答:
You could also use a langrangian method to maximize under constraint, you can define, assuming $\forall i, p_i>0 $ :

$$\mathcal{L}(\lambda,p_1,...,p_n)=log(V)+\lambda\left(\sum_{i=0}^{n}p_i-1 \right) $$
So you have the system :
$$\frac{\partial\mathcal{L}}{\partial\lambda}(\lambda,p_1,...,p_n)=\left(\sum_{i=0}^{n}p_i-1 \right)=0 \\ \forall i \in \{1...n \}, \frac{\partial\mathcal{L}}{\partial p_i}(\lambda,p_1,...,p_n)=\lambda+\frac{1}{p_i}=0 $$
So you can conclude that $p_1=p_2=...=p_n$, and the condition $\sum_{i=0}^{n}p_i=1 $ gives you the final result :

$$\forall i \in\{1...n \}, p_i=\frac{1}{n} $$

Title: maximizing $p_1p_2\cdots p_n$ where $p_j$ are probabilities

Body:
Let $V = p_1 p_2 \cdots p_n$ where $p_i \in [0,1]$ and $\sum p_i = 1$, so the $p_i$ define a discrete probability distribution. Now I'd like to find $p$ that maximizes $V$. Intuitively it makes sense that the maximum must be at $p_i = 1/n$ for all $i$, and it is easy to verify that the gradient of $V$ vanishes there. But now I'd like to prove the reverse: If $V$ is maximal then $p_i = 1/n$.

With Lagrangian mulitpliers it is rather straightforward, but I've tried solving the system you get by setting the gradient to zero but without much success. Can anyone spot a simplification or a nice argument that would let me continue there?

Here's my attempt so far:
First I just plugged in the $\sum p_i = 1$ (and for ease of writing $m:= n-1$, and we assume that the index now goes up to $m$). So

$$V = \underbrace{\Big(\prod_i p_i \Big)}_{=:P}\Big(1 - \underbrace{\sum_i p_i}_{=:S}\Big)$$

Actually it should be possible to prove that we have only one local maximum even without the restriction of $p_i \in [0,1]$:

$$0 \overset{!}{=} \frac{\partial V}{\partial p_j} = \frac{P}{p_j} (1-S) + P(p_j-S) $$

assuming all $p_i \neq 0$ (if some $p_i=0$ we can discard it an restart the problem with less variables) we can simplify that to

$$ S-1 = p_j(p_j-S)$$

(Note that this is a system of equations that holds for all $j=1,\ldots,m$, and here $S$ and $P$ depend on $p_j$.)

Now at that point I thought maybe we can use some fixed point iteration argument but didn't manage to find anything. So I'd love to hear if anyone could expand this argument, but I'm also interested in other elegant ways of solving this.


问:
Series Expansion of Exponential Function is Cauchy
Define $p_n (x) = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \dots + \frac{x^n}{n!}$.
Show that $\{ p_n (x) \}_{n=0}^{\infty}$ is a Cauchy sequence.

I'd just like to know the facts required to show this. We know that if a function $f: A \mapsto \mathbb{R}$, $A \subseteq \mathbb{R}$, is uniformly continuous on $A$ then, if $(x_n)$ is a Cauchy sequence in $A$, then $(f(x_n))$ is also a Cauchy sequence. Is this a relevant fact? Do I need to find some sequence, say $y_n$, that is Cauchy, and then find an explicit function that maps the terms of $y_n$ to the terms of $p_n(x)$?

Is there an easier way to do this?

Edit: I need to expand on this... the norm is $||p|| =$max$_{[0,1]} | p(x)|$. So for $p_n$ and $p_m$ the difference is $\sum _{ i = \min (n,m)} ^{\max (n,m)} \frac {x^i}{i!}$. The maximum term is the first term, because I've shown this to be a decreasing sequence. Now, is it still bounded by the geometric series with the factor 1/2?

答: HINT: For sufficiently big $n$ (say $n>2|x|$) you can compare with a geometric series with a suitable factor (say $\frac12$).

ContentLicense: CC BY-SA 3.0


问:
Summation Simplification for Poisson Product
I have a function like the following:

$$f(u) = \sum^u_{v=0} \frac{e^{-\theta}\theta^{u-v}}{(u-v)!}*\frac{e^{-\lambda}\lambda^v}{v!} $$

I am trying to get the answer of
$$\frac{e^{-(\theta + \lambda)}(\theta + \lambda)}{u!},$$
but I am unsure how to get to this expression. How can I simplify $f(u)$? Thanks!

ContentLicense: CC BY-SA 4.0

Body:
$$f(u)=\sum_{v=0}^{u} \frac{e^{-t}t^{u-v}}{(u-v)!} ~ \frac{e^{-s} s^v}{v!}=\frac{e^{-t-s} t^u}{u!} \sum_{v=0}^{u}\frac{u! (s/t)^v}{v!(u-v)!}=\frac{t^ue^{-(s-t)}}{u!} \sum_{v=0}^{u} {u \choose v} (s/t)^v.$$
$$\implies f(u)=\frac{t^ue^{-(s+t)}}{u!} (1+s/t)^u=\frac{e^{-(s+t)}}{u!} (s+t)^u$$


问:
If $(X_k)_{k\in\mathbb N}$, show that $\sum_{k=1}^n(X_k-\text E[X_k])$ converges almost surely as $n\to\infty$
Let $\nu$ be a $\sigma$-finite measure on $\mathbb R$ with $$\int_{(-1,\:1)}\nu({\rm d}x)x^2<\infty\tag1$$ and $\nu(\{0\})=0$, $$I_k:=\left(-\frac1k,-\frac1{k+1}\right]\cup\left[\frac1k,\frac1{k+1}\right)$$ and $$\nu_k(B):=\nu(B\cap I_k)\;\;\;\text{for }B\in\mathcal B(\mathbb R)$$ for $k\in\mathbb N$ and $(X_k)_{k\in\mathbb N}$ be a real-valued independent process on some probability space with $$\operatorname E[X_k]=\int\nu_k({\rm d}x)x=\int_{I_k}\nu({\rm d}x)\tag2$$ and $$\operatorname{Var}[X_k]=\int\nu_k({\rm d}x)x^2=\int_{I_k}\nu({\rm d}x^2)\tag3$$ for all $k\in\mathbb N$ and $$M_n:=\sum_{k=1}^n\left(X_k-\operatorname E[X_k]\right)\;\;\;\text{for }n\in\mathbb N.$$

It's easy to see that $M$ is a martingale with $$\operatorname E\left[M_n^2\right]=\sum_{k=1}^n\operatorname{Var}[X_k]\;\;\;\text{for all }n\in\mathbb N\tag5$$ and hence $$\sup_{n\in\mathbb N}\operatorname E\left[M_n^2\right]=\sum_{k\in\mathbb N}\operatorname{Var}[X_k]=\int_{(-1,\:1)}\nu({\rm d}x)x^2<\infty\tag6.$$

Why can we conclude that $(M_n)_{n\in\mathbb N}$ is almost surely convergent as $n\to\infty$?

答: The martingale convergence theorem, for $L^2$-bounded martingales.


问:
Question about combinatorics in case the arrangement is not important.
So, I know that the number of combinations of $n$ objects taken $r$ at a time (where the order does not matter) is given by $\frac{n!}{(n - r)!}$ divided by $r!$. Based on my understanding, I would divide the product of given formula $\frac{n!}{(n-r)!}$ by just $r$. I don't understand why it should be divided by $r!$. Can someone explain? For example, if we were to select two managers from five people then the possible ways to do it would be $5\cdot4 \cdot \frac{1}{2} = 10$. Also, I only partly understand the reason you have to divide it by anything when the order of the selected objects is not important. How to better understand the case when the order is not important?

答:
Imagine we are choosing $3$ elements from $\{1,2,3,4,5\}$, and order does matter. Then there are $\frac{5!}{2!} = 60$ choices, shown below.

\begin{array}{cccccc}
   123 & 132 & 213 & 231 & 312 & 321 \\
   124 & 142 & 214 & 241 & 412 & 421 \\
   125 & 152 & 215 & 251 & 512 & 521 \\
   134 & 143 & 314 & 341 & 413 & 431 \\
   135 & 153 & 315 & 351 & 513 & 531 \\
   145 & 154 & 415 & 451 & 514 & 541 \\
   234 & 243 & 324 & 342 & 423 & 432 \\
   235 & 253 & 325 & 352 & 523 & 532 \\
   245 & 254 & 425 & 452 & 524 & 542 \\
   345 & 354 & 435 & 453 & 534 & 543
\end{array}
The way I've written them, each row has $3! = 6$ different permutations of the same $3$ elements: the first row has $3!$ permutations of $123$, the second row has $3!$ permutations of $124$, and so on. Every subset of $3$ elements has its row.

If we are choosing $3$ elements from $\{1,2,3,4,5\}$, and order doesn't matter, then each of the rows in this table only corresponds to a single choice. So we should divide $60$ by $3!$ (the number of elements in each row) to get the number of rows.

This generalizes.


问:
If $X\sim \textbf{Laplace}(0,\lambda)$. Prove that $|X|\sim \textbf{Exp}(\lambda)$
How can I prove this statement?

If $X\sim \textbf{Laplace}(0,\lambda^{-1})$. Prove that $|X|\sim \textbf{Exp}(\lambda)$

My approach: Since that $X\sim \textbf{Laplace}(0,\lambda)$, so the density function is $$f_{X}(x)=\frac{1}{2\lambda^{-1}}e^{-\frac{|x-0|}{\lambda^{-1}}}, x\in \mathbb{R}, \lambda>0$$
Now, let $Y=|X|$, so we have that $Y$ is $\nearrow$ in $[0,\infty)$ and $y=x$, $x\in [0,\infty)$.

So, by transformation theorem and taking $h^{-1}(y)=y$, we have that $$f_{Y}(y)=\frac{1}{2\lambda^{-1}}e^{-\frac{|y|}{\lambda^{-1}}},\quad x\geq 0 $$
and similar, we have that $$f_{Y}(y)=\frac{1}{2\lambda^{-1}}e^{-\frac{|-y|}{\lambda^{-1}}},\quad x<0$$
So, we have that $$f_{|X|}(x)=\frac{1}{2\lambda^{-1}}e^{-\frac{|x|}{\lambda^{-1}}}, x\in \mathbb{R}, \lambda>0$$

But, I think that I can't conclude with that information that $|X|\sim \textbf{Exp}(\lambda)$.

答:
The PDF of $Y,f_Y(y)$ has to depend on $y$, not $x$. Since $Y=|X|$ is not a bijection, you run into an issue.

The distribution function of $Y$ is$$F_Y(y)=P(Y\le y)=P(|X|\le y)=\begin{cases}0,&y\le0\\P(-y\le X\le y),&y>0\end{cases}$$The PDF of $Y$ is $0$ for $y<0$. Using $P(-y\le X\le y)=F_X(y)-F_X(-y)$ where $F_X(x)$ is the distribution function of $X$, we get for $y>0$,$$f_Y(y)=f_X(y)+f_x(-y)=2f_x(y)$$which is the PDF of exponential distribution.


问:
For $\forall$ x $\in$ [0,1], $\lim\limits_{n\to+\infty} f(x+n)=0$ ($n\in \mathbb{N}$). Prove that $\lim\limits_{x\to+\infty}f(x)=0$.
I encountered a proof question:

Suppose $f(x)$ is uniformly continuous in $[0,+\infty)$. For $\forall$ x $\in$ [0,1], $\lim\limits_{n\to+\infty} f(x+n)=0$ ($n\in \mathbb{N}$). Prove that $\lim\limits_{x\to+\infty}f(x)=0$.

The question is whether the uniform continuity of $f(x)$ is necessary. When I use $x-[x]$ to represent $x\in [0,1]$, use $[x]$ to represent $n$, then $n\to +\infty$ means $[x] \to +\infty$, which can deduce $x\to+\infty$, then complete the proof. But condition uniform continuity of $f(x)$ is not used. Can someone point out the loopholes of my proof?

答:
It is possible to find a counterexample which shows that even if $f$ is continuous, and, for all $x\in[0,1]$, $\lim_{n\to\infty}f(x+n)=0$, this is not enough to conclude $\lim_{x\to\infty}f(x)= 0$ (hence, you need something stronger than continuity to conclude). Consider a function $f$ defined as follows: For $n\ge 2$, let $I_n=[1/n+n-1/n^2, 1/n+n+1/n^2]$, and let $f$ be a continous function s.t.:
$$
f(x)=
\begin{cases}
0 & x\notin \bigcup_n I_n\\
>0 & x\in \bigcup_n I_n\\
1 & x\in \bigcup_n \{1/n +n\}
\end{cases}
$$

This function doesn't converge to 0. However:

For $x=0, x+n\notin I_n=[(1/n-1/n^2)+n, (1/n+1/n^2)+n]$ for any $n$, thus $f(x+n)=0$ for all $n$. Same for $x > 3/4$.

For $x\in (0, 3/4]$, there will be some instant $n_0$ s.t. for all $n\ge n_0$, $x>1/n + 1/n^2$, thus, $x+n\notin I_n=[(1/n-1/n^2)+n, (1/n+1/n^2)+n]$ for all $n\ge n_0$. Hence, $f(x+n)=0$ for all $n\ge n_0$

In short, for all $x\in[0,1]$, $\lim_{n\to\infty}f(x+n)=0$.

As for errors in your proof. You say that, if you take a sufficiently large $x$, then $x=(x-[x]) + [x]$ with $[x]$ large, and because $\lim_{n\to\infty}f(x-[x]+n)=0$, $f(x)$ will be small. The problem I'd say is that for some such $x$, the sequence $(f(x-[x]+n))_n$ may fall to zero much later than at the instant $[x]$


问:
Homeomorphisms with quotient spaces
I'm finding some problems in solving this exercise: let $X=[0,1] \times [0,1]$ and $A= \left \{ 0 \right \} \times [0,1]$.

Show that $X/A=\left ([0,1] \times [0,1]  \right )/ \left ( \left \{ 0 \right \} \times [0,1]  \right )$ is homeomorphic to a triangle.

Show that $X/A$ is homeomorphic to $X$.

Regarding point 1, my goal is to answer without using compactness, writing a continuos and bijective function $f: X \rightarrow Y$, where $Y$ is a triangle, constant on the equivalente classes of $X/A$. This implies that the function $[f]: X/A \rightarrow Y$ is continuos and bijective too. Since the composition $f= [f] \circ \pi$ is commutative (where $\pi$ is the natural projection) I'll have to show that also $f^{-1}$ is continuos, and then $[f]^{-1}$ is continuos too. Hence $[f]$ is an homeomorphism.

How to formalise all this? I need a general method, so that I can solve point 2 by myself.

答:
Hint : Try to prove that the function $g : X \rightarrow X$ defined by
$$g(x,y)=(x,xy)$$

defines a injection $\tilde{g} : X/A \rightarrow X$, and show that $\tilde{g}$ is a homeomorphism between $X/A$ and the triangle of vertices $(0,0)$, $(1,0)$ and $(1,1)$.

Title: Homeomorphisms with quotient spaces


问:
Differentiation of integrals
I have a question I'm struggling with for quite a few hours now. For most of you its probably kindergarten mathematics: Can someone please explain exactly why the first-order condition of
$$
\max \left\{\frac{1}{1-\beta} \left[ \int_{0}^{N(t)}x_{\nu}(t)^{1 - \beta} d\nu \right]L^{\beta} - w(t)L -  \int_{0}^{N(t)} p_{\nu}(t) x_{\nu}(t)d\nu \right\}
$$
with respect to $ x_{\nu}(t)$ is $x_{\nu}(t) = Lp_{\nu} (t)^{-\frac{1}{\beta}}$. (I know how the basic differentiation rules but with the integrals I'm no sure why we can "throw them away"... Would be very happy about help! Thanks!

答:
I assume that "the first-order condition of $\max\{f\}$" refers to the equation $\frac{\partial f}{\partial x} = 0$. With that said, we have (using the Leibniz integral rule)
$$
\frac{\partial}{\partial x}\left[\frac{1}{1-\beta} \left[ \int_{0}^{N}x_{\nu}^{1 - \beta} d\nu \right]L^{\beta} - wL -  \int_{0}^{N} p_{\nu} x_{\nu}d\nu\right] = \\
\frac{1}{1-\beta} \left[\int_{0}^{N} \frac{\partial}{\partial x} x_{\nu}^{1 - \beta} d\nu \right]L^{\beta} - \frac{\partial}{\partial x} wL -  \int_{0}^{N} \frac{\partial}{\partial x}p_{\nu} x_{\nu}d\nu =\\
L^{\beta}\int_{0}^{N} x_{\nu}^{- \beta} d\nu  -  \int_{0}^{N} p_{\nu} d\nu = \\
\int_0^N [L^\beta x_\nu^{- \beta} - p_\nu]d\nu.
$$
So, we are apparently interested in quantities of $x_\nu$ for which $\int_0^N [L^\beta x_\nu^{- \beta} - p_\nu]d\nu = 0$. Now, we would need more context to know exactly what is meant by saying that your equation is "the first order condition". However, it is certainly true that if $[L^\beta x_\nu^{- \beta} - p_\nu] = 0$ for all $0 \leq \nu \leq N$, then the integral must be zero. So, setting this quantity to zero guarantees that the first order condition holds, but is not equivalent to the first order condition.

Rearranging this equation yields
$$
[L^\beta x_\nu^{- \beta} - p_\nu] = 0\\
L^\beta x_\nu^{- \beta} = p_\nu\\
x_\nu^{- \beta} = L^{-\beta}p_\nu\\
x_{\nu} = L p_\nu^{-1/\beta}.
$$

Title: Differentiation of integrals


问:
Show that a graph with 9 vertices has at least five vertices of degree 6, or at least six vertices of degree 5.
Let $G = (V, E)$ be a graph with nine vertices, such that each vertex has degree $5$ or $6$. Show that $G$ has at least five vertices of degree $6$, or at least six vertices of degree $5$.

My friend and I have been working on this question for the past two days and we have nothing. Any pointers or help?

答: Hint: The sum of all vertex degrees is twice the number of edges.


问:
To show that there are $3$ numbers between $n^2$ and $(n+1) ^2$ such that they satisfy a certain property.
Here is the problem

Show that for any natural number $n$, one can find three distinct
natural numbers $a, b, c$ between $n^2$ and $(n + 1)^2$ such that $a^2 + b^2$ is divisible
by $c$.

It seems like here are several ways to express $a^2 + b^2$.
\begin{array}
 aa^2 + b^2 &= (a - b) ^2 +  2ab \\
 &= (a + b) ^2 - 2ab \\
 &= \frac { (a + b) ^2 + (a - b)^2 }{2} \\
 &= 2b^2 + (a + b)(a - b)
\end{array}
What the problem is literally saying: Find algebraic expressions $X,Y$ and $Z$ such that $n^2 \lt X,Y,Z \lt(n + 1)^2$ and $Y^2 + Z^2 = AX$, for some expression $A$. (Hence, $X|Y^2 + Z^2$)

In the original problem, $X$ is $c$, $Y$ and $Z$ are $b$ and $a$.

People with enough experience might know the answer at once, but is it possible to actually find the expressions from identities/relations? For example the relations that I have found above?

From my trying it is very difficult because one needs to search for expression that are restricted between $n^2$ and $(n + 1)^2$, and then the factoring. What I tried was just plugging random expressions to $Y$ and $Z$ and trying to find a suitable $X$ that suffices the factoring, but to no avail.

Please help, and thanks for that!

答:
(Note: In the fixed olympiad question, $n\geq 2$, and $ n^2 < a , b, c < (n+1)^2$.)

There's a bunch of wishful thinking in solving this problem, which is a good healthy approach to take esp when solving Olympiad problems which typically have a "nice solution" but we have no idea what it is (IE "My problem is that how to find them" from the comments).

I've indicated this via "reasonable to assume". Of course, if things don't work out, then we have to revisit these "assumptions".

For $n=2$, the only solutions are $(6,7,5)$ and $(6,8,5)$.

For $n=3$, the only solutions are $(11,13,10)$, $(12,14,10)$, $(10, 11, 13)$, $(10, 15, 13)$.

At the point, it is reasonable to assume that we must have $c = n^2 + 1$. I like this assumption because it focuses what I'm investigating. We can reconsider what happens if this doesn't turn out to be true.

For $n=4, c = 17$, the only solutions are $(18, 21, 17)$ and $(20, 22, 17)$.

It is reasonable to assume that we have (a/several) family of solutions, and so let's see how we can split them.

It is reasonable to assume that the "smaller solution" $(6,7,5) \sim (11,13,10) \sim (18, 21, 17)$ is a family, which we can observe is $ (n^2 + 2, n^2+n+1, n^2 + 1)$.

It is likewise reasonable to assume that the "larger solution"  $(6,8,5) \sim (12, 14, 10) \sim (20, 22, 17)$ is a family, which we can observe is $(n^2 + n, n^2 + n + 2, n^2 + 1)$.

Finally, we verify that these families actually work.

Notes

Looking at the other solutions for $n=3$, it might be reasonable to assume that we have solutions for $c = n^2 + n + 1$ when $n \geq 3$. However, checking $n =4, 5 $ doesn't yield such a solution. This is a good example of where wishful thinking doesn't work out, and so we have to change something else.

In fact, for $n = 4$, the only solutions are those that we found. So if we didn't initially jump to the reasonable assumption that $c = n^2 +1$ after $n=3$, then we almost certainly should have after checking $n=4$


问:
Is there an analytical solution for the following
Is there an analytical solution to this equation:

$A. dM/dt + B. dN/dt = C$

Initial condition,At $t = 0$ ; $M(0) = N(0) = const$

$A,B,C$ are constants and $M(t)$ and $N(t)$.

答:
$$A. dM/dt + B. dN/dt = C$$
If $A,B,C$ are constants:
$$A. dM + B. dN = Cdt$$
Integrate:
$$AM+BN=Ct+K$$
Where $K$ is a constant $(K=AM(0)+BN(0))$. You have a relation between $M$ and $N$ but you need more information if you need $M$ or $N$ as a function only of the variable $t$.


问:
Definition of finite dimensional representation of a compact group
Consider the following fragment:

One asks that the map $\pi: G \to GL(V)$ is continuous. What does this mean? If $V$ is finite-dimensional, then we can view $GL(V)$ as matrices and then I guess it means that all the matrix entries are continuous?

What does continuity mean when $V$ is a Banach space (not necessarily finite-dimensional)? Which topology is put on $GL(V)$? Is it pointwise convergence topology?

答:
One of the most important representations of the additive group $\mathbb R$ takes place in the Hilbert space
$L^2(\mathbb R)$ and is given by
$$
\pi _s(\xi )|_t =  \xi (t-s), \quad \forall \xi \in L^2(\mathbb R),\ \forall t,s\in \mathbb R.
$$
This representation is not continuous with the norm topology on operators, but it is continuous with the strong
(i.e. pointwise) topology.  For that reason, most results about representations of topological groups assume strong continuity!

For a general locally compact group $G$,  there exists a left invariant measure $\mu $ on $G$ (called the Haar measure),  so we
may consider a similar  representation, this time  of $G$ on $L^2(G, \mu )$, given  by
$$
  \pi _s(\xi )|_t =  \xi (s^{-1}t), \quad \forall \xi \in L^2(G, \mu ),\ \forall t,s\in G.
  $$
Unless $G$ is discrete, this representation is not norm-continuous but again it is strongly continuous.


问:
How can I prove $c=\sup(S)$ satisfies $f(c)=d$?
I believe this is an alternative way to prove intermediate value theorem. Do not know if it is harder.

Let $D=[a,b]$ be a closed interval, where $a<b$ are real numbers. Let $f \colon D \rightarrow \mathbb{R}$ be a continuous function such that $f(a)<f(b)$. Consider $d \in \mathbb{R}$ such that $f(a)<d<f(b)$ and consider the set $S=\{x \in [a,b] \colon f(x) \leq d\}$.

We know that $S$ is not empty since $f(a) < d \implies a \in S$. We also know $S$ is bounded since it is a subset of $D$ ($S\subseteq D$). Thus by Axiom of Completeness, $\sup(S)$ exists.

How can I prove $c=\sup(S)$ satisfies $f(c)=d$? (without IVT obviously)

答:
Can we have $f(c)<d$? If that happens, then $c<b$ (because $f(b)\geqslant d$). Take $\delta>0$ such that$$|x-c|<\delta\wedge x\in[a,b]\implies\bigl|f(x)-f(c)\bigr|<d-f(c).$$Then, if $x_0\in(c,c+\delta)\cap[a,b]$ (this set is not empty, since $c<b$), you have\begin{align}f(x_0)&=f(x_0)-f(c)+f(c)\\&\leqslant\bigl|f(x_0)-f(c)\bigr|+f(c)\\&<d-f(c)+f(c)\\&=d.\end{align}But then $x_0\in S$ and $x_0>c$, which is impossible, since $c=\sup S$.

What about having $f(c)>d$? Then, take $\delta>0$ such that$$|x-c|<\delta\wedge x\in[a,b]\implies\bigl|f(x)-f(c)\bigr|<f(c)-d.$$Then, if $x_0\in(c-\delta,c)\cap[a,b]$,\begin{align}f(x_0)&=f(x_0)-f(c)+f(c)\\&\geqslant-\bigl|f(x_0)-f(c)\bigr|+f(c)\\&>-\bigl(f(c)-d\bigr)+f(c)\\&=d.\end{align}So, there is no element of $S$ in $(c-\delta,c)\cap[a,b]$, which is impossible, again because $c=\sup S$.

So, since we don't have $f(c)\ne d$, we must have $f(c)=d$.


问:
shortest distance between a point and a line segment
Okay, I know what you're thinking : this question has been asked a billion times before. But I have done my research, and sadly, the only answers I have found were presented in computer code, which I never learned, or involve too many vectors and constants that might seem a bit redundant (I'm sorry if I'm offending anybody. I'm a bit rusty on geometry, since it's been a while).

Basically, you have 3 coordinates that you know:
1) (px,py) is the coordinate of the point
2) (vx1, vy1) is the coordinate of one end of the line segment
3) (vx2, vy2) is the coordinate of the other end

Is there a way to find the shortest distance between this point and that segment? Please give me an answer that's relatively simple, using basic geometry/ algebra to explain. I love you all!

答:
I was never all that good with linear algebra, so someone please tell me if this is incorrect, but here goes...

First, note that if we have a line segment defined by vectors $\mathbf{P_0}$ and $\mathbf{P_1}$, then let $\mathbf{M} = \mathbf{P_1} - \mathbf{P_0}$. Then every point on the line can be represented as $\mathbf{P_0} + \mathbf{M}t$ for some $t$. If we restrict $0 \leq t \leq 1$ then we get all of the points on the line segment.

Going back to your problem, we have points $\mathbf{P_0}$ and $\mathbf{P_1}$ that comprise the endpoints of a line segment, and we have a point $\mathbf{Q}$. We want to find the unique $t$ that minimizes the distance between $\mathbf{P_0} + \mathbf{M}t$ and $\mathbf{Q}$, with the restriction that $0 \leq t \leq 1$. Let's also assume that $\mathbf{Q}$ is not on the line segment (this will be important later).

First, let's make this simpler by assuming that $\mathbf{P_0}$ is at the origin. So We simply want to minimize the distance between $\mathbf{Q}$ and $\mathbf{M}t$. Note that this distance is

$$D(t) = \sqrt{(\mathbf{Q} - \mathbf{M}t) \cdot (\mathbf{Q} - \mathbf{M}t)}$$

Because dot products are both distributive and commutative, we can expand this out to

$$D(t) = \sqrt{\mathbf{Q} \cdot \mathbf{Q} - 2\mathbf{Q} \cdot \mathbf{M}t + \mathbf{M} \cdot \mathbf{M}t^2}$$

In order to find the value of $t$ that minimizes this equation, let's take the derivative with respect to $t$:

$$D'(t) = \dfrac{1}{2} \times \dfrac{1}{\sqrt{\mathbf{Q} \cdot \mathbf{Q} - 2\mathbf{Q} \cdot \mathbf{M}t + \mathbf{M} \cdot \mathbf{M}t^2}} \times (-2\mathbf{Q} \cdot \mathbf{M} + 2 \mathbf{M} \cdot \mathbf{M}t) = 0$$

First, let's note that because we assumed that $\mathbf{Q}$ is not colinear with $\mathbf{P_0}$ and $\mathbf{P_1}$, the distance between $\mathbf{Q}$ and $\mathbf{M}t$ is positive for all $t$, so the denominator in this equation is always a positive.

Let's simplify the equation to
$$\dfrac{-\mathbf{Q} \cdot \mathbf{M} + \mathbf{M} \cdot \mathbf{M}t}{\sqrt{\mathbf{Q} \cdot \mathbf{M}t - 2\mathbf{Q} \cdot \mathbf{Q} + \mathbf{M} \cdot \mathbf{M}t^2}} = 0$$

We can multiply both sides by that ugly denominator, and the 0 makes it disappear! This leaves us with
$$-\mathbf{Q} \cdot \mathbf{M} + \mathbf{M} \cdot \mathbf{M}t = 0$$
$$t = \dfrac{\mathbf{Q} \cdot \mathbf{M}}{\mathbf{M} \cdot \mathbf{M}}$$

Note that technically we would want to take the second derivative to confirm that this is a minimum and not a maximum. But right now I am feeling lazy, and also intuitively I don't see how this function could be anything other than concave up.

If $t < 0$, then this tells us that to find the closest point we would have to extend the line segment past $\mathbf{P_0}$. So truncate $t \leftarrow 0$ and output $\mathbf{M} \times 0 = \mathbf{P_0}$ (i.e., the origin) as the closest point.

If $t > 1$, then this tells us that to find the closest point we would have to extend the line segment past $\mathbf{P_1}$. So we truncate $t \leftarrow 1$ and output $\mathbf{M} \times 1 = \mathbf{P_1}$ as the closest point.

Otherwise, we output $\mathbf{M}t$ as the closest point.

Ok, so what happens if $\mathbf{P_0}$ is not at the origin? Easy! We simply translate all three points until $\mathbf{P_0}$ is at the origin. We compute $\mathbf{P_0'}, \mathbf{P_1'}, \mathbf{Q'}$ such that $\mathbf{P_0'}$ is at the origin, and the other two points have been translated by the same amount in each dimension. We compute $\mathbf{M} = \mathbf{P_1} - \mathbf{P_0}$ the optimal value of $t = \dfrac{\mathbf{Q'} \cdot \mathbf{M}}{\mathbf{M} \cdot \mathbf{M}}$ Then, we just plug in this value of $t$ with our original points, and output $\mathbf{P_0} + \mathbf{M}t$

If this is incorrect, someone please let me know and I will gladly take this down ASAP.


问:
What is the difference between the rings $O_{(p)}$ and $O\left[\frac{1}{p} \right]$?
Let $K \supset \mathbb{Q}_p$ be the $p$-adic field. Let $O$ be its ring of integers. Let us denote by $O_{(p)}$ be the localisation of $O$ at the prime/maximal ideal $(p)$. Consider the another ring $O\left[\frac{1}{p} \right]$ obtained by inverting the prime $p$.

My question:

What is the difference between the rings $O_{(p)}$ and $O\left[\frac{1}{p} \right]$ ?

Edit: If $O=\mathbb{Z}_p$, then $O\left[\frac{1}{p} \right]=\mathbb{Q}_p$. But $O_{(p)}=(O \setminus \{(p)\})^{-1} O \subset \mathbb{Q}_p$.

答:
For $K/\Bbb{Q}_p$ a finite extension and $O$ its ring of integers then $(p)$ doesn't have to be a prime ideal, for example with $K=\Bbb{Q}_p(p^{1/2}),O=\Bbb{Z}_p[p^{-1/2}]$, the unique maximal ideal is $(p^{1/2})$. This is what we mean with "an uniformizer $\pi_K$": an element generating the maximal ideal.

We always have $(O-(\pi_K))^{-1}O = O$ and $O[p^{-1}]=K$.


问:
Is this DFA correct for the language $\{w \mid w \in \{ab\}^*\}$?
Language: $\{w \mid w \in \{ab\}^*\}$

答: your DFA will only accept string which starts from ab .so to make it work make every state will be a final state.</p>

Body: Language: $\{w \mid w \in \{ab\}^*\}$


问:
Intuitive way of constructing a graph with a large chromatic number?
I want to construct a graph with  a relatively large chromatic number (10). My only problem is that I can't have K9 (or the complete graph with 9 vertices) as an induced subgraph. Also My edges are at most 6n-12 (where n is the number of vertices). Without dwelling on the details what is one intuitive way I can go about doing this?</p>

答:
I couldn't get clarification via a comment due to insufficient reputation, but I assume the "or" in the parentheses in the second sentence is a typo, since the complete graph on 9 vertices is commonly denoted as $K_9$.

An easy way to do this would be to find a graph $G$ with chromatic number 5 and no induced $K_5$ as subgraph. Now to construct the graph of chromatic number 10 with no $K_9$, simply copy G two times and add all possible edges between the two copies of $G$ (such that when ignoring the edges within the copies of $G$, the vertices form a biclique).

To prove that this new graph has no induced $K_9$, consider that any set of 9 vertices must take at least 5 from one of the copies of $G$. However, $G$ does not contain a $K_5$, hence these 9 vertices cannot form a $K_9$.

To prove that it has chromatic number 10, consider that the colors used in one copy of $G$ must be completely disjoint from those used in the other copy, since we added all possible edges between them. Hence any coloring with less than 10 colors would imply a coloring with less than 5 colors for $G$.

The edge constraint can be easily circumvented by adding isolated vertices until it is satisfied.

All that remains is to construct a graph $G$ with chromatic number 5 and no induced $K_5$. Here is a paper that constructs such a graph with 11 vertices at the start of section 2: https://doi.org/10.1002/jgt.3190190111 (note that they construct one that does not contain an induced $K_4$, which is stronger). There is probably a simpler construction for such a graph $G$, but this certainly works.


问:
Showing a summation of powers of a prime power $q$
I calculated that the number of rank $k$ matrices in $\mathbb{F}_q^{n \times m}$ (where we may assume $n \leq m$) for a prime power $q$ is equal to

$$W_{n,m}(k) := \prod_{i = 1}^k \frac{(q^m - q^{i-1})(q^n - q^{i-1})}{(q^k - q^{i-1})}$$

And so summing all $W_{n,m}(k)$ from rank $k = 0$ to full rank $k = n$ should gives us all matrices of $\mathbb{F}_q^{n \times m}$

$$\sum_{k = 0}^n W_{n,m}(k) = q^{nm}$$

Can we also show this equality using pure arithmetic and combinatorial tricks, i.e. using just the formulas for $W_{n,m}$?

答:
I managed to solve this, but it took a number of pages. The full answer can be found here.

In short: one can show that the following $q$-analogue of Pascal's rule holds:
$$\prod_{i=1}^k \frac{q^{n+1} - q^{i-1}}{q^k - q^{i-1}} = q^k \cdot \prod_{i=1}^k \frac{q^{n} - q^{i-1}}{q^k - q^{i-1}} + \prod_{i=1}^{k-1} \frac{q^{n} - q^{i-1}}{q^{k-1} - q^{i-1}}$$

Then, this allows you to show that
$$
W_{n+1,m}(k) = q^k\cdot W_{n,m}(k) + (q^m - q^{k-1})\cdot W_{n,m}(k-1)
$$
by multiplying both sides with $\prod (q^m - q^{i-1})$.

Lastly, use induction on $n$ to get the final equality.

(Fun challenge, try to find the linear algebra interpretation of each statement)


问:
How to find expected value and variance of $a^X$, where a is positive.
How to find the expected value and variance of $a^X$, where a is positive. $X$ is a Poisson random variable with $\lambda$. This is very interesting, $X$ is at the exponential position. I only know the moment generating function, should we let $Y=a^X$, then write $E[Y]$ as sum of the value of $Y$ times the corresponding probability? But it seems very hard to discuss.

For variance, we can use the formula, but we still need to find $E[a^X]$ first.

答:
$Y=a^X$, so $Y$ takes the value $a^n$ with probability $f_X(n)$ where $f_X(m)$ is the PMF of $X$.

You can use $E[g(X)]=\sum_{x=0}^\infty g(x)f_X(x)$ or if you know the MGF of $X$ i.e $M_X(t)=E[e^{tX}]$, then simply substitute $t=\ln a$ into it which will give you $E[(e^{\ln a})^X]=E[a^X]$.

For the variance, you will additionally need $E[Y^2]=E[(a^2)^X]$. Once again substitute $t=\ln(a^2)=2\ln a$ into the MGF.


问:
From $3$ chocolate bars we can make at most $5$ chocolate rabbits. What is the greatest number of rabbits that can be made from $16$ bars?
I'm using some questions from the 2019 International Math Contest to help my students prepare for the Mathcounts competition. While the IMC gives me the answers, I like to give my students the worked solutions so that they learn how to get to the answer. I'm stuck on this one, though:

From three chocolate bars we can make at most 5 chocolate rabbits and have some leftover. What is the greatest number of chocolate rabbits that can be made from sixteen chocolate bars (maybe with some leftover)?

The answer that was given by the IMC was $31$ rabbits. I just can't seem to get there.

I first set up a straight up proportion:
$$\frac{3\;\text{bars}}{5\;\text{rabbits}} = \frac{16\;\text{bars}}{x\;\text{rabbits}}$$ and wound up with $26+\frac23$ rabbits, which is not correct.

So, I thought they must be making more rabbits from the leftovers, but I'm still not sure how they are getting to $31$ rabbits.

Please help!

Given that it takes $3$ bars to make $5$ rabbits, how can you make $31$ rabbits from $16$ bars?

答:
FWIW the phrasing (as given here) isn't clear, and I'd argue that this is a badly written question.

The idea they are going for is that the infimum (but not minimum) amount of chocolate that a rabbit uses is 0.5 chocolate bar, and so with 16 bars, we can produce 31 rabbits.

If you're unfamiliar with the concept of infimum, then it's that

A chocolate rabbit uses $ 0.5 + x $ bars of chocolate, where $ 0 < x \leq 0.1$.

Hence, from 16 bars of chocolate, we can produce $\frac{16}{0.5 + x}$ rabbits, and we get that $ 26 \leq \lfloor \frac{16}{0.5+x} \rfloor \leq 31 $.

So the maximum number of rabbits is 31.

But obviously, we don't know what $x$ is, nor is it well defined, and I agree that any answer from 26 to 31 is reasonable (given that it's reasonable to assume $x$ should be some fixed value).

The way I would rephrase this question for clarity is:

From three chocolate bars, we can make at most 5 chocolate rabbits with some leftover. What is the greatest number of chocolate rabbits that we could have made from sixteen chocolate bars (maybe with some leftover)?


问:
Probability of gaining more than $\$20$ in a two person die game

Jack and Jill gamble on a roll of a die, as follows. If the die shows $1$ or $2$ spots, Jack gives Jill $\$1$. If the die shows $5$ or $6$ spots, Jill gives Jack $\$1$. If the die shows $3$ or $4$ spots, no money changes hands. Suppose Jack and Jill play this game $400$ times. The chance that Jill’s net gain is more than $\$20$ is closest to:

a) $10\%$

b) $20\%$

c) $30\%$

d) $40\%$

Addition Rule: $\Pr(A\text{ or }B)= \Pr(A)+\Pr(B)$

$$\Pr(\text{Jack 1 or 2}, + \text{\$1 to Jill})=( 1/6)+(1/6)=2/6$$
$$\Pr(\text{Jill 5 or 6, + \$1 to Jack})=(1/6)+(1/6)=2/6$$
$$\Pr(\text{3 or 4; \$0 both Jack & Jill})= (1/6)+(1/6)=2/6$$

$n=400$

$\Pr(\text{net gain >\$20})$?

Body:
The probability of $a$ times \$-1, $b$ times \$0, and $c$ times \$1 is:
$$
\Pr(a, b, c) = \left(\frac{1}{3}\right)^{400} \frac{400!}{a!\,b!\,c!} \qquad (a+b+c=400)
$$
The probability to gain more than \$20 is:
$$
\begin{align}
& \sum_{a=0}^{400}\ \sum_{c=21+a}^{400-a}\ \Pr(a,\ 400-a-c,\ c) \\
&=\ \left(\frac{1}{3}\right)^{400} 400!\
\sum_{a=0}^{400}\ \frac{1}{a!}\ \sum_{c=21+a}^{400-a}\ \frac{1}{(400-a-c)!\ c!} \\
&\approx\  0.10
\end{align}
$$


答:
I think the intended question is as follows. Each rabbit contains a certain amount $x$ of chocolate. Each bar contains a certain amount $y$ of chocolate. $3$ bars is enough chocolate for $5$ rabbits, but not $6$. For some value $k$, $15$ bars is enough for $k$ rabbits, but not $k+1$.

What is the largest possible value of $k$, given this information? In other words, how large can you make $k$ if you are allowed to choose the values $x$ and $y$, provided they have to satisfy the given condition?

 With this interpretation, you know $y<2x$ and choosing $y=1.99$ and $x=1$ means you have enough chocolate for $31$ rabbits. Since $y<2x$ you can't make $32$.

This is different from what it sounds like at first reading: what is the largest number of rabbits you can make for any such values of $x$ and $y$. (If someone else gets to choose $x$ and $y$, they can prevent you from making more than $26$ as you say.)

Title: From $3$ chocolate bars we can make at most $5$ chocolate rabbits. What is the greatest number of rabbits that can be made from $16$ bars?

Body:
I'm using some questions from the 2019 International Math Contest to help my students prepare for the Mathcounts competition. While the IMC gives me the answers, I like to give my students the worked solutions so that they learn how to get to the answer. I'm stuck on this one, though:

From three chocolate bars we can make at most 5 chocolate rabbits and have some leftover. What is the greatest number of chocolate rabbits that can be made from sixteen chocolate bars (maybe with some leftover)?

The answer that was given by the IMC was $31$ rabbits. I just can't seem to get there.

I first set up a straight up proportion:
$$\frac{3\;\text{bars}}{5\;\text{rabbits}} = \frac{16\;\text{bars}}{x\;\text{rabbits}}$$ and wound up with $26+\frac23$ rabbits, which is not correct.

So, I thought they must be making more rabbits from the leftovers, but I'm still not sure how they are getting to $31$ rabbits.

Please help!

Given that it takes $3$ bars to make $5$ rabbits, how can you make $31$ rabbits from $16$ bars?


答:
Say you need exactly  $x$ chocolate bars to make one rabbit.  What do we know about $x$?

Well, with $3$ bars we can make $\frac 3x$ rabbits and we know that we can make $5$ rabbits but not $6$.  Thus $$5≤\frac 3x<6$$

Which implies that  $$\frac 36<x≤\frac 35$$

If $x$ were to equal $\frac 36$ (which it can't quite) then with $16$ bars we could make $$16\times \frac 63=32$$

But, since $x$ must be at least slightly greater than $\frac 36$ we can't quite arrange to make $32$, so $31$ is the best we can manage.

To stress, we don't know that we can make $31$.  That's just an upper bound.  The lower bound, the number we know we can make, is $$\big \lfloor 16\times \frac 53\big \rfloor=26$$


问:
Why is $\int x^2e^{x^2}\,d(x^2)$ not proper notation?
So, as the title states. My question is about mathematical notation. Whenever my lecturer in calculus one computes an integral they do some in my opinion very redundant steps.

When they attempt to compute for example $\int x^3e^{x^2}\,dx$, they will write down $u = x^2; du=2xdx$ and proceed to turn the integral into $\frac{1}{2}\int ue^u\,du$.

I don't see why it is required to hide this simple mathematical truth that $d(x^2)=2xdx$ behind a substitution, yet they tells me that not doing so would technically be incorrect.

By the way, I am not asking this to proof my lecturer wrong, since they are an epic person and when I told them that I found it easier without the substitution I was given permision to do so. I am merely asking why it would be "technically incorrect" to do so.

答:
Students can feel more comfortable when they see what is substituted as a single entity. But of course

$$\int x^3e^{x^2}dx=\frac12\int x^2e^{x^2}d(x^2)=\frac12\int ue^{u}du$$

are equivalent and technically correct.

You can even do the integral mentally if you want (but make sure your instructor believes you).

Extra trick:

Write down $x^2$ on a label and read it loud "$u$".


问:
How to prove that $\frac{|x+y+z|}{1+|x+y+z|} \le \frac{|x|}{1+|y|+|z|}+\frac{|y|}{|1+|x|+|z|}+\frac{|z|}{1+|x|+|y|}$
I need to prove $\frac{|x+y+z|}{1+|x+y+z|} \le \frac{|x|}{1+|y|+|z|}+\frac{|y|}{1+|x|+|z|}+\frac{|z|}{1+|x|+|y|}$. I've tried to use triangle inequality or to explore the form of $(a+b+c)^2$ but it won't get me anywhere. I would be grateful for some suggestions.

Body:
Let's check whether: $$S(N) = \frac{|\sum_{i=0}^{N} x_i|}{1+|\sum_{i=0}^{N} x_i|}\le \sum_{i=0}^{N} \frac{|x_i|}{1+ |\sum_{j=0;j \ne i}^{N} x_j|}$$ for all natural numbers $N$ and all natural numbers $i<N+1$ is true.

The inequation $\frac{|x+y+z|}{1+|x+y+z|} \le \frac{|x|}{1+|y|+|z|} + \frac{|y|}{1+|x|+|z|}+\frac{|z|}{1+|x|+|y|}$ is the special case for $i=2$ with $x=x_0, y=x_1, z=x_2$

For $N=0$ we have $\frac{|x_0|}{1+|x_0|} \le \frac{|x_0|}{1+ 0}$ for all $x_0$

Let's suppose for a certain $N$ that we have  $$S(N) = \frac{|\sum_{i=0}^{N} x_i|}{1+|\sum_{i=0}^{N} x_i|} \le \sum_{i=0}^{N} \frac{|x_i|}{1+ |\sum_{j=0;j \ne i}^{N} x_j|}$$ for all natural numbers $N$ and all natural numbers $i<N+1$.

and prove this inequality for $S(N+1)$ which means let's prove the following:

$$S(N+1) \le \sum_{i=0}^{N+1} \frac{|x_i|}{1+ |\sum_{j=0;j \ne i}^{N+1} x_j|}$$  all natural numbers $i<N+2$.

We first will use $\frac{|a+b|}{1+|a+b|} \le \frac{|a|}{1+|a|} + \frac{|b|}{1+|b|}$ like proved in this link:Prove $\frac{|a+b|}{1+|a+b|}<\frac{|a|}{1+|a|}+\frac{|b|}{1+|b|}$.

for $a = x_{N+1}$ and $ b= \sum_{i=0}^{N} x_i$

$$\frac{|x_{N+1}+\sum_{i=0}^{N} x_i|}{1+|x_{N+1}+\sum_{i=0}^{N}x_i|} \le \frac{|x_{N+1}|}{1+|x_{N+1}|} + \frac{|\sum_{i=0}^{N} x_i|}{1+|\sum_{i=0}^{N} x_i|}$$

means that

$$\frac{|x_{N+1}+\sum_{i=0}^{N} x_i|}{1+|x_{N+1}+\sum_{i=0}^{N}x_i|} \le \frac{|x_{N+1}|}{1+|x_{N+1}|} + S(N)$$

means that $$ \frac{|x_{N+1}+\sum_{i=0}^{N} x_i|}{1+|x_{N+1}+\sum_{i=0}^{N}x_i|} \le \frac{|x_{N+1}|}{1+|x_{N+1}|} + \sum_{i=0}^{N} \frac{|x_i|}{1+ |\sum_{j=0;j \ne i}^{N} x_j|}$$

means that

$$S(N+1) \le \sum_{i=0}^{N+1} \frac{|x_i|}{1+ |\sum_{j=0;j \ne i}^{N+1} x_j|}$$

Proved!


问:
Identity Element of Subtraction (Solution Verification)
This is about an exercise from Norman L. Biggs Discrete Mathematics. One is asked to check which binary operations are valid when $*$ represents the usual $-$ of arithmetic. Now, Biggs says that $-$ has an identity value. But, using the definition:

G3: There is an element $e \in G$ such that $$e*x=x*e=x$$ for all $x \in G$. Now, $e-x=x \iff e=2x$ and $x-e=x \iff e=0$. As $2x≠0$ for all $x \in G$

I answered that subtraction did not have an identity element. What am I missing?

答:
Subtraction is neither associative nor commutative, and there is no element $e$ such that for all $x$: $$e -x = x - e = x$$

However, there is a right-identity, namely $0$.


问:
Why are plurigenera birational invariants?
Let $X$ be a variety over an algebraically closed field $k$ and let $\Omega_X^n=\bigwedge_{i=1}^n\Omega_X$, where $\Omega_X$ is the canonical bundle.

I'm trying to understand the well-known fact that plurigenera $P_n:=h^0(X,\Omega_X^n)$ are birational invariants.

For convenience, I'll write invertible sheaves as divisors, so $\Omega_X$ will be replace by the canonical divisor $K_X$.

Let $\varphi:X\dashrightarrow Y$ be a birational map. My fist idea was to prove that for every divisor $D\in\text{Div}(Y)$ we should have $h^0(X,\varphi^*D)=h^0(Y,D)$, which is the same as finding an isomorphism of $k$-vector spaces $\Gamma(Y,D)\simeq \Gamma(X,\varphi^*D)$, which I'm not sure can be done.

Maybe this is true when $D=K_Y^n$, but I'm also not sure. For example, suppose $X,Y$ are smooth surfaces and $\varphi$ is the blowup at a point with exceptional divisor $E$ and $D=K_Y$. In this case $K_X=\varphi^*K_Y+E$, and I don't know what to do with $E$.

Any help will be appreciated. Thank you!

答:
I had a quick look at Beauville's book "Complex Algebraic surfaces" where a proof for surfaces is given Proposition 3.20, I think I got a feeling for the crux of argument. Lets restrict to surfaces first, note that it is only neccesary to check it for point blow-ups.

Let $Y \rightarrow X $ be the a blow-up at $p \in X$. Note that there is a zariski open subset in $Y$ that is isomorphic to $X \setminus p$. Suppose we have a differential form on $Y$, then we get a form on $X \setminus p$. This section can be extended to a rational section on $X$ (by a general fact which holds for any coherent sheaf). but the singularities of a rational section are always a divisor, so it is in fact a section. Then one may check that the morphism defined in this was is in fact injective, i.e. if two differential forms agree on a zariski open then they agree everywhere. This is just a fancy version of the identity theorem from complex analysis.

Now for higher dimensional, there is a theorem that birational varieties may be blown up along sequences of smooth centres with codimension at least $2$ to get a common variety. Then because the codimension is atleast $2$ we can apply the above argument; i.e. producing a rational section then seeing it has to be an actual section.


问:
math notation question regarding location within ordered set
if $\{x_i\}$ for $i=1...N$ are a set of ordered values such that $x_1\ge x_2 \ge ... \ge x_N$ how do I notate most efficiently for a given number $y$ where $x_1 \ge y \ge x_N$ the maximum subscript $j$ such that $x_j \ge y \ge x_{j+1}$

thanks in advance

答: I would write $$j^* = \max \{j : x_j \geq y \geq x_{j+1}\}.$$  Of course, there's also nothing wrong with the way you defined it in words.


问:
Which its the subgroup generated by -1 respect to +?
I have the question:

Which its the cyclic subgroup of $\mathbb{Z}$ generated by -1 respect to +? I know that its all $\mathbb{Z}$ but I don´t understad why?

Help, please.

答:
To my taste, a more positive way to think about "subgroup generated by subset $S$ of group $G$" is not to declare that it is the collection of all expressions involving elements of $S$ (and their inverses!), but, rather, that it is the "smallest subgroup of $G$ containing the set $S$". That there exists a smallest such subgroup, and that it is unique, is by observing that the intersection of all subgroups containing $S$ is a subgroup, contains $S$, and is a subgroup of any subgroup containing $S$.

So, in the case at hand, since "the subgroup generated by $-1$" (or, similarly, by $1$) first of all needs to be a subgroup, it has to contain additive inverses... so if it contains $-1$ it must contain $+1$, and vice versa. "Rules" about allowable expressions are secondary.


问:
Finding the smallest number $a \geq 0$ such that $a \equiv 7^{83} \bmod 11$.

Find the smallest number $a \geq 0$ such that $a \equiv 7^{83} \bmod 11$.

I know that I am supposed to come to the conclusion $a = 2$, but I don't understand how I reach this conclusion using Fermat's or Euler's theorem.

I have the order $83$ to $7$ in $\mathbb{Z}_{11}$, but I don't know what I write as an answer, because writing just $a = 2$ is not enough to show that I have understood how I came to the answer without using a calculator to just calculate that $7^{83} \bmod 11 = 2$.

答:
According to Fermat's Little Theorem $$a^{p-1} \equiv 1 \pmod p$$ for prime $p\nmid a$.

Therefore, $$7^{83} =\color{red}{(7^{11-1})}^8\cdot 7^3 \equiv \color{red}1\cdot 7^3 \equiv 343 \equiv 2\pmod {11}$$ Hence, $a=2$.


问:
"Let" in math texts
I'm trying to do the following exercise for my Real Analysis class:

Let $p$ be a given natural number. Give an example of a sequence $\left(x_{n}\right)$ that is not a Cauchy sequence, but that satisfies $\lim \left|x_{n+p}-x_{n}\right|=0$

However, I am in constant doubt in regard to the "let" word in math texts. Can I choose, say $p = 1$, or when one says "let" I am supposed to stick with $p \in \mathbb{N}$ and nothing more? What kind of "control" do I have over $p$?

For instance, when one says "Given $\epsilon \gt 0$", I usually see things like: Let $\epsilon = \frac{\epsilon}{2}$ so one can finish a certain argument.

Can someone help me? I need to solve this doubt once and for all, it bothers me very often.

答: You have no choice about $p$: you have to find an answer that works no matter what $p$ you are given. For example, $$x_n=\cos\frac{2\pi n}p$$would work for any given nonzero $p$, while $x_n=n$ would do in the trivial case $p=0$.

Title: "Let" in math texts


问:
Does the question mean she will do all 3 sports on one day in the week, or if she will do all 3 sports in a week, but individually on different days
Jo is going on an 8-day activity holiday. Each day she can choose one of the water sports: kayaking or sailing, or land-based sports. She never does different water sports on consecutive days. She also wants to try all three options on at least one day of her holiday. How many different schedules are possible</p>

答:
Simply put. She has 8 slots (marked as X below):

X X X X X X X X

1st condition: She must do all sports for at least 1 day (a complicated way of saying she has to do all sports at least once):

2nd condition: She can't do different consecutive water sports (a complicated way of saying she can't do kayaking or sailing in a row (KS or SK). So example solutions are as below:

K L S L K L K L

This fits both conditions


问:
Derivative of cross product equation
Recently I got a problem that equated the time derivative of a cross product $$\frac{d}{dt}(\vec{P}\times\vec{Q})$$ with a function of time(like $t + t^2$).

Ex. $$\frac{d}{dt}(\vec{P}\times\vec{Q})=5t-6t^2$$

My question is, how can you have an equation with a cross product derivative (which is itself a vector) with a function of time?

答:
As written the expression makes no sense, maybe the two vectors are in a fixed plane such that also the direction vector for the cross product is a fixed $\hat k$, in that case we should write

$$\frac{d}{dt}(\vec{P}\times\vec{Q})=\frac{d}{dt}\left[\left(\frac52t^2-2t^3+c\right)\hat k\right]=\left(5t-6t^2\right)\hat k$$

and as a notation we could obmit the $\hat k$.

Title: Derivative of cross product equation


问:
Proof Explanation on proposition for root spaces of a finite dimensional Lie algebra
Introduction to Lie Algebras K. Erdmann and M. Wildon page 100:

Proposition 10.9: Let $\alpha \in \Phi$, where $\Phi$ is the collection of all roots for Lie algebra $L$ with Cartan Subalgebra $H$. The root space $L_{\pm \alpha}$ are 1 dimensional. Moreover, the only multiples of $\alpha$ which lie in $\Phi$ are ${\pm \alpha}$. Let $h_\alpha$ be the semisimple basis element of $sl(\alpha) \cong sl(2,\mathbb{C})$

(I bolded the parts of the proof I am having trouble understanding)

proof:
If $c\alpha$ is a root, then $h_\alpha$ takes $c\alpha(h_\alpha)=2c$ as an eigenvalue. As the eigenvalues of $h_\alpha$ are integral, either $c \in \mathbb{Z}$ or $c \in \mathbb{Z} + \frac{1}{2}$. To rule out the unwanted values for $c$, we consider the root string module:

$M = H \oplus (\oplus_{c\alpha} L_{c\alpha})$

Let $K = ker\alpha \subset H$ (it is possible that $K=H$).

By an example on the previous page, $K \oplus sl(\alpha)$ is an $sl(\alpha)$-submodule of $M$. By Weyl's Theorem, modules for $sl(\alpha)$ are completely reducible, so we may write:

$M = K \oplus sl(\alpha) \oplus W$

Where $W$ is a complementary submodule.

If either of the conclusions of the proposition are false, then $W$ is non-zero. Let $V \cong V_s$ be an irreducible submodule of $W$. If $s$ is even, then it follows from the classification of chapter 8 that $V$ contains an $h_\alpha$-eigenvector with eigenvalue 0. Call this eigenvector $v$. The zero-eigenspace of $h_\alpha$ on $M$ is $H$, which is contained in $K \oplus sl(\alpha)$. Hence $v \in (K \oplus sl(\alpha)) \cap V = 0$, which is a contradiction.

Before considering the case where $s$ is odd, we pursue another consequence of this argument. Suppose that $2 \alpha \in \Phi$. Then $h_\alpha$ has $2\alpha(h_\alpha)=4$ as an eigenvalue. As the eigenvalues of $h_\alpha$ on $K \oplus sl(\alpha)$ are $0$ and $\pm 2$, the only way this could happen is if $W$ contains an irreducible submodule $V_s$ with $s$ even, which we just saw is impossible.

Now suppose that $s$ is odd. Then $V$ must contain and $h_\alpha$-eigenvector with eigenvalue $1$. As $\alpha(h_\alpha)=2$, this implies that $\frac{1}{2}\alpha$ is a root of $L$. But then both $\frac{1}{2}\alpha$ and $\alpha$ are roots of $L$, which contradicts the previous paragraph.

答:
"If either of the conclusions ...":
If the dimension of $L_\alpha$ is $\ge 2$, the eigenspace to the eigenvalue $2$ of $h_\alpha$ on $M$ has dimension $\ge 2$. But its eigenspace to the eigenvalue $2$ on $K\oplus sl(\alpha)$ only has dimension ...

If there is $c\neq \pm1$ with $L_{c\alpha} \neq 0$, then there is a nonzero $x\in L_{c\alpha} \subseteq M$ with $[h_\alpha, x] = 2cx$. But the only eigenvalues of $h_\alpha$ on $K\oplus sl(\alpha)$ are ...

"which is contained in ...": How large can $dim(H)-dim(K)$ at most be? And $sl(\alpha)$ contains the element $h_\alpha$ which is in $H$ but not in $K$.

"which contradicts the previous paragraph": If $\alpha$ and $\frac12 \alpha$ are both roots, then for the root $\beta := \frac12 \alpha$ we also have $2\beta$ as a root. But the previous paragraph showed that for any root, its double is not a root.


问:
Show that a class of structures is existentially axiomatisable iff it is closed under taking superstructures
A sentence is called existential if it's of the form $\exists x_1 \cdots \exists x_n \varphi(x_1, \cdots, x_n)$, where $\varphi$ is quantifier-free formula.

I'm trying to prove a lemma left as an exercise in my lecture notes that says

Let $C$ be an axiomatizable class. Then the following conditions are equivalent:
(i) $C$ is $\exists$-axiomatizable;
(ii) If $A \in C$ and $A \leq B$ then $B \in C$.

One direction is simple - if $C$ is $\exists$-axiomatizable then it follows easily by the fact that if $A \models \varphi(a_1, \cdots , a_n)$ for an existential formula $\varphi(v_1, \cdots , v_n)$, then $B \models \varphi(a_1, \cdots , a_n)$.

I'm really not sure how to go about proving the other direction.

I've proved the equivalent version for universal axiomatisation - i.e. that $C$ is $\forall$-axiomatizable iff $B \in C$ and $A \leq B$ then $A \leq C.$ To do that, I showed that Th$(C) \cup $Th$_∃(A)$ is finitely satisfiable (where $A\models$ Th$_\forall$(C) - i.e. $A$ is a model of the universal sentences in the theory of $C$) where Th$_∃(A)$ are the existential sentences of the theory of $A$.  It followed from that, and some other results I have, that Th$(C)\cup$Diag$(A)$ was satisfiable, from which it followed there is a model $B$ of Th$(C)$ such that $A \leq B$, which meant by the assumption that $A\in C$ and hence (since $A\models$ Th$_\forall$(C)) that $C$ was universally axiomatisable.

I'm really unsure how I'd go about taking a similar approach for the existential case - I haven't proved any similar results like the one that took me from Th$(C)\cup$Diag$(A)$ being satisfiable to there being a $B$ like I have here, and even if I did I'm not sure how I'd apply a similar kind of result - since in this case I have to start with a smaller model $A$ instead of a bigger one.

Any advice or suggestions you could offer would be much appreciated.

答:
The following general lemma is useful for proving axiomatizability by sentences of a particular form.

Separation Lemma: Let $\Delta$ be a class of sentences which contains $\bot$ and is closed under $\vee$ (up to logical equivalence). Suppose $T$ is a theory such that for any $M\models T$ and $N\not\models T$, there is some $\varphi\in \Delta$ such that $M\models \varphi$ and $N\not\models \varphi$. Then $T$ is axiomatizable by $\Delta$-sentences.

Proof: Fix some $N\not\models T$. For every model $M\models T$, there is some $\varphi_M\in \Delta$ such that $M\models \varphi_M$ and $N\not\models \varphi_M$. Thus $T\cup \{\lnot\varphi_M\mid M\models T\}$ is inconsistent. By compactness, there are finitely many $M_1,\dots,M_k$ such that $T\cup \{\lnot \varphi_{M_1},\dots,\lnot\varphi_{M_k}\}$ is inconsistent. So $T\models \bigvee_{i=1}^k\varphi_{M_i}$. By our assumption on $\Delta$, this disjunction is equivalent to a sentence in $\Delta$, which we call $\psi_N$. Note that $N\not\models \psi_N$.

Now $T' = \{\psi_N\mid N\not\models T\}$ is a $\Delta$-axiomatization of $T$. Indeed, $T\models \psi_N$ for all $\psi_N\in T'$, so every model of $T$ is a model of $T'$. And if $N$ is not a model of $T$, then $N\not\models \psi_N$, so $N$ is not a model of $T'$. $\square$

Rephrasing: To prove that $T$ is $\Delta$-axiomatizable, it suffices to show that if $M\models T$ and every $\Delta$-sentence true in $M$ is also true in $N$, then $N\models T$.

Now let's take $\Delta$ to be the class of existential sentences and assume that the class of models of $T$ is closed under superstructure. Let $M\models T$ and assume that every existential sentence true in $M$ is  true in $N$. We would like to show that $N\models T$.

Since the class of models of $T$ is closed under superstructure and elementary equivalence, it suffices to embed $M$ in a model $N'$ elementarily equivalent to $N$. So we look at $\text{Th}(N)\cup \text{Diag}(M)$. By compactness, this is consistent just in case $\text{Th}(N)\cup \{\theta(a_1,\dots,a_n)\}$ is consistent whenever $\theta$ is a conjunction of atomic and negated atomic formulas and $M\models \theta(a_1,\dots,a_n)$. But then $M\models \exists x_1,\dots,x_n \theta(x_1,\dots,x_n)$, so also $N\models \exists x_1,\dots,x_n \theta(x_1,\dots,x_n)$, and interpreting the constants $a_i$ as witnesses in $N$, we have $N\models \text{Th}(N)\cup \{\theta(a_1,\dots,a_n)\}$.


问:
Is there an analytical solution for the following
Is there an analytical solution to this equation:

$A. dM/dt + B. dN/dt = C$

Initial condition,At $t = 0$ ; $M(0) = N(0) = const$

$A,B,C$ are constants and $M(t)$ and $N(t)$.

答:
Hint:

Write as

$$\frac {d(AM+BN)}{dt}=C$$

and you have the solution (only available in terms of $AM+BN$).


问:
connected but not path-connected sets
Let $S\in[0,1]^2$ be defined as follows:
$$(x,y)\in S \Leftrightarrow \begin{cases}y\in[0, \frac{1}{q}], &x=\frac{p}{q}\in\mathbb{Q} \text{ with } q \text{ even}\\
y\in[1-\frac{1}{q}, 1], &x=\frac{p}{q}\in\mathbb{Q} \text{ with } q \text{ odd}\\
y\in[0,1], &\text{otherwise}\end{cases}$$
, where for every $x\in\mathbb{Q}$, the fractional representation $\frac{p}{q}$ is lowest in terms of $q>0$. Prove that $A$ is connected, but not pathwise connected.

I found that proving the connectedness of $S$ directly from definition is hard. So I'm trying to prove it using the property of connectedness that if $A\subset S\subset \bar{A}$, where $A$ is connected, then $S$ is connected. However, I'm having trouble finding this connected subset $A$. Any suggestions?

答:
Suppose we had two non-empty clopen sets $U, V$ partitioning $S$. Each set of the form $\{x\} \times [0, 1] \subseteq S$, where $x \in [0, 1] \setminus \Bbb{Q}$, must be entirely contained in $U$ or entirely contained in $V$, since each such subset is connected.

Let $A = \{x \in [0, 1] \setminus \Bbb{Q} : \{x\} \times [0, 1] \subseteq U\}$ and similarly define $B$ in relation to $V$. I claim that $A$ and $B$ are clopen in $[0, 1] \setminus \Bbb{Q}$. To see $A$ is open, suppose $x \in A$. Then $(x, 0) \in U$, which is open, hence a ball $B_S((x, y); r)$ exists in $U$. It's straight forward to verify that $B_A(x; r) \subseteq A$. Similarly, $B$ is open, and given both partition $[0, 1] \setminus \Bbb{Q}$, $A$ and $B$ are both clopen.

Due to the density of the irrationals, and the fact that $U, V$ are open and non-empty, we can also verify that $A$ and $B$ are both non-empty.

Consider now the closures of $A$ and $B$ in $[0, 1]$. The density of the irrationals imply that the closures of these sets cover $[0, 1]$. By the connectedness of $[0, 1]$, there must exist some point in their intersection. That is, there exists some $x$ that is simultaneously the limit (in $[0, 1]$) of a sequence $a_n \in A$ and a sequence $b_n \in B$. As both $A$ and $B$ are closed and disjoint, we must have $x \in \Bbb{Q}$.

Now, if $x$ has even lowest denominator, then consider the sequences $(a_n, 0)$ and $(b_n, 0)$. If $x$ has odd lowest denominator, then consider the sequences $(a_n, 1)$ and $(b_n, 1)$. In either case, each sequence respectively belongs to $A$ and $B$, converging to a point in $S$. As $U$ and $V$ are closed, this would produce a point in $U \cap V$, contradicting $U$ and $V$ being disjoint, and proving connectedness.

Title: connected but not path-connected sets


问:
Does $\left\{\mathrm{e}^{\mathrm{i} \frac{2\pi n}{L} x}\right\}_{n \in \mathbb{Z}}$ form a complete orthogonal basis in the interval $[-L/2,L/2]$?
$\mathrm{e}^{\mathrm{i}2\pi k x}$ and $\mathrm{e}^{\mathrm{i}2\pi k' x}$ are orthogonal in the following sense provided $k \ne k'$,

\begin{equation}
\langle \mathrm{e}^{\mathrm{i}2\pi k x}, \mathrm{e}^{\mathrm{i}2\pi k' x}\rangle = \int_{\infty}^{\infty} \mathrm{e}^{-\mathrm{i}2\pi k x} \, \mathrm{e}^{\mathrm{i}2\pi k' x} \, \mathrm{d}x = 0 \iff k \ne k'
\end{equation}

Therefore, $\mathrm{e}^{\mathrm{i}2\pi k x}$ forms a complete orthogonal basis over the interval $(-\infty,\infty)$ for all $k \in \mathbb{R}$

However, I'm interested in finding an orthogonal basis over the open interval $[-L/2,L/2]$, orthogonal such that

\begin{equation}
\int_{-L/2}^{L/2} \overline{\phi_j(x)} \, \phi_k(x) \, \mathrm{d}x = 0 \iff j \ne k
\end{equation}

I found an orthogonal set by considering $\mathrm{e}^{\mathrm{i}2\pi k x}$, and removing terms not orthogonal to (to choose a starting point) $1$, yielding the basis

\begin{equation}
\left\{\mathrm{e}^{\mathrm{i} \frac{2\pi n}{L}x}\right\}_{n \in \mathbb{Z}}
\end{equation}

Is this a complete orthogonal basis?

It seem that starting from a complete basis over a large interval and culling only what isn't orthogonal to a starting basis function seems promising to forming another complete basis, but I'm not convinced by this argument very much. Especially since this choice of basis suggests that the function is periodic with period $L$. While it seems that any arbitrary function could be constructed from infinitely many terms in the basis, it seems to fall short in that (due to the interval being open) the value of any function at $x=-L/2$ is equal to that ay $x=L/2$ due to this periodicity, $f(-L/2) = f(L/2)$. Surely, if the basis was complete, these values could be made independent. Is it the case that the basis is complete except for one basis function. If so, what is this elusive function?

答:
This is not really on the topic of the question, but the situation on the unbounded domain only intuitively works like you said. When written down carefully, the situation is much more subtle than that, and these subtleties have serious consequences in subjects like spectral theory.

Anyway, the situation in the bounded domain does work like you said. And yes, you have the counterintuitive situation that, given a $L^2$ function $f$ on $[-L/2,L/2]$, you can find a sequence of continuous and periodic functions $f_n$ which converge to $f$ in $L^2$ even if $f$ itself isn't periodic. (If you're not acquainted with the term $L^2$, replace the first instance with "continuous" and the second instance with "in the mean square sense").

The intuitive way that I interpret this is that $f$ on $[-L/2,L/2]$ is being understood as its periodic extension when we look at it from the point of view of Fourier series, so if it is actually not periodic then the periodic extension has a jump, near which the Fourier series misbehaves (but not so severely as to spoil the $L^2$ convergence). You can look up the Gibbs phenomenon to get some visualization of this.


问:
Suppose that a statistical model is given by the family of $N(μ, σ^2)$ where $θ = (μ, σ^2) ∈ R^1 × R^+$ is unknown. Calculate $\psi(μ, σ^2)$
Let's assume that we have a statistical model that is given by the family of $N(μ, σ^2)$ distributions where $θ = (μ, σ^2) ∈ R^1 × R^+$ is unknown. If our interest is in making inferences about the first quartile of the true distribution, then determine $ψ(μ, σ^2)$.

$\textbf{SOLUTION}$: We have that $ψ(μ, σ^2) = x_{0.25} = μ + σz_{0.25}$ where $\Phi(z_{0.25}) = 0.25$.

How? And What does that solution mean? What is $z_{0.25}$ and what is $\Phi(z_{0.25})$?

答:
The first quartile ($x_{0.25}$) of the true distribution satisfies:
$$
P(X \le x_{0.25}) = 0.25.
$$

Rescaling gives
$$
P \left ( \frac{X-\mu}{\sigma} \le \frac{x_{0.25} - \mu}{\sigma} \right) = 0.25
$$
Note that $\frac{X-\mu}{\sigma} $ is equal in distribution to a standard normal $Z\sim N(0,1)$, with CDF $\Phi$, so we can equivalently write
$$
\Phi \left ( \frac{x_{0.25} - \mu}{\sigma}\right) = 0.25
$$
Rearranging yields
$$
x_{0.25} = \sigma \Phi^{-1}(0.25) + \mu = \sigma z_{0.25} + \mu
$$
where $\Phi^{-1}(0.25) := z_{0.25}$ is the quartile of the standard normal.

If we want to do inference, we should come up with an estimator. The natural thing to do is MLE estimation. Note that the thing we are trying to estimate is a function of the parameters $\theta$, and it is a fact that the MLE of a function of parameters is the function of the MLEs of those parameters, so an estimate here would be
$$
\hat{x}_{0.25} = \hat{\sigma} z_{0.25} + \hat{\mu}
$$
where $\hat{\sigma} = \frac{1}{n} \sum_{i=1}^n (X_i - \bar{X})^2$ is the MLE estimate of the variance, and $\hat{\mu} = \bar{X} = \frac{1}{n} \sum_{i=1}^n X_i$ is the MLE estimate of the mean


问:
How many units are in the ring $(\Bbb Z/15\Bbb Z) \times (\Bbb Z/125\Bbb Z)$
How many units are in the ring $(\Bbb Z/15\Bbb Z) \times (\Bbb Z/125\Bbb Z)$

I am not sure how to count the units exactly because for smaller numbers I could make a multiplication chart and see which ones are equal to one. But $125$ is too large to do that.

答: Consider Euler phi function and compute it at 125. So, $\phi(125) = 100$ is the number of units in $\frac{\mathbb{Z}}{125\mathbb{Z}}$ and similarly, $\phi(15) = 8$ is the number of units in $\frac{\mathbb{Z}}{15\mathbb{Z}}$. Now, use the counting techinque to compute the units is the product.


问:
Resolution of $2\mathbf{\hat{i}}+3\mathbf{\hat{j}}$ along $\mathbf{\hat{i}}+\mathbf{\hat{j}}$ and $\mathbf{\hat{i}}-\mathbf{\hat{j}}$
I was solving this problem:

Resolve $2\mathbf{\hat{i}}+3\mathbf{\hat{j}}$ along $\mathbf{\hat{i}}+\mathbf{\hat{j}}$ and $\mathbf{\hat{i}}-\mathbf{\hat{j}}$.

I proceeded as below:

Let $\mathbf{A}$, $\mathbf{a}$ and $\mathbf{b}$ be the given vectors in the same order. Let
$$\mathbf{A}=\lambda\mathbf{a}+\mu\mathbf{b}$$
Putting the values and rearranging,
$$2\mathbf{\hat{i}}+3\mathbf{\hat{j}}= (\lambda+\mu)\mathbf{\hat{i}}+(\lambda-\mu)\mathbf{\hat{j}}$$
Then, noting that two vectors are equal iff their magnitudes as well as directions are equal, first we equate the directions and taking $\tan$ of both sides.
$$\frac 3 2 = \frac{\lambda-\mu}{\lambda+\mu}$$
Thus, $\lambda = -5\mu$.

Now, taking magnitudes and squaring,
$$13 = 2(\lambda^2+\mu^2)$$
Using our previous equation and doing some work, we get
$$\lambda=\pm \frac{5}{\sqrt{2}}\textrm{ and } \mu = \mp \frac{1}{\sqrt{2}}$$
However, I find no way of choosing which sign.

The book says the answer is $\frac{5}{\sqrt{2}}$ and $-\frac{1}{\sqrt{2}}$.

I know there is the way to take components by using dot product. But since the book has not introduced it yet, I wonder if there's a way I showed above.

Edit: As Jeea pointed out, the correct answers are
$$\lambda=\pm \frac{5}{2}\textrm{ and } \mu = \mp \frac{1}{2}$$
However, this worsens the situation, this time even the answers are different, let alone sign.

答:
Another way to solve this is using matrix techniques:

$$\begin{bmatrix}2\\3\end{bmatrix} = \lambda \begin{bmatrix}1\\1\end{bmatrix} + \mu \begin{bmatrix}1\\-1\end{bmatrix}
\\ = \begin{bmatrix}1&1\\1&-1\end{bmatrix}\begin{bmatrix}\lambda\\\mu\end{bmatrix}$$

So take inverse and premultiply and we get $\begin{bmatrix}\lambda\\\mu\end{bmatrix} = \begin{bmatrix}5/2\\-1/2\end{bmatrix}$


答: Hint: If $A$ and $B$ are commutative rings with unity, then $(A \times B)^\times = A^\times \times B^\times$, where $R^\times$ is the group of units of the ring $R$.

Body:
How many units are in the ring $(\Bbb Z/15\Bbb Z) \times (\Bbb Z/125\Bbb Z)$

I am not sure how to count the units exactly because for smaller numbers I could make a multiplication chart and see which ones are equal to one. But $125$ is too large to do that.


问:
$-2\arcsin\left(\sqrt{1-x}\right)$ and $\arcsin\left(2x-1\right)$ don't differ by a constant?
When you take the derivative of $-2\arcsin\left(\sqrt{1-x}\right)$ and $\arcsin\left(2x-1\right)$, you get the same result. Which would mean that $\arcsin\left(2x-1\right) + 2\arcsin\left(\sqrt{1-x}\right)$ is a constant. But using the addition and subtraction formulas, you can simplify the result to $\arcsin\left(2x-1\right) + \sin^{-1}\left(2\sqrt{1-x}\sqrt{x}\right) = \sin^{-1}\left(\left(2x-1\right)\sqrt{4x-3}+4\left(1-x\right)\sqrt{x}\right)$, at least for $x$ within $(0.5,1)$. That right hand side is looking not close to a constant. What's my mistake?

答:
Let

$\begin{align}y = -2\arcsin\sqrt{1-x} &\Rightarrow \arcsin\sqrt{1-x} = \frac{-y}{2} \\&\Rightarrow 1-x = \sin^2\left(\frac{-y}{2}\right) = \frac{1-\cos(\pm y)}{2}   ~~~(\text{ as cos is even })\\&\Rightarrow \cos(\pm y) = 2x-1 \\& \Rightarrow\sin(\frac{\pi}2\pm y) = 2x-1 \\ &\Rightarrow \pm y = \arcsin(2x-1) -\frac{\pi}{2} \\ &\Rightarrow y = \begin{cases}\arcsin(2x-1) -\frac{\pi}{2} \\ \text{or} \\-\arcsin(2x-1) +\frac{\pi}{2}\end{cases}\end{align}$

Of these two we can find that only the first one is permissible.

$-1\le x \le 1 \Rightarrow 0\le1- x \le 1 \Rightarrow 0 \le \arcsin\sqrt{1-x}\le\frac{\pi}{2} \Rightarrow \boxed{0\le -y\le \pi} \text{ or } \boxed{-\pi\le y\le0}$

Considering the first case,

$-\frac{\pi}{2}\le \arcsin(2x-1)\le\frac{\pi}{2} \Rightarrow -\pi \le\arcsin(2x-1) - \frac{\pi}{2} \le 0 \\ \color{blue}{\text{  satisfies the required range of y}}$

Considering the second case,

$-\frac{\pi}{2}\le -\arcsin(2x-1)\le\frac{\pi}{2} \Rightarrow 0 \le \frac{\pi}{2}-\arcsin(2x-1)  \le\pi \\ \color{blue}{\text{  doesn't satisfy the required range of y}}$

So,

$ -2\arcsin\sqrt{1-x} = \arcsin(2x-1) -\frac{\pi}{2}$


问:
Problems on topological and metric spaces.
I need help on the following problems:

Q1: Consider the map $f:(\Bbb R\times\Bbb R,\tau)\to(\Bbb R,\gamma)$ given by $f(x\times y)=|x-y|$,where $\tau$ is the product topology and $\gamma$ is the  usual topology. Find $f(1),f(0),f^{-1}(0)$ and $f^{-1}[0,1)$.

For this problem, I don’t know how to use the given definition to find say $f(1)$, also I find it difficult to figure out $f^{-1}$.

Another one i need help on: I read somewhere that the set of rational numbers $\Bbb Q$ as a subspace of $\Bbb R$ (with usual topology) does not have discrete topology since the point $0$ is not an open set in $\Bbb Q$. ¨

Why is this argument true? I thought that $0$ was rational. Thanks.

答:
The number $1$ is not (at least not without further conventions) an element of $\mathbb R\times \mathbb R$, hence I understand why you have problems finding out what $f(1)$ should be.

For $f^{-1}$, you need to find all $(x,y)$ such that $|x-y|$ ...

For the totally unrelated about $\mathbb Q$, note that $\frac1n\to 0$.


问:
Show that two compact self-adjoint operators in Hilbert space must share at least one eigenvector [Stein Chapter 4 Exercise 35]
I am working on Stein Real Analysis, Chapter 4, Exercise 35 (a), which is an invariant of the spectral theorem. The exercise is stated as follows:

If $T_{1}$ and $T_{2}$ are two linear self-adjoint and compact operators in $\mathcal{H}$ that commute, i.e. $T_{1}T_{2}=T_{2}T_{1}$, show that there exists an orthonormal basis for $\mathcal{H}$ which consists of eigenvectors for both $T_{1}$ and $T_{2}$.

Some similar post is here: If $S, T \in B(X)$ are self-adjoint, compact, and commuting bounded linear operators on a Hilbert space $X$, they can be simultaneously diagonalized. I have some attempts but I got stuck in the first step (I think I finished all the remaining proof). My idea is to adapt Stein's proof of spectral theorem in the page 193 of the book.

That is, we let $\mathcal{S}$ be the closure of the linear subspace of $\mathcal{H}$ spanned by all common eigenvectors of $T_{1}$ and $T_{2}$. Argue that $\mathcal{S}$ is not empty.

Our goal is then to prove $\mathcal{S}=\mathcal{H}$. It not, since $\mathcal{H}=\mathcal{S}\oplus\mathcal{S}^{\bot},$ $\mathcal{S}^{\bot}$ would be non-empty. We will then have reached a contradiction once we show that $\mathcal{S}^{\bot}$ contains a common eigenvector of $T_{1}$ and $T_{2}$.

However, what should I do to argue that $\mathcal{S}$ is not empty? I understand that for any compact, non-zero, self-adjoint operator, either $\|T\|$ or $-\|T\|$ is an eigenvalue, but I do not know how to show two such operators share at least one eigenvalue to ensure $\mathcal{S}$ not empty.

If I can show this, then here is the remaining proof:

Firstly, we note that both $T_{1}$ and $T_{2}$ respect the decomposition $\mathcal{H}=\mathcal{S}\oplus\mathcal{S}^{\bot}$. In other words, if $f\in\mathcal{S}$, then $T_{i}f\in\mathcal{S}$, which follows from the definitions. Also, if $g\in\mathcal{S}^{\bot}$, then $T_{i}g\in\mathcal{S}^{\bot}$ also holds. This is because $T_{i}$ is self-adjoint and maps $\mathcal{S}$ to itself, and therefore $$\langle T_{i}g,f\rangle=\langle g, T_{i}f\rangle =0\ \ \text{as long as}\ \ g\in\mathcal{S}^{\bot}\ \ \text{and}\ \ f\in\mathcal{S}.$$

Then, by the spectral theorem for one operator, $T_{1}$ must have an eigenvector in $\mathcal{S}^{\bot}$ with the eigenvalue denoted by $\lambda$. Let $E_{\lambda}$ be the eigenspace of $\lambda$ as a subspace of $\mathcal{S}^{\bot}$. Then, for any $v\in E_{\lambda}$, we have $$T_{1}(T_{2}v)=T_{2}(T_{1}v)=T_{2}(\lambda v)=\lambda (T_{2}v),$$ which implies that $T_{2}v\in E_{\lambda}$.

As $T_{2}$ fixes $E_{\lambda}$, it has at least one eigenvector in $E_{\lambda}$. This eigenvector is exactly the common eigenvector of $T_{1}$ and $T_{2}$, but it belongs to $\mathcal{S}^{\bot}$. The contradiction has been reached.

Is my proof for the remaining part correct?

The stein's version of spectral theorem is as follows:

Spectral Theorem. Suppose $T$ is a compact self-adjoint operator on a Hilbert space $\mathcal{H}$. Then there exists an orthonormal basis $\{\varphi_{k}\}_{k=1}^{\infty}$ of $\mathcal{H}$ that consists of eigenvectors of $T$. Moreover, if $T\varphi_{k}=\lambda_{k}\varphi_{k},$ then $\lambda_{k}\in\mathbb{R}$ and $\lambda_{k}\rightarrow 0$ as $k\rightarrow\infty$.

Thank you!

答:
Fix an eigenvalue $\lambda$ of $T_1$ and let $E_1$ be the eigenspace of $T_1$ corresponding to $\lambda_1.$ Observe that $E_1$ is invariant under $T_2$ because if $v\in E_1$ then $T_1(T_2(v))=T_2(\lambda v)=\lambda T_2(v)$ that is $T_2(v)\in E_1.$

Now restrict $T_2$ to $E_1$ and let $f\in E_1$ be an eigenvector of the restriction of $T_2.$ Note that $f$ is then an eigenvector of $T_2$ itself, not just the restriction. And, obviously $f$ is eigenvector of $T_1$ because $f\in E_1.$ Therefore, $f$ is a common eiegnvector of $T_1$ and $T_2.$


问:
Prove that $R = \Bbb C \left [X,Y,Z \right ]/ \left \langle 2X^2Y - 3XY^3 + Z \right \rangle$ is a unique factorization domain.

Let $R = \Bbb C \left [X,Y,Z \right ]/ \left \langle 2X^2Y - 3XY^3 + Z \right \rangle.$ Show that $R$ is a unique factorization domain.

How do I show that? Can anybody please help me proving that?

Thanks in advance.

EDIT $:$ Let $S = \Bbb C \left [X,Y \right ].$ Then $\alpha : = 2X^2Y - 3XY^3 \in S.$Therefore $$R = S \left [Z \right ] / \left \langle Z  + \alpha \right \rangle \cong S$$ since $\alpha \in S.$ Hence $R$ is isomorphic to $\Bbb C \left [X,Y \right ]$ as rings and since $\Bbb C \left [X,Y \right ]$ is a unique factorization domain (by Gauss's theorem) it follows that $R$ is also a unique factorization domain, as required.

Can anybody please check my reasoning above? Thanks in advance.

答: Your reasoning is entirely correct.</p>

Body: Your reasoning is entirely correct.


答:
I'm sure you've already proved $\int_{-L/2}^{L/2}e^{i2\pi(n-m)x/L}dx=L\delta_{mn}$ for $m,\,n\in\Bbb Z$, so the $u_n(x):=\tfrac{1}{\sqrt{L}}e^{i2\pi nx/L}$ form an orthonormal basis of the vector space they span. The hard part is how "diverse" a space of functions that is.

The way you've tried to write an arbitrary $f$ as a linear combination of these sounds familiar, so let's say $\sum_{n\in\Bbb Z}f_nu_n(x)$ is the attempted decomposition of $f(x)$ for some complex sequence $\{f_n\}_{n\in\Bbb Z}$. (You might have only tried e.g. $n\in\Bbb N$, but you can probably adapt around that for the sake of this discussion.) But this raises a few issues:

By definition, the space the $u_n$ span comprises only those $f$ for which we can choose all but finitely many of the $f_n$ to be zero, and such functions provably satisfy $f(x+L)\equiv f(x)$ (just induct on the number of non-zero $f_n$);

In at least some cases where infinitely many $f_n$ are nonzero, the function $\tilde{f}(x):=\sum_{n\in\Bbb Z}f_nu_n(x)$ makes sense, e.g. as $\lim_{N\to\infty}\sum_{n=-N}^Nf_nu_n(x)$;

However, even when $\tilde{f}$ is an OK function it may not be the $f$ you aimed for. For example, even if I started with an $f$ that obtains $\tilde{f}=f$, you could construct a new function $g$, differing from $f$ at some but only countably points in $[-L/2,\,L/2]$. Then $\tilde{g}\equiv\tilde{f}\equiv f\not\equiv g$.

Don't let the periodicity of the $u_n$ make this mysterious. The aforementioned proof by induction can't be extended to the infinite-terms case. If you think about it, this isn't all that different from what happens with Taylor series:

In the inner product $\langle h_1,\,h_2\rangle:=\oint_{|z|=1}\frac{h_1(z)h_2^\ast(z)dz}{2\pi iz}$, the $z^n$ are orthonormal (with $n\in\Bbb Z$, but let's just look at $n\ge0$ so I don't need to talk about Laurent series);

These monomials span the polynomials;

Lots of non-polynomial Taylor series converge to a sensible function, but maybe not the one you expected.

Title: Does $\left\{\mathrm{e}^{\mathrm{i} \frac{2\pi n}{L} x}\right\}_{n \in \mathbb{Z}}$ form a complete orthogonal basis in the interval $[-L/2,L/2]$?

Body:
$\mathrm{e}^{\mathrm{i}2\pi k x}$ and $\mathrm{e}^{\mathrm{i}2\pi k' x}$ are orthogonal in the following sense provided $k \ne k'$,

\begin{equation}
\langle \mathrm{e}^{\mathrm{i}2\pi k x}, \mathrm{e}^{\mathrm{i}2\pi k' x}\rangle = \int_{\infty}^{\infty} \mathrm{e}^{-\mathrm{i}2\pi k x} \, \mathrm{e}^{\mathrm{i}2\pi k' x} \, \mathrm{d}x = 0 \iff k \ne k'
\end{equation}

Therefore, $\mathrm{e}^{\mathrm{i}2\pi k x}$ forms a complete orthogonal basis over the interval $(-\infty,\infty)$ for all $k \in \mathbb{R}$

However, I'm interested in finding an orthogonal basis over the open interval $[-L/2,L/2]$, orthogonal such that

\begin{equation}
\int_{-L/2}^{L/2} \overline{\phi_j(x)} \, \phi_k(x) \, \mathrm{d}x = 0 \iff j \ne k
\end{equation}

I found an orthogonal set by considering $\mathrm{e}^{\mathrm{i}2\pi k x}$, and removing terms not orthogonal to (to choose a starting point) $1$, yielding the basis

\begin{equation}
\left\{\mathrm{e}^{\mathrm{i} \frac{2\pi n}{L}x}\right\}_{n \in \mathbb{Z}}
\end{equation}

Is this a complete orthogonal basis?

It seem that starting from a complete basis over a large interval and culling only what isn't orthogonal to a starting basis function seems promising to forming another complete basis, but I'm not convinced by this argument very much. Especially since this choice of basis suggests that the function is periodic with period $L$. While it seems that any arbitrary function could be constructed from infinitely many terms in the basis, it seems to fall short in that (due to the interval being open) the value of any function at $x=-L/2$ is equal to that ay $x=L/2$ due to this periodicity, $f(-L/2) = f(L/2)$. Surely, if the basis was complete, these values could be made independent. Is it the case that the basis is complete except for one basis function. If so, what is this elusive function?


问:
Why is $\int x^2e^{x^2}\,d(x^2)$ not proper notation?
So, as the title states. My question is about mathematical notation. Whenever my lecturer in calculus one computes an integral they do some in my opinion very redundant steps.

When they attempt to compute for example $\int x^3e^{x^2}\,dx$, they will write down $u = x^2; du=2xdx$ and proceed to turn the integral into $\frac{1}{2}\int ue^u\,du$.

I don't see why it is required to hide this simple mathematical truth that $d(x^2)=2xdx$ behind a substitution, yet they tells me that not doing so would technically be incorrect.

By the way, I am not asking this to proof my lecturer wrong, since they are an epic person and when I told them that I found it easier without the substitution I was given permision to do so. I am merely asking why it would be "technically incorrect" to do so.

答:
The notation is proper.  The Riemann-Sieltjes for a real-valued function $f(x)$ for $x\in [a,b]$ with respect to a real-valued function $g(x)$ is written

$$I=\int_a^b f(x) dg(x)$$

If $g$ is differentiable on $[a,b]$, then $I$ can be written as

$$I=\int_a^b f(x)\frac{dg(x)}{dx}\,dx$$

However, $g(x)$ need not be differentiable and, in fact, can even be discontinuous.

In the case of interest, $g(x)=x^2$ and $f(x)=x^2e^{x^2}$.  We have, therefore

$$\int_a^b x^2e^{x^2}d(x^2)=2\int_a^b x^3 e^{x^2}\,dx$$


问:
does the countinous function over continuous function is continuous?
If I have a function $g:[0,1]\rightarrow [0,1]$ which is continuous and $f:\Omega \rightarrow [0,1]$ which is right continuous. Is $g(f)$ continuous or just right continuous? Here $\Omega$ is the event space and $f$ is a random variable.

答: In general it will only be right-continuous as we can take $g(x)=x$ and then $g\circ f=f$.


问:
Is every ideal in $K[X]$ generated by monomials $x_1^{d_1}x_2^{d_2}\cdots x_n^{d_n}$, where each $d_i$ is a power of $p$, finitely generated?
Let $K$ be a field of characteristic $p$, where $p$ is a prime. Consider the ring $K[X]$ of the polynomials in the variables commutative and $X=\{x_1,x_2,... \}$. Is every ideal in $K[X]$ generated by monomials $x_1^{d_1}x_2^{d_2}\cdots x_n^{d_n}$, where each $d_i$ is a power of $p$, finitely generated?

答: Of course not; consider the ideal $(x_1^p,x_2^p,x_3^p,\ldots)$.


问:
Curious about the differing limits of integration when solving an ODE
This is one of those things that I just haven't thought about but always used. It came up when I was looking over this old problem to solve $x′′+ϕ(x)=0$ with IC $x(0)=x0>0$ and $x′(0)=0$. But even solving the most straight forward ODE involves integrating both sides to different limits, and that is I just decided, after years of use, seems strange.

So for the problem that made me stop and think about it, we let $y=x′$ to form a system and have $dy/dx=−ϕ(x)/y→ydy=−ϕ(x)dx$. And here is where I feel like some details are missing that I am unsure about. We integrate this equation to get:

$.5y2=−∫_0^xϕ(z)dz+C$

So the LHS was integrated from 0 to y and the RHS from 0 to x. But why should I expect that these will be equal when y need not equal x? I'm a bit chagrined that I never stopped to think about such a basic aspect of ODEs. I suspect there is an easy explanation. Can you help me? Thanks

答:
You're actually integrating both sides of the   equation $y{dy\over dx}=-\phi(x)$ with respect to $x$ over an interval $[0,t]$, say.

That is, you write  $$\int_0^t y{dy\over dx}\,dx=\int_0^t -\phi(x)\,dx.$$  The substitution $u=y$ in the integral on the left hand side of the above then transforms the equation  to
$$\int_{y(0)}^{y(t)} u\,du =\int_0^t -\phi(x)\,dx. $$  This gives the same equation as what you have (with $t$ used as the independent variable, rather than $x$). In the end the  "shortcut"  method you describe works.

Note in your example, you have $y(0)=0$.  In general, the limits of integration  are determined from the initial conditions . On the "$x$-side", from $x_0$ to $x$; on the "$y$-side", from $y(x_0)$ to $y$.


问:
Analyze the function $f \left( x \right) =\arcsin \left( x \right) -\arcsin \left( \sqrt{1-x^{2}} \right)$
The problem

The task is to find the intervals for which the function $f \left( x \right) =\arcsin \left( x \right) -\arcsin \left( \sqrt{1-x^{2}} \right)$ is constant and for which it takes the form of:

$f \left( x \right) =a \cdot \arcsin \left( x \right) +b, \\a,b=const, a \neq 0$

Constants a,b are also to be calculated.

What I have done so far

I've defined the functions domain, calculated its derivative and the derivatives domain.
$D_{f} x \in \left\langle -1,1\right\rangle\\
f' \left( x \right) =\frac{\sqrt{x^{2}}+x}{\sqrt{x^{2}-x^{4}}}\\
D_{f'} x \in \left(-1,0\right) \cup \left(0,1\right)\\ f' \left( x \right) =0 \Leftrightarrow \sqrt{x^{2}}+x=0\\ \left|x\right|=-x \Leftrightarrow x \le 0$

Which indicates that our function is constant for x'es from the interval (-1,0).
That's where I currently am. I don't know how to go about the second part of the task, that is finding the intervals for which the function behaves as:
$f \left( x \right) =a \cdot \arcsin \left( x \right) +b$
and calculating a,b.

Any help, greatly appreciated.

答:
Starting from what the OP found for the derivative of $f(x)=\arcsin(x)-\arcsin(\sqrt{1-x^2})$, we have

$$f'(x)={\sqrt{x^2}+x\over\sqrt{x^2-x^4}}=
\begin{cases}
0&\text{for }x\lt0\\
\displaystyle{2\over\sqrt{1-x^2}}&\text{for }x\gt0
\end{cases}$$

It follows that $f(x)$, which is continuous on $[-1,1]$, is constant on the interval $[-1,0]$ and equal to $2\arcsin(x)$ plus that constant on $[0,1]$. Since $f(0)=\arcsin(0)-\arcsin(1)=-\arcsin(1)=-\pi/2$, we get

$$\arcsin(x)-\arcsin\left(\sqrt{1-x^2}\right)=
\begin{cases}
-\displaystyle{\pi\over2}&\text{for }x\le0\\
2\arcsin(x)-\displaystyle{\pi\over2}&\text{for }x\ge0
\end{cases}$$


问:
How to prove that partial derivatives are zero for local max and min in 3d
I am confused on where to start on this proof. I read the proof for 2d and 1d(extreme value theorem) and it makes sense, but I cannot visualize it in 3d.

Show that if $(x_0,y_0,z_0)$ is a local minimum or maximum point of $w=f(x,y,z)$, then $ \frac{\partial w}{\delta x} $,$ \frac{\partial w}{\delta y} $,$ \frac{\partial w}{\delta z} $ are all zero at $(x_0,y_0,z_0)$.

答:
let $(x_0,y_0,z_0)$ be a minimum then $F(x)=f(x,y_0,z_0)$ has a minimum which means

$F'(x)=f_x(x,y,z)=0$


问:
Almost sure convergence problem
I'm working on a problem in which I should prove "almost sure" convergence for a sequence of random variables. I'm using Borel-Cantelli lemma to prove it. Here is the question and my solution - I appreciate if you can give me feedback and whether you have any other solution to this problem.

Assumptions:

$X_n$ is a sequence of random variables

$Pr(|X_n| > \epsilon)\le c^n$ for sufficiently large n, where $ 0 < c < 1$ and for $\forall \epsilon > 0$

Goal: to show:

$$X_n \rightarrow^{a.s} 0 \text{ a.s. stands for almost sure convergence}$$

My solution:

As $n \rightarrow \infty$, we can write:

$$\sum_{j = 1}^{\infty} Pr(|X_n| > \epsilon) \le \frac{c}{1-c} < \infty$$

Then by the application of Borel-Cantelli lemma, we can conclude:

$Pr\{|X_n| > \epsilon, i.o\} = 0 \Rightarrow 1 - Pr\{|X_n| > \epsilon, i.o\} = Pr\{|X_n| \le \epsilon, i.o\} \Rightarrow Pr\{|X_n| \le \epsilon, i.o\} = 1 $
$\Rightarrow X_n \rightarrow^{a.s} 0$

What do you think about this solution?

答:
Careful, the complement of the set $\{|X_n|\leqslant\varepsilon, \mbox{i.o.}\}$ is not $\{|X_n|>\varepsilon, \mbox{i.o.}\}$ (it is $\{\exists m\mid \forall n\geqslant m,|X_n|>\varepsilon\}$.

So the application of Borel-Cantelli's lemma yields a set $\Omega_{\varepsilon}\subset \Omega$ of measure $1$ for which if $\omega\in\Omega'$, then there is an integer $N=N(\omega)$ such that $|X_n|\leqslant \varepsilon$ whenever $n\geqslant N$.

Take $\Omega':=\bigcap_{q\in\Bbb Q,q>0}\Omega_q$: there is pointwise convergence on $\Omega'$ to $0$ and $\mathbb P(\Omega')=1$.

Since we just need the convergence of $\mathbb P(|X_n|>\varepsilon)$, we actually proved that

If for each $\varepsilon>0$, the series $\sum_{n=1}^{+\infty}\mathbb P(|X_n|>\varepsilon)$ is convergent, then $X_n\to 0$ almost surely.

Title: Almost sure convergence problem


问:
Question based on application of Cauchy's Inequality
This is question 9 of textbook Punnusamy and Silvermann on page 273( section Cauchy's Inequality).

I have done (a) but I don't know how to do (b). Problem is that I am not sure on how to use inequality here. I am not sure on how should I proceed so that I get term involving $(1-r) ^2 $ in denominator.

Please give hint on how to attempt.

Many Thanks!!

答:
Recall from Cauchy's formula that $f(w)=\frac{1}{2i\pi}\int_{C}\frac{f(\zeta)}{\zeta-w}d\zeta$ for $|w|<1.$ Use this to obtain
$$f(z_2)-f(z_1)=\frac{1}{2i\pi}\int_{C}f(\zeta)\frac{z_2-z_1}{(\zeta-z_1)(\zeta-z_2)}d\zeta.$$

Dividing both sides by $(z_2-z_1),$ putting an absolute value we easily see that $$\left|\frac{f(z_2)-f(z_1)}{(z_2-z_1)}\right|\le \frac{1}{(1-r)^2},$$

where we used the fact that $|f(\zeta)|\le 1$ and $|(\zeta-z_1)(\zeta-z_2)|\ge (1-r)^2.$


问:
Suggestions for a reading list in Statistics and intro to machine learning
I have built a reading list based on recommendations and suggestions found on several posts from this community. My intention is to use this reading list to acquire a good understanding about statistics and also about some topics on machine learning. Being my final goal to start a PhD in Statistics (in around 2 years).

I'm familiar with basic calculus and basic Linear Algebra (almost all the concepts covered by Gilbert Strang books, even though in a practical level). Furthermore, I'm familiar with basic concepts of probability and statistics. Additionally, I'm highly proficient in R and Python (and I have some experience using numpy, matplotlib, pandas and scikit-learn).

For this reason, I would appreciate further suggestions and recommendations, in order to improve this reading list.

The books on this reading list range from introductory to more advanced level, being the following:

Gilbert Strang - Calculus

Daniel J. Velleman - How to Prove It: A Structured Approach

Gilbert Strang - Linear Algebra and Its Applications

Larry Wasserman - All of Statistics

James, Witten, Hastie and Tibshirani - An Introduction to Statistical Learning

Kreyszig - Introductory Functional Analysis with Applications [chapters 1 to 3]

Golub and Van Loan - Matrix Computations

Stein and Shakarchi - Real Analysis: Measure Theory, Integration, and Hilbert Spaces

Jan R. Magnus - Matrix Differential Calculus with Applications in Statistics and Econometrics

Fitzpatrick - Advanced Calculus

Wakerly, Mendenhall and Scheaffer - Mathematical Statistics with Applications

Casella and Berger - Statistical Inference

Boyd and Vandenberghe - Convex Optimization

Hastie, Tibshirani and Friedman - The Elements of Statistical Learning: Data Mining, Inference, and Prediction

Gelman et al. - Bayesian Data Analysis

Kevin Murphy - Machine Learning: A Probabilistic Perspective

Click this link to see a visual representation of the previous reading list

Thanks in advance!

Edit 1: Added Hans Engler Feedback

Edit 2: Added All of Statistics as suggested by Wanshan

Edit 3: Added paf feedback

答: I recommend you the book Practical Econometrics with Python. The book links theory with practical examples in Python from most basic topics like OLS to advanced topics like VARMA.</p>


问:
Set of Discontinuities for Thomae's function in $\mathbb{R}^2$
For this question, Label:

Part A: Is $f(x,y)$ integrable? Question $3$-$7$ from Spivak's Calculus on Manifolds

Part B: Prove $f:[0,1]\times[0,1]→\mathbb{R}$ is integrable.

For part A and B question, I believe that the set of points of discontinuity is exactly equal to $[0,1]\times [0,1]\cap \mathbb{Q}\times \mathbb{Q}$ because part A is the extension of Thomae's funtion to $\mathbb{R}^2$ and part B is produced by taking a cross product of Thomae's funtion.

But when I use the sequential continuity definition, I get a larger set.

For Eg: For this part A; Let $p_n$ be a sequence of rational points converging to $\frac{1}{\sqrt{2}}$ but the sequence $f\left(p_n,\frac{1}{2}\right)=\frac{1}{2}$ doesn't converge to zero i.e. $f\left(\frac{1}{\sqrt{2}},\frac{1}{2}\right)=0$

Is there something wrong with the method? Else if the set is larger, what does the set look like?

Any Help will be appreciated. Thanks in advance.

答:
\begin{equation}
  f(x,y)=\begin{cases}
    0, & \text{if $x$ is irrational}.\\
    0, & \text{if $x$ is rational, $y$ is irrational}. \\
    1/q, & \text{if $x$ is rational, $y=p/q$ in lowest terms}.
  \end{cases}
\end{equation}

Let $\epsilon>0.$ Suppose $(a,b)\subseteq [0,1]\times [0,1]$ with $b$ irrational. Then, $f(a,b)=0.$ Choose an integer $N$ large enough so that  $1/N<\epsilon$ and consider $I\times Q=\{(x,k/n):x\in [0,1],\ \ 1\leq n\leq N, \ 0\le k\le n\}.$

Now $Q$ is a finite set, so as $b$ is irrational,  $\underset{y\in Q}\inf|y-b|=\delta>0.$ Now, if $(x,y)\in I\times (b-\delta,b+\delta),$ and if $x$ is either irrational or $x$ is rational and $y$ is irrational then $f(x,y)=0$. On the other hand, if $x,y$ are rational with $y=p/q$ then $f(x,y)=1/q<\epsilon$ because by construction, $y\notin Q$ so $q>N\Rightarrow 1/q<1/N<\epsilon.$ It follows that $f$ is continuous at all points of the form $(a,b):b$ irrational.

Now since $m(I\times (\mathbb Q\cap I))=0$ we have that the set of possible discontinuities of $f$ has measure zero, and so $f$ is Riemann integrable on $I\times I.$

But if you're going to use measure theory anyway, why not just note that $f$ is zero off a set of measure zero? Or better, do the integral from scratch using partitions: for any partition $P,\ \underline S(f,P)=0.$ And if we take the partition $P=\{(0,1/q,\cdots, q-1/q,1)\times [0,1]\}$, then $\overline S(f,P)=\sum_{k=1}^q\left(\frac{1}{q}\right)\left(\frac{1}{q}\right)=\left(\frac{1}{q^2}\right)\sum_{k=1}^q1=\frac{1}{q}\to 0.$


问:
The jumping times $\tau_k$ of a $\mathcal F$-poisson process $N$ are $\mathcal F$-stopping times and $N_t=\sum_k1_{\left\{\:\tau_k\:\le\:t\:\right\}}$
Let $(\Omega,\mathcal A,\operatorname P)$ be a probability space, $(\mathcal F_t)_{t\ge0}$ be a filtration on $(\Omega,\mathcal A,\operatorname P)$ and $(N_t)_{t\ge0}$ be a $\mathcal F$-Poisson process on $(\Omega,\mathcal A,\operatorname P)$ with intensity $\lambda>0$, i.e.

$N$ is an $\mathbb N_0$-valued $\mathcal F$-adapted process on $(\Omega,\mathcal A,\operatorname P)$;

$N_0=0$;

$N_t-N_s$ and $\mathcal F_s$ are independent for all $t\ge s\ge0$;

$N_t-N_s$  is Poisson distributed with parameter $\lambda(t-s)$ for all $t\ge s\ge0$.

Assuming $N$ is almost surely right-continuous, I was able to show that

$N$ is almost surely nondecreasing

and assuming that $N$ is almost surely càdlàg, I was able to show that$^1$

$\operatorname P\left[\forall t\ge0:\Delta N_t\in\{0,1\}\right]=1$.

Assume $N$ is (surely) càdlàg. Let $\tau_0:=0$ and $$\tau_k:=\inf\left\{t>\tau_{k-1}:\Delta N_t\ne0\right\}$$ for $k\in\mathbb N$.

Are we able to show that

$\operatorname P\left[\forall n\in\mathbb N_0:\exists t\ge0:N_t=n\right]=1$;

$\tau_k$ is a $\mathcal F$-stopping time for all $k\in\mathbb N$;

$\tau_k$ is almost surely finite for all $k\in\mathbb N$.

$N_t=\sum_{k\in\mathbb N_0}1_{\left\{\:\tau_k\:\le\:t\:\right\}}$ for all $t\ge0$ almost surely.

All these claims are intuitively trivial, but I really worry about some of the technical details we need to prove them rigorously. For example, does (8.) really hold or do we need to replace $\mathcal F$ with the right-continuous filtration generated by it?

For (7.), I've tried to consider $$\operatorname P\left[N_t<n\right]=\sum_{k=0}^{n-1}\operatorname P\left[N_t=k\right]=e^{-\lambda t}\sum_{k=0}^{n-1}\frac{(\lambda t)^k}{k!}\tag{11}$$ for all $t\ge0$ and $n\in\mathbb N$. $(11)$ would tend to $0$ as $t\to\infty$ if $t^ke^{-\lambda t}\xrightarrow{t\to\infty}0$, but does this really hold? We clearly got $t^ke^{-\lambda t}=e^{k\ln t-\lambda t}$, but now the only useful inequality I'm aware of is $\ln t\le t-1$ for all $t>0$, which is not enough to conclude $k\ln t-\lambda t\xrightarrow{t\to\infty}-\infty$.

$^1$ As usual, if $x:[0,\infty)\to\mathbb R$ is càdlàg, then $x(t-):=\lim_{s\to t-}x(s)$ and $\Delta x(t):=x(t)-x(t-)$ for $t\ge0$.

答:
From Thm 3.13 in Chapter $1$ of Karatzas-Shreve, a submartingale $X_t$ for a filtration satisfying the usual conditions has an RCLL (cadlag) modification that remains a submartingale under the same filtration , if and only if the function $t \to EX_t$ is continuous. One easily checks that this is the case for a Poisson process, and therefore the Poisson process, under the filtration being right-continuous and complete, has a cadlag modification that is still a submartingale. WLOG we will now assume that we are dealing with the modification, and this also answers the question about whether the filtration can be anything : it cannot.

Let $N'_t$ be the modification of $N_t$. Then $P[N'_t = N_t] = 1$ for all $t \geq 0$, is the definition of being a modification. We must show that $N'_t$ is a Poisson process if $N_t$ is, otherwise we can't work with it.

To do this, first put $t=0$ so $N'_0 = 0$ a.s. for starters. Throw out the null set where this doesn't happen. Now, for any rational $q$, we have $P(N'_q = N_q) = 1$, and so in particular $P(N'_q \in \mathbb N_0) = 1$. Throw out all the countably many null sets (so in total probability zero) for each positive rational, to get that $P(N'_q \in \mathbb N_0 \forall q \in \mathbb Q^+) = 0$. But we are already within the realm of right-continuity, since $N_q'$ is RCLL : this shows that $P(N_r' \in \mathbb N_0 \forall r \in \mathbb R) = 1$ since the limit of a sequence of integers, if it exists (it does by continuity) is an integer. Thus $N'_t$ is a.s. $\mathbb N_0$ valued.

Now, consider $t > s \geq 0$ and the filtration $\mathcal F_s$ along with $N'_t - N'_s$.We want to show independence. Let $A \in \mathcal F_s$. We know that $N'_t - N'_s = N_t - N_s$ almost surely, and therefore $1_A(N'_t - N'_s) = 1_A(N_t - N_s)$ almost surely. Taking expectations, $E[1_A(N'_t - N'_s)] = E[1_A(N_t - N_s)]=E[1_A]E[N_t - N_s] = E[1_A]E[N'_t - N'_s]$, so independence follows.

Finally, the distribution doesn't change under a.s. equivalence so the Poisson condition is trivial. This shows that $N'_t$ is also a Poisson process under this filtration which has cadlag paths. We let $N_t = N'_t$ from now on WLOG.

For $5$, note that for each pair of positive rationals $q<q'$ we have $N_{q'} \geq N_q$ almost surely by the Poisson condition. For a right-continuous process, we know that if $N_r < N_s$ for $r >s$ then there are rationals $p,q$ such that $N_p > N_q$, simply by taking rationals close enough to the right of $r$ and $s$ respectively. It follows then that $N$ has a.s. non-decreasing paths.

For $6$, you say you have done it so I will assume it done.

Finally onto the points you mention.

For $7$ , write for $n \in \mathbb N$, $N_n = (N_{n} - N_{n-1}) + (N_{n-1} - N_{n-2}) + ... + (N_2-N_1) + N_1$. So $N_n$ is a sum of $n$ iid $Poi(\lambda)$ random variables. By $SLLN$(verify the condition, it is easy) we get $\frac{N_n}{n} \to E[Poi(\lambda)] > 0$ a.s. , and in particular $\lim_{t \to \infty} N_t = \infty$ a.s.. Finally, as $N_t$ has a.s. only jumps in the range $\{0,1\}$, we have that the range of $N_t$ is $\mathbb N_0$ a.s., which is the assertion.

For $8$ we must go by induction. $\tau_1$ is a stopping time since $\{\tau_1 \leq t\} = \{N_t \geq 1\} \in \mathcal F_t$. Now, if $\tau_k$ is a stopping time, then the two events equated below can be seen to differ  symmetrically only by a null set (on which cadlag is violated), so by completeness that $\tau_{k+1}$ is a stopping time follows.
$$
\{\tau_{k+1} > t\} = \{\tau_k > t\} \cup \left(\cup_{0<q \in \mathbb Q<t}\left(\{\tau_{k} \leq q\} \cap \left(\cap_{q<q' \in \mathbb Q<t} \{N_q = N_{q'}\}\right)\right)\right)
$$

For $9$, the range of $N_t$ is $\mathbb N_0$ a.s., and for all such elements of the sample space, $\tau_k$ is finite, so this follows.

For $10$, we know that a.s., the set of jumps of $N_t$ is in $\{0,1\}$, hence fixing a $t$ we know that on this set we have $P(N_t \geq k)= P(\tau_k \leq t)$, since every jump is of size one. A similar identity $P(\sum_{k \in \mathbb N} 1_{\tau_k \leq t} \geq k) = P(\tau_k\geq t)$ is more obvious. So,$N_t$ and $\sum_{k \in \mathbb N} 1_{\tau_k \leq t}$ are modifications of each other, but both are cadlag (that the second is cadlag is left as an exercise)! It is a standard exercise(go via rationals) to then prove they are indistinguishable, as desired.