问:
Defining isometric functors?
Does such a thing exist? This would require a metric or distance between two objects of a generic category, but I haven't heard of this.

Thoughts: the required compositionality (transitivity) of a category means that the minimum distance between objects is a bit trivial: you can always short-cut. Perhaps one can define a less trivial notion of distance on the transitive reduction of a category, by which I mean:

For any category $C$ perform the following sequence,

take the preorder associated with $C$

construct a partial order on the quotient set of the equivalence, C / ~, the set of all equivalence classes of ~. (Let any two distinct members of the same equivalence class have zero distance from one another).

examine the minimum (undirected?) distance between any two objects on transitive reduction of this partial order.

(In graph theory, the transitive reduction of a directed graph D has the same reachability relation as D. Equivalently, D and its transitive reduction should have the same transitive closure as each other, and its transitive reduction should have as few edges as possible among all graphs with this property. )

Application

Consider functors F, G from the category with two objects and one non-identity morphism
$X \xrightarrow{f} Y$
to the category with three objects
$X' \xrightarrow{f'} Y'\xrightarrow{g'} Z' $.

Let the functor F take $f$ to $f'$, and the functor G take $f$ to $g' \circ f'$. Both F and G are fully faithful but intuitively F is more "veridical" or "isometric". Fullness and faithfulness do not say it all.  It is also important to consider what the distance would be, had the target category had an additional (non-composite) arrow $X' \xrightarrow{q} Z'$, and G instead mapped $f$ to $q$.

答: No, there is no notion of distance between two objects in an arbitrary category, though you’re incorrect that you can always shortcut in a category. (Consider the category $a\to b$, with only identity morphisms pictured.) Picking a transitive reduction gives a badly ill defined notion of distance, though in principle there’s nothing stopping you if such a reduction exists. There is a kind of enriched category that generalizes metric spaces, in which the hom-set is replaced with a real number and associativity is replaced with the triangle inequality.

Title: Defining isometric functors?


问:
Is there a geometric proof of the de Moivre's formula?
I am trying to understand (intuitively) this formula here:
$$e^{ic} = cos (c) + i.sin (c)$$

I understand the infinite sum (traditional) approach 1, but I am looking for something more geometric, maybe because of the involvement of the trig functions. I found another approach 2 which uses the fact that (assuming $f(x)=e^{ix}$):
$$f'(x)=i.e^{ix}=i(g(x)+i.h(x))=i.g(x)-h(x) $$

Since the pair of functions $g$ and $h$ for which $h' = g$ and $g' = -h$ happen to be sine and cosine, we get the proof.

I know that exponentiation involving imaginary numbers are more easily dealt with using the power series, but is there a more visual approach to this?

答:
Complex numbers admit the matrix representation $x+yi=\left(\begin{array}{cc}
x & -y\\
y & x
\end{array}\right)$. They can then be seen as a generalization of $2\times2$ rotation matrices that also including scaling. Writing rotation matrices as something exponentiated then just means anticlockwise rotations by $\theta,\,\phi$ compose to an anticlockwise rotation by $\theta+\phi$. Why the base is $e^{i}$ is the hard part, which I think needs Taylor series, albeit seen again in terms of matrices.

ContentLicense: CC BY-SA 4.0


问:
Uniformly $L^{q}$ bounded and convergence in probability implies $L^{p}$ convergence
Let $(X_{n})_{n\in \mathbb N}$ be some random variables such that $\sup\limits_{n \in \mathbb N}E[\lvert X_{n}\rvert ^{q}]<\infty$ and $X_{n}\xrightarrow{\mathbb P}X.$

Show that $X_{n} \xrightarrow{L^{p}}X$ for $p < q$

My idea:

By Fatou's Lemma, since convergence in probability implies a.s. convergence of a subsequence then w.l.o.g let $(X_{n})_{n \in \mathbb N}$ be the sequence.

Then
$$E[\lvert X\rvert^{q}]=E[\liminf\limits_{n \to \infty}\lvert X_{n}\rvert^{q}]\leq \liminf\limits_{n \to \infty}E[\lvert X_{n}\rvert^{q}] \leq \sup\limits_{n \in \mathbb N} E[\lvert X_{n}\rvert ^{q}]<\infty$$

and thus $\sup\limits_{n \in \mathbb N}E[\lvert X-X_{n}\rvert^{q}]<\infty$

Furthermore let $\varepsilon > 0$ and by Hölder:

$$E[\lvert X-X_{n}\rvert ^{p}]=E[\lvert X-X_{n}\rvert ^{p},\lvert X-X_{n}\rvert\leq \varepsilon]+E[\lvert X-X_{n}\rvert ^{p},\lvert X-X_{n}\rvert> \varepsilon]\\ \leq \varepsilon^{p}+E[\lvert X-X_{n}\rvert^{q},\lvert X-X_{n}\rvert> \varepsilon]^{\frac{p}{q}}E[\lvert X-X_{n}\rvert^{\frac{pq}{q-p}},\lvert X-X_{n}\rvert> \varepsilon]^{\frac{q-p}{q}}\\\leq \varepsilon^{p}+\left(\sup\limits_{m \in \mathbb N}E[\lvert X-X_{m}\rvert^{q}]\right)E[\lvert X-X_{n}\rvert^{\frac{pq}{q-p}},\lvert X-X_{n}\rvert> \varepsilon]^{\frac{q-p}{q}}$$

I do not know how to bound $\lvert X-X_{n}\rvert^{\frac{pq}{q-p}}1_{\{\lvert X-X_{n}\rvert> \varepsilon\}}$ that I need to prove the proposition. Any ideas?

答:
I dont quite understand the step where you used Hölder's Inequality, i would use it as follows:

\begin{align}
E|X-X_n|^p&\leq \epsilon^p+E\bigl(|X-X_n|^p\cdot \mathbb{1}_{|X-X_n|>\epsilon}\bigr)\\
&=\epsilon^p+\int |X-X_n|^p\cdot \mathbb{1}_{|X-X_n|>\epsilon}\\
&\leq \epsilon^p+\biggl(\int |X-X_n|^q\biggr)^{p/q}\cdot \biggl(\int\mathbb{1}_{|X-X_n|>\epsilon}\biggr)^{1-p/q}\\
&\leq \epsilon ^p+P(|X-X_n|>\epsilon)^{1-p/q}\bigl(\sup_{n}E|X-X_n|^q\bigr)^{p/q}\\
&=\epsilon^p+C\cdot P(|X-X_n|>\epsilon)^{1-p/q}
\end{align}
where $C=\bigl(\sup_{n}E|X-X_n|^q\bigr)^{p/q}<\infty$. So now, using $P(|X-X_n|>\epsilon)\to 0$ as $n\to \infty$ we get
$$\limsup_{n\to \infty}E|X_n-X|^p\leq \epsilon^p$$


答:
It seems you are still having trouble.

The following is a proof by induction fleshing out the ideas from previous users:

It can be seen that it is possible that

$P(1) = 1$ for $n= 1$ and

$P(n) < n$ for all $n > 1$.

The case for $n=2$ is clear since $P(2) = P(1) + \frac{1}{2} + \frac{1}{3} = 1+ \frac{1}{2} + \frac{1}{3}   < 2$

Suppose $P(n) < n$  for some $n > 1$.

Then $P(n+1) = P(n) + \frac {1}{2^n} + \frac {1}{2^n+1}+...+\frac {1}{2^{n+1}-1} < n + \frac {1}{2^n+1}+...+\frac {1}{2^{n+1}-1} $.

Notice that $2^{n} + 2^{n} = 2^{n+1}$ so that it may be clear to you that there are $2^{n}$ of these terms. Then since the terms are decreasing,

$\frac {1}{2^n} + \frac {1}{2^n+1}+...+\frac {1}{2^{n+1}-1} < \frac{1}{2^{n}} +...+\frac {1}{2^n}<1. $

Therefore $P(n+1) = P(n) + \frac {1}{2^n} + \frac {1}{2^n+1}+...+\frac {1}{2^{n+1}-1} < n + \frac {1}{2^n+1}+...+\frac {1}{2^{n+1}-1} = n+1$.

Then this implies that $P(100) < 100$, $(A)$.

Title: For a positive integer $ n$, $ P(n)=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}.....+\frac{1}{2^n-1}$, then

Body:
Four given options are:
$$A) P(100)<100$$ $$B) P(100)>100$$
$$C) P(200)<100$$ $$D) P(100)=150$$

Despite much try I am unable to find which option holds true. I am just unable to understand the question.
$$1+(1/2+1/3)+(1/4+1/5+1/6+1/7)+(1/8+....+1/15)+....+1/{(2^n)-1}$$
= $$1+(1/2+1/2²-1)+(1/4+1/5+1/6+1/2³-1)+.....+1/{(2^n) -1}$$
What to do next ?


问:
Compute the spectral measure of an operator $(T\mathbf{a})_{n}=a_{n+1}-a_{n}$ on the space of square summable sequences
Consider the space $\mathcal{H}$ of square summable sequences $\mathbf{a}=\{a_{n}\}_{n=-\infty}^{\infty}$. The operator is defined by $$(T\mathbf{a})_{n}=a_{n+1}-a_{n}.$$

I want to compute the spectral resolution of this operator. From this website, https://encyclopediaofmath.org/wiki/Spectral_resolution, the spectrum measure $\mu$ can define resolution by $\mu((-\infty, t))$. So I guess the key is to get the spectral measure of this operator.

Edit 1: I think I got at least the spectrum

Given $\Psi\in\ell^{2}(\mathbb{Z})$, we can define $\hat{\Psi}\in L^{2}([0,2\pi))$ by $\hat{\Psi}(\xi):=\sum_{n\in\mathbb{Z}}e^{in\xi}\Psi_{n},$ with $$\Psi_{n}=\dfrac{1}{2\pi}\int_{0}^{2\pi}e^{in\xi}d\xi.$$

Then, the map $$U:\ell^{2}(\mathbb{Z})\longrightarrow L^{2}([0,2\pi),\dfrac{d\xi}{2\pi}),\ \ \Psi\longrightarrow_{U}\hat{\Psi},$$ is unitary.

We want to know what does $UTU^{-1}$ give us. Take $f(\xi)\in L^{2}([0,2\pi))$, then $$(TU^{-1}f)(n)=\dfrac{1}{2\pi}\int_{0}^{2\pi}(e^{-i(n+1)\xi}-e^{-in\xi})f(\xi)d\xi=\dfrac{1}{2\pi}\int_{0}^{2\pi}e^{-in\xi}(e^{-i\xi}-1)f(\xi)d\xi,$$ which implies that $$(UTU^{-1})(\xi)=(e^{-i\xi}-1)f(\xi).$$ Now we have the form of our operator in Fourier space, we are equipped to determine its spectrum.

Look at the resolvent:
\begin{align*}
R(T):=\{\lambda:(\lambda-T)\ \text{is invertible}\}&=\{\lambda:\lambda-e^{-i\xi}+1\neq 0\ \ \text{for all}\ \ \xi\in [0,2\pi)\}\\
&=\{\lambda:\lambda\neq e^{-i\xi}-1\ \ \ \text{for all}\ \ \xi\in [0,2\pi)\}\\
&=\mathbb{C}\setminus \{|\lambda+1|=1\}.
\end{align*}

This implies that the spectrum is $$\sigma(T)= \{|\lambda+1|=1\}.$$ We denote this set to be $\mathbb{S}_{1}^{*}(-1,1)$.

Am I correct?

Edit 2: I possibly get the spectral measure. But it confuses me

Now, for $f\in C(\mathbb{S}^{*}(-1,1))$ in Fourier space $f(T)$ is just multiplication by $f(e^{-i\xi}-1)$. Fix $\Psi\in\ell^{2}(\mathbb{Z})$, and let $\hat{\Psi}$ be the corresponding function in $L^{2}([0,2\pi))$. Then
\begin{align*}
\int f(T)d\mu_{\Psi}=\langle \Psi|f(T)\psi\rangle_{\ell^{2}}&=\langle U^{-1}U\Psi|f(T)U^{-1}U\Psi\rangle_{\ell{^2}}\\
&=\Big\langle U\Psi\Bigg|\Big(Uf(T)U^{-1}\Big)U\Psi\rangle_{L^{2}}\\
&=\langle \hat{\Psi}|f(e^{-i\xi}-1)\hat{\Psi}\rangle_{L^{2}}\\
&=\dfrac{1}{2\pi}\int_{0}^{2\pi}\overline{\hat{\Psi}}f(e^{-i\xi}-1)\hat{\Psi}d\xi\\
&=\dfrac{1}{2\pi}\int_{0}^{2\pi}f(e^{-i\xi}-1)|\hat{\Psi}(\xi)|^{2}d\xi.
\end{align*}

Write $\lambda:=e^{-i\xi}-1$, then $d\xi=\frac{d\lambda}{-e^{-i\xi}}$ and $\xi=i\log(\lambda+1)$ where $\log$ is taking over the principal branch (it's okay if just a general $\log$ since $\xi\in [0,2\pi]$). Then,
\begin{align*}
\int fd\mu_{\Psi}&=\dfrac{1}{2\pi}\int_{\mathbb{S}_{1}^{*}(-1,0)}f(\lambda)|\hat{\Psi}(i\log(\lambda+1))|^{2}\dfrac{d\lambda}{-e^{-i\xi}},
\end{align*}
therefore, $$d\mu_{\Psi}=|\hat{\Psi}(i\log(\lambda+1))|^{2}\dfrac{d\lambda}{-e^{-i\xi}}\Bigg|_{\mathbb{S}_{1}^{*}(-1,0)}.$$

According to this website, https://encyclopediaofmath.org/wiki/Spectral_resolution, the spectrum measure $\mu$ can define resolution by $\mu((-\infty, t))$. But you see my spectrum density is restricted to the unit circle. I am not sure how to go back to the real line.

Are the above things correct? How can I proceed?

答:
By definition a spectral resolution  on a Hilbert space $H$,
defined on a measurable space $(X, \mathscr B)$, is a function
$$
  P:\mathscr B \mapsto  B(H),
  $$
such that

$P(E)$ is a projection for each $E$ in $\mathscr B$,

$P(\emptyset) = 0$, and $P(X) = I$,

whenever $\{E_n\}_{n\in \mathbb N}$ is a pairwise disjoint collection of measurable sets, one has that
$$
  P\left(\bigcup_{n\in \mathbb N}E_n\right) = \sum_{n\in \mathbb N}P(E_n),
  $$
the sum converging in the weak operator topology.

A well known form of the spectral theorem for bounded normal operators
asserts that, given any such operator $T$, there exists a spectral resolution $P$ defined on the Borel subsets of $\sigma (T)$ such
that
$$
  T=\int_{\sigma (T)} \lambda  \, dP(\lambda ).
  \tag 1
  $$
This integral is supposed to be interpreted in the weak sense, meaning that for every $\xi $ and $\eta $ in $H$, one
has that
$$
  \langle T\xi , \eta \rangle =\int_{\sigma (T)} \lambda  \, d P_{\xi , \eta }(\lambda ),
  $$
where $P_{\xi , \eta }$ is the actual (complex) measure on $\sigma (T)$, defined for every Borel set $E$ by
$$
  P_{\xi , \eta }(E)=  \langle P(E)\xi , \eta \rangle.
  $$

When $T$ is a self-adjoint operator, it is well known that $\sigma (T)\subseteq \mathbb R$, so its spectral resolution $P$ gives
rise to a so-called cummulative distributiuon, namely
$$
  \nu :t\in \mathbb R \mapsto  P(\sigma(T)\cap (-\infty ,t)) \in  B(H).
  $$
The importance of $\nu $ is that the integral in (1) can be expressed as a sort of Stieltjes integral
$$
  T=\int_{-\infty }^\infty  \lambda  \, d\nu (\lambda ).
  $$

For that reason, some authors prefer to emphasize $\nu $ over $P$, perhaps also because students have not yet learned any measure
theory at this point, but have already seen the Riemann-Stieltjes integral.  However I think  this is
not such a good idea because it has no counterpart for normal operators.

Finally,  if  $H=L^2(X, \mathscr B, \mu )$,  and $T$ is the so called multiplication operator defined on $H$ by
$$
  T(\xi )|_x = \varphi (x)\xi (x),  \quad \forall \xi \in  H,  \quad \forall x\in  X,
  $$
where $\varphi $ is a fixed bounded measurable complex valued function on $X$,  one may prove that the spectrum of $T$ is the essential range of
$\varphi $, and the spectral resolution provided by the spectral theorem is such that, for every $E\subseteq \sigma (T)$,
$P(E)$
is the orthogonal projection on the subspace $L_2(\varphi^{-1}(E))$.

Regarding the case in point,  we have that  $X=[0, 2π)$  and $φ(t) = e^{-it}+1$,  so the spectral resolution may be  obtained
as above.


问:
Why is it that $[\alpha^n] \ge \alpha^n/2$ for $n \ge 1$?
I'm reading through the proof of Etemadi's law of large numbers, and one of the inequalities that are used states that

$[\alpha^n] \ge \alpha^n/2$ for $n \ge 1$.

I seem to be perplexed by this inequality. Is this proven inductively? How would one make sense of the double bracket that would occur?

答:
Note the inequality is not true if $\alpha^n\lt1$, so presumably there is a tacit assumption that $\alpha\ge1$. In any event, we may as well prove that $\lfloor x\rfloor\gt x/2$ for all $x\ge1$ (and then substitute $x=\alpha^n$). Since $\lfloor x\rfloor\gt x-1$ always holds, we have, for $x\ge1$,

$$2\lfloor x\rfloor=\lfloor x\rfloor+\lfloor x\rfloor\gt(x-1)+1=x$$

so $\lfloor x\rfloor\gt x/2$.


问:
Perfect squares using 20 1's, 20 2's and 20 3's.
How many perfect squares can be formed using 20  1's, 20  2's and 20  3's. This is a recent exam question, which I had no clue how to solve?

There is some kind of trick here, since time allotted to solve it was just 4 minutes.

ContentLicense: CC BY-SA 3.0

Body:
How about this:  $$\begin {array} &1&2&3&1&2&3&1&2&3&1&2&3&1&2&3&1\\
2&&&&&&&&&&&&&&&2\\
3&&&&&&&&&&&&&&&3\\
1&&&&&&&&&&&&&&&1\\
2&&&&&&&&&&&&&&&2\\
3&&&&&&&&&&&&&&&3\\
1&&&&&&&&&&&&&&&1\\
2&&&&&&&&&&&&&&&2\\
3&&&&&&&&&&&&&&&3\\
1&&&&&&&&&&&&&&&1\\
2&&&&&&&&&&&&&&&2\\
3&&&&&&&&&&&&&&&3\\
1&&&&&&&&&&&&&&&1\\
2&&&&&&&&&&&&&&&2\\
3&&&&&&&&&&&&&&&3\\
1&2&3&1&2&3&1&2&3&1&2&3&1&2&3&1\\
\end {array}$$

If done in a properly spaced font, it looks like a perfect square to me.  The sum along each side is the same


问:
submatrix of signed incidence matrix of a graph containing a cycle

Let $G=(V,E)$ be a (simple) graph. Write $V = \{v_1,\cdots, v_n\}$ and $E = \{e_1,\cdots, e_m\}.$ Let $I(G)$ be the incidence matrix of $G$ (i.e. a $|V|\times |E|$ matrix whose entries $(i,j)$ are $1$ if vertex $i$ is incident with edge $j$ and $0$ otherwise). Let $S(G)$ be any signed incidence matrix of $G$, obtained by arbitrarily switching one of the two $1$'s in each column to a $-1$ (say edges start at the rows with $-1$ in the column corresponding to those edges).

Prove that if $F = \{f_1,\cdots, f_k\}$ is a subset of $E(G) = E,$ then $F$ corresponds to a linearly dependent set of columns of $S(G)$, denoted $S_F$, iff the subgraph $G_F = (V_F, F)$ of $G$ contains a cycle, where $V_F = \{v \in V : v\cap e \neq \emptyset\}$ for some $e\in F$.

I think I know how to prove that reverse implication, but I'm not sure how to prove the forward one. Below is what I've come up with so far.

Now suppose $F $ corresponds to a linearly dependent set of columns of $S(G).$ We construct a cycle in the graph $G_F$ by induction. For the base case, note that if $F$ is linearly dependent, we necessarily have that $|F| \geq 3$. Clearly it doesn't hold if $|F| = 1$, as $\{f_1\}$ is linearly independent, and if $|F| = 2,$ then the columns corresponding to $f_1$ and $f_2$ are linearly independent because otherwise the two edges $f_1$ and $f_2$ would have the same endpoints, which contradicts the fact that they are distinct edges (indeed a componentwise matching of the columns corresponding to $f_1$ and $f_2$ with the zero vector shows that if $c_1$ and $c_2$ are constants so that $c_1 col(f_1) + c_2 col(f_2) = 0, c_1 = c_2 = 0$, where $col(f_i)$ is the column index corresponding to $f_i$). So $|F|\geq 3,$ and we can find constants $y_1,\cdots, y_k,$ not all zero, so that $y_1 col(f_1)+\cdots y_k col(f_k) = 0.$ Each row of $S(G_F)$ must have two or $0$ nonzero entries, as otherwise there would be at least one row with only one nonzero entry in a column $f_i$, so  $y_i$ would need to be zero, and we know that the two remaining vectors are linearly independent by the above reasoning. So the base case holds. Now assume for some $k\geq 3$ that if $F$ corresponds to a linearly dependent set of columns, the subgraph $G_F$ contains a cycle. Let $F' = \{f'_1,\cdots, f'_{k+1}\}$ be a set of cardinality $k+1$ so that $F'$ corresponds to a linearly dependent set of columns of $S(G).$ Thus by definition we can find a column $c_{f'_e}$ that is a nontrivial linear combination of other columns, say columns $c_{f'_1},\cdots, c_{f'_l}.$

The problem I'm having is that removing a column may very well result in a linearly independent set of columns, which means I can't use the inductive hypothesis.

答:
I think it is easier to show this directly, rather than via induction. If $S_F$ is linearly dependent, then we can choose vectors $c_1, \dots, c_k \in S_F$, corresponding to edges $f_1, \dots, f_k \in F$, such that there exists a linear combination $\sum \alpha_i c_i = 0$. We may assume that $\alpha_i \neq 0$ for all $i$; otherwise, we just leave out the corresponding vector.

Now in the subgraph of $G$ induced by $f_1, \dots f_k$ every vertex must have degree at least two, for otherwise the corresponding coordinate could not sum to $0$ in the above linear combination. (You have made essentially the same observation in the base case of your induction proof idea.) But a graph with minimum degree two always contains a cycle; or said another way, a graph without a cycle – a forest – always contains a vertex of degree one – a leaf. (Unless it has no edges at all, in which case the minimum degree is even lower.)


问:
definition of finitely generated $A$ algebra, $A$ a ring
Let $A$ be a ring.
I'm a bit confused with definition of $A$ algebras and I would appreciate any clarification.
Let $A \subset B$ rings and suppose $B$ is a finitely generated $A$ algebra.
Does this mean there exist $x_1, .., x_n \in B$ for some $n$ such that
$B = A[x_1, .., x_n]$ ?

This is not clear to me looking at the definition in Wikipedia https://en.wikipedia.org/wiki/Finitely_generated_algebra for example.

答:
No. It only means that there exists $a_1, ... a_n \in B$ such that for every $x \in B$, there is a polynomial $f \in A[x_1, ... x_n]$ such that $f(a_1, ... a_n) = x$.

In other words, the "evaluation at $a_1, ... a_n$" homomorphism $A[x_1, ... x_n] \rightarrow B$ is surjective. By the first isomorphism theorem, this implies $B \cong A[x_1, ... x_n] / I$ for some ideal $I$ of $A[x_1, ... x_n]$. This ideal is the kernel of the evaluation homomorphism.


问:
How To Solve $\frac{1}{X}\bigg\lfloor-\frac{3}{2}+\frac{1}{2}\sqrt{8X+9}\bigg\rfloor=\frac{1}{N}$ over the integers.
Question: If $X$ and $N$ are positive numbers. How would I solve for $X$ in the following equation:

$$
\frac{1}{X}\bigg\lfloor-\frac{3}{2}+\frac{1}{2}\sqrt{8X+9}\bigg\rfloor=\frac{1}{N} \label{a}\tag{1}
$$

?

For example If $N=7$ then $X \in \{63,70,77\}.$ If $N=6$ then $X\in\{42,48,54\}.$ Careful with $N=4.$ The floor part of the LHS of $\ref{a}$ yields the sequence:
$$
0,1,1,1,2,2,2,2,3,3,3,3,3,4,4,4,4,4,4,5,5,5,5,5,5,5,\ldots;
$$
Here is a Wolfram Alpha "script" in the case $N=7$. You can change $7$ to another integer say $4$ etc..

```<code>Solve[Floor[-3/2 + (1/2) Sqrt[8 X + 9]]/X == 1/7, X, Integers]
</code>```

答:
Multiply by $X$ on both sides to get $$\left\lfloor-\frac{3}{2}+\frac{1}{2}\sqrt{8X+9}\right\rfloor=\frac{X}{N}$$

This shows that $X$ must be a multiple of $N$ and that $$\frac{X}{N} \le -\frac{3}{2}+\frac{1}{2}\sqrt{8X+9} < \frac{X}{N}+1$$

Multiply by $2$, add $3$, and square it to get $$\left(\frac{2X}{N}+3\right)^2 \le 8X+9 < \left(\frac{2X}{N}+5\right)^2$$

Expanding it, there are two inequalities: $$\frac{4X^{2}}{N^{2}}+\left(\frac{12}{N}-8\right)X \le 0$$ $$\frac{4X^{2}}{N^{2}}+\left(\frac{20}{N}-8\right)X+16 > 0$$

Assuming that $X, N > 0$, the first inequality means that $$X\le2N^{2}-3N \tag 1$$ and the second one means that $N<\frac{9}{2}$ or $$X<\frac{2N^{2}-5N-N\sqrt{9-20N+4N^{2}}}{2} \tag 2$$ or $$X>\frac{2N^{2}-5N+N\sqrt{9-20N+4N^{2}}}{2} \tag 3$$

$(2)$ would never be true because for $N \ge 5$, the upper bound on $X$ would be less than $N$ and $X$ has to be a positive multiple of $N$.

If $N < \frac{9}{2}$ (i.e. is in $\{ 1, 2, 3, 4 \}$), then $X = kN$, where $k$ is an integer in $[1, 2N-3]$. Otherwise, $X = kN$, where $k$ is an integer in $$\bigg(\frac{2N-5+\sqrt{9-20N+4N^{2}}}{2}, 2N-3\bigg]$$

Because $2N-5 - $ (the lower bound) $\le 1$ for $N \ge 5$, this means that this simplifies to $k \in \{2N-5, 2N-4, 2N-3\} \to X \in \{N(2N-5), N(2N-4), N(2N-3)\}$.


问:
Is it a standard to say that $a \oplus a_{\small 1}=0$ or $a \veebar a_{\small 1}=0$?
I am trying to express the following:

$a$ or $a_{\small 1}=0$ but only one of them equals zero.

so if $a=0$ then $a_{\small 1}\neq 0$ and if $a\neq 0$ then $a_{\small 1}=0$.

And I'm looking for a standard way to say that, from my little background in logic, I remember the logical operator xor which truth table fit my purposes. So is it ok to use any of the following notations?

$$a \oplus a_{\small 1}=0$$

$$a \veebar a_{\small 1}=0$$

答:
Both symbols are variously used to denote XOR

But it would need to be expressed thusly:

$$(a = 0) \oplus (a_1 = 0)$$

or

$$(a = 0) \veebar (a_1 = 0)$$

And NOT, e.g., $$a \oplus a_1 = 0$$


问:
If $\mathbb{K}$ is an extension of $\mathbb{Q}$ of degree $15$, which of these elements cannot belong to $\mathbb{K}$?
If $\mathbb{K}$ is an extension of $\mathbb{Q}$ of degree $15$, which of these elements cannot belong to $\mathbb{K}$ ?

I know that it has to do with minimal polynomial but I am unsure of how to determine them.

答:
The minimal polynomials are (A) $x^2+144$, (B) $x^{15}-3$, (C) $x^3-7$, and (D) $x^5-10$,

of degrees $2$, $15$, $3$, and $5$, respectively.

If $\alpha\in \Bbb K$, then $\mathbb Q(\alpha)$ is a subfield of $\mathbb K$ and therefore its degree is a factor of $[\mathbb K:\mathbb Q]$.

Only one of these degrees is not a factor of $15$, and that is the element that cannot belong to $\mathbb K$.


问:
Compute the following two cardinals: $\mathfrak{c}^{\mathfrak{c}^\mathfrak{c}}$ and $n^{\aleph _0} \aleph_0 ^n$
I'm really not sure how to go about this, for the first cardinal we essentially end up with $\mathfrak{c}^{2^\mathfrak{c}}$ but I have no idea where to go from there or even if that's the right step.

For the second one however I know that $\aleph_0 ^n=\aleph _0$, but I'm not sure about the other half. If I can prove that has to be = to $\aleph_0$ as well would that mean the proof is done as we'd end up with $\aleph_0 \aleph_0$, which we know is $\aleph_0$?

答:
Since $(A^B)^C\cong A^{B\times C}$, we have that $$\mathfrak c^{\mathfrak c^{\mathfrak c}}=\left(2^{\aleph_0}\right)^{\left(2^{\aleph_0}\right)^{2^{\aleph_0}}}=2^{\left(\aleph_0\cdot2^{\left(\aleph_0\cdot 2^{\aleph_0}\right)}\right)}=2^{\aleph_0\cdot2^{2^{\aleph_0}}}=2^{2^{2^{\aleph_0}}}$$

Your assertion on $\aleph_0^n$ is essentially correct (though you forgot to mention that $\aleph_0^0=1$). However for $2\le n<\aleph_0$ we have that $n^{\aleph_0}=2^{\aleph_0}>\aleph_0$. Therefore,
$$n^{\aleph_0}\cdot\aleph_0^n=\begin{cases}0&\text{if }n=0\\ \aleph_0&\text{if }n=1\\ 2^{\aleph_0}&\text{if }n\ge2\end{cases}$$


问:
Limit of convex increasing function
If $f$ is strictly increasing and strictly convex (or $f'>0$ & $f''>0$), then $$\lim_{x\rightarrow∞}{f(x)}=∞$$

Is this statement true?

If this statement is true, how can I prove?

答:
vadim123's answer is clinical, I like it. A more traditional proof would be:

For $\forall x_0$, consider for $a>0$, $f(x_0 + a)$'s Taylor expansion at $x_0$:
$$
f(x_0+a) = f(x_0) + af'(\xi_0)
$$
for some $\xi_0 \in (x_0,x_0+a)$. Since $f''>0$ implies $f'$ is strictly increasing, we have $f'(\xi_0) > f'(x_0)= b >0$, this gives us:
$$
f(x_0+a) > f(x_0) + ab.
$$
Now consider $f(x_0 + 2a)$'s Taylor expansion at $(x_0+a)$, similar argument as above gives:
$$
f(x_0+2a) = f(x_0+a) + af'(\xi_1)> f(x_0) + ab + af'(\xi_1) >  f(x_0) + 2ab
$$
Deduction yields:
$$
f(x_0 + na)> f(x_0) + nab.
$$
Back to $f$, for $a>0$ fixed:
$$
\lim_{x\to \infty} f(x)= \lim_{n\to \infty} f(x_0+ na) \geq \lim_{n\to \infty} \big( f(x_0) + nab\big) = \infty.
$$

Title: Limit of convex increasing function


问:
Taylor's Expansion II

Find the second degree Taylor polynomial for

$f(x,y) = e^{xy+2x+y+2}$

centered at $(x_0, y_0) = (-3, -2)$.

Hi, I'm wondering if anyone can help me with this Taylor Expansion question as I'm unsure on how to tackle it.

I know that I am supposed to get the answer of $-3-2y+2(y+2)^2+(x+3)(y+2)$

So far I have tried to take this approach:

finding $f(x,y)$ and $f(-3,-2)$:

$f(x,y)=e^{xy+2x+y+2}$

and hence $f(-3,-2)= e^0=1$

I have also found $f_x(x,y),f_y(x,y),f_{xx}(x,y),f_{yy}(x,y),f_{xy}(x,y)$ and the value of each using the boundaries given:

$f_x=0, f_y=-2, f_{xx}=0, f_{yy}=4, f_{xy}=1$

but here is where I get stuck. I know that there is a remainder theorem for the double differentiated terms but I'm not sure if this is what I should be looking at.

Any help is super appreciated.

答:
Since you are only asked for the second degree Taylor polynomial, the remainder term is not necessary here.

Recall the definition:

$$P_2(x,y) = f + f_x(x+3)+f_y(y+2) + \frac {f_{xx}}2(x+3)^2+f_{xy}(x+3)(y+2)+\frac {f_{yy}}2(y+2)^2$$

where all $f$ and all its derivatives are evaluated at the center $(-3,-2)$.

Hence

$$P_2(x,y) = 1-2(y+2)+(x+3)(y+2)+2(y+2)^2$$

which is equivalent to the answer.

Title: Taylor's Expansion II


问:
Real symmetric matrix of rank $n-1$ has a submatrix of rank $n-1$
Let $A$ be a symmetric real matrix of dimension $n \times n$ and rank $n-1$. Prove that there is a $k \in \{1,2,...n\}$ such that on deletion of the $k$th row and column the resulting matrix has rank $n-1$.

I think we would have to use adjugate of the matrix here since that is the space of all $(n-1) \times (n-1)$ submatrices, but I am not very sure of how to proceed

答:
Here is another (slightly) different way to use the fact that the adjugate is a rank $1$ symmetric matrix. If it has rank one and furthermore there does not exist some $k$ such that the determinant of the $(k, k)$th minor is nonzero, then the diagonal of the adjugate matrix (call it $A$) is all zeroes.

We know a rank one matrix is simply a matrix whose rows are all multiples of the same non-trivial vector (with at least one row being a nonzero multiple). Suppose the $i$th row of $A$ were nonzero. Then there exists some $j \neq i$ such that $A_{ij}$ (recall our assumption is that the diagonal elements are all zero). As the adjoint is symmetric, $A_{ji} \neq 0$. However, the $j$th row cannot be a multiple of the $i$th row (why?), a contradiction. So there cannot be a symmetric matrix of rank $1$ whose diagonal elements are all zero, and hence we conclude that at least one element of $A$'s diagonal (say the $k$th element) must be nonzero. Removing the $k$th row and column, of course, will yield an invertible $(n - 1) \times (n - 1)$ matrix.


问:
Shift function so it passes through a specific point.
I have a function that rappresents the average of a series of points.
This function is an approximation and once I have this, I need to shift it so it passes through a single given point.

So basically, take the given function, and shift it horizontally and vertically so it passes through a point.

the function is: y = b0+(b1*x)
where b0 = 1864, b1=-1.93, x=610

the point the function must pass through is: 558/778

Could you please explain how the above function needs to be edited in order to pass through the given point?

答:
Let $f\colon\mathbb R\to\mathbb R$ be any function and $(x_0,y_0)$ be any point in $\mathbb R^2$. The function $g\colon\mathbb R\to\mathbb R$ given by
$$
g(x) = f(x-x_0)+y_0-f(0)
$$
has as its graph the graph of $f$ where the point $(0, f(0))$ is shifted to $(x_0, y_0)$.

When $f(x) = b_0+b_1 x$, this yields
$$
g(x) = b_1(x-x_0)+y_0.
$$


问:
If $|G|=p^nq$, then $G$ contains a unique normal subgroup of index $q$
I'm trying to prove if $|G|=p^nq$ with $p\gt q$, primes, then $G$ contains a unique normal subgroup of index $q$.

I know by the first Sylow theorem that G has a Sylow p-subgroup $P$ with $[G:P]=q$. My problem is prove that it's unique.

I need help

Thanks in advance

答:
There is a Lemma saying for a subgroup $H$ of a finite $G$, if $[G:H]=q$ where $q$ is smallest prime dividing $|G|$, then $H$ is normal.

In our case, there is a subgroup of index $q$, which is some Sylow $p$-subgroup $P$. Notice that $p>q$, which implies $q$ is the smallest prime dividing $|G|$. So P is normal in G. Now any other subgroup of index $q$ must be a Sylow $p$-subgroup too, and must equal $P$ because $P$ is the unique Sylow $p$-subgroup.

It seems we have a stronger case about any subgroup of index q, not just normal ones.


问:
Solving this functional equation involving fixed polynomials
Let $f, g \in \mathbb R[x]$ be fixed polynomials of the same degree $\deg f = \deg g = n$.

Can you solve the functional equation $(f\circ y)(x) = g(x-y(x))$ explicitly for $y: \mathbb R\to \mathbb R$?

答:
Possible solution:

Let $f=g$ and let $y(x)=\frac{x}2$.

Then, $y(x)=x-y(x)$.


问:
Product of a subnormal subgroup and a normal subgroup
We all know that

(1) the product of a subgroup and a normal subgroup is a subgroup

(2) the product of two normal subgroups is a normal subgroup

Now my question is: Suppose $G$ is a group, $H$ is a subnormal subgroup of $G$ and $K$ is a normal subgroup of $G$ . Is $HK$ also a normal subgroup of $G$? I have no idea about it.

答:
No, $HK$ need not be normal (though it will be subnormal; see below)

A simple counterexample can be found with $p$-groups, since every subgroup of a $p$-group is subnormal. Let $G$ be the group
$$G = \langle x,y\mid x^{p^2}=y^{p^2}=[y,x]^{p^2}=[y,x,x]=[y,x,y]=e\rangle.$$
Elements of $G$ are of the form $x^ay^b[y,x]^c$, with multiplication given by
$$\left(x^ay^b[y,x]^c\right)\left(x^{\alpha}y^{\beta}[y,x]^{\gamma}\right) = x^{a+\alpha}y^{b+\beta}[y.x]^{c+\gamma+\alpha b}.$$

You can also think of the group as the multiplicative group of $3\times 3$ upper triangular matrices with coefficients in $\mathbb{Z}/p^2\mathbb{Z}$ and ones in the diagonal. The element $x$ corresponds to the matrix that also has a $1$ in the $(2,3)$ entry (and zeroes off the diagonal elsehwere); $y$ to the element that has a $1$ in the $(1,2)$ entry, and $[y,x]$ the element that has a $1$ in the $(1,3)$ entry.

The subgroup $K$ generated by $[y,x]^p$ is central, hence normal. The subgroup $H$ generated by $y$ is subnormal because $G$ is a $p$-group. The subgroup $HK$ consists of elements of the form $y^b[y,x]^{pc}$, $0\leq b\lt p^2$, $0\leq c\lt p$, and is abelian. But this subgroup is not normal, since it contains $y$, but it does not contain $x^{-1}yx = y[y,x]$.

It will, however, be subnormal. If $H\triangleleft N_1\triangleleft N_2\triangleleft\cdots\triangleleft N_k=G$ is a subnormal series for $H$, and $K$ is normal, then $HK\triangleleft N_1K\triangleleft\cdots\triangleleft N_kK=G$ is a subnormal series for $HK$.

To verify this, say $A\triangleleft B$. Then $AK$ and $BK$ are subgroups, and $AK\leq BK$. Given $ax\in AK$, $a\in A$, $x\in K$; and $by\in BK$, $b\in B$, $y\in K$, then $(by)^{-1}ax(by) = y^{-1}(b^{-1}ab)(b^{-1}xb)y$. Now, $b^{-1}xb\in K$, since $K$ is normal; $b^{-1}ab\in A$, since $A\triangleleft B$. So this element lies in $KAK$. But since $K$ is normal, $KA=AK$, so $KAK=AK$. Thus, $(by)^{-1}(ax)(by)\in AK$, proving $AK\triangleleft BK$, as claimed.


问:
Show: $\log(1-\frac{1}{n}) + \epsilon > - \frac{1}{n}$ for all but finitely many $n$
I am trying to show that, for $\epsilon > 0$, we have $\log(1-\frac{1}{n}) + \epsilon >  - \frac{1}{n}$
for all but finitely many $n$. I was thinking about using Borel-Cantelli lemma but I don't know where I should start. I would appreciate any suggestions.

答:
There is a power series,
$$\log(1-x) = -\Big( x + \frac{x^2}{2} + \frac{x^3}{3} + \cdots \Big)
$$
which converges for all $-1 < x < 1$. It can be derived by integrating the geometric series $1/(1-x) = 1+x+x^2+\cdots$.

Then for $0 < x < 1$,
$$
0 < \frac{x^2}{2} + \frac{x^3}{3} + \cdots < \frac{x^2}{2}\Big( 1 + x + x^2 + \cdots\Big) = \frac{x^2}{2(1-x)}.$$
It should now be plain that the right hand side converges to zero as $x \to 0$ and we can find some $x_0$ that ensures it is less than $\varepsilon$ for all $x < x_0$.  Accordingly for all $n > 1/x_0$ the same holds with $x$ replaced by $1/n$.  For such $n$, it follows
$$
-\log\Big(1-\frac{1}{n}\Big) - \frac{1}{n} < \varepsilon
$$
and noting that all but finaitely many $n$ will exceed $1/x_0$, you have your result.


问:
Finding HCF with the Euclidean Algorithm
Using the Euclidean algorithm, find $\mathrm{hcf}(86, 100)$, and use this to find integers $s, t$ such that
$\mathrm{hcf}(86, 100) = 86 · s + 100 · t$.

I have that the HCF is 2 but have forgotten what to do to find $s$ and $t$, any help would be appreciated.

Body:
We will use the Euclidean Algorithm to find $\gcd(86, 100)$. We have:

$$\gcd(86, 100)$$
$$\gcd(86, 100 - 86)$$
$$\gcd(86 - 6(100 - 86), 100 - 86) = \gcd(7(86) - 6(100), 100 - 86)$$
$$\gcd(7(86) - 6(100), (100 - 86) -6(7(86) - 6(100))) =\gcd(7(86) - 6(100), 37(100) - 43(86))$$

Note that $7(86) - 6(100) = 37(100) - 43(86) = 2$. Thus:

$$\boxed{\gcd(86, 100) = 2, (s, t) = (7, -6)}$$


问:
Prove that $x^2\equiv a\pmod {2^m}$ has exactly 4 solutions if $a\equiv 1\pmod8$.
An excercise 11.6.17 of "Number Theory: A Lively Introduction With Proofs Applications and Stories" involves prooving the following theorem:

Let $a,m\in \Bbb Z, m \geq 3$. The equation $x^2\equiv a\pmod {2^m}$ has exactly 4 solutions if $a\equiv 1\pmod8$ and no solutions if $a\not\equiv 1\pmod8$.

There is hint for this problem with the outline of the proof I used to follow:

Show that if $x\in \Bbb Z$ - odd, then $x^2\equiv 1\pmod8$. This one is easy to show as $8|x^2 -1 =(x - 1)(x + 1)$ for odd $x$.

Using 1. show that $x^2\equiv a\pmod{2^m}$ has no solutions if $a\not\equiv1\pmod8$. That's also clear for me as if $x^2\equiv 1\pmod8$ then so does $a\equiv 1\pmod8$.

Now use induction on $m$ to show that $x^2\equiv a\pmod{2^m}$ has exactly 4 solutions if $a\equiv 1\pmod8$. I'm starting from $m=3$ and showing directly that the entire solution set is $\{1, 3, 5, 7\}$.

Then by induction hypothesis assume that $x^2\equiv a\pmod{2^m}$ has exactly 4 solutions. Considering $x_1$ as solution it's easy to show that $x_1 + 2^{m-1}$ is also a solution (no problem with that), so the entire set of solutions is $\{x_1, x_1 + 2^{m-1}, x_2, x_1 + 2^{m-1}\}$.

Show that one of $x_1$ or $x_1 + 2^{m-1}$ solves $x^2\equiv a\pmod{2^{m+1}}$ (and so for $x_2$ or $x_2 + 2^{m-1}$).

Here is the step I'm stuck on. While I can see it clearly on numerical examples, I have no idea how to show it in general. I've tried substuting $x_1$ and $x_1 + 2^{m-1}$ in $x^2\equiv a\pmod{2^{m+1}}$ but it doesn't help much. But if I assume it's true, then I can finish the proof as follows:

Since we have two solutions for $x^2\equiv a\pmod{2^{m+1}}$ from 5. we can get the other two by adding $2^m$ as in 4.

Then since every solution of $x^2\equiv a\pmod{2^{m+1}}$ also solves $x^2\equiv a\pmod{2^{m}}$, and the last has exactly 4 solutions by induction hypothesis, it means that $x^2\equiv a\pmod{2^{m+1}}$ itself has exactly 4 solution. Thus, induction hypothesis is true and we are done.

So I'm asking for help with step 5. Thank you in advance for any thoughts.

BTW. this problem is from section that regards Jacobi Symbol. But actually I'm not sure how Jacobi Symbol could be applied here since we working with the even modulus.

Body:
Since $x_1$ is odd and $m \ge 3$:

$$(x_1 + 2^{m-1})^2 = x_1^2+2^mx_1+2^{2m-2} \equiv x_1^2+2^mx_1 \equiv x_1^2+2^m \pmod {2^{m+1}}$$

We also have $x_1^2 \equiv (x_1 + 2^{m-1})^2 \equiv a \pmod {2^m}$. Hence $x_1^2 \equiv a$ or $a+2^m \pmod {2^{m+1}}$.

If $x_1^2 \equiv a \pmod {2^{m+1}}$ we are done.

If $x_1^2 \equiv a+2^m \pmod {2^{m+1}}$ then

$$(x_1 + 2^{m-1})^2 \equiv x_1^2+2^m  \equiv a+2^m+2^m \equiv a \pmod {2^{m+1}}$$

Hence either $x_1$ or $x_1 + 2^m$ solves $x^2 \equiv a \pmod {2^{m+1}}$.


答:
If you use the Euclidean algorithm to find the hcf, reversing the steps will find you a solution.  Thus start with

$$86 s_1 + 100 t_1 = 2$$

The first step is to write $100 = 86 + 14$, so this becomes
$$ 86 (s_1 + t_1) + 14 t_1 = 2$$
which you rewrite as $86 s_2 + 14 t_1 = 2$, where $s_2 = s_1 + t_1$.  Then write $86 = 6 \cdot 14 + 2$, and so
$$ 2 s_2 + 14 t_2 = 2 $$
where $t_2 = t_1 + 6 s_2$.  At this point you can see that $s_2 = 1$, $t_2 = 0$ is a solution. Now go back and calculate $t_1$ and $s_1$.

Title: Finding HCF with the Euclidean Algorithm

Body:
Using the Euclidean algorithm, find $\mathrm{hcf}(86, 100)$, and use this to find integers $s, t$ such that
$\mathrm{hcf}(86, 100) = 86 · s + 100 · t$.

I have that the HCF is 2 but have forgotten what to do to find $s$ and $t$, any help would be appreciated.


答:
Indeed we have $86=2\cdot43$ and $100=2^{2}\cdot5^2$, so $\text{hcf}(86,100)=2.$

Next we apply the Euclidean algorithm $$100=1\cdot 86 + 14$$
$$86=6\cdot 14+2$$
$$14=2\cdot7+0$$

and we stop since we have $0$ remainder and indeed $\text{hcf}(100,86)=2$. Then we have $100-1\cdot86= 14$ from the first equality above and substituting into the second gives $$86=6\cdot 14+2=6\cdot(100-1\cdot86)+2=6\cdot100-6\cdot86+2$$

and re-arranging we obtain $$-6\cdot100+7\cdot86=2$$


问:
Product of N independent uniform random variables
We have N independent random variables of $X_1 , X_2 , ... X_N$ with uniform distribution $U(0,1)$. the distribution of $Y$ as
$$
Y = \frac{1}{X_1X_2X_3 ... X_N}
$$
is desired.

To solve this problem, I write:
$$
T = \ln(Y) = - \sum_{i=1}^N \ln(X_i) = \sum_{i=1}^N Z_i
$$
where $ Z_i =-\ln(X_i) $. Then the distribution of $Z_i$ random variables could be written as
$$
f_{Z_i}(Z_i) = \frac{f_{X_i}(e^{-Z_i})}{|e^{Z_i}|} = e^{-Z_i}f_{X_i}(e^{-Z_i})
$$
But I do not know how to solve it any further. Would be helpful to utilize its Characteristic function in order to rewrite the sum of natural-log sum of random variables? Or Should I solve it in a different way?

Thanks for your help in advance.

答:
As you noted,

$$logY=\Sigma_i (-logX_i)$$

$$W_i=-logX_i\sim Exp(1)$$

Thus $log(Y)=\Sigma_i W_i\sim Gamma(n;1)$

Now, knowing the distribution of $logY$ it is easy to find the requested density

$$f_Y(y)=\frac{log^{n-1}y}{\Gamma(n)y^2}\cdot \mathbb{1}_{(1;+\infty)}(y)$$

P.S.: I used the notation $log$ but Natural logarithm is meant


问:
Application of Mean Value/Rolle's Theorem?
Question

Let $f$ be a differentiable function such that $f'$ is continuous on $[0, 1]$ and $M$ the maximum value of $|f'(x)|$ on $[0, 1]\ $. Prove that, if $f(0) = f(1) = 0$, then $$\int^{1}_{0} |f(x)| \mathrm {d}x \leq \frac M 4\ .$$

I have stared long and hard at this question for quite some time, but I am not sure where to even begin. Since I see that the endpoints are given, I think of MVT, but since they are also equal, I think of Rolle's Theorem too. These are just my intuitions - they could be wrong. Moreover, the moduli of $f'(x)$ and $f(x)$ are throwing me off. Any hints/suggestions on how to approach this question would be greatly appreciated :)

P.S. I am only taking an introductory calculus module in college, so the tools at my disposal are, roughly, IVT, MVT and Rolle's Theorem.

答:
The rigourous proof given above proves the inequality, but I wish to give a little handwaiving argument that might give insight into why this should be true.

The problem is effectively about maximizing the area under a function in the interval $[-1,1]$ that satisfies the given conditions.

So in this interval we know that the functions end points are at $x=0$ and $x=1$ intersecting the $x$ axis.

In the interval $(0,1) $ The function has infinite possibilities, but it has to be differentiable, $f'(x)$ has to be continuous and satisfy $max |f' (x)|=M $ .

Given these 3 conditions we need to maximize the area under the curve.


问:
Is the map $z\mapsto z^2$ a fibration?
Let  $f:\mathbb C \to \mathbb C$ be defined as $f(z) = z^2$.

Is this a fibration? Or at least a Serre fibration?

I am not sure how to approach this.

答:
Connor's comment sums up the situation.

Two fibres of a Hurewicz fibration over points in the same path-component of the base are homotopy equivalent.

Thus $f:\mathbb{C}\rightarrow\mathbb{C}$, $z\mapsto z^2$, is not a Hurewicz fibration, since the origin has a unique preimage, whilst all other points have two.

A similar statement holds for Serre fibrations when homotopy equivalent is replaced by weakly homotopy equivalent. Thus $f$ is not even a Serre fibration.

On the other hand, by restriction we obtain $g:\mathbb{C}\setminus\{0\}\rightarrow\mathbb{C}\setminus\{0\}$, $z\mapsto z^2$, and this map is locally-trivial. Since $\mathbb{C}\setminus\{0\}$ is paracompact $T_2$, this map is thus a Hurewicz fibration. The fact that $g$ is locally trivial is easy to see. The easiest way to do so is to use polar coordinates to identify $\mathbb{C}\setminus\{0\}\cong S^1\times(0,\infty)$ and notice that $g$ is the product of the two-fold covering projection $S^1\rightarrow S^1$, $z\mapsto z^2$, and the homeomorphism $(0,1)\rightarrow (0,1)$, $t\mapsto t^2$.

Title: Is the map $z\mapsto z^2$ a fibration?


问:
All number fields with absolute value of discriminant $\le 20$

I need to find all number fields with absolute value of discriminant $\le 20$.

Using Minkovsky's theorem I understood that it should be quadric or cubic extension. The case of quadric is very easy. As for cubic case I understood that it should have 2 complex embeddings and his ideal class group should be trivial. But I don't know to how find all of them.

答:
I'm going to assume you mean that the absolute value of the discriminant has to be $\le 20$. You seem happy with the quadratic case, and have correctly narrowed down the cubic fields to complex cubic fields.

Hint. The trick here is that the smallest value the discriminant of a cubic field can take (in absolute value) is $23$, namely $\mathbb{Q}[x]/(x^3-x^2+1)$ whose discriminant is $-23$, so there are in fact no cubic fields of discriminant $|\Delta|\le 20$.


问:
Find the $[E_1 : F_1]$ for $F_1 = \mathbb{Q}(\sqrt 3 i)$ and splitting field $E_1$
$Q)$ $f(x) = x^6 -3, F_1 = \mathbb{Q}(\sqrt 3 i)$. Let the $E_1 = SF(f / F_1)$.($E_1$ means splitting field of the $f$ over $F_1 $ ). Find the $[E_1 : F_1]$

My answer is $6$. Let me explain why I get that result.

Firstly, All the roots of the $f$ are $\{\alpha \omega^i \vert 0 \leq i \leq5 \}$ $\Rightarrow$  $SF(f/\mathbb Q) = \mathbb{Q}(\alpha, \omega)$(Here the $\alpha = \sqrt[6] 3, w= e^{2 \pi i \over 6}$)

I easily checked the $F_1 \subset \mathbb{Q}(\alpha, \omega)$. By the way, the $E_1$ is a splitting field over the $F_1$ so does the $\mathbb{Q}(\alpha, \omega)$, I.e. both are splitting field over the $F_1$ for $f$, therefore We can get $\mathbb{Q}(\alpha, \omega) = E_1$. Surely $[E_1 : \mathbb{Q}] = [E_1 : \mathbb{Q}(\omega) ][\mathbb{Q}(\omega) : \mathbb{Q}] = 6 \cdot \phi(6) = 12 $

Since the $[E_1 : \mathbb{Q}] = [E_1 : F_1][F_1 : \mathbb{Q}]$ and $[F_1 : \mathbb{Q}] =2$, So the answer is $6$

But the answer was $12$  What did I wrong?

I guess questioner of this question who claim the answer is $12$ makes a wrong conclusion. Because, I can't find my any mistakes in my solution yet. What do you think about that? Is my solution right?

答:
Your solution looks fine to me, so I would assume the other person made an error. To remove just about any remaining doubt, the following Sage code verifies that the degree is indeed $6$:

```<code>A.<x> = QQ[]
F.<w> = NumberField(x^2 + 3)   # F = Q(w), where w^2 + 3 = 0
B.<y> = F[]
E.<a> = NumberField(y^6 - 3)   # E = F(a), where a^6 - 3 = 0
E.relative_degree()            # [E : F] equals 6
E.is_galois_relative()         # is true, indicating that E/F is a Galois extension
</code>```

Body:
Your solution looks fine to me, so I would assume the other person made an error. To remove just about any remaining doubt, the following Sage code verifies that the degree is indeed $6$:

A.<x> = QQ[]
F.<w> = NumberField(x^2 + 3)   # F = Q(w), where w^2 + 3 = 0
B.<y> = F[]
E.<a> = NumberField(y^6 - 3)   # E = F(a), where a^6 - 3 = 0
E.relative_degree()            # [E : F] equals 6
E.is_galois_relative()         # is true, indicating that E/F is a Galois extension


问:
Does the magnitude of a nonzero determinant give any information about how close to singular the matrix is?
As title says, I am wondering i the magnitude of a nonzero determinant give me information about the singularity of the matrix.

Can someone prove (or help me to understand how i could) prove this to be true or false?

答:
No, it does not.  The determinant of a matrix is not invariant under scaling since for an $n$ by $n$ matrix,

$\det (\alpha A)=\alpha^{n} \det(A)$

Thus for example, if $n=1000$, and $\alpha=0.1$,

$\det(I)=1$

$\det(\alpha I)=1.0 \times 10^{-1000}$

A more appropriate measure of the singularity of a matrix is the condition number

$\kappa(A)=\| A \| \| A^{-1} \|$

Different norms can be used here, but when the euclidean norm is used, it can be shown that

$\kappa(A)=\sigma_{1}/\sigma_{n}$

where $\sigma_{1}$ and $\sigma_{n}$ are the largest and smallest singular values of $A$.


问:
Prove that $\mathscr{A}(\alpha)+\mathscr{B}^*(\beta)=\xi$, $\mathscr{B}(\alpha)=\eta$.

Suppose that $\mathcal{L}$ is an $n$-dimensional real Euclid space with inner product $(*,\,*)_1$; $\mathcal{M}$ is an $m$-dimensional real Euclid space with inner product $(*,\,*)_2$. $\mathscr{A}$ is a linear transformation on $\mathcal{L}$; $\mathscr{B}$ is a linear map from $\mathcal{L}$ to $\mathcal{M}$. Let $\pi$ be an orthogonal projection from $\mathcal{L}$ into (not necessarily onto) $\ker(\mathscr{B})$ such that $\pi\circ\mathscr{A}\big|_{\ker(\mathscr B)}$ is a linear isomorphism. Furthermore, there is a constant $\lambda_0>0$ such that
$$
\inf_{0\ne\beta\in\mathcal{M}}\;\;\sup_{0\ne\alpha\in\mathcal{L}}\frac{|(\beta,\,\mathscr{B}(\alpha))_2|}{|\alpha|_1|\beta|_2}\geq\lambda_0. \tag{1}
$$
Show that for all $\xi\in\mathcal L$ and $\eta\in\mathcal M$, there exists a unique $\alpha\in\mathcal{L}$, $\beta\in\mathcal{M}$ such that $\mathscr{A}(\alpha) + \mathscr{B}^*(\beta)=\xi$, $\mathscr{B}(\alpha)=\eta$, where $(\mathscr{B}^*(\beta),\,\alpha)_1 = (\beta,\,\mathscr{B}(\alpha))_2$.

From the inequality $(1)$, I observed that $\mathscr{B}$ must be surjective.

Also, the prove the uniqueness, we should prove that if $\alpha_1$, $\beta_1$ and $\alpha_2$, $\beta_2$ both satisfy the conclusion, then it must be that $\alpha_1=\alpha_2$ and $\beta_1=\beta_2$.
Since $\alpha_2-\alpha_1\in\ker(\mathscr{B})$, and $\pi\circ\mathscr{A}\big|_{\ker(\mathscr B)}$ is an isomorphism, to prove $\alpha_2=\alpha_1$ it suffices to prove that $\pi\circ\mathscr{A}(\alpha_2-\alpha_1)=0$.

What I have got are listed above.

答:
The inequality (1) not only implies surjectivity for $\mathscr{B}$ but also implies $\mathscr{B}$ is a bijection from $\ker(B)^{\perp}$ to $\mathcal{M}$  because from this inf-sup condition we get
\begin{align}
\;\sup_{0\ne\alpha\in\mathcal{L}}\frac{|(\beta,\,\mathscr{B}(\alpha))_2|}{|\alpha|_1}
\sup_{0\ne\alpha\in\mathcal{L}}\frac{|(\mathscr{B}^{*}(\beta),\,\alpha)_1|}{|\alpha|_1}\geq\lambda_0 |\beta|_2 \quad \forall \beta \in \mathcal{M}.
\end{align}
In other words $|\mathscr{B}^{*}(\beta)|_1\geq \lambda_0|\beta|_2$. Now this inequality implies $\mathscr{B}^{*}$ is inyective and has closed range. Then $R(\mathscr{B}^{*})=\ker((\mathscr{B}^{*})^{*})^{\perp}=\ker \mathscr{B}^{\perp}$.
Hence, $\mathscr{B}^{*}:\mathcal{M} \to \ker \mathscr{B}^{\perp} $ is a linear isomorphism. Also $R(\mathscr{B})=\ker(\mathscr{B}^{*})^{\perp}=\{0\}^{\perp}=\mathcal{M}$. Then $\mathscr{B}:\ker{\mathscr{B}}^{\perp}\to\mathcal{M}$ is a linear isomorfism. Hence, there is only one $\alpha \in \ker{\mathscr{B}}^{\perp}$ with
\begin{align}
\mathscr{B}(\alpha) = \eta \tag{1}
\end{align}
$\pi\circ\mathscr{A}:\ker{\mathscr{B}} \to \ker{\mathscr{B}}$ is a bijection. Hence, exist $\alpha_0 \in \ker{\mathscr{B}}$, such that, $\pi\circ\mathscr{A}(\alpha_0)=\pi(\xi-\mathscr{A}(\alpha))$.
Then $\mathscr{A}(\alpha_0+\alpha)- \xi \in \ker{\mathscr{B}}^{\perp}$, now exist a unique $\beta \in \mathcal{M}$
\begin{align}
\mathscr{B}^{*}(\beta)=\xi-\mathscr{A}(\alpha_0+\alpha) \tag{2}
\end{align}
Now note $(\alpha+\alpha_0,\beta)\in \mathcal{L}\times \mathcal{M}$, solves (1) and (2). Hence, the problem has solution.
Now for the uniqueness let $(\alpha_0,\beta_0)$ and $(\alpha_1,\beta_1)$ two solutions in $\mathcal{L}\times \mathcal{M}$, doing some calculations we get
\begin{align}
\mathscr{A}(\alpha_1-\alpha_0)+\mathscr{B}^{*}(\beta_1-\beta_0) &= 0 \tag{3}\\
\mathscr{B}(\alpha_1-\alpha_0) &= 0   \tag{4}
\end{align}
Note $\mathscr{B}^{*}(\beta_1-\beta_0) \in \ker \mathscr{B}^{\perp}$. Appliying $\pi$ on (3), we get $\pi\circ \mathscr{A}(\alpha_1-\alpha_0)=0$, so $\alpha_1=\alpha_0$. Finally because $\mathscr{B}^{*}:\mathcal{M} \to \ker \mathscr{B}^{\perp} $ is a bijection and $\mathscr{B}^{*}(\beta_1-\beta_0)=0$ we get $\beta_1=\beta_2$.

Note: This theorem is well-know as Babuska-Brezzi Theorem, see [1] for example of applications of this result.

[1] GATICA, Gabriel N. A simple introduction to the mixed finite element method. Theory and Applications. Springer Briefs in Mathematics. Springer, London, 2014.


问:
Path-connected space
So I have this following question: Find a compact path-connected space $X \subset \mathbb{C}$ such that $X \backslash \{ O \}$ has an infinity of arc components.

I though about using an infinity of line segments (of length 1 for instance) which all have O as an extremity. However, I cannot get myself convinced that this set is compact.

So is this example fine, and if so how to prove it ? And else, do you have another example ?

Thank you for your help.

答:
The following is a standard example: let $q_n$ be an enumeration of the rational numbers in $[0,2\pi)$. Define $S_n$ as the segment with endpoints $(0,\frac{e^{iq_n}}{2^n})$. Define $S=\cup S_n$.

It is easy to see that $S$ is path connected and bounded. We claim that it is closed (compactness then follows from Heine-Borel):

To prove it, let $\{x_n\}\subset S$ be a converging sequence. Now, either $\{\arg(x_n)\}$ is a finite set (and we are done, since $S_n$ is closed and so $S_{n_1}\cup \dots \cup S_{n_k}$ is closed and so $\lim_n x_n\in S$), or it is not. In this case, however, $\lim_n |x_n|=0$, and so $x_n\to 0\in S$. Thus $S$ is closed.

We now claim that $S-\{0\}$ has a countable number of connected components. Indeed, $S_n-\{0\}$ is a connected subset of $S-\{0\}$, and it is maximal: to see this, let $X$ be the connected component of $S-\{0\}$ containing $S_n-\{0\}$. Its image through $\arg$ must be connected and contained in $\mathbb{Q}\cap [0,2\pi)$, and so it must be constant, which implies $X=S_n-\{0\}$.

One can also construct a compact set $X$ such that $X-\{(0,0)\}$ has an uncountable number of components: let $C$ be the cantor set, and $C_x$ the segment joining $(x_k,1)\in C\times \{1\}$ and the origin. Then $K=\cup_{x\in C} C_x$ is compact and path connected, but removing the origin gives us $K-\{(0,0)\}=\sqcup_{x\in C}C_x-\{(0,0)\}$, a set with $|C|=2^{\aleph_0}$ connected components

Title: Path-connected space


问:
time derivative of involved position-derivative
I am having a fair bit of trouble figuring out this differential. It's probably relatively straightforward but I can't wrap my head around it. I'm trying to work out:

$$
\frac{d}{dt}\left(-c^2\sqrt{1-\dot{x}^2\frac{1}{c^2}}\right)
$$
Where c is just a constant and $\dot{x} = \frac{dx}{dt}$

Any help would be greatly appreciated!

Body:
Use the chain rule twice.

Let's consider the $1$-dimensional case first. Since $\frac{d}{d\dot{x}}(1-\dot{x}^2/c^2)=-2\dot{x}/c^2$,$$\frac{d}{d\dot{x}}[-c^2(1-\dot{x}^2/c^2)^{1/2}]=\frac{\dot{x}}{(1-\dot{x}^2/c^2)^{1/2}},$$so$$\frac{d}{dt}[-c^2(1-\dot{x}^2/c^2)^{1/2}]=\frac{\dot{x}\ddot{x}}{(1-\dot{x}^2/c^2)^{1/2}}.$$

With a bit of work, you can prove the vector generalization is$$\frac{d}{dt}[-c^2(1-\dot{x}^2/c^2)^{1/2}]=\frac{\dot{x}\cdot\ddot{x}}{(1-\dot{x}^2/c^2)^{1/2}}.$$


问:
Distributing $m$ balls into $n$ urns with no urn left empty.
If $m \geq n$, how many different ways are there of distributing $m$ indistinguishable balls into $n$ distinguishable urns with no urn left empty? I have no idea how to even start with this so any help would be greatly appreciated. Thanks in advance.</p>

答:
Are you familiar with how to calculate the number of distributions of $m$ indistinguishable balls to $n$ distinguishable urns without any restrictions? If so, here's a hint:

Hint: Distribute 1 ball to each of the $n$ urns before doing anything. Then the number of distributions of $m$ indistinguishable balls to $n$ distinguishable urns with no urn left empty is the same as the number of distributions of $m-n$ indistinguishable balls to $n$ distinguishable urns with no restrictions.

If you need more help on the second calculation, feel free to post and I'll post more information!


问:
Sum of digits of 9(x+1) always equal to 9
For $x=1,2,....,9$ we know $9(x+1)$ is a two digits number $A+B$.
Prove sum of digits($A+B$) always equal to 9.

答:
Since $x$ only runs from $1$ to $9$, the simplest way is just to check them all quickly:

$$18, 27, 36, 45, 54, 63, 72, 81, 90.$$

Yes, it is true; this concludes the proof.


答:
Just check all the cases if you are comfortable with that.

Otherwise:

If $n$ is a positive integer divisible by $9$ then show that the sum of digits of $n$ must be divisible by $9$.

Finally notice that since $9$ is the only non-zero one digit number which is a multiple of $9$, so if you continuously keep on summing the digits to eventually reach to a one digit number, then this number should be $9$, since at each step you will have a multiple of $9$ obtained by summing the digits.

This will tell you that if you start with any multiple of $9$ and keep summing the digits till you get a one digit number (after which you cannot proceed any further) then this must be $9.$

Title: Sum of digits of 9(x+1) always equal to 9

Body:
For $x=1,2,....,9$ we know $9(x+1)$ is a two digits number $A+B$.
Prove sum of digits($A+B$) always equal to 9.


答:
HINT: To fulfill the "no urn left empty" clause, you must distribute $n$ balls into $n$ urns, one in each.

You can then distribute the remaining $(m-n)$ balls into $n$ urns however you like.

Title: Distributing $m$ balls into $n$ urns with no urn left empty.

Body: If $m \geq n$, how many different ways are there of distributing $m$ indistinguishable balls into $n$ distinguishable urns with no urn left empty? I have no idea how to even start with this so any help would be greatly appreciated. Thanks in advance.</p>


问:
Why when rearranging this equation (under limit) gives different answer?
When I try to solve this by using l'hopital rule

$\lim_{x \to 0} \large \frac{(2e)^x-x\ln2-x-1}{xtanx} = k$

$\lim_{x \to 0} \large \frac{\ln(2e)(2e)^x-\ln2-1}{tanx+xsec^2 x} = k$

$\lim_{x \to 0} \large \frac{(\ln2e)^2(2e)^x}{sec^2 x+sec^2 x x+ 2xsec^2 xtanx} = k$

I get
$k=\frac{(ln2+1)^2}{2}$ as the answer

But when I rearrange the equation as

$\large \frac{\lim_{x \to 0}\frac{(2e)^x-1}{x}-\ln2-1}{\lim_{x \to 0}tanx} = k
$
$\small ,since \lim_{x \to 0} \frac{(2e)^x-1}{x}=\lim_{x \to 0} (2e)^x\ln2e$

$\lim_{x \to 0}\large \frac{(2e)^x\ln2e -\ln2e -1}{tanx} =k, $

$\lim_{x \to 0} \small (2e)^x(\ln2e)^2=k$

And then use l'hopital  to evaluate the limit I get the answer $\small (\ln2e+1)^2 =k$

(The upper one is correct)

Where have I made the mistake?

Body:
$$L=\lim_{x\to 0} \frac{(2e)^x-x\ln 2-x-1}{x \tan x}$$
$(2e)^x=1+x\ln (2e)+x^2 \ln^2(2e)/2+...$, \tan x=x+x^3/3+..
$$L=\lim_{x\to 0}\frac {1+x\ln(2e)+x^2\ln^2(2e)/2-x\ln 2-x-1}{x^2+x^4/3}$$
$$L=\lim_{x \to 0} \frac{x^2 \ln^2(2e)/2}{x^2+x^4/3}=\frac{1}{2}\ln^2 (2e).$$


问:
Calculating matrix exponential

Given matrix $$M = \begin{pmatrix} 7i& -6-2i\\6-2i&-7i\end{pmatrix}$$ how do I calculate matrix exponential $e^M$?

I know I can use that $e^A=Pe^DP^{-1}$ where $D=P^{-1}AP$. I computed the characteristic polynomial of the above matrix as

$$P(\lambda)=\lambda^2+89$$

Is there an easier way to do this than trying to compute the diagonalized matrix?

答:
You can also use Euclidean divisions to compute the exponential once you have the characteristic polynomial :

$$\forall n\in\mathbb{N}^*,\exists(P_n,R_n)\in\mathbb{R}[X], \deg(R)\leq1, X^n=P_n(X)\times(X^2+89)+R_n(X) $$

Let $R_n(X)=a_nX+b_n$, you get :
$$M^n=a_nM+b_n1 $$

and you get the values of $a_n,b_n$ with :
$$(i\sqrt{89})^n=a_ni\sqrt{89}+b_n\\
(-i\sqrt{89})^n=-a_ni\sqrt{89}+b_n$$

So you get :
$$b_n=\frac{(i\sqrt{89})^n+(-i\sqrt{89})^n}{2}\\
a_n=\frac{(i\sqrt{89})^n-(-i\sqrt{89})^n}{2i\sqrt{89}} $$

And then you use the definition :
$$e^M=\sum_{n=0}^{\infty}\frac{M^n}{n!}=\sum_{n=0}^{\infty}\frac{a_nM+b_n1}{n!} $$

And then you can conclude from here.

Title: Calculating matrix exponential


答:
Simplify dividing by $c$ from the given
$$ -c \frac{d}{dt}\sqrt{c^2- \dot x^{^2}}$$

The derivative of $\sqrt u$ is $\dfrac{1}{2\sqrt u},\text{so that is using Chain Rule }$

$$= -c \cdot \frac{-2 \dot x \ddot x}{2\sqrt{c^2- \dot x^{^2}}}$$

$$=  \frac{c \dot x \ddot x}{\sqrt{c^2- \dot x^{^2}}}.$$

Title: time derivative of involved position-derivative

Body:
I am having a fair bit of trouble figuring out this differential. It's probably relatively straightforward but I can't wrap my head around it. I'm trying to work out:

$$
\frac{d}{dt}\left(-c^2\sqrt{1-\dot{x}^2\frac{1}{c^2}}\right)
$$
Where c is just a constant and $\dot{x} = \frac{dx}{dt}$

Any help would be greatly appreciated!


Body:
$y:=-c^2(1-(x')^2/c^2)^{1/2}$, and let $y\not =0;$

$y^2=c^4(1-(x')^2/c^2);$

Chain rule:

$2yy'=c^4(-2(x')(x'')/c^2);$

$y'=(-c^2)\dfrac{(x')(x'')}{(-c^2)(1-(x')^2/c^2)^{1/2}},$

$y'=\dfrac{(x')(x'')}{(1-(x')^2/c^2)^{1/2}}.$


问:
Interesting binomial coefficients sum: $\sum\limits_{i = 0}^b(-1)^i\binom{b}{i}\binom{a+b-i-1}{a-i}$
Any idea on whether or not $$\sum_{i = 0}^b(-1)^i\binom{b}{i}\binom{a+b-i-1}{a-i}$$
has a closed formula on $a$ and $b$ (and on what it is, in case it does)?

It is supposed that $b \le a$.

答:
if $a=0,$ you get $1.$

Assume now $a>0.$ Notice that if $i=0$ you are evaluating $\binom{a+b-1}{b-1}$ this is the number of ways to write $a$ as a sum of $b$ $(x_1,\cdots ,x_b)$ non-negative integers by the stars and bars technique.

Consider now the following sets $$A_i = \{(x_1,\cdots ,x_b):\sum _{j=1}^bx_j=a , \, x_i\geq 1\},$$
notice that at all possible ways have to be such that at least one of the $x_j\geq 1$ hence, by the principle of inclusion-exclusion,
$$\binom{a+b-1}{b-1}=\left |\bigcup _{i=1}^b A_i\right |=\sum _{i=1}(-1)^{i-1}\sum _{Y\in \binom{[b]}{i}}\left |\bigcap _{y\in Y}A_y\right |.$$
Notice, further, that $|\cap _{y\in Y}A_y|=\binom{a-|Y|+b-1}{b-1}$ because all the $x_y\geq 1$ so you can redefine $z_y=x_y-1$ and pass the $|Y|$ 1's to the other side of the equation. So you get that
$$\binom{a+b-1}{b-1}=\sum _{i=1}^b(-1)^{i-1}\binom{b}{i}\binom{a-i+b-1}{b-1},$$
but passing the RHS to the LHS gives your identity.


答:
Suppose we have $$9x=AB=10A+B$$
Then we have $$9(x+1)=10A+B=9A+(A+B)$$.

This gives us $$A+B=9(x+1)-9A=A+B=9(x+1-A). $$

Thus the sum of digits of $9(x+1)$ is $9(x+1-A)$. Since $9(x+1-A)$ is divisible by $9$, then so is the sum of digits of $9(x+1)$


问:
What is the probability of each cell being "True" in this table?
Say I have a table with the following values.
The following things must be true:

Each column must have 3 Trues

Each row must have 1 True

```<code>                   |     1     |      2    |      3    |     4     |     5     |     6     |      7
---------------------------------------------------------------------------------------------------------
         Apple     |    False  |    False  |    False  |    False  |    False  |    False  |     True
         Banana    |    False  |  -------  |  -------  |    False  |    False  |  -------  |    False
         Car       |     True  |    False  |    False  |    False  |    False  |    False  |    False
         Dinosaur  |    False  |  -------  |  -------  |  -------  |    False  |  -------  |    False
         Elephant  |    False  |  -------  |  -------  |  -------  |    False  |  -------  |    False
         Fig       |     True  |    False  |    False  |    False  |    False  |    False  |    False
         Giraffe   |    False  |    False  |     True  |    False  |    False  |    False  |    False
         Hound     |    False  |  -------  |  -------  |  -------  |    False  |    False  |  -------
         Ice       |    False  |  -------  |  -------  |    False  |    False  |  -------  |  -------
         Jay       |    False  |    False  |    False  |    False  |    False  |     True  |    False
         Kangaroo  |    False  |    False  |    False  |    False  |    False  |  -------  |  -------
         Leopard   |    False  |    False  |    False  |    False  |    False  |     True  |    False
         Moose     |    False  |    False  |    False  |    False  |     True  |    False  |    False
         Numbat    |    False  |    False  |  -------  |  -------  |    False  |    False  |  -------
         Octopus   |    False  |    False  |    False  |     True  |    False  |    False  |    False
         Pig       |    False  |    False  |    False  |    False  |    False  |  -------  |  -------
         Quinoa    |    False  |  -------  |  -------  |  -------  |    False  |    False  |  -------
         Rhino     |     True  |    False  |    False  |    False  |    False  |    False  |    False
         Snake     |    False  |    False  |    False  |    False  |     True  |    False  |    False
         Tapeworm  |    False  |    False  |    False  |    False  |     True  |    False  |    False
         Unicorn   |    False  |    False  |    False  |     True  |    False  |    False  |    False
</code>```

The "-------" mark unknown cells.

Is it possible to figure out the probability of any unknown cell being true?

Or, figure out the number of solutions this table could have, then examine each solution and count the number of "True" occurrences for any given unknown cell, and then divide that count by the number of solutions?

答:
This is equivalent to putting the $21$ items into $7$ bins such that each bin contains exactly $3$ items. A True in a cell indicates that the item in this cells row is put into the bin corresponding to the cells column.

Since the bins $1$ and $5$ are already full and some items are already placed, we can simplify the problem to this:

```<code>           |   bin 2   |   bin 3   |   bin 4   |   bin 6   |  bin  7
           |  3 items  |  2 items  |   1 item  |   1 item  |  2 items
-------------------------------------------------------------------------
 Banana    |  -------  |  -------  |    False  |  -------  |    False
 Dinosaur  |  -------  |  -------  |  -------  |  -------  |    False
 Elephant  |  -------  |  -------  |  -------  |  -------  |    False
 Hound     |  -------  |  -------  |  -------  |    False  |  -------
 Ice       |  -------  |  -------  |    False  |  -------  |  -------
 Kangaroo  |    False  |    False  |    False  |  -------  |  -------
 Numbat    |    False  |  -------  |  -------  |    False  |  -------
 Pig       |    False  |    False  |    False  |  -------  |  -------
 Quinoa    |  -------  |  -------  |  -------  |    False  |  -------
</code>```

At this point I can't reach any further conclusions or simplifications. You might want to use a computer to find all valid solutions. Then you can calculate all the probabilities you like using the method described in your question.

Body:
This is equivalent to putting the $21$ items into $7$ bins such that each bin contains exactly $3$ items. A True in a cell indicates that the item in this cells row is put into the bin corresponding to the cells column.

Since the bins $1$ and $5$ are already full and some items are already placed, we can simplify the problem to this:

           |   bin 2   |   bin 3   |   bin 4   |   bin 6   |  bin  7
           |  3 items  |  2 items  |   1 item  |   1 item  |  2 items
-------------------------------------------------------------------------
 Banana    |  -------  |  -------  |    False  |  -------  |    False
 Dinosaur  |  -------  |  -------  |  -------  |  -------  |    False
 Elephant  |  -------  |  -------  |  -------  |  -------  |    False
 Hound     |  -------  |  -------  |  -------  |    False  |  -------
 Ice       |  -------  |  -------  |    False  |  -------  |  -------
 Kangaroo  |    False  |    False  |    False  |  -------  |  -------
 Numbat    |    False  |  -------  |  -------  |    False  |  -------
 Pig       |    False  |    False  |    False  |  -------  |  -------
 Quinoa    |  -------  |  -------  |  -------  |    False  |  -------

At this point I can't reach any further conclusions or simplifications. You might want to use a computer to find all valid solutions. Then you can calculate all the probabilities you like using the method described in your question.


问:
Finding constraints on right-hand side to yield feasible constrained linear problem
I have the following constrained linear system:

$$
Ax = y \\
Cx \ge b
$$

where
$$
y\in \mathbb{R}^3 \\
x \in \mathbb{R}^n \\
b \in \mathbb{R}^m \\
$$
Also, $n$ and $m$ are typically greater than 3, e.g. say 4 and 10 respectively.

For my specific application, the inequality constraints are always such that they create a convex polygon, in case that is of any use.

Is it possible to find a set of linear constraints on $y$, i.e.
$$
Dy \ge g
$$
such that the original system has solutions?

Algebraic solutions are preferred, but numerical solutions are also of interest.

答:
So I spent some time on the problem and arrived at the following:

Assuming the 3 rows of $A$ are not linearly dependent, we can first partition the unknowns $x$ into subsets $x_1 \in \mathbb{R}^3$ and $x_2 \in \mathbb{R}^{k}$, where $k=n-3$. This leads to a corresponding partitioning of the matrices:

$$
Ax =
\begin{pmatrix}
A_1 & A_2
\end{pmatrix}
\begin{pmatrix}
x_1\\
x_2
\end{pmatrix} = y\\
Cx =
\begin{pmatrix}
C_1 & C_2
\end{pmatrix}
\begin{pmatrix}
x_1\\
x_2
\end{pmatrix} \ge b\\
$$
Furthermore, we can introduce the auxiliary variables $\lambda = x_2$, by augmenting the equality system with $k$ additional equalities, yielding
$$
\begin{pmatrix}
A_1 & A_2\\
0 & \mathbb{I}_{k \times k}
\end{pmatrix}
\begin{pmatrix}
x_1\\
x_2
\end{pmatrix} =
\begin{pmatrix}
y\\
\lambda
\end{pmatrix} .
$$
Now, the block matrix can be inverted yielding
$$
\begin{pmatrix}
x_1\\
x_2
\end{pmatrix} =
\begin{pmatrix}
A_1^{-1} & -A_1^{-1}A_2\\
0 & \mathbb{I}_{k \times k}
\end{pmatrix}
\begin{pmatrix}
y\\
\lambda
\end{pmatrix} .
$$
This can now be plugged into the system of inequalities,
$$
\begin{pmatrix}
C_1 & C_2
\end{pmatrix}
\begin{pmatrix}
x_1\\
x_2
\end{pmatrix} =
\begin{pmatrix}
C_1 & C_2
\end{pmatrix}
\begin{pmatrix}
A_1^{-1} & -A_1^{-1}A_2\\
0 & \mathbb{I}_{k \times k}
\end{pmatrix}
\begin{pmatrix}
y\\
\lambda
\end{pmatrix} \ge b.
$$
Simplifying, we get
$$
\begin{pmatrix}
C_1 A_1^{-1} & C_2 -C_1 A_1^{-1}A_2
\end{pmatrix}
\begin{pmatrix}
y\\
\lambda
\end{pmatrix} \ge b\\
C_1 A_1^{-1}y +(C_2 -C_1 A_1^{-1}A_2)\lambda \ge b
$$
or
$$
C_1 A_1^{-1}y \ge b + (C_1 A_1^{-1}A_2 - C_2)\lambda \, .
$$
These are now the sought inequalities for $y$ for which there are solutions to the original problem. However, we now have the parameters $\lambda$ left, which are allowed to take on any value as long as all the inequalities hold simultaneously for a given $y$. In other words, $y$ is a solution if there exists a $\lambda$ for which the inequalities hold.

I'm convinced it should be possible to get rid of the parametrization and just have fixed inequalities, but I haven't gotten anywhere with that yet.  Any help with solving that part would be greatly appreciated.


问:
how can I show that this function is uniformly continuous on $(1,0) $ and $(1,\infty)$
How can I show that $\sqrt x$ is uniformly continuous on the intervals: $(0,1)$ and $(1,\infty)?$

答: Hint : For the first one, show that $\sqrt{x}$ is even uniformly continuous on $[0,1]$. For the second one, show that $\sqrt{x}$ is lipschitz on the given interval.


问:
Cone over a variety is a variety?
Let $\mathbb{F}$ be a field (possibly algebraically closed) and let $V\subseteq\mathbb{F}^n$ be an affine variety. Now define the cone over $V$ as follows: $$\mathrm{Cone}(V):=\{\lambda\mathbf{p}: \mathbf{p}\in V, \lambda\in\mathbb{F}\}.$$ I want to show that $\mathrm{Cone}(V)\subseteq\mathbb{F}^n$ is also an affine variety, but I'm having trouble. Any advice would be appreciated.

答:
So an affine variety is the shared zero set among some collection of polynomials $\{f_1, \dots, f_m\}\subset \mathbb{F}[x_1, \dots, x_n]$. We can show that $Cone(V)$ is an affine variety in $\mathbb{F}^{n+1}$ by taking the $f_i$ and transforming them into homogenous polynomials, i.e. polynomials $p$ satisfying $$p(\lambda x_1, \dots, \lambda x_n)=\lambda^{deg(p)}p(x_1, \dots, x_n)$$
We can homogenize any non-homogenous polynomial by appending a coordinate and doing a sort of "renormalization" on inputs for the original. Specifically, take $\lambda$ to be a new coordinate and define
$$\widetilde{f_i}(\lambda,x_1, \dots, x_n)=\lambda^{deg(f_i)}f_i(\tfrac{x_1}{\lambda}, \dots, \tfrac{x_n}{\lambda})$$
Given some $(a_1, \dots, a_n) \in V$, we have
$$\widetilde{f_i}(\lambda, \lambda a_1, \dots, \lambda a_n)=\lambda^{deg(f_i)}f_i(\tfrac{\lambda a_1}{\lambda}, \dots, \tfrac{\lambda a_n}{\lambda})=\lambda^{deg(f_i)}f_i(a_1, \dots, a_n)=0$$

Which means $Cone(V)$ is an affine variety in $\mathbb{F}^{n+1}$ (or, can be thought of as a projective variety in $\mathbb{P}^n$).

Title: Cone over a variety is a variety?


问:
Why are the cosets of the Bruhat decomposition independent of the representation of the Weyl group chosen?
Why are the cosets of the Bruhat decomposition independent of the representation of the Weyl group chosen?

For the sake of concreteness, let $G=SL_n$. Let $B_+$ the subgroup of upper triangular matrices and $B_-$ the subgroup of lower triangular matrices and let $T = B_+ \cap B_-$. Then the Weyl group $W$ of $SL_n$ be can canonically identified with $\frac{N(T)}{T}$, and furthermore we obtain two BRuhat decomposition's:

$SL_n = \bigsqcup_{w \in W} B_+wB_+=\bigsqcup_{v \in W} B_-vB_-$

The Weyl group here is being understood as the quotient group $\frac{N(T)}{T}$. Can somebody help me understand why the double coset $B_+wB_+$ is independent of the reprentative chosen for $w$ from the quotient group? Thanks

答: If you choose a different representative for $w$ in the quotient group, that amounts to multiplying $w$ by some element $t \in T$.  Since $T = B_+ \cap B_-$, then $t B_+ = B_+$ and $t B_- = B_-$, i.e, the $t$ can be absorbed into the $B_+$ or $B_-$ of the double coset.


问:
Prove that a solution to differential equation $x'=f(x)$ is bounded with $\left| f(x)-\cos(x)\right|\leq1$.
Let $x: \mathbb{R} \to \mathbb{R}$ be a solution to a Cauchy problem
$$x'=f(x),\quad x(0)=x_0,$$
where $x_0\in\mathbb{R},f\in\mathcal{C}^1\left(\mathbb{R}\right)$ and $f$ is such that
$$\forall_{x\in\mathbb{R}} \left|f\left(x\right)-\cos x\right|\leq 1.$$
Prove that for any $x_0 \in \mathbb{R}$ function $x$ is bounded.

Edit: My partial solution had an error in it and it didn't work. What I've tried was to bound $\int\frac{dx}{f(x)}$ but the $f(x)$ isn't always positive nor negative so it doesn't work here.

答:
Do you know about Picard iteration/the Picard-Lindelöf theorem?

It tells you that a solution $\lambda$ of $x' = f(x)$ satisfies $\lambda(t) = x_0 + \displaystyle\int_{0}^t f(\lambda(s)) \mathrm{d}s$ (where $f$ is locally Lipschitzean, so it's true when $f \in \mathcal{C}^1$).

Then, you have $|f(x) - \cos x| \leq 1 \implies |f(x)| \leq 1 + |\cos x| \leq 2$. So in particular:

$|\lambda(t)| \leq x_0 + \displaystyle \int_{0}^t  2 \mathrm{d}s = x_0 + 2t$. This doesn't ensure boundedness of $\lambda$ yet, but it makes sure that it is defined globally. Indeed, the only way for a solution to a Cauchy problem to not be defined globally, is if the solution blows up in finite time (ie, it goes to infinity at some finite value, which explains why you can't extend it any more).

You can skip this step if you're not interested in the domain of your solution, and all you want to show is that it's bounded (whatever the domain).

Now that you know $\lambda$ is defined globally, all you need to do is check that it stays bounded.

A very useful fact, when $f$ is $\mathcal{C}^1$ is that solutions cannot cross.

Now, the condition $|f(x) - \cos x| \leq 1$ tells you that $f$ is always within a distance $1$ of $\cos$. This, in particular, tells you that $f$ will have to be $0$ infinitely many times, and you can find zeros as big as you want! (Look at it graphically if you can't see why, it has to do with the fact cosine will oscillate infinitely many times between $1$ and $-1$.)

But if $x^*$ is a zero of $f$, then $x \mapsto x^*$ is a constant solution of your differential equation. So you can find two zeros of $f$, $x_1^*$ and $x_2^*$, such that $x_1^* \leq x_0 \leq x_2^*$.
In particular, since $\lambda$ cannot cross other solutions, it will never be able to leave that box:

So $\lambda$ is bounded.


问:
Describe $\sigma$-algebra subsets of $[0;1]$ section generated by..

Describe $\sigma$-algebra subsets of $[0;1]$ section generated by following sets system that consists of all single-point sets.

However I do have several ideas, I'm rather new to this topic and not really sure whether I'm right, my ideas are the following:

In this task we need to list all the possible elements of $\sigma$-algebra given, so for any point we may take it and may not, so $\{\{x\}, \{[0;1] \setminus x \} | \forall x \in [0;1] \}$.
Is it right?

Any response/solutions is greatly appreciated.

答:
You want to describe a $\sigma$-algebra generated by
$$\mathcal{E} = \left\{\left\{x\right\}:x\in[0,1]\right\}.$$
Let's call that $\sigma(\mathcal{E})$. By the definition of $\sigma$-algebra, $\sigma(\mathcal{E})$ needs to contain the empty set, be closed under complements, and closed under countable union.

Right now, your collection is not a $\sigma$-algebra because it is not closed under countable unions. For example, the set
$$\left\{\frac1n:n\in\mathbb{N}\right\} = \bigcup_n \left\{\frac1n\right\}$$
should be in $\sigma(\mathcal{E})$ but it is not in your collection you described.

For the actual characterization of $\mathcal{M}$, I think what you would end up with is the so-called "countable co-countable" $\sigma$-algebra. That is, we have
$$\mathcal{M}= \left\{ E\subseteq [0,1]:E\text{ is countable or }[0,1]\setminus E\text{ is countable}\right\}$$
as the $\sigma$-algebra generated by single point sets. You should prove the two following claims as exercises (they are pretty straightforward)

Claim 1: $ \mathcal{M}$ is in fact a $\sigma$-algebra.

Claim 2: $\mathcal{E}\subseteq \mathcal{M}$.

From the above claims, it follows that $\mathcal{M}\supseteq \sigma(\mathcal{E})$ because $\sigma(\mathcal{E})$ is the smallest $\sigma$-algebra containing $\mathcal{E}$. Lastly, proving that (also straightforward) $\mathcal{M}\subseteq \sigma(\mathcal{E})$ will give you $\mathcal{M} = \sigma(\mathcal{E})$.


问:
Curvature of a non-compact complete surface
Assume $\Sigma$ is a non compact, complete surface. Assume the integral $$\int_{\Sigma}K$$ is convergent, where K is the Gauss curvature of $\Sigma$. Is it always true that $$\frac{1}{2\pi}\int_{\Sigma}K$$ is an integer?</p>

答: No. Start with a one-nappe circular cone. Cut off the tip and replace with a subset of a sphere, resulting in what is now a $C^1$ infinite surface. Smooth near the circle where they join, so the result is at least $C^2.$ The result has zero curvature except near the spherical part. Since we did not specify the central angle of the cone, we get whatever we like in the way of total Gauss curvature, up to an effective maximum given by a hemisphere and an infinite cylinder.</p>


问:
Numerically computing an oscillatory integral
I want to numerically compute integrals of the form
$$F(s) = \frac{1}{2\pi} \int_0^A \frac{e^{isu}}{1 + B\cos(r u)}\,du$$
for fixed $A,r>0, 1 > B >0$ for a finite number of values of $s\in \mathbb{R}$. I am familiar with the Filon method to deal with the highly oscillating case ($|s|\gg 1$). However, I am also concerned with the evaluations with slow oscillations ($0 < |s| \ll 1$), and I'm not sure if Filon method will be enough as it failed in several instances. Is it enough to do a naive approach of applying Filon method for large frequencies, otherwise use the default integral command?

答:
You might write this as a series in powers of $B$, which converges since $0 < B < 1$:

$$ F(s) = \frac{1}{2\pi}\sum_{k=0}^\infty (-B)^k \int_0^A e^{isu}  \cos(ru)^k\; du $$
where each term can be computed in closed form.


问:
Prove that in $C([0,1])$, $||x||= \text{sup}_{t \in [0,1]}|tx(t)|$ the norms $||.||$ and $||.||_{\infty}$ are not equivalent
In $C([0,1]),||.||)$  $with ||x||= \text{sup}_{t \in [0,1]}|tx(t)|$ prove that $||.||$ and $||.||_{\infty}$ are not equivalent norms

Try:

In $0 \leq t \leq 1 $   $|tf(t)| \leq |f(t)|$ so $||f(t)|| \leq ||f(t)||_{\infty}$

Now the question is if there exists a $c>0$ such that  $||f(t)||_{\infty} \leq c||f(t)||$

We take a sequence defined as follows:

$$ f_n(t)= \begin{cases} \frac{1}{ t} && ,\frac{1}{n} \leq t \leq 1 \\ n &&, 0 \leq t \leq \frac{1}{n}  \end{cases}   $$

$||f_n||_{\infty}= \text{sup}_{t \in [0,1]} |f_n(t)|=n$
$||f_n||=\text{sup}_{t \in [0,1]} |t f_n(t)|=\text{max} \{\text{sup}_{0 \leq t \leq \frac{1}{n}} |t f_n(t)|,\text{sup}_{\frac{1}{n} \leq t \leq 1} |t f_n(t)| \}=\text{max} \{\text{sup}_{0 \leq t \leq \frac{1}{n}} nt,\text{sup}_{\frac{1}{n} \leq t \leq 1} \frac{1}{t}t \}= \text{max} \{1,1\}=1 $

Then $\frac{||f_n||_{\infty}}{||f_n||} \to \infty$, when $n \to \infty$. So the norms are not equivalent

My questions are, in addition to knowing if this is correct :

In this approach I don't need the sequence to be Cauchy, right?.  in order to prove that no such c exists , why do we consider a sequence in the first place? what's the idea?

Why do they take the limit of $\frac{||f_n||_{\infty}}{||f_n||} \to \infty$ at the end, how  does this proves that no constant c exists that makes the norms equivalent ?

答:
What you did is fine and it proves indeed that there is no constant $c$ so that we always have$$\|f\|_\infty\leqslant c\|f\|.\tag1$$Note that $(1)$ is equivalent to $\frac{\|f\|_\infty}{\|f\|}\leqslant c$. So, proving that $(1)$ doesn't hold for some $c$ and for every $f\in C\bigl([0,1]\bigr)$ is equivalent to asserting that for every $c$ there is some $f_c\in C\bigl([0,1]\bigr)$ such that $\frac{\|f_c\|_\infty}{\|f_c\|}>c$. And, in order no prove that, it is enough to prove that, for every $n\in\Bbb N$ there is some $f_n\in C\bigl([0,1]\bigr)$ such that $\frac{\|f_n\|_\infty}{\|f_n\|}>c$. So, it is natural to take a sequence $(f_n)_{n\in\Bbb N}$ such that $\lim_{n\to\infty}\frac{\|f_n\|_\infty}{\|f_n\|}=\infty$.

And, no, it doesn't have to be a Cauchy sequence.


问:
Question about Protter's proof that a Cadlag, locally square integrable local martingale is a semimartingale
This is Corollary $1$ in Chapter $2$ of Protter's Stochastic Integration and Differential Equations.

Theorem 8 states that each $L^2$ martingale (martingales $X$ such that $X_0 = 0$ and $E[X_\infty^2]<\infty$) with cadlag paths is a total semimartingale.

The Corollary to Theorem 6 states that : If $X$ is a process and there exists a sequence $T_n$ of stopping times increasing to $\infty$ a.s. such that $X^{T_n}$ ( or $X^{T_n} 1_{\{T_n>0\}}$) is a semimartingale for each $n$ then $X$ is a semimartingale.

But I don't see how the proof is so straightforward here.

First, let $X$ be a cadlag, locally square integrable local martingale.

Then does this mean that $X$ is locally a square integrable martingale, i.e. we have a fundamental sequence $T_n$ such that $X^{T_n} 1_{\{T_n >0\}}$ is a square integrable martingale?  I have seen several questions about this on StackExchange, but no real answer to this.

It seems like we would need this condition but then taking $X^n := X^{T_n} 1_{\{T_n >0\}}$, square integrability just means that $E[(X^n_t)^2]<\infty$ for each $t$. We don't get $L^2$ boundedness over all $t$ from this. So how do we use the corollary to Theorem 6 here?

My attempt: $X$ being a semimartingale is by definition in the text, $X^t$ being a total semimartingale for each $t \ge 0$.

Assuming that being locally square integrable local martingale is the same as being locally a square integrable martingale, we can find a fundamental sequence $T_n$ such that $X^{T_n} 1_{\{T_n > 0\}}$ is a square integrable martingale. For convenience denote $M := X^{T_n} 1_{\{T_n > 0\}}$. Then we have $E[(M_t^n)^2]<\infty$ for all $t \ge 0$.

Now in order to apply the Corollary to Theorem 6, we would need to show that $M$ is a semimartingale. And to do this we need to use Theorem 8.

Thus, we are done if we show that for each $s \ge 0$, $M^s$ is a $L^2$ - martingale.

Now, for each $s \ge 0$, $M^s_t = X_{T_n \wedge s \wedge t} 1_{\{T_n > 0\}}$. Since a stopped martingale is a martingale, $M^s$ is still a martingale. Moreover, $s$ and $t$ are not random, so square integrability gives us that $M^s$ is a $L^2$-bounded uniformly integrable martingale. Hence, by Theorem 8, $M^s$ is a total semimartingale, hence $M$ is a semimartingale. Finally, the Corollary to Theorem 6 applies. QED.

This proof has been bothering me for a long time now. I think my final argument assuming a single fundamental sequence that makes $X$, a locally square integrable local martingale, into a locally square integrable martingale, is correct but I don't know how to show this part. I would greatly appreciate any help.

答:
Let $X$ be cadlag and a locally square-integrable local martingale, with $X_0=0$ for simplcity. Because $X$ is locally square integrable, there is a localizing sequence $(T'_n)$ such that $E[(X^{T_n'}_t)^2]<\infty$ for each $t>0$ and each $n$. Because $X$ is a local martingale
there is a localizing sequence $(T''_n)$ such that $(X^{T_n''})$ is a uniformly integrable martingale, for each $n$. Define $T_n:=T_n'\wedge T_n''\wedge n$. Then $(T_n)$ is an increasing sequence of stopping times with a.s. limit $\infty$.  Moreover, $X^{T_n}$ is for each $n$ both square integrable and a U.I. martingale. In fact, because $T_n\le n$, $X^{T_n}$ is an $L^2$ martingale. Thus by Theorem 8, each $X^{T_n}$ is a semimartingale. Finally, by the Corollary to Theorem 6, $X$ is a semimartingale.


问:
Non-reflexive module isomorphic to its double dual

Could you give me an example of a non-reflexive module isomorphic to its double dual?

I found an example here but I cannot understand it, do you have any simpler examples?

By this question we should only use not finitely generated modules.

答:
What about $l^1(\mathbb{R})$ as an $\mathbb{R}$ vector space?

The natural map $J_1\colon l^\infty \to (l^1)^\vee$, which reads a bounded sequence as the multiplication by said sequence, is an isomorphism, while the dual map $J_2\colon l^1 \to (l^\infty)^\vee$ is not. We then get, composing the natural bidual map with the traspose of $J_1$, that
$$l^1 \to (l^1)^{\vee\vee} \to (l^\infty)^{\vee}$$
is, again, $J_2$! Hence $l^1$ cannot be reflexive. Still, we can find bases of the same cardinality for the two spaces.


问:
How can I prove, the sequence converges to the given limit?
Prove that the sequence $ {f(n)}$ defined by $$0<f(1)<f(2) \\ f(n+2) = [f(n+1)\cdot f(n)]^{1/2}$$ for $n \geq 1$ converges to $$[f(1) \cdot  {f(2)}^2]^{1/3}$$

答:
Hint: take the logarithm of $f(n)$ and turn it into a linear difference equation.

After that there's a shortcut to solve the equation, see
Finding explicit formula for recursive relation


答:
Your matrix $M$ is diagonalizable with eigenvalues $\pm i\sqrt{89}$. This means that $e^M = p(M)$ where $p \in \Bbb{C}[x]$ is the unique polynomial of degree less than $2$ such that
$$p(i\sqrt{89}) = e^{i\sqrt{89}}, \quad p(-i\sqrt{89}) = e^{-i\sqrt{89}}.$$

Using Lagrange interpolation formula, we see that
$$p(x) = \frac{x+i\sqrt{89}}{2i\sqrt{89}}e^{i\sqrt{89}}-\frac{x-i\sqrt{89}}{2i\sqrt{89}}e^{-i\sqrt{89}} = \frac{\sin\sqrt{89}}{\sqrt{89}}x+ \cos\sqrt{89}$$
so
$$e^M = p(M) = \frac{\sin\sqrt{89}}{\sqrt{89}}M+ \cos\sqrt{89}I = \left(
\begin{array}{cc}
 \frac{7 i \sin \sqrt{89}}{\sqrt{89}} +\cos\sqrt{89}& -\frac{(6+2 i) \sin \sqrt{89}}{\sqrt{89}} \\
 \frac{(6-2 i) \sin \sqrt{89}}{\sqrt{89}} &  -\frac{7 i \sin \sqrt{89}}{\sqrt{89}} +\cos\sqrt{89} \\
\end{array}
\right).$$

Title: Calculating matrix exponential

Body:
Given matrix $$M = \begin{pmatrix} 7i& -6-2i\\6-2i&-7i\end{pmatrix}$$ how do I calculate matrix exponential $e^M$?

I know I can use that $e^A=Pe^DP^{-1}$ where $D=P^{-1}AP$. I computed the characteristic polynomial of the above matrix as

$$P(\lambda)=\lambda^2+89$$

Is there an easier way to do this than trying to compute the diagonalized matrix?


问:
Finding probability of an absolute value function
$X$ and $Y$ are continuous random variables, and $Y=|X|$

I want to find $P(Y \leq y) = P(|X| \leq y)$

So I attempted the following

$$\begin{align*}
P(|X| \leq y) & = P(|X|<y) && \text{ Since } X \text{ is a cont. r.v. } P(X=y)=0 \\
& = P(X<y)+P(X>-y)
\end{align*}$$

However, my in my textbook's solution,
$$P(|X| \leq y) = 1 - P(|X|>y) = 1 - (P(X>y)+P(X<-y))$$

I was wondering if these two equations are equal. If we rearrange my textbook's solution,
$$\begin{align*}
P(|X| \leq y) & = (1 - P(X>y)) - P(X<-y) \\
& = P(X<y)-P(X<-y)
\end{align*}$$

Therefore, we check if $$P(X>-y) = -P(X<-y)$$

$-P(X<-y) = -(1 -P(X>-y)) = -1 + P(X>-y)$

We can see that
$$-1+P(X>-y)\neq P(X>-y)$$

So I'm not sure if my calculations are wrong, but why does the book do $P(|X| \leq y) = 1 - P(|X|>y)$ instead of directly calculating $P(|X| \leq y) = P(X<y)+P(X>-y)$ and why are these approaches different?

Is it wrong that I assumed $P(|X| = y) = 0$?

答:
Here is an easy way to solve the problem

$$F_Y(y)=\mathbb{P}[|X|\leq y]=\mathbb{P}[-y \leq X\leq y]=F_X(y)-F_X(-y)$$

Example:

Let $X\sim U[-1;1]$ . Find the distribution of $Y=|X|$

$$F_Y(y)=F_X(y)-F_X(-y)=\frac{y+1}{2}-\frac{1-y}{2}=y$$

Thus $Y\sim U[0;1]$

EDIT: answering to your comment,

$$\mathbb{P}[-y\leq X\leq y]=\underbrace{\mathbb{P}[-\infty\leq X\leq y]}_{=\mathbb{P}[X\leq y]=F_X(y)}-\underbrace{\mathbb{P}[-\infty\leq X\leq -y]}_{=\mathbb{P}[X\leq -y]=F_X(-y)}$$


问:
Constraint optimization question with linear function being constrained
I have a multivariable function $F(x, y) = a(y^2+x)$ representing the total manufacturing output of some process. $x$ and $y$ are variables representing the quantity of resources of each type input into the process and $a$ is a constant which depends on the machine used.

The idea is that we are using the process $F$ which accepts quantities of two types of resources ($x$ and $y$), of which $y$ is far superior (it contributes exponentially to the number of products produced whereas $x$ contributes linearly).

Now, $x$ is cheaper to procure than $y$. So the total cost of a manufacturing run is given by:

$C = tx + wy$

Where $t$ is the cost per unit of $x$ and $w$ is the cost per unit of $y$.

My goal is to find the ideal combination of $x$ and $y$ to fulfill a customer's order of $P$ products. That is, the cheapest combination of $x$ and $y$ which will yield $P$ as the value of $F(x, y)$.

How can I approach this a problem?

Edit:

Thanks to the tip about using Lagrangian multiplier, I now have the following:

$P = ay^2 + ax$ <-- The constraint function

$C = tx + wy$

Let us define the Lagrangian:

$\mathscr{L}(x, y, \lambda)=tx + wy -\lambda(ay^2+ax-P)$

$=tx + wy -a \lambda y^2 -a \lambda x +\lambda P$

We take the partial derivatives and set to zero to get the critical points of the Lagrangian:

$0 = \frac{\partial \mathscr{L}}{\partial x} = t-a \lambda$    $\dagger$

$0 = \frac{\partial \mathscr{L}}{\partial y} = w-2a \lambda y$   $\ddagger$

And we have our constraint function, with $P$ substituted for the output:

$0 = -ay^2 - ax + P$

So the derivative of the Lagrangian at the critical points is given by:

$\nabla \mathscr{L} = \begin{bmatrix} t-a \lambda \cr w-2a \lambda y \cr -ax - ay^2 + P\end{bmatrix} = \begin{bmatrix} 0\cr 0 \cr 0 \end{bmatrix}$

From here I am unsure how to interpret this result to find the minimum of $C$ in terms of $x$ and $y$.

When I try to work out the values for $x$ and $y$ I run into a problem: using the derivative in $\ddagger$ I can get $y$ in terms of $\lambda$ but because the derivative in $\dagger$ does not have a $x$ term, I don't see how I can get $x$ in terms of $\lambda$. I am sure I am missing something here.

答:
This is an equality constrained minimization problem.

You want to minimize $C(x,y) = tx + wy$ given that $F(x,y) = a(y^2+x) = P$ or $G(x,y)=a(y^2+x)-P = 0$, which is the equality constraint.

This type of problem can be solved using Lagrange Multipliers, in which we define a new function $\mathscr{L}(x,y)$, called the Lagrangian.

$$\mathscr{L}(x,y,\lambda) = C(x,y) - \lambda G(x,y)$$

We then minimize this new function. To do so, we must obtain the gradient of the Lagrangian and then calculate the point $P=(x,y,\lambda)$ which makes the gradient be zero.

$$\nabla \mathscr{L} = \begin{bmatrix} t-a \lambda \cr w-2a \lambda y \cr -ax - ay^2 + P\end{bmatrix} = \begin{bmatrix} 0\cr 0 \cr 0 \end{bmatrix}$$

Now we solve for $x, y$ and $\lambda$.
From the first term of the gradient we get:

$$\lambda = \frac{t}{a}$$
Substituting this in the second term, we get:

$$y=\frac{w}{2t}$$

Finally we can solve for $x$ with the third entry of the the gradient:

$$x=\frac{P}{a}-\frac{w^2}{4t^2}$$

We now know that the point $P_c=(\frac{P}{a}-\frac{w^2}{4t^2},\frac{w}{2t},\frac{t}{a})$ is a critical point of the Lagrangian. But to tell whether it's a minimum or a maximum, we use the discriminant, which is the determinant of the hessian matrix, defined as:

$$H(x,y, \lambda)=\begin{bmatrix}
{\frac{\partial^2 \mathscr{L}}{\partial^2 x}} & {\frac{\partial^2 \mathscr{L}}{\partial x \partial y}} & {\frac{\partial^2 \mathscr{L}}{\partial x \partial \lambda}}
\\ {\frac{\partial^2 \mathscr{L}}{\partial y \partial x}} & {\frac{\partial^2 \mathscr{L}}{\partial^2 y}} & {\frac{\partial^2 \mathscr{L}}{\partial y \partial \lambda}} \\ {\frac{\partial^2 \mathscr{L}}{\partial \lambda \partial x}} & {\frac{\partial^2 \mathscr{L}}{\partial \lambda \partial y}} & {\frac{\partial^2 \mathscr{L}}{\partial^2 \lambda}} \end{bmatrix}$$

Calculating all these derivatives, gives us:

$$H(x,y, \lambda)=\begin{bmatrix}
{0} & {0} & {-a}
\\ {0} & {-2a\lambda} & {-2ay} \\ {-a} & {-2ay} & {0} \end{bmatrix}$$

The discriminant is then:

$$D(x,y,\lambda)=|H(x,y,\lambda)|$$
$$D(x,y,\lambda)=2a^3\lambda$$

Now, the point we found will definitely be a minimum if the discriminant is greater than zero on it.

$$D(P_c)=2ta^2$$

Since $a^2$ is always positive, it then depends on the value of $t$ whether the point is a minimum. If $t>0$ then the point $P_c$ will be the minimum, if $t<0$ then it will be a maximum, and if $t=0$ or $a=0$ then the hessian test is not enough.


问:
Soft question about analytic continuation of a (family) of curves.
A lot of times in math one considers not just one object but a family of objects, or a space of objects.

In set theory one often deals with collections of numbers that form a set. In functional analysis one can consider function spaces. In group theory one often cares about understanding the discrete collection of symmetries of an object.

I'm wondering if that's done at all with regards to analytic continuation. That is, does one ever consider a family of real functions and continue each and every one of them to the complex plane?

Is there anything one can learn by doing this as opposed to analytically continuing just one candidate function.

To be more specific, if you have $f(x)=1/x$ of a real variable you could analytically continue this to a function of a complex variable $f(z)=1/z$. But what if you have an infinite collection, like $f_k(x)=k/x$ for $k\in \Bbb R.$ This can be done by partitioning the real plane into a disjoint union. Then all the analytic continuations would differ by a constant multiple.

I am highly skeptical that there is any point in continuing each and every curve in the space because I think any one curve will capture the essence, the information that is needed. But I want to hear from more knowledgeable people.

In practice does one analytically continue just one function in the family, or analytically continue the space of all the curves in the family?

答:
Dirichlet L functions are an incredibly important tool in Analytic Number theory and they were first used by Dirichlet in his proof of the Prime Number Theorem in Arithmetic progressions. A key step of his proof is analytically continuing the function

$$\prod_{\chi\mod k}L(\chi,s)$$

to $\Re(s)\geq1$, where the product is taken over all Dirichlet characters $\chi$ with period $k$. I will note that his proof relies on showing that this function must have a pole at $\chi=1$ and thus can't be analytically be continued and instead must be meromorphically continued, but multiplying by $(s-1)$ yields a nice analytic continuation of a family of functions (any choice of $k$).


问:
6th grade question
I got this problem from some 6-th grade kid:

If $n=(a+b)(b+c)(c+a)$, where $a,b,c \in \mathbb{N}^*$, then prove that

$$\frac{2^{n+1}+7^{n-1}}{15} \in \mathbb{N}$$

It has been some time since I tackled problems like this one so I am a bit rusty on the subject.

I tried to first see if expanding the expression of $n$ would give rise to something that can related to the expression that has to be proven. I'll put it below although it is just basic algebra.

$$n = 2abc + a^2(b+c)+ b^2(a+c) + c^2(a+b)$$

Having no hint from that result I went to the expression that has to be proven. I wrote it as

$$14\cdot 2^n + 7^n = k\cdot 105$$

where $k\in \mathbb{N}^*$. One hint came from this, but it did not deliver a solution.

I noticed that $15=2^3 + 7$ from which I can write $15 \cdot k = k(2^3 + 7)$. From this, for $k = 1$ and $k=7$, I could check by hand if I can find a value for $n$ such that it satisfies the equation that has to be proven. I got the values $n=1$ and $n=2$. However, those values cannot be written in as it is required by the first relation.

Next I thought that putting numbers for the values $a,b,c$ could give some clues. Unfortunately brute force with a calculator fails since the smallest value for $n$ is $8$, and the next one is $18$.

Note 1: This last step was also meant to check if by pure luck I can find a value for $n$ where the equation does not hold, thus solving the problem like a boss by mentioning "The problem is incorrect or it has a typo".

Note 2: I am interested in a solution that is fit for a 6-th grader, however any other level (e.g. undergraduate) should be good (I'm not a mathematician, but an engineer).

答:
This answer is perfectly suitable for a $6$-grader.

First of all, note that $n$ is always even. To accept it, note that $n$ can be odd if and only if each one of $a+b$, $b+c$ and $c+a$ must be odd. If any of $a$, $b$ and $c$ is odd, then the other two must be even, implying their sum is even. Hence, $n$ is definitely even. If this sounds confusing, remember

$$\begin{align}
\mathrm{odd}\times \mathrm{odd}&=\mathrm{odd}\\
\mathrm{odd}\times \mathrm{even}&=\mathrm{even}\\
\mathrm{even}\times \mathrm{even}&=\mathrm{even}\end{align}$$

and

$$\begin{align} \mathrm{odd}+\mathrm{even}&=\mathrm{odd}\\
\mathrm{odd}+\mathrm{odd}&=\mathrm{even}\\
\mathrm{even}+\mathrm{even}&=\mathrm{even}\end{align}$$

Now, comes the second part of the question. Note that
$$\frac{2^{n+1}+7^{n-1}}{15}\in\mathbb{N}\equiv 15|(2^{n+1}+7^{n-1})$$
(If you don't know the statement means that $2^{n+1}+7^{n-1}$ is exactly divisible by $15$.)

Assume, for simplicity, $x=2^{n+1}+7^{n-1}$.

To prove this, we shall do this in two parts - a number is divisible by $15$ if and only if it is divisible by both $3$ and $5$.

A number is divisible by $5$ if and only if it ends in $5$ or $0$.

Thus, $x$ is always divisible by $5$.

Now, let's observe the remainder left by $2^{n+1}$ when divided by $3$.

Thus, $2^{n+1}$ always leaves a remainder of $2$ when divided by $3$.

Now, consider $7^{n-1}$.

Together, $x$ will make a remainder of $3$, which when divided by $3$ gives remainder $0$.

Thus, $x$ is exactly divisible by $3$.

Altogether, we can say that $x$ is divisible by $15$, as required.

Hope this helps :)

Title: 6th grade question