答:
Strict answer: The description

$\frac1{1-n}$ for $n\in\Bbb N$

does not define a sequence. It defines something very sequence-like, but the fact that it's undefined for $n=1$ means it isn't a sequence. Until this is remedied, there is no point in going further.

Intuitive answer: The convergence of a sequence cares not about what happens for the first few terms (for any reasonable definition of "few").

Title: Does the sequence $\frac{1}{1-n}$ converge?

Body:
New to analysis here. I recognise this sequence converges to $0$, but surely it's unbounded since it's undefined at $n = 1$, which is a contradiction. Could someone just point out the problem in my thought process?

Thanks


问:
Proof function with two variables is continuous $f(x,y) = 2xy/(x^2+y^2)$
I have this function:

$f(x,y)= \frac{2xy}{x^2+y^2}$ if $(x,y)$ is not $(0,0)$.

Now I want to prove that this function is continuous. I have tried using the epsilon delta method, but I don't really know if it works with a two variables function. The $|f(x)-f(a)|$ seems to get very complicated, because of the $(x,y)-(a_1,a_2)$. How should I approach this proof?

答:
Your function is continuous as a composition of continuous functions on $\mathbb{R}^2 \backslash \{(0,0)\}$ as DonAntonio suggested.

Real Analysis: Continuity of a Composition Function

$z(x,y) = \frac{x}{y}$,

$g(x,y) = 2xy$

$h(x,y) = x^2+y^2$

Then $f(x,y) = z(g(x,y),h(x,y))$

ContentLicense: CC BY-SA 4.0


问:
How to solve $\frac{\partial V}{\partial t} + x + \frac{\partial V}{\partial x}- \frac{1}{2} \frac{1}{\left(\frac{\partial V}{\partial x}\right)} = 0$
I'm solving this problem of optimal control using the dynamic programming:
$$
\begin{cases}
\min \displaystyle \int_0^2(x-u)dt + x(2) \\
\dot x = 1+u^2 \\
x(0) = 1
\end{cases}
$$

Then solving the Bellman-Hamilton-Jacobi equation I found the following PDE:
$$\frac{\partial V}{\partial t} + x + \frac{\partial V}{\partial x}- \frac{1}{4} \frac{1}{\left(\frac{\partial V}{\partial x}\right)} = 0$$

The problem gives an hint: In order to solve BHJ equation, we suggest to find the solution in the family of functions $\mathcal{F} = \{V(t,x) = A +Bt + Ct^2 + D\log(3-t) + E(3-t)x\}$ where $A,B,C,D,E$ are all real costants.

My question is:

How can someone derives that all the solution of the BHJ equation are in that family of functions? In other words, how could I manage to solve the BHJ without any hint?

答:
I find that this problem can be more easily solved using Pontryagin's maximum principle, which gives the following Hamiltonian

$$
H(t,x,\lambda) = x - u + \lambda\,(1 + u^2),
$$

such that

$$
\dot{\lambda} = -1, \quad u = \frac{1}{2\,\lambda}
$$

and from the terminal cost is follows that $\lambda(2) = 1$. In this case solving for the co-state as a function of time is easy, namely $\lambda(t) = 3 - t$ and thus

$$
u(t) = \frac{1}{2\,(3 - t)}. \tag{1}
$$

Note that this solution is still independent of $x(0)$.

When formulating the PDE one also obtains that

$$
u = \frac{1}{2\,V_x}, \tag{2}
$$

with $V_x$ shorthand notation for the partial derivative of $V(t,x)$ with respect to $x$. Equating $(2)$ to $(1)$ yields

$$
V_x = 3 - t. \tag{3}
$$

Therefore, the final expression for $V(t,x)$ should be of the form

$$
V(t,x) = x\,(3 - t) + U(t), \tag{4}
$$

with $U(t)$ a yet unknown function of only $t$ and no $x$. Substituting $(4)$ together with $(3)$ in the PDE yields

\begin{align}
0 &= -x + \dot{U} + x + 3 - t - \frac{1}{4} \frac{1}{3 - t}, \\
&= \dot{U} + 3 - t - \frac{1}{4} \frac{1}{3 - t}.
\end{align}

From this it becomes hopefully clear where the function family comes from. Namely, the last term is derived from $(3)$, the second order polynomial comes from integrating $t - 3$ and the logarithm from integrating $(3 - t)^{-1}$.

I am not sure how one could spot this family of functions without Pontryagin's maximum principle. Though, I suspect this is also why the exercise gave the function family, because solving nonlinear PDE's is hard.


问:
Prove that using Fundamental Theorem of Calculus

$\forall x \in \mathbb{R}$ Prove that
$$\displaystyle\int_{0}^{x}\left|t\right|\mathrm{d}t=\dfrac{x|x|}{2}$$

First I distinguish the t value with knowing that absolute value act differently for $t \ge 0$ and $t< 0$. So:

$$\displaystyle |t| = \begin{cases} t &\text{if }t \ge 0 \\ -t &\text{if }t < 0 \end{cases}$$
So if we integrate:

$$\displaystyle \int|t|\mathrm{d}t = \begin{cases} \dfrac{t^2}{2} &\text{if } t>0 \\ -\dfrac{t^2}{2} &\text{if }t < 0 \end{cases}$$

And then I plugged the variables. But I didn't get the result. Thanks.

答: If $x\geqslant 0$, then\begin{align}\int_0^x|t|\,\mathrm dt&=\int_0^xt\,\mathrm dt\\&=\frac{x^2}2\\&=\frac{x|x|}2\end{align}and if $x\leqslant0$, then\begin{align}\int_0^x|t|\,\mathrm dt&=-\int_x^0-t\,\mathrm dt\\&=\int_x^0t\,\mathrm dt\\&=-\frac{x^2}2\\&=\frac{x|x|}2.\end{align}


问:
Find the Laurent expansion of $(1-z)e^{1/z}$ - When can we use Taylor series to find Laurent series?
I'm currently taking a course in mathemathical tools, where we are covering complex analysis (Note that this course is not very rigorous and we cover complex analysis in only 4 lectures). The Laurent series have been introduced as:

Laurent series$$
\begin{array}{l}
f(z)=\sum_{n=-\infty}^{\infty} a_{n}\left(z-z_{0}\right)^{n} \\
a_{n}=\frac{1}{2 \pi i} \oint_{C} \frac{f\left(z^{\prime}\right) d z^{\prime}}{\left(z^{\prime}-z_{0}\right)^{n+1}}
\end{array}
$$

However, I noticed that the above integral isn't used in the solutions to any of the problems regarding laurent series. So I'm trying to understand why we can avoid using it.
What I understood so far, is that the difference between the Taylor and Laurent series, is that the Laurent series also contains negative powers. Where it's not possible to expand a Taylor series around a point $z_0$ where f is not analytically, the opposite applies for the Laurent series.
In the case we expand f around a point where f is analytically, the Laurent series and Taylor expansion will be the same.
If that is the correct understanding, please help me understand the solution to the following problem:

Find the Laurent series for $f(z)=(1-z)e^{1/z}$ about $z=0$

What I thought should be my approach is using the formula above solving the contour integral, since $f(z)$ is non-analytically at $z=0$. However, the solution uses directly that $e^{1/z}=\sum_{n=0}^{\infty} \frac{z^{-n}}{n !}$. So here comes my first question:

Why is the Laurent series of $e^{1/z}=\sum_{n=0}^{\infty} \frac{z^{-n}}{n !}$.

It seems to me, that they have just substituted $z \rightarrow 1/z$ in the Taylor series for $e^z$. But $e^{1/z}$ is not analytic at $z=0$, so the Taylor expansion around that point shouldn't exist? And if that is not the Taylor series, how do we know it's the Laurent series?

After this they multiply the two expressions together:
$(z-1) e^{1 / z}=(z-1) \sum_{n=0}^{\infty} \frac{z^{-n}}{n !}=z-\sum_{n=1}^{\infty}\left(\frac{n}{n+1}\right) \frac{z^{-n}}{n !}$

Is that because $z-1$ is a polynomial and thus is analytically around $z=0$ and is it's own Taylor series and Laurent series. So we can just find the Laurent series of each factor and multiply them together to find the final Laurent series?

Fx: Consider $h(z)=f(z)g(z)$, if we want to find the Laurent series around $z=z_0$. If the Laurent series of f(z) around $z_0$ is $\sum_{n=-\infty}^{\infty} a_{n}\left(z-z_{0}\right)^{n}$ and of $g(z)$ is $\sum_{n=-\infty}^{\infty} b_{n}\left(z-z_{0}\right)^{n}$. Is the Laurent series of $h(z)$ then:
$h(z)=(\sum_{n=-\infty}^{\infty} a_{n}\left(z-z_{0}\right)^{n}) (\sum_{n=-\infty}^{\infty} b_{n}\left(z-z_{0}\right)^{n})$?

答:
Let us first briefly try to understand the differences between the Laurent and the Taylor series. You have correctly observed that the Taylor series is only defined in a region in which the function is analytic. To be a little bit more specific: If $f$ is analytic at $z_0$, its Taylor series converges uniformly in the biggest circle centered around $z_0$ in which $f$ is analytic, i.e. the radius of the circle is the distance to the closest singularity. The Laurent series is however define around points were $f$ need not be analytic. In fact, we are interested in Laurent series presicely because they have this property. So suppose $f$ has an isolated singularity at $z_0$, then we can expand $f$ in a Laurent series which converges uniformly on a punctured circle around $z_0$ or an annulus. As you have observed the most apparent difference between the two series is that the Laurent series contains negative powers. This really means that is the sum of two series, one converging in an area $|z-z_0| < R$ and another convering in an area $|z-z_0| > r$. The sum will converge in the annulus $r < |z-z_0| < R$. Intuitively, the negative powers "make up for" the behaviour of $f$ close to the singularity.

However, if $f$ was actually analytic at $z_0$, the Laurent series will equal the Taylor series. You can easily see this by studying your expression for $a_n$. If $f$ is analytic at $z_0$ this equal $0$ for all $n<0$.

Now, to the computational aspect of your question. We usually ever use the integral formula for $a_n$, simply because it is tedious and unnecessary. It is used in proofs, but rarely to find Laurent expansions. It is a legitimate question why you can simply substitute $1/z$ for $z$ in the Taylor series of $e^z$. Think of it this way: Let $w \in \mathbb{C}$. You know that $e^w = \sum_{k=0}^{\infty} \frac{w^k}{k!}$. Now, let $z \neq 0$. $\frac{1}{z}$ is a complex number, so substituting $w = \frac{1}{z}$ is completely legitimate, and we get

$$e^{\frac{1}{z}} = \sum_{k=0}^{\infty} \frac{1}{z^k k!}.$$

Notice that this works whever $z \neq 0$, and that the series has now turned into a Laurent series. We claim that it converges for all $z \neq 0$, which follows immediately from how we constructed the series. The "annulus" in which the series converges is $\{ z \in \mathbb{C} \, | \, |z| > 0 \} = \mathbb{C} \setminus \{ 0 \}$. Using this we can easily find the Laurent series you are asked to find:

$$(1-z)e^{1/z}=e^{1/z}-ze^{1/z}=\sum_{k=0}^{\infty} \frac{1}{z^k k!}-z \sum_{k=0}^{\infty} \frac{1}{z^k k!} = \sum_{k=0}^{\infty} \frac{1}{z^k k!} - \sum_{k=0}^{\infty} \frac{1}{z^{k-1} k!}$$

Rearranging this, we get

$$(1-z)e^{1/z} = -z + \sum_{k=0}^{\infty} \bigg(\frac{1}{k!}-\frac{1}{(k+1)!} \bigg) \frac{1}{z^k}.$$

At last, how do we know that this is the Laurent series and not some other series converging to $(1-z)e^{1/z}$ ? Well, this is a theorem one would usually prove in an introductory course in complex analysis. A series representation of a function of the form

$$\sum_{n=-\infty}^{\infty} a_n z^n$$

is unique.


问:
Cardinality != Density?
I was in a discussion where I argued that the density of two sets of the same cardinality could be different in respect to the infinite range of non-negative integers. Does cardinality imply that any set of ${\aleph_0}$ has equal density to any other set of the same ${\aleph_0}$?

Also, does cardinality imply equivalence in set size? Or is a formal ranking of different tiers which cannot be compared?

Ok, got that so far. Expanding question now:

Densities can be different but the count is the same. Infinite == Infinite iff cardinality is equivalent. Why doesn't a different density imply infinite set != infinite set even if cardinality is equivalent?

答:
I think that it is not completely clear to you what cardinality and density means.

Firstly cardinality is the "rawest" notion of size. For two sets to have the same cardinality they need only a bijection, which is to say "There exists a table with two columns, one for each set, and all the elements appear exactly once in that column."

The notion which you describe is called Dedekind infinite, which is to have the same cardinality as a proper subset.

As for density, this can be treated in several aspects:

A dense ordered set is a set such that for every $x,y$ there is $z$ which is strictly between them. The rationals are such set, the positive integers are not.

If there is a first and/or last point (for example all the non-negative rationals which are less or equal to one) then it is customary to exclude the end points from the density criteria and say that there is a first/last point as well.

Topological dense sets: if $A$ is a topological space then $D\subseteq A$ is dense in $A$ if and only if its intersection with every open set is nonempty. The rationals are dense in the real numbers, in this sense, since every non empty interval contains a rational point inside it.

If the topological space is endowed with a linear order topology (e.g. the real numbers) then a dense set in the topological sense is a dense order as well (although not vice versa, consider all the real numbers versus the rationals between $-1$ to $1$)

The real numbers give example of a very large space (its cardinality is continuum) with a very small (countable) subset which is dense (topologically).

The third notion of size is measure, which roughly translates to volume. A large set is a set of positive measure (or equal measure to the measure of the entire space).

The Lebesgue measure is a way to determine the volume of subsets of the real numbers in the way we want it to work, that is if we just shift around the set it will not change its volume and if we stretch it the volume will increase as the factor we stretched by.

One can build a set which is of Lebesgue measure $0$ (i.e. has no volume at all), not dense at any point (that is if a point is outside the set then it has an open interval which does not intersect this set) and yet it is of the cardinality of the continuum. This can be generalized to have any volume that we want as well.

Therefore the notions are hardly related, if we have a very small (in cardinality) set which is dense (large topologically) and another set which is very large in cardinality but topologically speaking is very very small.

To sum up, there are many different ways to measure how big a set is and it gets more and more complex with every new technique you acquire (filters, cofinality, and more). They may or may not be related or correlated, but the case is usually that we want a new way of "sizing" up sets, mostly because the ones we have are insufficient or cumbersome for the task at hand.

ContentLicense: CC BY-SA 3.0

Title: Cardinality != Density?


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

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

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

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

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

Seeing no pattern, I did the same once more:

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

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

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

Edit:

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

答:
To proceed

Actually we can compute $\lim a_n$ explicitly, since
$$
\frac{c_{n+1}}{c_n}=\frac{1}{c-\frac{1}{c_n}} \text{ converges to } 1 \quad \Leftrightarrow \quad c_n \text{ converges to } \frac{1}{c-1}  $$
which is also equivalent to that
$ a_n \text{ converges to } \frac{1}{c-1}-1 $,
where $c=\frac{c_2}{c_3}+\frac{1}{c_2}$.

We know$\ c_2=a_2+1=\frac{(1+a_1)(1+b_1)}{b_1},c_3=a_3+1=\frac{(+a_2)(1+b_2)}{b_2}=\frac{(1+a_1)^2(1+b_1)^2}{b_1(1+b_1+a_1b_1)}$

Now we get $c=\frac{1+2b_1+a_1b_1}{(1+a_1)(1+b_1)}$ and
$$
a_n\to \frac{1}{c-1}-1=\frac{1+2a_1+a_1b_1}{b_1-a_1}
$$
Edit:

Thanks to Mihai Dicu, once we notice that $\frac{1}{1+a_{n+1}}-\frac{1}{1+b_{n+1}}=\frac{1}{1+a_{n}}-\frac{1}{1+b_{n}}=\frac{1}{1+a_{1}}-\frac{1}{1+b_{1}}$, it is rather easy to find the limit. From my previous answer, if $b_1>a_1>0$, we can actually show that both $a_n$ and $b_n$ are stictly increasing, and $b_n\to +\infty$. Therefore,
$$
\frac{1}{1+a_n}=\frac{1}{1+b_n}+\frac{1}{1+a_1}-\frac{1}{1+b_1}>\frac{1}{1+a_1}-\frac{1}{1+b_1}>0 $$
which shows that ${a_n}$ is bounded, and thus its limits exists.
$$
\lim_{n\to \infty}b_n=+\infty \Rightarrow \lim_{n\to \infty}\frac{1}{1+a_n}=\frac{1}{1+a_1}-\frac{1}{1+b_1}=\frac{b_1-a_1}{(1+a_1)(1+b_1)}
$$
which is equivalent to
$$
\lim_{n\to \infty}a_n=\frac{(1+a_1)(1+b_1)}{b_1-a_1}-1=\frac{1+2a_1+a_1b_1}{b_1-a_1}
$$


问:
implicit functions in higher Dimensions
For Instance: Given the function: $\varphi(x) = x\,\varphi(x)^2+2\,x^2\,e^{\varphi(x)}$.

It has to be shown that it exist a function $\varphi: U \to\mathbb{R}$ that solves it.

In this case $f(x,y) = x\,y^2+2\,x^2\,e^{y}-y\quad$ hence $D_yf(x_0,y_0) = -1$ where $(x_0,y_0) = (0,0)$.

So it is solvable.

Now I ask myself how a function $\varphi(x)$ that depends on more variables $y_1,y_2$ would look like.

In general if there is a function $f(x,y_1,y_2) = \left(\begin{array}{c}x+y_1+{y_2}^2\\x^2-y_2 \end{array}\right)$. How to state a function $\varphi(x)$ ?

For me, this means rearranging for y. But if there are 2 of them? Something like $\varphi(x)=\left(\begin{array}{c}y_1 \\y_2 \end{array}\right)$ ?

答:
Dini's theorem in higher dimension states that, given a function $\underline f:\Omega\subseteq\mathbb R^m\times\mathbb R^n\to\mathbb R^n$, where $\Omega$ is an open set and $\underline f\in\mathcal C^1(\Omega)$, if $(\underline x_0,\underline {y_0})\in\Omega $ such that $\underline f(\underline x_0,\underline {y_0})=0$ and $J\underline f|_{\underline y}(\underline x_0,\underline {y_0})$ is invertible$\implies$

$\exists U(\underline x_0)$ and $V(\underline f(\underline{y_0}))$, with $\overline{U(\underline x_0)\times V(\underline f(\underline{y_0}))}\subseteq\Omega$ such that

$$(1)\text{ The set }Z_{\underline f}=\{\underline x\in\mathbb \Omega\:\underline f(\underline x)=\underline 0\}\text{ resticted to }U(\underline x_0)\times V(\underline f(\underline{y_0}))$$$$\text{ coincides with the graph of the implicit function }\underline \psi: U(\underline x_0)\to V(\underline f(\underline{y_0}))$$
$$(2)\text{ }\underline\psi\in\mathcal C^1(U(\underline x_0))\text{ and }J\underline\psi|_{\underline y}(\underline x)=-(J\underline f|_{\underline y})^{-1}(J\underline f|_{\underline x})(\underline x,\underline\psi(\underline x))\forall \underline x\in U(\underline x_0)$$

In order to see how it works, consider the function $G:\mathbb R^5\to\mathbb R^2$ such that
$$G(u,v,x,y,z)=\begin{pmatrix}ux^2-yz^2+v\cos y-1\\xe^y+vz-\cos u-2 \end{pmatrix}$$
and try to verify that $G(u,v,x,y,z)=(0,0)$ implicitly defines a function $(u,v)=\psi (x,y,z)$ in a neighbourhood of the point $(0,1,2,0,1)$.

Hint: if $JG=\begin{pmatrix}x^2&\cos y&2ux&-z^2-v\sin y&-2yz\\\sin u&z&e^y&xe^y&v \end{pmatrix}$ how can you see if $G$ satisfies the Hp of Dini's theorem?

Title: implicit functions in higher Dimensions


问:
Countable set of finite sets is countable (without the axiom of choice).
So i have the infinite family of sets:
$$\{A_0, A_1, A_2 \dots\}$$
and $\forall i \in \omega,\; A_i$ is finite. I need to show that $\cup A_i$ is countable  ($\exists f: \cup A_i \rightarrow \omega $ and f is bijection).
I seem to understand it intuitively, but I can't prove it. I also don't have to use the axiom of choice. Any tips ?

答:
This statement cannot be proven without appeal to axiom of choice. To be precise, it is consistent with ZF axioms that there is a countable family of sets $A_0,A_1,A_2,\dots$ such that each $A_i$ has two elements, and yet $\bigcup_{i\in\omega}A_i$ is not countable. Families of sets like that are sometimes known as "Russell's socks", because of the relation to (in)famous analogy about how you can't pick out socks from infinitely many pairs of socks without using the axiom of choice.

You can find this result and more interesting results concerning Russell's socks in this paper by Herrlich and Tachtsis.


答:
1

But $e^{1/z}$ is not analytic at $z=0$, so the Taylor expansion around that point shouldn't exist?

No. Not being analytic at point $z_0$ does not necessarily mean that Taylor expansion around $z_0$ does not exist. Consider the example $\frac{\sin z}{z}$ at $0$. The singularity is removable. Though $z=0$ is not a removable singularity of $e^{1/z}$.

On the other hand, you are looking for the Laurent series of $e^{1/z}$, not the "Taylor series".

All you do in this case is to look at the Taylor series (or definition, depending on how you define the exponential function) of $e^w$ at $w=0$:
$$
e^w=\sum_{n=0}^\infty \frac{w^n}{n!}\tag{1}
$$
This expansion is valide for any $w\in\mathbb{C}$. So you can do the substitution $w=\frac{1}{z}$ for any $z\ne 0$.

How do you know this is the Laurent series in your definition? The substitution above gives you the coefficients $a_n$. You can verify that
$$
a_n=\frac{1}{2\pi i}\oint_C\frac{f(z)}{z^{n+1}}dz
$$
with $f(z)=e^{1/z}$.

Or, you can simply use the uniqueness of the coefficients.

2

Think about what finding Laurent series of $g(z):=(1-z)e^{1/z}$ about $z=0$ really means. What you really looking for is a double-sided series
$$
\sum_{n=-\infty}^\infty a_nz^n\tag{2}
$$
such that at every $z$ with $0<|z|<R$ (for some $R$), (2) is equal to $g(z)$.

On the one hand, you can use your integral definition to find the coefficients. On the other hand, you can directly find the coefficients so that
$$
g(z)=\sum_{n=-\infty}^\infty a_nz^n
$$
at every $z$ with $0<|z|<R$. Since the coefficients of the Laurent series are unique, you would have the same answer no matter which way you take.

Title: Find the Laurent expansion of $(1-z)e^{1/z}$ - When can we use Taylor series to find Laurent series?

Body:
I'm currently taking a course in mathemathical tools, where we are covering complex analysis (Note that this course is not very rigorous and we cover complex analysis in only 4 lectures). The Laurent series have been introduced as:

Laurent series$$
\begin{array}{l}
f(z)=\sum_{n=-\infty}^{\infty} a_{n}\left(z-z_{0}\right)^{n} \\
a_{n}=\frac{1}{2 \pi i} \oint_{C} \frac{f\left(z^{\prime}\right) d z^{\prime}}{\left(z^{\prime}-z_{0}\right)^{n+1}}
\end{array}
$$

However, I noticed that the above integral isn't used in the solutions to any of the problems regarding laurent series. So I'm trying to understand why we can avoid using it.
What I understood so far, is that the difference between the Taylor and Laurent series, is that the Laurent series also contains negative powers. Where it's not possible to expand a Taylor series around a point $z_0$ where f is not analytically, the opposite applies for the Laurent series.
In the case we expand f around a point where f is analytically, the Laurent series and Taylor expansion will be the same.
If that is the correct understanding, please help me understand the solution to the following problem:

Find the Laurent series for $f(z)=(1-z)e^{1/z}$ about $z=0$

What I thought should be my approach is using the formula above solving the contour integral, since $f(z)$ is non-analytically at $z=0$. However, the solution uses directly that $e^{1/z}=\sum_{n=0}^{\infty} \frac{z^{-n}}{n !}$. So here comes my first question:

Why is the Laurent series of $e^{1/z}=\sum_{n=0}^{\infty} \frac{z^{-n}}{n !}$.

It seems to me, that they have just substituted $z \rightarrow 1/z$ in the Taylor series for $e^z$. But $e^{1/z}$ is not analytic at $z=0$, so the Taylor expansion around that point shouldn't exist? And if that is not the Taylor series, how do we know it's the Laurent series?

After this they multiply the two expressions together:
$(z-1) e^{1 / z}=(z-1) \sum_{n=0}^{\infty} \frac{z^{-n}}{n !}=z-\sum_{n=1}^{\infty}\left(\frac{n}{n+1}\right) \frac{z^{-n}}{n !}$

Is that because $z-1$ is a polynomial and thus is analytically around $z=0$ and is it's own Taylor series and Laurent series. So we can just find the Laurent series of each factor and multiply them together to find the final Laurent series?

Fx: Consider $h(z)=f(z)g(z)$, if we want to find the Laurent series around $z=z_0$. If the Laurent series of f(z) around $z_0$ is $\sum_{n=-\infty}^{\infty} a_{n}\left(z-z_{0}\right)^{n}$ and of $g(z)$ is $\sum_{n=-\infty}^{\infty} b_{n}\left(z-z_{0}\right)^{n}$. Is the Laurent series of $h(z)$ then:
$h(z)=(\sum_{n=-\infty}^{\infty} a_{n}\left(z-z_{0}\right)^{n}) (\sum_{n=-\infty}^{\infty} b_{n}\left(z-z_{0}\right)^{n})$?


问:
$\lim_{x \to 0 }|x| = 0$
Given $\epsilon > 0$ we want to $\delta > 0$ such as $|x - 0| < \delta \Rightarrow |f(x) - f(0)| < \epsilon $.

How
\begin{align}
|x| =
\begin{cases}
x, \text{ if } x > 0\\
-x, \text{ if } x < 0
\end{cases}
\end{align}

From desigualities we have

if  $x > 0$ : $|x - 0| < \epsilon$

and if $x < 0 : $

$|-x - 0 | < \epsilon \Rightarrow |-1||x - 0| < \epsilon \Rightarrow |x - 0| < \epsilon$

Because this, we take it $\delta = \epsilon$.

So, $|x - 0|<\delta \Rightarrow |x - 0|<\epsilon$.

Therefore, for definition $\lim_{x \to 0 }|x| = 0$

It's right?

With the hints I rewrited like this:
Given $\epsilon > 0$ we want to $\delta > 0$ such as $|x - 0| < \delta
> \Rightarrow ||x| - 0| < \epsilon $.

As $||x| - 0| = |x|$ we have that $|x| < \epsilon$ so we take it
$\delta = \epsilon$.

So, $|x - 0|<\delta \Rightarrow ||x| -
0|<\epsilon$.

Therefore, for definition $\lim_{x \to 0 }|x| = 0$

答:
You are looking at the wrong definition. What you need should be the definition of $\lim_{x\to 0}f(x)=0$, not $\lim_{x\to 0}f(x)=f(0)$. The latter is the continuity of $f$ at $x=0$. The function $f(x)=|x|$ is indeed continuous at $0$ though.

The estimate you want should be
$$
||x|-0|<\epsilon.
$$

You don't need to discuss two cases. Simply observe that
$$
||x|-0|=|x|
$$

Title: $\lim_{x \to 0 }|x| = 0$


问:
$G/Z$ cannot be isomorphic to quaternion group

Let $Q_8$ be the quaternion group. Show that $G/Z$ can never be isomorphic to $Q_8$, where $Z$ is the center of $G$.

Hint: if $G/Z\cong Q_8$, show that $G$ has two abelian subgroups of index $2$.

I'm trying to prove the hint by looking at the canonical homomorphism $\pi:G\rightarrow G/Z$. Well, $Q_8$ has some abelian subgroups of index $2$, such as $H=\{1,-1,i,-i\}$, but that doesn't seem to map to abelian subgroups of index $2$ of $G$. (We know that $\pi(ab)=\pi(a)\pi(b)=\pi(b)\pi(a)=\pi(ba)$ for $a,b$ in $\pi^{-1}(H)$, but that only yields $aba^{-1}b^{-1}\in Z$, not $aba^{-1}b^{-1}=1$.)

答:
$Q_8$ has cyclic subgroups of index $2$, such as $\langle i\rangle$ and $\langle j\rangle$.
Their inverse images in $G$ are of index $2$ and their respective centres contains at least the centre $Z$ of $G$. Hence their quotient by their centre is cyclic and finally $\pi^{-1}(\langle i\rangle)$ and $\pi^{-1}(\langle j\rangle)$ are abelian (cf. Jykri Lahtonen's comment).
Select $g\in G$ with $\pi(g)=-1$. Then $g\in \pi^{-1}(\langle i\rangle)\cap \pi^{-1}(\langle j\rangle)$, hence it commutes with all elements of $\pi^{-1}(\langle i\rangle)$ and all elements of $\pi^{-1}(\langle j\rangle)$, therefore also with all elements of $\pi^{-1}(\langle i,j\rangle)=\pi^{-1}(Q_8)=G$, contradicting $g\notin Z$.</p>


问:
Prove that if $f^{(k)}(f^{-1}(x))$ exists, and is nonzero, then $(f^{-1})^{(k)}(x)$ exists.
I think I found an error and just want to confirm.

This is from Calculus by Michael Spivak, 3rd edition, Chapter 12 (Inverse Functions), problem 21.

12-21. Prove that if $f^{(k)}(f^{-1}(x))$ exists, and is nonzero, then $(f^{-1})^{(k)}(x)$ exists.

This appears to be false. For example, take $f(x) = x^3$.

In this case
$$f^{-1}(x) = x^{\frac{1}{3}}$$
$$f(0) = f^{-1}(0) = 0$$
$$f^{(3)}(0) = 6$$

If the statement is true, $(f^{-1})^{(3)}(0)$ would exist. However, $(f^{-1})'(0)$ doesn't exist (because $f'(0) = 0$) so higher order derivatives cannot.

I think the question should instead say "Prove that if $f^{(k)}(f^{-1}(x))$ exists, and $f'(f^{-1}(x))$ is nonzero, then $(f^{-1})^{(k)}(x)$ exists.

Does that sound about right to y'all?

Edit: As discussed with Paul Frost, the revised version of the statement

Prove that if $f^{(k)}(f^{-1}(x))$ exists, and $f'(f^{-1}(x))$ is nonzero, then $(f^{-1})^{(k)}(x)$ exists

is missing one other thing.

We need $f$ to be one-one on some interval containing $f^{-1}(x)$.

For cases $k\geq 2$, $f$ one-one is implied by $f'(f^{-1}(x))≠0$, and the existence of $″(f^{-1}(x))$ (which gives us the existence of $f'$ for all points in some interval containing $f^{-1}(x)$ and continuity of $f'$ at $f^{-1}(x)$). We can prove the (corrected) statement for $k≥2$. However, we don't know if $f$ is one-one for the case $k=1$. As problem 11-63 shows, $f'(f^{-1}(x))\neq 0$ by itself isn't enough to guarantee $f$ is increasing (or decreasing) over some interval containing $f^{-1}(x)$

But that's ok. The higher derivative cases are the ones we're interested in anyway, so we can take $k = 2$ as the base case and use induction from there.

The corrected corrected statement reads:
"Prove that if $f^{(k)}(f^{-1}(x))$ exists, and $f'(f^{-1}(x))$ is nonzero, then $(f^{-1})^{(k)}(x)$ exists for all $k\geq 2$, $k\in \mathbb{N}$

For the $k=1$ case, we have theorem 12-5.

答:
Your counterexample is correct; Spivak seems to be a bit unattentive. He also does not say anything about $f$ except that $f^{(k)}(f^{-1}(x_0))$ exists (and is $\ne 0$, which you have shown to be inadequate). The existence of $f^{(k)}(f^{-1}(x_0))$ means more precisely that (a) there is an open neigborhood $U$ of $f^{-1}(x_0)$ on which $f$ is $(k-1)$-times differentiable, and (b) $f^{(k)}(f^{-1}(x_0))$ exists.

In Theorems 12.4 and 12.5 Spivak states for an injective function $f$ which is differentiable at $f^{-1}(x_0)$:

If $f'(f^{-1}(x_0)) = 0$, then $f^{-1}$ is not differentiable at $x_0$.

If $f'(f^{-1}(x_0)) \ne 0$, then $f^{-1}$ is differentiable at $x_0$ and $(f^{-1})'(x_0) = \dfrac{1}{f'(f^{-1}(x_0))}$.

Thus the minimal requirement for the existence of $(f^{-1})'(x_0)$, and hence for the existence of $(f^{-1})^{(k)}(x_0)$ for some $k \ge 1$, is
$$f'(f^{-1}(x_0)) \ne 0 \tag{1}.$$

Now assume that $f^{(k)}(f^{-1}(x_0))$ exists (with any value of $f^{(k)}(f^{-1}(x_0))$) and $(1)$ is satisfied. We shall (a) construct an open neigborhood $V$ of $x_0$ such that $f^{-1}$ is $(k-1)$-times differentiable on $V$ (b) show that $(f^{-1})^{(k)}(x_0))$ exists.

For $k = 1$ nothing remains to do; 2. applies. We now consider $k > 1$.

Since $f''(f^{-1}(x_0))$ exists, $f'$ is continuous at $f^{-1}(x_0)$, thus $f'(y) \ne 0$ for all $y$ in some open neigborhood $U' \subset U$ of $f^{-1}(x_0)$. For $x \in V = f(U')$ we have $f^{-1}(x) \in U'$, thus by 2. $f^{-1}$ is differentiable at $x$ with derivative  $(f^{-1})'(x) =  \dfrac{1}{f'(f^{-1}(x))}$. Clearly $x_0 = f(f^{-1}(x_0)) \in V$. Since $f^{-1}$ is continuous (Theorem 12.3), $V = f(U') = (f^{-1})^{-1}(U')$ is open. Thus $V$ is an open neigborhood of $x_0$.

We shall now show by induction on $r$ that $(f^{-1})^{(r)}(x_0)$ exists on $V$ for all $r < k$. The base case $r=1$ has been treated above. Let us now consider $r=2$, the details of the general case are left to you.

Quotient rule and chain rule show that $(f^{-1})''(x)$ exists for all $x \in V$, where
$$(f^{-1})''(x) = -\dfrac{f''(f^{-1}(x))}{(f'(f^{-1}(x)))^3} .$$
Induction produces for all $x \in V$ a formula for $(f^{-1})^{(r)}(x)$ as a fraction whose denominator is a power of $f'(f^{-1}(x))$ and whose nominator has summands which are products of powers of derivatives $f^{(i)}(f^{-1}(x))$ with $i = 1,\ldots,r$.

This shows that  $(f^{-1})^{(k-1)}(x_0)$ exists on $V$. Clearly we can once more differentiate at $x_0$.


问:
How do I solve this non linear equation?
I will like to know how to solve this non linear equation.
$$\frac{\tan 6a}{\tan 2a}=\frac{3}{2}$$
Thank you for your time.

答: Using the angle addition formula for $\tan(\theta + 2 \theta)$, $\tan 3 \theta = \frac{3 \tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta}$. Thus $\frac{\tan 3 \theta}{\tan \theta} = \frac{3 - \tan^2 \theta}{1 - 3 \tan^2 \theta} = \frac{2}{3}$ which is now a quadratic equation in terms of $\tan \theta$, where $\theta = 2a$.


问:
Probability of 4 drivers in the same group
Yesterday I asked a question that a colleague quickly answered @mjw but some things were not clear and the question was closed. I will reformulate it and count on everyone's support again.

The problem is this: I have a go-kart group in which 32 drivers participate. In each stage, the drivers are divided into 2 groups (GroupA and GroupB). The names of the 32 drivers (D1, D2, ..., D32) are drawn one by one and distributed in the two groups according to the example below:

1º - I take a name randomly from a box and place it in the "A" group;

2º - I take a name randomly from a box and place it in the "B" group;

3º - I take a name randomly from a box and place it in the "A" group;

...

I will perform these steps until all the drivers are distributed in the groups as shown. Each group will have 16 drivers.

Question: What are the chances of drivers "D1", "D2", "D3", and "D4" being in the same group (GroupA or GroupB) considering that all drivers can be chosen equally?

I tried to think about solving this problem using the inclusion-exclusion theory, or conditional probability, but I failed. If I'm not wrong, the number of possible combinations is

$\binom{32}{16}$

But how can I consider the division into 2 groups?

答:
Let $D_1$ be in any of the two groups. Then the probability that

$D_2$ is in the same group as $D_1$ is $\displaystyle \frac{15}{31}$. This is because there are still $15$ spots in the group and $31$ remaining drivers to choose from.

Similarly,

that $D_3$ is also in the same group is $\displaystyle \frac{14}{30}$.

that $D_4$ is also in the same group is $\displaystyle \frac{13}{29}$.

So probability $P$ that $D_1, D_2, D_3, D_4$ are all in the same group is

$ = \displaystyle \frac{15}{31} \cdot \frac{14}{30} \cdot \frac{13}{29} = \frac{91}{899}$

Or you can also state it as $P = \displaystyle \frac{2 \cdot {28 \choose 12}} {32 \choose 16} = \frac{91}{899}$

(the numerator is number of ways of having $D_1, D_2, D_3, D_4$ drivers together in either group $A$ or group $B$ and rest $12$ drivers in the group chosen from remaining $28$ drivers).


问:
Fundamental solution of heat equation: Show $\lvert v(x,t)-v(x,t')\rvert\leq \left\lVert\frac{d^2}{dx^2}u_0\right\rVert_{L^\infty}\lvert t-t'\rvert$
Let
$$
v(x,t)=\frac{1}{\sqrt{4\pi t}}\int_\mathbb{R} e^{-(x-y)^2/(4t)}u_0(y)\, dy
$$
be the fundamental solution of the heat equation.

Would like to prove that
$$
\lvert v(x,t)-v(x,t')\rvert\leq \left\lVert\frac{d^2}{dx^2}u_0\right\rVert_{L^\infty}\lvert t-t'\rvert.
$$

The only way I see is to use integration by parts.

Substituting $z=\frac{y-x}{2\sqrt{t}}$, I get

$$
v(x,t)=\frac{1}{\sqrt{\pi}}\int_\mathbb{R}e^{-z^2} u_0(2\sqrt{t}z+x)\, dz
$$

Integrating by parts twice gives me
$$
v(x,t)=\left[\frac{\textrm{erf}(z)}{2}\cdot u_0(2\sqrt{t}z+x)\right]_{-\infty}^\infty - \sqrt{t}\left[\left(z\cdot\textrm{erf}(z)+\frac{e^{-z^2}}{\sqrt{\pi}}\right)\cdot u_0'(2\sqrt{t}z+x)\right]_{-\infty}^\infty+2t\int_\mathbb{R} \left(z\cdot\textrm{erf}(z)+\frac{e^{-z^2}}{\sqrt{\pi}}\right)u_0''(2\sqrt{t}z+x)\, dz
$$
and, analogously,
$$
v(x,t')=\left[\frac{\textrm{erf}(z)}{2}\cdot u_0(2\sqrt{t'}z+x)\right]_{-\infty}^\infty - \sqrt{t'}\left[\left(z\cdot\textrm{erf}(z)+\frac{e^{-z^2}}{\sqrt{\pi}}\right)\cdot u_0'(2\sqrt{t'}z+x)\right]_{-\infty}^\infty+2t'\int_\mathbb{R} \left(z\cdot\textrm{erf}(z)+\frac{e^{-z^2}}{\sqrt{\pi}}\right)u_0''(2\sqrt{t'}z+x)\, dz
$$

Looks terrible. However, can one use this to get the desired result?

答:
I have not tried to go through the entire argument myself, but have you considered using the fact that

$$|v(x, t)- v(x, t')| = |\int_{t'}^t v_{\tau}(x, \tau) \, d \tau| = |\int_{t'}^t v_{xx} (x, \tau) \, d \tau| \leq \lVert v_{xx}(x, \tau) \rVert_{\infty} |t-t'|$$

and try to relate $\lVert v_{xx}(x, \tau) \rVert_{\infty}$ to $\lVert u_0 ''(x) \rVert_{\infty}$?


问:
Analytic formula for integral $I_p(\theta) := \int_0^{2\pi}\cos(t)^p\cos(t-\theta)^pdt$
Let $p$ be a nonnegative integer and $\theta \in [0, \pi]$.

Question. What is an analytic formula for the integral $I_p(\theta) := \int_0^{2\pi}\cos(t)^p\cos(t-\theta)^pdt$ ?

Note. My ultimate goal is to compute $I_p(1)$, $I_p(0)$, $I_p'(0)$, and $I_p''(0)$.

My guess is everything can be succinctly expressed in terms of special functions (gamma, beta, etc.).

答:
Amazingly, I found this kind of integrals in my old (65 years old !) cookbook.

The idea is to write
$$\cos (t) \cos (t-\theta )=\frac{1}{2} (\cos (\theta )+\cos (2 t-\theta ))$$ and to use the binomial expansion when you raise it to power $p$.

$$I_p(\theta)= \int_0^{2\pi}\big[\cos(t)\,\cos(t-\theta)\big]^p\,dt=  4^{1-p}\,\pi \,J_p(\theta)$$

Now, the first results
$$\left(
\begin{array}{cc}
 p & J_p(\theta) \\
 1 & \cos (\theta ) \\
 2 & 2+\cos (2 \theta ) \\
 3 & 9 \cos (\theta )+\cos (3 \theta ) \\
 4 & 18+16 \cos (2 \theta )+\cos (4 \theta ) \\
 5 & 100 \cos (\theta )+25 \cos (3 \theta )+\cos (5 \theta ) \\
 6 & 200+225 \cos (2 \theta )+36 \cos (4 \theta )+\cos (6 \theta )\\
 7 & 1225 \cos (\theta )+441 \cos (3 \theta )+49 \cos (5 \theta )+\cos (7 \theta )
   \\
 8 &2450+ 3136 \cos (2 \theta )+784 \cos (4 \theta )+64 \cos (6 \theta )+\cos (8 \theta
   ) \\
 9 & 15876 \cos (\theta )+7056 \cos (3 \theta )+1296 \cos (5 \theta )+81 \cos (7
   \theta )+\cos (9 \theta )
\end{array}
\right)$$ The patterns (for odd and even values of $p$) seem to be quite clear.


问:
Theory of bijection with no finite cycle has trivial associated geometry
I am working through the following exercise in Tent and Ziegler's A Course in Model Theory.

Exercise 5.7.2. Show that the theory of an infinite set equipped with a
bijection without finite cycles is strongly minimal and that the associated
geometry is trivial.

I think I've managed to show that this theory has QE and hence (having no relation symbol) can only define finite or cofinite sets, and hence is strongly minimal. What I don't understand is what is meant by "the associated geometry is trivial". I am guessing this refers to the pregeometry $(M, \text{cl})$, where for each $A\subseteq M$, $\text{cl}(A)=\text{acl}^M(A)\cap M$. But what does it mean for it to be trivial and how do we prove it? Thank you!

答:
Whenever we have a strongly minimal theory, we get a pregeometry on any model $M$ of that theory: $(M,\text{acl})$. There's no reason to write $\text{acl}^M(A)\cap M$, because $\text{acl}(A) = \text{acl}^M(A)\subseteq M$. Maybe what you were thinking of is that if $D$ is a strongly minimal set definable in a model $M$, then the pregeometry is $(D,\text{cl})$, where for $A\subseteq D$, $\text{cl}(A) = \text{acl}^M(A)\cap D$.

The meaning of "associated geometry" is explained in Section C.1 (bottom of p. 205). If  $(X,\text{cl})$ is a pregeometry, let $X^\bullet = X\setminus \text{cl}(\emptyset)$. Define an equivalence relation $\sim$ on $X^\bullet$ by $x\sim y$ iff $\text{cl}(x) = \text{cl}(y)$. Finally, let $X' = X^\bullet/{\sim}$, and define $\text{cl}'$ on $X'$ by $\text{cl}'(A/{\sim}) = \text{cl}(A)^\bullet/{\sim}$. That is, $\text{cl}'$ of a set $B$ of equivalence classes contains the equivalence class of every point which is in the $\text{cl}$-closure of a set of representatives for $B$  but not in $\text{cl}(\emptyset)$. Then $(X',\text{cl}')$ is a geometry, the geometry associated to $(X,\text{cl})$.

The meaning of "trivial" is explained on p. 80 in passing: a geometry is trivial if $\text{cl}(A) = A$ for all sets $A$.

A word about terminology: Tent and Ziegler provide another definition of "trivial" in the context of pregeometries. On p. 207, we read that a pregeometry is trivial if $\text{cl}(A\cup B) =  \text{cl}(A) \cup \text{cl}(B)$ for all sets $A$ and $B$. Some people (including me) prefer call such  a pregeometry disintegrated to distinguish it from the more trivial meaning of trivial given above. But the reason for the terminology is that a pregometry is trivial (in the disintegrated sense) if and only if its associated geometry is trivial (in the truly trivial sense). This is a good exercise to strengthen your understanding of the associated geometry construction.

As for how to solve the exercise from Tent and Ziegler: Now that you know what the words mean, I think it's better to try to solve it yourself. If you get stuck, drop me a comment.


问:
fractional power of a fraction - $\log(s/s_o)^{2/3}$
I have an expression $\log(s/s_o)^{2/3}$

I would like to rewrite this as a difference of logs using the rules of logs.  If there was no exponent I could rewrite this as $\log(s)-\log(s_o)$. But the fractional exponent is confusing me.

I know that if I was dealing with a whole number, e.g. $\log(x)^{2/3}$ then I could just rewrite as $\sqrt[3]{\log(x)^2}$.

I don't understand how to combine these two rules here?

How can I rewrite $\log(s/s_o)^{2/3}$ as a difference of logs?

答: $$\log\left(\frac s{s_0}\right)^{2/3}=\frac23\log\frac s{s_0}\stackrel{\text{assuming}\,s,\,s_0>0}=\frac23\left(\log s-\log s_0\right)$$


Title: fractional power of a fraction - $\log(s/s_o)^{2/3}$

Body:
I have an expression $\log(s/s_o)^{2/3}$

I would like to rewrite this as a difference of logs using the rules of logs.  If there was no exponent I could rewrite this as $\log(s)-\log(s_o)$. But the fractional exponent is confusing me.

I know that if I was dealing with a whole number, e.g. $\log(x)^{2/3}$ then I could just rewrite as $\sqrt[3]{\log(x)^2}$.

I don't understand how to combine these two rules here?

How can I rewrite $\log(s/s_o)^{2/3}$ as a difference of logs?

Body: $$\log\left(\frac{s}{s_0}\right)^{2/3}=\frac{2}{3}[\log s-\log s_0]=\frac{2}{3}\log s-\frac{2}{3}\log s_0$$


问:
Prove that equation $x^5-5x^3+4x-1=0$ has exactly 5 roots

Prove that equation $x^5-5x^3+4x-1=0$ has exactly 5 root.

By using intermediate value theorem, I can show that $x^5-5x^3+4x-1=0$ has at least one root in each following intervals: $(-2,-1.5),\ (-1.5,-1),\ (-1,0.5),\ (0.5,1),\ (1,3)$. So , it has exactly 5 roots.

But I wonder that there is some logical ways we can find intervals that contain roots without guessing?

答:
If you rewrite the equation as

$$x^4-5x^2+4={1\over x}$$

and then sketch the curves $y=x^4-5x^2+4=(x^2-1)(x^2-4)$ and $y=1/x$ sufficiently carefully, you can see that the curves cross at two points in the interval $(-2,-1)$, two more points in the interval $(0,1)$, and one final point in the interval $(2,\infty)$.

The intersections in $(0,1)$ are the hardest to see (you might miss them if your sketch is too sketchy, as mine was at first), but are easily confirmed by comparing the values of $y$ at $x=1/2$, since the hyperbola clearly lies above the polynomial curve at $x=1$ and as $x\to0^+$. Doing so tells you one root lies in the interval $(0,1/2)$ and the other in the interval $(1/2,1)$. Similarly, comparing the $y$ values at $x=-3/2$ splits $(-2,-1)$ into $(-2,-3/2)$ and $(-3/2,-1)$, each with one root. And finally, it's easy to see that $(3^2-1)(3^2-4)\gt1/3$, so the largest root lies in the interval $(2,3)$.

Note, this curve-sketching approach works as nicely as it does only because of the special form of the equation, in particular the nice factorization of $x^4-5x^2+4$, which makes that curve's general shape easy to draw. It's not an approach you can always rely on to be helpful, but it's worth keeping in mind for occasions when it is.


问:
Find $M(x, y)$ for which $M(x, y) + (3xy^2 + 20x^2y^3)y' = 0$ is an exact differential equation.
Title says it all.

Find $M(x, y)$ so that $M(x, y) + (3xy^2 + 20x^2y^3)y' = 0$ is an exact differential equation.

I tried to use the $d M(x,y)/dy = d N(x,y)/dx$ formula but got stuck. I guess we must cancel the derivatives by applying an integral in both sides but I'm not sure how.

答:
hint

Your condition can be written as

$$M(x,y)dx+(3xy^2+20x^2y^3)dy=0=dU$$

with
$$\frac{\partial U}{\partial y}=3xy^2+20x^2y^3$$
so

$$U=xy^3+5x^2y^4+C(x)$$
and

$$M(x,y)=\frac{\partial U}{\partial x}=...$$


答:
Just a hint:

Write $y'$ as $\frac{dy}{dx}$ till the ODE looks like

$$M(x,y)dx+(3xy^2+20x^2y^3)dy=0.$$

Now, find the possible function for $M(x,y)$ such that $M_y=N_x.$

Title: Find $M(x, y)$ for which $M(x, y) + (3xy^2 + 20x^2y^3)y' = 0$ is an exact differential equation.

Body:
Title says it all.

Find $M(x, y)$ so that $M(x, y) + (3xy^2 + 20x^2y^3)y' = 0$ is an exact differential equation.

I tried to use the $d M(x,y)/dy = d N(x,y)/dx$ formula but got stuck. I guess we must cancel the derivatives by applying an integral in both sides but I'm not sure how.


问:
Does $ax \equiv b \pmod m$ have exactly one solution for groups in general?
In the context of the chinese remainder theorem there is a lemma:

Lemma 1: Let $a, b, m \in \mathbb Z$ with $m \geq 2$ and $\gcd(a,m)=1$.

Then the equation $ax \equiv b \pmod m$ has exactly one solution
in $\mathbb Z_m$.

I am trying to contrast that with the following statement about groups:

Lemma 2:

Let $G = (S, \circ)$ be a group, thus subject to the group axioms

G1 $\circ$ is associative

G2 There exists a neutral element $e \in S$

G3 Every element $a \in S$ has an inverse $a^{-1}$

From these axioms it can be derived that $\circ$ is surjective, i.e.
that

$$\exists x : a \circ x = b \text{ and } \exists y : y \circ a = b$$

Now what I am interested in is how to prove the second lemma.
My first thought was to use the proof of the first lemma as a guide, but that fails almost immediately:
The proof for the first lemma uses results from the euclidean algorithm, in particular that $\exists f, g: af + mg = \gcd(a, m) = 1$, which only works for euclidean rings but we only have a group.

My second observation is that Lemma 2 is a weaker statement than Lemma 1 since L2 only asserts existence, but not uniqueness of $x$ and $y$.

Long story short, I have three questions:

How do I prove Lemma 2 with only the 3 group axioms?

Is it really true that Lemma 2 is a weaker version of Lemma 1, or is there really no connection between those two lemmas.

Is there a way to prove Lemma 1 for groups in general, or is it only possible for euclidean rings?

For reference, I got both lemmas from the book "Diskrete Strukturen Band 1", second edition by Steger.

答:
They are special cases of the following (in the commutative case).

Lemma $ $ TFAE for $\,a\in M\,$ a commutative monoid

$(1)\ \  \forall\ b\in M\!:\ ax=b\,$ has a unique root $\,x\in M$

$(2)\ \  \forall\ b\in M\!:\ ax=b\,$ has a root $\,x\in M$

$(3)\ \ a\,$ has an inverse $\,a^{-1}\in M$

Proof $\ (1\!\Rightarrow\! 2)\ $ Clear. $\ (2\!\Rightarrow\! 3)\ $ Let $\,b =  1.\ $ $\ (3\!\Rightarrow\! 1)\ \  ax = b\!\iff\! x = a^{-1}b\,$ by scaling by $\,a^{\pm 1}$.


问:
Extrapolating a limit from finite numerical data
Simply for fun, I'm wondering if there are techniques (rigourous or not) for estimating a sequential limit from finitely many terms

(under certain niceness assumptions, e.g., there is only negligible transient behaviour).

For definiteness, consider the following setting:

A toy problem

I have two sequences, $(a_n)_{n=1}^∞$ and $(b_n)_{n=1}^∞$ for which I have only been able to calculate the first 8 terms.

Here's what is known:

$a_n \leq b_n$ for all $n$.

$(a_n)$ increasing.

$(b_n)$ decreasing.

$(b_n-a_n) → 0$.

I.e., they converge and have the same limit, sandwiched between the two sequences.

Numerically,

$(a_n)_{n=1}^8 \approx (0.7767, 0.7863, 0.7940, 0.7995, 0.8034, 0.8062, 0.8081,
  0.8095)$.

$(b_n)_{n=1}^8 \approx (0.8760, 0.8582, 0.8443, 0.8350, 0.82895, 0.8248, 0.8222,
  0.8202)$.

Here is a plot:

So the shared limit is likely between 0.81 and 0.82.

The question: if a gun was placed to your head and you were told to guess the limit, what would you guess, and why?

(Or a probability density, if you prefer.)

Naive idea

One naive idea, focusing on each sequence individually, would be to try to approximate each data set with a converging curve. E.g., for some $N$, choose $(λ_n)_{n=1}^N$ so that

$$ a_n \approx \sum_{k=0}^N λ_k n^{-k}$$

in some fashion, and take $λ_0$ as the candidate limit. (And similarly for $(b_n)$.)

There are plenty of problems with this, with the arbitrariness of polynomial terms in the sum and the choice of fitting.

With the probably-poor choice of least-squares fitting, the following happens with the estimate when we vary $N$:

The estimates from each sequence mostly disagree!

Most estimates lie outside the admissible window.

In particular, for $N≥7$, the value bifurcates and settles down onto a totally wrong estimate.

These estimates are depicted in the following graph. (For larger $N$, it continues to settle down to the wrong value.)

Subquestion (optional): Can this naive method be improved, e.g., with a different choice of fitting functions and fitting method?

答:
You're looking for series/sequence acceleration, such as Aitken's method. If you interpret another sequence as

$$x_{2n}=a_n,\quad x_{2n-1}=b_n$$

then you may also get accelerated convergence by applying the Van Wijngaarden transformation, which essentially suggests you should be looking at the averages of $a_n$ and $b_n$.

For your data set, the averages of the last three $a_n$ and $b_n$ are

$$\left(\frac{a_n+b_n}2\right)_{n=6}^8=(0.81550, 0.81515, 0.81485)$$

and the Aitken acceleration applied to those terms yields $0.81305$.


问:
Close-form for triple integral $ \int_0^c \int_0^b \int_0^a \sqrt{x^2+y^2+z^2} dx dy dz$
I am able to work out the double integral

$$\int_0^b \int_0^a \sqrt{x^2+y^2} dx dy $$
with brute-force (i.e. integrating $x$, then $y$) to arrive at the close-form result

$$\frac13ab\sqrt{a^2+b^2} +\frac16a^3\sinh^{-1}\frac ba +\frac16 b^3 \sinh^{-1}\frac ab$$

which has the expected parity between $a$ and $b$. However, it gets unwieldy to tackle the triple-integral extension $$\int_0^c \int_0^b \int_0^a \sqrt{x^2+y^2+z^2} dx dy dz$$
this way and I am unable to slug it out. Does anyone know the corresponding close-form expression for the triple version?

Body:
You can refer to J.M. Borwein's box integral

\begin{gather*}
\int_0^c\int_0^b\int_0^a\dfrac{1}{\sqrt{x^2+y^2+z^2}}{\rm\,d}x{\rm\,d}y{\rm\,d}z\\
=
\begin{array}{r}
ab\ln\left(\frac{c}{\sqrt{a^2+b^2}}+\sqrt{1+\frac{c^2}{a^2+b^2}}\right)\\
+ac\ln\left(\frac{b}{\sqrt{a^2+c^2}}+\sqrt{1+\frac{b^2}{a^2+c^2}}\right)
+bc\ln\left(\frac{a}{\sqrt{b^2+c^2}}+\sqrt{1+\frac{a^2}{b^2+c^2}}\right)\\
-\frac{a^2}{2}\arctan\left(\frac{bc}{a\sqrt{a^2+b^2+c^2}}\right)-\frac{b^2}{2}\arctan\left(\frac{ac}{b\sqrt{a^2+b^2+c^2}}\right)-\frac{c^2}{2}\arctan\left(\frac{ab}{c\sqrt{a^2+b^2+c^2}}\right)
\end{array}\\
\\
\\
\int_0^c\int_0^b\int_0^a\sqrt{x^{\overset{\,}{2}}+y^2+z^2}{\rm\,d}x{\rm\,d}y{\rm\,d}z\\
=
\begin{array}{r}
\frac{abc}{4}\sqrt{a^2+b^2+c^2}+\frac{ab\left(a^2+b^2\right)}{6}\ln\left(\frac{c}{\sqrt{a^2+b^2}}+\sqrt{1+\frac{c^2}{a^2+b^2}}\right)\\
+\frac{ac\left(a^2+c^2\right)}{6}\ln\left(\frac{b}{\sqrt{a^2+c^2}}+\sqrt{1+\frac{b^2}{a^2+c^2}}\right)
+\frac{bc\left(b^2+c^2\right)}{6}\ln\left(\frac{a}{\sqrt{b^2+c^2}}+\sqrt{1+\frac{a^2}{b^2+c^2}}\right)\\
-\frac{a^4}{12}\arctan\left(\frac{bc}{a\sqrt{a^2+b^2+c^2}}\right)-\frac{b^4}{12}\arctan\left(\frac{ac}{b\sqrt{a^2+b^2+c^2}}\right)-\frac{c^4}{12}\arctan\left(\frac{ab}{c\sqrt{a^2+b^2+c^2}}\right)
\end{array}\\
\\
\\
\int_0^c\int_0^b\int_0^a\bigg(x^2+y^2+z^2\bigg)^{\frac{3}{2}}{\rm\,d}x{\rm\,d}y{\rm\,d}z\\
=
\begin{array}{r}
\frac{2abc\left(a^2+b^2+c^2\right)}{15}\sqrt{a^2+b^2+c^2}+\frac{ab\left(9a^4+10a^2b^2+9b^4\right)}{120}\ln\left(\frac{c}{\sqrt{a^2+b^2}}+\sqrt{1+\frac{c^2}{a^2+b^2}}\right)\\
+\frac{ac\left(9a^4+10a^2c^2+9c^4\right)}{120}\ln\left(\frac{b}{\sqrt{a^2+c^2}}+\sqrt{1+\frac{b^2}{a^2+c^2}}\right)\\
+\frac{bc\left(9b^4+10b^2c^2+9c^4\right)}{120}\ln\left(\frac{a}{\sqrt{b^2+c^2}}+\sqrt{1+\frac{a^2}{b^2+c^2}}\right)\\
-\frac{a^6}{30}\arctan\left(\frac{bc}{a\sqrt{a^2+b^2+c^2}}\right)-\frac{b^6}{30}\arctan\left(\frac{ac}{b\sqrt{a^2+b^2+c^2}}\right)-\frac{c^6}{30}\arctan\left(\frac{ab}{c\sqrt{a^2+b^2+c^2}}\right)
\end{array}
\end{gather*}


问:
Prove $\log x!$ is $\Omega (xlogx)$
Find a positive real number $C$ and a nonnegative real number $x_o$ such that

$Cx$$\log x$ $\leq$ $\log x!$ for all real numbers $x > x_o$.

I tried to expand $\log x!$ into $\log 1 + \log2 +\log3 +....\log x$. But how do I choose $C$ and $x_o$ so the above inequality hold?

Any hints would be appreciated.

答:
For $k\ge 2$, we have $\log k\ge \int_{k-1}^{k}\log t\,dt$. Summing from $k=2$ to $n$, we find  that
$$\log 2 +\log 3 + \log 4+ \cdots +\log n \ge \int_1^n \log t\,dt=n\log n-n.$$

If $n\ge 9$, then $n\log n-n \gt \frac{1}{2}n\log n$.

Title: Prove $\log x!$ is $\Omega (xlogx)$


问:
$h_n \to h$ uniformly+ $\operatorname{supp}(h_n)$ compact for each $n\in \mathbb{N}$ $\Rightarrow$ $\operatorname{supp}(h)$ is compact?
Let $h_n, h: \mathbb{R}^{d} \to \mathbb{R}$, with $h_n \in C_c(\mathbb{R}^{d})$ for each $n \in \mathbb{N}$ such that $(h_n)$ converges uniformly to $h$. Suppose that $\operatorname{supp}(h_n)$ is compact for each $n\in \mathbb{N}$.

My question: Is it true that $\operatorname{supp}(h)$ is compact?

That would be to justify a passage in Rudin's book Fourier Analysis on Groups. (See figure below, where $h_n=f_n \ast g_n$, $h=f \ast g$ and $G$ is a locally compact abelian group, in particular, $G=\mathbb{R}^{d}$)

答:
No, let $h$ be defined by $h(x)=\frac{1}{1+x^2}$ (or any other function that goes to zero for $x\to\pm\infty$). Let $g_n$ be suitable cut-off functions, i.e. \begin{align}g_n(x)=\begin{cases}1 ~~&\text{if }|x|\leq n,\\ 0 &\text{if } |x|\geq n+1\end{cases}\end{align}
and $0\leq g_n(x)\leq 1$ for $n<|x|<n+1$. Now set $h_n=g_nh$. We have $\|h_n-h\|_\infty\leq h(n)\to0$ as $n\to\infty$, i.e. $h_n$ converges uniformly to $h$, the supports of $h_n$ are compact but the support of $h$ is not.

Note that Rudin says '$f*g\in C_0$' and not $C_c$. Sometimes $C_0$ denotes the subspace of functions vanishing at infinity. In this case the statement would be correct, but I am not sure what $C_0(G)$ means for LCA groups.


问:
Let $f:\mathbb{N}\to \mathbb{N}$ be bijection such that $\sum_{n=1}^{\infty} \frac{f(n)}{n^2}< \infty$
Let $f:\mathbb{N}\to \mathbb{N}$ be bijection such that

$$\sum_{n=1}^{\infty} \frac{f(n)}{n^2}< \infty$$

Then numbers of such bijections is zero, or one or finitely many but more than one or infinite such bijections.

I think there will be no such bijection . Am i right ? if yes how to prove that ? any hint

答:
There is no such $f$. Indeed, for every $N$, you have
$$\sum_{n=N+1}^{3N} \frac{f(n)}{n^2} \geq \sum_{n=N+1}^{3N} \frac{f(n)}{(3N)^2} \geq \frac{N^2}{9N^2}$$

because the $f(n)$ for $n \in [N+1, 3N]$ are $2N$ distinct integers, so at least $N$ of them are greater than $N$.

So
$$\sum_{N+1}^{3N} \frac{f(n)}{n^2} \geq \frac{1}{9}$$

so the series is not Cauchy, so it diverges.


问:
Prove that: If there exists $a \in \mathbb R$ such that $f'(a) > 0$, then $\lim_{x \to \infty}f(x) = \infty$
Let $f$ be differentiable in $\mathbb R$, with $f'$ differentiable and $f'(x) \geq 0,f''(x) \geq 0$ for every $x \in \mathbb R$.

How do I rigorously prove that (using Lagranges Mean Value Theorem):

If there exists $a \in \mathbb R$ such that $f'(a) > 0$, then $\lim_{x \to \infty}f(x) = \infty$

I know that if f is monotonic increasing and $f'>m>0$ for $x>a$ (where $f'(a)=m>0$) then $\lim_{x \to \infty} f(x) > \lim_{x \to \infty} mx = \infty$

But I want something more rigorouse.

Thanks for any idea or direction

答:
Another approch is to note that $f'(x) \geq f'(a) >0$ for all $x\geq a$ because $f''$ is non-negative. Thus $f$ is strictly increasing in $[a, \infty) $.

It follows that either $f(x) \to L$ or $f(x) \to \infty $ as $x\to\infty $. But if $f(x) \to L$ then (via LMVT) $f' (c) =f(x+1)-f(x)\to 0$ which contradicts $f'(c) \geq f'(a) >0$. Hence we must have $f(x) \to\infty$.


问:
Transfroming 2-factor to another 2 factor graph.
I'm reading Petersen's DIE THEORIE DER REGULAREN GRAPHS.

There is interesting statement at chapter 7, page 6.

Given 2-factor F.
You can transfrom F to any other 2-factor applying following operation many times.

Erase 2 edges ab, cd of F. Create 2 edges ac, bd or ad, bc.

The proof looks short. However, unfortunetly, the paper is written in German. I could not understand the proof with dictionary.

Does anyone explain the proof to me?

答:
I don't think the statement you are asking about is even claimed on p.6 (p.198). Later, on p.7 (p.199) Petersen claims that the edge swap operation lets us turn any graph into any other graph with the same degree sequence. (Maybe he is only claiming this for regular graphs, since I think for Petersen only regular graphs exist...?) This implies the statement you want.

More is true in that regard: even if we do not allow multigraphs to appear in intermediate steps (as Petersen does) then we can turn any graph into any other graph with the same degree sequence, regular or not. This is a theorem proved independently by Havel and by Hakimi, and I go into some detail about how to prove it here. (This gets me out of reading more German, which is not my favorite language for doing math in either.)

Meanwhile, on the page you're looking at, Petersen claims that

If we start with a graph that has a $2$-factorization, and apply this operation, we get another graph that has a $2$-factorization.

There are two cases to check here, which Petersen does. First, if $ab$ and $cd$ are in the same $2$-factor, then deleting them and replacing them by $ac$ and $bd$ leaves degrees in that $2$-factor unchanged, so it is still a $2$-factor, and the $2$-factorization is still preserved.

Second, if $ab$ and $cd$ are in different $2$-factors, the union of those $2$-factors is $4$-regular: a $4$-factor, I suppose, though that's not a very modern thing to say. After applying the edge swap, the $4$-factor is preserved, and by Petersen's preceding theorem, that $4$-factor can be factored into two $2$-factors, restoring the $2$-factorization of the whole graph.


问:
Find a function discontinuous everywhere but $\ x = 0$
Find a function discontinuous everywhere but $\ x = 0$
After considering modifications to the Dirichlet function however I have not got anywhere. Is there a method to find such function or is it purely intuition with a bit of creativity?

答:
$f: \Bbb R \to \Bbb R$ defined by:
$$f(x) = \begin{cases} x & x \in \Bbb Q\\
                       -x & x \in \Bbb R\setminus \Bbb Q\\
\end{cases}$$

is only continuous at $x=0$ (source "counterexamples in analysis" p. 22)


问:
Why is it important to have a classifying space for principal $G$-bundles over a based space $X$?
For a group $G$ its classifying space $BG$ can be thought of as a space with the property that $G∼ΩBG$ (the based loop space of $BG$).

This actually works more generally for spaces equipped with a multiplication that is 'highly homotopically associative'.

An important use for $BG$ is classifying principal $G$-bundles over a based space $X$: there is a bijection between the set of based homotopy classes $[X,BG]$ and the set of isomorphism classes of principal $G$-bundles.

Ok, but it is not clear to me why it is useful to have a bijection with the set of based homotopy classes $[X,BG]$.

答:
The space $BG$ has a non-trivial cohomology $H^*(BG;A)$ with coefficients in a group $A$. Now any fiber bundle over $X$ is equivalent to a homotopy class of maps $X\rightarrow BG$. But from this we get a map $H^*(BG;A)\rightarrow H^*(X;A)$. If this map in cohomology is non-trivial, the fiber bundle is non-trivial. An element in the image of this map is called a characteristic class for the fiber bundle. Stiefel-Whitney, Chern, and more exotic things are all examples of characteristic classes.

Sometimes one also might be able to explicitly compute the set $[X,BG]$ by other means (for example obstruction theory). Then you know all bundles! A good example is $BS^1=CP^\infty$. Then we know that all principal bundles $S^1$ are classified by $[X,\mathbb CP^\infty]$. But this set coincides with $H^2(X;\mathbb{Z})$ as $CP^\infty$ is also a $K(Z,2)$. One of the miracles of topology.


问:
Specific example of the universal property of $V \otimes W$
I'm trying to familiarize myself with the definition of tensors, so I was wondering if I understood the definition in terms of the universal property.

Consider a bilinear $B : V \times W \rightarrow U$. That is, $B(\sum \alpha_i v_i, \sum \beta_j w_j) = \sum_{i,j} \alpha_i \beta_j B(v_i,w_j) \in U$.

Then I can quotient out the desired equivalence relation to get $T : V \times W \rightarrow V \otimes W$, which is of the form $T(\sum \alpha_i v_i, \sum \beta_j w_j) = \sum_{i,j} \alpha_i \beta_j (v_i \otimes w_j)$. Then the unique linear map $\ell$ is of the form:

$$\sum \alpha_i \beta_j (v_i \otimes w_j) \mapsto \sum_{k=1}^{ij} \gamma_k \ell((v \otimes w)_k)$$

where $k$ ranges over every ordered pair $(i,j)$, and is $\gamma_k = \alpha_i\beta_j$, and similarly for each basis of $V \otimes W$.

Is this correct? Or am I way off?

答:
I think you have the right idea, but you're overcomplicating it a bit.

By the universal property of $V\otimes W$, there is a unique linear map $\ell : V\otimes W \to U$ such that $B = \ell\circ T$. For $(v, w) \in V\times W$ we have $T(v, w) = v\otimes w$, so

$$\ell(v\otimes w) = \ell(T(v, w)) = (\ell\circ T)(v, w) = B(v, w).$$

Note that a generic element of $V\otimes W$ is not of the form $v\otimes w$, but it is a sum of such elements. For $\sum_{i=1}^nv_i\otimes w_i \in V\otimes W$, we have $$\ell\left(\sum_{i=1}^nv_i\otimes w_i\right) = \sum_{i=1}^nB(v_i, w_i)$$

as $\ell$ is linear.


问:
Finding the volume of a solid through 2$\pi$ radians
With reference to this question

Curve $C$ with the equation:
$$y = (2x-1)^{\frac{3}{4}}, \quad x \ge\frac{1}{2}$$

The finite region $S$, is bounded by the curve $C$, the $x$-axis. The $y$-axis and the line $y = 8$. This region is rotated through $2\pi$ radians about the $x$-axis.

Find the exact value of the volume of the solid generated.

Here's what I have tried:
$\mathbf{k}$ or $\mathbf{x}$ was found to be at $\frac{17}{2}$

$$\int\limits^{17/2}_{1/2}(2x-1)^{3/4}dx=\frac{256}{7}$$

I'm wondering if I approached this correctly?

答:
As you've done, let's look at region S:

Now you can obtain the volume of region S when rotated around the y-axis by $2\pi$ by rotating the rectangular region (represented by the blue lines) which will form a cylinder, and then subtracting the volume of the curve when rotated around the x-axis.

Cylinder volume
The radius of the cylinder will be the $r = 8$ (since the rectangle is bounded at a height of $8$). The height will be $h=17/2 - 1/2 = 8$ since that is the width of the bounded rectangle.
So using the formula for the volume of a cylinder $V = {\pi}{r^2}h$, we obtain:

$V_1 = 512{\pi}$

Curve C volume
Using the volumes of revolution formula (you can learn more about this here)

$V = \pi\int_b^a{(y(x))^2 dx}$, we arrive at the integral:
$\pi\int^{17/2}_{1/2} (2x-1)^{3/2} dx$ which can be done by either inspection (what differentiates to $(2x-1)^{3/2}$?) or by using a substition (something like $u = 2x-1$). Once this is done and you have plugged in the limits, you obtain:

$V_2 = \frac{1024\pi}{5}$

Final result
Then the region S can be obtained by $V_s = V_1 - V_2$, which gives the result:
$V_s = 307.2\pi$


问:
Simplifying an equality of products of matrices
I have five matrices $A$, $B$, $C$, $D$, $E$, and I know that $ABC = DEC$. Can I conclude that $AB = DE$?</p>

答:
If $C$ is invertible, you can right multiply both sides of the equation by $C^{-1}$. One way to check if $C$ is invertible is to calculate if it has a nonzero determinant, assuming $C$ is a square matrix.

If $C$ is not square, this problem is more difficult, and you must check if there is a matrix $C'$ such that $CC' = I$, where $I$ is the identity matrix.


问:
Einstein says that this equation is true, but WolframAlpha says it's not always true. Who is right?
Einstein says
$$\cos\mathrm{i}x=\frac1{\sqrt{1-\left(\mathrm{i}\tan\mathrm{i}x\right)^2}},$$
but WolframAlpha says that this isn't true for $x=\pm2$ and $x=\pm9/5$. What's happening?

From page 34 of The Meaning of Relativity:

$$v=\mathrm{i}\tan\mathrm{i}\theta,\quad\mathrm{i}\sin\mathrm{i}\theta=\dfrac{v}{\sqrt{1-v^2}}\quad\text{and}\quad \cos\mathrm{i}\theta=\dfrac1{\sqrt{1-v^2}}.$$

Otherwise, the functions do appear to be identical.

答:
Einstein is right assuming real x. $$\cos (ix) = \frac{1}{\sqrt{1-(i\tan(ix))^2}} = \frac{1}{\sqrt{1+\tan^2(ix)}} = \frac{1}{\frac{1}{\cos(ix)}}~.$$

As $\cos(ix)\gt 0$ this is an identity for all real x.

OwnerDisplayName: Expert Nonexpert


问:
How to show that $\lim_{(x,y) \to (0,0)}\frac{x^{2}y^{2}}{\left|x\right|^{3}+\left|y\right|^{3}}=0$
How to show that $$\lim_{(x,y) \to (0,0)}\frac{x^{2}y^{2}}{\left|x\right|^{3}+\left|y\right|^{3}}=0$$
I know that

$$\left|\frac{x^{2}y^{2}}{\left|x\right|^{3}+\left|y\right|^{3}}\right|\le\frac{x^{2}y^{2}}{\left|x\right|^{3}}=\frac{y^{2}}{\left|x\right|}$$

But now we have a problem with $\left|x\right|$ and I cannot find a lower bound other than $0$.

Also$$\left|\frac{x^{2}y^{2}}{\left|x\right|^{3}+\left|y\right|^{3}}\right|\le\frac{x^{2}y^{2}}{\left|x^{3}+y^{3}\right|}=\frac{\left|x+y\right|}{\left|x^{2}-xy+y^{2}\right|}\le\frac{\left|x\right|+\left|y\right|}{\left|x^{2}-xy+y^{2}\right|}$$

But I'm not sure if that helps.

If we show that $\lim_{(x,y) \to (0,0)}\frac{y^{2}}{\left|x\right|}=0$ then we are done.

答:
You can directly apply the general statement:

If $a, b \ge 0$, $c, d > 0$ and $\frac{a}{c} + \frac{b}{d} > 1$ then
$$\lim_{(x,y)\rightarrow (0,0)}\frac{\left|x\right|^a\left|y\right|^b}{\left|x\right|^c + \left|y\right|^d} = 0$$

You can also work it out in an ad hoc way.

Let $u=|x|^3$ and $v=|y|^3$
and note that
$$
u^{2/3}v^{2/3}=u^{1/3}(u^{1/3}v^{2/3})\le u^{1/3}(\frac13u+\frac23 v)\tag{1}
\le u^{1/3}(u+v)
$$
and thus
$$
|f(x,y)|=\frac{u^{2/3}v^{2/3}}{u+v}\le u^{1/3}=|x|
$$
In the first inequality of (1), Young's inequality is used.


问:
Calculate $\iint_D 2x*cos(2(x-y)x) $ where D is a rectangle with corners at (π/6 ,2+ π /6 ), ( π/6,4+ π/6), (π /4,4+ π/4) and (π/4,2+ π/4)
I have been trying to calculate the double integral:  $\iint_D 2x*cos(2(x-y)x) $ where D is a rectangle with corners at  (π/6 ,2+ π /6 ), ( π/6,4+ π/6),  (π /4,4+ π/4)  and  (π/4,2+ π/4) for a very long time now, but I get stuck and wonder if there is something I'm missing in my solution? It seems very complicated to integrate this function, have I missed out on any simplifying variable substitution (I tried one as you can see below but that didn't make it much easier) or have I approached the problem in the wrong way from the beginning? Down below are my calculations (the figure is very unproportional):

Thanks in advance!

/Nick

答:
First of all, the region is a parallelogram and not a rectangle. The parallelogram is made of two vertical lines $x = \frac{\pi}{6}$ and $x = \frac{\pi}{4}$ and you can work out the other two lines as

$y = x + 2$ and $y = x + 4$.

Now change of variable is not necessary if you integrate wrt $y$ first as integral is much simpler then.

$\displaystyle \int_{\pi/6}^{\pi/4} \bigg[\int_{x+2}^{x+4} 2 x \cos(2x^2-2xy) \ dy \bigg] \ dx$

$\displaystyle = \int_{\pi/6}^{\pi/4} \big[\sin (2xy-2x^2)\big]_{x+2}^{x+4} \ dx$

$\displaystyle = \int_{\pi/6}^{\pi/4} (\sin 8x - \sin 4x) \ dx$

Can you take it from here?


答:
It is not just the fact that there is a bijection $[X, BG] \to \operatorname{Prin}_G(X)$, but also the form of the bijection which is useful. There is a universal principal $G$-bundle $EG \to BG$ and the bijection is given by $[f] \mapsto f^*EG$; that is, every principal $G$-bundle $P \to X$ is obtained by pulling back the universal principal $G$-bundle by some map $f : X \to BG$ (called a classifying map for $P$).

This description can be used to prove general facts about principal $G$-bundles by proving them 'universally', i.e. for the universal principal $G$-bundle. For example, you can prove the following fact by working universally: for a real line bundle $\ell$, we have $c_1(\ell\otimes_{\mathbb{R}}\mathbb{C}) = \beta(w_1(\ell))$ where $c_1$ and $w_1$ are the first Chern and Stiefel-Whitney classes respectively, and $\beta$ is the Bockstein.

Title: Why is it important to have a classifying space for principal $G$-bundles over a based space $X$?

Body:
For a group $G$ its classifying space $BG$ can be thought of as a space with the property that $G∼ΩBG$ (the based loop space of $BG$).

This actually works more generally for spaces equipped with a multiplication that is 'highly homotopically associative'.

An important use for $BG$ is classifying principal $G$-bundles over a based space $X$: there is a bijection between the set of based homotopy classes $[X,BG]$ and the set of isomorphism classes of principal $G$-bundles.

Ok, but it is not clear to me why it is useful to have a bijection with the set of based homotopy classes $[X,BG]$.


问:
Difference between "Gaussian distribution" and "random variable whose distribution is a Gaussian distribution."
The question was a question to me in a comment to a post of mine yesterday here.

I can't see any difference, and thought that "Gaussian distribution" implicitly meant "distribution [of a random variable]".

I realize this is a soft-question, and if there isn't any substance to the topic, I'll delete it. However, I tend to think there is something I am not aware of, because the user asking me this follow-up question proceeded to tell me that these two concepts would correspond to convolutions $N(2\mu, 4 \sigma^2)$ for the distribution of the sum of two $X,Y \sim N(\mu,\sigma^2)$; and $N(2 \mu, 2\sigma^2)$ for $N(\mu,\sigma^2)$ with itself.

I am not asking for the actual convolutions, but perhaps the set up of the integral, and the difference between both concepts.

答:
The obvious difference is:

"Gaussian distribution" refers to the distribution

"a random variable whose distribution is Gaussian" refers to the random variable.

Recall in Kolmogorov's axiomatization of probability theory (which most mathematicians use) one starts with a probability space $(\Omega,\mathcal{F},\mathbb{P})$ and defines

A distribution is a probability measure $\mathbb{S}$ on some model measure space $(\Omega',\mathcal{F}')$

A (real-valued) random variable is a measurable function $\Omega\to\mathbb{R}$.

The point @LeanderTilstedKristensen made is that you can obviously add two real-valued functions together, but if you add two probability measures you get a finite measure with total measure $2$ and so not a probability measure.  Hence you will need to define the map "adding distributions"
$$
\mathcal{M}^+_1(\Omega,\mathcal{F})\times\mathcal{M}^+_1(\Omega,\mathcal{F})\to\mathcal{M}^+_1(\Omega,\mathcal{F})
$$


问:
Calculating a limit using Landau notation
I'm trying to do a problem but I can't figure out how to do it. We have to calculate a limit using Landau notation. I'm in the french schooling system so I'm not sure if this is what it is called but I mean domination, preponderance and equivalence of functions around a certain value writen with the symbols O , o and ~ respectively. I also know some basics of how to use Taylor Young developpements but was explicitly told not to use Hopital's rule.

Anyway, with all of that out of the way, here is the limit:

$$\lim_{t \to 1}{(\frac{a+t}{a+1})^{\frac{t}{1-t}}} $$

with a as constant
Thank you for your time

Body:
Using that $\log(1 + x) = x + o(x)$ when $x\to 0$
$$
\log\lim_{t\to 1}{\left(\frac{a+t}{a+1}\right)^{\frac{t}{1-t}}} =
\lim_{t\to 1}\log{\left(\frac{a+t}{a+1}\right)^{\frac{t}{1-t}}} =$$
$$
\lim_{t\to 1}\frac{t}{1-t}\,\log{\left(\frac{a+t}{a+1}\right)} =
\lim_{t\to 1}\frac{t}{1-t}\,\log{\left(\frac{(a+1) + (t-1)}{a+1}\right)} =
$$
$$
\lim_{t\to 1}\frac{t}{1-t}\,\log{\left(1 + \frac{t-1}{a+1}\right)} =
\lim_{t\to 1}\frac{t}{1-t}\,\left(\frac{t-1}{a+1} + o(t-1)\right) =\cdots
$$


问:
Kernel, Basis and Column Space
If a matrix $A$ has row reduced form

$A=$$\begin{pmatrix} 1&3&0&3 \\ 0&0&1&-1 \\ 0&0&0&0 \end{pmatrix}$

then the kernel of the linear map $T : \mathbb{R}^4 → \mathbb{R}^3$ defined by $T(x) = Ax$ has basis $$(−3,1,0,0), (−3, 0, 1, 1).$$

How do I prove or disprove the claim regarding the kernel?

Any help is appreciated guys. Also, how can I show that columns 2 and 4 forms a basis for the column space?

答:
Use the rank-nullity theorem, which says that

$\dim R^4 = \dim ker(A) + \dim rank (A)$

Here $\dim ker(A) = 4 - 2 = 2$. Your two vectors are linearly independent and so form a basis of $ker(A)$.

Title: Kernel, Basis and Column Space


问:
Find $c\in\mathbb{R}$, and $f$ such that $\int_c^xt\cdot f(t)\,\mathrm{d}x=x\sin (x)+\cos (x)+\frac{x^3}3$

Find a number and a function that satisfies:
$$\displaystyle\int_{c}^{x}t\cdot f(t)\,\mathrm{d}x=x\sin (x)+\cos (x)+\dfrac{x^3}{3}$$

I tried taking derivative both sides so I got:

$$\left(\displaystyle\int_{0}^{x}t\cdot f(t)\,\mathrm{d}t-\displaystyle\int_{0}^{c}t \cdot f(t)\,\mathrm{d}t\right)^\prime=x \cos(x)+x^2 $$

It's clear that the right part of the integral will be $0$ because it's constant:

$$\left( \displaystyle\int_{0}^{x} t \cdot f(t)\,\mathrm{d}t \right)^{\prime}=x \cos(x)+x^2$$

But I didn't do rest. Because I don't know how to derive the integral with two functions. I know with one function like $\int_{0}^{x} f(t)\,\mathrm{d}t$. Thanks!

答:
You can still think of $t\mapsto tf(t)$ as a function, and differentiate as you normally would using the fundamental theorem of calculus:
$$\frac{d}{dx}\left( \int_{0}^{x} t \cdot f(t)\,\mathrm{d}t \right) = xf(x).$$


问:
Calculate big-$\Theta$ for $T(x) = \log(x2x!)$
$T(x) = \log(x2x!)$

use the property of log, $\log(x2x!)$ is equivalent to $\log(2x) + \log(x!)$

My approach is to prove big-$O$ and big-$\Omega$ for $T(x)$,then big-$\Theta$ just follows.

If I want to calculate big-$O$ and big-$\Omega$ for $T(x)$, can I treat $\log(2x)$ as a constant and ignore it since it's growth rate is so slow compare to $\log(x!)$?

答:
You may simply write
$$
\log(x!) = \log(x)+ \log(x-1)+ \cdots+ \log(1) < x\log(x).
$$

Then,
$$
\log(2x)+ \log(x!) < (x+2) \log(x),
$$

which is $O(x \log(x))$ as $x \to \infty$.


问:
Meaning of derivative of a function with respect to another function
In chain rule we use,$\frac{dy}{dx}=\frac{dy}{dz} \times \frac{dz}{dx}$. My question is what is meant by $\frac{dy}{dz}$.When we find derivative,we find it with respect to an independent vairable meaning if $x$ changes by some amount,how much does $f(x)$ changes.But how much $g(f(x))$ changes by a change in $f(x)$ doesn't seem to make sense to me.Also derivatives are all about slopes of a graph.Now we will we draw the graph of $g(f(x))$ vs $f(x)$ since we can't place $f(x)$ on real axis as it can't achieve all real numbers for it being a dependent variable.So could someone please shed some light on what we are actually doing?

答: Consider this example: let's say we have a variable $V$ representing volume, and another variable $T$ representing temperature. Now the rate of change of volume with respect to temperature is given by $\frac{dV}{dT}$. If we further suppose that $T$ depends on another variable $t$ representing time, then the rate of change of volume with respect to time can be found by $$\frac{dV}{dt}=\frac{dV}{dT}\cdot \frac{dT}{dt}$$


问:
How to calculate limits of sequences?

Calculate the limits of the following sequences
$$ a_n = \frac{n+4n-n^5}{n^3+3n^5-2n}$$
$$ b_n =  \sqrt{n^2+4} - \sqrt{n^2+2} $$

I know the formulas to transform sequences to find out its limit but I do not understand to which point one has to transform a sequence to see its limit. I mean the limit of $a_n$ seems to be $- \frac{1}{3}$ and the limit of $b_n$ seems to be $0$ by just looking at them. I did not calculate anything. So I wonder to which point I have to transform sequences to "see" their limits. Can anyone explain me what the goal is when it comes to calculate limits of sequences?

答:
For the first one, divide the numerator and the denominator by the dominant power of the numerator that is $n^5$ in this case:
$$\begin{align}
\lim a_n &= \lim\frac{n+4n-n^5}{n^3+3n^5-2n} \\
&= \lim \frac{\color{blue}{5/n^3} - 1}{\color{blue}{1/n^2} + 3 - \color{blue}{2/n}} \\
&=\frac{-1}{3}\end{align}$$
because all blue terms tend to $0$ as $n$ goes to infinity.

For the second one, here is the hint:
$$b_n =  \sqrt{n^2+4} - \sqrt{n^2+2} = \frac{2}{\sqrt{n^2+4} +\sqrt{n^2+2}}$$

Title: How to calculate limits of sequences?


答:
Numerically, the most dominant term in the numerator of $a_n$ is $-x^5$ and in the denominator it is $3x^5$, so the $$\lim_{n \to \infty} a_n=-\frac{1}{3}.$$

Next $$L=\lim_{n \to \infty}b_n= \lim_{n \to\infty}[\sqrt{n^2+4}-\sqrt{n^2+2}]= \lim_{n\to \infty} n[(1+4/n^2)^{1/2}-(1+2/n^2)^{1/2}].$$
Use binomial expansion $(1+z)^k =1+kz+k(k-1)z^2/2+.... |z|<<1$
Then $$L=\lim_{n\to \infty} n[1+2/n^2+...-1-1/n^2-...]=\lim_{n \to \infty}n/n^2= \lim_{n \to \infty}1/n=0$$

Title: How to calculate limits of sequences?

Body:
Calculate the limits of the following sequences
$$ a_n = \frac{n+4n-n^5}{n^3+3n^5-2n}$$
$$ b_n =  \sqrt{n^2+4} - \sqrt{n^2+2} $$

I know the formulas to transform sequences to find out its limit but I do not understand to which point one has to transform a sequence to see its limit. I mean the limit of $a_n$ seems to be $- \frac{1}{3}$ and the limit of $b_n$ seems to be $0$ by just looking at them. I did not calculate anything. So I wonder to which point I have to transform sequences to "see" their limits. Can anyone explain me what the goal is when it comes to calculate limits of sequences?


问:
Random values with the same distribution
My question relates to the very basics of probability theory.
I can’t understand why if two functions that have the same distribution are not equal (apart from the set of measure zero).

For example, the central limit theorem. There are the countable set of the “same” functions and their sum converges to another function...

Is there any geometric answer (or picture) why the functions with the same distribution are not equal?

答:
Functions with the same distribution can be very different--they may never equal actually, like the example offered by Henry, which works pretty well here. If we let 1 be the head of the coin and -1 be the tail, then X which is the outcome of a fair coin flip has the same distribution as Y=-X: both assigning probability 0.5 to head and tail equally, but they are never the same.

We can define two independent random variables(actually as many as you want) on one single probability space, which I think answers your last question--to distinguish two tosses of a fair coin.

Take [0,1] with Borel sets for example, and let the probability measure be Lebesgue measure(i.e. $\lambda[a,b]=b-a$ as usual). Let
$$X(t)=\begin{cases}1, t\in[0,0.5]\\-1,t\in(0.5,1]\end{cases},Y(t)=\begin{cases}1,t\in[0,0.25)\cup[0.5,0.75),\\-1,t\in[0.25,0.5)\cup[0.75,1]\end{cases}.$$

Then you can check easily that they are independent by definition, while having the same distribution like coin tosses: one half for each cases. Continue dividing the intervals we can define a sequence of i.i.d coin tosses(which is a good exercise for you to check the mutual independence).


问:
What is the domain of $y$?
We have the differential equation $y'=\frac{3x^2}{3y^2-4}$. We want to find the domain of $y$. From the equation we have that $y^{3}-4y=x^{3}+C$, for $C\in \mathbb{R}$. From this relation, how can we find the domain of $y$ ?

答:
From $y'=\frac{3x^2}{3y^2-4}$ we can see denominator of the fraction can't be zero so $3y^2-4\neq0$  hence $y\neq\pm\frac{2}{\sqrt3}$.

So the domain of $y$ is $\mathbb{R}-\{\pm\frac{2}{\sqrt3}$}

Title: What is the domain of $y$?


问:
A trigonometric function comparing with $0$

Short Version

We need to solve this problem:

Prove that function $$f(x) = 2\sin\left(\frac{\pi}{x+\frac{1}{x}+1}\right)-\sin\left(\frac{\pi}{x^{2}+x+1}\right)-\sin\left(\frac{\pi}{\frac{1}{x^{2}}+\frac{1}{x}+1}\right)$$ is always larger than to $0$ on $(0,1)$, and is always smaller than $0$ on $(-1,0)$.

Long Version

Me and my friends are working on an extended problem from another forum:

Prove that function $$f(x) = 2\sin\left(\frac{\pi}{x+\frac{1}{x}+1}\right)-\sin\left(\frac{\pi}{x^{2}+x+1}\right)-\sin\left(\frac{\pi}{\frac{1}{x^{2}}+\frac{1}{x}+1}\right)$$ is always larger than or equal to $0$ on $(0,+\infty)$, and is always smaller than or equal to $0$ on $(-\infty,0)$.

Despite our best effort, we failed to solve it. Our initial opinion is that to show that $f'(x)$ is always smaller than or larger than $0$ before or after a constant since we are able to show that $$\lim_{x\to\pm\infty} = 0$$ However we failed.

Every approch or guidance to the problem, if mentioned, will be appreciated. Thanks in advance.

UPD: Calculating $f'(x)$ and determining whether it is positive / negetive throughout is not necessary!

OK, after long thoughts I decided to post an answer to the problem.

Actually, $f(x) = f\left(\frac{1}{x}\right)$ is a very helpful tool.

Via calculating $f'(x)$ we can discover that $f(x)$ is always positive on $(0, 1)$ and always negative on $(-1, 0)$.

Hence, let $g:(0,1)\to(1,+\infty)$ become $g(x)=\frac 1 x$, we know that if $f(x)>0$, then $f(g(x))>0$, and hence the question is proved.

Similarly, the other part can be proved.

Some questioned the inequality, well, you can check the link: https://www.desmos.com/calculator/clfhhlq5zi

Some asked for the proof of "$f(x)$ is always positive on $(0, 1)$"

Firstly, given $$
f'(x)=\frac{\pi  (2 x+1) \cos \left(\frac{\pi }{x^2+x+1}\right)}{\left(x^2+x+1\right)^2}-\frac{2 \pi  \left(1-\frac{1}{x^2}\right) \cos \left(\frac{\pi }{x+\frac{1}{x}+1}\right)}{\left(x+\frac{1}{x}+1\right)^2}+\frac{\pi  \left(-\frac{2}{x^3}-\frac{1}{x^2}\right) \cos \left(\frac{\pi }{\frac{1}{x^2}+\frac{1}{x}+1}\right)}{\left(\frac{1}{x^2}+\frac{1}{x}+1\right)^2}
$$

Concluding from demos graph, $f'(x)$ is negative from a constant, $x_0$, and returns to $0$ at $1$. This conclusion can also be proven by pluging in $f'(x)$.

Hence, $f(1)$ is a minimum point, since $f'(1) = 0$. However, $f(1)>0$ and, also, $f(x_0)>0$, we can now conclude that $\forall x\in(0,1)$, $f(x)>0$.

This, well, is not a rigorous proof, but it seems convincing to me and I hope someone can help me finish this.

Therefore, the problem is not ended since we had a little problem when it comes to $(0,1)$.

答:
Edit: A slightly different approach is described as follows:

Denote $A = \frac{\pi}{x+\frac{1}{x}+1}$, $B = \frac{\pi}{x^{2}+x+1}$ and $C = \frac{\pi}{\frac{1}{x^{2}}+\frac{1}{x}+1}$.
Clearly, $A + B + C = \pi$.

We have
\begin{align}
f(x) &= 2\sin A - \sin B - \sin C\\
&= 2\sin (B + C) - 2 \sin\frac{B+C}{2} \cdot \cos \frac{B-C}{2}\\
&= 2\sin \frac{B+C}{2} \cdot \left(2\cos \frac{B+C}{2} - \cos \frac{B-C}{2}\right).
\end{align}
Thus, if suffices to prove that
$$2\cos \frac{B+C}{2} - \cos \frac{B-C}{2} \ge 0.$$

Similarly, we may use bounds. Omitted.

Proof of $f(x) \ge 0$ on $(0, \infty)$:

Since $f(x) = f(1/x)$ for all $x > 0$, we only need to prove the case $x \ge 1$.

First we give the following auxiliary results (Facts 1 through 5). The proofs are easy and thus omitted.

Fact 1: $\sin u \le u$ for all $u \ge 0$.

Fact 2: $\sin u \ge \frac{2}{\pi}u$ for all $u \in [0, \pi/2]$.

Fact 3: $\sin u \ge \frac{\sqrt{3}}{2} + \frac{1}{2}(u - \pi/3) - \frac{\sqrt{3}}{4}(u - \pi/3)^2$ on $[0, \pi/3]$.

Fact 4: $\sin u \le \frac{\sqrt{3}}{2} + \frac{1}{2}(u - \pi/3) - \frac{\sqrt{3}}{6}(u - \pi/3)^2$ on $[0, \pi/3]$.

Fact 5: $\sin u \le \frac{\sqrt{3}}{2} + \frac{1}{2}(u - \pi/3) - \frac{\sqrt{3}}{4}(u - \pi/3)^2$ on $[\pi/3, \pi]$.

Now, we split into two cases:

$x \ge 8$

It suffice to prove that
$$2\sin\left(\frac{\pi}{x+\frac{1}{x}+1}\right)-\sin\left(\frac{\pi}{x^{2}+x+1}\right)-\sin\left(
\pi - \frac{\pi}{\frac{1}{x^{2}}+\frac{1}{x}+1}\right) \ge 0.$$

By Facts 1-2, it suffices to prove that
$$2\cdot \frac{2}{\pi} \cdot \frac{\pi}{x+\frac{1}{x}+1} - \frac{\pi}{x^{2}+x+1} - \left(\pi - \frac{\pi}{\frac{1}{x^{2}}+\frac{1}{x}+1}\right) \ge 0$$
that is
$$\frac{(4 - \pi)x - 2\pi}{x^2+x+1} \ge 0$$
which is true.

$1\le x < 8$

Denote $A = \frac{\pi}{x+\frac{1}{x}+1}$, $B = \frac{\pi}{x^{2}+x+1}$ and $C = \frac{\pi}{\frac{1}{x^{2}}+\frac{1}{x}+1}$.

Note that $A \in [0, \pi/3]$, $B \in [0, \pi/3]$ and $C \in [\pi/3, \pi)$.

By Facts 3-5, it suffices to prove that
\begin{align}
&2\cdot \left(\frac{\sqrt{3}}{2} + \frac{1}{2}(A - \pi/3) - \frac{\sqrt{3}}{4}(A - \pi/3)^2\right)\\
&\quad - \left(\frac{\sqrt{3}}{2} + \frac{1}{2}(B - \pi/3) - \frac{\sqrt{3}}{6}(B - \pi/3)^2\right)\\
&\quad - \left(\frac{\sqrt{3}}{2} + \frac{1}{2}(C - \pi/3) - \frac{\sqrt{3}}{4}(C - \pi/3)^2\right) \ge 0
\end{align}
that is
$$\frac{\pi (x-1)^2\Big[(8\pi \sqrt{3} - 54)x^2 + (32\pi \sqrt{3} - 54)x + 5\pi \sqrt{3} - 54\Big]}{108(x^2+x+1)^2} \ge 0$$
which is true.

We are done.


问:
Infinite series, injective function and rearrangement inequality
Let $f:\mathbb{N^*}\to\mathbb{N^*}$ an injective function. Show that the following infinite series $$\sum_{i=1}^{\infty}\frac{f(n)}{n^2}$$ is divergent. I am supposed to deal with this using the rearrangement inequality. But I don't know how!Could some explain this to me step by step? I really want to understand...</p>

答:
Here is a direct proof without the use of the rearrangement inequality.

For every $N$, you have
$$\sum_{n=N+1}^{3N} \frac{f(n)}{n^2} \geq \sum_{n=N+1}^{3N} \frac{f(n)}{(3N)^2} \geq \frac{N^2}{9N^2}$$

because the $f(n)$ for $n \in [N+1, 3N]$ are $2N$ distinct integers, so at least $N$ of them are greater than $N$.

So
$$\sum_{N+1}^{3N} \frac{f(n)}{n^2} \geq \frac{1}{9}$$

so the series is not Cauchy, so it diverges.


问:
Closure of operators
Let $X$ and $Y$ Banach. We say that the linear operator $A:\mathcal{D}(A)\subseteq X\rightarrow Y$ admits a closure if there's a linear operator $B:\mathcal{D}(B)\subseteq X\rightarrow Y$ such that $\mathcal{D}(A)\subseteq \mathcal{D}(B)$, $B|_{\mathcal{D}(A)} = A$
and $\mathcal{G}(B)= \overline{\mathcal{G}(A)}$, where $\mathcal{G}(Z)$ is the graph of $Z$.

I want to prove that $A$ admits a closure if and only if every sequence $\{x_n\}_{n\in\mathbb{N}}\subseteq \mathcal{D}(A)$ such that $(x_n,A(x_n))\xrightarrow{n\rightarrow\infty}(0,y)$, with $y\in Y$, satisfy that $y=0$.

I proved ($\Rightarrow$), but I have had problems in ($\Leftarrow$), because I don't know how to define the operator $B$. Please, somebody can help me?

Thanks in advance.

答:
The operator $B$ is defined as follows:

If $x \in \mathcal{D}(A)$, then $Bx := Ax$.

If $x \notin \mathcal{D}(A)$ happens to be the limit of a sequence $\{x_n\}_n \subset \mathcal{D}(A)$ such that $\{Ax_n\}_n$ converges in $Y$---note that such a sequence need not exist in general, in which case we can't define $B$ on $x$---then $Bx := \lim_{n\to\infty}Ax_n$.

The point, then, is to check that $B$ is actually well-defined in the second case. So, suppose that $x \notin \mathcal{D}(A)$ is the limit both of $\{x_n\}_n \subset \mathcal{D}(A)$ and of $\{x_n^\prime\}_n \subset \mathcal{D}(A)$, such that $\{Ax_n\}_n$ and $\{Ax_n^\prime\}_n$ converge in $Y$. Then $$\{(x_n-x_n^\prime,A(x_n-x_n^\prime))\}_n = \{(x_n-x_n^\prime,Ax_n - Ax_n^\prime)\}_n \to_{n\to\infty} \left(0,\lim_{n\to\infty}Ax_n - \lim_{n\to\infty}Ax_n^\prime\right);$$
what can you now conclude?

Title: Closure of operators


答: For $b_n$, Consider $f(x)=\sqrt{x}$  on $[n^2+2, n^2+4]$. Note that $f$ satisfies the hypothesis of mean value theorem. Then there exist $c \in (n^2+2, n^2+4) $ such that $$\frac{1}{2\sqrt{c}}=f'(c)=\frac{\sqrt{n^2+4}-\sqrt{n^2+2}}{2}$$ So $$0<\sqrt{n^2+4}-\sqrt{n^2+2}=\frac{1}{\sqrt{c}}<\frac{1}{\sqrt{n^2+2}}$$The result follows by Squeeze theorem!

Title: How to calculate limits of sequences?


答:
Easier: if $(x,y) = (r\cos\theta,r\sin\theta)$,
$$
\left|\frac{x^2 y^2}{|x|^3+|y|^3}\right|
= \frac{r^4\cos^2\theta\sin^2\theta}{r^3(|\cos\theta|^3 + |\sin\theta|^3)}\le\frac{r}{m} = \frac{\sqrt{x^2 + y^2}}{m},
$$
with
$$m = \min_{\theta\in[0,2\pi]}(|\cos\theta|^3 + |\sin\theta|^3) > 0.$$
(Why $> 0$)?

Title: How to show that $\lim_{(x,y) \to (0,0)}\frac{x^{2}y^{2}}{\left|x\right|^{3}+\left|y\right|^{3}}=0$

Body:
How to show that $$\lim_{(x,y) \to (0,0)}\frac{x^{2}y^{2}}{\left|x\right|^{3}+\left|y\right|^{3}}=0$$
I know that

$$\left|\frac{x^{2}y^{2}}{\left|x\right|^{3}+\left|y\right|^{3}}\right|\le\frac{x^{2}y^{2}}{\left|x\right|^{3}}=\frac{y^{2}}{\left|x\right|}$$

But now we have a problem with $\left|x\right|$ and I cannot find a lower bound other than $0$.

Also$$\left|\frac{x^{2}y^{2}}{\left|x\right|^{3}+\left|y\right|^{3}}\right|\le\frac{x^{2}y^{2}}{\left|x^{3}+y^{3}\right|}=\frac{\left|x+y\right|}{\left|x^{2}-xy+y^{2}\right|}\le\frac{\left|x\right|+\left|y\right|}{\left|x^{2}-xy+y^{2}\right|}$$

But I'm not sure if that helps.

If we show that $\lim_{(x,y) \to (0,0)}\frac{y^{2}}{\left|x\right|}=0$ then we are done.


答:
A probability distribution is a measure on some probability space $\Omega$. Such a probability distribution $\lambda$ takes a subset $A$ of $\Omega$ and assigns a probability
$\lambda(A)$, which is a number between $0$ and $1$. In particular the Gaussian distribution $N(\mu,\sigma^2)$ is the measure defined on $\mathbb{R}$ as
$$\lambda(A):= \int_A \frac{1}{\sqrt{2\pi} \sigma} e^{-\frac{(x-\mu)^2}{2\sigma^2}} \: dx.\quad \text{(for $A \subseteq \mathbb{R}$)}.$$
A random variable on the other hand, is a function $X: \Omega \rightarrow \mathbb{R}$, where $\Omega$ is a probability space (or sometimes called sample space), which itself has a probability measure, that we will call $\mathbb{P}$. A random variable $X$ induces a measure $\mathbb{P}_X$ on $\mathbb{R}$ defined by the formula
$$\mathbb{P}_X(A): = \mathbb{P}(X\in A) = \mathbb{P}(\{ \omega \in \Omega \: | \: X(\omega) \in A\}).$$
This measure is called the distribution of $X$ and in particular $X$ has a Gaussian distribution if
$$\mathbb{P}_X(A):= \int_A \frac{1}{\sqrt{2\pi} \sigma} e^{-\frac{(x-\mu)^2}{2\sigma^2}} \: dx.$$
It seems to be the case, that you believe that the distribution of $X+Y$ is always the convolution of $\mathbb{P}_X$ with $\mathbb{P}_Y$, but this is wrong, you need to assume that $X$ and $Y$ are independent. The point of the example that i commented in your previous post, was that $X$ is usually not independent of $X$ and thus the distribution of $X+X=2X$ is NOT the convolution of $\mathbb{P}_X$ with itself.

To be more explicit the example was: Let $X \sim N(\mu,\sigma^2)$ and let $Y=X$. Then $X+Y = 2X$, and by basic computations we see that $E[2X] = 2\mu$ and $Var(2X) = 4Var(X) = 4 \sigma^2$. Since the sum is Gaussian, we see that $X+Y \sim N(2\mu , 4\sigma^2)$, however if we instead had taken $Y \sim N(\mu,\sigma^2)$ to be independent of $X$, then $X+Y \sim N(2\mu , 2\sigma^2)$. In both cases we have $X,Y \sim N(\mu,\sigma^2)$, but the resulting distribution of $X+Y$ is not the same. This is also what i refer to by saying "you actually have to properly define what it means to add distributions".

Title: Difference between "Gaussian distribution" and "random variable whose distribution is a Gaussian distribution."

Body:
The question was a question to me in a comment to a post of mine yesterday here.

I can't see any difference, and thought that "Gaussian distribution" implicitly meant "distribution [of a random variable]".

I realize this is a soft-question, and if there isn't any substance to the topic, I'll delete it. However, I tend to think there is something I am not aware of, because the user asking me this follow-up question proceeded to tell me that these two concepts would correspond to convolutions $N(2\mu, 4 \sigma^2)$ for the distribution of the sum of two $X,Y \sim N(\mu,\sigma^2)$; and $N(2 \mu, 2\sigma^2)$ for $N(\mu,\sigma^2)$ with itself.

I am not asking for the actual convolutions, but perhaps the set up of the integral, and the difference between both concepts.


问:
Find $M(x, y)$ for which $M(x, y) + (3xy^2 + 20x^2y^3)y' = 0$ is an exact differential equation.
Title says it all.

Find $M(x, y)$ so that $M(x, y) + (3xy^2 + 20x^2y^3)y' = 0$ is an exact differential equation.

I tried to use the $d M(x,y)/dy = d N(x,y)/dx$ formula but got stuck. I guess we must cancel the derivatives by applying an integral in both sides but I'm not sure how.

答:
I hope that this is right, if not please correct me.

If $M(x,y)$ is an exact equation, it's true that:

$\frac{dM(x,y)}{dy}=\frac{dN(x,y)}{dx}$

Then:

$\frac{dM(x,y)}{dy}= \frac{d(3xy^2+20x^2y^3)}{dx}$

$\frac{dM(x,y)}{dy}= 3y^2+40xy^3$

$dM(x,y)= (3y^2+40xy^3)dy$

$\int{dM(x,y)}= \int{(3y^2+40xy^3)dy}$

Therefore:

$M(x,y)= y^3+10xy^4+c$


答:
First of all, when involving two variables, you should clarify the variable which you are differentiating. Following from your question, because of the fundamental theorem of calculus, we have:
$$
\frac{d\left( \displaystyle\int_{0}^{x} t \cdot f(t)\mathrm{d}t \right)}{dx}=x \cos(x)+x^2
$$
Implies:
$$
xf(x)=x\cos(x)+x^2 \implies f(x)=\cos(x)+x
$$
Let's go back to the beginning. If we substitute the value of $f$, we get:
Implies:

$$
\left( \displaystyle\int_{c}^{x} t \cdot (\cos(t)+t)\mathrm{d}t \right)=x\sin (x)+\cos (x)+\dfrac{x^3}{3}
$$
After doing all the calculations, we will have:
$$
c\sin (c)+\cos (c)+\dfrac{c^3}{3}=0
$$
I used an external program to solve this, but you can use Bolzano's Theorem to see that the equation has a solution for some $c\in [-1,0]$

Title: Find $c\in\mathbb{R}$, and $f$ such that $\int_c^xt\cdot f(t)\,\mathrm{d}x=x\sin (x)+\cos (x)+\frac{x^3}3$

Body:
Find a number and a function that satisfies:
$$\displaystyle\int_{c}^{x}t\cdot f(t)\,\mathrm{d}x=x\sin (x)+\cos (x)+\dfrac{x^3}{3}$$

I tried taking derivative both sides so I got:

$$\left(\displaystyle\int_{0}^{x}t\cdot f(t)\,\mathrm{d}t-\displaystyle\int_{0}^{c}t \cdot f(t)\,\mathrm{d}t\right)^\prime=x \cos(x)+x^2 $$

It's clear that the right part of the integral will be $0$ because it's constant:

$$\left( \displaystyle\int_{0}^{x} t \cdot f(t)\,\mathrm{d}t \right)^{\prime}=x \cos(x)+x^2$$

But I didn't do rest. Because I don't know how to derive the integral with two functions. I know with one function like $\int_{0}^{x} f(t)\,\mathrm{d}t$. Thanks!


问:
integrating a vector over a sphere
I have the following triple integral in spherical coordinates $(r,\theta,\phi)$: $$\int_0^{2\pi}\int_0^\pi\int_0^RCr^3\hat\theta\cdot r^2dr\sin{\theta}d\theta d\phi$$

How do I handle the $\hat\theta$? If I ignore it, I get $\frac{2}{3}\pi CR^6$. So is my answer the vector $\frac{2}{3}\pi CR^6\hat\theta$? Do I need to integrate the unit vector $\hat\theta$? If so, how?

答: Yes, you need to integrate this, as $\hat \theta$ is a function of $\theta, \phi$.  A straightforward way to attack the problem is to write $\hat \theta$ as a linear combination of $\hat x, \hat y, \hat z$, but with the components expressed in terms of $r, \theta, \phi$.</p>

Title: integrating a vector over a sphere


Body:
$$M(x,y)dx+(3xy^2+20x^2y^3)dy=0$$
$$M(x,y)dx+xd(y^3)+5x^2d(y^4)=0$$
$$M_1(x,y)dx+xd(y^3)+M_2(x,y)dx+5x^2d(y^4)=0$$
$$(\color{red}{y^3dx}+xd(y^3))+(\color{red}{5y^4dx^2}+5x^2d(y^4))=0$$
$$d(y^3x)+5d(y^4x^2)=0$$
$$y^3x+5y^4x^2=C$$
There are many other solutions since you can write $M(x,y)$ as $M_1(x,y)dx+M_2(x,y)dx+f(x)dx$.
So that:
$$M(x,y)=y^3+10y^4x+f(x)$$