问:
$\sum_{\Im(\rho)>0}\frac{1-2\ \Re(\rho)}{|\frac{1}{2}-\rho|^2}=0 $ iff the Riemann Hypothesis is true.
Question  Prove that Riemann Hypothesis is equivalent to the statement that $$\sum_{\Im(\rho)>0}\frac{1-2\ \Re(\rho)}{|\frac{1}{2}-\rho|^2}=0 $$ where $\rho$ are the non trivial zeros of Riemann zeta function.

Attempt
Assume that RH is true.

Then, $\Re(\rho)=\frac{1}{2}$

Hence, $$\sum_{\Im(\rho)>0}\frac{1-2\ \Re(\rho)}{|\frac{1}{2}-\rho|^2}=0 $$
Conversely Assume, $$\sum_{\Im(\rho)>0}\frac{1-2\ \Re(\rho)}{|\frac{1}{2}-\rho|^2}=0 $$
Assume on the contrary that there exists some $\rho_0$ such that $\Re(\rho_0)\neq \frac{1}{2}$. Since $0<\Re(\rho)<1$, thus let, $0<\Re(\rho_0)<\frac{1}{2}$.
How to prove the converse part?.

Edit
$$\sum_{\Im(\rho)>0}\frac{1-2\ \Re(\rho)}{|\frac{1}{2}-\rho|^2}=0 $$
Since $\rho$ has $\Im(\rho)>0$ , so $1-\bar{\rho}$ has $\Im(1-\bar{\rho})>0$.
$$\sum_{\Im(\rho)>0}\frac{1-2\ \Re(\rho)}{|\frac{1}{2}-\rho|^2} +    \sum_{\Im(\rho)>0}\frac{1-2\ \Re(1-\bar{\rho})}{|\frac{1}{2}-(1-\bar{\rho}|^2}
  =0 $$
$$\sum_{\Im(\rho)>0}\frac{1-2\ \Re(\rho)}{|\frac{1}{2}-\rho|^2} +    \sum_{\Im(\rho)>0}\frac{2\Re(\rho)-1}{|\frac{1}{2}-{\rho}|^2}
  =0 $$
What to do next?

答:
The claim in the OP is not true as $\sum_{\Im(\rho)>0}\frac{1-2\ \Re(\rho)}{|\frac{1}{2}-\rho|^2}=0$ unconditionally (the identity holds if RH true and if RH false)

By symmetry whenever $\rho$ is a root with $\Im \rho >0$, so is $1-\bar \rho$; but $\frac{1}{|1/2-\rho|^2}=\frac{1}{|\bar \rho -1/2|^2}$, while $\Re \rho + \Re (1-\bar \rho)=1$, so in the case of RH false we group the terms with $\Im \rho >0$ as above $\rho, 1-\bar \rho$ when they are distinct (in other the roots off the critical line with positive imaginary part) together and we get $0$, while the critical line zeroes give $0$ trivially as noted in the original post.


问:
Show that the $p$-series $\sum_{n=1}^\infty \frac{1}{n^p}$ is convergent iff $p \gt 1$.
Show that the $p$-series $\sum_{n=1}^\infty \frac{1}{n^p}$ is convergent iff $p \gt 1$.

Attempt:
For the right direction:
Let $\sum_{n=1}^\infty \frac{1}{n^p}$ is convergent. Then, $\lim\limits_{n\to \infty} \frac{1}{n^p} = 0$.
In particular, $\lim\limits_{n\to \infty} \frac{1}{n^p} = 0$ for $p \gt 0$. But, for $0 \lt p \le 1$, the series $\sum_{n=1}^\infty \frac{1}{n^p}$ is divergent. Hence, we must have $p \gt 1$. $\Box$

Proof that series $\sum_{n=1}^\infty \frac{1}{n^p}$ is divergent:
We know that $n^p \le n$ for all positive integers $n$ and $0 \lt p \le 1$.
Then,
\begin{equation*}
\frac{1}{n} \le \frac{1}{n^p}.
\end{equation*}
Now, since the partial sums of the harmonic series are not bounded, this
inequality shows that the partial sums of the $p-$series are not bounded when $0 \lt p \le 1$. Hence, the $p-$series diverges for theses values of $p. \Box$

For the left direction:
Let $p \gt 1$. Let $f(x) = \frac{1}{x^p}$ for all $x \in [1.\infty)$. Then, $f$ is a positive, continous, and decreasing sequence. Hence, we can apply the Integral Test here. Notice that
\begin{align*}
\int_1^\infty \frac{1}{x^p} &= \lim\limits_{b \to \infty} \int_1^b \frac{1}{x^p} \\
&= \lim\limits_{b\to \infty} \left[\frac{1}{1-p}x^{1-p}\right]_1^b \\
&= \lim\limits_{b \to \infty} \left( \frac{1}{1-p}b^{1-p} - \frac{1}{1-p} \right) \\
&= \frac{1}{p-1} \lim\limits_{b \to \infty} \left(1 - \frac{1}{b^{p-1}} \right) \\
&= \frac{1}{p-1}.
\end{align*}
Thus,
\begin{equation*}
\int_1^\infty \frac{1}{x^p}dx
\end{equation*}
is convergent. Consequently, $\int_1^\infty \frac{1}{n^p}dn$ is convergent. By the Integral Test,
we have that
\begin{equation*}
\sum_{n=1}^\infty \frac{1}{n^p}
\end{equation*}
is convergent. $\Box$

Am I correct, especially for the right direction ?

答:
The left direction is correct. An alternative way to prove this is using the Cauchy condensation test.

The right direction is not correct. You say: for $0 < p \le 1$ the series $\sum_n 1/n^p$ is divergent but you did not prove this!

Alternatively, you can also use the integral test or the Cauchy condensation test for this direction.

ContentLicense: CC BY-SA 4.0


问:
Localisation of an ideal
This should be quite easy, but somehow I can't find the proof.
Let $P\neq Q$ be two maximal ideals in the commutative ring $R$. Then $P_Q=R_Q$.

($P_Q$ is the localisation of the R-module $P$ at $Q$ and $R_Q$ is the localisation of R at Q)

答:
Since $P$ and $Q$ are distinct maximal ideals, $P$ is not contained in $Q$ and thus there exists $x \in P \cap (R \setminus Q)$.  This element becomes a unit in the localization, so the localized ideal contains a unit and is thus the entire localized ring $R_Q$.

This is a special case of basic results on pushing forward and pulling back ideals under a localization map: see e.g. $\S 7.2$ of my commutative algebra notes for more details.  (Or see any other commutative algebra text, of course.)

ContentLicense: CC BY-SA 3.0

Title: Localisation of an ideal


问:
Can we put a constant sequence ? for example : $U_n = a$ where a is a real number
i found a question that asks to say whether the proposition is True or False and prove it if it is True and give a counter example if it is false, so they said if a sequence is converge to 0 then it's monotonous so i think it's False and i wanted to give as counter example this sequence $U_n = 0$

答:
You are correct in that $U_{n}$ does indeed converge to $0$. However, your alleged counter example in fact is a positive example, because your sequence is indeed monotonous by definition:

$U_{n} monotonous :\iff (\forall n \in \mathbb{N}:U_{n} \leq U_{n+1} \lor  \forall n \in \mathbb{N}: U_{n+1} \leq U_{n}) $

In other words, the statement talks about monotonous, not strictly monotonous. A constant series is both monotonically increasing as well as monotonically decreasing.

Try to find a series that is indeed neither but still converges. Hint: it may involve $(-1)^n$. Spoiler:

Consider $U_{n}=\frac{(-1)^{n}}{n}$


问:
What is the interior of this set? (weak topology)
Here the $\mathbb R$ is a set of real numbers. Let the topological space, $T$ of wich basis are $B =\{(a,b] \vert a,b\in \mathbb R\} $

Define $f_1, f_2 : (\mathbb R,T_1) \rightarrow (\mathbb R,T)$  by $f_1(x) =x^2$ and $f_2(x) = -x^2$ respectively. Here the $(\mathbb R ,T_1)$  be the smallest topological spaces defined by both of the $f_1$ and $f_2$ are continuous on $\mathbb R$. Find the interior of the $[0,1)$ in $(\mathbb R, T_1)$

In my solution all the $a$ to $e$  are positive real number. In the $f_1$ case, the open set $G_1$ can be $\{0\}$ and $[-b,-a) \cup (a,b] $ considering the $f_1^{-1}((-a,0])$ and $f_1^{-1}((a^2,b^2])$  set in order.  Vice versa when the $f_2$ case with the same method, open set $G_2$ would be $(-c,c)$ and $(-c,-d] \cup [d,c)$ considering the $f_2^{-1}((-c^2,0])$ and $f_2^{-1}((-c^2,-d^2])$ in sequence.

Consequently, the $T$ should have a subbasis as $G_1$ and $G_2$. The open set form of the $T $ are like the below

(1) $\{0\}$ considering the intersection of $G_1 =\{0\}$ and $G_2 = (-c,c)$

(2) $(-e,-d) \cup (d,e)$ considering the intersection of $G_1 = [-b,-a) \cup (a,b] $ and $G_2= (-c,-d] \cup [d,c) $

So my answer is "interior of the $[0,1) $ is $\{0\}$". But the answer was $\phi$ in my workbook. What did I wrong?

答:
A base for $\mathcal{T}_1$ is given by all sets of the form

$$f_1^{-1}[O]\cap f_2^{-1}[O'] \text{, where } O,O' \in \mathcal{T}$$

So $\{0\} = f_1^{-1}[(-1,0]] \cap f_2^{-1}[(-1,0]] = \{0\}\cap (-1,1)$ is indeed an open neighbourhood of $0$, so the interior of $[0,1)$ contains at least that. So your book's answer of $\emptyset$ is not correct.


问:
Given a set $A$, define relation $R$ on $\mathcal{P}(A)$ by $\{(U,V) \in \mathcal{P}(A) \times \mathcal{P}(A) \colon \dots \}$. Is $R$ transitive?
Given a set $A$, define relation $R$ on $\mathcal{P}(A)$  by
$$R = \{(U,V) \in \mathcal{P}(A) \times \mathcal{P}(A) \colon (U \cap V \neq \emptyset) \lor (U \cup V = \emptyset)\}.$$

I want to check if $R$ is a transitive relation.

My scratch work so far:

Let $A = \{a, b, c, d, e\}$ then $\mathcal{P}(A)=\{\{\emptyset\}, \{a\},\{b\},\{c\},\{d\},\{e\},\dots,\{a,b,c,d,e\}\}$.

So I know I want to prove that $URV \land VRW \Longrightarrow VRW$ for transitivity.

If I let $(U,V)=(\{a\},\{a,b\})$ then $(U \cap V) = \{a\} \neq \emptyset$, so that  $URV$,

$(V,W)=(\{a,b\},\{b\})$ then $(V \cap W) = \{b\} \neq \emptyset$, so that  $VRW$

Then $(U \cap V)\cap(V \cap W) = \emptyset$ right? Which means $URV∧VRW⟹V\not RW$.  Am I missing a key concept? Is this enough to prove it's not transitive? I know there exists subsets I can pull to show transitivity but if I show one isn't transitive, it should show the overall relation isn't transitive right?

Any feedback on this is appreciated. Thanks.

答:
Your argument is fine, I’m just writing the solution because you seem to have some little mistakes in your writing (I will pointed out)

To show that a relation $R$ is transitive, we need to show that

\begin{align*}
\forall x,y,z \, ,\, xRy \wedge yRz \implies xRz. \tag{1}
\end{align*}

In this case, if we let $A = \{a,b\}$, $U = \{a\}$, $V = \{a,b\}$ and $W = \{b\}$, then we have that

\begin{align*}
U \cap V = \{a\} \neq \emptyset \implies URV \\
V \cap W = \{b\} \neq \emptyset \implies VRW
\end{align*}

Although, we have that

\begin{align*}
U \cap W = \emptyset \implies U \not R W \tag{2}
\end{align*}

Therefore, you have that $R$ is not a transitive relation. $\square$

At $(1)$, when in your question you write the definition of a transitive relation, you say that for all $U,V,W \subseteq A$ we have that $URV \wedge VRW \implies VRW$ which is wrong, it should be $URV \wedge VRW \implies URW$.

At $(2)$, you check the intersection $(U \cap V) \cap (V \cap W)$ which is wrong. You only need to check the intersection between $U$ and $W$ (So, the intersection $U \cap W$) in order to say something about $URW$.

Finally, I just want to say that your thought is right, your arguments is correct. In order to show that $R$ is not transitive you only need to work out a counter example, as you well did. Continuation of a good work.


问:
Why the convergency of Cauchy product requires at least one of the series to be abs. convergent while the product of two convergent seq. converges?
In baby rudin, Theorem 3.3(c) states,

suppose $\{s_n\}$,$\{t_n\}$ are complex sequences and $\lim_{n\rightarrow \infty}s_n=s$, $\lim_{n\rightarrow \infty}t_n=t$, then $\lim_{n\rightarrow \infty}s_nt_n=st$

here we don't require $\{s_{n+1}-s_n\}$ or $\{t_{n+1}-t_n\}$ to be absolutely convergent.
But then Theorem 3.50 states,

if

$\sum_{n=0}^\infty a_n$ converges absolutely,

$\sum_{n=0}^{\infty}a_n=A$ and $\sum_{n=0}^{\infty}b_n=B$,

$c_n=\sum_{k=0}^{n}a_kb_{n-k}$,

then $\sum_{n=0}^\infty c_n=AB$

which means $\lim_{n\rightarrow \infty}A_nB_n=AB$, where $A_n=\sum_{k=0}^na_k$ and $B_n=\sum_{k=0}^nb_k$.

But if we consider $\{A_n\}$ and $\{B_n\}$ as two sequences, it doesn't require $\sum_{n=0}^\infty a_n$ to be absolutely convergent to make sure $\lim_{n\rightarrow \infty}A_nB_n=AB$ according to Theorem 3.3(c). I'm quite confused about this. Any answer or hint would be greatly appreciated!

答:
Yes, Theorem 3.3(c) proves that $$\lim_{n\to\infty} \left(\sum_{k=0}^n a_k\right)\left(\sum_{k=0}^n b_k\right) = AB$$ but that's not the Cauchy product:

$$\left(\sum_{k=0}^n a_k\right)\left(\sum_{k=0}^n b_k\right) \ne \sum_{k=0}^n\sum_{i=0}^k a_i b_{k-i}$$


问:
Showing that a radical function is continuous
I want to show that $f(x) = x^{1/5}$ is continuous for all $x \in \mathbb R$.
My try:
Fix $\epsilon > 0, c \in \mathbb R $ and try to find $\delta > 0$ such that $|x-c| < \delta \Rightarrow |f(x) -f(c) | < \epsilon$ .
The case for $c = 0$ is quite straightforward. When I use the identity to expand $a^5 - b^5 $, I get: $$|x^{1/5} - c^{1/5}| = \frac {|x-c|}{|x^{4/5} + c^{4/5} + (xc)^{2/5} + x^{1/5}c^{3/5} + x^{3/5} c^{1/5} |} \leq \frac{|x-c|}{|x^{1/5}c^{3/5} + x^{3/5} c^{1/5}|}$$ because $$ x^{4/5} + c^{4/5} + (xc)^{2/5} \geq 0$$ for all reals.
But I am stuck from here..
Any help would be great

Body:
Consider $\delta<\dfrac c2$ so that $x>\dfrac c2$.

Then the long denominator is at least

$$(2^{-4/5}+2^{-3/5}+2^{-2/5}+2^{-1/5}+1)\,c^{4/5}$$

and

$$|x^{1/5}-c^{1/5}|<\frac{|x-c|}{c^{4/5}}<\frac{\delta}{c^{4/5}}.$$

So you can take

$$\delta=\min(\frac c2,c^{4/5}\epsilon).$$


问:
Notation of max?
What does the following notation mean?

$$V^{(k)}(s) = \max_{a \in A} \sum_{s' \in S}T(s,a,s') [R(s,a,s') + \gamma\cdot V^{(k-1)}(s')]$$

Does it mean max the value of the function with a element of A. What will be the result of the Function a element of A? What do they mean with a element of A max?

答: $\max_{a ∈ A} f(a)$ is just the maximum of the set $\{f(a): a ∈ A\}$, i.e. the largest of the values $f(a)$ where $a$ ranges the set $A$. In other notation, it's $\max f[A]$.

Title: Notation of max?


答:
Good question! I'll try to simplify the situation for you. Suppose, $A = \{1,2,3,4\}$. Also suppose there is a function $f: A\to\mathbb R$, such that $$f(1) = 0, f(2) = 3, f(3) = -2, f(4) = 1$$
I shall explain what $$\max_{a\in A} f(a)$$ means! Well, it essentially asks you to iterate over set $A$, compute the value $f(a)$, and return back the maximum value you see in the process. In our case, $\max_{a\in A} f(a) = 3$. Another useful notion is that of $\text{argmax}$. $$\operatorname{argmax}\limits_{a\in A} f(a) = 2$$
$\text{argmax}_{a\in A}f(a)$ returns the element(s) from set $A$, for which $f(a) $ is maximum. On the other hand, as you saw, $\max_{a\in A}$ returns the maximum value itself! Hope that clarifies the notation.

P.S. Isn't that a Bellman equation in Reinforcement Learning? Super cool, have fun!

Title: Notation of max?

Body:
What does the following notation mean?

$$V^{(k)}(s) = \max_{a \in A} \sum_{s' \in S}T(s,a,s') [R(s,a,s') + \gamma\cdot V^{(k-1)}(s')]$$

Does it mean max the value of the function with a element of A. What will be the result of the Function a element of A? What do they mean with a element of A max?


问:
Need references for study of fundamental group of 3-manifold
Can anyone provide references on the study of the fundamental group of 3-manifolds?

The references should be geared for a physicist that has good knowledge of differential geometry and group theory, but is not very proficient in the latter.

答:
I've been studying fundamental groups of 3-manifolds lately, and one of my main references has been Hempel's book 3-Manifolds (Amer. Math. Soc., Vol. 349, 2004), which although somewhat heavy in its content, discusses a lot about the group-theoretic properties of fundamental groups of 3-manifolds.

Otherwise, there's Aschenbrenner et. al.'s 3-Manifold Groups (Zürich: European Math. Soc., Vol. 20, 2015) which is a nice survey of various things known, which in itself contains a myriad of references about the subject.

If you want the differential geometry side of things, Hempel deals with $I$-bundles over surfaces and the like, and if you want the knot theory side of things, then either Rolfsen's Knots and Links (Amer. Math. Soc., Vol. 346, 2003) is a very accessible text, as is Prasolov & Sossinsky's Knots, Links, Braids and 3-Manifolds (Amer. Math. Soc., No. 154, 1997).

There is much overlap in the above texts, and in my opinion Hempel is the most intense read, but the most rewarding, while Prasolov & Sossinsky give a very hands-on approach with lots of beautiful pictures.


问:
Interviews of famous modern mathematicians
I was wondering, are there any good collections of interviews of famous modern mathematicians? It can be text interviews, or audio or video recordings.

I am not sure what exactly I mean by "modern". It just feels that mathematics in the last, say, 20 years is different than before in the sense that it is getting more and more abstract, and different mathematical fields become further away from each other. So it seems that now more people are trying to find concrete (or at least more concrete) applications of advanced theories, and also to relate different mathematical subjects, other than to solve hard problems within a field. (This may be completely wrong, this is just an impression I have now ). Therefore, it seems that modern mathematicians have a different approach to math, and it is very interesting to read about their vision of math in interviews.

I have read some very nice interviews by Manin, Kontsevich, Voevodsky, Arnold and some others, but I have found them accidentally. So if you have links to some websites collecting interviews with modern mathematicians, I would really appreciate that!

Thank you very much! I hope this question is appropriate here!

Edit: I have found this very nice collection of interviews, collected by E.Dynkin.

答:
Interview  with  You tuber Richard Borcherds (Fields Medalist)

Interview Date Mar $17, 2021$

video link


答:
After the last step you mentioned the denominator i.e.

$$|x^{1/5}c^{3/5}+x^{3/5}c^{1/5}|=|x^{1/5}||c^{1/5}||c^{2/5}+x^{2/5}|\ge |x^{1/5}||c^{3/5}|$$

as both terms in the last modulus are positive

Now for $c\ne 0\ \exists \delta\ |x-c|<\delta\Rightarrow\ x\in [c/2,3c/2]$ in whatever order the set is applicable.

Using this assumption the net expression can be written as $\le\frac{|x-c|}{|(c/2)^{1/5}\times c^{3/5}|}$ following which we can easily continue to show that the function is continuous using epsilon-delta definition.

Title: Showing that a radical function is continuous

Body:
I want to show that $f(x) = x^{1/5}$ is continuous for all $x \in \mathbb R$.
My try:
Fix $\epsilon > 0, c \in \mathbb R $ and try to find $\delta > 0$ such that $|x-c| < \delta \Rightarrow |f(x) -f(c) | < \epsilon$ .
The case for $c = 0$ is quite straightforward. When I use the identity to expand $a^5 - b^5 $, I get: $$|x^{1/5} - c^{1/5}| = \frac {|x-c|}{|x^{4/5} + c^{4/5} + (xc)^{2/5} + x^{1/5}c^{3/5} + x^{3/5} c^{1/5} |} \leq \frac{|x-c|}{|x^{1/5}c^{3/5} + x^{3/5} c^{1/5}|}$$ because $$ x^{4/5} + c^{4/5} + (xc)^{2/5} \geq 0$$ for all reals.
But I am stuck from here..
Any help would be great


问:
Relation between stability of fixed points of discrete and continuous dynamical systems
Let $f:\mathbb R^d\to\mathbb R^d$ be a diffeomorphism, $x_0$ be a fixed point of $$\frac{\rm d}{{\rm d}t}\varphi_t(x)=f\left(\varphi_t(x)\right)\tag1$$ (i.e. $f(x_0)=0$) and $A:={\rm D}f(x_0)$.

It is a well-known result that if the largest eigenvalue $\lambda_1$ of $A$ is negative, then $x_0$ is asymptotically stable.

Now, on the other hand, if $x_0$ is a fixed point of the discrete system $$x_n=f(x_{n-1})\tag2,$$ we may consider the linearization $$h_n=Ah_{n-1}\tag3$$ and it is well-known that $x_0$ is asymptotically stable if $|\lambda_1|<1$.

How are these stability results related?

I've read that we somehow need to "integrate" $(2)$ over time and then use that if $\lambda$ is an eigenvalue of $A$, then $e^\lambda$ is an eigenvalue of $e^A$, but I neither know how precisely we need to transform $(2)$, nor why $e^A$ is occurring then.

答:
The link between the two results comes from linear system analysis.

1. From continuous-time dynamic system(CTDS) to discrete-time dynamic system(DTDS), linear case

Consider the state-space representation of a CTDS,
$\dot{x}=Ax,x(0)=x_0$ where $x,x_0\in\mathbb{R}^d,A\in\mathbb{R}^{d\times d}.$ We can explicitly solve this:
$$x(t)=e^{At}x(0).$$

Let $x_n=x(nT)$ where $T$ is a positive constant, we have
$$x_{n+1}=e^{AT}x_n=e^Te^Ax_n.$$ That is to say, we obtain a DTDS by sampling a CTDS. This is how $e^A$ occurs.

2. Stability and stability criteria of the two linear systems

Let $x_0$ be a fixed point of the CTDS, then $Ax_0=0$. Since $$e^A=\sum_{n=0}^\infty\frac{A^n}{n!},$$ we have that $$e^Ax_0=0.$$

Hence $x_0$ is also a fixed point of the DTDS.

Using the definition for stability, we can easily check that the (in)stability of the CTDS at $x_0$ is equivalent to the (in)stability of the DTDS at $x_0$ and the factor $e^{T}$ doesn't matter.

As you said, if $λ$ is an eigenvalue of $A$, then $e^λ$ is an eigenvalue of $e^A$, therefore all eigenvalue of $A$ is negative is equivalent to all eigenvalue of $e^A$ is inside the unit circle.

3. Linearization keeps the stability invariant

See Hartman–Grobman theorem.

Edit 1:

The full logic goes like this: the stability of a nonlinear system (continuous-case or discrete-case) at some equilibrium point can be partly inferred by analyzing the linear system obtained by linearization at that point (Hartman–Grobman theorem for the continuous-case and Banach fixed-point theorem for the discrete-case). Therefore, we can focus on the linearized version of the two systems, and the stability analysis still holds.

To put it simply, We are actually dealing with four systems here:
$$\dot{x}=f(x),\tag{1}$$
$$\dot{x}=Ax,A:={\textrm D}f(x_0),\tag{1'}$$
$$x_n=f(x_{n-1}),\tag{2}$$
$$x_n=Ax_{n-1},A:={\textrm D}f(x_0),\tag{2'}$$
System (1') and (2') are the linearized version of systems (1) and (2), respectively. As you mentioned, if the largest eigenvalue $\lambda_1$ of $A$ is negative, then (1') and (1) are both asymptotically stable at $x_0$. Similarly, if $|\lambda_1|<1$, then (2') and (2) are both asymptotically stable at $x_0$.

Since (1') is a linear time-invariant system, we can easily explicitly solve it: $x(t)=e^{At}x(0)$. By sampling, we get another DTDS:
$$x_{n+1}=e^Te^Ax_n\tag{3}$$

It is by connecting (1') with (2') using (3) that we get the relationship of the two criteria. The results are shown above.


问:
Are both values of cos15° that I obtained equal and if yes, why do I get such different-looking values?
I found the value of $\cos{15}^\circ$ using 2 methods.

Method 1: using $\cos{(a-b)} = \cos{a}\cos{b}+\sin{a}\sin{b}$

$\cos{(45-30)^\circ} = \cos{45^\circ}\cos{30^\circ}+\sin{45^\circ}\sin{30^\circ}$

$\cos{15^\circ} = \frac{1}{\sqrt{2}}\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\frac{1}{2}$

$\cos{15^\circ} = \frac{\sqrt{3}+1}{2\sqrt{2}}$

Method 2: using $\cos{2\alpha} = 2\cos^2{\alpha}-1$

$\cos{(2_\dot{}15)^\circ} = 2\cos^2{15^\circ}-1$

$\cos{30^\circ} = 2\cos^2{15^\circ}-1$

substituting $\cos{15^\circ}$ with $x$ and $\cos{30^\circ}$ with $\frac{\sqrt{3}}{2}$

$\frac{\sqrt{3}}{2} = 2x^2-1$

$2x^2-1-\frac{\sqrt{3}}{2} = 0$

solving for $x$, we get $x = \frac{\pm\sqrt{\sqrt{3}+2}}{2}$ and therefore $\cos{15^\circ} = \frac{\pm\sqrt{\sqrt{3}+2}}{2}$

Note: Only the positive value out of the two values we have got can be the correct value of $\cos{15\circ}$ since $\cos{15\circ}$ cannot be a negative value. So going forward I shall only refer to the positive value obtained in Method 2 as the solution from Method 2.

Now my question is: Are both answers equal? If yes, why do we even get these two answers?

The reason I ask equal and not correct is because the universal answer everywhere on the internet and in books is that  $\cos{15^\circ} = \frac{\sqrt{3}+1}{2\sqrt{2}}$

So I am confused as to how you can get completely different-looking answers. Now a quick use of the calculator reveals that the decimal approximation of both answers is the same i.e. 0.96592... . So if that is the case then why do we get such different-looking solutions?

答:
To see that $$\sqrt {\sqrt 3 +2}=\frac {\sqrt 3 +1}{\sqrt 2}$$

simply square both sides.

Since both terms are positive we see that our desired equality is equivalent to $$\sqrt 3+ 2 =\frac {3+1+2\sqrt 3}{2}=2+\sqrt 3$$

and we are done.

As a general remark, algebraic relations can be very complicated and unintuitive.  It is certainly a good idea to compare numerical expressions, as you did, as sanity checks.


问:
Why is there this recurring pattern in the derivative of $x\cdot e^{-k\cdot x}$?
I was doing an exercise for which I needed to take the derivative(s) of the function

$$f_k(x)=x\cdot e^{-k\cdot x}$$

Doing this a few times, I noticed some sort of recurring pattern. I concluded that one can write the $n$-th  derivative of the function as

$$\frac{\partial^n}{\partial x^n}f_k(x)=k^{n-1}\cdot e^{-kx}((-1)^nkx-(-1)^n\cdot n)$$

or perhaps, to improve readability

$$\frac{\partial^n}{\partial x^n}f_k(x)=k^{n-1}\cdot e^{-kx}(-kx+n)$$
for odd $n$ and

$$\frac{\partial^n}{\partial x^n}f_k(x)=k^{n-1}\cdot e^{-kx}(kx-n)$$
for even $n$.

While I know how to compute each derivative, I do not understand why the derivative follows this pattern-like structure. Is there any intuitive idea behind this?

Clarification: By "intuitive", I mean anything that does not purely rely on blindly computing each integral. A very beautiful way to represent derivatives visually is shown in this 3blue1brown video; I don't know if anything like this is possible to explain the above described pattern. Basically, I am looking for something which actually explains why there is a pattern instead of just noticing it.

Note: This is not the actual homework assignment, it just came up when doing it.

答:
Introduce an auxiliary variable $\lambda$, and put
$$f(x,\lambda):=xe^{\lambda x}={\partial\over\partial\lambda}\bigl(e^{\lambda x}\bigr)\ .$$
The exchangeability of partial derivatives then implies
$${\partial^n\over\partial x^n}f(x,\lambda)={\partial\over\partial\lambda}{\partial^n\over\partial x^n}\bigl(e^{\lambda x}\bigr)={\partial\over\partial\lambda}\bigl(\lambda^n e^{\lambda x}\bigr)=\lambda^{n-1}(n+\lambda x)e^{\lambda x}\ .$$
When $\lambda:=-k$ we have your formulas.


答:
I present an answer in three different levels of analysis (quite literally): a first, dirtier, more cautious approach; a second, slicker, more daring approach; and finally, a generalized version of the second approach that hopefully offers some insight into why it worked in the first place. You can choose which one you like best!

1. Dirty proof. Let $\varepsilon > 0$ and $c > 0$. Then, if $x<c$, we have
$$|x^{1/5} - c^{1/5}| = c^{1/5} \left|1-\left(\frac xc\right)^{1/5}  \right| \leqslant c^{1/5} \left| 1 - \frac xc \right| = c^{-4/5}|x-c|; $$
taking $\delta_1 := c^{4/5}\varepsilon$ we obtain $|x^{1/5}-c^{1/5}| \leqslant \varepsilon$ for all $x$ with $|x-c|\leqslant \delta_1$.

If instead $c<x$, then by the same argument
$$|x^{1/5} - c^{1/5}| \leqslant x^{-4/5}|x-c| \leqslant\begin{cases}
x^{-1}|x-c| & x \in (0,1). \\
x|x-c| & x > 1,
\end{cases} $$
Suppose $x >1$ and define $\delta_>'$ in such a way that, a posteriori, it works as before: it's not too hard to guess $\delta_2 := -\frac c 2 + \frac 1 2 \sqrt{c^2+4\varepsilon}$. Then, seeing as $|x-c|\leqslant \delta_2$ would imply $x \leqslant c+\delta2$, and thus
$$\begin{split}
|x^{1/5} - c^{1/5}| &\leqslant (c+ \delta_2)\delta_2 = \left(\frac c 2 + \frac 1 2 \sqrt{c^2+4\varepsilon} \right)\left(-\frac c 2 + \frac 1 2 \sqrt{c^2+4\varepsilon} \right) \\
&= \frac 1 4 (c^2+4\varepsilon) - \frac 1 4 c^2 = \varepsilon.
\end{split} $$
Similarly, if $x \in (0,1)$, we can define $\delta_3:=\frac c 2 + \frac 1 2 \sqrt{c^2- 4\varepsilon}$, provided $\varepsilon < c^2/4$ (which can always be arranged). In that situation we also see that $|x-c|\leqslant\delta_3$ implies $c-\delta_3\leqslant x$, and thus again $|x^{1/5}-c^{1/5}| \leqslant (c-\delta_3)\delta_3= \varepsilon$ (try it).

Continuity on $(0,+\infty)$ follows by choosing $\delta := \min\{\delta_1,\delta_2,\delta_3\}$ and from the arbitrariness of $\varepsilon$ and $c$.

If $c=0$ the relevant inequality follows immediately with $\delta:=\varepsilon^5$, so the function is continuous at $0$; finally, continuity on $(-\infty,0)$ follows from the oddity relation $(-c)^{1/5} = -(c^{1/5})$.

(Notice that our choice of $\delta$ depends on $\varepsilon$ and $c$: this doesn't have to be so. See below!)

2. Prophetic proof. Let $\varepsilon > 0$ and $c\neq 0$. Suppose an angelic being manifests itself to you and gifts you with the knowledge of a $\delta$ which does not depend on $c$: $\delta := \varepsilon^5$. Indeed, this works. As before, suppose $c<x$ (i.e. $0<c/x<1$), so that $$\frac cx \leqslant  \left(\frac{c}{x}\right)^{1/5}, \implies 1- \left(\frac{c}{x}\right)^{1/5}\leqslant1-\left(\frac{c}{x}\right),$$ and similarly
$$\left[1-\left(\frac{c}{x}\right)\right]^{1/5}\geqslant 1-\left(\frac{c}{x}\right).  $$
Hence
$$
\begin{split}
|x^{1/5} - c^{1/5}| &= \frac{|x^{1/5}-c^{1/5}|}{|x-c|^{1/5}} |x-c|^{1/5} = \frac{c^{1/5}\left|1-\left(\frac{c}{x}\right)^{1/5}\right|}{c^{1/5}\left|1-\left(\frac{c}{x}\right) \right|^{1/5}} |x-c|^{1/5} \\
&\leqslant \left|\frac{1-\left(\frac{c}{x}\right)}{1-\left(\frac{c}{x}\right)} \right| |x-c|^{1/5} = |x-c|^{1/5};
\end{split}
$$
we can exchange the role of $x$ and $c$ to obtain the same inequality for the case $x<c$ (although we need to require $x\neq 0$, which is not a problem). So, as long as $|x-c|\leqslant \delta=\varepsilon^5$, we have the desired $|x^{1/5}-c^{1/5}|\leqslant\varepsilon$.

The angel's $\delta$ also works for $c=0$, as we've already shown, so continuity (uniform continuity!) on the whole of $\mathbb R$ follows from the arbitrariness of $\varepsilon$.

(But how did the angel know that this $\delta$ would work? See below!)

3. Transcended proof. Let $0<\beta<1$, and $f : x \mapsto x^\beta$ with $x \in (0,+\infty)$.
We can repeat the same argument as in the second proof, replacing $1/5$ with $\beta$, to conclude that
$$\begin{split}
\sup_{x\neq c}\frac{|f(x)-f(c)|}{|x-c|^\beta}&=\sup_{x\neq c} \frac{|x^{\beta} - c^{\beta}|}{|x-c|^{\beta}} \leqslant  1 < + \infty.
\end{split}$$
This is equivalent to saying that there is a finite constant $a>0$ (in this case the optimal choice is $a=1$) such that
$$|f(x)-f(c)|\leq a|x-c|^\beta; $$
any function of this kind is said to be $\beta$-Hölder continuous (with constant $a>0$) on $(0,+\infty)$ – written $f \in C^{0,\beta}(0,+\infty)$. (This is a generalization of Lipschitz continuity, which stems from a similar condition but with exponent $\beta=1$.)

On the domain $(-\infty,0)$, the function $f$ is defined only when $\beta$ is a rational number that has an odd denominator when expressed in minimal terms (let this be the case from now on). The $\beta$-Hölder continuity extends to this domain because of the parity relation $f(-x)= \pm f(x)$ (sign depending on the parity of the numerator of $\beta$). Also, clearly we see that, with the optimal choice of Hölder constant $a=1$,
$$|f(x)-f(0)| = |x^{\beta}| \leq |x-0|^{\beta}, $$
so we actually have that $f\in C^{0,\beta}(\mathbb R)$.

Does $\beta$-Hölder continuity with constant $a>0$ on $\mathbb R$ imply regular old continuity? Yes, because it implies uniform continuity: given an $x \in \mathbb R$ and a $\varepsilon>0$, we can take $\delta := (\varepsilon/a)^{1/\beta}$ (the angel's $\delta$!), so that for all $y$ with $|x-c|< \delta$ we have
$$|f(x)-f(c)|\leq a|x-c|^\beta < a\delta^\beta = \varepsilon.$$


问:
Non-Recursive Fundamental Recurrence Formulas
Is there a non-recursive version of the fundamental recurrence formulas for continued fractions? I am trying to compute $A_{1000}$, and it is taking me an extremely long time.

By the way, I am expanding $\sqrt2=1+1/(2+\cdots)$.

答:
I'm guessing this is about the 57th problem from Project Euler, and feel that it might be more of a question on programming and creating an optimised algorithm.

You can try to set up better recurrences. Let $d_t$ and $n_t$ denote the $t$-th denominator and numerator respectively. Then you'll find the following recurrence relations.
$$
\begin{align*}
d_t&=d_{t-1}+n_{t-1} \\
n_t&=2d_{t-1}+n_{t-1} \\
\Rightarrow n_t&=d_t+d_{t-1}
\end{align*}
$$
You can also find recurrence relations which are exclusive of one another (so the denominator recurrence relation contains no terms dependent upon the numerator and vice versa).
$$
\begin{align*}
d_t&=2d_{t-1}+d_{t-2} \\
n_t&=2n_{t-1}+n_{t-2}
\end{align*}
$$
Given that $d_0=n_0=1$, and $d_1=2$, $n_1=3$, the denominators actually form the sequence of Pell numbers, and so there is a closed form for the denominator that I'm familiar with.
$$
d_{t}=\frac{(1+\sqrt{2})^{t+1}-(1-\sqrt{2})^{t+1}}{2\sqrt{2}}
$$
You can also calculate a closed form for the numerator.
$$
n_{t}=\frac{(1+\sqrt{2})^{t+1}+(1-\sqrt{2})^{t+1}}{2}
$$


问:
How would I prove this statement using an indirect proof?
So I am new to discrete math and I am learning about proofs. I have learned a little about contrapositive and contradiction proofs.

This is the example I have trouble with $$\forall k, l \in \Bbb{Z} \, , \, kl \text{ is even} \implies k \text{ is even } \vee l \text{ is even}.$$

Suppose $k$ is odd and $l$ is odd, then there exist integers $a$ and $b$ such that $k = 2a + 1$, so $kl = (ab??)$

What I have?

答:
$$\forall k, l \in \Bbb{Z} \, , \, kl \text{ is even} \implies k \text{ is even } \vee l \text{ is even}.$$

$1)$ A direct proof.

When proving a proposition of the form $p \implies q$ using the method of direct proof, you assume that $p$ is true and, with this assumption, you show (directly) that $q$ must also be true.

Proof. Let $k,l \in \Bbb{Z}$. Suppose that $kl$ is even. Then, $kl = 2j$ for some $j \in \Bbb{Z}$. If $k$ is even, then there is nothing to prove (because $k$ is even $\vee$ $l$ is even is true). Suppose that $k$ is not even. Then, $k$ must be an odd integer. Hence, $k = 2q + 1$ for some $q \in \Bbb{Z}$. Then

\begin{align*}
kl = 2j & \implies (2q+1)l = 2j \\
& \implies 2ql +l = 2j \\
& \implies l = 2j - 2ql \\
& \implies l = 2(j-ql).
\end{align*}

Since, $j-ql \in \Bbb{Z}$ we have that $l$ is an even number and this ends the proof.

$2)$ A contrapositive proof.

When proving a proposition of the form $p \implies q$ using the method of contrapositive proof, you assume $q$ to be false and, with this assumption, you show that $p$ must also be false. This works because for any two propositions we have that
\begin{align*}
(p \implies q) \iff (\sim q \implies \sim p)
\end{align*}

Proof. Let $k,l \in \Bbb{Z}$. Suppose that it is not true that $k$ is even or $l$ is even. Then (by applying the De Morgan Law), we conclude that $k$ and $l$ are not even. So, $k$ and $l$ are odd integers. Then, $k = 2j+1$ and $l = 2p + 1$ for some $j,p \in \Bbb{Z}$. Hence,

\begin{align*}
kl = (2j+1)(2p+1) = 4jp + 2j + 2p + 1 = 2(2jp + j + p) + 1.
\end{align*}

Since $2jp+j+p \in \Bbb{Z}$, then $kl$ is an odd integer. So, $kl$ is not even. This ends the proof and result is proved by contrapositive.

$3)$ A reductio ad absurdum proof (proof by contradiction).

When proving a proposition of the form $p \implies q$ using the method of proof by contradiction, we assume that the proposition $p \implies q$ is false (so we assume that $p$ is true and $q$ is false) and then we derive a contradiction. This contradiction comes from assuming that $p \implies q$ is false and therefore we conclude that $p \implies q$ must be true.

Proof: Let $k,l \in \Bbb{Z}$ such that $kl$ is even. Suppose that $k$ and $l$ are odd integers. Then, $k = 2j$ and $l = 2p+1$ for some $j,p \in \Bbb{Z}$. Hence

\begin{align*}
kl = (2j+1)(2p+1) = 4jp + 2j + 2p + 1 = 2(2jp + j + p) + 1.
\end{align*}

Since $2jp+j+p \in \Bbb{Z}$, then $kl$ is an odd integer. But this is absurd, because it contradicts the fact that $k,l \in \Bbb{Z}$ are such that $kl$ is even. The contradiction came from assuming that $k$ and $l$ are odd integers. Therefore, $k$ is even or $l$ is even.


问:
Question on parallelograms and area - please comfirm!
Alice drew parallelogram ABCD and Bob chose a point P on its diagonal AC. Then Alice drew lines EF and  GH through P parallel to the sides of the parallelogram as shown on the figure. Is it true that the areas of parallelograms DHPF and BGPE are equal to each other regardless of the choice of the original parallelogram ABCD and the choice of the point P on its diagonal?

I then came up with the following.

My solution:

No: point P will move $x$ units right and $x$ units up, so the height of each parallelogram will increase by
$a$ and the base will decrease by $a$. If we multiply the side lengths of one of the parallelograms together:
$$(h-a)(b+a)=bh\Rightarrow bh+ah-ab-a^2=bh \Rightarrow ah-ab-a^2=0 \Rightarrow \color{DarkRed}{h-b-a=0}$$

$\color{DarkRed}{h-b-a=0}$ does not have to be true for example if $h=8,b=4,a=2$

However, I am not sure if my values for the height, base and $a$ are

realistic

possible

actually mean something

and I can't

prove it

disprove it

Would you mind telling me how to prove/disprove it, and how?

Cheers!

答:
Triangles $ACD$ and $ABC$ are congruent, so is the pair $AGP$ and $APF$ as well as $PCH$ and $PCE$.  Enough?

Alternatively, let $a=\vec{AB}$, $b=\vec{AD}$, and $0<r<1$ such that $\vec{AP}=r(a+b)$. Then the parallelogram $FPHD$ is spanned by $(1-r)b$ and $ra$, the parallelogram $GBEP$ is spanned by $rb$ and $(1-r)a$. Now compare the cross products
$$(1-r)b\times ra\quad\text{and}\quad rb\times (1-r)a.$$

OR recall that the area of parallelogram ist the product of the two sides and the sin of the enclosed angle.


问:
Solve trivariate linear Diophantine equation: $ 2x +3y +4z = 5 $
This exercise is from Rosen's Elementary Number Theory:

Solve  $ 2x +3y +4z = 5  $ in the set of integers.

The solution given in the book is:

"Since $(2,3)=1$, we take $z$ to be any integer $t$ and solve the equation $2x +3y = 5-4z$, which leads to the solution $x=-5+3s-2t, y= 5-2s, z=t$.  ".

My approach was following:

$2x +3y = 5-4z$

Since $1= 2(-1) +3 \cdot 1 $, we have $ 5-4z = 2(-(5-4z)) + 3(5-4z) $ so particular solutions are $x_0 = -(5-4z), y_0 = 5-4z$  and then the solution is $x= (-5+4z)+3s, y= (5-4z)-2s, z=z $.

Is my approach correct, it is different from Rosen's? If it is correct, how can be proved that they are same?

Why $y$ in Rosen's solution does not depend on $t$?

Thanks in advance.

答:
Hint: Rosen has
$\begin{align} &\color{#c00}{(3,-2,0)}s\\ +\, &\color{#0a0}{(-2,0,1)}t\\ +\, &(-5,5,0)\end{align}$ vs.
$\begin{align} &(3,-2,0)s\\ +\, &\color{#90f}{(4,-4,1)}z\\ +\,&(-5,5,0)\end{align}\,$ in yours

But:  $\ 2\color{#c00}{(3,-2,0)} + \color{#0a0}{(-2,0,1)}\ \ =\ \ \color{#90f}{(4,-4,1)}\,$ so they are equivalent, i.e. the span doesn't change by the elementary row operation of adding $3$ times the first row to the second, since that's an invertible operation.

Remark $ $ See here for the general linear theory, and a formula / algorithm for the trivariate case.


问:
I dont know how to finish proving this statement using an indirect proof
The problem is

$$\forall j \in \Bbb{Z} \, , \, 4j − 1 \text{ is prime } \implies j \text{ is odd}.$$

I did: assume $j$ is even
let $j =2k$ $k \notin \Bbb{Z}$
$2k \notin \Bbb{Z}+
4^2k - 1$ not Prime

I dont know what to do next

答:
$$\forall j \in \Bbb{Z} \, , \, 4j − 1 \text{ is prime } \implies j \text{ is odd}.$$

This proposition if false, so all you to do is to present a counter example.

Proof. The proposition if false. For example, let $j = 2$. Then $j \in \Bbb{Z}$ is such that $4j-1 = 7$ is prime, but $2$ is not odd. $\square$

Why is it enough to present a counter example? I’ll explain.

If you want to prove that $\forall x, p(x)$ is false, you can do it by showing that its negation is true. By De Morgan’s laws, we have that the negation of this proposition is $\exists x \colon \neg p(x)$. And to show that this latter is true, you have to prove the existence of some object $x$ such that $\neg p(x)$ is true (i.e., such that $p(x)$ is false).


问:
Im having trouble with this question
Find an indexed collection $\{A_n\}_{n \in \Bbb{N}}$ of two distinct sets (that is, no two sets are equal) satisfying the given conditions

\begin{align*}
\bigcap_{n = 1}^{\infty} A_n = \{0\} && \text{and} && \bigcup_{n=1}^{\infty} A_n = [0,1].
\end{align*}

This the only problem that I am stuck, and any help will be greatly appreciated

Title: Im having trouble with this question

Body: Define $\{A_n\}_{n \in \Bbb{N}} = \{A_i = [0,\frac{1}{i}] \colon i \in \Bbb{N}\}$.


问:
Bertrand's postulate in another point of view
I was just wondering why can't we use prime number's theorem to prove Bertrand's postulate.
We know that if we show that for all natural numbers $n>2, \pi(2n)-\pi(n)>0$ we are done.
Why can't it be proven by just showing (By using the prime number's theorem) that for every natural numbers $n>2, \frac{2n}{ln(2n)}-\frac{n}{ln(n)}>0$?

答:
As per user60930's answer, we have via the prime number theorem that for any $\epsilon >0 $ however small there is some $x_0$ such that for $x > x_0$ there is always a prime between x and $x(1+\epsilon).$ By letting $\epsilon = 1/k$ we get any number of impressive results (k=1 is Bertrand's postulate.) The only issue, and a very thorny one, is how big to make $x_0$ so that all subsequent intervals contain at least one prime.

Here we either have to be quite ingenious (like Chebyshev-Erdos-Ramanujan or, say, Nagura) or we have to invoke results that come from more advanced methods. To prove Bertrand's theorem even using somewhat dated (1941) results like Rosser's, valid for $x > 55$ --

$$\frac{x}{\ln x +2}< \pi(x) < \frac{x}{\ln x - 4}$$

--we are really shooting fish in a barrel.

The intervals $$(\frac{x}{\ln x +2},\frac{x}{\ln x -4})$$ and $$(\frac{2x}{\ln 2x + 2},\frac{2x}{\ln 2x - 4})$$ are mutually exclusive beyond about $ x = 44052.9$ if I have calculated correctly.

In other words, $\pi(2x)$ is trapped on an interval that is numerically higher than the interval in which $\pi(x)$ is trapped. The difference is therefore positive and so we have proven Bertrand using the PNT in the first instance but in particular an understanding of the error of the PNT acquired using more advanced methods.

Could we prove Bertrand's theorem using the PNT alone? The problem is that the PNT is an asymptotic result which does not in itself furnish any idea of its error on shorter intervals (but see last edit below). We need an ancillary indication (like Rosser) of how big numbers must be for the result to consistently obtain.

Edit: it seems worth mentioning for comparison that Pierre Dusart's (1999) result,

$$\frac{x}{\ln x}(1+\frac{0.992}{\ln x}) < \pi(x) < \frac{x}{\ln x}(1+\frac{1.2762}{\ln x})$$

is an amazing improvement and is valid for $x\geq 599.$ Rosser used about 3.5 million zeros of the zeta function to derive his estimates, while Dusart uses work of Brent showing that 1,500,000,001 zeros lie on the critical line. Not sure what to make of that last zero but Mathematica says 1,500,000,001 is prime.

Further edit: This question and answer to the same effect: no one has used the PNT (alone) to prove Bertrand's postulate.

The first two proofs of the PNT (de la Vallee Poussin, Hadamard, ca. 1896) were done by showing that zeros of the zeta function cannot lie on the line Re(s) = 1.  De la Vallee Poussin's form for the relative error of the PNT came in a separate proof in 1899. So if we want to find an $x_0$ large enough to use the PNT to establish Bertrand's postulate, even for early versions of the PNT we are relying on separate theorems about the error.

It is interesting that Selberg's 'elementary' proof of the PNT, which avoids complex function theory, extends methods used by Chebyshev in proving Bertrand's postulate (1850) and later by Nagura in sharpening Bertrand to $(x,(1+1/5)x).$ We often read that the prime number theorem is "deeper" than Bertrand's postulate but the connection between Chebyshev's identity

$T(x) = \psi(x) + \psi(x/2)+\psi(x/3)+\psi(x/4)...$

used in his proof of Bertrand's postulate, and the version of the prime number theorem proved by Selberg in 1948, a connection described in detail by Edwards in Riemann's Zeta Function at 281-297, suggests that the assertion should be weighed carefully.


问:
$arg(z) $ for negative angle ilustrated
so my task is to illustrate this in a coordinate system

$-(\pi/2) < arg(z-1+i) < \pi/2$

or this -(pi/2)<arg(z)<3pi/4

I know how to do it for positive angles but how can you illustrate it for negative (in this case for both positive and negative)?
Also, excuse me for not using all the fancy formating, it's my first time here on this website.

Thanks you for help.

答:
The set of complex numbers with argument between $\;-\cfrac\pi2\;$ and $\;\cfrac\pi2\;$ is the whole right side complex semiplane $\;H\;$. Thus, we want complex $\;z=x+iy\,,\,\,x,y\in\Bbb R\;$ , such that:

$$-\frac\pi2<\arg(z-1+i)<\frac\pi2\iff z-1+i=(x-1)+(y+1)i\in H\iff$$

$$x-1>0\iff x>1$$

You can now graph the set of complex numbers fulfilling the condition.

Title: $arg(z) $ for negative angle ilustrated


问:
Resultant Understanding
The resultant of two polynomials
$f\left(x\right)=a_{n}x^{n}+\ldots+a_{0},g\left(x\right)=b_{n}x^{m}+\ldots+b_{0}\in K\left[x\right]$
over a field K defined as $\det R\left(f,g\right)$
when $R(f,g)$ is the sylvester matrix of $f,g$:

$$R\left(f,g\right)=\begin{pmatrix}a_{n} & a_{n-1} & \cdots & a_{0} & 0 & \cdots & \cdots & 0\\
0 & a_{n} & a_{n-1} & \cdots & a_{0} & 0 & \cdots & 0\\
\cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots\\
b_{m} & b_{m-1} & \cdots & b_{0} & 0 & \cdots & \cdots & 0\\
0 & b_{m} & b_{m-1} & \cdots & b_{0} & 0 & \cdots & 0\\
\cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots
\end{pmatrix}$$

Why the rows of $R(f,g)$ represent the coefficients of $f,xf,\ldots,x^{m-1}f,g,xg,\ldots,x^{n-1}g$?

For example, how can we recognize the second row as coefficients of $xf$?

答: You are reading the rows backwards, it's $x^{m-1}f, x^{m-2}f, \dots, f, x^{n-1}g, \dots, g$.

Title: Resultant Understanding


问:
Show that a norm is continuous in 0 (with the 2-norm )
In this proof I want to use the definition of continuity with the 2-norm.
Let N : $\mathbb{R}^n \longrightarrow \mathbb{R}$ be another norm.
I want to show that N is continuous in 0.

To do that I took a sequence $(x_k)$ included in $\mathbb{R}^n$ such that $lim_{k \rightarrow \infty} x_k = 0$
but then I don't know how to conclude that $lim_{k \rightarrow \infty}  N(x_k) = 0 = N(0)$

(Here, I can't use the fact that all norms are equivalent.)

Thank you by advance for your help.

答:
Salut Chloé?, je vais répondre en anglais pour pas qu'ils suppriment la réponse

Consider an arbitrary $x \in \mathbb R^n$,  $N(x) = N(\sum_{i=1}^n\lambda_i e_i)$ where $\lambda_i$ are scalars and $e_i$ vectors in $\mathbb R^n$'s basis.  Now use triangular inequality and then the fameux théorème des gendarmes.


问:
Topology of uniform convergence on elements of $\gamma$
Let $\gamma$ be a cover of space $X$ and consider $C_\gamma (X)$ of all continuous functions on $X$ with values in the discrete space $D=\{0,1\}$ endowed with the topology of uniform convergence on elements of $\gamma$. What does "topology of uniform convergence on elements of $\gamma$" mean?</p>

答:
In general we have a metric co-domain $(R,d)$, so we consider (continuous) functions from
$X$ to $R$, and we have a cover $\gamma$ of $X$.
A subbase for the topology of uniform convergence on elements of $\gamma$ is given by sets of the form $S(A, f, \epsilon)$, for all $f \in C(X,R)$, $A \in \gamma$, $\epsilon>0$ real, and

$S(A, f, \epsilon) = \{g \in C(X,R): \forall x \in A: \, d(f(x),g(x)) < \epsilon \}$

For the cover of singletons we get the pointwise topology, and for the cover $\{X\}$ we get the uniform metric, and also the cover by all compact sets is used (topology of compact convergence).

In your case we have $\{0,1\}$ as codomain, so we can just consider all $S(A,f,1)$ as subbasic sets, and those are all functions that exactly coincide with $f$ on $S$, due to the discreteness of the codomain.


问:
Is there a general notion of definability/expressibility?
If we want to talk about uncountable sets, then we can distinguish between objects in the set that we can actually "specify" and those that we can't. An example of objects that can be specified are $5$, $\sup_xf(x)$, and Chaitin's constant. I can't give any examples of objects that can't be specified, since I would have specified them in doing so.

This notion is formalized in some specific contexts: E.g. a definable real number, or a definable set (of natural numbers). Based on a few quick google searches I could only find these two examples. I'd just like to get a pointer to some literature that I can read about it.

Is there a general notion of "definability" of arbitrary mathematical objects?

In particular, in classical mathematics. In constructive mathematics the notion of definability collapses to computability afaik.

I'm particularly interested if it's in type-theoretic foundations.

I can imagine there are some difficulties related to logical paradoxes when we try to define a notion of "definability" in full generality?

答:
For any structure $\mathfrak{M}$ there is a general notion of (first-order) definable subset/function/relation/element of $\mathfrak{M}$, provided by the (first-order) formulas in the language of $\mathfrak{M}$. We can also talk about definability with parameters, and so on. See here for a quick summary of this.

Now note the "first-order" bit above. This refers to the specific choice of first-order logic as our framework for making and evaluating definitions. There are other logics out there - second-order logic, infinitary logic(s), etc. - and each logic gives rise to its own definability notion. If you're interested, I strongly recommend the (legally freely available online!) wonderful collection Model-theoretic logics.

As to meta-issues, for any "reasonable" logic $\mathcal{L}$ there is a general obstacle (usually phrased for first-order logic, but much more generally applicable than that) to talking about the $\mathcal{L}$-definable phenomena in a structure $\mathfrak{M}$ within that structure itself. The existence of pointwise-definable models of set theory provides a particularly powerful example of how strong this limitation truly is! However, this doesn't prevent us from performing this analysis from outside $\mathcal{M}$. So we just have to be careful what statement is being made where.


问:
Ask unbiased estimator of $\sigma ^2$ in normal distribution when either $\mu$ known or $\mu$ unknown？
If $\mu$ is unknown, then $\frac{1}{n-1}  \sum_{i=1}^n (X_i - \overline X)^2$ is the unbiased estimator of $\sigma ^2$.

However, if $\mu$ is known, then $\frac{1}{n}  \sum_{i=1}^n (X_i - \mu)^2$ is the unbiased estimator of $\sigma ^2$.

I am very confused. From introductory statistics class, I know that given any random population, $E(S^2)$ is always equal to $\sigma ^2$. Hence, for a normal distribution, I think the sample variance = $\frac{1}{n-1}  \sum_{i=1}^n (X_i - \overline X)^2$ should always be unbiased.

答:
Of course if the mean is a known value it is self evident that it is better to use it instead of using an estimation of the mean...to understand if the "natural" estimator for $\sigma^2$ is biased / unbiased it is enough to calculate its expectation

$$\frac{1}{n}E(\Sigma_iX_i^2-2\mu\Sigma_iX_i+n\mu^2)=E(X^2)-2\mu^2+\mu^2=\sigma^2+\mu^2-2\mu^2+\mu^2=\sigma^2$$

Furthermore observe that you are in a Gaussian model, thus it is important to realize that

$$\frac{\Sigma_i(X_i-\overline{X}_n)^2}{\sigma^2}\sim \chi_{(n-1)}^2$$

while

$$\frac{\Sigma_i(X_i-\mu)^2}{\sigma^2}\sim \chi_{(n)}^2$$


问:
Change of order of integration
Can anyone please help me in identifying the change of order in the following integration $$\int\limits_0^\infty  {\int\limits_{\frac{{t - z}}{h}}^\infty  {f\left( {h,t,u,z} \right)du \,dz} }, $$ if I substitute for $v={\frac{{t - z}}{h}},$ where $f()$ is any function, $0\le t \le \infty$ and $h\ge 0$.

Thanks

答:
From drawing a picture of the region in question, swapping the order of integration gives two integrals

$$\int_{-\infty}^{\frac{t}{h}} \int_{t-hu}^\infty \cdots + \int_{\frac{t}{h}}^\infty \int_0^\infty \cdots$$

Or more consicsely

$$\int_{-\infty}^\infty \int_{\max\left(t-hu,0\right)}^\infty \cdots$$

Title: Change of order of integration


问:
Verifying field automorphisms
I am trying to compute the Galois group of $p(x) = x^4 + x^3 + x^2 + x + 1$ over $\mathbb{Q}$. I more or less know how to do it, but I am hung up on one detail. The idea is that $p(x)$ factors as $(x - \omega_1)(x - \omega_2)(x - \omega_3)(x - \omega_4)$, where $\omega_k = e^{\frac{2\pi i}{5}k}$ for $k = 1, .., 4$. That means the splitting field is $\mathbb{Q}(\omega)$, where $\omega = \omega_1$. Since $\omega$ is a root of $p(x)$, then the minimal polynomial $m_\omega(x) \in \mathbb{Q}[x]$ has degree at most deg$(p(x)) = 4$ (otherwise it wouldn't be minimal). Therefore the degree of the extension $[\mathbb{Q}(\omega) : \mathbb{Q}] \leq 4$. Since $p(x)$ separates into distinct linear factors over $\mathbb{Q}(\omega)$, the degree of the extension is exactly equal to the size of the Galois group. So $|\text{Gal}(\mathbb{Q}(\omega)/\mathbb{Q})| \leq 4$. We can define maps $\sigma_j : \omega \mapsto \omega^j$ that extend linearly to the rest of $\mathbb{Q}(\omega)$ for $j = 1, 2, 3, 4$, and the point is supposed to be that these four maps are supposed to define the entire Galois group, since there are at most four elements, and we have found four.

But my question is, how do we actually see that these maps really are automorphisms? If we actually knew what the degree of the extension was, then we could explicitly write elements of $\mathbb{Q}(\omega)$ as linear combinations of powers of $\omega$, so that we could verify that each $\sigma_j$ was an automorphism. But as it is, we don't know what the degree of the extension is, other than it is no more than $4$. I can of course just check with a computer to figure out what the degree is, but that defeats the point. How can I be sure that $\omega$ doesn't satisfy some polynomial in $\mathbb{Q}[x]$ of degree $3$ or less?

答:
Suppose that the degree were less than $4$. Then $\omega$ would satisfy a monic minimal polynomial $f$ of degree less than $4$. Hence it suffices to check that $p(x)$ is irreducible. There are a few ways.

Firstly, notice that none of the roots you wrote down are rational, hence $p(x)$ has no linear factors. Thus, if $p(x)$ is reducible it must be a product of quadratic factors. Say
\begin{align*}
p(x) &= (x^2 + ax + b)(x^2 + cx + d)
\end{align*}
comparing coefficients will give you a system of equations to solve, and you will deduce a contradiction.

A perhaps better way is to consider $p(x + 1)$ (which has a root if and only if $p(x)$ does) and notice that this is Eisenstein for the prime $5$, hence irreducible.

Edit: In response to the question in the comments. Suppose that $\alpha$ is a root of a polynomial $f(x)$ over a field $F$. Then I claim that the minimal polynomial $p(x)$ of $\alpha$ divides $f(x)$. Why? By the division algorithm
$$f(x) = q(x)p(x) + r(x)$$
where $r$ has degree less than $p$. But $p$ is the minimal polynomial hence $r(\alpha)$ is nonzero, or $r(x) = 0$. But $f(\alpha) = p(\alpha) = 0$, so it must be the second case, and $p$ divides $f$.

Title: Verifying field automorphisms


问:
Rank of a matrix if one of the diagonal elements is $0$ during elementary row operations
I'm required to find the rank of a square matrix A of dimension $m * m$. While doing elementary row operations I encountered a $0$ at a diagonal position. I tried all the possible row exchanges, but none of these could get a nonzero number at that location.

At this step what can one tell about the rank of the matrix? Is it $0$ or $m-1$ or something else? And what is the reasoning behind it?

答:
Let $A,B \in \mathcal{M}_{3 \times 3}(\mathbb{K})$ be defined as

\begin{align*}
A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 1 \end{bmatrix} && B = I_3
\end{align*}

Matrix $A$ is obtained from $B$ by performing row operations, but yet $\text{rank}(A) = \text{rank}(B) = 3$.

But there is something of interest. If the question was

What does it mean for the rank of a matrix if we obtain a zero at the diagonal in the row reduced form?

And from this you can conclude that (for example) if $A$ is an $n\times n$ matrix and $m$ is the number of zeros that you find in this situation, then $\text{rank}(A) = n -m$.


问:
Help by solving simple equation with logarithm.
I know I have to know it, but somehow I have difficulty with solving the logarithmic equations.
In the script, I found one and can't go further.
I would be glad if someone could just write me the between steps, how to come from this equation:

$\frac{1}{(\frac{1}{F(E2)}-1)*e^{\frac{E1-E2}{kT}}+1}$

to this equation

$\frac{1}{e^\frac{E1-E2 + kT*ln[(\frac{1}{F(E2)}-1)]}{kT}+1}$

Body:
$\frac{1}{(\frac{1}{F(E2)}-1)\cdot e^{\frac{E1-E2}{kT}}+1}\overset{(1)}{=}\frac{1}{e^{\ln (\frac{1}{F(E2)}-1)}\cdot e^{\frac{E1-E2}{kT}}+1}=\frac{1}{e^{\ln (\frac{1}{F(E2)}-1)+\frac{E1-E2}{kT}}+1}=\frac{1}{ e^{ \frac{kT \ln (\frac{1}{F(E2)}-1)+E1-E2}{kT} +1}}=\frac{1}{ e^{ \frac{E1-E2+kT \ln (\frac{1}{F(E2)}-1)}{kT} +1}}$

(1) since $x=e^{\ln (x)},\ x>0$


问:
A general rule for L'Hôpital's Rule
In my text there are as many as $11$ theorems dealings with the different cases of L'Hôpital's Rule and due to the quantitative heaviness I often forget rules during problem session. Isn't there a general rule for L'Hôpital's Rule which is easy to remind so that I won't need to spend useless time recalling result?</p>

答:
I think of it as a sort of "algorithm". Let's say you want to calculate the limit of

$$\lim _{x \to \infty} \frac{f(x)}{g(x)}$$

If this limit tends to either $\frac{0}{0}$ or $\pm\frac{\infty}{\infty}$, then we apply the rule and differentiate $f$ and $g$ on their own. By this, I mean you don't use the quotient rule, but you just do $\frac{f'(x)}{g'(x)}$. Now, you calculate the limit of this. Again, if you get the same thing, you differentiate again.

Let's consider an example:

Evaluate

$$\lim_{x \to 0} {\frac{2\sin (x) -\sin (2x)}{x-\sin (x)}}.$$

Here, we have $f(x) = 2 \sin(x) - \sin(2x)$ and $g(x) = x - \sin (x)$. Straight away, you notice that as $x \to 0$, we get the fraction tending to $\frac{0}{0}$. So we apply L'Hopitals rule:

$$f'(x) = 2\cos (x) -2\cos (2x)$$
$$g'(x) = 1-\cos (x)$$

So now, we are calculating

$$\lim_{x\to 0}{\frac{2\cos (x) -2\cos (2x)}{1-\cos (x)}}$$

again, this goes to

$$\frac{2 - 2}{1 - 1} \to \frac{0}{0}$$

and so we apply L'Hopitals again. Now we have:

$$f''(x) = -2\sin (x) +4\sin (2x)$$
$$g''(x) = \sin (x)$$

Now, calulating the limit gives

$$ \lim_{x\to 0}{\frac{-2\sin (x) +4\sin (2x)}{\sin (x)}} \to \frac{0}{0}$$

and so AGAIN, we apply the rule. Differetiating the third time gives us

$$f'''(x) = -2\cos (x) +8\cos (2x)$$
$$g'''(x) = \cos (x)$$

So now taking the limit gives

$$ \lim_{x\to 0}{\frac{-2\cos (x) +8\cos (2x)}{\cos (x)}} \to {\frac{-2 +8}{1}} \to 6.$$

Title: A general rule for L'Hôpital's Rule


问:
A "viscosity" solution of ODE arising from switching systems
$\newcommand{\d}[1]{\underline{\mathrm{#1}}}$

$\color{red}{\d{Setting}}$

Let $f : \mathbb R \to \{0,1\}$ be a function and let $x_0 \in (0,1)$. I have two fixed distinct real numbers $A_0,A_1$ and define $g(x) = A_{f(x)}x$. Basically, $f$ assigns to each $x$ a number $A_0$ or $A_1$, and then we get $g(x)$ by multiplying by $A$.

Consider the initial value problem
$$
\dot{W}(x) = g(W(x)) \quad W(0) = x_0
$$

for what $f$ can I assert existence and uniqueness of $W$ as a "solution" for any initial point $x_0$? I'd like existence and uniqueness only till $f$ exists $[0,1]$ for the first time.

$\color{blue}{\d{Motivation}}$

This arises in the notion of switching linear systems. Imagine a system where a particle is at a point initially, say the particle is a feather. At every point in the space, there is a fan, which blows in a certain direction, and dictates a local behaviour of the motion of the feather like a linear ODE. Now we find when the feather has a unique valid movement in this configuration.

Apart from feathers in fans, my particular application is in epidemic theory, where the trajectory of a epidemic can be controlled with an on-off switch, with different linear dynamics in the on and off case.

$\color{fuchsia}{\d{Example}}$

Of course, if $f$ is constant then the solution exists as a strong solution, and is unique given the initial condition. Let's take a first non-trivial example. Fix a $y \in (0,1)$ and let $f(x) = -1$ if $x < y$ and $f(x) = 1$ otherwise.

Now, let $x_0$ be any point. Note that the Lipschitz continuity condition fails for $f$ at $y$, in fact even continuity fails. So I'm not expecting a strong solution anyway : a solution to this won't even be differentiable, looking at the behaviour at $y$.

However, I'd like to think that a solution of some kind exists as follows : say for example that $x_0 < y$. Then , by using the existence/uniqueness theorem for $(0,y)$ I can get the solution on this interval, which will be $W(z) = e^{-z} + (x_0-1)$. This solution, by continuity of the weak solution, will fix a unique value at $y$ i.e. $W(y) = e^{-y} + (x_0-1)$. Then, using the uniqueness/existence condition in $[y,1]$ we get an extension of the solution to $[0,1]$, uniquely (I can write down the explicit answer but that's not important). We call this solution as $W$.

I read up the notion of a viscosity solution which matches heavily with my construction (For the definition, see here or see Fleming and Soner, chapter 2). I can't go for a weak solution, since that just neglects the role of $y$ in the above construction, and you can just work piecewise. I'd like the pieces to be "joined together" continuously, and a viscosity solution is ensured continuous by construction.

$\color{green}{\d{Questions}}$

According to Wikipedia , a viscosity solution seems to be defined only for certain forms of PDE that arise from the HJB. I believe that this falls in that category of "degenerate elliptic" PDE because it is a first order PDE so the condition is vacuous, but I need confirmation on this one because the switching and discontinuity of $g$ is getting to me.

Is the function I wrote above the unique viscosity solution of the ODE described? I'd like an easy way to see that it is, but the definitions are getting to me. Let me try : so the situation is $V(0) = x_0$ and $$F(x, V(x), \dot{V}(x)) = \begin{cases}
\dot{V}(x) - V(x) & x \geq y \\
\dot{V}(x) + V(x) & x < y
\end{cases} $$
To check that the $W$ constructed is a sub and super solution we only need to see the point $y$ since at the other points differentiability is present. If $\phi$ is differentiable at $y$ and $\phi \geq W$ in a neighbourhood of $y$ then $\phi(t) - \phi(y) \geq W(t) - W(y)$ for all $t$ in a neighbourhood of $y$. Now we can divide by $y$ and let $t \to u$ from left and right and get that $|y| \geq \phi'(y)$, so $W$ is a subsolution. We should get $W$ being a super solution symmetrically : if $\phi \leq W$ in a neighbourhood of $y$ then the inequalities reverse which basically leads to the reverse conclusion $|y| \leq \phi'(y)$ and we are done. (Kindly point out casual errors, I don't expect any)

If $f$ has only finitely many points of discontinuity then I expect a unique viscosity solution to exist. How can I be expected to work if $f$ had infinitely many points of discontinuity? Or at a limit point of the discontinuities? Could the solution be definable in the viscous sense here and be unique?

Essentially, I'm trying to find the largest class of $f$ for which I can find a viscosity solution. I need the class to be large enough because I want a minimizer of a certain functional of that class, which I can't really study unless I know something about this class.

答:
In control theory, one generally recognizes 2 different cases.

The switchings are somehow spaced in time, either because there is some timing scheme or device, or because the continuous trajectories move away from the discontinuities of the right-hand side of the differential equation, which define the switching surfaces. In this case existence is not really a problem, because the solutions in each interval where Lipschitz conditions hold can be patched together, final condition of one interval to initial condition of the next.

Switchings occur with infinite frequency, because trajectories on both sides of the switching surfaces lead back towards it. In this case the switching surfaces are called "sliding modes" and are studied under the label VSS, for variable structure systems. Classical solutions do not exist, and the concept employed is often called a Filippov solution, after the author of a monograph titled "Differential Equations with Discontinuous Righthand Sides".

I hope this is helpful, although it's not a very formal answer. You may be on to a good point about the connection with viscosity solutions. Those are used in a different context, partial differential equations, and I'm not sure how the 2 things relate.

(I have to add that in my opinion sliding modes do not often lead to realistic models of physical phenomena or useful control design methods. This is the personal opinion of an engineer, and many researchers disagree.)


答:
The intuition is that if you have to estimate $\mu$ then $\overline{X}$ will be the minimizer of

$$f(m) = \sum_{i = 1}^n (X_i - m)^2.$$

To see this, set the derivative equal to $0$:

$$f'(m) = \sum_{i = 1}^n 2(X_i - m) = 0 \iff \left(\sum_{i = 1}^n X_i\right) - nm = 0.$$

But that means that if $\mu$ is the true mean

$$ \sum_{i = 1}^n (X_i - \overline{X})^2 \le \sum_{i = 1}^n (X_i - \mu)^2.$$

And so there is some bias introduced when we want to estimate $\sigma^2$, because $\sum_{i = 1}^n (X_i - \overline{X})^2$ is too small.

Well, it turns out that $\sum_{i = 1}^n (X_i - \overline{X})^2$ is too small by exactly a factor of $\frac{n - 1}{n}$.

In summary,

$$\mathbf{E}\left[ \frac{1}{n - 1} \sum_{i = 1}^n (X_i - \overline{X})^2 \right] = \mathbf{E}\left[ \frac{1}{n} \sum_{i = 1}^n (X_i - \mu)^2 \right] = \sigma^2.$$

Thus both
$$ \frac{1}{n - 1} \sum_{i = 1}^n (X_i - \overline{X})^2 \text{ and } \frac{1}{n} \sum_{i = 1}^n (X_i - \mu)^2$$

are unbiased estimators of $\sigma^2$. The second estimator is "better" in the sense that it's variance from $\sigma^2$ is smaller, but the catch is that one needs to know the true mean in order to use it.

Title: Ask unbiased estimator of $\sigma ^2$ in normal distribution when either $\mu$ known or $\mu$ unknown？

Body:
If $\mu$ is unknown, then $\frac{1}{n-1}  \sum_{i=1}^n (X_i - \overline X)^2$ is the unbiased estimator of $\sigma ^2$.

However, if $\mu$ is known, then $\frac{1}{n}  \sum_{i=1}^n (X_i - \mu)^2$ is the unbiased estimator of $\sigma ^2$.

I am very confused. From introductory statistics class, I know that given any random population, $E(S^2)$ is always equal to $\sigma ^2$. Hence, for a normal distribution, I think the sample variance = $\frac{1}{n-1}  \sum_{i=1}^n (X_i - \overline X)^2$ should always be unbiased.


问:
Linear regression. Lowering response maintaining equal independent variable.
I have put some data together and modelled the behaviour of the response ($y$) as function of three independent variables $x_1$, $x_2$ and $x_3$. A simple multi-linear regression. The model looks like:

$y = k + a*x_1 + b*x_2 + c*x_3 + e$

Up to this point everything is OK. But now I want to lower the response by a $15%$. Like reducing some commissions or costs. The only idea I came across is to multiply the responses by $0.85$ and readjust the whole model. Recalculate $a$, $b$ and $c$ with the new values. I have been trying to find another way of doing this without touching the data samples. Just changing and adjusting the coefficients $a$, $b$ and $c$. Does anybody know how this should be done? An idea you come across with would be okay.

答:
The regression coefficients are given by:

$\hat{\beta} = (X'X)^{-1}X'Y$

Thus, if you scale $Y$ values by $0.85$ then this is effectively equivalent to the following:

$Y_{\text{new}} = 0.85e' Y $

where,

$e$ is a vector of ones.

Thus, the new estimate is given by

$\hat{\beta}_{\text{new}} = (X'X)^{-1}X'  (0.85 e'Y)$

Thus, you get the following relationshiop:

$\hat{\beta}_{\text{new}} = 0.85 \hat{\beta}$


问:
If $f=u+iv$ is differentiable in a domain $D$ with either $u(x,y), v(x,y),$ or arg$f(z)$ constant in $D$, then $f(z)$ is constant in $D$.
Theorem: If $f=u+iv$ is differentiable in a domain $D$ with either $u(x,y), v(x,y),$ or arg$f(z)$ constant in $D$, then $f(z)$ is constant in $D$. Moreover, if$f$ and $g$ are two differentiable functions in a domain $D$, and Im$f(z)$=Im$g(z)$ on $D$, then $f(z)=g(z)$+real constant.

To prove the first part, suppose that $u(x,y)=c$ on $D$. Then $u_x=u_y=0$ on $D$, and by the Cauchy-Riemann equations, $v_y=v_x=0$. Therefore $v$ is constant on $D$. The same can be said of when $u(x,y)$ is constant in $D$.

However, I don't know how to prove this when arg$f(z)$ is constant in $D$. How can I use the Cauchy-Riemann equations in this case?

Also, for the second part, I can use the first part with Im$f(z)-g(z)=0$ on $D$, to easily conclude that $f(z)=g(z)$+constant, but I don't understand how I can show that the constant must be real.

I would greatly appreciate it if anyone explains the solution to the above questions.

答:
Let $\arg f(z)=\theta_0$, and then
$$f(z)=|f(z)|e^{\rm{i}\theta_0},\qquad \forall z\in D.$$
Let
$$F(z)=e^{-\rm{i}\theta_0}f(z)=|f(z)|\in \mathbb R,\qquad \forall z\in D,$$
then $F$ is real-valued holomorphic function,
which implies $f$ is constant.


问:
Show that $\text{arg}(f(z))$ is a constant $\Rightarrow$ $f(z)$ is constant in $D$.
The full question is as follows
Let $f(z)$ be an analytic function in a region $D$ and $f(z) \neq 0$ in $D$. Show that $\text{arg}(f(z))$ is a constant $\Rightarrow$ $f(z)$ is constant in $D$.

My approach would be to use Cauchy Riemann in terms of polar coordiantes.

In polar coordinates the Cauchy-Riemann equations become  $$\dfrac{du}{dr}=\dfrac{1}{r}\dfrac{dv}{d\theta} ~~,~~ \dfrac{dv}{dr} = -\dfrac{1}{r}\dfrac{du}{d\theta}$$

The derivative in polar version at a point $z$ whose polar coordinates are $(r,\theta)$ is then $$f^{'}(z) = e^{-i\theta}(\dfrac{du}{dr}+i\dfrac{dv}{dr}) = \dfrac{1}{r}e^{-i\theta}(\dfrac{dv}{d\theta}-i\dfrac{du}{d\theta})$$

So how do i go on from here?
Since $arg(f(z))$ is equivalent to the $\theta$ in question, can i just say that $v_\theta = u_\theta = 0$?

Any help would be appreciated.

答:
Let $\arg f(z)=\theta_0$, and then
$$f(z)=|f(z)|e^{\rm{i}\theta_0},\qquad \forall z\in D.$$
Let
$$F(z)=e^{-\rm{i}\theta_0}f(z)=|f(z)|\in \mathbb R,\qquad \forall z\in D,$$
then $F$ is real-valued holomorphic function $(\text{Im} F(z)=0)$,
which implies $f$ is constant.


问:
How to get daily APR from APY
I want to calculate the daily APR from an APY (Annual Percentage Yield) rate.

I know the APY formula is:

$$\mathrm{APY} = \left(1 + \frac{r}{n}\right)^n - 1,$$

which is for example in my case

$$20 = \left(1 + \frac{r}{365.25}\right)^{365.25} - 1.$$

My maths skills are far behind me, how can I calculate $r$?

Title: How to get daily APR from APY

Body:
\begin{alignat}{2}
20 &= \left(1 + \frac{r}{365.25}\right)^{365.25} - 1 &\\
21 &=\left(1 + \frac{r}{365.25}\right)^{365.25} &\text{(add $1$ to each side)}\\
21^{1/365.25}&=1 + \frac{r}{365.25}  &\text{(raise each side to the $1/365.25$ power)}\\
21^{1/365.25} - 1 &=\frac{r}{365.25}  &\text{(subtract $1$ from each side)}\\
365.25\left(21^{1/365.25} - 1\right) &= r  &\text{(multiply each side by $365.25$)}
\end{alignat}

So, we arrive at $$r = 365.25\left(21^{1/365.25} - 1\right) \approx 3.057.$$


答:
This is what I come up with to the best of my abilities:

Note that in order to be able to change the order of integration, Fubini-Tonelli's Theorem requires the existence of $\int \int |f|$  in the first place. If that integral is not finite, it does not exist and a change of order of integration is not generally allowed.

Assuming it exists, you are looking to map $\{(z,u) \mid u=\frac{t-z}{h}, z\in [0,\infty) \}$ to $\{(u,z) \mid z=t-hu, u\in (-\infty,\frac{t}{h}) \} \cup \{(u,z) \mid z\in[0,\infty), u\in [\frac{t}{h},\infty) \}$.

Thus change of order of integration entails that $\int_{0}^{\infty} \int_{\frac{t-z}{h}}^{\infty}f dudz $ becomes $\int_{-\infty}^{\frac{t}{h}} \int_{t-hu}^{\infty}f dzdu + \int_{\frac{t}{h}}^{\infty} \int_{0}^{\infty}f dzdu $.

How to arrive at this? The boring (and incomplete) answer, solve the initial inner lower limit for z (inverse function). The more intuitive answer: consider the set of points that is created by letting $z$ go from $0$ to $\infty$ and for each value of $z$ you you consider all values of $u$ such that $u \geq \frac{t-z}{h}$. You end up with the section above the graph of the function $u(z)=\frac{t-z}{h}$. This equates to the domain of integration in the initial problem.

Now reverse the interpretation and look at the set created by letting $u$ go from $-\infty$ to $\frac{t}{h}$, which is the image of $\left.u(z)\right|_{[0,\infty)}$. To arrive at the same set of points as before you want to consider the same graph, but in terms of $u$ instead of $z$. This leads to the function $z(u)=t-hu$ which satisfies i) $z(0)=t$ and ii) $z(\frac{t}{h})=0$. We just created the same linear function with inversion of variables. Obviously this is the inverse. However, this is not quite the full picture. We still lack the points for which $u\geq \frac{t}{h} \land z\geq 0$, which leads to the second integral.

You can see this most clearly by actually drawing the graphs, which I strongly suggest, because putting it into words can only do so much while seeing it visually is almost self-explanatory.

(Edit: I added the second integral which I realized my answer was missing after seeing Ninad Munshi's correct answer)

Title: Change of order of integration

Body:
Can anyone please help me in identifying the change of order in the following integration $$\int\limits_0^\infty  {\int\limits_{\frac{{t - z}}{h}}^\infty  {f\left( {h,t,u,z} \right)du \,dz} }, $$ if I substitute for $v={\frac{{t - z}}{h}},$ where $f()$ is any function, $0\le t \le \infty$ and $h\ge 0$.

Thanks


问:
Calculating gaussian integral
I have  a basic idea about Gaussian integral i.e $$\int_{-\infty}^{\infty}  e^{-x^2}\,dx = \sqrt {\pi}.$$

How can I calculate this integral using this basic idea:
$$\int_{-\infty}^{\infty} x^4e^{-4x^2}\,dx$$

答:
Hint:

Consider the integral

$$\int_{\mathbb{R}} e^{-\alpha x^2}\,dx.$$

You can calculate this as a function of $\alpha$ based on your known integral (do a simple change of variable to get it into your original form). What happens if you differentiate this twice with respect to $\alpha$? Can you take it from here?

Title: Calculating gaussian integral


问:
Does the converse of uniform continuity -> Preservance of Cauchy sequences hold?

We know that if a function $f$ is uniformly continuous on an interval $I$ and $(x_n)$ is a Cauchy sequence in $I$, then $f(x_n)$ is a Cauchy sequence as well.

Now, I would like to ask the following question:

The function $g:(0,1) \rightarrow \mathbb{R}$ has the following property: for every Cauchy sequence $(x_n)$ in $(0,1)$, $(g(x_n))$ is also a Cauchy sequence. Prove that g is uniformly continuous on $(0,1)$.

How do we go about doing it?

答:
Just a remark: a key point is that the open interval $(0,1)$ is not complete as a metric space.  Suppose you have a function $f: X \rightarrow \mathbb{R}$ where $X$ is complete.  Then $f$ is Cauchy-continuous -- i.e., if $\{x_n\}$ is a Cauchy
sequence in $X$, then $\{f(x_n)\}$ is Cauchy in $\mathbb{R}$ -- iff it is continuous.  To see that this is in general much weaker than being uniformly continuous, consider the example $f: \mathbb{R} \rightarrow \mathbb{R}$, $x \mapsto x^2$.

In the other direction, if $X$ is incomplete, but with compact completion, then every Cauchy continuous function $f: X \rightarrow \mathbb{R}$ extends to a continuous function $\overline{f}: \overline{X} \rightarrow \mathbb{R}$ on the completion $\overline{X}$.  Since $\overline{X}$ is compact, $\overline{f}$ is uniformly continuous, hence so is its restriction $f$.  This latter situation obtains in the question asked by the OP.


问:
Proving that brownian motion pass almost ever by zero in $]0,t[$
Consider $B=(B_0)_{t\geq 0}$ a real $\mathcal F_t$ - brownian motion starting at zero, in a probability space $(\Omega, \mathcal F, (\mathcal F_t)_{t\geq 0}, \mathbb P)$. Then, consider
$$ \Phi_t(x) :=  \mathbb P \left\{\exists s \in ]0, t[ : B_s + x= 0\right\}$$
Show that $\Phi_t(0)=1 \forall t \geq 0$.

I have tried to explore the fact that $\Phi_t$ is even, since $- B$ is also a brownian motion. However, I still don't see how to prove it.

Any advice will be appreciate. Thank's in advance.

答:
Of course, what is the best approach to this question depends very much on what you know and what you do not know and, unfortunately, you are silent about this, but anyway, here is a try, using only elementary tools...

Let $X_t=\mathrm e^{-B_t-t/2}$, then $(X_t)$ is a martingale. Let $T$ denote the first hitting time of $0$ by $B$. Let $x\gt0$. Then, conditionally on $B_0=x$, $(X_{t\wedge T})$ is uniformly integrable since $X_t\leqslant1$ almost surely when $t\leqslant T$. Conditionally on $B_0=x$,  $X_0=\mathrm e^{-x}$ and $X_T=\mathrm e^{-T/2}$, hence the stopping time theorem yields $E_x[\mathrm e^{-T/2}]=\mathrm e^{-x}$. By symmetry, for every $x\ne0$, $E_x[\mathrm e^{-T/2}]=\mathrm e^{-|x|}$.

We deduce from this the case when $B_0=0$ almost surely. Consider the first hitting time $T_t$ of $0$ after $t$, then $T\leqslant T_t$ and $B_t\ne0$ almost surely hence
$$
E_0[\mathrm e^{-T/2}]\geqslant\mathrm e^{-t/2}E_0[E_{B_t}[\mathrm e^{-T/2}]]=E_0[Y_t],\qquad Y_t=\mathrm e^{-|B_t|-t/2}.
$$
This holds for every $t\gt0$ and, when $t\to0$, $Y_t\to1$ and $|Y_t|\leqslant1$ almost surely, hence $E_0[Y_t]\to1$. Thus, $E_0[\mathrm e^{-T/2}]\geqslant1$. Since $\mathrm e^{-T/2}\leqslant1$ almost surely, $\mathrm e^{-T/2}=1$ almost surely, that is, $T=0$ almost surely.


问:
Multiple division in algebra and expressing as powers
I can’t really find any resources on this so if you have some please let me know, but I suppose it’s really specific I only need to know this for units.

If you have the following:

$$
p=\frac{x}{\dfrac{y}{q\cdot \frac{1}{z}}}
$$

Is this how you’d write them in powers?:

$$
p=x\cdot y^{-1}\cdot q^{-2}\cdot z^{-3}
$$

I think I’m wrong because when I see units written in this way it doesn’t seem to happen like that.

答: $$\dfrac{x}{\dfrac{y}{q\cdot\dfrac{1}{z}}} = \dfrac{x\cdot q\cdot\dfrac{1}{z}}{y} = \dfrac{xq}{zy}$$.


答:
Hint:

Decompose the integrand as

$$x^3\cdot xe^{-x^2}$$ and integrate by parts.

Then repeat.

Title: Calculating gaussian integral

Body:
I have  a basic idea about Gaussian integral i.e $$\int_{-\infty}^{\infty}  e^{-x^2}\,dx = \sqrt {\pi}.$$

How can I calculate this integral using this basic idea:
$$\int_{-\infty}^{\infty} x^4e^{-4x^2}\,dx$$


问:
If $3(x +7) - y(2x+9)$ is the same for all values of $x$, what number must $y$ be?
If we let $x = 0$.

\begin{align*}
3(0+7)-y(2(0)+9) \\
21-9y \\
\end{align*}

Then $9y$ should always equal $21$?
Solving for $y$ finds $\frac{7}{3}$.

But $3(x+7)-\frac{7}{3}(2(x)+9)$ does not have the same result for diffrent values of $x$.

Where am I going wrong?

Body:
$3 (x + 7) - y (2 x + 9)=21-9y$ for $x=0$ and

$3 (x + 7) - y (2 x + 9)=24 - 11 y$ for $x=1$

They are the same, so must be
$$21-9y=24-11y\to y=\frac32$$


问:
Transformation of a normal distribution- Find marginal distribution
I got these 2 models for hourly wage for 2 periods:

The hourly wage for period 1 is normally distributed with mean $µ$ and variance $σ^2$ so $Y_1 \sim N(\mu,\sigma^2)$.

And the hourly wage for period 2 is given by:
$$Y_2=\alpha+\beta Y_1+U$$ where $Y_1$ and $U$ are independent and $U \sim N(0,v^2)$.
Then we assume that $\beta \neq 0$ and let $\mu = 350$ and $\sigma^2=12365$ and $\alpha=350\cdot(1-\beta)$ and $v^2=12365 \cdot (1-\beta^2)$.

Now I have to find the marginal distributions of $Y_1$ and $Y_2$. But how can I found that? Normally I would integrate by $Y_1$ to find the marginal distribution of $Y_2$ and integrate by $Y_2$ to find the marginal distribution of $Y_1$. But here I dont have an expression with both $Y_1$ and $Y_2$?

答: No integrals are needed. Linear combinations of gaussians are still gaussian thus all you have to do is to calculate mean and variance of $Y_2$ and I explained you how to fo that in my  answer to your previous question


问:
Concurrency of perpendiculars through touch points of excircles with sides of a triangle
Let $ABC$ be a triangle. The ex-circles touch sides $BC, CA$ and $AB$ at points $U, V$ and $W$, respectively. Be $r_u$ a straight line that passes through $U$ and is perpendicular to $BC$, $r_v$ the straight line that passes through $V$ and is perpendicular to $AC$ and $r_w$ the straight line that passes through $W$ and is perpendicular to $AB$. Prove that the lines $r_u$, $r_v$ and $r_w$ pass through the same point.

Attempt: I think I saw a property somewhere in this problem: '' circuncenter of $ I_a I_b I_c $ and isogonal conjugates '', but I can't demonstrate it. Is there any way to solve this problem, or to solve it using the mentioned property? I really have no ideas for the problem

答:
You're thinking of correct property. Here are the hints :

Recall $I$ is orthocenter of $\triangle I_aI_bI_c$ where $A,B,C$ are the feet of altitudes.

Show that the three $r_i$'s are isogonal lines of those altitudes. For this it is enough to show that $\angle CI_bV=C/2$ and $\angle BI_bI_c=C/2$. As a further hint, use quadrilateral $AIBI_c$ to show the second equality.

Hence the $r_i$'s will concur at isogonal conjugate of the orthocenter $I$, which will be the circumcenter of $\triangle I_aI_bI_c.$


问:
Direct limit in Set
If we consider a directed direct system in the category of sets (order by inclusion), then the direct limit is just the union of all the sets appearing in the direct system.

Is this true more generally for a direct system in sets which is not directed? Or does the direct limit take a different form then?

答: The direct limit of a diagram consisting of two sets $A$ and $B$ with no arrows between them is the disjoint union of the two sets, not their union. Even if $A=B$, the direct limit is then $A\coprod A$, the disjoint union of $A$ with itself (which can be realized as the set $\{(a,t)\mid a\in A,t\in \{\star, *\}\}$, where $\star \ne *$ are any two symbols). </p>

Title: Direct limit in Set


答:
You are right in saying that what you have is incorrect. Note that $\frac{1}{\frac{1}{x}} \ne \frac{1}{x^2}= x^{-2}$. In fact, $\frac{1}{\frac{1}{x}} = x = x^1$.

Now to your problem: Let's simplify the above fraction by taking the bottom fraction and instead of dividing by it, we are going to multiply by its reciprocol: $$p=\frac{x}{\frac{y}{q*\frac{1}{z}}}$$ $$ = x * (\frac{q*\frac{1}{z}}{y})$$
Next, let's move the $z$ fraction out from the bigger fraction,
$$ = x * \frac{q}{y}* \frac{1}{z}$$
Finally, we decompose again, by knowing that $\frac{a}{b} = a * \frac{1}{b}$, give it a little more notation and we get
$$ = x * q * \frac{1}{y}*\frac{1}{z}$$ $$ = x^1*q^1*y^{-1}*z^{-1}$$

Title: Multiple division in algebra and expressing as powers

Body:
I can’t really find any resources on this so if you have some please let me know, but I suppose it’s really specific I only need to know this for units.

If you have the following:

$$
p=\frac{x}{\dfrac{y}{q\cdot \frac{1}{z}}}
$$

Is this how you’d write them in powers?:

$$
p=x\cdot y^{-1}\cdot q^{-2}\cdot z^{-3}
$$

I think I’m wrong because when I see units written in this way it doesn’t seem to happen like that.


答:
You have the function:

$$ f \left( x, y \right) = 3 \left( x + 7 \right) - y \left( 2 x + 9 \right) $$

Since the function is constant for $ x $ then it means:

$$ \nabla_{x} f \left( x, y \right) = 0 = 3 - 2 y \Rightarrow y = \frac{3}{2} $$

Another way thinking about it, is by definition the function with respect to $ x $ must be constant (Specific case of Linear).

Hence its coefficients (Which are given by the derivative since it's a specific case of Linear Function) must vanish (As seen in @Deepak answer).

Title: If $3(x +7) - y(2x+9)$ is the same for all values of $x$, what number must $y$ be?

Body:
If we let $x = 0$.

\begin{align*}
3(0+7)-y(2(0)+9) \\
21-9y \\
\end{align*}

Then $9y$ should always equal $21$?
Solving for $y$ finds $\frac{7}{3}$.

But $3(x+7)-\frac{7}{3}(2(x)+9)$ does not have the same result for diffrent values of $x$.

Where am I going wrong?


问:
Show that $f({x_1+x_2+\cdots x_n\over n})\le \frac 1n [f(x_1)+f(x_2)+\cdots f(x_n)]$
If $f''(x)\ge 0$ on $[a,b]$ prove that $f({x_1+x_2+\cdots x_n\over n})\le \frac 1n[f(x_1)+f(x_2)+\cdots f(x_n)]$

for any n points $x_1,x_2,...,x_n$ in $[a,b]$.

Applying MVT on $[x_1,{x_1+x_2\over 2}]$ & $[{x_1+x_2\over 2},x_2]$ and using the fact $f'$ is increasing on $[a,b]$, I could prove the proposition for $n=2$. But I couldn't use induction after that.

答:
This is a special caes of Jensen's inequality. Standard proofs are given on this wiki page.

It is also possible to prove this special case by a slightly unusual induction.

We let $P(n)$ denote the proposition for $n$. You have already shown that $P(2)$ is true.

Now the proof goes like this:

We show that, for $n \geq 2$, if $P(n)$ is true, then $P(2n)$ is true.

We show that, for $n \geq 2$, if $P(n)$ is true, then $P(n - 1)$ is true.

These two statements combined together will show that $P(n)$ is true for all $n$.

For 1, we have $$f(\frac{x_1 + \dots x_{2n}}{2n}) \leq \frac 1 2[f(\frac{x_1 + \dots + x_n}n) + f(\frac{x_{n + 1} + \dots + x_{2n}}n)] \leq \frac 1 {2n}[f(x_1) + \dots + f(x_{2n})]$$ where the first inequality uses $P(2)$ and the second inequality uses $P(n)$.

For 2, we put $x_n = \frac {x_1 + \dots + x_{n - 1}}{n - 1}$ in $P(n)$ and get $P(n - 1)$.


答:
We have : \begin{aligned}\int_{-\infty}^{+\infty}{x^{4}\,\mathrm{e}^{-4x^{2}}\,\mathrm{d}x}&=\underbrace{\left[-\frac{\mathrm{e}^{-4x^{2}}}{8}\times x^{3}\right]_{-\infty}^{+\infty}}_{0}+\frac{3}{8}\int_{-\infty}^{+\infty}{x^{2}\,\mathrm{e}^{-4x^{2}}\,\mathrm{d}x}\\ &=\frac{3}{8}\underbrace{\left[-\frac{\mathrm{e}^{-4x^{2}}}{8}\times x\right]_{-\infty}^{+\infty}}_{0}+\frac{3}{64}\int_{-\infty}^{+\infty}{\mathrm{e}^{-4x^{2}}\,\mathrm{d}x}\\ &=\frac{3}{128}\int_{-\infty}^{+\infty}{\mathrm{e}^{-\left(2x\right)^{2}}\times2\,\mathrm{d}x}\\ &=\frac{3}{128}\int_{-\infty}^{+\infty}{\mathrm{e}^{-y^{2}}\,\mathrm{d}y}\\ \int_{-\infty}^{+\infty}{x^{4}\,\mathrm{e}^{-4x^{2}}\,\mathrm{d}x}&=\frac{3\sqrt{\pi}}{128}\end{aligned}

Note : We did a first integration by parts setting $ u':x\mapsto x\,\mathrm{e}^{-4x^{2}} $, and $ v:x\mapsto x^{3} $, then a second integration by parts setting $ f'=u' $, and $ g=\mathrm{Id}_{\mathbb{R}} $.

Title: Calculating gaussian integral


问:
Prove $P(A-B)\subseteq(P(A)-P(B))\cup\{\emptyset\}$
Prove that $P(A-B)\subseteq(P(A)-P(B))\cup\{\emptyset\}$

Attempt: Suppose $X\in P(A-B)$ then $X\subseteq A-B$, which means $X\subseteq A$ and $X\cap B=\emptyset$. Then, $X\in P(A)$ and $X\notin P(B)$. Therefore, $X\in(P(A)-P(B))$. This leads to $X\in(P(A)-P(B))\cup\{\emptyset\}$.

Question: Is this correct? If not, can someone give me the answer or point my mistakes out? Thank you!

Note: $P$ is the power set, not probability.

答: $X \nsubseteq B$ might lead that some elements in $X$ are in $B$ but not all, but what we want is that none of the elements of $X$ are in $B$ and thus we need the second statement $X \cap B = \phi$.


答:
The easiest way to think about this is to realise you need the expression to be independent of the value of $x$. Which means the coefficient of the $x$ term has to vanish, leaving only a constant term.

That means $3-2y = 0 \implies y = \frac 32$.


问:
The Lattice of Open Sets Determines the Topology
In a Hausdorff space $X$, it's a known theorem that the lattice structure on the open sets $\mathcal{O}(X)$ determine the topology of $X$ up to homeomorphism (actually, the Hausdorff condition can even be related a little bit to sober spaces, but I don't want too many unfamiliar concepts).  Two questions:

Is there an elementary proof that doesn't get into 'lattice theory' proper, filters, sober spaces etc.?  The result is very intuitive but I'm not sure what the cleanest way to define the homeomorphism is.  Obviously we need to define when two lattices are isomorphic, but can we avoid other theory?

When is the same true of the lattice of closed sets?  Of the poset on the closed + connected subsets?

Thanks!

答:
Re: $1$, I don't see how you'd hope for a proof avoiding some lattice theory. After all, a homeomorphism crucially involves points, and the only way points are coded into the lattice of opens of a sober space is via filters. To do anything "point-flavored" with the lattice of opens, you need in some sense to talk about those filters.

Re: $2$, as noted in the comments the lattice of closed sets carries exactly the same information as the lattice of open sets (they're duals). Restricting attention to connected sets is a bit more interesting. Of course for most spaces we lose too much information, but among "nice" spaces there are some distinctions we can draw. For example, if $m\not=n$ then the lattice of connected open subsets of $\mathbb{R}^m$ is not homeomorphic to the lattice of connected open subsets of $\mathbb{R}^n$ (think about covering dimension).


问:
A question about finding maxima and minima
Today I see a problem very wired.

To find the maxima and minima of function
$$f(x,y) = x^3 - y^3 -12xy -63x + 63y$$

I tried to take the partial derivative with respect to x and y, and obtain:

$$\frac{\partial}{\partial x}f(x,y) = 3x^2 - 12y-63$$

$$\frac{\partial}{\partial y} = -3y^2 - 12x + 63$$

and when I tried to find the stationary point, I found that I don't know how to solve the system of equations.

答:
My first advice is to learn latex my friend, it's very helpfull!

Lets consider $$f(x, y) = x^3 - y^3 - 12xy -63x + 63y$$
Taking both partial derivate of $x$ and $y$ and making them equal $0$,
$$\frac{\partial}{\partial x} f(x, y) = 3x^2-12y-63 = 0, \quad (1)$$ and $$\frac{\partial}{\partial y} f(x, y) = -3y^2 -12x +63x = 0, \quad (2)$$
Clearing $y$ from equation $(1)$ and substituting in equation $(2)$, $$y=\frac{3x^2-63}{12} \Longrightarrow -3\Big(\frac{3x^2-63}{12}\Big)^2-12x+63=-\frac{3x^4}{16}+\frac{63x^2}{8}-12x+\Big(63-\frac{1323}{16}\Big)=0$$ which is a fourth degree polynomial, then $$x\in\{-7, -1, 3, 5\}$$ Using this set of points for $x$ on equation $(1)$ (or $(2)$, it's the same), were have the following set of stationary points, $$A = \{(-7, 7), (-1, -5), (3,-3), (5, 1)\}$$
Let's consider the following Theorem.
Let $$D=\frac{\partial^2}{\partial x^2} f(a, b) \cdot \frac{\partial^2}{\partial y^2} f(a, b) -\Big(\frac{\partial^2}{\partial x \partial y} f(a, b)\Big)^2, \quad (a, b) \in A$$ then,

$1)$ If $D>0$ and $\frac{\partial^2}{\partial x^2} f(a, b) > 0$, then $f$ has a minimum at $(a, b)$.

$2)$ If $D>0$ and $\frac{\partial^2}{\partial x^2} f(a, b) < 0$, then $f$ has a maximum at $(a, b)$.

$3)$ If $D<0$, then $f$ has a saddle point at $(a, b)$.

$4)$ If $D=0$, then we have no conclusion for $f$ at $(a, b)$.

Using this Theorem with the points of the set $A$, we conclude that, $(-7, 7)$ is maximum of $f$, $(3, -3)$ is minimum of $f$ and both $(-1, -5)$ and $(5, 1)$ are saddle points of $f$.


问:
Existence & uniqueness of radix representation by decrementing descent
I'm reading a proof in George E. Andrew's Number Theory of the Basis Representation Theorem (i.e. each positive integer $n$ has a unique representation in base $k$), and I'm having some difficulty. For reference, the argument in the proof runs like this:

Suppose we have a representation of $n$ as
$$n = a_0k^s + a_1k^{s-1} + ... + a_{s}.$$

If some of the trailing $a_i$ are $0$, then exclude them so that
$$n = a_0k^s + a_1k^{s-1} + ... + a_{s-t}k^t,$$
where now $a_0 \neq 0$ and $a_{s-t} \neq 0$. Then
$$
\begin{align*}
n-1 &=  a_0k^s + a_1k^{s-1} + ... + a_{s-t}k^t - 1\\
&= a_0k^s + a_1k^{s-1} + ... + (a_{s-t}-1)k^t + k^t - 1\\
&= a_0k^s + a_1k^{s-1} + ... + (a_{s-t}-1)k^t + \sum_{j=0}^{t-1}(k-1)k^j.
\end{align*}
$$

Let $b_k(n)$ denote the number of representations of $n$ in the base $k$. We just showed that for every representation of $n$, we can find one for $n-1$, so that
$$b_k(n) \leq b_k(n-1).$$

With this fact, the proof continues and concludes
$$1 \leq b_k(k^n) \leq b_k(n) \leq b_k(1) = 1, \ \ \ (1.1)$$

So we have existence and uniqueness. $\blacksquare$

My question: Part of the theorem's hypothesis says each $a_i$ is such that $0 \leq a_i \leq k-1$. I want to understand where the proof fails if you remove this requirement. If you allow $a_i$'s to be negative, then I see why the proof fails, because then $1$ can be expressed in numerous ways, so that $b_k(1) \neq 1$ (for example $1 = -(k-1)\cdot k^0 + 1\cdot k^1$). However, I don't see where the proof fails in the case where you require $a_i \geq 0$, but allow them to be arbitrarily large (i.e. $0 \leq a_i < \infty)$. In this case, $1$ still has a unique representation, so the problem lies somewhere in the left side of $(1.1)$ rather than the right side. I can see the issue with a particular example: if you allow arbitrarily large $a_i$, then you could have for instance $3 = 1\cdot3^1 = 3*3^0$, so that $b_3(3) = 2$, while $b_3(1) = 1$, so $b_3(3) > b_3(1) = 1$, which is the opposite of the inequality $b_k(n) \leq b_k(n-1)$ in the proof, but I don't see how the proof fails in this case. I'm rather confused, so insights would be appreciated.

答:
The issue is that different representations of $n$ can result in the same representation for $n-1$ if $a_i$ is allowed to be any nonnegative integer.
For example, allowing for such $a_i$ but following the process Andrews describes we find
$$\begin{array}{rcll}
10 = 1\cdot 10^1
    &\rightarrow&
    9 = (1-1) 10^1 + (10-1) 10^0
    = 9\cdot 10^0 &\\
10 = 10\cdot 10^0
    &\rightarrow&
    9 = (10-1) 10^0
    = 9\cdot 10^0. &\quad(!)
\end{array}$$
Thus, it does not follow that
$b_k(n)\le b_k(n-1)$.
It is critical to Andrews' argument that different representations of $n$ result in different representations for $n-1$.
This only occurs if $0\le a_i\le k-1$.


问:
Choosing an answer randomly
You are asked a question and you don't have a clue about it. But, luckily you are given $K$ ($K \ge 4$) possible answers and only one is the right one. Because you choose your answer randomly you would like to increase your chance of answering the question right. You have two options:

Choose an answer randomly.

Choose an answer randomly, exclude it and then choose another answer randomly from the other $K-1$ given answers.

So the question, would you have better chance if you choose option 2 than option 1? And if it's better, how many times would be the best option, and stopping before it become a bad idea (excluding large part of the answers).

Let's say. $K=4$. Then you'll have $25$% odds choosing the right answer with option 1.
If you exclude that answer you have $33$% or $0$% odds of choosing the right answer.

答:
These are two different procedures, whose outcome is selecting one of the $K$ answers.  All $K$ outcomes are equally likely, so the procedures are identical in outcome.  Procedure 1 is simpler and quicker, so is to be preferred.

Clarifying edit: Think of these procedures as "black boxes" that produce a guess.  It doesn't matter how the procedures work, they produce a guess.  By symmetry, none of the answers are any more likely than any of the others to be chosen, because the black box doesn't know the answer to the question.  Hence every such mechanism is identical to every other such mechanism.

Title: Choosing an answer randomly


答:
If we let $x=0$ then we do indeed find that the expression equals $21-9y$. But this does not mean that $21-9y$ must equal $0$; it only means that $21-9y$ is a constant. If, as Raffaele has suggested, we let $x=1$, then we find that the expression equals $24-11y$. Hence, $21-9y$ and $24-11y$ must be equal to the same constant, and so solving the problem boils down to solving the equation $21-9y=24-11y$:
\begin{align}
21 - 9y &= 24 - 11y &&\text{Subtract $21$ from both sides}\\
-9y &= 3 - 11y &&\text{Add $11y$ to both sides}\\
2y &= 3 &&\text{Divide both sides by $2$}\\
y &= \frac{3}{2} \, .
\end{align}