问:
Intuitive interpretation of Ricci Flow
What is the best way to interpret, explain or somehow visualize the basic idea behind formal definition of Ricci Flow?

I am familiar with the hackneyed expressions like

"Ricci Flow is a non-linear analogue for the heat equation which smoothens metric",

or

"Ricci Flow describes the deformation of the Riemannian metric tensor on manifold".

However, I was looking for something similar to the balloon-under-pressure interpretation of the mean curvature flow,  in particular of the surface tension flow:

Ultimately, I am looking for something that would make Ricci Flow concept clear for undergraduate students without diving too deep into technicalities of Ricci tensor and volume forms.

答:
I'm late to the question, but this writeup by Terry Tao is the best intuitive, high level source on the subject that I know. I'll summarize it in my own words.

First we need to describe the notion of a Riemannian metric, since this is what is evolving under the Ricci flow. The Riemannian metric $g$ is a way to describe the shape of a manifold, by defining the lengths, areas and angles of that manifold locally. For example, standing on the surface of the earth, you only have a perspective of distances, angles and area with respect to that surface rather than the larger space $\mathbb{R}^3$.

The Ricci curvature of a point $x \in M$ tells us how "non-Euclidean" a Riemannian manifold is at $x$. Specifically, we can measure this by considering the area of a sector of a circle having angle $\theta$ and radius $r$. In Euclidean space, this area is $\frac{1}{2}\theta r^2$ and in a Riemannian manifold $M$, we can denote it by $|A(x,r,\theta,v)|$. The Ricci curvature of a surface is the (scaled) difference between these:
$$ \text{Ric}(x)(v,v) = \lim_{r\rightarrow 0} \lim_{\theta \rightarrow 0} \frac{\frac{1}{2} \theta r^2 - |A(x,r,\theta,v)| }{\theta r^4/24}.$$

This is the formula for a 2-dimensional manifold (surface). More generally, the Ricci curvature measures the difference between the Euclidean volume and the manifold volume.

(Here is an image of how the area of a triangle differs in spherical, Euclidean, and hyperbolic space.)

The Ricci flow an intrinsic flow, meaning that it describes the evolution of a manifold using only its Riemannian metric (recalling that this is a local description of the shape at every point). It decreases the Ricci curvature in time. This means that a surface becomes "more Euclidean" at every point as the Ricci flow evolves it.
Specifically, the Ricci flow is
$$\frac{dg}{dt} = -2\text{Ric}$$
meaning that the Riemannian metric itself evolves, at a rate proportional to its negative Ricci curvature. This resembles mean curvature flow, in which a manifold evolves so that its mean curvature at every point "smooths out". The difference is that the Ricci is an intrinsic flow, measured by the local metric, as opposed to MCF which is extrinsic, where curvature is measured from the ambient Euclidean space.

As for your balloon-under-pressure interpretation I'm not sure. This probably has to do with the volume measured under the Riemannian metric. You might want to check out the surface tension viewpoint of minimal surfaces (surfaces that are stationary under MCF).

Title: Intuitive interpretation of Ricci Flow


问:
If $\lim_{n\to\infty}\sum_{i=1}^{n}w_i=1$, then $\lim_{n\to\infty}\sum_{i=1}^{n}w_i^{2}=0$?
If $\lim_{n\to\infty}\sum_{i=1}^{n}w_i=1$, then $\lim_{n\to\infty}\sum_{i=1}^{n}w_i^{2}=0$?

If it is true, then can you help prove it?

ContentLicense: CC BY-SA 4.0

Body: This is not true. $$\sum_1^\infty (\frac{1}{2})^n = 1,$$ but $$\sum_1^\infty (\frac{1}{2})^{2n} = \sum_1^\infty (\frac{1}{4})^n = \frac{1}{3} \neq 0.$$


问:
Solving a Linear Congruence
I've been trying to solve the following linear congruence with not much success:

$19\equiv 21x\pmod{26}$

If anyone could point me to the solution I'd be grateful, thanks in advance

答: HINT: $21\equiv-5\pmod{26}$, so $21\cdot5\equiv-25\equiv1\pmod{26}$, and $21\cdot5\cdot19\equiv1\cdot19\pmod{26}$.</p>

Title: Solving a Linear Congruence

ContentLicense: CC BY-SA 3.0


答: Hint: $26 = 2 \cdot 13$ and the Chinese remainder theorem. Modulo $2$ we have to solve $1 \cong x \pmod 2$, that is $x = 2k + 1$ for some $k$, now solve $19 \cong 42k + 21 \pmod{13}$.

Title: Solving a Linear Congruence

Body:
I've been trying to solve the following linear congruence with not much success:

$19\equiv 21x\pmod{26}$

If anyone could point me to the solution I'd be grateful, thanks in advance


问:
Why do you we say a is congruent to b mod n and not equal?
When talking about modulos why is that we use congruency and not equality?

From the accepted answer in the post What is the difference between congruency and equality? It states that:

... two figures are equal if they have the same points.

Congruent figures have the same shape and size (informally) but
possibly different points.

Which confuses me since I think 7 mod 4 results in the same value or the same point as 3.

答:
This is mostly a question of taste; but there's very, very good reason for that taste. Indeed, it's important to disambiguate things thoroughly to make sure we're correct when it concerns technical details.

Generally, the core factor where you can say "7 is equal to 3" without the fear mistaken is being in the right context, namely, being in the space $\mathbb{Z}/4\mathbb{Z}$. In this "four hour clock", $7$ and $3$ are indeed equal, but it's not the $7$ and $3$ that you know from $\mathbb{Z}$, it's one element, generally written with a hat, like so: $\hat{-1} = \hat3 = \hat 7$. The way we write this element refers to $\mathbb{Z}$, since we construct $\mathbb{Z}/4\mathbb{Z}$ from $\mathbb{Z}$; so it helps our intuition. But I could just as well forbid you from writing in a way that refers to $\mathbb{Z}$, and decide to write $\mathbb{Z}/4\mathbb{Z} = \{ 0, \alpha, \beta, \gamma \}$.

In the context of $\mathbb{Z}$ itself, $3$ and $7$ are NEVER equal. They are only EQUIVALENT, relative to the equivalence relation of congruence modulo 4. From any equivalence relation, you can make an algebraic quotient, and in the resulting space, the elements now become equal: $7$ and $3$ are both mapped to a single element $\gamma$, and their results are then equal. But you must never lose track of the space/context in which things are happening, since this is ripe for misuse and mistake.


问:
Show that $U = int(cl(U))$
Let $U \subset X$ and $V := X  \backslash  U$. Show that $U = int(cl(U))$ only if, $V = cl(int(V))$.
I found this: $cl(U)=U \cup int(cl(U))$, but i don't know if it can be used here. Is it true that $U \subset int(cl(U)) $? If it is then $U \cup int(cl(U)) = int(cl(U))$ ?. And also i found that $cl(U) = U $ if $U$ is open, but how to check if this set is open?

答:
We need the following well known lemma:

Lemma: For any $A$ we have $$int(A)=cl(A^c)^c.$$

Now
$$U=int(cl(U))= cl(cl(U)^c)^c \Leftrightarrow U^c=cl(cl(U)^c)=cl(cl(U^{cc})^c)=cl(int(U^c)).
$$

where second equality follows from Lemma applied with $A=cl(U)$, and last equality follows from Lemma applyied with  $A=U^c$.

Title: Show that $U = int(cl(U))$


问:
Theorem for some structure that it cannot be axiomatized in any language?
Godel’s theorem states that there cannot be an enumerable axiomatization of arithmetic $(\mathbb N,0,1,+)$ in first order logic. By implication, there cannot be any axiomatization in any language that is complete.

But this doesn’t mean that arithmetic cannot be axiomatized at all. E.g. we have second order Peano axioms that characterizes arithmetic up to isomorphism. It just means that we cannot derive all propositions from this axiomatization (which is possible because second-order logic is not complete).

My question is: is there a different theorem, similar in spirit to Godel’s, for some mathematical structure that can be constructed explicitly, that there cannot be an axiomatization that characterizes it up to isomorphism, in any language? (I.e. not first order, or second order, or higher order, or anything else that I don’t know about).

Intuitively, I would guess there might be, because: There is only a countable amount of possible enumerable axiomatizations in all languages combined, since there is only a countable amount of Turing machines. On the other hand there is an uncountable amount of structures. So there must be some structure that cannot be axiomatized. My question is whether there is an explicit constructable structure where we know that it cannot be.

答:
If I understand the question correctly, you can always construct a halting problem.  You can make a second-order system first order by importing the second-order concepts as first-order axioms, and then you just have the halting problem with your additional axioms.

For instance, let's say that you have a set of operations A.  You can construct a halting problem based on A.  Now, let's say that you have another set of operations B, which includes all of A plus a halting oracle for A.  This set of operations will then itself be subject to a halting problem.

I think Turing covers this in "Systems of Logic Based on Ordinals".


问:
There were $107$ people at the wedding and $11$ tables, which could accommodate $10$ people each.

There were $107$ people at the wedding and $11$ tables, which could accommodate $10$ people each. This means there were about $11^{107}$ possible seating plans.

I read that in a book but didn't really understand the solution. I mean the solution doesn't consider the fact that each table can have a max of $10$ people and it actually doesn't care about the capacity of the table. Can anyone explain it to me?

答:
$11^{107} \approx 2.7 \times 10^{111}$ is the number of ways you allocate each person to a table, not a seat.  But you could end up with more than $10$ people to a table in that count.

$\frac{110!}{10!^{11}}\approx 1.1 \times 10^{106}$ would be the way of allocating $110$ people to tables with $10$ at each table.  But there will be $110-107 =3$ empty spaces, so I think that number is perhaps $\frac{110\times 109 \times 108}{{11 \choose 1} \times 10\times 9 \times 8+{11 \choose 2}{2\choose 1}\times 10^2\times 9 +{11 \choose3}10^3 } \approx 4.76$ times too high so perhaps about $2.3 \times 10^{105}$ is the true figure.

So $11^{107}$ is too big on this basis, perhaps out by a factor of a million, but this is not that dramatic compared with other possible changes to the calculation: you would get a smaller number if individual tables did not matter, dividing by $11!\approx 4 \times 10^7$, while if you wanted to allocate seats as well as tables too you would get Ross Millikan's much higher numbers.


问:
Question about $\sum \frac{1}{1+(a_n)^2}$ and its properties
If there is a random sum $\sum a_n$ that is absolutely convergent (so $\sum |a_n|$ converges), would $\sum \frac{1}{1+(a_n)^2}$ converge as well?

Presently I have no clue how this is supposed to work, but I got some ideas. Since this is any $a_n$ that is absolutely convergent, then it is also convergent in general. If it's convergent, then it cannot be something like $2^n$ or $n^4$ and the like. So if you reciprocate it (put a fraction on it) then it most likely converges too.

答:
The answer is that $\sum \frac 1 {1 + a_n^2}$ will never converge.

The reason is that, for a series $\sum x_n$ to converge, whether absolutely or not, a necessary condition is that the limit $\lim_{n \rightarrow \infty} x_n$ is $0$.

Now if $\lim a_n$ is zero, then we would have $\lim \frac 1 {1 + a_n^2} = 1$ and hence $\sum \frac 1 {1 + a_n^2}$ doesn't converge.


问:
Show that $f''(\xi)≥8$

Let $f:[0,1]\rightarrow\mathbb{R}$ be a $\mathcal{C}^{2}([0,1])$ function such that $f(0)=f(1)=1$ and such that the minimum of $f$ is negative. Prove that there must exist a point $\xi\in [0,1]$ such that $f''(\xi)≥8$

Maybe I can use the fact that there exists maximum and minimum in $[0,1]$ so there are two critical points. But I don't know how to proceed. Can someone give me any advice? Thanks before!

答:
Assume $f'(x)<8$ for all $x\in(0,1)$.
At ist minimum point $x_\min$, we have
$$f(x_\min)<0,\qquad f'(x_\min)=0. $$
Then for $x>x_\min$,
$$ f'(x)=f'(x)-f'(x_\min)=\int_{x_\min}^xf''(t)\,\mathrm dt<8(x-x_\min)$$
whereas for $x<x_\min$,
$$ f'(x)=f'(x)-f'(x_\min)=-\int_x^{x_\min}f''(t)\,\mathrm dt>8(x-x_\min).$$
By integrating again,
$$\tag1 1<f(1)-f(x_\min)=\int_{x_\min}^1f'(x)\,\mathrm dx<8\int_{x_\min}^1(x-x_\min)\,\mathrm dx=4(1-x_\min)^2$$
and
$$\tag2 1<f(0)-f(x_\min)=-\int_0^{x_\min}f'(x)\,\mathrm dx<8\int_0^{x_\min}(x_\min-x)\,\mathrm dx=4x_\min^2.$$
Multiply $(1)$ and $(2)$ and use AM-GM inequality:
$$ 1<16\left(\sqrt{(1-x_\min)(x_\min)}\right)^4<16\left(\frac{(1-x_\min)+x_\min}2\right)^4=1,$$
contradiction.

Title: Show that $f''(\xi)≥8$


问:
Tricky Proof in Stochastic Processes/ Probability Theory
Say we have a probability space $(\Omega, \mathscr{F}, \mathbb{P}$), and $\mathbb{F} = (\mathscr{F}_t)_{t \geq 0}$ is a filtration of sub-$\sigma$-algebras of $\mathscr{F}$ satisfying the standard conditions. I've been told to consider a one-dimensional $\mathbb{F}$-Wiener process $(W_t)_{t \geq 0}$ and also the process $(X_t)_{t \geq 0}$ given by: $$X_t = f(W_t) - \frac{1}{2} \int_{0}^{t} f''(W_s) ds$$ for every $t > 0$, where $f: \mathbb{R} \mapsto \mathbb{R}$ is a $C^2$ function.

Furthermore, I've been told to assume that there exists a constant $C > 0$ such that $|f''(x)| \leq C$ for all $x \in \mathbb{R}$.

Show that $(X_t)_{t \geq 0}$ is a martingale (with respect to $\mathbb{F}$).

I've been quite stumped by this proof for a while. How could I demonstrate this idea?

答:
Applying Ito's formula to $f$, we have
\begin{align*}
f(W_t) &= f(W_0) + \int_0^t f'(W_s)dW_s + \frac 12 \int_0^t f''(W_s)ds
\end{align*}
so
\begin{align*}
X_t = f(W_t)-\frac 12 \int_0^t f''(W_s)ds &= f(W_0) + \int_0^t f'(W_s)dW_s
\end{align*}
is a (local) martingale.  To prove it is a true martingale, use the fact that $|f''(x)| \le C$ to show that
\begin{align*}
\mathbb{E}\left[\int_0^t |f'(W_s)|^2 ds \right] < \infty
\end{align*}
for all $t \ge 0$.


答: If $19 \equiv 21 x \mod 26$ then we must have $19=21x+26y$ for some integer $y$. You could use the Euclidean Algorithm to solve this problem, finding integers $s,t$ such that $\gcd(21,26) = 21s+26t$</p>

Title: Solving a Linear Congruence


问:
Evaluate $\int_{0}^{\pi} \frac{x\coth x-1}{x^2}dx$
I've been trying to evaluate certain series recently, and I found that
$$\sum_{r=1}^{\infty}\frac{1}{r}\arctan\frac{1}{r}=\frac{\pi}{2}\int_{0}^{\pi} \frac{x\coth x-1}{x^2} \, dx$$
Therefore, I would like to know how to evaluate $$\int_{0}^{\pi} \frac{x\coth x-1}{x^2} \, dx$$
The integral is approximately equal to $0.895004...$ but even Wolfram Alpha gives no closed form solution. Any ideas? Part of the problem is that the indefinite integral has no elementary antiderivative. I've tried using 'Feynman's trick' on this integral too but without success.

Thank you for your help.

答:
Another method, if you'd like:

Consider the continued fraction of $\coth x$ which is

$$\coth x=\dfrac{1}{x}\left\{1+\dfrac{x^2}{3+\dfrac{x^2}{5+\dfrac{x^2}{7+\dfrac{x^2}{\ddots}}}}\right\}$$
Considering the third iteration, that is:

$$\dfrac{1}{x}+\dfrac{x}{3+\dfrac{x^2}{5+\dfrac{x^2}{7}}}=\dfrac{1}{x}+\dfrac{5x+\dfrac{x^3}{7}}{15+\dfrac{10x^2}{7}}=\dfrac{105+45x^2+x^4}{5x(21+2x^2)}$$
we may write

$$\int_0^{\pi}\dfrac{105+45x^2+x^4}{5x^2(21+2x^2)}-\dfrac{5(21+2x^2)}{5x^2(21+2x^2)}dx= \int_0^{\pi}\dfrac{35+x^2}{105+10x^2}dx\approx 0.89\color{red}{6287}$$
which was only the third iteration.


问:
How to find the smallest side of a triangle when the interior angles are unknown?
I am confused on how to find the answer for this problem. So far what I believe would apply is the triangle inequality but I'm not sure on how to use it.

The figure $ABC$ is a triangle so as $BDC$. It is known that the
  length of $AB = 8$ inches and $\angle BAD = 2\,\angle BCD$ and
  $\angle\,DBC = 3 \angle BCD$. Find the smallest whole number that $BD$
  can attain.

The alternatives given in my book are:

$\begin{array}{ll}
1.&7\,\textrm{inches}\\
2.&4\,\textrm{inches}\\
3.&8\,\textrm{inches}\\
4.&5\,\textrm{inches}\\
5.&6\,\textrm{inches}\\
\end{array}$

So far the only thing I could come up with was to "guess" it might be $4$ since, $\angle BDA = 4\omega$ could it be that since $BD$ is opposing $2\omega$ hence that side might be $4$ inches. But that's how far I went. Needless to say that I'm unsure if that's a right way to go. Can somebody help me with this question?. Please try to include some drawing or sketch of the situation as I'd like to know what can I do to solve this problem relying mostly in geometry?.

答:
Drawing Issocoles Triangles was inspired and leads to an easy analysis

$\angle ADB = 4\omega$ as its external to Triangle $BED$

$BD$ at max lies along $AD$ when $\angle ADB =120°$

$BD$ min when $\angle ADB = 90°$

$2\omega = 45° \text{ & } AD = BD\\
8^2 = 2BD^2\\
\therefore BD = 4 \sqrt 2 ≈ 5.656$

$BD$ again nearly reaches max just before triangle collapses when $AD \implies 16 \text{ and } \angle ADB \implies 0°$

Smallest whole number is $6$


答:
$$\frac{26}{21}=1+\frac5{21}=1+\frac1{\frac{21}5}=1+\frac1{4+\frac15}$$

The previous convergent of $\frac{26}{21}$ is $1+\frac14=\frac54$

Using Convergent property (Theorem #$3$ here) of continued fraction, $21\cdot5-26\cdot4=1\implies 21^{-1}\equiv 5\pmod {26} $

Title: Solving a Linear Congruence


问:
Is the union of all $l^p$ spaces meagre in $l^\infty$?
Is the union of all $l^p$ spaces meagre in $l^\infty$? i.e. is $ \bigcup_{p=1}^\infty l^p$ meagre?

I am revisiting this variety of math after a long break so help is appreciated. Please correct anything I get wrong too.

I know that a space is meagre if it can be written as a countable union of nowhere dense sets, and that a set is nowhere dense if it is closed and has no interior points.

My attempts/ideas:

(1) I seem to remember that $l^1 \subset l^2 \subset l^3 \subset l^4 \subset ...\subset l^\infty$, but maybe I'm wrong about this? I can't seem to find it anywhere or prove it either. Is this true? If this were the case then it would be enough to show that $\bigcup_{p=1}^\infty l^p=l^\infty$ is meagre. Then I'm stuck again... how do I show that $l^\infty$ is meagre?

(2) There is already a countable union (of $l^p$s) here (do I have to show that it is countable? This is obvious isn't it?), so that makes me think that it would be good if I could show that each $l^p$ is nowhere dense. How could I do this?

Any critiques/ideas?

答:
Complementing Jonas answer, the sequence with terms $ x_k = \dfrac{1}{\ln(k+1)}$ for all $k\geq 1$ is in $c_0$, but it's not in any of the $l^{p}$ sets. Suppose, by contradiction, that $(x_k)_{k\in\mathbb{N}} \in l^{p}$. The idea to prove it is using Hölder's inequality and get the inequality, for each $k\in\mathbb{N}$,
$$\sum^{k}_{i=1} \dfrac{1}{ (1+k) \ln(k+1)}\leq \left(\sum^{k}_{i=1} \dfrac{1}{ (1+k)^{q}}\right)^{1/q}\left( \sum^{k}_{i=1} \dfrac{1}{\ln(k+1)^p} \right)^{1/p}$$ with $\dfrac{1}{p}+\dfrac{1}{q}=1$, which would, erroneously, mean that the sequence of terms $ y_k = \dfrac{1}{(k+1)\ln(k+1)}$ belongs to $l^{1}.$

The argument to derive that $(y_k)_{k\in\mathbb{N}}$ is not in $l^{1}$ is to realize that, for each $k\in\mathbb{N}$,
$$
\int^{k}_{1} \dfrac{1}{(\tau+1)\ln(\tau+1)} d\tau = \ln(\ln(k+1))- \ln(\ln(2)) \leq \sum^{k}_{i=1} \dfrac{1}{(k+1)\ln(k+1)},
$$ using the integral test for convergence.


问:
Is gradient descent nothing other than discretized gradient flow?
I can find definitions of the gradient flow of a scalar field $f$ as
$$\frac{d \xi}{dt} = - \nabla_\xi f$$
in here and here.

Gradient descent can be used to find a minimum in $f$ and can be written as
$$\xi_{i+1} = \xi_{i} - \lambda \nabla_\xi f,$$
where $\lambda$ is a constant scalar.

As far as I understand, gradient descent is nothing other than discretized gradient flow. So I set $\lambda = 1$ to obtain
$$\frac{d \xi}{dt} = \lim_{\Delta t \rightarrow 0} \frac{\xi_{i+1} - \xi_i}{\Delta t} = - \lim_{\Delta t \rightarrow 0} \frac{\nabla_\xi f}{\Delta t} = - \frac{d}{dt} \nabla_\xi f \stackrel{???}{=} - \nabla_\xi f = \frac{d \xi}{dt}.$$

This does not make much sense to me... Shouldn’t both $\frac{d \xi}{dt}$ give the same? What mistake am I making? Could anyone help me, please?

答:
I think the confusion might stem from the ambiguity in notation. The definition of gradient flow is $\frac{d \xi}{dt}(t_0) = - \nabla f(\xi(t_0))$, where the right-hand side is the gradient of $f$ evaluated at $\xi(t_0)$. (This is closer to the notation used in the second link.)

Further, the discrete step is $\xi_{i+1} = \xi_i - \lambda \nabla f(\xi_i)$. This is simply a rearrangement of $$-\nabla f(\xi(t_i)) = \frac{d\xi}{dt}(t_i) \approx \frac{\xi(t_i + \lambda) - \xi(t_i)}{\lambda},$$
where $t_i$ is such that $\xi_i = \xi(t_i)$.
In particular, $\lambda$ is the parameter that should be driven to zero to see the asymptotic correspondence with gradient flow; you cannot just set it at $\lambda=1$.


答:
What you are missing is that $\lambda$ "is" the time scale $\Delta t$. So you should rather interpret the first formula as
$$\xi_{t+\Delta t}:=\xi_t - (\Delta t)\nabla_{\xi_t}f,$$
and with this you get what you want.

Title: Is gradient descent nothing other than discretized gradient flow?

Body:
I can find definitions of the gradient flow of a scalar field $f$ as
$$\frac{d \xi}{dt} = - \nabla_\xi f$$
in here and here.

Gradient descent can be used to find a minimum in $f$ and can be written as
$$\xi_{i+1} = \xi_{i} - \lambda \nabla_\xi f,$$
where $\lambda$ is a constant scalar.

As far as I understand, gradient descent is nothing other than discretized gradient flow. So I set $\lambda = 1$ to obtain
$$\frac{d \xi}{dt} = \lim_{\Delta t \rightarrow 0} \frac{\xi_{i+1} - \xi_i}{\Delta t} = - \lim_{\Delta t \rightarrow 0} \frac{\nabla_\xi f}{\Delta t} = - \frac{d}{dt} \nabla_\xi f \stackrel{???}{=} - \nabla_\xi f = \frac{d \xi}{dt}.$$

This does not make much sense to me... Shouldn’t both $\frac{d \xi}{dt}$ give the same? What mistake am I making? Could anyone help me, please?


问:
$\displaystyle\lim_{z\to 0} |zf(z)| = 0$ implies $z = 0$ is a removable singularity.
Given that $f$ is analytic around $z = 0$ and that $\displaystyle\lim_{z\to 0} |zf(z)| = 0$, we can conclude that $f(z)$ has a removable singularity at $z = 0$.

I think that we should deduce that $f(z)$ is bounded around $z = 0$ and use Riemann to conclude that it has a removable singularity. However, I have no idea how to show that this must be bounded. Am I on the right track?

答:
Let $g(z) = zf(z)$. By assumption $g$ is bounded near $z=0$, so $g$ has a removable singularity at $z=0$. Hence
$$f(z) = \frac{g(z)}{z},$$
on a punctured neighbourhood of $z=0$. Since $g(0) = 0$, it follows that $g(z) = zh(z)$ for some analytic function $h$, i.e.
$$f = h$$
on a punctured neighbourhood of $z=0$, which shows that $f$ is bounded near $0$.</p>


问:
Calculate volume of intersection two non-aligned cuboids
I'm on the lookout to finding an algorithm that calculates the volume of the intersection between two cuboids. Most answers so far assume all the axes of the two cuboids in question are nicely aligned. Not the case in my problem. See figure below

Finding out what corners of the two cuboid are in each others space is not that difficult. Also determining some boundary points on the cuboids faces shouldn't be problematic. The hard part where I get lost is to determine what kind of shape the intersection actually is and how to calculate it's volume. I think it's safe to assume the intersection will be convex. Any suggestions?

答:
The intersection of convex sets is always convex.  Divide the intersection into disjoint tetrahedra and use the well-known formula for the volume of a tetrahedron, given the vertices.  It seems to me that you should be able to do this by selecting any point in the interior of the intersection to be the apex of all the tetrahedra.  Then triangulate each face of the intersection.  The vertices of the triangles, together with the distinguished interior point, give all the tetrahedra.

I haven't really thought about how to program this, but it doesn't sound very hard.


问:
Can you find the determinant of a matrix given its row echelon form?
I believe I stumbled on a resource for quickly computing the determinant of a matrix by converting it to row echelon form then acting upon the diagonal elements, either they're sum or product. However, I cannot find the specific link/resource, and so I'm no longer sure that this is a valid way of reaching the determinant.

Edit: B is an upper-triangular matrix, so detB is the product of its diagonal entries so perhaps my question should be, is it better/easier to just use the upper triangular matrix than to work with row echelon form?

答:
You need to keep track of how many row swaps you have done because this multiplies the determinant by $-1$.

You can use elementary row operation matrices. The ones that correspond to adding/subtracting a row to another one have determinant one. Swapping rows has determinant $-1$. Hence, your determinant is the determinant of the matrix in row echelon from multiplied by $(-1)^n$, where $n$ is the number of row swaps you have done.

EDIT: I was using the wrong definition of row echelon form. One also has to keep track of the values you get in the diagonal before you divide to get $1$, and multiply it at the end as well.


问:
Dominated convergence theorem for stochastic processes integrated with respect to Lebesgue measure
I consider a filtered probability space $(\Omega, \mathcal{F}, P)$ with filtration $(\mathcal{F}_t)$ on which $(X_t)$ is an $(\mathcal{F}_t)$-adapted stochastic process. In addition to this, $(X^n_t)$ is a sequence of stochastic processes where for each $n$, $(X_t^n)$ is  $(\mathcal{F}_t)$-adapted. I want to show that for every fixed $t \geq 0$
$$
\lim_{n \to \infty} \int_0^t \vert X_s^n -X_s \vert^2 ds = 0, \quad \text{P-a.s}
$$
My questions is, does their exist a dominated convergence result which can be used in this context? More precisely, is it enough if I can show that for every fixed $t \geq 0$
$$
 \lim_{n \to \infty} \vert X_t^n - X_t \vert^2  = 0, \quad \text{P-a.s}
$$
and for every $t \geq 0$ there exist a stochastic variable $Y_t$ with $E[\vert Y_t \vert ]< \infty$ such that for every $n$
$$
\vert X_t^n - X_t \vert^2 \leq Y_t \quad \text{P-a.s}
$$
I'm looking for a reference.

答:
unfortunately I don't know a reference, but I can give you a simple proof

I think we need an additional hypothesis: $X^n,X$ need to have a.s. càdlàg (or càglàd) paths for all $n$. For simlicity we assume it to hold for all $\omega\in\Omega$ and not only a.s. (otherwise intersect all the following sets with an appropriate set). Another additional hypothesis would be to hold your two equation/inequation not only for all $t\geq0$ a.s., but a.s. for all $t\geq0$ (that's a significant difference!). The proof is very similar of the given one below.

By assumption for every $t\geq0$ there is a set $A_t\subset\Omega$, $P(A_t^\complement)=0$ s.t.
$$\forall \omega\in A_t: \lim_{n\to\infty}|X^n_t(\omega)-X_t(\omega)|^2=0\tag{1}$$ and a set $B_t\subset\Omega$, $P(B_t^\complement)=0$ s.t.
$$\forall \omega\in B_t\forall n\in\Bbb N: |X^n_t(\omega)-X_t(\omega)|^2\leq Y_t(\omega).\tag{2}$$ Now consider the set
$$M:=\bigcap_{t\in\Bbb Q_{\geq0}}(A_t\cap B_t).
$$
Note: as $\left(A\cap B\right)^{\complement}=A^{\complement}\cup B^{\complement}$ and we have a countable intersection, $P(M^\complement)=0$ holds.

The trick is now to consider the integral path-wise. Let $\omega\in M$ be fixed. As the processes are càdlàg (or càglàd) there is only a countable set $S\subset\Bbb R$ on which $X^n(\omega),X(\omega)$ $(n\in\Bbb N)$ have jumps. So consider function $f_n(t):=|X^n_t(\omega)-X_t(\omega)|^2\cdot1_{S^\complement}(t)$ instead of $|X^n_t(\omega)-X_t(\omega)|^2$. By continuity on $S^\complement $ the properties $(1)$ and $(2)$ hold for $f_n$ for all $t$ and not only on $\Bbb Q_{\geq0} $. For $f_n$ you can use the 'traditional' dominated convergence theorem for
$$\lim_{n\to\infty}\int_0^tf_n(s)\Bbb ds=0$$
and as $S$ is a Lebesgue null-set $\int_0^tf_n(s)\Bbb ds=\int_0^t \vert X_s^n(\omega) -X_s(\omega) \vert^2\Bbb ds$ holds.

Alltogether we found a set $M$, $P(M^\complement)=0$ s.t. for all $\omega\in M$
$$\lim_{n\to\infty}\int_0^t\vert X_s^n(\omega) -X_s(\omega) \vert^2\Bbb ds=0$$
holds.


问:
Is a weakly differentiable, $L^{2}$ function over $\mathbb{R}_{+}$, bounded?
Consider $u \in W^{1,2}(\mathbb{R}_{+})$. Is there always a bounded representative of $u$?

I have come across this sort of question multiple times, but I just cannot wrap my head around it. This is connected, at least from my point of view and from my experience, to this more generic question: given a function $u \in L^p(\mathbb{R})$, for $p \neq \infty$, under which additional hypothesis (if any) is $u$ bounded?

I have often found myself with integrability hypotheses over unbounded sets such as $\mathbb{R}$ and have hoped to deduce some boundedness, but I can't think clearly about it.

答:
We have the continuity of the embedding $W^{1,2}(0,1)$ into $C([0,1])$. That  is there is $c>0$ such that for all $u$
$$
\|u\|_{L^\infty(0,1)} \le c \|u\|_{W^{1,2}(0,1)} .
$$
This is true for all intervals of length $1$, so we can sum this up:
$$
\sum_{n=-\infty}^\infty \|u\|_{L^\infty(n,n+1)}^2 \le c^2 \sum_{n=-\infty}^\infty \|u\|_{W^{1,2}(n,n+1)}^2 = c^2 \|u\|_{W^{1,2}(\mathbb R)}^2.
$$
Hence, the series on the left is converging, in particular $\|u\|_{L^\infty(\mathbb R)} \le  c \|u\|_{W^{1,2}(\mathbb R)}$.
Moreover, the series tells us that
$$
\lim_{|n|\to\infty}\|u\|_{L^\infty(n,n+1)}=0,
$$
i.e., $u$ kind of vanishes at infinity.


问:
Characterization of line graphs
I'm confused about how you to recognize line graphs.

From the Wikipedia page on the subject, I read that

A graph is a line graph if and only if it does not contain one of
  these nine graphs as an induced subgraph:

Which explains why this graph is not a line graph as it induces a claw, which is one of the forbidden subgraph.

However here's a graph and its subgraph (in green). This line graph also contains a claw, if you look at the vertex (4,3) which is adjacent to (1,3), (1,4) and (4,5).

What am I missing?

答: When we say induced subgraph, we mean you choose some vertices and take all edges between them. So when you choose (4,3), (1,3), (1,4), and (4,5) you get a diamond (with an edge across its middle), not the claw.

Title: Characterization of line graphs


问:
How does Cramer's rule work?
I know Cramer's rule works for 3 linear equations. I know all steps to get solutions. But I don't know why (how) Cramer's rule gives us solutions?

Why do we get $x=\frac{\Delta_1}\Delta$ and $y$ and $z$ in the same way?

I want to know how these steps give us solutions?

答:
Let $ A $ be an $ n \times n $ invertible matrix, and $ b \in \mathbb{R}^{n} $. We're looking for the $ x \in \mathbb{R}^n $ such that $ Ax = b $.

Let's write columns of $ A $ as $ A_1, \ldots, A_n $ (since $ A $ is invertible, these form a basis of $ \mathbb{R}^n$). Now $ Ax = b $ becomes $ x_1 A_1 + \ldots + x_n A_n = b $. Notice $ x_k $ is that scalar $ t $ such that $ b - t A_k $ lies in the span of $ \{ A_i : i \neq k \} $ [ We can visualise this for $ n \leq 3 $ ].

So $ \det(A_1, \ldots, b - x_k A_k, \ldots, A_n) = 0 $, from which $ x_k = \frac{\det(A_1, \, \ldots \, , \, b \, , \, \ldots \, , A_n)}{\det(A_1, \, \ldots \, , A_n)} $ [ Here the $ b $ in numerator appears in $k^{th}$ position ].

Title: How does Cramer's rule work?


问:
Maximize $f(x,y) = x^{2} - y^{2}$ subject to $g(x,y) = 1 - x^{2} - y^{2} = 0$
$$\begin{array}{ll} \text{maximize} & f(x,y) := x^{2} - y^{2}\\ \text{subject to} & g(x,y) := 1 - x^{2} - y^{2} = 0\end{array}$$

I tried solving this question using Lagrange multiplier and I had set up the following Lagrangean,

$$ L = x^{2} - y^{2} - \lambda (1 - x^{2} - y^{2}) $$

I got the following first-order conditions:

$$\frac{\partial L}{\partial x} = x +  \lambda x \\ \frac{\partial L}{\partial y} = \lambda y - y $$

I am not sure how to solve after this.
When I equate the FOC to 0 I get x= 0 and y=0 but if I keep x= 1 and y = 0  a larger value for f(x,y) is obtained which still satisifes the constraint.

答:
As your constraint is unit circle $x^2+y^2=1$, it is easy to see that you have maxima  when the objective function is a hyperbola $x^2 - y^2 = 1$ tangent to the circle and you have minima when it is hyperbola $x^2-y^2 = -1$, again tangent to the circle.

As far as Lagrange Multiplier method, you have

$x(\lambda+1) = 0$

$y(\lambda - 1) = 0$

$1 - x^2 - y^2 = 0$

From first equation, $\lambda = -1$ is a possible solution. Plugging into second, $y = 0$ and from third, $x = \pm1$.

Also from second, $\lambda = 1$ is a possible solution. That gives you $x = 0$ and $y = \pm1$.

That leads to maxima of $f(x, y) = x^2-y^2$ as $1$ and minima as $-1$.

You also have $x = 0$ from the first equation and $y = 0$ from the second but they both lead to the same set of solutions.


问:
Stopping time for irreducible recurrent finite Markov chains in continuous time
Let $(X_t)_t$ be a continuous-time Markov chain with values in a finite set $E$.

Assume that $\forall x,y \in E, \forall t > 0, \quad \mathbb{P}(X_t = y | X_0 = x) > 0$.

How do you prove that $\forall x,y \in E, \quad \mathbb{E}[\inf \{ t > 0, X_t = y \} | X_0 = x ] < +\infty$?

The method I found consists in considering the discrete-time Markov chain $(X_n)_{n \in \mathbb{N}}$ but there should be a continuous-time method.

答:
I write $\Bbb E^x$ and $\Bbb P^x$ to indicate the initial condition $X_0=x$.

Define $u(x):=\Bbb E^x\left[\int_0^\infty e^{-t} 1_{\{X_t=y\}}dt\right]$. Then $\delta:=\inf\{u(x):x\in E\}>0$. Choose $b>0$ so large that $e^{-b}<\delta/2$. Writing $T_y:=\inf\{t>0:X_t=y\}$, you have
$$
\eqalign{
v(x) &\le\Bbb E^x\left[\int_0^b e^{-t}1_{\{X_t=y\}}dt\right]+e^{-b}\cr
&=\Bbb E^x\left[\int_0^b e^{-t}1_{\{X_t=y\}}1_{\{T_y\le b\}}dt\right]+e^{-b}\cr
&\le\Bbb E^x\left[\int_0^b e^{-t}1_{\{T_y\le b\}}dt\right]+e^{-b}\cr
&=(1-e^{-b})\Bbb P^x\left[T_y\le b\right]+e^{-b}\cr
&\le\Bbb P^x\left[T_y\le b\right]+e^{-b}\cr
}
$$
It follows that
$$
\Bbb P^x[T_y>b]\le 1-\delta/2,\qquad\forall x.
$$
Noting that $\{T_y>2b\} =\{T_y>b\}\cap\{\inf\{t>b:X_t=y\}>b\}$, and using the Markov property at time $b$:
$$
\eqalign{
\Bbb P^x[T_y>2b]
&=\Bbb E^x\left[ 1_{\{T_y>b\}}\Bbb P^{X_b}[T_y>b]\right]\cr
&\le\Bbb E^x\left[ 1_{\{T_y>b\}}\cdot(1-\delta/2)\right]\cr
&\le(1-\delta/2)^2.
}
$$
Repeating this you see that
$$
\Bbb P^x[T_y>nb]\le(1-\delta/2)^n,\qquad n=1,2,\ldots, x\in E.
$$
It follows easily from this that there is a constant $C>0$ such that
$$
\Bbb P^x[T_y>t]\le Ce^{-\rho t},\qquad t>0, x\in E,
$$
where $\rho:=-[\log(1-\delta/2)]/b$. This implies that $T_y$ has finite moments of all orders; indeed
$$
\Bbb E^x[\exp(\alpha T_y)]<\infty,\qquad x\in E,
$$
provided $\alpha<\rho$.


问:
How can I find the surface area of a normal chicken egg?
This morning, I had eggs for breakfast, and I was looking at the pieces of broken shells and thought "What is the surface area of this egg?" The problem is that I have no real idea about how to find the surface area.

I have learned formulas for circles, and I know the equation for an ellipse; however, I don't know how to apply that.

The only idea I can think of is to put an egg on a sheet of paper and trace it, and then measure the outline drawn, and then try to find an equation for that ellipse and rotate that about the $x$-axis.  Now, my problem is how I can find the equation of the ellipse from the graph, and will my tracing method really be the edge of the egg?  Also, can I use the standard surface area integral? Will I have to use some techniques to solve the integral that are not covered in the AP BC Calculus?

There has to be a better method for finding the surface area. Please, help me understand how to find the surface area of an egg; i.e., how to use my mathematical knowledge for something other than passing exams.

答:
Joining the frenzy of ways to do it, here is a physics-lab-like method:

Take some millimeter paper

Break the shell in almost-flat pieces and paint them with ink (sink it in ink).

With some tweezers take the pieces, make sure there is no spilling ink, and put them on the millimeter paper, ink face down, with any comfortable distance between pieces

Cover the paper with another piece of paper and press smoothly to even out the rounded corners of the shell

let the ink dry, take the shells and count the squares covered with ink.

There will be some approximation, given that it's hard to estimate half-squares, but I don't see why it's worse than approximating the thickness of a shell as being uniform.

EDIT

This may be formulated as hands-on, distructive "triangulation". Though it may not fit the accurate mathematical definition of the notion.


问:
Given $n \in N$, prove that $f: [0, \infty) \rightarrow [0, \infty) f(x) = x^n$ is increasing without calculus (derivates)
I have to prove this using $x_1 < x_2 \rightarrow f(x_1) < f(x_2)$.

I have no idea on how to do this, but what I have tried so far is:

Consider $x_1, x_2  \in [0, \infty)$ such that $x_1 < x_2 \rightarrow f(x_1) < f(x_2)$

$\rightarrow x_1^n < x_2^n$

I do not know what to do next

Body:
You have

$$x_2^n - x_1^n = (x_2-x_1) (x_2^{n-1} + x_2^{n-2} x_1 + \cdots + x_2 x_1^{n-2}+ x_1^{n-1}) \gt 0$$ for $x_2 \gt x_1$.


问:
Clarifying Polygonal Paths in Complex Analysis
Introduction to Complex Analysis by H. A. Priestley (Revised ed.) shows a diagram (page 40) which is a polygonal path in a set $G$ (link below). Their proof claims we can polygonally pave our way from $z$ to $a$ by using the fact that open discs centred at $z$ enclose a point $w$ such that the line segment $[z,w]\subset D(z;r),r>0$. I am confident in that notion, but maybe it's their diagram which confuses me.

It displays large line segments paving through the set $G$ which if you were to draw open discs centred from the original point, the open disc is then not contained by $G$. I assume they could have used smaller open discs, therefore, more $w$s to get to $a$.

Any clarification on this topic would help me out a lot. Thank you very much.

答:
H. A. Priestley, Introduction to Complex Analysis, Revised Edition (Oxford University Press 1990), p.39f.:

Let $G_1 = \{z \in G \colon\ \text{there exists a polygonal path in } G \text{ with endpoints } a \text{ and } z\}$ and let $G_2 = G \setminus G_1.$ We require $G_1 = G.$ We shall prove that each of $G_1$ and $G_2$ is open. Connectedness of $G$ will then imply that one of these sets is empty. This cannot be $G_1,$ since $a \in G_1.$

We now establish our claim that $G_1$ and $G_2$ are open. For any $z \in G,$ we can find $r$ such that $D(z; r) \subseteq G.$ For each $w \in D(z; r),$ $[z, w] \subseteq G.$ It follows that $z$ can be joined to $a$ by a polygonal path in $G$ if and only if $w$ can be (see Fig. 3.3). Hence, for $k = 1, 2,$ $z \in G_k$ implies $D(z;r) \subseteq G_k.$

The diagram illustrates the implication that if $z \in G_1$ then $w \in G_1.$ A similar diagram could be drawn to illustrate the converse implication that if $w \in G_1$ then $z \in G_1;$ but that would be overkill. Instead, just imagine that the penultimate segment of the polygonal path from $a$ terminates at $w$ rather than $z.$


问:
The biggest $n$ so the complementary of any subset with $n$ elements of $A=\{1, 2, 3, …, 2003\}$ has at least one pair of consecutive numbers
I am asked to determine the biggest natural number $n$ such that the complementary of any subset with $n$ elements of the set $A=\{1, 2, 3, …, 2003\}$ contains at least one pair of consecutive numbers.

My progress so far:

I know that the number of susets is $2^{2003}$. We may ask: how many subsets of ${1,2,...,2003}$ have no two consecutive numbers ?
Well, the subsets are interpreted as $n$-words from the alphabet ${0,1}$. Let an be the number of words with no consecutive ones. Then, a word can start from $0$ and proceed in $a_{n−1}$ ways or start with $10$ and proceed in $an−2$ ways. Therefore, $an=a_{n−1}+a_{n−2}$, $a_1=2$,$a_2=3$. So, $a_n$=$F_{n+2}$.
So, in our case, we have $F(2005)$. However, this still does not answer our question.

I saw Largest value of $n$ such that the complementary set of any $n$-element subset contains at least two elements that are coprime. Does it apply here?
Any help, please?

答:
The answer comes from insight, not calculation. There are sufficient hints in the comments, but to flesh those ideas out:

$A$ contains $1001$ even numbers and $1002$ odd numbers. If you remove all of the even numbers, the remaining subset of $1002$ odd numbers has no two consecutive numbers. If you add back one even number, it falls between two odd numbers, so it creates two distinct pairs of consecutive numbers: $2n-1,2n$ and $2n,2n+1$. So removing one odd number in an attempt to offset the added even number is insufficient to remove all pairs of consecutive numbers. In general, you have to remove at least one more odd number than the number of even numbers you restore in order to avoid consecutive pairs.

So a subset with $1002$ odd numbers plus $1$ even number (total of $1003$ elements) must have at least one pair of consecutive numbers. Therefore, the largest $n$ is $2003-1003=1000$. That is to say, you can remove $n=1001$ or more elements in some manner that leaves all non-consecutive numbers. If you remove fewer than $1001$ elements, there will always be at least $1003$ remaining elements, which perforce must feature at least one pair of consecutive numbers.


答:
Say $x,y \in [0,\infty) $ and $x\lt y$. Then $$f(x) \lt f(y) \\ \iff x^n \lt y^n \\ $$

Since $\ln x$ is an increasing function, $$\iff n\ln x \lt n \ln y \\ \iff \ln x \lt \ln y \\ \iff x\lt y $$

But that’s of course true.

Title: Given $n \in N$, prove that $f: [0, \infty) \rightarrow [0, \infty) f(x) = x^n$ is increasing without calculus (derivates)

Body:
I have to prove this using $x_1 < x_2 \rightarrow f(x_1) < f(x_2)$.

I have no idea on how to do this, but what I have tried so far is:

Consider $x_1, x_2  \in [0, \infty)$ such that $x_1 < x_2 \rightarrow f(x_1) < f(x_2)$

$\rightarrow x_1^n < x_2^n$

I do not know what to do next


问:
Constructing self-complementary graphs
How does one go about systematically constructing a self-complementary graph, on say 8 vertices?

[Added: Maybe everyone else knows this already, but I had to look up my guess to be sure it was correct: a self-complementary graph is a simple graph which is isomorphic to its complement.  --PLC]

答: Systematically is easy; systematically and efficiently, I don't know. It's easy to work out how many edges such a graph must have, that's a start. There's also some information at http://oeis.org/A000171</p>

Title: Constructing self-complementary graphs


问:
Question related to Erdös-Szekeres Happy Ending problem
After reading the papers of Erdös and Szekeres regarding the "Happy Ending problem", I am wondering if there is some research / paper regarding the existence of a given number of points $F_2(n)$ in general position such that every subset of $n$ points is a convex n-gon.

For instance, $F_2(3)=3$, as every set of three points forms a triangle, which is always convex. I am not sure if $F_2(n>3)$ does even exist, I have not found an example of such a set of points. I guess that it does not exist, but I am not sure of how it would be possible to prove it.

If we generalize the question to three dimensions, regarding the existence of a given number of points $F_3(n)$ in general position such that every subset of $n$ points is a convex polyhedron, I get that $F_3(4)=4$, because every set of four points in general position in three dimensions form a convex polyhedron.

Thus, it can be conjectured that, for a given $d$ number of dimensions, the only given number of points $F_d(n)$ such that every subset of $n$ points is a given convex polytope is $F_d(n=d+1)=d+1$. Also, it seems trivial that necessarily $F_d(n<d+1)$ does not exist, as it is impossible to form any convex $d$-polytope with $d$ or less points.

If you could provide some research / paper regarding this question, on how to prove the conjecture, or provide an example of some $F_d(n\neq d+1)$, it would be welcomed.

Thanks in advance!

答:
It seems to me that the answer is no: in the plane take three points not on a line and a fourth point inside the triangle. No matter how many points you add, the initial four points do not form a convex 4-gon.

This generalizes to all dimensions.


Title: Maximize $f(x,y) = x^{2} - y^{2}$ subject to $g(x,y) = 1 - x^{2} - y^{2} = 0$

Body:
$$\begin{array}{ll} \text{maximize} & f(x,y) := x^{2} - y^{2}\\ \text{subject to} & g(x,y) := 1 - x^{2} - y^{2} = 0\end{array}$$

I tried solving this question using Lagrange multiplier and I had set up the following Lagrangean,

$$ L = x^{2} - y^{2} - \lambda (1 - x^{2} - y^{2}) $$

I got the following first-order conditions:

$$\frac{\partial L}{\partial x} = x +  \lambda x \\ \frac{\partial L}{\partial y} = \lambda y - y $$

I am not sure how to solve after this.
When I equate the FOC to 0 I get x= 0 and y=0 but if I keep x= 1 and y = 0  a larger value for f(x,y) is obtained which still satisifes the constraint.

Body:
Elementary option

Constraint: $y^2=1-x^2;$

$f(x,y)=x^2-y^2=2x^2-1;$

1)Min: at $x=0,$ and $y=\pm1:$

$f_{min}(x,y)=-1;$

2)Max: at $x=\pm1$ and $y=0:$

$f_{max}(x,y)=1.$


问:
Absolute convergence of $\sum\limits_{k=0}^{\infty}2^{-(k+1)}\sum\limits_{j=0}^{k}2^ja_j$

Let be $\sum\limits_{n=0}^{\infty}a_n$ an absolutely convergent series and $b_k:=2^{-(k+1)}\sum\limits_{j=0}^{k}2^ja_j$. Show that $\sum\limits_{k=0}^{\infty}b_k$ converges absolutely.

答:
I just want to share another answer which uses Cauchy's product rule (https://en.wikipedia.org/wiki/Cauchy_product#:~:text=In%20mathematics%2C%20more%20specifically%20in,French%20mathematician%20Augustin%20Louis%20Cauchy.).

We know that $\sum\limits_{j=0}^{\infty}a_j$ and $\sum\limits_{k=0}^{\infty}2^{-k-1}$ are both absolute convergent. Therefore we are allowed to apply Cauchy's product rule:

$1\cdot\sum\limits_{j=0}^{\infty}a_j=\sum\limits_{k=0}^{\infty}2^{-k-1}\sum\limits_{j=0}^{\infty}a_j=\sum\limits_{k=0}^{\infty}\sum\limits_{j=0}^{k}a_j2^{-(k-j)-1}=\sum\limits_{k=0}^{\infty}\sum\limits_{j=0}^{k}2^{j-(k+1)}a_j=\sum\limits_{k=0}^{\infty}b_k$.

Hence, $\sum\limits_{k=0}^{\infty}b_k$ converges absolutely and $\sum\limits_{k=0}^{\infty}b_k=\sum\limits_{j=0}^{\infty}a_j$.


问:
Find the least $n$ such that $f(n)/g(n)=4/7$

Let $f(n)$ and $g(n)$ be functions satisfying $$f(n)=\cases{\sqrt{n} \hspace{50pt} \sqrt{n} \in \mathbb{N} \\ 1+f(n+1) \hspace{14pt} \text{otherwise}} $$
$$g(n)=\cases{\sqrt{n} \hspace{50pt} \sqrt{n} \in \mathbb{N} \\ 2+g(n+2) \hspace{14pt} \text{otherwise}}$$
for positive integers $n$. Find the least positive value of $n$ such that $\dfrac{f(n)}{g(n)}=\dfrac47$.

I found the values for $f$ and $g$ till $n=9$, I noticed that $f(n)=g(n)$ happened at $n=1,2,4,5,7,9$ and for some values it also happened that if $f(a)=b$, then $f(b)=a$. I don't know if that is useful in anyway. I can't see how to approach the problem.

Please give some hints.

Body:
I can not believe the number of arithmetic errors I made.  All inexcusable. but.

Let $n > 0$ so there is a $k$ so that $(k-1)^2 < k \le n^2$ and let $j = k^2 - n$.

Then $$f(n) = f(k^2-j)= 1 + f(k^2-j+1) = 2+f(k^2 - j + 2) =.... =j + f(k^2 -j + j) = j + f(k^2) = j + k$$. (That holds if $j=0$.)

Now similarly if $j$ is even then $g(n) = j +k$ as well.  But if $j$ is odd then.

$$g(k^2 - j) = (j-1) + g(k^2 - 1) = (j+1) + g(k^2 + 1) = (j + 2k + 1) + g(k^2+ 2k + 1) = (j+2k +1) + g((k+1)^2) = (j+2k +1) + (k+1) = j + 3k + 2$$.

So if $\frac {f(n)}{g(n)} \ne 1$ we must have $j$ is odd and  $\frac {f(n)}{g(n)} = \frac {j+k}{j + 3k + 2}$.

So we need $\frac {j+k}{j + 3k + 2} = \frac {4M}{7M}$ and

$j + k = 4M; j + 3k + 2 = 7M$ and $j$ is odd.  And ... arithmetic the bane of mathematicians.

$j= 4M - k$ (so $k$ is odd)

$4M - k + 3k + 2 = 4m + 2k + 2 = 7M$ so $k = \frac {3M-2}2$.

A bit of manipulation we can see than $M$ is divisible by $4$ as $k$ is odd. ($\frac {3M-2}2 = 2w -1\implies 3M=4m$).

Plugging in $M=4w$ we get

$k = \frac {12w-2}2 = 6w -1$ and $j = 16M -k = 10w + 1$.

But we must also have that $j < k^2 - (k+1)^2 = 2k-1$ so $10w + 1 < 12w -3$ so $w > 2$.

Now $w = 3$ and we have $k=17$ and $j = 31$ and $n = 17^2 - 31 > 17^2 -34 +1 = 16^2$.

And $f(17^2 - 31) = 31 + f(17^2) = 48 = 4\cdot 12$.

And $$g(17^2 -31) = 30 + g(17^2 -1)=32 + g(17^2+1)=32+34 + g(17^2 + 34 + 1)=66+g(18^2) = 84 = 7\cdot 12$$.

So $n = 17^2 - 31 = 258$ is the smallest such $n$.

(To minimimize $(k-1)^2 < n=k^2 -j < k^2$ where $n = k^2 -j=(6w-1)^2-(10w + 1)=36w^2 -20w=4w(6w-5)$ we must minimize $w$.)


问:
$\sum_{n=1}^\infty 1/n^{1+\epsilon}$ Converges for all $\epsilon >0$
If $\epsilon>0$, Prove that the sum$$\sum_{n=1}^\infty \frac{1}{n^{1+\epsilon}}$$ converges.

This problem is from my calculus book, and Since this problem follows some other problems like: Show that $\sum_{n=1}^\infty 1/n^2$ And $\sum_{n=1}^\infty 1/n^3$ converges, i tried to use the same method that i used to solve these problems, but it didn’t work.

The method is very simple, all what you have to do is to find a series that you know that it definitely converges and you also know that this series Is less or equal to your series.

The series i used to prove that the above series converges is $$\sum_{n=0}^\infty \frac{1}{2^n}$$
But again it didn’t help me in this $\sum_{n=1}^\infty \frac{1}{n^{1+\epsilon}}$

答: Since this is a decreasing positive sequence, the Cauchy Condensation test is applicable. The condensed series becomes $\displaystyle\sum_{n=1}^\infty 2^{-n\epsilon}$ in this case, which is a GP with common ratio less than $1$.


问:
Decide whether statement (3) is true if statements (1) and (2) are both true.
I am trying to solve the following problem from the book Reading, Writing, and Proving: A Closer Look at Mathematics 2nd ed. on page 44 Problem 4.20)b)

Decide whether statement $(3)$ is true if statements $(1)$ and $(2)$ are both true. Give reasons for your answers.

The three statements are:

$(1)$ If Susie goes to the ball in the red dress, I will stay home.

$(2)$ Susie went to the ball in the green dress.

$(3)$ I did not stay home.

The following is my attempted solution:

The universe for all variables is the set of all humans.

Let $B(x)$ denotes the statement: $x$ goes to the ball.

Let $R(x)$ denotes the statement: $x$ wears a red dress.

Let $S(x)$ denotes the statement: $x$ stays home.

Each of the statement $(1),(2)$ is expressed symbolically as:

\begin{align}
&\exists x,(B(x) \wedge R(x)) \rightarrow \exists y,(y \ne x \wedge S(y)). &\text{(1)}\\
&B(x) \wedge \lnot R(x) &\text{(2)}
\end{align}

Because the statement $(2)$ makes the statement $(1)$ vacuously true, $(3)$ is True or False and not both.

I am not sure whether my answer is correct or not.

Reference:

Daepp, U., & Gorkin, P. (2011). Reading, writing, and proving: A closer look at mathematics. In Reading, writing, and proving: A closer look at mathematics (2nd ed., p. 44). New York: Springer.

答: The third statement does not follow from the first two. If "Susie goes to the ball in the red dress" is A, and  "I will stay home" is B, then of course $\bar{B}\implies \bar{A}$, where the bar denotes negation. But you cannot deny the hypothesis to conclude that $B$ is False, so  $\bar{B}$ cannot be deduced from $\bar{A}$. The wording of this question makes it appear as if you should answer it with either "(3) is false" or "(3) is true", but I don't believe this is what the author intended.


答: In this case you should use the p-series test. The p-series test states that the infinite series $$\sum_{n=1}^\infty \frac{1}{n^p}$$ converges only if p > 1. For p < 1 or p = 1 (harmonic series), this series diverges. In your case, you have p = $\epsilon$ + 1. From there it's pretty straight-forward. Because $\epsilon$ by definition has to be greater than zero, it insures that p must be greater than 1. The p=series test therefore confirms that the series converges.

Title: $\sum_{n=1}^\infty 1/n^{1+\epsilon}$ Converges for all $\epsilon >0$

Body:
If $\epsilon>0$, Prove that the sum$$\sum_{n=1}^\infty \frac{1}{n^{1+\epsilon}}$$ converges.

This problem is from my calculus book, and Since this problem follows some other problems like: Show that $\sum_{n=1}^\infty 1/n^2$ And $\sum_{n=1}^\infty 1/n^3$ converges, i tried to use the same method that i used to solve these problems, but it didn’t work.

The method is very simple, all what you have to do is to find a series that you know that it definitely converges and you also know that this series Is less or equal to your series.

The series i used to prove that the above series converges is $$\sum_{n=0}^\infty \frac{1}{2^n}$$
But again it didn’t help me in this $\sum_{n=1}^\infty \frac{1}{n^{1+\epsilon}}$


问:
AF Space subsets
Let $(X, \tau)$ be the Arens-Fort space. If we pick $X$ \ $\{0,0\}$, can we say $X$ \ $\{0,0\}$ is sequentially closed but not closed?

答: The main fact you need: there is no sequence $(x_n)$ in Arens-Fort space such that $x_n \neq (0,0)$ for all $n$, that converges to $(0,0)$, so $(0,0)$ is not in the sequential closure of $A:= X\setminus \{(0,0)\}$. So indeed $A$ is not closed, but it does equal its sequential closure, so is sequentially closed.

Title: AF Space subsets


问:
If $(a,b,c)=1$, is there $n\in \mathbb Z$ such that $(a,b+nc)=1$?
In the book Lectures on modular forms, one finds the statement at page 8 that

If $(a,b,c)=1$ then there is $n\in \mathbb Z$ such that $(a,b+nc)=1$.

I know that, if $(a,b)=1$, then we can take $n=0$. But, if $(a,b)\not=1$, then what could we do? Further, I tried to look at the linear combinations of $a, b, c$ which are $=1$, but to no avail have I discovered anything. The main difficulty I encountered is that the coefficient of $b$ might not be $1$.

Any hint is well-appreciated.

Thanks in advance.

答:
(Overkill proof)

We know that $\gcd(a,\gcd(b,c))=1$. Let $b+c=\gcd(b,c)P+\gcd(b,c)Q=\gcd(b,c)(P+Q)$ where $b=\gcd(b,c)P$ and $c=\gcd(b,c)Q$. Note that $\gcd(P,Q)=1$ (since we divided by the greatest common divisor). By Dirichlet's Theorem on primes in arithmetic progression there are infinitely many primes of the form $P \mod Q$ or in other words there are infinitely many primes $\pi=P+nQ$. Obviously there is a $\pi$ such that $\gcd(a, \pi)=1$ and $\gcd(a,\gcd(b,c)\pi)=1$. Well $\gcd(b,c)\pi=\gcd(b,c)(P+nQ)=b+nc$.


答:
$$\sum_{n=1}^\infty \frac{1}{n^{1+\epsilon}}\leq\int_{1}^\infty \frac{1}{n^{1+\epsilon}}\,\mathrm{d}n=\frac{1}{\varepsilon}-\lim_{n\to\infty}\frac{n^{-\varepsilon}}{\varepsilon}$$
Only converges for $\varepsilon>0$.


问:
Determining the Hausdorff axiom for the quotient space of \mathbb{R}_K.
I'm undergraduated student studying basic topology. While solving the exercises in the Munkres-Topology, one asked me to prove that $\mathbb{R}^*_K$ is not Hausdorff. (section 22, problem 6-(a))

$\mathbb{R}_K$ is a topological space on $\mathbb{R}$ where the basis is given by the form of $(a,b)$ and $(a,b) - K$  ($K = \{ \frac{1}{n} | n \in \mathbb{N} \}$), and $\mathbb{R}^*_K$ is defined by the quotient space obtained from $\mathbb{R}_K$ where the entire set $K$ is considered as a point.

Every solution that I could find on the internet were solving this problem by the following logic: there are no disjoint neighborhoods of $0$ and $K$ in $\mathbb{R}^*_K$, since every open set $(a,b)$ containing $K$ must intersect with $(c,d)$ or $(c,d)-K$ that contains $0$. For instance, $\prod_{i=1}^{\infty} U_i $ can be a neighborhood of $K$ where $U_i = (\dfrac{1}{i} - \epsilon_i , \dfrac{1}{i} + \epsilon_i)$, which (I think that) behavior is totally different from just $(a,b)$.

I can fully understand the statement "every open set $(a,b)$ containing $K$ must intersect with neighborhood containing $0$", but I think that $(a,b)$ just indicates the basis (kind of... I haven't learned any information about basis in the quotient topology) so only dealing about $(a,b)$ cannot fully describe every open set containing $K$.

I'll be very thankful if someone tells me a reason that treating only $(a,b)$ are enough for the solution. Or if it isn't, some guidelines for this problem will be very helpful.

Thank you in advance.

答:
$\Bbb R_K$ was invented to be a Hausdorff space with a closed set $K$ such that $0 \notin K$ but $0$ and $K$ do not have disjoint open neighbourhoods (showing it's not regular). Assume this fact for this exercise.

Now we form the quotient by identifying $K$ to a point (and leaving all points outside $K$ in their own class). The quotient map $q: \Bbb R_K \to X$ (where $X$ is that quotient, so the set of classes $\{\{x\}\mid x \notin K\} \cup \{K\}$) is closed and continuous, $\Bbb R_K$ is Hausdorff but $X$ is not:

Suppose we could separate the points $K$ and $\{0\}$ in $X$ by disjoint open sets $U$ and $V$ respectively. This would imply that $q^{-1}[U]$ is an open neighbourhood of $K$ and $q^{-1}[V]$ of $0$, and disjoint still. But this contradicts the basic fact about $\Bbb R_K$ from the start. So $\{0\}$ and $K$ cannot be separated in $X$ so $X$ is not Hausdorff.


问:
Prove $q = \begin{cases} [{{\frac{a}{b}}}], & b>0 \\[2ex] [\frac{a}{b}]+1 & b<0 \end{cases}$.
Prove if $$a = bq+r$$     $$0\le r<|b|$$
then
$$q =
\begin{cases}
[{{\frac{a}{b}}}],  & b>0  \\[2ex]
[\frac{a}{b}]+1 &  b<0
\end{cases}$$

I have that $q\leq\frac{a}{|b|}< q+1$
but unable to continue from here.

答: Hint : ${a \over b} = q + {r \over b}$, so $\lfloor {a \over b} \rfloor = \lfloor q + {r \over b} \rfloor = q + \lfloor {r \over b} \rfloor$. Now, look at the value of $\lfloor {r \over b} \rfloor$ depending on the sign of $b$.


答:
For any number $n$, consider the distance to the next perfect square.  This is either even (including $0$) or odd.  If it's even... that is, if $n=x^2-2k$, with $k<x$... then $f(n)=g(n)=x+2k$.  So for the ratio to be $4/7$ (to be anything other than $1$, really), you must have $$n=x^2-2k+1=(x+1)^2-2(k+x),$$ with $k<x$.  In that case, $f(n)=x+2k-1$, but $g(n)=3x+2k+1$.  Then
$$
\frac{4}{7}=\frac{f(n)}{g(n)}=\frac{x+2k-1}{3x+2k+1}\implies12x+8k+4=7x+14k-7\implies5x=6k-11.
$$
There will be infinitely many solutions to this that satisfy $k<x$; all must have $k\equiv 1$ (mod $5$), so you can just check $k=1,6,11,\ldots$  The smallest is $(x,k)=(17,16)$, followed by $(23,21)$, $(29,26)$, and so on; generally $(x,k)=(17+6q, 16+5q)$ for $q\ge 0$.  These correspond to infinitely many solutions for $n=x^2-2k+1$: $n=258, 488, 790, \ldots$  The general solution is $$n=(17+6q)^2-2(16+5q)+1=258+194q+36q^2.$$

But by far the easiest approach is just to code it up: in Python,

```<code>>>> def f(n): return getSqrt(n) or 1 + f(n+1)

>>> def g(n): return getSqrt(n) or 2 + g(n+2)

>>> [n for n in xrange(1,10000) if 7*f(n) == 4*g(n)]
[258, 488, 790, 1164, 1610, 2128, 2718, 3380, 4114, 4920, 5798, 6748, 7770, 8864]
</code>```

assuming that you've defined getSqrt(n) to return $\sqrt{n}$ when $n$ is a perfect square and otherwise return None.

Title: Find the least $n$ such that $f(n)/g(n)=4/7$

Body:
Let $f(n)$ and $g(n)$ be functions satisfying $$f(n)=\cases{\sqrt{n} \hspace{50pt} \sqrt{n} \in \mathbb{N} \\ 1+f(n+1) \hspace{14pt} \text{otherwise}} $$
$$g(n)=\cases{\sqrt{n} \hspace{50pt} \sqrt{n} \in \mathbb{N} \\ 2+g(n+2) \hspace{14pt} \text{otherwise}}$$
for positive integers $n$. Find the least positive value of $n$ such that $\dfrac{f(n)}{g(n)}=\dfrac47$.

I found the values for $f$ and $g$ till $n=9$, I noticed that $f(n)=g(n)$ happened at $n=1,2,4,5,7,9$ and for some values it also happened that if $f(a)=b$, then $f(b)=a$. I don't know if that is useful in anyway. I can't see how to approach the problem.

Please give some hints.

Body:
For any number $n$, consider the distance to the next perfect square.  This is either even (including $0$) or odd.  If it's even... that is, if $n=x^2-2k$, with $k<x$... then $f(n)=g(n)=x+2k$.  So for the ratio to be $4/7$ (to be anything other than $1$, really), you must have $$n=x^2-2k+1=(x+1)^2-2(k+x),$$ with $k<x$.  In that case, $f(n)=x+2k-1$, but $g(n)=3x+2k+1$.  Then
$$
\frac{4}{7}=\frac{f(n)}{g(n)}=\frac{x+2k-1}{3x+2k+1}\implies12x+8k+4=7x+14k-7\implies5x=6k-11.
$$
There will be infinitely many solutions to this that satisfy $k<x$; all must have $k\equiv 1$ (mod $5$), so you can just check $k=1,6,11,\ldots$  The smallest is $(x,k)=(17,16)$, followed by $(23,21)$, $(29,26)$, and so on; generally $(x,k)=(17+6q, 16+5q)$ for $q\ge 0$.  These correspond to infinitely many solutions for $n=x^2-2k+1$: $n=258, 488, 790, \ldots$  The general solution is $$n=(17+6q)^2-2(16+5q)+1=258+194q+36q^2.$$

But by far the easiest approach is just to code it up: in Python,

>>> def f(n): return getSqrt(n) or 1 + f(n+1)

>>> def g(n): return getSqrt(n) or 2 + g(n+2)

>>> [n for n in xrange(1,10000) if 7*f(n) == 4*g(n)]
[258, 488, 790, 1164, 1610, 2128, 2718, 3380, 4114, 4920, 5798, 6748, 7770, 8864]

assuming that you've defined getSqrt(n) to return $\sqrt{n}$ when $n$ is a perfect square and otherwise return None.


问:
A functional equation problem: $ \frac { f \left( f ( y ) ^ 2 + x y \right) } { f ( y ) } = f ( x ) + y $

Let $ \mathbb R ^ + $ denote the set of the positive real numbers. Find all functions $ f : \mathbb R ^ + \to \mathbb R ^ + $ satisfying
$$ \frac { f \left( f ( y ) ^ 2 + x y \right) } { f ( y ) } = f ( x ) + y $$
for all $ x , y \in \mathbb R ^ + $.

I am very thankful for any solution, please help!

I tried to set $ x = y = 1 $, $ x = y = 2 $, $ x = 1 $, $ y = 2 $, so on, but this problem is more difficult.

答:
$ \def \R {\mathbb R ^ +} $
You can show that the only $ f : \R \to \R $ satisfying
$$ f \left( x y + f ( y ) ^ 2 \right) = \big( f ( x ) + y \big) f ( y ) \tag 0 \label 0 $$
for all $ x , y \in \R $ is the identity function. It's straightforward to verify that the identity function is a solution. We try to prove the converse. The messiest part is showing $ f ( 1 ) = 1 $, and after that, everything goes rather smoothly. Before going further, let's make the simple observation that $ f $ must be injective: if $ f ( x ) = f ( y ) $, then by \eqref{0} we have
$$ x = \frac { f \left( y x + f ( x ) ^ 2 \right) } { f ( x ) } - f ( y ) = \frac { f \left( x y + f ( y ) ^ 2 \right) } { f ( y ) } - f ( x ) = y \text . $$

Showing $ f ( 1 ) = 1 $

Letting $ a = f ( 1 ) $ and plugging $ y = 1 $ in \eqref{0} we get
$$ f \left( x + a ^ 2 \right) = a \big( f ( x ) + 1 \big) \text . \tag 1 \label 1 $$
Assuming $ a > 1 $, \eqref{1} gives
$$ f \left( x + n a ^ 2 \right) = a ^ n f ( x ) + \frac { a ^ { n + 1 } - a } { a - 1 } \tag 2 \label 2 $$
by induction on the positive integer $ n $. Setting $ x = y = 1 + n a ^ 2 $ in \eqref{0} and using \eqref{2} we get
$$ f \left( \left( 1 + n a ^ 2 \right) ^ 2 + \left( \frac { a ^ { n + 2 } - a } { a - 1 } \right) ^ 2 \right) = \frac { a ^ { n + 2 } - a } { a - 1 } \left( \frac { a ^ { n + 2 } - a } { a - 1 } + 1 + n a ^ 2 \right) \text . \tag 3 \label 3 $$
Also, letting $ m = 2 n + \left\lfloor n ^ 2 a ^ 2 + \left( \frac { a ^ { n + 1 } - 1 } { a - 1 } \right) ^ 2 \right\rfloor $ and $ b = \left( 1 + n a ^ 2 \right) ^ 2 + \left( \frac { a ^ { n + 2 } - a } { a - 1 } \right) ^ 2 - m a ^ 2 $, we can see that $ m $ is a positive integer and $ b \in \R $, and thus we can use \eqref{2} to get
$$ f \left( \left( 1 + n a ^ 2 \right) ^ 2 + \left( \frac { a ^ { n + 2 } - a } { a - 1 } \right) ^ 2 \right) = a ^ m f ( b ) + \frac { a ^ { m + 1 } - a } { a - 1 } \text . $$
But this means that the right-hand side of \eqref{3} must be greater than $ \frac { a ^ { m + 1 } - a } { a - 1 } $, which can't happen for large enough $ n $. Thus the assumption $ a > 1 $ is absurd, and we must have $ a \le 1 $. Now, if $ a < 1 $, we can set $ x = 1 - a ^ 2 $ and $ y = 1 $ in \eqref{0} to get $ a = a \Big( f \left( 1 - a ^ 2 \right) + 1 \Big) $, which contradicts $ a , f \left( 1 - a ^ 2 \right) \in \R $. Thus the assumption $ a < 1 $ is absurd, too, and we must have $ a = 1 $.

Showing $ f = \operatorname {Id} _ { \R } $

Using \eqref{1} and induction on the positive integer $ n $, we get
$$ f ( x + n ) = f ( x ) + n \text , \tag 4 \label 4 $$
which in particular shows $ f ( n ) = n $ for any positive integer $ n $. Hence, letting $ y = n $ in \eqref{0} we have
$$ f ( n x ) = n f ( x ) \tag 5 \label 5 $$
for any positive integer $ n $. Letting $ x = \frac n y $ in \eqref{0} and using \eqref{4} and \eqref{5}, we get
$$ f \left( f ( y ) ^ 2 \right) - y f ( y ) = n \Bigg( f ( y ) f \left( \frac 1 y \right) - 1 \Bigg) \text , $$
and since the left-hand side doesn't depend on $ n $, both sides must be equal to $ 0 $. In particular, we have
$$ f \left( f ( y ) ^ 2 \right) = y f ( y ) \text . \tag 6 \label 6 $$
Letting $ x = \frac { f ( y ) ^ 2 } y $ in \eqref{0} and using \eqref{5} and \eqref{6} we get
$$ f \left( \frac { f ( y ) ^ 2 } y \right) = y \text . \tag 7 \label 7 $$
Using \eqref{4} and \eqref{7}, we can see that
$$ f \left( \frac { f ( y ) ^ 2 + 2 f ( y ) + 1 } { y + 1 } \right) = f \left( \frac { \big( f ( y ) + 1 \big) ^ 2 } { y + 1 } \right) = f \left( \frac { f ( y + 1 ) ^ 2 } { y + 1 } \right) \\
= y + 1 = f \left( \frac { f ( y ) ^ 2 } y \right) + 1 = f \left( \frac { f ( y ) ^ 2 } y + 1 \right) = f \left( \frac { f ( y ) ^ 2 + y } y \right) \text . $$
By injectivity of $ f $, we have
$$ \frac { f ( y ) ^ 2 + 2 f ( y ) + 1 } { y + 1 } = \frac { f ( y ) ^ 2 + y } y \text , $$
or equivalently
$$ y f ( y ) ^ 2 + 2 y f ( y ) + y = y f ( y ) ^ 2 + y ^ 2 + f ( y ) ^ 2 + y \text , $$
which can be simplified to $ \big( f ( y ) - y \big) ^ 2 = 0 $, and thus $ f $ is the identity function.


问:
Gamma distribution with rate
I'm trying to solve this question:

Let T be a random variable with Gamma(r=7, LAMBDA) distribution, where r is the shape parameter and LAMBDA the rate parameter. What is P(T > E[T])?

I'm trying to plug this into a gamma distribution graph calculator online, but I'm not sure what exactly to put in the place of lambda since it's not given? Also, what exactly is E(T) in this case?

答:
Refer to the Wikipedia article for the gamma distribution.  Specifically, look at the second column of the table on the right-hand side, which shows that the probability density function for shape $\alpha$ and rate $\beta$ is given by

$$f_X(x) = \frac{\beta^\alpha}{\Gamma(\alpha)} x^{\alpha-1} e^{-\beta x}, \quad x > 0.$$  Under this parametrization, the mean or expected value is $$\operatorname{E}[X] = \alpha/\beta.$$

Rewriting this in terms of your variables, $X = T$, $\alpha = r = 7$, $\beta = \lambda$,

$$F_T(t) = \frac{\lambda^7}{6!} t^6 e^{-\lambda t}, \quad t > 0, \\ \operatorname{E}[T] = 7/\lambda.$$  You want to compute $$\Pr[T > 7/\lambda].$$  You can do this either by writing the integral directly:
$$\Pr[T > 7/\lambda] = \int_{t = 7/\lambda}^\infty \frac{\lambda^7}{6!} t^6 e^{-\lambda t} \, dt, \tag{1}$$
or you can define a new, transformed random variable, say $W = \lambda T$, whose density is given by $$f_W(w) = \frac{1}{\lambda} f_T(w/\lambda) = \frac{1}{\lambda} \cdot \frac{\lambda^7}{6!} (w/\lambda)^6 e^{- \lambda (w/\lambda)} = e^{-w} \frac{w^6}{6!}.$$  And now you can see that $$\Pr[T > 7/\lambda] = \Pr[W > 7] = \int_{w=7}^\infty e^{-w} \frac{w^6}{6!} \, dw, \tag{2}$$ and the $\lambda$ drops out entirely.  You could also have seen this by performing the variable substitution $w = \lambda t$, $dw = \lambda \, dt$ in Equation $(1)$.  This is why $\lambda$ is not given, because the answer doesn't depend on it.

I leave the computation of the integral as an exercise.

Title: Gamma distribution with rate


问:
Dividing Divergent and Convergent Sequences
Let $a_n$ be a diverging sequence and $b_n$ be a converging sequence. With that in mind does $\frac{a_n}{b_n}$ always diverge?

If so, what is the proof for this?

答: Let $b_n$ converge to $l_1$, and suppose $\frac{a_n}{b_n}$ converges to $l_2$. Then by the Algebraic Limit Theorem we have$$\lim_{n\to \infty}b_n\cdot \frac{a_n}{b_n} = l_1l_2.$$ But $$b_n \cdot \frac{a_n}{b_n} =a_n,$$ which is divergent, thus giving us a contradiction. Therefore $\frac{a_n}{b_n}$ is divergent.


问:
Find the value of the constants $a$ and $b$ such that the following limit is satisfied.
Find the value of the constants $a$ and $b$ such that the following limit is satisfied.

$\lim_{x\to a}\dfrac{bx-1}{x-a}=\frac12 $.

$\lim_{x\to a}\dfrac{bx+9}{x-a}=a-6 $.

$\lim_{x\to a}\dfrac{ax+b}{x^2+3x-4}=4$.

I desperately need help with these 3 problems as I do not understand them. Thanks!

答:
For $(bx-1)/(x-a)\to\frac12$ as $x\to a$ recognize that $bx-1=\frac12(x-a)$. Reduce and equate coefficients:$$bx-1=\frac12(x-a)\\bx-1=\frac12x-\frac{a}2\\\implies b=\frac12,a=2$$

Similarly, for our second limit, we see that $bx+9=(a-6)(x-a)$, reduce, and equate coefficients:$$bx+9=(a-6)x-(a-6)a$$Immediately we see that $b=a-6$ and that $-9=(a-6)a=a^2-6a$. Moving terms to our right-hand side, we have a factorizable quadratic equation in $a$:$$a^2-6a+9=0\\(a-3)^2=0\\\implies a=3$$... and thus $b=-3$.

Can you use a similar approach to tackle our third limit?

Recognize that our denominator is easily factorizable to yield $x^2+3x-4=(x+4)(x-1)$. Note that since there are no singularities unless $x=a$ and since we have a finite limit and the numerator is only linear we know that cannot be true, we may "distribute" our limit safely to yield:$$\frac{\lim\limits_{x\to a}(ax+b)}{\lim\limits_{x\to a}(x^2+3x-4)}=4\\\frac{a^2+b}{(a+4)(a-1)}=4\\\implies a^2+b=4(a+4)(a-1)$$Expand our right-hand side to yield $4(a+4)(a-1)=4(a^2+3a-4)=4a^2+12a-16$, yielding $$a^2+b=4a^2+12a-16$$Equating coefficients we find $a=0$ and $b=-16$.


问:
Help with Problem 10 in Chapter 7 of Lang's Algebra
I have no idea how to proceed with the following problem from Lang's Algebra:

Let $n$ be a positive integer and let $\zeta,\zeta'$ be primitive $n$-th roots of unity.

a) Show that $\frac{1-\zeta}{1-\zeta'}$ is an algebraic integer.

b) If $n\geq 6$ is divisible by at least two primes, show that $1-\zeta$ is a unit in the ring $\mathbb{Z}[\zeta]$.

This has been asked twice before, here and here.

However, I don't fully understand the approach given in either. In the first, I don't understand why $1-\zeta^{p^k}$ is a unit mod $p$ if $\zeta^{p^k}$ is an $m$-th root of unity. In the second, I don't understand how the substitution advised in the comments helps to solve the problem. This is probably supposed to be obvious but please help, I don't get it.

Thanks.

答:
For a): Fix some primitive $n$-th root of unity $\zeta$. Then all other primitive $n$-th roots of unity are of the form $\zeta^k$ with $(k,n)=1$. If this is the case, then there is $k'$ such that $kk'\equiv1\mod n$. Now write $\zeta'=\zeta^{k}$ and $\zeta=(\zeta^k)^{k'}$ and note the following:

$$\frac{1-\zeta}{1-\zeta'}=\frac{1-(\zeta^k)^{k'}}{1-\zeta^k}=\sum_{i=0}^{k'-1}(\zeta^k)^i\in\mathbb Z[\zeta]$$

This proves a) as $\zeta$ is an algebraic integers and hence all its powers and sums of these.

For b): This is best done using a norm argument (as mentioned in one of the linked posts). Let $G=\operatorname{Gal}(\mathbb Q(\zeta)/\mathbb Q)$. Define the norm

$$\mathbf{N}_{\mathbb Q}^{\mathbb Q(\zeta)}\colon\mathbb Q(\zeta)\to\mathbb Q,\,\alpha\mapsto\prod_{\sigma\in G}\sigma(\alpha)\, .$$

Restricted to $\mathbb Z[\zeta]$ the image will be contained in $\mathbb Z$ and $\alpha\in\mathbb Z[\zeta]$ is a unit iff $\mathbf{N}_{\mathbb Q}^{\mathbb Q(\zeta)}(\alpha)=\{\pm1\}$. So computing the norm of $1-\zeta$ in the given cases will suffices for showing that $1-\zeta$ is a unit. Note that $\sigma(\zeta)$ just ranges over the different primitive roots of unity and so

$$\mathbf{N}_{\mathbb Q}^{\mathbb Q(\zeta)}(1-\zeta)=\prod_{\sigma\in G}\sigma(\alpha)=\prod_{(k,n)=1}(1-\zeta^k)$$

Claim. $\displaystyle\prod_{(k,n)=1}(1-\zeta^k)=1$

Proof. Let $\Phi(X)=\frac{X^n-1}{X-1}=X^{n-1}+\cdots+1$ and denote by $\Phi_d(X)$ the $d$-th cyclotomic polynomial. Then

$$\Phi(X)=\prod_{\substack{d\mid n\\d>1}}\Phi_d(X)$$

Note that our polynomial is $\Phi_n(X)$. Now, $\Phi(1)=n$ and for prime $p$ also $\Phi_p(1)=p$. Moreover, $\Phi_{p^r}(1)=p$ for all $r\ge1$. To see this, consider

$$\frac{X^{p^r}-1}{X^{p^{r-1}}-1}=X^{(p-1)p^{r-1}}+\cdots+1$$

and note that this polynomial divides $X^{p^r}-1$ and is of degree $(p-1)p^{r-1}$. But $\varphi(p^r)=(p-1)p^{r-1}$ is the exact number of primitive $p^r$-th roots of unity. Hence $\Phi_{p^r}(X)=X^{(p-1)p^{r-1}}+\cdots+1$ and $\Phi_{p^r}(1)=p$.

We return to $\Phi(1)=n$ written as

$$n=\Phi(1)=\prod_{\substack{d\mid n\\d>1}}\Phi_d(1)\ .$$

Now, $n$ factors $p_1^{r_1}\cdots p_s^{r_s}$ for some distinct primes $p_1,\dots,p_s$. But these factors already occur only considering the prime power factors. Hence the remaining factors, that is $\Phi_d(1)$ for $d>1$ and $d\ne p_i^{r_i}$ have to be $\pm1$. Inductively we see that $\Phi_d(1)=1$ for all $d$ as described, and so in particular $\Phi_n(1)=1$. $\square$

(This argument can be found in Lang's Algebraic Number Theory who credits Bass for it.)

The assumption that $n$ is divisible by at least two distinct primes is used subtly: it ensures that $n$ is not a prime power (or prime) in which case the above argument would collapse as would have no factors like $d$ in the end. As shown the product evaluates to $p$ in case of primes powers.


问:
Show holomorphic function in unit disk $D$ s.t. $|f(z)| < M$, $ \forall z \in D$ and $f(0) = 0$ is s.t. $|f(z)| \leq M|z|$
To solve this problem I thought about using the function $g(z) = \frac{f(z)}{z}$ because I know it is also holomorphic, and then to apply on $g(z)$ the Maximum Modulus Principle, which I recall below:

If $f$ is holomorphic in a domain D and continuous on it, then $|f(0)| \leq \max_{z \in D} |f(z)|$

I know it can be done this way, but I don't understand how to apply this Principle since $g(0)$ is actually not defined.. any clue?

答:
Define $g(z):=\begin{cases} \frac{f(z)}{z}&\text{if $z\neq 0$}\\ f'(0) & \text{if $z=0$}\end{cases}$.  Now, note that since $f(0)=0$, we have
\begin{align}
\lim\limits_{z\to 0}g(z)=\lim_{z\to 0}\frac{f(z)}{z}=\lim_{z\to 0}\frac{f(z)-f(0)}{z-0}=f'(0)=g(0).
\end{align}
Hence, $g$ is clearly holomorphic on $0<|z|<1$, and it is continuous at $0$. Under this situation, it actually follows that $g$ is also holomorphic at the origin (for example, using Morera's theorem, or by Cauchy's inequalities applied to the Laurent expansion of $g$). Now can you continue?


问:
Piecewise function notation and applications
I would like to define a piecewise function to imitate some physical phenomenon, which, for example, occurs once in a year (at June for a whole month long). Say, frequency/domain (integers) of the func is representing months, and January is at 0.

How to say mathematically, that y=0 if x is not a member of positive Z which are divisible by smth (to imitate all months except for June) and y=1 if x is a member of pos. Z which are divisible by smth (to imitate June)?

For me here the problem also is that I cannot easily say that it is y=1 for all integers divisible by 6-1 (6 as a June and -1 to compensate for starting with January at 0) because it would also cover next January etc.

答:
Since $0$ corresponds to January, numbers that correspond to June are $5$, $17$, $22$, etc. That sequence increases by $12$ each time, so the difference between any two of them is some multiple of $12$. And because of this, if we divide any of them by $12$, we'll get an integer quotient representing the number of years, and a remainder of $5$.

One understandable, but nonstandard way to write your function would be something like:
$$f(n)=\begin{cases}
1 & \text{ if the remainder of }n\div12\text{ is }5\\
0 & \text{ otherwise}
\end{cases}$$

There is a special language and notation for dealing with things like this, in the words of so-called "modular arithmetic". Two integers whose difference is a multiple of $12$ are said to be "congruent modulo $12$", and we use the notation $n_1\equiv n_2\pmod {12}$ in math.

A more standard notation would be something like:
$$f(n)=\begin{cases}
1 & \text{ if }n\equiv5\pmod{12}\\
0 & \text{ otherwise}
\end{cases}$$

In some mathematics subfields, functions that are $1$ or $0$ in this way are common, and there are special notations like the Iverson bracket or indicator functions. But  these may not be universally understood.

Finally, in programming-related contexts, the calculation of the remainder is emphasized more than the equivalence, so they would be more likely to write things in terms of the modulo operation which outputs the remainder.


问:
Given $n \in N$, prove that $f: [0, \infty) \rightarrow [0, \infty) f(x) = x^n$ is increasing without calculus (derivates)
I have to prove this using $x_1 < x_2 \rightarrow f(x_1) < f(x_2)$.

I have no idea on how to do this, but what I have tried so far is:

Consider $x_1, x_2  \in [0, \infty)$ such that $x_1 < x_2 \rightarrow f(x_1) < f(x_2)$

$\rightarrow x_1^n < x_2^n$

I do not know what to do next

答:
We can use the result $a\le b\implies ax\le bx\quad\forall x\ge 0$

Note that since $0\le x_1<x_2$ by hypothesis then $x_2\neq 0$,

So we can divide and get $\  0\le a=\dfrac{x_1}{x_2}<1$ and multiplying by $a$ then $0\le a^2\le a<1$

And we can proceed by induction on $n$ to show $a^n\le a^{n-1}<1$, which we can retransform back into $x_1^n<x_2^n$.


答:
For the first one, note that denominator is 0 when $x=a$ so for the limit to exist, the numerator must also equal 0 when $x=a$. This should give you b in terms of a. Now either factorizing of using L'Hopital's Rule, find what the limit would be in terms of a and then solve for the limit to equal 2.

Similar work for the second question.

For the third question consider the cases when a makes the denominator 0 and when it makes the denominator non-zero.

Body:
Find the value of the constants $a$ and $b$ such that the following limit is satisfied.

$\lim_{x\to a}\dfrac{bx-1}{x-a}=\frac12 $.

$\lim_{x\to a}\dfrac{bx+9}{x-a}=a-6 $.

$\lim_{x\to a}\dfrac{ax+b}{x^2+3x-4}=4$.

I desperately need help with these 3 problems as I do not understand them. Thanks!


问:
Finding Groups and Subgroups Such That Some Relations (and Homomorphism) Hold
Find groups $G$ and $G'$ and a homomorphism $f : G \to G'$ so that there are subgroups $U_1, U_2 \leq G$ (with $U_1 \neq U_2$) and $U'_1, U'_2 \leq G'$ for which following diagram is true.

In particular, this means that

$f(U_2) = U'_1 = f(U_1)$;

$f^{-1} (U'_1) = U_2 = f^{-1}(U'_2)$; and

there is no subgroup $U_3 \leq G$ so that $f(U_3)=U'_2.$

I thought about $f:(\mathbb{Z}, +)\rightarrow (\mathbb{Z}, +)$ with $x \mapsto 2x$ (which is a homomorphism).

Then, I chose $U_1=(\mathbb{Z}_e,+)=U'_1,$ which is the subgroup of even integers, $U_2=(\mathbb{Z},+)=U'_2,$ which is clearly the (trivial) subgroup of all integers.

Since the function only maps to even numbers, there is no subgroup $U \leq G$ so that $f(U)=U'_2$.

And since the inverse image of the odd numbers has no element, it follows that $f^{-1} (U'_1) = U_2 = f^{-1}(U'_2)$

Lastly, $f(U_1) \overset{!}{=} U'_1,$ but I'm not too sure about that since I would need all the odd numbers to truly map to every element of $U'_1.$

Could anyone give me a hint how to resolve the last problem or give me a better example?

Body:
Take $G = \mathbb Z_{12}$ and $G' = \mathbb Z_{18}$.

Define $f \colon G \to G'$ by $f(1_G) = 6_{G'}$ (and extend).

Then $f$ is a homomorphism with $\operatorname{im} f = \langle 6_{G'} \rangle = \{0, 6, 12\}$.

Let $U_2' = G' = \mathbb Z_{18}$, $U_2 = G = \mathbb Z_{12}$. Note that $f^{-1}(U_2') = U_2$.

But $f$ is not surjective, so $\not\exists U_3 \le G$ such that $f(U_3) = U_2'$.

Let $U_1 = \langle 2_G \rangle = \{0, 2, 4, 6, 8, 10\} \le \mathbb Z_{12}$.

Let $U_1' = \langle 6_{G'} \rangle = \{0, 6, 12\} \le \mathbb Z_{18}$.

Note that $f(U_2) = \operatorname{im} f = U_1'$.

But also, $f(U_1) = \{f(0), f(2), f(4), f(6), f(8), f(10)\} = \{0, 12, 6, 0, 12, 6\} = \{0, 6, 12\} = U_1'$.


问:
Unknown sequence with some values known
I know that $n$ and $m$ are independant of eachother, and $f(n,m)$ is symmetrical. I can see different sequences within it, but is there a definite function $f: \mathbb{N}^2 \to \mathbb{N}$ for all natural numbers, or does it have to be a composite function?

$$
\begin{array}{c|lcr}
n,m & 1 & 2 & 3 & 4 & 5 & 6 & 7\\
\hline
1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 &\cdots \\
2 &   & 6 & 10 & 15 & 21 & 28 & 36     \\
3 &   &   & 20 & 35 & 56 & 84 & 120     \\
4 &   &   &    & 70 & 126 & 210 & 330&\\
5 &   &   &    &    & 252 & 462  & 792  &\\
6 &   &   &    &    &     & 924  & 1716  & \\
\end{array}
$$

答: It appears that the solution I was looking for is indeed related to the binomial cefficent. The actual formula is: $ f(n,m) = \left( \array{n\\ m+n} \right) $, or when simplified $\displaystyle f(n,m) = \frac{(n+ m)!}{n!\ m!}$

Title: Unknown sequence with some values known


问:
There were $107$ people at the wedding and $11$ tables, which could accommodate $10$ people each.

There were $107$ people at the wedding and $11$ tables, which could accommodate $10$ people each. This means there were about $11^{107}$ possible seating plans.

I read that in a book but didn't really understand the solution. I mean the solution doesn't consider the fact that each table can have a max of $10$ people and it actually doesn't care about the capacity of the table. Can anyone explain it to me?

答:
Let $a(n,t)$ be the number of ways to assign $n$ distinguishable people to $t$ indistinguishable tables, each with capacity $c$.  Without the capacity, this is a Stirling number of the second kind.  By conditioning on the number $k$ of people at the table that contains the first person, we obtain
$$a(n,t) =
\begin{cases}
1 &\text{if $n = 0$} \\
0 &\text{if $t = 0$}\\
\sum_{k=1}^{\min(c,n)} \binom{n-1}{k-1} a(n-k,t-1) &\text{otherwise}
\end{cases}
$$
For $c=10$, we have $a(107,11)=5.8152\times 10^{97}$.

Let $b(n,t)$ be the number of ways to seat $n$ distinguishable people around $t$ indistinguishable tables, each with capacity $c$.  Without the capacity, this is a Stirling number of the first kind.  By conditioning on the number $k$ of people at the table that contains the first person, we obtain
$$b(n,t) =
\begin{cases}
1 &\text{if $n = 0$} \\
0 &\text{if $t = 0$}\\
\sum_{k=1}^{\min(c,n)} (k-1)! \binom{n-1}{k-1} a(n-k,t-1) &\text{otherwise}
\end{cases}
$$
For $c=10$, we have $b(107,11)=1.2132\times 10^{156}$.  If rotations are not equivalent, multiply each summand by $k!$ instead of $(k-1)!$.

For both $a(n,t)$ and $b(n,t)$, if the tables are instead distinguishable, multiply the final results by $t!$.


问:
Is there any example such that $\lim\limits_{k \to \infty} S_{2k} < \infty$ but $\lim\limits_{k \to \infty} S_k = \infty$?
Let $S_k = \sum\limits_{n=1}^{k}a_n$ be a series.
Is there any example that
$\lim\limits_{k\rightarrow\infty}S_{2k}$ exists but $\sum\limits_{n=1}^{\infty}a_n$ is convergent.

答:
Yes. Counter-example is easy to construct.

We try to choose $(a_{n})$ such that $S_{n}=\begin{cases}
n, & \mbox{if }n\mbox{ is odd}\\
0, & \mbox{if }n\mbox{ is even}
\end{cases}.$ This is possible by solving equations. For example, $a_{1}=S_{1}=1$,
$a_{2}=S_{2}-S_{1}=-1$, $a_{3}=S_{3}-S_{2}=3$, etc...

Clearly $\lim_{n\rightarrow\infty}S_{2n}=0$ while $\lim_{n\rightarrow\infty}S_{n}$
does not exist.


问:
Different results obtained for $\int e^{-x}\sin x \mathrm dx$ using formula and using by parts.
So I have a very simple integral at hand which can be evaluated by parts directly, and the answer so obtained matches with the correct antiderivative.

$$\int e^{-x}\sin x\mathrm dx$$

My question arises when I try to use the formula for integrals of the type $\int e^{ax}\sin(bx+c)\mathrm dx$. Although the formula itself is derived from two applications of by parts technique, the result finally obtained for the said integral differs in sign. Here's how:

$$\begin{aligned}\int e^{ax}\sin (bx+c)\mathrm dx &= \frac{e^{ax}}{a^2+b^2}\left(a\sin(bx+c)-b\cos(bx+c)\right)\\ & = \frac{e^{ax}}{r}\sin(bx+c-\phi) \ , \ r=\sqrt{a^2+b^2} \ , \ \phi=\arctan\left(\frac{b}{a}\right)\end{aligned}$$

Plugging in $a=-1$, $b=1$ and $c=0$ for the integral at hand gives $1/\sqrt{2}\cdot \sin(x+\pi/4)$ which is negative of the correct antiderivative. What mistake am I making? Any pointers are appreciated. Thanks.

答: You are calculating $\phi=\arctan (-1)=-\pi/4$. But it could also be $3\pi/4$ which I think will give you the answer you want. Check carefully the restrictions on the parameters in whatever source you used for the result.


答:
Assumed that $n \in \Bbb{Z^+}.$

Alternative approach that is based on the binomial theorem:

$(a + b)^n = \sum_{k=0}^n \binom{n}{k}a^{n-k}b^k.$

$x_2 > x_1 \geq 0 \implies \exists b \in \Bbb{R^+}$ such that
$x_2 = (x_1 + b)$.

Therefore,

$$(x_2)^n - (x_1)^n = (x_1 + b)^n - (x_1)^n = \sum_{k=1}^n \binom{n}{k}(x_1)^{n-k}b^k > 0.$$

Note that the above summation skips the term corresponding to $k=0$.  Further, regardless of whether $x_1 = 0$ or $x_1 > 0$, all of the terms are non-negative, and the very last term (which is $b^n$) must be positive.


问:
A space where no paths are homotopic
I was just thinking about algebraic topology and was wondering: It is possible to find a Hausdorff space where no paths are homotopic?

Excluding trivial examples.

答: The answer is no. I presume by excluding trivial examples you are asking for a space $X$ with two distinct points $x,y$ such that there is a path $\gamma$ from $x$ to $y$ (note that $x,y$ being distinct is not so important, if there is a nonconstant loop at $x$ automatically this is true by just picking $y$ to be the midpoint). Given any such space, there is a homotopy of paths from the constant loop at $x$ to $\gamma^{-1} \circ \gamma$. Clearly these paths are unequal, since their images are different.

Title: A space where no paths are homotopic