问:
A product identity involving the gamma function
I have reduced this problem (thanks @Mhenni) to the following (which needs to be proved):

$$\prod_{k=1}^n\frac{\Gamma(3k)\Gamma\left(\frac{k}{2}\right)}{2^k\Gamma\left(\frac{3k}{2}\right)\Gamma(2k)}=\prod_{k=1}^n\frac{2^k(1+k)\Gamma(k)\Gamma\left(\frac{3(1+k)}{2}\right)}{(1+3k)\Gamma(2k)\Gamma\left(\frac{3+k}{2}\right)}.$$

As you see it's quite a mess. Hopefully one can apply some gamma-identities and cancel some stuff out. I have evaluated both products for large numbers and I know that the identity is true, I just need to learn how to manipulate those gammas.

答:
Put all gamma functions to one side. Then

$\Gamma(2k)$ cancels out.

Using gamma function duplication formula, one can replace
$$\frac{1}{1+k}\Gamma\left(\frac{k}{2}\right)\Gamma\left(\frac{k+3}{2}\right)=\frac12\Gamma\left(\frac{k}{2}\right)\Gamma\left(\frac{k}{2}+\frac12\right)=2^{-k}\sqrt{\pi}\,\Gamma(k).$$

Similarly,
$$\frac{1}{1+3k}\Gamma\left(\frac{3k}{2}\right)\Gamma\left(\frac{3k+3}{2}\right)=
\frac12\Gamma\left(\frac{3k}{2}\right)\Gamma\left(\frac{3k}{2}+\frac12\right)=
2^{-3k}\sqrt{\pi}\,\Gamma(3k).
$$

The identity follows immediately.


问:
$C(K)$ is reflexive if and only if $K$ is finite
Let $C(K)$ be the set of all continuous complex valued functions on a compact Hausdorff space $K$. Is it true that $K$ must be finite if $C(K)$ is reflexive?
To me it seems true, but I don't know how to prove it. As $C(K)$ is reflexive then we have canonical isometry onto $C(K)^{**}$. How does that help?

答:
There is another different answer to you question in the special case that you allow $C(K)$ for $K = \mathbb T$, the torus, to be equipped with the weak topology.

The following is taken from the article Exact Support Recovery for Sparse Spikes Deconvolution by
Vincent Duval and Gabriel Peyré, from subsection 2.1 Topology of Radon Measures:

$M(\mathbb T)$ can be defined as the dual space of $C(\mathbb T)$ endowed with the uniform norm $\| \cdot \|_{\infty}$. It is naturally a Banach space when endowed  with the dual norm (also called total variation norm)
$$
\| m \|_{\text{TV}}
:= \sup_{\substack{f \in C(\mathbb T) \\ \| f \|_{\infty} = 1}} \int_{\mathbb T} f dm.
$$
In that case the dual of $M(\mathbb T)$ is a complicated space, and it is strictly larger than $C(\mathbb T)$, as $C(\mathbb T)$ is not reflexive.
However, if we endow $M(\mathbb T)$ with its weak$^*$ topology (i.e. the coarsest topology such that elements of $C(\mathbb T)$ defined continuous linear functionals on $M(\mathbb T)$), then $M(\mathbb T)$ is a locally convex space whose dual is $C(\mathbb T)$.
Thus if we endow $C(\mathbb T)$ with its weak and $M(\mathbb T)$ with its weak$^*$-topology, both have symmetrical roles, on is the dual of the other, and conversely.

ContentLicense: CC BY-SA 4.0


问:
Solving two ODEs that reference each other
I'm an engineer not a mathmatician. I'm trying to learn the maths as I work through an interesting problem I'm working on. I don't know how to proceed because I don't even know which area of maths I need to solve the following ODEs. The problem is they keep referencing each other in a recursive manner.

$$\frac{dq(t)}{dt} = \frac{W(t)}{ \alpha + q(t) }$$
$$\frac{dW(t)}{dt} = \frac{1}{\alpha + q(t)}$$

Note $\alpha$ is a constant and $q(0) = 0$ and $W(0)$ will be some constant e.g. 2.

I cannot solve the ODE by separation of variable because I can't integrate the functions as they keep expanding each other recursively.

I would very much like to learn how to solve the problem not just the solution to this problem. If you could suggest to me where to study next please. At the moment it reminds me of induction but I'm not sure if that applies to continuous systems.

Thank-you very much in advance!

答:
Take the quotient of both equations to get $dq=WdW$ or $q=\frac12W^2+C$. The constant $C$ depends on the solution, but is constant along each solution. Now insert into the second equation and solve
$$
(α+C+\tfrac12W^2)dW=dt
$$
This gives an implicit cubic equation for $W$.


问:
An interesting stopping time problem using PGF's
This is a question I thought of the other day:

Suppose $U_1, U_2, U_3, \cdots \overset{\text{iid}}{\sim} \text{Uniform}(0, 1)$ and let $N = \min\{n: \sum_{i=1}^{n}U_i \ge 1\}$. Compute $\mathbb{P}(N \in 2\mathbb{Z})$ (i.e. probability $N$ is even). How about $\mathbb{P}(N \in 3 \mathbb{Z})$? In general, $\mathbb{P}(N \in k\mathbb{Z})$. Similarly, what is $\mathbb{E}[N|N \in k\mathbb{Z}]$?

I'll post the answer below, but challenge yourself before looking!

答:
Here is a more general solution for any stopping point $0 <c \le 1$ that nests the special case of $c = 1$ posed in the question.

As set out at Random sums of iid Uniform random variables, by simple geometric construction, the cdf of the number $N$ of draws required to exceed $c$, for $0<c \le 1$, is:  $P(N \leq n) = 1 - \large \frac{c^n}{n!}$.

Then the pmf of $N$ is $f(n)$:

The probability that $N$ is even  (using Mathematica/mathStatica here to quickly do the nitty gritties), is:

Then, the pmf of $N$, conditional on $N$ being even is:

Now that we have the conditional pmf, it is straightforwards to derive any conditional moments. The mean of $N$, conditional on $N$ being even is:

In the special case when $c = 1$ (as considered by the OP), the solution simplifies to: $\frac{1+e^2}{2 (e-1)} \approx 2.44112$

Similarly, $\text{Var}(N)$ conditional on $N$ being even is simply:

The same approach works just as well for multiples of 3 or 4 etc


问:
How to Taylor expand a scalar quantity about a unit sphere?
Consider a vector field $\mathbf{v} (r,\theta)$ expressed in the system of (axisymmetric) spherical coordinates with $r$ denoting the radial distance and $\theta$ the polar angle.
We consider a surface $S$ defined by $r = 1 + \epsilon f(\theta)$ where $\epsilon \ll 1$.
We denote by $\mathbf{n}$ the vector normal to that surface.
Specifically,
$$
\mathbb{n} = \mathbb{e}_r - \epsilon f'(\theta) \, \mathbb{e}_\theta \, ,
$$
wherein $\mathbb{e}_r$ and $\mathbb{e}_\theta$ are the basis unit vectors.

Using perturbation analysis, the field $\mathbb{v}$ can be expressed to leading order as
$$
\mathbb{v} = \mathbb{v}^{(0)} + \epsilon \mathbb{v}^{(1)} \, .
$$
Accordingly,
$$
\mathbb{v} \cdot \mathbb{n} =
v_r^{(0)} + \epsilon\left( v_r^{(1)} - f'(\theta) v_\theta^{(0)} \right) \,
$$
wherein $v_r := \mathbb{v} \cdot \mathbb{e}_r$ and $v_\theta := \mathbb{v} \cdot \mathbb{e}_\theta$.

I was wondering how one can Taylor expand the dot product $\mathbb{v} \cdot \mathbb{n}$ about the unit sphere $r=1$.
Specifically,

$$
\bigg. \mathbb{v} \cdot \mathbb{n} \bigg|_{r = 1+\epsilon f(\theta)}
=
[\bigg. \mathbb{v}^{(0)} \cdot \mathbb{e}_r + \epsilon \left( \text{something} \right) ] \bigg|_{r = 1}
+ \mathcal{O}(\epsilon^2) \, .
$$

Any help is highly appreciated!

Thank you

答:
We may view $\vec{v} \cdot \hat{n}$ as a scalar function $g=g(r,\theta)$, and then expand this function in the argument $r$ by regular Taylor series

$$
\vec{v} \cdot \hat{n} \bigg\vert_{r=1}=g(r,\theta) \bigg\vert_{r=1} =g(1,\theta)+(r-1)\partial_rg(1,\theta)+\mathcal{O}((r-1)^2)
$$

With (a mess of arguments, to keep track of what is a function of what)

$$
g(r,\theta)=v_r^{(0)}(r,\theta)+\varepsilon \left(v_r^{(1)}(r,\theta)-f'(\theta) v_{\theta}^{(0)}(r,\theta) \right) \\
\partial_rg(1,\theta)= \partial_rv_r^{(0)}(1,\theta)+\varepsilon \left(\partial_r v_r^{(1)}(1,\theta)-f'(\theta) \partial_r v_{\theta}^{(0)}(1,\theta) \right)
$$

Substituting back

$$
g(r,\theta) \bigg\vert_{r=1} =v_r^{(0)}(1,\theta) + \varepsilon v_r^{(1)}(1,\theta)-\varepsilon f'(\theta)v_{\theta}^{(0)} + (r-1)\left( \partial_r v_r^{(0)}(1,\theta)+\varepsilon \partial_rv_r^{(1)}(1,\theta)-\varepsilon f'(\theta)\partial_r v_\theta ^{(0)}\right)
$$

Collecting powers of $\varepsilon$, and setting $(r-1)=\varepsilon f(\theta)$ we have

$$
g(\varepsilon,\theta) \bigg\vert_{r=1} =v_r^{(0)}(1,\theta)+\varepsilon \left[ v_r^{(1)}(1,\theta)-f'(\theta)v_\theta ^{(0)}(1,\theta)+f(\theta) \partial_r v_r^{(0)}(1,\theta) \right] +\mathcal{O}(\varepsilon^2)
$$

Just to check this is reasonable, I chose some arbitrary functions for $v$ and $f$, and plotted the result above versus the actual functions near $r=1$ across a few values of $\theta$:

We see it does what we want: produce a linear approximant in $\varepsilon$ at $r=1$ for arbitrary $\theta$.


问:
Trigonometry/Geometry triangle help
In right triangle $ABC$, ($\angle BAC = 90$), $D$ is found on side $AC$, $BD$ is the angle bisector of $\angle ABC$ and $BC= \sqrt3 BD$.

Find the value of $\angle CBD$.

With regards to the point $E$ (found on $AC$), establish whether the areas of the circles which circumscribe the triangles $BDE$ and $BCE$ are equivalent.

(Not connected to number 2): Given that $CE=2DE$, find the value of the angle $\angle BEC$.

Note: I tried to call $\angle ABC =\alpha$ and thereby bisect it. I found different trigonometric relationships, but I could not work out how to solve for either the values of the sides nor the angles.

答:
(1) By sine law,

$${\sin({\pi\over 2}-x)\over \sin({\pi\over 2}-2x)}=\sqrt3$$

$$\sin({\pi\over 2}-x)- \sqrt{3}\sin({\pi\over 2}-2x)=0$$

$$\cos(x)-\sqrt{3}\cos(2x)=0$$

$$\cos(x)-\sqrt{3}(2\cos^2(x)-1)=0$$

$$6\cos^2(x)-\sqrt{3}\cos(x)-3=0$$

$$\cos(x)={\sqrt{3}\pm\sqrt{75}\over 12}={\sqrt{3}\pm5\sqrt{3}\over 12}={1\over2}\sqrt{3},-{1\over3}\sqrt{3}$$

Since $0<x<90$ degrees, we take the positive result and $x$ is therefore $30$ degrees.

(2) $BC>BE>BD$ as given.

Angle $C$ correspond to a larger arc in the circumcircle of $\triangle BCE$ than in $\triangle BCD$. This means the circumcircle of $\triangle BCE$ is larger than the circumcircle of $\triangle BCD$.

Angle $BDC$ correspond to a larger arc in the circumcircle of $\triangle BCD$ than in $\triangle BDE$. This means the circumcircle of $\triangle BCD$ is larger than the circumcircle of $\triangle BDE$.

Combining both results, the circumcircle of $\triangle BCE$ is larger than the circumcircle of $\triangle BDE$.

(3) From (1) we already knows $\triangle ABC$ is a $30,60,90$ triangle. This means $CD=2AD$ therefore $CD:DE:AD=4:2:3$. Since $AC:AB=\sqrt{3}$ we know $AE:AB={5\over 9}\sqrt{3}$ and $AB:AE={3\over 5}\sqrt{3}$. Therefore $\angle AEB=\arctan({3\over 5}\sqrt{3})$ and $\angle BEC=\pi - \arctan({3\over 5}\sqrt{3})$

Title: Trigonometry/Geometry triangle help


问:
Derivative of $\exp[(aX+Y)t]$ w.r.t $a$ where $X,Y$ are matrices of dimension $n\times n$
Let $X,Y\in\mathbb{R}^{n\times n}$ and define
$$
F(a,t) = \exp[(aX+Y)t].
$$
Then, compute
$$
\frac{\partial F}{\partial a} = ?
$$

My attempt: Let $M_a = aX + Y$, then
$$
\frac{\partial F}{\partial a} = \frac{\partial}{\partial a} ( I_n + M_a t + M_a^2 t^2 / 2! + M_a^3 t^3 / 3! + \dots ).
$$
Since $\partial M_a^k /\partial a = \sum_{j=0}^{k-1} M_a^j X M_a^{k-1-j}$, we obtain
$$
\frac{\partial F}{\partial a} = \sum_{k=1}^\infty \sum_{j=0}^{k-1} M_a^j X M_a^{k-1-j} \frac{t^k}{k!}.
$$
The question, therefore, is that whether it's possible to write this power series as a function?

答:
$\def\m#1{\left[\begin{array}{c}#1\end{array}\right]}\def\p#1#2{\frac{d #1}{d #2}}$Given a scalar function, like $f(x)=\exp(x)$, and the matrices
$$\eqalign{
M = M(a)&= aXt + Yt,\qquad &\dot M = \dot M(a) = \p{M}{a} = Xt \\
F = F(a) &= f(M),
&\dot F = \dot F(a) = \p{F}{a} \\
}$$
Evaluating the function with a block-triangular argument yields
$$\eqalign{
B \;=\; f\left(\m{M&\dot M\\{0}&M}\right) \;=\; \m{F&\dot F\\0&F} \\
}$$
Block-matrix analogs of the standard basis vectors
$$\eqalign{
e_1 &= \m{1\\0} \quad &e_2 = \m{0\\1} \\
E_1 &= e_1\otimes I_n \qquad &E_2 = e_2\otimes I_n \\
}$$
can be used to extract the upper-right block, which one might call a closed-form solution
$$\eqalign{
\dot F = E_1^T B\, E_2 \\
}$$


问:
How to integrate $\frac{1}{x\ln(1+x)}$?
I tried to evaluate the indefinite integral

$$\int{\frac{dx}{x\ln(1+x)}}.$$

Even Wolfram Alpha can't give any result in terms of standard mathematical functions. Here are some related integrals:

$$\int{\frac{\ln(x)}{\ln(1+x)}}\,dx$$

$$\int{\frac{Li(x)}{1+x}}\,dx,$$

Unfortunately, I'm not able to solve any of them. Is there any special function related with this kind of integrals?

答:
$$I=\int\frac{dx}{x\ln(1+x)}$$
let $u=\ln(1+x)$ so $du=\frac{1}{x+1}dx\Rightarrow dx=e^udu$ and $x=e^u-1$ so:
$$I=\int\frac{e^udu}{(e^u-1)u}=\int\frac{du}{u}+\int\frac{du}{(e^u-1)u}$$
the first integral is obviously quite easy and you can note that:
$$\frac{1}{(e^u-1)}=\frac{e^{-u}}{1-e^{-u}}$$
which might look familiar to you as the sum of a geometric series, try this out and see if it comes to anything :)

Title: How to integrate $\frac{1}{x\ln(1+x)}$?


问:
Calculate $\iint_D \frac{y}{x}  $
I am trying to calculate $\iint_D \frac{y}{x} $ where D is restricted by 3<_3x+y<_5 and by 4x<_y<_7x. I started by switching variables so that s=3x+y and t=$\frac{y}{x}$. However that gives me a very weird Jacobian determinant; $\frac{x^3}{3x^3+y}$, so maybe I shouldn't set t to $\frac{y}{x}$? Does anyone have any suggestion on how to go about this? Here are my calculations this far:

Thanks in advance!

/Nick

答:
Pierre's answer is a good one, but for completeness I'll continue on the track you started. From your picture you have the derivatives in the wrong entries of your matrix so the Jacobian is not correct. Instead it should be

$$J = \frac{x^2}{3x+y}$$

To isolate $x$ substitute $y=xt$ into $s$

$$3x+xt = s \implies x =\frac{s}{3+t}$$

which means the Jacobian is

$$\frac{\frac{s^2}{(3+t)^2}}{s} = \frac{s}{(3+t)^2} $$

and the integral becomes

$$\int_3^5\int_4^7 \frac{st}{(3+t)^2}dtds = \left(\int_3^5 sds\right)\left(\int_4^7\frac{1}{3+t}-\frac{3}{(3+t)^2}dt\right)$$

$$=8\cdot \left(\log\frac{10}{7} -\frac{9}{70}\right)$$

Title: Calculate $\iint_D \frac{y}{x}  $


问:
How to compute $\int^{\infty}_{0} t^{(\frac1n-1)}\cos t \,\mathrm{d}t$?
How to calculate the below integral?
$$
\int^{\infty}_{0} \frac{\cos t}{t^{1-\frac{1}{n}}} \textrm{d}t = \frac{\pi}{2\sin(\frac{\pi}{2n})\Gamma(1-\frac{1}{n})}
$$
where $n\in \mathbb{N}$.</p>

答: This is just the definition of the Incomplete gamma function. The actual value can be found using the recurrence relation gotten by integration by parts as well as the definition of the Gamma function.</p>

ContentLicense: CC BY-SA 3.0


问:
Find the volume of $V\subset \mathbb{R^3}$ limited by a plane and a paraboloid

Let $V=\{(x,y,z)\in\mathbb{R^3}: x^2+y^2\le z, z\le x+2\}$

Then the volume of V is:

(A) Vol(V) = $\frac{75}{8}\pi$

(B) Vol(V) = $\frac{81}{32}\pi$

(C) Vol(V) = $\frac{9}{4}\pi$

(D) Vol(V) = $\frac{45}{8}\pi$

I already understand how the figure sits on 3D space, is limited by a paraboloid on the bottom and a plane on the top, which is inclined. I guessed it would be easier if I used cylindrical or spherical coordinates but I'm having a hard time with the inclined plane. I would appreciate any hint in or guided resolution in order to understand what thing I'm missing. Thank you.

答:
Start with $x^2\leq z\leq x+2.$ So $x^2-x-2\leq 0,$ and $x\in[-1,2].$

Given $x,$ $z\in [x^2,x+2].$

Given $x$ and $z,$ $y\in [-\sqrt{z-x^2},\sqrt{z-x^2}].$

So this gives the integral:

$$\int_{-1}^2 \int_{x^2}^{x+2}\int_{-\sqrt{z-x^2}}^{\sqrt{z-x^2}}dy\,dz\,dx$$


问:
CRC polynomial must include $X^0$?
Must every polynomial include $x^0$, for example $G(x) =\dots+ x^0$?

I could not find any resource which elaborates on it.
From computer networks:

Both the high- and low-order bits of the generator must be 1

答: Yes, it's a basic property of the GF (generator polymonial of cyclic codes, on which CRC are based), its independent coefficient cannot be zero. That's explained in every textbook, see eg here, proof in page 3.</p>

Title: CRC polynomial must include $X^0$?


问:
$z$- score probability

The proportion of people who like playing basketball is 2%, so a
student randomly sample 1000, but find the proportion is 4%, if the
true proportion is 2%. What the probability that this student detect
at least 4% in random sample?

I used the formula, but found the $z$ score is $4.51$ which is impossible I think.

答:
Your z-score is correct. We are conducting a test on a sample proportion, so the zscore under the hypotheses

$$H_0:p=.02\\
H_A:p>.02$$

is

$$\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}}}=\frac{.04-.02}{\sqrt{\frac{.02(.98)}{1000}}}\\
=4.517$$

This is indeed a high zscore, but it's not impossible. You might think it is strange because most tables in books don't have z score that go this high in the tables. It's true. You must use a computer program if your table doesn't carry z scores this high, or you could just report the probability as "approximately zero" should also be fine. The probability of detecting at least .04/getting a zscore of at least 4.517 is

```<code>> 1-pnorm(4.517)
[1] 3.136095e-06
</code>```

Title: $z$- score probability

Body:
Your z-score is correct. We are conducting a test on a sample proportion, so the zscore under the hypotheses

$$H_0:p=.02\\
H_A:p>.02$$

is

$$\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}}}=\frac{.04-.02}{\sqrt{\frac{.02(.98)}{1000}}}\\
=4.517$$

This is indeed a high zscore, but it's not impossible. You might think it is strange because most tables in books don't have z score that go this high in the tables. It's true. You must use a computer program if your table doesn't carry z scores this high, or you could just report the probability as "approximately zero" should also be fine. The probability of detecting at least .04/getting a zscore of at least 4.517 is

> 1-pnorm(4.517)
[1] 3.136095e-06


问:
A fair coin is tossed 100 times. What is the probability that more than 55 heads are observed?

A fair coin is tossed 100 times. What is the probability that more than 55 heads are observed?

I need a clarification on how to use binomial distribution formula in this problem. What i got was:

$$x = 55, \quad \mu = 50, \quad SD = 5$$
$$ z = (55-50)/5 \implies z = 0.84 \implies 1- z = 0.16$$
$$p(x) = 0.16$$

Is this the right answer?

答:
Considering $n = 100$ is very large we are going to approximate this as normal distribution

$$\mu = np = 100*0.5 =50$$
$$\sigma =\sqrt{npq} = \sqrt{100*0.5*0.5} = 5$$

Using the mean and standard deviation we can calculate a Z-score:

$$z = \frac{x - \mu}{\sigma} = \frac{55 - 50}{5} = 1$$
$$P(z > 1) \approx 0.16$$


问:
Can we prove any statement in the first order logic operating only in the space of prenex normal forms?
Let's assume that we have a set of statements written in first order logic (axioms and maybe some proven theorems). Now we want to prove that a given "new" statement is true.

My naive approach to the prove would be like follows:

Take all statements that are used as inputs (axioms and proven theorems) and connect them with and (conjunction) so that we get just one large statement (sentence in FOL).

Bring this "large" statement to a prenex normal form using simple deterministic rules (for example as explained here).

Similarly, bring the statement that you want to prove to the prenex normal form as well.

Now, use formulas like:

$\forall x. P(x) \implies \exists x . P(x)$

to show that the "output" expression (written in the prenex normal form) follows from the "input" expression (also written in the prenex normal form). In other words, we will use only transformations that preserve the prenex normal form. So, all our derivations will be in the space of the prenex normal form.

So, now my questions is if such an approach would work. I guess that it is not the case, because it would be too simple. So, the second part of the question is: Why this approach would not work? Where can if fail?

答:
That's a famous problem, called the Entscheidungsproblem. "is first order logic decidable?", that is: if we're given a sentence is there an effective method for determining whether or not it is valid?

Turing and Church proved (the first with his machines the second with lambda calculus) that it is not the case.

Roughly:

By the Church-Turing thesis every computable function is Turing-computable.
Also consider the Halting Problem: the function $h(e,w)$ that halts with output $1$ if the Turing-machine described
by $e$ halts on input $w$ and outputs $0$ otherwise, is not Turing-computable.

The strategy is to reduce the decision problem to the Halting problem.

So we create two Turing-machines:
For every input $w$ and a Turing Machine $M$;
one sentence is $\tau (M,w)$ which represents the instructions of $M$ and the input $w$ another sentence is $\alpha (M,w)$ expressing "M eventually halts" such that:
$$\vDash \tau (M,w) \to \alpha (M,w) \;\; \text{iff} \;\; \text{M halts for input}\, w$$

Proving that reduces the decision problem to the halting problem, all that remains to do is to code first order logic into numbers so that Turing machines can operate on them.


问:
Sum of $(\tan{12°}/\cos{24°})+(\tan{24°}/\cos{48°})+(\tan{48°}/\cos{96°})+(\tan{96°}/\cos{192°})$
$$(\tan{12°}/\cos{24°})+(\tan{24°}/\cos{48°})+(\tan{48°}/\cos{96°})+(\tan{96°}/\cos{192°})$$

it should equals 0 - zero. But I don't know how?

Can anyone help me with this? Thank You!

Body:
Hint:
$${\tan(x)\over\cos(2x)}={2\sin(x)\cos(x)\over 2\cos^2(x)\cos(2x)}={\sin(2x)\over (\cos(2x)+1)\cos(2x)}=\sin(2x)({1\over \cos(2x)}-{1\over \cos(2x)+1})$$

$$=\tan(2x)-{\sin(2x)\over\cos(2x)+1}=\tan(2x)-{2\sin(x)\cos(x)\over 2\cos^2(x)}=
\tan(2x)-\tan(x)$$


问:
For what $x$ values does the integral have a minimum

Question

My Understanding

I'm really confused because I flipped the function to solve it (turned the integral negative and made lower value upper value...) and I got $2$ as the answer, but when I checked it, it was wrong. I know it can't be $0$ either. So it's $-2$? Or can there be multiple minimum points? Because $-2$ is a minimum when we don't flip the function. But $2$ is the minimum when we flip it. I'm really confused.

答:
Suppose that the problem was changed to minimizing $h(x) = \int_x^2 g(t)dt$ and further suppose that you were (temporarily) only considering values of $x$ such that $x < 1$.  It is clear that

$h(-6) = 0$.

Because of the symmetry around $g(t)$, and because the graph of $g(t)$ for the area between $(-2 \leq t \leq 2)$ is completely positive, it is clear that there is no value for $x < 1$ such that $h(x) < 0.$

Furthermore, when $x$ is constrained to $x < 1$, minimizing $f(x)$ is equivalent to minimizing $h(x)$.  That is, $f(x)$ will have a minimum at $(x = x_0 < 1) \iff h(x)$ is similarly minimized at $(x = x_0 < 1).$

Consequently, the problem is reduced to considering only those values of $x$ such that $x > 1$.  Further, values of $x$ such that $1 \leq x \leq 2$ can be considered at the end of this answer.

Given that, it is clear that minimizing $f(x)$ for $x > 2$ is (again) similar to minimizing $h(x)$ for $x > 2$.  Therefore,  the following considerations apply, when $x>2$.

$\int_x^2 g(t)dt = -\int_2^x g(t)dt.$

The graph of $g(t)$ for the region $2 \leq x \leq 6$ is completely negative.

Therefore, it is immediate that for $x > 2$, the minimum achievable value for $h(x)$ will be $0 = h(10) = -\int_2^{10} g(t)dt.$

Consider $\int_2^1 g(t)dt = -\int_ 1^2 g(t)dt < 0$.

You can see by the symmetry of the graph of $g(t)$, that just as $h(-6) = h(2) = h(10),~f(-6) = f(2) = f(10).$


问:
How can $ zi(e^{iw} + e^{-iw}) = e^{iw} - e^{-iw} $ become $e^{iw}(1-iz) = e^{-iw}(1+iz)$?
I need to get from this from:

$$ zi(e^{iw} + e^{-iw}) = e^{iw} - e^{-iw} $$

Into this other form:

$$e^{iw}(1-iz) = e^{-iw}(1+iz)$$

I've tried doing the sums but the answer is nowhere from where I'm trying to get. I've tried as well multiplying by $(iz)^{-1}$ but that got me nowhere as well.

Thanks for the help.

答:
It's just moving terms from one side of the equality to the other. We have
$$ zi (e^{iw}+e^{-iw}) = e^{iw}- e^{-iw}$$
$$ zi e^{iw}+ zi e^{-iw} = e^{iw}- e^{-iw}$$
$$ zi e^{iw} - e^{iw} =- zi e^{-iw} - e^{-iw}$$
$$ -(1-zi ) e^{iw}=- (1+zi) e^{-iw} $$
$$ (1-zi ) e^{iw}= (1+zi) e^{-iw} $$


问:
Confusion with vector spaces
If we consider $\mathbb{R}^2$ space with cartesian coordinates, then $\vec{V_{c1}}+\vec{V_{c2}} = \overrightarrow{V_{c1}+V_{c2}}$ implies we add components of the vectors. This idea of adding vectors should be invariant to the basis used but if we consider adding vectors using polar coordinates the way we add components is completely different compared to using cartesian coordinates.

Therefore, does this mean when we define a vector space it's basis specific and we must specify how actual addition/scaling operations are implemented under the basis OR are the implementations of the operations overlooked and by specifying that these operations exist we merely mean that we can put the "tail" of one vector on the "head" of another.

答:
The vector space axioms tell how addition and scalar multiplication in a vector space must behave. The Euclidean plane with addition and scalar multiplication defined using the usual coordinate system satisfies those axioms. With those definitions the arithmetic agrees with your intuition about how vectors should behave geometrically.

Those axioms were chosen to capture the idea of "linearity".

There is no reason to expect that arbitrary other coordinate systems will allow you to calculate vector sums one coordinate at a time. In fact, the ones that do are precisely those defined by applying an invertible linear transformation. In vector space terms that's choosing a different basis.

Title: Confusion with vector spaces


问:
Locus of center of tilted rectangle that travels around an ellipse
In order to determine the stochastic visibility (https://link.springer.com/book/10.1007/978-1-4612-2690-1) in Poisson fields where the center of blockages is randomly distributed according to a Poisson point process. We are interested in the following scenario in which a rectangular blockage can obstruct the Fresnel zone between a transmitter and a receiver.

Consider an ellipse with equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$,

and a rectangle of width W and length L tilted at an angle $\theta$ with respect to the horizontal axis (Ox).

This rectangle is shifted on the outer surface of the ellipse for fixed theta as illustrated in the figure.

The movement of the rectangle can be described alternatively as follows: One corner of the rectangle traces part of the ellipse until one side becomes tangent, at this point the rectangle slides along the tangent line, and the process restarts. The path of the rectangle's center seems to consist of four translated arcs of the ellipse connected by four line segments.

Any tips to obtain the parametric equations of the curve drawn by the center of the rectangle are appreciated?

答: Yes, the path is formed by four arcs of ellipse (when the rectangle doesn't slide) and four segments joining them (when the rectangle slides). If the rectangle doesn't slide, its center is translated by the same vector with respect to the contact point (red arrows below) and it then describes an ellipse (red) which is a translated copy of the original ellipse (green). When the rectangle slides its center describes a segment (green in figure below), bur after sliding the vector (blue) between contact point and center has changed, so the center describes a different translated ellipse. And so on.


问:
What is the metric tensor on the n-sphere (hypersphere)?
I am considering the unit sphere (but an extension to one of radius $r$ would be appreciated) centered at the origin. Any coordinate system will do, though the standard angular one (with 1 radial and $n-1$ angular coordinates) would be preferable.

I know that on the 2-sphere we have $ds^2 = d\theta^2+\sin^2(\theta)d\phi^2$ (in spherical coordinates) but I'm not sure how this generalizes to $n$ dimensions.

Added note: If anything can be discovered only about the determinant of the tensor (when presented in matrix form), that would also be quite helpful.

答:
I will define the metric of $S^{n-1}$ via pullback of the Euclidean metric on ${\mathbb{R}}^{n}$.

To start with we take $n$-dimension Cartesian co-ordinates:
$(x_1,x_2......x_n)$.
The metric here is $g_{ij }= \delta_{ij}$, where $\delta$ is the Kronecker delta.

We specify the surface patches of  $S^{n-1}$ by the parametrization  $f$:
$$x_1=r{\cos{\phi_1}},$$

$$x_p=r{\cos{\phi_p}}{\Pi_{m=1}^{p-1}}{\sin{\phi_{m}}},$$

$$x_n=r{\prod_{m=1}^{n-1}}{\sin{\phi_{m}}},$$

Where $r$ is the radius of the hypersphere and the angles have the usual range.

We see that the pullback of the Euclidean metric $g'_{ab} = (f^*g)_{ab}$ is the metric tensor of the hypersphere. Its components are:

$$g'_{ab} = g_{ij} {\frac{\partial{x_i}}{\partial{\phi_a}}} {\frac{\partial{x_j}}{\partial{\phi_b}}} = {\frac{\partial{x_i}}{\partial{\phi_a}}}{\frac{\partial{x_i}}{\partial{\phi_b}}}$$

We get $2$ cases here:

i) $a>b$ or $b>a$, For these components one obtains a series of terms with alternating signs which vanishes, $g'_{ab}=0$ and thus all off-diagonal components of the tensor vanish.

ii) $a=b$,

$$g'_{11}={r^2}$$

$$g'_{aa} ={r^2} \prod_{m=1}^{a-1} \sin^2{\phi_{m}}$$

where $2\leq a\leq {n-1}$

The determinant is very straightforward to calculate:

$$ \det{(g'_{ab})} = {r^2} \prod_{m=1}^{n-1} g'_{mm}$$

Finally, we can write the metric of the hypersphere as:

$$g' = {r^2} \, d\phi_{1}\otimes d\phi_{1} +  {r^2} \sum_{a=2}^{n-1} \left( \prod_{m=1}^{a-1} \sin^2{\phi_{m}} \right) d\phi_{a} \otimes d\phi_{a} $$


问:
Going from the unit-counit definition of the dual pair to the Hom-adjunction
In "Categories and Sheaves" by Kashiwara and Schapira on the page 101, the definition of the dual pair in tensor category is given and in the next theorem it is shown that if $(X,Y)$ is a dual pair, then for every pair of objects $W, Z$ we have $$Hom(Z, W\otimes X)\cong Hom(Z\otimes Y, W)$$
He claims that the maps $$Hom(Z, W\otimes X)\ni u\mapsto \bigg(Z\otimes Y\xrightarrow{u\otimes Id_Y}W\otimes X\otimes Y\xrightarrow{Id_W\otimes \eta}W\otimes 1\cong W\bigg)\in Hom(Z\otimes Y, W)$$
and $$Hom(Z\otimes Y, W)\ni v\mapsto \bigg(Z\cong Z\otimes 1\xrightarrow{Id_Z\otimes\epsilon}Z\otimes Y\otimes X\xrightarrow{v\otimes Id_X}W\otimes X\bigg)\in Hom(Z, W\otimes X)$$
are mutually inverse. However, I was unable to show that (of course I know that unit-counit equations should somehow be involved, but I can't make any use of them) or find any other resource which tells about the proof of that fact something more than that "it is an easy diagram chase". For sure I am missing something obvious. Any help with that problem would be highly appreciated.

答:
Here's a nice way to see this:

Let $C$ be a monoidal category. Then the functor $ad: C\to Fun(C,C)$ given by $X\mapsto X\otimes -$ has a canonical monoidal structure, where $Fun(C,C)$ is (strict) monoidal with composition.

In fact, the monoidal structure just comes from the associators: $ad(X)\circ ad(Y) = X\otimes (Y\otimes -) \cong (X\otimes Y)\otimes -$

The fact that this does in fact define a monoidal structure on $ad$ uses the pentagon identity (this is why it's there in a sense)

In particular, a dual pair $(X,Y)$ gives you functors $F,G: C\to C$ given by $X\otimes -,Y\otimes-$, with a unit $\eta: id \to GF$ (because $id\cong 1\otimes -$) and co-unit $\epsilon : FG\to id$ satisfying the triangle identities. In other words, this defines an adjunction.

Now the fact that adjunctions induce bijections as you suggest is well-known, this is e.g. proposition 1.4 on this nLab page

The point is simply the following: start with $f: Fa\to b$. Apply $G$ to that, to get $GFa \to Gb$, and precompose with $\eta_a$ to get $a\to Gb$. Now apply $F$ to that and postcompose with $\epsilon_b$: you want to prove that you get $f$ back.

By definition, the last map is the composite $Fa\to FGFa \overset{FGf}\to FGb \to b$. But $\epsilon$ is a natural transformation, so you have the following commutative diagram :

$$\require{AMScd}\begin{CD}Fa @>>> FGFa @>{FGf}>> FGb \\
& @V{\epsilon_a}VV @V{\epsilon_b}VV \\
& & Fa @>{f}>> b\end{CD}$$

Now, by definition, the new map you get is the composite "right-right-down". By commutativity of the diagram, this is "right-down-right". By the triangle identity, this is the same as just $f$, because "right-down" is $id_{Fa}$.

This proves that the composite $\hom(Fa,b)\to \hom(a,Gb)\to \hom(Fa,b)$ is the identity, and of course the dual version shows that the other composite is also the identity, so they are inverse bijections.


问:
If I start walking on the border of a bounded domain, will I end up where I started?
If I am on the border of a bounded domain with continuous border (or maybe a manifold?) and I start moving in some direction, will I eventually end up where I started?

And if so, will I eventually reach the point where I started moving in the starting direction?

答: No. To see this first we note that if we continuously rotate points on a circle by some irrational multiple of $2\pi$ that no matter how many times we do it we will never return any point to its starting location. Now extrude the circle into a cylinder and glue the ends together to form a torus. Choose a point on the circles that were glued together and a direction that would wind around the cylinder so the path ends at a point which is an irrational multiple of $2\pi$ away from the starting location on the circle. This straight path will never reach its initial location.


问:
A formula satisfiable from an infinite sequence of formulas

Let $\lbrace\varphi_n\rbrace_{n\geq 1}$ be a sequence of semantically inequivalent formulas s.t. $\vDash \varphi_{n+1} \to \varphi_{n}$. Is there a formula $\psi$ s.t. $(\lbrace\varphi_n\rbrace_{n\geq 1}\vDash \psi) \land (\forall n\geq 1.\psi\vDash \varphi_{n})$?

I think there is no such $\psi$ but I can't formulate the proof quite yet.
By assuming such $\psi$ exists and using compactness theorem we can have a finite subset $\Delta \subset \lbrace\varphi_n\rbrace_{n\geq 1}$ s.t. $\forall n\geq 1.\Delta \vDash \varphi_{n}$ but I can't see where does a contradiction suppose to come from.

答:
Below I'm assuming that "syntactically inequivalent" is a typo for "semantically inequivalent," since otherwise as GVT commented above the problem is trivial.

In fact, there can never be such a $\psi$ - essentially, no nontrivial infinite conjunction is ever expressible by a first-order sentence.

To see this, suppose otherwise and consider the theory $$T=\{\neg\psi\}\cup\{\varphi_i:i\in\mathbb{N}\}.$$ If this is finitely satisfiable, then by compactness it is satisfiable and any model of $T$ witnesses $\{\varphi_i:i\in\mathbb{N}\}\not\models\psi$. So we just need to show that $T$ is finitely satisfiable.

Suppose $T_0\subset T$ is finite. WLOG, $T_0$ has the form $\{\neg\psi\}\cup\{\varphi_i: i\le n\}$ for some $n$. By assumption on the $\varphi_i$s, this is semantically equivalent to the sentence $\neg\psi\wedge\varphi_n$. Now if $\neg\psi\wedge\varphi_n$ were unsatisfiable, this would mean $\varphi_n\models\psi$. But since $\psi\models\varphi_i$ for every $i\in\mathbb{N}$ this would give e.g. $\varphi_n\models\varphi_{n+1}$, which - since $\varphi_{n+1}\models\varphi_n$ by assumption - would imply $\varphi_n\equiv\varphi_{n+1}$, contradicting the assumed inequivalence.

(Note that I'm writing e.g. "$\varphi_{n+1}\models\varphi_n$" instead of "$\models\varphi_{n+1}\rightarrow\varphi_n$." Of course, these are equivalent, and I find the former easier to think about.)


问:
Antiderivative of $e^{x^n}$
How do you prove the following identity ?
$$\int \exp(x^n)dx = - \frac{\Gamma(1/n, -x^n)}{n \cdot i^{2/n}} + C$$

where $\Gamma$ is the Incomplete gamma function and $i$ the imaginary unit (initially, $i^{2/n}$ was written $(-1)^{1/n}$).

答:
Define
$$
\Gamma(a,z) = \int_z^\infty e^{-t} t^{a-1}\;dt,\qquad a > 0.
$$
So
$$
\Gamma\left(\frac{1}{n},-x^n\right) = \int_{-x^n}^\infty e^{-t} t^{1/n-1}\;dt
$$
Differentiate, using the fundamental theorem of calculus and the chain rule:
\begin{align}
\frac{d}{dx} \Gamma\left(\frac{1}{n},-x^n\right)
&= - e^{-(-x^n)}\;(-x^n)^{1/n-1}\;(-n x^{n-1})
= e^{x^n} (-1)^{1/n-1} (x^n)^{1/n-1} (-n) x^{n-1}
\\ &= n e^{x^n} (-1)^{1/n} x^{1-n} x^{n-1}
= n e^{x^n} i^{2/n}
\end{align}

Therefore,
$$
\int e^{x^2}\;dx = \frac{1}{n i^{2/n}} \Gamma\left(\frac{1}{n},-x^n\right)
+C$$

Title: Antiderivative of $e^{x^n}$


问:
Why is my result not matching exactly with the solution of the problem?
I am reading the following problem:

A farmer sells 10 tons of potatoes every year. He also grows enough
seed potatoes for next year's planting. If his yield is exactly 20
times what he sows how many tons of potatoes does he need to plant to
ensure a perpetual supply?

My approach:

Let $p$ be the seeds for regular potatoes.

Let $s$ be the seeds for seed potatoes.

Let $plant = p + s$ i.e. $plant$ is what the farmer sows.

We know that the yield is $20*plant = 20(p+s) = 20p + 20s $

We need to have $10$ tons of potatoes to sell and enough seed potatoes to be able to plant enough potatoes next year in order to have a perpetual supply.

Hence:

$20p = 10\space tons \equiv p = \frac{10\space tons}{20} \approx \frac{10000\space kg}{20} = 500\space kg$

Since we need enough seed potatoes to sow $500kg$ potatoes for next year we need:

$20s = 500\space kg \equiv s=\frac{500}{20} = 25 \space kg$ seed potatoes.

So total plant is $500 + 25 = 525kg$

So to review since the yield is $20x$ the sow we have:

$500\space kg *20 = 10000\space kg \approx 10 \space tons$ to sell

$25\space kg * 20 = 500\space kg$ to plant.

So it seems to check out.

The solution now is different and states:

Let $x$ be the weight of seed potatoes the farmer plants per year.

Then we know that:

$20x = 10 + x \equiv 19x = 10 \equiv x = \frac{10}{19} \space tons = 0.526 \space tons$

I see that my approach was needlessly convoluted as I spitted the types of potato seeds but it seems to me that the final result should match. I.e. my total is $500 + 25 = 525\space kg = 0.525 \space tons$ but it is not really the same as the solution's result ($0.526\space tons$).

So what are the errors in my reasoning since I should be getting $526\space kg$ and not $525kg$ as I have

答:
Your error is you didn't take into account the seed potatoes the farmer will have to plant next year. So $s$ is not $25$ kg, but a little larger.

That is, the farmer must plant enough to produce $10$ tons and plant again a quantity plant. So we have:

$$\mathrm{plant}=\frac{10\;\mathrm{tons}}{20}+\frac{\mathrm{plant}}{20}$$

Here, $\frac{\mathrm{plant}}{20}$ is what must be sowed to be able to plant again the quantity $plant$ next year.

$$\frac{19}{20}\mathrm{plant}=500\;\mathrm{kg}$$

$$\mathrm{plant}=\frac{20}{19}\times500\;\mathrm{kg}\simeq 526.3\;\mathrm{kg}$$


Title: Antiderivative of $e^{x^n}$

Body:
How do you prove the following identity ?
$$\int \exp(x^n)dx = - \frac{\Gamma(1/n, -x^n)}{n \cdot i^{2/n}} + C$$

where $\Gamma$ is the Incomplete gamma function and $i$ the imaginary unit (initially, $i^{2/n}$ was written $(-1)^{1/n}$).

Body:
\begin{align*}
\frac{d}{dx}\left(-\frac{\Gamma(1/n,-x^n)}{n\cdot (-1)^{1/n}}+C\right)&=-\frac{1}{n(-1)^{1/n}}\frac{d}{dx}\int_{-x^n}^\infty t^{1/n-1}e^{-t}dt\\
&= -\frac{1}{n(-1)^{1/n}}\left((-x^n)^{1/n-1}\cdot e^{x^n}\right)\frac{d}{dx}(x^n)\\
&= -\frac{1}{n(-1)^{1/n}} \left((-1)^{1/n-1}\cdot x^{1-n}\cdot e^{x^n}\right)(nx^{n-1})\\
&= \frac{1}{n(-1)^{1/n}} \left((-1)^{1/n}\cdot x^{-(n-1)}\cdot e^{x^n}\right)(nx^{n-1})\\
&=\exp(x^n)
\end{align*}


问:
What is a plane figure covering all sets of diameter $1$?
Problem : (1) Show that there is a plane figure $F$ of least area
which is capable of covering any plane figure of unit diameter.

(2) Try to guess what is $F$.

Proof of (1) : Define $\mathcal{H}$ to be set of all compact convex set in unit
ball. Here $\mathcal{H}$ is a compact metric space and an area function on $\mathcal{H}$ is continuous, which is crucial in the proof of well-known isoperimetric problem.

We will mimic that if we can do : Note that a set of diameter $\leq 1$ can be covered by a convex set of diameter $1$, which can be provved.

If $F'$ is cover for all convex set of diameter $1$, then we define $\mathcal{I}$ to be a set of all such $F'$.

Clealy $\mathcal{I}$ is compact. But we don't know whether the area function is continuous or not.

Proof of (2) : Consider a regular $(2n+1)$-gon $Q_{2n+1}=(p_1\cdots p_{2n+1})$ s.t. $$|p_1p_n|=|p_1p_{n+1}| =1 $$

Each triangle $p_1p_np_{n+1}$ define a circular sector $A$ in unit-ball.

Hence we attach $(2n+1)$-copies of $A$ to $Q_{2n+1}$ so that we have $Q_{2n+1}'$. I guess that $F$ is from overlapping all such $Q_{2n+1}'$.

答:
Here is a sketch of a proof for (1). (Take all sets to be closed in this argument, to avoid some convergence/boundary issues.)

I will assume for this proof that we can place some bound on an optimal region; say, that it is contained inside a closed disc of radius $10^{100}$. This seems patently obvious given that we are looking for a cover of shapes that have diameter $1$, but I haven't found a great argument for this yet; if I see a proof, I will edit it in. Suggestions welcome in the comments.

First, note that we can restrict to convex sets of unit diameter, since taking the convex hull doesn't increase the diameter.

Now consider the set of convex polygons with rational coordinates of diameter $\le1$ - this is a countable set, so we can enumerate them $P_1, P_2,\ldots$.

Observe that any (compact) cover $F$ for all $P_i$ in our list is a cover for all unit-diameter convex sets. This is because, given convex $S$ of unit diameter, the space of possible placements of scaled copies of $S$ is a compact set. But any copy of $S$ scaled by a factor less than $1$ fits inside one of our $P_i$ and therefore inside our cover. So the continuous function on our space of placements measuring the scale of $S$ has a supremum of $1$, and by compactness attains it somewhere in the set.

So how do we find a cover of minimal area for the $P_i$? Let $A$ be the infimum of possible areas; this infimum obviously exists, since a circle of area $\pi$ suffices as a cover.

We place $P_1, P_2,\ldots$ on top of each other iteratively via an inductive process: after placing $P_1,\ldots,P_n$, we choose $P_n$ such that for all $\epsilon>0$, there is a placement of all further $P_i$'s which attains area at most $A+\epsilon$. (Intuitively, we place each new polygon so as to never "box ourselves into a corner" and place a suboptimal lower bound on the area of our cover.)

Why can we choose such a location for $P_n$? By the inductive hypothesis, there is one such placement for every $\epsilon>0$, but by our boundedness assumption, we can restrict our attention to a compact set of possibilities, so one placement will attain this. (The base case is trivial; any placement of $P_1$ will do.)

Repeating this process, we end up with a series of sets ordered by inclusion: $P_1\subset (P_1\cup P_2)\subset (P_1\cup P_2\cup P_3)\subset\ldots$, each of which has area at most $A$. Thus, their union also has area at most $A$, as does its closure. The result gives us a cover $F$ of area exactly $A$ for all planar sets of unit diameter.

Edit: Actually, I'm not sure I can justify that the area of the closure is equal to the area of $A$. It seems obvious enough, but there are some pathological cases in which it fails. Suggestions for how to remedy this are welcome, but in the meantime this proof is certainly incomplete.

In the case where $F$ is convex, this is Lebesgue's universal covering problem; it is an open problem, and the best-known solutions thus far are not particularly obvious or guessable. So I would be quite surprised if (2) can be answered very well.

In practice, I don't think any non-convex covers are known that improve on the best-known convex case? I'm not sure if much work has been done on that side of things, though.


问:
Uniformly continuous function on bounded open interval is bounded
Let $f(x)$ be uniformly continuous on a bounded open interval $a<x<b$. Show that $f$ is bounded (i.e. $\exists M$ such that $|f(x)|\le M \ \forall x\in (a,b)$).

To be honest, I have no idea how to solve thus problem. I tried to pass directly by the definition of uniformly continuous function and extract something, and I passed by cases distinction ($f$ monotonic or not), but I still can't conclude. Intuitively I see why it is true but can't find a good approach to this problem. If someone could give a hint, I would appreciate it. Thanks in advance

答: Assume it is not bounded. Then there is a sequence $(x_n)_{n=1}^\infty\subseteq (a,b)$ such that $|f(x_n)|>n$ for all $n\in\mathbb{N}$. Since this sequence is bounded, it must have some Cauchy subsequence $(x_{n_k})$. Now use uniform continuity to show that $f(x_{n_k})$ is also Cauchy (this easily follows from the definition), and this will be a contradiction.


答: If $f$ was not bounded, there would be a sequence $\{c_n\}$ converging to let say $c$ such that $\{\vert f(c_n) \vert\}$ is unbounded. Derive a contradiction to uniform continuity using a subsequence of $\{a_n\}$.

Title: Uniformly continuous function on bounded open interval is bounded

Body:
Let $f(x)$ be uniformly continuous on a bounded open interval $a<x<b$. Show that $f$ is bounded (i.e. $\exists M$ such that $|f(x)|\le M \ \forall x\in (a,b)$).

To be honest, I have no idea how to solve thus problem. I tried to pass directly by the definition of uniformly continuous function and extract something, and I passed by cases distinction ($f$ monotonic or not), but I still can't conclude. Intuitively I see why it is true but can't find a good approach to this problem. If someone could give a hint, I would appreciate it. Thanks in advance


答: If $f$ is uniformly continous then $\lim_{x\to a_+}f(x) =A$ and $\lim_{x\to b_-} f(x) =B $ exist moreover if you define $ f(a) =A $ and $f(b)=B$ then you obtain a continous function on $[a,b]. $


问:
Find the volume of $V\subset \mathbb{R^3}$ limited by a plane and a paraboloid

Let $V=\{(x,y,z)\in\mathbb{R^3}: x^2+y^2\le z, z\le x+2\}$

Then the volume of V is:

(A) Vol(V) = $\frac{75}{8}\pi$

(B) Vol(V) = $\frac{81}{32}\pi$

(C) Vol(V) = $\frac{9}{4}\pi$

(D) Vol(V) = $\frac{45}{8}\pi$

I already understand how the figure sits on 3D space, is limited by a paraboloid on the bottom and a plane on the top, which is inclined. I guessed it would be easier if I used cylindrical or spherical coordinates but I'm having a hard time with the inclined plane. I would appreciate any hint in or guided resolution in order to understand what thing I'm missing. Thank you.

答:
$V=\{(x,y,z)\in\mathbb{R^3}: x^2+y^2\le z, z\le x+2\}$

You can either use cylindrical coordinates or cartesian coordinates. Spherical coordinates is not the right way to go for this when you have a choice.

Let's first look at the intersection of the paraboloid and the plane.

$z = x + 2 = x^2 + y^2 \implies x^2 - x + y^2 = 2$

You can rewrite it as $(x-\frac{1}{2})^2 + y^2 = \frac{9}{4}$

Hence we use the substitution as below -

$x = \frac{1}{2} + r \cos \theta, y = r \sin \theta$

$0 \leq r \leq \frac{3}{2}, 0 \leq \theta \leq 2\pi$

For the bound of $z$, as you said it is bound below by the paraboloid with $z = x^2 + y^2 = (\frac{1}{2} + r \cos \theta)^2 + r^2 \sin^2\theta$, and above by the plane $z = x + 2 = \frac{1}{2} + r \cos \theta + 2$

So the integral to find the desired volume becomes,

$V = \displaystyle \int_0^{2\pi} \int_0^{3/2} \int_{\frac{1}{4} + r^2 + r \cos \theta}^{\frac{5}{2} + r \cos\theta} r \ dz \ dr \ d\theta$

which is a lot more straightforward integration than if we used $x = r \cos\theta, y = r\sin\theta$.


问:
Find all functions $f:\mathbb{R} \to \mathbb{R}$ such that $f(x^2 + y + f(y)) = 2y + (f(x))^2$

Find all functions $f:\mathbb{R} \to \mathbb{R}$ such that $$f(x^2 + y + f(y)) = 2y + (f(x))^2$$

I tried something like this but that inner $f(y)$ makes things worse to my opinion. Any help is appreciated.

答:
Replacing $x$ with $-x$ in the original equation, we get $f(-x)^2 = f(x)^2$ for any $x$. Equivalently, $f(x)^2 = f(|x|)^2$ holds for all $x$.

We have $f(x^2 + y + f(y)) \geq 2y$. For any $y$, if $f(y) \leq 0$, then putting $x = \sqrt{-f(y)}$ in the inequality gives $f(y) \geq 2y$, hence $y \leq 0$.

In other words, if $y > 0$, then $f(y) > 0$.

For any $y \neq 0$, we have $f(y)^2 = f(|y|)^2 \neq 0$ by 1 and 2. That is, if $f(y) = 0$, then $y = 0$.

Putting $y = -f(x)^2 / 2$ in the original equation, we get $f(\frac{x^2}2 + f(-\frac{x^2}2)) = 0$, which implies $f(-\frac{x^2}2) = -\frac{x^2}2$ by 3. Since $x$ is arbitrary, we see that $f(x) = x$ for any $x \leq 0$.

Combining 4 with 1 and 2, we get $f(x) = x$ for all $x$.


答:
Explanation of your method

I noticed why your answer doesn't work. Your answer involves a recursive procedure of growing seeds to replace the seeds that have been used, that is you reasoned as follows: I grew enough seeds to sell 10 tons, this is 1/2 ton. Now I need to grow enough for the 1/2 ton, this is 1/40 ton. But I need to grow enough for the 1/40 ton, this is 1/800 tons. You cannot stop because if you do, you will run out. So the actual calculation looks like this:

$$\frac 1 2+\frac 1{40}+\frac 1 {800}+...=\frac 1 2\left(1+\frac 1 {20}+\frac 1 {20^2}+...\right)\\
=\frac 1 2\sum_{i=0}^\infty(1/20)^i\text{ infinite geometric series}\\
=\frac 1 2\left(\frac 1{1-1/20}\right)\\
=\frac 1 2\cdot \frac{20}{19}\\
=\frac{10}{19}$$

Explanation of the given solution

The solution says $20p>10+p$ because you will output 20p tons of potatoes. You need this to be at least 10 tons plus p for next year (allowing you to plant p again next year).

```<code>> 10/19
[1] 0.5263158
</code>```

Indeed,

$$\frac{10}{19}*20=10+\frac {10}{19}$$

so if you plant 10/19 tons, you will get 10 tons of potatoes as well as get 10/19 "back", in one fell swoop if I may add.

Title: Why is my result not matching exactly with the solution of the problem?

Body:
I am reading the following problem:

A farmer sells 10 tons of potatoes every year. He also grows enough
seed potatoes for next year's planting. If his yield is exactly 20
times what he sows how many tons of potatoes does he need to plant to
ensure a perpetual supply?

My approach:

Let $p$ be the seeds for regular potatoes.

Let $s$ be the seeds for seed potatoes.

Let $plant = p + s$ i.e. $plant$ is what the farmer sows.

We know that the yield is $20*plant = 20(p+s) = 20p + 20s $

We need to have $10$ tons of potatoes to sell and enough seed potatoes to be able to plant enough potatoes next year in order to have a perpetual supply.

Hence:

$20p = 10\space tons \equiv p = \frac{10\space tons}{20} \approx \frac{10000\space kg}{20} = 500\space kg$

Since we need enough seed potatoes to sow $500kg$ potatoes for next year we need:

$20s = 500\space kg \equiv s=\frac{500}{20} = 25 \space kg$ seed potatoes.

So total plant is $500 + 25 = 525kg$

So to review since the yield is $20x$ the sow we have:

$500\space kg *20 = 10000\space kg \approx 10 \space tons$ to sell

$25\space kg * 20 = 500\space kg$ to plant.

So it seems to check out.

The solution now is different and states:

Let $x$ be the weight of seed potatoes the farmer plants per year.

Then we know that:

$20x = 10 + x \equiv 19x = 10 \equiv x = \frac{10}{19} \space tons = 0.526 \space tons$

I see that my approach was needlessly convoluted as I spitted the types of potato seeds but it seems to me that the final result should match. I.e. my total is $500 + 25 = 525\space kg = 0.525 \space tons$ but it is not really the same as the solution's result ($0.526\space tons$).

So what are the errors in my reasoning since I should be getting $526\space kg$ and not $525kg$ as I have

Body:
Explanation of your method

I noticed why your answer doesn't work. Your answer involves a recursive procedure of growing seeds to replace the seeds that have been used, that is you reasoned as follows: I grew enough seeds to sell 10 tons, this is 1/2 ton. Now I need to grow enough for the 1/2 ton, this is 1/40 ton. But I need to grow enough for the 1/40 ton, this is 1/800 tons. You cannot stop because if you do, you will run out. So the actual calculation looks like this:

$$\frac 1 2+\frac 1{40}+\frac 1 {800}+...=\frac 1 2\left(1+\frac 1 {20}+\frac 1 {20^2}+...\right)\\
=\frac 1 2\sum_{i=0}^\infty(1/20)^i\text{ infinite geometric series}\\
=\frac 1 2\left(\frac 1{1-1/20}\right)\\
=\frac 1 2\cdot \frac{20}{19}\\
=\frac{10}{19}$$

Explanation of the given solution

The solution says $20p>10+p$ because you will output 20p tons of potatoes. You need this to be at least 10 tons plus p for next year (allowing you to plant p again next year).

> 10/19
[1] 0.5263158

Indeed,

$$\frac{10}{19}*20=10+\frac {10}{19}$$

so if you plant 10/19 tons, you will get 10 tons of potatoes as well as get 10/19 "back", in one fell swoop if I may add.


问:
understanding a proof about the set of nowhere differentiable functions

Show that the set of nowhere differentiable functions is residual in $\mathcal{C}[0,1]$.

It suffices to show its complement is first category. Below is part of a proof of this statement. In the proof below, it's not too hard to show that $A = \cup_{k,l} A_{k,l}$ and that each $A_{k,l}$ is closed.

The part I don't really understand is the claim "By our choice of $\omega$ we can choose $x\in [0,1]$ with $0<|x-a| < \frac{1}k$ such that ..." Why does this $x$ exist? I tried explicitly finding this $x$, but to no avail.

答: Recall that $\frac{2\pi}\omega<\min\left\{\frac1k,\frac{r}{2(\ell+m)}\right\}$. Let $x_0=a-\frac{2\pi}\omega$ and $x_1=a+\frac{2\pi}\omega$. If $x_0<x<x_1$, then $0<|x-a|<\frac1k$, and $|x-a|<\frac{r}{2(\ell+m)}$. Moreover, $\sin\omega x_0=0=\sin\omega x_1$, so $\sin\omega x$ goes through at least one full cycle as $x$ ranges over $[x_0,x_1]$, and there must be points $x_2,x_3\in(x_0,x_1)$ such that $\sin\omega x_2=1$ and $\sin\omega x_3=-1$. Let $x=x_2$ if $\sin\omega a\le 0$, and let $x=x_3$ otherwise.


问:
How can I prove that if $a$ and $b$ are integers , and if $a\mid b$ then $b\ge a$

How can I prove that if $a$ and $b$ are integers , and if $a\mid b$ then $b\ge a$

I tried doing this, From $$a\mid b \implies $$ a number $k$ exists ,where $b=ak$ Then I did $$ak>a,$$ removed $a$ from each side and got $$ k>1 $$ which didn't really make sense and didn't show anything. So that's basically all I have got. Thank you in advance.

答:
If you assume $a$ and $b$ to be positive integers, then since $a|b$ so $\dfrac ba$ is an integer greater than or equal to $1$, we have $b\ge a$.

Just put the equality sign also and you are done since $k\ge 1\implies ak=b\ge a $.


问:
Principle of inclusion exclusion and counting surjective functions
Let $f$ be a function from $A \rightarrow B$.

$|A| = 4, |B| = 3$

The number of surjective functions by applying the principle of inclusion exclusion is given by: $3^{4} - \binom{3}{1} 2^{4} + \binom{3}{2}1^{4}$.

The rationale is that we begin with the set of all possible functions, and subtract off the functions with one element in the codomain that is not in the range, and add back the functions with two elements in the codomain not in the range.

However, I don't understand why we are adding back $\binom{3}{2}1^{4}$.

If $\binom{3}{1}2^{4}$ already include functions with two elements from the codomain not in the range, then wouldn't it be done already, as that's all the non-surjective functions?

答:
Let the elements of $B$ be $a$, $b$, and $c$. For a set $S\subseteq B$, let $f_S$ be the number of functions whose range is a subset of $S$. The formula that you're suggesting can be expressed like this:

$f_B-f_{\{a,b\}}-f_{\{a,c\}}-f_{\{b,c\}}.$

To see the problem with this equation, consider the function that sends everything to $a$. That function is not surjective so we want it to make zero contribution to our counting. It is counted by the terms $f_B, f_{\{a,b\}}$, and $f_{\{a,c\}}$, so its total contribution to this formula is $1-1-1+0=-1$. We counted this function twice when we were getting rid of bad functions, so we need to add it back once to cancel out the negative contribution. The same holds for the function that sends everything to $b$ and the function that sends everything to $c$. Hence, the formula we want looks like

$f_B-f_{\{a,b\}}-f_{\{a,c\}}-f_{\{b,c\}}+f_{\{a\}}+f_{\{b\}}+f_{\{c\}}.$

This is counted by the first formula you give, which is why it's correct.


问:
Conditions for a ring to be isomorphic to the product of rings.
Let $R,R_1,\dots,R_n$ be rings. Show that $R\cong R_1 \times \cdots\times R_n$ if and only if there exist ideals $I_1,\dots,I_n$ of $R$ such that

(a) $I_i\cong R_i$ for all $i$

(b) $R= I_1 +\cdots+I_n$

(c) $I_i \cap (I_1+\cdots+I_{i-1}+I_{i+1}+\cdots+I_n)=0$ for all $i$.

Any advice would be greatly appreciated ! Thank you.

答:
One direction is easy: if $R\cong R_1\times\dots\times R_n$, finding ideals satisfying (a), (b) and (c) is trivial.

Let's do the converse. Conditions (b) and (c) say that $R=I_1\oplus I_2\oplus\dots\oplus I_n$ as abelian groups. This is a standard argument, so I'll not talk about it here.

In particular, there is an additive isomorphism
$f\colon I_1\times I_2\times\dots\times I_n\to R$ given by
$$
f(x_1,x_2,\dots,x_n)=x_1+x_2+\dots+x_n.
$$
We need to show that $f$ is also a homomorphism with respect to the multiplication defined componentwise on $I_1\times I_2\times\dots\times I_n$.

Now, if $i\ne j$, $r\in I_i$ and $s\in I_j$, we have $rs=0$. Indeed, $rs\in I_i\cap I_j=\{0\}$ because of condition (c). Thus, if $x=(x_1,x_2,\dots,x_n)$ and $y=(y_1,y_2,\dots,y_n)$ are elements of $I_1\times I_2\times\dots\times I_n$, we have
$$
f(xy)=x_1y_1+x_2y_2+\dots+x_ny_n
$$
by definition, while
$$
f(x)f(y)=\biggl(\sum_{i=1}^n x_i\biggr)\biggl(\sum_{i=1}^n y_i\biggr)
=\sum_{i,j=1}^n x_iy_j=\sum_{i=1}^n x_iy_i
$$
as was to be proved.


问:
What is this formula for mean called and where can I find the derivation
$\overline{x}= \frac{\sum{x-c}}{n}+ c$ I'm solving a question that requires this formula and I have done it but I dont understand what the C stands for here

答:
Normally we define the mean by $$\bar{x} = \frac{\sum_{i=1}^{n} x_{i}}{n}.$$  Now, if we wanted to, we could add and subtract by any number $c$ in each term of the sum, to get
\begin{align}\bar{x} &= \frac{\sum_{i=1}^{n} (x_{i} - c + c)}{n}\\
&= \frac{\sum_{i=1}^{n} (x_{i}-c) + \sum_{i=1}^{n}c}{n}\\
&= \frac{\sum_{i=1}^{n} (x_{i}-c) + cn}{n}\\
&= \frac{\sum_{i=1}^{n}(x_{i}-c)}{n} + c.
\end{align}


问:
Problem in undertsanding the definition of Neighborhood (and it's followed corollaries) in Metric Space Topology
Mtric Sets : (Topology)

(1) Definition : Let $(M,d)$ be any metric set. Let $p$ be a fixed
point in M, and let ϵ be any positive real number. We define a
'neighborhhod $N(p,ϵ)$ of $p$, with radius $ϵ$ as follows :
$N(p,ϵ) = \{q \in M : d(p,q) < ϵ\}$.

(2) Let $A$ be any set with metric defined by $d(p,q) = 0$ if $p=q$, and
$d(p,q)=1$ if $p \neq q$. Then $N(p,ϵ) = p$ if $ϵ \leq 1$ and $N(p,ϵ) = A$.

(3) A 'deleted neighborhood $N^{′}(p,ϵ)$ is defined as follows :
$N^{′}(p,ϵ) = {q, 0,d(p,q)<ϵ}$. Note that, $N^{′}(p,ϵ)$ has precisely
one point less than $N(p,ϵ)$ that is, $N^{′}(p,ϵ) = N(p,ϵ) - {p}$.

I'm unable to undertsand these 3 points in terms of Metric Space topology, I want logical (with Cantors) explanation of these 3 points.

Edit :

(1) I did undertsnad the first definitons more or less, since it's the key point I want some detailed (geometrical wise) explanation of this definiton along with the rest 2 points.

Yes, I tried to think about it, but I'm not convinced enough by my way of thinking, I'm missing the core understanding of it.

(2) Yes, I thought in terms of $\mathbb{R}^2$. $N(p,ϵ)$ is the set of points insde a circel with center $p=(a,b)$  and radius $ϵ$, described as : $\{x,y : (x-a)^2+(y-b)^2 < ϵ^2\}$.

(3) So, the point (1) I undertsnad in the sense of given meaning but I can't undertsand when I use the same key definition to understand the points (2) and (3).

答:
For (1), maybe instead of thinking of an arbitrary metric space, just think of $\mathbb{R}^2$ with the usual Euclidean metric, i.e. just the 'distance' that you're used to in everyday life. More intuitively, think about an infinitely large sheet of paper where you can measure distance with a ruler. First thing you do now is draw any point on this paper and call it $p$. For convenience, let's fix $\epsilon = 1$ but note that any positive radius would be possible. So with that analogy in mind, what is now $N(p,1)$? We have that

$$N(p,1) = \{x \in \mathbb{R}^2 | d(p,x) < 1\},$$

so all points $x$ on your sheet of paper that are less than the distance 1 away from your initial point $p$. If you actually draw this on a piece of paper it should be clear to you that with the distance we are used to this leads to a circle of radius 1 around $p$, just without the border at the outside (these points would be exactly at the distance 1 from $p$.)

The thing now is, we can actually use this definition for other metrics, or 'distance functions' than the one we are used to in everyday life. For instance we can try to see what happens if you apply this reasoning to the metric you have defined in (2). Be aware that this doesn't work like a normal 'infinite sheet of paper' anymore, so you might have to adapt your intuition a bit.


问:
Field of meromorphic functions over Riemann surface of genus g.
I want to study the field $\mathcal{M}(\Sigma_g)$ as Galois extension over $\mathcal{M}(S^2)$ when we have the information about the Galois group of this extension. Can we say the degree of extension of this extension or anything related to $\mathcal{M}(\Sigma_g)$?

答:
The field of meromorphic functions on a Riemann surface depends not only on the genus of the surface $S$ but also its complex structure. Every closed Riemann surface is a smooth projective curve over $\newcommand{\CC}{\mathbb{C}}\CC$. (This is a consequence of the Riemann existence theorem; more generally, Chow's theorem generalizes this to higher-dimensional closed complex manifolds.) In other words, every meromorphic function on a closed Riemann surface is rational.

By the theory of algebraic curves (see, for example, Hartshorne's Algebraic Geometry, section I.6), the category of smooth projective curves over $\CC$ with surjective regular maps is equivalent to the category of function fields of transcendence degree $1$ over $\CC$ with $\CC$-homomorphisms. This tells us two relevant things:

Two smooth projective curves have isomorphic function fields if and only if the curves themselves are isomorphic, i.e., there's a regular (locally polynomial) map between them with regular inverse.

Embeddings of $\CC(t)$, the function field of the complex projective line $\mathbb{P}^1$, into the function field of a smooth projective curve $X$ correspond to surjective regular maps from $X$ to $\mathbb{P}^1$, and the degree of the function field as a field extension of $\CC(t)$ is the degree of the regular map.

In general, given a smooth projective curve $X$, there are many surjective regular maps $X \to \mathbb{P}^1$. The smallest degree of such a map is called the gonality of $X$, and is an important geometric invariant of the curve.

The relation between gonality and genus is somewhat complicated. For example:

A smooth projective curve $X$ over $\CC$ has gonality 1 if and only $X$ has genus 0.

If $X$ has genus 1 or 2, then $X$ has gonality 2. (A curve of gonality 2 is called hyperelliptic, or elliptic in the special case of genus 1. Elliptic curves are a huge area of study in their own right.)

If $X$ has genus 3, then $X$ is either hyperelliptic (and is the completion of the affine curve given by an equation $y^2 = f(x)$ with $\deg(f) = 8$), or is given by a quartic equation in $\mathbb{P}^2$. In the latter case, $X$ has gonality 3 (such a curve is called trigonal).

The Galois groups can vary as well: Among the genus 3 curves of gonality 3, there are the Picard curves, given in an affine patch by an equation of the form $y^3 = f(x)$ with $\deg(f) = 4$, whose function fields are cyclic cubic extensions of $\CC(t)$. But a "generic" genus 3 curve's function field is an $S_3$ cubic extension of $\CC(t)$.

The picture gets increasingly complicated as the genus increases. In general, for genus $\geq 3$, the gonality $\operatorname{gon}(C)$ and the genus $g(C)$ are related by the inequality
$$\operatorname{gon}(C) \leq \left\lfloor \frac{g(C) + 3}{2} \right\rfloor,$$
which is an equality for "generic" curve of any given genus (more precisely: the space of curves of a $g$ that have smaller gonality form a lower-dimensional subvariety of the moduli space of all curves of genus $g$). And of course many different Galois groups can occur among curves of a given gonality, just as many different Galois groups occur among number fields of a given degree.

This is an active area of research with plenty of open problems; there are many interesting questions you can ask about the gonality of algebraic curves. One other keyword that's useful for finding more about this is Brill–Noether theory.

EDIT: I should also mention that, unlike in the number field case, the degree over $\CC(t)$ isn't an intrinsic property of the function field but rather depends on the embedding. (Gonality is only concerned with the least possible degree.) So, for example, the function field $K$ of the elliptic curve with Weierstrass equation $y^2 = x^3 + 1$ (elliptic curve 36.a4 in the LMFDB) is isomorphic to the quadratic extension $\CC(t)(\sqrt{t^3 + 1})$ by projection onto the $x$-coordinate, but also to the cubic extension $\CC(t)(\sqrt[3]{t^2 - 1})$ by projection onto the $y$-coordinate, and many other realizations are possible. (As an even simpler example, every nonzero rational function gives an embedding of $\CC(t)$ into itself.)


问:
How can I prove that the limit doesn't exist?
I have to study the following limit

$$\lim_{(x,y)\to(0,0)}\frac{1-\cos(\sqrt{|xy|})}{x}.$$

I think that this limit does not exist, so I'm trying to prove it. First, I discovered that, if $x=y$, then, the limit is equal to zero. Is there any other variable changing that I can use?

答:
hint

The limit exists and it is zero.
$$(\forall x,y\in \Bbb R)\;\;$$
$$ 1-\cos(\sqrt{|xy|})=2\sin^2(\frac{\sqrt{|xy|}}{2})$$
$$\le \frac 12|xy|$$

because
$$(\forall X\in \Bbb R)\;\;|\sin(X)|\le |X|$$

thus, if $ x\ne 0$,
$$|f(x,y)|\le \frac 12|y|$$


问:
Check proof for a statement of linear independence involving 5 matrices in $M_2(\mathbb{R})$
Let $A,B,C,D,E \in M_2(\mathbb{R})$

I'm asked to prove or disprove that if the set $A = \{EA,EB,EC,ED\}$ is linearly independent so the set $\{A,B,C,D\}$ is linearly independent.

I was having troubles with matrices algebra but this is my shot, I'm not sure I'm allowed to factorise $E$ but here's it:

The set $\{EA,EB,EC,ED\}$ is linearly independent if and only if:
$$\alpha\cdot EA + \beta\cdot EB + \gamma\cdot EC + \delta\cdot ED = 0$$
$$ \Rightarrow \alpha = \beta = \gamma = \delta = 0 $$

Thus, if we factor out $E$ out of the first equation we get that
$$E\cdot(\alpha\cdot A + \beta\cdot B + \gamma\cdot C + \delta\cdot D) = 0$$
We know that $A$ is linearly independent so $E \neq 0$, thus we get that:
$$ \alpha\cdot A + \beta\cdot B + \gamma\cdot C + \delta\cdot D = 0 $$
We also know that
$$\alpha = \beta = \gamma = \delta = 0$$

Thus, $\{A,B,C,D\}$ is linearly independent.

$\blacksquare$

答:
The correct argument goes like this. Assume
$$
a A + b B + c C + d D = 0,
$$
with $a, b, c, d \in \Bbb{R}$.

You have to prove that this implies $a = b = c = d = 0$.

Multiply by $E$ on the left to get
$$
a E A + b  E B + c E C + d E D = 0.
$$
By assumption $EA, EB, EC, ED$ are linearly independent, so this implies $a = b = c = d = 0$, done.


问:
How can the chain rule be explained more rigorously?
The 'proof' for the chain rule that is often used at school is unsatisfying for me because it treats derivatives as fractions:

$$
\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}
$$

However, the more rigorous proofs that are used in University are unfathomable to me because they are intended for people with a much greater level of background knowledge. Is there a way I can think of the chain rule (perhaps not a rigorous proof) that acknowledges that derivatives are a shorthand for limit expressions, but does not use esoteric notation or complicated methods?

For reference, here is a list of things that I do and don't know:

I know how to differentiate from first principles using $$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$

Apart from the chain rule, I know the product rule and the quotient rule (but again, I don't know the proofs for these rules)

I know some limit laws (e.g. the quotient law for limits)

I don't have a rigorous  understanding of limits, but I think I have a good intuitive grasp of them

Similarly, I have an intuitive understanding of continuous vs. discontinuous functions (continuous = not lifting your pen off the page), but I have not been taught the formal definition for continuity

Thank you for reading.

答:
The chain rule helps us differentiate the composition of functions. It is often the case that the variables $y$ and $x$ can be linked by an intermediate variable $u$. If $y=f(u)$, where $u=g(x)$, then $y=f(g(x))$, meaning that $y$ and $x$ are linked through the composite function $f \circ g$. Here is an informal argument, using differentials, that
$$
\frac{dy}{dx}=f'(g(x))g'(x) \, .
$$
If $y=f(g(x))$, then by definition
$$
dy=f(g(x+dx))-f(g(x)) \, .
$$
To make things a little easier to look at, let $u=g(x)$ and $du=g(x+dx)-g(x)$. Then we have
\begin{align}
dy &= f(u+du)-f(u) \\
&= f(u)+f'(u)du - f(u)\\
&= f'(u)du \\
\end{align}
and so
$$
\frac{dy}{dx}=f'(u)\frac{du}{dx}=f'(g(x))g'(x) \, .
$$


问:
Is it possible for the graph of a multivariable function to be a curve instead of a surface?
The graph of a function, let's say for example $f: \mathbb{R}^2  \rightarrow \mathbb{R} $ is a subset of the Cartesian product of the domain and the codomain, in this case of $\mathbb{R}^3$. When speaking about the graph of functions like $f$ the term that is always associated is that one of surface. However i was wondering if there is a possibility that the graph of a function of this type is instead a curve, that can be represented as a vector function $\vec{f}$. For example, a function $F(x, y)$ with the property that the condition of each level set is only satisfied by one point.

$$ F(x, y) = c$$

I imagine that the previous example is not possible, nonetheless i haven't been able to prove it. I don't know if the question is rather trivial, and i have a misconception about the concept of the  graph of a function.

答:
It's a well known result that $\mathbb{R}^2$ has the same cardinality as $\mathbb{R}$, so there is a bijection $h:\mathbb{R}^2\longrightarrow\mathbb{R}$. Then, for each $c\in\mathbb{R}$ there exists one, and only one point $(x,y)\in\mathbb{R}^2$ such that $h(x,y)=c$.

Anyways, the graph of such a function is obviously very difficult to imagine. So probably if you ask almost any condition over $F$ such as continuity, monotony, or whatever, that won't be possible.


问:
Non-compact manifolds with everywhere positive curvature
What are examples of non-compact complete Riemannian manifolds with everywhere positive curvature?

Can you give examples of 2-dimensional surfaces in $\mathbb{R}^3$ with this property?

Note that by Bonnet-Myers theorem, if the curvature is bounded from below by a positive number, then the manifold is compact, so the curvature must decrease to 0.

答: Think about a paraboloid like $z=x^2+y^2$. The idea is that the curvature is positive but tends towards zero as we take $x^2+y^2=r^2$ to infinity. This is a good calculation to work out as an example.


问:
Is my Interpretation of $L^{1}$ and $L^{\infty}$ correct in this case
Let $X:=[0,1]$ and $\mathcal{A}:=\{A \subseteq [0,1]: A \operatorname{or} A^{c} \operatorname{countable}\}$ and $\mu$ be the counting measure.

I am asked to characterize both $L^{1}(X,\mathcal{A}, \mu)$ as well as $L^{\infty}(X,\mathcal{A}, \mu)$

My ideas:

for $L^{\infty}(X,\mathcal{A}, \mu)$ it is clear that in order for $f \in L^{\infty}(X,\mathcal{A},\mu)$: $\sup\limits_{x \in [0,1]}\vert f(x)\vert<\infty$ since every point has positive measure $1$.

for $L^{1}(X,\mathcal{A}, \mu)$, I want to say that:

$f \in L^{1}(X,\mathcal{A}, \mu)\iff f1_{A} \in L^{1}(A,\mathcal{A},\mu) \wedge f1_{A^{c}}=0$ for all $A \subseteq [0,1]$ that are countable

Am I missing something?

答:
We first characterize measurable functions: A function $f:[0,1]\to\mathbb{C}$ is measurable if and only if there exists a sequence $(x_n)\subset [0,1]$ such that $f$ is constant when restricted to $[0,1]\setminus\{x_n:n\geq1\}$. I don't think one can explain this better than the accepted answer in this post.

For $L^\infty(\mu)$: as you observed, if $f\in L^\infty(\mu)$ and $x\in [0,1]$, since $\mu(\{x\})=1>0$, we must have that $|f(x)|<\|f\|_\infty$, so $\sup_{x\in[0,1]}|f(x)|<\infty$. On the other hand, any $f$ that satisfies $\sup_{x\in [0,1]}|f(x)|<\infty$ is obviously an element of $L^\infty(\mu)$.

Now for $L^1(\mu)$. Let $f:[0,1]\to\mathbb{C}$ be a measurable function, so there exists a constant $c\in\mathbb{C}$ and a sequence $(x_n)\subset[0,1]$ so that $f=c$ on $[0,1]\setminus\{x_n:n\geq1\}$. Now
$$\int_{[0,1]}|f|d\mu=\int_{[0,1]\setminus\{x_n:n\geq1\}}|f|d\mu+\int_{\{x_n:n\geq1\}}|f|d\mu=$$

$$\int_{[0,1]\setminus\{x_n:n\geq1\}}|f|d\mu+\sum_{n=1}^\infty\int_{\{x_n\}}|f|d\mu=$$ $$=\int_{[0,1]\setminus\{x_n:n\geq1\}}|f|d\mu+\sum_{n=1}^\infty|f(x_n)|=|c|\cdot\mu([0,1]\setminus\{x_n:n\geq1\})+\sum_{n=1}^\infty|f(x_n)|=$$
$$=|c|\cdot\infty+\sum_{n=1}^\infty|f(x_n)| $$

Now this quantity on the last line of the equation is finite if and only if $c=0$ and $\sum_n|f(x_n)|<\infty$, i.e. $f=0$ everywhere except a countable set of points in $[0,1]$ on which $f$ assumes values that form a sequence of $\ell^1$.


答:
Your proof depends on the following conjecture:

For any matrices $V,W$, if $VW=0$ and $V\ne0$, then $W=0$.

This conjecture, unfortunately, is only true if $V$ is invertible.

Instead of showing that:

If $\{EA,EB,EC,ED\}$ is linearly independent, then $\{A,B,C,D\}$ is linearly independent.

we will prove its contrapositive:

If $\{A,B,C,D\}$ is linearly dependent, then $\{EA,EB,EC,ED\}$ is linearly dependent.

Suppose $\{A,B,C,D\}$ is linearly dependent. Then for some $a,b,c,d\in \mathbb{R}$ (at least one of which is nonzero), we have:
$$aA+bB+cC+dD=0$$

Left multiplying each side of the equation by $E$ yields:
$$a(EA)+b(EB)+c(EC)+d(ED)=0$$
where at least one of $a,b,c,d\in \mathbb{R}$ is nonzero. Thus, by definition, $\{EA,EB,EC,ED\}$ is linearly dependent, as desired.

Title: Check proof for a statement of linear independence involving 5 matrices in $M_2(\mathbb{R})$

Body:
Let $A,B,C,D,E \in M_2(\mathbb{R})$

I'm asked to prove or disprove that if the set $A = \{EA,EB,EC,ED\}$ is linearly independent so the set $\{A,B,C,D\}$ is linearly independent.

I was having troubles with matrices algebra but this is my shot, I'm not sure I'm allowed to factorise $E$ but here's it:

The set $\{EA,EB,EC,ED\}$ is linearly independent if and only if:
$$\alpha\cdot EA + \beta\cdot EB + \gamma\cdot EC + \delta\cdot ED = 0$$
$$ \Rightarrow \alpha = \beta = \gamma = \delta = 0 $$

Thus, if we factor out $E$ out of the first equation we get that
$$E\cdot(\alpha\cdot A + \beta\cdot B + \gamma\cdot C + \delta\cdot D) = 0$$
We know that $A$ is linearly independent so $E \neq 0$, thus we get that:
$$ \alpha\cdot A + \beta\cdot B + \gamma\cdot C + \delta\cdot D = 0 $$
We also know that
$$\alpha = \beta = \gamma = \delta = 0$$

Thus, $\{A,B,C,D\}$ is linearly independent.

$\blacksquare$


问:
Solve $-(x+1)y' - (x+1)y = 0$
I must solve the ode: $$-(x+1)y' - (x+1)y = 0$$
where $y=y(x)$. However, because I have two non-constant coefficients I do not know the method needed to solve this. Can someone please tell me the method needed?

答:
Divide both sides by $-(x+1)$ in order to get
\begin{align*}
y' + y = 0 & \Longleftrightarrow e^{x}y' + e^{x}y = 0\\\\
& \Longleftrightarrow (e^{x}y)' = 0\\\\
& \Longleftrightarrow e^{x}y = c\\\\
& \Longleftrightarrow y(x) = ce^{-x}
\end{align*}

Hopefully this helps!

Title: Solve $-(x+1)y' - (x+1)y = 0$


问:
What is the difference between the small inductive dimension, the large inductive dimension and the lebesgue covering dimension?
I am studying fractals autodidactically (I have never had any topology discipline) and according to Mandelbrot: a fractal has a Hausdorff dimension that exceeds the topological dimension. I understood what Hausdorff's dimension is: it is the value of $s$ (unique) for which a jump between infinity and zero occurs:
$$ \dim_H(X):=\sup\{s:\mathcal{H}^s(F)=\infty\}:=\inf\{s:\mathcal{H}^s(F)=0\} $$
However, I have been trying to understand what the topological dimension is for several days and three concepts always appear to me: the small inductive dimension, the large inductive dimension and the lebesgue covering dimension. I do not understand the definition and the difference between the three and to understand the concept of fratal I just need to understand the concept of topological dimension. Could you explain the concept to me intuitively and with examples and then the mathematical formalism, please?

答:
First off, the term "fractal" does not have a universally agreed upon meaning.  I have written about this before on Math SE, so I won't go into that again.  I will point out that Mandelbrot walked back from the definition you provide in the second edition of The Fractal Geometry of Nature, so even he was not entirely wed to that notion.

Inductive Dimensions

Regarding the meat of your question, you cite three different topological notions of dimension: the small inductive dimension, the large inductive dimension, and the Lebesgue covering dimension.  For the sake of completeness, let us recall the definitions (note: this presentation assumes familiarity with some of the basics of point-set topology, i.e. the definitions of open and closed sets, limit points, boundaries, etc—these definitions are cribbed from James Robinson's Dimensions, Embeddings, and Attractors):

Definition: The small inductive dimension, denoted $\DeclareMathOperator{ind}{ind}\ind(\cdot)$, is defined as follows:

$\ind(\varnothing) = -1$,

$\ind(X) \le n$ if for every point $p \in X$, $p$ has "arbitrarily small" neighborhoods $U$ with $\ind(\partial U) \le n-1$, where $\partial U$ denotes the boundary of $U$, and

$\ind(X) = n$ if $\ind(X) \le n$ but it is not true that $\ind(X) \le n-1$.

The large inductive dimension, denoted $\DeclareMathOperator{Ind}{Ind}\Ind(\cdot)$ is defined similarly, with (2) replaced by

$\Ind(X) \le n$ if for every closed set $A \subseteq X$ and each open set $V \subseteq X$ which contains $A$ there exists an open set $U \subseteq X$ such that
$$ A\subseteq \overline{U} \subseteq V \qquad\text{and}\qquad \Ind(\partial U) \le n-1. $$

Both of these definitions are pretty gnarly, but they are trying to get at the same essential idea:  we can understand the dimension of a set by looking at the boundaries of subsets; the dimension of a space should be greater than the dimension of the boundary of an open subset of the space.

Thinking in a very Euclidean way, an open set in $\mathbb{R}^2$ is (more or less) a disk.  The boundary of a disk is a circle, so $\mathbb{R}^2$ should be one dimension greater than a circle.  An open set in the circle is (roughly) an interval, and the boundary of an interval consists of two points.  Thus the circle should have dimension one greater than a set containing two points.  The boundary of a discrete set is empty, and so a collection of discrete points should have dimension one greater than the emptyset.  Thus

\begin{align}
\DeclareMathOperator{Dim}{dim}
\Dim(\mathbb{R}^2)
&= 1 + \Dim(\text{a circle}) \\
&= 1 + (1 + \Dim(\text{two points})) \\
&= 1 + (1 + (1 + \Dim(\varnothing))) \\
&= 1 + (1 + (1 + -1))) \\
&= 2,
\end{align}
where $\Dim$ denotes some appropriate notion of dimension.

The precise definitions of the small and large inductive dimensions try to capture this idea in a more general way, with slight variations in the way in which the open sets of concern are chosen.  As is often the case, the jump from intuitive to precise requires a great deal of work.

Covering Dimension

The Lebesgue covering dimension approaches things slightly differently:

Definition:  The order of an open covering of a topological space is the largest $n$ such that there exist $n+1$ sets in the covering with non-empty intersection.  An open covering $\mathscr{V}$ is a refinement of an open covering $\mathscr{U}$ if every member of $\mathscr{V}$ is contained in some element of $\mathscr{U}$.

A set $A$ has Lebesgue covering dimension less than or equal to $n$, denoted $\Dim_{L}(A) \le n$, if every covering has a refinement of order less than or equal to $n$.  $\dim_{L}(A) = n$ if $\dim_{L}(A) \le n$, but not $\Dim_{L}(A) \le n-1$.

Again, there is a pretty intuitive observation about Euclidean space running around behind the scenes.  Imagine trying to cover $\mathbb{R}$ by open intervals, and then throwing away as many sets as you can.  What you will likely end up with is a collection of intervals which overlap "just a little" at their ends, see the figure.

No matter how gnarly the initial cover is, I can always find a way to "throw away" enough sets so that the remaining sets don't overlap too much.  Specifically, no point on the real line will be contained in more than two of the covering intervals.

By contrast, if you attempt to cover the plane $\mathbb{R}^2$ by disks such that no point is contained in more than two disks, you will rapidly find that this is impossible—there must be some points contained in three disks.

Further Reading

Robinson, James C., Dimensions, embeddings, and attractors, Cambridge Tracts in Mathematics 186. Cambridge: Cambridge University Press (ISBN 978-0-521-89805-8/hbk). xii, 205 p. (2011). ZBL1222.37004.

Engelking, Ryszard, Dimension theory. A revised and enlarged translation of ”Teoria wymiaru”, Warszawa 1977, by the author, North-Holland Mathematical Library. Vol. 19. Amsterdam, Oxford, New York: North-Holland Publishing Company. Warszawa: PWN - Polish Scientific Publishers. X, 314p. $ 44.50; Dfl. 100.00 (1978). ZBL0401.54029.

Hurewicz, W.; Wallman, H., Dimension theory., 165 p. Princeton University Press (1941). ZBL67.1092.03.


问:
can a sum of n greater than 2 random variables be uniform?
conjecture posed here in one of the answers.

The sum of $ n \geq 3 $ independent r.v.'s distributed on $ [-\alpha, \alpha] $ cannot be uniform on $ [-n\alpha, n\alpha] $.

Is this conjecture true if the random variables are allowed to be distributed on $[-n\alpha, n\alpha]$?

答:
The modifed conjecture is false.

Let:

$U$ be $\text{Unif}(-\frac{3\alpha}4,\frac{3\alpha}4)$.

$X$ take the values $-\frac{3\alpha}4$ and $\frac{3\alpha}4$ each with probability 0.5.

$Y$ take the values $-\frac{6\alpha}4$ and $\frac{6\alpha}4$ each with probability 0.5.

Then,

$U+X\text~\text{Unif}(-\frac{6\alpha}4,\frac{6\alpha}4)$ and $U+X+Y\text~\text{Unif}(-3\alpha,3\alpha)$.

In general, it can be shown by induction that if:

$U$ is $\text{Unif}(-\beta,\beta)$

$X_1$ takes the values $-\beta$ and $\beta$ each with probability 0.5.

$X_2$ takes the values $-2 \beta$ and $2\beta$ each with probability 0.5.

$X_3$ takes the values $-2^2 \beta$ and $2^2 \beta$ each with probability 0.5.

...

$X_{n-1}$ takes the values $-2^{n-2}\beta$ and $2^{n-2}\beta$ each with probability 0.5.

Then,

$$U+\sum_{i=1}^{n-1} X_i\text{~Unif}\left(-2^{n-1}\beta,2^{n-1}\beta\right)$$

Therefore, if we take $\beta=n 2^{1-n}\alpha$, then the sum of the $n$ random variables will be uniform on $ [-n\alpha, n\alpha]$.


答:
Unless I'm overlooking something, the graph in ${\mathbb R}^3$ of $z = f(x,y),$ where

$$ f(x,y) \; = \; \sqrt{x^2 + y^2 - 1} \; + \; \sqrt{1 - x^2 - y^2} $$

is the circle in the $xy$-plane with radius $1$ and center $(0,0,0).$

Note that the domain of $f(x,y)$ is $x^2 + y^2 = 1$ (a circle in the $xy$-plane, not a cylinder in $xyz$-space), and for each point $(a,b)$ such that $a^2 + b^2 = 1$ we have $f(a,b) = 0.$

Title: Is it possible for the graph of a multivariable function to be a curve instead of a surface?

Body:
The graph of a function, let's say for example $f: \mathbb{R}^2  \rightarrow \mathbb{R} $ is a subset of the Cartesian product of the domain and the codomain, in this case of $\mathbb{R}^3$. When speaking about the graph of functions like $f$ the term that is always associated is that one of surface. However i was wondering if there is a possibility that the graph of a function of this type is instead a curve, that can be represented as a vector function $\vec{f}$. For example, a function $F(x, y)$ with the property that the condition of each level set is only satisfied by one point.

$$ F(x, y) = c$$

I imagine that the previous example is not possible, nonetheless i haven't been able to prove it. I don't know if the question is rather trivial, and i have a misconception about the concept of the  graph of a function.


问:
Modular operation on polynomial rings
I originally started modular arithmetic by the following:

$1 \bmod 2$

$1/2$ is $0.5$

$0$ times $2$ is $0$

$1-0=1$

$1$ equals $1 \bmod 2$.

Is it the same way to compute a quotient of a polynomial ring such as $\dfrac{\mathbb{C}[x_{1},\dots,x_{n}]}{x^{2}+y^{2}-z^{2}}$?

$x^3+2xy^2-2xz^2+x \bmod \; x^2+y^2-z^2$

$\dfrac{x^3+2xy^2-2xz^2+x}{x^2+y^2-z^2}$

$\dfrac {(x^3+2xy^2-2xz^2+x) \times (x^3+2xy^2-2xz^2+x)}{(x^2+y^2-z^2)}$

$x^3+2xy^2-2xz^2+x - \dfrac {(x^3+2xy^2-2xz^2+x) \times (x^3+2xy^2-2xz^2+x)}{x^2+y^2-z^2}$

答:
You can think of two numbers being congruent modulo $n$ as follows:

$ a \equiv b \text{ (mod } n)  \iff a - b$ is a multiple of $n$

In your example we get:

$1 - 1 = 0$ which is indeed divisible by $2$, in fact it's divisible by any number $n$.

For another example, suppose we wish to calculate $40$ (mod $17$), we could proceed as you did:

$\frac{40}{17} \text{ is } 2.35294...$

$2$ times $17$ is $34$

$40 -34 = 6$.

so $40 \equiv 6$ (mod $17$).

This is correct but somehow it feels, for lack of a better word, clunky. Instead, we can approach this from a slightly different point of view.

We observe that $40 = 2(17) + 6$

$\implies 40 - 6 = 2(17)$, which is a multiple of 17

$\implies 40 \equiv 6$ (mod $17)$.

We have in essence done the same thing both times, but in the latter calculation it is clearer how we "extract" the multiples of $17$ from $40$.

This latter way of thinking carries over nicely to other quotient rings, such as the polynomial ring you are interested in. That is

$f(x) \equiv g(x) \; \big(\text{mod }h(x)\big) \iff f(x) - g(x) = p(x)h(x)$ for some $p(x) \in \mathbb{C}[x_1,...x_n]$. ie. $f(x) - g(x)$ is a multiple of $h(x)$. We now want to calculate

$x^3+2xy^2-2xz^2+x$ $\big($mod  $x^2+y^2-z^2 \big)$.

As we did in our above example, we wish to "extract" any multiples of $x^2+y^2-z^2$ from $x^3+2xy^2-2xz^2+x$.

We notice that $x^3+2xy^2-2xz^2 + x = x(2x^2+2y^2-2z^2) - x^3 + x = 2x(x^2+y^2-z^2) - x^3 + x$

$\implies (x^3+2xy^2-2xz^2+x) -(- x^3 + x) = 2x(x^2+y^2-z^2)$ is a multiple of $x^2+y^2-z^2$

$\implies x^3+2xy^2-2xz^2+x \equiv - x^3 + x \text{ (mod }x^2+y^2-z^2)$.

One final thing to note is that there are other equivalent ways to state this, for example, we could have instead said that

$x^3+2xy^2-2xz^2 + x = x(x^2+y^2-z^2) + xy^2 - xz^2 + x$

$\implies x^3+2xy^2-2xz^2 + x \equiv xy^2 - xz^2 + x \text{ (mod }x^2+y^2-z^2)$

and this is of course fine too.

Title: Modular operation on polynomial rings


问:
Series/inequalities proof
Suppose that $\{x_m\}$ is a sequence of positive, non-zero numbers such that $\sum_{n=1}^{\infty} {x_n}$ diverges, and let $S_n = x_1 + x_2 + ... + x_n$. Prove the following:

(i) $\sum_{n=1}^{\infty} \frac{x_n}{1 + x_n}$ diverges.

(ii) For all $m, n \geq 1$, $\sum_{j=1}^{n} \frac{x_{m + j}}{S_{m + j}} \geq 1 - \frac{S_m}{S_{m + n}}$

(iii) For all $n \geq 2$, $\frac{x_n}{S_n^2} \leq \frac{S_n - S_{n-1}}{S_nS_{n-1}}$

My attempt:

(i) We check using the convergence/divergence tests:

Divergence test: Consider the limit: $\lim_{n\to\infty} \frac{x_n}{1 + x_n}$. Does this limit $\neq 0$? Inconclusive.

Direct Comparison Test: Is $x_n \leq \frac{x_n}{1 + x_n}$? This only happens if $1 \leq \frac{1}{1 + x_n} \implies 1 \geq 1 + x_n$. This is false for $n \geq 1$ as $x_n$ is non-negative.

Ratio test: Consider the limit: $\lim_{n\to\infty} |\frac{\frac{x_{n+1}}{1 + x_{n+1}}}{\frac{x_{n}}{1 + x_{n}}}| = |\frac{x_{n+1}(1 + x_n)}{(1 + x_{n+1})x_n}|$. Inconclusive.

Root test: Consider the limit: $\lim_{n\to\infty} {|\frac{x_{n}}{(1 + x_{n})}|}^{1/n}$. Inconclusive.

(ii) No idea how to do this one.

(iii) Consider the elementary sum $S_n - S_{n-1}$:

$$S_n - S_{n-1} = (x_1 + x_2 + ... + x_n) - (x_1 + x_2 + ... + x_{n-1}) = x_n$$

Notice that because $x_n$ are non-negative, we have that

$$S_n \geq S_{n-1} \implies S_n^2 \geq S_nS_{n-1} \implies \frac{1}{S_n^2} \leq \frac{1}{S_nS_{n-1}} \implies \frac{x_n}{S_n^2} \leq \frac{x_n}{S_nS_{n-1}} \implies \frac{x_n}{S_n^2} \leq \frac{S_n - S_{n-1}}{S_nS_{n-1}}$$

For the ones I attempted, are they correct? Despite trying for the last few hours to use what I have learned thus far, I still cannot figure out how to go about proving (i) and (ii) ones? Any hints or assistance are much appreciated.

答:
For (i) separate two cases: If $x_n>1/2$ infinitely often then $x_n/(1+x_n)$ does not tend to zero. Otherwise $x_n/(1+x_n) \ge  x_n/(3/2)$ for all large enough $n$.
For (ii) replace $S_{m+j}$ on LHS by $S_{m+n}$ to reduce the sum.

Title: Series/inequalities proof


问:
Ultraweak convergence on matrix over a von Neumann algebra
Let $N$ be a von Neumann algebra and $n\geq1$be an integer. Let $x_\lambda=[x_{i, j}^\lambda]_{i,j} $ be a net in $M_n(N)$. Is it true that $(x_\lambda)_\lambda$ converges ultraweakly to zero in $M_n(N)$ if and only if each entry-net $(x_{i,j}^\lambda)_\lambda$ converges to 0 in the ultraweak topology of $N$?

I was trying to prove that the maps $N\to M_n(N)$ sending an element of $N$ to the matrix with 0 everywhere except the $i, j$ position and the element in the $i, j$ position are ultraweakly continuous and the same for the maps $M_n(N) \to N$ sending a matrix to one of its entries, but I could not do it. Any hint would be appreciated.

答:
I believe the simplest answer goes as follows.

First of all, note that this is easily deduced when $N=\mathcal{B(H)}$ because we have a nice description of the predual of $\mathcal{B(H)}$: the trace-class operators, or, if you prefer,
$$\mathcal{B(H)}_*=\overline{\{\phi\in(\mathcal{B(H)},\|\cdot\|)^*:\phi\text{ is WOT continuous}\}}^{\|\cdot\|_{\mathcal{B(H)}^*}}$$
I actually think that this description is more practical here than the trace class operators. Now the $*$-isomorphism $M_n(\mathcal{B(H)})\cong\mathcal{B(H^{(n)})})$ can be used to obtain the result.

Now let $N$ be an arbitrary von Neumann algebra. Then choose a concrete representation of $N$, i.e. find a faithful representation $\pi:N\to\mathcal{B(H)}$ so that $\pi(N)$ is SOT-closed in $\mathcal{B(H)}$. Then $\pi(N)$ is ultraweakly closed: Indeed, if $(\pi(x_\lambda))\subset\pi(N)$ is a net such that $\pi(x_\lambda)\to y$ ultraweakly, then $\pi(x_\lambda)\to y$ in (WOT) because the ultraweak topology is stronger than the weak topology. But $\pi(N)$ is a convex set and (SOT) closed, thus it is also (WOT) closed, so $y\in\pi(N)$. My point is, $\pi(N)\subset\mathcal{B(H)}$ is an inclusion of von Neumann algebras. Since $N\cong\pi(N)$ and $*$-isomorphisms are topological homeomorphisms with respect to the ultraweak topologies, we may simplify this to $N\subset\mathcal{B(H)}$ as von Neumann algebras. But now $M_n(N)\subset M_n(\mathcal{B(H)})$ as von Neumann algebras and the result follows from the one we proved in the special case of $\mathcal{B(H)}$.

This was hinted to me on MO by Nik Weaver, I deleted my post there because on second thought it was way too elementary for math overflow.


问:
There exists $(z_n)_{n\geq 1}$ monotone $z_n\to x_0$ such that $\sum |\phi(z_i)-\phi(z_{i-1})|=\infty$
I'm reading "Measure and integral" Wheeden & Zygmund. One of its exercises on bounded variation, asked me to prove :

If $V[\phi; a,b]=\infty$. Show there exists a point $z\in[a,b]$ and a monotone sequence $(z_n)_{n\geq 1}$ such that $z_n\to z$ and such that $\sum_{i=1}^{\infty} |\phi(z_i)-\phi(z_{i-1})|=\infty$

The variation of a function $\phi$ on $[a,b]$ is defined as: $$V[\phi;a,b]=\sup_{\Gamma}S_\Gamma=\sup_{\Gamma}\sum_{i=1}^m|\phi(x_i)-\phi(x_{i-1})|,\ \mathrm{where}\ \Gamma=\{x_0,\cdots,x_m\}\  \mathrm{it's\ a \ partition \ of\  } [a,b]  $$

When $V[\phi;a,b]<\infty$ it is say that $\phi$ is of bounded variation on $[a,b]$. In order to prove this result, I already have demostrated the following two previous exercises:

If $V[\phi;a,b]=\infty$ there is a point $z\in [a,b]$ such that either $V[\phi;I]=\infty$ for every subinterval $I$ of $[a,b]$ having $z$ as left-hand endpoint or $V[\phi;I]=\infty$ if the interval has   $z$ as right-hand endpoint.

Let $\phi$ is finite on $[a,b]$ and of bounded variation on every interval $[a+\varepsilon,b]$, $\varepsilon>0$, with $V[\phi;a+\varepsilon,b]\leq M<\infty$. Then $V[\phi;a,b]<\infty.$

Attempt: The idea is to use the previous two results. Let $z$ the point found, such that $V[\phi;z,z+\varepsilon]=\infty$ for every $\varepsilon>0$. Let $M>0$ as large as you want. Since variation in $[z,z+\varepsilon]$ is infinite, i.e., $\sup_{\Gamma} S_\Gamma=\infty$ there should be $\Gamma_{\varepsilon}=\{\tilde{z}_0,\cdots,\tilde{z}_m\}$ partition of $[z,z+\varepsilon]$ such that $S_{\Gamma_\varepsilon}=\sum_{i=1}^m|\phi(\tilde{z_i})-\phi(\tilde{z}_{i-1})|>M$. Since we want $(z_n)$ to be monotone, I was thinking in $z_0=\tilde{z}_m,z_1=\tilde{z}_{m-1},\cdots,z_{m-1}=\tilde{z}_1$, now since we want $z_n\to z=z_0$, I thought $z_k=({z_{k-1}+z})/2$, for all $k\geq m$ which works. I did all that, because I wanted to take advantage of $S_{\Gamma_\varepsilon}>M$, because $$\sum_{i=1}^\infty |\phi(z_i)-\phi(z_{i-1})|\geq S_{\Gamma_\varepsilon}>M $$

At this point I have two serious questions:

Can I conclude that since $$\sum_{i=1}^\infty |\phi(z_i)-\phi(z_{i-1})|>M $$
where $M$ was arbitrarily chosen, that it is infinite? I asked this because If I change $M$ then the partition $\Gamma_\varepsilon$ chance as well, and therefore the sequence, in order to have the "$>M$".

The reason why $$\sum_{i=1}^\infty |\phi(z_i)-\phi(z_{i-1})|\geq S_{\Gamma_\varepsilon} \ \ (**) $$ is based on a question that I have asked here: Triangle inequality infinite terms $|a-b|\leq \sum_{i=1}^\infty|x_{i}-x_{i-1}|$ but the main difference is that, as $i\to \infty$, we have $z_i\to z$ but there is no reason for $\phi(z_i)\to \phi(z)$. If $\phi(z_i)\to \phi(z)$ were true, then ( ** ) is clearly true.

If someone can bring solutions is ok, but I'm also interested on understand if I made mistakes, so I can correct my thinking. This is an independent learning.

答:
I will post the answer, not occured to me before, for if someone will need it in future. The above approach was not fine.

Suppose W.L.O.G that $V[\phi,I]=\infty$ for any subinterval having $z$ as left-hand endpoint (see results above for the existence). There is an $n_0$ such that $z+\frac{1}{n_0}<b$ (to have the subinterval inside $[a,b]$) so for all $n\geq n_0$ there is $\Gamma_{n}$ partition of $[z,z+\frac{1}{n}]$ such that $S_{\Gamma_n}>n$, it follows that $S_{\Gamma_n}\to\infty$ as $n\to\infty$. Consider $$\Gamma = \bigcup_{n\geq n_0}\Gamma_n $$ now let's define $b\geq z_0>z_1>\cdots>z_{n_0-1}>z+\frac{1}{n_0} $ and for $n\geq n_0$ define $$z_n=\max\left\lbrace \Gamma\ \ \  \backslash \ \ \bigcup_{0\leq k\leq n-1} \{z_k\} \right\rbrace$$ for definition, we have for example $z_{n_0}=z+\frac{1}{n_0}$ and $z_n$ monotone decreasing. Indeed, for each $\varepsilon>0$ there is, $M$ and $\tilde{M}$ with $0<M\leq \tilde{M}$ such that $\frac{1}{M}<\varepsilon$ and $z_{\tilde{M}}=z+\frac{1}{M}$ (this is true because for a fix $k$ we have $z_k\leq z+\frac{1}{k}$ most of the time '<' ) so if $n\geq \tilde{M}$ we have $$|z_n-z|=z_n-z\leq z+\frac{1}{M}-z<\varepsilon $$ which proves that $z_n\to z$. This is our candidate for sequence. It remains only to show that $\sum_i|\phi(z_i)-\phi(z_{i-1})|=\infty$. For this notice that, $$ \sum_{i=0}^\infty |\phi(z_i)-\phi(z_{i-1})|\geq S_{\Gamma_n}>n,\ \ \forall n
 $$
taking limits as $n\to\infty$ we get the desired result.
My answer comes after this idea: Is $\max$ a valid order for $\cup_{n\in\mathbb{N}}\Gamma_n$, each $\Gamma_n$ is finite.


问:
Simplest explanation of a limit for a calc I student
I had a calculus I student come to my office hours with the following problem:
Suppose $\lim_{x\to a}\frac{f(x)}{(x-a)^2}=L$. Find $\lim_{x\to a}\frac{f(x)}{x-a}$. I sifted through my real analysis bag of tricks to find a solution for them but unfortunately it seemed to go beyond the scope of their class. If anyone has any pedagogical advice on how to explain this problem to a calculus I student, that would be great!

答:
This is just an elementary limit law.  You've assumed that $\lim_{x \to a} \frac{f(x)}{(x-a)^2}$ exists and is equal to $L$. Then
$$
\lim_{x \to a} \frac{f(x)}{x-a} = \lim_{x \to a} \left(\frac{f(x)}{(x-a)^2} \cdot (x-a) \right)= \lim_{x \to a} \frac{f(x)}{(x-a)^2} \cdot \lim_{x \to a}(x-a) = L\cdot 0 =0.
$$