9} k $.

Is it correct or I am missing something?

I don't understand my book shows a different solution.

答:
$$y'= 6(\cos3x+\cos6x)= 6 \cos 3x (1+2 \sin 3x)$$

This is simply not true. You are implying that:

$$\cos6x=2\cos3x\sin3x$$

..which is plain wrong.


答:
Check your calculation of $y'$. The derivative of $$2\sin 3x + \sin 6x$$ is
$$6\cos 3x + 6\cos 6x.$$

Title: maximum and minimum of the function $ y=2 \sin3x+\sin 6x$

Body:
I have to find maximum and minimum of the function
$$ y=2 \sin3x+\sin 6x$$

$y'= (\cos3x+\cos6x)= 6 \cos 3x (1+2 \sin 3x)=0 \Rightarrow \cos3x=0  \lor \sin3x=- \frac{1}{2}  \Rightarrow 3x= \frac{\pi}{2} \lor 3x= \frac{3\pi}{2} \lor 3x= \frac{7\pi}{6}\lor 3x= \frac{11\pi}{6}$.
Analyzing the sign of the derivative we can say that in $[0,2\pi]$:

$y'>0 \Rightarrow 0< 3x< \frac{\pi}{2} \lor \frac{7}{6} \pi<3x < \frac{3}{2}\pi \lor \frac{11}{6} \pi<3x <2 \pi $.

For $3x= \frac{\pi}{2}$ and $3x= \frac{3}{2} \pi$ there are two maxima.

For $3x= \frac{7}{6}\pi$ and $3x=  \frac{11}{6}\pi$ there are two maxima.

The first two positions can be summarized in $3x= \frac{\pi}{2}+k\pi \Rightarrow x= \frac{\pi}{6}+\frac{k \pi}{3}= \frac{\pi+2k \pi}{6}= \frac {1+2k}{6} \pi$.

The other two disequalities can be summarized in $3x= \frac{\pi}{2}+ \frac{2 \pi}{3} k \Rightarrow x=  \frac{\pi}{6}+ \frac{2 \pi}{9} k $.

Is it correct or I am missing something?

I don't understand my book shows a different solution.


问:
distance between two circle's centres
Two circles of equal radius intersect at points C and D.The centres of the two circles are points A and B respectively.If their radius is 10 units,the area of triangle ABC is 40,then how do we find the distance between A and B?</p>

答:
Draw a figure.

You have to find the length of the third side of an isosceles triangle with two sides length = 10 and area= 40.

This is equivalent to finding the length of twice one of the short sides of a right angled triangle with hypotenuse 10 and area 20.

So there are two solutions.

Introduce notation.

If $p, q$ are the lengths of the short sides in the right-angled triangle (so you are looking for AB = $2p$ or $2q$) then you have to solve
$$ p^2 + q^2 = 10^2, \mbox{ (Pythagoras)}$$
$$ \frac{pq}{2} = 20. \mbox{ (area triangle)}$$

If you solve this quadratic equation you get e.g.
$$p = \sqrt{80} \mbox{ or } p = \sqrt{20},$$
$$q = \frac{20}{p}.$$

Edit: corrected formula for area of triangle and solution for $p$.

Title: distance between two circle's centres


问:
4 balls with different colors, there are 3 identical boxes, how many ways?
4 balls with different colors, there are 3 identical boxes. the box can be empty. of course, all balls must be used. how many ways to distribute the balls?

The striling number of the second kind S(n,k) counts the number of ways to partition a set of n objects into k non-empty subsets. so my answer is,
S(4,1)+S(4,2)+S(4,3)
but I don't know how to get numeric answer from here like how do I find what S(4,1) is?

答: You can put all balls in $1$ box in $1$ way. Then you can put $1$ ball in a box and the other $3$ balls in a second box in $4$ ways, or pair the $4$ balls off in two boxes in $3$ ways. Finally you can make ${4\choose2}=6$ pairs of balls for one box and put the remaining balls in two more boxes. In all you have $1+4+3+6=14$ ways, corresponding with $S(4,1)+S(4,2)+S(4,3)=14$.


问:
What is the meaning of the term "direction along a level curve g(x,y) = c"?
I am studying the topic of Lagrange multiplier and I'm at the point of learning why the method is true.

Provided that we want to minimize/maximize function $f(x,y)$ subject to the constrain of function $g(x,y) = c$.

There is a statement in the proof "At constrained min/max point, in any direction along the level curve $g(x,y) = c$, the rate of change of function $f(x,y)$ must be $0$"

Could you please explain me what does "direction along the level curve $g(x,y) = c$" really mean ? Can this term be understood as the direction of the tangent vector to the level curve $g(x,y) = c$ ?

The reason I ask this question is that, say, if we have $g(x,y) = c = 3 = x^2+y^2$ which is a circle, then innocently, I imagine the "direction along this curve" is an "arc" but it sounds not true.

Thank you very much for the clarification!

答:
Yes, this is the direction of the tangent at a point. You don't consider an arc because a vector represents a single direction at a single point. In other words, it is the instantaneous direction when you move along the curve.

Note that a smooth curved can be closely approximated as a sequence of line segments.


问:
Prove that $f(\overline C ) \subset \overline{ f(C)}$ .
Prove that $f: A \to B$ is continuous iff its graph is compact where $A$ is compact and $A,B$ are metric spaces.

My attempt: I have already proved it. But somehow i am not satisfied with my proof. Implies part is Ok. But converse part i want to prove f is continuous by using $f(\overline C) \subset \overline{ f(C)}$.

For this i took $C\subset E$ arbitrary. Let $T=\{(x, f(x)):\, x  \in C\}$, $F=\{(x, f(x)): x \in \overline{C}\}$. Now $\overline T$ is closed in graph(f),  hece it is compact. I cant proceed further. Please give me hint.

答:
As noted in the comments, I don't think what you're asking to prove will actually solve the desired question.  So I will use sequences to prove your question.  So suppose your graph is compact.  Note that since we are dealing with Metric spaces then we have the Bolzano-Weierstrass property (in the sense that every sequence in our graph has a convergent subsequence).

Let $C \subset f(A)$ be closed and let $x_n \in f^{-1}(C)$ with $x_n \rightarrow x \in A$.  Since our graph is compact then the sequence $(x_n, f(x_n))$ has a convergent subsequence in the graph of $f$, i.e. $(x_{n_k}, f(x_{n_k})) \rightarrow (x,y)$, which must be in the graph of $f$ since our graph is closed.  But then $(x,y) = (x,f(x))$, and $f(x_{n_k}) \rightarrow f(x)$ so then $f(x) \in C$.  Finally, this means that $x \in f^{-1}(C)$, thus $f^{-1}(C)$ is closed and $f$ is continuous.


问:
Calculate a point from the end of a line, on a particular angle and distance
Considering a cartesian plane; There is a line from point A to B, at any angle from the horizontal axis. We need to draw another line from point B to C. The distance d of the new line is given as well as the angle a it will have from the first line. We need to find the x and y coordinates of the point C.

Explanation:
Lets suppose I have point A on the cartesian plane as such { Ax : 10 , Ay : 10 }. I have another point B such that { Bx : 20 , By : 20}. Now I need to find a point C in terms of Cx and Cy. While I have the length of the line BC or the magnitude of the vector BC; that is d = 20. And also the Angle AC that is a = 135. The Image attached also clarifies it further.

I found a solution to draw a line at 90 degrees from the end of a line, but it doesnt gives a solution for other angles. I also found a solution to find a point from a particular point at an angle relative to the X-axis, but it doesnt gives a solution for the angle being relative to a line instead.
Image Attached

答:
Following an anticlockwise angular convention.

Let angle between AB and BC be  $\beta $ and let the slope of AB be $\phi$

$$ xC= xB + d \cos ( \pi + \phi -\beta) $$

$$ yC= yB + d \sin ( \pi + \phi -\beta) $$


问:
Show that a statement is true by induction for $x,y > 0$ and $n \in N$.
I ve been assigned the following problem:

How can we prove by induction that for every $x,y > 0$ and $n \in N$ it is true that:
$$\frac{x^n+y^n}{2} \geq \bigg(\frac{x+y}{2}\bigg)^n$$

I have established a base of $n=1$ so that $\frac{x+y}{2} = \frac{x+y}{2}$ and have assumed that the statement is true for any natural $k$ but am having problem proving that it is true for $k+1$.

Any help is appreciated!

答:
Wlog, let $x+y=1$ and the inequality becomes
$$ x^n+(1-x)^n\ge \frac{1}{2^{n-1}}. $$
Let $$ f(x)=x^n+(1-x)^n $$
and then
$$ f'(x)=n(x^{n-1}-(1-x)^{n-1}),f''(x)=n(n-1)(x^{n-2}+(1-x)^{n-2}).$$
Setting $f'(x)=0$ gives $x=\frac12$ and $f''(\frac12)>0$. Therefore $f(x)$ attains a local minimum $\frac{1}{2^{n-1}}$ at $x=\frac12$. Note $f(0)=f(1)=1$ and thus $f(x)$ attain the global minimum $\frac{1}{2^{n-1}}$ at $x=\frac12$; Namely
$$ x^n+(1-x)^n\ge \frac{1}{2^{n-1}} $$
and the equal sign "=" holds iff $x=y\frac12$.


问:
Calculating the matrix exponential in this case?
I am wondering how in general one can calculate the matrix exponential. For example, I am working in the Lie algebra $\mathfrak{su}(3)$. I have two matrices $$ m_1 = \begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} $$ and $$m_4 = \begin{pmatrix} 0 & i & 0 \\ i & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}. $$ Their linear combination $\alpha m_1 + \beta m_4$ can be written as $$ X := \alpha m_1 + \beta m_4 = \begin{pmatrix} 0 & - \overline{z} & 0 \\ z & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} $$ where $z = \alpha + i \beta$.

Now I wish to obtain a closed formula (if possible) for the matrix exponential $e^{tX}$ (where $t$ is some real number), perhaps in terms of sines and cosines? How does one in general calculate the matrix exponential like this?

Body:
Notice that $X^3 = -|z|^2 X$. With that in mind, we can write
$$
\exp(tX) = \sum_{n=0}^{\infty}\frac{(tX)^n}{n!}
= I + \sum_{k=1}^\infty \frac{(tX)^{2k}}{(2k)!}
+ \sum_{k=0}^\infty \frac{(tX)^{2k+1}}{(2k+1)!}\\
=
I + \left(\sum_{k=1}^\infty \frac{(-1)^k(|z|t)^{2k}}{(2k)!}\right) \frac{X^2}{|z|^2}
+ \left(\sum_{k=0}^\infty \frac{(-1)^k(|z|t)^{2k+1}}{(2k+1)!}\right) X
\\
= I  + \sin(t|z|) X + \frac{\cos(t|z|) - 1}{|z|^2} X^2.
$$


问:
derive the least squares estimate b1 from the normal equations
I am stuck on the following question*:

derive the least squares estimate:

$$  b1 = \frac{\sum X_iY_i - \frac{\sum X_i \sum Y_i}{n}}
          {\sum X_i^2 - \frac{( \sum X_i)^2}{n}} $$

from the normal equations:

$$ (i)   \sum Y_i = nb_0 +  b_1 \sum X_i    $$
$$ (ii) \sum X_iY_i = b_0 \sum X_i + b_1 \sum X_i^2$$

I get:

$$ b1 = \frac { \sum X_i Y_i -  \frac {\sum X_i \sum Y_i}{n} }
              {(\sum X_i)^2 - \frac {(\sum X_i)^2}{n}} $$

I think I am there except that I have $(\sum X_i)^2$ instead of $\sum X_i^2$ in the demonimator. I take the latter to mean the sum (X * X) rather than the sum(X) * sum(X). However I cannot see how you would get sum (X * X)  from the original equations. Any help would be much appreciated.

Unless (X * X) is interpreted the matrix muliplication (X * X') where X is a column vector in which case $( \sum X_i)^2$ and $\sum X_i^2$ are equal.

(Please see below for working)

I first solve for $b_0$ with $i - (\frac {ii} {\sum X_i})$:

$(\frac {ii} {\sum X_i})$:    $  \frac{\sum X_iY_i}{\sum X_i} = b_0  + b_1 \sum X_i$

$(i - \frac {ii} {\sum X_i})$:   $  \sum Y_i -  \frac{\sum X_iY_i}{\sum X_i} = nb_0 - b_0$

rearranging yeilds:

$$  b_o = \frac {\sum X_i \sum Y_i - \sum X_i Y_i}{\sum X_i (n -1)} $$

then substituting into (i) (rearranged to isolate b1) and cancelling:

$$ b1 = \frac { \sum Y_i - n \frac { \sum X_i \sum Y_i - \sum X_i Y_i} {\sum X_i(n-1)}} { \sum X_i} $$

$$ b1 \sum X_i = \frac { \sum X_i \sum Y_i (n-1) - n \sum X_i \sum Y_i - n\sum X_i Y_i}                      {\sum X_i(n-1)} $$

$$ b1 = \frac {(n \sum X_i \sum Y_i  - \sum X_i \sum Y_i)  - (n \sum X_i \sum Y_i - n\sum X_i Y_i)}                      {\sum X_i \sum X_i(n-1)} $$

$$ b1 = \frac {n \sum X_i Y_i -  \sum X_i \sum Y_i }
              {\sum X_i \sum X_i(n-1)} $$

$$ b1 = \frac {n \sum X_i Y_i -  \sum X_i \sum Y_i }
              {n(\sum X_i)^2 - (\sum X_i)^2} $$

multiply both sides by $\frac {\frac{1}{n}}{\frac{1}{n}}$

$$ b1 = \frac { \sum X_i Y_i -  \frac {\sum X_i \sum Y_i}{n} }
              {(\sum X_i)^2 - \frac {(\sum X_i)^2}{n}} $$

*q2.30 from Applied Linear Statistical Models, Neter et al 3rd edition.

答:
Your error occurs when you divide $(ii)$ by $\sum X_i$. The result should be:
$$\frac{\sum X_iY_i}{\sum X_i} = b_0  + b_1 \frac{\sum X_i^2}{\sum X_i}
$$ because, as you mentioned, $\sum X_i^2$ does not mean $(\sum X_i)(\sum X_i)$.


问:
Uniqueness of ring product
Sorry for my bad English.

Let $A_1,\dots,A_n,B_1,\dots,B_m$ be commutative integral domains.

We assume $A_1\times \cdots A_n\cong B_1\times \cdots \times B_m$.

Now, is  the following correct?

$n=m$ and there is  $\sigma\in S_n$ such that $A_i\cong B_{\sigma(i)}$?

We have no idea this is correct or false, help me thanks.

答:
It's true under the slightly weaker assumption that each of the $A_i$ and $B_i$ have no nontrivial idempotents; a commutative ring $R$ satisfying this property is called connected, because it's equivalent to $\text{Spec } R$ being connected in the Zariski topology. Conceptually the idea is that, if we write $R = \prod A_i \cong \prod B_i$, then $\text{Spec } A_i, \text{Spec } B_i$ must be two lists of the connected components of $\text{Spec } R$ (as an affine scheme), so must be the same up to permutation.

I mention all this by way of motivation; it's possible to give a completely elementary proof in terms of idempotents without knowing any algebraic geometry (although I find it a little unmotivated without the geometric picture), as follows. Let $e = \prod A_i$ be an idempotent. The condition $e^2 = e$ gives that each component of $e$ is either $0$ or $1$, so $\prod_{i=1}^n A_i$ has exactly $2^n$ idempotents; this already gives $n = m$.

A nonzero idempotent is primitive if it can't be written as the sum of two other nonzero idempotents. The primitive idempotents of $\prod A_i$ are exactly those equal to $1$ in exactly one index and equal to $0$ otherwise, and an isomorphism $\prod A_i \cong \prod B_i$ sends primitive idempotents to primitive idempotents. Given an idempotent $e$ in a commutative ring $R$ we can consider the quotient of $R$ by $1 - e$, which can naturally be identified with $eR$; the quotients corresponding to primitive idempotents in $\prod A_i$ recover each of the factor rings $A_i$, so an isomorphism $\prod A_i \cong \prod B_i$ sends each $A_i$ to some $B_{\sigma(i)}$ as desired.

Title: Uniqueness of ring product


问:
Use Cauchy's Residue Theorem to solve for $\displaystyle{\frac{1}{2\pi i} \oint_{|z|=1} \frac{\cos \left(e^{-z}\right)}{z^2} dz}$
I know how to solve for the basic about this.  But in this problem

$\displaystyle{\frac{1}{2\pi i} \oint_{|z|=1} \frac{\cos \left(e^{-z}\right)}{z^2} dz}$,

I don't know how to start.  Can somebody help me or guide me about this? Or just give me a hint.  Thanks.  I would appreciate the help.  Thanks.

答:
You could also do this using the residue theorem directly, as the headline asks for. I wish to emphazise that I am aware that the calulation is completely analogous, and that I simply show this as an alternative since it appears from the comments and the headline that the asker might be more comfortable with the residue theorem as the starting point here. I urge him however, to think about why these two approaches are equivalent in this case.

Observe that $f(z) = \frac{\cos(e^{-z})}{z^2}$ has a pole of order 2 at $z = 0$ with residue

$$\text{Res}(0) = \lim_{z \rightarrow 0} \frac{d}{dz} z^2 f(z) = \lim_{z \rightarrow 0} \sin(e^{-z})e^{-z} = \sin(1).$$ By the residue theorem we then have

$$\frac{1}{2\pi i} \int_{|z|=1} \frac{\cos(e^{-z})}{z^2} dz = \sin(1) $$


问:
Showing a polynomial is not a subspace
I'm working on a question that asks me to show that the set $S = \{(x,y,z) \in \mathbb R^3 : x + 2y + 3z = 1\} $ is not a subspace of $\mathbb R^3$ under the regular addition and scalar multiplication. I understand that to prove a subspace I have to show that it is closed under addition and under scalar multiplication. Or I show that there exists a 0 such that 0 + u = u and there exists a negative of u such that u + (-u) = 0.

I can't figure out a way to show that one of them doesn't work. Could someone help?

答:
If $W$ is a subspace of $V \implies$ $\forall w_1,w_2 \in W$ the vector $w_1+w_2 \in W$.

Let $W=S=\{(x_1,x_2,x_3):x_1+2x_2+3x_3=1\}$ and $V=\mathbb R^3$.

If we now take $w_1=(2,1,-1)$ and $w_2=(0,-1,1)$, they belong to $S$ but $w_1+w_2=(2,0,0)\notin S$.

This shows that $S$ can't be a subspace of $\mathbb R^3.$

Title: Showing a polynomial is not a subspace


问:
Rotating the graph of a function.
The problem is to rotate a function f(x) by 90 degrees and obtain a new rotated function g(x). I've tried to use the rotation matrix and the formula g(x,y) =(f(xcos(a) - y sin(o)) - xsin(o) ) / cos(o) and they seem to work, but I don't really end up with a form of g(x).
Is there even a way to do that, because on the internet I only read that the rotated graph isn't a function anymore?

In the attached pictures the problem is visualized. In the first figure the original function f(x) and in the second picture the wished rotated function is shown.

original function f(x)

rotated function g(x)

答:
You don't give your $f(x)$ so it is impossible to be precise, but it appears that $f(x) = -2 \tanh (2 x)$:

in which the "rotated" version is simply the inverse function (with a minus sign), i.e., $g(x) = 1/2 \tanh^{-1} (x/2)$:

Title: Rotating the graph of a function.


问:
Integral of $\int \frac{\sin(x)dx}{3-\cos(x)}$
I am trying to solve this integral and I need your suggestions.

I don't know if its OK to set $3-\cos(x)$ as $t$ $\rightarrow dt = \sin(x)dx$ or just take $\cos(x)$ and set it as $t$
$$\int \frac{\sin(x)dx}{3-\cos(x)}$$
Thanks

答:
Hint: if $\,f\,$ is a derivable function, then

$$\int\frac{f'(x)}{f(x)}dx=\log|f(x)|+C(=\text{ a constant})$$


问:
Composition of Riemann integrable function and C1 function is Riemann integrable
Let $f$ be a Riemann integrable function on $[a,b]$ and $g: [\alpha, \beta] \to [a, b]$ be a continuously differentiable function. Then how can I prove that $f \circ g$ is Riemann integrable? I think this is an easy task, but I can't get it. Maybe uniform continuity of $f$(+uniform continuity of $f'$) seems to be crucial, but I can't proceed any more.

There are no more assumptions about $f$ other than stated. According to the textbook containing this(Krantz, Real Analysis and Foundations, 4th ed, sec 7.3. It is an elementary real analysis textbook), this is an 'easier result' than proving that 'composition $\phi \circ f$ of continuous function $\phi$ on compact interval and Riemann integral function $f$ on $[a,b]$ is Riemann integrable'. I would appreciate answers not using measure theory if possible.

答:
This is false. Let $\alpha=a=0, \beta = b = 1.$ Define $f=\chi_{(0,1]}.$ Then $f$ is Riemann integrable on $[0,1].$

To define $g,$ we use the following fact: Given a closed subset $E\subset \mathbb R,$ there is a $C^\infty$ function $g:\mathbb R\to [0,1]$ such that $g=0$ on $E$ and $g>0$ everywhere else.

With that in mind, choose a nowhere dense compact set $K\subset [0,1],$ with $m(K)>0.$ Use $E=K$ in the last paragraph to obtain $g\in C^\infty$ with the properties above.

Claim: $f\circ g$ is discontinuous at each $x\in K.$

Proof: Let $x\in K.$ Because $K$ is nowhere dense, there exists a sequence $x_n \in [0,1]\setminus K$ with $x_n\to x.$ We then have $g(x_n)>0$ for all $n.$ Hence $f\circ g(x_n) = 1$ for all $n.$ But $x\in K$ implies $g(x)=0,$ which gives $f\circ g(x)=0.$ This gives the claim.

Now if $f\circ g$ were Riemann integrable on $[0,1],$ then its set of discontinuities would have measure $0.$ But the claim implies $f\circ g$ is discontinuous on a set of positive measure, namely $K.$ This is a contradiction, proving $f\circ g$ is not Riemann integrable


问:
Proving a polynomial splits fully in a certain field
I have a polynomial $f(x)=x^6+3$, and define $a$ a root of $f(x)$. Then construct the field $\mathbb{Q}(a)$. I need to prove that $g(x)=x^3+3$ splits completely over $F(a)$. It hints at the fact that $a^2$ is a root of $g(x)$ and $a^3$ is a root of $x^2+3$. However I'm at a complete loss of how to use this information. I can gather that $(x-a^2)(x^2+a^2x+a^4)$ is a factorization of $g(x)$, but I don't think this is the right approach.

Any hints would be appreciated, thanks in advance.

答:
The roots of $g$ are $a^2, \zeta a^2$, and $\zeta^2 a^2$ where $\zeta$ is a third root of unity, so it suffices to show that $\zeta \in \mathbb{Q}(a)$. Recall that $\zeta = \frac{-1 + \sqrt{-3}}{2}$. Since $a^6 = -3$, then $a^3 = \sqrt{-3}$ (this is the second part of your hint), so
$$
\zeta = \frac{-1 + \sqrt{-3}}{2} = \frac{-1 + a^3}{2} \in \mathbb{Q}(a) \, .
$$