问:
Solving Math Problems in your head
Recently, I have read about chess players who can not only visualize some moves, but actually "see" a chess board that is not there and play moves (e.g. Nakamura, Carlsen etc.).

Now I wonder whether the same is possible to achieve for mathematics. I recently tried solving some problems in my head (e.g. $\int \frac{1}{x(x-1)x(x-2)}dx$) and I have some partial success: I get to the solution (and it is correct), however, I am considerably slower than I am on paper and I also noticed the following fact: It is not like I am writing on a nonexistent paper or anything, but rather, I remember some partial solutions and try my best to not forget them.

Let me illustrate that on an example: I want to do the following polynomial division: $(x^3-5x^2+3x+1):(x-1)=x^2-4x-1$. Whilst trying to get the $-4x$ part of the solution, I need to repeat $x^2$ (first part of the solution) as to not forget it. This wouldn't happen when solving it on paper as it is just written there and you basically can't forget it. Of course this isn't a problem when doing such easy tasks, but it might evolve into a gigantic problem when actually attempting to solve geometric problems or something like that.

How do I evade my problem? Is it just practice? If so, what is the most efficient way for practicing? Just as a reminder: My end goal is to feel (almost) no difference between solving something on paper and in my head -- quite similar to chess: The top players (GMs) can play games in their head nearly as well as regularly over the board.

Thank you for your answers!

答:
It's matter of practice and using the right algorithms. With practice there is a short term effect that will make you better at this over a period of weeks to months. But I've noticed that I have continued to improve on much longer time scales. I can do much more complex computations in my head than I could do when I was 20 years old.

Let's start with the right algorithms. In the division example you mentioned, I think you are using long division, but you should use synthetic division. I can easily solve the division problem in my head, because I do the division as follows. I first divide the polynomial by $x$, this yields:

$$x^2 - 5 x + 3$$

where I've dropped the fractional part. Then we just add to the coefficient of $x$ the coefficient of $x^2$. If we were dividing by $x-a$, we would add $a$ times the coefficient of $x^2$ to the coefficient of $x$. So, we then get the intermediary result:

$$x^2 - 4 x + 3$$

And then we need to add the coefficient of $x$ which is $-4$ to the constant term of $3$, this gets us to the result:

$$x^2 - 4 x - 1$$

Usually there are many different methods to do computations, the methods that are taught are usually what's most convenient for working things out on paper, not for mental computations.

Lifestyle and diet

Another important factor that's usually ignored in these sorts of questions is lifestyle and diet. When I was 20 I was not eating the sort of extremely healthy diet I'm eating now. I now eat a mainly whole food, plant-based diet. I do eat meat and fish, but in small quantities. This sort of a diet gives me much more fiber than a normal diet, and it's now known that fiber is important for the brain by feeding the microbiome, see here. I get 100 grams of fiber a day, while most people get less than 20. I also get 150 grams of protein a day while the RDA for me would be just 60 grams and would on a normal diet be met by meat and dairy. About 100 grams of that protein comes from only the dry bread, vegetables, pasta, potatoes I eat.

Changing my diet happened gradually over the years. It's not possible to change a normal diet to the sort of diet I'm eating in just a few weeks, as your intestines won't be able to process the sheer volume of food and the  massive amount of fiber in it.

I also exercise a lot, about 1 hour of fast running per day. This burns about 1000 Kcal and therefore need to eat 1000 kcal more. All that extra food contains a massive amount of nutrients that will also benefit my brain.

Another thing I do that most people don't do is that I take a large dose of vitamin D. I take 10,000 IU on 5 days per week, so my intake is about 7,000 IU/day. This is much higher than the RDA, but it's what I would get from the Sun if I were outside all day long in a country at a latitude consistent with my skin color.

Now, many brilliant mathematicians who are very good at working out things mentally do not stick to a healthy lifestyle. Here genetics may play a role. The human body is extremely robust, it can function quite well on a suboptimal diet. But on the long term the body will have to compensate for that, it will adjust gene expression and switch to an alternative mode of operation. The consequences for that for your energy levels and your long term health will then be dependent on your genetics.

So, there are people with the right genes whose academic performance is not affected all that much by a suboptimal lifestyle, they end up becoming the new professors while the rest of the population will end up not pursuing an academic career.

Sleep

Getting enough and good quality sleep is obviously also important. One thing I noticed is that when I was about 20 years old, I could not read while dreaming. So, in a dream, I could see letters but I was unable to read. But this gradually changed. I now do dream about seeing texts and being able to read. This may be important for the brain to develop the ability to efficiently process texts and formulas mentally.

Title: Solving Math Problems in your head


问:
Repeating period in non binary numeral systems
I came across this question -> Repeating period in binary conversion. I want to know if there is a way to calculate repeating period in numeral systems with base other than 2.

For example:

Number $0.1_{10}$ is $0.01212121212121..._4$ so there are 2 repeating ciphers .

Number $0.1_{10}$ is $0.02222222222222222222...._5$ so there is only 1 repeating cipher .

Also I noticed that the repeating periods are repeating :

number 0.1

base 2 repeating period : 4

base 3 repeating period : 4

base 4 repeating period : 2

base 5 repeating period : 1

base 6 repeating period : 1

base 7 repeating period : 4

base 8 repeating period : 4

base 9 repeating period : 2

base 10 repeating period : 1

base 11 repeating period : 1

So if there is way to calculate even these repeatings ? Thanks.

答:
As you will have seen from the other post, if the base $b$ is coprime to $10$ then the period $p$ for the number $0.1_{10}$ will be the least power of $b$ such that $b^p\equiv 1\pmod {10}.$ Note that
$$3^4\equiv 1\pmod {10}$$
$$7^4\equiv 1\pmod {10}$$
$$9^2\equiv 1\pmod {10}$$
$$11^1\equiv 1\pmod {10}$$
This will give us a pattern which repeats every $10$ bases since $13\equiv 3\pmod {10}$, $15\equiv 5\pmod {10}$ and so on.

If the base $b$ is even but coprime to $5$ then the period $p$ will be the least power of $b$ such that $b^p\equiv 1\pmod {5}.$ Note that
$$2^4\equiv 1\pmod {5}$$
$$4^2\equiv 1\pmod {5}$$
$$6^1\equiv 1\pmod {5}$$
$$8^4\equiv 1\pmod {5}$$
Again this will give us a pattern which repeats every $10$ bases.

If the base $b$ is odd but but a multiple $5$ then the period $p=1$ since $b^p\equiv 1\pmod {2}.$ The period is also $1$ if $b$ is a multiple of $10$.

This gives you the pattern you have noticed with overall period  being $5$.

ContentLicense: CC BY-SA 4.0


问:
A is a matrix of integers , prove that A+I is invertible
Question:

$2 \le d \in \Bbb Z$

Let $A \in M_n(\Bbb Q) s.t$ All of it's elements are integers divisible by d.

Prove that $I+A$ is invertible.

What I thought:

I thought of using the determinant of A and extracting $d^2$ from it, but I can't use that on det(A+I)... tried also to calculate the determinant for n=2, and got a polynomial of 2nd degree (With no idea on how to prove that it doesn't equal 0).

答:
There's no need of doing an induction; when you consider your matrix $I+A$ modulo $d$ it's just the identity, so
$$
\det(I+A)\equiv 1\pmod{d}
$$
and it can't be $0$.

Let's do it with more details. Your matrices are in $\def\Z{\mathbb{Z}}M_n(\Z)$. We have a ring homomorphism $\delta\colon\Z\to\Z/d\Z$ that easily extends to a homomorphism $\delta_n\colon M_n(\Z)\to M_n(\Z/d\Z)$, applying $\delta$ to each entry of the matrix.

What can we say about $\delta(\det A)$ and $\det(\delta_n(A))$? They are equal for any matrix $A$:
$$
\det A=\sum_{\sigma}(-1)^{\mathrm{sgn}(\sigma)}
  a_{1,\sigma(1)}a_{2,\sigma(2)}\dots a_{n,\sigma(n)}
$$
so applying $\delta$ to it is exactly the same as doing
$$
\sum_{\sigma}(-1)^{\mathrm{sgn}(\sigma)}
  \delta(a_{1,\sigma(1)})\delta(a_{2,\sigma(2)})\dots \delta(a_{n,\sigma(n)})
$$
($\sigma$ runs over the permutations of $\{1,2,\dots,n\}$).

If now the matrix $A$ has all of its entries in $d\Z$, $\delta_n(A)=0$ (the null matrix in $M_n(\Z/n\Z)$); thus
$$
\delta(\det(I+A))=
\det(\delta_n(I+A))=
\det I=1
$$
(the unity in $\Z/dZ$), which means exactly
$$
\det(I+A)\equiv 1\pmod{d}
$$
Thus $\det(I+A)\ne0$, so the matrix is invertible in $M_n(\mathbb(Q)$.

ContentLicense: CC BY-SA 3.0


问:
Relation between total variation and KS distance between measures on $[0,1]^d$
Let $P$ and $Q$ be two probability measures on the space $[0,1]^d$, $d \in \{1, 2, \ldots \}$, endowed with the $L_\infty$ norm and the corresponding Borel $\sigma$-field, $\mathcal{B}$. Let
$$F_P(\mathbf{u})=P([\mathbf{0},\mathbf{u}]), \, \quad F_Q(\mathbf{u})=Q([\mathbf{0},\mathbf{u}]),$$
denote the distribution functions associated to $P$ and $Q$, respectively. Then, we have that
$$
d_{KS}(F_P,F_Q):=\sup_{\mathbf{u}\in[0,1]^d}
|F_P(\mathbf{u})-F_Q(\mathbf{u})| \leq \sup_{B \in \mathcal{B}}|P(B)-Q(B)|=:d_{TV}(P,Q).
$$
My question is the following: assume $F_P$ and $F_Q$ are Lipschitz continuous, then does (some form of) converse inequality also hold true?

I was reasoning in this way: since $P$ and $Q$ are regular, for every $B \in \mathcal{B}$ and $\epsilon>0$ there exist closed sets $C_{B,\epsilon}^{(P)},C_{B,\epsilon}^{(Q)}$ and open sets  $O_{B,\epsilon}^{(P)},O_{B,\epsilon}^{(Q)}$ such that $O_{B,\epsilon}^{(\bullet)} \subset B \subset C_{B,\epsilon}^{(\bullet)}$ and
$$
P(C_{B,\epsilon}^{(P)}\setminus O_{B,\epsilon}^{(P)})\leq \epsilon, \quad
Q(C_{B,\epsilon}^{(Q)}\setminus O_{B,\epsilon}^{(Q)})\leq \epsilon.
$$
Whence,
$
|P(B)-Q(B)| \leq 2 \epsilon + |P(O_{B,\epsilon}^{(P)})-Q(O_{B,\epsilon}^{(Q)})|.
$
Yet, from now on it is not clear how to proceed. Maybe cover
each open set with uniform metric-balls $\{B_1^\bullet,\ldots,B_{m_\bullet}^\bullet\}$ of radius $\delta$? Herein , we could maybe exploit the covering number inequality $m_\bullet \leq (3d/\delta)^d$. Observe that each ball is of the form
$$
B_i^\bullet=\times_{j=1}^d(u_{i,j}^\bullet-\delta,u_{i,j}^\bullet+\delta),
$$
where $\mathbf{u}_i^\bullet=(u_{i,1}^\bullet, \ldots, u_{i,d}^\bullet) \in [0,1]^d.
$
In particular, by absolute continuity, we could choose $\delta$ such that $$
|F_P(\mathbf{u}_i^Q+\delta \mathbf{1})-F_Q(\mathbf{u}_i^Q-\delta \mathbf{1})|\leq \epsilon',
\quad |F_Q(\mathbf{u}_i^P+\delta \mathbf{1})-F_P(\mathbf{u}_i^P-\delta \mathbf{1})|\leq \epsilon'
$$
for some arbitrarily small $\epsilon'>0$. But still it is not evident to me that this could lead to a suitable upperbound encompassing $d_{KS}(F_p,F_Q)$. Do you have any clue?

答:
The reverse inequality does not hold for $d=1$ even when the CDFs are Lipschitz.

A) Infinite family of counterexamples:

Let $U_\epsilon$ denote the uniform measure supported on $A_\epsilon=[\tfrac{1-\epsilon}{2},\tfrac{1+\epsilon}{2}]\subseteq[0,1]$ for $0<\epsilon\leq 1$. Its CDF is continuous and even Lipchitz since the derivative $U_\epsilon$ is bounded. The CDF is given by,
$$
F_{\epsilon}(x)
=
\begin{cases}
0 & x \in \left(-\infty,\tfrac{1-\epsilon}{2}\right],\\
\tfrac{1}{\epsilon}(x-\tfrac{1}{2})+\tfrac{1}{2} & x\in \left[\tfrac{1-\epsilon}{2},\tfrac{1+\epsilon}{2}\right],\\
1 & x\in \left[\tfrac{1+\epsilon}{2},\infty\right).
\end{cases}
$$

First note that,
$$
d_{TV}(U_\epsilon,U_1)\!=\!\sup_{B\in\mathcal B} |U_\epsilon(B)-U_1(B)|\!\geq\! |U_\epsilon(A_\epsilon)-U_1(A_\epsilon)|
\!=\!1-\epsilon.
$$

Next note that since $|F_1-F_\epsilon|$ achieves a maximum value at $x^*_{\pm}=\tfrac{1\pm\epsilon}{2}$. In particular, $|F_1(x^*_-)-F_\epsilon(x^*_-)|=F_1(x^*_-)=x^*_-$ and so,
$$
d_{KS}(F_\epsilon,F_1)
= \sup_{t\in[0,1]} |F_\epsilon(t)-F_1(t)|
= F_1(x^*_-)
= \frac{1-\epsilon}{2}.
$$

We can make $d_{KS}(U_\epsilon,U_1)<d_{TV}(U_\epsilon,U_1)$ occur by demanding $\tfrac{1-\epsilon}{2}<1-\epsilon$, or equivalently, $\epsilon<1$. Thus, if we choose $P=U_\epsilon$ and $Q=U_1$, where $0<\epsilon<1$, then the CDFs are Lipschitz and $d_{KS}(F_P,F_Q)<d_{TV}(P,Q)$.

B) Counterexample with both measures supported on [0,1]:

To produce these counterexamples, we can simply perturb the previous ones. Consider the mixture of uniform measures $W_\epsilon = \epsilon U_1 + (1-\epsilon) U_\epsilon$. Note that $\operatorname{supp} W_\epsilon = [0,1]$ for all $0<\epsilon\leq 1$. Moreover, it's CDF is given by
$$
G_\epsilon(x)=W_\epsilon([0,x])= \epsilon F_1(x) + (1-\epsilon) F_\epsilon(x).
$$

Note that $|W_\epsilon-W_1| = (1-\epsilon)|U_\epsilon-U_1|$. Similarly, $|G_\epsilon-G_1| = (1-\epsilon)|F_\epsilon-F_1|$. Thus,
$$
\begin{align*}
d_{TV}(W_\epsilon,W_1) &= (1-\epsilon)d_{TV}(U_\epsilon,U_1), \\
d_{KS}(G_\epsilon,G_1) &= (1-\epsilon)d_{KS}(F_\epsilon,F_1).
\end{align*}
$$

By using our results in A), we obtain
$$
\begin{align*}
d_{TV}(W_\epsilon,W_1) &= (1-\epsilon)d_{TV}(U_\epsilon,U_1)\geq
(1-\epsilon)^2, \\
d_{KS}(W_\epsilon,W_1) &= (1-\epsilon)d_{KS}(F_\epsilon,F_1)=
\frac{(1-\epsilon)^2}{2}.
\end{align*}
$$

Thus, $d_{KS}(W_\epsilon,W_1)<d_{TV}(W_\epsilon,W_1)$ since $\tfrac{1}{2}(1-\epsilon)^2<(1-\epsilon)^2$ holds for all $\epsilon\neq 1$.


问:
The topological dual for the weak topology is the topological dual.
Let $E$ be Banach space and $E'$ its topological dual. The weak topology on $E$ is define to be the coarsest topology (with the least amount of open sets) on $E$ such that all element of $E'$ are continuous. Now if we denote by $\widetilde{E'}$ the topological dual of $E$ for the weak topology, we clearly have $E' \subset \widetilde{E'}$. In my functional analysis course, we said that these  sets has to be algebrically the same but I don't really understand from where comes the reverse inclusion. Any help ?

答:
The comment of Ruy is correct. Since norm convergence is stronger for elements in $E$ than weak convergence, every weakly continuous functional is norm continuous.

For the reverse, if $\psi:E \to \mathbb{C}$ is a norm continuous functional and $a_\alpha \to a$ converge weakly, we have to see that $\psi(a_\alpha) \to \psi(a)$. But that follows from the fact that $a_\alpha$ converge weakly.


问:
Given a recursion $a_{n+ 1}= \sqrt{na_{n}+ 2n+ 1}$ with $a_{1}\geq 1.$ Prove that $a_{n}\sim n\,{\rm as}\,n\rightarrow\infty$

Given a recursion $a_{n+ 1}= \sqrt{na_{n}+ 2n+ 1}$ with $a_{1}\geq 1.$ Prove that
$$a_{n}\sim n\,{\rm as}\,n\rightarrow\infty$$

Let $b_{n}= \dfrac{a_{n}}{n},$ we need to prove $\lim b_{n}= 1,$ then we find a value $\beta\in\left ( 0, 1 \right )$ satisfying $\left | b_{n+ 1}- 1 \right |\leq\beta\left | b_{n}- 1 \right |$ so that
$$\beta\rightarrow 0\,{\rm aut}\,n\rightarrow\infty$$
I can't predict the relationship among them, I need the the help to go to the induction, thank you.

答:
You have
$$\begin{aligned}
a_{n+1} &=\sqrt{na_n +2n+1}\\
&= \sqrt{n^2+2n+1+(n a_n -n^2)}\\
&=(n+1) \sqrt{1+\left(\frac{n}{n+1}\right)^2\left(\frac{a_n}{n}-1\right)}
\end{aligned}$$

Hence denoting $b_n =\frac{a_n}{n}$

$$\begin{aligned}
b_{n+1} -1
&=\sqrt{1+\left(\frac{n}{n+1}\right)^2\left(b_n-1\right)}-1
\end{aligned}$$

As for $x\gt 0$

$$\begin{aligned}
\sqrt{1+x}-1 &=\frac{x}{\sqrt{1+x}+1}\\
&\le \frac{1}{2}x
\end{aligned}$$

you get

$$\begin{aligned}0 &\le b_{n+1}-1\\
&\le \frac{1}{2} \left(\frac{n}{n+1}\right)^2 (b_n-1)\\
&\le \frac{1}{2}(b_n-1)
\end{aligned}$$ and can conclude to the desired result.


问:
$\frac{t_1 + y}{s_1} \leq \frac{t_2 + y}{s_2}$ and $\frac{t_1}{s_1} \leq \frac{t_2}{s_2}$
Consider the following inequality
$\frac{t_1 + y}{s_1} \leq \frac{t_2 + y}{s_2}$ where $y$ is a non-negative integer and others are positive integers, does this imply $\frac{t_1}{s_1} \leq \frac{t_2}{s_2}$ and if yes how? if not why not?

答:
$$\frac{t_1 + y}{s_1} \leq \frac{t_2 + y}{s_2}\implies \frac{t_1 }{s_1} \leq \frac{t_2}{s_2}+y\left(\frac1{s_2}-\frac1{s_1}\right)$$

If $s_2\ge s_1$, then $y(1/s_2-1/s_1)\le0$ which gives $t_1/s_1\le t_2/s_2$.

If $s_2< s_1$, then for $y\ne0$ we have $y(1/s_2-1/s_1)>0$ and the result is not necessarily true unless $y=0$. For example, take $t_1=s_1=8,t_2=2,s_2=4,y=64$.


问:
Explanation of application of superposition theorem for homogenious linear ODE
\begin{equation}
x^{\prime}+a(t)x+b(t)=0 \, \, \, \text{Denote by $(1)$}
\end{equation}

\begin{equation}
x^{\prime}+a(t)x=0 \,\,\, \text{Denote by $(2)$}
\end{equation}

Particular solution: When the solution $x(t)$ of say $(1)$ passes throught a given point say
$(t_0,x_0)$ i.e. $x_0=x(t_0)$ then the solution is particular.

Example:

For example the solution of $(2)$ is $\int_{x_0}^x{du}=x_0-\int_{t_0}^{t}{a(s)x(s)ds}$ and write
$x=x(t,t_0,x_0)$
The problem here is that I dont understand why the solution of $(2)$ looks of these form, because the solution is given by
$$x^{\prime}=-a(t)x$$
$$\frac{dx}{dt}=-a(t)x $$
$$ \frac{dx}{x}=-a(t)dt$$
$$ \log(x)=-\int{a(t)dt}+C$$
$$x=Ke^{-\int{a(t)dt}}$$ and when then by the condition $x_0=x(t_0)$
$$ x_0=Ke^{-\int_{t_0}^{t}{a(s)ds}}$$ but from here I don´t know how to get the form of $(2)$ in the example.

Someone can explain me how is that the author get the form of $(2)$? it was very helpful.

My first goal is understand the following theorem

Theorem:Let us assume that $x(t,t_0,x_0)$ is solution of $(2)$ then $p(t)=\int_{t_0}^{t}x(t,s,b(s))ds$ is solution of $(1)$ and have general solution $y(t)=x(t,t_0,x_0)+\int_{t_0}^{t}x(t,s,b(s))ds$.

I really I don´t understand how to apply the theorem to solve the equation
$y^{\prime}-\frac{2}{t}y=t+\frac{1}{t} \sin (\frac{1}{t^2})$
I know how to get the solution without this method (in fact is $y=t^2 \log(t)+\frac{t^2}{2} \cos (\frac{1}{t^2})+t^2 c$), but when I try to apply the method I fail.

If someone can help me to understand the theorem above was vwet useful.

答:
Actually,
$$
\int_{x_0}^x\,du = x_0 - \int_{t_0}^t a(s)x(s)\,ds
$$
is not a solution. It is an integral equation equivalent to your differential equation $(2)$.

The theorem says that $y(t) = \int_{t_0}^t x(t; s, b(s))\,ds$ is a particular solution of equation $(1)$ which satisfies $y(t_0) = 0$.

The proof is very simple:
$$
\begin{align*}
&\left[\int_{t_0}^t x(t; s, b(s))\,ds\right]' + a(t)\int_{t_0}^t x(t; s, b(s))\,ds \\&= x(t; t, b(t)) + \int_{t_0}^t x'(t; s, b(s))\,ds + \int_{t_0}^t a(t) x(t; s, b(s))\,ds \\&= b(t) - \int_{t_0}^t a(t) x(t; s, b(s))\,ds + \int_{t_0}^t a(t) x(t; s, b(s))\,ds = b(t).
\end{align*}
$$

In fact, this theorem is just a more formal approach to the method of variation of constants.


问:
A convexity inequality
The function $\mathbb{R}^n\to \mathbb{R}$, $x\mapsto |x|^p$ (where $p>1$) is convex and thus the inequality
$$|y|^p-|x|^p\ge p(y-x)\cdot x |x|^{p-2}$$ is valid. In some lecture notes of Peter Lindqvist, it is remarked that this inequality can be strengthened to
$$|y|^p-|x|^p\ge p(y-x)\cdot x |x|^{p-2} + C(p) |y-x|^p$$ (of course $C(p)>0$) at least for $p>2$.
Does anyone know a proof of this inequality?</p>

答:
(I'm expanding my comment.)

This is a two-dimensional problem. One may assume $x=(1,0)$, $y=(1+\alpha, \beta)$ with $\sqrt{\alpha^2+\beta^2}=:r$. Then
$$|y|^p-|x|^p=((1+\alpha)^2 +\beta^2)^{p/2}-1\ ,\qquad p(y-x)\cdot x |x|^{p-2}=p\alpha\ .$$
It follows that we have to prove an inequality of the form
$$(1+2\alpha+r^2)^{p/2}\geq 1 + p\alpha + Cr^p \qquad\qquad (1)$$
for a suitable $C$, and we may assume $p\geq2$.

Putting $r:=0$ in (1) the statement reads $(1+2\alpha)^{p/2}\geq 1 + p\alpha$, and this is true for $p\geq2$ by Bernoulli's inequality. Now the derivative  of the left side of (1) with respect to $r$ is
$${p\over 2}(1+2\alpha +r^2)^{p/2 -1}\ 2r\geq {p\over 2}r^{p-2}\  2r=pr^{p-1}\ ,$$
and the derivative of the right side of (1) with respect to $r$ is $Cp r^{p-1}$. So if $C=1$ the left side of (1) grows faster with $r$ than the right side. It follows that (1) is true with $C=1$.

Title: A convexity inequality


问:
How to show $\mathbb{Z}[w]/(2,w) \simeq \mathbb{Z}_2$?
Let $\mathbb{Z}[w]=\mathbb{Z}[\frac{1+\sqrt{-15}}{2}]$ be the quadratic integers.

I want to show that $\mathbb{Z}[w]/(2,w) \simeq \mathbb{Z}_2$. It seems very clear, but how can I show the isomorphism rigorously by isomorphism? I found somewhat similar using 3rd isomorphism theorem(here). But I think there will be much easier way.

I tried $\mathbb{Z}[w]/(2,w) \simeq (\mathbb{Z}(X)/(f(X))/(2+f(X),X+f(X))$ but I'm not sure then it is isomorphic to $\mathbb{Z}(X)/(2,X)\simeq \mathbb{Z}_2$.

I also tried $\mathbb{Z}[X] \to \mathbb{Z}[w] \to \mathbb{Z}[w]/(2,w)$, but not sure what the kernel is. Obviousely $2,w$ goes to $0$ but isn't it possible that $1 \mapsto 0$?

答:
Hint: Define:

$$
\varphi(a + b \omega) = \begin{cases} 0 &: a \text{ even} \\
1 &: a \text{ odd}
\end{cases}
$$

Show that this is a homomorphism from $\Bbb Z[\omega]$ to $\Bbb Z/2\Bbb Z$ and find its kernel. Apply the first isomorphism theorem.


问:
Why $\sum_{i=1}^{k} \frac{1}{p^i} = \frac{p^{-k}-1}{1-p}$?
I have the statement:

$$\sum_{i=1}^{k} \frac{1}{p^i} = \frac{p^{-k}-1}{1-p}$$

without a clear step explanation why.

I've tried expanding the LHS and rewriting the RHS, but I don't see it yet.

$$\sum_{i=1}^{k} \frac{1}{p^i}= ? =\frac{p^{-k}-1}{1-p}$$

I'd appreciate an intermediary step "?" to help understand the result.

答:
You simply apply the formula for the sum of a geometric series:
$$x+x^2+\dots+x^k=x\bigl(1+x+\dots+x^{k-1}\bigr)=x\,\frac{x^k-1}{x-1},$$
substitute $\dfrac1p$ to $x$ in this formula and simplify.


答:
This is a geometric series. Hint: write
$$S=a+ar+ar^2+...+ar^N$$
and then multiply it by the common factor $r$ to obtain
$$rS=ar+ar^2+...+ar^{N+1}.$$
What you get if you substract one equation from the another one?

Title: Why $\sum_{i=1}^{k} \frac{1}{p^i} = \frac{p^{-k}-1}{1-p}$?

Body:
I have the statement:

$$\sum_{i=1}^{k} \frac{1}{p^i} = \frac{p^{-k}-1}{1-p}$$

without a clear step explanation why.

I've tried expanding the LHS and rewriting the RHS, but I don't see it yet.

$$\sum_{i=1}^{k} \frac{1}{p^i}= ? =\frac{p^{-k}-1}{1-p}$$

I'd appreciate an intermediary step "?" to help understand the result.


问:
Integrating a set of measure 0 but unbounded values
I am currently working on learning some measure theory, and I recently came across this question, posed to me in my book:

Show that if $f \geq 0$ and $\int_{\Omega} f d\mu = 0$, then $f=0$ almost everywhere

If I assume the additional condition that $0 \leq f \leq M$, this is an easy question, but my question is this: Let's say that I had $f = g + h$ such that $g$ is some nice function on a set $ E \subset \Omega$ of nonzero measure, and $h$ is a function such that the values of $h$ are all infinite but $\mu(E^c) = 0$, where $E^c \subset \Omega$. Would the value of $\int_{E^c} h d\mu = 0$ still? Or does the idea of assigning infinite values not make sense in this case? Any help would be appreciated! Thanks!

答:
The integral of a function $f\colon \Omega \to [0,\infty]$ is usually defined as the supremum of all $\int_\Omega \varphi d\mu$, where $\varphi$ is any simple function with $\varphi \leq f$.

I think that usually simple functions are only allowed to take finite values (i.e. in $\mathbb{R}$), hence the problem with $0\cdot \infty$ wouldn't occur here. So, if $f$ is almost everywhere zero, all the $\int_\Omega \varphi d\mu$ are zero and the supremum as well.


答:
Quoting from wikipedia:

$$\sum_{k=1}^{n} ar^{k-1} = a\ \frac{1-r^n}{1-r}.$$

To derive this formula, first write a general geometric series as:

$$\sum_{k=1}^{n} ar^{k-1} = ar^0+ar^1+ar^2+ar^3+\cdots+ar^{n-1}.$$

We can find a simpler formula for this sum by multiplying both sides
of the above equation by $1 − r,\ $ and we'll see that:

\begin{align} (1-r) \sum_{k=1}^{n} ar^{k-1} & = (1-r)(ar^0 +
 ar^1+ar^2+ar^3+\cdots+ar^{n-1}) \\  & = ar^0 +
 ar^1+ar^2+ar^3+\cdots+ar^{n-1} - ar^1-ar^2-ar^3-\cdots-ar^{n-1} - ar^n
 \\  & = a - ar^n \end{align}

since all the other terms cancel. If $r\neq1,\ $ we can rearrange the
above to get the convenient formula for a geometric series that
computes the sum of n terms:

$$\sum_{k=1}^{n} ar^{k-1} = a\ \frac{1-r^n}{1-r}.$$


问:
Finding minimum value of an integral
For $0<\lambda<0.5$ we define the integral:
$$ \int_{0}^{\infty}{\frac{1}{1+y}\cdot y^{-2\lambda}}dy $$ We have to find the minimum value of this integral as $\lambda$ varies in the interval $(0,0.5)$.
What I did was first put $y=1/x$ to reduce it to $$\int_{0}^{\infty}\frac{x^{2\lambda-1}}{1+x}dx $$ Now I defined $$I(\alpha)=\int_{0}^{\infty}e^{-\alpha(x+1)}\frac{x^{2\lambda-1}}{1+x}dx $$ and differenciate w.r.t $\alpha$. Since $I(\infty)=0$, I integrated the expression from $0$ to $\infty$. The end result was that $$I(0)=(-2\lambda)!(2\lambda-1)!$$ Now how do I find the minimum value of the above expression. I do not know how to deal with $\Gamma$ or Beta functions... according to the book, the value is $$\pi\over \sin(2\pi \lambda)$$ How are these two equivalent?

答:
A Second Approach Not Using Beta or Gamma Functions

The function
$$
f(\lambda)=\int_0^\infty\frac{y^{-2\lambda}}{1+y}\,\mathrm{d}y\tag1
$$
is convex:
$$
\begin{align}
f''(\lambda)
&=\int_0^\infty\frac{4\log(y)^2y^{-2\lambda}}{1+y}\,\mathrm{d}y\tag{2a}\\[6pt]
&\ge0\tag{2b}
\end{align}
$$
Furthermore, by substituting $y\mapsto1/y$, we see that
$$
\begin{align}
f'(1/4)
&=\int_0^\infty\frac{-2\log(y)\,y^{-1/2}}{1+y}\,\mathrm{d}y\tag{3a}\\
&=\int_0^\infty\frac{2\log(y)\,y^{-1/2}}{1+y}\,\mathrm{d}y\tag{3b}\\[6pt]
&=0\tag{3c}
\end{align}
$$
Therefore, the minimum occurs at $\lambda=\frac14$. To evaluate at $\lambda=\frac14$, we can substitute $y\mapsto y^2$:
$$
\begin{align}
\int_0^\infty\frac{y^{-1/2}}{1+y}\,\mathrm{d}y
&=2\int_0^\infty\frac{\mathrm{d}y}{1+y^2}\tag{4a}\\
&=\left.2\arctan(y)\vphantom{\int}\right]_0^\infty\tag{4b}\\[5pt]
&=\pi\tag{4c}
\end{align}
$$

Original Answer Showing How to Apply the Beta Function
$$
\begin{align}
\int_0^\infty\frac{y^{-2\lambda}}{1+y}\,\mathrm{d}y
&=\Gamma(1-2\lambda)\Gamma(2\lambda)\tag1\\
&=\pi\csc(2\pi\lambda)\tag2
\end{align}
$$
Explanation:

$(1)$: Beta function integral

$(2)$: Euler's reflection formula

What is the maximum of $\sin(2\pi\lambda)$ on $\left[0,\frac12\right]$?

How To Evaluate the Integral

This answer uses contour integration to evaluate a generalization of the integral, which shows that  the integral is $\pi\csc(2\pi\lambda)$. It also derives Euler's Reflection Formula.

Title: Finding minimum value of an integral


答:
First, from $a_1\ge 1$, one can easily obtain $a_n \ge n$, by induction argument.

Second, using induction again, it is easy to prove that $a_n \le 3a_1n$, for $n\ge 1$.

Hence $1\le b_n \le 3a_1$, where $b_n:=\frac{a_n}{n}$. To show that $a_n \sim n$, it suffices to show $\limsup_{n \rightarrow \infty}b_n\le 1$. Here is the argument:

Fix $N>1$, we have $\frac{N+1}{N}b_{N+1}=\sqrt{b_N+\frac{2N+1}{N^2}}$, and hence $$b_{N+1}<\sqrt{b_N+\frac{3}{N}}\le \sqrt{b_N+\frac{3}{N}b_N}=\left((1+\frac{3}{N})b_N\right)^{1/2}.$$

From iteration, it is easy to see that $$b_{N+k}< \left(1+\frac{3}{N}\right)^{\sum_{i=1}^k1/2^i}b_N^{1/2^k}.$$

Hence from the above, and $b_n \le 3a_1n$, we have that $$\limsup_{n \rightarrow \infty}b_n=\limsup_{k\rightarrow \infty}b_{N+k}<1+3/N.$$

Then let $N\rightarrow \infty$, we have $\limsup_{n\rightarrow \infty}b_n\le 1$. This finishes the proof.

Title: Given a recursion $a_{n+ 1}= \sqrt{na_{n}+ 2n+ 1}$ with $a_{1}\geq 1.$ Prove that $a_{n}\sim n\,{\rm as}\,n\rightarrow\infty$

Body:
Given a recursion $a_{n+ 1}= \sqrt{na_{n}+ 2n+ 1}$ with $a_{1}\geq 1.$ Prove that
$$a_{n}\sim n\,{\rm as}\,n\rightarrow\infty$$

Let $b_{n}= \dfrac{a_{n}}{n},$ we need to prove $\lim b_{n}= 1,$ then we find a value $\beta\in\left ( 0, 1 \right )$ satisfying $\left | b_{n+ 1}- 1 \right |\leq\beta\left | b_{n}- 1 \right |$ so that
$$\beta\rightarrow 0\,{\rm aut}\,n\rightarrow\infty$$
I can't predict the relationship among them, I need the the help to go to the induction, thank you.


答:
You don't need $\beta \rightarrow 0$, you just need that for sufficiently large $n$ $$\beta < 1$$ Then $$|b_{n+M}-1|<\beta^n|b_M-1| \rightarrow 0$$

You can check that
$$ b_{n+1} = \sqrt{\frac{b_nn^2 +2n+1}{(n+1)^2}}{} = \sqrt{(b_n-1)\frac{n^2}{(n+1)^2}+1}$$
that is
$$ b_{n+1}-1 = \sqrt{(b_n-1)\frac{n^2}{(n+1)^2}+1} - 1= \frac{(b_n-1)\frac{n^2}{(n+1)^2}}{\sqrt{(b_n-1)\frac{n^2}{(n+1)^2}+1} +1} $$
Since $a_1 \ge 1$, then $b_1 -1 \ge 0$ and by induction it's easy to show that $b_n -1 \ge 0$. We have then
$$ \frac{\frac{n^2}{(n+1)^2}}{\sqrt{(b_n-1)\frac{n^2}{(n+1)^2}+1} +1} \le  \frac{\frac{n^2}{(n+1)^2}}{\sqrt{1}+1} = \frac{n^2}{2(n+1)^2} \le \frac12 $$
Therefore
$$b_{n+1} -1  \le \frac12(b_n-1)$$
As showed before, it is enough to prove that $b_{n+1}-1 \rightarrow 0$.


问:
Adam Bashforth predictor corrector method second order differential equation
I'm trying to numerical solve a 2nd order differential equation, I know the usual way of doing this is using Runge-Kutta's Method. But then I wonder if I could do it with Adam-Bashford and Adam-Bashford predictor-corrector. To my surprise, although theoretically, it is possible to do so, I didn't end up finding any implementation nor straight forward formula, so I wrote this:

```<code># AB
Y[i + 1] = Y[i] + h*(55.0 * U[i] - 59.0 * U[i-1] + 37.0 * U[i-2] - 9.0 * U[i-3])/24.0
U[i + 1] = U[i] + h*(55.0 * dudx(X[i],Y[i],U[i]) - 59.0 * dudx(X[i-1],Y[i-1],U[i-1]) + 37.0 * dudx(X[i-2],Y[i-2],U[i-2]) - 9.0 * dudx(X[i-3],Y[i-3],U[i-1]))/24.0
#AB predictor-corrector
Y[i + 1] = Y[i] + h*(55.0 * U[i] - 59.0 * U[i-1] + 37.0 * U[i-2] - 9.0 * U[i-3])/24.0
U[i + 1] = U[i] + h*(55.0 * dudx(X[i],Y[i],U[i]) - 59.0 * dudx(X[i-1],Y[i-1], U[i-1]) + 37.0 * dudx(X[i-2],Y[i-2], U[i-2]) - 9.0 * dudx(X[i-3],Y[i-3],U[i-3]))/24.0

Y[i + 1] = Y[i] + h*(9.0 * U[i+1] + 19.0 * U[i] - 5.0 * U[i-1] + U[i-2])/24.0
U[i + 1] = U[i] + h*(9.0 * dudx(X[i+1], Y[i + 1], U[i+1]) + 19.0 * dudx(X[i],Y[i], U[i]) - 5.0 * dudx(X[i-1],Y[i-1], U[i-1]) + dudx(X[i-2],Y[i-2], U[i-2]))/24.0
</code>```

where $u=\frac{dy}{dx}$ and $dudx=\frac{d^2 y}{dx^2}$. I believed that this would result in a better estimate of the solution. But it seems AB is less accurate than rk4 and ABP-C is just a little better than rk4, Have I done something wrong? If not, Is there any way that someone can explain why this happens?

This is the result I got for the equation:
$$
\left\{\begin{array}{l}
y''+y'-6y=0\\
y(0)=1, y'(0)=2
\end{array}\right.
$$
where $y(x)=e^{2x}$ with this different methods:

答:
There is one error and one non-standard feature in your formulas, and thus apparently also in your computations. In the AB4 formula for the Y component, the last index should be i-3 as the ones directly before it, not i-1. In the repetition as predictor for AM4 you have that correct. In the corrector you are using the freshly computed value for Y[i+1] in the computation of U[i+1]. This will produce a difference to vector-based standard implementations. This is basically the difference between a Gauß-Seidel scheme and a Jacobi scheme. If one were to continue the corrector iteration until numerical convergence, there should be no differences above floating point noise.

To check if you implemented the method correctly, you can compare the errors when dividing the step size by N=2,4,8 etc. You should observe a reduction in error by a factor $16$ in each step. For a more visual confirmation, divide the step size by N=10, 100, etc. Then the number of correct digits should increase by 4 in each step. I computed such tables below, they show that clearly in the columns for N=10. It should be true in general that the columns have about 2, 6 and 10 correct digits after the dot, respectively, RK4 and AM4 partially a little more, AB4 a little less.

Runge-Kutta 4

x
N=1
N=10
N=100

0.0
[ 1.000000000000,  2.000000000000]
[ 1.000000000000,  2.000000000000]
[ 1.000000000000,  2.000000000000]

0.1
[ 1.221400000000,  2.442800000000]
[ 1.221402757840,  2.442805515680]
[ 1.221402758160,  2.442805516320]

0.2
[ 1.491817960000,  2.983635920000]
[ 1.491824696859,  2.983649393718]
[ 1.491824697641,  2.983649395282]

0.3
[ 1.822106456344,  3.644212912688]
[ 1.822118798957,  3.644237597914]
[ 1.822118800390,  3.644237600781]

0.4
[ 2.225520825779,  4.451041651557]
[ 2.225540926158,  4.451081852316]
[ 2.225540928492,  4.451081856984]

0.5
[ 2.718251136606,  5.436502273212]
[ 2.718281824895,  5.436563649789]
[ 2.718281828459,  5.436563656917]

0.6
[ 3.320071938250,  6.640143876501]
[ 3.320116917512,  6.640233835024]
[ 3.320116922736,  6.640233845472]

0.7
[ 4.055135865379,  8.110271730758]
[ 4.055199959400,  8.110399918800]
[ 4.055199966844,  8.110399933688]

0.8
[ 4.952942945974,  9.905885891948]
[ 4.953032414003,  9.906064828007]
[ 4.953032424394,  9.906064848788]

0.9
[ 6.049524514213, 12.099049028426]
[ 6.049647450134, 12.099294900267]
[ 6.049647464411, 12.099294928823]

1.0
[ 7.388889241659, 14.777778483319]
[ 7.389056079552, 14.778112159104]
[ 7.389056098929, 14.778112197857]

Adams-Bashford 4

x
N=1
N=10
N=100

0.0
[ 1.000000000000,  2.000000000000]
[ 1.000000000000,  2.000000000000]
[ 1.000000000000,  2.000000000000]

0.1
[ 1.221400000000,  2.442800000000]
[ 1.221402748820,  2.442805497640]
[ 1.221402758159,  2.442805516318]

0.2
[ 1.491817960000,  2.983635920000]
[ 1.491824670099,  2.983649340199]
[ 1.491824697638,  2.983649395276]

0.3
[ 1.822106456344,  3.644212912688]
[ 1.822118747045,  3.644237494090]
[ 1.822118800384,  3.644237600769]

0.4
[ 2.225359751835,  4.450719503670]
[ 2.225540839268,  4.451081678535]
[ 2.225540928483,  4.451081856965]

0.5
[ 2.717819501390,  5.435639002780]
[ 2.718281690082,  5.436563380165]
[ 2.718281828444,  5.436563656888]

0.6
[ 3.319281371915,  6.638562743829]
[ 3.320116717817,  6.640233435634]
[ 3.320116922715,  6.640233845429]

0.7
[ 4.053852018449,  8.107704036897]
[ 4.055199672699,  8.110399345399]
[ 4.055199966813,  8.110399933626]

0.8
[ 4.950979883921,  9.901959767841]
[ 4.953032011560,  9.906064023120]
[ 4.953032424351,  9.906064848702]

0.9
[ 6.046643715383, 12.093287430766]
[ 6.049646894750, 12.099293789499]
[ 6.049647464353, 12.099294928705]

1.0
[ 7.384781911467, 14.769563822935]
[ 7.389055323232, 14.778110646464]
[ 7.389056098849, 14.778112197697]

Adams-Moulton 4-1

x
N=1
N=10
N=100

0.0
[ 1.000000000000,  2.000000000000]
[ 1.000000000000,  2.000000000000]
[ 1.000000000000,  2.000000000000]

0.1
[ 1.221400000000,  2.442800000000]
[ 1.221402758700,  2.442805517401]
[ 1.221402758160,  2.442805516321]

0.2
[ 1.491817960000,  2.983635920000]
[ 1.491824699412,  2.983649398824]
[ 1.491824697642,  2.983649395283]

0.3
[ 1.822106456344,  3.644212912688]
[ 1.822118803910,  3.644237607820]
[ 1.822118800391,  3.644237600782]

0.4
[ 2.225527878319,  4.451055756639]
[ 2.225540934448,  4.451081868896]
[ 2.225540928493,  4.451081856986]

0.5
[ 2.718268691144,  5.436537382288]
[ 2.718281837757,  5.436563675514]
[ 2.718281828460,  5.436563656920]

0.6
[ 3.320104159474,  6.640208318947]
[ 3.320116936565,  6.640233873130]
[ 3.320116922738,  6.640233845476]

0.7
[ 4.055188406071,  8.110376812142]
[ 4.055199986754,  8.110399973508]
[ 4.055199966847,  8.110399933694]

0.8
[ 4.953023229681,  9.906046459362]
[ 4.953032452400,  9.906064904800]
[ 4.953032424398,  9.906064848797]

0.9
[ 6.049642249362, 12.099284498725]
[ 6.049647503123, 12.099295006245]
[ 6.049647464417, 12.099294928835]

1.0
[ 7.389057076415, 14.778114152830]
[ 7.389056151712, 14.778112303424]
[ 7.389056098937, 14.778112197874]


问:
Calculate the limit of two interrelated sequences?
I'm given two sequences:

$$a_{n+1}=\frac{1+a_n+a_nb_n}{b_n},b_{n+1}=\frac{1+b_n+a_nb_n}{a_n}$$

as well as an initial condition $a_1=1$, $b_1=2$, and am told to find: $\displaystyle \lim_{n\to\infty}{a_n}$.

Given that I'm not even sure how to approach this problem, I tried anyway. I substituted $b_{n-1}$ for $b_n$ to begin the search for a pattern. This eventually reduced to:

$$a_{n+1}=\frac{a_{n-1}(a_n+1)+a_n(1+b_{n-1}+a_{n-1}b_{n-1})}{1+b_{n-1}+a_{n-1}b_{n-1}}$$

Seeing no pattern, I did the same once more:

$$a_{n+1}=\frac{a_{n-2}a_{n-1}(a_n+1)+a_n\left(a_{n-2}+(a_{n-1}+1)(1+b_{n-2}+a_{n-2}b_{n-2})\right)}{a_{n-2}+(a_{n-1}+1)(1+b_{n-2}+a_{n-2}b_{n-2})}$$

While this equation is atrocious, it actually reveals somewhat of a pattern. I can sort of see one emerging - though I'm unsure how I would actually express that. My goal here is generally to find a closed form for the $a_n$ equation, then take the limit of it.

How should I approach this problem? I'm totally lost as is. Any pointers would be very much appreciated!

Edit:

While there is a way to prove that $\displaystyle\lim_{n\to\infty}{a_n}=5$ using $\displaystyle f(x)=\frac{1}{x-1}$, I'm still looking for a way to find the absolute form of the limit, $\displaystyle\frac{1+2a+ab}{b-a}$.

答:
It's obvious that $a_n$ and $b_n$ are in the same situation, so their limits highly depend on the initial values. Following are some points we can obtain from $a_1=1$ and $b_1=2$:

$a_n>0$ and $b_n>0\ ;a_{n+1}-a_{n}=\frac{1+a_n}{b_n}>0$ and similarly $b_{n+1}-b_n>0$. Therefore, $\{a_n\}$ and $\{b_n\}$ are strictly increasing sequences;

$b_{n+1}-a_{n+1}=\frac{(b_n-a_n)+(b_n^2-a_n^2)+a_n\cdot b_n(b_n-a_n)}{a_n\cdot b_n}$, and thus by induction $b_n>a_n$ for every $n$;

$b_{n+1}-b_{n}=\frac{1+b_n}{a_n}>\frac{b_n}{a_n}>1$, which implies $b_n$ increases to $+\infty$;

$\frac{a_{n+1}}{a_n}=1+\frac{1}{b_n}+\frac{1}{a_n\cdot b_n}$ converges to $1$ as $n\to \infty$.

Now we prove $\lim a_n$ exists and find its closed form. To show the existence, it suffices to show $\{a_n\}$ is bounded. First assume $a_n$ increases to infinity, and we will derive a contradiction with the last point listed above.
From the fact
$$b_n(a_{n+1}+1)=(1+a_n)(1+b_n)=a_n(b_{n+1}+1)$$
Denote $c_n:=a_n+1$ and $d_n:=b_n+1$, then we obtain
$$
\begin{cases}
\frac{c_n-1}{c_n}=\frac{d_n}{d_{n+1}}\\
\frac{d_n-1}{d_n}=\frac{c_n}{c_{n+1}}
\end{cases}
\Rightarrow
\begin{cases}
\frac{d_{n+1}}{d_n}=\frac{c_n}{c_n-1}\\
\frac{c_{n+1}}{c_n}=\frac{d_n}{d_n-1}
\end{cases}
$$
For $n\ge 2$,
$d_n=d_1\cdot \frac{d_2}{d_1}\cdots \frac{d_n}{d_{n-1}}=d_1\cdot \frac{c_1}{c_1-1}\cdots \frac{c_{n-1}}{c_{n-1}-1}$, which implies
$$
d_1(1-\frac{c_n}{c_{n+1}})=(1-\frac{1}{c_1})\cdots (1-\frac{1}{c_{n-1}})
$$
In fact,$$\frac{c_{n+1}}{c_n}=\frac{d_n}{d_n-1}=\frac{d_1\cdot \frac{c_1}{c_1-1}\cdots \frac{c_{n-1}}{c_{n-1}-1}}{d_1\cdot \frac{c_1}{c_1-1}\cdots \frac{c_{n-1}}{c_{n-1}-1}-1} \Rightarrow \frac{c_{n}}{c_{n+1}}=1-\frac{1}{d_1}((1-\frac{1}{c_1})\cdots (1-\frac{1}{c_{n-1}}))$$

Together with $d_1(1-\frac{c_{n+1}}{c_{n+2}})=(1-\frac{1}{c_1})\cdots (1-\frac{1}{c_{n}})$, we get $(1-\frac{1}{c_n})(1-\frac{c_n}{c_{n+1}})=1-\frac{c_{n+1}}{c_{n+2}}$. That is
$$
\frac{c_{n+1}}{c_{n+2}}-\frac{c_{n}}{c_{n+1}}=\frac{1}{c_{n}}-\frac{1}{c_{n+1}}
$$
Hence, for $n\ge 2$
$$
\frac{c_{n}}{c_{n+1}}-\frac{c_2}{c_3}=\frac{1}{c_2}-\frac{1}{c_{n}} $$
which is equivalent to
$$\frac{c_{n}}{c_{n+1}}+\frac{1}{c_{n}}=c(constant):=\frac{c_2}{c_3}+\frac{1}{c_2}=\frac{7}{6} \\ (c_1=2,d_1=3;c_2=3,d_2=6;c_3=\frac{18}{5})
$$
Now it is clear that $$\frac{c_{n+1}}{c_{n}}=\frac{c_n}{\frac{7}{6}c_n-1}=\frac{1}{\frac{7}{6}-\frac{1}{c_n}} $$
Well, the problem has been reduced to solve $c_n(=a_n+1)$, and it's you can use the same method to solve $b_n$, and I would like to leave this open to you, but note as I mentioned before if $a_n\to +\infty$, then $c_n\to +\infty$ and $$
\frac{c_{n+1}}{c_{n}}(=\frac{\frac{a_{n+1}}{a_n}+\frac{1}{a_n}}{1+\frac{1}{a_n}})=\frac{1}{\frac{7}{6}-\frac{1}{c_n}} \text{converges to } \frac{6}{7} \text{instead of } 1
$$ this contradicts the last point.


问:
Can we show that $(x_n)$ is relatively compact if and only if $(\langle x_n,\;\cdot\;\rangle)$ restricted to a closed ball is relatively compact?
Let $E$ be a Banach space, $\iota:E\to E''$ denote the canonical injection of $E$ into $E''$, i.e. $$\iota x:=\langle x,\;\cdot\;\rangle\;\;\;\text{for }x\in E,$$ $\sigma_c(E',E)$ denote the topology of compact convergence on $E$ and $(x_n)_{n\in\mathbb N}$.

Are we able to show that $(x_n)_{n\in\mathbb N}$ is relatively compact if and only if there is a $r>0$ such that $(\left.\iota x_n\right|_{B^{E'}_r(0)})_{n\in\mathbb N}$ is relatively compact with respect to the topology of uniform convergence on $C(B^{E'}_r(0),\sigma_c(E',E))$?

The claim can be found in the proof of Linde's book Probability in Banach Spaces - Stable and Infinitely Distributions, Lemma 2.3.6:$^1$

He also seems to conclude (most probably using the Arzelà-Ascoli theorem) that "$(\left.\iota x_n\right|_{B^{E'}_r(0)})_{n\in\mathbb N}$ is relatively compact with respect to the topology of uniform convergence on $C(B^{E'}_r(0),\sigma_c(E',E))$" follows from "$(\left.\iota x_n\right|_{B^{E'}_r(0)})_{n\in\mathbb N}$ is $\sigma_c(E',E)$-equicontinuous". But this is only true, if $(\left.\iota x_n\right|_{B^{E'}_r(0)})_{n\in\mathbb N}$ is pointwise bounded, which should be the case if (and only if) $(x_n)_{n\in\mathbb N}$ is bounded. Why is that the case?

$^1$ $V_\delta:=\{\varphi\in E':\left\|\varphi\right\|_{E'}\le\delta\}$ is endowed with the restriction of the topology $\sigma_c(E',E)$ and $\delta_x$ denotes the Dirac measure concentrated at $x$.

答:
Peter Morfe's answer answered my question, which is why I've accepted it. However, since the whole stuff with the $\sigma_c(E',E)$-topology on $E'$ confused me and it turned out that everything is way more elementary than I thought, I will reformulate the result, which can be proved in the way Peter described in his answer.

Let

$\mathbb F=\mathbb R$ or $\mathbb F=\mathbb C$;

$X$ be a normed $\mathbb F$-vector space;

$\langle\;\cdot\;,\;\cdot\;\rangle$ denote the canonical duality pairing between $X$ and $X'$;

$\iota$ denote the canonical embedding of $X$ into $X''$, i.e. $$\iota:X\to X''\;,\;\;\;x\mapsto\langle x,\;\cdot\;\rangle;$$

$V_r:=\overline B^{X'}_r(0)$ for $r>0$.

Remember that if $V$ is any set, then $$B(V):=\left\{f:V\to\mathbb F\mid\left\|f\right\|_\infty<\infty\right\}$$ equipped with $\left\|\;\cdot\;\right\|_\infty$ is a $\mathbb F$-Banach space.

Proposition: Let $(x_n)_{n\in\mathbb N}\subseteq X$ and $$f_n:=\iota(x_n)\;\;\;\text{for }n\in\mathbb N.$$

If $(x_n)_{n\in\mathbb N}$ is relatively compact, then $(f_n)_{n\in\mathbb N}$ is relatively compact in $\iota(X)$. In particular, $\left(\left.f_n\right|_{V_r}\right)_{n\in\mathbb N}$ is relatively compact in $B(V_r)$ for all $r>0$.

If $X$ is complete and $\left(\left.f_n\right|_{V_r}\right)_{n\in\mathbb N}$ is relatively compact in $B(V_r)$ for some $r>0$, then $(x_n)_{n\in\mathbb N}$ is relatively compact.

Now the result asked for in the question is a simple corollary: If $r>0$ and $\tau$ is any compact topology on $V_r$, then $$C(V_r,\tau):=\left\{f:V_r\to\mathbb F\mid f\text{ is }\tau\text{-continuous}\right\}$$ is a closed subspace of $B(V_r)$. Thus, if $\iota(X)\subseteq C(V_r,\tau)$, then $\left(\left.f_n\right|_{V_r}\right)_{n\in\mathbb N}$ being relatively compact in $B(V_r)$ implies that $\left(\left.f_n\right|_{V_r}\right)_{n\in\mathbb N}$ is relatively compact in $C(V_r,\tau)$.


问:
Prove that $\sqrt2$ is the least upper bound
Consider the set $A=\{\,x\in\mathbb{Q} : x^2<2\,\}.$

I want to prove that $\sqrt2$ is the least upper bound for $A$.

To do this, I think I need to prove that if I take any $\alpha\in\mathbb{R}$ with $\alpha<\sqrt2$, then there exists $y\in A$ with $y>\alpha$.

I'm not sure how to choose $y\in A$ to satisfy this property.

答:
Let $0\lt d\lt\sqrt2$, where $d\in\mathbb{Q}$.

Then we can prove the existence of a rational between $d$ and $\sqrt2$ by noting that such a rational, $\frac{a}{b}$, satisfies:

$$d\lt \frac{a}{b}\lt \sqrt2$$
$$bd\lt a\lt b\sqrt2$$

and we can make $b$ as large as required in order to find such an $a$.


问:
What types of holes $\pi_1(X)$ can detect?
I begin studying algebraic topology, and I'm wondering what type of hole $\pi_1$ can detect ? More precisely what are the limits of $\pi_1$ in the study of holes ?
I mean, $\mathbb{S}^2$ has a hole while $\pi_1(\mathbb{S}^2) = 0$ and $\mathbb{S}^1\times [0,1]$ as the same type of hole than $\mathbb{S}^1$ (through $\pi_1$) so we can have hole in every dimension through $\pi_1$...
After I know there is $\mathbb{R}\mathbb{P}(n)$ which give strange homology group (and I don't imagine how wierd are the homotopy groups). So I don't see what type of hole it detects and it's links with dimension if there is one (only the fact that loops in $\mathbb{R}^n$ are 1D manifold...)

答: The first homotopy group $ \pi_1(X) $ detects whether all loops (that is the images of a circle) can be continously deformed to a point. You can do it on a sphere $S^2$ (because you can contract the loop moving it towards any chosen pole), but you cannot always do it on a  cylinder $S^1\times[0,1]$.


答:
I'll expand my comment.

WLOG one may assume that $\alpha>0$. If $\alpha\le 0$ it is clear that $y:=\frac{1}{2}\in A$ and $\alpha<\frac12<\sqrt2$.

Assume $0<\alpha<\sqrt2$ is an upper bound of $A$. Archimedean property (or the density of rationals) guarantees the existence of $y\in\Bbb Q$ such that $\alpha<y<\sqrt2$, so (since $0<y<\sqrt2$) $y^2<2$ and $y\in A$.

Title: Prove that $\sqrt2$ is the least upper bound

Body:
Consider the set $A=\{\,x\in\mathbb{Q} : x^2<2\,\}.$

I want to prove that $\sqrt2$ is the least upper bound for $A$.

To do this, I think I need to prove that if I take any $\alpha\in\mathbb{R}$ with $\alpha<\sqrt2$, then there exists $y\in A$ with $y>\alpha$.

I'm not sure how to choose $y\in A$ to satisfy this property.


问:
Stereographic Projection: Cartography Applications
Compared to the Mercator's, which is also conformal, how does the Stereographic projection help in areas such as navigation? Or any application besides simply mapping polar areas, although I would prefer answers on the polar aspect.

Additionally, from Wikipedia:

"When the projection is centered at the Earth's north or south pole, it has additional desirable properties: It sends meridians to rays emanating from the origin and parallels to circles centered at the origin".

Why are these desirable qualities? How would it benefit one in any hypothetical situation?

This may not be strictly math related, but any ideas where the stereographic projections' properties could be utilized to solve a problem would still be appreciated.

Edit: Any suitable scenario where its properties can be of use would suffice, it can be very bizarre, but I would prefer it to be related to navigation or locations on Earth.

答:
Cartography benefit is significant. Also easy to calculate mappings.

If we consider a small patch like a state on the globe  its geographical  mapping from sphere surface onto a plane at south pole  SPp has an approximately constant magnification. But the angles are preserved (conformal but not isometric). It is useful for GIS and GPS assisted navigation on land.

If the earth's curvature can be neglected.. its small area map is represented on flat SPp tangent plane faithfully.

Corresponding  angles do not change in magnitude except for $\phi \to \pi -\phi $  reckoned from meridian. Large loxodromes are mapped to logarithmic spirals along rhumb lines, very useful in ship navigation.

Small circles ( radius small compared to earth's radius ) project to perfect circles on SPp.

When entire Lat/Long lines map is slid by $ 90^{\circ}$ ( so N-S poles axis is along a diameter of equator plane)  projection from north pole point light NPl  casts a bipolar coordinate map or net.

Properties of circle inversions is useful and interesting.

3d printed surface models cast interesting angle preserving shadow projections on SPp cast from NPl.


问:
Proving number of partitions of $n$ to $3$ parts at most.
I have an exercise, to prove that the number of partitions of $n$ to at most $3$ integers is $\frac{(n+3)^2}{12}$ rounded. I tried to prove by induction but I don't know how.</p>

答:
The number of partitions of $n$ into at most $k$ integers is equal to the number of partitions of $n$ into integers no bigger than $k$.  Let's call this number $p(n,k)$.  In general $$p(n,k)=p(n,k-1)+p(n-k,k)$$ starting with $p(0,0)=1$ and $p(n,0)=0$ for $n\not=0$ and zero if $n$ or $k$ are negative.

For given $k$, patterns are going to repeat every $k!$ terms, so we can find and show by induction that $p(m,1)=1$ and $p(2m,2)=p(2m+1,2)=m+1$ for $m \ge 0$ and thus

$p(6m+0,3)= 3m^2+3m+1 = \frac{((6m+0)+3)^2}{12} + \frac1{4}$

$p(6m+1,3)= 3m^2+4m+1 = \frac{((6m+1)+3)^2}{12} - \frac1{3}$

$p(6m+2,3)= 3m^2+5m+2 = \frac{((6m+2)+3)^2}{12} - \frac1{12}$

$p(6m+3,3)= 3m^2+6m+3 = \frac{((6m+3)+3)^2}{12} + 0$

$p(6m+4,3)= 3m^2+7m+4 = \frac{((6m+4)+3)^2}{12} - \frac1{12}$

$p(6m+5,3)= 3m^2+8m+5 = \frac{((6m+5)+3)^2}{12} - \frac1{3}$

Since all the fractions on the right hand side are smaller than $\frac12$, this  leads to the conclusion that you can round and say $$p(n,3)= \left[\frac{(n+3)^2}{12}\right].$$


答:
Denoting $\,\Bbb Z_p:=$ the prime field of characteristic $\,p\,$ , try to follow and prove the following:

$$\Bbb Z[w]/(2,w)\cong\left(\Bbb Z[w]/(2)\right)/\left((2,w)/(2)\right)\cong\Bbb Z_2[w]/(w)\cong\Bbb Z_2$$

Title: How to show $\mathbb{Z}[w]/(2,w) \simeq \mathbb{Z}_2$?

Body:
Let $\mathbb{Z}[w]=\mathbb{Z}[\frac{1+\sqrt{-15}}{2}]$ be the quadratic integers.

I want to show that $\mathbb{Z}[w]/(2,w) \simeq \mathbb{Z}_2$. It seems very clear, but how can I show the isomorphism rigorously by isomorphism? I found somewhat similar using 3rd isomorphism theorem(here). But I think there will be much easier way.

I tried $\mathbb{Z}[w]/(2,w) \simeq (\mathbb{Z}(X)/(f(X))/(2+f(X),X+f(X))$ but I'm not sure then it is isomorphic to $\mathbb{Z}(X)/(2,X)\simeq \mathbb{Z}_2$.

I also tried $\mathbb{Z}[X] \to \mathbb{Z}[w] \to \mathbb{Z}[w]/(2,w)$, but not sure what the kernel is. Obviousely $2,w$ goes to $0$ but isn't it possible that $1 \mapsto 0$?


问:
Naming a property of continua
Let me define a continuum $X$ to be nice if for each pair $a,b\in X$ of distinct points, there are subcontinua $M,N\subseteq X$ such that: $a\in M\setminus N$; $b\in N\setminus M$; and $M\cup N=X$.  (This is a very strong form of aposyndesis.) If a continuum is both locally connected (base consisting of connected open sets) and colocally connected (base consisting of open sets with connected complements), then it is clearly nice.  My question is: does this property have an established name in the continuum-theoretic literature?

答:
Yes, this is called freely decomposable and actually it's equivalent to aposyndesis as a global property of metric continua.  Your "colocally connected" is a stronger form of a property called semi-locally connected (slc), where each point has a neighborhood base whose elements have only finitely many components in their complements.  The slc property is also equivalent to freely decomposable as a global property.

Let $x, y \in X$ be distinct, where $X$ is a continuum.

Assume $X$ is freely decomposable, i.e. there are subcontinua $A \cup B = X$ with $x \in A \setminus B$ and $y \in B \setminus A$.  If $x \notin \text{int}(A)$ then every neighborhood of $x$ intersects $X \setminus A \subset B \implies x \in \overline{B} = B$, a contradiction.  Thus $A$ is a neighborhood of $x$ and $A \subset X \setminus \lbrace y \rbrace$, so that $X$ is aposyndetic at $x$ wrt $y$.  But $x, y$ were arbitrary, so $X$ is aposyndetic.

Now assume $X$ is aposyndetic and $U$ is a neighborhood of $x$.  Since $X$ is aposyndetic at $y$ wrt $x$ for every $y \notin U$ there is a neighborhood continuum $K_y$ of $y$ not containing $x$, and $X \setminus U$ is contained in $\cup K_y$.  Since $X \setminus U$ is compact it's contained in finitely many such continua, say $K_1, K_2, \dots, K_n$.  Then $V = X \setminus (\cup K_j)$ is a neighborhood of $x$ with $V \subset U$ and $X \setminus V = K_1 \cup \cdots \cup K_n$ has at most $n$ components.  Therefore $X$ is slc at $x$ since $U$ was an arbitrary neighborhood of $x$.  But $x$ was arbitrary, so $X$ is slc.

(note that the above implies: If $X$ is aposyndetic at $y$ wrt $x$ for every $y \in X \setminus \lbrace x \rbrace$ then $X$ is slc at $x$.  As local properties slc is not equivalent to aposyndesis, even for planar curves)

Now assume that $X$ is slc and we show that $X$ is freely decomposable (this is the hardest part).  Let $U$ be a neighborhood of $x$ with $y \notin \overline{U}$ such that $X \setminus U$ has only finitely many components $C_1, C_2, \dots, C_k$ with $y \in C_1$.  Since $y \notin \overline{U}$ we have $y \in \text{int}(C_1)$.  Note that since $U$ was open each $C_j$ is closed.

Since $X$ is slc at $y$ pick an open neighborhood $V \subset C_1$ of $y$ such that $X \setminus V$ has only finitely many components $D_1, D_2, \dots, D_n$ with $x \in D_1$ (similarly, $x \in \text{int}(D_1)$ and the $D_j$'s are closed).  Then $\mathcal{A} = \lbrace C_1, \dots, C_k, D_1, \dots, D_n \rbrace$ is a finite cover of $X$ by closed, connected sets.

Then $C_1 \cup D_2, \cup \cdots \cup D_n$ is a neighborhood continuum of $y$ since $C_1$ is a neighborhood and $C_1 \cup D_i$ is connected for any $i$ by the boundary bumping theorem.  Similarly, $D_1 \cup C_2 \cup \cdots \cup C_n = D_1$ since $U$ (and thus $U \cup C_2 \cup \cdots \cup C_n$ by bbp) is contained in $D_1$ since $D_1$ is a component of $X \setminus V$.  Then these are a decomposition of $X$ between $x$ and $y$ as desired.

The above proofs work fine for Hausdorff continua.  Below is the triangle basin $T$.  $T$ is locally connected, and thus aposyndetic, at $v$, but is not slc at $v$.  It's slc at $x$ but not aposyndetic at $x$ with respect to $v$.

Title: Naming a property of continua


问:
Question about sequence
Let $N_0\in \mathbb{N}.$ If a sequence of complex numbers $\{F_N\}_{N \in \mathbb{N}}$ has the following properties:
$$\lim_{N \rightarrow \infty} |F_N|^{1/N}=0$$ and for all $N \geq N_0$,
$$|F_N|\leq N\sum_{k=N+1}^{\infty} |F_k|,\quad \quad \quad  $$
then there exists $N_1\in \mathbb{N}$ such that $F_N=0$ for all $N \geq N_1.$

I would like to know if this is true or false.

Title: Question about sequence

Body:
No. Let's try $F_{N} = \frac{1}{(N-1)!}$.

Clearly Nth root of $F_N$ tends to $0$.

Also $$N*(\sum_{k=N+1}^{\infty}F_{k}) > N*(F_{N+1})= N*\frac{1}{N!} = F_{N}$$


问:
Why in tensors is the identity $\int{A_{\nu}\partial_{\mu}}\partial^{\nu}A^{\mu} = \int A_{\mu}\partial^{\mu}\partial^{\nu}A_{\nu}$ valid
I am trying to understand the  and there is a step in the Derivation of something and there is a step which I cannot understand.

Its basically a lack of understanding of tensors so any help would be appreciated:

I am unsure about this line $$\int{A_{\nu}\partial_{\mu}}\partial^{\nu}A^{\mu} = \int A_{\mu}\partial^{\mu}\partial^{\nu}A_{\nu}$$

$A_{\mu}$ is a real free vector by the way.

I thought it may have something to do with using the metric tensor and have tried the following

$$\int{A_{\nu}\partial_{\mu}}\partial^{\nu}A^{\mu} = \int A_{\nu}\partial^s g_{s \mu} \partial^{\nu}A^{\mu} = \int A_{\nu}\partial^s  \partial^{\nu}A_{s} = \int A_{\nu}\partial^\mu  \partial^{\nu}A_{\mu} $$

however

$$\int A_{\nu}\partial^\mu  \partial^{\nu}A_{\mu}  \neq  \int A_{\mu}\partial^{\mu}\partial^{\nu}A_{\nu}$$

Any help in understanding where I went wrong and how to get the first equation I stated woul be appreciated.

答:
Assuming that $A$ has continuous derivatives (or is a distribution), the partial derivatives commute, so
$$
A_\nu \partial_\mu \partial^\nu A^\mu
= A_\nu \partial^\nu \partial_\mu A^\mu
= (A_\nu \partial^\nu) (\partial_\mu A^\mu)
= \{ \mu \leftrightarrow \nu \}
= (A_\mu \partial^\mu) (\partial_\nu A^\nu)
\\
= \{ \text{see-saw rule} \}
= (A_\mu \partial^\mu) (\partial^\nu A_\nu)
= A_\mu \partial^\mu \partial^\nu A_\nu
.
$$


答:
As far as I understand you initially operate in flat Minkowski space with $g_{\mu\nu}=g^{\mu\nu}=diag(1,-1,-1,-1)$
In this case $\partial_\mu=\frac{\partial}{\partial{x}^\mu}$

You have a summation over repeated indices: $\partial^\mu{A}_\mu=g^{\mu\lambda}\partial_\lambda{A}_\mu=\partial_\mu{A}^\mu=\partial_0A^0+\partial_1A^1+\partial_2A^2+\partial_3A^3=\partial_0A_0-\partial_1A_1-\partial_2A_2-\partial_3A_3$

Index $\mu$ here is a dumb index of summation- you may can write $\partial^\delta{A}_\delta$ instead.

Next, in Minkowski space $\partial_\mu\partial_\nu=\partial_\nu\partial_\mu$ (ordinary derivatives). Renaming you summation indices and swapping derivatives you get ${A_{\nu}\partial_{\mu}}\partial^{\nu}A^{\mu}={A_{\mu}\partial^{\mu}}\partial^{\nu}A_{\nu}$

In curved space we have to switch to covariant derivatives $\partial_\mu{A}^\nu\to{A}^\nu;\mu$. In general ${A}^\nu;\nu;\mu\neq{A}^\nu;\mu;\nu$

Title: Why in tensors is the identity $\int{A_{\nu}\partial_{\mu}}\partial^{\nu}A^{\mu} = \int A_{\mu}\partial^{\mu}\partial^{\nu}A_{\nu}$ valid

Body:
I am trying to understand the  and there is a step in the Derivation of something and there is a step which I cannot understand.

Its basically a lack of understanding of tensors so any help would be appreciated:

I am unsure about this line $$\int{A_{\nu}\partial_{\mu}}\partial^{\nu}A^{\mu} = \int A_{\mu}\partial^{\mu}\partial^{\nu}A_{\nu}$$

$A_{\mu}$ is a real free vector by the way.

I thought it may have something to do with using the metric tensor and have tried the following

$$\int{A_{\nu}\partial_{\mu}}\partial^{\nu}A^{\mu} = \int A_{\nu}\partial^s g_{s \mu} \partial^{\nu}A^{\mu} = \int A_{\nu}\partial^s  \partial^{\nu}A_{s} = \int A_{\nu}\partial^\mu  \partial^{\nu}A_{\mu} $$

however

$$\int A_{\nu}\partial^\mu  \partial^{\nu}A_{\mu}  \neq  \int A_{\mu}\partial^{\mu}\partial^{\nu}A_{\nu}$$

Any help in understanding where I went wrong and how to get the first equation I stated woul be appreciated.


问:
$A: B_1 \rightarrow B_2$ be a mapping, where $B_1$, $B_2$ are two Banach Spaces.Comsequently Hilbert
Let $A: B_1 \rightarrow B_2$ be a mapping, where $B_1$, $B_2$ are two Banach Spaces.

Then the adjoint of $A$, $A^*$, is defined by

$A^*:B_2^* \rightarrow B_1^*$ s.t for all $f \in B_1^*, f:B_1 \rightarrow \mathbb{R}$ is linear.

$\langle Ax{,}y\rangle$=$\langle x{,}A^*y\rangle$

My Question is: why when dealing with Hilbert Spaces $H_1$ and $H_2$, we simply say: IF $A:H_1 \rightarrow H_2$, THEN $A^*:H_2 \rightarrow H_1$, without taking into consideration $H_2^*$ and $H_1^*$ ?

答:
The reason behind this is that every Hilbert space is isometrically isomorphic to its dual by the Riesz representation theorem, stating that any functional $\phi$ on a Hilbert space $H$ is of the form $\phi(\cdot)=\langle\cdot,z\rangle$ for some $z\in H$. Using this one-to-one correspondence between $H^*$ and $H$, one can see that if $A:H_1\to H_2$ is a bounded linear operator, then the "Hilbert space adjoint: $A^*:H_2\to H_1$ which is defined by $\langle Ax,y\rangle=\langle x,A^*y\rangle$ for all $x,y$ is "essentially the same" as the "Banach space adjoint: $A^*:H_2^*\to H_1^*$.

If $\sigma:H_1\xrightarrow{\cong}H_1^*$ and $\rho:H_2\xrightarrow{\cong}H_2^*$ are the isometric isomorphisms, then $\sigma^{-1}\circ A^{*,Ban}\circ\rho=A^{*,Hil}$.

The only essential difference of the two adjoints is that in the Hilbert setting we have that $(\lambda A)^*=\bar{\lambda}A^*$ while on the Banach setting we have that $(\lambda A)^*=\lambda\cdot A^*$.

Related: Hilbert space adjoint vs Banach space adjoint?


问:
Is there any sequence that is in the range of $[0,1]$ that converges to something outside of this range?
Is there any sequence that is in the range of $\cdots\cup[0,1]\cup[2,3]\cup\cdots$  that converges to something outside of this range?  I thought about taking a sequence such as $ \{\frac{n}{n+1}\}$ but this clearly converges to $1$. In any case I haven't been able to find a sequence that converges outside of this range

答:
Let $x_n$ tend to a limit $x$. Then, for sufficiently large $n$, $|x_n-x|<0.5$.

Then all these $x_n$ are in one of the sets $[2n,2n+1]$, which is a closed set. Therefore $x$ is also in this set.


问:
Given a recursion $a_{n+ 1}= \sqrt{na_{n}+ 2n+ 1}$ with $a_{1}\geq 1.$ Prove that $a_{n}\sim n\,{\rm as}\,n\rightarrow\infty$

Given a recursion $a_{n+ 1}= \sqrt{na_{n}+ 2n+ 1}$ with $a_{1}\geq 1.$ Prove that
$$a_{n}\sim n\,{\rm as}\,n\rightarrow\infty$$

Let $b_{n}= \dfrac{a_{n}}{n},$ we need to prove $\lim b_{n}= 1,$ then we find a value $\beta\in\left ( 0, 1 \right )$ satisfying $\left | b_{n+ 1}- 1 \right |\leq\beta\left | b_{n}- 1 \right |$ so that
$$\beta\rightarrow 0\,{\rm aut}\,n\rightarrow\infty$$
I can't predict the relationship among them, I need the the help to go to the induction, thank you.

答:
You can prove by induction on $n$ that
$$
n \le a_n  \le n + a_1 .
$$
Indeed, for $n=1$, these inequalities hold. To move from $n$ to $n+1$, note that
$$
a_{n + 1}  \ge \sqrt {n \cdot n + 2n + 1}  = \sqrt {(n + 1)^2 }  = n + 1
$$
and
\begin{align*}
a_{n + 1}  & \le \sqrt {n(n + a_1 ) + 2n + 1}  = \sqrt {(n + 1)^2  + na_1 } \\ & \le \sqrt {(n + 1)^2  + 2(n + 1)a_1  + a_1^2 }  = n + 1 + a_1 .
\end{align*}


问:
Is $(-1)^{\varepsilon}=1+i\pi \varepsilon$?
Let $\varepsilon$ be defined as in the dual numbers. Using the standard matrix representation. I tried $(-1)^{\left(
\begin{array}{cc}
 0 & 1 \\
 0 & 0 \\
\end{array}
\right)}$ and got the following result: $\left(
\begin{array}{cc}
 1 & i \pi  \\
 0 & 1 \\
\end{array}
\right)$. But complex numbers should have different representations in matrix form, so I onder if my interpretation is correct.

答:
No,
$$ (-1)^\epsilon = (e^{i\pi})^{\epsilon} = e^{i\pi\epsilon} = \cos(\pi\epsilon) + i \sin(\pi\epsilon)$$
or, at the very least this is one of the ways that $(-1)^\epsilon$ can be defined. You can get other results using $-1 = e^{i\pi+2n\pi}$, $n\in\mathbb{N}\setminus\{0\}$; the function $z^\epsilon$ is a so-called a multi-valued function on a complex plane and can be defined in amny ways.


问:
Strong duality by Slater's condition
In Section 5.3.2 of Boyd, Vandenberghe: Convex Optimization, strong duality is proved under the assumption that ker(A^T)={0} for the linear map describing the equality constraint, though it is remarked that this condition is not necessary. Where can I find a proof of the statement without the assumption on the kernel?</p>

答:
Define $\mathcal{A}$ as in Boyd's (5.37). Then we can think of $\mathcal{A}$ as an epigraph of $\mathcal{G}$ (again from Boyd).
Let $\tilde{\mathbf{x}}\in\text{ri}\,D$ with $f_i(\tilde{\mathbf{x}})<0,\,i=1,\dots,m$ and $A\tilde{\mathbf{x}}-\mathbf{b}=\tilde{\mathbf{y}}$. Also denote by $F=(f_1,\dots,f_m)^T$ the vectors of inequality constraints. Since the vector $(F(\tilde{\mathbf{x}}), \tilde{\mathbf{y}}, f_0(\tilde{\mathbf{x}}))\subseteq\mathbb{R}^m\times\mathbb{R}^p\times\mathbb{R}$ is on the relative boundary of $\mathcal{A}=\text{epi}(\mathcal{G})$, it follows by the supporting hyperplane theorem that there exists a separating hyperplane between $(F(\tilde{\mathbf{x}}), \tilde{\mathbf{y}}, f_0(\tilde{\mathbf{x}}))$ and $\mathcal{A}$, meaning that there exists a nonzero vector $(\mathbf{p},\mathbf{q},-\alpha)\subseteq\mathbb{R}_+^m\times\mathbb{R}^p\times\mathbb{R}$ for which \begin{equation}\tag{1}
\mathbf{p}^TF(\tilde{\mathbf{x}})-\mathbf{q}^T\tilde{\mathbf{y}}-\alpha f_0(\tilde{\mathbf{x}})\geq\mathbf{p}^TF(\mathbf{x})+\mathbf{q}^T\mathbf{y}-\alpha t,\quad \forall (F(\mathbf{x}),\mathbf{y},t)\in\mathcal{A}.
\end{equation}
Since $(F(\tilde{\mathbf{x}}), \tilde{\mathbf{y}}, f_0(\tilde{\mathbf{x}}))$ is on the relative boundary of $\mathcal{A}$ it follows that $(F(\tilde{\mathbf{x}}), \tilde{\mathbf{y}}, f_0(\tilde{\mathbf{x}})+1)\in \mathcal{A}$ as well. Hence we can plug $\mathbf{x}=\tilde{\mathbf{x}},\mathbf{y}=\tilde{\mathbf{y}}$ and $t=f_0(\tilde{\mathbf{x}})+1$ to the equation above and get
$$
\mathbf{p}^TF(\tilde{\mathbf{x}})-\mathbf{q}^T\tilde{\mathbf{y}}-\alpha f_0(\tilde{\mathbf{x}})\geq\mathbf{p}^TF(\tilde{\mathbf{x}})+\mathbf{q}^T\tilde{\mathbf{y}}-\alpha (f_0(\tilde{\mathbf{x}})+1),\quad \iff \alpha\geq 0
$$
After showing that $\alpha\geq 0$ we need to prove that $\alpha\neq 0$. Consider $L=\text{aff}(\mathcal{A})$. Since $\mathcal{A}\subseteq L$ and a supporting hyperplane $(\mathbf{p},\mathbf{q},-\alpha)$ by definition does not contain $\mathcal{A}$, it does not contain $L$ as well. Now consider the affine combination $\mathbf{s}_{\lambda}=(1-\lambda)\tilde{\mathbf{x}}+\lambda\mathbf{s},\;\lambda\in\mathbb{R}$, and $\mathbf{r}_{\lambda}$ is the corresponding vector for $A\mathbf{s}_{\lambda}$. The points in this line all lie in the affine set $L$ and $\mathbf{s}_{\lambda}\to\tilde{\mathbf{x}}, \mathbf{r}_{\lambda}\to\tilde{\mathbf{y}}$ as $\lambda\to 0$. Since $\tilde{\mathbf{x}}\in\text{ri}\,D$, for small values of $|\lambda|$ the points $\mathbf{s}_{\lambda}$ lie in $D$. Hence we can plug $\mathbf{s}_{\lambda}$ into $(1)$ and get (after some  algebra)
$$\mathbf{p}^T(F(\tilde{\mathbf{x}})-F(\mathbf{s}_{\lambda}))+\lambda\mathbf{q}^T(\tilde{\mathbf{y}}-\mathbf{r})-\alpha(f_0(\tilde{\mathbf{x}})-f_0(\mathbf{s}_{\lambda}))\geq 0.\tag{2}$$
Inequality $(2)$ must hold for sufficiently small positive and negative values of $\lambda$. Consider the case when $\alpha=0$. Since $\tilde{\mathbf{x}}\in\text{ri}\,D$ then $f_i(\tilde{\mathbf{x}})<0$ and for $\mathbf{s}_{\lambda}\in D\setminus\text{ri}\,D$ we have $f_i(\mathbf{s}_{\lambda})=0$. This implies that
$$
\mathbf{p}^T\overbrace{(F(\tilde{\mathbf{x}})-F(\mathbf{s}_{\lambda}))}^{<0}+\lambda\mathbf{q}^T(\tilde{\mathbf{y}}-\mathbf{r})\geq 0.\tag{3}
$$
Then either $\mathbf{p}=\mathbf{0}$ or $\mathbf{q}=\mathbf{0}$, but never both, since $(\mathbf{0},\mathbf{0},0)$ cannot be a supporting hyperplane. If $\mathbf{p}=\mathbf{0}$ then $\lambda\mathbf{q}^T(\tilde{\mathbf{y}}-\mathbf{r})\geq 0$ cannot hold true for positive and negative (small) values of $\lambda$, unless $\mathbf{q}=0$. If $\mathbf{q}=0$ then again $\mathbf{p}^T(F(\tilde{\mathbf{x}})-F(\mathbf{s}_{\lambda}))\geq 0$ can hold for only if $\mathbf{p}=0$. However if $\alpha>0$ there is no problem. Take for example the case where $\mathbf{p}=0$. Then the inequality
$$
\lambda\mathbf{q}^T(\tilde{\mathbf{y}}-\mathbf{r})-\alpha(f_0(\tilde{\mathbf{x}})-f_0(\mathbf{s}_{\lambda}))\geq 0
$$
will hold. This is because if we analyze the convex function $f_0(\mathbf{s}_{\lambda})$ we notice it's monotone decreasing in $\lambda<0$ (it increases as $\lambda$ moves away from zero in the negative direction), hence offsetting the flip of sign from $\lambda\mathbf{q}^T(\tilde{\mathbf{y}}-\mathbf{r})$.

Overall we have shown that there exists a non-vertical supporting hyperplane to $\mathcal{A}$ at its boundary, and therefore strong duality holds.

Title: Strong duality by Slater's condition


问:
Concerning left nullspace
There's a question that makes me reduce a matrix A to echelon form, which has a row of zeros. Then knowing the combinations of the rows which make the zero row in the echelon form of A, we can work out the vector(s) in the null space.

My question is why does this work? As in we know row3 - 2row2 + row1 gives a zero row in the echelon form of A. but why does this mean the null space contains the vector (1, -2, 1) which are the coefficients of the above row combination?

答: Take a look at the intermediate steps while going from the first matrix to the second one. The original matrix us$$\left[\begin{array}{rrr|l}1&2&3&b_1\\4&5&6&b_2\\7&8&9&b_3\end{array}\right].\tag1$$Then you add the first line times $-4$ to the second one and you add the first line times $-7$ to the third one, getting$$\left[\begin{array}{rrr|l}1&2&3&b_1\\0&-3&-6&b_2-4b_1\\0&-6&-12&b_3-7b_1\end{array}\right].$$Then you add the seconf line times $-2$ to the third line and you get$$\left[\begin{array}{rrr|l}1&2&3&b_1\\0&-3&-6&b_2-4b_1\\0&0&0&b_1-2b_2+b_3\end{array}\right].$$That $b_1-2b_2+b_3$ at the lower right corner tells you how the bottom line was obtained from the lines of $(1)$: it's the first line minus twice the second line plus the third line. But then, in particular,$$(0,0,0)=(1,2,3)-2(4,5,6)+(7,8,9).$$

Title: Concerning left nullspace


答:
If you apply the triangle inequality
$$
|f(x)-f(y)|  \leq |x-y| + |\tan^{-1}x - \tan^{-1}y|
$$
then  the right-hand side is already $> |x-y|$ for $x\ne y$, so that will not lead to the desired  result.

Better use the mean-value theorem: If $x \ne y$ then
$$
 f(x) - f(y) = f'(c) (x-y)
$$
for some $c$ between $x$ and $y$. Compute the derivative and show that $|f'(c)| < 1$ for all $c$ in the domain of $f$.

Title: Show $|f(x) - f(y)| < |x-y|$ for $f(x) = \frac{\pi}{2} + x - \tan^{-1}x$

Body:
Given the function $f : (-\frac{\pi}{2},\frac{\pi}{2}) \to \mathbb{R}$:
$$f(x) = \frac{\pi}{2} + x - \tan^{-1}x$$

I need to check whether $|f(x) - f(y)| < |x-y|$.

Now $$|f(x)-f(y)| = |(x-y)-(\tan^{-1}x - \tan^{-1}y)| \\
\leq |x-y| + |\tan^{-1}x - \tan^{-1}y| \\
= |x-y| + |\tan^{-1}(\frac{x-y}{1+xy})|\leq|x-y|+|\frac{x-y}{1+xy}| \, .
$$
What can I do next?


问:
Solving an ODE with square
I want to solve the IVP:

$$
\left\{
\begin{array}{c}
x'=\frac{x^2}{t^2}-2 \\
x(1)=1
\end{array}
\right.
$$

This is what I got for now:

I solved the homogenous ODE to $x(t)=t$.

Then I variated the constant for which I got: $c'(t)=\frac{c^2-c-2}{t}$.

For this, I repeated the steps and got $c=\frac{1}{1-Kt}$ and variated the constant $K$ again.

But that's where I'm stuck and think I did something wrong because I can't solve it.

If someone could help me it'd be very nice.

Title: Solving an ODE with square

Body:
Hint

Set $y(t)=tx(t).$ Then $$ty'(t)+y(t)=y(t)^2-1\iff \frac{y'(t)}{y(t)^2-y(t)-1}=\frac{1}{t}.$$


问:
Does $\mathbb Q(\sqrt{-2})$ contain a square root of $-1$?
This isn't a homework question but one I found online.

Does $\mathbb Q(\sqrt{-2})$ contain a square root of $-1$?

We just started doing field theory in my class and I want extra practice, but I have no idea how to even start this problem. It's no, right? I'm not sure why.

答:
Use contradiction.

Start by assuming $\sqrt(-1)$ is in the set $\mathbb{Q}(\sqrt(-1))$

so there is an element in $\mathbb Q(\sqrt(-2))=\{a+b\sqrt(-2)| a,b\in\mathbb Q\}$ such that its square is equal to -1


问:
Confusion on notation in the definition of reflexivity.
This is a very silly question i assume to most of you but how would I read this

$\forall x \in X : (x,x) \in R$

I don't quite know what the colon indicates and this definition is used to define reflexivity .

Thanks in advance.

答:
Normally the colon is read as "such that"

∀ ∈  :(,) ∈

For all $x$ in {the set} X such that $(x,x)$ {$x$ maps to itself} in {the binary relation} R {which is a subset of X}.


答:
You could also consider this as a Riccati equation and employ the parametrization $x=-2\frac{u}{u'}$ so that
$$
\frac{4u^2}{t^2u'^2}-2 = x'=-2+2\frac{uu''}{u'^2}
\implies
t^2u''-4u=0,~~ u(1)=1,~u'(1)=-2.
$$
This now is an Euler-Cauchy equation with basis solutions $t^m$ where $m$ is the roots of the characteristic equation $m(m-1)-4=0$. Then
$$
x(t)=-2\frac{C_1t^{m_1}+C_2t^{m_2}}{m_1C_1t^{m_1-1}+m_2C_2t^{m_2-1}}.
$$

Title: Solving an ODE with square

Body:
I want to solve the IVP:

$$
\left\{
\begin{array}{c}
x'=\frac{x^2}{t^2}-2 \\
x(1)=1
\end{array}
\right.
$$

This is what I got for now:

I solved the homogenous ODE to $x(t)=t$.

Then I variated the constant for which I got: $c'(t)=\frac{c^2-c-2}{t}$.

For this, I repeated the steps and got $c=\frac{1}{1-Kt}$ and variated the constant $K$ again.

But that's where I'm stuck and think I did something wrong because I can't solve it.

If someone could help me it'd be very nice.


问:
Lyapunov function for a non linear 3 dimensional system
How it is possible to find a Lyapunov function for the following system?
\begin{cases}
  \dot {x_1}=x_2+x_3 & \\
         \dot {x_2}=-\sin x_1-x_3 & \\
 \dot {x_3}=-\sin x_1+x_2 &
\end{cases}
I have tried with $V(x)=\frac{1}{2}x^TPx+\int_0^{x_1} \sin{(s)} ds$ but I can't find out the right elements of P in order to obtain $\dot{V}<0$. Can you help me?

答:
This system is not asymptotically stable. Near the origin we have $\sin x\approx x$ so the dynamics are practically

$$
\cases{
\dot x_1 = x_2 + x_3\\
\dot x_2 = -x_1 - x_3\\
\dot x_3 = -x_1 + x_2
}
$$

or

$$
\cases{
x_1\dot x_1 = x_1x_2 + x_1x_3\\
x_2\dot x_2 = -x_1x_2 - x_2x_3\\
x_3\dot x_3 = -x_1x_3 + x_2x_3
}
$$

and after adding up

$$
x_1^2+x_2^2+x_3^2 = C
$$

those orbits represent a center at the origin.


问:
Matrix norm differentiation problem.
Consider $J = \|X - AS\|^2$, where $A \in \mathbb{R}^{N \times R}$, $S \in \mathbb{R}^{R \times M}$.

We need to calculate $\dfrac{d J}{d S}$.

I've done the following:

$J = \operatorname{tr}((X - AS)(X - AS)^\top) = \operatorname{tr}((X-AS) (X^\top - S^\top A^\top)) = \operatorname{tr}(XX^\top - ASX^\top - XS^\top A^\top + ASS^\top A^\top)$ and hence

$$\dfrac{dJ}{dS} = \dfrac{-\operatorname{tr}(ASX) - \operatorname{tr}(XS^\top A^\top) + \operatorname{tr}(ASS^\top A^\top)}{dS}$$

$\dfrac{d\operatorname{tr}(ASX)}{dS} = \dfrac{d\operatorname{tr}(SXA)}{dS} = (XA)^\top = A^\top X^\top$

$\dfrac{d\operatorname{tr}(XS^\top A^\top)}{dS} = \dfrac{d\operatorname{tr}((XS^\top A^\top)^\top}{dS} = \dfrac{d\operatorname{tr}(ASX^\top)}{dS} = \dfrac{\operatorname{tr}(SX^\top A)}{dS} = (X^\top A)^\top = A^\top X$

$\dfrac{d\operatorname{tr}(AS(AS)^\top)}{dS}$, so $$\displaystyle \dfrac{d\sum_{n,m}(AS)_{n,m} ((AS)^\top)_{m,n}}{d s_{i,j}} = \dfrac{d\sum_{n,m}(AS)^2_{n,m}}{d s_{i,j}} = \dfrac{d \sum_{n,m} \left(\sum_{q} a_{n,q} s_{q,m}\right)^2}{d s_{i,j}} = \dfrac{d\sum_{n,m} \sum_{q}a_{n,q}s_{q,m} \sum_{l}a_{n,l} s_{l,m}}{ds_{i,j}} =$$ $$\sum_{n,m} \sum_{q} \sum_{l} a_{n,q} s_{q,m} a_{n,l} \delta_{l}^{i} \delta_{m}^{j} = \sum_{n} \sum_{q} a_{n,q}s_{q,j} a_{n, i} = \sum_{n} A^\top_{i,n}(AS)_{n,j} = (A^\top A S)_{i,j}$$

Hence we have: $\dfrac{dJ}{dS} = A^\top (AS - X - X^\top)$. Unfortunately, I'm very bad at matrix calculus and do not use some kind of properties. I'm interested in whether my approach is OK?

答:
We have, using $\operatorname{tr}(C+C^T)=2\operatorname{tr}(C)$
$$\begin{align}
d_SJ(v)&=\operatorname{tr}\bigl((-Av)^T(X-AS)+(X-AS)^T(-Av)\bigr)\\
&=\operatorname{tr}\bigl((X-AS)^T(-Av)+((X-AS)^T(-Av))^T \bigr)\\
&=-2\operatorname{tr}\bigl((X-AS)^TAv\bigr)\\
&=-2\operatorname{tr}\bigl(\bigl(A^T(X-AS)\bigr)^Tv\bigr)\\
&=-2\langle (A^T(X-AS)),v\rangle.
\end{align}
$$
Hence $$d_SJ=-2(A^T(X-AS)).$$

EDIT (even simpler) Using chain and product rule:
$$d_SJ(v)=2\langle X-SA,-vA\rangle=-2\langle (A^T(X-AS)),v\rangle.$$

Title: Matrix norm differentiation problem.


问:
Questions about domain and range of composite functions

Let $f: \mathbb R \to \mathbb Z, \ g: \mathbb R \setminus \{0\} \to \mathbb  R$ be defined by $f(x) = \lfloor x\rfloor, g(x) = \displaystyle{\frac 1x}$. Note that in this case, $g \circ f$ is not defined ($\color{red}{\text{why?}}$). However, $f \circ g: \mathbb R \setminus \{0\} \to \mathbb Z$ is defined. You can $\color{blue}{\text{verify}}$ that the composite is given by $$(f \circ g)(x) = \left\lfloor\frac 1x\right\rfloor =\begin{cases}0 \text{ if } x \in (1, \infty) \\  n \text{ if } x \in \color{orange}{\left(\frac{1}{n + 1},\frac 1n\right]_{n \in \mathbb N}} \\  -(n + 1) \text{ if } x \in \color{orange}{\left(-\frac 1n, -\frac{1}{n + 1}\right]_{n \in \mathbb N}} \\ -1 \text{ if } x \in (-\infty, -1)\end{cases}$$

How do we know that we have defined $(f \circ g)(x)$ for all $x \in \mathbb R \setminus \{0\}$? $\color{green}{\text{Recall }\bigcup_n [1/n,1) = (0,1)}.$

My questions:

Answer to the $\color{red}{\text{question in red}}$ above: $g \circ f$ is not defined as $0$ in the range of $f$, but not in the domain of $g$. Is that correct?

$\color{blue}{\text{Suggestion in blue}}$ above. I'll only consider one branch to see if I can do it. Suppose $\displaystyle{-\frac 1n < x \le -\frac{1}{n + 1}}$. Then $\displaystyle{-n > \frac 1x \ge  -(n + 1)} \iff \left\lfloor \frac 1x \right\rfloor = -(n + 1)$ by definition of $\lfloor \cdot \rfloor?$

$\color{green}{\text{Hint in green}}$ above. If $x \in \mathbb Q$, then $g(x) \in \mathbb N.$ If $x
\in \mathbb N$, then $g(x) \ne 0$. Is this what the mysterious(?) $\color{green}{\text{hint in green}}$ above alluding to?

How do they know the given partition of the domain of $f \circ g$ covers all the necessary cases? In particular, how did they know to choose the $\color{orange}{\text{intervals in orange}}$?

Thanks.

答:
Red and Blue: They look good to me.

Green: You're given a piecewise function that's defined and equal to $f \circ g$ on $(-\infty, -1) \cup (-\frac 1n, -\frac{1}{n + 1})_{n \in \mathbb N} \cup (\frac{1}{n + 1},\frac{1}{n})_{n \in \mathbb N}\cup (1, \infty)$, but how are you sure that these intervals make up all of $\mathbb{R}/\{0\}$?

If you can show that your piecewise function has been defined on $[1/n,1)_{n \in \mathbb N}$, the hint will tell you that you've indeed defined it on $(0,1)$. From there, proving that the function is defined on all of $\mathbb{R}/\{0\}$ is straightforward.

Orange: I believe it takes a bit of intuition and trial and error. That's what I would have done at least. The value of $\left\lfloor x \right\rfloor$ seems to change if and only if $x$ "crosses" an integer, which means that $\left\lfloor\frac{1}{x}\right\rfloor$ should change whenever $x$ "crosses" $\frac{1}{n}$ for some $n$. This suggests for us to use the partition that was given. Now all we need to do is figure out the details (endpoints) and verify if it works.


问:
minimum of functional for $u$ in $H^1$
Let $\Omega \subset \mathbb{R^n}$ for $n \ge 3$ be a bounded domain with smooth boundary.
Let $$I(u)= \int_{\Omega}|\nabla{u}|^2 +\int_{\Omega}u^2+\beta\int_{\Omega}|u|^p$$ with $\beta >0$, $2<p<2^*=\frac{2n}{n-2}$ and $u \in H^{1}_0(\Omega)$. Show that there exists a $u_0 \in H^{1}_0(\Omega)$ such that $I(u_0)=\min I(u)$  for $u \in H^{1}_0(\Omega)$.

I understand that this means that $u_0$ minimizes the functional but have not reached the sophistication required to prove this. I'd like to see why this is true.
Would the idea of using a minimizing sequence work? and if so how?

答:
I am gonna apply the direct method here. I will drop the $\beta>0$, as it is irrelevant in this context.

$\textbf{Step 1}:$ Lower bound

$$
I(u)=||u||_{H^1(\Omega)}^2+||u||_{L^p(\Omega)}^p \geq 0
$$
$\textbf{Step 2}:$Compactness

First of all, let us note that the Rellich embedding gives $u \in H^1(\Omega) \implies u \in L^p(\Omega)$. Your exponents match perfectly, as $p$ is smaller than the respective Sobolev conjugate.

Let $\{ u_k \} \subset H^1_0(\Omega)$ be a minizining sequence. Then we have
$$
||u||_{H^1(\Omega)}^2+||u||_{L^p(\Omega)}^p \leq C
$$
for some $C>0$. In particular, there exists a subsequence (not relabeled) which satsifies
$$
u_k \rightharpoonup u \; \textrm{ in } H^1(\Omega)   \\
u_k \to u \; \textrm{ in } L^p(\Omega)
$$
i.e. weakly in $H^1(\Omega)$ and strongly in $L^p(\Omega)$. The first one is simply a property of the Hilbert space you are working on - bounded sequences in Hilbert spaces have weakly convergent subsequences. The second one is again a consquence of the Rellich embedding, which is also compact in your setting. The fact that $u \in H^1_0(\Omega)$ follows from the fact that closed, convex sets are also weakly closed.

$\textbf{Step 3}:$ Conclusion

The $H^1(\Omega)$-norm (squared) is weakly lower semicontinuous and we have even strong $L^p(\Omega)$ convergence for our minimizing sequence. Therefore, we see that our functional is lower-semicontinuous w.r.t. to our sequence. We then conclude:
$$
\inf I(\cdot) \leq I(u) \leq \liminf I(u_k)=\inf I(\cdot)  \implies
I(u)=\min I(\cdot)
$$
I hope this gives you an idea of what to do, there are still some details to fill in. If you have any questions, feel free to ask.

Title: minimum of functional for $u$ in $H^1$


问:
Find the dimensions of some kernel subspaces of $\operatorname{Hom}(V, W)$.
I am a bit unconfident with the following proof that I made to find the dimensions of the subspace below,

$$
S =\{T\in\operatorname{Hom}(V, W): T(v)=0\}
$$
where $V$ and $W$ are finite-dimensional vector spaces over F of dimensions $n$ and $m$ respectively and $v\in V$ be a non-zero vector. What I did is to define the following,
$$
U:\operatorname{Hom}(V,W)\rightarrow W
$$
where $T\mapsto T(v)$. And show $U$ is a linear transformation, then using the Dimension Theorem to find the $\operatorname{dim}(S)$ since $S=\operatorname{ker}(U)$. To show $U$ is linear is not so difficult, but I was confused at the end that was whether if $\operatorname{Im}(U)=W$. I know $\operatorname{Im}(U)\subset$ W, but if $\operatorname{Im}(U)\neq W$, then I can't derive the conclusion such that $\operatorname{dim}(S)= nm-m$.

Additionally, I also have the following question that is to find the possible dimensions of the subspace below,
$$
C =\{T\in\operatorname{Hom}(V,W):T(v_1)=0,T(v_2)=0,T(v_3)=0\}
$$
where $v_1,v_2,v_3\in V$. Basically, I show this in a similar way to the previous one. Let
$$
\Phi:\operatorname{Hom}(V, W)\rightarrow W\times W\times W
$$
which is defined differently depending on the independence of $\{v_1, v_2, v_3\}$, but to show all of them are linear is not so difficult either. For example, if $v_1,v_2,v_3$ are linearly independent then $\Phi:T\mapsto (T(v_1),T(v_2),T(v_3))$. Since $C=\operatorname{ker}(\Phi)$ and if $\operatorname{Im}(\Phi)=W\times W\times W$, then using Dimension Theorem $\operatorname{dim}(C)=nm-3m$. Similarily, depending on the dependence between $v_1, v_2$ and $v_3$, I can also derive two possible dimensions that are $\operatorname{dim}(C)=nm-2m$ or $\operatorname{dim}(C)=nm-m$.

I wonder how to end the proof (e.g. how to tell $\operatorname{Im}(U)=W$?), and whether if my guesses to the second question is correct or not.

PS: I am a newcomer to linear algebra so if I made anything unclear or wrong in the proof, please help me to point it out since this is helpful for me to improve my skill when writing a proof.

Thanks in advance.

答:
Any element in Hom(V, W) is a map from V to W.

v is a fixed element in V.

Thus for any w∈W, we can define a Tw: V -> W which map v to w regardless of what Tw map other elements of V to

This Tw always exists since it belongs to the set Hom(V, W)

This implies that Image(U)=W


问:
How to think about $\sum_{k\leq t}k=\frac{t^2}{2}+\mathcal{O}(t)$
I am looking for different ways to think of why the below equality holds:

$\sum_\limits{k\leq t} k=\frac{t^2}{2}+\mathcal{O}(t)$

答:
So, using the standard formula $$\sum_{k=1}^{t}k=\frac{t(t+1)}{2}=\frac{t^2}{2}+\frac{t}{2}$$
we can see that the question is expecting you to see the link between the $t/2$ term and $O(t)$. So what does $O(t)$ mean? We say a function $f(t)$ is $O(t)$ if there is a (positive) constant $C$ such that
$$
|f(t)|\leq Ct\ .
$$
Here, our $f(t)=t/2$ and we can take any $C\geq 1/2$.

More generally, we say $f(x)=O\big(g(x)\big)$ if there is a constant $C$ (normally positive, depending on context) with
$$
|f(x)|\leq Cg(x)
$$
for all $x>0$ (again, depending on context).

Here are a few examples:

$$
\begin{array}{rcll}
f(x)&=&O(f(x))& \text{always true, just take }C=1\\
x&=&O(x^n)&\text{for any }n\geq 1\text{ and }x\geq 1\\
x^n&=&O(e^x)&\text{for any }n>0\text{, so long as it is understood }x\geq 0\\
\sin x&=&O(1)&\text{in fact, any bounded function is }O(1)
\end{array}
$$


问:
ODE - existence and uniqueness of a solution
Let $p(t)$ be a polynomial with no real roots.
I have to prove that the following ODE-IVP:
$$\begin{cases}
y'=\frac{1}{p(xy)}=f(x,y) \\
y(0)=0
\end{cases}
$$

Has a unique solution in $\mathbb{R}$:

$$f_y(x,y)=-\frac{p'(xy)x}{(p(xy))^2}$$

I tried to show that $f_y(x,y)$ is bounded at $J\times(\infty,\infty)$ for all closed intervals $J$, but found out that when $p(t)=t^2+t+1$,
$$f_y(x,y)=-\frac{2x^2y+x}{((xy)^2+xy+1)^2},$$

and in that case, $f_y(x,y)$ is not bounded for small values of $y$.

Maybe I should use the Lipschitz condition?

Thanks.

答:
As $p$ is a polynomial without a real root, $f(x,y)$ is  locally uniformly Lipschitz continuous in $y$ and continuous in $x$. Therefore, the IVP has a unique maximal solution satisfying $y(0)=0$ according to Picard-Lindelöf theorem.

Remains to prove that the solution is global on $\mathbb R$. For this, just remind that if a solution has a bounded maximal interval, it explodes in this interval. It’s not the case here as for all $x \in \mathbb R$

$$\vert y(x) \vert \le \frac{\vert x \vert}{k}$$ where $k \gt 0$ is such that $\vert p(t) \vert \gt k$ for all $t \in \mathbb R$. The existence of $k$ is a consequence of the fact that $p$ has no real root.


答:
As you say, $f(\epsilon)=f’(0)\epsilon$ for any differentiable function $f$ defined on a real neighborhood of $0$, when $\epsilon^2=0$. To compute $(-1)^\epsilon$ you need to determine $f$. The matrix expression doesn’t make anything easier here.

Generally, $(-1)^x=e^{x\log(-1)}$ and $\log(-1)=i(2k+1)\pi$ is multivalued. So there is a choice $f_k(x)=e^{xi(2k+1)\pi}$ of function here for each integer. Making a choice here is called choosing a branch cut of $\log$. We have $f_k’(0)=(2k+1)i\pi$, so $f_k(\epsilon)=(2k+1)i\pi\epsilon$. Choosing the branch cut $k=0$ recovers your calculation, but it is one of infinitely many correct answers, as almost always happens with complex exponentials.

Title: Is $(-1)^{\varepsilon}=1+i\pi \varepsilon$?

Body:
Let $\varepsilon$ be defined as in the dual numbers. Using the standard matrix representation. I tried $(-1)^{\left(
\begin{array}{cc}
 0 & 1 \\
 0 & 0 \\
\end{array}
\right)}$ and got the following result: $\left(
\begin{array}{cc}
 1 & i \pi  \\
 0 & 1 \\
\end{array}
\right)$. But complex numbers should have different representations in matrix form, so I onder if my interpretation is correct.


问:
Stereographic Projection: Cartography Applications
Compared to the Mercator's, which is also conformal, how does the Stereographic projection help in areas such as navigation? Or any application besides simply mapping polar areas, although I would prefer answers on the polar aspect.

Additionally, from Wikipedia:

"When the projection is centered at the Earth's north or south pole, it has additional desirable properties: It sends meridians to rays emanating from the origin and parallels to circles centered at the origin".

Why are these desirable qualities? How would it benefit one in any hypothetical situation?

This may not be strictly math related, but any ideas where the stereographic projections' properties could be utilized to solve a problem would still be appreciated.

Edit: Any suitable scenario where its properties can be of use would suffice, it can be very bizarre, but I would prefer it to be related to navigation or locations on Earth.

答:
You consider stereographic projection from North pole.

But I would like to stress the fact that, when the center of projection is done from a point $E$ on the equator, the following pattern is generated:

(Figure taken from this article) explaining the following kind of map where meridians and parallels are "rendered" as two families of orthogonal circles:

with the advantage of minimal deformations along the equator (Figure taken from Oxford Advanced Atlas 1936).
This technique had been already used by Mercator in this wonderful map:

(a testimony of the fact that Mercator was mastering other projections than the one he is famous for).

An interesting article about stereographic projection, its history and an application to another scientific field (geology) can be found here


问:
Functions and Infinity
I have been reading a lot about real analysis and infinite sets lately. Infinite sets I mean the most common ones ($\mathbb{N}$ and $\mathbb{R}$). I have come across many problems and there are two which look very similar in terms of solving technique but I could not seem to figure them out.

Problem 1: Is there a function ($f:$ $\mathbb{R} \rightarrow  \mathbb{R}$) whose limit at every point is $ \infty $ ?

Problem 2: Can a function ($f:$ $\mathbb{R} \rightarrow  \mathbb{R}$) have continuum many local min/max values ?
(We are looking for local extrema values not points, so the constant function only has 1 in this regard)

So both problems ask questions about some properties maximum cardinality. Obviously countably finite is achieveable for both questions ($\tan{x}$ and $x\sin{(1/x)}$). I think the answer is negative for both questions.

For Problem 1 I think all values have to be 'large' for this to happen, namely $\infty$ but could not exactly prove this. I also tried to make a bijection with rational numbers on how many points can have a limit at infinity, but also could not finish.

For problem 2 I tried bijection with rationals but I could not make it so that for every rational I pair it with one Local min/max.

I feel like the thought process should be roughly similar for both but I could not find it. I may be completely on the wrong foot either by guessing that the answer must be negative.

Can someone tell me if my ideas can be finished, or if not then what way would work for proving these, and also if my guess about the answer is right.

答:
Here is a proof for Problem 1:

Assume that the limit of $f$ at every point is $\infty$.
Then, for every $M\in \mathbb{Z}$, the set
$$\{x\mid f(x) \leq M\}$$
cannot have any accumulation point (*), which implies that it is countable (**).
But on the other hand
$$\bigcup_{M\in\mathbb{Z}} \{x\mid f(x) \leq M\} = \mathbb{R}$$
which is a contradiction since the LHS is countable.

(*) If the set $\{x\mid f(x) \leq M\}$ had an accumulation point $x^*$, then there would exist a convergent sequence $x_n \to x^*$ with $f(x_n) \leq M$, but this contradicts $\lim_{x\to x^*} f(x) = \infty$.

(**) If a set $A\subseteq \mathbb{R}$ is uncountable but has no accumulation point, then for some $K\in \mathbb{N}$ the set $A\cap[-K,K]$ would be uncountable and have no accumulation point, contradicting Bolzano-Weierstrass.

For problem 2, you are looking at strict local extrema, right? (Else a constant function would have extrema everywhere)

For pairing the local extrema: it might work to consider the a local extremum $x$ as global extremum of the function restricted to some interval $(x-\varepsilon, x+\varepsilon)$.
Then (by making it smaller if necessary) you can have the interval bounds be rational.
So $x$ is global extremum of $f$ on some interval $(a,b)$, $a,b \in \mathbb{Q}$. Since strict global min/max are unique, we can have at most one such $x$ for any interval with rational bounds. But there are only countably many such intervals, hence the number of strict local extrema of $f$ must be countable.

So the idea would be to pair local min/max to a pair of rational numbers, instead of a single rational number.

Title: Functions and Infinity


问:
Euclidean norm between the difference of two matrices $| B^{i+1} - B^{i} |_2$
So I am curious what/if anything the euclidean distance between two matrices means? I have an algorithm that computes the solution of this matrix $B$ iteratively and one of the questions it asks is if $| B^{i+1} - B^{i} |_2 < \epsilon$ then a First Order, Second Order Necessary condition has been met or even a Second Order Sufficient Condition.

I would appreciate some intuitive insight if possible.

Thanks

答:
Informally, you can think of the condition $\|B^{i+1} - B^i\|_2 \le \epsilon$ as "$B$ have not changed much in the last iteration". When $\|B^{i+1} -B^i\|_2 = 0$, i.e. $B^{i+1} = B^i$, your method have converged (and with suitable assumptions, you reached optimum). $\|B^{i+1} - B^i\|_2 \le \epsilon$ is "close enough" for convergance in practice (when $\epsilon$ is small enough).

For example, assume you apply a first-order gradient descent method on a problem $\min_B f(B)$, where $f$ is convex and differentiable. Specifically, assume that you run $B^{i+1} = B^{i} - \eta \nabla f(B^i)$ with constant step size $\eta$ until the condition $\|B^{i+1} - B^i\|_2 \le \epsilon$ is met. In this scenario, $\|B^{i+1} - B^i\|_2 = \eta \|\nabla f(B^i)\|_2$, hence the condition is $\|\nabla f(B^i)\|_2 \le \frac{\epsilon}{\eta}$. Since $f$ is convex and differentiable, $\|\nabla f(B^i)\|_2 = 0$ is a sufficient condition for optimality. Demanding $\|B^{i+1} - B^i\|_2 \le \epsilon$ is now equivalent to "the gradient is very small now, so we are close enough to the minimum".


问:
Showing an inequality in the real positive numbers
I want to show that $\forall c>0$:

$$0<\left(1-\frac{(wl)^2}{r^2+(wl)^2}\right)^2+\left(\frac{wrl}{r^2+(wl)^2}\right)^2<(1-w^2lc)^2+(wrc)^2\tag1$$

Where all variables are bigger than zero.

My work. If we simplify the middle term I get: $$\frac{r^2}{r^2+(wl)^2}$$ So we need to show $$0<\frac{r^2}{r^2+(wl)^2}<(1-w^2lc)^2+(wrc)^2$$

But what can I conclude from this?

答:
Let $X=r^2+w^2l^2$. The expression is clearly greater than $0$, so consider just the RH inequality.

The central terms can be simplified:
$$\left(1-\frac{(wl)^2}{r^2+(wl)^2}\right)^2+\left(\frac{wrl}{r^2+(wl)^2}\right)^2=\left(\frac{r^2}{X}\right)^2+\left(\frac{wrl}{X}\right)^2=\frac{r^2}{X}=1-\frac{w^2l^2}{X}.$$
The RH tems can also be simplified:
$$1-2w^2lc+w^4l^2c^2+w^2r^2c^2=1-2w^2lc+w^2c^2X$$

We now need to prove $$-\frac{w^2l^2}{X}<-2w^2lc+w^2c^2X$$

This simplifies to
$$w^2(l^2-2lcX+c^2X^2)>0.$$

The LHS is $w^2(l-cX)^2$ and so the result is true unless $l=c(r^2+w^2l^2)$.


问:
Prove, that the multiplicative group of a finite field is cyclic.
I have almost completed the proof, I have shown, that there exists an element of order e, where e is min{n:$g^n=1$} and $g\in G$ where $G$ is the multiplicative group of a finite field. Now I noticed, that the polynomial $x^e-1$ is solved by all $g\in G$. I am finished if I can show, that e=q-1 where q=$|G|$. However I only know that q-1$\leq$e. My idea is, that if q-1 <e, e >min{n:$g^n=1$}, so e has to be q-1 wich means $G$ is cyclic. Is the reasoning for q-1=e sufficient?

答:
Hint: Suppose $q-1<e$:

By division with remainder, $e = p(q-1) + r$ with remainder $0\leq r<q-1$.
Then $1=g^e = (g^{q-1})^p g^r = g^r$ for all $g$.


问:
$0<f(1)<f(3)<f(2)$ Limits, Continuity, Maxima and minima

Given the function $f(x)$ that fulfills the following $0<f(1)<f(3)<f(2)$ determine if the statements are true or false.

if the function is defined in the interval $[1,3]$ then we will conclude that $\lim \limits_{x \to2}f(x)>0$

if the function is continuous in $[1,3]$ we can conclude that there is a point $c$ in the interval $(1,3)$ that is $f'(c)=0$

if the function is differentiable in $[1,3]$ we can conclude that $f'(2)=0$

if the function is defined in $[1,3]$ then there is no Vertical asymptotes

if the function is differentiable in $[1,3]$ we can conclude that $x=2$ is a local maxima or global maxima

What I tried:

I think that it is true according to the information given $0<f(1)<f(3)<f(2)$ and $\lim \limits_{x \to2}f(x)$ is $f(2)$ and they are all greater than zero.

In part 2 I thought about two things, the first thing was that the statement is not true because even if it is continuous it does not mean that the function is necessarily differentiable so that makes it false. But the second thing I thought about was the Extreme value theorem but even with that it still does not mean that $f′(c)=0$ so my guess is that the statement is false(?)

I also believe that this is false, this rang a bell for Rolle's theorem at first we need it to be differentiable in $(1,3)$ and continuous in $[1,3]$ and we need $f(a)=f(b)$ and we don't have that here according to $0<f(1)<f(3)<f(2)$ but I am not sure.

This seems true to me because they said it is defined in $[1,3]$

Also seems true according to Extreme value theorem and the information given $0<f(1)<f(3)<f(2)$.

I usually have more information on what I tried on my posts but honestly, on this question I found it very hard to even begin with, I would very much appreciate tips and hints on how to approach these questions.

答:
One can construct a function $f$ such that $f(x_0)$ is defined but $\lim_{x\to x_0} f(x)$ is not defined. Consider $sign(x)$ at point $x_0=0$. There are also functions for which $\lim_{x\to x_0} f(x)$ defined but is not equal to $f(x_0)$. Consider $f$ where $f(0) = 0$ and $f(x) = 1$ for every $x \neq 0$ at point $x_0=0$.

You are correct. Continuity of $f$ does not imply differentiability. Though if we assume $f$ to be differentiable, the statement is true. Apply the intermediate value theorem to interval $[1,2]$: let $a \in [1,2]$ be such that $f(a)=f(3)$. Then apply Rolle's theorem to interval $[a, 3]$.

One can easily provide a contradicting example. Just draw three points $(1, f(1))$, $(2, f(2))$, $(3, f(3))$ on a plane such that $0<f(1)<f(3)<f(2)$ and draw a curve through them that doesn't have a horizontal tangent at point $x=2$. For a rigorous construction, a parabola $y=ax^2+bx+c$ with $a<0$ and its vertex $x_0 \in (2,3)$ might be a good choice -- the details are up to you.

Think of a function $f$ that is defined at $[1,2)\cup (2,3]$ and $\lim_{x\to 2} f(x)=\pm \infty$. You then obviously have a vertical asymptote at point $x=2$. Setting $f(2)=0$ does not do anything to asymptote.

An example contradicting 3. will also probably contradict 5.


问:
Does half-life mean something can never completely decay?
Caffeine has a half-life of approximately six hours. I understand this to mean that every six hours, the amount of caffeine in the body is half of what it was six hours prior. Does that mean that caffeine never completely leaves the body? It just keeps reducing to half after a fixed time ad infintum.

I guess a broader statement of my question is: When calculating the half life of a thing, is it true that it will never reach zero? And how is this used/reconciled in a more practical setting like when calculating the amount of caffeine in the body?

答:
Any chemical or radioactivity has a level below which it cannot be detected, so you are safe if you are drinking coffee.

Meanwhile, half-lives are an approximation that works very well when the number of particles involved is essentially uncountable by us. If you get down to ten atoms remaining that might decay, there is nothing dangerous going on. And I just made up the number ten, an actual physicist could give specifics about any such issue that worries you.